Solution manual university phyiscs 12e Young

1 UNITS, PHYSICAL QUANTITIES AND VECTORS 1.1. 1.2. IDENTIFY: Convert units from mi to km and from km to ft. SET UP: ...

Author: Hugh D. Young | Roger A. Freedman | and Lewis Ford (Paperback - Aug 24 | 2007)

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1

UNITS, PHYSICAL QUANTITIES AND VECTORS

1.1.

1.2.

IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. = 2.54 cm , 1 km = 1000 m , 12 in. = 1 ft , 1 mi = 5280 ft . ⎛ 5280 ft ⎞⎛ 12 in. ⎞⎛ 2.54 cm ⎞⎛ 1 m ⎞⎛ 1 km ⎞ EXECUTE: (a) 1.00 mi = (1.00 mi) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ 2 ⎟⎜ 3 ⎟ = 1.61 km ⎝ 1 mi ⎠⎝ 1 ft ⎠⎝ 1 in. ⎠⎝ 10 cm ⎠⎝ 10 m ⎠

⎛ 103 m ⎞⎛ 102 cm ⎞ ⎛ 1 in. ⎞⎛ 1 ft ⎞ 3 (b) 1.00 km = (1.00 km) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 3.28 × 10 ft 1 km 1 m 2.54 cm 12 in. ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. IDENTIFY: Convert volume units from L to in.3 . SET UP: 1 L = 1000 cm3 . 1 in. = 2.54 cm 3

⎛ 1000 cm3 ⎞ ⎛ 1 in. ⎞ 3 0.473 L × ⎜ ⎟×⎜ ⎟ = 28.9 in. . ⎝ 1 L ⎠ ⎝ 2.54 cm ⎠ EVALUATE: 1 in.3 is greater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in cm3 , which is 473 cm3 . IDENTIFY: We know the speed of light in m/s. t = d / v . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is v = 3.00 × 108 m/s . 1 ft = 0.3048 m . 1 s = 109 ns . 0.3048 m EXECUTE: t = = 1.02 × 10−9 s = 1.02 ns 3.00 × 108 m/s EVALUATE: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km = 1.86 × 105 mi . IDENTIFY: Convert the units from g to kg and from cm3 to m3 . SET UP: 1 kg = 1000 g . 1 m = 1000 cm . EXECUTE:

1.3.

1.4.

3

EXECUTE:

1.5.

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3 . IDENTIFY: Convert volume units from in.3 to L. SET UP: 1 L = 1000 cm3 . 1 in. = 2.54 cm . EXECUTE:

1.6.

( 327 in. ) × ( 2.54 cm in.) × (1 L 1000 cm ) = 5.36 L 3

3

3

EVALUATE: The volume is 5360 cm3 . 1 cm3 is less than 1 in.3 , so the volume in cm3 is a larger number than the volume in in.3 . IDENTIFY: Convert ft 2 to m 2 and then to hectares. SET UP: 1.00 hectare = 1.00 × 104 m 2 . 1 ft = 0.3048 m . EXECUTE:

1.7.

g ⎛ 1 kg ⎞ ⎛ 100 cm ⎞ 4 kg × 11.3 ⎟×⎜ ⎟ = 1.13 × 10 3 ⎜ cm ⎝ 1000 g ⎠ ⎝ 1 m ⎠ m3

⎛ 43,600 ft 2 ⎞ ⎛ 0.3048 m ⎞ The area is (12.0 acres) ⎜ ⎟⎜ ⎟ ⎝ 1 acre ⎠ ⎝ 1.00 ft ⎠

2

⎛ 1.00 hectare ⎞ = 4.86 hectares . ⎜ 4 2 ⎟ ⎝ 1.00 × 10 m ⎠

EVALUATE: Since 1 ft = 0.3048 m , 1 ft 2 = (0.3048) 2 m 2 . IDENTIFY: Convert seconds to years. SET UP: 1 billion seconds = 1 × 109 s . 1 day = 24 h . 1 h = 3600 s . EXECUTE:

⎛ 1 h ⎞⎛ 1 day ⎞ ⎛ 1 y ⎞ 1.00 billion seconds = (1.00 × 109 s ) ⎜ ⎟ = 31.7 y . ⎟⎜ ⎟⎜ ⎝ 3600 s ⎠⎝ 24 h ⎠ ⎝ 365 days ⎠ 1-1

1-2

1.8.

Chapter 1

EVALUATE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d , which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.

⎛ 0.125 mi ⎞⎛ 1 fortnight ⎞⎛ 1 day ⎞ furlongs fortnight ) ⎜ ⎟ = 67 mi/h ⎟⎜ ⎟⎜ ⎝ 1 furlong ⎠⎝ 14 days ⎠ ⎝ 24 h ⎠ EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi = 1.609 km . 1 gallon = 3.788 L. EXECUTE:

1.9.

(180,000

⎛ 1.609 km ⎞⎛ 1 gallon ⎞ (a) 55.0 miles/gallon = (55.0 miles/gallon) ⎜ ⎟⎜ ⎟ = 23.4 km/L . ⎝ 1 mi ⎠⎝ 3.788 L ⎠ 1500 km 64.1 L (b) The volume of gas required is = 64.1 L . = 1.4 tanks . 23.4 km/L 45 L/tank EVALUATE: 1 mi/gal = 0.425 km/L . A km is very roughly half a mile and there are roughly 4 liters in a gallon, EXECUTE:

1.10.

so 1 mi/gal ∼ 24 km/L , which is roughly our result. IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg . EXECUTE:

ft ⎛ mi ⎞ ⎛ 1h ⎞ ⎛ 5280 ft ⎞ (a) ⎜ 60 ⎟ ⎜ ⎟ ⎜ ⎟ = 88 h ⎠ ⎝ 3600s ⎠ ⎝ 1mi ⎠ s ⎝

⎛ ft ⎞ ⎛ 30.48cm ⎞ (b) ⎜ 32 2 ⎟ ⎜ ⎟ ⎝ s ⎠ ⎝ 1ft ⎠

m ⎛ 1m ⎞ ⎜ ⎟ = 9.8 2 100 cm s ⎝ ⎠ 3

g ⎞ ⎛ 100 cm ⎞ ⎛ 1 kg ⎞ ⎛ 3 kg (c) ⎜1.0 3 ⎟ ⎜ ⎟ = 10 3 ⎟ ⎜ cm 1 m 1000 g m ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1.11.

EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation 32 ft/s 2 = 9.8 m/s 2 is accurate to only two significant figures. IDENTIFY: We know the density and mass; thus we can find the volume using the relation density = mass/volume = m / V . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: Density = 19.5 g/cm3 and mcritical = 60.0 kg. For a sphere V = 34 π r 3 . EXECUTE:

⎛ 60.0 kg ⎞⎛ 1000 g ⎞ 3 V = mcritical / density = ⎜ ⎟ = 3080 cm . 3 ⎟⎜ 19.5 g/cm 1.0 kg ⎝ ⎠⎝ ⎠ r=3

1.12.

EVALUATE: The density is very large, so the 130 pound sphere is small in size. IDENTIFY: Use your calculator to display π × 107 . Compare that number to the number of seconds in a year. SET UP: 1 yr = 365.24 days, 1 day = 24 h, and 1 h = 3600 s.

⎛ 24 h ⎞ ⎛ 3600 s ⎞ 7 7 7 (365.24 days/1 yr) ⎜ ⎟⎜ ⎟ = 3.15567... × 10 s ; π × 10 s = 3.14159... × 10 s 1 day 1 h ⎝ ⎠ ⎝ ⎠ The approximate expression is accurate to two significant figures. EVALUATE: The close agreement is a numerical accident. IDENTIFY: The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. 10 m EXECUTE: (a) = 1.1 × 10−3%. 890 × 103 m (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE: In this case a very small percentage error has disastrous consequences. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. EXECUTE:

1.13.

1.14.

3V 3 3 ( 3080 cm3 ) = 9.0 cm . = 4π 4π

Units, Physical Quantities and Vectors

1-3

SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE: (a) (12 mm ) × ( 5.98 mm ) = 72 mm 2 (two significant figures)

5.98 mm = 0.50 (also two significant figures) 12 mm (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm. IDENTIFY and SET UP: In each case, estimate the precision of the measurement. EXECUTE: (a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%. (b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3 × 10−3%. (c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 2.8 × 10−2%. EVALUATE: The percent errors are those due only to the limit of precision of the measurement. IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area. SET UP: The length could be as large as 5.11 cm and the width could be as large as 1.91 cm. 0.07 cm 2 EXECUTE: The area is 9.69 ± 0.07 cm2. The fractional uncertainty in the area is = 0.72%, and the 9.69 cm 2 0.01 cm 0.01 cm = 0.20% and = 0.53%. The sum of these fractional uncertainties in the length and width are 5.10 cm 1.9 cm fractional uncertainties is 0.20% + 0.53% = 0.73% , in agreement with the fractional uncertainty in the area. EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities. SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target variables and from these we get the uncertainty. EXECUTE: (a) The volume of a disk of diameter d and thickness t is V = π (d / 2) 2 t. (b)

1.15.

1.16.

1.17.

The average volume is V = π (8.50 cm/2) 2 (0.50 cm) = 2.837 cm3 . But t is given to only two significant figures so the answer should be expressed to two significant figures: V = 2.8 cm3 . We can find the uncertainty in the volume as follows. The volume could be as large as V = π (8.52 cm/2) 2 (0.055 cm) = 3.1 cm 3 , which is 0.3 cm3 larger than the average value. The volume could be as small as V = π (8.52 cm/2) 2 (0.045 cm) = 2.5 cm3 , which is 0.3 cm3 smaller than the average value. The

1.18.

uncertainty is ±0.3 cm3 , and we express the volume as V = 2.8 ± 0.3 cm3 . (b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm = 170. By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8.52 cm/0.045 cm = 190. The smallest possible value of the ratio is 8.48 / 0.055 = 150. Thus the uncertainty is ±20 and we write the ratio as 170 ± 20. EVALUATE: The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much less, so the percentage uncertainty in the volume and in the ratio should be about 10%. IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 3 × 108 people, so 2 × 108 cars. EXECUTE: ( Number of cars × miles/car day ) / ( mi/gal ) = gallons/day

( 2 ×10

8

1.19.

cars × 10000 mi/yr/car × 1 yr/365 days ) / ( 20 mi/gal ) = 3 × 108 gal/day

EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S. IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs. 1 in. = 2.54 cm . 1 y = 12 months . EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. ⎛ 1 in. ⎞ 3 (b) 200 m = (2.00 × 104 cm) ⎜ ⎟ = 7.9 × 10 inches . This is much greater than the height of a person. 2.54 cm ⎝ ⎠ (c) 200 cm = 2.00 m = 79 inches = 6.6 ft . Some people are this tall, but not an ordinary man.

1-4

1.20.

Chapter 1

(d) 200 mm = 0.200 m = 7.9 inches . This is much too short. ⎛ 1y ⎞ (e) 200 months = (200 mon) ⎜ ⎟ = 17 y . This is the age of a teenager; a middle-aged man is much older than this. ⎝ 12 mon ⎠ EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. IDENTIFY: The number of kernels can be calculated as N = Vbottle / Vkernel . SET UP: Based on an Internet search, Iowan corn farmers use a sieve having a hole size of 0.3125 in. ≅ 8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth as 3 mm. We must also apply the conversion factors 1 L = 1000 cm3 and 1 cm = 10 mm. EXECUTE: The volume of the kernel is: Vkernel = (10 mm )( 6 mm )( 3 mm ) = 180 mm3 . The bottle’s volume is: 3 3 Vbottle = ( 2.0 L ) ⎣⎡(1000 cm3 ) (1.0 L ) ⎦⎤ ⎡⎣(10 mm ) (1.0 cm ) ⎤⎦ = 2.0 × 106 mm3 . The number of kernels is then

1.21.

1.22.

N kernels = Vbottle / Vkernels ≈ ( 2.0 × 106 mm3 ) (180 mm3 ) = 11,000 kernels . EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since these dimensions vary amongst the different available types of corn, acceptable answers could range from 6,500 to 20,000. IDENTIFY: Estimate the number of pages and the number of words per page. SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). EXECUTE: An estimate for the number of words is about 106 . EVALUATE: We can expect that this estimate is accurate to within a factor of 10. IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to find the volume in m3 breathed in a year. ⎛ 24 h ⎞⎛ 60 min ⎞ 5 2 SET UP: Assume 10 breaths/min . 1 y = (365 d) ⎜ ⎟⎜ ⎟ = 5.3 × 10 min . 10 cm = 1 m so ⎝ 1 d ⎠⎝ 1 h ⎠ 106 cm3 = 1 m3 . The volume of a sphere is V = 34 π r 3 = 16 π d 3 , where r is the radius and d is the diameter. Don’t forget to account for four astronauts. ⎛ 5.3 × 105 min ⎞ 4 3 EXECUTE: (a) The volume is (4)(10 breaths/min)(500 × 10−6 m3 ) ⎜ ⎟ = 1 × 10 m / yr . 1y ⎝ ⎠ 1/ 3

⎛ 6V ⎞ (b) d = ⎜ ⎟ ⎝ π ⎠

1.23.

1.24.

1/ 3

⎛ 6[1 × 104 m3 ] ⎞ =⎜ ⎟ π ⎝ ⎠

= 27 m

EVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute. 1 y = 365 days . 1 day = 24 h , 1 h = 60 min . Use 80 years for the lifetime. ⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞ 8 EXECUTE: The number of blinks is (10 per min) ⎜ ⎟⎜ ⎟ (80 y/lifetime) = 4 × 10 ⎟⎜ ⎝ 1 h ⎠ ⎝ 1 day ⎠⎝ 1 y ⎠ EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. ⎛ 60 min ⎞⎛ 24 h ⎞⎛ 365 days ⎞⎛ 80 yr ⎞ 9 EXECUTE: N beats = ( 75 beats/min ) ⎜ ⎟ ⎟⎜ lifespan ⎟ = 3 × 10 beats/lifespan yr ⎝ 1 h ⎠ ⎜⎝ 1 day ⎟⎜ ⎠⎝ ⎠⎝ ⎠ 9 ⎛ 1 L ⎞⎛ 1 gal ⎞ ⎛ 3 × 10 beats ⎞ 7 Vblood = ( 50 cm3 /beat ) ⎜ ⎟⎜ ⎟ = 4 × 10 gal/lifespan 3 ⎟⎜ ⎝ 1000 cm ⎠⎝ 3.788 L ⎠ ⎝ lifespan ⎠ EVALUATE: This is a very large volume.


1.25.

1-5

IDENTIFY: Estimation problem SET UP: Estimate that the pile is 18 in. × 18 in. × 5 ft 8 in.. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is V = 18 in. × 18 in. × 68 in. = 22,000 in.3 . Convert to cm3 :

V = 22,000 in.3 (1000 cm3 / 61.02 in.3 ) = 3.6 × 105 cm3 . The density of gold is 19.3 g/cm3 , so the mass of this volume of gold is

m = (19.3 g/cm3 )(3.6 × 105 cm3 ) = 7 × 106 g. The monetary value of one gram is $10, so the gold has a value of ($10 / gram)(7 × 106 grams) = $7 × 107 , or about

1.26.

$100 × 106 (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3 . Convert m3 to L. SET UP: Estimate the diameter of a drop to be d = 2 mm . The volume of a spherical drop is V = 34 π r 3 = 16 π d 3 . 103 cm3 = 1 L . 1000 cm3 = 2 × 105 4 × 10−3 cm3 EVALUATE: Since V ∼ d 3 , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year. SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) EXECUTE: They eat a total of 104 pizzas. EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each. IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill. SET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance from the earth to the moon is 3.8 × 108 m. EXECUTE:

1.27.

1.28.

V = 16 π (0.2 cm)3 = 4 × 10−3 cm3 . The number of drops in 1.0 L is

⎛ 3.8 × 108 m ⎞⎛ 103 mm ⎞ 12 12 N bills = ⎜ ⎟ = 3.8 × 10 bills ≈ 4 × 10 bills ⎟⎜ ⎝ 0.1 mm/bill ⎠ ⎝ 1 m ⎠ EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is significantly less – roughly 1 billion dollars. IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the surface area of a single dollar bill. SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or 3,380,000 mi 2 . This estimate is within 10 percent of the actual area, 3,794,083 mi 2 . The population is roughly 3.0 × 108 while the area of a dollar bill, as measured with a ruler, is approximately 6 18 in. by 2 85 in. EXECUTE:

1.29.

AU.S. = ( 3,380,000 mi 2 ) [( 5280 ft ) / (1 mi )] ⎣⎡(12 in.) (1 ft )⎦⎤ = 1.4 × 1016 in.2 = ( 6.125 in.)( 2.625 in.) = 16.1 in.2

EXECUTE:

Abill

2

2

Total cost = N bills = AU.S. Abill = (1.4 × 1016 in.2 ) (16.1 in.2 / bill ) = 9 × 1014 bills

1.30.

Cost per person = (9 × 1014 dollars) /(3.0 × 108 persons) = 3 × 106 dollars/person EVALUATE: The actual cost would be somewhat larger, because the land isn’t flat. IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative orientation of the two displacements. SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel. EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30. EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m.

Figure 1.30

1-6

1.31.

Chapter 1

IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. " " " " " " " " SET UP: Call the three displacements A , B , and C . The resultant displacement R is given by R = A + B + C . " EXECUTE: The vector addition diagram is given in Figure 1.31. Careful measurement gives that R is 7.8 km, 38# north of east . EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km + 4.0 km + 3.1 km .

Figure 1.31 1.32.

IDENTIFY: Draw the vector addition diagram, so scale. " " SET UP: The two vectors A and B are specified in the figure that accompanies the problem. " " " " EXECUTE: (a) The diagram for C = A + B is given in Figure 1.32a. Measuring the length and angle of C gives C = 9.0 m and an angle of θ = 34° . " " " " (b) The diagram for D = A − B is given in Figure 1.32b. Measuring the length and angle of D gives D = 22 m and an angle of θ = 250° . " " " " " " (c) − A − B = −( A + B ) , so − A − B has a magnitude of 9.0 m (the same as A + B ) and an angle with the + x axis " " of 214° (opposite to the direction of A + B ). " " " " " " (d) B − A = −( A − B ) , so B − A has a magnitude of 22 m and an angle with the + x axis of 70° (opposite to the " " direction of A − B ). " " EVALUATE: The vector − A is equal in magnitude and opposite in direction to the vector A .

Figure 1.32 1.33.

IDENTIFY: Since she returns to the starting point, the vectors sum of the four displacements must be zero. " " " " SET UP: Call the three given displacements A , B , and C , and call the fourth displacement D . " " " " A+ B + C + D = 0. " EXECUTE: The vector addition diagram is sketched in Figure 1.33. Careful measurement gives that D is144 m, 41# south of west.


1-7

" " " " D is equal in magnitude and opposite in direction to the sum A + B + C .

EVALUATE:

Figure 1.33 1.34.

1.35.

IDENTIFY and SET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding horizontal and vertical components. EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m. (b) Figure 1.34 gives components −15.6 km, 15.6 km . (c) Figure 1.34 gives components 3.82 cm, − 5.07 cm . EVALUATE: The signs of the components depend on the quadrant in which the vector lies.

Figure 1.34 " " IDENTIFY: For each vector V , use that Vx = V cosθ and Vy = V sin θ , when θ is the angle V makes with the + x

axis, measured counterclockwise from the axis. " " " " SET UP: For A , θ = 270.0° . For B , θ = 60.0° . For C , θ = 205.0° . For D , θ = 143.0° . EXECUTE: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m . Dx = −7.99 m ,

Dy = 6.02 m . The signs of the components correspond to the quadrant in which the vector lies. A IDENTIFY: tan θ = y , for θ measured counterclockwise from the + x -axis. Ax " " SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. EVALUATE: 1.36.

EXECUTE: (a) tan θ = (b) tan θ = (c) tan θ = (d) tan θ =

1.37.

Ay AX Ay Ax Ay Ax Ay Ax

=

−1.00 m = −0.500 . θ = tan −1 ( −0.500 ) = 360° − 26.6° = 333° . 2.00 m

=

1.00 m = 0.500 . θ = tan −1 ( 0.500 ) = 26.6° . 2.00 m

=

1.00 m = −0.500 . θ = tan −1 ( −0.500 ) = 180° − 26.6° = 153° . −2.00 m

=

−1.00 m = 0.500 . θ = tan −1 ( 0.500 ) = 180° + 26.6° = 207° −2.00 m

EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of θ . IDENTIFY: Find the vector sum of the two forces. SET UP: Use components to add the two forces. Take the + x -direction to be forward and the + y -direction to be upward.

1-8

Chapter 1

The second force has components F2 x = F2 cos32.4° = 433 N and F2 y = F2 sin 32.4° = 275 N. The

EXECUTE:

first force has components F1x = 725 N and F1 y = 0.

Fx = F1x + F2 x = 1158 N and Fy = F1 y + F2 y = 275 N

1.38.

The resultant force is 1190 N in the direction 13.4° above the forward direction. EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less than the sum of their magnitudes. IDENTIFY: Find the vector sum of the three given displacements. SET UP: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are: $ $ $ A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east .

Rx = Ax + Bx + C x = 0 + 4.0 km + ( 3.1 km ) cos ( 45# ) = 6.2 km ; Ry = Ay + By + C y =

EXECUTE:

2.6 km + 0 + (3.1 km) ( sin45# ) = 4.8 km ; R = Rx2 + Ry2 = 7.8 km ; θ = tan −1 ⎡⎣( 4.8 km ) ( 6.2 km )⎤⎦ = 38# ; $ R = 7.8 km, 38# north of east. This result is confirmed by the sketch in Figure 1.38. " EVALUATE: Both Rx and Ry are positive and R is in the first quadrant.

1.39.

Figure 1.38 " " " IDENTIFY: If C = A + B , then Cx = Ax + Bx and C y = Ay + By . Use C x and C y to find the magnitude and " direction of C . SET UP: From Figure 1.34 in the textbook, Ax = 0 , Ay = −8.00 m and Bx = + B sin 30.0° = 7.50 m ,

By = + B cos30.0° = 13.0 m . " " " EXECUTE: (a) C = A + B so Cx = Ax + Bx = 7.50 m and C y = Ay + By = +5.00 m . C = 9.01 m .

1.40.

Cy

5.00 m and θ = 33.7° . Cx 7.50 m " " " " " " (b) B + A = A + B , so B + A has magnitude 9.01 m and direction specified by 33.7° . " " " D −21.0 m (c) D = A − B so Dx = Ax − Bx = −7.50 m and Dy = Ay − By = −21.0 m . D = 22.3 m . tan φ = y = and Dx −7.50 m " φ = 70.3° . D is in the 3rd quadrant and the angle θ counterclockwise from the + x axis is 180° + 70.3° = 250.3° . " " " " " " (d) B − A = −( A − B ) , so B − A has magnitude 22.3 m and direction specified by θ = 70.3° . EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.32. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors. " " SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. tan θ =

EXECUTE: (b)

=

(a)

⎛ 5.20 ⎞ (−8.60 cm) 2 + (5.20 cm) 2 = 10.0 cm, arctan ⎜ ⎟ = 148.8° (which is 180° − 31.2° ). ⎝ −8.60 ⎠

⎛ −2.45 ⎞ (−9.7 m) 2 + (−2.45 m) 2 = 10.0 m, arctan ⎜ ⎟ = 14° + 180° = 194°. ⎝ −9.7 ⎠

⎛ −2.7 ⎞ (7.75 km)2 + (−2.70 km) 2 = 8.21 km, arctan ⎜ ⎟ = 340.8° (which is 360° − 19.2° ). ⎝ 7.75 ⎠ EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agree with our sketches. (c)


1.41.

1-9

IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum. SET UP:

A = 3.25 km B = 4.75 km C = 1.50 km

Figure 1.41a

" " " Select a coordinate system where + x is east and + y is north. Let A, B and C be the three displacements of the " " " " " professor. Then the resultant displacement R is given by R = A + B + C . By the method of components, Rx = Ax + Bx + C x and Ry = Ay + By + C y . Find the x and y components of each vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. EXECUTE: Ax = 0, Ay = +3.25 km Bx = −4.75 km, By = 0 Cx = 0,

C y = −1.50 km

Rx = Ax + Bx + C x Rx = 0 − 4.75 km + 0 = −4.75 km Ry = Ay + By + C y Ry = 3.25 km + 0 − 1.50 km = 1.75 km Figure 1.41b

R = Rx2 + Ry2 = ( −4.75 km) 2 + (1.75 km) 2 R = 5.06 km R 1.75 km = −0.3684 tan θ = y = Rx −4.75 km θ = 159.8° Figure 1.41c

1.42.

The angle θ measured counterclockwise from the + x -axis. In terms of compass directions, the resultant displacement is 20.2° N of W. " EVALUATE: Rx < 0 and Ry > 0, so R is in 2nd quadrant. This agrees with the vector addition diagram. " " " " IDENTIFY: Add the vectors using components. B − A = B + (− A) . " " " " " " SET UP: If C = A + B then Cx = Ax + Bx and C y = Ay + By . If D = B − A then Dx = Bx − Ax and Dy = By − Ay . EXECUTE: (a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + ( −3.75 cm) = −1.50 cm. (b) Using Equations (1.7) and (1.8),

⎛ −1.50 ⎞ (5.40cm) 2 (−1.50 cm) 2 = 5.60 cm, arctan ⎜ ⎟ = 344.5° ccw. ⎝ +5.40 ⎠

1-10

Chapter 1

(c) Similarly, 4.10 cm − (1.30 cm ) = 2.80 cm, 23.75 cm − ( 2.25 cm ) = 26.00 cm.

⎛ −6.00 ⎞ (2.80cm) 2 + (−6.00cm) 2 = 6.62 cm, arctan ⎜ ⎟ = 295° (which is 360° − 65° ). ⎝ 2.80 ⎠ EVALUATE: We can draw the vector addition diagram in each case and verify that our results are qualitatively correct. " " " " IDENTIFY: Vector addition problem. A − B = A + (− B ). " " SET UP: Find the x- and y-components of A and B. Then the x- and y-components of the vector sum are " " calculated from the x- and y-components of A and B. EXECUTE: Ax = A cos(60.0°) (d)

1.43.

Ax = (2.80 cm)cos(60.0°) = +1.40 cm Ay = A sin(60.0°) Ay = (2.80 cm)sin(60.0°) = +2.425 cm Bx = B cos( −60.0°) Bx = (1.90 cm)cos(−60.0°) = +0.95 cm By = B sin(−60.0°) By = (1.90 cm)sin(−60.0°) = −1.645 cm Note that the signs of the components correspond to the directions of the component vectors. Figure 1.43a " " " (a) Now let R = A + B. Rx = Ax + Bx = +1.40 cm + 0.95 cm = +2.35 cm.

Ry = Ay + By = +2.425 cm − 1.645 cm = +0.78 cm. R = Rx2 + Ry2 = (2.35 cm) 2 + (0.78 cm) 2 R = 2.48 cm R +0.78 cm tan θ = y = = +0.3319 Rx +2.35 cm θ = 18.4° Figure 1.43b EVALUATE:

" " " The vector addition diagram for R = A + B is " R is in the 1st quadrant, with Ry < Rx , in agreement with our calculation.

Figure 1.43c


" " " (b) EXECUTE: Now let R = A − B. Rx = Ax − Bx = +1.40 cm − 0.95 cm = +0.45 cm. Ry = Ay − By = +2.425 cm + 1.645 cm = +4.070 cm.

R = Rx2 + Ry2 = (0.45 cm) 2 + (4.070 cm) 2

R = 4.09 cm R 4.070 cm = +9.044 tan θ = y = Rx 0.45 cm θ = 83.7°

Figure 1.43d " " " EVALUATE: The vector addition diagram for R = A + − B is

( )

" R is in the 1st quadrant, with Rx < Ry , in agreement with our calculation.

Figure 1.43e (c) EXECUTE:

" " " " B − A= − A−B " " " " B − A and A − B are equal in magnitude and opposite in direction. R = 4.09 cm and θ = 83.7° + 180° = 264°

(

Figure 1.43f

)

1-11

1-12

Chapter 1

" " " EVALUATE: The vector addition diagram for R = B + − A is

( )

" R is in the 3rd quadrant, with Rx < Ry , in agreement with our calculation.

Figure 1.43g 1.44.

" " IDENTIFY: The velocity of the boat relative to the earth, vB/E , the velocity of the water relative to the earth, vW/E , " " " " and the velocity of the boat relative to the water, vB/W , are related by vB/E = vB/W + v W/E . " " SET UP: vW/E = 5.0 km/h , north and vB/W = 7.0 km/h , west. The vector addition diagram is sketched in Figure 1.44. v 5.0 km/h 2 2 2 EXECUTE: vB/E = vW/E + vB/W and vB/E = (5.0 km/h) 2 + (7.0 km/h) 2 = 8.6 km/h . tan φ = W/E = and vB/W 7.0 km/h

φ = 36° , north of west. EVALUATE: Since the two vectors we are adding are perpendicular we can use the Pythagorean theorem directly to find the magnitude of their vector sum.

Figure 1.44 1.45.

" " " " IDENTIFY: Let A = 625 N and B = 875 N . We are asked to find the vector C such that A + B = C = 0 . SET UP: Ax = 0 , Ay = −625 N . Bx = (875 N)cos30° = 758 N , By = (875 N)sin 30° = 438 N . EXECUTE:

C x = −( Ax + Bx ) = −(0 + 758 N) = −758 N . C y = −( Ay + By ) = −( −625 N + 438 N) = +187 N . Vector

" C y 187 N C and its components are sketched in Figure 1.45. C = Cx2 + C y2 = 781 N . tan φ = and φ = 13.9° . = C x 758 N " C is at an angle of 13.9° above the − x -axis and therefore at an angle 180° − 13.9° = 166.1° counterclockwise from the + x -axis . " " " EVALUATE: A vector addition diagram for A + B + C verifies that their sum is zero.

Figure 1.45


1.46.

1-13

IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of components. " " " SET UP: The two vectors A and B and their resultant C are shown in Figure 1.46. Let + y be in the direction of the resultant. A = B . EXECUTE: C y = Ay + By . 372 N = 2 A cos 43.0° and A = 254 N . EVALUATE: The sum of the magnitudes of the two forces exceeds the magnitude of the resultant force because only a component of each force is upward.

Figure 1.46 1.47.

IDENTIFY: Find the components of each vector and then use Eq.(1.14). SET UP: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m . Dx = −7.99 m ,

Dy = 6.02 m .

1.48.

" " " EXECUTE: A = (−8.00 m) ˆj ; B = (7.50 m)iˆ + (13.0 m) ˆj ; C = (−10.9 m) iˆ + (−5.07 m) ˆj ; " D = (−7.99 m)iˆ + (6.02 m) ˆj . EVALUATE: All these vectors lie in the xy-plane and have no z-component. " IDENTIFY: The general expression for a vector written in terms of components and unit vectors is A = Ax iˆ + Ay ˆj " " " SET UP: 5.0 B = 5.0(4iˆ − 6 ˆj ) = 20i − 30 j

EXECUTE: (a) Ax = 5.0 , Ay = −6.3 (b) Ax = 11.2 , Ay = −9.91 (c) Ax = −15.0 , Ay = 22.4 (d) Ax = 20 , Ay = −30 1.49.

EVALUATE: The components are signed scalars. IDENTIFY: Use trig to find the components of each vector. Use Eq.(1.11) to find the components of the vector sum. Eq.(1.14) expresses a vector in terms of its components. SET UP: Use the coordinates in the figure that accompanies the problem. " EXECUTE: (a) A = ( 3.60 m ) cos70.0°iˆ + ( 3.60 m ) sin 70.0° ˆj = (1.23 m ) iˆ + ( 3.38 m ) ˆj " B = − ( 2.40 m ) cos 30.0°iˆ − ( 2.40 m ) sin 30.0° ˆj = ( −2.08 m ) iˆ + ( −1.20 m ) ˆj " " " (b) C = ( 3.00 ) A − ( 4.00 ) B = ( 3.00 )(1.23 m ) iˆ + ( 3.00 )( 3.38 m ) ˆj − ( 4.00 )( −2.08 m ) iˆ − ( 4.00 )( −1.20 m ) ˆj = (12.01 m)iˆ + (14.94) ˆj (c) From Equations (1.7) and (1.8), 2 2 ⎛ 14.94 m ⎞ C = (12.01 m ) + (14.94 m ) = 19.17 m, arctan ⎜ ⎟ = 51.2° ⎝ 12.01 m ⎠

EVALUATE: 1.50.

C x and C y are both positive, so θ is in the first quadrant.

IDENTIFY: Find A and B. Find the vector difference using components. SET UP: Deduce the x- and y-components and use Eq.(1.8). " EXECUTE: (a) A = 4.00iˆ + 3.00 ˆj; Ax = +4.00; Ay = +3.00

A = Ax2 + Ay2 = (4.00) 2 + (3.00) 2 = 5.00

1-14

Chapter 1

" B = 5.00iˆ − 2.00 ˆj; Bx = +5.00; By = −2.00

B = Bx2 + By2 = (5.00) 2 + (−2.00) 2 = 5.39

" " EVALUATE: Note that the magnitudes of A and B are each larger than either of their components. " " EXECUTE: (b) A − B = 4.00iˆ + 3.00 ˆj − 5.00iˆ − 2.00 ˆj = (4.00 − 5.00) iˆ + (3.00 + 2.00) ˆj " " A − B = −1.00iˆ + 5.00 ˆj " " " (c) Let R = A − B = −1.00iˆ + 5.00 ˆj. Then Rx = −1.00, Ry = 5.00.

(

)

R = Rx2 + Ry2 R = ( −1.00) 2 + (5.00) 2 = 5.10. tan θ =

EVALUATE: 1.51.

1.52.

1.53.

5.00 = −5.00 Rx −1.00 θ = −78.7° + 180° = 101.3°. =

Figure 1.50 " Rx < 0 and Ry > 0, so R is in the 2nd quadrant.

IDENTIFY: A unit vector has magnitude equal to 1. SET UP: The magnitude of a vector is given in terms of its components by Eq.(1.12). EXECUTE: (a) iˆ + ˆj + kˆ = 12 + 12 + 12 = 3 ≠ 1 so it is not a unit vector.

" " (b) A = Ax2 + Ay2 + Az2 . If any component is greater than +1 or less than −1, A > 1 , so it cannot be a unit " vector. A can have negative components since the minus sign goes away when the component is squared. " 1 2 2 (c) A = 1 gives a 2 ( 3.0 ) + a 2 ( 4.0 ) = 1 and a 2 25 = 1 . a = ± = ±0.20 . 5.0 EVALUATE: The magnitude of a vector is greater than the magnitude of any of its components. " " " " " " IDENTIFY: If vectors A and B commute for addition, A + B = B + A . If they commute for the scalar product, " " " " A⋅ B = B ⋅ A . " " SET UP: Express the sum and scalar product in terms of the components of A and B . " " " " EXECUTE: (a) Let A = Ax iˆ + Ay ˆj and B = Bx iˆ + By ˆj . A + B = ( Ax + Bx ) iˆ + ( Ay + By ) ˆj . " " " " " " B + A = ( Bx + Ax ) iˆ + ( By + Ay ) ˆj . Scalar addition is commutative, so A + B = B + A . " " " " " " " " A ⋅ B = Ax Bx + Ay By and B ⋅ A = Bx Ax + By Ay . Scalar multiplication is commutative, so A ⋅ B = B ⋅ A . " " (b) A × B = ( Ay Bz − Az By ) iˆ + ( Az Bx − Ax Bz ) ˆj + ( Ax By − Ay Bx ) kˆ . " " B × A = ( By Az − Bz Ay ) iˆ + ( Bz Ax − Bx Az ) ˆj + ( Bx Ay − By Ax ) kˆ . Comparison of each component in each vector

product shows that one is the negative of the other. " " " " EVALUATE: The result in part (b) means that A × B and B × A have the same magnitude and opposite direction. " " IDENTIFY: A ⋅ B = AB cos φ " " " " " " SET UP: For A and B , φ = 150.0° . For B and C , φ = 145.0° . For A and C , φ = 65.0° . " " EXECUTE: (a) A ⋅ B = (8.00 m)(15.0 m)cos150.0° = −104 m 2 " " (b) B ⋅ C = (15.0 m)(12.0 m)cos145.0° = −148 m 2 " " (c) A ⋅ C = (8.00 m)(12.0 m)cos65.0° = 40.6 m 2 When φ < 90° the scalar product is positive and when φ > 90° the scalar product is negative. " " IDENTIFY: Target variables are A ⋅ B and the angle φ between the two vectors. " " SET UP: We are given A and B in unit vector form and can take the scalar product using Eq.(1.19). The angle φ can then be found from Eq.(1.18). EVALUATE:

1.54.

Ry


1.55.

EXECUTE: " " (a) A = 4.00iˆ + 3.00 ˆj , B = 5.00iˆ − 2.00 ˆj; A = 5.00, B = 5.39 " " A ⋅ B = (4.00iˆ + 3.00 ˆj ) ⋅ (5.00iˆ − 2.00 ˆj ) = (4.00)(5.00) + (3.00)( −2.00) = 20.0 − 6.0 = +14.0. " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector written in terms of unit vectors. " " ⎛ −22 ⎞ EXECUTE: (a) A ⋅ B = −22, A = 40, B = 13, and so φ = arccos ⎜ ⎟ = 165° . ⎝ 40 13 ⎠

" " (b) A ⋅ B = 60, A = 34, B = 136,

1.56.

1-15

⎛

60 ⎞ ⎟ = 28° . ⎝ 34 136 ⎠

φ = arccos ⎜

" " (c) A ⋅ B = 0 and φ = 90° . " " " " " " EVALUATE: If A ⋅ B > 0 , 0 ≤ φ < 90° . If A ⋅ B < 0 , 90° < φ ≤ 180° . If A ⋅ B = 0 , φ = 90° and the two vectors are perpendicular. " " " " " " IDENTIFY: A ⋅ B = AB cos φ and A × B = AB sin φ , where φ is the angle between A and B . " " " " " " SET UP: Figure 1.56 shows A and B . The components A% of A along B and A⊥ of A perpendicular to B are " " " " shown in Figure 1.56a. The components of B% of B along A and B⊥ of B perpendicular to A are shown in Figure 1.56b. " " EXECUTE: (a) From Figures 1.56a and b, A% = A cos φ and B% = B cos φ . A ⋅ B = AB cos φ = BA% = AB% . " " (b) A⊥ = A sin φ and B⊥ = B sin φ . A × B = AB sin φ = BA⊥ = AB⊥ . " " " " " EVALUATE: When A and B are perpendicular, A has no component along B and B has no component along " " " " " " " " A and A ⋅ B = 0 . When A and B are parallel, A has no component perpendicular to B and B has no component " " " perpendicular to A and A × B = 0 .

1.57.

Figure 1.56 " " IDENTIFY: A × D has magnitude AD sin φ . Its direction is given by the right-hand rule. SET UP: φ = 180° − 53° = 127° " " " " EXECUTE: A × D = (8.00 m)(10.0 m)sin127° = 63.9 m 2 . The right-hand rule says A × D is in the

− z -direction (into the page). EVALUATE: 1.58.

" " " " The component of D perpendicular to A is D⊥ = D sin 53.0° = 7.00 m . A × D = AD⊥ = 63.9 m 2 ,

which agrees with our previous result. " " IDENTIFY: Target variable is the vector A × B , expressed in terms of unit vectors. " " SET UP: We are given A and B in unit vector form and can take the vector product using Eq.(1.24). " " EXECUTE: A = 4.00iˆ + 3.00 ˆj , B = 5.00iˆ − 2.00 ˆj

1-16

Chapter 1

" " A × B = 4.00iˆ + 3.00 ˆj × 5.00iˆ − 2.00 ˆj = 20.0iˆ × iˆ − 8.00iˆ × ˆj + 15.0 ˆj × iˆ − 6.00 ˆj × ˆj " " But iˆ × iˆ = ˆj × ˆj = 0 and iˆ × ˆj = kˆ , ˆj × iˆ = − kˆ , so A × B = −8.00kˆ + 15.0 − kˆ = −23.0kˆ. " " The magnitude of A × B is 23.0. " " EVALUATE: Sketch the vectors A and B in a coordinate system where the xy-plane is in the plane of the paper and the z-axis is directed out toward you.

(

) (

)

( )

1.59.

Figure 1.58 " " By the right-hand rule A × B is directed into the plane of the paper, in the − z -direction. This agrees with the above calculation that used unit vectors. IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude. SET UP: φ = 120.0° . " " EXECUTE: (a) The direction of A× B is into the page (the − z -direction ). The magnitude of the vector product

is AB sin φ = ( 2.80 cm )(1.90 cm ) sin120# = 4.61 cm 2 .

" " (b) Rather than repeat the calculations, Eq. (1.23) may be used to see that B × A has magnitude 4.61 cm2 and is in the + z -direction (out of the page). EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is Cz = Ax By − Ay Bx = ( 2.80 cm ) cos 60.0° ( −1.90 cm ) sin 60° − ( 2.80 cm ) sin 60.0° (1.90 cm ) cos60.0° = −4.61 cm 2 .

1.60.

This gives the same result. IDENTIFY: Area is length times width. Do unit conversions. SET UP: 1 mi = 5280 ft . 1 ft 3 = 7.477 gal . EXECUTE:

(a) The area of one acre is

1 8

mi × 801 mi =

mi 2 , so there are 640 acres to a square mile.

1 640

2

1.61.

⎛ 1 mi 2 ⎞ ⎛ 5280 ft ⎞ 2 (b) (1 acre ) × ⎜ ⎟×⎜ ⎟ = 43,560 ft 640 acre 1 mi ⎝ ⎠ ⎝ ⎠ (all of the above conversions are exact). ⎛ 7.477 gal ⎞ 5 (c) (1 acre-foot) = ( 43,560 ft 3 ) × ⎜ ⎟ = 3.26 × 10 gal, which is rounded to three significant figures. 3 1 ft ⎝ ⎠ EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-foot is much larger than a gallon. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from this the radius. SET UP: The earth has mass mE = 5.97 × 1024 kg and radius rE = 6.38 × 106 m . The volume of a sphere is

V = 34 π r 3 . ρ = 1.76 g/cm3 = 1760 km/m3 . EXECUTE:

(a) The planet has mass m = 5.5mE = 3.28 × 1025 kg . V = 1/ 3

⎛ 3V ⎞ r =⎜ ⎟ ⎝ 4π ⎠

3

ρ

=

3.28 × 1025 kg = 1.86 × 1022 m3 . 1760 kg/m3

1/ 3

⎛ 3[1.86 × 10 m ] ⎞ =⎜ ⎟ 4π ⎝ ⎠ 22

m

= 1.64 × 107 m = 1.64 × 104 km

(b) r = 2.57 rE EVALUATE: Volume V is proportional to mass and radius r is proportional to V 1/ 3 , so r is proportional to m1/ 3 . If the planet and earth had the same density its radius would be (5.5)1/ 3 rE = 1.8rE . The radius of the planet is greater than this, so its density must be less than that of the earth.


1.62.

IDENTIFY and SET UP:

(b)

Unit conversion.

(a) f = 1.420 × 109 cycles/s, so

EXECUTE:

1-17

1 s = 7.04 × 10−10 s for one cycle. 1.420 × 109

3600 s/h = 5.11 × 1012 cycles/h 7.04 × 10−10 s/cycle

(c) Calculate the number of seconds in 4600 million years = 4.6 × 109 y and divide by the time for 1 cycle:

(4.6 × 109 y)(3.156 × 107 s/y) = 2.1 × 1026 cycles 7.04 × 10−10 s/cycle

1.63.

⎛ 4.60 × 109 ⎞ 4 (d) The clock is off by 1 s in 100,000 y = 1 × 105 y, so in 4.60 × 109 y it is off by (1 s) ⎜ ⎟ = 4.6 × 10 s 5 ⎝ 1 × 10 ⎠ (about 13 h). EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer. IDENTIFY: The number of atoms is your mass divided by the mass of one atom. SET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the mass of one H2O molecule: 18.015 u × 1.661 × 10227 kg/u = 2.992 × 10226 kg/molecule.

( 70 kg ) / ( 2.992 × 10226 kg/molecule ) = 2.34 × 1027

EXECUTE:

1.64.

molecules. Each H 2O molecule has 3 atoms, so

there are about 6 × 1027 atoms. EVALUATE: Assuming carbon to be the most common atom gives 3 × 1027 molecules, which is a result of the same order of magnitude. IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume. SET UP: The volume of a sphere of radius r is V = 34 π r 3 . The volume of a cylinder of radius r and length l is

V = π r 2l . The density of water is 1000 kg m3 . (a) Estimate the volume as that of a sphere of diameter 10 cm: V = 5.2 × 10−4 m3 .

EXECUTE:

m = ( 0.98 ) (1000 kg m3 )( 5.2 × 10−4 m3 ) = 0.5 kg .

(b) Approximate as a sphere of radius r = 0.25 μm (probably an over estimate): V = 6.5 × 10−20 m3 .

m = ( 0.98 ) (1000 kg m3 )( 6.5 × 10−20 m3 ) = 6 × 10−17 kg = 6 × 10−14 g . (c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V = π r 2l = 2.8 × 10−7 m3 .

m = ( 0.98 ) (1000 kg m3 )( 2.8 × 10−7 m3 ) = 3 × 10−4 kg = 0.3 g .

EVALUATE: 1.65.

IDENTIFY: SET UP: EXECUTE:

The mass is directly proportional to the volume. Use the volume V and density ρ to calculate the mass M: ρ =

M M ,so V = . V ρ

The volume of a cube with sides of length x is x3 . The volume of a sphere with radius R is 34 π R 3 . (a) x3 =

0.200 kg = 2.54 × 10−5 m3 . x = 2.94 × 10−2 m = 2.94 cm . 7.86 × 103 kg/m3

4 3 π R = 2.54 × 10−5 m 3 . R = 1.82 × 10−2 m = 1.82 cm . 3 EVALUATE: 34 π = 4.2 , so a sphere with radius R has a greater volume than a cube whose sides have length R. IDENTIFY: Estimate the volume of sand in all the beaches on the earth. The diameter of a grain of sand determines its volume. From the volume of one grain and the total volume of sand we can calculate the number of grains. SET UP: The volume of a sphere of diameter d is V = 16 π d 3 . Consulting an atlas, we estimate that the continents (b)

1.66.

have about 1.45 × 105 km of coastline. Add another 25% of this for rivers and lakes, giving 1.82 × 105 km of coastline. Assume that a beach extends 50 m beyond the water and that the sand is 2 m deep. 1 billion = 1 × 109 . EXECUTE: (a) The volume of sand is (1.82 × 108 m)(50 m)(2 m) = 2 × 1010 m3 . The volume of a grain is

V = 16 π (0.2 × 10−3 m)3 = 4 × 10−12 m 3 . The number of grains is

2 × 1010 m3 = 5 × 1021 . The number of grains of sand 4 × 10−12 m3

is about 1022 . (b) The number of stars is (100 × 109 )(100 × 109 ) = 1022 . The two estimates result in comparable numbers for these two quantities.

1-18

1.67.

Chapter 1

EVALUATE: Both numbers are crude estimates but are probably accurate to a few powers of 10. IDENTIFY: The number of particles is the total mass divided by the mass of one particle. SET UP: 1 mol = 6.0 × 1023 atoms . The mass of the earth is 6.0 × 1024 kg . The mass of the sun is 2.0 × 1030 kg .

The distance from the earth to the sun is 1.5 × 1011 m . The volume of a sphere of radius R is 34 π R 3 . Protons and neutrons each have a mass of 1.7 × 10−27 kg and the mass of an electron is much less. ⎛ 6.0 × 1023 atoms ⎞ 50 mole (a) (6.0 × 1024 kg) × ⎜ ⎟ = 2.6 × 10 atoms. −3 kg × 14 10 mole ⎠ ⎝ (b) The number of neutrons is the mass of the neutron star divided by the mass of a neutron: (2)(2.0 × 1030 kg) = 2.4 × 1057 neutrons. (1.7 × 10−27 kg neutron) EXECUTE:

(c) The average mass of a particle is essentially 32 the mass of either the proton or the neutron, 1.7 × 10−27 kg. The total number of particles is the total mass divided by this average, and the total mass is the volume times the average density. Denoting the density by ρ ,

1.68.

4 π R 3 ρ (2π )(1.5 × 1011 m)3 (1018 kg m3 ) M 3 = = = 1.2 × 1079. −27 2 mave 1.7 10 kg × mp 3 Note the conversion from g/cm3 to kg/m3. EVALUATE: These numbers of particles are each very, very large but are still much less than a googol. " " " " " " " " " " IDENTIFY: Let D be the fourth force. Find D such that A + B + C + D = 0 , so D = − A + B + C . " SET UP: Use components and solve for the components Dx and Dy of D .

(

EXECUTE:

)

Ax = + A cos30.0° = +86.6 N, Ay = + A cos30.0° = +50.00 N .

Bx = − B sin 30.0° = −40.00 N, By = + B cos30.0° = +69.28 N . Cx = +C cos53.0° = −24.07 N, C y = −C sin 53.0° = −31.90 N . Then Dx = −22.53 N , Dy = −87.34 N and D = Dx2 + Dy2 = 90.2 N . tan α = Dy / Dx = 87.34 / 22.53 . α = 75.54° .

φ = 180° + α = 256° , counterclockwise from the + x-axis. EVALUATE:

" As shown in Figure 1.68, since Dx and Dy are both negative, D must lie in the third quadrant.

Figure 1.68 1.69.

IDENTIFY: We know the magnitude and direction of the sum of the two vector pulls and the direction of one pull. We also know that one pull has twice the magnitude of the other. There are two unknowns, the magnitude of the smaller pull and its direction. Ax + Bx = Cx and Ay + By = C y give two equations for these two unknowns. " " " " " SET UP: Let the smaller pull be A and the larger pull be B . B = 2 A . C = A + B has magnitude 350.0 N and is northward. Let + x be east and + y be north. Bx = − B sin 25.0° and By = B cos 25.0° . C x = 0 , C y = 350.0 N . " " A must have an eastward component to cancel the westward component of B . There are then two possibilities, as " " sketched in Figures 1.69 a and b. A can have a northward component or A can have a southward component. EXECUTE: In either Figure 1.69 a or b, Ax + Bx = Cx and B = 2 A gives (2 A)sin 25.0° = A sin φ and φ = 57.7° . In

Figure 1.69a, Ay + By = C y gives 2 A cos 25.0° + A cos57.7° = 350.0 N and A = 149 N . In Figure 1.69b, 2 A cos 25.0° − A cos57.7° = 350.0 N and A = 274 N . One solution is for the smaller pull to be 57.7° east of north. In this case, the smaller pull is 149 N and the larger pull is 298 N. The other solution is for the smaller pull to be 57.7° east of south. In this case the smaller pull is 274 N and the larger pull is 548 N.


1-19

" EVALUATE: For the first solution, with A east of north, each worker has to exert less force to produce the given resultant force and this is the sensible direction for the worker to pull.

Figure 1.69 1.70.

IDENTIFY: Find the vector sum of the two displacements. " " " " " " " SET UP: Call the two displacements A and B , where A = 170 km and B = 230 km . A + B = R . A and B are as shown in Figure 1.70. EXECUTE: Rx = Ax + Bx = (170 km) sin 68° + (230 km) cos 48° = 311.5 km .

Ry = Ay + By = (170 km) cos 68° − (230 km) sin 48° = −107.2 km . R = Rx2 + Ry2 =

( 311.5 km ) + ( −107.2 km ) 2

2

= 330 km . tanθR =

Ry Rx

=

107.2 km = 0.344 . 311.5 km

θR = 19° south of east . EVALUATE:

1.71.

" Our calculation using components agrees with R shown in the vector addition diagram, Figure 1.70.

Figure 1.70 " " " " " " " IDENTIFY: A + B = C (or B + A = C ). The target variable is vector A. " SET UP: Use components and Eq.(1.10) to solve for the components of A. Find the magnitude and direction of " A from its components. EXECUTE: (a) Cx = Ax + Bx , so Ax = Cx − Bx

C y = Ay + By , so Ay = C y − By Cx = C cos 22.0° = (6.40 cm)cos 22.0° C x = +5.934 cm C y = C sin 22.0° = (6.40 cm)sin 22.0° C y = +2.397 cm Bx = B cos(360° − 63.0°) = (6.40 cm)cos 297.0° Bx = +2.906 cm By = B sin 297.0° = (6.40 cm)sin 297.0° By = −5.702 cm Figure 1.71a (b) Ax = Cx − Bx = +5.934 cm − 2.906 cm = +3.03 cm

Ay = C y − By = +2.397 cm − ( −5.702) cm = +8.10 cm

1-20

Chapter 1

(c)

A = Ax2 + Ay2 A = (3.03 cm)2 + (8.10 cm) 2 = 8.65 cm tan θ =

Ay

Ax θ = 69.5°

1.72.

=

8.10 cm = 2.67 3.03 cm

Figure 1.71b " " EVALUATE: The A we calculated agrees qualitatively with vector A in the vector addition diagram in part (a). IDENTIFY: Add the vectors using the method of components. SET UP: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m . EXECUTE:

(a) Rx = Ax + Bx + C x = −3.4 m . Ry = Ay + By + C y = −0.07 m . R = 3.4 m . tan θ =

−0.07 m . −3.4 m

θ = 1.2° below the − x-axis . (b) S x = Cx − Ax − Bx = −18.4 m . S y = C y − Ay − By = −10.1 m . S = 21.0 m . tan θ =

Sy

=

−10.1 m . θ = 28.8° −18.4 m

Sx below the − x-axis . " " EVALUATE: The magnitude and direction we calculated for R and S agree with our vector diagrams.

Figure 1.72 1.73.

IDENTIFY: Vector addition. Target variable is the 4th displacement. SET UP: Use a coordinate system where east is in the + x -direction and north is in the + y -direction. " " " " Let A, B, and C be the three displacements that are given and let D be the fourth unmeasured displacement. " " " " " " Then the resultant displacement is R = A + B + C + D. And since she ends up back where she started, R = 0. " " " " " " " " 0 = A + B + C + D, so D = − A + B + C

(

)

Dx = −( Ax + Bx + Cx ) and Dy = −( Ay + By + C y ) EXECUTE:

Ax = −180 m, Ay = 0 Bx = B cos315° = (210 m)cos315° = +148.5 m By = B sin 315° = (210 m)sin 315° = −148.5 m Cx = C cos 60° = (280 m)cos60° = +140 m C y = C sin 60° = (280 m)sin 60° = +242.5 m Figure 1.73a

Dx = −( Ax + Bx + Cx ) = −( −180 m + 148.5 m + 140 m) = −108.5 m


1-21

Dy = −( Ay + By + C y ) = −(0 − 148.5 m + 242.5 m) = −94.0 m D = Dx2 + Dy2 D = (−108.5 m) 2 + (−94.0 m) 2 = 144 m −94.0 m = 0.8664 Dx −108.5 m θ = 180° + 40.9° = 220.9° " ( D is in the third quadrant since both Dx and Dy are negative.) tan θ =

Dy

=

Figure 1.73b " " The direction of D can also be specified in terms of φ = θ − 180° = 40.9°; D is 41° south of west. EVALUATE: The vector addition diagram, approximately to scale, is

" Vector D in this diagram agrees qualitatively with our calculation using components.

Figure 1.73c 1.74.

IDENTIFY: Solve for one of the vectors in the vector sum. Use components. SET UP: Use coordinates for which + x is east and + y is north. The vector displacements are: $ $ $ A = 2.00 km, 0°of east; B = 3.50 m, 45° south of east; and R = 5.80 m, 0° east EXECUTE: Cx = Rx − Ax − Bx = 5.80 km − ( 2.00 km ) − ( 3.50 km )( cos 45° ) = 1.33 km ; C y = Ry − Ay − By 2 2 = 0 km − 0 km − ( −3.50 km )( sin 45° ) = 2.47 km ; C = (1.33 km ) + ( 2.47 km ) = 2.81 km ; θ = tan −1 ⎡⎣( 2.47 km ) (1.33 km )⎤⎦ = 61.7° north of east. The vector addition diagram in Figure 1.74 shows good

qualitative agreement with these values. EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive.

Figure 1.74 1.75.

IDENTIFY: The sum of the vector forces on the beam sum to zero, so their x components and their y components " sum to zero. Solve for the components of F . SET UP: The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The " 100-N pull makes an angle of 30.0° + 40.0° = 70.0° with the horizontal. F and the 100-N pull have been replaced by their x and y components. EXECUTE: (a) The sum of the x-components is equal to zero gives Fx + (100 N)cos70.0° = 0 and Fx = −34.2 N . " The sum of the y-components is equal to zero gives Fy + (100 N)sin 70.0° − 124 N = 0 and Fy = +30.0 N . F and

its components are sketched in Figure 1.75b. F = Fx2 + Fy2 = 45.5 N . tan φ =

Fy Fx

=

" 30.0 N and φ = 41.3° . F is 34.2 N

directed at 41.3° above the − x -axis in Figure 1.75a. " (b) The vector addition diagram is given in Figure 1.75c. F determined from the diagram agrees with " F calculated in part (a) using components.

1-22

Chapter 1

" EVALUATE: The vertical component of the 100 N pull is less than the 124 N weight so F must have an upward component if all three forces balance.

Figure 1.75 1.76.

" " " " " " IDENTIFY: The four displacements return her to her starting point, so D = − ( A + B + C ) , where A , B and " " C are in the three given displacements and D is the displacement for her return. START UP: Let + x be east and + y be north. EXECUTE:

(a) Dx = −[(147 km ) sin 85° + (106 km ) sin167° + (166 km ) sin 235°] = −34.3 km .

Dy = −[(147 km ) cos85° + (106 km ) cos167° + (166 km ) cos 235°] = +185.7 km . D = (−34.3 km) 2 + (185.7 km) 2 = 189 km .

1.77.

" ⎛ 34.3 km ⎞ (b) The direction relative to north is φ = arctan ⎜ ⎟ = 10.5° . Since Dx < 0 and Dy > 0 , the direction of D ⎝ 185.7 km ⎠ is 10.5° west of north. EVALUATE: The four displacements add to zero. " IDENTIFY and SET UP: The vector A that connects points ( x1 , y1 ) and ( x2 , y2 ) has components Ax = x2 − x1 and Ay = y2 − y1 . ⎛ 200 − 20 ⎞ (a) Angle of first line is θ = tan −1 ⎜ ⎟ = 42°. Angle of second line is 42° + 30° = 72°. ⎝ 210 − 10 ⎠ Therefore X = 10 + 250 cos 72° = 87 , Y = 20 + 250 sin 72° = 258 for a final point of (87,258). (b) The computer screen now looks something like Figure 1.77. The length of the bottom line is 258 − 200 ⎞ 2 2 ( 210 − 87 ) + ( 200 − 258) = 136 and its direction is tan −1 ⎛⎜ ⎟ = 25° below straight left. ⎝ 210 − 87 ⎠ EVALUATE: Figure 1.77 is a vector addition diagram. The vector first line plus the vector arrow gives the vector for the second line. EXECUTE:

Figure 1.77


1.78.

1-23

" " " IDENTIFY: Let the three given displacements be A , B and C , where A = 40 steps , B = 80 steps and " " " " " " C = 50 steps . R = A + B + C . The displacement C that will return him to his hut is − R . SET UP: Let the east direction be the + x -direction and the north direction be the + y -direction. EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.78. (b) Rx = ( 40 ) cos 45° − ( 80 ) cos 60° = −11.7 and Ry = ( 40 ) sin 45° + ( 80 ) sin 60° − 50 = 47.6. The magnitude and direction of the resultant are

⎛ 47.6 ⎞ (−11.7) 2 + (47.6) 2 = 49, arctan ⎜ ⎟ = 76° , north of west. ⎝ 11.7 ⎠

" We know that R is in the second quadrant because Rx < 0 , Ry > 0 . To return to the hut, the explorer must take 49 steps in a direction 76° south of east, which is 14° east of south. " EVALUATE: It is useful to show Rx , Ry and R on a sketch, so we can specify what angle we are computing.

Figure 1.78 1.79.

IDENTIFY: Vector addition. One vector and the sum are given; find the second vector (magnitude and direction). " " SET UP: Let + x be east and + y be north. Let A be the displacement 285 km at 40.0° north of west and let B be the unknown displacement. " " " " A + B = R where R = 115 km, east " " " B = R− A Bx = Rx − Ax , By = Ry − Ay EXECUTE:

Ax = − A cos 40.0° = −218.3 km, Ay = + A sin 40.0° = +183.2 km

Rx = 115 km, Ry = 0 Then Bx = 333.3 km, By = −183.2 km. B = Bx2 + By2 = 380 km;

tan α = By / Bx = (183.2 km)/(333.3 km)

α = 28.8°, south of east Figure 1.79

1.80.

" " EVALUATE: The southward component of B cancels the northward component of A. The eastward component " " of B must be 115 km larger than the magnitude of the westward component of A. IDENTIFY: Find the components of the weight force, using the specified coordinate directions. SET UP: For parts (a) and (b), take + x direction along the hillside and the + y direction in the downward direction and perpendicular to the hillside. For part (c), α = 35.0° and w = 550 N . EXECUTE: (a) wx = w sin α (b) wy = w cos α (c) The maximum allowable weight is w = wx ( sin α ) = ( 550 N ) ( sin35.0° ) = 959 N . EVALUATE: The component parallel to the hill increases as α increases and the component perpendicular to the hill increases as α decreases.

1-24

1.81.

Chapter 1

IDENTIFY: Vector addition. One force and the vector sum are given; find the second force. SET UP: Use components. Let + y be upward.

" B is the force the biceps exerts.

Figure 1.81a " " " " E is the force the elbow exerts. E + B = R, where R = 132.5 N and is upward. Ex = Rx − Bx , E y = Ry − By EXECUTE:

Bx = − B sin 43° = −158.2 N, By = + B cos 43° = +169.7 N, Rx = 0, Ry = +132.5 N

Then Ex = +158.2 N, E y = −37.2 N E = Ex2 + E y2 = 160 N;

tan α = E y / Ex = 37.2 /158.2

α = 13°, below horizontal Figure 1.81b " " EVALUATE: The x-component of E cancels the x-component of B. The resultant upward force is less than the " upward component of B, so E y must be downward. 1.82.

IDENTIFY: Find the vector sum of the four displacements. SET UP: Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction, and the z-axis vertical. The first displacement is then (−30 m) kˆ , the second is (−15 m) ˆj , the third is (200 m) iˆ,

and the fourth is (100 m) ˆj. EXECUTE: (a) Adding the four displacements gives (−30 m) kˆ + ( −15 m) ˆj + (200 m) iˆ + (100 m) ˆj = (200 m) iˆ + (85 m) ˆj − (30 m) kˆ. (b) The total distance traveled is the sum of the distances of the individual segments: 30 m + 15 m + 200 m + 100 m = 345 m. The magnitude of the total displacement is: D = Dx2 + Dy2 + Dz2 = (200 m) 2 + (85 m) 2 + ( −30 m ) = 219 m. 2

1.83.

EVALUATE: The magnitude of the displacement is much less than the distance traveled along the path. IDENTIFY: The sum of the force displacements must be zero. Use components. " " " " " SET UP: Call the displacements A , B , C and D , where D is the final unknown displacement for the return " " " " " " " from the treasure to the oak tree. Vectors A , B , and C are sketched in Figure 1.83a. A + B + C + D = 0 says Ax + Bx + Cx + Dx = 0 and Ay + By + C y + Dy = 0 . A = 825 m , B = 1250 m , and C = 1000 m . Let + x be eastward

and + y be north. EXECUTE:

(a) Ax + Bx + Cx + Dx = 0 gives Dx = −( Ax + Bx + Cx ) = −(0 − [1250 m]sin30.0° +[1000 m]cos40.0°) = −141 m .

Ay + By + C y + Dy = 0 gives Dy = −( Ay + By + C y ) = −( −825 m + [1250 m]cos30.0° + [1000 m]sin 40.0°) = −900 m . " The fourth displacement D and its components are sketched in Figure 1.83b. D = Dx2 + Dy2 = 911 m . tan φ =

Dx Dy

=

141 m and φ = 8.9° . You should head 8.9° west of south and must walk 911 m. 900 m


1-25

" (b) The vector diagram is sketched in Figure 1.83c. The final displacement D from this diagram agrees with the " vector D calculated in part (a) using components. " " " " EVALUATE: Note that D is the negative of the sum of A , B , and C .

1.84.

Figure 1.83 " " IDENTIFY: If the vector from your tent to Joe’s is A and from your tent to Karl’s is B , then the vector from " " Joe’s tent to Karl’s is B − A . SET UP: Take your tent's position as the origin. Let + x be east and + y be north. EXECUTE: The position vector for Joe’s tent is ([21.0 m]cos 23° ) iˆ − ([21.0 m]sin 23° ) ˆj = (19.33 m) iˆ − (8.205 m) ˆj.

The position vector for Karl's tent is ([32.0 m]cos 37° ) iˆ + ([32.0 m]sin 37° ) ˆj = (25.56 m) iˆ + (19.26 m) ˆj. The difference between the two positions is (19.33 m − 25.56 m ) iˆ + ( −8.205 m − 19.25 m ) ˆj = −(6.23 m)iˆ − (27.46 m) ˆj. The magnitude of this vector is the distance between the two tents: D =

1.85.

( −6.23 m )

2

+ ( −27.46 m ) = 28.2 m 2

EVALUATE: If both tents were due east of yours, the distance between them would be 32.0 m − 21.0 m = 17.0 m . If Joe’s was due north of yours and Karl’s was due south of yours, then the distance between them would be 32.0 m + 21.0 m = 53.0 m . The actual distance between them lies between these limiting values. " " IDENTIFY: In Eqs.(1.21) and (1.27) write the components of A and B in terms of A, B, θ A and θ B . SET UP: From Appendix B, cos(a − b) = cos a cos b + sin a sin b and sin(a − b) = sin a cos b − cos a sin b . EXECUTE:

(a) With Az = Bz = 0 , Eq.(1.21) becomes

Ax Bx + Ay By = ( A cos θ A )( B cos θB ) + ( A sin θ A )( B sin θB ) Ax Bx + Ay By = AB ( cos θ Acos θB + sin θ Asin θB ) = AB cos ( θ A − θB ) = AB cos φ , where the expression for the cosine of the difference between two angles has been used. " (b) With Az = Bz = 0 , C = Cz kˆ and C = C z . From Eq.(1.27), C = Ax By − Ay Bx = ( A cos θ A )( B sin θB ) − ( A sin θ A )( B cos θ A ) C = AB cos θ A sin θB − sin θ A cos θB = AB sin ( θB − θ A ) = AB sin φ , where the expression for the sine of the

1.86.

difference between two angles has been used. EVALUATE: Since they are equivalent, we may use either Eq.(1.18) or (1.21) for the scalar product and either (1.22) or (1.27) for the vector product, depending on which is the more convenient in a given application. IDENTIFY: Apply Eqs.(1.18) and (1.22). SET UP: The angle between the vectors is 20° + 90° + 30° = 140°. " " EXECUTE: (a) Eq. (1.18) gives A ⋅ B = ( 3.60 m )( 2.40 m ) cos 140° = −6.62 m 2 . (b) From Eq.(1.22), the magnitude of the cross product is ( 3.60 m )( 2.40 m ) sin 140° = 5.55 m 2 and the direction,

from the right-hand rule, is out of the page (the + z -direction ). " " EVALUATE: We could also use Eqs.(1.21) and (1.27), with the components of A and B .

1-26

1.87.

1.88.

Chapter 1

Compare the magnitude of the cross product, AB sin φ , to the area of the parallelogram. " " SET UP: The two sides of the parallelogram have lengths A and B. φ is the angle between A and B . EXECUTE: (a) The length of the base is B and the height of the parallelogram is A sin φ , so the area is AB sin φ . This equals the magnitude of the cross product. " " " " (b) The cross product A × B is perpendicular to the plane formed by A and B , so the angle is 90° . EVALUATE: It is useful to consider the special cases φ = 0° , where the area is zero, and φ = 90° , where the parallelogram becomes a rectangle and the area is AB. IDENTIFY: Use Eq.(1.27) for the components of the vector product. SET UP: Use coordinates with the + x -axis to the right, + y -axis toward the top of the page, and + z -axis out of IDENTIFY:

the page. Ax = 0 , Ay = 0 and Az = −3.50 cm . The page is 20 cm by 35 cm, so Bx = 20 cm and By = 35 cm . " " " " " " EXECUTE: A × B = 122 cm 2 , A × B = −70 cm 2 , A × B = 0.

(

1.89.

)

(

x

)

y

(

)

z

EVALUATE: From the components we calculated the magnitude of the vector product is 141 cm 2 . B = 40.3 cm and φ = 90° , so AB sin φ = 141 cm 2 , which agrees. " " " " IDENTIFY: A and B are given in unit vector form. Find A, B and the vector difference A − B. " " " " " " " " SET UP: A = −2.00i + 3.00 j + 4.00k , B = 3.00i + 1.00 j − 3.00k Use Eq.(1.8) to find the magnitudes of the vectors. EXECUTE:

(a) A = Ax2 + Ay2 + Az2 = (−2.00) 2 + (3.00) 2 + (4.00) 2 = 5.38

B = Bx2 + By2 + Bz2 = (3.00) 2 + (1.00) 2 + (−3.00) 2 = 4.36 " " (b) A − B = ( −2.00iˆ + 3.00 ˆj + 4.00kˆ ) − (3.00iˆ + 1.00 ˆj − 3.00kˆ ) " " A − B = ( −2.00 − 3.00) iˆ + (3.00 − 1.00) ˆj + (4.00 − ( −3.00))kˆ = −5.00iˆ + 2.00 ˆj + 7.00kˆ. " " " (c) Let C = A − B, so Cx = −5.00, C y = +2.00, Cz = +7.00

1.90.

1.91.

1.92.

C = Cx2 + C y2 + C z2 = (−5.00) 2 + (2.00) 2 + (7.00) 2 = 8.83 " " " " " " " " B − A = −( A − B ), so A − B and B − A have the same magnitude but opposite directions. EVALUATE: A, B and C are each larger than any of their components. IDENTIFY: Calculate the scalar product and use Eq.(1.18) to determine φ . SET UP: The unit vectors are perpendicular to each other. EXECUTE: The direction vectors each have magnitude 3 , and their scalar product is (1)(1) + (1)( −1) + (1)( −1) = 21, so from Eq. (1.18) the angle between the bonds is ⎛ −1 ⎞ ⎛ 1⎞ arccos ⎜ ⎟ = arccos ⎜ − 3 ⎟ = 109°. 3 3 ⎝ ⎠ ⎝ ⎠ EVALUATE: The angle between the two vectors in the bond directions is greater than 90° . IDENTIFY: Use the relation derived in part (a) of Problem 1.92: C 2 = A2 + B 2 + 2 AB cosφ , where φ is the angle " " between A and B . SET UP: cos φ = 0 for φ = 90° . cos φ < 0 for 90° < φ < 180° and cos φ > 0 for 0° < φ < 90° . " " EXECUTE: (a) If C 2 = A2 + B 2 , cos φ = 0, and the angle between A and B is 90° (the vectors are perpendicular). " " (b) If C 2 < A2 + B 2 , cosφ < 0, and the angle between A and B is greater than 90° . " " (c) If C 2 > A2 + B 2 , cosφ > 0, and the angle between A and B is less than 90°. EVALUATE: It is easy to verify the expression from Problem 1.92 for the special cases φ = 0 , where C = A + B , and for φ = 180° , where C = A − B . " " " " " IDENTIFY: Let C = A + B and calculate the scalar product C ⋅ C . " " " " " SET UP: For any vector V , V ⋅V = V 2 . A ⋅ B = AB cos φ . " " EXECUTE: (a) Use the linearity of the dot product to show that the square of the magnitude of the sum A + B is " " " " " " " " " " " " " " " " " " " " A+ B ⋅ A+ B = A ⋅ A + A ⋅ B + B ⋅ A + B ⋅ B = A ⋅ A + B ⋅ B + 2 A ⋅ B = A2 + B 2 + 2 A ⋅ B

(

)(

)

= A2 + B 2 + 2 AB cos φ


1-27

(b) Using the result of part (a), with A = B, the condition is that A2 = A2 + A2 + 2 A2cos φ , which solves for 1 = 2 + 2cos φ , cos φ = − 12 , and φ = 120°. 1.93.

EVALUATE: The expression C 2 = A2 + B 2 + 2 AB cos φ is called the law of cosines. IDENTIFY: Find the angle between specified pairs of vectors. " " A⋅ B SET UP: Use cos φ = AB " ˆ EXECUTE: (a) A = k (along line ab) " B = iˆ + ˆj + kˆ (along line ad )

A = 1, B = 12 + 12 + 12 = 3 " " A ⋅ B = kˆ ⋅ iˆ + ˆj + kˆ = 1 " " A⋅ B = 1/ 3; φ = 54.7° So cos φ = AB " (b) A = iˆ + ˆj + kˆ (along line ad ) " B = ˆj + kˆ (along line ac)

(

A = 12 + 12 + 12 = 3; B = 12 + 12 = 2 " " A ⋅ B = iˆ + ˆj + kˆ ⋅ iˆ + ˆj = 1 + 1 = 2 " " A⋅ B 2 2 So cos φ = = = ; φ = 35.3° AB 3 2 6 EVALUATE: Each angle is computed to be less than 90°, in agreement with what is deduced from Fig. 1.43 in the textbook. " " " " IDENTIFY: The cross product A × B is perpendicular to both A and B . " " SET UP: Use Eq.(1.27) to calculate the components of A × B . EXECUTE: The cross product is ⎡ ⎛ 6.00 ⎞ ˆ 11.00 ˆ ⎤ (−13.00) iˆ + (6.00) ˆj + (−11.00) kˆ = 13 ⎢ −(1.00) iˆ + ⎜ k ⎥ . The magnitude of the vector in square ⎟ j− ⎝ 13.00 ⎠ 13.00 ⎦ ⎣

(

1.94.

)

brackets is

)(

)

1.93, and so a unit vector in this direction is ⎡ −(1.00) iˆ + (6.00 /13.00) ˆj − (11.00 /13.00) kˆ ⎤ ⎢ ⎥. 1.93 ⎣⎢ ⎦⎥

The negative of this vector,

1.95.

1.96.

⎡ (1.00) iˆ − (6.00 /13.00) ˆj + (11.00 /13.00) kˆ ⎤ ⎢ ⎥, 1.93 ⎣⎢ ⎦⎥ " " is also a unit vector perpendicular to A and B . EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular to both vectors is perpendicular to this plane. " " " IDENTIFY and SET UP: The target variables are the components of C . We are given A and B. We also know " " " " A ⋅ C and B ⋅ C , and this gives us two equations in the two unknowns C x and C y . " " " " EXECUTE: A and C are perpendicular, so A ⋅ C = 0. AxC x + AyC y = 0, which gives 5.0C x − 6.5C y = 0. " " B ⋅ C = 15.0, so −3.5Cx + 7.0C y = 15.0 We have two equations in two unknowns C x and C y . Solving gives Cx = 8.0 and C y = 6.1 " " " EVALUATE: We can check that our result does give us a vector C that satisfies the two equations A ⋅ C = 0 and " " B ⋅ C = 15.0. IDENTIFY: Calculate the magnitude of the vector product and then use Eq.(1.22). SET UP: The magnitude of a vector is related to its components by Eq.(1.12).

1-28

Chapter 1

EXECUTE:

" " A× B " " A × B = AB sinθ . sinθ = = AB

( −5.00 ) + ( 2.00 ) ( 3.00 )( 3.00 ) 2

2

= 0.5984 and

θ = sin −1 ( 0.5984 ) = 36.8°.

1.97.

" " EVALUATE: We haven't found A and B , just the angle between them. " " " " " " (a) IDENTIFY: Prove that A ⋅ B × C = A × B ⋅ C .

(

) (

)

SET UP: Express the scalar and vector products in terms of components. EXECUTE: " " " " " " " " " A ⋅ B × C = Ax B × C + Ay B × C + Az B × C

(

)

(

)

(

x

)

(

y

)

z

" " " A ⋅ B × C = Ax ( By Cz − BzC y ) + Ay ( BzC x − BxCz ) + Az ( BxC y − By Cx )

(

)

"

"

"

"

"

"

"

( A× B) ⋅C = ( A× B) C + ( A× B) x

x

y

" " C y + A × B Cz

(

)

z

" " " A × B ⋅ C = ( Ay Bz − Az By ) Cx + ( Az Bx − Ax Bz ) C y + ( Ax By − Ay Bx ) Cz " " " " " " Comparison of the expressions for A ⋅ B × C and A × B ⋅ C shows they contain the same terms, so " " " " " " A ⋅ B × C = A × B ⋅ C. " " " " " " (b) IDENTIFY: Calculate A× B ⋅ C , given the magnitude and direction of A, B, and C . " " " " SET UP: Use Eq.(1.22) to find the magnitude and direction of A × B. Then we know the components of A × B " and of C and can use an expression like Eq.(1.21) to find the scalar product in terms of components. EXECUTE: A = 5.00; θ A = 26.0°; B = 4.00, θ B = 63.0° " " A × B = AB sin φ . " " The angle φ between A and B is equal to φ = θ B − θ A = 63.0° − 26.0° = 37.0°. So " " " " A × B = (5.00)(4.00)sin 37.0° = 12.04, and by the right hand-rule A × B is in the + z -direction. Thus " " " A × B ⋅ C = (12.04)(6.00) = 72.2 " " " " " " EVALUATE: A × B is a vector, so taking its scalar product with C is a legitimate vector operation. A × B ⋅ C

(

(

(

1.98.

) (

)

)

(

(

)

(

)

)

)

(

)

is a scalar product between two vectors so the result is a scalar. IDENTIFY: Use the maximum and minimum values of the dimensions to find the maximum and minimum areas and volumes. SET UP: For a rectangle of width W and length L the area is LW. For a rectangular solid with dimensions L, W and H the volume is LWH. EXECUTE: (a) The maximum and minimum areas are ( L + l )(W + w ) = LW + lW + Lw,

( L − l )(W − w ) = LW − lW − Lw, where the common terms wl have been omitted. The area and its uncertainty are then WL ± (lW + Lw), so the uncertainty in the area is a = lW + Lw. (b) The fractional uncertainty in the area is

1.99.

a lW + Wl l w = = + , the sum of the fractional uncertainties in the A WL L W

length and width. (c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the v lWH + LwH + LWh l w h volume is = = + + , the sum of the fractional uncertainties in the length, width and V LWH L W H height. EVALUATE: The calculation assumes the uncertainties are small, so that terms involving products of two or more uncertainties can be neglected. IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to the receiver. SET UP: Add the x-components and the y-components.


1-29

EXECUTE: The receiver's position is [( +1.0 + 9.0 − 6.0 + 12.0 ) yd]iˆ + [( −5.0 + 11.0 + 4.0 + 18.0 ) yd] ˆj = (16.0 yd ) iˆ + ( 28.0 yd ) ˆj .

The vector from the quarterback to the receiver is the receiver's position minus the quarterback's position, or 2 2 (16.0 yd ) iˆ + ( 35.0 yd ) ˆj , a vector with magnitude (16.0 yd ) + ( 35.0 yd ) = 38.5 yd . The angle is

1.100.

⎛ 16.0 ⎞ arctan ⎜ ⎟ = 24.6° to the right of downfield. ⎝ 35.0 ⎠ EVALUATE: The vector from the quarterback to receiver has positive x-component and positive y-component. IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the distance between two objects is the magnitude of this vector. Use the scalar product to find the angle between two vectors. " SET UP: If object A has coordinates ( x A , y A ) and object B has coordinates ( xB , yB ) , the vector rAB from A to B has x-component xB − x A and y-component yB − y A . EXECUTE: (a) The diagram is sketched in Figure 1.100. (b) (i) In AU,

(0.3182) 2 + (0.9329) 2 = 0.9857.

(ii) In AU, (1.3087) 2 + ( −0.4423) 2 + (−0.0414) 2 = 1.3820. (iii) In AU (0.3182 − 1.3087) 2 + (0.9329 − (−0.4423))2 + (0.0414)2 = 1.695. (c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1.18) and (1.21), ⎛ (−0.3182)(1.3087 − 0.3182) + ( −0.9329)( −0.4423 − 0.9329) + (0) ⎞ ⎟ = 54.6°. (0.9857)(1.695) ⎝ ⎠

φ = arccos ⎜

(d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o. EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that on this date Mars was farther from the earth than it is from the Sun.

Figure 1.100 1.101.

IDENTIFY: Draw the vector addition diagram for the position vectors. " SET UP: Use coordinates in which the Sun to Merak line lies along the x-axis. Let A be the position vector of " " Alkaid relative to the Sun, M is the position vector of Merak relative to the Sun, and R is the position vector for Alkaid relative to Merak. A = 138 ly and M = 77 ly . " " " EXECUTE: The relative positions are shown in Figure 1.101. M + R = A . Ax = M x + Rx so

Rx = Ax − M x = (138 ly)cos 25.6° − 77 ly = 47.5 ly . Ry = Ay − M y = (138 ly)sin 25.6° − 0 = 59.6 ly . R = 76.2 ly is the distance between Alkaid and Merak. Rx 47.5 ly = and θ = 51.4° . Then φ = 180° − θ = 129° . R 76.2 ly The concepts of vector addition and components make these calculations very simple.

(b) The angle is angle φ in Figure 1.101. cosθ = EVALUATE:

Figure 1.101

1-30

1.102.

Chapter 1

" " " IDENTIFY: Define S = Aiˆ + Bˆj + Ckˆ . Show that r ⋅ S = 0 if Ax + By + Cz = 0 . SET UP: Use Eq.(1.21) to calculate the scalar product. " " EXECUTE: r ⋅ S = ( xiˆ + yˆj + zkˆ ) ⋅ ( Aiˆ + Bˆj + Ckˆ ) = Ax + By + Cz " " " " If the points satisfy Ax + By + Cz = 0, then r ⋅ S = 0 and all points r are perpendicular to S . The vector and plane are sketched in Figure 1.102. EVALUATE: If two vectors are perpendicular their scalar product is zero.

Figure 1.102

2

MOTION ALONG A STRAIGHT LINE

2.1.

IDENTIFY:

The average velocity is vav-x =

Δx . Δt

Let + x be upward. 1000 m − 63 m EXECUTE: (a) vav-x = = 197 m/s 4.75 s 1000 m − 0 (b) vav-x = = 169 m/s 5.90 s

SET UP:

63 m − 0 = 54.8 m/s . When the velocity isn’t constant the 1.15 s average velocity depends on the time interval chosen. In this motion the velocity is increasing. Δx IDENTIFY: vav-x = Δt SET UP: 13.5 days = 1.166 × 105 s . At the release point, x = +5.150 × 106 m . EVALUATE:

2.2.

x2 − x1 5.150 × 106 m = = −4.42 m/s 1.166 × 106 s Δt (b) For the round trip, x2 = x1 and Δx = 0 . The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. IDENTIFY: Target variable is the time Δt it takes to make the trip in heavy traffic. Use Eq.(2.2) that relates the average velocity to the displacement and average time. Δx Δx so Δx = vav-x Δt and Δt = . SET UP: vav-x = Δt vav-x EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities: EXECUTE:

2.3.

For the first 1.15 s of the flight, vav-x =

(a) vav-x =

Δx = vav-x Δt = (105 km/h)(1 h/60 min)(140 min) = 245 km. Now use vav-x for heavy traffic to calculate Δt ; Δx is the same as before:

Δt =

Δx 245 km = = 3.50 h = 3 h and 30 min. vav-x 70 km/h

The trip takes an additional 1 hour and 10 minutes. EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is (105/ 70)(140 m) = 210 min. 2.4.

Δx . Use the average speed for each segment to find the time traveled Δt in that segment. The average speed is the distance traveled by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m + 280 m = 480 m . 200 m 280 m EXECUTE: (a) The eastward run takes time = 40.0 s and the westward run takes = 70.0 s . The 5.0 m/s 4.0 m/s 480 m = 4.4 m/s . average speed for the entire trip is 110.0 s Δx −80 m (b) vav-x = = = −0.73 m/s . The average velocity is directed westward. Δt 110.0 s IDENTIFY:


2-1

2-2

2.5.

2.6.

Chapter 2

EVALUATE: The displacement is much less than the distance traveled and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. IDENTIFY: When they first meet the sum of the distances they have run is 200 m. SET UP: Each runs with constant speed and continues around the track in the same direction, so the distance each runs is given by d = vt . Let the two runners be objects A and B. 200 m EXECUTE: (a) d A + d B = 200 m , so (6.20 m/s)t + (5.50 m/s)t = 200 m and t = = 17.1 s . 11.70 m/s (b) d A = v At = (6.20 m/s)(17.1 s) = 106 m . d B = vBt = (5.50 m/s)(17.1 s) = 94 m . The faster runner will be 106 m from the starting point and the slower runner will be 94 m from the starting point. These distances are measured around the circular track and are not straight-line distances. EVALUATE: The faster runner runs farther. IDENTIFY: To overtake the slower runner the first time the fast runner must run 200 m farther. To overtake the slower runner the second time the faster runner must run 400 m farther. SET UP: t and x0 are the same for the two runners. (a) Apply x − x0 = v0 xt to each runner: ( x − x0 )f = (6.20 m/s)t and ( x − x0 )s = (5.50 m/s)t .

EXECUTE:

( x − x0 )f = ( x − x0 )s + 200 m gives (6.20 m/s)t = (5.50 m/s)t + 200 m and t =

200 m = 286 s . 6.20 m/s − 5.50 m/s

( x − x0 )f = 1770 m and ( x − x0 )s = 1570 m .

2.7.

(b) Repeat the calculation but now ( x − x0 )f = ( x − x0 )s + 400 m . t = 572 s . The fast runner has traveled 3540 m. He has made 17 full laps for 3400 m and 140 m past the starting line in this 18th lap. EVALUATE: In part (a) the fast runner will have run 8 laps for 1600 m and will be 170 m past the starting line in his 9th lap. IDENTIFY: In time tS the S-waves travel a distance d = vStS and in time tP the P-waves travel a distance d = vPtP . SET UP:

tS = tP + 33 s

d d 1 1 ⎛ ⎞ − = + 33 s . d ⎜ ⎟ = 33 s and d = 250 km . vS vP ⎝ 3.5 km/s 6.5 km/s ⎠ EVALUATE: The times of travel for each wave are tS = 71 s and tP = 38 s . EXECUTE:

2.8.

IDENTIFY: SET UP:

Δx . Use x(t ) to find x for each t. Δt x (0) = 0 , x(2.00 s) = 5.60 m , and x(4.00 s) = 20.8 m The average velocity is vav-x =

(a) vav-x =

EXECUTE:

5.60 m − 0 = +2.80 m/s 2.00 s

20.8 m − 0 = +5.20 m/s 4.00 s 20.8 m − 5.60 m (c) vav-x = = +7.60 m/s 2.00 s EVALUATE: The average velocity depends on the time interval being considered. (a) IDENTIFY: Calculate the average velocity using Eq.(2.2). Δx SET UP: vav-x = so use x(t ) to find the displacement Δx for this time interval. Δt EXECUTE: t = 0 : x = 0 t = 10.0 s: x = (2.40 m/s 2 )(10.0 s) 2 − (0.120 m/s3 )(10.0 s)3 = 240 m − 120 m = 120 m. (b) vav-x =

2.9.

Δx 120 m = = 12.0 m/s. Δt 10.0 s (b) IDENTIFY: Use Eq.(2.3) to calculate vx (t ) and evaluate this expression at each specified t. Then vav-x =

dx = 2bt − 3ct 2 . dt EXECUTE: (i) t = 0 : vx = 0 SET UP:

vx =

(ii) t = 5.0 s: vx = 2(2.40 m/s 2 )(5.0 s) − 3(0.120 m/s3 )(5.0 s)2 = 24.0 m/s − 9.0 m/s = 15.0 m/s. (iii) t = 10.0 s: vx = 2(2.40 m/s 2 )(10.0 s) − 3(0.120 m/s3 )(10.0 s)2 = 48.0 m/s − 36.0 m/s = 12.0 m/s.

Motion Along a Straight Line

(c) IDENTIFY: SET UP:

2-3

Find the value of t when vx (t ) from part (b) is zero.

vx = 2bt − 3ct 2

vx = 0 at t = 0. vx = 0 next when 2bt − 3ct 2 = 0 EXECUTE:

2.10.

2.11.

2b = 3ct so t =

2b 2(2.40 m/s 2 ) = = 13.3 s 3c 30(.120 m/s3 )

EVALUATE: vx (t ) for this motion says the car starts from rest, speeds up, and then slows down again. IDENTIFY and SET UP: The instantaneous velocity is the slope of the tangent to the x versus t graph. EXECUTE: (a) The velocity is zero where the graph is horizontal; point IV. (b) The velocity is constant and positive where the graph is a straight line with positive slope; point I. (c) The velocity is constant and negative where the graph is a straight line with negative slope; point V. (d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III. EVALUATE: The sign of the velocity indicates its direction. Δx IDENTIFY: The average velocity is given by vav-x = . We can find the displacement Δt for each constant Δt velocity time interval. The average speed is the distance traveled divided by the time. SET UP: For t = 0 to t = 2.0 s , vx = 2.0 m/s . For t = 2.0 s to t = 3.0 s , vx = 3.0 m/s . In part (b),

vx = −3.0 m/s for t = 2.0 s to t = 3.0 s . When the velocity is constant, Δx = vx Δt . EXECUTE: (a) For t = 0 to t = 2.0 s , Δx = (2.0 m/s)(2.0 s) = 4.0 m . For t = 2.0 s to t = 3.0 s , Δx = (3.0 m/s)(1.0 s) = 3.0 m . For the first 3.0 s, Δx = 4.0 m + 3.0 m = 7.0 m . The distance traveled is also 7.0 m. Δx 7.0 m = = 2.33 m/s . The average speed is also 2.33 m/s. Δt 3.0 s (b) For t = 2.0 s to 3.0 s, Δx = ( −3.0 m/s)(1.0 s) = −3.0 m . For the first 3.0 s, Δx = 4.0 m + (−3.0 m) = +1.0 m . The dog runs 4.0 m in the + x -direction and then 3.0 m in the − x -direction, so the distance traveled is still 7.0 m. Δx 1.0 m 7.00 m = = 0.33 m/s . The average speed is = 2.33 m/s . vav-x = Δt 3.0 s 3.00 s EVALUATE: When the motion is always in the same direction, the displacement and the distance traveled are equal and the average velocity has the same magnitude as the average speed. When the motion changes direction during the time interval, those quantities are different. Δv IDENTIFY and SET UP: aav,x = x . The instantaneous acceleration is the slope of the tangent to the vx versus Δt t graph. EXECUTE: (a) 0 s to 2 s: aav,x = 0 ; 2 s to 4 s: aav,x = 1.0 m/s 2 ; 4 s to 6 s: aav,x = 1.5 m/s 2 ; 6 s to 8 s: The average velocity is vav-x =

2.12.

aav,x = 2.5 m/s 2 ; 8 s to 10 s: aav,x = 2.5 m/s 2 ; 10 s to 12 s: aav,x = 2.5 m/s 2 ; 12 s to 14 s: aav,x = 1.0 m/s 2 ; 14 s to 16 s: aav,x = 0 . The acceleration is not constant over the entire 16 s time interval. The acceleration is constant between 6 s and 12 s. (b) The graph of vx versus t is given in Fig. 2.12. t = 9 s : ax = 2.5 m/s 2 ; t = 13 s : ax = 1.0 m/s 2 ; t = 15 s : ax = 0 .

2-4

Chapter 2

EVALUATE: The acceleration is constant when the velocity changes at a constant rate. When the velocity is constant, the acceleration is zero.

Figure 2.12 2.13.

Δvx . Δt SET UP: Assume the car is moving in the + x direction. 1 mi/h = 0.447 m/s , so 60 mi/h = 26.82 m/s , 200 mi/h = 89.40 m/s and 253 mi/h = 113.1 m/s . EXECUTE: (a) The graph of vx versus t is sketched in Figure 2.13. The graph is not a straight line, so the acceleration is not constant. 26.82 m/s − 0 89.40 m/s − 26.82 m/s (b) (i) aav-x = = 12.8 m/s 2 (ii) aav-x = = 3.50 m/s 2 (iii) 2.1 s 20.0 s − 2.1 s 113.1 m/s − 89.40 m/s aav-x = = 0.718 m/s 2 . The slope of the graph of vx versus t decreases as t increases. This is 53 s − 20.0 s consistent with an average acceleration that decreases in magnitude during each successive time interval. EVALUATE: The average acceleration depends on the chosen time interval. For the interval between 0 and 53 s, 113.1 m/s − 0 aav-x = = 2.13 m/s 2 . 53 s IDENTIFY:

The average acceleration for a time interval Δt is given by aav-x =

Figure 2.13


2.14.

2-5

Δvx . ax (t ) is the slope of the vx versus t graph. Δt SET UP: 60 km/h = 16.7 m/s 16.7 m/s − 0 0 − 16.7 m/s EXECUTE: (a) (i) aav-x = = 1.7 m/s 2 . (ii) aav-x = = −1.7 m/s 2 . 10 s 10 s (iii) Δvx = 0 and aav-x = 0 . (iv) Δvx = 0 and aav-x = 0 . IDENTIFY:

aav-x =

(b) At t = 20 s , vx is constant and ax = 0 . At t = 35 s , the graph of vx versus t is a straight line and

ax = aav-x = −1.7 m/s 2 .

2.15.

EVALUATE: When aav-x and vx have the same sign the speed is increasing. When they have opposite sign the speed is decreasing. dx dv and ax = x to calculate vx (t ) and ax (t ). IDENTIFY and SET UP: Use vx = dt dt dx 2 = 2.00 cm/s − (0.125 cm/s )t EXECUTE: vx = dt dv ax = x = −0.125 cm/s 2 dt (a) At t = 0, x = 50.0 cm, vx = 2.00 cm/s, ax = −0.125 cm/s 2 . (b) Set vx = 0 and solve for t: t = 16.0 s. (c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0 s. The turtle returns to the starting point after 32.0 s. (d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. Set x = 60.0 cm and solve for t: t = 6.20 s and t = 25.8 s. At t = 6.20 s, vx = +1.23 cm/s.

At t = 25.8 s, vx = −1.23 cm/s. Set x = 40.0 cm and solve for t: t = 36.4 s (other root to the quadratic equation is negative and hence nonphysical). At t = 36.4 s, vx = −2.55 cm/s. (e) The graphs are sketched in Figure 2.15.

Figure 2.15

2.16.

EVALUATE: The acceleration is constant and negative. vx is linear in time. It is initially positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the − x -direction. IDENTIFY: Use Eq.(2.4), with Δt = 10 s in all cases. SET UP: vx is negative if the motion is to the right.

( ( 5.0 m/s ) − (15.0 m/s ) ) / (10 s ) = −1.0 m/s (b) ( ( −15.0 m/s ) − ( −5.0 m/s ) ) / (10 s ) = −1.0 m/s (c) ( ( −15.0 m/s ) − ( +15.0 m/s ) ) / (10 s ) = −3.0 m/s EXECUTE:

(a)

2

2

2

In all cases, the negative acceleration indicates an acceleration to the left. Δv IDENTIFY: The average acceleration is aav-x = x Δt SET UP: Assume the car goes from rest to 65 mi/h (29 m/s) in 10 s. In braking, assume the car goes from 65 mi/h to zero in 4.0 s. Let + x be in the direction the car is traveling. 29 m/s − 0 EXECUTE: (a) aav-x = = 2.9 m/s 2 10 s 0 − 29 m/s (b) aav-x = = −7.2 m/s 2 4.0 s EVALUATE:

2.17.

2-6

2.18.

Chapter 2

(c) In part (a) the speed increases so the acceleration is in the same direction as the velocity. If the velocity direction is positive, then the acceleration is positive. In part (b) the speed decreases so the acceleration is in the direction opposite to the direction of the velocity. If the velocity direction is positive then the acceleration is negative, and if the velocity direction is negative then the acceleration direction is positive. EVALUATE: The sign of the velocity and of the acceleration indicate their direction. Δv IDENTIFY: The average acceleration is aav-x = x . Use vx (t ) to find vx at each t. The instantaneous acceleration Δt dvx . is ax = dt SET UP: vx (0) = 3.00 m/s and vx (5.00 s) = 5.50 m/s . EXECUTE:

(a) aav-x =

Δvx 5.50 m/s − 3.00 m/s = = 0.500 m/s 2 Δt 5.00 s

dvx = (0.100 m/s3 )(2t ) = (0.200 m/s3 )t . At t = 0 , ax = 0 . At t = 5.00 s , ax = 1.00 m/s 2 . dt (c) Graphs of vx (t ) and ax (t ) are given in Figure 2.18. (b) ax =

ax (t ) is the slope of vx (t ) and increases at t increases. The average acceleration for t = 0 to

EVALUATE:

t = 5.00 s equals the instantaneous acceleration at the midpoint of the time interval, t = 2.50 s , since ax (t ) is a linear function of t.

Figure 2.18 2.19.

(a) IDENTIFY and SET UP: EXECUTE:

vx is the slope of the x versus t curve and ax is the slope of the vx versus t curve.

t = 0 to t = 5 s : x versus t is a parabola so ax is a constant. The curvature is positive so ax is

positive. vx versus t is a straight line with positive slope. v0 x = 0. t = 5 s to t = 15 s : x versus t is a straight line so vx is constant and ax = 0. The slope of x versus t is positive so vx is positive. t = 15 s to t = 25 s: x versus t is a parabola with negative curvature, so ax is constant and negative. vx versus t is a straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s. t = 25 s to t = 35 s: x versus t is a straight line so vx is constant and ax = 0. The slope of x versus t is negative so vx is negative. t = 35 s to t = 40 s: x versus t is a parabola with positive curvature, so ax is constant and positive. vx versus t is a straight line with positive slope. The velocity reaches zero at t = 40 s.


2-7

The graphs of vx (t ) and ax (t ) are sketched in Figure 2.19a.

Figure 2.19a (b) The motions diagrams are sketched in Figure 2.19b.

Figure 2.19b

2.20.

EVALUATE: The spider speeds up for the first 5 s, since vx and ax are both positive. Starting at t = 15 s the spider starts to slow down, stops momentarily at t = 20 s, and then moves in the opposite direction. At t = 35 s the spider starts to slow down again and stops at t = 40 s. dx dv and ax (t ) = x IDENTIFY: vx (t ) = dt dt d n n −1 SET UP: (t ) = nt for n ≥ 1 . dt EXECUTE: (a) vx (t ) = (9.60 m/s 2 )t − (0.600 m/s6 )t 5 and ax (t ) = 9.60 m/s 2 − (3.00 m/s 6 )t 4 . Setting vx = 0 gives

t = 0 and t = 2.00 s . At t = 0 , x = 2.17 m and ax = 9.60 m/s 2 . At t = 2.00 s , x = 15.0 m and ax = −38.4 m/s 2 . (b) The graphs are given in Figure 2.20.

2-8

Chapter 2

EVALUATE:

For the entire time interval from t = 0 to t = 2.00 s , the velocity vx is positive and x increases.

While ax is also positive the speed increases and while ax is negative the speed decreases.

Figure 2.20 2.21.

IDENTIFY: Use the constant acceleration equations to find v0 x and ax . (a) SET UP: The situation is sketched in Figure 2.21.

x − x0 = 70.0 m t = 7.00 s vx = 15.0 m/s v0 x = ? Figure 2.21

2( x − x0 ) 2(70.0 m) ⎛v +v ⎞ − vx = − 15.0 m/s = 5.0 m/s. Use x − x0 = ⎜ 0 x x ⎟ t , so v0 x = t 7.00 s 2 ⎝ ⎠ v −v 15.0 m/s − 5.0 m/s = 1.43 m/s 2 . (b) Use vx = v 0 x + axt , so ax = x 0 x = t 7.00 s EVALUATE: The average velocity is (70.0 m)/(7.00 s) = 10.0 m/s. The final velocity is larger than this, so the EXECUTE:

2.22.

antelope must be speeding up during the time interval; v0 x < vx and ax > 0. IDENTIFY: Apply the constant acceleration kinematic equations. SET UP: Let + x be in the direction of the motion of the plane. 173 mi/h = 77.33 m/s . 307 ft = 93.57 m . EXECUTE: (a) v0 x = 0 , vx = 77.33 m/s and x − x0 = 93.57 m . vx2 = v02x + 2ax ( x − x0 ) gives ax =

vx2 − v02x (77.33 m/s) 2 − 0 = = 32.0 m/s 2 . 2( x − x0 ) 2(93.57 m)

2( x − x0 ) 2(93.57 m) ⎛v +v ⎞ (b) x − x0 = ⎜ 0 x x ⎟ t gives t = = = 2.42 s v + v 0 + 77.33 m/s 2 ⎝ ⎠ 0x x

2.23.

EVALUATE: Either vx = v0 x + axt or x − x0 = v0 xt + 12 axt 2 could also be used to find t and would give the same result as in part (b). IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Assume the ball starts from rest and moves in the + x -direction. EXECUTE: (a) x − x0 = 1.50 m , vx = 45.0 m/s and v0 x = 0 . vx2 = v02x + 2ax ( x − x0 ) gives

ax =

vx2 − v02x (45.0 m/s) 2 = = 675 m/s 2 . 2( x − x0 ) 2(1.50 m)

2( x − x0 ) 2(1.50 m) ⎛v +v ⎞ (b) x − x0 = ⎜ 0 x x ⎟ t gives t = = = 0.0667 s v0 x + vx 45.0 m/s 2 ⎝ ⎠ v 45.0 m/s EVALUATE: We could also use vx = v0 x + axt to find t = x = = 0.0667 s which agrees with our ax 675 m/s 2 previous result. The acceleration of the ball is very large.


2.24.

2-9

IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Assume the ball moves in the + x direction. EXECUTE: (a) vx = 73.14 m/s , v0 x = 0 and t = 30.0 ms . vx = v0 x + axt gives

vx − v0 x 73.14 m/s − 0 = = 2440 m/s 2 . t 30.0 × 10−3 s ⎛ v + v ⎞ ⎛ 0 + 73.14 m/s ⎞ −3 (b) x − x0 = ⎜ 0 x x ⎟ t = ⎜ ⎟ (30.0 × 10 s) = 1.10 m 2 ⎝ 2 ⎠ ⎝ ⎠ ax =

EVALUATE:

2.25.

x − x0 = 12 (2440 m/s 2 )(30.0 × 10−3 s) 2 = 1.10 m , which agrees with our previous result. The acceleration of the ball is very large. IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set ax equal to its maximum allowed value. SET UP:

Let + x be the direction of the initial velocity of the car. ax = −250 m/s 2 . 105 km/h = 29.17 m/s .

vx2 − v02x 0 − (29.17 m/s) 2 = = 1.70 m . 2ax 2(−250 m/s 2 ) EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your 1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while stopping. IDENTIFY: Apply constant acceleration equations to the motion of the car. SET UP: Let + x be the direction the car is moving. vx2 (20 m s) 2 EXECUTE: (a) From Eq. (2.13), with v0 x = 0, ax = = = 1.67 m s 2 . 2( x − x0 ) 2(120 m) EXECUTE:

2.26.

We could also use x − x0 = v0 xt + 12 axt 2 to calculate x − x0 :

v0 x = +29.17 m/s . vx = 0 . vx2 = v02x + 2ax ( x − x0 ) gives x − x0 =

(b) Using Eq. (2.14), t = 2( x − x0 ) v x = 2(120 m) (20 m s) = 12 s.

2.27.

(c) (12 s)(20 m s) = 240 m. EVALUATE: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far. Δv IDENTIFY: The average acceleration is aav-x = x . For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) Δt apply. SET UP: Assume the shuttle travels in the + x direction. 161 km/h = 44.72 m/s and 1610 km/h = 447.2 m/s . 1.00 min = 60.0 s Δv 44.72 m/s − 0 = 5.59 m/s 2 EXECUTE: (a) (i) aav-x = x = Δt 8.00 s 447.2 m/s − 44.72 m/s (ii) aav-x = = 7.74 m/s 2 60.0 s − 8.00 s ⎛ v + v ⎞ ⎛ 0 + 44.72 m/s ⎞ (b) (i) t = 8.00 s , v0 x = 0 , and vx = 44.72 m/s . x − x0 = ⎜ 0 x x ⎟ t = ⎜ ⎟ (8.00 s) = 179 m . 2 ⎝ 2 ⎠ ⎝ ⎠

(ii) Δt = 60.0 s − 8.00 s = 52.0 s , v0 x = 44.72 m/s , and vx = 447.2 m/s .

2.28.

⎛ v + v ⎞ ⎛ 44.72 m/s + 447.2 m/s ⎞ 4 x − x0 = ⎜ 0 x x ⎟ t = ⎜ ⎟ (52.0 s) = 1.28 × 10 m . 2 2 ⎝ ⎠ ⎝ ⎠ EVALUATE: When the acceleration is constant the instantaneous acceleration throughout the time interval equals the average acceleration for that time interval. We could have calculated the distance in part (a) as x − x0 = v0 xt + 12 axt 2 = 12 (5.59 m/s 2 )(8.00 s) 2 = 179 m , which agrees with our previous calculation. IDENTIFY: Apply the constant acceleration kinematic equations to the motion of the car. SET UP: 0.250 mi = 1320 ft . 60.0 mph = 88.0 ft/s . Let + x be the direction the car is traveling. EXECUTE:

ax =

(a) braking: v0 x = 88.0 ft/s , x − x0 = 146 ft , vx = 0 . vx2 = v02x + 2ax ( x − x0 ) gives

vx2 − v02x 0 − (88.0 ft/s) 2 = = −26.5 ft/s 2 2( x − x0 ) 2(146 ft)

Speeding up: v0 x = 0 , x − x0 = 1320 ft , t = 19.9 s . x − x0 = v0 xt + 12 axt 2 gives ax =

2( x − x0 ) 2(1320 ft) = = 6.67 ft/s 2 t2 (19.9 s) 2

2-10

Chapter 2

(b) vx = v0 x + axt = 0 + (6.67 ft/s 2 )(19.9 s) = 133 ft/s = 90.5 mph

vx − v0 x 0 − 88.0 ft/s = = 3.32 s ax −26.5 ft/s 2 EVALUATE: The magnitude of the acceleration while braking is much larger than when speeding up. That is why it takes much longer to go from 0 to 60 mph than to go from 60 mph to 0. IDENTIFY: The acceleration ax is the slope of the graph of vx versus t. (c) t =

2.29.

SET UP:

The signs of vx and of ax indicate their directions.

EXECUTE: to the left.

(a) Reading from the graph, at t = 4.0 s , vx = 2.7 cm/s , to the right and at t = 7.0 s , vx = 1.3 cm/s ,

8.0 cm/s = −1.3 cm/s 2 . The acceleration is constant and equal to 6.0 s 1.3 cm/s 2 , to the left. It has this value at all times. (c) Since the acceleration is constant, x − x0 = v0 xt + 12 axt 2 . For t = 0 to 4.5 s, (b) vx versus t is a straight line with slope −

x − x0 = (8.0 cm/s)(4.5 s) + 12 (−1.3 cm/s 2 )(4.5 s) 2 = 22.8 cm . For t = 0 to 7.5 s, x − x0 = (8.0 cm/s)(7.5 s) + 12 (−1.3 cm/s 2 )(7.5 s) 2 = 23.4 cm (d) The graphs of ax and x versus t are given in Fig. 2.29. EVALUATE:

⎛v +v ⎞ In part (c) we could have instead used x − x0 = ⎜ 0 x x ⎟ t . ⎝ 2 ⎠

Figure 2.29 2.30.

IDENTIFY: Use the constant acceleration equations to find x, v0 x , vx and ax for each constant-acceleration segment of the motion. SET UP: Let + x be the direction of motion of the car and let x = 0 at the first traffic light. ⎛ v + v ⎞ ⎛ 0 + 20 m/s ⎞ EXECUTE: (a) For t = 0 to t = 8 s : x = ⎜ 0 x x ⎟ t = ⎜ ⎟ (8 s) = 80 m . 2 ⎝ 2 ⎠ ⎝ ⎠

vx − v0 x 20 m/s = = +2.50 m/s 2 . The car moves from x = 0 to x = 80 m . The velocity vx increases linearly t 8s from zero to 20 m/s. The acceleration is a constant 2.50 m/s 2 . Constant speed for 60 m: The car moves from x = 80 m to x = 140 m . vx is a constant 20 m/s. ax = 0 . This ax =

60 m + 8 s = 11 s . 20 m/s Slowing from 20 m/s until stopped: The car moves from x = 140 m to x = 180 m . The velocity decreases linearly 2(40 m) ⎛v +v ⎞ = 4 s . vx2 = v02x + 2ax ( x − x0 ) gives from 20 m/s to zero. x − x0 = ⎜ 0 x x ⎟ t gives t = 20 m/s + 0 ⎝ 2 ⎠ interval starts at t = 8 s and continues until t =

ax =

−(20.0 m/s) 2 = −5.00 m/s 2 This segment is from t = 11 s to t = 15 s . The acceleration is a 2(40 m)

constant −5.00 m/s 2 . The graphs are drawn in Figure 2.30a. (b) The motion diagram is sketched in Figure 2.30b.


2-11

! ! ! ! EVALUATE: When a and v are in the same direction, the speed increases ( t = 0 to t = 8 s ). When a and v are in opposite directions, the speed decreases ( t = 11 s to t = 15 s ). When a = 0 the speed is constant t = 8 s to t = 11 s .

Figure 2.30a-b 2.31.

(a) IDENTIFY and SET UP: The acceleration ax at time t is the slope of the tangent to the vx versus t curve at time t. EXECUTE: At t = 3 s, the vx versus t curve is a horizontal straight line, with zero slope. Thus ax = 0.

At t = 7 s, the vx versus t curve is a straight-line segment with slope

45 m/s − 20 m/s = 6.3 m/s 2 . 9 s −5 s

Thus ax = 6.3 m/s 2 . At t = 11 s the curve is again a straight-line segment, now with slope

−0 − 45 m/s = −11.2 m/s 2 . 13 s − 9 s

Thus ax = −11.2 m/s 2 . EVALUATE:

ax = 0 when vx is constant, ax > 0 when vx is positive and the speed is increasing, and ax < 0

when vx is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval t = 0 to t = 5 s the acceleration is constant and equal to zero. For the time interval t = 5 s to t = 9 s the acceleration is constant and equal to 6.25 m/s 2 . For the interval t = 9 s to t = 13 s the acceleration is constant and equal to −11.2 m/s 2 . EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. v0 x = 20 m/s ax = 0 t = 5 s x − x0 = ? x − x0 = v0 xt (ax = 0 so no

1 2

axt 2 term)

x − x0 = (20 m/s)(5 s) = 100 m; this is the distance the officer travels in the first 5 seconds. During the interval t = 5 s to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time t = 0 and ends at time t = 5 s. (Note that the acceleration is not constant over the entire t = 0 to t = 9 s interval.) v0 x = 20 m/s ax = 6.25 m/s 2 t = 4 s x0 = 100 m x − x0 = ? x − x0 = v0 xt + 12 axt 2 x − x0 = (20 m/s)(4 s) + 12 (6.25 m/s 2 )(4 s) 2 = 80 m + 50 m = 130 m. Thus x − x0 + 130 m = 100 m + 130 m = 230 m.

2-12

Chapter 2

At t = 9 s the officer is at x = 230 m, so she has traveled 230 m in the first 9 seconds. During the interval t = 9 s to t = 13 s the acceleration is again constant. The constant acceleration formulas can be applied for this 4 second interval but not for the whole t = 0 to t = 13 s interval. To use the equations restart our clock so this interval begins at time t = 0 and ends at time t = 4 s. v0 x = 45 m/s (at the start of this time interval) ax = −11.2 m/s 2 t = 4 s x0 = 230 m x − x0 = ? x − x0 = v0 xt + 12 axt 2 x − x0 = (45 m/s)(4 s) + 12 ( −11.2 m/s 2 )(4 s) 2 = 180 m − 89.6 m = 90.4 m. Thus x = x0 + 90.4 m = 230 m + 90.4 m = 320 m. At t = 13 s the officer is at x = 320 m, so she has traveled 320 m in the first 13 seconds. EVALUATE: The velocity vx is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval Δt is vav-x = Δx / Δt. For t = 0 to 5 s,

2.32.

vav-x = 20 m/s. For t = 0 to 9 s, vav-x = 26 m/s. For t = 0 to 13 s, vav-x = 25 m/s. These results are consistent with Fig. 2.33. IDENTIFY: In each constant acceleration interval, the constant acceleration equations apply. SET UP: When ax is constant, the graph of vx versus t is a straight line and the graph of x versus t is a parabola. When ax = 0 , vx is constant and x versus t is a straight line. EXECUTE: The graphs are given in Figure 2.32. EVALUATE: The slope of the x versus t graph is vx (t ) and the slope of the vx versus t graph is ax (t ) .

Figure 2.32 2.33.

(a) IDENTIFY: The maximum speed occurs at the end of the initial acceleration period. SET UP: ax = 20.0 m/s 2 t = 15.0 min = 900 s v0 x = 0 vx = ?

vx = v0 x + axt EXECUTE: vx = 0 + (20.0 m/s 2 )(900 s) = 1.80 × 104 m/s (b) IDENTIFY: Use constant acceleration formulas to find the displacement Δx. The motion consists of three constant acceleration intervals. In the middle segment of the trip ax = 0 and vx = 1.80 × 104 m/s, but we can’t directly find the distance traveled during this part of the trip because we don’t know the time. Instead, find the distance traveled in the first part of the trip (where ax = +20.0 m/s 2 ) and in the last part of the trip (where

ax = −20.0 m/s 2 ). Subtract these two distances from the total distance of 3.84 × 108 m to find the distance traveled in the middle part of the trip (where ax = 0). first segment SET UP: x − x0 = ? t = 15.0 min = 900 s ax = +20.0 m/s 2 v0 x = 0 x − x0 = v0 xt + 12 axt 2 EXECUTE:

x − x 0 = 0 + 12 (20.0 m/s 2 )(900 s)2 = 8.10 × 106 m = 8.10 × 103 km

second segment SET UP: x − x0 = ? t = 15.0 min = 900 s ax = −20.0 m/s 2 v0 x = 1.80 × 104 m/s x − x0 = v0 xt + 12 axt 2 EXECUTE: x − x0 = (1.80 × 104 s)(900 s) + 12 (−20.0 m/s 2 )(900 s) 2 = 8.10 × 106 m = 8.10 × 103 km (The same distance as traveled as in the first segment.)


2-13

Therefore, the distance traveled at constant speed is 3.84 × 108 m − 8.10 × 106 m − 8.10 × 106 m = 3.678 × 108 m = 3.678 × 105 km. 3.678 × 108 m = 0.958. The fraction this is of the total distance is 3.84 × 108 m (c) IDENTIFY: We know the time for each acceleration period, so find the time for the constant speed segment. SET UP: x − x0 = 3.678 × 108 m vx = 1.80 × 104 m/s ax = 0 t = ? x − x0 = v0 xt + 12 axt 2 x − x0 3.678 × 108 m = = 2.043 × 104 s = 340.5 min. v0 x 1.80 × 104 m/s The total time for the whole trip is thus 15.0 min + 340.5 min + 15.0 min = 370min. EVALUATE: If the speed was a constant 1.80 × 104 m/s for the entire trip, the trip would take (3.84 × 108 m)/(1.80 × 104 m/s) = 356 min. The trip actually takes a bit longer than this since the average velocity is EXECUTE:

2.34.

t=

less than 1.80 × 108 m/s during the relatively brief acceleration phases. IDENTIFY: Use constant acceleration equations to find x − x0 for each segment of the motion. SET UP: Let + x be the direction the train is traveling. EXECUTE: t = 0 to 14.0 s: x − x0 = v0 xt + 12 axt 2 = 12 (1.60 m/s 2 )(14.0 s) 2 = 157 m . At t = 14.0 s , the speed is vx = v0 x + axt = (1.60 m/s 2 )(14.0 s) = 22.4 m/s . In the next 70.0 s, ax = 0 and x − x0 = v0 xt = (22.4 m/s)(70.0 s) = 1568 m . For the interval during which the train is slowing down, v0 x = 22.4 m/s , ax = −3.50 m/s 2 and vx = 0 . vx2 − v02x 0 − (22.4 m/s) 2 = = 72 m . 2a x 2(−3.50 m/s 2 ) The total distance traveled is 157 m + 1568 m + 72 m = 1800 m . EVALUATE: The acceleration is not constant for the entire motion but it does consist of constant acceleration segments and we can use constant acceleration equations for each segment. IDENTIFY: vx (t ) is the slope of the x versus t graph. Car B moves with constant speed and zero acceleration. Car A moves with positive acceleration; assume the acceleration is constant. SET UP: For car B, vx is positive and ax = 0 . For car A, ax is positive and vx increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.35a. (b) The two cars have the same position at times when their x-t graphs cross. The figure in the problem shows this occurs at approximately t = 1 s and t = 3 s . (c) The graphs of vx versus t for each car are sketched in Figure 2.35b. (d) The cars have the same velocity when their x-t graphs have the same slope. This occurs at approximately t=2s. (e) Car A passes car B when x A moves above xB in the x-t graph. This happens at t = 3 s . vx2 = v02x + 2ax ( x − x0 ) gives x − x0 =

2.35

(f) Car B passes car A when xB moves above x A in the x-t graph. This happens at t = 1 s . EVALUATE:

When ax = 0 , the graph of vx versus t is a horizontal line. When ax is positive, the graph of

vx versus t is a straight line with positive slope.


IDENTIFY: Apply the constant acceleration equations to the motion of each vehicle. The truck passes the car when they are at the same x at the same t > 0 .

2-14

Chapter 2

The truck has ax = 0 . The car has v0 x = 0 . Let + x be in the direction of motion of the vehicles. Both

SET UP:

vehicles start at x0 = 0 . The car has aC = 3.20 m/s 2 . The truck has vx = 20.0 m/s . EXECUTE:

(a) x − x0 = v0 xt + 12 axt 2 gives xT = v0Tt and xC = 12 aCt 2 . Setting xT = xC gives t = 0 and v0T = 12 aCt , so

2v0T 2(20.0 m/s) = = 12.5 s . At this t, xT = (20.0 m/s)(12.5 s) = 250 m and x = 12 (3.20 m/s 2 )(12.5 s) 2 = 250 m . aC 3.20 m/s 2 The car and truck have each traveled 250 m. (b) At t = 12.5 s , the car has vx = v0 x + axt = (3.20 m/s 2 )(12.5 s) = 40 m/s . t=

(c) xT = v0Tt and xC = 12 aCt 2 . The x-t graph of the motion for each vehicle is sketched in Figure 2.36a. (d) vT = v0T . vC = aCt . The vx -t graph for each vehicle is sketched in Figure 2.36b. EVALUATE: When the car overtakes the truck its speed is twice that of the truck.


IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Take + y to be downward, so the motion is in the + y direction. 19,300 km/h = 5361 m/s , 1600 km/h = 444.4 m/s , and 321 km/h = 89.2 m/s . 4.0 min = 240 s . EXECUTE: (a) Stage A: t = 240 s , v0 y = 5361 m/s , v y = 444.4 m/s . v y = v0 y + a y t gives

v y − v0 y

444.4 m/s − 5361 m/s = −20.5 m/s 2 . 240 s Stage B: t = 94 s , v0 y = 444.4 m/s , v y = 89.2 m/s . v y = v0 y + a y t gives ay =

t

v y − v0 y

=

89.2 m/s − 444.4 m/s = −3.8 m/s 2 . 94 s Stage C: y − y0 = 75 m , v0 y = 89.2 m/s , v y = 0 . v y2 = v02y + 2a y ( y − y0 ) gives ay =

ay =

t

=

v y2 − v02y

2( y − y0 ) upward.

=

0 − (89.2 m/s) 2 = −53.0 m/s 2 . In each case the negative sign means that the acceleration is 2(75 m)

⎛ v + v y ⎞ ⎛ 5361 m/s + 444.4 m/s ⎞ (b) Stage A: y − y0 = ⎜ 0 y ⎟t = ⎜ ⎟ (240 s) = 697 km . 2 ⎠ ⎝ 2 ⎠ ⎝

2.38.

⎛ 444.4 m/s + 89.2 m/s ⎞ Stage B: y − y0 = ⎜ ⎟ (94 s) = 25 km . 2 ⎝ ⎠ Stage C: The problem states that y − y0 = 75 m = 0.075 km . The total distance traveled during all three stages is 697 km + 25 km + 0.075 km = 722 km . EVALUATE: The upward acceleration produced by friction in stage A is calculated to be greater than the upward acceleration due to the parachute in stage B. The effects of air resistance increase with increasing speed and in reality the acceleration was probably not constant during stages A and B. IDENTIFY: Assume an initial height of 200 m and a constant acceleration of 9.80 m/s 2 . SET UP: Let + y be downward. 1 km/h = 0.2778 m/s and 1 mi/h = 0.4470 m/s .


EXECUTE:

2-15

(a) y − y0 = 200 m , a y = 9.80 m/s 2 , v0 y = 0 . v y2 = v02y + 2a y ( y − y0 ) gives

v y = 2(9.80 m/s 2 )(200 m) = 60 m/s = 200 km/h = 140 mi/h .

2.39.

(b) Raindrops actually have a speed of about 1 m/s as they strike the ground. (c) The actual speed at the ground is much less than the speed calculated assuming free-fall, so neglect of air resistance is a very poor approximation for falling raindrops. EVALUATE: In the absence of air resistance raindrops would land with speeds that would make them very dangerous. IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground, a y = g , downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP:

At the maximum height v y = 0.

v y = 0 y − y0 = 0.440 m a y = −9.80 m/s 2 v0 y = ? v y2 = v02y + 2a y ( y − y0 ) EXECUTE: (b) SET UP:

v0 y = −2a y ( y − y0 ) = −2(−9.80 m/s 2 )(0.440 m) = 2.94 m/s When the flea has returned to the ground y − y0 = 0.

y − y0 = 0 v0 y = +2.94 m/s a y = −9.80 m/s 2 t = ? y − y0 = v0 yt + 12 a yt 2 EXECUTE: EVALUATE: 2.40.

With y − y0 = 0 this gives t = −

2v0 y ay

=−

2(2.94 m/s) = 0.600 s. −9.80 m/s 2

We can use v y = v0 y + a y t to show that with v0 y = 2.94 m/s, v y = 0 after 0.300 s.

IDENTIFY: Apply constant acceleration equations to the motion of the lander. SET UP: Let + y be positive. Since the lander is in free-fall, a y = +1.6 m/s 2 . EXECUTE:

v0 y = 0.8 m/s , y − y0 = 5.0 m , a y = +1.6 m/s 2 in v y2 = v02y + 2a y ( y − y0 ) gives

v y = v02y + 2a y ( y − y0 ) = (0.8 m/s) 2 + 2(1.6 m/s 2 )(5.0 m) = 4.1 m/s .

2.41.

EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to gravity on earth is much larger than on the moon. IDENTIFY: Apply constant acceleration equations to the motion of the meterstick. The time the meterstick falls is your reaction time. SET UP: Let + y be downward. The meter stick has v0 y = 0 and a y = 9.80 m/s 2 . Let d be the distance the

meterstick falls. EXECUTE:

(a) y − y0 = v0 yt + 12 a yt 2 gives d = (4.90 m/s 2 )t 2 and t =

d . 4.90 m/s 2

0.176 m = 0.190 s 4.90 m/s 2 EVALUATE: The reaction time is proportional to the square of the distance the stick falls. IDENTIFY: Apply constant acceleration equations to the vertical motion of the brick. SET UP: Let + y be downward. a y = 9.80 m/s 2 (b) t =

2.42.

EXECUTE:

(a) v0 y = 0 , t = 2.50 s , a y = 9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 = 12 (9.80 m/s 2 )(2.50 s) 2 = 30.6 m . The

building is 30.6 m tall. (b) v y = v0 y + a yt = 0 + (9.80 m/s 2 )(2.50 s) = 24.5 m/s (c) The graphs of a y , v y and y versus t are given in Fig. 2.42. Take y = 0 at the ground.

2-16

Chapter 2

EVALUATE:

⎛ v + vy ⎞ 2 2 We could use either y − y0 = ⎜ 0 y ⎟ t or v y = v0 y + 2a y ( y − y0 ) to check our results. 2 ⎝ ⎠

Figure 2.42 2.43.

IDENTIFY: When the only force is gravity the acceleration is 9.80 m/s 2 , downward. There are two intervals of constant acceleration and the constant acceleration equations apply during each of these intervals. SET UP: Let + y be upward. Let y = 0 at the launch pad. The final velocity for the first phase of the motion is the initial velocity for the free-fall phase. EXECUTE: (a) Find the velocity when the engines cut off. y − y0 = 525 m , a y = +2.25 m/s 2 , v0 y = 0 .

v y2 = v02y + 2a y ( y − y0 ) gives v y = 2(2.25 m/s 2 )(525 m) = 48.6 m/s . Now consider the motion from engine cut off to maximum height: y0 = 525 m , v0 y = +48.6 m/s , v y = 0 (at the maximum height), a y = −9.80 m/s 2 . v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y 2a y

=

0 − (48.6 m/s) 2 = 121 m and 2(−9.80 m/s 2 )

y = 121 m + 525 m = 646 m . (b) Consider the motion from engine failure until just before the rocket strikes the ground: y − y0 = −525 m ,

a y = −9.80 m/s 2 , v0 y = +48.6 m/s . v y2 = v02y + 2a y ( y − y0 ) gives v y = − (48.6 m/s) 2 + 2( −9.80 m/s 2 )(−525 m) = −112 m/s . Then v y = v0 y + a y t gives t=

v y − v0 y ay

=

−112 m/s − 48.6 m/s = 16.4 s . −9.80 m/s 2

(c) Find the time from blast-off until engine failure: y − y0 = 525 m , v0 y = 0 , a y = +2.25 m/s 2 .

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(525 m) = = 21.6 s . The rocket strikes the launch pad ay 2.25 m/s 2

21.6 s + 16.4 s = 38.0 s after blast off. The acceleration a y is +2.25 m/s 2 from t = 0 to t = 21.6 s . It is −9.80 m/s 2 from t = 21.6 s to 38.0 s . v y = v0 y + a y t applies during each constant acceleration segment, so the graph of v y versus t is a straight line with positive slope of 2.25 m/s 2 during the blast-off phase and with negative slope of −9.80 m/s 2 after engine failure. During each phase y − y0 = v0 yt + 12 a yt 2 . The sign of a y determines the curvature of y (t ) . At t = 38.0 s the rocket has returned to y = 0 . The graphs are sketched in Figure 2.43. EVALUATE:

In part (b) we could have found the time from y − y0 = v0 yt + 12 a yt 2 , finding v y first allows us to

avoid solving for t from a quadratic equation.

Figure 2.43


2.44.

2-17

IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take + y upward. a y = −9.80 m/s 2 . The initial velocity of the sandbag equals the velocity of the

balloon, so v0 y = +5.00 m/s . When the balloon reaches the ground, y − y0 = −40.0 m . At its maximum height the sandbag has v y = 0 . (a) t = 0.250 s : y − y0 = v0 yt + 12 a yt 2 = (5.00 m/s)(0.250 s) + 12 (−9.80 m/s 2 )(0.250 s) 2 = 0.94 m . The

EXECUTE:

sandbag is 40.9 m above the ground. v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(0.250 s) = 2.55 m/s . t = 1.00 s : y − y0 = (5.00 m/s)(1.00 s) + 12 (−9.80 m/s 2 )(1.00 s) 2 = 0.10 m . The sandbag is 40.1 m above the ground. v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(1.00 s) = −4.80 m/s . (b) y − y0 = −40.0 m , v0 y = 5.00 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives

−40.0 m = (5.00 m/s)t − (4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (5.00 m/s)t − 40.0 m = 0 and

)

(

1 5.00 ± (−5.00) 2 − 4(4.90)(−40.0) s = (0.51 ± 2.90) s . t must be positive, so t = 3.41 s . 9.80 (c) v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(3.41 s) = −28.4 m/s t=

(d) v0 y = 5.00 m/s , a y = −9.80 m/s 2 , v y = 0 . v y2 = v02y + 2a y ( y − y0 ) gives

y − y0 =

v y2 − v02y 2a y

=

0 − (5.00 m/s) 2 = 1.28 m . The maximum height is 41.3 m above the ground. 2(−9.80 m/s 2 )

(e) The graphs of a y , v y , and y versus t are given in Fig. 2.44. Take y = 0 at the ground . EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed.

Figure 2.44 2.45.

IDENTIFY: (a) SET UP:

The balloon has constant acceleration a y = g , downward. Take the + y direction to be upward.

t = 2.00 s, v0 y = −6.00 m/s, a y = −9.80 m/s 2 , v y = ? EXECUTE: (b) SET UP: EXECUTE: (c) SET UP:

v y = v0 y + a yt = −6.00 m/s + ( −9.80 m/s 2 )(2.00 s) = −25.5 m/s y − y0 = ? y − y0 = v0 yt + 12 a yt 2 = (−6.00 m/s)(2.00 s) + 12 (−9.80 m/s 2 )(2.00 s) 2 = −31.6 m y − y0 = −10.0 m, v0 y = −6.00 m/s, a y = −9.80 m/s 2 , v y = ?

v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

v y = − v02y + 2a y ( y − y0 ) = − (−6.00 m/s) 2 + 2( −9.80 m/s 2 )(−10.0 m) = −15.2 m/s

(d) The graphs are sketched in Figure 2.45.

Figure 2.45 EVALUATE: The speed of the balloon increases steadily since the acceleration and velocity are in the same direction. v y = 25.5 m/s when y − y0 = 31.6 m, so v y is less than this (15.2 m/s) when y − y0 is less (10.0 m).

2-18

2.46.

Chapter 2

IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the egg. SET UP: Take + y to be upward. At the maximum height, v y = 0 . (a) y − y0 = −50.0 m , t = 5.00 s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives

EXECUTE:

y − y0 1 −50.0 m 1 − 2 ayt = − 2 ( −9.80 m/s 2 )(5.00 s) = +14.5 m/s . t 5.00 s (b) v0 y = +14.5 m/s , v y = 0 (at the maximum height), a y = −9.80 m/s 2 . v y2 = v02y + 2a y ( y − y0 ) gives v0 y =

y − y0 =

v y2 − v02y 2a y

=

0 − (14.5 m/s) 2 = 10.7 m . 2(−9.80 m/s 2 )

(c) At the maximum height v y = 0 . (d) The acceleration is constant and equal to 9.80 m/s 2 , downward, at all points in the motion, including at the maximum height. (e) The graphs are sketched in Figure 2.46. v y − v0 y −14.5 m/s = = 1.48 s . The egg has EVALUATE: The time for the egg to reach its maximum height is t = ay −9.8 m/s 2

returned to the level of the cornice after 2.96 s and after 5.00 s it has traveled downward from the cornice for 2.04 s.

Figure 2.46 2.47.

Use the constant acceleration equations to calculate ax and x − x 0 .

IDENTIFY:

vx = 224 m/s, v0 x = 0, t = 0.900 s, ax = ?

(a) SET UP:

vx = v0 x + axt vx − v0 x 224 m/s − 0 = = 249 m/s 2 t 0.900 s (b) ax / g = (249 m/s 2 ) /(9.80 m/s 2 ) = 25.4 ax =

EXECUTE:

(c) x − x0 = v0 xt + 12 axt 2 = 0 + 12 (249 m/s 2 )(0.900 s) 2 = 101 m (d) SET UP: Calculate the acceleration, assuming it is constant: t = 1.40 s, v0 x = 283 m/s, vx = 0 (stops), ax = ?

vx = v0 x + axt

2.48.

vx − v0 x 0 − 283 m/s = = −202 m/s 2 t 1.40 s ax / g = (−202 m/s 2 ) /(9.80 m/s 2 ) = −20.6; ax = −20.6 g If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is only 20.6g. But if the acceleration is not constant it is certainly possible that at some point the instantaneous acceleration could be as large as 40g. EVALUATE: It is reasonable that for this motion the acceleration is much larger than g. IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the boulder. SET UP: Take + y to be upward. EXECUTE:

ax =

EXECUTE:

(a) v0 y = +40.0 m/s , v y = +20.0 m/s , a y = −9.80 m/s 2 . v y = v0 y + a y t gives

t=

v y − v0 y ay

=

20.0 m/s − 40.0 m/s = +2.04 s . −9.80 m/s 2


(b) v y = −20.0 m/s . t =

v y − v0 y ay

=

2-19

−20.0 m/s − 40.0 m/s = +6.12 s . −9.80 m/s 2

(c) y − y0 = 0 , v0 y = +40.0 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives t = 0 and

t=−

2v0 y ay

2(40.0 m/s) = +8.16 s . −9.80 m/s 2

=−

(d) v y = 0 , v0 y = +40.0 m/s , a y = −9.80 m/s 2 . v y = v0 y + a y t gives t =

v y − v0 y ay

=

0 − 40.0 m/s = 4.08 s . −9.80 m/s 2

2

(e) The acceleration is 9.80 m/s , downward, at all points in the motion. (f) The graphs are sketched in Figure 2.48. EVALUATE: v y = 0 at the maximum height. The time to reach the maximum height is half the total time in the air,

so the answer in part (d) is half the answer in part (c). Also note that 2.04 s < 4.08 s < 6.12 s . The boulder is going upward until it reaches its maximum height and after the maximum height it is traveling downward.

Figure 2.48 2.49.

IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall and acceleration due to gravity and setting up a ratio involving time of fall and acceleration due to gravity. SET UP: Let g En be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the

common height from which the object is dropped. Let + y be downward, so y − y0 = h . v0 y = 0 2 y − y0 = v0 yt + 12 a yt 2 gives h = 12 gtE2 and h = 12 g En tEn . Combining these two equations gives

EXECUTE:

2

2.50.

2

⎛t ⎞ ⎛ 1.75 s ⎞ 2 gt = g t and g En = g ⎜ E ⎟ = (9.80 m/s 2 ) ⎜ ⎟ = 0.0868 m/s . t 18.6 s ⎝ ⎠ ⎝ En ⎠ EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use Eqs.(2.17) and (2.18). Use the values of vx and of x at t = 1.0 s to evaluate v0 x and x0 . 2 E

2 En En

SET UP: EXECUTE:

1

∫ t dt = n + 1t n

n +1

, for n ≥ 0 . t

(a) vx = v0 x + ∫ α tdt = v0 x + 12 α t 2 = v0 x + (0.60 m/s3 )t 2 . vx = 5.0 m/s when t = 1.0 s gives 0

v0 x = 4.4 m/s . Then, at t = 2.0 s , vx = 4.4 m/s + (0.60 m/s3 )(2.0 s) 2 = 6.8 m/s . t 1 (b) x = x0 + ∫ (v0 x + 12 α t 2 ) dt = x0 + v0 xt + α t 3 . x = 6.0 m at t = 1.0 s gives x0 = 1.4 m . Then, at t = 2.0 s , 0 6 1 x = 1.4 m + (4.4 m/s)(2.0 s) + (1.24 m/s3 )(2.0 s)3 = 11.8 m . 6 (c) x (t ) = 1.4 m + (4.4 m/s)t + (0.20 m/s3 )t 3 . vx (t ) = 4.4 m/s + (0.60 m/s3 )t 2 . ax (t ) = (1.20m/s3 )t . The graphs are sketched in Figure 2.50.

2-20

Chapter 2

We can verify that ax =

EVALUATE:

dvx dx and vx = . dt dt

Figure 2.50 2.51.

ax = At − Bt 2 with A = 1.50 m/s3 and B = 0.120 m/s 4 (a) IDENTIFY: SET UP:

Integrate ax (t ) to find vx (t ) and then integrate vx (t ) to find x(t ). t

vx = v0 x + ∫ ax dt 0

t

vx = v0 x + ∫ ( At − Bt 2 ) dt = v0 x + 12 At 2 − 13 Bt 3

EXECUTE:

0

At rest at t = 0 says that v0 x = 0, so vx = 12 At 2 − 13 Bt 3 = 12 (1.50 m/s3 )t 2 − 13 (0.120 m/s 4 )t 3 vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3 SET UP:

t

x − x0 + ∫ vx dt

EXECUTE:

0

x = x0 + ∫

t 0

(

1 2

At 2 − 13 Bt 3 ) dt = x0 + 16 At 3 − 121 Bt 4

At the origin at t = 0 says that x0 = 0, so x = 16 At 3 − 121 Bt 4 = 16 (1.50 m/s3 )t 3 − 121 (0.120 m/s 4 )t 4 x = (0.25 m/s3 )t 3 − (0.010 m/s 4 )t 4 dx dv and ax (t ) = x . dt dt dvx dvx (b) IDENTIFY and SET UP: At time t, when vx is a maximum, = 0. (Since ax = , the maximum velocity dt dt is when ax = 0. For earlier times ax is positive so vx is still increasing. For later times ax is negative and vx is decreasing.) dv EXECUTE: ax = x = 0 so At − Bt 2 = 0 dt One root is t = 0, but at this time vx = 0 and not a maximum. EVALUATE:

We can check our results by using them to verify that vx (t ) =

A 1.50 m/s3 = = 12.5 s B 0.120 m/s 4 At this time vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3 gives The other root is t =

vx = (0.75 m/s3 )(12.5 s) 2 − (0.040 m/s 4 )(12.5 s)3 = 117.2 m/s − 78.1 m/s = 39.1 m/s. 2.52.

EVALUATE: For t < 12.5 s, ax > 0 and vx is increasing. For t > 12.5 s, ax < 0 and vx is decreasing. IDENTIFY: a (t ) is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2.52. EXECUTE: (a) Slope = a = 0 for t ≥ 1.3 ms . (b) 1 hmax = Area under v-t graph ≈ ATriangle + ARectangle ≈ (1.3 ms) (133 cm/s ) + (2.5 ms − 1.3 ms)(133 cm s) ≈ 0.25 cm 2 133cm s 2 (c) a = slope of v-t graph. a (0.5 ms) ≈ a (1.0 ms) ≈ = 1.0 × 105 cm s . 1.3ms a (1.5 ms) = 0 because the slope is zero.


2-21

1 (d) h = area under v-t graph. h(0.5 ms) ≈ ATriangle = (0.5 ms) ( 33 cm/s ) = 8.3 × 10−3 cm . 2 1 h(1.0 ms) ≈ ATriangle = (1.0 ms)(100 cm s) = 5.0 × 10−2 cm . 2 1 h(1.5 ms) ≈ ATriangle + ARectangle = (1.3 ms) (133 cm/s ) (0.2 ms)(1.33) = 0.11 cm 2 EVALUATE: The acceleration is constant until t = 1.3 ms , and then it is zero. g = 980 cm/s 2 . The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering.

Figure 2.52 2.53.

(a) IDENTIFY and SET UP: The change in speed is the area under the ax versus t curve between vertical lines at t = 2.5 s and t = 7.5 s. EXECUTE: This area is 12 (4.00 cm/s 2 + 8.00 cm/s 2 )(7.5 s − 2.5 s) = 30.0 cm/s This acceleration is positive so the change in velocity is positive. (b) Slope of vx versus t is positive and increasing with t. The graph is sketched in Figure 2.53.

Figure 2.53 EVALUATE: 2.54.

2.55.

The calculation in part (a) is equivalent to Δvx = ( aav-x )Δt. Since ax is linear in t,

aav-x = (a0 x + ax ) / 2. Thus aav-x = 12 (4.00 cm/s 2 + 8.00 cm/s 2 ) for the time interval t = 2.5 s to t = 7.5 s. IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that first 10 mi 10 miles is = 1.25 h . 8 mi/h 20 mi = 5.0 h . The second 10 mi must EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of 4 mi/h 10 mi be covered in 5.0 h − 1.25 h = 3.75 h . This corresponds to an average speed of = 2.7 mi/h . 3.75 h 20 mi = 1.67 h . The second 10 mi must be (b) An average speed of 12 mi/h for 20 mi gives a total time of 12 mi/h 10 mi covered in 1.67 h − 1.25 h = 0.42 h . This corresponds to an average speed of = 24 mi/h . 0.42 h 20 mi = 1.25 h . But 1.25 h was already spent (c) An average speed of 16 mi/h for 20 mi gives a total time of 16 mi/h during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. dx dv and ax = x . IDENTIFY: vx (t ) = dt dt d n SET UP: (t ) = nt n −1 , for n ≥ 1 . dt EXECUTE: (a) vx (t ) = (9.00 m/s3 )t 2 − (20.0 m/s 2 )t + 9.00 m/s . ax (t ) = (18.0 m/s 3 )t − 20.0 m/s 2 . The graphs are sketched in Figure 2.55.

2-22

Chapter 2

(b) The particle is instantaneously at rest when vx (t ) = 0 . v0 x = 0 and the quadratic formula gives

1 (20.0 ± (20.0) 2 − 4(9.00)(9.00)) s = 1.11 s ± 0.48 s . t = 0.63 s and t = 1.59 s . These results agree with the 18.0 vx -t graphs in part (a).

t=

(c) For t = 0.63 s , ax = (18.0 m/s3 )(0.63 s) − 20.0 m/s 2 = −8.7 m/s 2 . For t = 1.59 s , ax = +8.6 m/s 2 . At t = 0.63 s

the slope of the vx -t graph is negative and at t = 1.59 s it is positive, so the same answer is deduced from the vx (t ) graph as from the expression for ax (t ) . 20.0 m/s 2 = 1.11 s . 18.0m/s3 (e) When the particle is at its greatest distance from the origin, vx = 0 and ax < 0 (so the particle is starting to move back toward the origin). This is the case for t = 0.63 s , which agrees with the x-t graph in part (a) . At t = 0.63 s , x = 2.45 m . (f) The particle’s speed is changing at its greatest rate when ax has its maximum magnitude. The ax -t graph in part (d) vx (t ) is instantaneously not changing when ax = 0 . This occurs at t =

(a) shows this occurs at t = 0 and at t = 2.00 s . Since vx is always positive in this time interval, the particle is speeding up at its greatest rate when ax is positive, and this is for t = 2.00 s . The particle is slowing down at its greatest rate when ax is negative and this is for t = 0 . EVALUATE:

Since ax (t ) is linear in t, vx (t ) is a parabola and is symmetric around the point where vx (t ) has its

minimum value ( t = 1.11 s ). For this reason, the answer to part (d) is midway between the two times in part (c).

Figure 2.55 2.56.

IDENTIFY:


Δx . The average speed is the distance traveled divided by the Δt

elapsed time. SET UP: Let + x be in the direction of the first leg of the race. For the round trip, Δx ≥ 0 and the total distance traveled is 50.0 m. For each leg of the race both the magnitude of the displacement and the distance traveled are 25.0 m. Δx 25.0 m EXECUTE: (a) vav-x = = = 1.25 m/s . This is the same as the average speed for this leg of the race. 20.0 s Δt (b) vav-x =

Δx 25.0 m = = 1.67 m/s . This is the same as the average speed for this leg of the race. Δt 15.0 s

(c) Δx = 0 so vav-x = 0 .

50.0 m = 1.43 m/s . 35.0 s EVALUATE: Note that the average speed for the round trip is not equal to the arithmetic average of the average speeds for each leg. IDENTIFY: Use information about displacement and time to calculate average speed and average velocity. Take the origin to be at Seward and the positive direction to be west. distance traveled (a) SET UP: average speed = time EXECUTE: The distance traveled (different from the net displacement ( x − x0 ) ) is 76 km + 34 km = 110 km. (d) The average speed is

2.57.

Find the total elapsed time by using vav-x = Seward to Auora: t =

Δx x − x0 = to find t for each leg of the journey. Δt t

x − x0 76 km = = 0.8636 h vav-x 88 km/h


2-23

x − x0 −34 km = = 0.4722 h vav-x −72 km/h Total t = 0.8636 h + 0.4722 h = 1.336 h. 110 km = 82 km/h. Then average speed = 1.336 h Δx (b) SET UP: vav-x = , where Δx is the displacement, not the total distance traveled. Δt Auora to York: t =

42 km = 31 km/h. l.336 h EVALUATE: The motion is not uniformly in the same direction so the displacement is less than the distance traveled and the magnitude of the average velocity is less than the average speed. IDENTIFY: The vehicles are assumed to move at constant speed. The speed (mi/h) divided by the frequency with which vehicles pass a given point (vehicles/h) is the total space per vehicle (the length of the vehicle plus space to the next vehicle). SET UP: 96 km/h = 96 × 103 m/h 96 × 103 m/h EXECUTE: (a) The total space per vehicle is = 40 m/vehicle . Since the average length of a 2400 vehicles/h vehicle is 4.6 m, the average space between vehicles is 40 m − 4.6 m = 35 m . 96 × 103 m/h (b) The frequency of vehicles (vehicles/h) is = 7000 vehicles/h . (4.6 + 9.2) m/vehicle EVALUATE: The traffic flow rate per lane would nearly triple. Note that the traffic flow rate is directly proportional to the traffic speed. Δv v −v (a) IDENTIFY: Calculate the average acceleration using aav-x = x = x 0 x Use the information about the time Δt t and total distance to find his maximum speed. SET UP: v0 x = 0 since the runner starts from rest. For the whole trip he ends up 76 km − 34 km = 42 km west of his starting point. vav-x =

2.58.

2.59.

t = 4.0 s, but we need to calculate vx , the speed of the runner at the end of the acceleration period. EXECUTE: For the last 9.1 s − 4.0 s = 5.1 s the acceleration is zero and the runner travels a distance of d1 = (5.1 s)vx (obtained using x − x0 = v0 xt + 12 axt 2 ) During the acceleration phase of 4.0 s, where the velocity goes from 0 to vx , the runner travels a distance ⎛v +v ⎞ v d 2 = ⎜ 0 x x ⎟ t = x (4.0 s) = (2.0 s)vx 2 ⎝ 2 ⎠ The total distance traveled is 100 m, so d1 + d 2 = 100 m. This gives (5.1 s)vx + (2.0 s)vx = 100 m. vx =

100 m = 14.08 m/s. 7.1 s

vx − v0 x 14.08 s − 0 = = 3.5 m/s 2 . t 4.0 s (b) For this time interval the velocity is constant, so aav − x = 0.

Now we can calculate aav-x : aav-x = EVALUATE:

Now that we have vx we can calculate d1 = (5.1 s)(14.08 m/s) = 71.9 m and

d 2 = (2.0 s)(14.08 m/s) = 28.2 m. So, d1 + d 2 = 100 m, which checks. vx − v0 x , where now the time interval is the full 9.1 s of the race. t We have calculated the final speed to be 14.08 m/s, so (c) IDENTIFY and SET UP:

aav-x =

14.08 m/s = 1.5 m/s 2 . 9.1 s EVALUATE: The acceleration is zero for the last 5.1 s, so it makes sense for the answer in part (c) to be less than half the answer in part (a). (d) The runner spends different times moving with the average accelerations of parts (a) and (b). IDENTIFY: Apply the constant acceleration equations to the motion of the sled. The average velocity for a time Δx interval Δt is vav-x = . Δt aav-x =

2.60.

2-24

Chapter 2

SET UP: Let + x be parallel to the incline and directed down the incline. The problem doesn’t state how much time it takes the sled to go from the top to 14.4 m from the top. 25.6 m − 14.4 m EXECUTE: (a) 14.4 m to 25.6 m: vav-x = = 5.60 m/s . 25.6 to 40.0 m: 2.00 s 40.0 m − 25.6 m 57.6 m − 40.0 m vav-x = = 7.20 m/s . 40.0 m to 57.6 m: vav-x = = 8.80 m/s . 2.00 s 2.00 s (b) For each segment we know x − x0 and t but we don’t know v0 x or vx . Let x1 = 14.4 m and x2 = 25.6 m . For

x −x ⎛v +v ⎞ x −x this interval ⎜ 1 2 ⎟ = 2 1 and at = v2 − v1 . Solving for v2 gives v2 = 12 at + 2 1 . Let x2 = 25.6 m and t t ⎝ 2 ⎠ ⎛v +v ⎞ x −x x3 = 40.0 m . For this second interval, ⎜ 2 3 ⎟ = 3 2 and at = v3 − v2 . Solving for v2 gives t ⎝ 2 ⎠ x − x v2 = − 12 at + 3 2 . Setting these two expressions for v2 equal to each other and solving for a gives t 1 1 a = 2 [( x3 − x2 ) − ( x2 − x1 )] = [(40.0 m − 25.6 m) − (25.6 m − 14.4 m)] = 0.80 m/s 2 . t (2.00 s) 2 Note that this expression for a says a =

vav-23 − vav-12 , where vav-12 and vav-23 are the average speeds for successive t

2.00 s intervals. (c) For the motion from x = 14.4 m to x = 25.6 m , x − x0 = 11.2 m , ax = 0.80 m/s 2 and t = 2.00 s . x − x0 1 11.2 m 1 − 2 axt = − (0.80 m/s 2 )(2.00 s) = 4.80 m/s . t 2.00 s 2 (d) For the motion from x = 0 to x = 14.4 m , x − x0 = 14.4 m , v0 x = 0 , and vx = 4.8 m/s . x − x0 = v0 xt + 12 axt 2 gives v0 x =

2( x − x0 ) 2(14.4 m) ⎛v +v ⎞ x − x0 = ⎜ 0 x x ⎟ t gives t = = = 6.0 s . v0 x + vx 4.8 m/s ⎝ 2 ⎠ (e) For this 1.00 s time interval, t = 1.00 s , v0 x = 4.8 m/s , ax = 0.80 m/s 2 .

x − x0 = v0 xt + 12 axt 2 = (4.8 m/s)(1.00 s) + 12 (0.80 m/s 2 )(1.00 s) 2 = 5.2 m .

2.61.

EVALUATE: With x = 0 at the top of the hill, x (t ) = v0 xt + 12 axt 2 = (0.40 m/s 2 )t 2 . We can verify that t = 6.0 s gives x = 14.4 m , t = 8.0 s gives 25.6 m, t = 10.0 s gives 40.0 m, and t = 12.0 s gives 57.6 m. IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. Since vx is always positive the motion is always in the + x direction and the total distance moved equals the magnitude

of the displacement. The acceleration ax is the slope of the vx versus t graph. SET UP:

For the t = 0 to t = 10.0 s segment, v0 x = 4.00 m/s and vx = 12.0 m/s . For the t = 10.0 s to

12.0 s segment, v0 x = 12.0 m/s and vx = 0 . EXECUTE:

⎛ v + v ⎞ ⎛ 4.00 m/s + 12.0 m/s ⎞ (a) For t = 0 to t = 10.0 s , x − x0 = ⎜ 0 x x ⎟ t = ⎜ ⎟ (10.0 s) = 80.0 m . For 2 ⎝ 2 ⎠ ⎝ ⎠

⎛ 12.0 m/s + 0 ⎞ t = 10.0 s to t = 12.0 s , x − x0 = ⎜ ⎟ (2.00 s) = 12.0 m . The total distance traveled is 92.0 m. 2 ⎝ ⎠ (b) x − x0 = 80.0 m + 12.0 m = 92.0 m (c) For t = 0 to 10.0 s, ax =

ax =

12.0 m/s − 4.00 m/s = 0.800 m/s 2 . For t = 10.0 s to 10.2 s, 10.0 s

0 − 12.0 m/s = −6.00 m/s 2 . The graph of ax versus t is given in Figure 2.61. 2.00 s


2-25

EVALUATE: When vx and ax are both positive, the speed increases. When vx is positive and ax is negative, the speed decreases.

Figure 2.61 2.62.

IDENTIFY: Since light travels at constant speed, d = ct SET UP: The distance from the earth to the sun is 1.50 × 1011 m . The distance from the earth to the moon is 3.84 × 108 m . c = 186,000 mi/s . EXECUTE:

⎛ 365 14 d ⎞ ⎛ 24 h ⎞⎛ 3600 s ⎞ 15 (a) d = ct = (3.0 × 108 m/s)(1 y) ⎜ ⎟⎜ ⎟⎜ ⎟ = 9.5 × 10 m 1 y 1 d 1 h ⎝ ⎠⎝ ⎠ ⎝ ⎠

(b) d = ct = (3.0 × 108 m/s)(10−9 s) = 0.30 m (c) t =

d 1.5 × 1011 m = = 500 s = 8.33 min c 3.0 × 108 m s

(d) t =

d 2(3.84 × 108 m) = = 2.6 s 3.0 × 108 m s c

d 3 × 109 mi = = 16,100 s = 4.5 h c 186,000 mi s EVALUATE: The speed of light is very large but it still takes light a measurable length of time to travel a large distance. IDENTIFY: Speed is distance d divided by time t. The distance around a circular path is d = 2π R , where R is the radius of the circular path. SET UP: The radius of the earth is RE = 6.38 × 106 m . The earth rotates once in 1 day = 86,400 s . The radius of (e) t =

2.63.

the earth’s orbit around the sun is 1.50 × 1011 m and the earth completes this orbit in 1 year = 3.156 × 107 s . The speed of light in vacuum is c = 3.00 × 108 m/s . d 2π RE 2π (6.38 × 106 m) EXECUTE: (a) v = = = = 464 m/s . t t 86,400 s 2π R 2π (1.50 × 1011 m) = = 2.99 × 104 m/s . t 3.156 × 107 s d 2π RE 2π (6.38 × 106 m) (c) The time for light to go around once is t = = = = 0.1336 s . In 1.00 s light would go c c 3.00 × 108 m/s 1.00 s = 7.49 times . around the earth 0.1336 s EVALUATE: All these speeds are large compared to speeds of objects in our everyday experience. IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. For t = 0 to 5.0 s, vx is positive and the ball moves in the + x direction. For t = 5.0 s to 20.0 s, vx is negative and the (b) v =

2.64.

ball moves in the − x direction. The acceleration ax is the slope of the vx versus t graph. SET UP:

For the t = 0 to t = 5.0 s segment, v0 x = 0 and vx = 30.0 m/s . For the t = 5.0 s to t = 20.0 s segment,

v0 x = −20.0 m/s and vx = 0 .

2-26

Chapter 2

EXECUTE:

⎛ v + v ⎞ ⎛ 0 + 30.0 m/s ⎞ (a) For t = 0 to 5.0 s, x − x0 = ⎜ 0 x x ⎟ t = ⎜ ⎟ (5.0 m/s) = 75.0 m . The ball travels a 2 ⎝ 2 ⎠ ⎝ ⎠

⎛ −20.0 m/s + 0 ⎞ distance of 75.0 m. For t = 5.0 s to 20.0 s, x − x0 = ⎜ ⎟ (15.0 m/s) = −150.0 m . The total distance 2 ⎝ ⎠ traveled is 75.0 m + 150.0 m = 225.0 m . (b) The total displacement is x − x0 = 75.0 m +( − 150.0 m) = −75.0 m . The ball ends up 75.0 m in the negative xdirection from where it started. 30.0 m/s − 0 0 − ( −20.0 m/s) (c) For t = 0 to 5.0 s, ax = = 6.00 m/s 2 . For t = 5.0 s to 20.0 s, ax = = +1.33 m/s 2 . 5.0 s 15.0 s The graph of ax versus t is given in Figure 2.64. (d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity doesn't change instantaneously. So, no, the actual graph of vx (t ) is not really vertical at 5.00 s. EVALUATE:

For t = 0 to 5.0 s, both vx and ax are positive and the speed increases. For t = 5.0 s to 20.0 s, vx is

negative and ax is positive and the speed decreases. Since the direction of motion is not the same throughout, the displacement is not equal to the distance traveled.

Figure 2.64 2.65.

IDENTIFY and SET UP: Apply constant acceleration equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find x − x0 for the first 5.0 s. EXECUTE: For the first 5.0 s of the motion, v0 x = 0, t = 5.0 s.

vx = v0 x + axt gives vx = ax (5.0 s). This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: v0 x = ax (5.0 s), t = 5.0 s, x − x0 = 150 m. x − x0 = v0 xt + 12 axt 2 gives 150 m = (25 s 2 )ax + (12.5 s 2 )ax and ax = 4.0 m/s 2 Use this ax and consider the first 5.0 s of the motion: x − x0 = v0 xt + 12 axt 2 = 0 + 12 (4.0 m/s 2 )(5.0 s) 2 = 50.0 m. EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact, x − x0 is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go four times as far. In 10.0 s we calculated it went 50 m + 150 m = 200 m, which is four times 50 m. 2.66.

IDENTIFY: Apply x − x0 = v0 xt + 12 axt 2 to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train x0 = 0 and for

the caboose of the freight train x0 = 200 m . For the freight train vF = 15.0 m/s and aF = 0 . For the passenger train vP = 25.0 m/s and aP = −0.100 m/s 2 . EXECUTE:

(a) x − x0 = v0 xt + 12 axt 2 for each object gives xP = vPt + 12 aPt 2 and xF = 200 m + vFt . Setting

xP = xF gives vPt + 12 aPt 2 = 200 m + vFt . (0.0500 m/s 2 )t 2 − (10.0 m/s)t + 200 m = 0 . The

(

)

1 +10.0 ± (10.0) 2 − 4(0.0500)(200) s = (100 ± 77.5) s . The collision occurs at 0.100 t = 100 s − 77.5 s = 22.5 s . The equations that specify a collision have a physical solution (real, positive t), so a collision does occur.

quadratic formula gives t =


2-27

(b) xP = (25.0 m/s)(22.5 s) + 12 (−0.100 m/s 2 )(22.5 s) 2 = 537 m . The passenger train moves 537 m before the collision. The freight train moves (15.0 m/s)(22.5 s) = 337 m . (c) The graphs of xF and xP versus t are sketched in Figure 2.66. EVALUATE: The second root for the equation for t, t = 177.5 s is the time the trains would meet again if they were on parallel tracks and continued their motion after the first meeting.

Figure 2.66 2.67.

IDENTIFY: Apply constant acceleration equations to the motion of the two objects, you and the cockroach. You catch up with the roach when both objects are at the same place at the same time. Let T be the time when you catch up with the cockroach. SET UP: Take x = 0 to be at the t = 0 location of the roach and positive x to be in the direction of motion of the two objects. roach: v0 x = 1.50 m/s, ax = 0, x0 = 0, x = 1.20 m, t = T you: v0 x = 0.80 m/s, x0 = −0.90 m, x = 1.20 m, t = T , ax = ?

Apply x − x0 = v0 xt + 12 axt 2 to both objects: EXECUTE: roach: 1.20 m = (1.50 m/s)T , so T = 0.800 s. you: 1.20 m − (−0.90 m) = (0.80 m/s)T + 12 axT 2 2.10 m = (0.80 m/s)(0.800 s) + 12 ax (0.800 s) 2 2.10 m = 0.64 m + (0.320 s 2 )ax ax = 4.6 m/s 2 . ⎛v +v ⎞ Your final velocity is vx = v0 x + axt = 4.48 m/s. Then x − x0 = ⎜ 0 x x ⎟ t = 2.10 m, which checks. ⎝ 2 ⎠ You have to accelerate to a speed greater than that of the roach so you will travel the extra 0.90 m you are initially behind. IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together. SET UP: Each car has moved 100 m when they hit. 100 m = 10 s . During this time the grasshopper travels a distance of EXECUTE: The time until the cars hit is 10 m/s (15 m/s)(10 s) = 150 m . EVALUATE: The grasshopper ends up 100 m from where it started, so the magnitude of his final displacement is 100 m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction. IDENTIFY: Apply constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.69a Let d be the distance that the auto initially is behind the truck, so x0 (auto) = −d and x0 (truck) = 0. Let T be the EVALUATE:

2.68.

2.69.

time it takes the auto to catch the truck. Thus at time T the truck has undergone a displacement x − x0 = 40.0 m, so is at x = x0 + 40.0 m = 40.0 m. The auto has caught the truck so at time T is also at x = 40.0 m.

Figure 2.69a

2-28

Chapter 2

(a) SET UP: Use the motion of the truck to calculate T: x − x0 = 40.0 m, v0 x = 0 (starts from rest), ax = 2.10 m/s 2 , t = T

x − x0 = v0 xt + 12 axt 2 Since v0 x = 0, this gives t =

2( x − x0 ) ax

2(40.0 m) = 6.17 s 2.10 m/s 2 (b) SET UP: Use the motion of the auto to calculate d: x − x0 = 40.0 m + d , v0 x = 0, ax = 3.40 m/s 2 , t = 6.17 s EXECUTE:

T=

x − x0 = v0 xt + 12 axt 2 EXECUTE: d + 40.0 m = 12 (3.40 m/s 2 )(6.17 s) 2 d = 64.8 m − 40.0 m = 24.8 m (c) auto: vx = v0 x + axt = 0 + (3.40 m/s 2 )(6.17 s) = 21.0 m/s

truck: vx = v0 x + axt = 0 + (2.10 m/s 2 )(6.17 s) = 13.0 m/s (d) The graph is sketched in Figure 2.69b.

Figure 2.69b

2.70.

EVALUATE: In part (c) we found that the auto was traveling faster than the truck when they come abreast. The graph in part (d) agrees with this: at the intersection of the two curves the slope of the x-t curve for the auto is greater than that of the truck. The auto must have an average velocity greater than that of the truck since it must travel farther in the same time interval. IDENTIFY: Apply the constant acceleration equations to the motion of each car. The collision occurs when the cars are at the same place at the same time. SET UP: Let + x be to the right. Let x = 0 at the initial location of car 1, so x01 = 0 and x02 = D . The cars collide

when x1 = x2 . v0 x1 = 0 , ax1 = ax , v0 x 2 = −v0 and ax 2 = 0 . EXECUTE: 1 2

(a) x − x0 = v0 xt + 12 axt 2 gives x1 = 12 axt 2 and x2 = D − v0t . x1 = x2 gives

axt 2 + v0t − D = 0 . The quadratic formula gives t =

so t =

(

)

(

)

1 2

axt 2 = D − v0t .

1 −v0 ± v02 + 2ax D . Only the positive root is physical, ax

1 −v0 + v02 + 2ax D . ax

(b) v1 = axt = v02 + 2ax D − v0 (c) The x-t and vx -t graphs for the two cars are sketched in Figure 2.70.


EVALUATE:

2-29

In the limit that ax = 0 , D − v0t = 0 and t = D / v0 , the time it takes car 2 to travel distance D. In the

limit that v0 = 0 , t =

2D , the time it takes car 1 to travel distance D. ax

Figure 2.70

The average speed is the distance traveled divided by the time. The average velocity is vav-x =

Δx . Δt

2.71.

IDENTIFY:

2.72.

SET UP: The distance the ball travels is half the circumference of a circle of diameter 50.0 cm so is 1 π d = 12 π (50.0 cm) = 78.5 cm . Let + x be horizontally from the starting point toward the ending point, so 2 Δx equals the diameter of the bowl. 1 π d 78.5 cm EXECUTE: (a) The average speed is 2 = = 7.85 cm/s . t 10.0 s Δx 50.0 cm = = 5.00 cm/s . (b) The average velocity is vav-x = Δt 10.0 s EVALUATE: The average speed is greater than the magnitude of the average velocity, since the distance traveled is greater than the magnitude of the displacement. IDENTIFY: ax is the slope of the vx versus t graph. x is the area under the vx versus t graph. SET UP: The slope of vx is positive and decreasing in magnitude. As vx increases, the displacement in a given amount of time increases. EXECUTE: The ax -t and x-t graphs are sketched in Figure 2.72. EVALUATE: vx is the slope of the x versus t graph. The x (t ) graph we sketch has zero slope at t = 0 , the slope is always positive, and the slope initially increases and then approaches a constant. This behavior agrees with the vx (t ) that is given in the graph in the problem.

Figure 2.72 2.73.

IDENTIFY: Apply constant acceleration equations to each vehicle. SET UP: (a) It is very convenient to work in coordinates attached to the truck. Note that these coordinates move at constant velocity relative to the earth. In these coordinates the truck is at rest, and the initial velocity of the car is v0 x = 0. Also, the car’s acceleration in these coordinates is the same as in coordinates fixed to the earth. EXECUTE: First, let’s calculate how far the car must travel relative to the truck: The situation is sketched in Figure 2.73.

Figure 2.73

2-30

Chapter 2

The car goes from x0 = −24.0 m to x = 51.5 m. So x − x0 = 75.5 m for the car. Calculate the time it takes the car to travel this distance: ax = 0.600 m/s 2 , v0 x = 0, x − x0 = 75.5 m, t = ? x − x0 = v0 xt + 12 axt 2 t=

2( x − x0 ) 2(75.5 m) = = 15.86 s ax 0.600 m/s 2

It takes the car 15.9 s to pass the truck. (b) Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. In these coordinates v0 x = 20.0 m/s for the car. Take the origin to be at the initial position of the car. v0 x = 20.0 m/s, ax = 0.600 m/s 2 , t = 15.86 s, x − x0 = ? x − x0 = v0 xt + 12 axt 2 = (20.0 m/s)(15.86 s) + 12 (0.600 m/s 2 )(15.86 s) 2 x − x0 = 317.2 m + 75.5 m = 393 m. (c) In coordinates fixed to the earth: vx = v0 x + axt = 20.0 m/s + (0.600 m/s 2 )(15.86 s) = 29.5 m/s

2.74.

EVALUATE: In 15.9 s the truck travels x − x0 = (20.0 m/s)(15.86 s) = 317.2 m. The car travels 392.7 m − 317.2 m = 75 m farther than the truck, which checks with part (a). In coordinates attached to the truck, ⎛v +v ⎞ for the car v0 x = 0, vx = 9.5 m/s and in 15.86 s the car travels x − x0 = ⎜ 0 x x ⎟ t = 75 m, which checks with ⎝ 2 ⎠ part (a). IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use t dv ax (t ) = x and x = x0 + ∫ vx (t )dt . 0 dt 1 n +1 SET UP: ∫ t n dt = t for n ≥ 0 . n +1 EXECUTE:

t

(a) x (t ) = x0 + ∫ [α − β t 2 ]dt = x0 + α t − 13 β t 3 . x = 0 at t = 0 gives x0 = 0 and 0

dvx = −2 β t = −(4.00 m/s3 )t . dt (b) The maximum positive x is when vx = 0 and ax < 0 . vx = 0 gives α − β t 2 = 0 and x (t ) = α t − 13 β t 3 = (4.00 m/s)t − (0.667 m/s3 )t 3 . ax (t ) =

t=

2.75.

α 4.00 m/s = = 1.41 s . At this t, ax is negative. For t = 1.41 s , β 2.00 m/s3

x = (4.00 m/s)(1.41 s) − (0.667 m/s 3 )(1.41 s)3 = 3.77 m . EVALUATE: After t = 1.41 s the object starts to move in the − x direction and goes to x = −∞ as t → ∞ . a (t ) = α + β t , with α = −2.00 m/s 2 and β = 3.00 m/s3 (a) IDENTIFY and SET UP: EXECUTE:

Integrage ax (t ) to find vx (t ) and then integrate vx (t ) to find x(t ).

t

t

0

0

vx = v0 x + ∫ ax dt = v0 x + ∫ (α + β ) dt = v0 x + α t + 12 β t 2

t

t

0

0

x = x0 + ∫ vx dt = x0 + ∫ (v0 x + α t + 12 β t 2 ) dt = x0 + v0 xt + 12 α t 2 + 16 β t 3

At t = 0, x = x0 . To have x = x0 at t1 = 4.00 s requires that v0 xt1 + 12 α t12 + 16 β t13 = 0. Thus v0 x = − 16 β t12 − 12 α t1 = − 16 (3.00 m/s3 )(4.00 s) 2 − 12 (−2.00 m/s 2 )(4.00 s) = −4.00 m/s. (b) With v0 x as calculated in part (a) and t = 4.00 s,

v0 = v0 x + α t + 12 β t 2 = −4.00 s + (−2.00 m/s 2 )(4.00 s) + 12 (3.00 m/s3 )(4.00 s) 2 = +12.0 m/s. EVALUATE: ax = 0 at t = 0.67 s. For t > 0.67 s, ax > 0. At t = 0, the particle is moving in the − x -direction and is speeding up. After t = 0.67 s, when the acceleration is positive, the object slows down and then starts to move in the + x -direction with increasing speed.


2.76.

2-31

IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let + y be downward. The egg has v0 y = 0 and a y = 9.80 m/s 2 . At the height of the professor’s head,

the egg has y − y0 = 44.2 m . EXECUTE:

2.77.

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(44.2 m) = = 3.00 s . The professor walks a distance ay 9.80 m/s 2

x − x0 = v0 xt = (1.20 m/s)(3.00 s) = 3.60 m . Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is (9.80 m/s 2 )(3.00s) = 29.4 m/s . It is traveling much faster than the professor. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let + y be upward. At the maximum height, v y = 0 . When the rock returns to the surface, y − y0 = 0 . EXECUTE:

(a) v y2 = v02y + 2a y ( y − y0 ) gives a y H = − 12 v02y , which is constant, so aE H E = aM H M .

⎛a ⎞ ⎛ 9.80 m/s 2 ⎞ HM = HE ⎜ E ⎟ = H ⎜ = 2.64 H . 2 ⎟ ⎝ 3.71 m/s ⎠ ⎝ aM ⎠ (b) y − y0 = v0 yt + 12 a yt 2 with y − y0 = 0 gives a y t = −2v0 y , which is constant, so aETE = aMTM .

2.78.

⎡a ⎤ TM = TE ⎢ E ⎥ = 2.64T . ⎣ aM ⎦ EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let + y be upward. a y = −9.80 m/s 2 . y − y0 = 0 when the ball returns to its original

position. 5.50 m = 2.20 s to reach the table and an equal time to return. For the ball, 2.50 m/s y − y0 = 0 , t = 4.40 s and a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives

EXECUTE:

(a) It takes her

v0 y = − 12 a y t = − 12 (−9.80 m/s 2 )(4.40 s) = 21.6 m/s . (b) Find y − y0 when t = 2.20 s . y − y0 = v0 yt + 12 a yt 2 = (21.6 m/s)(2.20 s) + 12 (−9.80 m/s 2 )(2.20 s) 2 = 23.8 m

2.79.

EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. (a) IDENTIFY: Use constant acceleration equations, with a y = g , downward, to calculate the speed of the diver

when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the + y -direction to be downward. y − y0 = +21.3 m, a y = +9.80 m/s 2 , v0 y = 0 (since diver just steps off), v y = ? v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

v y = + 2a y ( y − y0 ) = + 2(9.80 m/s 2 )(31.3 m) = +20.4 m/s.

We know that v y is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use y − y0 = v0 yt + 12 a yt 2 to find t = 2.085 s. The diver gains 9.80 m/s of speed each second, so has v y = (9.80 m/s 2 )(2.085 s) = 20.4 m/s when she reaches the water, which checks. (b) IDENTIFY: Calculate the initial upward velocity needed to give the diver a speed of 25.0 m/s when she reaches the water. Use the same coordinates as in part (a). SET UP: v0 y = ?, v y = +25.0 m/s, a y = +9.80 m/s 2 , y − y0 = +21.3 m

v y2 = v02y + 2a y ( y − y0 )

2-32

Chapter 2

v0 y = − v y2 − 2a y ( y − y0 ) = − (25.0 m/s) 2 − 2(9.80 m/s 2 )(21.3 m) = −14.4 m/s

EXECUTE:

(v0 y is negative since the direction of the initial velocity is upward.) EVALUATE: One way to decide if this speed is reasonable is to calculate the maximum height above the platform it would produce: v0 y = −14.4 m/s, v y = 0 (at maximum height), a y = +9.80 m/s 2 , y − y0 = ?

v y2 = v02y + 2a y ( y − y0 ) y − y0 = 2.80.

v y2 − v02y

2a y

=

0 − (−14.4 s) 2 = −10.6 m 2(+9.80 m/s)

This is not physically attainable; a vertical leap of 10.6 m upward is not possible. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the windowsill to the top of the window. SET UP: Let + y be downward. Throughout the motion a y = +9.80 m/s 2 . EXECUTE:

Motion past the window: y − y0 = 1.90 m , t = 0.420 s , a y = +9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives

y − y0 1 1.90 m 1 − 2 a yt = − (9.80 m/s 2 )(0.420 s) = 2.466 m/s . This is the velocity of the flowerpot when it is 0.420 s 2 t at the top of the window. Motion from the windowsill to the top of the window: v0 y = 0 , v y = 2.466 m/s , a y = +9.80 m/s 2 . v0 y =

v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y

=

2a y

(2.466 m/s) 2 − 0 = 0.310 m . The top of the window is 0.310 m 2(9.80 m/s 2 )

below the windowsill. EVALUATE:

It takes the flowerpot t =

v y − v0 y ay

=

2.466 m/s = 0.252 s to fall from the sill to the top of the 9.80 m/s 2

window. Our result says that from the windowsill the pot falls 0.310 m + 1.90 m = 2.21 m in 0.252 s + 0.420 s = 0.672 s . y − y0 = v0 yt + 12 a yt 2 = 12 (9.80 m/s 2 )(0.672 s) 2 = 2.21 m , which checks. 2.81.

IDENTIFY: For parts (a) and (b) apply the constant acceleration equations to the motion of the bullet. In part (c) neglect air resistance, so the bullet is free-fall. Use the constant acceleration equations to establish a relation between initial speed v0 and maximum height H. SET UP: For parts (a) and (b) let + x be in the direction of motion of the bullet. For part (c) let + y be upward, so

a y = − g . At the maximum height, v y = 0 . EXECUTE:

ax =

(a) x − x0 = 0.700 m , v0 x = 0 , vx = 965 m/s . vx2 = v02x + 2ax ( x − x0 ) gives

vx2 − v02x (965 m/s) 2 − 0 a = = 6.65 × 105 m/s 2 . x = 6.79 × 104 , so ax = (6.79 × 104 ) g . g 2( x − x0 ) 2(0.700 m)

2( x − x0 ) 2(0.700 m) ⎛v +v ⎞ = = 1.45 ms . (b) x − x0 = ⎜ 0 x x ⎟ t gives t = v0 x + vx 0 + 965 m/s ⎝ 2 ⎠ (c) v y2 = v02y + 2a y ( y − y0 ) and v y = 0 gives

v02y y − y0

= −2a y , which is constant.

2 v01 v2 = 02 . H1 H 2

2

⎛ v2 ⎞ ⎛ 1v ⎞ H 2 = H1 ⎜ 022 ⎟ = H ⎜ 2 01 ⎟ = H / 4 . ⎝ v01 ⎠ ⎝ v01 ⎠ v2 − v2 −(965 m/s) 2 EVALUATE: H = y 0 y = = 47.5 km . Rifle bullets fired vertically don't actually reach such a 2a y 2( −9.80 m/s 2 ) 2.82.

large height; it is not an accurate approximation to ignore air resistance. IDENTIFY: Assume the firing of the second stage lasts a very short time, so the rocket is in free-fall after 25.0 s. The motion consists of two constant acceleration segments. SET UP: Let + y be upward. After t = 25.0 s , a y = −9.80 m/s 2 . EXECUTE:

(a) Find the height of the rocket at t = 25.0 s : v0 y = 0 , a y = +3.50 m/s 2 , t = 25.0 s .

y − y0 = v0 yt + 12 a yt 2 = 12 (3.50 m/s)(25.0 s) 2 = 1.0938 × 103 m . Find the displacement of the rocket from firing of the


2-33

second stage until the maximum height is reached: v0 y = 132.5 m/s , v y = 0 (at maximum height), a y = −9.80 m/s 2 . v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y

2a y

=

0 − (132.5 m/s) 2 = 896 m . The total height is 2(−9.80 m/s 2 )

1094 m + 896 m = 1990 m . (b) v0 y = +132.5 m/s , a y = −9.80 m/s 2 , y − y0 = −1094 m . y − y0 = v0 yt + 12 a yt 2 gives −1093.8 m = (132.5 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 33.7 s as the positive root. The rocket returns to the launch pad 33.7 s after the second stage fires. (c) v y = v0 y + a yt = +132.5 m/s + ( −9.80 m/s 2 )(33.7 s) = −198 m/s . The rocket has speed 198 m/s as it reaches the

2.83.

launch pad. EVALUATE: The speed when the rocket returns to the launch pad is greater than 132.5 m/s. When the rocket returns to the height where the second stage fired, its velocity is 132.5 m/s downward and it continues to speed up during the rest of the descent. Take positive y to be upward. (a) IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand. SET UP: v0 y = 0, v y = ?, a y = 45.0 m/s 2 , y − y0 = 0.640 m v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

v y = 2a y ( y − y0 ) = 2(45.0 m/s 2 )(0.640 m) = 7.59 m/s

(b) IDENTIFY: Consider the motion of the shot from the point where he releases it to its maximum height, where v = 0. Take y = 0 at the ground.

y0 = 2.20 m, y = ?, a y = −9.80 m/s 2 (free fall), v0 y = 7.59 m/s

SET UP:

(from part (a), v y = 0 at maximum height) v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

y − y0 =

v y2 − v02y

2a y

=

0 − (7.59 m/s) 2 = 2.94 m 2(−9.80 m/s 2 )

y = 2.20 m + 2.94 m = 5.14 m. (c) IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. Take y = 0 at the ground. y0 = 2.20 m, y = 1.83 m, a y = −9.80 m/s 2 , v0 y = +7.59 m/s, t = ? y − y0 = v0 yt + 12 a yt 2

SET UP: EXECUTE:

1.83 m − 2.20 m = (7.59 m/s)t + 12 ( −9.80 m/s 2 )t 2

= (7.59 m/s)t − (4.90 m/s 2 )t 2

4.90t 2 − 7.59t − 0.37 = 0, with t in seconds. Use the quadratic formula to solve for t: 1 7.59 ± (7.59) 2 − 4(4.90)(−0.37) = 0.774 ± 0.822 t= 9.80 t must be positive, so t = 0.774 s + 0.822 s = 1.60 s EVALUATE: Calculate the time to the maximum height: v y = v0 y + a yt , so

(

)

t = (v y − v0 y ) / a y = −(7.59 m/s)/( − 9.80 m/s 2 ) = 0.77 s. It also takes 0.77 s to return to 2.2 m above the ground, for a

2.84.

total time of 1.54 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his head a little later than 1.54 s after being thrown; the answer of 1.60 s in part (c) makes sense. IDENTIFY: The teacher is in free-fall and falls with constant acceleration 9.80 m/s 2 , downward. The sound from her shout travels at constant speed. The sound travels from the top of the cliff, reflects from the ground and then travels upward to her present location. If the height of the cliff is h and she falls a distance y in 3.0 s, the sound must travel a distance h + (h − y ) in 3.0 s. SET UP: EXECUTE:

Let + y be downward, so for the teacher a y = 9.80 m/s 2 and v0 y = 0 . Let y = 0 at the top of the cliff. (a) For the teacher, y = 12 (9.80 m/s 2 )(3.0 s) 2 = 44.1 m . For the sound, h + ( h − y ) = vst .

h = 12 (vst + y ) = 12 ([340 m/s][3.0 s] + 44.1 m) = 532 m , which rounds to 530 m. (b) v y2 = v02y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(532 m) = 102 m/s .

2-34

Chapter 2

EVALUATE: 2.85.

She is in the air for t =


v y − v0 y ay

=

102 m/s = 10.4 s and strikes the ground at high speed. 9.80 m/s 2

Let + y be upward. Each ball moves with constant acceleration a y = −9.80 m/s 2 . In

parts (c) and (d) require that the two balls be at the same height at the same time. EXECUTE: (a) At ceiling, v y = 0, y − y0 = 3.0 m, a y = −9.80 m/s 2 . Solve for v0 y . v y2 = v02y + 2a y ( y − y0 ) gives v0 y = 7.7 m/s. (b) v y = v0 y + a yt with the information from part (a) gives t = 0.78 s. (c) Let the first ball travel downward a distance d in time t. It starts from its maximum height, so v0 y = 0.

y − y0 = v0 y t = 12 a yt 2 gives d = (4.9 m/s 2 )t 2

The second ball has v0 y = 32 (7.7 m/s) = 5.1 m/s. In time t it must travel upward 3.0 m − d to be at the same place as the first ball. y − y0 = v0 yt + 12 a yt 2 gives 3.0 m − d = (5.1 m/s)t − (4.9 m/s 2 )t 2 .

2.86.

We have two equations in two unknowns, d and t. Solving gives t = 0.59 s and d = 1.7 m. (d) 3.0 m − d = 1.3 m EVALUATE: In 0.59 s the first ball falls d = (4.9 m/s 2 )(0.59 s) 2 = 1.7 m, so is at the same height as the second ball. IDENTIFY: The helicopter has two segments of motion with constant acceleration: upward acceleration for 10.0 s and then free-fall until it returns to the ground. Powers has three segments of motion with constant acceleration: upward acceleration for 10.0 s, free-fall for 7.0 s and then downward acceleration of 2.0 m/s 2 . SET UP: Let + y be upward. Let y = 0 at the ground. EXECUTE: (a) When the engine shuts off both objects have upward velocity v y = v0 y + a yt = (5.0 m/s 2 )(10.0 s) = 50.0 m/s and are at y = v0 yt + 12 a yt 2 = 12 (5.0 m/s 2 )(10.0 s) 2 = 250 m . For the helicopter, v y = 0 (at the maximum height), v0 y = +50.0 m/s , y0 = 250 m , and a y = −9.80 m/s 2 . v y2 = v02y + 2a y ( y − y0 ) gives y =

v y2 − v02y

2a y

+ y0 =

0 − (50.0 m/s) 2 + 250 m = 378 m , which rounds to 380 m. 2( −9.80 m/s 2 )

(b) The time for the helicopter to crash from the height of 250 m where the engines shut off can be found using v0 y = +50.0 m/s , a y = −9.80 m/s 2 , and y − y0 = −250 m . y − y0 = v0 yt + 12 a yt 2 gives

−250 m = (50.0 m/s)t − (4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (50.0 m/s)t − 250 m = 0 . The quadratic formula gives

(

)

1 50.0 ± (50.0) 2 + 4(4.90)(250) s . Only the positive solution is physical, so t = 13.9 s . Powers therefore 9.80 has free-fall for 7.0 s and then downward acceleration of 2.0 m/s 2 for 13.9 s − 7.0 s = 6.9 s . After 7.0 s of free-fall he is at y − y0 = v0 yt + 12 a yt 2 = 250 m + (50.0 m/s)(7.0 s) + 12 ( −9.80 m/s 2 )(7.0 s) 2 = 360 m and has velocity t=

vx = v0 x + axt = 50.0 m/s + ( −9.80 m/s 2 )(7.0 s) = −18.6 m/s . After the next 6.9 s he is at y − y0 = v0 yt + 12 a yt 2 = 360 m + (−18.6 m/s)(6.9 s) + 12 ( −2.00 m/s 2 )(6.9 s) 2 = 184 m . Powers is 184 m above the

2.87.

ground when the helicopter crashes. EVALUATE: When Powers steps out of the helicopter he retains the initial velocity he had in the helicopter but his acceleration changes abruptly from 5.0 m/s 2 upward to 9.80 m/s 2 downward. Without the jet pack he would have crashed into the ground at the same time as the helicopter. The jet pack slows his descent so he is above the ground when the helicopter crashes. IDENTIFY: Apply the constant acceleration equations to his motion. Consider two segments of the motion: the last 1.0 s and the motion prior to that. The final velocity for the first segment is the initial velocity for the second segment. SET UP: Let + y be downward, so a y = +9.80 m/s 2 . EXECUTE:

Motion from the roof to a height of h / 4 above ground: y − y0 = 3h / 4 , a y = +9.80 m/s 2 , v0 y = 0 .

v y2 = v02y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 3.834 h y − y0 = h / 4 , a y = +9.80 m/s 2 , v0 y = 3.834 h

m / s . Motion from height of h / 4 to the ground:

m / s , t = 1.00 s . y − y0 = v0 yt + 12 a yt 2 gives


h = 3.834 h 4

m + 4.90 m . Let h = u 2 and solve for u. 14 u 2 − 3.834u

(

u = 2 3.834 ± ( −3.834) 2 + 4.90

)

2-35

m − 4.90 m = 0 .

m . Only the positive root is physical, so u = 16.52

m and h = u 2 = 273 m ,

which rounds to 270 m. The building is 270 m tall. EVALUATE:

2.88.

With h = 273 m the total time of fall is t =

2h = 7.46 s . In 7.47 s − 1.00 s = 6.46 s Spider-Man ay

falls a distance y − y0 = 12 (9.80 m/s 2 )(6.46 s) 2 = 204 m . This leaves 69 m for the last 1.0 s of fall, which is h / 4 . IDENTIFY: Apply constant acceleration equations to the motion of the rock. Sound travels at constant speed. SET UP: Let tfall be the time for the rock to fall to the ground and let ts be the time it takes the sound to travel from the impact point back to you. tfall + ts = 10.0 s . Both the rock and sound travel a distance d that is equal to the height of the cliff. Take + y downward for the motion of the rock. The rock has v0 y = 0 and a y = 9.80 m/s 2 . EXECUTE:

(a) For the rock, y − y0 = v0 yt + 12 a yt 2 gives tfall =

2d . 9.80 m/s 2

d = 10.0 s . Let α 2 = d . 0.00303α 2 + 0.4518α − 10.0 = 0 . α = 19.6 and d = 384 m . 330 m/s (b) You would have calculated d = 12 (9.80 m/s 2 )(10.0 s) 2 = 490 m . You would have overestimated the height of the cliff. It actually takes the rock less time than 10.0 s to fall to the ground. EVALUATE: Once we know d we can calculate that tfall = 8.8 s and ts = 1.2 s . The time for the sound of impact to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it strikes the ground. (a) IDENTIFY: Let + y be upward. The can has constant acceleration a y = − g . The initial upward velocity of the

For the sound, ts =

2.89.

can equals the upward velocity of the scaffolding; first find this speed. SET UP: y − y0 = −15.0 m, t = 3.25 s, a y = −9.80 m/s 2 , v0 y = ? EXECUTE:

y − y0 = v0 yt + 12 a yt 2 gives v0 y = 11.31 m/s

Use this v0 y in v y = v0 y + a y t to solve for v y : v y = −20.5 m/s (b) IDENTIFY: Find the maximum height of the can, above the point where it falls from the scaffolding: SET UP: v y = 0, v0 y = +11.31 m/s, a y = −9.80 m/s 2 , y − y0 = ? EXECUTE:

2.90.

v y2 = v02y + 2a y ( y − y0 ) gives y − y0 = 6.53 m

The can will pass the location of the other painter. Yes, he gets a chance. EVALUATE: Relative to the ground the can is initially traveling upward, so it moves upward before stopping momentarily and starting to fall back down. IDENTIFY: Both objects are in free-fall. Apply the constant acceleration equations to the motion of each person. SET UP: Let + y be downward, so a y = +9.80 m/s 2 for each object. EXECUTE:

(a) Find the time it takes the student to reach the ground: y − y0 = 180 m , v0 y = 0 , a y = 9.80 m/s 2 .

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(180 m) = = 6.06 s . Superman must reach the ground in ay 9.80 m/s 2

6.06 s − 5.00 s = 1.06 s : t = 1.06 s , y − y0 = 180 m , a y = +9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives y − y0 1 180 m 1 − 2 a yt = − (9.80 m/s 2 )(1.06 s) = 165 m/s . Superman must have initial speed v0 = 165 m/s . 1.06 s 2 t (b) The graphs of y-t for Superman and for the student are sketched in Figure 2.90. (c) The minimum height of the building is the height for which the student reaches the ground in 5.00 s, before Superman jumps. y − y0 = v0 yt + 12 a yt 2 = 12 (9.80 m/s 2 )(5.00 s) 2 = 122 m . The skyscraper must be at least v0 y =

122 m high.

2-36

Chapter 2

165 m/s = 369 mi/h , so only Superman could jump downward with this initial speed.

EVALUATE:

Figure 2.90 2.91.

IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground. SET UP: Let + y be upward. At the instant that the canister is released, it has the same velocity as the rocket.

After it is released, the canister has a y = −9.80 m/s 2 . At its maximum height the canister has v y = 0 . EXECUTE:

(a) Find the speed of the rocket when the canister is released: v0 y = 0 , a y = 3.30 m/s 2 ,

y − y0 = 235 m . v y2 = v02y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 2(3.30 m/s 2 )(235 m) = 39.4 m/s . For the

motion of the canister after it is released, v0 y = +39.4 m/s , a y = −9.80 m/s 2 , y − y0 = −235 m . y − y0 = v0 yt + 12 a yt 2 gives −235 m = (39.4 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 12.0 s as the

positive solution. Then for the motion of the rocket during this 12.0 s, y − y0 = v0 yt + 12 a yt 2 = 235 m + (39.4 m/s)(12.0 s) + 12 (3.30 m/s 2 )(12.0 s) 2 = 945 m . (b) Find the maximum height of the canister above its release point: v0 y = +39.4 m/s , v y = 0 , a y = −9.80 m/s 2 .

v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y

2a y

=

0 − (39.4 m/s) 2 = 79.2 m . After its release the canister travels 2(−9.80 m/s 2 )

upward 79.2 m to its maximum height and then back down 79.2 m + 235 m to the ground. The total distance it travels is 393 m. EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is v y = v0 y + a yt = 39.4 m/s + (3.30 m/s 2 )(12.0 s) = 79.0 m/s . We can calculate its height at this instant by v y2 = v02y + 2a y ( y − y0 ) with v0 y = 0 and v y = 79.0 m/s . y − y0 =

2.92.

v y2 − v02y

2a y

=

(79.0 m/s) 2 = 946 m , which agrees 2(3.30 m/s 2 )

with our previous calculation. IDENTIFY: Both objects are in free-fall and move with constant acceleration 9.80 m/s 2 , downward. The two balls collide when they are at the same height at the same time. SET UP: Let + y be upward, so a y = −9.80 m/s 2 for each ball. Let y = 0 at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H. y0 A = 0 , y0B = H . EXECUTE:

(a) y − y0 = v0 yt + 12 a yt 2 applied to each ball gives y A = v0t − 12 gt 2 and yB = H − 12 gt 2 . y A = yB gives

v0t − 12 gt 2 = H − 12 gt 2 and t =

H . v0

(b) For ball A at its highest point, v yA = 0 and v y = v0 y + a y t gives t =

part (a) gives

H v0 v2 = and H = 0 . g v0 g

EVALUATE:

In part (a), using t =

less than

v0 . Setting this equal to the time in g

⎛ gH ⎞ H in the expressions for y A and yB gives y A = yB = H ⎜1 − 2 ⎟ . H must be v0 ⎝ 2v0 ⎠

2v02 in order for the balls to collide before ball A returns to the ground. This is because it takes ball A g


2-37

2v0 2v 2 2v 2 to return to the ground and ball B falls a distance 12 gt 2 = 0 during this time. When H = 0 the g g g two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before they collide. IDENTIFY and SET UP: Use vx = dx / dt and ax = dvx / dt to calculate vx (t ) and ax (t ) for each car. Use these equations to answer the questions about the motion. dx dv EXECUTE: x A = α t + β t 2 , v Ax = A = α + 2 β t , a Ax = Ax = 2 β dt dt dx dv xB = γ t 2 − δ t 3 , vBx = B = 2γ t − 3δ t 2 , aBx = Bx − 2γ − 6δ t dt dt (a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger v0 x . time t =

2.93.

EXECUTE:

At t = 0, v Ax = α and vBx = 0. So initially car A moves ahead.

(b) IDENTIFY and SET UP:

Cars at the same point implies x A = xB .

αt + βt = γ t − δ t EXECUTE: One solution is t = 0, which says that they start from the same point. To find the other solutions, 2

2

3

divide by t: α + β t = γ t − δ t 2

δ t 2 + ( β − γ )t + α = 0

)

(

)

(

1 1 −( β − γ ) ± ( β − γ ) 2 − 4δα = +1.60 ± (1.60) 2 − 4(0.20)(2.60) = 4.00 s ± 1.73 s 2δ 0.40 So x A = xB for t = 0, t = 2.27 s and t = 5.73 s. t=

EVALUATE: Car A has constant, positive ax . Its vx is positive and increasing. Car B has v0 x = 0 and ax that is initially positive but then becomes negative. Car B initially moves in the + x -direction but then slows down and finally reverses direction. At t = 2.27 s car B has overtaken car A and then passes it. At t = 5.73 s, car B is moving in the − x-direction as it passes car A again. d ( xB − x A ) . If this (c) IDENTIFY: The distance from A to B is xB − x A . The rate of change of this distance is dt d ( xB − x A ) distance is not changing, = 0. But this says vBx − v Ax = 0. (The distance between A and B is neither dt decreasing nor increasing at the instant when they have the same velocity.) SET UP: v Ax = vBx requires α + 2 β t = 2γ t − 3δ t 2 EXECUTE:

3δ t 2 + 2( β − γ )t + α = 0

)

(

(

1 1 3.20 ± 4(−1.60)2 − 12(0.20)(2.60) −2( β − γ ) ± 4( β − γ ) 2 − 12δα = 6δ 1.20 t = 2.667 s ± 1.667 s, so v Ax = vBx for t = 1.00 s and t = 4.33 s. t=

)

EVALUATE: At t = 1.00 s, v Ax = vBx = 5.00 m/s. At t = 4.33 s, v Ax = vBx = 13.0 m/s. Now car B is slowing down while A continues to speed up, so their velocities aren’t ever equal again. (d) IDENTIFY and SET UP: a Ax = aBx requires 2β = 2γ − 6δ t EXECUTE: EVALUATE: 2.94.

t=

γ − β 2.80 m/s 2 − 1.20 m/s 2 = = 2.67 s. 3δ 3(0.20 m/s3 )

At t = 0, aBx > a Ax , but aBx is decreasing while a Ax is constant. They are equal at t = 2.67 s but

for all times after that aBx < a Ax . IDENTIFY: The apple has two segments of motion with constant acceleration. For the motion from the tree to the top of the grass the acceleration is g, downward and the apple falls a distance H − h . For the motion from the top of the grass to the ground the acceleration is a, upward, the apple travels downward a distance h, and the final speed is zero. SET UP: Let + y be upward and let y = 0 at the ground. The apple is initially a height H + h above the ground. EXECUTE:

(a) Motion from y0 = H + h to y = H : y − y0 = − H , v0 y = 0 , a y = − g . v y2 = v02y + 2a y ( y − y0 ) gives

v y = − 2 gH . The speed of the apple is

2 gH as it enters the grass.

2-38

Chapter 2

(b) Motion from y0 = h to y = 0 : y − y0 = − h , v0 y = − 2 gH . v y2 = v02y + 2a y ( y − y0 ) gives

ay =

v y2 − v02y

2( y − y0 )

=

−2 gH gH = . The acceleration of the apple while it is in the grass is gH / h , upward. h 2( −h)

(c) Graphs of y-t, v y -t and a y -t are sketched in Figure 2.94. EVALUATE: increases.

The acceleration a produced by the grass increases when H increases and decreases when h

Figure 2.94 2.95.

IDENTIFY: Apply constant acceleration equations to the motion of the two objects, the student and the bus. SET UP: For convenience, let the student's (constant) speed be v0 and the bus's initial position be x0 . Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as

functions of time are then x1 = v0t and x2 = x0 + (1 2)at 2 . EXECUTE:

) ( (5.0 m s) − 2(0.170 m s )(40.0 m) ) = 9.55 s and 49.3 s .

(a) Setting x1 = x2 and solving for the times t gives t =

(

1 v0 ± v02 − 2ax0 . a

1 2 2 (5.0 m s) ± (0.170 m s 2 ) The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance v0t = (5 m s)(9.55 s) = 47.8 m. t=

(b) The speed of the bus is (0.170 m/s 2 )(9.55 s) = 1.62 m/s . (c) The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown in Figure 2.95a. (d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus's velocity is ( 0.170 m s 2 ) ( 49.3 s ) = 8.38 m s. (e) No; v02 < 2ax0 , and the roots of the quadratic are imaginary. When the student runs at 3.5 m s, Figure 2.95b shows that the two lines do not intersect: (f) For the student to catch the bus, v02 > 2ax0 . and so the minimum speed is

2 ( 0.170 m s 2 ) ( 40 m s ) = 3.688 m s. She would be running for a time distance ( 3.688 m s ) ( 21.7 s ) = 80.0 m.

3.69 m s = 21.7 s, and covers a 0.170 m/s 2

However, when the student runs at 3.688 m s, the lines intersect at one point, at x = 80 m , as shown in Figure 2.95c. EVALUATE: The graph in part (c) shows that the student is traveling faster than the bus the first time they meet but at the second time they meet the bus is traveling faster. t2 = ttot − t1

Figure 2.95


2.96.

2-39

Apply y − y0 = v0 yt + 12 a yt 2 to the motion from the maximum height, where v0 y = 0 . The time spent

IDENTIFY:

above ymax / 2 on the way down equals the time spent above ymax / 2 on the way up. SET UP:

Let + y be downward. a y = g . y − y0 = ymax / 2 when he is a distance ymax / 2 above the floor. The time from the maximum height to ymax / 2 above the floor is given by ymax / 2 = 12 gt12 . The time

EXECUTE:

2 and the time from a height of ymax / 2 to the floor is . from the maximum height to the floor is given by ymax = 12 gt tot

t1 = t2

ymax / 2 ymax − ymax / 2

EVALUATE: 2.97.

=

1 = 2.4 . 2 −1

The person spends over twice as long above ymax / 2 as below ymax / 2 . His average speed is less

above ymax / 2 than it is when he is below this height. IDENTIFY: Apply constant acceleration equations to both objects. SET UP: Let + y be upward, so each ball has a y = − g . For the purpose of doing all four parts with the least 1 1 2 repetition of algebra, quantities will be denoted symbolically. That is, let y1 = h + v0t − gt 2 , y2 = h − g ( t − t0 ) . 2 2 In this case, t0 = 1.00 s . EXECUTE: 1 2

(a) Setting y1 = y2 = 0, expanding the binomial ( t − t0 ) and eliminating the common term 2

gt 2 yields v0t = gt0t − 12 gt02 . Solving for t: t =

⎞ gt02 1 t ⎛ = 0⎜ ⎟. gt0 − v0 2 ⎝ 1 − v0 /( gt0 ) ⎠ 1 2

Substitution of this into the expression for y1 and setting y1 = 0 and solving for h as a function of v0 yields, after

( 12 gt0 − v0 ) 2 ( gt0 − v0 )

2

some algebra, h = 12 gt02

. Using the given value t0 = 1.00 s and g = 9.80 m s 2 ,

2 ⎛ 4.9 m s − v0 ⎞ h = 20.0 m = ( 4.9 m ) ⎜ ⎟ . ⎝ 9.8 m s − v0 ⎠ This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative square root before solving for v0 , and yields 8.2 m s. The graph of y versus t for each ball is given in Figure 2.97. (b) The above expression gives for (i), 0.411 m and for (ii) 1.15 km. (c) As v0 approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If v0 > 9.8 m s, the first ball can never catch the second ball. (d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer

(on its way down) to the top of the roof when the second ball is released. If v0 < 4.9 m s, the first ball will already have passed the roof on the way down before the second ball is released, and the second ball can never catch up. EVALUATE: Note that the values of v0 in parts (a) and (b) are all greater than vmin and less than vmax .

Figure 2.97 2.98.

IDENTIFY: Apply constant acceleration equations to the motion of the boulder. SET UP: Let + y be downward, so a y = + g .

2-40

Chapter 2

EXECUTE:

(a) Let the height be h and denote the 1.30-s interval as Δt ; the simultaneous equations

h = 12 gt 2 , 23 h = 12 g (t − Δt ) 2 can be solved for t. Eliminating h and taking the square root,

t 3 = , and t − Δt 2

Δt , and substitution into h = 12 gt 2 gives h = 246 m. 1 − 2/3 (b) The above method assumed that t > 0 when the square root was taken. The negative root (with Δt = 0) gives an answer of 2.51 m, clearly not a “cliff”. This would correspond to an object that was initially near the bottom of this “cliff” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer. EVALUATE: For the first two-thirds of the distance, y − y0 = 164 m , v0 y = 0 , and a y = 9.80 m/s 2 . t=

v y = 2a y ( y − y0 ) = 56.7 m/s . Then for the last third of the distance, y − y0 = 82.0 m , v0 y = 56.7 m/s and a y = 9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives (4.90 m/s 2 )t 2 + (56.7 m/s)t − 82.0 m = 0 . t=

(

)

1 −56.7 + (56.7) 2 + 4(4.9)(82.0) s = 1.30 s , as required. 9.8

3

MOTION IN TWO OR THREE DIMENSIONS

3.1.

IDENTIFY and SET UP: Use Eq.(3.2), in component form. Δx x2 − x1 5.3 m − 1.1 m = = = 1.4 m/s EXECUTE: ( vav ) x = Δt t2 − t1 3.0 s − 0

( vav ) y =

Δy y2 − y1 −0.5 m − 3.4 m = = = −1.3 m/s Δt t2 − t1 3.0 s − 0 tan α =

( vav ) y ( vav ) x

=

−1.3 m/s = −0.9286 1.4 m/s

α = 360° − 42.9° = 317° vav =

( vav ) x + ( vav ) y 2

2

vav = (1.4 m/s) 2 + (−1.3 m/s) 2 = 1.9 m/s Figure 3.1

3.2.

! EVALUATE: Our calculation gives that vav is in the 4th quadrant. This corresponds to increasing x and decreasing y. ! IDENTIFY: Use Eq.(3.2), written in component form. The distance from the origin is the magnitude of r . SET UP: At time t1 , x1 = y1 = 0 . EXECUTE:

(a) x = (vav-x )Δt = ( −3.8m s)(12.0 s) = −45.6 m and y = (vav-y )Δt = (4.9m s)(12.0 s) = 58.8 m .

(b) r = x 2 + y 2 = (−45.6 m) 2 + (58.8 m) 2 = 74.4 m. ! ! EVALUATE: Δr is in the direction of vav . Therefore, Δx is negative since vav-x is negative and Δy is positive

since vav-y is positive.

3.3.

! (a) IDENTIFY and SET UP: From r we can calculate x and y for any t. Then use Eq.(3.2), in component form. ! EXECUTE: r = ⎡⎣ 4.0 cm + ( 2.5 cm/s 2 ) t 2 ⎤⎦ iˆ + ( 5.0 cm/s ) tˆj ! At t = 0, r = ( 4.0 cm ) iˆ. ! At t = 2.0 s, r = (14.0 cm ) iˆ + (10.0 cm ) ˆj.

Δx 10.0 cm = = 5.0 cm/s. Δt 2.0 s Δy 10.0 cm = = = 5.0 cm/s. Δt 2.0 s

( vav ) x = ( vav ) y

3-1

3-2

Chapter 3

vav =

( vav ) x + ( vav ) y

tan α =

2

( vav ) y ( vav ) x

2

= 7.1 cm/s

= 1.00

θ = 45°. Figure 3.3a

! EVALUATE: Both x and y increase, so vav is in the 1st quadrant. ! ! (b) IDENTIFY and SET UP: Calculate r by taking the time derivative of r (t ). ! ! dr = ⎡⎣5.0 cm/s 2 ⎤⎦ t iˆ + ( 5.0 cm/s ) ˆj EXECUTE: v = dt t = 0 : vx = 0, v y = 5.0 cm/s; v = 5.0 cm/s and θ = 90°

(

)

t = 1.0 s: vx = 5.0 cm/s, v y = 5.0 cm/s; v = 7.1 cm/s and θ = 45° t = 2.0 s: vx = 10.0 cm/s, v y = 5.0 cm/s; v = 11 cm/s and θ = 27° (c) The trajectory is a graph of y versus x. x = 4.0 cm + (2.5 cm/s 2 )t 2 , y = (5.0 cm/s)t For values of t between 0 and 2.0 s, calculate x and y and plot y versus x.

Figure 3.3b 3.4.

3.5.

EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory. ! ! IDENTIFY: v = dr/dt . This vector will make a 45° -angle with both axes when its x- and y-components are equal. d (t n ) SET UP: = nt n −1 . dt ! EXECUTE: v = 2btiˆ + 3ct 2 ˆj . vx = v y gives t = 2b 3c . ! EVALUATE: Both components of v change with t. IDENTIFY and SET UP: Use Eq.(3.8) in component form to calculate ( aav ) x and ( aav ) y .

Motion in Two or Three Dimensions 3-3

EXECUTE:

(a) The velocity vectors at t1 = 0 and t2 = 30.0 s are shown in Figure 3.5a.

Figure 3.5a (b) ( aav ) x =

( aav ) y =

Δv y Δt

Δvx v2 x − v1x −170 m/s − 90 m/s = = = −8.67 m/s 2 Δt t2 − t1 30.0 s =

v2 y − v1 y t2 − t1

=

40 m/s − 110 m/s = −2.33 m/s 2 30.0 s

(c)

a=

( aav ) x + ( aav ) y

tan α =

2

( aav ) y ( aav ) x

2

=

= 8.98 m/s 2

−2.33 m/s 2 = 0.269 −8.67 m/s 2

α = 15° + 180° = 195° Figure 3.5b EVALUATE: 3.6.

! The changes in vx and v y are both in the negative x or y direction, so both components of aav are in

the 3rd quadrant. IDENTIFY: Use Eq.(3.8), written in component form. 2 2 2 2 SET UP: ax = (0.45m s )cos31.0° = 0.39m s , a y = (0.45m s )sin 31.0° = 0.23m s EXECUTE:

(a) aav-x =

Δv Δvx 2 and vx = 2.6 m s + (0.39 m s )(10.0 s) = 6.5 m s . aav-y = y and Δt Δt 2

v y = −1.8 m s + (0.23 m s )(10.0 s) = 0.52 m s . ⎛ 0.52 ⎞ (b) v = (6.5m s) 2 + (0.52 m s) 2 = 6.48m s , at an angle of arctan ⎜ ⎟ = 4.6° above the horizontal. ⎝ 6.5 ⎠ ! ! (c) The velocity vectors v1 and v2 are sketched in Figure 3.6. The two velocity vectors differ in magnitude and direction. ! EVALUATE: v1 is at an angle of 35° below the + x -axis and has magnitude v1 = 3.2 m/s , so v2 > v1 and the ! ! direction of v2 is rotated counterclockwise from the direction of v1 .

Figure 3.6 3.7.

IDENTIFY and SET UP: Use Eqs.(3.4) and (3.12) to find vx , v y , ax , and a y as functions of time. The magnitude ! ! and direction of r and a can be found once we know their components.

3-4

Chapter 3

EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are given in Figure 3.7a.

Figure 3.7a

dx dy = α vy = = −2 β t dt dt dv dv a y = x = 0 a y = y = −2 β dt dt ! ! Thus v = aiˆ − 2 β tˆj a = −2 β ˆj (b) vx =

(c) velocity: At t = 2.0 s, vx = 2.4 m/s, v y = −2(1.2 m/s 2 )(2.0 s) = −4.8 m/s

v = vx2 + v y2 = 5.4 m/s −4.8 m/s = −2.00 2.4 m/s vx α = −63.4° + 360° = 297°

tan α =

vy

=

Figure 3.7b

acceleration: At t = 2.0 s, ax = 0, a y = −2(1.2 m/s 2 ) = −2.4 m/s 2 a = ax2 + a y2 = 2.4 m/s 2 tan β =

ay ax

=

−2.4 m/s 2 = −∞ 0

β = 270° Figure 3.7c

! EVALUATE: (d) a has a component a" in the same direction as ! v , so we know that v is increasing (the bird is speeding up.) ! ! a also has a component a⊥ perpendicular to v , so that the ! direction of v is changing; the bird is turning toward the − y -direction (toward the right) Figure 3.7d


3.8.

! ! v is always tangent to the path; v at t = 2.0 s shown in part (c) is tangent to the path at this t, conforming to this ! ! general rule. a is constant and in the − y -direction; the direction of v is turning toward the − y -direction. ! ! ! IDENTIFY: The component a⊥ of a perpendicular to the path is related to the change in direction of v and the ! ! ! component a" of a parallel to the path is related to the change in the magnitude of v . ! ! ! SET UP: When the speed is increasing, a" is in the direction of v and when the speed is decreasing, a" is opposite ! to the direction of v . When v is constant, a" is zero and when the path is a straight line, a⊥ is zero. EXECUTE: The acceleration vectors in each case are sketched in Figure 3.8a-c. ! EVALUATE: a⊥ is toward the center of curvature of the path.

Figure 3.8a-c 3.9.

IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal. SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position of the book, at the point where it leaves the table top.

x-component: ax = 0, v0 x = 1.10 m/s, t = 0.350 s y-component: a y = −9.80 m/s 2 , v0 y = 0, t = 0.350 s

Figure 3.9a

Use constant acceleration equations for the x and y components of the motion, with ax = 0 and a y = − g . EXECUTE:

(a) y − y0 = ?

y − y0 = v0 yt + 12 a yt 2 = 0 + 12 ( −9.80 m/s 2 )(0.350 s) 2 = −0.600 m. The table top is 0.600 m above the floor. (b) x − x0 = ?

x − x0 = v0 xt + 12 axt 2 = (1.10 m/s)(0.350 s) + 0 = 0.358 m. (c) vx = v0 x + axt = 1.10 m/s (The x-component of the velocity is constant, since ax = 0.)

v y = v0 y + a yt = 0 + ( −9.80 m/s 2 )(0.350 s) = −3.43 m/s

v = vx2 + v y2 = 3.60 m/s −3.43 m/s = = −3.118 vx 1.10 m/s α = −72.2° ! Direction of v is 72.2° below the horizontal

tan α =

Figure 3.9b

vy

3-6

Chapter 3

(d) The graphs are given in Figure 3.9c

Figure 3.9c EVALUATE:

In the x-direction, ax = 0 and vx is constant. In the y-direction, a y = −9.80 m/s 2 and v y is

downward and increasing in magnitude since a y and v y are in the same directions. The x and y motions occur

3.10.

independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it travels horizontally. IDENTIFY: The bomb moves in projectile motion. Treat the horizontal and vertical components of the motion separately. The vertical motion determines the time in the air. SET UP: The initial velocity of the bomb is the same as that of the helicopter. Take + y downward, so ax = 0 , a y = +9.80 m/s 2 , v0 x = 60.0 m/s and v0 y = 0 . EXECUTE:

(a) y − y0 = v0 yt + 12 a yt 2 with y − y0 = 300 m gives t =

2( y − y0 ) 2(300 m) = = 7.82 s . ay 9.80 m/s 2

(b) The bomb travels a horizontal distance x − x0 = v0 xt + 12 axt 2 = (60.0 m/s)(7.82 s) = 470 m . (c) vx = v0 x = 60.0 m/s . v y = v0 y + a yt = (9.80 m/s 2 )(7.82 s) = 76.6 m/s . (d) The graphs are given in Figure 3.10. (e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m directly above the bomb at impact. EVALUATE: The initial horizontal velocity of the bomb doesn’t affect its vertical motion.

Figure 3.10 3.11.

IDENTIFY: Each object moves in projectile motion. SET UP: Take + y to be downward. For each cricket, ax = 0 and a y = +9.80 m/s 2 . For Chirpy, v0 x = v0 y = 0 . For

Milada, v0 x = 0.950 m/s , v0 y = 0

3.12.

EXECUTE: Milada's horizontal component of velocity has no effect on her vertical motion. She also reaches the ground in 3.50 s. x − x0 = v0 xt + 12 axt 2 = (0.950 m/s)(3.50 s) = 3.32 m EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that the time is the same for both. IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time she falls 9.00 m vertically. SET UP: Take + y downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0 . EXECUTE:

Time to fall 9.00 m: y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(9.00 m) = = 1.36 s . ay 9.80 m/s 2

Speed needed to travel 1.75 m horizontally during this time: x − x0 = v0 xt + 12 axt 2 gives x − x0 1.75 m = = 1.29 m/s . t 1.36 s EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has traveled horizontally farther than 1.75 m. IDENTIFY: The car moves in projectile motion. The car travels 21.3 m − 1.80 m = 19.5 m downward during the time it travels 61.0 m horizontally. SET UP: Take + y to be downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0 . v0 = v0 x =

3.13.


Use the vertical motion to find the time in the air: 2( y − y0 ) 2(19.5 m) = = 1.995 s y − y0 = v0 yt + 12 a yt 2 gives t = ay 9.80 m/s 2

EXECUTE:

Then x − x0 = v0 xt + 12 axt 2 gives v0 = v0 x =

x − x0 61.0 m = = 30.6 m/s . t 1.995 s

(b) vx = 30.6m s since ax = 0 . v y = v0 y + a yt = −19.6m s . v = vx2 + v y2 = 36.3m s . 3.14.

EVALUATE: We calculate the final velocity by calculating its x and y components. IDENTIFY: The marble moves with projectile motion, with initial velocity that is horizontal and has magnitude v0 . Treat the horizontal and vertical motions separately. If v0 is too small the marble will land to the

left of the hole and if v0 is too large the marble will land to the right of the hole. SET UP:

Let + x be horizontal to the right and let + y be upward. v0 x = v0 , v0 y = 0 , ax = 0 , a y = −9.80 m/s 2

Use the vertical motion to find the time it takes the marble to reach the height of the level ground; 2( y − y0 ) 2( −2.75 m) y − y0 = −2.75 m . y − y0 = v0 yt + 12 a yt 2 gives t = = = 0.749 s . The time does not depend ay −9.80 m/s 2

EXECUTE:

on v0 . Minimum v0 : x − x0 = 2.00 m , t = 0.749 s . x − x0 = v0 xt + 12 axt 2 gives v0 =

x − x0 2.00 m = = 2.67 m/s . t 0.749 s

3.50 m = 4.67 m/s . 0.749 s EVALUATE: The horizontal and vertical motions are independent and are treated separately. Their only connection is that the time is the same for both. IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has magnitude v0 . Maximum v0 : x − x0 = 3.50 m and v0 =

3.15.

The height h of the table and v0 are the same; the acceleration due to gravity changes from g E = 9.80 m/s 2 on earth to g X on planet X. SET UP: Let + x be horizontal and in the direction of the initial velocity of the marble and let + y be upward.

v0 x = v0 , v0 y = 0 , ax = 0 , a y = − g , where g is either g E or g X . EXECUTE:

Use the vertical motion to find the time in the air: y − y0 = − h . y − y0 = v0 yt + 12 a yt 2 gives t =

Then x − x0 = v0 xt + 12 axt 2 gives x − x0 = v0 xt = v0

2h . x − x0 = D on earth and 2.76D on Planet X. g

gE = 0.131g E = 1.28 m/s 2 . (2.76) 2 EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor, and it travels farther horizontally. IDENTIFY: The football moves in projectile motion. SET UP: Let + y be upward. ax = 0 , a y = − g . At the highest point in the trajectory, v y = 0 . ( x − x0 ) g = v0 2h , which is constant, so D g E = 2.76 D g X . g X =

3.16.

2h . g

v0 y

16.0m s = 1.63 s . 2 9.80m s (b) Different constant acceleration equations give different expressions but the same numerical result: v2 1 gt 2 = 12 v y 0t = 0 y = 13.1 m . 2 2g (c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s (d) ax = 0 , so x − x0 = v0 xt = (20.0 m s)(3.27 s) = 65.3 m . (e) The graphs are sketched in Figure 3.16. EXECUTE:

(a) v y = v0 y + a y t . The time t is

g

=

3-8

Chapter 3

EVALUATE:

When the football returns to its original level, vx = 20.0 m/s and v y = −16.0 m/s .

Figure 3.16 3.17.

IDENTIFY: The shell moves in projectile motion. SET UP: Let + x be horizontal, along the direction of the shell's motion, and let + y be upward. ax = 0 ,

a y = −9.80 m/s 2 . EXECUTE:

(a) v0 x = v0 cos α 0 = (80.0 m/s)cos60.0° = 40.0 m/s , v0 y = v0 sin α 0 = (80.0 m/s)sin60.0° = 69.3 m/s .

(b) At the maximum height v y = 0 . v y = v0 y + a y t gives t = (c) v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y 2a y

=

v y − v0 y ay

=

0 − 69.3 m/s = 7.07 s . −9.80 m/s 2

0 − (69.3 m/s) 2 = 245 m . 2(−9.80 m/s 2 )

(d) The total time in the air is twice the time to the maximum height, so x − x0 = v0 xt + 12 axt 2 = (40.0 m/s)(14.14 s) = 566 m . (e) At the maximum height, vx = v0 x = 40.0 m/s and v y = 0 . At all points in the motion, ax = 0 and

a y = −9.80 m/s 2 . EVALUATE:

The equation for the horizontal range R derived in Example 3.8 is R =

v02 sin 2α 0 . This gives g

(80.0 m/s) 2 sin(120.0°) = 566 m , which agrees with our result in part (d). 9.80 m/s 2 IDENTIFY: The flare moves with projectile motion. The equations derived in Example 3.8 can be used to find the maximum height h and range R. v 2 sin 2 α 0 v 2 sin 2α 0 and R = 0 . SET UP: From Example 3.8, h = 0 2g g R=

3.18.

EXECUTE:

(a) h =

(125 m/s) 2 (sin 55.0°) 2 (125 m/s) 2 (sin110.0°) . = 535 m R = = 1500 m . 2(9.80 m/s 2 ) 9.80 m/s 2

⎛ 9.80 m/s 2 ⎞ (b) h and R are proportional to 1/ g , so on the Moon, h = ⎜ (535 m) = 3140 m and 2 ⎟ ⎝ 1.67 m/s ⎠

3.19.

⎛ 9.80 m/s 2 ⎞ R=⎜ (1500 m) = 8800 m . 2 ⎟ ⎝ 1.67 m/s ⎠ EVALUATE: The projectile travels on a parabolic trajectory. It is incorrect to say that h = ( R / 2) tan α 0 . IDENTIFY: The baseball moves in projectile motion. In part (c) first calculate the components of the velocity at this point and then get the resultant velocity from its components. SET UP: First find the x- and y-components of the initial velocity. Use coordinates where the + y -direction is upward, the + x -direction is to the right and the origin is at the point where the baseball leaves the bat.

v0 x = v0 cos α 0 = (30.0 m/s)cos36.9° = 24.0 m/s v0 y = v0 sin α 0 = (30.0 m/s)sin 36.9° = 18.0 m/s

Figure 3.19a

Use constant acceleration equations for the x and y motions, with ax = 0 and a y = − g .


EXECUTE: (a) y-component (vertical motion): y − y0 = +10.0 m/s, v0 y = 18.0 m/s, a y = −9.80 m/s 2 , t = ?

y − y0 = v0 y + 12 a yt 2 10.0 m = (18.0 m/s)t − (4.90 m/s 2 )t 2 (4.90 m/s 2 )t 2 − (18.0 m/s)t + 10.0 m = 0 1 ⎡ Apply the quadratic formula: t = 9.80 18.0 ± ( −18.0 ) − 4 ( 4.90 )(10.0 ) ⎤ s = (1.837 ± 1.154 ) s ⎢⎣ ⎥⎦ The ball is at a height of 10.0 above the point where it left the bat at t1 = 0.683 s and at t2 = 2.99 s. At the earlier time the ball passes through a height of 10.0 m as its way up and at the later time it passes through 10.0 m on its way down. (b) vx = v0 x = +24.0 m/s, at all times since ax = 0. 2

v y = v0 y + a y t t1 = 0.683 s: v y = +18.0 m/s + (−9.80 m/s 2 )(0.683 s) = +11.3 m/s. ( v y is positive means that the ball is traveling upward at this point. t2 = 2.99 s: v y = +18.0 m/s + (−9.80 m/s 2 )(2.99 s) = −11.3 m/s. ( v y is negative means that the ball is traveling downward at this point.) (c) vx = v0 x = 24.0 m/s Solve for v y : v y = ?, y − y0 = 0 (when ball returns to height where motion started), a y = −9.80 m/s 2 , v0 y = +18.0 m/s v y2 = v02y + 2a y ( y − y0 ) v y = −v0 y = −18.0 m/s (negative, since the baseball must be traveling downward at this point) ! Now that have the components can solve for the magnitude and direction of v . v = vx2 + v y2 v=

( 24.0 m/s )

2

+ ( −18.0 m/s ) = 30.0 m/s 2

−18.0 m/s = vx 24.0 m/s α = −36.9°, 36.9° below the horizontal tan α =

vy

Figure 3.19b

3.20.

The velocity of the ball when it returns to the level where it left the bat has magnitude 30.0 m/s and is directed at an angle of 36.9° below the horizontal. EVALUATE: The discussion in parts (a) and (b) explains the significance of two values of t for which y − y0 = +10.0 m. When the ball returns to its initial height, our results give that its speed is the same as its initial speed and the angle of its velocity below the horizontal is equal to the angle of its initial velocity above the horizontal; both of these are general results. IDENTIFY: The shot moves in projectile motion. SET UP: Let + y be upward. EXECUTE: (a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and 2 − g = −9.80 m s vertically downward. (b) The x-component of velocity is constant at vx = (12.0 m s)cos51.0° = 7.55 m s . The y-component is 2

v0 y = (12.0 m s)sin 51.0° = 9.32 m s at release and v y = v0 y − gt = (10.57 m s) − (9.80 m s )(2.08 s) = −11.06 m s when the shot hits. (c) x − x0 = v0 xt = (7.55 m s)(2.08 s) = 15.7 m . (d) The initial and final heights are not the same. (e) With y = 0 and v0 y as found above, Eq.(3.18) gives y0 = 1.81m . (f) The graphs are sketched in Figure 3.20.

3-10

Chapter 3

EVALUATE:

When the shot returns to its initial height, v y = −9.32 m/s . The shot continues to accelerate

downward as it travels downward 1.81 m to the ground and the magnitude of v y at the ground is larger than 9.32 m/s.

Figure 3.20 3.21.

IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take positive y to be upward. The quarter moves in projectile motion, with ax = 0, and a y = − g . It travels vertically for the time it

takes it to travel horizontally 2.1 m. v0 x = v0 cos α 0 = (6.4 m/s)cos60° v0 x = 3.20 m/s v0 y = v0 sin α 0 = (6.4 m/s)sin 60° v0 y = 5.54 m/s Figure 3.21 (a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels through the air: t = ?, x − x0 = 2.1 m, v0 x = 3.2 m/s, ax = 0

x − x0 = v0 xt + 12 axt 2 = v0 xt , since ax = 0 x − x0 2.1 m = = 0.656 s v0 x 3.2 m/s SET UP: Now find the vertical displacement of the quarter after this time: y − y0 = ?, a y = −9.80 m/s 2 , v0 y = +5.54 m/s, t = 0.656 s EXECUTE:

t=

y − y0 + v0 yt + 12 a yt 2 EXECUTE: (b) SET UP:

y − y0 = (5.54 m/s)(0.656 s) + 12 (−9.80 m/s 2 )(0.656 s) 2 = 3.63 m − 2.11 m = 1.5 m. v y = ?, t = 0.656 s, a y = −9.80 m/s 2 , v0 y = +5.54 m/s

v y = v0 y + a y t EXECUTE: EVALUATE:

3.22.

v y = 5.54 m/s + (−9.80 m/s 2 )(0.656 s) = −0.89 m/s.

! The minus sign for v y indicates that the y-component of v is downward. At this point the quarter

has passed through the highest point in its path and is on its way down. The horizontal range if it returned to its original height (it doesn’t!) would be 3.6 m. It reaches its maximum height after traveling horizontally 1.8 m, so at x − x0 = 2.1 m it is on its way down. IDENTIFY: Use the analysis of Example 3.10. d SET UP: From Example 3.10, t = and ydart = (v0 sin α 0 )t − 12 gt 2 . v0 cos α 0 EXECUTE:

Substituting for t in terms of d in the expression for ydart gives ⎛ ⎞ gd ydart = d ⎜ tan α 0 − 2 ⎟. 2 α 2 v cos 0 0 ⎠ ⎝

Using the given values for d and α 0 to express this as a function of v0 , ⎛ 26.62 m 2 s 2 ⎞ y = (3.00 m) ⎜ 0.90 − ⎟. v02 ⎝ ⎠ (a) v0 = 12.0 m/s gives y = 2.14 m . (b) v0 = 8.0 m/s gives y = 1.45 m .


3.23.

(c) v0 = 4.0 m/s gives y = −2.29 m . In this case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal distance. EVALUATE: For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c) the dart moves in a parabola and returns to the ground before it reaches the x-coordinate of the monkey. IDENTIFY: Take the origin of coordinates at the roof and let the + y -direction be upward. The rock moves in

projectile motion, with ax = 0 and a y = − g . Apply constant acceleration equations for the x and y components of the motion. SET UP:

v0 x = v0 cos α 0 = 25.2 m/s v0 y = v0 sin α 0 = 16.3 m/s

Figure 3.23a (a) At the maximum height v y = 0.

a y = −9.80 m/s 2 , v y = 0, v0 y = +16.3 m/s, y − y0 = ? v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

y − y0 =

v y2 − v02y 2a y

=

0 − (16.3 m/s) 2 = +13.6 m 2(−9.80 m/s 2 )

(b) SET UP: Find the velocity by solving for its x and y components. vx = v0 x = 25.2 m/s (since ax = 0 )

v y = ?, a y = −9.80 m/s 2 , y − y0 = −15.0 m (negative because at the ground the rock is below its initial position), v0 y = 16.3 m/s v y2 = v02y + 2a y ( y − y0 ) v y = − v02y + 2a y ( y − y0 ) ( v y is negative because at the ground the rock is traveling downward.) EXECUTE:

v y = − (16.3 m/s) 2 + 2(−9.80 m/s 2 )( −15.0 m) = −23.7 m/s

Then v = vx2 + v y2 = (25.2 m/s) 2 + (−23.7 m/s) 2 = 34.6 m/s. (c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air: t = ?, v y = −23.7 m/s (from part (b)), a y = −9.80 m/s 2 , v0 y = +16.3 m/s EXECUTE:

t=

v y − v0 y ay

=

−23.7 m/s − 16.3 m/s = +4.08 s −9.80 m/s 2

SET UP: Can use this t to calculate the horizontal range: t = 4.08 s, v0 x = 25.2 m/s, ax = 0, x − x0 = ? EXECUTE:

x − x0 = v0 xt + 12 axt 2 = (25.2 m/s)(4.08 s) + 0 = 103 m

(d) Graphs of x versus t, y versus t, vx versus t, and v y versus t:

Figure 3.23b

3-12

Chapter 3

EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally. With v0 y = +16.3 m/s the time it takes the rock to return to the level of the roof ( y = 0) is t = 2v0 y / g = 3.33 s. The 3.24.

time in the air is greater than this because the rock travels an additional 15.0 m to the ground. IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m horizontally in 3.00 s. SET UP: Let + y be upward. ax = 0 , a y = −9.80 m/s 2 . v0 x = v0 cosθ 0 , v0 y = v0 sin θ 0 . EXECUTE:

(a) x − x0 = v0 xt + 12 axt 2 gives x − x0 = v0 (cosθ 0 )t and cosθ 0 =

45.0 m = 0.600 ; θ 0 = 53.1° (25.0 m/s)(3.00 s)

(b) At the highest point vx = v0 x = (25.0 m/s)cos53.1° = 15.0 m/s , v y = 0 and v = vx2 + v y2 = 15.0 m/s . At all points

in the motion, a = 9.80 m/s 2 downward. (c) Find y − y0 when t = 3.00s : y − y0 = v0 yt + 12 a yt 2 = (25.0 m/s)(sin53.1°)(3.00 s) + 12 (−9.80 m/s 2 )(3.00 s)2 = 15.9 m vx = v0 x = 15.0 m/s , v y = v0 y + a yt = (25.0 m/s)(sin53.1°) − (9.80m/s 2 )(3.00 s) = −9.41 m/s , and v = vx 2 + v y 2 = (15.0 m/s) 2 + (−9.41 m/s) 2 = 17.7 m/s EVALUATE: The acceleration is the same at all points of the motion. It takes the water v 20.0 m/s t = − 0y = − = 2.04 s to reach its maximum height. When the water reaches the building it has passed ay −9.80 m/s 2 3.25.

its maximum height and its vertical component of velocity is downward. IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the balloon. Use constant acceleration equations for the x and y components of its motion. Take + y to be upward. EXECUTE: (a) Use the vertical motion of the rock to find the initial height. t = 6.00 s, v0 y = +20.0 m/s, a y = +9.80 m/s 2 , y − y0 = ? y − y0 = v0 yt + 12 a yt 2 gives y − y0 = 296 m (b) In 6.00 s the balloon travels downward a distance y − y0 = (20.0 s)(6.00 s) = 120 m. So, its height above ground when the rock hits is 296 m − 120 m = 176 m. (c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the

rock and the basket is 176 m, so the rock is (176 m) 2 + (90 m) 2 = 198 m from the basket when it hits the ground. (d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the basket. Just before the rock hits the ground, its vertical component of velocity is v y = v0 y + a yt =

3.26.

20.0 m/s + (9.80 m/s 2 )(6.00 s) = 78.8 m/s, downward, relative to the ground. The basket is moving downward at 20.0 m/s, so relative to the basket the rock has downward component of velocity 58.8 m/s. (e) horizontal: 15.0 m/s; vertical: 78.8 m/s EVALUATE: The rock has a constant horizontal velocity and accelerates downward IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have y − y0 = 25.0 m when it has x − x0 = 60.0 m . SET UP: EXECUTE:

Let + y be upward. ax = 0 , a y = − g . v0 x = v0 cos 43° , v0 y = v0 sin 43° . (a) horizontal motion: x − x0 = v0 xt so t =

60.0 m . (v0 cos 43°)t

vertical motion: y − y0 = v0 yt + 12 a yt 2 gives 25.0m = (v0 sin 43.0°)t + 12 ( −9.80m/s 2 )t 2 . Solving these two simultaneous equations for v0 and t gives v0 = 3.26 m/s and t = 2.51 s . (b) v y when shell reaches cliff:

v y = v0 y + a yt = (32.6 m/s) sin 43.0° − (9.80 m/s 2 )(2.51 s) = −2.4 m/s The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. v EVALUATE: The shell reaches its maximum height at t = − 0 y = 2.27 s , which confirms that at t = 2.51 s it has ay 3.27.

passed its maximum height and is on its way down when it strikes the edge of the cliff. IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity of the airplane.


Take + y to be upward. ax = 0 , a y = − g .

SET UP:

EXECUTE: Use the vertical motion to find the time it takes the suitcase to reach the ground: v0 y = v0 sin23°, a y = −9.80 m/s 2 , y − y0 = −114 m, t = ? y − y0 = v0 yt + 12 a yt 2 gives t = 9.60 s .

The distance the suitcase travels horizontally is x − x0 = v0 x = (v0 cos23.0°)t = 795 m . EVALUATE: 3.28.

An object released from rest at a height of 114 m strikes the ground at t =

2( y − y0 ) = 4.82 s . The −g

suitcase is in the air much longer than this since it initially has an upward component of velocity. IDENTIFY: Determine how arad depends on the rotational period T. 4π 2 R . T2 EXECUTE: For any item in the washer, the centripetal acceleration will be inversely proportional to the square of the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that arad =

SET UP:

3.29.

the new period T ′ is given in terms of the previous period T by T ′ = T / 3 . EVALUATE: The rotational period must be decreased in order to increase the rate of rotation and therefore increase the centripetal acceleration. IDENTIFY: Apply Eq. (3.30). SET UP: T = 24 h . 4π 2 (6.38 × 106 m) = 0.034 m/s 2 = 3.4 × 10−3 g . EXECUTE: (a) arad = ((24 h)(3600 s/h)) 2 4π 2 (6.38 × 106 m) = 5070 s =1.4 h. 9.80 m/s 2 1 EVALUATE: arad is proportional to 1/T 2 , so to increase arad by a factor of = 294 requires that T be 3.4 × 10−3 1 24 h multiplied by a factor of . = 1.4 h . 294 294 IDENTIFY: Each blade tip moves in a circle of radius R = 3.40 m and therefore has radial acceleration arad = v 2 / R . (b) Solving Eq. (3.30) for the period T with arad = g , T =

3.30.

550 rev/min = 9.17 rev/s , corresponding to a period of T =

SET UP:

(a) v =

EXECUTE:

1 = 0.109 s . 9.17 rev/s

2π R = 196 m/s . T

v2 = 1.13 × 104 m/s 2 = 1.15 × 103 g . R 4π 2 R EVALUATE: arad = gives the same results for arad as in part (b). T2 IDENTIFY: Apply Eq.(3.30). SET UP: R = 7.0 m . g = 9.80 m/s 2 . (b) arad =

3.31.

(a) Solving Eq. (3.30) for T in terms of R and arad ,

EXECUTE:

T = 4π R / arad = 4π 2 (7.0 m)/(3.0)(9.80 m/s 2 ) = 3.07 s . 2

(b) arad = 10 g gives T = 1.68 s . EVALUATE: 3.32.

IDENTIFY: SET UP:

When arad increases, T decreases. Each planet moves in a circular orbit and therefore has acceleration arad = v 2 / R .

The radius of the earth’s orbit is r = 1.50 × 1011 m and its orbital period is T = 365 days = 3.16 × 107 s .

For Mercury, r = 5.79 × 1010 m and T = 88.0 days = 7.60 × 106 s . EXECUTE:

(a) v =

2π r = 2.98 × 104 m/s T

v2 = 5.91 × 10−3 m/s 2 . r (c) v = 4.79 × 104 m/s , and arad = 3.96 × 10−2 m/s 2 . (b) arad =

3-14

3.33.

3.34.

Chapter 3

EVALUATE: Mercury has a larger orbital velocity and a larger radial acceleration than earth. IDENTIFY: Uniform circular motion. ! dv ! SET UP: Since the magnitude of v is constant. vtan = = 0 and the resultant acceleration is equal to the radial dt component. At each point in the motion the radial component of the acceleration is directed in toward the center of the circular path and its magnitude is given by v 2 / R. v 2 (7.00 m/s) 2 EXECUTE: (a) arad = = = 3.50 m/s 2 , upward. R 14.0 m (b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the circle is downward. The acceleration at this point in the motion is 3.50 m/s 2 , downward. (c) SET UP: The time to make one rotation is the period T, and the speed v is the distance for one revolution divided by T. 2π R 2π R 2π (14.0 m) EXECUTE: v = so T = = = 12.6 s T v 7.00 m/s EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in the ! motion directed toward the center of the circular path. The acceleration is perpendicular to v and is nonzero ! because the direction of v changes. IDENTIFY: The acceleration is the vector sum of the two perpendicular components, arad and atan . ! SET UP: atan is parallel to v and hence is associated with the change in speed; atan = 0.500 m/s 2 . EXECUTE:

(a) arad = v 2 / R = (3 m/s) 2 /(14 m) = 0.643 m/s 2 .

a = ((0.643 m/s 2 ) 2 + (0.5 m/s 2 ) 2 )1/ 2 = 0.814 m/s 2 , 37.9° to the right of vertical. (b) The sketch is given in Figure 3.34.

Figure 3.34 3.35.

v2 . The speed in rev/s is 1/T , R where T is the period in seconds (time for 1 revolution). The speed v increases with R along the length of his body but all of him rotates with the same period T. SET UP: For his head R = 8.84 m and for his feet R = 6.84 m . IDENTIFY:

Each part of his body moves in uniform circular motion, with arad =

EXECUTE:

(a) v = Rarad = (8.84 m)(12.5)(9.80 m/s 2 ) = 32.9 m/s

(b) Use arad =

T = 2π

4π 2 R . Since his head has arad = 12.5 g and R = 8.84 m , T2

R 8.84 m R 4π 2 (6.84 m) 1.688 s a = 2π = . Then his feet have = = = 94.8 m/s 2 = 9.67 g . rad 12.5(9.80 m/s 2 ) (1.688 s) 2 arad T2

The difference between the acceleration of his head and his feet is 12.5 g − 9.67 g = 2.83 g = 27.7 m/s 2 . (c)

1 1 = = 0.592 rev/s = 35.5 rpm T 1.69 s

His feet have speed v = Rarad = (6.84 m)(94.8 m/s 2 ) = 25.5 m/s ! ! IDENTIFY: The relative velocities are vS/F , the velocity of the scooter relative to the flatcar, vS/G , the scooter ! ! ! ! relative to the ground and vF/G , the flatcar relative to the ground. vS/G = vS/F + vF/G . Carry out the vector addition by drawing a vector addition diagram. ! ! ! ! ! SET UP: vS/F = vS/G − vF/G . vF/G is to the right, so −vF/G is to the left. EXECUTE: In each case the vector addition diagram gives EVALUATE:

3.36.


3.37.

(a) 5.0 m/s to the right (b) 16.0 m/s to the left (c) 13.0 m/s to the left. EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative to the ! ! flatcar, since in that case the two velocities vS/F and vF/G are in the same direction and their magnitudes add. IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length divided by the velocity of the woman relative to the ground. SET UP: Let W stand for the woman, G for the ground, and S for the sidewalk. Take the positive direction to be the direction in which the sidewalk is moving. The velocities are vW/G (woman relative to the ground), vW/S (woman relative to the sidewalk), and vS/G (sidewalk relative to the ground). Eq.(3.33) becomes vW/G = vW/S + vS/G .

The time to reach the other end is given by t = EXECUTE:

distance traveled relative to ground vW/G

(a) vS/G = 1.0 m/s

vW/S = +1.5 m/s vW/G = vW/S + vS/G = 1.5 m/s + 1.0 m/s = 2.5 m/s. t=

35.0 m 35.0 m = = 14 s. vW/G 2.5 m/s

(b) vS/G = 1.0 m/s

vW/S = −1.5 m/s

3.38.

3.39.

3.40.

vW/G = vW/S + vS/G = −1.5 m/s + 1.0 m/s = −0.5 m/s. (Since vW/G now is negative, she must get on the moving sidewalk at the opposite end from in part (a).) −35.0 m −35.0 m = = 70 s. t= vW/G −0.5 m/s EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion of the sidewalk. IDENTIFY: Calculate the rower’s speed relative to the shore for each segment of the round trip. SET UP: The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream. EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). 1.5 km 1.5 km + = 1.47 h = 88.2 min. The total time the rower takes is 6.8 km/h 1.2 km/h EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h. The rower spends more time at the slower speed. IDENTIFY: Apply the relative velocity relation. ! ! SET UP: The relative velocities are vC/E , the canoe relative to the earth, vR/E , the velocity of the river relative to ! the earth and vC/R , the velocity of the canoe relative to the river. ! ! ! ! ! ! ! EXECUTE: vC/E = vC/R + vR/E and therefore vC/R = vC/E − vR/E . The velocity components of vC/R are

−0.50 m/s + (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south, for a velocity relative to the river of 0.36 m/s, at 52.5° south of west. EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of the canoe relative to the earth. IDENTIFY: Use the relation that relates the relative velocities. ! SET UP: The relative velocities are the velocity of the plane relative to the ground, vP/G , the velocity of the plane ! ! ! ! relative to the air, vP/A , and the velocity of the air relative to the ground, vA/G . vP/G must due west and vA/G must ! ! ! be south. vA/G = 80 km/h and vP/A = 320 km/h . vP/G = vP/A + vA/G . The relative velocity addition diagram is given in Figure 3.40. v 80 km/h EXECUTE: (a) sin θ = A/G = and θ = 14° , north of west. vP/A 320 km/h 2 2 (b) vP/G = vP/A − vA/G = (320 km/h) 2 − (80.0 km/h) 2 = 310 km/h .

3-16

Chapter 3

EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward component and also a component that is northward, opposite to the wind direction.

Figure 3.40 3.41.

IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time displacement) depends on his velocity relative to the earth so we must solve for this velocity. (a) SET UP: View the motion from above.

The velocity vectors in the problem are: ! vM/E , the velocity of the man relative to the earth ! vW/E , the velocity of the water relative to the earth ! vM/W , the velocity of the man relative to the water The rule for adding these velocities is ! ! ! vM/E = vM/W + vW/E Figure 3.41a

! ! The problem tells us that vW/E has magnitude 2.0 m/s and direction due south. It also tells us that vM/W has magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.41b This diagram shows the vector addition ! ! ! vM/E = vM/W + vW/E ! ! and also has vM/W and vW/E in their specified directions. Note that the vector diagram forms a right triangle. Figure 3.41b 2 2 2 The Pythagorean theorem applied to the vector addition diagram gives vM/E = vM/W + vW/E .

vM/W 4.2 m/s = = 2.10; θ = 65°; or vW/E 2.0 m/s φ = 90° − θ = 25°. The velocity of the man relative to the earth has magnitude 4.7 m/s and direction 25° S of E. (b) This requires careful thought. To cross the river the man must travel 800 m due east relative to the earth. The ! man’s velocity relative to the earth is vM/E . But, from the vector addition diagram the eastward component of vM/E EXECUTE:

2 2 vM/E = vM/W + vW/E = (4.2 m/s) 2 + (2.0 m/s) 2 = 4.7 m/s tan θ =

equals vM/W = 4.2 m/s. Thus t =

x − x0 800 m = = 190 s. vx 4.2 m/s

! (c) The southward component of vM/E equals vW/E = 2.0 m/s. Therefore, in the 190 s it takes him to cross the river the distance south the man travels relative to the earth is y − y0 = v yt = (2.0 m/s)(190 s) = 380 m.

3.42.

EVALUATE: If there were no current he would cross in the same time, (800 m) /(4.2 m/s) = 190 s. The current carries him downstream but doesn’t affect his motion in the perpendicular direction, from bank to bank. IDENTIFY: Use the relation that relates the relative velocities. ! ! SET UP: The relative velocities are the water relative to the earth, vW/E , the boat relative to the water, vB/W , and ! ! ! the boat relative to the earth, vB/E . vB/E is due east, vW/E is due south and has magnitude 2.0 m/s. vB/W = 4.2 m/s . ! ! ! vB/E = vB/W + vW/E . The velocity addition diagram is given in Figure 3.42.


EXECUTE:

! v 2.0 m/s (a) Find the direction of vB/W . sin θ = W/E = . θ = 28.4° , north of east. vB/W 4.2 m/s

2 2 (b) vB/E = vB/W − vW/E = (4.2 m/s) 2 − (2.0 m/s) 2 = 3.7 m/s

800 m 800 m = = 216 s . vB/E 3.7 m/s EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.41. In the direction straight across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s. (c) t =

Figure 3.42 3.43.

IDENTIFY: Relative velocity problem in two dimensions. ! ! (a) SET UP: vP/A is the velocity of the plane relative to the air. The problem states that vP/A has magnitude 35 m/s and direction south. ! ! vA/E is the velocity of the air relative to the earth. The problem states that vA/E is to the southwest ( 45° S of W) and has magnitude 10 m/s. ! ! ! The relative velocity equation is vP/E = vP/A + vA/E .

Figure 3.43a EXECUTE:

(b) (vP/A ) x = 0, (vP/A ) y = −35 m/s

(vA/E ) x = −(10 m/s)cos 45° = −7.07 m/s, (vA/E ) y = −(10 m/s)sin 45° = −7.07 m/s (vP/E ) x = (vP/A ) x + (vA/E ) x = 0 − 7.07 m/s = −7.1 m/s (vP/E ) y = (vP/A ) y + (vA/E ) y = −35 m/s − 7.07 m/s = −42 m/s (c)

vP/E = (vP/E ) 2x + (vP/E ) 2y vP/E = (−7.1 m/s) 2 + (−42 m/s) 2 = 43 m/s tan φ =

(vP/E ) x −7.1 = = 0.169 (vP/E ) y −42

φ = 9.6°; ( 9.6° west of south) Figure 3.43b EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition must be done using components. The wind adds both southward and westward components to the velocity of the plane relative to the ground.

3-18

3.44.

Chapter 3

IDENTIFY: Use Eqs.(2.17) and (2.18). SET UP: At the maximum height v y = 0 . EXECUTE:

(a) vx = v0 x +

α

γ

α

β

γ

t 3 , v y = v0 y + β t − t 2 , and x = v0 xt + t 4 , y = v0 y t + t 2 − t 3 . 3 2 12 2 6

γ

(b) Setting v y = 0 yields a quadratic in t , 0 = v0 y + β t − t 2 , which has as the positive solution 2 1⎡ t = β + β 2 + 2v0γ ⎤ = 13.59 s . Using this time in the expression for y(t) gives a maximum height of 341 m. ⎦ γ⎣ (c) The path of the rocket is sketched in Figure 3.44. (d) y = 0 gives 0 = v0 yt +

β

2

γ

γ

β

t 2 − t 3 and t 2 − t − v0 y = 0 . The positive solution is t = 20.73 s . For this t, 6 6 2

x = 3.85 × 104 m . EVALUATE: The graph in part (c) shows the path is not symmetric about the highest point and the time to return to the ground is less than twice the time to the maximum height.

Figure 3.44 3.45.

IDENTIFY:

! ! ! dr ! dv and a = v= dt dt

d n (t ) = nt n −1 . At t = 1.00 s , ax = 4.00 m/s 2 and a y = 3.00 m/s 2 . At t = 0 , x = 0 and y = 50.0 m . dt dx dv EXECUTE: (a) vx = = 2 Bt . ax = x = 2 B , which is independent of t. ax = 4.00 m/s 2 gives B = 2.00 m/s 2 . dt dt dv dy = 3Dt 2 . a y = y = 6 Dt . a y = 3.00 m/s 2 gives D = 0.500 m/s 2 . x = 0 at t = 0 gives A = 0 . y = 50.0 m at vy = dt dt t = 0 gives C = 50.0 m . ! ! (b) At t = 0 , v = 0 and v = 0 , so v = 0 . At t = 0 , a = 2 B = 4.00 m/s 2 and a = 0 , so a = (4.00 m/s 2 )iˆ . SET UP:

x

x

y

y

(c) At t = 10.0 s , vx = 2(2.00 m/s )(10.0 s) = 40.0 m/s and v y = 3(0.500 m/s )(10.0 s) 2 = 150 m/s . 2

3

v = vx2 + v y2 = 155 m/s .

3.46.

! (d) x = (2.00 m/s 2 )(10.0 s) 2 = 200 m , y = 50.0 m + (0.500 m/s 3 )(10.0 s)3 = 550 m . r = (200 m)iˆ + (550 m) ˆj . EVALUATE: The velocity and acceleration vectors as functions of time are ! ! v (t ) = (2 Bt )iˆ + (3Dt 2 ) ˆj and a (t ) = (2 B) iˆ + (6 Dt ) ˆj . The acceleration is not constant. ! t! ! ! dv IDENTIFY: r = r0 + ∫ v (t )dt and a = . 0 dt SET UP: At t = 0 , x0 = 0 and y0 = 0 . ! ! β γ (a) Integrating, r = (α t − t 3 )iˆ + ( t 2 ) ˆj . Differentiating, a = (−2 β t ) iˆ + γ ˆj . 3 2 (b) The positive time at which x = 0 is given by t 2 = 3α β . At this time, the y-coordinate is EXECUTE:

γ 3αγ 3(2.4 m/s)(4.0 m/s 2 ) = = 9.0 m . y = t2 = 2 2β 2(1.6 m/s3 ) EVALUATE:

The acceleration is not constant.


3.47.

IDENTIFY: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the incline determines the initial velocity and initial position for the projectile motion. SET UP: For motion along the incline let + x be directed up the incline. vx2 = v02x + 2ax ( x − x0 ) gives

vx = 2(1.25 m/s 2 )(200 m) = 22.36 m/s . When the projectile motion begins the rocket has v0 = 22.36 m/s at 35.0° above the horizontal and is at a vertical height of (200.0 m)sin 35.0° = 114.7 m . For the projectile motion let + x be horizontal to the right and let + y be upward. Let y = 0 at the ground. Then y0 = 114.7 m , v0 x = v0 cos35.0° = 18.32 m/s , v0 y = v0 sin 35.0° = 12.83 m/s , ax = 0 , a y = −9.80 m/s 2 . Let x = 0 at point A, so x0 = (200.0 m)cos35.0° = 163.8 m . EXECUTE:

y − y0 =

(a) At the maximum height v y = 0 . v y2 = v02y + 2a y ( y − y0 ) gives

v y2 − v02y 2a y

=

0 − (12.83 m/s) 2 = 8.40 m and y = 114.7 m + 8.40 m = 123 m . The maximum height above 2( −9.80 m/s 2 )

ground is 123 m. (b) The time in the air can be calculated from the vertical component of the projectile motion: y − y0 = −114.7 m , v0 y = 12.83 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives (4.90 m/s 2 )t 2 − (12.83 m/s)t − 114.7 m . The

3.48.

(

)

1 12.83 ± (12.83) 2 + 4(4.90)(114.7) s . The positive root is t = 6.32 s . Then 9.80 x − x0 = v0 xt + 12 axt 2 = (18.32 m/s)(6.32 s) = 115.8 m and x = 163.8 m + 115.8 m = 280 m . The horizontal range of the rocket is 280 m. EVALUATE: The expressions for h and R derived in Example 3.8 do not apply here. They are only for a projectile fired on level ground. IDENTIFY: The person moves in projectile motion. Use the results in Example 3.8 to determine how T, h and D depend on g and set up a ratio. 2v sin α 0 v 2 sin 2 α 0 , the maximum height is h = 0 and the SET UP: From Example 3.8, the time in the air is t = 0 g 2g quadratic formula gives t =

horizontal range (called D in the problem) is D =

v02 sin 2α 0 . The person has the same v0 and α 0 on Mars as on g

the earth. EXECUTE:

hg =

⎛g ⎞ ⎛ gE ⎞ tg = 2v0 sin α 0 , which is constant, so tE g E = tM g M . tM = ⎜ E ⎟ tE = ⎜ ⎟ tE = 2.64tE . g ⎝ M⎠ ⎝ 0.379 g E ⎠

⎛g ⎞ v02 sin 2 α 0 , which is constant, so hE g E = hM g M . hM = ⎜ E ⎟ hE = 2.64hE . Dg = v02 sin 2α 0 , which is constant, 2 ⎝ gM ⎠

⎛g ⎞ so DE g E = DM g M . DM = ⎜ E ⎟ DE = 2.64 DE . ⎝ gM ⎠ EVALUATE: All three quantities are proportional to 1/ g so all increase by the same factor of g E / g M = 2.64 . 3.49.


3.50.

The range for a projectile that lands at the same height from which it was launched is R =

v02 sin 2α . g

The maximum range is for α = 45° . Assuming α = 45° , and R = 50 m , v0 = gR = 22 m/s .

EVALUATE: We have assumed that debris was launched at all angles, including the angle of 45° that gives maximum range. IDENTIFY: The velocity has a horizontal tangential component and a vertical component. The vertical component v2 of acceleration is zero and the horizontal component is arad = x R SET UP: Let + y be upward and + x be in the direction of the tangential velocity at the instant we are considering.

3-20

Chapter 3

EXECUTE:

(a) The bird’s tangential velocity can be found from

vx =

circumference 2π (8.00 m) 50.27 m = = = 10.05 m/s time of rotation 5.00 s 5.00 s

Thus its velocity consists of the components vx = 10.05 m/s and v y = 3.00 m/s . The speed relative to the ground is then v = vx2 + v y2 = 10.5 m/s .

3.51.

(b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the horizontal direction, toward v 2 (10.05 m/s) 2 the center of its spiral path–and has magnitude arad = x = = 12.6 m/s 2 . r 8.00 m 3.00 m/s (c) Using the vertical and horizontal velocity components θ = tan −1 = 16.6° . 10.05 m/s EVALUATE: The angle between the bird’s velocity and the horizontal remains constant as the bird rises. IDENTIFY: Take + y to be downward. Both objects have the same vertical motion, with v0 y and a y = + g . Use

constant acceleration equations for the x and y components of the motion. SET UP: Use the vertical motion to find the time in the air: v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 25 m, t = ? EXECUTE:

3.52.

y − y0 = v0 yt + 12 a yt 2 gives t = 2.259 s

During this time the dart must travel 90 m, so the horizontal component of its velocity must be x − x0 90 m v0 x = = = 40 m/s t 2.25 s EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle velocity greater than 40 m/s. IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air. SET UP: Take + y upward. v0 x = 15.0 m/s , v0 y = +10.0 m/s , ax = 0 , a y = −9.80 m/s 2 . EXECUTE:

(a) Use the vertical motion to find the time in the air: y − y0 = v0 yt + 12 a yt 2 with y − y0 = −30.0 m

gives −30.0 m = (10.0 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t=

(

)

1 +10.0 ± ( −10.0) 2 − 4(4.9)( −30) s . The positive solution is t = 3.70 s . During this time she travels a 2(4.9)

horizontal distance x − x0 = v0 xt + 12 axt 2 = (15.0 m/s)(3.70 s) = 55.5 m . She will land 55.5 m south of the point where she drops from the helicopter and this is where the mats should have been placed. (b) The x-t, y-t, vx -t and v y -t graphs are sketched in Figure 3.52. 2(30.0 m) = 2.47 s . 9.80 m/s 2 She is in the air longer than this because she has an initial vertical component of velocity that is upward. EVALUATE:

If she had dropped from rest at a height of 30.0 m it would have taken her t =

Figure 3.52 3.53.

IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant acceleration equations for the x and y components of motion.


SET UP:

Take the origin of coordinates at the point where the canister is released. Take + y to be upward. The initial velocity of the canister is the velocity of the plane, 64.0 m/s in the + x-direction. Figure 3.53

Use the vertical motion to find the time of fall: t = ?, v0 y = 0, a y = −9.80 m/s 2 , y − y0 = −90.0 m (When the canister reaches the ground it is 90.0 m below the origin.) y − y0 = v0 yt + 12 a yt 2 EXECUTE:

Since v0 y = 0, t =

2( y − y0 ) 2(−90.0 m) = = 4.286 s. −9.80 m/s 2 ay

SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this time: x − x0 = ?, ax − 0, v0 x = 64.0 m/s,

3.54.

EXECUTE: x − x0 = v0t + 12 at 2 = (64.0 m/s)(4.286 s) + 0 = 274 m. EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at constant speed. IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment plus the distance the ship moves while the equipment is in the air. SET UP: For the motion of the equipment take + x to be to the right and + y to be upwards. Then ax = 0 ,

a y = −9.80 m/s 2 , v0 x = v0 cos α 0 = 7.50 m/s and v0 y = v0 sin α 0 = 13.0 m/s . When the equipment lands in the front of the ship, y − y0 = −8.75 m . EXECUTE:

Use the vertical motion of the equipment to find its time in the air: y − y0 = v0 yt + 12 a yt 2 gives

)

(

1 13.0 ± (−13.0) 2 + 4(4.90)(8.75) s . The positive root is t = 3.21 s . The horizontal range of the 9.80 equipment is x − x0 = v0 xt + 12 axt 2 = (7.50 m/s)(3.21 s) = 24.1 m . In 3.21 s the ship moves a horizontal distance (0.450 m/s)(3.21 s) = 1.44 m , so D = 24.1 m + 1.44 m = 25.5 m . t=

v02 sin 2α 0 from Example 3.8 can't be used because the starting and ending points g of the projectile motion are at different heights. IDENTIFY: Projectile motion problem. Take the origin of coordinates at the point where the ball leaves the bat, and take + y to be upward. v0 x = v0 cos α 0 EVALUATE:

3.55.

The equation R =

v0 y = v0 sin α 0 , but we don’t know v0 . Figure 3.55

Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding equation for the vertical displacement. The time t is the same for both components, so this will give us two equations in two unknowns ( v0 and t).

3-22

Chapter 3

(a) SET UP: y-component: a y = −9.80 m/s 2 , y − y0 = −0.9 m, v0 y = v0 sin 45°

y − y0 = v0 y + t 12 a yt 2 EXECUTE: −0.9 m = (v0 sin 45°)t + 12 (−9.80 m/s 2 )t 2 SET UP: x-component: ax = 0, x − x0 = 188 m, v0 x = v0 cos 45°

x − x0 = v0 xt + 12 axt 2 EXECUTE:

t=

x − x0 188 m = v0 x v0 cos 45°

Put the expression for t from the x-component motion into the y-component equation and solve for v0 . (Note that sin 45° = cos 45°. )

⎛ 188 m ⎞ 2 ⎛ 188 m ⎞ −0.9 m = (v0 sin 45°) ⎜ ⎟ − (4.90 m/s ) ⎜ ⎟ ⎝ v0 cos 45° ⎠ ⎝ v0 cos 45° ⎠

2

2

⎛ 188 m ⎞ 4.90 m/s ⎜ ⎟ = 188 m + 0.9 m = 188.9 m ⎝ v0 cos 45° ⎠ 2

2

2 2 ⎛ 188 m ⎞ 4.90 m/s ⎛ v0 cos 45° ⎞ 4.90 m/s = 42.8 m/s , v0 = ⎜ ⎟ ⎜ ⎟ = 188.9 m ⎝ cos 45° ⎠ 188.9 m ⎝ 188 m ⎠

(b) Use the horizontal motion to find the time it takes the ball to reach the fence: SET UP: x-component: x − x0 = 116 m, ax = 0, v0 x = v0 cos 45° = (42.8 m/s)cos 45° = 30.3 m/s, t = ?

x − x0 = v0 xt + 12 axt 2 x − x0 116 m = = 3.83 s v0 x 30.3 m/s SET UP: Find the vertical displacement of the ball at this t: y-component: y − y0 = ?, a y = −9.80 m/s 2 , v0 y = v0 sin 45° = 30.3 m/s, t = 3.83 s EXECUTE:

t=

y − y0 = v0 yt + 12 a yt 2 EXECUTE:

y − y0 = (30.3 s)(3.83 s) + 12 (−9.80 m/s 2 )(3.83 s) 2

y − y0 = 116.0 m − 71.9 m = +44.1 m, above the point where the ball was hit. The height of the ball above the ground is 44.1 m + 0.90 m = 45.0 m. It’s height then above the top of the fence is 45.0 m − 3.0 m = 42.0 m. EVALUATE: With v0 = 42.8 m/s, v0 y = 30.3 m/s and it takes the ball 6.18 s to return to the height where it was

3.56.

hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 , a y = − g . EXECUTE: The equations of motions are y = (v0 sin α )t − 12 gt 2 and x = (v0 cos α )t . When the water goes in the tank for the minimum velocity, y = 2 D and x = 6 D . When the water goes in the tank for the maximum velocity,

y = 2 D and x = 7 D . In both cases, sin α = cos α = 2 / 2. To reach the minimum distance: 6 D =

2 2 v0t , and 2 D = v0t − 12 gt 2 . Solving the first equation for t gives 2 2 2

⎛ 6D 2 ⎞ 6D 2 . Substituting this into the second equation gives 2 D = 6 D − 12 g ⎜⎜ t= ⎟⎟ . Solving this for v0 gives v0 ⎝ v0 ⎠ v0 = 3 gD .


To reach the maximum distance: 7 D =

2 2 v0t , and 2 D = v0t − 12 gt 2 . Solving the first equation for t gives 2 2 2

t=

⎛ 7D 2 ⎞ 7D 2 . Substituting this into the second equation gives 2 D = 7 D − 12 g ⎜⎜ ⎟⎟ . Solving this for v0 gives v0 ⎝ v0 ⎠

v0 = 49 gD / 5 = 3.13 gD , which, as expected, is larger than the previous result.

3.57.

EVALUATE: A launch speed of v0 = 6 gD = 2.45 gD is required for a horizontal range of 6D. The minimum speed required is greater than this, because the water must be at a height of at least 2D when it reaches the front of the tank. IDENTIFY: The equations for h and R from Example 3.8 can be used. v 2 sin 2 α 0 v 2 sin 2α 0 SET UP: h = 0 and R = 0 . If the projectile is launched straight up, α 0 = 90° . 2g g EXECUTE:

(a) h =

v02 and v0 = 2 gh . 2g

(b) Calculate α 0 that gives a maximum height of h when v0 = 2 2 gh . h =

8 gh sin 2 α 0 = 4h sin 2 α 0 . sin α 0 = 12 and 2g

α 0 = 30.0° .

(2 (c) R = EVALUATE:

3.58.

)

2

2 gh sin 60.0° g

= 6.93h .

v02 2h 2h sin(2α 0 ) = so R = . For a given α 0 , R increases when h increases. For α 0 = 90° , 2 g sin α 0 sin 2 α 0

R = 0 and for α 0 = 0° , h = 0 and R = 0 . For α 0 = 45° , R = 4h . IDENTIFY: To clear the bar the ball must have a height of 10.0 ft when it has a horizontal displacement of 36.0 ft. The ball moves as a projectile. When v0 is very large, the ball reaches the goal posts in a very short time and the acceleration due to gravity causes negligible downward displacement. SET UP: 36.0 ft = 10.97 m ; 10.0 ft = 3.048 m . Let + x be to the right and + y be upward, so ax = 0 , a y = − g , v0 x = v0 cos α 0 and v0 y = v0 sin α 0 EXECUTE:

(a) The ball cannot be aimed lower than directly at the bar. tan α 0 =

(b) x − x0 = v0 xt + 12 axt 2 gives t =

10.0 ft and α 0 = 15.5° . 36.0 ft

x − x0 x − x0 = . Then y − y0 = v0 yt + 12 a yt 2 gives v0 x v0 cos α 0

⎛ x − x0 ⎞ 1 ( x − x0 ) 2 1 ( x − x0 ) 2 . y − y0 = (v0 sin α 0 ) ⎜ = ( x − x0 ) tan α 0 − g 2 ⎟− g 2 2 2 v0 cos 2 α 0 ⎝ v0 cos α 0 ⎠ 2 v0 cos α 0 v0 =

( x − x0 ) g 10.97 m 9.80 m/s 2 = = 12.2 m/s cos α 0 2[( x − x0 ) tan α 0 − ( y − y0 )] cos 45.0° 2[10.97 m − 3.048 m]

EVALUATE:

3.59.

With the v0 in part (b) the horizontal range of the ball is R =

v02 sin 2α 0 = 15.2 m = 49.9 ft . The ball g

reaches the highest point in its trajectory when x − x0 = R / 2 , so when it reaches the goal posts it is on its way down. IDENTIFY: Apply Eq.(3.27) and solve for x. SET UP: The change in height is y = − h . EXECUTE: (a) We get a quadratic equation in x, the solution to which is x=

v02 cos α 0 ⎡ 2 2 gh ⎤ v0 cos α 0 ⎡ v sin α 0 + v02 sin 2 α 0 + 2 gh ⎤ . ⎢ tan α 0 + 2 ⎥= ⎣ 0 ⎦ g v0 cos α 0 ⎦ g ⎣

If h = 0 , the square root reduces to v0 sin α 0 , and x = R .

3-24

Chapter 3

(b) The expression for x becomes x = (10.2 m)cos α 0 + [sin 2 α 0 + sin 2 α 0 + 0.98] . The graph of x as a function of

α 0 is sketched in Figure 3.59. The angle α 0 = 90° corresponds to the projectile being launched straight up, and there is no horizontal motion. If α 0 = 0 , the projectile moves horizontally until it has fallen the distance h. (d) The graph shows that the maximum horizontal distance is for an angle less than 45° . EVALUATE: For α 0 = 45° the x and y components of the initial velocity are equal. For α 0 < 45° the x component of the initial velocity is less than the y component. Height comes from the initial position and less vertical component of initial velocity is needed for the maximum range.

Figure 3.59 3.60.

IDENTIFY: The snowball moves in projectile motion. In part (a) the vertical motion determines the time in the air. In part (c), find the height of the snowball above the ground after it has traveled horizontally 4.0 m. SET UP: Let + y be downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 cosθ 0 = 5.36 m/s , v0 y = v0 sin θ 0 = 4.50 m/s . EXECUTE:

(a) Use the vertical motion to find the time in the air: y − y0 = v0 yt + 12 a yt 2 with y − y0 = 14.0 m gives

14.0 m = (4.50 m/s)t + (4.9 m/s 2 )t 2 . The quadratic formula gives t =

(

)

1 −4.50 ± (4.50) 2 − 4(4.9)( −14.0) s . 2(4.9)

The positive root is t = 1.29 s . Then x − x0 = v0 xt + 12 axt 2 = (5.36 m/s)(1.29 s) = 6.91 m . (b) The x-t, y-t, vx -t and v y -t graphs are sketched in Figure 3.60. (c) x − x0 = v0 xt + 12 axt 2 gives t =

x − x0 4.0 m = = 0.746 s . In this time the snowball travels downward a v0 x 5.36 m/s

distance y − y0 = v0 yt + 12 a y t 2 = 6.08 m and is therefore 14.0 m − 6.08 m = 7.9 m above the ground. The snowball passes well above the man and doesn’t hit him. EVALUATE: If the snowball had been released from rest at a height of 14.0 m it would have reached the ground 2(14.0 m) in t = = 1.69 s . The snowball reaches the ground in a shorter time than this because of its initial 9.80 m/s 2 downward component of velocity.

Figure 3.60 3.61.

(a) IDENTIFY and SET UP:

Use the equation derived in Example 3.8: ⎛ 2v sin α 0 ⎞ R = (v0 cos α 0 ) ⎜ 0 ⎟ g ⎝ ⎠


Call the range R1 when the angle is α 0 and R2 when the angle is 90° − α . ⎛ 2v sin α 0 ⎞ R1 = (v0 cos α 0 ) ⎜ 0 ⎟ g ⎝ ⎠ ⎛ 2v sin(90° − α 0 ) ⎞ R2 = (v0 cos(90° − α 0 )) ⎜ 0 ⎟ g ⎝ ⎠ The problem asks us to show that R1 = R2 . EXECUTE: We can use the trig identities in Appendix B to show: cos(90° − α 0 ) = cos(α 0 − 90°) = sin α 0 sin(90° − α 0 ) = − sin(α 0 − 90°) = −( − cos α 0 ) = + cos α 0 ⎛ 2v cos α 0 ⎞ ⎛ 2v0 sin α 0 ⎞ Thus R2 = (v0 sin α 0 ) ⎜ 0 ⎟ = (v0 cos α 0 ) ⎜ ⎟ = R1. g g ⎝ ⎠ ⎝ ⎠ v02 sin 2α 0 Rg (0.25 m)(9.80 m/s 2 ) . so sin 2α 0 = 2 = g v0 (2.2 m/s) 2 This gives α = 15° or 75°. EVALUATE: R = (v02 sin 2α 0 ) / g , so the result in part (a) requires that sin 2 (2α 0 ) = sin 2 (180° − 2α 0 ), which is true. (b) R =

3.62.

(Try some values of α 0 and see!) IDENTIFY: Mary Belle moves in projectile motion. SET UP: Let + y be upward. ax = 0 , a y = − g . EXECUTE:

(a) Eq.(3.27) with x = 8.2 m , y = 6.1 m and α 0 = 53° gives v0 = 13.8 m/s .

(b) When she reached Joe Bob, t =

8.2 m = 0.9874 s . vx = v0 x = 8.31 m/s and v y = v0 y + a yt = +1.34 m/s . v0 cos53°

v = 8.4 m/s , at an angle of 9.16° . (c) The graph of vx (t ) is a horizontal line. The other graphs are sketched in Figure 3.62. (d) Use Eq. (3.27), which becomes y = (1.327) x − (0.071115 m −1 ) x 2 . Setting y = −8.6 m gives x = 23.8 m as the positive solution.

Figure 3.62 3.63.

(a) IDENTIFY:

Projectile motion. Take the origin of coordinates at the top of the ramp and take + y to be upward. The problem specifies that the object is displaced 40.0 m to the right when it is 15.0 m below the origin. Figure 3.63

We don’t know t, the time in the air, and we don’t know v0 . Write down the equations for the horizontal and vertical displacements. Combine these two equations to eliminate one unknown. SET UP: y-component: y − y0 = −15.0 m, a y = −9.80 m/s 2 , v0 y = v0 sin 53.0° y − y0 = v0 yt + 12 a yt 2 EXECUTE:

−15.0 m = (v0 sin 53.0°)t − (4.90 m/s 2 )t 2

3-26

Chapter 3

SET UP: x-component: x − x0 = 40.0 m, ax = 0, v0 x = v0 cos53.0°

x − x0 = v0 xt + 12 axt 2 EXECUTE:

40.0 m = (v0t )cos53.0°

40.0 m = 66.47 m. cos53.0° Use this to replace v0t in the first equation: The second equation says v0t =

−15.0 m = (66.47 m)sin 53° − (4.90 m/s 2 )t 2 t=

(66.46 m)sin 53° + 15.0 m 68.08 m = = 3.727 s. 2 4.90 m/s 4.90 m/s 2

Now that we have t we can use the x-component equation to solve for v0 : v0 = EVALUATE:

40.0 m 40.0 m = = 17.8 m/s. t cos53.0° (3.727 s)cos53.0°

Using these values of v0 and t in the y = y0 = v0 y + 12 a y t 2 equation verifies that y − y0 = −15.0 m.

(b) IDENTIFY: v0 = (17.8 m/s) / 2 = 8.9 m/s This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: SET UP: y − y0 = −100 m; v0 y = +v0 sin 53.0° = 7.11 m/s; a y = −9.80 m/s 2

y − y0 = v0 yt + 12 a yt 2 gives −100 = 7.11t − 4.90t 2 EXECUTE:

(

1 4.90t 2 − 7.11t − 100 = 0 and t = 9.80 7.11 ± (7.11) 2 − 4(4.90)(−100)

)

t = 0.726 s ± 4.57 s so t = 5.30 s. The horizontal distance he travels in this time is x − x0 = v0 xt = (v0 cos53.0°)t = (5.36 m/s)(5.30 s) = 28.4 m.

3.64.

He lands in the river a horizontal distance of 28.4 m from his launch point. EVALUATE: He has half the minimum speed and makes it only about halfway across. IDENTIFY: The rock moves in projectile motion. SET UP: Let + y be upward. ax = 0 , a y = − g . Eqs.(3.22) and (3.23) give vx and v y . EXECUTE: Combining equations 3.25, 3.22 and 3.23 gives v 2 = v02 cos 2 α 0 + (v0 sin α 0 − gt ) 2 = v02 (sin 2 α 0 + cos 2 α 0 ) − 2v0 sin α 0 gt + ( gt ) 2 .

1 v 2 = v02 − 2 g (v0 sin α 0t − gt 2 ) = v02 − 2 gy , where Eq.(3.21) has been used to eliminate t in favor of y. For the case 2 of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y = − h into the above expression, yielding v = v02 + 2 gh , which is independent of α 0 . EVALUATE: This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long as v02 − 2 gy > 0 . 3.65.


Take + y to be upward. The rocket moves with projectile motion, with v0 y = +40.0 m/s

and v0 x = 30.0 m/s relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. EXECUTE:

(a) v y = 0 (at maximum height), v0 y = +40.0 m/s, a y = −9.80 m/s 2 , y − y0 = ?

v y2 = v02y + 2a y ( y − y0 ) gives y − y0 = 81.6 m (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. v0 y = 40 m/s, a y = −9.80 m/s 2 , v y = −40 m/s, t = ?

v y = v0 y + a yt gives t = 8.164 s Then x − x0 = v0 xt = (30.0 m/s)(8.164 s) = 245 m.


(d) Relative to the ground the rocket has initial velocity components v0 x = 30.0 m/s and v0 y = 40.0 m/s, so it is

traveling at 53.1° above the horizontal. (e) (i)

Figure 3.65a

Relative to the cart, the rocket travels straight up and then straight down (ii)

Figure 3.65b

3.66.

Relative to the ground the rocket travels in a parabola. EVALUATE: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart. IDENTIFY: The ball moves in projectile motion. SET UP: The woman and ball travel for the same time and must travel the same horizontal distance, so for the ball v0 x = 6.00 m/s . v0 x 6.00 m/s = and θ 0 = 72.5° . v0 20.0 m/s (b) Relative to the ground the ball moves in a parabola. The ball and the runner have the same horizontal component of velocity, so relative to the runner the ball has only vertical motion. The trajectories as seen by each observer are sketched in Figure 3.66. EVALUATE: The ball could be thrown with a different speed, so long as the angle at which it was thrown was adjusted to keep v0 x = 6.00 m/s . EXECUTE:

(a) v0 x = v0cosθ 0 . cosθ 0 =

Figure 3.66 3.67.

IDENTIFY: The boulder moves in projectile motion. SET UP: Take + y downward. v0 x = v0 , v0 y = 0 . ax = 0 , a y = +9.80 m/s 2 . (a) Use the vertical motion to find the time for the boulder to reach the level of the lake: 2( y − y0 ) 2(20 m) y − y0 = v0 yt + 12 a yt 2 with y − y0 = +20 m gives t = = = 2.02 s . The rock must travel ay 9.80 m/s 2

EXECUTE:

x − x0 100 m = = 49.5 m/s t 2.02 s (b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of y − y0 = 45 m .

horizontally 100 m during this time. x − x0 = v0 xt + 12 axt 2 gives v0 = v0 x =

t=

3.68.

2( y − y0 ) 2(45 m) = = 3.03 s and x − x0 = v0 xt = (49.5 m/s)(3.03 s) = 150 m . The rock lands ay 9.80 m/s 2

150 m − 100 m = 50 m beyond the foot of the dam. EVALUATE: The boulder passes over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake. IDENTIFY: The bagels move in projectile motion. Find Henrietta’s location when the bagels reach the ground, and require the bagels to have this horizontal range.

3-28

Chapter 3

SET UP:

Let + y be downward and let x0 = y0 = 0 . ax = 0 , a y = + g . When the bagels reach the ground,

y = 43.9 m . EXECUTE: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels to 1 1 fall 43.9 m from rest. Get the time to fall: y = gt 2 , 43.9 m = (9.80 m/s 2 )t 2 and t = 2.99 s . So, she has been 2 2 jogging for 9.00 s + 2.99 s = 12.0 s . During this time she has gone x = vt = (3.05 m/s)(12.0 s) = 36.6 m . Bruce

3.69.

must throw the bagels so they travel 36.6 m horizontally in 2.99 s. This gives x = vt . 36.6 m = v(2.99 s) and v = 12.2 m/s . (b) 36.6 m from the building. EVALUATE: If v > 12.2 m/s the bagels land in front of her and if v < 12.2 m/s they land behind her. There is a range of velocities greater than 12.2 m/s for which she would catch the bagels in the air, at some height above the sidewalk. IDENTIFY: The shell moves in projectile motion. To find the horizontal distance between the tanks we must find the horizontal velocity of one tank relative to the other. Take + y to be upward. (a) SET UP: The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the vertical motion of the shell to find the time the shell is in the air: v0 y = v0 sin α = 43.4 m/s, a y = −9.80 m/s 2 , y − y0 = 0 (returns to initial height), t = ? EXECUTE:

3.70.

y − y0 = v0 yt + 12 a yt 2 gives t = 8.86 s

SET UP: Consider the motion of one tank relative to the other. EXECUTE: Relative to tank #1 the shell has a constant horizontal velocity v0 cos α = 246.2 m/s. Relative to the ground the horizontal velocity component is 246.2 m/s + 15.0 m/s = 261.2 m/s. Relative to tank #2 the shell has horizontal velocity component 261.2 m/s − 35.0 m/s = 226.2 m/s. The distance between the tanks when the shell was fired is the (226.2 m/s)(8.86 s) = 2000 m that the shell travels relative to tank #2 during the 8.86 s that the shell is in the air. (b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels 310 m, in the same direction. Therefore, their separation increases by 310 m − 133 m = 177 m. So, the separation becomes 2180 m (rounding to 3 significant figures). EVALUATE: The retreating tank has greater speed than the approaching tank, so they move farther apart while the shell is in the air. We can also calculate the separation in part (b) as the relative speed of the tanks times the time the shell is in the air: (35.0 m/s − 15.0 m/s)(8.86 s) = 177 m. IDENTIFY: The object moves with constant acceleration in both the horizontal and vertical directions. SET UP: Let + y be downward and let + x be the direction in which the firecracker is thrown. EXECUTE:

The firecracker’s falling time can be found from the vertical motion: t =

2h g

.

The firecracker’s horizontal position at any time t (taking the student’s position as x = 0 ) is x = vt − 12 at 2 . x = 0 when cracker hits the ground, so t = 2v / a . Combining this with the expression for the falling time gives

3.71.

2v 2h 2v 2 g = and h = 2 . a g a EVALUATE: When h is smaller, the time in the air is smaller and either v must be smaller or a must be larger. ! ! IDENTIFY: The velocity vT/G of the tank relative to the ground is related to the velocity vR/G of the rocket relative ! ! ! ! to the ground and the velocity vT/R of the tank relative to the rocket by vT/G = vT/R + vR/G . SET UP: Let + y be upward and take y = 0 at the ground. Let + x be in the direction of the horizontal component of the tank's motion. Once the tank is released it has ax = 0 , a y = −9.80 m/s 2 , relative to the ground. EXECUTE:

(a) For the rocket v y = v0 y + a yt = (1.75m/s 2 )(22.0 s) = 38.5 m/s and vx = 0 . The rocket has speed 38.5

m/s at the instant when the fuel tank is released. ! (b) (i) The rocket's path is vertical, so relative to the crew member vT/R-x = +25.0 m/s and vT/R-y = 0 . (ii) vR/G is ! vertical and vT/R is horizontal, so vT/G-x = +25.0 m/s and vT/G − y = +38.5 m/s . (c) (i) The tank initially moves horizontally, at an angle of zero. (ii) tan α 0 =

vT/G-y vT/G-x

=

38.5 m/s and α 0 = 57.0° . 25.0 m/s


(d) Consider the motion of the tank, in the reference frame of the technician on the ground. At the instant the tank is released the rocket at a height y − y0 = v0 yt + 12 a yt 2 = 12 (1.75 m/s 2 )(22.0 s) 2 = 423.5 m . So, for the tank

y0 = 423.5 m , v0 y = 38.5 m/s and a y = −9.80 m/s 2 . v y = 0 at the maximum height. v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

3.72.

v y2 − v02y 2a y

=

0 − (38.5 m/s) 2 = 75.6 m . y = 423.5 m + 75.6 m = 499 m . The tank reaches a height of 499 m 2(9.80 m/s 2 )

above the launch pad. EVALUATE: Relative to the crew member in the rocket the jettisoned tank has an acceleration of 1.75 m/s 2 + 9.80 m/s 2 = 11.5 m/s 2 , downward. Relative to the rocket the tank follows a parabolic path, but with zero initial vertical velocity and with a downward acceleration that has magnitude greater than g. ! ! IDENTIFY: The velocity vR/G of the rocket relative to the ground is related to the velocity vS/G of the secondary ! rocket relative to the ground and the velocity vS/R of the secondary rocket relative to the rocket by ! ! ! vS/G = vS/R + vR/G . SET UP: Let + y be upward and let y = 0 at the ground. Let + x be in the direction of the horizontal component of the secondary rocket's motion. After it is launched the secondary rocket has ax = 0 and a y = −9.80 m/s 2 , relative to the ground. EXECUTE: (a) (i) vS/R-x = (12.0 m/s)cos53.0° = 7.22 m/s and vS/R-y = (12.0 m/s)sin 53.0° = 9.58 m/s . (ii) vR/G-x = 0 and vR/G-y = 8.50 m/s . vS/G-x = vS/R-x + vR/G-x = 7.22 m/s and vS/G-y = vS/R-y + vR/G-y = 9.58 m/s + 8.50 m/s = 18.1 m/s . (b) vS/G = (vS/G-x ) 2 + (vS/G-y ) 2 = 19.5 m/s . tan α 0 =

vS/G-y vS/G-x

=

18.1 m/s and α 0 = 68.3° . 7.22 m/s

(c) Relative to the ground the secondary rocket has y0 = 145 m , v0 y = +18.1 m/s , a y = −9.80 m/s 2 and v y = 0 (at

the maximum height). v y2 = v02y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02y 2a y

=

0 − (18.1 m/s) 2 = 16.7 m . 2( −9.80 m/s 2 )

y = 145 m + 16.7 m = 162 m . EVALUATE:

3.73.

The secondary rocket reaches its maximum height in time t =

v y − v0 y ay

=

−18.1 m/s = 1.85 s after it −9.80 m/s 2

is launched. At this time the primary rocket has height 145 m + (8.50 m/s)(1.85 s) = 161 m , so is at nearly the same height as the secondary rocket. The secondary rocket first moves upward from the primary rocket but then loses vertical velocity due to the acceleration of gravity. IDENTIFY: The original firecracker moves as a projectile. At its maximum height it's velocity is horizontal. The ! ! velocity vA/G of fragment A relative to the ground is related to the velocity vF/G of the original firecracker relative to ! ! ! ! the ground and the velocity vA/F of the fragment relative to the original firecracker by vA/G = vA/F + vF/G . Fragment B obeys a similar equation. SET UP: Let + x be along the direction of the horizontal motion of the firecracker before it explodes and let + y be upward. Fragment A moves at 53.0° above the + x direction and fragment B moves at 53.0° below the + x direction. Before it explodes the firecracker has ax = 0 and a y = −9.80 m/s 2 EXECUTE: The horizontal component of the firecracker's velocity relative to the ground is constant (since ax = 0 ), so vF/G-x = (25.0 m/s)cos30.0° = 21.65 m/s . At the time of the explosion, vF/G-y = 0 . For fragment A,

vA/F-x = (20.0 m/s)cos53.0° = 12.0 m/s and vA/F-y = (20.0 m/s)sin 53.0° = 16.0 m/s . vA/G-x = vA/F-x + vF/G-x = 12.0 m/s + 21.65 m/s = 33.7 m/s . vA/G-y = vA/F-y + vF/G-y = 16.0 m/s . tan α 0 =

3.74.

vA/G-y vA/G-x

=

16.0 m/s and α 0 = 25.4° . The calculation for fragment B is the same, except vA/F-y = −16.0 m/s . 33.7 m/s

The fragments move at 25.4° above and 25.4° below the horizontal. EVALUATE: As the initial velocity of the firecracker increases the angle with the horizontal for the fragments, as measured from the ground, decreases. IDENTIFY: The grenade moves in projectile motion. 110 km/h = 30.6 m/s . The horizontal range R of the grenade must be 15.8 m plus the distance d that the enemy's car travels while the grenade is in the air.

3-30

Chapter 3

For the grenade take + y upward, so ax = 0 , a y = − g . Let v0 be the magnitude of the velocity of the

SET UP:

grenade relative to the hero. v0 x = v0 cos 45° , v0 y = v0 sin 45° . 90 km/h = 25 m/s ; The enemy’s car is traveling away from the hero’s car with a relative velocity of vrel = 30.6 m/s − 25 m/s = 5.6 m/s . EXECUTE:

y − y0 = v0 yt + 12 a yt 2 with y − y0 = 0 gives t = −

R = v0 xt = v0 (cos 45°)t =

2v0 y ay

=

2v0 sin 45° 2v0vrel . d = vrelt = . g g

2v02 sin 45° cos 45° v02 v2 2vrel v0 + 15.8 m . = . R = d + 15.8 m gives that 0 = g g g g

v02 − 2vrelv0 − (15.8 m)g = 0 . v02 − 7.92v0 − 154.8 = 0 . The quadratic formula gives v0 = 17.0 m/s = 61.2 km/h . The grenade has velocity of magnitude 61.2 km/h relative to the hero. Relative to the hero the velocity of the grenade has components v0 x = v0 cos 45° = 43.3 km/h and v0 y = v0 sin 45° = 43.3 km/h . Relative to the earth the velocity of the grenade has components vEx = 43.3 km/h + 90 km/h = 133.3 km/h and vEy = 43.3 km/h . The magnitude of the velocity relative to the earth is vE = vE2x + vE2y = 140 km/h . EVALUATE:

The time the grenade is in the air is t =

2v0 sin 45° 2(17.0 m/s)sin 45° = = 2.45 s . During this time g 9.80 m/s 2

the grenade travels a horizontal distance x − x0 = (133.3 km/h)(2.45 s)(1 h / 3600 s) = 90.7 m , relative to the earth,

3.75.

and the enemy’s car travels a horizontal distance x − x0 = (110 km/h)(2.45 s)(1 h / 3600 s) = 74.9 m , relative to the earth. The grenade has traveled 15.8 m farther. IDENTIFY and SET UP: Use Eqs. (3.4) and (3.12) to get the velocity and acceleration components from the position components. EXECUTE: x = R cos ωt , y = R sin ωt (a) r = x 2 + y 2 = R 2 cos 2 ωt + R 2 sin 2 ωt = R 2 (sin 2 ωt + cos 2 ωt ) = R 2 = R,

since sin 2 ωt + cos 2 ωt = 1. dx dy (b) vx = = − Rω sin ωt , v y = = Rω cos ωt dt dt ! ! v ⋅ r = vx x + v y y = (− Rω sin ωt )( R cos ωt ) + ( Rω cos ωt )( R sin ωt ) ! ! ! ! v ⋅ r = R 2ω (− sin ωt cos ωt + sin ωt cos ωt ) = 0, so v is perpendicular to r . (c) ax =

ay =

dv y dt

dvx = − Rω 2 cos ωt = −ω 2 x dt = − Rω 2 sin ωt = −ω 2 y

a = ax2 + a y2 = ω 4 x 2 + ω 4 y 2 = ω 2 x 2 + y 2 = Rω 2 . ! ! a = ax iˆ + a y ˆj = −ω 2 ( xiˆ + yˆj ) = −ω 2 r .

! ! Since ω 2 is positive this means that the direction of a is opposite to the direction of r . (d) v = vx2 + v y2 = R 2ω 2 sin 2 ωt + R 2ω 2 cos 2 ωt = R 2ω 2 (sin 2 ωt + cos 2 ωt ). v = R 2ω 2 = Rω.

3.76.

(e) a = Rω 2 , ω = v / R, so a = R(v 2 / R 2 ) = v 2 / R. EVALUATE: The rock moves in uniform circular motion. The position vector is radial, the velocity is tangential, and the acceleration is radially inward. IDENTIFY: All velocities are constant, so the distance traveled is d = vB/Et , where vB/E is the magnitude of the ! ! velocity of the boat relative to the earth. The relative velocities vB/E , vS/W (boat relative to the water) ! ! ! ! and v W/E (water relative to the earth) are related by vB/E = vB/W + v W/E .

Let + x be east and let + y be north. vW/E-x = +30.0 m/min and vW/E-y = 0 . vB/W = 100.0 m/min . The ! direction of vB/W is the direction in which the boat is pointed or aimed. SET UP:

EXECUTE:

(a) vB/W-y = +100.0 m/min and vB/W-x = 0 . vB/E-x = vB/W-x + vW/E-x = 30.0 m/min and

vB/E-y = vB/W-y + vW/E-y = 100.0 m/min . The time to cross the river is t =

y − y0 400.0 m = = 4.00 min . vB/E-y 100.0 m/min


x − x0 = (30.0 m/min)(4.00 min) = 120.0 m . You will land 120.0 m east of point B, which is 45.0 m east of point C. The distance you will have traveled is

(400.0 m) 2 + (120.0 m) 2 = 418 m .

! 75.0 m (b) vB/W is directed at angle φ east of north, where tan φ = and φ = 10.6° . 400.0 m vB/W-x = (100.0 m/min)sin10.6° = 18.4 m/min and vB/W-y = (100.0 m/min)cos10.6° = 98.3 m/min .

vB/E-x = vB/W-x + vW/E-x = 18.4 m/min + 30.0 m/min = 48.4 m/min . vB/E-y = vB/W-y + vW/E-y = 98.3 m/min . t=

y − y0 400.0 m = = 4.07 min . x − x0 = (48.4 m/min)(4.07 min) = 197 m . You will land 197 m downstream vB/E-y 98.3 m/min

from B, so 122 m downstream from C. ! (c) (i) If you reach point C, then vB/E is directed at 10.6° east of north, which is 79.4° north of east. We don't know ! ! the magnitude of vB/E and the direction of vB/W . In part (a) we found that if we aim the boat due north we will land ! east of C, so to land at C we must aim the boat west of north. Let vB/W be at an angle φ of north of west. The relative velocity addition diagram is sketched in Figure 3.76. The law of sines says

sin θ sin 79.4° = . vW/E vB/W

⎛ 30.0 m/min ⎞ sin θ = ⎜ ⎟ sin 79.4° and θ = 17.15° . Then φ = 180° − 79.4° − 17.15° = 83.5° . The boat will head ⎝ 100.0 m/min ⎠ 83.5° north of west, so 6.5° west of north. vB/E-x = vB/W-x + vW/E-x = −(100.0 m/min)cos83.5° + 30.0 m/min = 18.7 m/min .

vB/E-y = vB/W-y + vW/E-y = −(100.0 m/min)sin83.5° = 99.4 m/min . Note that these two components do give the ! direction of vB/E to be 79.4° north of east, as required. (ii) The time to cross the river is t=

y − y0 400.0 m = = 4.02 min . (iii) You travel from A to C, a distance of vB/E-y 99.4 m/min

(400.0 m)2 + (75.0 m) 2 = 407 m .

(iv) vB/E = (vB/E-x ) 2 + (vB/E-y ) 2 = 101 m/min . Note that vB/Et = 406 m , the distance traveled (apart from a small difference due to rounding). EVALUATE: You cross the river in the shortest time when you head toward point B, as in part (a), even though you travel farther than in part (c).

Figure 3.76 3.77.

IDENTIFY:

vx = dx / dt , v y = dy / dt , ax = dvx / dt and a y = dv y / dt .

d (sin ωt ) d (cos ωt ) = ω cos(ωt ) and = −ω sin(ωt ) . dt dt EXECUTE: (a) The path is sketched in Figure 3.77. (b) To find the velocity components, take the derivative of x and y with respect to time: vx = Rω (1 − cos ωt ), and SET UP:

v y = Rω sin ωt. To find the acceleration components, take the derivative of vx and v y with respect to time: ax = Rω 2 sin ωt, and a y = Rω 2 cos ωt. (c) The particle is at rest (v y = vx = 0) every period, namely at t = 0, 2π / ω, 4π / ω,.... At that time,

x = 0 , 2πR, 4πR,...; and y = 0. The acceleration is a = Rω 2 in the + y - direction.

3-32

Chapter 3 1/ 2

2 2 (d) No, since a = ⎡( Rω 2 sin ωt ) + ( Rω 2 cos ωt ) ⎤ = Rω 2 . The magnitude of the acceleration is the same as for ⎢⎣ ⎥⎦ uniform circular motion. EVALUATE: The velocity is tangent to the path. v0 x is always positive; v y changes sign during the motion.

Figure 3.77 3.78.

IDENTIFY: At the highest point in the trajectory the velocity of the projectile relative to the earth is horizontal. ! ! The velocity vP/E of the projectile relative to the earth, the velocity vF/P of a fragment relative to the projectile, and ! ! ! ! the velocity vF/E of a fragment relative to the earth are related by vF/E = vF/P + vP/E . SET UP: Let + x be along the horizontal component of the projectile motion. Let the speed of each fragment relative to the projectile be v. Call the fragments 1 and 2, where fragment 1 travels in the + x direction and fragment 2 is in the − x-direction , and let the speeds just after the explosion of the two fragments relative to the earth be v1 and v2 . Let vp be the speed of the projectile just before the explosion.

vF/E-x = vF/P-x + vP/E-x gives v1 = vp + v and −v2 = vp − v . Both fragments start from the same height with

EXECUTE:

zero vertical component of velocity relative to the earth, so they both fall for the same time t, and this is also the same time as it took for the projectile to travel a horizontal distance D, so vpt = D . Since fragment 2 lands at A it travels a horizontal distance D as it falls and v2t = D . −v2 = +vp − v gives v = vp + v2 and vt = vpt + v2t = 2 D . Then v1t = vpt + vt = 3D . This fragment lands a horizontal distance 3D from the point of explosion and hence 4D from A.

3.79.

EVALUATE: Fragment 1, that is ejected in the direction of the motion of the projectile travels with greater speed relative to the earth than the fragment that travels in the opposite direction. v 2 4π 2 R IDENTIFY: arad = = . All points on the centrifuge have the same period T. R T2 60 s/min SET UP: The period T in seconds is related to n, the number of revolutions per minute, by n = . T a a a 4π 2 EXECUTE: (a) rad = 2 , which is constant. rad,1 = rad,2 . Let R1 = R , so arad,1 = 5.00 g and let R2 = R / 2 . R T R1 R2

⎛R ⎞ arad,2 = arad,1 ⎜ 2 ⎟ = (5.00 g )(1/ 2) = 2.50 g . ⎝ R1 ⎠ 4π 2 R arad 4π 2 R ⎛ 60 s/min ⎞ 2 2 2 (b) T = ⎜ and a = gives a = 4 π Rn /(60 s/min) . = , which is constant. rad rad ⎟ n n 2 (60 s/min) 2 T2 ⎝ ⎠

arad,1 n12

=

n2 = n1

arad,2 n22

. Let arad,1 = 5.00 g , so n1 = n and arad,2 = 5 g Mercury = 5(0.378) g . Then

arad,2 arad,1

EVALUATE:

=n

5(0.378) g = 0.615n . 5.00 g

The radial acceleration is less for points closer to the rotation axis. Since g Mercury < g , a smaller

rotation rate is required to produce 5g Mercury than to produce 5g . 3.80.

IDENTIFY: Use the relation that relates the relative velocities. ! ! SET UP: The relative velocities are the raindrop relative to the earth, vR/E , the raindrop relative to the train, vR/T , ! ! ! ! ! ! and the train relative to the earth, vT/E . vR/E = vR/T + vT/E . vT/E is due east and has magnitude 12.0 m/s. vR/T is ! 30.0° west of vertical. vR/E is vertical. The relative velocity addition diagram is given in Figure 3.80. ! ! ! EXECUTE: (a) vR/E is vertical and has zero horizontal component. The horizontal component of vR/T is −vT/E , so is 12.0 m/s westward. vT/E 12.0 m/s vT/E 12.0 m/s (b) vR/E = = = 20.8 m/s . vR/T = = = 24.0 m/s . tan 30.0° tan30.0° sin 30.0° sin30.0°


EVALUATE: The speed of the raindrop relative to the train is greater than its speed relative to the earth, because of the motion of the train.

Figure 3.80 3.81.

IDENTIFY: Relative velocity problem. The plane’s motion relative to the earth is determined by its velocity relative to the earth. SET UP: Select a coordinate system where + y is north and + x is east. The velocity vectors in the problem are: ! vP/E , the velocity of the plane relative to the earth. ! vP/A , the velocity of the plane relative to the air (the magnitude vP/A is the air speed of the plane and the direction ! of vP/A is the compass course set by the pilot). ! vA/E , the velocity of the air relative to the earth (the wind velocity). ! ! ! The rule for combining relative velocities gives vP/E = vP/A + vA/E . (a) We are given the following information about the relative velocities: ! vP/A has magnitude 220 km/h and its direction is west. In our coordinates is has components (vP/A ) x = −220 km/h

and (vP/A ) y = 0.

! From the displacement of the plane relative to the earth after 0.500 h, we find that vP/E has components in our coordinate system of

(vP/E ) x = −

120 km = −240 km/h (west) 0.500 h

20 km = −40 km/h (south) 0.500 h With this information the diagram corresponding to the velocity addition equation is shown in Figure 3.81a. (vP/E ) y = −

Figure 3.81a ! We are asked to find vA/E , so solve for this vector: ! ! ! ! ! ! vP/E = vP/A + vA/E gives vA/E = vP/E − vP/A . EXECUTE: The x-component of this equation gives (vA/E ) x = (vP/E ) x − (vP/A ) x = −240 km/h − (−220 km/h) = −20 km/h. The y-component of this equation gives (vA/E ) y = (vP/E ) y − (vP/A ) y = −40 km/h.

3-34

Chapter 3

! Now that we have the components of vA/E we can find its magnitude and direction.

vA/E = (vA/E ) 2x + (vA/E ) 2y vA/E = ( −20 km/h) 2 + (−40 km/h) 2 = 44.7 km/h 40 km/h = 2.00; φ = 63.4° 20 km/h The direction of the wind velocity is 63.4° S of W, or 26.6° W of S. tan φ =

Figure 3.81b EVALUATE: The plane heads west. It goes farther west than it would without wind and also travels south, so the wind velocity has components west and south. ! ! ! (b) SET UP: The rule for combining the relative velocities is still vP/E = vP/A + vA/E , but some of these velocities have different values than in part (a). ! vP/A has magnitude 220 km/h but its direction is to be found. ! vA/E has magnitude 40 km/h and its direction is due south. ! The direction of vP/E is west; its magnitude is not given. ! ! ! The vector diagram for vP/E = vP/A + vA/E and the specified directions for the vectors is shown in Figure 3.81c.

Figure 3.81c

3.82.

The vector addition diagram forms a right triangle. v 40 km/h EXECUTE: sin φ = A/E = = 0.1818; φ = 10.5°. vP/A 220 km/h The pilot should set her course 10.5° north of west. EVALUATE: The velocity of the plane relative to the air must have a northward component to counteract the wind and a westward component in order to travel west. IDENTIFY: Both the bolt and the elevator move vertically with constant acceleration. SET UP: Let + y be upward and let y = 0 at the initial position of the floor of the elevator, so y0 for the bolt is 3.00 m. EXECUTE: (a) The position of the bolt is 3.00 m + (2.50 m/s)t − (1/ 2)(9.80 m/s 2 )t 2 and the position of the floor is (2.50 m/s)t. Equating the two, 3.00 m = (4.90 m/s 2 )t 2 . Therefore, t = 0.782 s . (b) The velocity of the bolt is 2.50 m/s − (9.80 m/s 2 )(0.782 s) = −5.17 m/s relative to Earth, therefore, relative to an observer in the elevator v = −5.17 m/s − 2.50 m/s = −7.67 m/s. (c) As calculated in part (b), the speed relative to Earth is 5.17 m/s. (d) Relative to Earth, the distance the bolt traveled is (2.50 m/s)t − (1/ 2)(9.80 m/s 2 )t 2 = (2.50 m/s)(0.782 s) − (4.90 m/s 2 )(0.782 s) 2 = −1.04 m . EVALUATE:

As viewed by an observer in the elevator, the bolt has v0 y = 0 and a y = −9.80 m/s 2 , so in 0.782 s it

falls − 12 (9.80 m/s 2 )(0.782 s) 2 = −3.00 m . 3.83.

IDENTIFY: In an earth frame the elevator accelerates upward at 4.00 m/s 2 and the bolt accelerates downward at 9.80 m/s 2 . Relative to the elevator the bolt has a downward acceleration of 4.00 m/s 2 + 9.80 m/s 2 = 13.80 m/s 2 . In either frame, that of the earth or that of the elevator, the bolt has constant acceleration and the constant acceleration equations can be used. SET UP: Let + y be upward. The bolt travels 3.00 m downward relative to the elevator. EXECUTE:

(a) In the frame of the elevator, v0 y = 0 , y − y0 = −3.00 m , a y = −13.8 m/s 2 .

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2( −3.00 m) = = 0.659 s . ay −13.8 m/s 2


(b) v y = v0 y + a y t . v0 y = 0 and t = 0.659 s . (i) a y = −13.8 m/s 2 and v y = −9.09 m/s . The bolt has speed 9.09 m/s

when it reaches the floor of the elevator. (ii) a y = −9.80 m/s 2 and v y = −6.46 m/s . In this frame the bolt has speed 6.46 m/s when it reaches the floor of the elevator. (c) y − y0 = v0 yt + 12 a yt 2 . v0 y = 0 and t = 0.659 s . (i) a y = −13.8 m/s 2 and y − y0 = 12 (−13.8 m/s 2 )(0.659 s) 2 = −3.00 m . The bolt falls 3.00 m, which is correctly the distance between the floor and roof of the elevator. (ii) a y = −9.80 m/s 2 and y − y0 = 12 (−9.80 m/s 2 )(0.659 s) 2 = −2.13 m . The bolt falls

3.84.

2.13 m. EVALUATE: In the earth's frame the bolt falls 2.13 m and the elevator rises 1 (4.00 m/s 2 )(0.659 s) 2 = 0.87 m during the time that the bolt travels from the ceiling to the floor of the elevator. 2 ! ! IDENTIFY: The velocity vP/E of the plane relative to the earth is related to the velocity vP/A of the plane relative to ! ! ! ! the air and the velocity vA/E of the air relative to the earth (the wind velocity) by vP/E = vP/A + vA/E . SET UP:

Let + x be to the east. With no wind vP/A = vP/E =

5550 km = 840.9 km/h . vA/E-x = +225 km/h . The 6.60 h

distance between A and B is 2775 km. EXECUTE: vP/E-x = vP/A-x + vA/E-x . For the trip A to B, vP/A-x = +840.9 km/h and 2775 km = 2.60 h . For the trip B to 1065.9 km/h = −840.9 km/h + 225 km/h = −615.9 km/h and the travel time is

vP/E-x = 840.9 km/h + 225 km/h = 1065.9 km/h and the travel time is t AB = A, vP/A-x = −840.9 km/h and vP/E-x

−2775 km = 4.51 h . The total time for the round trip will be t = t AB + t BA = 7.11 h . −615.9 km/h EVALUATE: The round trip takes longer when the wind blows, even though the plane travels with the wind for 1065.9 km/h + 615.9 km/h = 840.9 km/h , one leg of the trip. The arithmetic average of the speeds for each leg is 2 the same speed when there is no wind. But the plane spends more time traveling at the slower speed relative to the ground and the average speed is less than the arithmetic average of the speeds for each half of the trip. IDENTIFY: Relative velocity problem. SET UP: The three relative velocities are: ! vJ/G , Juan relative to the ground. This velocity is due north and has magnitude vJ/G = 8.00 m/s. ! vB/G , the ball relative to the ground. This vector is 37.0° east of north and has magnitude vB/G = 12.00 m/s. ! vB/J , the ball relative to Juan. We are asked to find the magnitude and direction of this vector. ! ! ! ! ! ! The relative velocity addition equation is vB/G = vB/J + vJ/G , so vB/J = vB/G − v J/G . The relative velocity addition diagram does not form a right triangle so we must do the vector addition using components. Take + y to be north and + x to be east.

t BA =

3.85.

EXECUTE:

vB/Jx = + vB/G sin 37.0° = 7.222 m/s

vB/Jy = + vB/G cos37.0° − vJ/G = 1.584 m/s

3.86.

These two components give vB/J = 7.39 m/s at 12.4° north of east. EVALUATE: Since Juan is running due north, the ball’s eastward component of velocity relative to him is the same as its eastward component relative to the earth. The northward component of velocity for Juan and the ball are in the same direction, so the component for the ball relative to Juan is the difference in their components of velocity relative to the ground. IDENTIFY: (a) The ball moves in projectile motion. When it is moving horizontally, v y = 0 . SET UP: EXECUTE:

Let + x be to the right and let + y be upward. ax = 0 , a y = − g . (a) v0 y = 2 gh = 2(9.80 m/s 2 )(4.90 m) = 9.80 m/s.

(b) v0 y / g = 1.00 s . (c) The horizontal component of the velocity of the ball relative to the man is

(10.8 m/s) 2 − (9.80 m/s) 2 = 4.54 m/s , the horizontal component of the velocity relative to the hoop is 4.54 m/s + 9.10 m/s = 13.6 m/s , and the man must be 13.6 m in front of the hoop at release.

3-36

Chapter 3

⎛ 9.80 m/s ⎞ (d) Relative to the flat car, the ball is projected at an angle θ = tan −1 ⎜ ⎟ = 65°. Relative to the ground the ⎝ 4.54 m/s ⎠ ⎛ ⎞ 9.80 m/s angle is θ = tan −1 ⎜ ⎟ = 35.7° . 4.54 m/s 9.10 m/s + ⎝ ⎠ EVALUATE: In both frames of reference the ball moves in a parabolic path with ax = 0 and a y = − g . The only

3.87.

difference between the description of the motion in the two frames is the horizontal component of the ball’s velocity. IDENTIFY: The pellets move in projectile motion. The vertical motion determines their time in the air. SET UP: v0 x = v0 cos1.0° , v0 y = v0 sin1.0° . ⎛ 2v sin1.0° ⎞ . x − x0 = v0 xt gives x − x0 = (v0 cos1.0°)⎜ 0 ⎟ = 80 m . g ⎝ ⎠ (b) The probability is 1000 times the ratio of the area of the top of the person’s head to the area of the circle in ⎛ π (10 × 10−2 m) 2 ⎞ −3 which the pellets land. (1000) ⎜ ⎟ = 1.6 × 10 . 2 ⎝ π (80 m) ⎠ (c) The slower rise will tend to reduce the time in the air and hence reduce the radius. The slower horizontal velocity will also reduce the radius. The lower speed would tend to increase the time of descent, hence increasing the radius. As the bullets fall, the friction effect is smaller than when they were rising, and the overall effect is to decrease the radius. EVALUATE: The small angle of deviation from the vertical still causes the pellets to spread over a large area because their time in the air is large. IDENTIFY: Write an expression for the square of the distance ( D 2 ) from the origin to the particle, expressed as a EXECUTE:

3.88.

(a) t =

2v0 y g

function of time. Then take the derivative of D 2 with respect to t, and solve for the value of t when this derivative is zero. If the discriminant is zero or negative, the distance D will never decrease. SET UP: D 2 = x 2 + y 2 , with x (t ) and y (t ) given by Eqs.(3.20) and (3.21).

3.89.

EXECUTE: Following this process, sin −1 8/ 9 = 70.5°. EVALUATE: We know that if the object is thrown straight up it moves away from P and then returns, so we are not surprised that the projectile angle must be less than some maximum value for the distance to always increase with time. IDENTIFY: The baseball moves in projectile motion. SET UP: Use coordinates where the x-axis is horizontal and the y-axis is vertical. EXECUTE: (a) The trajectory of the projectile is given by Eq. (3.27), with α 0 = θ + φ, and the equation describing the incline is y = x tan θ. Setting these equal and factoring out the x = 0 root (where the projectile is on the

incline) gives a value for x0 ; the range measured along the incline is ⎡ 2v 2 ⎤ ⎡ cos 2 (θ + φ) ⎤ x / cos θ = ⎢ 0 ⎥ [tan(θ + φ) − tan θ ] ⎢ ⎥ . ⎣ g ⎦ ⎣ cos θ ⎦ (b) Of the many ways to approach this problem, a convenient way is to use the same sort of substitution, involving double angles, as was used to derive the expression for the range along a horizontal incline. Specifically, write the above in terms of α = θ + φ, as

⎡ 2v02 ⎤ 2 R=⎢ ⎥ [sin α cos α cosθ − cos α sin θ ] . 2 ⎣ g cos θ ⎦

The dependence on α and hence φ is in the second term. Using the identities sin α cos α = (1/ 2)sin 2α and cos 2 α = (1/ 2)(1 + cos 2α ), this term becomes (1/ 2)[cos θ sin 2α − sin θ cos 2α − sin θ ] = (1/ 2)[sin(2α − θ ) − sin θ ] .

3.90.

This will be a maximum when sin(2α − θ ) is a maximum, at 2α − θ = 2φ + θ = 90°, or φ = 45° − θ / 2. EVALUATE: Note that the result reduces to the expected forms when θ = 0 (a flat incline, φ = 45° and when θ = −90° (a vertical cliff), when a horizontal launch gives the greatest distance). IDENTIFY: The arrow moves in projectile motion. SET UP: Use coordinates that for which the axes are horizontal and vertical. Let θ be the angle of the slope and let φ be the angle of projection relative to the sloping ground.


EXECUTE:

The horizontal distance x in terms of the angles is ⎛ gx ⎞ 1 tan θ = tan(θ + φ ) − ⎜ 2 ⎟ . 2 ⎝ 2v0 ⎠ cos (θ + φ )

Denote the dimensionless quantity gx / 2v02 by β ; in this case

β=

(9.80 m/s 2 )(60.0 m)cos30.0° = 0.2486. 2(32.0 m/s) 2

The above relation can then be written, on multiplying both sides by the product cosθ cos(θ + φ ), sin θ cos(θ + φ ) = sin(θ + φ )cosθ −

β cosθ , cos(θ + φ )

β cosθ . The term on the left is sin((θ + φ ) − θ ) = sin φ , so the result cos(θ + φ ) of this combination is sin φ cos(θ + φ ) = β cosθ . and so sin(θ + φ )cosθ − cos(θ + φ )sin θ =

Although this can be done numerically (by iteration, trial-and-error, or other methods), the expansion sin a cos b = 12 (sin( a + b) + sin( a − b)) allows the angle φ to be isolated; specifically, then 1 (sin(2φ + θ ) + sin(−θ )) = β cosθ , with the net result that sin(2φ + θ ) = 2 β cosθ + sin θ . 2 (a) For θ = 30°, and β as found above, φ = 19.3° and the angle above the horizontal is θ + φ = 49.3°. For level ground, using β = 0.2871, gives φ = 17.5°. (b) For θ = −30°, the same β as with θ = 30° may be used (cos30° = cos(−30°)), giving φ = 13.0° and φ + θ = −17.0°. EVALUATE:

For θ = 0 the result becomes sin(2φ ) = 2 β = gx / v02 . This is equivalent to the expression

v02 sin(2α 0 ) derived in Example 3.8. g ! IDENTIFY: Find Δv and use this to calculate the magnitude and direction of the average acceleration. v Δt SET UP: In a time Δt, the velocity vector has moved through an angle (in radians) Δφ = (see Figure 3.28 in R ! the textbook). By considering the isosceles triangle formed by the two velocity vectors, the magnitude Δv is seen R=

3.91.

to be 2v sin(φ / 2) .

EXECUTE:

! Δv v # ⎛ vΔt ⎞ 10 m/s sin([1.0 / s]Δt) aav = = 2 sin ⎜ ⎟= Δt Δt ⎝ 2 R ⎠ Δt

Using the given values gives magnitudes of 9.59 m/s 2 , 9.98 m/s 2 and 10.0 m/s 2 . The changes in direction of the vΔt and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle of R π + Δθ = π / 2 + 1/ 2 rad(or 118.6°), the average acceleration vector with the original velocity vector is 2 π / 2 + 0.1 rad(or 95.7°), and π / 2 + 0.05 rad(or 92.9°).

velocity vectors are given by Δθ =

3.92.

EVALUATE: The instantaneous acceleration magnitude, v 2 / R = (5.00 m/s) 2 /(2.50 m ) = 10.0 m/s 2 is indeed # approached in the limit at Δt → 0. Also, the direction of aav approaches the radially inward direction as Δt → 0 . IDENTIFY: The rocket has two periods of constant acceleration motion. SET UP: Let + y be upward. During the free-fall phase, ax = 0 and a y = − g . After the engines turn on, ax = (3.00 g )cos30.0° and a y = (3.00 g )sin 30.0° . Let t be the total time since the rocket was dropped and let T be the time the rocket falls before the engine starts. EXECUTE: (i) The diagram is given in Figure 3.92a. (ii) The x-position of the plane is (236 m/s)t and the x-position of the rocket is (236 m/s)t + (1/ 2)(3.00)(9.80 m/s 2 )cos30°(t − T ) 2 . The graphs of these two equations are sketched in Figure 3.92b.

3-38

Chapter 3

(iii) If we take y = 0 to be the altitude of the airliner, then y (t ) = −1/ 2 gT 2 − gT (t − T ) + 1/ 2(3.00)(9.80 m/s 2 )(sin 30°)(t − T ) 2 for the rocket. The airliner has constant y. The graphs are sketched in Figure 3.92b. In each of the Figures 3.92a-c, the rocket is dropped at t = 0 and the time T when the motor is turned on is indicated. By setting y = 0 for the rocket, we can solve for t in terms of T: 0 = −(4.90 m/s 2 )T 2 − (9.80 m/s 2 )T (t − T ) + (7.35 m/s 2 )(t − T ) 2 . Using the quadratic formula for the variable x = t − T we find x = t − T =

(9.80 m/s 2 )T + (9.80 m/s 2T ) 2 + (4)(7.35 m/s 2 )(4.9)T 2 , or t = 2.72 T . Now, 2(7.35 m/s 2 )

using the condition that xrocket − xplane = 1000 m , we find (236 m/s)t + (12.7 m/s 2 )(t − T )2 − (236 m/s)t = 1000 m, or (1.72T ) 2 = 78.6 s 2 . Therefore T = 5.15 s. EVALUATE: During the free-fall phase the rocket and airliner have the same x coordinate but the rocket moves downward from the airliner. After the engines fire, the rocket starts to move upward and its horizontal component of velocity starts to exceed that of the airliner.

Figure 3.92 3.93.

IDENTIFY: Apply the relative velocity relation. SET UP: Let vC/W be the speed of the canoe relative to water and vW/G be the speed of the water relative to the ground. EXECUTE: (a) Taking all units to be in km and h, we have three equations. We know that heading upstream vC/W − vW/G = 2 . We know that heading downstream for a time t , (vC/W + vW/G )t = 5. We also know that for the bottle vW/G (t + 1) = 3. Solving these three equations for vW/G = x, vC/W = 2 + x, therefore (2 + x + x)t = 5 or ⎛3 ⎞ (2 + 2 x )t = 5. Also t = 3/ x − 1, so (2 + 2 x) ⎜ − 1⎟ = 5 or 2 x 2 + x − 6 = 0. The positive solution is ⎝x ⎠ x = vW/G = 1.5 km/h.

(b) vC/W = 2 km/h + vW/G = 3.5 km/h. EVALUATE: When they head upstream, their speed relative to the ground is 3.5 km/h − 1.5 km/h = 2.0 km/h . When they head downstream, their speed relative to the ground is 3.5 km/h + 1.5 km/h = 5.0 km/h . The bottle is moving downstream at 1.5 km/s relative to the earth, so they are able to overtake it.

4

NEWTON’S LAWS OF MOTION

4.1.

IDENTIFY: Consider the vector sum in each case. ! ! ! ! SET UP: Call the two forces F1 and F2 . Let F1 be to the right. In each case select the direction of F2 such that ! ! ! F = F1 + F2 has the desired magnitude. EXECUTE: (a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. The two vectors and their sum are sketched in Figure 4.1a. (b) The forces form the sides of a right isosceles triangle, and the angle between them is 90° . The two vectors and their sum are sketched in Figure 4.1b. (c) For the sum to have zero magnitude, the forces must be antiparallel, and the angle between them is 180° . The two vectors are sketched in Figure 4.1c. EVALUATE: The maximum magnitude of the sum of the two vectors is 2F, as in part (a).

Figure 4.1 4.2.

IDENTIFY: Add the three forces by adding their components. SET UP: In the new coordinates, the 120-N force acts at an angle of 53° from the − x -axis, or 233° from the + x -axis, and the 50-N force acts at an angle of 323° from the + x -axis. EXECUTE: (a) The components of the net force are Rx = (120 N)cos 233° + (50 N)cos323° = −32 N

Ry = (250 N) + (120 N)sin 233° + (50 N)sin 323° = 124 N.

4.3.

4.4.

⎛ 124 ⎞ (b) R = Rx2 + Ry2 = 128 N, arctan ⎜ ⎟ = 104° . The results have the same magnitude as in Example 4.1, and the ⎝ −32 ⎠ angle has been changed by the amount (37°) that the coordinates have been rotated. EVALUATE: We can use any set of coordinate axes that we wish to and can therefore select axes for which the analysis of the problem is the simplest. IDENTIFY: Use right-triangle trigonometry to find the components of the force. SET UP: Let + x be to the right and let + y be downward. EXECUTE: The horizontal component of the force is (10 N)cos 45° = 7.1 N to the right and the vertical component is (10 N)sin 45° = 7.1 N down. EVALUATE: In our coordinates each component is positive; the signs of the components indicate the directions of the component vectors. IDENTIFY: Fx = F cosθ , Fy = F sin θ .

SET UP: Let + x be parallel to the ramp and directed up the ramp. Let + y be perpendicular to the ramp and directed away from it. Then θ = 30.0° .

4-1

4-2

Chapter 4

EXECUTE: (a) F =

Fx 60.0 N = = 69.3 N. cosθ cos30°

(b) Fy = F sin θ = Fx tan θ = 34.6 N. EVALUATE: We can verify that Fx2 + Fy2 = F 2 . The signs of Fx and Fy show their direction. 4.5.

IDENTIFY: Vector addition. ! SET UP: Use a coordinate system where the + x -axis is in the direction of FA , the force applied by dog A. The forces are sketched in Figure 4.5. EXECUTE:

FAx = +270 N, FAy = 0 FBx = FB cos60.0° = (300 N)cos60.0° = +150 N FBy = FB sin 60.0° = (300 N)sin 60.0° = +260 N

Figure 4.5a ! ! ! R = FA + FB

Rx = FAx + FBx = +270 N + 150 N = +420 N Ry = FAy + FBy = 0 + 260 N = +260 N R = Rx2 + Ry2 R = (420 N) 2 + (260 N) 2 = 494 N tan θ =

Ry

Rx θ = 31.8°

= 0.619

Figure 4.5b

4.6.

EVALUATE: The forces must be added as vectors. The magnitude of the resultant force is less than the sum of the magnitudes of the two forces and depends on the angle between the two forces. IDENTIFY: Add the two forces using components. ! SET UP: Fx = F cosθ , Fy = F sin θ , where θ is the angle F makes with the + x axis. EXECUTE: (a) F1x + F2 x = (9.00 N)cos120° + (6.00 N)cos(233.1°) = −8.10 N F1 y + F2 y = (9.00 N)sin120° + (6.00 N)sin(233.1°) = +3.00 N.

4.7.

4.8.

4.9.

(b) R = Rx2 + Ry2 = (8.10 N) 2 + (3.00 N) 2 = 8.64 N. ! EVALUATE: Since Fx < 0 and Fy > 0 , F is in the second quadrant. ! ! IDENTIFY: Apply ∑ F = ma . SET UP: Let + x be in the direction of the force. EXECUTE: ax = Fx / m = (132 N)/ (60 kg) = 2.2 m / s 2 . EVALUATE: The acceleration is in the direction of the force. ! ! IDENTIFY: Apply ∑ F = ma .

SET UP: Let + x be in the direction of the acceleration. EXECUTE: Fx = max = (135 kg)(1.40 m/s 2 ) = 189 N. EVALUATE: The net force must be in the direction of the acceleration. ! ! IDENTIFY: Apply ∑ F = ma to the box. SET UP: Let + x be the direction of the force and acceleration.

∑F

x

= 48.0 N .

Newton’s Laws of Motion

4.10.

∑F

48.0 N x = = 16.0 kg . 3.00 m/s 2 ax EVALUATE: The vertical forces sum to zero and there is no motion in that direction. IDENTIFY: Use the information about the motion to find the acceleration and then use

EXECUTE:

∑F

x

= max gives m =

4-3

calculate m. SET UP: Let + x be the direction of the force.

∑F

x

∑F

x

= max to

= 80.0 N .

EXECUTE: (a) x − x0 = 11.0 m , t = 5.00 s , v0 x = 0 . x − x0 = v0 xt + 12 axt 2 gives ax =

2( x − x0 ) 2(11.0 m) ∑ Fx = 80.0 N = 90.9 kg . = = 0.880 m/s 2 . m = 2 2 t (5.00 s) 0.880 m/s 2 ax

(b) ax = 0 and vx is constant. After the first 5.0 s, vx = v0 x + axt = (0.880 m/s 2 )(5.00 s) = 4.40 m/s .

4.11.

x − x0 = v0 xt + 12 axt 2 = (4.40 m/s)(5.00 s) = 22.0 m . EVALUATE: The mass determines the amount of acceleration produced by a given force. The block moves farther in the second 5.00 s than in the first 5.00 s. IDENTIFY and SET UP: Use Newton’s second law in component form (Eq.4.8) to calculate the acceleration produced by the force. Use constant acceleration equations to calculate the effect of the acceleration on the motion. EXECUTE: (a) During this time interval the acceleration is constant and equal to ax =

Fx 0.250 N = = 1.562 m/s 2 m 0.160 kg

We can use the constant acceleration kinematic equations from Chapter 2. x − x0 = v0 xt + 12 axt 2 = 0 + 12 (1.562 m/s 2 )(2.00 s) 2 , so the puck is at x = 3.12 m. vx = v0 x + axt = 0 + (1.562 m/s 2 )(2.00 s) = 3.13 m/s. (b) In the time interval from t = 2.00 s to 5.00 s the force has been removed so the acceleration is zero. The speed stays constant at vx = 3.12 m/s. The distance the puck travels is x − x0 = v0 xt = (3.12 m/s)(5.00 s − 2.00 s) = 9.36 m. At the end of the interval it is at x = x0 + 9.36 m = 12.5 m. In the time interval from t = 5.00 s to 7.00 s the acceleration is again ax = 1.562 m/s 2 . At the start of this interval

v0 x = 3.12 m/s and x0 = 12.5 m. x − x0 = v0 xt + 12 axt 2 = (3.12 m/s)(2.00 s) + 12 (1.562 m/s 2 )(2.00 s) 2 . x − x0 = 6.24 m + 3.12 m = 9.36 m. Therefore, at t = 7.00 s the puck is at x = x0 + 9.36 m = 12.5 m + 9.36 m = 21.9 m.

vx = v0 x + axt = 3.12 m/s + (1.562 m/s 2 )(2.00 s) = 6.24 m/s

4.12.

EVALUATE: The acceleration says the puck gains 1.56 m/s of velocity for every second the force acts. The force acts a total of 4.00 s so the final velocity is (1.56 m/s)(4.0 s) = 6.24 m/s. ! ! IDENTIFY: Apply ∑ F = ma . Then use a constant acceleration equation to relate the kinematic quantities.

SET UP: Let + x be in the direction of the force. EXECUTE: (a) ax = Fx / m = (140 N) /(32.5 kg) = 4.31 m/s 2 . (b) x − x0 = v0 xt + 12 axt 2 . With v0 x = 0, x = 12 at 2 = 215 m .

4.13.

(c) vx = v0 x + axt . With v0 x = 0, vx = axt = 2 x / t = 43.0 m/s . EVALUATE: The acceleration connects the motion to the forces. IDENTIFY: The force and acceleration are related by Newton’s second law. SET UP: ∑ Fx = max , where ∑ Fx is the net force. m = 4.50 kg . EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value. ∑ Fx = max = (4.50 kg)(10.0 m/s2 ) = 45.0 N . This maximum force occurs between 2.0 s and 4.0 s. (b) The net force is constant when the acceleration is constant. This is between 2.0 s and 4.0 s. (c) The net force is zero when the acceleration is zero. This is the case at t = 0 and t = 6.0 s . EVALUATE: A graph of ∑ Fx versus t would have the same shape as the graph of ax versus t.

4-4

4.14.

Chapter 4

IDENTIFY: The force and acceleration are related by Newton’s second law. ax =

dvx , so ax is the slope of the dt

graph of vx versus t.

SET UP: The graph of vx versus t consists of straight-line segments. For t = 0 to t = 2.00 s , ax = 4.00 m/s 2 . For t = 2.00 s to 6.00 s, ax = 0 . For t = 6.00 s to 10.0 s, ax = 1.00 m/s 2 .

∑F

x

= max , with m = 2.75 kg .

∑F

x

is the net force.

EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value. ∑ Fx = max = (2.75 kg)(4.00 m/s2 ) = 11.0 N . This maximum occurs in the interval t = 0 to t = 2.00 s . (b) The net force is zero when the acceleration is zero. This is between 2.00 s and 6.00 s. (c) Between 6.00 s and 10.0 s, ax = 1.00 m/s 2 , so ∑ Fx = (2.75 kg)(1.00 m/s 2 ) = 2.75 N . 4.15.

EVALUATE: The net force is largest when the velocity is changing most rapidly. IDENTIFY: The net force and the acceleration are related by Newton’s second law. When the rocket is near the ! surface of the earth the forces on it are the upward force F exerted on it because of the burning fuel and the ! downward force Fgrav of gravity. Fgrav = mg . SET UP: Let + y be upward. The weight of the rocket is Fgrav = (8.00 kg)(9.80 m/s 2 ) = 78.4 N . EXECUTE: (a) At t = 0 , F = A = 100.0 N . At t = 2.00 s , F = A + (4.00 s 2 ) B = 150.0 N and 150.0 N − 100.0 N = 12.5 N/s 2 . 4.00 s 2 (b) (i) At t = 0 , F = A = 100.0 N . The net force is B=

∑F

y

ay =

m

∑F

y

=

y

= F − Fgrav = 100.0 N − 78.4 N = 21.6 N .

21.6 N = 2.70 m/s 2 . (ii) At t = 3.00 s, F = A + B(3.00 s) 2 = 212.5 N . 8.00 kg

= 212.5 N − 78.4 N = 134.1 N . a y =

∑F

y

m

=

134.1 N = 16.8 m/s 2 . 8.00 kg

212.5 N = 26.6 m/s 2 . 8.00 kg EVALUATE: The acceleration increases as F increases. IDENTIFY: Use constant acceleration equations to calculate ax and t. Then use

(c) Now Fgrav = 0 and 4.16.

∑F

∑F

y

= F = 212.5 N . a y =

!

!

∑ F = ma to calculate the

net force. SET UP: Let + x be in the direction of motion of the electron. EXECUTE: (a) v0 x = 0 , ( x − x0 ) = 1.80 × 10−2 m , vx = 3.00 × 106 m/s . vx2 = v02x + 2ax ( x − x0 ) gives

ax =

vx2 − v02x (3.00 × 106 m/s) 2 − 0 = = 2.50 × 1014 m/s 2 2( x − x0 ) 2(1.80 × 10−2 m)

(b) vx = v0 x + axt gives t = (c)

4.17.

∑F

x

= max = (9.11 × 10−31 kg)(2.50 × 1014 m/s 2 ) = 2.28 × 10−16 N .

EVALUATE: The acceleration is in the direction of motion since the speed is increasing, and the net force is in the direction of the acceleration. IDENTIFY and SET UP: F = ma. We must use w = mg to find the mass of the boulder. EXECUTE:

4.18.

vx − v0 x 3.00 × 106 m/s − 0 = = 1.2 × 10−8 s 2.50 × 1014 m/s 2 ax

m=

w 2400 N = = 244.9 kg g 9.80 m/s 2

Then F = ma = (244.9 kg)(12.0 m/s 2 ) = 2940 N. EVALUATE: We must use mass in Newton’s second law. Mass and weight are proportional. ! ! IDENTIFY: Apply ∑ F = ma .

SET UP:

m = w / g = (71.2 N) /(9.80 m/s 2 ) = 7.27 kg .

Fx 160 N = = 22.0 m/s 2 m 7.27 kg EVALUATE: The weight of the ball is a vertical force and doesn’t affect the horizontal acceleration. However, the weight is used to calculate the mass. EXECUTE:

ax =


4.19.

4.20.

4-5

IDENTIFY and SET UP: w = mg. The mass of the watermelon is constant, independent of its location. Its weight differs on earth and Jupiter’s moon. Use the information about the watermelon’s weight on earth to calculate its mass: w 44.0 N = 4.49 kg. EXECUTE: w = mg gives that m = = g 9.80 m/s 2 On Jupiter’s moon, m = 4.49 kg, the same as on earth. Thus the weight on Jupiter’s moon is

w = mg = (4.49 kg)(1.81 m/s 2 ) = 8.13 N. EVALUATE: The weight of the watermelon is less on Io, since g is smaller there. IDENTIFY: Weight and mass are related by w = mg . The mass is constant but g and w depend on location. SET UP: On earth, g = 9.80 m/s 2 . EXECUTE: (a)

4.21.

w w w = m , which is constant, so E = A . wE = 17.5 N , g E = 9.80 m/s 2 , and wA = 3.24 N . g gE gA

⎛w ⎞ ⎛ 3.24 N ⎞ 2 2 gA = ⎜ A ⎟ gE = ⎜ ⎟ (9.80 m/s ) = 1.81 m/s . w 17.5 N ⎝ ⎠ ⎝ E⎠ w 17.5 N = 1.79 kg . (b) m = E = g E 9.80 m/s 2 EVALUATE: The weight at a location and the acceleration due to gravity at that location are directly proportional. IDENTIFY: Apply ∑ Fx = max to find the resultant horizontal force.

SET UP: Let the acceleration be in the + x direction. EXECUTE: ∑ Fx = max = (55 kg)(15 m/s 2 ) = 825 N . The force is exerted by the blocks. The blocks push on the

4.22.

4.23.

sprinter because the sprinter pushes on the blocks. EVALUATE: The force the blocks exert on the sprinter has the same magnitude as the force the sprinter exerts on the blocks. The harder the sprinter pushes, the greater the force on him. ! ! IDENTIFY: ∑ F = ma refers to forces that all act on one object. The third law refers to forces that a pair of objects exert on each other. SET UP: An object is in equilibrium if the vector sum of all the forces on it is zero. A third law pair of forces have the same magnitude regardless of the motion of either object. EXECUTE: (a) the earth (gravity) (b) 4 N; the book (c) no, these two forces are exerted on the same object (d) 4 N; the earth; the book; upward (e) 4 N, the hand; the book; downward (f) second (The two forces are exerted on the same object and this object has zero acceleration.) (g) third (The forces are between a pair of objects.) (h) No. There is a net upward force on the book equal to 1 N. (i) No. The force exerted on the book by your hand is 5 N, upward. The force exerted on the book by the earth is 4 N, downward. (j) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (k) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (l) One, only the gravity force. (m) No. There is a net downward force of 5 N exerted on the book. EVALUATE: Newton’s second and third laws give complementary information about the forces that act. IDENTIFY: Identify the forces on the bottle. SET UP: Classify forces as contact or noncontact forces. The noncontact force is gravity and the contact forces come from things that touch the object. Gravity is always directed downward toward the center of the earth. Air resistance is always directed opposite to the velocity of the object relative to the air. EXECUTE: (a) The free-body diagram for the bottle is sketched in Figure 4.23a

The only forces on the bottle are gravity (downward) and air resistance (upward).

Figure 4.23a (b)

4-6

Chapter 4

w is the force of gravity that the earth exerts on the bottle. The reaction to this force is w′, force that the bottle exerts on the earth

Figure 4.23b Note that these two equal and opposite forces produce very different accelerations because the bottle and the earth have very different masses. Fair is the force that the air exerts on the bottle and is upward. The reaction to this force is a downward force Fair′

4.24.

4.25.

that the bottle exerts on the air. These two forces have equal magnitudes and opposite directions. EVALUATE: The only thing in contact with the bottle while it is falling is the air. Newton’s third law always deals with forces on two different objects. IDENTIFY: The reaction forces in Newton’s third law are always between a pair of objects. In Newton’s second law all the forces act on a single object. SET UP: Let + y be downward. m = w / g . EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational ∑ Fy = a gives force that the passenger exerts on the earth, upward and also of magnitude 650 N. y m 650 N − 620 N ay = = 0.452 m/s 2 . The passenger’s acceleration is 0.452 m / s 2 , downward. (650 N)/(9.80 m/s 2 ) EVALUATE: There is a net downward force on the passenger and the passenger has a downward acceleration. IDENTIFY: Apply Newton’s second law to the earth. SET UP: The force of gravity that the earth exerts on her is her weight, w = mg = (45 kg)(9.8 m/s 2 ) = 441 N. By Newton’s 3rd law, she exerts an equal and opposite force on the earth. ! ! ! Apply ∑ F = ma to the earth, with ∑ F = w = 441 N, but must use the mass of the earth for m.

EXECUTE:

4.26.

a=

w 441 N = = 7.4 × 10−23 m/s 2 . m 6.0 × 1024 kg

EVALUATE: This is much smaller than her acceleration of 9.8 m/s 2 . The force she exerts on the earth equals in magnitude the force the earth exerts on her, but the acceleration the force produces depends on the mass of the object and her mass is much less than the mass of the earth. ! IDENTIFY and SET UP: The only force on the ball is the gravity force, Fgrav . This force is mg , downward and is independent of the motion of the object. EXECUTE: The free-body diagram is sketched in Figure 4.26. The free-body diagram is the same in all cases. EVALUATE: Some forces, such as friction, depend on the motion of the object but the gravity force does not.

Figure 4.26 4.27.

IDENTIFY: Identify the forces on each object. SET UP: In each case the forces are the noncontact force of gravity (the weight) and the forces applied by objects ! that are in contact with each crate. Each crate touches the floor and the other crate, and some object applies F to crate A. EXECUTE: (a) The free-body diagrams for each crate are given in Figure 4.27. FAB (the force on mA due to mB ) and FBA (the force on mB due to mA ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them.


4-7

EVALUATE: Crate B is accelerated by FBA and crate A is accelerated by the net force F − FAB . The greater the total weight of the two crates, the greater their total mass and the smaller will be their acceleration.

4.28.

4.29.

4.30.

Figure 4.27 IDENTIFY: The surface of block B can exert both a friction force and a normal force on block A. The friction force is directed so as to oppose relative motion between blocks B and A. Gravity exerts a downward force w on block A. SET UP: The pull is a force on B not on A. EXECUTE: (a) If the table is frictionless there is a net horizontal force on the combined object of the two blocks, and block B accelerates in the direction of the pull. The friction force that B exerts on A is to the right, to try to prevent A from slipping relative to B as B accelerates to the right. The free-body diagram is sketched in Figure 4.28a. f is the friction force that B exerts on A and n is the normal force that B exerts on A. (b) The pull and the friction force exerted on B by the table cancel and the net force on the system of two blocks is zero. The blocks move with the same constant speed and B exerts no friction force on A. The free-body diagram is sketched in Figure 4.28b. EVALUATE: If in part (b) the pull force is decreased, block B will slow down, with an acceleration directed to the left. In this case the friction force on A would be to the left, to prevent relative motion between the two blocks by giving A an acceleration equal to that of B.

Figure 4.28 IDENTIFY: Since the observer in the train sees the ball hang motionless, the ball must have the same acceleration as the train car. By Newton’s second law, there must be a net force on the ball in the same direction as its acceleration. ! SET UP: The forces on the ball are gravity, which is w, downward, and the tension T in the string, which is directed along the string. EXECUTE: (a) The acceleration of the train is zero, so the acceleration of the ball is zero. There is no net horizontal force on the ball and the string must hang vertically. The free-body diagram is sketched in Figure 4.29a. (b) The train has a constant acceleration directed east so the ball must have a constant eastward acceleration. There must be a net horizontal force on the ball, directed to the east. This net force must come from an eastward ! component of T and the ball hangs with the string displaced west of vertical. The free-body diagram is sketched in Figure 4.29b. EVALUATE: When the motion of an object is described in an inertial frame, there must be a net force in the direction of the acceleration.

Figure 4.29 IDENTIFY: Identify the forces for each object. Action-reaction pairs of forces act between two objects. SET UP: Friction is parallel to the surfaces and is directly to oppose relative motion between the surfaces. EXECUTE: The free-body diagram for the box is given in Figure 4.30a. The free body diagram for the truck is given in Figure 4.30b. The box’s friction force on the truck bed and the truck bed’s friction force on the box form an action-reaction pair. There would also be some small air-resistance force action to the left, presumably negligible at this speed.

4-8

Chapter 4

EVALUATE: The friction force on the box, exerted by the bed of the truck, is in the direction of the truck's acceleration. This friction force can't be large enough to give the box the same acceleration that the truck has and the truck acquires a greater speed than the box.

Figure 4.30 4.31.

4.32.

IDENTIFY: Identify the forces on the chair. The floor exerts a normal force and a friction force. SET UP: Let + y be upward and let + x be in the direction of the motion of the chair. EXECUTE: (a) The free-body diagram for the chair is given in Figure 4.31. (b) For the chair, a y = 0 so ∑ Fy = ma y gives n − mg − F sin 37° = 0 and n = 142 N . ! EVALUATE: n is larger than the weight because F has a downward component.

Figure 4.31 ! ! IDENTIFY: Identify the forces on the skier and apply ∑ F = ma . Constant speed means a = 0 . SET UP: Use coordinates that are parallel and perpendicular to the slope. EXECUTE: (a) The free-body diagram for the skier is given in Figure 4.32. (b) ∑ Fx = max with ax = 0 gives T = mg sin θ = (65.0 kg)(9.80 m / s 2 )sin 26.0° = 279 N . EVALUATE: T is less than the weight of the skier. It is equal to the component of the weight that is parallel to the incline.

4.33.

Figure 4.32 ! ! ! IDENTIFY: ∑ F = ma must be satisfied for each object. Newton’s third law says that the force FC on T that the car ! exerts on the truck is equal in magnitude and opposite in direction to the force FT on C that the truck exerts on the car. ! SET UP: The only horizontal force on the car is the force FT on C exerted by the truck. The car exerts a force ! ! FC on T on the truck. There is also a horizontal friction force f that the highway surface exerts on the truck. Assume the system is accelerating to the right in the free-body diagrams. EXECUTE: (a) The free-body diagram for the car is sketched in Figure 4.33a (b) The free-body diagram for the truck is sketched in Figure 4.33b.


4-9

! (c) The friction force f accelerates the system forward. The tires of the truck push backwards on the highway surface as they rotate, so by Newton’s third law the roadway pushes forward on the tires. ! EVALUATE: FT on C and FC on T each equal the tension T in the rope. Both objects have the same acceleration a . T = mC a and f − T = mT a , so f = ( mC + mT ) a . The acceleration of the two objects is proportional to f.

Figure 4.33 4.34.

IDENTIFY: Use a constant acceleration equation to find the stopping time and acceleration. Then use ! ! ∑ F = ma to calculate the force. ! SET UP: Let + x be in the direction the bullet is traveling. F is the force the wood exerts on the bullet. ⎛v +v ⎞ EXECUTE: (a) v0 x = 350 m/s , vx = 0 and ( x − x0 ) = 0.130 m . ( x − x0 ) = ⎜ 0 x x ⎟ t ⎝ 2 ⎠ gives t =

2( x − x0 ) 2(0.130 m) = = 7.43 × 10−4 s . v0 x + vx 350 m/s

(b) vx2 = v02x + 2ax ( x − x0 ) gives ax =

∑F

x

4.35.

vx2 − v02x 0 − (350 m/s) 2 = = −4.71× 105 m/s 2 2( x − x0 ) 2(0.130 m)

= max gives − F = max and F = − max = −(1.80 × 10−3 kg)( − 4.71 × 105 m/s 2 ) = 848 N .

EVALUATE: The acceleration and net force are opposite to the direction of motion of the bullet. IDENTIFY: Vector addition problem. Write the vector addition equation in component form. We know one vector and its resultant and are asked to solve for the other vector. ! ! SET UP: Use coordinates with the + x -axis along F1 and the + y -axis along R; as shown in Figure 4.35a. F1x = +1300 N, F1 y = 0 Rx = 0, Ry = +1300 N

Figure 4.35a ! ! ! ! ! ! F1 + F2 = R, so F2 = R − F1 EXECUTE:

F2 x = Rx − F1x = 0 − 1300 N = −1300 N

F2 y = Ry − F1 y = +1300 N − 0 = +1300 N ! The components of F2 are sketched in Figure 4.35b. F2 = F22x + F22y = (−1300 N) 2 + (1300 N) 2 F = 1840 N F +1300 N tan θ = 2 y = = −1.00 F2 x −1300 N θ = 135°

4.36.

Figure 4.35b ! ! The magnitude of F2 is 1840 N and its direction is 135° counterclockwise from the direction of F1. ! ! ! EVALUATE: F2 has a negative x-component to cancel F1 and a y-component to equal R. IDENTIFY: Use the motion of the ball to calculate g, the acceleration of gravity on the planet. Then w = mg . SET UP: Let + y be downward and take y0 = 0 . v0 y = 0 since the ball is released from rest.

4-10

Chapter 4

1 2 1 gt gives 10.0 m = g (2.2 s) 2 . g = 4.13 m / s 2 and then 2 2 wX = mg X = (0.100 kg)(4.03 m / s 2 ) = 0.41 N . EVALUATE: g on Planet X is smaller than on earth and the object weighs less than it would on earth. IDENTIFY: If the box moves in the + x -direction it must have a y = 0, so ∑ Fy = 0.

EXECUTE: Get g on X: y =

4.37.

The smallest force the child can exert and still produce such motion is a force that makes the y-components of all three forces sum to zero, but that doesn’t have any x-component.

Figure 4.37 ! ! ! F1 and F2 are sketched in Figure 4.37. Let F3 be the force exerted by the child.

SET UP:

∑F

y

= ma y implies F1 y + F2 y + F3 y = 0, so F3 y = −( F1 y + F2 y ).

EXECUTE:

F1 y = + F1 sin 60° = (100 N)sin 60° = 86.6 N

F2 y = + F2 sin( −30°) = − F2 sin 30° = −(140 N)sin 30° = −70.0 N Then F3 y = −( F1 y + F2 y ) = −(86.6 N − 70.0 N) = −16.6 N; F3 x = 0 The smallest force the child can exert has magnitude 17 N and is directed at 90° clockwise from the + x -axis shown in the figure. (b) IDENTIFY and SET UP: Apply ∑ Fx = max . We know the forces and ax so can solve for m. The force exerted by the child is in the − y -direction and has no x-component.

EXECUTE:

F1x = F1 cos 60° = 50 N

F2 x = F2 cos30° = 121.2 N

∑F = F ∑F m= x

1x

+ F2 x = 50 N + 121.2 N = 171.2 N

171.2 N x = = 85.6 kg 2.00 m/s 2 ax Then w = mg = 840 N.

EVALUATE: In part (b) we don’t need to consider the y-component of Newton’s second law. a y = 0 so the mass

∑ Fy = ma y equation. ! ! Use ∑ F = ma to calculate the acceleration of the tanker and then use constant acceleration

doesn’t appear in the

4.38.

IDENTIFY:

kinematic equations. SET UP: Let + x be the direction the tanker is moving initially. Then ax = − F / m .

EXECUTE:

vx2 = v02x + 2ax ( x − x0 ) says that if the reef weren't there the ship would stop in a distance of x − x0 = −

v02x v02 mv 2 (3.6 × 107 kg)(1.5 m / s) 2 = = 0 = = 506 m, 2ax 2( F / m) 2 F 2(8.0 × 104 N)

so the ship would hit the reef. The speed when the tanker hits the reef is found from vx2 = v02x + 2ax ( x − x0 ) , so it is v = v02 − (2 Fx / m) = (1.5 m/s) 2 −

4.39.

2(8.0 × 104 N)(500 m) = 0.17 m/s, (3.6 × 107 kg)

and the oil should be safe. EVALUATE: The force and acceleration are directed opposite to the initial motion of the tanker and the speed decreases. IDENTIFY: We can apply constant acceleration equations to relate the kinematic variables and we can use Newton’s second law to relate the forces and acceleration. (a) SET UP: First use the information given about the height of the jump to calculate the speed he has at the instant his feet leave the ground. Use a coordinate system with the + y -axis upward and the origin at the position when his feet leave the ground.


4-11

v y = 0 (at the maximum height), v0 y = ?, a y = −9.80 m/s 2 , y − y0 = +1.2 m v y2 = v02y + 2a y ( y − y0 )

EXECUTE:

v0 y = −2a y ( y − y0 ) = −2( −9.80 m/s 2 )(1.2 m) = 4.85 m/s

(b) SET UP: Now consider the acceleration phase, from when he starts to jump until when his feet leave the ground. Use a coordinate system where the + y -axis is upward and the origin is at his position when he starts his jump. EXECUTE: Calculate the average acceleration: v y − v0 y

4.89 m/s − 0 = = 16.2 m/s 2 0.300 s t (c) SET UP: Finally, find the average upward force that the ground must exert on him to produce this average upward acceleration. (Don’t forget about the downward force of gravity.) The forces are sketched in Figure 4.39. EXECUTE: 890 N m = w/ g = = 90.8 kg 9.80 m/s 2 ∑ Fy = ma y (aav ) y =

Fav − mg = m(aav ) y Fav = m( g + (aav ) y ) Fav = 90.8 kg(9.80 m/s 2 + 16.2 m/s 2 ) Fav = 2360 N

Figure 4.39

4.40.

This is the average force exerted on him by the ground. But by Newton’s 3rd law, the average force he exerts on the ground is equal and opposite, so is 2360 N, downward. EVALUATE: In order for him to accelerate upward, the ground must exert an upward force greater than his weight. IDENTIFY: Use constant acceleration equations to calculate the acceleration ax that would be required. Then use

∑F

x

= max to find the necessary force.

SET UP: Let + x be the direction of the initial motion of the auto. v02x . The force F is directed opposite to the EXECUTE: vx2 = v02x + 2ax ( x − x0 ) with vx = 0 gives ax = − 2( x − x0 ) motion and ax = −

F . Equating these two expressions for ax gives m F =m

4.41.

v02x (12.5 m / s) 2 = (850 kg) = 3.7 × 106 N. 2( x − x0 ) 2(1.8 × 10−2 m)

EVALUATE: A very large force is required to stop such a massive object in such a short distance. IDENTIFY: Apply Newton’s second law to calculate a. (a) SET UP: The free-body diagram for the bucket is sketched in Figure 4.41.

The net force on the bucket is T − mg , upward.

Figure 4.41

4-12

Chapter 4

(b) EXECUTE:

∑F

y

a=

4.42.

= ma y gives T − mg = ma T − mg 75.0 N − (4.80 kg)(9.80 m/s 2 ) 75.0 N − 47.04 N = = = 5.82 m/s 2 . m 4.80 kg 4.80 kg

EVALUATE: The weight of the bucket is 47.0 N. The upward force exerted by the cord is larger than this, so the bucket accelerates upward. ! ! IDENTIFY: Apply ∑ F = ma to the parachutist. ! SET UP: Let + y be upward. Fair is the force of air resistance. EXECUTE: (a) w = mg = (55.0 kg)(9.80 m/s 2 ) = 539 N (b) The free-body diagram is given in Fig. 4.42.

∑F

∑F

y

= Fair − w = 620 N − 539 N = 81 N . The net force is upward.

81 N y = = 1.5 m/s 2 , upward. 55.0 kg m EVALUATE: Both the net force and the acceleration are upward. Since her velocity is downward and her acceleration is upward, her speed decreases.

(c) a y =

Figure 4.42 4.43.

IDENTIFY: Use Newton’s 2nd law to relate the acceleration and forces for each crate. (a) SET UP: Since the crates are connected by a rope, they both have the same acceleration, 2.50 m/s 2 . (b) The forces on the 4.00 kg crate are shown in Figure 4.43a.

EXECUTE: ∑ Fx = max T = m1a = (4.00 kg)(2.50 m/s 2 ) = 10.0 N.

Figure 4.43a (c) SET UP: Forces on the 6.00 kg crate are shown in Figure 4.43b

The crate accelerates to the right, so the net force is to the right. F must be larger than T.

Figure 4.43b (d) EXECUTE:

∑F

x

= max gives F − T = m2 a

F = T + m2 a = 10.0 N + (6.00 kg)(2.50 m/s 2 ) = 10.0 N + 15.0 N = 25.0 N


4-13

EVALUATE: We can also consider the two crates and the rope connecting them as a single object of mass m = m1 + m2 = 10.0 kg. The free-body diagram is sketched in Figure 4.43c.

∑F

x

= max

F = ma = (10.0 kg)(2.50 m/s 2 ) = 25.0 N This agrees with our answer in part (d).

Figure 4.43c 4.44.

4.45.

IDENTIFY: Apply Newton's second and third laws. SET UP: Action-reaction forces act between a pair of objects. In the second law all the forces act on the same object. EXECUTE: (a) The force the astronaut exerts on the cable and the force that the cable exerts on the astronaut are an action-reaction pair, so the cable exerts a force of 80.0 N on the astronaut. (b) The cable is under tension. F 80.0 N = 0.762 m / s 2 . (c) a = = m 105.0 kg (d) There is no net force on the massless cable, so the force that the shuttle exerts on the cable must be 80.0 N (this is not an action-reaction pair). Thus, the force that the cable exerts on the shuttle must be 80.0 N. F 80.0 N (e) a = = = 8.84 × 10−4 m / s 2 . m 9.05 × 104 kg EVALUATE: Since the cable is massless the net force on it is zero and the tension is the same at each end. IDENTIFY and SET UP: Take derivatives of x (t ) to find vx and ax . Use Newton’s second law to relate the acceleration to the net force on the object. EXECUTE: (a) x = (9.0 × 103 m/s 2 )t 2 − (8.0 × 104 m/s3 )t 3 x = 0 at t = 0 When t = 0.025 s, x = (9.0 × 103 m/s 2 )(0.025 s) 2 − (8.0 × 104 m/s3 )(0.025 s)3 = 4.4 m. The length of the barrel must be 4.4 m. dx (b) vx = = (18.0 × 103 m/s 2 )t − (24.0 × 104 m/s3 )t 2 dt At t = 0, vx = 0 (object starts from rest). At t = 0.025 s, when the object reaches the end of the barrel, vx = (18.0 × 103 m/s 2 )(0.025 s) − (24.0 × 104 m/s3 )(0.025 s) 2 = 300 m/s

(c)

∑F

x

= max , so must find ax .

dvx = 18.0 × 103 m/s 2 − (48.0 × 104 m/s3 )t dt (i) At t = 0, ax = 18.0 × 103 m/s 2 and ∑ Fx = (1.50 kg)(18.0 × 103 m/s 2 ) = 2.7 × 104 N. ax =

(ii) At t = 0.025 s, ax = 18 × 103 m/s 2 − (48.0 × 104 m/s3 )(0.025 s) = 6.0 × 103 m/s 2 and

∑F

x

4.46.

= (1.50 kg)(6.0 × 103 m/s 2 ) = 9.0 × 103 N.

EVALUATE: The acceleration and net force decrease as the object moves along the barrel. ! ! IDENTIFY: Apply ∑ F = ma and solve for the mass m of the spacecraft. SET UP: w = mg . Let + y be upward. EXECUTE: (a) The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward. When it is speeding up, the acceleration is downward. (b) In each case the net force is in the direction of the acceleration. Speeding up: w > F and the net force is downward. Slowing down: w < F and the net force is upward.

4-14

Chapter 4

(c) Denote the y-component of the acceleration when the thrust is F1 by a1 and the y-component of the acceleration when the thrust is F2 by a1. a y = +1.20 m/s 2 and a2 = −0.80 m/s 2 . The forces and accelerations are then related by F1 − w = ma1 , F2 − w = ma2 . Dividing the first of these by the second to eliminate the mass gives F1 − w a1 = , and solving for the weight w gives F2 − w a2 w=

a1F2 − a2 F1 . Substituting the given numbers, with + y upward, gives a1 − a2 w=

4.47.

(1.20 m / s 2 )(10.0 × 103 N) − (−0.80 m / s 2 )(25.0 × 103 N) = 16.0 × 103 N. 1.20 m / s 2 − (−0.80 m / s 2 )

EVALUATE: The acceleration due to gravity at the surface of Mercury did not need to be found. IDENTIFY: The ship and instrument have the same acceleration. The forces and acceleration are related by Newton’s second law. We can use a constant acceleration equation to calculate the acceleration from the information given about the motion. ! SET UP: Let + y be upward. The forces on the instrument are the upward tension T exerted by the wire and the ! downward force w of gravity. w = mg = (6.50 kg)(9.80 m/s 2 ) = 63.7 N EXECUTE: (a) The free-body diagram is sketched in Figure 4.47. The acceleration is upward, so T > w . 2( y − y0 ) 2(276 m) = = 2.45 m/s 2 . y − y0 = 276 m , t = 15.0 s , v0 y = 0 . y − y0 = v0 yt + 12 a yt 2 gives a y = t2 (15.0 s) 2

∑F

y

= ma y gives T − w = ma and T = w + ma = 63.7 N + (6.50 kg)(2.45 m/s 2 ) = 79.6 N .

EVALUATE: There must be a net force in the direction of the acceleration.

Figure 4.47 4.48. 4.49.

If the rocket is moving downward and its speed is decreasing, its acceleration is upward, just as in Problem 4.47. The solution is identical to that of Problem 4.47. ! ! IDENTIFY: Apply ∑ F = ma to the gymnast.

SET UP: The upward force on the gymnast gives the tension in the rope. The free-body diagram for the gymnast is given in Figure 4.49. EXECUTE: (a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so the tension must equal the weight; T = mg . (b) No motion is no acceleration, so the tension is again the gymnast’s weight. ! ! (c) T − w = T − mg = ma = m a (the acceleration is upward, the same direction as the tension), so T = m( g + a ) . ! (d) T − w = T − mg = ma = − m a (the acceleration is downward, the opposite direction to the tension), so ! T = m( g − a ) . EVALUATE: When she accelerates upward the tension is greater than her weight and when she accelerates downward the tension is less than her weight.

Figure 4.49


4.50.

4-15

! ! Apply ∑ F = ma to the elevator to relate the forces on it to the acceleration.

IDENTIFY:

The free-body diagram for the elevator is sketched in Figure 4.50.

(a) SET UP:

The net force is T − mg (upward).

Figure 4.50

Take the + y -direction to be upward since that is the direction of the acceleration. The maximum upward acceleration is obtained from the maximum possible tension in the cables. EXECUTE: ∑ Fy = ma y gives T − mg = ma T − mg 28,000 N − (2200 kg)(9.80 m/s 2 ) = = 2.93 m/s 2 . m 2200 kg (b) What changes is the weight mg of the elevator. T − mg 28,000 N − (2200 kg)(1.62 m/s 2 ) a= = = 11.1 m/s 2 . m 2200 kg EVALUATE: The cables can give the elevator a greater acceleration on the moon since the downward force of gravity is less there and the same T then gives a greater net force. IDENTIFY: He is in free-fall until he contacts the ground. Use the constant acceleration equations and ! ! apply ∑ F = ma . a=

4.51.

SET UP:

Take + y downward. While he is in the air, before he touches the ground, his acceleration

is a y = 9.80 m/s 2 . EXECUTE:

(a) v0 y = 0 , y − y0 = 3.10 m , and a y = 9.80 m/s 2 . v y2 = v02y + 2a y ( y − y0 ) gives

v y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(3.10 m) = 7.79 m/s (b) v0 y = 7.79 m/s , v y = 0 , y − y0 = 0.60 m . v y2 = v02y + 2a y ( y − y0 ) gives

v y2 − v02y

0 − (7.79 m/s) 2 = −50.6 m/s 2 . The acceleration is upward. 2( y − y0 ) 2(0.60 m) ! (c) The free-body diagram is given in Fig. 4.51. F is the force the ground exerts on him. ∑ Fy = ma y gives mg − F = −ma . F = m( g + a) = (75.0 kg)(9.80 m/s2 + 50.6 m/s2 ) = 4.53 × 103 N , upward. ay =

=

F 4.53 × 103 N = = 6.16 , so F = 6.16w . w (75.0 kg)(9.80 m/s 2 )

! By Newton's third law, the force his feet exert on the ground is − F . EVALUATE: The force the ground exerts on him is about six times his weight.

4.52.

IDENTIFY:

Figure 4.51 ! ! Apply ∑ F = ma to the hammer head. Use a constant acceleration equation to relate the motion to the

acceleration. SET UP: Let + y be upward. EXECUTE: (a) The free-body diagram for the hammer head is sketched in Figure 4.52. (b) The acceleration of the hammer head is given by v y2 = v02y + 2a y ( y − y0 ) with v y = 0 , v0 y = −3.2 m/s 2 and y − y0 = −0.0045 m . a y = v02y / 2( y − y0 ) = (3.2 m / s) 2 / 2(0.0045 cm) = 1.138 × 103 m / s 2 . The mass of the hammer

4-16

Chapter 4

head is its weight divided by g , (4.9 N) /(9.80 m / s 2 ) = 0.50 kg , and so the net force on the hammer head is (0.50 kg)(1.138 × 103 m / s 2 ) = 570 N. This is the sum of the forces on the hammer head: the upward force that the nail exerts, the downward weight and the downward 15-N force. The force that the nail exerts is then 590 N, and this must be the magnitude of the force that the hammer head exerts on the nail. (c) The distance the nail moves is 0.12 m, so the acceleration will be 4267 m / s 2 , and the net force on the hammer head will be 2133 N. The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N. EVALUATE: For the shorter stopping distance the acceleration has a larger magnitude and the force between the nail and hammer head is larger.

4.53.

IDENTIFY:

Figure 4.52 ! ! Apply ∑ F = ma to some portion of the cable.

SET UP: The free-body diagrams for the whole cable, the top half of the cable and the bottom half are sketched in Figure 4.53. The cable is at rest, so in each diagram the net force is zero. EXECUTE: (a) The net force on a point of the cable at the top is zero; the tension in the cable must be equal to the weight w. (b) The net force on the cable must be zero; the difference between the tensions at the top and bottom must be equal to the weight w, and with the result of part (a), there is no tension at the bottom. (c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be half the weight, w / 2 . Equivalently, the net force on the upper half of the cable must be zero. From part (a) the tension at the top is w, the weight of the top half is w / 2 and so the tension in the cable at the middle must be w − w / 2 = w / 2 . (d) A graph of T vs. distance will be a negatively sloped line. EVALUATE: The tension decreases linearly from a value of w at the top to zero at the bottom of the cable.

Figure 4.53 4.54.

IDENTIFY: Note that in this problem the mass of the rope is given, and that it is not negligible compared to the ! ! other masses. Apply ∑ F = ma to each object to relate the forces to the acceleration. (a) SET UP:

The free-body diagrams for each block and for the rope are given in Figure 4.54a.

Figure 4.54a


4-17

Tt is the tension at the top of the rope and Tb is the tension at the bottom of the rope. EXECUTE: (b) Treat the rope and the two blocks together as a single object, with mass m = 6.00 kg + 4.00 kg + 5.00 kg = 15.0 kg. Take + y upward, since the acceleration is upward. The free-body diagram is given in Figure 4.54b. ∑ Fy = ma y

F − mg = ma F − mg m 200 N − (15.0 kg)(9.80 m/s 2 ) a= = 3.53 m/s 2 15.0 kg

a=

Figure 4.54b (c) Consider the forces on the top block (m = 6.00 kg), since the tension at the top of the rope (Tt ) will be one of these forces.

∑F

y

= ma y

F − mg − Tt = ma Tt = F − m( g + a ) T = 200 N − (6.00 kg)(9.80 m/s 2 + 3.53 m/s 2 ) = 120 N Figure 4.54c

Alternatively, can consider the forces on the combined object rope plus bottom block (m = 9.00 kg):

∑F

y

= ma y

Tt − mg = ma Tt = m( g + a ) = 9.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 120 N, which checks Figure 4.54d (d) One way to do this is to consider the forces on the top half of the rope (m = 2.00 kg). Let Tm be the tension at the midpoint of the rope.

∑F

y

= ma y

Tt − Tm − mg = ma Tm = Tt − m( g + a ) = 120 N − 2.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 93.3 N Figure 4.54e

To check this answer we can alternatively consider the forces on the bottom half of the rope plus the lower block taken together as a combined object (m = 2.00 kg + 5.00 kg = 7.00 kg):

∑F

y

= ma y

Tm − mg = ma Tm = m( g + a ) = 7.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 93.3 N, which checks Figure 4.54f

4-18

4.55.

Chapter 4

EVALUATE: The tension in the rope is not constant but increases from the bottom of the rope to the top. The tension at the top of the rope must accelerate the rope as well the 5.00-kg block. The tension at the top of the rope is less than F; there must be a net upward force on the 6.00-kg block. ! ! IDENTIFY: Apply ∑ F = ma to the barbell and to the athlete. Use the motion of the barbell to calculate its

acceleration. SET UP: Let + y be upward. EXECUTE: (a) The free-body diagrams for the baseball and for the athlete are sketched in Figure 4.55. (b) The athlete’s weight is mg = (90.0 kg)(9.80 m / s 2 ) = 882 N . The upward acceleration of the barbell is found from y − y0 = v0 yt + 12 a yt 2 . a y =

2( y − y0 ) 2(0.600 m) = = 0.469 m/s 2 . The force needed to lift the barbell is given t2 (1.6 s) 2

by Flift − wbarbell = ma y . The barbell’s mass is (490 N) / (9.80 m/s 2 ) = 50.0 kg , so Flift = wbarbell + ma = 490 N + (50.0 kg)(0.469 m / s 2 ) = 490 N + 23 N = 513 N . The athlete is not accelerating, so Ffloor − Flift − wathlete = 0 . Ffloor = Flift + wathlete = 513 N + 882 N = 1395 N . EVALUATE: Since the athlete pushes upward on the barbell with a force greater than its weight the barbell pushes down on him and the normal force on the athlete is greater than the total weight, 1362 N, of the athlete plus barbell.

4.56.

IDENTIFY:

Figure 4.55 ! ! Apply ∑ F = ma to the balloon and its passengers and cargo, both before and after objects are

dropped overboard. SET UP: When the acceleration is downward take + y to be downward and when the acceleration is upward take + y to be upward. EXECUTE: (a) The free-body diagram for the descending balloon is given in Figure 4.56. L is the lift force. (b) ∑ Fy = ma y gives Mg − L = M ( g / 3) and L = 2Mg / 3 . (c) Now + y is upward, so L − mg = m( g / 2) , where m is the mass remaining. L = 2 Mg / 3 , so m = 4M / 9 . Mass 5M / 9 must be dropped overboard. EVALUATE: In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the total weight.

Figure 4.56


4.57.

IDENTIFY: SET UP:

! ! Apply ∑ F = ma to the entire chain and to each link.

m = mass of one link. Let + y be upward.

EXECUTE:

(a) The free-body diagrams are sketched in Figure 4.57. Ftop is the force the top and middle links

exert on each other. Fmiddle is the force the middle and bottom links exert on each other. (b) (i) The weight of each link is mg = (0.300 kg)(9.80 m / s 2 ) = 2.94 N . Using the free-body diagram for the whole chain:

a=

Fstudent − 3mg 12 N − 3(2.94 N) 3.18 N = = = 3.53 m / s 2 3m 0.900 kg 0.900 kg

(ii) The top link also accelerates at 3.53 m / s 2 , so Fstudent − Ftop − mg = ma . Ftop = Fstudent − m( g + a ) = 12 N − (0.300 kg)(9.80 m/s 2 + 3.53 m/s 2 ) = 8.0 N . EVALUATE:

The force exerted by the middle link on the bottom link is given by Fmiddle − mg = ma and

Fmiddle = m( g + a ) = 4.0 N . We can verify that with our results

4.58.

y

= ma y is satisfied for the middle link.

Figure 4.57 ! ! ! ! ! IDENTIFY: Calculate a from a = d 2 r / dt 2 . Then Fnet = ma . SET UP: w = mg EXECUTE: Differentiating twice, the acceleration of the helicopter as a function of time is ! ! a = (0.120 m / s3 )tiˆ − (0.12 m / s 2 )kˆ and at t = 5.0s , the acceleration is a = (0.60 m / s 2 )iˆ − (0.12 m / s 2 )kˆ . The force is then ! ! w ! (2.75 × 105 N) ⎡ (0.60 m / s 2 )iˆ − (0.12 m / s 2 )kˆ ⎤⎦ = (1.7 × 104 N)iˆ − (3.4 × 103 N)kˆ F = ma = a = g (9.80 m / s 2 ) ⎣ EVALUATE:

4.59.

∑F

IDENTIFY:

The force and acceleration are in the same direction. They are both time dependent. d 2x Fx = max and ax = 2 . dt

d n (t ) = nt n −1 dt EXECUTE: The velocity as a function of time is vx (t ) = A − 3Bt 2 and the acceleration as a function of time is SET UP:

4.60.

ax (t ) = −6 Bt , and so the force as a function of time is Fx (t ) = ma (t ) = −6mBt . EVALUATE: Since the acceleration is along the x-axis, the force is along the x-axis. t ! ! ! ! ! IDENTIFY: a = F / m . v = v0 + ∫ a dt . 0

v0 = 0 since the object is initially at rest. ! 1 t ! 1⎛ k ⎞ EXECUTE: v (t ) = ∫ F dt = ⎜ k1tiˆ + 2 t 4 ˆj ⎟ . 0 m m⎝ 4 ⎠ ! ! EVALUATE: F has both x and y components, so v develops x and y components. IDENTIFY: Follow the steps specified in the problem. dv dv dv dv SET UP: The chain rule for differentiating says = = v. dt dx dt dx SET UP:

4.61.

4-19

4-20

Chapter 4

dv cannot be integrated with respect to time, as the unknown dt C dv function v(t ) is part of the integrand. The equation must be separated before integration; that is, − dt = 2 and m v Ct 1 1 − =− + , m v v0

EXECUTE:

(a) The equation of motion, −Cv 2 = m

where v0 is the constant of integration that gives v = v0 at t = 0 . Note that this form shows that if v0 = 0 , there is −1

dx ⎛ 1 Ct ⎞ no motion. This expression may be rewritten as v = =⎜ + ⎟ , dt ⎝ v0 m ⎠ m ⎡ Ctv0 ⎤ which may be integrated to obtain x − x0 = ln ⎢1 + . C ⎣ m ⎥⎦ To obtain x as a function of v, the time t must be eliminated in favor of v; from the expression obtained after the m ⎛v ⎞ Ctv0 v0 first integration, = − 1 , so x − x0 = ln ⎜ 0 ⎟ . C ⎝v⎠ m v dv dv (b) Applying the chain rule, ∑ F = m = mv . Using the given expression for the net force, dt dx ⎛v⎞ C m ⎛v ⎞ C dv ⎛ dv ⎞ −Cv 2 = ⎜ v ⎟ m . − dx = . Integrating gives − ( x − x0 ) = ln ⎜ ⎟ and x − x0 = ln ⎜ 0 ⎟ . m v C m v ⎝ dx ⎠ ⎝v⎠ ⎝ 0⎠ EVALUATE: If C is positive, our expression for v (t ) shows it decreases from its value of v0 . As v decreases, so does the acceleration and therefore the rate of decrease of v. 4.62.

IDENTIFY:

t

t

0

0

x = ∫ vx dt and vx = ∫ ax dt , and similar equations apply to the y-component.

SET UP: In this situation, the x-component of force depends explicitly on the y-component of position. As the ycomponent of force is given as an explicit function of time, v y and y can be found as functions of time and used in

the expression for ax (t ) . EXECUTE:

a y = (k3 / m)t , so v y = (k3 / 2m)t 2 and y = ( k3 / 6m)t 3 , where the initial conditions v0 y = 0, y0 = 0 have

been used. Then, the expressions for ax , vx and x are obtained as functions of time: ax =

k1 k2 k3 3 + t , m 6m 2

k1 kk k kk t + 2 32 t 4 and x = 1 t 2 + 2 3 2 t 5 . m 24m 2m 120m ! ⎛ k1 2 k2 k3 5 ⎞ ˆ ⎛ k3 3 ⎞ ˆ ! ⎛k kk ⎞ ⎛ k ⎞ t + t ⎟i + ⎜ t ⎟ j and v = ⎜ 1 t + 2 32 t 4 ⎟ iˆ + ⎜ 3 t 2 ⎟ ˆj . In vector form, r = ⎜ 2 120m ⎠ ⎝ 6m ⎠ 24m ⎠ ⎝ 2m ⎠ ⎝m ⎝ 2m EVALUATE: ax depends on time because it depends on y, and y is a function of time. vx =

5

APPLYING NEWTON’S LAWS

5.1.

IDENTIFY:

a = 0 for each object. Apply

∑F

y

= ma y to each weight and to the pulley.

SET UP: Take + y upward. The pulley has negligible mass. Let Tr be the tension in the rope and let Tc be the tension in the chain. EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a. ∑ Fy = ma y gives Tr = w = 25.0 N . (b) The free-body diagram for the pulley is given in Figure 5.1b. Tc = 2Tr = 50.0 N . EVALUATE: The tension is the same at all points along the rope.

Figure 5.1a, b 5.2.

5.3.

IDENTIFY:

! ! Apply ∑ F = ma to each weight.

SET UP: Two forces act on each mass: w down and T ( = w) up. EXECUTE: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w. EVALUATE: The tension is the same in all three cases. IDENTIFY: Both objects are at rest and a = 0 . Apply Newton’s first law to the appropriate object. The maximum tension Tmax is at the top of the chain and the minimum tension is at the bottom of the chain. SET UP: Let + y be upward. For the maximum tension take the object to be the chain plus the ball. For the minimum tension take the object to be the ball. For the tension T three-fourths of the way up from the bottom of the chain, take the chain below this point plus the ball to be the object. The free-body diagrams in each of these three cases are sketched in Figures 5.3a, 5.3b and 5.3c. mb+c = 75.0 kg + 26.0 kg = 101.0 kg . mb = 75.0 kg . m is

the mass of three-fourths of the chain: m = 34 (26.0 kg) = 19.5 kg . EXECUTE:

(a) From Figure 5.3a,

∑F

y

= 0 gives Tmax − mb+c g = 0 and Tmax = (101.0 kg)(9.80 m/s 2 ) = 990 N .

∑ F = 0 gives T − m g = 0 and T = (75.0 kg)(9.80 m/s ) = 735 N . (b) From Figure 5.3c, ∑ F = 0 gives T − (m + m ) g = 0 and T = (19.5 kg + 75.0 kg)(9.80 m/s ) = 926 N . From Figure 5.3b,

2

min

y

min

b

2

y

b

5-1

5-2

Chapter 5

EVALUATE:

The tension in the chain increases linearly from the bottom to the top of the chain.

Figure 5.3a–c 5.4.

IDENTIFY: Apply Newton’s 1st law to the person. Each half of the rope exerts a force on him, directed along the rope and equal to the tension T in the rope. SET UP: (a) The force diagram for the person is given in Figure 5.4

T1 and T2 are the tensions in each half of the rope. Figure 5.4 EXECUTE:

∑F

x

=0

T2 cosθ − T1 cosθ = 0 This says that T1 = T2 = T (The tension is the same on both sides of the person.)

∑F

y

=0

T1 sin θ + T2 sin θ − mg = 0 But T1 = T2 = T , so 2T sin θ = mg T=

mg (90.0 kg)(9.80 m/s 2 ) = = 2540 N 2sin θ 2sin10.0°

(b) The relation 2T sin θ = mg still applies but now we are given that T = 2.50 × 104 N (the breaking strength) and are asked to find θ .

sin θ =

5.5.

mg (90.0 kg)(9.80 m/s 2 ) = = 0.01764, θ = 1.01°. 2T 2(2.50 × 104 N)

EVALUATE: T = mg /(2sin θ ) says that T = mg / 2 when θ = 90° (rope is vertical). T → ∞ when θ → 0 since the upward component of the tension becomes a smaller fraction of the tension. ! ! IDENTIFY: Apply ∑ F = ma to the frame. SET UP: Let w be the weight of the frame. Since the two wires make the same angle with the vertical, the tension is the same in each wire. T = 0.75w . EXECUTE: The vertical component of the force due to the tension in each wire must be half of the weight, and w 3w cosθ this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical. = 2 4 and θ = arccos 23 = 48° . EVALUATE:

If θ = 0° , T = w / 2 and T → ∞ as θ → 90° . Therefore, there must be an angle where T = 3w / 4 .

Applying Newton’s Laws

5.6.

5-3

IDENTIFY: Apply Newton’s 1st law to the car. The forces are the same as in Example 5.5. SET UP: The free-body diagram is sketched in Figure 5.6.

EXECUTE: ∑ Fx = max

T cos α − n sin α = 0 T cos α = n sin α ∑ Fy = ma y

n cos α + T sin α − w = 0 n cos α + T sin α = w

Figure 5.6

⎛ cos α ⎞ The first equation gives n = T ⎜ ⎟. ⎝ sin α ⎠ Use this in the second equation to eliminate n: ⎛ cos α ⎞ ⎜T ⎟ cos α + T sin α = w ⎝ sin α ⎠ Multiply this equation by sin α : T (cos 2 α + sin 2 α ) = w sin α

5.7.

T = w sin α (since cos 2 α + sin 2 α = 1 ). ⎛ cos α ⎞ ⎛ cos α ⎞ Then n = T ⎜ ⎟ = w sin α ⎜ ⎟ = w cos α . ⎝ sin α ⎠ ⎝ sin α ⎠ EVALUATE: These results are the same as obtained in Example 5.5. The choice of coordinate axes is up to us. Some choices may make the calculation easier, but the results are the same for any choice of axes. ! ! IDENTIFY: Apply ∑ F = ma to the car. SET UP: Use coordinates with + x parallel to the surface of the street. EXECUTE: ∑ Fx = 0 gives T = w sin α . F = mg sinθ = (1390 kg)(9.80 m/s 2 )sin17.5° = 4.10 × 103 N .

5.8.

EVALUATE: The force required is less than the weight of the car by the factor sin α . IDENTIFY: Apply Newton’s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the cable. SET UP: The force diagram for the wrecking ball is sketched in Figure 5.8.

Figure 5.8 EXECUTE: (a) ∑ Fy = ma y

TB cos 40° − mg = 0 mg (4090 kg)(9.80 m/s 2 ) = = 5.23 × 104 N cos 40° cos 40° (b) ∑ Fx = max TB =

TB sin 40° − TA = 0 TA = TB sin 40° = 3.36 × 104 N EVALUATE:

If the angle 40° is replaces by 0° (cable B is vertical), then TB = mg and TA = 0.

5-4

5.9.

Chapter 5

Apply

IDENTIFY:

!

!

∑ F = ma to the object and to the knot where the cords are joined.

Let + y be upward and + x be to the right.

SET UP: EXECUTE:

(a) TC = w, TA sin30° + TB sin 45° = TC = w, and TA cos30° − TB cos 45° = 0. Since sin 45° = cos 45°,

adding the last two equations gives TA (cos30° + sin 30°) = w, and so TA = TB = TA

w = 0.732 w. Then, 1.366

cos30° = 0.897 w. cos 45°

(b) Similar to part (a), TC = w, − TA cos60° + TB sin 45° = w, and TA sin 60° − TB cos 45° = 0.

Adding these two equations, TA = EVALUATE: 5.10.

w sin 60° = 2.73w, and TB = TA = 3.35w. cos 45° (sin 60° − cos 60°)

In part (a), TA + TB > w since only the vertical components of TA and TB hold the object against

gravity. In part (b), since TA has a downward component TB is greater than w. IDENTIFY: Apply Newton’s first law to the car. SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp. EXECUTE: (a) The free-body diagram for the car is given in Figure 5.10. The vertical weight w and the tension T in the cable have each been replaced by their x and y components. sin 25.0° sin 25.0° (b) ∑ Fx = 0 gives T cos31.0° − w sin 25.0° = 0 and T = w = (1130 kg)(9.80 m/s 2 ) = 5460 N . cos31.0° cos31.0° (c) ∑ Fy = 0 gives n + T sin 31.0° − w cos 25.0° = 0 and

n = w cos 25.0°− T sin 31.0° = (1130 kg)(9.80 m/s 2 )cos 25.0° − (5460 N)sin 31.0° = 7220 N EVALUATE: We could also use coordinates that are horizontal and vertical and would obtain the same values of n and T.

Figure 5.10 5.11.

IDENTIFY: Since the velocity is constant, apply Newton’s first law to the piano. The push applied by the man must oppose the component of gravity down the incline. ! SET UP: The free-body diagrams for the two cases are shown in Figures 5.11a and b. F is the force applied by the man. Use the coordinates shown in the figure. EXECUTE: (a) ∑ Fx = 0 gives F − w sin11.0° = 0 and F = (180 kg)(9.80 m/s 2 )sin11.0° = 337 N . (b)

∑F

y

= 0 gives n cos11.0° − w = 0 and n =

w . cos11.0°

w ⎛ ⎞ F =⎜ ⎟ sin11.0° = w tan11.0° = 343 N . ° cos11.0 ⎝ ⎠

∑F

x

= 0 gives F − n sin11.0° = 0 and


5-5

EVALUATE: A slightly greater force is required when the man pushes parallel to the floor. If the slope angle of the incline were larger, sin α and tan α would differ more and there would be more difference in the force needed in each case.

Figure 5.11a, b 5.12.

IDENTIFY: Apply Newton’s 1st law to the hanging weight and to each knot. The tension force at each end of a string is the same. (a) Let the tensions in the three strings be T, T ′, and T ′′, as shown in Figure 5.12a.

Figure 5.12a SET UP:

The free-body diagram for the block is given in Figure 5.12b. EXECUTE: ∑ Fy = 0

T′ − w = 0 T ′ = w = 60.0 N

Figure 5.12b SET UP:

The free-body diagram for the lower knot is given in Figure 5.12c. EXECUTE: ∑ Fy = 0

T sin 45° − T ′ = 0 T′ 60.0 N = = 84.9 N T= sin 45° sin 45°

Figure 5.12c

5-6

Chapter 5

(b) Apply

∑F

x

∑F

x

= 0 to the force diagram for the lower knot:

=0

F2 = T cos 45° = (84.9 N)cos 45° = 60.0 N SET UP: The free-body diagram for the upper knot is given in Figure 5.12d. EXECUTE: ∑ Fx = 0

T cos 45° − F1 = 0 F1 = (84.9 N)cos 45° F1 = 60.0 N Figure 5.12d

Note that F1 = F2 . EVALUATE:

Applying

∑F

y

= 0 to the upper knot gives T ′′ = T sin 45° = 60.0 N = w. If we treat the whole

system as a single object, the force diagram is given in Figure 5.12e.

∑F ∑F

x

= 0 gives F2 = F1 , which checks

y

= 0 gives T ′′ = w, which checks

Figure 5.12e 5.13.

IDENTIFY: Apply Newton’s first law to the ball. The force of the wall on the ball and the force of the ball on the wall are related by Newton’s third law. SET UP: The forces on the ball are its weight, the tension in the wire, and the normal force applied by the wall. 16.0 cm To calculate the angle φ that the wire makes with the wall, use Figure 5.13a. sin φ = and φ = 20.35° 46.0 cm EXECUTE: (a) The free-body diagram is shown in Figure 5.13b. Use the x and y coordinates shown in the figure. w (45.0 kg)(9.80 m/s 2 ) = 470 N ∑ Fy = 0 gives T cosφ − w = 0 and T = cosφ = cos 20.35° (b)

∑F

x

= 0 gives T sin φ − n = 0 . n = (470 N)sin 20.35° = 163 N . By Newton’s third law, the force the ball

exerts on the wall is 163 N, directed to the right. ⎛ w ⎞ EVALUATE: n = ⎜ ⎟ sin φ = w tan φ . As the angle φ decreases (by increasing the length of the wire), T ⎝ cos φ ⎠ decreases and n decreases.

5.14.

Figure 5.13a, b ! ! IDENTIFY: Apply ∑ F = ma to each block. a = 0 . SET UP:

Take + y perpendicular to the incline and + x parallel to the incline.


5-7

EXECUTE: The free-body diagrams for each block, A and B, are given in Figure 5.14. (a) For B, ∑ Fx = max gives T1 − w sin α = 0 and T1 = w sin α . (b) For block A, (c)

∑F

y

∑F

x

= max gives T1 − T2 − w sin α = 0 and T2 = 2w sin α .

= ma y for each block gives nA = nB = w cos α .

(d) For α → 0 , T1 = T2 → 0 and nA = nB → w . For α → 90° , T1 = w , T2 = 2w and nA = nB = 0 . EVALUATE: The two tensions are different but the two normal forces are the same.

Figure 5.14a, b 5.15.

IDENTIFY: Apply Newton’s first law to the ball. Treat the ball as a particle. SET UP: The forces on the ball are gravity, the tension in the wire and the normal force exerted by the surface. The normal force is perpendicular to the surface of the ramp. Use x and y axes that are horizontal and vertical. EXECUTE: (a) The free-body diagram for the ball is given in Figure 5.15. The normal force has been replaced by its x and y components. mg (b) ∑ Fy = 0 gives n cos35.0° − w = 0 and n = = 1.22mg . cos35.0° (c) ∑ Fx = 0 gives T − n sin 35.0° = 0 and T = (1.22mg )sin 35.0° = 0.700mg . EVALUATE: Note that the normal force is greater than the weight, and increases without limit as the angle of the ramp increases towards 90° . The tension in the wire is w tan φ , where φ is the angle of the ramp and T also increases without limit as φ → 90° .

Figure 5.15 5.16.

IDENTIFY: Apply Newton’s second law to the rocket plus its contents and to the power supply. Both the rocket and the power supply have the same acceleration. SET UP: The free-body diagrams for the rocket and for the power supply are given in Figures 5.16a and b. Since the highest altitude of the rocket is 120 m, it is near to the surface of the earth and there is a downward gravity force on each object. Let + y be upward, since that is the direction of the acceleration. The power supply has

mass mps = (15.5 N) /(9.80 m/s 2 ) = 1.58 kg

5-8

Chapter 5

EXECUTE:

a= (b)

(a)

∑F

y

= ma y applied to the rocket gives F − mr g = mr a .

F − mr g 1720 N − (125 kg)(9.80 m/s 2 ) = = 3.96 m/s 2 . 125 kg mr

∑F

y

= ma y applied to the power supply gives n − mps g = mps a .

n = mps ( g + a) = (1.58 kg)(9.80 m/s 2 + 3.96 m/s 2 ) = 21.7 N . EVALUATE: The acceleration is constant while the thrust is constant and the normal force is constant while the acceleration is constant. The altitude of 120 m is not used in the calculation.

Figure 5.16a, b 5.17.

IDENTIFY: Use the kinematic information to find the acceleration of the capsule and the stopping time. Use Newton’s second law to find the force F that the ground exerted on the capsule during the crash. SET UP: Let + y be upward. 311 km/h = 86.4 m/s . The free-body diagram for the capsule is given in Figure 15.17. EXECUTE: y − y0 = −0.810 m , v0 y = −86.4 m/s , v y = 0 . v y2 = v02y + 2a y ( y − y0 ) gives

ay = (b)

v y2 − v02y 2( y − y0 )

∑F

y

=

0 − (−86.4 m/s) 2 = 4610 m/s 2 = 470 g . 2( −0.810) m

= ma y applied to the capsule gives F − mg = ma and

F = m( g + a) = (210 kg)(9.80 m/s 2 + 4610 m/s 2 ) = 9.70 × 105 N = 471w. ⎛ v + vy ⎞ 2( y − y0 ) 2(−0.810 m) = = 0.0187 s (c) y − y0 = ⎜ 0 y ⎟ t gives t = v + v − 86.4 m/s 2 + 0 2 ⎝ ⎠ y 0y EVALUATE: The upward force exerted by the ground is much larger than the weight of the capsule and stops the capsule in a short amount of time. After the capsule has come to rest, the ground still exerts a force mg on the capsule, but the large 9.00 × 105 N force is exerted only for 0.0187 s.

Figure 5.17 5.18.

IDENTIFY: Apply Newton’s second law to the three sleds taken together as a composite object and to each individual sled. All three sleds have the same horizontal acceleration a. SET UP: The free-body diagram for the three sleds taken as a composite object is given in Figure 5.18a and for each individual sled in Figure 5.18b-d. Let + x be to the right, in the direction of the acceleration. mtot = 60.0 kg . EXECUTE:

a=

(a)

∑F

x

= max for the three sleds as a composite object gives P = mtot a and

P 125 N = = 2.08 m/s 2 . mtot 60.0 kg


(b)

∑F

x

5-9

= max applied to the 10.0 kg sled gives P − TA = m10 a and

TA = P − m10 a = 125 N − (10.0 kg)(2.08 m/s 2 ) = 104 N .

∑F

x

= max applied to the 30.0 kg sled gives

TB = m30 a = (30.0 kg)(2.08 m/s ) = 62.4 N . 2

EVALUATE:

If we apply

x

= max to the 20.0 kg sled and calculate a from TA and TB found in part (b), we get

TA − TB 104 N − 62.4 N = = 2.08 m/s 2 , which agrees with the value we calculated in part (a). m20 20.0 kg

TA − TB = m20 a . a =

5.19.

∑F

Figure 5.18a–d ! ! IDENTIFY: Apply ∑ F = ma to the load of bricks and to the counterweight. The tension is the same at each end

of the rope. The rope pulls up with the same force (T ) on the bricks and on the counterweight. The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude. (a) SET UP: The free-body diagrams for the bricks and counterweight are given in Figure 5.19.

Figure 5.19

∑F

= ma y to each object. The acceleration magnitude is the same for the two objects. ! For the bricks take + y to be upward since a for the bricks is upward. For the counterweight take + y to be ! downward since a is downward. bricks: ∑ Fy = ma y (b) EXECUTE:

Apply

y

T − m1 g = m1a counterweight:

∑F

y

= ma y

m2 g − T = m2 a Add these two equations to eliminate T: (m2 − m1 ) g = (m1 + m2 ) a ⎛ m − m1 ⎞ ⎛ 28.0 kg − 15.0 kg ⎞ 2 2 a =⎜ 2 ⎟g = ⎜ ⎟ (9.80 m/s ) = 2.96 m/s ⎝ 15.0 kg + 28.0 kg ⎠ ⎝ m1 + m2 ⎠ (c) T − m1 g = m1a gives T = m1 (a + g ) = (15.0 kg)(2.96 m/s 2 + 9.80 m/s 2 ) = 191 N As a check, calculate T using the other equation. m2 g − T = m2 a gives T = m2 ( g − a) = 28.0 kg(9.80 m/s 2 − 2.96 m/s 2 ) = 191 N, which checks.

5-10

5.20.

Chapter 5

EVALUATE: The tension is 1.30 times the weight of the bricks; this causes the bricks to accelerate upward. The tension is 0.696 times the weight of the counterweight; this causes the counterweight to accelerate downward. If m1 = m2 , a = 0 and T = m1 g = m2 g . In this special case the objects don’t move. If m1 = 0, a = g and T = 0; in this special case the counterweight is in free-fall. Our general result is correct in these two special cases. IDENTIFY: In part (a) use the kinematic information and the constant acceleration equations to calculate the ! ! ! ! acceleration of the ice. Then apply ∑ F = ma . In part (b) use ∑ F = ma to find the acceleration and use this in

the constant acceleration equations to find the final speed. SET UP: Figures 5.20a and b give the free-body diagrams for the ice both with and without friction. Let + x be directed down the ramp, so + y is perpendicular to the ramp surface. Let φ be the angle between the ramp and the horizontal. The gravity force has been replaced by its x and y components. EXECUTE: (a) x − x0 = 1.50 m , v0 x = 0 , vx = 2.50 m/s . vx2 = v02x + 2ax ( x − x0 ) gives ax =

vx2 − v02x (2.50 m/s) 2 − 0 = = 2.08 m/s 2 . 2( x − x0 ) 2(1.50 m)

∑F

x

= max gives mg sin φ = ma and sin φ =

a 2.08 m/s 2 = . g 9.80 m/s 2

φ = 12.3° . (b)

∑F

a=

mg sin φ − f (8.00 kg)(9.80 m/s 2 )sin12.3° − 10.0 N = = 0.838 m/s 2 . m 8.00 kg

x

= max gives mg sin φ − f = ma and

Then x − x0 = 1.50 m , v0 x = 0 , ax = 0.838 m/s 2 and vx2 = v02x + 2ax ( x − x0 ) gives

vx = 2ax ( x − x0 ) = 2(0.838 m/s 2 )(1.50 m) = 1.59 m/s EVALUATE:

5.21.

IDENTIFY:

With friction present the speed at the bottom of the ramp is less.

Figure 5.20a, b ! ! Apply ∑ F = ma to each block. Each block has the same magnitude of acceleration a.

SET UP: Assume the pulley is to the right of the 4.00 kg block. There is no friction force on the 4.00 kg block, the only force on it is the tension in the rope. The 4.00 kg block therefore accelerates to the right and the suspended block accelerates downward. Let + x be to the right for the 4.00 kg block, so for it ax = a , and let + y be

downward for the suspended block, so for it a y = a . (a) The free-body diagrams for each block are given in Figures 5.21a and b. T 10.0 N = = 2.50 m/s 2 . (b) ∑ Fx = max applied to the 4.00 kg block gives T = (4.00 kg)a and a = 4.00 kg 4.00 kg EXECUTE:

(c)

∑F

m=

y

= ma y applied to the suspended block gives mg − T = ma and

T 10.0 N = = 1.37 kg . g − a 9.80 m/s 2 − 2.50 m/s 2

(d) The weight of the hanging block is mg = (1.37 kg)(9.80 m/s 2 ) = 13.4 N . This is greater than the tension in the rope; T = 0.75mg .


5-11

EVALUATE: Since the hanging block accelerates downward, the net force on this block must be downward and the weight of the hanging block must be greater than the tension in the rope. Note that the blocks accelerate no matter how small m is. It is not necessary to have m > 4.00 kg , and in fact in this problem m is less than 4.00 kg.

Figure 5.21a, b 5.22.

(a) Consider both gliders together as a single object, apply

IDENTIFY:

!

!

∑ F = ma , and solve for a. Use a in a

constant acceleration equation to find the required runway length. ! ! (b) Apply ∑ F = ma to the second glider and solve for the tension Tg in the towrope that connects the two gliders. SET UP:

In part (a), set the tension Tt in the towrope between the plane and the first glider equal to its maximum

value, Tt = 12,000 N . EXECUTE: (a) The free-body diagram for both gliders as a single object of mass 2m = 1400 kg is given in Figure 5.22a.

∑F

x

= max gives Tt − 2 f = (2m)a and a =

Tt − 2 f 12,000 N − 5000 N = = 5.00 m/s 2 . Then 2m 1400 kg

ax = 5.00 m/s 2 , v0 x = 0 and vx = 40 m/s in vx2 = v02x + 2ax ( x − x0 ) gives ( x − x0 ) =

vx2 − v02x = 160 m . 2a x

(b) The free-body diagram for the second glider is given in Figure 5.22b. ∑ Fx = max gives Tg − f = ma and T = f + ma = 2500 N + (700 kg)(5.00 m/s2 ) = 6000 N . EVALUATE:

We can verify that

∑F

x

= max is also satisfied for the first glider.

Figure 5.22a, b 5.23.

IDENTIFY:

The maximum tension in the chain is at the top of the chain. Apply

!

!

∑ F = ma to the composite

object of chain and boulder. Use the constant acceleration kinematic equations to relate the acceleration to the time. SET UP: Let + y be upward. The free-body diagram for the composite object is given in Figure 5.23. T = 2.50 wchain . mtot = mchain + mboulder = 1325 kg . EXECUTE:

(a)

∑F

y

= ma y gives T − mtot g = mtot a . a =

⎛ 2.50[575 kg] ⎞ a =⎜ − 1⎟ (9.80 m/s 2 ) = 0.832 m/s 2 . ⎝ 1325 kg ⎠

⎞ T − mtot g 2.50mchain g − mtot g ⎛ 2.50mchain = =⎜ − 1⎟ g mtot mtot ⎝ mtot ⎠

5-12

Chapter 5

(b) Assume the acceleration has its maximum value: a y = 0.832 m/s 2 , y − y0 = 125 m and v0 y = 0 .

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(125 m) = = 17.3 s ay 0.832 m/s 2

EVALUATE: The tension in the chain is T = 1.41 × 104 N and the total weight is 1.30 × 104 N . The upward force exceeds the downward force and the acceleration is upward.

5.24.

Figure 5.23 ! ! Apply ∑ F = ma to the composite object of elevator plus student ( mtot = 850 kg ) and also to the

IDENTIFY:

student ( w = 550 N ). The elevator and the student have the same acceleration. SET UP: Let + y be upward. The free-body diagrams for the composite object and for the student are given in Figure 5.24a and b. T is the tension in the cable and n is the scale reading, the normal force the scale exerts on the student. The mass of the student is m = w / g = 56.1 kg . EXECUTE:

ay =

(a)

∑F

y

= ma y applied to the student gives n − mg = ma y .

n − mg 450 N − 550 N = = −1.78 m/s 2 . The elevator has a downward acceleration of 1.78 m/s 2 . m 56.1 kg 670 N − 550 N = 2.14 m/s 2 . 56.1 kg

(b) a y =

(c) n = 0 means a y = − g . The student should worry; the elevator is in free-fall. (d)

∑F

y

= ma y applied to the composite object gives T − mtot g = mtot a . T = mtot (a y + g ) . In part (a),

T = (850 kg)(−1.78 m/s 2 + 9.80 m/s 2 ) = 6820 N . In part (c), a y = − g and T = 0 . EVALUATE: In part (b), T = (850 kg)(2.14 m/s 2 + 9.80 m/s 2 ) = 10,150 N . The weight of the composite object is 8330 N. When the acceleration is upward the tension is greater than the weight and when the acceleration is downward the tension is less than the weight.

5.25.

Figure 5.24a, b ! ! Apply ∑ F = ma to the puck. Use the information about the motion to calculate the acceleration. The

IDENTIFY:

table must slope downward to the right. SET UP: Let α be the angle between the table surface and the horizontal. Let the + x -axis be to the right and parallel to the surface of the table. EXECUTE: ∑ Fx = max gives mg sin α = max . The time of travel for the puck is L / v0 , where L = 1.75 m and v0 = 3.80 m/s . x − x0 = v0 xt + 12 axt 2 gives ax =

2 x 2 xv02 a 2 xv02 = 2 , where x = 0.0250 m . sin α = x = . 2 t L g gL2

⎛ 2(2.50 × 10−2 m)(3.80 m / s) 2 ⎞ ⎟ = 1.38° . ⎜ ( 9.80 m / s 2 ) (1.75 m) 2 ⎟ ⎝ ⎠

α = arcsin ⎜


5.26.

5-13

EVALUATE: The table is level in the direction along its length, since the velocity in that direction is constant. The angle of slope to the right is small, so the acceleration and deflection in that direction are small. ! dv ! IDENTIFY: Acceleration and velocity are related by a y = y . Apply ∑ F = ma to the rocket. dt ! SET UP: Let + y be upward. The free-body diagram for the rocket is sketched in Figure 5.26. F is the thrust force. EXECUTE: (a) v y = At + Bt 2 . a y = A + 2 Bt . At t = 0 , a y = 1.50 m/s 2 so A = 1.50 m/s 2 . Then v y = 2.00 m/s at

t = 1.00 s gives 2.00 m/s = (1.50 m/s 2 )(1.00 s) + B (1.00 s) 2 and B = 0.50 m/s3 . (b) At t = 4.00 s , a y = 1.50 m/s 2 + 2(0.50 m/s3 )(4.00 s) = 5.50 m/s 2 . (c)

∑F

y

= ma y applied to the rocket gives T − mg = ma and

T = m(a + g ) = (2540 kg)(9.80 m/s 2 + 5.50 m/s 2 ) = 3.89 × 104 N . T = 1.56 w . (d) When a = 1.50 m/s 2 , T = (2540 kg)(9.80 m/s 2 + 1.50 m/s 2 ) = 2.87 × 104 N EVALUATE: During the time interval when v(t ) = At + Bt 2 applies the magnitude of the acceleration is increasing, and the thrust is increasing.

Figure 5.26 5.27.

IDENTIFY: Consider the forces in each case. There is the force of gravity and the forces from objects that touch the object in question. SET UP: A surface exerts a normal force perpendicular to the surface, and a friction force, parallel to the surface. EXECUTE: The free-body diagrams are sketched in Figure 5.27a-c. EVALUATE: Friction opposes relative motion between the two surfaces. When one surface is stationary the friction force on the other surface is directed opposite to its motion.

Figure 5.27a–c 5.28.

IDENTIFY:

f s ≤ μs n and f k = μk n . The normal force n is determined by applying

!

!

∑ F = ma to the block.

Normally, μ k ≤ μs . f s is only as large as it needs to be to prevent relative motion between the two surfaces. SET UP: Since the table is horizontal, with only the block present n = 135 N . With the brick on the block, n = 270 N . EXECUTE: (a) The friction is static for P = 0 to P = 75.0 N . The friction is kinetic for P > 75.0 N . (b) The maximum value of f s is μs n . From the graph the maximum f s is f s = 75.0 N , so max fs 75.0 N f 50.0 N = = 0.556 . f k = μk n . From the graph, f k = 50.0 N and μ k = k = = 0.370 . n 135 N n 135 N (c) When the block is moving the friction is kinetic and has the constant value f k = μk n , independent of P. This is

μs =

why the graph is horizontal for P > 75.0 N . When the block is at rest, f s = P since this prevents relative motion. This is why the graph for P < 75.0 N has slope +1. (d) max fs and f k would double. The values of f on the vertical axis would double but the shape of the graph would be unchanged. EVALUATE: The coefficients of friction are independent of the normal force.

5-14

5.29.

Chapter 5

(a) IDENTIFY: Constant speed implies a = 0. Apply Newton’s 1st law to the box. The friction force is directed opposite to the motion of the box. ! SET UP: Consider the free-body diagram for the box, given in Figure 5.29a. Let F be the horizontal force applied by the worker. The friction is kinetic friction since the box is sliding along the surface. EXECUTE: ∑ Fy = ma y

n − mg = 0 n = mg So f k = μk n = μ k mg Figure 5.29a

∑F

x

= max

F − fk = 0 F = f k = μk mg = (0.20)(11.2 kg)(9.80 m/s 2 ) = 22 N (b) IDENTIFY: Now the only horizontal force on the box is the kinetic friction force. Apply Newton’s 2nd law to the box to calculate its acceleration. Once we have the acceleration, we can find the distance using a constant acceleration equation. The friction force is f k = μ k mg , just as in part (a). SET UP: The free-body diagram is sketched in Figure 5.29b. EXECUTE: ∑ Fx = max

− f k = max − μk mg = max ax = − μ k g = −(0.20)(9.80 m/s 2 ) = −1.96 m/s 2 Figure 5.29b

Use the constant acceleration equations to find the distance the box travels: vx = 0, v0 x = 3.50 m/s, ax = −1.96 m/s 2 , x − x0 = ? vx2 = v02x + 2ax ( x − x0 ) vx2 − v02x 0 − (3.50 m/s) 2 = = 3.1 m 2ax 2( −1.96 m/s 2 ) EVALUATE: The normal force is the component of force exerted by a surface perpendicular to the surface. Its ! ! magnitude is determined by ∑ F = ma . In this case n and mg are the only vertical forces and a y = 0, so n = mg . x − x0 =

5.30.

Also note that f k and n are proportional in magnitude but perpendicular in direction. ! ! IDENTIFY: Apply ∑ F = ma to the box. SET UP: Since the only vertical forces are n and w, the normal force on the box equals its weight. Static friction is as large as it needs to be to prevent relative motion between the box and the surface, up to its maximum possible value of f smax = μs n . If the box is sliding then the friction force is f k = μk n . EXECUTE: (a) If there is no applied force, no friction force is needed to keep the box at rest. (b) f smax = μs n = (0.40)(40.0 N) = 16.0 N . If a horizontal force of 6.0 N is applied to the box, then f s = 6.0 N in the opposite direction. (c) The monkey must apply a force equal to f smax , 16.0 N. (d) Once the box has started moving, a force equal to f k = μk n = 8.0 N is required to keep it moving at constant velocity. EVALUATE: μ k < μs and less force must be applied to the box to maintain its motion than to start it moving.


5.31.

5-15

! ! Apply ∑ F = ma to the crate. f s ≤ μs n and f k = μk n .

IDENTIFY:

SET UP: Let + y be upward and let + x be in the direction of the push. Since the floor is horizontal and the push is horizontal, the normal force equals the weight of the crate: n = mg = 441 N . The force it takes to start the crate

moving equals max fs and the force required to keep it moving equals f k 313 N 208 N = 0.710 . f k = 208 N , so μ k = = 0.472 . 441 N 441 N (b) The friction is kinetic. ∑ Fx = max gives F − f k = ma and F = f k + ma = 208 + (45.0 kg)(1.10 m/s 2 ) = 258 N . max f s = 313 N , so μs =

EXECUTE:

(c) (i) The normal force now is mg = 72.9 N . To cause it to move, F = max fs = μs n = (0.710)(72.9 N) = 51.8 N .

F − f k 258 N − (0.472)(72.9 N) = = 4.97 m/s 2 m 45.0 kg EVALUATE: The kinetic friction force is independent of the speed of the object. On the moon, the mass of the crate is the same as on earth, but the weight and normal force are less. ! ! IDENTIFY: Apply ∑ F = ma to the box and calculate the normal and friction forces. The coefficient of kinetic (ii) F = f k + ma and a =

5.32.

fk . n Let + x be in the direction of motion. ax = −0.90 m/s 2 . The box has mass 8.67 kg.

friction is the ratio SET UP: EXECUTE:

The normal force has magnitude 85 N + 25 N = 110 N. The friction force, from FH − f k = ma is 28 N = 0.25. 110 N The normal force is greater than the weight of the box, because of the downward component of the

f k = FH − ma = 20 N − (8.67 kg)(−0.90 m/s 2 ) = 28 N . μ k =

5.33.

EVALUATE: push force. IDENTIFY: Apply

!

!

∑ F = ma to the composite object consisting of the two boxes and to the top box. The friction

the ramp exerts on the lower box is kinetic friction. The upper box doesn’t slip relative to the lower box, so the friction between the two boxes is static. Since the speed is constant the acceleration is zero. SET UP: Let + x be up the incline. The free-body diagrams for the composite object and for the upper box are 2.50 m given in Figures 5.33a and b. The slope angle φ of the ramp is given by tan φ = , so φ = 27.76° . Since the 4.75 m boxes move down the ramp, the kinetic friction force exerted on the lower box by the ramp is directed up the incline. To prevent slipping relative to the lower box the static friction force on the upper box is directed up the incline. mtot = 32.0 kg + 48.0 kg = 80.0 kg . EXECUTE:

∑F

x

(a)

∑F

y

= ma y applied to the composite object gives ntot = mtot g cos φ and f k = μk mtot g cos φ .

= max gives f k + T − mtot g sin φ = 0 and

T = (sin φ − μk cosφ )mtot g = (sin 27.76° − [0.444]cos 27.76°)(80.0 kg)(9.80 m/s 2 ) = 57.1 N . The person must apply a force of 57.1 N, directed up the ramp. (b) ∑ Fx = max applied to the upper box gives f s = mg sin φ = (32.0 kg)(9.80 m/s 2 )sin 27.76° = 146 N , directed up the ramp. EVALUATE:

For each object the net force is zero.

Figure 5.33a, b

5-16

5.34.

Chapter 5

IDENTIFY:

Use

!

!

∑ F = ma

to find the acceleration that can be given to the car by the kinetic friction force. Then

use a constant acceleration equation. SET UP: Take + x in the direction the car is moving. EXECUTE: (a) The free-body diagram for the car is shown in Figure 5.34.

∑F

x

∑F

y

= ma y gives n = mg .

= max gives − μk n = max . − μk mg = max and ax = − μ k g . Then vx = 0 and vx2 = v02x + 2ax ( x − x0 ) gives

( x − x0 ) = −

v02x v2 (29.1 m/s) 2 = + 0x = = 54.0 m . 2a x 2 μ k g 2(0.80)(9.80 m/s 2 )

(b) v0 x = 2 μ k g ( x − x0 ) = 2(0.25)(9.80 m/s 2 )(54.0 m) = 16.3 m/s EVALUATE:

For constant stopping distance

v02x

μk

is constant and v0 x is proportional to

μk . The answer to

part (b) can be calculated as (29.1 m/s) 0.25/ 0.80 = 16.3 m/s .

Figure 5.34 5.35.

5.36.

IDENTIFY: For a given initial speed, the distance traveled is inversely proportional to the coefficient of kinetic friction. SET UP: From Table 5.1 the coefficient of kinetic friction is 0.04 for Teflon on steel and 0.44 for brass on steel. 0.44 EXECUTE: The ratio of the distances is = 11 . 0.04 EVALUATE: The smaller the coefficient of kinetic friction the smaller the retarding force of friction, and the greater the stopping distance. IDENTIFY: Constant speed means zero acceleration for each block. If the block is moving the friction force the ! ! tabletop exerts on it is kinetic friction. Apply ∑ F = ma to each block. SET UP: The free-body diagrams and choice of coordinates for each block are given by Figure 5.36. m A = 4.59 kg and mB = 2.55 kg . EXECUTE:

(a)

∑F

y

= ma y with a y = 0 applied to block B gives mB g − T = 0 and T = 25.0 N .

ax = 0 applied to block A gives T − f k = 0 and f k = 25.0 N . n A = mA g = 45.0 N and μ k =

∑F

x

= max with

f k 25.0 N = = 0.556 . nA 45.0 N

(b) Now let A be block A plus the cat, so mA = 9.18 kg . n A = 90.0 N and f k = μk n = (0.556)(90.0 N) = 50.0 N .

∑F

x

= max for A gives T − f k = m Aax .

∑F

y

= ma y for block B gives mB g − T = mB a y . ax for A equals a y for B,

so adding the two equations gives mB g − f k = ( mA + mB )a y and a y = The acceleration is upward and block B slows down.

mB g − f k 25.0 N − 50.0 N = = −2.13 m/s 2 . mA + mB 9.18 kg + 2.55 kg


EVALUATE:

5-17

The equation mB g − f k = ( mA + mB )a y has a simple interpretation. If both blocks are considered

together then there are two external forces: mB g that acts to move the system one way and f k that acts oppositely. The net force of mB g − f k must accelerate a total mass of mA + mB .

5.37.

IDENTIFY:

Figure 5.36 ! ! Apply ∑ F = ma to each crate. The rope exerts force T to the right on crate A and force T to the left

on crate B. The target variables are the forces T and F. Constant v implies a = 0. SET UP: The free-body diagram for A is sketched in Figure 5.37a EXECUTE: ∑ Fy = ma y

nA − mA g = 0 nA = mA g f kA = μ k n A = μ k m A g Figure 5.37a

∑F

x

= max

T − f kA = 0 T = μk mA g SET UP: The free-body diagram for B is sketched in Figure 5.37b. EXECUTE: ∑ Fy = ma y

nB − mB g = 0 nB = mB g f kB = μk nB = μ k mB g Figure 5.37b

∑F

x

= max

F − T − f kB = 0 F = T + μk mB g Use the first equation to replace T in the second: F = μ k mA g + μ k mB g . (a) F = μ k (mA + mB ) g (b) T = μk mA g EVALUATE:

We can also consider both crates together as a single object of mass (mA + mB ).

this combined object gives F = f k = μk ( mA + mB ) g , in agreement with our answer in part (a).

∑F

x

= max for

5-18

5.38.

Chapter 5

f = μr n . Apply

IDENTIFY: SET UP:

!

!

∑ F = ma to the tire.

n = mg and f = ma .

v 2 − v02 , where L is the distance covered before the wheel’s speed is reduced to half its original L a v 2 − v 2 v02 − 14 v02 3 v02 = = speed and v = v0 / 2 . μ r = = 0 . 2 Lg 2 Lg 8 Lg g ax =

EXECUTE:

3 (3.50 m / s) 2 = 0.0259 . 8 (18.1 m)(9.80 m / s 2 )

Low pressure, L = 18.1 m and High pressure, L = 92.9 m and EVALUATE: 5.39.

SET UP:

μ r is inversely proportional to the distance L, so Apply

IDENTIFY:

3 (3.50 m / s) 2 = 0.00505 . 8 (3.50 m / s) 2

!

!

μr1 L2 = . μr2 L1

∑ F = ma to the box. Use the information about sliding to calculate the mass of the box.

f k = μk n , f r = μr n and n = mg . Without the dolly: n = mg and F − μk n = 0 ( ax = 0 since speed is constant).

EXECUTE:

m=

F

μk g

=

160 N = 34.74 kg (0.47) (9.80 m s 2 )

With the dolly: the total mass is 34.7 kg + 5.3 kg = 40.04 kg and friction now is rolling friction, f r = μr mg . F − μr mg = ma . a = EVALUATE: 5.40.

F − μr mg = 3.82 m s 2 . m

f k = μk mg = 160 N and f r = μr mg = 4.36 N , or,

f r μr = . The rolling friction force is much less fk μk

than the kinetic friction force. ! ! IDENTIFY: Apply ∑ F = ma to the truck. For constant speed, a = 0 and Fhoriz = f r . SET UP: EXECUTE:

f r = μr n = μ r mg . Let m2 = 1.42m1 and μ r2 = 0.81μr1 . Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and Fnet in

the horizontal direction must be zero. Therefore f r = μr n = Fhoriz = 200 N before the weight and pressure changes are made. After the changes, (0.81μr ) (1.42n) = Fhoriz , because the speed is still constant and Fnet = 0 . We can (0.81μr )(1.42n) F = horiz and (0.81) (1.42) (200 N) = Fhoriz = 230 N . μr n 200 N EVALUATE: The increase in weight increases the normal force and hence the friction force, whereas the decrease in μ r reduces it. The percentage increase in the weight is larger, so the net effect is an increase in the friction force. ! ! IDENTIFY: Apply ∑ F = ma to each block. The target variables are the tension T in the cord and the

simply divide the two equations:

5.41.

acceleration a of the blocks. Then a can be used in a constant acceleration equation to find the speed of each block. The magnitude of the acceleration is the same for both blocks. SET UP: The system is sketched in Figure 5.41a.

For each block take a positive coordinate direction to be the direction of the block’s acceleration.

Figure 5.41a


5-19

block on the table: The free-body is sketched in Figure 5.41b. EXECUTE: ∑ Fy = ma y n − mA g = 0 n = mA g f k = μk n = μ k mA g Figure 5.41b

∑F

x

= max

T − f k = mAa T − μk mA g = mAa SET UP: hanging block: The free-body is sketched in Figure 5.41c. EXECUTE: ∑ Fy = ma y

mB g − T = mB a T = mB g − mB a Figure 5.41c (a) Use the second equation in the first

mB g − mB a − μk mA g = m Aa (mA + mB ) a = (mB − μk mA ) g a=

( mB − μ k mA ) g (1.30 kg − (0.45)(2.25 kg))(9.80 m/s 2 ) = = 0.7937 m/s 2 mA + mB 2.25 kg + 1.30 kg

SET UP: Now use the constant acceleration equations to find the final speed. Note that the blocks have the same speeds. x − x0 = 0.0300 m, ax = 0.7937 m/s 2 , v0 x = 0, vx = ?

vx2 = v02x + 2ax ( x − x0 ) EXECUTE:

vx = 2ax ( x − x0 ) = 2(0.7937 m/s 2 )(0.0300 m) = 0.218 m/s = 21.8 cm/s.

(b) T = mB g − mB a = mB ( g − a ) = 1.30 kg(9.80 m/s 2 − 0.7937 m/s 2 ) = 11.7 N

Or, to check, T − μk m A g = m Aa T = mA (a + μ k g ) = 2.25 kg(0.7937 m/s 2 + (0.45)(9.80 m/s 2 )) = 11.7 N, which checks. EVALUATE:

5.42.

The force T exerted by the cord has the same value for each block. T < mB g since the hanging block

accelerates downward. Also, f k = μk mA g = 9.92 N. T > f k and the block on the table accelerates in the direction of T. ! ! IDENTIFY: Apply ∑ F = ma to the box. When the box is ready to slip the static friction force has its maximum possible value, f s = μs n . SET UP: Use coordinates parallel and perpendicular to the ramp. EXECUTE: (a) The normal force will be wcos θ and the component of the gravitational force along the ramp is wsin θ . The box begins to slip when w sin θ > μs w cos θ , or tan θ > μs = 0.35, so slipping occurs at θ = arctan(0.35) = 19.3° . (b) When moving, the friction force along the ramp is μ k w cos θ , the component of the gravitational force along the ramp is w sin θ , so the acceleration is

( w sin θ − wμk cos θ ) m = g (sin θ − μk cos θ ) = 0.92 m s 2 . (c) Since v0 x = 0 , 2ax = v 2 , so v = (2ax)1 2 , or v = [(2)(0.92m s 2 )(5 m)]1 2 = 3 m/s . EVALUATE: When the box starts to move, friction changes from static to kinetic and the friction force becomes smaller.

5-20

5.43.

Chapter 5

(a) IDENTIFY:

Apply

!

!

∑ F = ma

to the crate. Constant v implies a = 0. Crate moving says that the friction is

kinetic friction. The target variable is the magnitude of the force applied by the woman. SET UP: The free-body diagram for the crate is sketched in Figure 5.43. EXECUTE: ∑ Fy = ma y

n − mg − F sin θ = 0 n = mg + F sin θ f k = μk n = μ k mg + μ k F sin θ

Figure 5.43

∑F

x

= max

F cosθ − f k = 0 F cosθ − μk mg − μk F sin θ = 0 F (cosθ − μk sin θ ) = μ k mg

μk mg cosθ − μk sin θ (b) IDENTIFY and SET UP: F=

“start the crate moving” means the same force diagram as in part (a), except that μs mg μ k is replaced by μs . Thus F = . cosθ − μs sin θ

cosθ 1 F → ∞ if cosθ − μs sin θ = 0. This gives μs = = . sin θ tan θ ! EVALUATE: F has a downward component so n > mg . If θ = 0 (woman pushes horizontally), n = mg and EXECUTE:

F = f k = μk mg . 5.44.

IDENTIFY:

Apply

!

!

∑ F = ma to the box.

SET UP: Let + y be upward and + x be horizontal, in the direction of the acceleration. Constant speed means a = 0 . EXECUTE: (a) There is no net force in the vertical direction, so n + F sin θ − w = 0, or n = w − F sin θ = mg − F sin θ. The friction force is f k = μk n = μ k (mg − F sin θ ). The net horizontal force

is F cos θ − f k = F cos θ − μk ( mg − F sin θ ) , and so at constant speed, F= (b) Using the given values, F = EVALUATE: 5.45.

IDENTIFY:

μk mg cos θ + μk sin θ

(0.35)(90 kg)(9.80m s 2 ) = 290 N . (cos 25° + (0.35)sin 25°)

If θ = 0° , F = μ k mg . ! ! Apply ∑ F = ma to each block.

SET UP: For block B use coordinates parallel and perpendicular to the incline. Since they are connected by ropes, blocks A and B also move with constant speed. EXECUTE: (a) The free-body diagrams are sketched in Figure 5.45. (b) The blocks move with constant speed, so there is no net force on block A; the tension in the rope connecting A and B must be equal to the frictional force on block A, μ k = (0.35) (25.0 N) = 9 N. (c) The weight of block C will be the tension in the rope connecting B and C; this is found by considering the forces on block B. The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the component of the weight along the ramp, the friction on block B and the tension in the second rope. Thus, the weight of block C is wC = 9 N + wB (sin 36.9° + μk cos36.9°) = 9 N + (25.0 N)(sin 36.9° + (0.35)cos 36.9°) = 31.0 N

The intermediate calculation of the first tension may be avoided to obtain the answer in terms of the common weight w of blocks A and B, wC = w( μk + (sin θ + μ k cos θ )), giving the same result. (d) Applying Newton’s Second Law to the remaining masses (B and C) gives: a = g ( wC − μ k wB cos θ − wB sin θ ) ( wB + wC ) = 1.54m s 2 .


5-21

EVALUATE: Before the rope between A and B is cut the net external force on the system is zero. When the rope is cut the friction force on A is removed from the system and there is a net force on the system of blocks B and C.

Figure 5.45 5.46.


The derivative of v y gives a y as a function of time, and the integral of v y gives y as a

function of time. EXECUTE: Differentiating Eq. (5.10) with respect to time gives the acceleration ⎛k⎞ a = vt ⎜ ⎟ e − ( k m )t = ge −( k m )t , where Eq. (5.9), vt = mg k , has been used. Integrating Eq. (5.10) with respect to time ⎝m⎠ with y0 = 0 gives t ⎡ ⎛m⎞ ⎤ ⎛m⎞ ⎡ m ⎤ y = ∫ vt [1 − e− ( k m ) t ] dt = vt ⎢t + ⎜ ⎟ e − ( k m )t ⎥ − vt ⎜ ⎟ = vt ⎢t − (1 − e− ( k m ) t ) ⎥ . 0 ⎝k⎠ ⎣ k ⎦ ⎣ ⎝k⎠ ⎦

EVALUATE: 5.47.

We can verify that dy / dt = v y .

IDENTIFY and SET UP: EXECUTE: (b) vt =

Apply Eq.(5.13).

(a) Solving for D in terms of vt , D =

mg (80 kg) (9.80 m s 2 ) = = 0.44 kg m. vt2 (42 m s) 2

mg (45 kg)(9.80 m s 2 ) = = 42 m s. D (0.25 kg m)

EVALUATE:

vt is less for the daughter since her mass is less. ! ! Apply ∑ F = ma to the ball. At the terminal speed, f = mg .

5.48.

IDENTIFY:

5.49.

SET UP: The fluid resistance is directed opposite to the velocity of the object. At half the terminal speed, the magnitude of the frictional force is one-fourth the weight. EXECUTE: (a) If the ball is moving up, the frictional force is down, so the magnitude of the net force is (5/4)w and the acceleration is (5/4)g, down. (b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is (3/4)g, down. EVALUATE: The frictional force is less than mg in each case and in each case the net force is downward and the acceleration is downward. ! ! IDENTIFY: Apply ∑ F = ma to one of the masses. The mass moves in a circular path, so has acceleration

v2 , directed toward the center of the path. R SET UP: In each case, R = 0.200 m . In part (a), let + x be toward the center of the circle, so ax = arad . In part (b) arad =

let + y be toward the center of the circle, so a y = arad . + y is downward when the mass is at the top of the circle ! and + y is upward when the mass is at the bottom of the circle. Since arad has its greatest possible value, F is in ! the direction of arad at both positions. EXECUTE:

(a)

∑F

x

= max gives F = marad = m

v2 . F = 75.0 N and v = R

FR (75.0 N)(0.200 m) = = 3.61 m/s . m 1.15 kg

(b) The free-body diagrams for a mass at the top of the path and at the bottom of the path are given in figure 5.49. At the top, ∑ Fy = ma y gives F = marad − mg and at the bottom it gives F = mg + marad . For a given rotation rate

and hence value of arad , the value of F required is larger at the bottom of the path. (c) F = mg + marad so

v2 F = − g and R m

⎛ 75.0 N ⎞ ⎛F ⎞ v = R ⎜ − g ⎟ = (0.200 m) ⎜ − 9.80 m/s 2 ⎟ = 3.33 m/s ⎝m ⎠ ⎝ 1.15 kg ⎠

5-22

Chapter 5

! EVALUATE: The maximum speed is less for the vertical circle. At the bottom of the vertical path F and the weight are in opposite directions so F must exceed marad by an amount equal to mg. At the top of the vertical path F and mg are in the same direction and together provide the required net force, so F must be larger at the bottom.

Figure 5.49 5.50.

IDENTIFY: Since the car travels in an arc of a circle, it has acceleration arad = v 2 / R , directed toward the center of the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the minimum coefficient of friction that is required, set the static friction force equal to its maximum value, f s = μs n . Friction is static friction because the car is not sliding in the radial direction. SET UP: The free-body diagram for the car is given in Figure 5.50. The diagram assumes the center of the curve is to the left of the car. v2 v2 EXECUTE: (a) ∑ Fy = ma y gives n = mg . ∑ Fx = max gives μs n = m . μs mg = m and R R v2 (25.0 m/s) 2 μs = = = 0.290 gR (9.80 m/s 2 )(220 m) (b)

v2

μs

= Rg = constant , so

v12

μs1

=

v22

μs2

. v2 = v1

μs2 μ /3 = (25.0 m/s) s1 = 14.4 m/s . μs1 μs1

EVALUATE: A smaller coefficient of friction means a smaller maximum friction force, a smaller possible acceleration and therefore a smaller speed.

Figure 5.50 5.51.

We can use the analysis done in Example 5.23. As in that example, we assume friction is negligible. v2 SET UP: From Example 5.23, the banking angle β is given by tan β = . Also, n = mg / cos β . gR 65.0 mi/h = 29.1 m/s . IDENTIFY:

(29.1 m/s) 2 and β = 21.0° . The expression for tan β does not involve the mass (9.80 m/s 2 )(225 m) of the vehicle, so the truck and car should travel at the same speed. (1125 kg)(9.80 m/s 2 ) = 1.18 × 104 N and ntruck = 2ncar = 2.36 × 104 N , since mtruck = 2mcar . (b) For the car, ncar = cos21.0° EVALUATE: The vertical component of the normal force must equal the weight of the vehicle, so the normal force is proportional to m. IDENTIFY: The acceleration of the person is arad = v 2 / R , directed horizontally to the left in the figure in the ! ! 2π R . Apply ∑ F = ma to the person. problem. The time for one revolution is the period T = v EXECUTE:

5.52.

(a) tan β =


5-23

The person moves in a circle of radius R = 3.00 m + (5.00 m)sin 30.0° = 5.50 m . The free-body diagram ! is given in Figure 5.52. F is the force applied to the seat by the rod. v2 mg EXECUTE: (a) ∑ Fy = ma y gives F cos30.0° = mg and F = . ∑ Fx = max gives F sin 30.0° = m . cos30.0° R SET UP:

Combining these two equations gives v = Rg tan θ = (5.50 m)(9.80 m/s 2 ) tan 30.0° = 5.58 m/s . Then the period 2π R 2π (5.50 m) = = 6.19 s . v 5.58 m/s (b) The net force is proportional to m so in

is T =

!

!

∑ F = ma the mass divides out and the angle for a given rate of

rotation is independent of the mass of the passengers. EVALUATE: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod.

5.53.

IDENTIFY:

Figure 5.52 ! ! Apply ∑ F = ma to the composite object of the person plus seat. This object moves in a horizontal

circle and has acceleration arad , directed toward the center of the circle. SET UP: The free-body diagram for the composite object is given in Figure 5.53. Let + x be to the right, in the ! direction of arad . Let + y be upward. The radius of the circular path is R = 7.50 m . The total mass is (255 N + 825 N) /(9.80 m/s 2 ) = 110.2 kg . Since the rotation rate is 32.0 rev/min = 0.5333 rev/s , the period T is 1 = 1.875 s . 0.5333 rev/s EXECUTE:

∑F

x

∑F

y

= ma y gives TA cos 40.0° − mg = 0 and TA =

mg 255 N + 825 N = = 1410 N . cos 40.0° cos 40.0°

= max gives TA sin 40.0° + TB = marad and

4π 2 R 4π 2 (7.50 m) − TA sin 40.0° = (110.2 kg) − (1410 N)sin 40.0° = 8370 N . 2 T (1.875 s) 2 The tension in the horizontal cable is 8370 N and the tension in the other cable is 1410 N. EVALUATE: The weight of the composite object is 1080 N. The tension in cable A is larger than this since its vertical component must equal the weight. marad = 9280 N . The tension in cable B is less than this because part of the required inward force comes from a component of the tension in cable A. TB = m

Figure 5.53

5-24

5.54.

Chapter 5

IDENTIFY:

Apply

!

!

∑ F = ma to the button. The button moves in a circle, so it has acceleration a

rad

.

The situation is equivalent to that of Example 5.22. 2 v2 2π R EXECUTE: (a) μs = . Expressing v in terms of the period T, v = so μs = 4π2 R . A platform speed of T g T Rg SET UP:

4π 2 (0.150 m) = 0.269. (1.50 s) 2 (9.80 m s 2 ) (b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period (longer periods mean slower speeds, so the button may be moved further out) and so is inversely proportional to 2 ⎛ 40.0 ⎞ the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m) ⎜ ⎟ = 0.067 m . ⎝ 60.0 ⎠

40.0 rev/min corresponds to a period of 1.50 s, so μs =

4π 2 R . The maximum radial acceleration that friction can give is μs mg . At the faster rotation T2 rate T is smaller so R must be smaller to keep arad the same.

EVALUATE:

5.55.

arad =

4π 2 R . T2 SET UP: R = 800 m . 1/T is the number of revolutions per second. EXECUTE: (a) Setting arad = g and solving for the period T gives IDENTIFY:

The acceleration due to circular motion is arad =

T = 2π

R 400 m = 2π = 40.1 s, g 9.80 m s 2

so the number of revolutions per minute is (60 s min) (40.1 s) = 1.5 rev min . (b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations, T ′ = (1.5 rev min) × 3.70 9.8 = 0.92 rev min. EVALUATE:

In part (a) the tangential speed of a point at the rim is given by arad =

v2 , so R

v = Rarad = Rg = 62.6 m/s ; the space station is rotating rapidly. 5.56.

2π R . The apparent weight of a person is the normal force exerted on him by the seat he is sitting v on. His acceleration is arad = v 2 / R , directed toward the center of the circle. SET UP: The period is T = 60.0 s. The passenger has mass m = w / g = 90.0 kg . IDENTIFY:

T=

v 2 (5.24 m/s) 2 2π R 2π (50.0 m) = 0.549 m/s 2 . = = 5.24 m/s . Note that arad = = R 50.0 m T 60.0 s (b) The free-body diagram for the person at the top of his path is given in Figure 5.56a. The acceleration is downward, so take + y downward. ∑ Fy = ma y gives mg − n = marad . EXECUTE:

(a) v =

n = m( g − arad ) = (90.0 kg)(9.80 m/s 2 − 0.549 m/s 2 ) = 833 N . The free-body diagram for the person at the bottom of his path is given in Figure 5.56b. The acceleration is upward, so take + y upward. ∑ Fy = ma y gives n − mg = marad and n = m( g + arad ) = 931 N . (c) Apparent weight = 0 means n = 0 and mg = marad . g =

v2 and v = gR = 22.1 m/s . The time for one R

2π R 2π (50.0 m) = = 14.2 s . Note that arad = g . v 22.1 m/s (d) n = m( g + arad ) = 2mg = 2(882 N) = 1760 N , twice his true weight. revolution would be T =


5-25

EVALUATE: At the top of his path his apparent weight is less than his true weight and at the bottom of his path his apparent weight is greater than his true weight.

5.57.

Figure 5.56a, b ! ! IDENTIFY: Apply ∑ F = ma to the motion of the pilot. The pilot moves in a vertical circle. The apparent weight ! is the normal force exerted on him. At each point arad is directed toward the center of the circular path. (a) SET UP: “the pilot feels weightless” means that the vertical normal force n exerted on the pilot by the chair on which the pilot sits is zero. The force diagram for the pilot at the top of the path is given in Figure 5.57a. EXECUTE: ∑ Fy = ma y

mg = marad g=

v2 R

Figure 5.57a

Thus v = gR = (9.80 m/s 2 )(150 m) = 38.34 m/s ⎛ 1 km ⎞⎛ 3600 s ⎞ v = (38.34 m/s) ⎜ 3 ⎟⎜ ⎟ = 138 km/h ⎝ 10 m ⎠⎝ 1 h ⎠ (b) SET UP: The force diagram for the pilot at the bottom of the path is given in Figure 5.57b. Note that the vertical normal force exerted on the pilot by the chair on which the pilot sits is now upward. EXECUTE: ∑ Fy = ma y

v2 R v2 n = mg + m R This normal force is the pilot’s apparent weight. n − mg = m

Figure 5.57b

w = 700 N, so m =

w = 71.43 kg g

3 ⎛ 1 h ⎞ ⎛ 10 m ⎞ v = (280 km/h) ⎜ ⎟ = 77.78 m/s ⎟⎜ ⎝ 3600 s ⎠ ⎝ 1 km ⎠

(77.78 m/s) 2 = 3580 N. 150 m EVALUATE: In part (b), n > mg since the acceleration is upward. The pilot feels he is much heavier than when at

Thus n = 700 N + 71.43 kg

rest. The speed is not constant, but it is still true that arad = v 2 / R at each point of the motion. 5.58.

! IDENTIFY: arad = v 2 / R , directed toward the center of the circular path. At the bottom of the dive, arad is upward. The apparent weight of the pilot is the normal force exerted on her by the seat on which she is sitting. SET UP: The free-body diagram for the pilot is given in Figure 5.58.

5-26

Chapter 5

EXECUTE: (b)

∑F

y

(a) arad =

v2 v2 (95.0 m/s) 2 = = 230 m . gives R = arad 4.00(9.80 m/s 2 ) R

= ma y gives n − mg = marad .

n = m( g + arad ) = m( g + 4.00 g ) = 5.00mg = (5.00)(50.0 kg)(9.80 m/s 2 ) = 2450 N EVALUATE: Her apparent weight is five times her true weight, the force of gravity the earth exerts on her.

5.59.

IDENTIFY:

Figure 5.58 ! ! Apply ∑ F = ma to the water. The water moves in a vertical circle. The target variable is the speed

v; we will calculate arad and then get v from arad = v 2 / R SET UP: Consider the free-body diagram for the water when the pail is at the top of its circular path, as shown in Figures 5.59a and b. The radial acceleration is in toward the center of the circle so at this point is downward. n is the downward normal force exerted on the water by the bottom of the pail. Figure 5.59a EXECUTE: ∑ Fy = ma y

n + mg = m

v2 R

Figure 5.59b

5.60.

At the minimum speed the water is just ready to lose contact with the bottom of the pail, so at this speed, n → 0. (Note that the force n cannot be upward.) v2 With n → 0 the equation becomes mg = m . v = gR = (9.80 m/s 2 )(0.600 m) = 2.42 m/s. R EVALUATE: At the minimum speed arad = g . If v is less than this minimum speed, gravity pulls the water (and bucket) out of the circular path. IDENTIFY: The ball has acceleration arad = v 2 / R , directed toward the center of the circular path. When the ball is at the bottom of the swing, its acceleration is upward. SET UP: Take + y upward, in the direction of the acceleration. The bowling ball has mass m = w / g = 7.27 kg . v 2 (4.20 m/s) 2 = = 4.64 m/s , upward. R 3.80 m (b) The free-body diagram is given in Figure 5.60. ∑ Fy = ma y gives T − mg = marad . EXECUTE:

(a) arad =

T = m( g + arad ) = (7.27 kg)(9.80 m/s 2 + 4.64 m/s 2 ) = 105 N


EVALUATE:

5-27

The acceleration is upward, so the net force is upward and the tension is greater than the weight.

Figure 5.60 5.61.

IDENTIFY:

! ! Apply ∑ F = ma to the knot.

SET UP: a = 0 . Use coordinates with axes that are horizontal and vertical. EXECUTE: (a) The free-body diagram for the knot is sketched in Figure 5.61. T1 is more vertical so supports more of the weight and is larger. You can also see this from ∑ Fx = max :

T2 cos 40° − T1 cos60° = 0 . T2 cos 40° − T1 cos60° = 0 . (b) T1 is larger so set T1 = 5000 N. Then T2 = T1 1.532 = 3263.5 N . ∑ Fy = ma y gives

T1 sin 60° + T2 sin 40° = w and w = 6400 N . EVALUATE: The sum of the vertical components of the two tensions equals the weight of the suspended object. The sum of the tensions is greater than the weight.

5.62.

IDENTIFY:

Figure 5.61 ! ! Apply ∑ F = ma to each object . Constant speed means a = 0 .

SET UP: The free-body diagrams are sketched in Figure 5.62. T1 is the tension in the lower chain, T2 is the tension in the upper chain and T = F is the tension in the rope. EXECUTE: The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w and the tension in the rope, which equals F, is w 2 . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w. EVALUATE: The pulley combination allows the worker to lift a weight w by applying a force of only w / 2 .

Figure 5.62 5.63.

IDENTIFY:

! ! Apply ∑ F = ma to the rope.

SET UP: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the force due to the tension in the rope are an action-reaction pair. EXECUTE: (a) The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of w 2 on the rope. Therefore, the downward force that the rope exerts at each end is

Tend sin θ = w 2 , so Tend = w (2sin θ ) = Mg (2sin θ ).

5-28

Chapter 5

(b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; Tend cos θ = Tmiddle , so Tmiddle = Mg cos θ (2sin θ ) = Mg (2tan θ ).

5.64.

(c) Mathematically speaking, θ ≠ 0 because this would cause a division by zero in the equation for Tend or Tmiddle . Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight. EVALUATE: The tension in the rope is not the same at all points along the rope. ! ! ! ! IDENTIFY: Apply ∑ F = ma to the combined rope plus block to find a. Then apply ∑ F = ma to a section of

the rope of length x. First note the limiting values of the tension. The system is sketched in Figure 5.64a.

At the top of the rope T = F At the bottom of the rope T = M ( g + a )

Figure 5.64a SET UP: Consider the rope and block as one combined object, in order to calculate the acceleration: The freebody diagram is sketched in Figure 5.64b. EXECUTE: ∑ Fy = ma y

F − ( M + m) g = ( M + m) a a=

F −g M +m

Figure 5.64b SET UP: Now consider the forces on a section of the rope that extends a distance x < L below the top. The tension at the bottom of this section is T ( x ) and the mass of this section is m( x / L). The free-body diagram is sketched in Figure 5.64c. EXECUTE: ∑ Fy = ma y

F − T ( x ) − m( x / L) g = m( x / L) a T ( x ) = F − m( x / L ) g − m ( x / L ) a Figure 5.64c

Using our expression for a and simplifying gives ⎛ ⎞ mx T ( x) = F ⎜1 − ⎟ ( ) L M + m ⎝ ⎠

5.65.

EVALUATE: Important to check this result for the limiting cases: x = 0 : The expression gives the correct value of T = F . x = L : The expression gives T = F ( M /( M + m)). This should equal T = M ( g + a ), and when we use the expression for a we see that it does. ! ! IDENTIFY: Apply ∑ F = ma to each block. SET UP:

Constant speed means a = 0 . When the blocks are moving, the friction force is f k and when they are at

rest, the friction force is f s . EXECUTE: (a) The tension in the cord must be m2 g in order that the hanging block move at constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so m2 g = ( m1 g sin α + μk m1 g cos α ) and m2 = m1 (sin α + μk cos α ) . (b) In this case, the friction force acts in the same direction as the tension on the block of mass m1 , so

m2 g = (m1 g sin α − μk m1 g cos α ) , or m2 = m1 (sin α − μk cos α ) .


5-29

(c) Similar to the analysis of parts (a) and (b), the largest m2 could be is m1 (sin α + μs cos α ) and the smallest m2

could be is m1 (sin α − μs cos α ) . EVALUATE: In parts (a) and (b) the friction force changes direction when the direction of the motion of m1 changes. In part (c), for the largest m2 the static friction force on m1 is directed down the incline and for the 5.66.

smallest m2 the static friction force on m1 is directed up the incline. IDENTIFY: The system is in equilibrium. Apply Newton’s 1st law to block A, to the hanging weight and to the knot where the cords meet. Target variables are the two forces. (a) SET UP: The free-body diagram for the hanging block is given in Figure 5.66a. EXECUTE: ∑ Fy = ma y

T3 − w = 0 T3 = 12.0 N Figure 5.66a SET UP:

The free-body diagram for the knot is given in Figure 5.66b. EXECUTE: ∑ Fy = ma y

T2 sin 45.0° − T3 = 0 T3 12.0 N = sin 45.0° sin 45.0° T2 = 17.0 N

T2 =

Figure 5.66b

∑F

x

= max

T2 cos 45.0° − T1 = 0 T1 = T2 cos 45.0° = 12.0 N SET UP: The free-body diagram for block A is given in Figure 5.66c. EXECUTE: ∑ Fx = max

T1 − f s = 0 f s = T1 = 12.0 N Figure 5.66c EVALUATE:

Also can apply

∑F

y

= ma y to this block: n − wA = 0 n = wA = 60.0 N

Then μs n = (0.25)(60.0 N) = 15.0 N; this is the maximum possible value for the static friction force. We see that f s < μs n; for this value of w the static friction force can hold the blocks in place. (b) SET UP: We have all the same free-body diagrams and force equations as in part (a) but now the static friction force has its largest possible value, f s = μs n = 15.0 N. Then T1 = fs = 15.0 N. EXECUTE: From the equations for the forces on the knot 15.0 N T2 cos 45.0° − T1 = 0 implies T2 = T1 / cos 45.0° = = 21.2 N cos 45.0° T2 sin 45.0° − T3 = 0 implies T3 = T2 sin 45.0° = (21.2 N)sin 45.0° = 15.0 N

And finally T3 − w = 0 implies w = T3 = 15.0 N. EVALUATE: Compared to part (a), the friction is larger in part (b) by a factor of (15.0/12.0) and w is larger by this same ratio.

5-30

5.67.

Chapter 5

IDENTIFY:

Apply

!

!

∑ F = ma to each block. Use Newton’s 3rd law to relate forces on A and on B.

SET UP: Constant speed means a = 0 . EXECUTE: (a) Treat A and B as a single object of weight w = wA + wB = 4.80 N . The free-body diagram for this

combined object is given in Figure 5.67a.

∑F

y

= ma y gives n = w = 4.80 N . f k = μk n = 1.44 N .

∑F

x

= max

gives F = f k = 1.44 N (b) The free-body force diagrams for blocks A and B are given in Figure 5.67b. n and f k are the normal and

friction forces applied to block B by the tabletop and are the same as in part (a). f kB is the friction force that A applies to B. It is to the right because the force from A opposes the motion of B. nB is the downward force that A exerts on B. f kA is the friction force that B applies to A. It is to the left because block B wants A to move with it. n A is the normal force that block B exerts on A. By Newton’s third law, f kB = f kA and these forces are in opposite

directions. Also, n A = nB and these forces are in opposite directions.

∑F

y

= ma y for block A gives n A = wA = 1.20 N , so nB = 1.20 N .

f kA = μ k nA = (0.300)(1.20 N) = 0.36 N , and f kB = 0.36 N.

∑F ∑F

x

= max for block A gives T = f kA = 0.36 N .

x

= max for block B gives F = f kB + f k = 0.36 N + 1.44 N = 1.80 N

EVALUATE: In part (a) block A is at rest with respect to B and it has zero acceleration. There is no horizontal force on A besides friction, and the friction force on A is zero. A larger force F is needed in part (b), because of the friction force between the two blocks.

5.68.

IDENTIFY:

Figure 5.67a–c ! ! Apply ∑ F = ma to the brush. Constant speed means a = 0. Target variables are two of the forces

on the brush. SET UP: Note that the normal force exerted by the wall is horizontal, since it is perpendicular to the wall. The kinetic friction force exerted by the wall is parallel to the wall and opposes the motion, so it is vertically downward. The free-body diagram is given in Figure 5.68. EXECUTE: ∑ Fx = max

n − F cos53.1° = 0 n = F cos53.1° f k = μk n = μ k F cos53.1° Figure 5.68

∑F

y

= ma y

F sin 53.1° − w − f k = 0 F sin 53.1° − w − μ k F cos53.1° = 0 F (sin 53.1° − μ k cos53.1°) = w F=

w sin 53.1° − μ k cos53.1°


120 N = 16.9 N sin 53.1° − μk cos53.1° sin 53.1° − (0.15)cos53.1° (b) n = F cos53.1° = (16.9 N)cos53.1° = 10.1 N EVALUATE: In the absence of friction w = F sin 53.1°, which agrees with our expression. IDENTIFY: The net force at any time is Fnet = ma . SET UP: At t = 0 , a = 62 g . The maximum acceleration is 140g at t = 1.2 ms . (a) F =

5.69.

5-31

w

=

EXECUTE: (a) Fnet = ma = 62mg = 62(210 × 10−9 kg)(9.80 m/s 2 ) = 1.3 × 10−4 N . This force is 62 times the flea’s weight. (b) Fnet = 140mg = 2.9 × 10−4 N .

5.70.

(c) Since the initial speed is zero, the maximum speed is the area under the ax -t graph. This gives 1.2 m/s. EVALUATE: a is much larger than g and the net external force is much larger than the flea's weight. ! ! IDENTIFY: Apply ∑ F = ma to the instrument and calculate the acceleration. Then use constant acceleration

equations to describe the motion. SET UP: The free-body diagram for the instrument is given in Figure 5.70. The instrument has mass m = w g = 1.531 kg . T − mg = 13.07 m s 2 . m v0 y = 0, v y = 330 m s, a y = 13.07 m s 2 , t = ? Then v y = v0 y + a y t gives t = 25.3 s . Consider forces on the

EXECUTE:

(a) For on the instrument, ∑ Fy = ma y gives T − mg = ma and a =

rocket; rocket has the same a y . Let F be the thrust of the rocket engines. F − mg = ma and F = m( g + a ) = (25,000 kg) (9.80 m s 2 + 13.07 m s 2 ) = 5.72 × 105 N . (b) y − y0 = v0 yt + 12 a yt 2 gives y − y0 = 4170 m. EVALUATE: The rocket and instrument have the same acceleration. The tension in the wire is over twice the weight of the instrument and the upward acceleration is greater than g.

5.71.

IDENTIFY:

Figure 5.70 ! ! a = dv / dt . Apply ∑ F = ma to yourself.

SET UP: The reading of the scale is equal to the normal force the scale applies to you. EXECUTE: The elevator’s acceleration is

a=

dv (t ) = 3.0 m s 2 + 2(0.20 m s3 )t = 3.0 m s 2 + (0.40 m s3 )t dt

At t = 4.0 s, a = 3.0 m s 2 + (0.40 m s3 )(4.0 s) = 4.6 m s 2 . From Newton’s Second Law, the net force on you is Fnet = Fscale − w = ma and Fscale = w + ma = (72 kg)(9.8 m s 2 ) + (72 kg)(4.6 m s 2 ) = 1040 N 5.72.

EVALUATE: a increases with time, so the scale reading is increasing. ! ! IDENTIFY: Apply ∑ F = ma to the passenger to find the maximum allowed acceleration. Then use a constant

acceleration equation to find the maximum speed. SET UP: The free-body diagram for the passenger is given in Figure 5.72. EXECUTE: ∑ Fy = ma y gives n − mg = ma . n = 1.6mg , so a = 0.60 g = 5.88 m s 2 . 2

y − y0 = 3.0 m, a y = 5.88 m s , v0 y = 0 so v y2 = v02y + 2a y ( y − y0 ) gives v y = 5.0 m s .

5-32

Chapter 5

EVALUATE:

A larger final speed would require a larger value of a y , which would mean a larger normal force on

the person.

5.73.

IDENTIFY:

Figure 5.72 ! ! Apply ∑ F = ma to the package. Calculate a and then use a constant acceleration equation to

describe the motion. SET UP: Let + x be directed up the ramp. EXECUTE: (a) Fnet = − mg sin 37° − f k = − mg sin 37° − μ k mg cos37° = ma and a = −(9.8 m s 2 )(0.602 + (0.30)(0.799)) = −8.25m s 2

Since we know the length of the slope, we can use vx2 = v02x + 2ax ( x − x0 ) with x0 = 0 and vx = 0 at the top. v02 = −2ax = −2( −8.25 m s 2 )(8.0 m) = 132 m 2 s 2 and v0 = 132 m 2 s 2 = 11.5 m s (b) For the trip back down the slope, gravity and the friction force operate in opposite directions to each other. Fnet = − mg sin 37° + μk mg cos37° = ma and a = g (− sin 37° + 0.30 cos37°) = (9.8 m s 2 )((−0.602) + (0.30)(0.799)) = −3.55 m s 2 .

Now we have v0 = 0, x0 = −8.0 m, x = 0 and v 2 = v02 + 2a ( x − x0 ) = 0 + 2(−3.55 m s 2 )( −8.0 m) = 56.8 m 2 s 2 , so

5.74.

v = 56.8 m 2 s 2 = 7.54 m s . EVALUATE: In both cases, moving up the incline and moving down the incline, the acceleration is directed down the incline. The magnitude of a is greater when the package is going up the incline, because mg sin 37° and f k are in the same direction whereas when the package is going down these two forces are in opposite directions. ! ! IDENTIFY: Apply ∑ F = ma to the hammer. Since the hammer is at rest relative to the bus its acceleration

equals that of the bus. SET UP: The free-body diagram for the hammer is given in Figure 5.74. EXECUTE: ∑ Fy = ma y gives T sin 74° − mg = 0 so T sin 74° = mg . ∑ Fx = max gives T cos74° = ma. Divide the a 1 and a = 2.8 m s 2 . = g tan74° EVALUATE: When the acceleration increases the angle between the rope and the ceiling of the bus decreases, and the angle the rope makes with the vertical increases.

second equation by the first:

5.75.

IDENTIFY:

Figure 5.74 ! ! Apply ∑ F = ma to the washer and to the crate. Since the washer is at rest relative to the crate, these

two objects have the same acceleration. SET UP: The free-body diagram for the washer is given in Figure 5.75. EXECUTE: It’s interesting to look at the string’s angle measured from the perpendicular to the top of the crate. This angle is θstring = 90° − angle measured from the top of the crate . The free-body diagram for the washer then leads to the following equations, using Newton’s Second Law and taking the upslope direction as positive: − mw g sin θslope + T sin θstring = mw a and T sin θstring = mw ( a + g sinθslope ) − mw g cosθslope + T cos θstring = 0 and T cosθstring = mw g cos θslope


Dividing the two equations: tanθstring =

5-33

a + g sin θslope g cos θslope

For the crate, the component of the weight along the slope is − mc g sin θslope and the normal force is mc g cos θslope . Using Newton’s Second Law again: − mc g sin θslope + μk mc g cos θslope = mc a . μ k =

a + g sin θslope g cos θslope

. This leads to the

interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic friction:

μ k = tan θstring = tan(90° − 68°) = tan 22° = 0.40 . EVALUATE:

In the limit that μ k → 0 , θstring → 0 and the string is perpendicular to the top of the crate.

As μ k increases, θstring increases.

5.76.

IDENTIFY:

Figure 5.75 ! ! Apply ∑ F = ma to yourself and calculate a. Then use constant acceleration equations to describe

the motion. SET UP: The free-body diagram is given in Figure 5.76. EXECUTE: (a) ∑ Fy = ma y gives n = mg cos α . ∑ Fx = max gives mg sin α − f k = ma . Combining these two equations, we have a = g (sin α − μk cos α ) = −3.094 m s 2 . Find your stopping distance: vx = 0, ax = −3.094 m s 2 , v0 x = 20 m s . vx2 = v02x + 2ax ( x − x0 ) gives x − x0 = 64.6 m, which is greater than 40 m. You don’t stop before you reach the hole, so you fall into it. (b) ax = −3.094 m s 2 , x − x0 = 40 m, vx = 0 . vx2 = v02x + 2ax ( x − x0 ) gives v0 x = 16 m s. EVALUATE: Your stopping distance is proportional to the square of your initial speed, so your initial speed is proportional to the square root of your stopping distance. To stop in 40 m instead of 64.6 m your initial speed must 40 m = 16 m/s . be (20 m/s) 64.6 m

5.77.

IDENTIFY:

Figure 5.76 ! ! Apply ∑ F = ma to each block and to the rope. The key idea in solving this problem is to recognize

that if the system is accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn’t accelerate.) SET UP: Take a positive coordinate direction for each object to be in the direction of the acceleration of that object. All three objects have the same magnitude of acceleration. EXECUTE: The Second Law equations for the three different parts of the system are: Block A (The only horizontal forces on A are tension to the right, and friction to the left): − μk mA g + TA = m Aa. Block B (The only vertical forces on B are gravity down, and tension up): mB g − TB = mB a. Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the portion of the rope that hangs vertically): mR d g + TB − TA = mR a. L

( )

5-34

Chapter 5

To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations: m + mR (d / L) − μ k mA − μ k mA g + mB g + mR d g = (m A + mB + mR )a, or a = g B . L ( mA + mB + mR )

( )

(a) When μ k = 0, a = g

mB + mR ( d / L) . As the system moves, d will increase, approaching L as a limit, and thus ( mA + mB + mR )

the acceleration will approach a maximum value of a = g

mB + mR . (m A + mB + mR )

(b) For the blocks to just begin moving, a > 0, so solve 0 = [mB + mR ( d / L) − μs m A ] for d. Note that we must use

static friction to find d for when the block will begin to move. Solving for d, d =

L ( μs mA − mB ) or mR

d = 1.0 m (0.25(2 kg) − 0.4 kg) = 0.63 m. 0.160 kg (c) When mR = 0.04 kg, d = 1.0 m (0.25(2 kg) − 0.4 kg) = 2.50 m . This is not a physically possible situation 0.04 kg since d > L. The blocks won’t move, no matter what portion of the rope hangs over the edge. EVALUATE: For the blocks to move when released, the weight of B plus the weight of the rope that hangs vertically must be greater than the maximum static friction force on A, which is μs n = 4.9 N . 5.78.

Apply Newton’s 1st law to the rope. Let m1 be the mass of that part of the rope that is on the table,

IDENTIFY:

and let m2 be the mass of that part of the rope that is hanging over the edge. ( m1 + m2 = m, the total mass of the rope). Since the mass of the rope is not being neglected, the tension in the rope varies along the length of the rope. Let T be the tension in the rope at that point that is at the edge of the table. SET UP: The free-body diagram for the hanging section of the rope is given in Figure 5.78a EXECUTE: ∑ Fy = ma y

T − m2 g = 0 T = m2 g Figure 5.78a SET UP:

The free-body diagram for that part of the rope that is on the table is given in Figure 5.78b. EXECUTE: ∑ Fy = ma y

n − m1 g = 0 n = m1 g Figure 5.78b

When the maximum amount of rope hangs over the edge the static friction has its maximum value: f s = μs n = μs m1 g

∑F

x

= max

T − fs = 0 T = μs m1 g Use the first equation to replace T: m2 g = μs m1 g m2 = μs m1 The fraction that hangs over is

5.79.

m2 μs m1 μs = = . m m1 + μs m1 1 + μs

EVALUATE: As μs → 0, the fraction goes to zero and as μs → ∞, the fraction goes to unity. IDENTIFY: First calculate the maximum acceleration that the static friction force can give to the case. Apply ! ! ∑ F = ma to the case.


5-35

(a) SET UP: The static friction force is to the right in Figure 5.79a (northward) since it tries to make the case move with the truck. The maximum value it can have is f s = μs N . EXECUTE: ∑ Fy = ma y

n − mg = 0 n = mg f s = μs n = μs mg Figure 5.79a

∑F

x

= max

f s = ma

μs mg = ma a = μs g = (0.30)(9.80 m/s 2 ) = 2.94 m/s 2 The truck’s acceleration is less than this so the case doesn’t slip relative to the truck; the case’s acceleration is a = 2.20 m/s 2 (northward). Then f s = ma = (30.0 kg)(2.20 m/s 2 ) = 66 N, northward. (b) IDENTIFY: Now the acceleration of the truck is greater than the acceleration that static friction can give the case. Therefore, the case slips relative to the truck and the friction is kinetic friction. The friction force still tries to keep the case moving with the truck, so the acceleration of the case and the friction force are both southward. The free-body diagram is sketched in Figure 5.79b. SET UP: EXECUTE: ∑ Fy = ma y n − mg = 0 n = mg f k = μ k mg = (0.20)(30.0 kg)(9.80 m/s 2 ) f k = 59 N, southward Figure 5.79b

fk 59 N = = 2.0 m/s 2 . The magnitude of the acceleration of the case is less m 30.0 kg than that of the truck and the case slides toward the front of the truck. In both parts (a) and (b) the friction is in the direction of the motion and accelerates the case. Friction opposes relative motion between two surfaces in contact. ! ! IDENTIFY: Apply ∑ F = ma to the car to calculate its acceleration. Then use a constant acceleration equation to f k = ma implies a =

EVALUATE:

5.80.

find the initial speed. SET UP: Let + x be in the direction of the car’s initial velocity. The friction force f k is then in the − x -direction . 192 ft = 58.52 m . EXECUTE: n = mg and f k = μk mg . ∑ Fx = max gives − μ k mg = max and ax = − μ k g = −(0.750)(9.80 m/s 2 ) = −7.35 m/s 2 . vx = 0 (stops), x − x0 = 58.52 m . vx2 = v02x + 2ax ( x − x0 ) gives v0 x = −2ax ( x − x0 ) = −2(−7.35 m/s 2 )(58.52 m) = 29.3 m/s = 65.5 mi/h . He was guilty. x − x0 =

EVALUATE:

vx2 − v02x v2 = − 0 x . If his initial speed had been 45 mi/h he would have stopped in 2a x 2ax

2

5.81.

⎛ 45 mi/h ⎞ ⎜ ⎟ (192 ft) = 91 ft . ⎝ 65.5 mi/h ⎠ ! ! IDENTIFY: Apply ∑ F = ma to the point where the three wires join and also to one of the balls. By symmetry the tension in each of the 35.0 cm wires is the same.

5-36

Chapter 5

The geometry of the situation is sketched in Figure 5.81a. The angle φ that each wire makes with the 12.5 cm and φ = 15.26° . Let TA be the tension in the vertical wire and let TB be the vertical is given by sin φ = 47.5 cm tension in each of the other two wires. Neglect the weight of the wires. The free-body diagram for the left-hand ball is given in Figure 5.81b and for the point where the wires join in Figure 5.81c. n is the force one ball exerts on the other. EXECUTE: (a) ∑ Fy = ma y applied to the ball gives TB cos φ − mg = 0 . SET UP:

TB =

mg (15.0 kg)(9.80 m/s 2 ) = = 152 N . Then cos φ cos15.26°

∑F

y

= ma y applied in Figure 5.81c gives TA − 2TB cos φ = 0 and

TA = 2(152 N)cos φ = 294 N . (b)

∑F

x

= max applied to the ball gives n − TB sin φ = 0 and n = (152 N)sin15.26° = 40.0 N .

EVALUATE:

5.82.

IDENTIFY:

TA equals the total weight of the two balls.

Figure 5.81a–c ! ! Apply ∑ F = ma to the box. Compare the acceleration of the box to the acceleration of the truck and

use constant acceleration equations to describe the motion. SET UP: Both objects have acceleration in the same direction; take this to be the + x -direction. EXECUTE: If the block were to remain at rest relative to the truck, the friction force would need to cause an acceleration of 2.20 m s 2 ; however, the maximum acceleration possible due to static friction is (0.19)(9.80 m s 2 ) = 1.86 m s 2 , and so the block will move relative to the truck; the acceleration of the box would be μ k g = (0.15)(9.80 m s 2 ) = 1.47 m s 2 . The difference between the distance the truck moves and the distance the box moves (i.e., the distance the box moves relative to the truck) will be 1.80 m after a time t=

5.83.

2Δx 2(1.80 m) = = 2.221 s. 2 atruck − abox (2.20 m s 2 − 1.47 m s )

In this time, the truck moves 12 atruck t 2 = 12 (2.20m s 2 ) (2.221 s) 2 = 5.43 m. EVALUATE: To prevent the box from sliding off the truck the coefficient of static friction would have to be μs = (2.20 m/s 2 ) / g = 0.224 . ! ! IDENTIFY: Apply ∑ F = ma to each block. Forces between the blocks are related by Newton’s 3rd law. The target variable is the force F. Block B is pulled to the left at constant speed, so block A moves to the right at constant speed and a = 0 for each block. SET UP: The free-body diagram for block A is given in Figure 5.83a. nBA is the normal force that B exerts on A. f BA = μk nBA is the kinetic friction force that B exerts on A. Block A moves to the right relative to B, and f BA opposes this motion, so f BA is to the left.


5-37

Note also that F acts just on B, not on A. EXECUTE: ∑ Fy = ma y

nBA − wA = 0 nBA = 1.40 N f BA = μk nBA = (0.30)(1.40 N) = 0.420 N Figure 5.83a

∑F

x

= max

T − f BA = 0 T = f BA = 0.420 N SET UP: The free-body diagram for block B is given in Figure 5.83b.

Figure 5.83b EXECUTE:

n AB is the normal force that block A exerts on block B. By Newton’s third law n AB and nBA are equal

in magnitude and opposite in direction, so n AB = 1.40 N. f AB is the kinetic friction force that A exerts on B. Block B moves to the left relative to A and f AB opposes this motion, so f AB is to the right. f AB = μk n AB = (0.30)(1.40 N) = 0.420 N. n and f k are the normal and friction force exerted by the floor on block B; f k = μk n. Note that block B moves to the left relative to the floor and f k opposes this motion, so f k is to the right.

∑F

y

= ma y

n − wB − n AB = 0 n = wB + n AB = 4.20 N + 1.40 N = 5.60 N Then f k = μk n = (0.30)(5.60 N) = 1.68 N.

∑F

x

= max

f AB + T + f k − F = 0 F = T + f AB + f k = 0.420 N + 0.420 N + 1.68 N = 2.52 N

5.84.

EVALUATE: Note that f AB and f BA are a third law action-reaction pair, so they must be equal in magnitude and opposite in direction and this is indeed what our calculation gives. ! ! IDENTIFY: Apply ∑ F = ma to the person to find the acceleration the PAPS unit produces. Apply constant

acceleration equations to her free-fall motion and to her motion after the PAPS fires. SET UP: We take the upward direction as positive. EXECUTE: The explorer’s vertical acceleration is −3.7 m s 2 for the first 20 s. Thus at the end of that time her vertical velocity will be v y = a yt = (−3.7 m s 2 )(20 s) = −74 m s. She will have fallen a distance ⎛ −74 m s ⎞ d = vavt = ⎜ ⎟ (20 s) = −740 m and will thus be 1200 m − 740 m = 460 m above the surface. Her vertical 2 ⎝ ⎠ velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as

5-38

Chapter 5

v y2 − v02y

0 − (−74 m s) 2 = 5.95 m s 2 , which is the vertical 2( y − y0 ) −460 m acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by y0 = 0, v y2 = v02y + 2a y ( y − y0 ) gives a y =

t=

v y − v0 y ay

=

=

0 − (−74 m s) = 12.4 s 5.95 m s 2

Using Newton’s Second Law for the vertical direction FPAPSv + mg = ma . This gives FPAPSv = ma − mg = m( a + g ) = (150 kg)(5.95 − (−3.7)) m s 2 = 1450 N , which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is ay =

5.85.

v y − v0 y t

=

0 − 33 m s 2 = 2.66 m s 2 12.4 s

and the force needed is FPAPSh = ma = (150 kg)(2.66 m s 2 ) = 400 N , since there are no other horizontal forces. EVALUATE: The acceleration produced by the PAPS must bring to zero both her horizontal and vertical components of velocity. ! ! IDENTIFY: Apply ∑ F = ma to each block. Parts (a) and (b) will be done together.

Figure 5.85a

Note that each block has the same magnitude of acceleration, but in different directions. For each block let the ! direction of a be a positive coordinate direction. SET UP: The free-body diagram for block A is given in Figure 5.85b. EXECUTE: ∑ Fy = ma y

TAB − mA g = mA a TAB = m A (a + g ) TAB = 4.00 kg(2.00 m/s 2 + 9.80 m/s 2 ) = 47.2 N Figure 5.85b SET UP:

The free-body diagram for block B is given in Figure 5.85b. EXECUTE: ∑ Fy = ma y

n − mB g = 0 n = mB g Figure 5.85c

f k = μ k n = μ k mB g = (0.25)(12.0 kg)(9.80 m/s 2 ) = 29.4 N

∑F

x

= max

TBC − TAB − f k = mB a TBC = TAB + f k + mB a = 47.2 N + 29.4 N + (12.0 kg)(2.00 m/s 2 ) TBC = 100.6 N


SET UP:

5-39

The free-body diagram for block C is sketched in Figure 5.85d. EXECUTE: ∑ Fy = ma y

mC g − TBC = mC a mC ( g − a ) = TBC mC =

TBC 100.6 N = = 12.9 kg g − a 9.80 m/s 2 − 2.00 m/s 2

Figure 5.85d EVALUATE:

If all three blocks are considered together as a single object and

!

!

∑ F = ma

is applied to this

combined object, mC g − mA g − μk mB g = ( mA + mB + mC ) a. Using the values for μ k , mA and mB given in the 5.86.

problem and the mass mC we calculated, this equation gives a = 2.00 m/s 2 , which checks. ! ! IDENTIFY: Apply ∑ F = ma to each block. They have the same magnitude of acceleration, a. SET UP: Consider positive accelerations to be to the right (up and to the right for the left-hand block, down and to the right for the right-hand block). EXECUTE: (a) The forces along the inclines and the accelerations are related by T − (100 kg)g sin 30° = (100 kg)a and (50 kg)g sin 53° − T = (50 kg)a, where T is the tension in the cord and a the mutual magnitude of acceleration. Adding these relations, (50 kg sin 53° − 100 kg sin 30°) g = (50 kg + 100 kg)a, or a = −0.067 g. Since a comes out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive accelerations were to the left, a would be +0.067 g . 2

5.87.

(b) a = 0.067(9.80 m s 2 ) = 0.658 m s . (c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the above relations involving T yields 424 N. EVALUATE: For part (a) we could have compared mg sin θ for each block to determine which direction the system would move. IDENTIFY: Let the tensions in the ropes be T1 and T2 .

Figure 5.87a

Consider the forces on each block. In each case take a positive coordinate direction in the direction of the acceleration of that block. SET UP: The free-body diagram for m1 is given in Figure 5.87b.

EXECUTE: ∑ Fx = max

T1 = m1a1

Figure 5.87b

5-40

Chapter 5

The free-body diagram for m2 is given in Figure 5.87c.

SET UP:

EXECUTE: ∑ Fy = ma y

m2 g − T2 = m2 a2 Figure 5.87c

This gives us two equations, but there are 4 unknowns ( T1 , T2 , a1 , and a2 ) so two more equations are required. SET UP: The free-body diagram for the moveable pulley (mass m) is given in Figure 5.87d.

EXECUTE: ∑ Fy = ma y

mg + T2 − 2T1 = ma

Figure 5.87d

But our pulleys have negligible mass, so mg = ma = 0 and T2 = 2T1. Combine these three equations to eliminate T1 and T2 : m2 g − T2 = m2 a2 gives m2 g − 2T1 = m2 a2 . And then with T1 = m1a 1 we have m2 g − 2m1a1 = m2 a2 . There are still two unknowns, a1 and a2 . But the accelerations a1 and a2 are related. In any time

SET UP:

interval, if m1 moves to the right a distance d, then in the same time m2 moves downward a distance d / 2. One of the constant acceleration kinematic equations says x − x0 = v0 xt + 12 axt 2 , so if m2 moves half the distance it must have half the acceleration of m1 : a2 = a1 / 2, or a1 = 2a2 . EXECUTE: This is the additional equation we need. Use it in the previous equation and get m2 g − 2m1 (2a2 ) = m2 a2 . a2 (4m1 + m2 ) = m2 g a2 =

m2 g 2m2 g and a1 = 2a2 = . 4m1 + m2 4m1 + m2

EVALUATE: 5.88.

IDENTIFY:

If m2 → 0 or m1 → ∞, a1 = a2 = 0. If m2 >> m1 , a2 = g and a1 = 2 g . ! ! Apply ∑ F = ma to block B, to block A and B as a composite object and to block C. If A and B slide

together all three blocks have the same magnitude of acceleration. SET UP: If A and B don’t slip the friction between them is static. The free-body diagrams for block B, for blocks A and B, and for C are given in Figures 5.88a-c. Block C accelerates downward and A and B accelerate to the right. In each case take a positive coordinate direction to be in the direction of the acceleration. Since block A moves to the right, the friction force f s on block B is to the right, to prevent relative motion between the two blocks. When C has its largest mass, f s has its largest value: f s = μs n . EXECUTE:

a = μs g .

∑F

x

∑F

x

= max applied to the block B gives f s = mB a . n = mB g and f s = μs mB g . μs mB g = mB a and

= max applied to blocks A + B gives T = mAB a = mAB μs g .

mC g − T = mC a . mC g − mAB μs g = mC μs g . mC =

∑F

y

= ma y applied to block C gives

mAB μs ⎛ 0.750 ⎞ = (5.00 kg + 8.00 kg) ⎜ ⎟ = 39.0 kg . 1 − μs ⎝ 1 − 0.750 ⎠


5-41

EVALUATE: With no friction from the tabletop, the system accelerates no matter how small the mass of C is. If mC is less than 39.0 kg, the friction force that A exerts on B is less than μs n . If mC is greater than 39.0 kg, blocks C and A have a larger acceleration than friction can give to block B and A accelerates out from under B.

Figure 5.88 5.89.

IDENTIFY: Apply the method of Exercise 5.19 to calculate the acceleration of each object. Then apply constant acceleration equations to the motion of the 2.00 kg object. SET UP: After the 5.00 kg object reaches the floor, the 2.00 kg object is in free-fall, with downward acceleration g. 5.00 kg − 2.00 kg EXECUTE: The 2.00-kg object will accelerate upward at g = 3 g 7, and the 5.00-kg object will 5.00 kg + 2.00 kg

accelerate downward at 3 g 7. Let the initial height above the ground be h0 . When the large object hits the ground, the small object will be at a height 2h0 , and moving upward with a speed given by v02 = 2ah0 = 6 gh0 7. The small object will continue to rise a distance v02 2 g = 3h0 7, and so the maximum height reached will be

5.90.

2h0 + 3h0 7 = 17h0 7 = 1.46 m above the floor , which is 0.860 m above its initial height. EVALUATE: The small object is 1.20 m above the floor when the large object strikes the floor, and it rises an additional 0.26 m after that. ! ! IDENTIFY: Apply ∑ F = ma to the box. SET UP: The box has an upward acceleration of a = 1.90 m/s 2 . EXECUTE: The floor exerts an upward force n on the box, obtained from n − mg = ma, or n = m( a + g ). The friction force that needs to be balanced is 2

μ k n = μk m(a + g ) = (0.32)(28.0 kg)(1.90 m s 2 + 9.80 m s ) = 105 N.

5.91.

EVALUATE: If the elevator wasn't accelerating the normal force would be n = mg and the friction force that would have to be overcome would be 87.8 N. The upward acceleration increases the normal force and that increases the friction force. ! ! IDENTIFY: Apply ∑ F = ma to the block. The cart and the block have the same acceleration. The normal force

exerted by the cart on the block is perpendicular to the front of the cart, so is horizontal and to the right. The friction force on the block is directed so as to hold the block up against the downward pull of gravity. We want to calculate the minimum a required, so take static friction to have its maximum value, f s = μs n. SET UP: The free-body diagram for the block is given in Figure 5.91. EXECUTE: ∑ Fx = max n = ma f s = μs n = μs ma Figure 5.91

∑F

y

= ma y

f s − mg = 0

μs ma = mg a = g / μs EVALUATE: An observer on the cart sees the block pinned there, with no reason for a horizontal force on it because the block is at rest relative to the cart. Therefore, such an observer concludes that n = 0 and thus f s = 0, and he doesn’t understand what holds the block up against the downward force of gravity. The reason for this

5-42

Chapter 5

difficulty is that

5.92.

!

!

∑ F = ma

does not apply in a coordinate frame attached to the cart. This reference frame is

accelerated, and hence not inertial. The smaller μs is, the larger a must be to keep the block pinned against the front of the cart. ! ! IDENTIFY: Apply ∑ F = ma to each block. SET UP: Use coordinates where + x is directed down the incline. EXECUTE: (a) Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i.e., the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, (4.00 kg)g (sin30° − (0.25)cos 30°) − T = (4.00 kg)a, or 11.11 N − T = (4.00 kg)a, and similarly for the larger, 15.44 N + T = (8.00 kg)a . Adding these two relations,

26.55 N = (12.00 kg)a, a = 2.21 m s 2 . (b) Substitution into either of the above relations gives T = 2.27 N. (c) The string will be slack. The 4.00-kg block will have a = 2.78 m s 2 and the 8.00-kg block will have

5.93.

a = 1.93 m s 2 , until the 4.00-kg block overtakes the 8.00-kg block and collides with it. EVALUATE: If the string is cut the acceleration of each block will be independent of the mass of that block and will depend only on the slope angle and the coefficient of kinetic friction. The 8.00-kg block would have a smaller acceleration even though it has a larger mass, since it has a larger μ k . ! ! IDENTIFY: Apply ∑ F = ma to the block and to the plank. SET UP: Both objects have a = 0 . EXECUTE: Let nB be the normal force between the plank and the block and n A be the normal force between the

block and the incline. Then, nB = w cos θ and n A = nB + 3w cos θ = 4w cos θ . The net frictional force on the block is

μ k (nA + nB ) = μ k 5w cosθ . To move at constant speed, this must balance the component of the block’s weight along the incline, so 3w sin θ = μk 5w cos θ , and μ k = 53 tan θ = 53 tan 37° = 0.452.

5.94.

EVALUATE: In the absence of the plank the block slides down at constant speed when the slope angle and coefficient of friction are related by tan θ = μk . For θ = 36.9° , μ k = 0.75 . A smaller μ k is needed when the plank is present because the plank provides an additional friction force. ! ! IDENTIFY: Apply ∑ F = ma to the ball, to m1 and to m2 SET UP: The free-body diagrams for the ball, m1 and m2 are given in Figures 5.94a-c. All three objects have the ! same magnitude of acceleration. In each case take the direction of a to be a positive coordinate direction. EXECUTE: (a) ∑ Fy = ma y applied to the ball gives T cosθ = mg . ∑ Fx = max applied to the ball gives

T sin θ = ma . Combining these two equations to eliminate T gives tan θ = a / g .

(b)

∑F

x

= max applied to m2 gives T = m2 a .

∑F

y

= ma y applied to m1 gives m1 g − T = m1a . Combining these

⎛ m1 ⎞ m1 250 kg = and θ = 9.46° . two equations gives a = ⎜ ⎟ g . Then tan θ = + m m m m 1500 kg + 2 ⎠ 1 2 ⎝ 1 (c) As m1 becomes much larger than m2 , a → g and tan θ → 1 , so θ → 45° . EVALUATE: The device requires that the ball is at rest relative to the platform; any motion swinging back and forth must be damped out. When m1 mg . The bananas also move up. (b) The bananas and monkey move with the same acceleration and the distance between them remains constant. (c) Both the monkey and bananas are in free fall. They have the same initial velocity and as they fall the distance between them doesn’t change. (d) The bananas will slow down at the same rate as the monkey. If the monkey comes to a stop, so will the bananas. EVALUATE: None of these actions bring the monkey any closer to the bananas.


5.100.

SET UP:

Follow the analysis that leads to Eq.(5.10), except now the initial speed is v0 y = 3mg / k = 3vt rather than

zero. EXECUTE:

The separated equation of motion has a lower limit of 3vt instead of 0; specifically, v

dv

∫ v−v

3 vt

5.101.

5-45

! ! IDENTIFY: Apply ∑ F = ma , with f = kv .

= ln t

⎛ v 1⎞ vt − v k ⎡1 ⎤ = ln ⎜ − ⎟ = − t , or v = 2vt ⎢ + e − ( k m )t ⎥ . −2vt m ⎣2 ⎦ ⎝ 2vt 2 ⎠

EVALUATE: As t → ∞ the speed approaches vt . The speed is always greater than vt and this limit is approached from above. ! ! IDENTIFY: Apply ∑ F = ma to the rock. SET UP: Equations 5.9 through 5.13 apply, but with a0 rather than g as the initial acceleration. EXECUTE: (a) The rock is released from rest, and so there is initially no resistive force and a0 = (18.0 N) (3.00 kg) = 6.00 m s 2 . (b) (18.0 N − (2.20 N ⋅ s m) (3.00 m s)) (3.00 kg) = 3.80 m s 2 . (c) The net force must be 1.80 N, so kv = 16.2 N and v = (16.2 N) (2.20 N ⋅ s m) = 7.36 m s. (d) When the net force is equal to zero, and hence the acceleration is zero, kvt = 18.0 N and

vt = (18.0 N) (2.20 N ⋅ s m) = 8.18 m s. (e) From Eq.(5.12), ⎡ ⎤ 3.00 kg y = (8.18 m s) ⎢(2.00 s) − (1 − e−((2.20 N⋅s m) (3.00 kg))(2.00 s) )⎥ = +7.78 m. ⋅ 2.20 N s m ⎣ ⎦ From Eq. (5.10), v = (8.18 m s)[1 − e − ((2.20 N ⋅s m) (3.00 kg))(2.00 s) ] = 6.29 m s. From Eq.(5.11), but with a0 instead of g, a = (6.00 m s 2 )e − ((2.20 N⋅s m) (3.00 kg))(2.00 s) = 1.38 m s 2 . (f) 1 −

v m = 0.1 = e − ( k m ) t and t = ln (10) = 3.14 s. k vt

EVALUATE:

5.102.

The acceleration decreases with time until it becomes zero when v = vt . The speed increases with

time and approaches vt as t → ∞ . ! ! dv dx IDENTIFY: Apply ∑ F = ma to the rock. a = and v = yield differential equations that can be integrated to dt dt give v(t ) and x (t ) . SET UP: The retarding force of the surface is the only horizontal force acting. v dv F F −kv1 2 dv dv k k t EXECUTE: (a) Thus a = net = R = = and 1 2 = − dt . Integrating gives ∫ 1 2 = − ∫ dt and v 0 m m m dt v m v m 0 12 2 2 v kt k t kt + 2v1 2 vv0 = − . This gives v = v0 − 0 . m 4m 2 m dx v1 2 kt k 2t 2 v1 2 ktdt k 2t 2 dt = v0 − 0 + and dx = v0 dt − 0 + For the rock’s position: . 2 dt m 4m m 4m 2 v1 2 kt 2 k 2t 3 + . Integrating gives x = v0t − 0 2m 12m 2 v1 2 kt k 2t 2 (b) v = 0 = v0 − 0 + . This is a quadratic equation in t; from the quadratic formula we can find the single m 2m 2 2mv10 2 . solution t = k (c) Substituting the expression for t into the equation for x: x = v0 ⋅ EVALUATE:

2mv10 2 v01 2 k 4m 2v0 k 2 8m3v03 2 2mv03 2 − ⋅ + ⋅ = 2 k 2m k 12m 2 k3 3k

The magnitude of the average acceleration is aav =

Fav = maav = 12 kv1/0 2 , which is

1 2

times the initial value of the force.

Δv v0 1 kv1/0 2 − = . The average force is 1/ 2 Δt (2mv0 / k ) 2 m

5-46

Chapter 5

5.103.


Apply

!

!

∑ F = ma to the object, with and without including the buoyancy force.

At the terminal speed vt , a = 0 . Without buoyancy, kvt = mg , so k =

⎛ 0.24 m s ⎞ B = mg − kvt = mg ⎜1 − ⎟ = mg 3 . ⎝ 0.36 m s ⎠ EVALUATE: At the terminal speed, B and f = kv together equal mg. The presence of B reduces the value of f required, so the presence of B reduces the terminal speed. IDENTIFY: The block has acceleration arad = v 2 / r , directed to the left in the figure in the problem. Apply ! ! ∑ F = ma to the block. upward buoyancy force B, so B + kvt = mg

5.104.

mg mg = . With buoyancy included there is the additional vt 0.36 s

SET UP:

.

The block moves in a horizontal circle of radius r = (1.25 m) 2 − (1.00 m) 2 = 0.75 m . Each string

1.00 m , so θ = 36.9° . The free-body diagram for the block is given in 1.25 m Figure 5.104. Let + x be to the left and let + y be upward.

makes an angle θ with the vertical. cosθ =

EXECUTE:

(a)

∑F

y

= ma y gives Tu cosθ − Tl cosθ − mg = 0 .

mg (4.00 kg)(9.80 m/s 2 ) = 80.0 N − = 31.0 N . cosθ cos36.9° v2 (b) ∑ Fx = max gives (Tu + Tl )sin θ = m . r Tl = Tu −

v=

r (Tu + Tl )sin θ (0.75 m)(80.0 N + 31.0 N)sin 36.9° = = 3.53 m/s . The number of revolutions per second is m 4.00 kg

v 3.53 m/s = = 0.749 rev/s = 44.9 rev/min . 2π r 2π (0.75 m) (c) If Tl → 0 , Tu cosθ = mg and Tu =

v=

v2 mg (4.00 kg)(9.80 m/s 2 ) = = 49.0 N . Tu sin θ = m . r cosθ cos36.9°

rTu sin θ (0.75 m)(49.0 N)sin 36.9° = = 2.35 m/s . The number of revolutions per minute is m 4.00 kg

⎛ 2.35 m/s ⎞ (44.9 rev/min) ⎜ ⎟ = 29.9 rev/min ⎝ 3.53 m/s ⎠ EVALUATE: The tension in the upper string must be greater than the tension in the lower string so that together they produce an upward component of force that balances the weight of the block.

5.105.

Figure 5.104 ! ! IDENTIFY: Apply ∑ F = ma to the falling object. SET UP:

Follow the steps that lead to Eq.(5.10), except now v0 y = v0 and is not zero.


(a) Newton’s 2nd law gives m

EXECUTE:

dv y dt

5-47

v

= mg − kv y , where

y t dv y k mg = vt . ∫ = − ∫ dt . This is the same m0 k v0 v y − vt

expression used in the derivation of Eq. (5.10), except the lower limit in the velocity integral is the initial speed v0 instead of zero. Evaluating the integrals and rearranging gives v = v0e − kt m + vt (1 − e − kt m ) . Note that at t = 0 this expression says v y = v0 and at t → α it says v y → vt . (b) The downward gravity force is larger than the upward fluid resistance force so the acceleration is downward, until the fluid resistance force equals gravity when the terminal speed is reached. The object speeds up until v y = vt . Take + y to be downward. The graph is sketched in Figure 5.105a. (c) The upward resistance force is larger than the downward gravity force so the acceleration is upward and the object slows down, until the fluid resistance force equals gravity when the terminal speed is reached. Take + y to be downward. The graph is sketched in Figure 5.105b. (d) When v0 = vt the acceleration at t = 0 is zero and remains zero; the velocity is constant and equal to the terminal velocity. EVALUATE: In all cases the speed becomes vt as t → ∞ .

Figure 5.105a, b 5.106.

! ! IDENTIFY: Apply ∑ F = ma to the rock. At the maximum height, v y = 0 . Let + y be upward. Suppress the y subscripts on v and a.

SET UP:

(a) To find the maximum height and time to the top without fluid resistance: v 2 = v02 + 2a ( y − y0 ) and

EXECUTE:

y − y0 =

v 2 − v02 0 − (6.0 m s) 2 v − v0 0 − 6.0 m s = = 1.84 m . t = = = 0.61 s . 2a 2( − 9.8 m s 2 ) a −9.8 m s 2

dv = mg − kv . We rearrange and integrate, taking dt downward as positive as in the text and noting that the velocity at the top of the rock’s flight is zero: 0 dv k −v −2.0 m s 0 ∫ v v − vt = − m t . ln(v − vt ) v = ln v − vt t = ln −6.0 m s − 2.0 m s = ln(0.25) = −1.386 From Eq.(5.9), m k = vt g = (2.0 m s 2 ) (9.8 m s 2 ) = 0.204 s, and t = − m (−1.386) = (0.204 s) (1.386) = 0.283 s k dx to the top. Equation 5.10 in the text gives us = vt (1 − e− ( k m )t ) = vt − vt e − ( k m )t . dt (b) Starting from Newton’s Second Law for this situation m

x

t

t

0

0

0

x = ∫ dx = ∫ vt dt − ∫ vt e− ( k m ) t dt = vtt +

5.107.

vt m − ( k m )t (e − 1) . k

x = (2.0 m s) (0.283 s) + (2.0 m s) (0.204 s)(e −1.387 − 1) = 0.26 m . EVALUATE: With fluid resistance present the maximum height is much less and the time to reach it is less. ! ! IDENTIFY: Apply ∑ F = ma to the car. SET UP: The forces on the car are the air drag force f D = Dv 2 and the rolling friction force μ r mg . Take the velocity to be in the + x -direction. The forces are opposite in direction to the velocity. EXECUTE:

(a) ∑ Fx = max gives − Dv 2 − μ r mg = ma . We can write this equation twice, once with v = 32 m s

and a = − 0.42 m s 2 and once with v = 24 m s and a = −0.30 m/s 2 . Solving these two simultaneous equations in the unknowns D and μ r gives μ r = 0.015 and D = 0.36 N ⋅ s 2 m 2 . (b) n = mg cos β and the component of gravity parallel to the incline is mg sin β , where β = 2.2°. For constant speed, mg sin 2.2° − μ r mg cos 2.2° − Dv 2 = 0. Solving for v gives v = 29 m s.

5-48

Chapter 5

(c) For angle β , mg sin β − μ r mg cos β − Dv 2 = 0 and v =

mg (sin β − μ r cosβ ) . The terminal speed for a falling D

object is derived from Dvt2 − mg = 0, so vt = mg D. v vt = sin β − μ r cos β . And since

μ r = 0.015, v vt = sin β − (0.015) cosβ . EVALUATE: 5.108.

SET UP:

In part (c), v → vt as β → 90° , since in that limit the incline becomes vertical. ! ! ∑ F = ma to the person and to the cart.

Apply

IDENTIFY:

The apparent weight, wapp , which is the same as the upward force on the person exerted by the car seat.

EXECUTE: (a) The apparent weight is the actual weight of the person minus the centripetal force needed to keep him moving in its circular path:

wapp = mg −

5.109.

⎡ mv 2 (12 m s) 2 ⎤ = (70 kg) ⎢(9.8 m s 2 ) − ⎥ = 434 N . R 40 m ⎦ ⎣

(b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to mv 2 = 0 . v = Rg = (40 m) (9.8 m/s 2 ) = 19.8 m s . The answer doesn’t exert any upward force on it: mg − R depend on the cart’s mass, because the centripetal force needed to hold it on the road is proportional to its mass and so to its weight, which provides the centripetal force in this situation. EVALUATE: At the speed calculated in part (b), the downward force needed for circular motion is provided by gravity. For speeds greater than this more, downward force is needed and there is no source for it and the cart leaves the circular path. For speeds less than this, less downward force than gravity is needed, so the roadway must exert an upward vertical force. (a) IDENTIFY: Use the information given about Jena to find the time t for one revolution of the merry-go-round. ! Her acceleration is arad , directed in toward the axis. Let F1 be the horizontal force that keeps her from sliding off. ! ! Let her speed be v1 and let R1 be her distance from the axis. Apply ∑ F = ma to Jena, who moves in uniform

circular motion. SET UP: The free-body diagram for Jena is sketched in Figure 5.109a EXECUTE: ∑ Fx = max F1 = marad F1 = m

v12 , v1 = R1

R1F1 = 1.90 m/s m

Figure 5.109a

The time for one revolution is t =

2π R1 m . Jackie goes around once in the same time but her speed = 2π R1 v1 R1F1

(v2 ) and the radius of her circular path ( R2 ) are different. ⎛ 1 ⎞ R1F1 R2 R1F1 2π R2 = 2π R2 ⎜ = . ⎟ t R1 m ⎝ 2π R1 ⎠ m ! ! IDENTIFY: Now apply ∑ F = ma to Jackie. She also moves in uniform circular motion. v2 =

SET UP:

The free-body diagram for Jackie is sketched in Figure 5.109b. EXECUTE: ∑ Fx = max

F2 = marad Figure 5.109b

F2 = m

v22 ⎛ m ⎞⎛ R22 ⎞ ⎛ R1F1 ⎞ ⎛ R2 ⎞ ⎛ 3.60 m ⎞ = ⎜ ⎟⎜ 2 ⎟ ⎜ ⎟ = ⎜ ⎟ F1 = ⎜ ⎟ (60.0 N) = 120.0 N R2 ⎝ R2 ⎠⎝ R1 ⎠ ⎝ m ⎠ ⎝ R1 ⎠ ⎝ 1.80 m ⎠

(b) F2 = m

v22 , so v2 = R2

F2 R2 (120.0 N)(3.60 m) = = 3.79 m/s m 30.0 kg


EVALUATE: 5.110.

5-49

Both girls rotate together so have the same period T. By Eq.(5.16), arad is larger for Jackie so the

force on her is larger. Eq.(5.15) says R1 / v1 = R2 / v2 so v2 = v1 ( R2 / R1 ); this agrees with our result in (a). ! ! IDENTIFY: Apply ∑ F = ma to the passenger. The passenger has acceleration arad , directed inward toward the center of the circular path. SET UP: The passenger’s velocity is v = 2π R t = 8.80 m s. The vertical component of the seat’s force must balance the passenger’s weight and the horizontal component must provide the centripetal force. mv 2 EXECUTE: (a) Fseat sin θ = mg = 833 N and Fseat cosθ = = 188 N . Therefore R tan θ = (833 N) (188 N) = 4.43; θ = 77.3° above the horizontal. The magnitude of the net force exerted by the seat (note that this is not the net force on the passenger) is Fseat = (833 N) 2 + (188 N) 2 = 854 N

5.111.

(b) The magnitude of the force is the same, but the horizontal component is reversed. v2 EVALUATE: At the highest point in the motion, Fseat = mg − m = 645 N . At the lowest point in the motion, R v2 Fseat = mg + m = 1021 N . The result in parts (a) and (b) lies between these extreme values. R ! ! IDENTIFY: Apply ∑ F = ma to the person. The person moves in a horizontal circle so his acceleration is

arad = v 2 / R, directed toward the center of the circle. The target variable is the coefficient of static friction between ⎛ 2π R ⎞ ⎛ 2π (2.5 m) ⎞ the person and the surface of the cylinder. v = (0.60 rev/s) ⎜ ⎟ = (0.60 rev/s) ⎜ ⎟ = 9.425 m/s ⎝ 1 rev ⎠ ⎝ 1 rev ⎠ (a) SET UP: The problem situation is sketched in Figure 5.111a.

Figure 5.111a

The free-body diagram for the person is sketched in Figure 5.111b. The person is held up against gravity by the static friction force exerted on him by the wall. The acceleration of the person is arad , directed in towards the axis of rotation. Figure 5.111b (b) EXECUTE:

∑F

y

To calculate the minimum μs required, take f s to have its maximum value, f s = μs n.

= ma y

f s − mg = 0

μs n = mg

∑F

x

= max

n = mv 2 / R Combine these two equations to eliminate n: μs mv 2 / R = mg

μs =

Rg (2.5 m)(9.80 m/s 2 ) = = 0.28 v2 (9.425 m/s) 2

5-50

Chapter 5

5.112.

v must be to keep the person from sliding down. For smaller μs the cylinder must rotate faster to make n larger enough. ! ! IDENTIFY: Apply ∑ F = ma to the combined object of motorcycle plus rider.

(c) EVALUATE:

No, the mass of the person divided out of the equation for μs . Also, the smaller μs is, the larger

SET UP: The object has acceleration arad = v 2 / r , directed toward the center of the circular path. EXECUTE: (a) For the tires not to lose contact, there must be a downward force on the tires. Thus, the v2 (downward) acceleration at the top of the sphere must exceed mg, so m > mg , and R

5.113.

v > gR = (9.80 m s 2 ) (13.0 m) = 11.3 m s. (b) The (upward) acceleration will then be 4g, so the upward normal force must be 5mg = 5(110 kg) (9.80 m s 2 ) = 5390 N. EVALUATE: At any nonzero speed the normal force at the bottom of the path exceeds the weight of the object. ! ! IDENTIFY: Apply ∑ F = ma to your friend. Your friend moves in the arc of a circle as the car turns. (a) Turn to the right. The situation is sketched in Figure 5.113a.

As viewed in an inertial frame, in the absence of sufficient friction your friend doesn’t make the turn completely and you move to the right toward your friend. Figure 5.113a (b) The maximum radius of the turn is the one that makes arad just equal to the maximum acceleration that static

friction can give to your friend, and for this situation f s has its maximum value f s = μs n. SET UP: The free-body diagram for your friend, as viewed by someone standing behind the car, is sketched in Figure 5.113b. EXECUTE: ∑ Fy = ma y

n − mg = 0 n = mg Figure 5.113b

∑F

x

= max

f s = marad

μs n = mv 2 / R μs mg = mv 2 / R R=

5.114.

v2 (20 m/s) 2 = = 120 m μs g (0.35)(9.80 m/s 2 )

EVALUATE: The larger μs is, the smaller the radius R must be. IDENTIFY: The tension F in the string must be the same as the weight of the hanging block, and must also provide the resultant force necessary to keep the block on the table in uniform circular motion. SET UP: The acceleration of the block is arad = v 2 / r , directed toward the hole.

v2 , so v = gr M m. r EVALUATE: The larger M is the greater must be the speed v, if r remains the same. EXECUTE:

Mg = F = m


5.115.

5-51

! ! IDENTIFY: Apply ∑ F = ma to the circular motion of the bead. Also use Eq.(5.16) to relate arad to the period of rotation T. SET UP: The bead and hoop are sketched in Figure 5.115a.

The bead moves in a circle of radius R = r sin β . The normal force exerted on the bead by the hoop is radially inward.

Figure 5.115a

The free-body diagram for the bead is sketched in Figure 5.115b. EXECUTE: ∑ Fy = ma y

n cos β − mg = 0 n = mg / cos β

∑F

x

= max

n sin β = marad Figure 5.115b

Combine these two equations to eliminate n: ⎛ mg ⎞ ⎜ ⎟ sin β = marad ⎝ cos β ⎠ sin β arad = cos β g arad = v 2 / R and v = 2π R / T , so arad = 4π 2 R / T 2 , where T is the time for one revolution. R = r sin β , so arad =

4π 2 r sin β T2

sin β 4π 2 r sin β = cos β T 2g This equation is satisfied by sin β = 0, so β = 0, or by Use this in the above equation:

1 4π 2 r T 2g = 2 , which gives cos β = 2 cos β T g 4π r (a) 4.00 rev/s implies T = (1/ 4.00) s = 0.250 s (0.250 s) 2 (9.80 m/s 2 ) and β = 81.1°. 4π 2 (0.100 m) (b) This would mean β = 90°. But cos90° = 0, so this requires T → 0. So β approaches 90° as the hoop rotates very fast, but β = 90° is not possible. (c) 1.00 rev/s implies T = 1.00 s (1.00 s) 2 (9.80 m/s 2 ) T 2g The cos β = 2 equation then says cos β = = 2.48, which is not possible. The only way to 4π 2 (0.100 m) 4π r ! ! have the ∑ F = ma equations satisfied is for sin β = 0. This means β = 0; the bead sits at the bottom of the hoop.

Then cos β =

5-52

Chapter 5

β → 90° as T → 0 (hoop moves faster). The largest value T can have is given by T 2 g/(4π 2 r ) = 1 so

EVALUATE:

5.116.

T = 2π r / g = 0.635 s. This corresponds to a rotation rate of (1/ 0.635) rev/s = 1.58 rev/s. For a rotation rate less than 1.58 rev/s, β = 0 is the only solution and the bead sits at the bottom of the hoop. Part (c) is an example of this. ! d 2x d2y ! IDENTIFY: ax = 2 and a y = 2 . Then apply ∑ F = ma to calculate the components of the net force. dt dt ! SET UP: The components of F determine its magnitude and direction. EXECUTE: (a) Differentiating twice, ax = −6 βt and a y = −2δ , so Fx = max = (2.20 kg) ( − 0.72 N s)t = −(1.58 N/s)t and Fy = ma y = (2.20 kg) ( − 2.00 m s 2 ) = −4.40 N . (b) The graph is given in Figure 5.116. (c) At t = 3.00 s, Fx = −4.75 N and Fy = −4.40 N, so F = (−4.75 N) 2 + (−4.40 N) 2 = 6.48 N at an angle of

(

)

arctan −4.40 = 223°. −4.75 EVALUATE: Fy is constant and negative. Fx is zero at t = 0 and becomes increasingly more negative as t increases.

Figure 5.116 5.117.

5.118.

IDENTIFY: The velocity is tangent to the path. The acceleration has a tangential component when the speed is changing and a radial component when the path is curving. ! ! ! SET UP: arad is toward the center of curvature of the path. atan is parallel to v when the speed is increasing and ! ! ! antiparallel to v when the speed is decreasing. The net force F is proportional to a . EXECUTE: The diagram is sketched in Figure 5.117. ! ! ! EVALUATE: v , a , and F all change during the motion.

Figure 5.117 ! ! ! IDENTIFY: Apply ∑ F = ma to the car. It has acceleration arad , directed toward the center of the circular path.

The analysis is the same as in Example 5.24. ⎛ ⎛ v2 ⎞ (12.0 m/s) 2 ⎞ EXECUTE: (a) FA = m ⎜ g + ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 + ⎟ = 61.8 N. R⎠ 5.00 m ⎠ ⎝ ⎝ SET UP:

⎛ ⎛ v2 ⎞ (12.0 m/s) 2 ⎞ (b) FB = m ⎜ g − ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 − ⎟ = −30.4 N. , where the minus sign indicates that the track R⎠ 5.00 m ⎠ ⎝ ⎝ pushes down on the car. The magnitude of this force is 30.4 N. EVALUATE: FA > FB . FA − 2mg .


5.119.

5-53

IDENTIFY: The analysis is the same as for Problem 5.96. SET UP: The speed is related to the period by v = 2π R T = 2π h(tan β ) / T , or T = 2π h(tan β ) / v . EXECUTE: The maximum and minimum speeds are the same as those found in Problem 5.96,

vmax = gh tan β

cos β + μs sin β cos β − μs sin β and vmin = gh tan β . sin β − μs cos β sin β + μs cos β

The minimum and maximum values of the period T are then Tmin = 2π

R . The result for v then tan β agrees with the result in Example 5.23, if we take into account that in this problem β is measured from the vertical whereas in Example 5.23 it is measured relative to the horizontal. ! ! IDENTIFY: Apply ∑ F = ma to the block and to the wedge. EVALUATE:

5.120.

h tan β sin β − μs cos β h tan β sin β + μs cos β and Tmax = 2π . g cos β + μs sin β g cos β − μs sin β

For μs = 0 the results for the speeds reduce to vmin = vmax = gh . h =

SET UP: For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be vertical and positive upward. The normal force between the block and the wedge is n; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A, and the components of acceleration of the block are ax and a y . EXECUTE:

(a) The equations of motion are then MA = − n sin α , max = n sin α and ma y = n cos α − mg . Note

that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A, ax , a y and n. Solution is possible with the imposition of the relation between A, ax and a y . An observer on the wedge is not in an inertial frame, and should not apply Newton’s laws, but the kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical acceleration of the block is a y , but the horizontal acceleration of the block is ax − A. To this observer, the block descends at an angle α , so the relation needed is

ay ax − A

= − tan α. At this point, algebra is unavoidable. A

possible approach is to eliminate ax by noting that ax = −

M A , using this in the kinematic constraint to eliminate m

a y and then eliminating n. The results are: A=

− gm ( M + m) tanα + ( M tan α )

ax =

gM ( M + m) tanα + ( M tan α )

ay =

− g ( M + m) tan α ( M + m) tanα + ( M tan α )

(b) When M >> m, A → 0, as expected (the large block won’t move). Also, g tan α ax → =g = g sin α cos α which is the acceleration of the block ( gsinα in this case), tan α + (1 tan α ) tan 2α + 1

with the factor of cos α giving the horizontal component. Similarly, a y → − g sin 2 α . (c) The trajectory is a spiral. EVALUATE: If m >> M , our general results give ax = 0 and a y = − g . The massive block accelerates straight 5.121.

downward, as if it were in free-fall. ! ! IDENTIFY: Apply ∑ F = ma to the block and to the wedge. SET UP:

From Problem 5.120, max = n sin α and ma y = n cos α − mg for the block. a y = 0 gives ax = g tan α .

EXECUTE: If the block is not to move vertically, both the block and the wedge have this horizontal acceleration and the applied force must be F = ( M + m)a = ( M + m) gtanα . EVALUATE: F → 0 as α → 0 and F → ∞ as α → ∞ .

5-54

Chapter 5

5.122.

IDENTIFY:

Apply

!

!

∑ F = ma .

SET UP: Let + x be directed up the ramp. EXECUTE: The normal force that the ramp exerts on the box will be n = wcosα − Tsinα . The rope provides a force of Tcosθ up the ramp, and the component of the weight down the ramp is wsinα . Thus, the net force up the ramp is

F = Tcosθ − w sin α − μ k ( w cos α − Tsinθ ) = T (cosθ + μ k sinθ ) − w(sin α + μ k cosα )

The acceleration will be the greatest when the first term in parentheses is greatest and this occurs when tan θ = μ k .

5.123.

EVALUATE: Small θ means F is more nearly in the direction of the motion. But θ → 90° means F is directed to reduce the normal force and thereby reduce friction. The optimum value of θ is somewhere in between and depends on μ k . When μ k = 0 , the optimum value of θ is θ = 0° . IDENTIFY: Use the results of Problem 5.44. d2 f df >0. SET UP: f ( x) is a minimum when = 0 and dx 2 dx EXECUTE: (a) F = μ k w /(cosθ + μ k sinθ ) (b) The graph of F versus θ is given in Figure 5.123. (c) F is minimized at tan θ = μ k . For μ k = 0.25 , θ = 14.0° . EVALUATE: Small θ means F is more nearly in the direction of the motion. But θ → 90° means F is directed to reduce the normal force and thereby reduce friction. The optimum value of θ is somewhere in between and depends on μ k .

5.124.

Figure 5.123 ! ! IDENTIFY: Apply ∑ F = ma to the ball. At the terminal speed, a = 0 . SET UP: For convenience, take the positive direction to be down, so that for the baseball released from rest, the acceleration and velocity will be positive, and the speed of the baseball is the same as its positive component of velocity. Then the resisting force, directed against the velocity, is upward and hence negative. EXECUTE: (a) The free-body diagram for the falling ball is sketched in Figure 5.124. (b) Newton’s Second Law is then ma = mg − Dv 2 . Initially, when v = 0, the acceleration is g, and the speed increases. As the speed increases, the resistive force increases and hence the acceleration decreases. This continues as the speed approaches the terminal speed. mg (c) At terminal velocity, a = 0, so vt = in agreement with Eq. (5.13). D dv g 2 2 = (vt − v ) . This is a separable equation and may be (d) The equation of motion may be rewritten as dt vt2

expressed as EVALUATE:

⎛ v ⎞ gt dv g 1 = 2 ∫ dt or arctanh ⎜ ⎟ = 2 . v = vt tanh ( gt vt ) . 2 −v vt vt ⎝ vt ⎠ vt x −x e −e tanh x = x . At t → 0 , tanh( gt / vt ) → 0 and v → 0 . At t → ∞ , tanh( gt / vt ) → ∞ and v → vt . e + e− x

∫v

2 t

Figure 5.124


5.125.

5-55

! ! IDENTIFY: Apply ∑ F = ma to each of the three masses and to the pulley B. SET UP: Take all accelerations to be positive downward. The equations of motion are straightforward, but the kinematic relations between the accelerations, and the resultant algebra, are not immediately obvious. If the acceleration of pulley B is aB , then aB = −a3 , and aB is the average of the accelerations of masses 1 and 2,

or a1 + a2 = 2aB = −2a3 . EXECUTE:

(a) There can be no net force on the massless pulley B, so TC = 2TA . The five equations to be solved

are then m1 g − TA = m1a1 , m2 g − TA = m2 a2 , m3 g − TC = m3a3 , a1 + a2 + 2a3 = 0 and 2TA − TC = 0 . These are five equations in five unknowns, and may be solved by standard means. The accelerations a1 and a2 may be eliminated by using 2a3 = −(a1 + a2 ) = −(2 g − TA ((1 m1 ) + (1 m2 ))). The tension TA may be eliminated by using TA = (1 2)TC = (1 2)m3 ( g − a3 ). Combining and solving for a3 gives a3 = g

−4m1m2 + m2 m3 + m1m3 . 4m1m2 + m2 m3 + m1m3

(b) The acceleration of the pulley B has the same magnitude as a3 and is in the opposite direction. (c) a1 = g −

a1 = g

TA T m = g − C = g − 3 ( g − a3 ). Substituting the above expression for a3 gives m1 2m1 2m1

4m1m2 − 3m2 m3 + m1m3 . 4m1m2 + m2 m3 + m1m3

4m1m2 − 3m1m3 + m2 m3 . 4m1m2 + m2 m3 + m1m3 (e), (f) Once the accelerations are known, the tensions may be found by substitution into the appropriate equation 4m1m2 m3 8m1m2 m3 of motion, giving TA = g , TC = g . 4m1m2 + m2 m3 + m1m3 4m1m2 + m2 m3 + m1m3 (d) A similar analysis (or, interchanging the labels 1 and 2) gives a2 = g

(g) If m1 = m2 = m and m3 = 2m, all of the accelerations are zero, TC = 2mg and TA = mg . All masses and pulleys are in equilibrium, and the tensions are equal to the weights they support, which is what is expected. EVALUATE: It is useful to consider special cases. For example, when m1 = m2 >> m3 our general result gives 5.126.

a1 = a2 = + g and a3 = − g . ! ! IDENTIFY: Apply ∑ F = ma to each block. The tension in the string is the same at both ends. If T < w for a

block, that block remains at rest. SET UP: In all cases, the tension in the string will be half of F. EXECUTE: (a) F 2 = 62 N, which is insufficient to raise either block; a1 = a2 = 0. (b) F 2 = 62 N. The larger block (of weight 196 N) will not move, so a1 = 0, but the smaller block, of weight

98 N, has a net upward force of 49 N applied to it, and so will accelerate upwards with a2 =

49 N = 4.9 m s 2 . 10.0 kg

(c) F 2 = 212 N, so the net upward force on block A is 16 N and that on block B is 114 N, so

16 N 114 N = 0.8 m s 2 and a2 = = 11.4 m s 2 . 20.0 kg 10.0 kg EVALUATE: The two blocks need not have accelerations with the same magnitudes. ! ! IDENTIFY: Apply ∑ F = ma to the ball at each position. a1 =

5.127.

SET UP: When the ball is at rest, a = 0 . When the ball is swinging in an arc it has acceleration component v2 arad = , directed inward. R EXECUTE: Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension force must balance the weight, so TA cos β = w or TA = w cos β . At point B, the ball is not in equilibrium; its speed is instantaneously 0, so there is no radial acceleration, and the tension force must balance the radial component of the weight, so TB = w cos β and the ratio (TB TA ) = cos 2 β . EVALUATE: At point B the net force on the ball is not zero; the ball has a tangential acceleration.

WORK AND KINETIC ENERGY

6.1.

6.2.

6

IDENTIFY: Apply Eq.(6.2). SET UP: The bucket rises slowly, so the tension in the rope may be taken to be the bucket’s weight. EXECUTE: (a) W = Fs = mgs = (6.75 kg) (9.80 m / s 2 )(4.00 m) = 265 J. (b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq.(6.2) gives the negative of the result of part (a), or −265 J . (c) The total work done on the bucket is zero. EVALUATE: When the force is in the direction of the displacement, the force does positive work. When the force is directed opposite to the displacement, the force does negative work. IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so W = Fs cos φ . SET UP: In part (a), when the cable pulls horizontally φ = 0° and when it pulls at 35.0° above the horizontal φ = 35.0° . In part (b), if the cable pulls horizontally φ = 180° . If the cable pulls on the car at 35.0° above the horizontal it pulls on the truck at 35.0° below the horizontal and φ = 145.0° . For the gravity force φ = 90° , since the force is vertical and the displacement is horizontal. EXECUTE: (a) When the cable is horizontal, W = (850 N)(5.00 × 103 m)cos 0° = 4.25 × 106 J . When the cable is

35.0° above the horizontal, W = (850 N)(5.00 × 103 m)cos35.0° = 3.48 × 106 J .

6.3.

(b) cos180° = − cos0° and cos145.0° = − cos35.0° , so the answers are −4.26 × 106 J and −3.48 × 106 J . (c) Since cos φ = cos90° = 0 , W = 0 in both cases. EVALUATE: If the car and truck are taken together as the system, the tension in the cable does no net work. IDENTIFY: Each force can be used in the relation W = F! s = ( F cosφ ) s for parts (b) through (d). For part (e), apply

the net work relation as Wnet = Wworker + Wgrav + Wn + W f . SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, Fworker = f k = μ k n. EXECUTE: (a) The magnitude of the force the worker must apply is:

Fworker = f k = μ k n = μ k mg = ( 0.25 ) ( 30.0 kg ) ( 9.80 m/s 2 ) = 74 N (b) Since the force applied by the worker is horizontal and in the direction of the displacement, φ = 0° and the work is:

Wworker = ( Fworker cos φ ) s = [( 74 N )( cos0° )] ( 4.5 m ) = +333 J (c) Friction acts in the direction opposite of motion, thus φ = 180° and the work of friction is:

W f = ( f k cos φ ) s = [( 74 N )( cos180° )] ( 4.5 m ) = −333 J (d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any work on the crate and Wgrav = Wn = 0.0 J. (e) Substituting into the net work relation, the net work done on the crate is:

Wnet = Wworker + Wgrav + Wn + W f = +333 J + 0.0 J + 0.0 J − 333 J = 0.0 J EVALUATE: 6.4.

The net work done on the crate is zero because the two contributing forces, Fworker and Ff , are equal in

magnitude and opposite in direction. IDENTIFY: The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies a = 0. We " " must use ∑ F = ma applied to the crate to find the forces acting on it.

6-1

6-2

Chapter 6

(a) SET UP:

The free-body diagram for the crate is given in Figure 6.4.

EXECUTE:

∑F

y

= ma y

n − mg − F sin 30° = 0 n = mg + F sin 30° f k = μ k n = μ k mg + F μ k sin 30° Figure 6.4

∑F

x

= max

F cos30° − f k = 0 F cos30° − μ k mg − μ k sin 30° F = 0

μk mg 0.25(30.0 kg)(9.80 m/s 2 ) = = 99.2 N cos30° − μ k sin 30° cos30° − (0.25)sin 30° (b) WF = ( F cos φ ) s = (99.2 N)(cos30°)(4.5 m) = 387 J F=

" " " ( F cos30° is the horizontal component of F ; the work done by F is the displacement times the component of F in the direction of the displacement.) (c) We have an expression for f k from part (a): f k = μ k ( mg + F sin 30°) = (0.250)[(30.0 kg)(9.80 m/s 2 ) + (99.2 N)(sin 30°)] = 85.9 N

φ = 180° since f k is opposite to the displacement. Thus W f = ( f k cos φ ) s = (85.9 N)(cos180°)(4.5 m) = −387 J (d) The normal force is perpendicular to the displacement so φ = 90° and Wn = 0. The gravity force (the weight) is

perpendicular to the displacement so φ = 90° and Ww = 0 (e) Wtot = WF + W f + Wn + Ww = +387 J + (−387 J) = 0 EVALUATE: Forces with a component in the direction of the displacement do positive work, forces opposite to the displacement do negative work and forces perpendicular to the displacement do zero work. The total work, obtained as the sum of the work done by each force, equals the work done by the net force. In this problem, Fnet = 0 since 6.5.

6.6.

a = 0 and Wtot = 0, which agrees with the sum calculated in part (e). IDENTIFY: The gravity force is constant and the displacement is along a straight line, so W = Fs cosφ . SET UP: The displacement is upward along the ladder and the gravity force is downward, so φ = 180.0° − 30.0° = 150.0° . w = mg = 735 N . EXECUTE: (a) W = (735 N)(2.75 m)cos150.0° = −1750 J . (b) No, the gravity force is independent of the motion of the painter. EVALUATE: Gravity is downward and the vertical component of the displacement is upward, so the gravity force does negative work. IDENTIFY and SET UP: WF = ( F cos φ ) s, since the forces are constant. We can calculate the total work by summing the work done by each force. The forces are sketched in Figure 6.6. EXECUTE:

W1 = F1s cos φ1

W1 = (1.80 × 106 N)(0.75 × 103 m)cos14° W1 = 1.31 × 109 J W2 = F2 s cos φ2 = W1 Figure 6.6

Wtot = W1 + W2 = 2(1.31× 109 J) = 2.62 × 109 J EVALUATE: Only the component F cos φ of force in the direction of the displacement does work. These " components are in the direction of s so the forces do positive work.

Work and Kinetic Energy

6.7.

6-3

IDENTIFY: All forces are constant and each block moves in a straight line. so W = Fs cosφ . The only direction the system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right. SET UP: Since the 12.0 N block moves at constant speed, a = 0 for it and the tension T in the string is T = 12.0 N . Since the 20.0 N block moves to the right at constant speed the friction force f k on it is to the left and f k = T = 12.0 N . EXECUTE:

(a) (i) φ = 0° and W = (12.0 N)(0.750 m)cos0° = 9.00 J . (ii) φ = 180° and

W = (12.0 N)(0.750 m)cos180° = −9.00 J . (b) (i) φ = 90° and W = 0 . (ii) φ = 0° and W = (12.0 N)(0.750 m)cos0° = 9.00 J . (iii) φ = 180° and W = (12.0 N)(0.750 m)cos180° = −9.00 J . (iv) φ = 90° and W = 0 .

6.8.

6.9.

(c) Wtot = 0 for each block. EVALUATE: For each block there are two forces that do work, and for each block the two forces do work of equal magnitude and opposite sign. When the force and displacement are in opposite directions, the work done is negative. IDENTIFY: Apply Eq.(6.5). SET UP: iˆ ⋅ iˆ = ˆj ⋅ ˆj = 1 and iˆ ⋅ ˆj = ˆj ⋅ iˆ = 0 " " EXECUTE: The work you do is F ⋅ s = ((30 N)iˆ − (40 N) ˆj ) ⋅ ((−9.0 m)iˆ − (3.0 m) ˆj ) " " F ⋅ s = (30 N)(−9.0 m) + (−40 N)( −3.0 m) = −270 N ⋅ m + 120 N ⋅ m = −150 J . " " EVALUATE: The x-component of F does negative work and the y-component of F does positive work. The total " work done by F is the sum of the work done by each of its components. IDENTIFY: Apply Eq.(6.2) or (6.3). " " SET UP: The gravity force is in the − y -direction , so Fmg ⋅ s = − mg ( y2 − y1 ) EXECUTE: (a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is − mg ( y2 − y1 ). When y1 = y2 , Wmg = 0 . (b) (i) Tension does no work. (ii) Let l be the length of the string. Wmg = − mg ( y2 − y1 ) = −mg (2l ) = −25.1 J

6.10.

EVALUATE: In part (b) the displacement is upward and the gravity force is downward, so the gravity force does negative work. IDENTIFY: K = 12 mv 2 SET UP: 65 mi/h = 29.1 m/s EXECUTE: (a) K = 12 (750 kg)(29.1 m/s) 2 = 3.18 × 105 J (b) K1 = 12 mv12 . K 2 = 12 mv22 , with v2 = v1 / 2 , so K 2 = 12 m(v1 / 2) 2 = 14 ( 12 mv12 ) = K1 / 4 . The change in kinetic energy is a

decrease of

3 4

K1 .

(c) K 2 = 12 K1 .

6.11.

K K K m = = constant , so 21 = 22 . v2 = v1 K 2 / K1 = (65 mi/h) v1 v2 v2 2

1 2

K1 / K1 = 46 mi/h .

EVALUATE: Since K ∼ v 2 , to have half the kinetic energy the speed must be less than half of the original speed. IDENTIFY: K = 12 mv 2 . Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic energy. SET UP: v = 12 km/s = 1.2 × 104 m/s . A 1.0 megaton bomb releases 4.184 × 1015 J of energy. EXECUTE: (a) K = 12 (1.4 × 108 kg)(1.2 × 104 m/s) 2 = 1.0 × 1016 J .

1.0 × 1016 J = 2.4 . The energy is equivalent to 2.4 one-megaton bombs. 4.184 × 1015 J EVALUATE: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large crater. IDENTIFY: K = 12 mv 2 . Use the equations for free-fall to find the speed of the weight when it reaches the ground. SET UP: Estimate that a person has speed 2 m/s when walking and 6 m/s when running. The mass of an electron is 9.11 × 10−31 kg . In part (c) take + y downward, so a y = +9.80 m/s 2 . Estimate a shoulder height of 1.6 m. (b)

6.12.

EXECUTE:

(a) Walking: K = 12 (75 kg)(2 m/s) 2 = 150 J . Running: K = 12 (75 kg)(6 m/s) 2 = 1400 J .

(b) K = 12 (9.11 × 10−31 kg)(2.19 × 106 m/s) 2 = 2.2 × 10−18 J . (c) v y2 = v02y + 2a y ( y − y0 ) gives v y = 2(9.80 m/s 2 )(1.6 m) = 5.6 m/s . K = 12 (1.0 kg)(5.6 m/s) 2 = 16 J .

6-4

Chapter 6

(d) v =

6.13.

2K 2(100 J) = = 2.6 m/s . Yes, this is reasonable. 30 kg m

EVALUATE: A walking speed of 2 m/s corresponds to walking a mile in about 13 min. A running speed of 6 m/s corresponds to running a 100 m dash in about 17 s. IDENTIFY: K = 12 mv 2 . Set up a ratio that relates K, m and v. SET UP: EXECUTE:

mp = 1836me (a) K p = K e gives meve2 = mpvp2 . ve = vp mp / me = V 1836 = 42.85V .

(b) vp = ve gives

6.14.

Kp mp

=

Ke . K p = K e ( mp / me ) = 1836 K . me

EVALUATE: The electron has less mass so must travel faster to have the same kinetic energy. And with equal speeds the proton has more kinetic energy. IDENTIFY: Only gravity does work on the watermelon, so Wtot = Wgrav . Wtot = ΔK and K = 12 mv 2 . SET UP: EXECUTE:

Since the watermelon is dropped from rest, K1 = 0 . (a) Wgrav = mgs = (4.80 kg)(9.80 m/s 2 )(25.0 m) = 1180 J

(b) Wtot = K 2 − K1 so K 2 = 1180 J . v =

2K2 2(1180 J) = = 22.2 m/s . 4.80 kg m

(c) The work done by gravity would be the same. Air resistance would do negative work and Wtot would be less than

Wgrav . The answer in (a) would be unchanged and both answers in (b) would decrease. 6.15.

EVALUATE: The gravity force is downward and the displacement is downward, so gravity does positive work. IDENTIFY: Wtot = K 2 − K1 . In each case calculate Wtot from what we know about the force and the displacement. SET UP: The gravity force is mg, downward. The friction force is f k = μ k n = μ k mg and is directed opposite to the displacement. The mass of the object isn't given, so we expect that it will divide out in the calculation. EXECUTE:

(a) K1 = 0 . Wtot = Wgrav = mgs . mgs = 12 mv22 and v2 = 2 gs = 2(9.80 m/s 2 )(95.0 m) = 43.2 m/s .

(b) K 2 = 0 (at the maximum height). Wtot = Wgrav = − mgs . − mgs = − 12 mv12 and

v1 = 2 gs = 2(9.80 m/s 2 )(525 m) = 101 m/s . (c) K1 = 12 mv12 . K 2 = 0 . Wtot = W f = − μ k mgs . − μ k mgs = − 12 mv12 . s = (d) K1 = 12 mv12 . K 2 = 12 mv22 . Wtot = W f = − μ k mgs . K 2 = Wtot + K1 .

1 2

v12

2μk g

=

(5.00 m/s) 2 = 5.80 m . 2(0.220)(9.80 m/s 2 )

mv22 = − μ k mgs + 12 mv12

v2 = v12 − 2 μk gs = (5.00 m/s) 2 − 2(0.220)(9.80 m/s 2 )(2.90 m) = 3.53 m/s . (e) K1 = 12 mv12 . K 2 = 0 . Wgrav = −mgy2 , where y2 is the vertical height. − mgy2 = − 12 mv12 and

(12.0 m/s) 2 v12 = = 7.35 m . 2 g 2(9.80 m/s 2 ) EVALUATE: In parts (c) and (d), friction does negative work and the kinetic energy is reduced. In part (a), gravity does positive work and the speed increases. In parts (b) and (e), gravity does negative work and the speed decreases. The vertical height in part (e) is independent of the slope angle of the hill. IDENTIFY: From the work-energy relation, W = Wgrav = ΔK rock . y2 =

6.16.

SET UP: As the rock rises, the gravitational force, F = mg , does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s, the work is negative. Let h be the vertical distance the rock travels. EXECUTE: (a) Applying Wgrav = K 2 − K1 we obtain − mgh = 12 mv22 − 12 mv12 . Dividing by m and solving for v1 ,

v1 = v22 + 2 gh . Substituting h = 15.0 m and v2 = 25.0 m/s, 2 v1 = ( 25.0 m/s ) + 2 ( 9.80 m/s 2 ) (15.0 m ) = 30.3 m/s

(b) Solve the same work-energy relation for h. At the maximum height v2 = 0 .

− mgh = 12 mv22 − 12 mv12 and h =

2 2 v12 − v22 ( 30.3 m/s ) − ( 0.0 m/s ) = = 46.8 m . 2 2g 2 ( 9.80 m/s )


6.17.

6-5

EVALUATE: Note that the weight of 20 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m. Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have an initial velocity of 30.3 m/s. As the rock moves upward gravity does negative work and this reduces the kinetic energy of the rock. IDENTIFY and SET UP: Apply Eq.(6.6) to the box. Let point 1 be at the bottom of the incline and let point 2 be at the skier. Work is done by gravity and by friction. Solve for K1 and from that obtain the required initial speed. EXECUTE:

Wtot = K 2 − K1

K1 = mv , K 2 = 0 1 2

2 0

Work is done by gravity and friction, so Wtot = Wmg + W f . Wmg = − mg ( y2 − y1 ) = −mgh W f = − fs. The normal force is n = mg cos α and s = h / sin α , where s is the distance the box travels along the incline. W f = −( μk mg cos α )(h / sin α ) = − μk mgh / tan α Substituting these expressions into the work-energy theorem gives − mgh − μk mgh / tan α = − 12 mv02 . Solving for v0 then gives v0 = 2 gh(1 + μk / tan α ).

6.18.

EVALUATE: The result is independent of the mass of the box. As α → 90°, h = s and v0 = 2 gh , the same as throwing the box straight up into the air. For α = 90° the normal force is zero so there is no friction. IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK .

Parallel to incline: force component W! = mg sin α , down incline; displacement s = h/ sin α , down incline.

SET UP:

Perpendicular to the incline: s = 0 . EXECUTE: (a) W|| = (mg sin α )(h / sin α ) = mgh . W⊥ = 0 , since there is no displacement in this direction. Wmg = W|| + W⊥ = mgh , same as falling height h.

6.19.

(b) Wtot = K 2 − K1 gives mgh = 12 mv 2 and v = 2 gh , same as if had been dropped from height h. The work done by gravity depends only on the vertical displacement of the object. When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction. When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller. (c) h = 15.0 m , so v = 2 gh = 17.1 s . EVALUATE: The acceleration and time of travel are different for an object sliding down an incline and an object in free-fall, but the final velocity is the same in these two cases. IDENTIFY: Wtot = K 2 − K1 with Wtot = W f . The car stops, so K 2 = 0 . In each case identify what is constant and set

up a ratio. SET UP: W f = − fs , so − fs = − 12 mv02 . EXECUTE:

6.20.

(a) v0b = 3v0 a . sa = D . f is constant.

2

⎛v ⎞ v2 v2 v02 2 f = = constant , so 0 a = 0b . sb = sa ⎜ 0b ⎟ = D(3) 2 = 9 D . sa sb s m ⎝ v0 a ⎠

⎛f ⎞ (b) f b = 3 f a . v0 is constant. fs = 12 mv02 = constant , so f a sa = fb sb . sb = sa ⎜ a ⎟ = D / 3 . ⎝ fb ⎠ EVALUATE: The stopping distance is proportional to the square of the initial speed. When the friction force increases, the stopping distance decreases. IDENTIFY and SET UP: Apply Eq.(6.6). The relation between the speeds v1 and v2 tells us the relation between K1 and K 2 . EXECUTE:

(a) W = K 2 − K1

K1 = 12 mv , K 2 = 12 mv22 2 1

v2 = 14 v1 gives that K 2 = 12 m ( 14 v1 ) = 161 ( 12 mv12 ) = 161 K1 2

15 W = K 2 − K1 = 161 K1 − K1 = − 16 K1 " (b) EVALUATE: K depends only on the magnitude of v not on its direction, so the answer for W in part (a) does not depend on the final direction of the electron’s motion. The electron slows down, so its kinetic energy decreases and the total work done on it is negative.

6-6

6.21.

Chapter 6

IDENTIFY:

Apply W = Fs cos φ and Wtot = ΔK .

SET UP: φ = 0° EXECUTE: From Equations (6.1), (6.5) and (6.6), and solving for F,

F=

6.22.

ΔK 12 m(v22 − v12 ) 12 (8.00 kg)((6.00 m / s) 2 − (4.00 m / s) 2 ) = = = 32.0 N. (2.50 m) s s

EVALUATE: The force is in the direction of the displacement, so the force does positive work and the kinetic energy of the object increases. IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work done by the foot on the ball. Then use Eq.(6.2) to find the distance over which this force acts. EXECUTE: Wtot = K 2 − K1

K1 = 12 mv12 = 12 (0.420 kg)(2.00 m/s) 2 = 0.84 J K 2 = 12 mv22 = 12 (0.420 kg)(6.00 m/s) 2 = 7.56 J Wtot = K 2 − K1 = 7.56 J − 0.84 J = 6.72 J

The 40.0 N force is the only force doing work on the ball, so it must do 6.72 J of work. WF = ( F cosφ ) s gives that s=

6.23.

W 6.72 J = = 0.168 m F cos φ (40.0 N)(cos0)

EVALUATE: The force is in the direction of the motion so positive work is done and this is consistent with an increase in kinetic energy. IDENTIFY: Apply Wtot = ΔK . SET UP: v1 = 0 , v2 = v . f k = μk mg and f k does negative work. The force F = 36.0 N is in the direction of the motion and does positive work. EXECUTE: (a) If there is no work done by friction, the final kinetic energy is the work done by the applied force, and solving for the speed,

v=

2W 2 Fs 2(36.0 N)(1.20 m) = = = 4.48 m / s. (4.30 kg) m m

(b) The net work is Fs − f k s = ( F − μk mg ) s , so

v=

6.24.

EVALUATE: The total work done is larger in the absence of friction and the final speed is larger in that case. IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK SET UP: The gravity force has magnitude mg and is directed downward. EXECUTE: (a) On the way up, gravity is opposed to the direction of motion, and so W = − mgs = −(0.145 kg)(9.80 m / s 2 )(20.0 m) = −28.4 J . (b) v2 = v12 + 2

6.25.

2( F − μ k mg ) s 2(36.0 N − (0.30)(4.30 kg)(9.80 m / s 2 ))(1.20 m) = = 3.61 m/s m (4.30 kg)

W 2(−28.4 J) = (25.0 m / s) 2 + = 15.3 m / s . m (0.145 kg)

(c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the way down, gravity will have done both negative and positive work on the ball, but the net work at this height will be the same. EVALUATE: As the baseball moves upward, gravity does negative work and the speed of the baseball decreases. (a) IDENTIFY and SET UP: Use Eq.(6.2) to find the work done by the positive force. Then use Eq.(6.6) to find the final kinetic energy, and then K 2 = 12 mv22 gives the final speed. EXECUTE:

Wtot = K 2 − K1 , so K 2 = Wtot + K1

K1 = mv = 12 (7.00 kg)(4.00 m/s) 2 = 56.0 J The only force that does work on the wagon is the 10.0 N force. This force is in the direction of the displacement so φ = 0° and the force does positive work: 1 2

2 1

WF = ( F cos φ ) s = (10.0 N)(cos0)(3.0 m) = 30.0 J Then K 2 = Wtot + K1 = 30.0 J + 56.0 J = 86.0 J. K 2 = 12 mv22 ; v2 =

2K2 2(86.0 J) = = 4.96 m/s m 7.00 kg


(b) IDENTIFY:

6-7

" " Apply ∑ F = ma to the wagon to calculate a. Then use a constant acceleration equation to calculate

the final speed. The free-body diagram is given in Figure 6.25. SET UP: EXECUTE:

∑F

x

= max

F = max ax =

F 10.0 N = = 1.43 m/s 2 m 7.00 kg

Figure 6.25

v22x = v12x + 2a2 ( x − x0 ) v2 x = v12x + 2ax ( x − x0 ) = (4.00 m/s) 2 + 2(1.43 m/s 2 )(3.0 m) = 4.96 m/s

6.26.

EVALUATE: This agrees with the result calculated in part (a). The force in the direction of the motion does positive work and the kinetic energy and speed increase. In part (b), the equivalent statement is that the force produces an acceleration in the direction of the velocity and this causes the magnitude of the velocity to increase. IDENTIFY: Apply Wtot = K 2 − K1 . SET UP: K1 = 0 . The normal force does no work. The work W done by gravity is W = mgh , where h = L sin θ is the vertical distance the block has dropped when it has traveled a distance L down the incline and θ is the angle the plane makes with the horizontal. 2K 2W EXECUTE: The work-energy theorem gives v = = = 2 gh = 2 gL sin θ . Using the given numbers, m m

6.27.

v = 2(9.80 m / s 2 )(0.75 m)sin 36.9° = 2.97 m / s. EVALUATE: The final speed of the block is the same as if it had been dropped from a height h. IDENTIFY: Wtot = K 2 − K1 . Only friction does work. SET UP: EXECUTE:

Wtot = W fk = − μ k mgs . K 2 = 0 (car stops). K1 = 12 mv02 . (a) Wtot = K 2 − K1 gives − μ k mgs = − 12 mv02 . s =

(b) (i) μ kb = 2 μ ka . sμ k =

v02

2μk g

.

⎛μ ⎞ v02 = constant so sa μ ka = sb μ kb . sb = ⎜ ka ⎟ sa = sa / 2 . The minimum stopping distance 2g ⎝ μ kb ⎠ 2

would be halved. (ii) v0b = 2v0 a .

⎛v ⎞ s 1 s s = = constant , so 2a = 2b . sb = sa ⎜ 0b ⎟ = 4 sa . The stopping distance v02 2 μ k g v0 a v0b ⎝ v0 a ⎠

would become 4 times as great. (iii) v0b = 2v0 a , μ kb = 2 μ ka .

sμk 1 sμ sμ = = constant , so a 2 ka = b 2 kb . 2 v0 2g v0 a v0b

2

6.28.

⎛ μ ⎞⎛ v ⎞ ⎛1⎞ sb = sa ⎜ ka ⎟⎜ 0b ⎟ = sa ⎜ ⎟ (2) 2 = 2sa . The stopping distance would double. ⎝2⎠ ⎝ μkb ⎠⎝ v0 a ⎠ EVALUATE: The stopping distance is directly proportional to the square of the initial speed and indirectly proportional to the coefficient of kinetic friction. IDENTIFY: The work that must be done to move the end of a spring from x1 to x2 is W = 12 kx22 − 12 kx12 . The force required to hold the end of the spring at displacement x is Fx = kx . SET UP: When the spring is at its unstretched length, x = 0 . When the spring is stretched, x > 0 , and when the spring is compressed, x < 0 . 2W 2(12.0 J) = 2.67 × 104 N/m . EXECUTE: (a) x1 = 0 and W = 12 kx22 . k = 2 = x2 (0.0300 m) 2 (b) Fx = kx = (2.67 × 104 N/m)(0.0300 m) = 801 N . (c) x1 = 0 , x2 = −0.0400 m . W = 12 (2.67 × 104 N/m)( −0.0400 m) 2 = 21.4 J .

Fx = kx = (2.67 × 104 N/m)(0.0400 m) = 1070 N . EVALUATE: When a spring, initially unstretched, is either compressed or stretched, positive work is done by the force that moves the end of the spring.

6-8

6.29.

Chapter 6


Use Eq.(6.8) to calculate k for the spring. Then Eq.(6.10), with x1 = 0, can be used to

calculate the work done to stretch or compress the spring an amount x2 . EXECUTE: Use the information given to calculate the force constant of the spring. F 160 N Fx = kx gives k = x = = 3200 N/m x 0.050 m (a) Fx = kx = (3200 N/m)(0.015 m) = 48 N Fx = kx = (3200 N/m)( − 0.020 m) = −64 N (magnitude 64 N) (b) W = 12 kx 2 = 12 (3200 N/m)(0.015 m) 2 = 0.36 J

6.30.

6.31.

W = 12 kx 2 = 12 (3200 N/m)( − 0.020 m) 2 = 0.64 J Note that in each case the work done is positive. EVALUATE: The force is not constant during the displacement so Eq.(6.2) cannot be used. A force in the + x direction is required to stretch the spring and a force in the opposite direction to compress it. The force Fx is in the same direction as the displacement, so positive work is done in both cases. IDENTIFY: The magnitude of the work can be found by finding the area under the graph. SET UP: The area under each triangle is 1/2 base × height . Fx > 0 , so the work done is positive when x increases during the displacement. EXECUTE: (a) 1/ 2 (8 m)(10 N) = 40 J . (b) 1/ 2 (4 m)(10 N) = 20 J . (c) 1/ 2 (12 m)(10 N) = 60 J . EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c). IDENTIFY: Use the work-energy theorem and the results of Problem 6.30. SET UP: For x = 0 to x = 8.0 m , Wtot = 40 J . For x = 0 to x = 12.0 m , Wtot = 60 J . EXECUTE:

(a) v =

(2)(40 J) = 2.83 m / s 10 kg

(2)(60 J) = 3.46 m / s . 10 kg " " EVALUATE: F is always in the + x -direction. For this motion F does positive work and the speed continually increases during the motion. (b) v =

6.32.

IDENTIFY:

x2

The force has only an x-component and the motion is along the x-direction, so W = ∫ Fx dx . x1

SET UP: x1 = 0 and x2 = 6.9 m . EXECUTE: The work you do with your changing force is x2

x2

x2

x1

x1

x1

W = ∫ F ( x)dx = ∫ (−20.0 N) dx − ∫ (3.0 N/m)xdx = (−20.0 N) x |xx12 −(3.0 N/m)( x 2 /2) |xx12

6.33.

W = −138 N ⋅ m − 71.4 N ⋅ m = −209 J . EVALUATE: The work is negative because the cow continues to move forward (in the + x -direction ) as you vainly attempt to push her backward. IDENTIFY: Apply Eq.(6.6) to the box. SET UP: Let point 1 be just before the box reaches the end of the spring and let point 2 be where the spring has maximum compression and the box has momentarily come to rest. EXECUTE: Wtot = K 2 − K1 K1 = 12 mv02 , K 2 = 0 Work is done by the spring force. Wtot = − 12 kx22 , where x2 is the amount the spring is compressed. − 12 kx22 = − 12 mv02 and x2 = v0 m / k = (3.0 m/s) (6.0 kg)/(7500 N/m) = 8.5 cm

6.34.

EVALUATE: The compression of the spring increases when either v0 or m increases and decreases when k increases (stiffer spring). IDENTIFY: The force applied to the springs is Fx = kx . The work done on a spring to move its end from x1 to x2 is

W = 12 kx22 − 12 kx12 . Use the information that is given to calculate k. SET UP:

When the springs are compressed 0.200 m from their uncompressed length, x1 = 0 and x2 = −0.200 m .

When the platform is moved 0.200 m farther, x2 becomes −0.400 m .


6-9

2W 2(80.0 J) = = 4000 N/m . Fx = kx = (4000 N/m)(−0.200 m) = −800 N . The 2 x − x1 (0.200 m) 2 − 0 magnitude of force that is required is 800 N. (b) To compress the springs from x1 = 0 to x2 = −0.400 m , the work required is (a) k =

EVALUATE:

2 2

W = 12 kx22 − 12 kx12 = 12 (4000 N/m)(−0.400 m) 2 = 320 J . The additional work required is 320 J − 80 J = 240 J . For

6.35.

x = −0.400 m , Fx = kx = −1600 N . The magnitude of force required is 1600 N. EVALUATE: More work is required to move the end of the spring from x = −0.200 m to x = −0.400 m than to move it from x = 0 to x = −0.200 m , even though the displacement of the platform is the same in each case. The magnitude of the force increases as the compression of the spring increases. " " IDENTIFY: Apply ∑ F = ma to calculate the μs required for the static friction force to equal the spring force. SET UP:

(a) The free-body diagram for the glider is given in Figure 6.35. EXECUTE:

∑F

y

= ma y

n − mg = 0 n = mg f s = μs mg Figure 6.35

∑F

x

= max

f s − Fspring = 0

μs mg − kd = 0 μs =

kd (20.0 N/m)(0.086 m) = = 1.76 mg (0.100 kg)(9.80 m/s 2 )


Apply

"

"

∑ F = ma

to find the maximum amount the spring can be compressed and still

have the spring force balanced by friction. Then use Wtot = K 2 − K1 to find the initial speed that results in this compression of the spring when the glider stops. EXECUTE: μs mg = kd

μs mg

(0.60)(0.100 kg)(9.80 m/s 2 ) = 0.0294 m 20.0 N/m k Now apply the work-energy theorem to the motion of the glider: Wtot = K 2 − K1 d=

=

K1 = 12 mv12 , K 2 = 0 (instantaneously stops) Wtot = Wspring + Wfric = − 12 kd 2 − μ k mgd (as in Example 6.8) Wtot = − 12 (20.0 N/m)(0.0294 m) 2 − 0.47(0.100 kg)(9.80 m/s 2 )(0.0294 m) = −0.02218 J Then Wtot = K 2 − K1 gives −0.02218 J = − 12 mv12 . v1 =

6.36.

2(0.02218 J) = 0.67 m/s 0.100 kg

EVALUATE: In Example 6.8 an initial speed of 1.50 m/s compresses the spring 0.086 m and in part (a) of this problem we found that the glider doesn’t stay at rest. In part (b) we found that a smaller displacement of 0.0294 m when the glider stops is required if it is to stay at rest. And we calculate a smaller initial speed (0.67 m/s) to produce this smaller displacement. IDENTIFY: For the spring, W = 12 kx12 − 12 kx22 . Apply Wtot = K 2 − K1 . SET UP: EXECUTE:

x1 = −0.025 m and x2 = 0 . (a) W = 12 kx12 = 12 (200 N / m)(−0.025 m) 2 = 0.060 J .

2W 2(0.060 J) = = 0.18 m / s. m (4.0 kg) EVALUATE: The block moves in the direction of the spring force, the spring does positive work and the kinetic energy of the block increases. (b) The work-energy theorem gives v2 =

6-10

6.37.

Chapter 6


The magnitude of the work done by Fx equals the area under the Fx versus x curve. The

work is positive when Fx and the displacement are in the same direction; it is negative when they are in opposite directions. EXECUTE: (a) Fx is positive and the displacement Δx is positive, so W > 0. W = 12 (2.0 N)(2.0 m) + (2.0 N)(1.0 m) = +4.0 J (b) During this displacement Fx = 0, so W = 0.

6.38.

(c) Fx is negative, Δx is positive, so W < 0. W = − 12 (1.0 N)(2.0 m) = −1.0 J (d) The work is the sum of the answers to parts (a), (b), and (c), so W = 4.0 J + 0 − 1.0 J = +3.0 J (e) The work done for x = 7.0 m to x = 3.0 m is +1.0 J. This work is positive since the displacement and the force are both in the − x-direction. The magnitude of the work done for x = 3.0 m to x = 2.0 m is 2.0 J, the area under Fx versus x. This work is negative since the displacement is in the − x -direction and the force is in the + x -direction. Thus W = +1.0 J − 2.0 J = −1.0 J EVALUATE: The work done when the car moves from x = 2.0 m to x = 0 is − 12 (2.0 N)(2.0 m) = −2.0 J. Adding this to the work for x = 7.0 m to x = 2.0 m gives a total of W = −3.0 J for x = 7.0 m to x = 0. The work for x = 7.0 m to x = 0 is the negative of the work for x = 0 to x = 7.0 m. IDENTIFY: Apply Wtot = K 2 − K1 . SET UP: K1 = 0 . From Exercise 6.37, the work for x = 0 to x = 3.0 m is 4.0 J. W for x = 0 to x = 4.0 m is also 4.0 J. For x = 0 to x = 7.0 m , W = 3.0 J . EXECUTE: (a) K = 4.0 J , so v = 2 K m = 2(4.0 J) (2.0 kg) = 2.00 m / s . (b) No work is done between x = 3.0 m and x = 4.0 m , so the speed is the same, 2.00 m/s.

6.39.

(c) K = 3.0 J , so v = 2 K / m = 2(3.0 J) /(2.0 kg) = 1.73 m / s . EVALUATE: In each case the work done by F is positive and the car gains kinetic energy. IDENTIFY and SET UP: Apply Eq.(6.6). Let point 1 be where the sled is released and point 2 be at x = 0 for part (a) and at x = −0.200 m for part (b). Use Eq.(6.10) for the work done by the spring and calculate K 2 . Then K 2 = 12 mv22

gives v2 . EXECUTE:

(a) Wtot = K 2 − K1 so K 2 = K1 + Wtot

K1 = 0 (released with no initial velocity), K 2 = 12 mv22 The only force doing work is the spring force. Eq.(6.10) gives the work done on the spring to move its end from x1 to x2 . The force the spring exerts on an object attached to it is F = − kx , so the work the spring does is

Wspr = − ( 12 kx22 − 12 kx12 ) = 12 kx12 − 12 kx22 . Here x1 = −0.375 m and x2 = 0. Thus Wspr = 12 (4000 N/m)( − 0.375 m) 2 − 0 = 281 J. K 2 = K1 + Wtot = 0 + 281 J = 281 J Then K 2 = 12 mv22 implies v2 =

2K2 2(281 J) = = 2.83 m/s. m 70.0 kg

(b) K 2 = K1 + Wtot

K1 = 0 Wtot = Wspr = 12 kx12 − 12 kx22 . Now x2 = 0.200 m, so Wspr = 12 (4000 N/m)( − 0.375 m) 2 − 12 (4000 N/m)( − 0.200 m) 2 = 281 J − 80 J = 201 J Thus K 2 = 0 + 201 J = 201 J and K 2 = 12 mv22 gives v2 =

6.40.

2K2 2(201 J) = = 2.40 m/s. m 70.0 kg

EVALUATE: The spring does positive work and the sled gains speed as it returns to x = 0. More work is done during the larger displacement in part (a), so the speed there is larger than in part (b). IDENTIFY: Fx = kx SET UP: When the spring is in equilibrium, the same force is applied to both ends of any segment of the spring. EXECUTE: (a) When a force F is applied to each end of the original spring, the end of the spring is displaced a distance x. Each half of the spring elongates a distance xh , where xh = x / 2 . Since F is also the force applied to each

⎛ x half of the spring, F = kx and F = kh xh . kx = kh xh and kh = k ⎜ ⎝ xh

⎞ ⎟ = 2k . ⎠


6-11

(b) The same reasoning as in part (a) gives kseg = 3k , where kseg is the force constant of each segment.

6.41.

EVALUATE: For half of the spring the same force produces less displacement than for the original spring. Since k = F / x , smaller x for the same F means larger k. IDENTIFY and SET UP: Apply Eq.(6.6) to the glider. Work is done by the spring and by gravity. Take point 1 to be where the glider is released. In part (a) point 2 is where the glider has traveled 1.80 m and K 2 = 0. There two points are shown in Figure 6.41a. In part (b) point 2 is where the glider has traveled 0.80 m. EXECUTE: (a) Wtot = K 2 − K1 = 0. Solve for x1 , the amount the spring is initially compressed.

Wtot = Wspr + Ww = 0 So Wspr = −Ww (The spring does positive work on the glider since the spring force is directed up the incline, the same as the direction of the displacement.) Figure 6.41a

The directions of the displacement and of the gravity force are shown in Figure 6.41b. Ww = ( w cos φ ) s = ( mg cos130.0°) s Ww = (0.0900 kg)(9.80 m/s 2 )(cos130.0°)(1.80 m) = −1.020 J (The component of w parallel to the incline is directed down the incline, opposite to the displacement, so gravity does negative work.) Figure 6.41b

Wspr = −Ww = +1.020 J 2Wspr

2(1.020 J) = 0.0565 m k 640 N/m (b) The spring was compressed only 0.0565 m so at this point in the motion the glider is no longer in contact with the spring. Points 1 and 2 are shown in Figure 6.41c. Wspr = 12 kx12 so x1 =

=

Wtot = K 2 − K1 K 2 = K1 + Wtot K1 = 0 Figure 6.41c

Wtot = Wspr + Ww From part (a), Wspr = 1.020 J and Ww = (mg cos130.0°) s = (0.0900 kg)(9.80 m/s 2 )(cos130.0°)(0.80 m) = −0.454 J Then K 2 = Wspr + Ww = +1.020 J − 0.454 J = +0.57 J.

6.42.

EVALUATE: The kinetic energy in part (b) is positive, as it must be. In part (a), x2 = 0 since the spring force is no longer applied past this point. In computing the work done by gravity we use the full 0.80 m the glider moves. IDENTIFY: Apply Wtot = K 2 − K1 to the brick. Work is done by the spring force and by gravity. SET UP:

At the maximum height. v = 0 . Gravity does negative work, Wgrav = − mgh . The work done by the spring

2

is kd , where d is the distance the spring is compressed initially. EXECUTE: The initial and final kinetic energies of the brick are both zero, so the net work done on the brick by the spring and gravity is zero, so (1 2)kd 2 − mgh = 0 , or 1 2

d = 2mgh / k = 2(1.80 kg)(9.80 m / s 2 )(3.6 m) /(450 N / m) = 0.53 m. The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length. But when the spring reaches its uncompressed length the brick has an upward velocity and leaves the spring. EVALUATE: Gravity does negative work because the gravity force is downward and the brick moves upward. The spring force does positive work on the brick because the spring force is upward and the brick moves upward.

6-12

6.43.

Chapter 6

IDENTIFY: SET UP:

Apply the relation between energy and power. W to solve for W, the energy the bulb uses. Then set this value equal to Use P = Δt

speed. EXECUTE:

mv 2 and solve for the

W = PΔt = (100 W)(3600 s) = 3.6 × 105 J

K = 3.6 × 105 J so v = 6.44.

1 2

2K 2(3.6 × 105 J) = = 100 m/ s 70 kg m

EVALUATE: Olympic runners achieve speeds up to approximately 36 m/s, or roughly one third the result calculated. IDENTIFY: Energy is power times time. SET UP: 1 W = 1 J/s . 1 yr = 3.16 × 107 s . EXECUTE: (b)

(a)

(1.0 × 1019 J / yr) = 3.2 × 1011 W. (3.16 × 107 s / yr)

3.2 × 1011 W = 1.1 kW/person. 3.0 × 108 folks

3.2 × 1011 W = 8.0 × 108 m 2 = 800 km 2 . (0.40)1.0 × 103 W / m 2 EVALUATE: The area in part (c) corresponds to a square about 28 km on a side, which is about 18 miles. The space required is not an impediment. ΔW . ΔW is the energy released. IDENTIFY: Pav = Δt SET UP: ΔW is to be the same. 1 y = 3.156 × 107 s . (c) A =

6.45.

Pav Δt = ΔW = constant , so Pav-sun Δtsun = Pav-m Δtm .

EXECUTE:

6.46.

⎛ Δt ⎞ ⎛ [2.5 × 105 y][3.156 × 107 s/y] ⎞ 13 Pav-m = Pav-sun ⎜ sun ⎟ = ⎜ ⎟ = 3.9 × 10 P . 0.20 s t Δ ⎠ ⎝ m ⎠ ⎝ EVALUATE: Since the power output of the magnetar is so much larger than that of our sun, the mechanism by which it radiates energy must be quite different. IDENTIFY: The thermal energy is produced as a result of the force of friction, F = μ k mg . The average thermal power is thus the average rate of work done by friction or P = F!vav . SET UP:

v2 + v1 ⎛ 8.00 m/s + 0 ⎞ =⎜ ⎟ = 4.00 m/s 2 2 ⎝ ⎠ P = Fvav = ⎡⎣( 0.200 ) ( 20.0 kg ) ( 9.80 m/s 2 ) ⎤⎦ ( 4.00 m/s ) = 157 W

vav =

EXECUTE: EVALUATE:

The power could also be determined as the rate of change of kinetic energy, ΔK t , where the time is

calculated from vf = vi + at and a is calculated from a force balance, 6.47.

1 hp = 746 W

EXECUTE:

6.48.

k

Use the relation P = F!v to relate the given force and velocity to the total power developed.

IDENTIFY: SET UP:

∑ F = ma = μ mg.

The total power is P = F!v = (165 N )( 9.00 m/s ) = 1.49 × 103 W. Each rider therefore contributes

Peach rider = (1.49 × 103 W ) / 2 = 745 W ≈ 1 hp. EVALUATE: The result of one horsepower is very large; a rider could not sustain this output for long periods of time. IDENTIFY and SET UP: Calculate the power used to make the plane climb against gravity. Consider the vertical motion since gravity is vertical. EXECUTE: The rate at which work is being done against gravity is P = Fv = mgv = (700 kg)(9.80 m/s 2 )(2.5 m/s) = 17.15 kW. This is the part of the engine power that is being used to make the airplane climb. The fraction this is of the total is 17.15 kW/75 kW = 0.23. EVALUATE: engine.

The power we calculate for making the airplane climb is considerably less than the power output of the


6.49.

6-13

ΔW . The work you do in lifting mass m a height h is mgh. Δt SET UP: 1 hp = 746 W EXECUTE: (a) The number per minute would be the average power divided by the work (mgh) required to lift one (0.50 hp) (746 W hp) box, = 1.41 s, or 84.6 min. (30 kg) (9.80 m s 2 ) (0.90 m) Pav =

IDENTIFY:

(100 W) = 0.378 s, or 22.7 min. (30 kg) (9.80 m s 2 ) (0.90 m) EVALUATE: A 30-kg crate weighs about 66 lbs. It is not possible for a person to perform work at this rate. IDENTIFY and SET UP: Use Eq.(6.15) to relate the power provided and the amount of work done against gravity in 16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers. EXECUTE: Find the total mass that can be lifted: Pt ΔW mgh Pav = = , so m = av Δt t gh (b) Similarly,

6.50.

⎛ 746 W ⎞ 4 Pav = (40 hp) ⎜ ⎟ = 2.984 × 10 W 1 hp ⎝ ⎠ m=

Pavt (2.984 × 104 W)(16.0 s) = = 2.436 × 103 kg gh (9.80 m/s 2 )(20.0 m)

This is the total mass of elevator plus passengers. The mass of the passengers is 2.436 ×103 kg − 600 kg = 1.836 ×103 kg. 1.836 × 103 kg = 28.2. 28 passengers can ride. 65.0 kg EVALUATE: Typical elevator capacities are about half this, in order to have a margin of safety. IDENTIFY: Calculate the gallons of gasoline consumed and from that the energy consumed. Find the time Δt for the ΔW , where ΔW is the energy consumed. trip and use Pav = Δt SET UP: 200 km = 124 mi 124 mi EXECUTE: (a) The gallons of gasoline consumed is = 4.13 gal . The energy consumed is 30 mi/gal

The number of passengers is 6.51.

(4.13 gal)(1.3 × 109 J/gal) = 5.4 × 109 J . ΔW 5.4 × 109 J 124 mi = 2.07 h = 7450 s . Pav = = = 7.2 × 105 W = 720 kW . 60 mi/h 7450 s Δt 720 × 103 W The rate of energy consumption is = 970 hp . 746 W/hp

(b) The time for the trip is EVALUATE: 6.52.

Apply P = F!v . F! is the force F of water resistance.

IDENTIFY: SET UP:

1 hp = 746 W . 1 km/h = 0.228 m/s F=

EXECUTE: EVALUATE: 6.53.

(0.70) P (0.70) (280,000 hp)(746 W hp) = = 8.1 × 106 N. v (65 km h) ((0.228 m/s) (1 km/h))

The power required depends on speed, because of the factor of v in P = F!v and also because the

resistive force increases with speed. IDENTIFY: To lift the skiers, the rope must do positive work to counteract the negative work developed by the component of the gravitational force acting on the total number of skiers, Frope = Nmg sin α . SET UP:

P = F!v = Fropev

EXECUTE:

Prope = Fropev = ⎣⎡ + Nmg ( cos φ ) ⎦⎤ v .

⎡ ⎛ 1 m/s ⎞ ⎤ Prope = ⎡⎣( 50 riders ) ( 70.0 kg ) ( 9.80 m/s 2 ) ( cos75.0º ) ⎤⎦ ⎢(12.0 km/h ) ⎜ ⎟ . ⎝ 3.60 km/h ⎠ ⎥⎦ ⎣ Prope = 2.96 × 104 W = 29.6 kW . EVALUATE: rest.

Some additional power would be needed to give the riders kinetic energy as they are accelerated from

6-14

6.54.

6.55.

Chapter 6

IDENTIFY: Relate power, work and time. SET UP: Work done in each stroke is W = Fs and Pav = W t . EXECUTE: 100 strokes per second means Pav = 100 Fs t with t = 1.00 s, F = 2mg and s = 0.010 m. Pav = 0.20 W. EVALUATE: For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per second, the average power output would be 7.0 × 105 W . This power output is very far beyond the capability of a person. IDENTIFY: For mass dm located a distance x from the axis and moving with speed v, the kinetic energy is K = 12 (dm)v 2 . Follow the procedure specified in the hint. SET UP: The bar and an infinitesimal mass element along the bar are sketched in Figure 6.55. Let M = total mass 2πx . and T = time for one revolution . v = T 1 M dx , so EXECUTE: K = ∫ (dm)v 2 . dm = 2 L 2

L 1 ⎛ M ⎞ ⎛ 2πx ⎞ 1 ⎛ M ⎞ ⎛ 4π 2 ⎞ K = ∫ ⎜ dx ⎟ ⎜ ⎟ ⎟ = ⎜ ⎟ ⎜ 2⎝ L ⎠ ⎝ T ⎠ 2⎝ L ⎠ ⎝ T2 ⎠ 0

1 ⎛ M ⎞ ⎛ 4π 2 ⎞ ⎛ L3 ⎞ 2 2 2 2 2 = x dx ⎟ ⎜ ⎟ = π ML T ⎜ ⎟ ⎜ ∫0 2⎝ L ⎠ ⎝ T2 ⎠ ⎝ 3 ⎠ 3 L

There are 5 revolutions in 3 seconds, so T = 3 5 s = 0.60 s 2 K = π 2 (12.0 kg) (2.00 m)2 (0.60 s) 2 = 877 J. 3 EVALUATE: If a point mass 12.0 kg is 2.00 m from the axis and rotates at the same rate as the bar, 2π (2.00 m) v= = 20.9 m/s and K = 12 mv 2 = 12 (12 kg)(20.9 m/s) 2 = 2.62 × 103 J . K for the bar is smaller by a factor of 0.60 s 0.33. The speed of a segment of the bar decreases toward the axis.

Figure 6.55 6.56.

IDENTIFY: Density is mass per unit volume, ρ = m / V , so we can calculate the mass of the asteroid. K = 12 mv 2 . Since the asteroid comes to rest, the kinetic energy it delivers equals its initial kinetic energy. 1 SET UP: The volume of a sphere is related to its diameter by V = π d 3 . 6

π

(320 m)3 = 1.72 × 107 m3 . m = ρV = (2600 kg/m3 )(1.72 × 107 m 3 ) = 4.47 × 1010 kg . 6 K = 12 mv 2 = 12 (4.47 × 1010 kg)(12.6 × 103 m/s) 2 = 3.55 × 1018 J .

EXECUTE:

(a) V =

3.55 × 1018 J = 56.5 devices . 6.28 × 1016 J EVALUATE: If such an asteroid were to hit the earth the effect would be catastrophic. IDENTIFY and SET UP: Since the forces are constant, Eq.(6.2) can be used to calculate the work done by each force. The forces on the suitcase are shown in Figure 6.57a. (b) The yield of a Castle/Bravo device is (1 s)(4.184 × 1015 J) = 6.28 × 1016 J .

6.57.

Figure 6.57a

In part (f ) , Eq.(6.6) is used to relate the total work to the initial and final kinetic energy. EXECUTE: (a) WF = ( F cos φ ) s " " Both F and s are parallel to the incline and in the same direction, so φ = 90° and WF = Fs = (140 N)(3.80 m) = 532 J


6-15

(b) The directions of the displacement and of the gravity force are shown in Figure 6.57b. Ww = ( w cos φ ) s φ = 115°, so Ww = (196 N)(cos115°)(3.80 m)

Ww = −315 J Figure 6.57b Alternatively, the component of w parallel to the incline is w sin 25°. This component is down the incline so its angle " with s is φ = 180°. Ww sin 25° = (196 Nsin 25°)(cos180°)(3.80 m) = −315 J. The other component of w, w cos 25°, is " perpendicular to s and hence does no work. Thus Ww = Ww sin 25° = −315 J, which agrees with the above. (c) The normal force is perpendicular to the displacement (φ = 90°), so Wn = 0. (d) n = w cos 25° so f k = μ k n = μ k w cos 25° = (0.30)(196 N)cos 25° = 53.3 N

W f = ( f k cos φ ) x = (53.3 N)(cos180°)(3.80 m) = −202 J (e) Wtot = WF + Ww + Wn + W f = +532 J − 315 J + 0 − 202 J = 15 J (f ) Wtot = K 2 − K1 , K1 = 0, so K 2 = Wtot 1 2

6.58.

mv22 = Wtot so v2 =

2Wtot 2(15 J) = = 1.2 m/s 20.0 kg m

EVALUATE: The total work done is positive and the kinetic energy of the suitcase increases as it moves up the incline. IDENTIFY: The work he does to lift his body a distance h is W = mgh . The work per unit mass is (W m) = gh. SET UP: The quantity gh has units of N/kg. EXECUTE: (a) The man does work, (9.8 N kg ) (0.4 m) = 3.92 J kg. (b) (3.92 J kg) (70 J kg ) × 100 = 5.6%. (c) The child does work (9.8 N kg)(0.2 m) = 1.96 J kg. (1.96 J kg) (70 J kg ) × 100 = 2.8%. (d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg

6.59.

instead of 3.92 J kg, the child should be able to do more chin-ups. EVALUATE: Since the child has arms half the length of his father’s arms, the child must lift his body only 0.20 m to do a chin-up. IDENTIFY: Apply the definitions of IMA and AMA given in the problem. SET UP: When the object moves a distance L along the ramp, it rises a vertical distance L sin α . 1 . EXECUTE: (a) sin = L, sout = L sin α, so IMA = sin α (b) If AMA = IMA, ( Fout Fin ) = ( sin sout ) and so ( Fout ) (sout ) = ( Fin ) (sin ) , or Wout = Win . (c) The pulley is sketched in Figure 6.59. W ( F )( s ) F F AMA (d) e = out = out out = out in = . Win ( Fin )( sin ) sin sout IMA EVALUATE:

Fin = w sin α and Fout = w . ( Fin )( sin ) = ( w sin α ) L . ( Fout )( sout ) = w(sin α L) . Therefore,

( Fin )( sin ) = ( Fout )( sout ) . A smaller force acting over a larger distance does the same amount of work as a larger force acting over a smaller distance.

6.60.

IDENTIFY: SET UP:

mB =

Apply

Figure 6.59 " F = m a to each block to find the tension in the string. Each force is constant and W = Fs cosφ . ∑

The free-body diagram for each block is given in Figure 6.60. m A =

12.0 N = 1.22 kg . g

20.0 N = 2.04 kg and g

6-16

Chapter 6

EXECUTE:

T − f k = m Aa . wB − T = mB a . wB − f k = ( mA + mB )a .

⎛ wB ⎞ ⎛ mA ⎞ ⎛ wA ⎞ fk = 0 . a = ⎜ ⎟ and T = wB ⎜ ⎟ = wB ⎜ ⎟ = 7.50 N . ⎝ mA + mB ⎠ ⎝ mA + mB ⎠ ⎝ wA + wB ⎠ 20.0 N block: Wtot = Ts = (7.50 N)(0.750 m) = 5.62 J . 12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 7.50 N)(0.750 m) = 3.38 J (b) f k = μ k wA = 6.50 N . a =

⎛ mA ⎞ ⎛ wA ⎞ wB − μ k wA . T = f k + ( wB − μ k wA ) ⎜ ⎟ = μ k wA + ( wB − μ k wA ) ⎜ ⎟. m A + mB ⎝ mA + mB ⎠ ⎝ wA + wB ⎠

T = 6.50 N + (5.50 N)(0.625) = 9.94 N . 20.0 N block: Wtot = (T − f k ) s = (9.94 N − 6.50 N)(0.750 m) = 2.58 J . 12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 9.94 N)(0.750 m) = 1.54 J . EVALUATE: Since the two blocks move with equal speeds, for each block Wtot = K 2 − K1 is proportional to the mass (or weight) of that block. With friction the gain in kinetic energy is less, so the total work on each block is less.

Figure 6.60 6.61.

IDENTIFY: K = 12 mv 2 . Find the speed of the shuttle relative to the earth and relative to the satellite. SET UP: Velocity is distance divided by time. For one orbit the shuttle travels a distance 2π R . 2

EXECUTE:

6.62.

(a)

2

⎛ 2π (6.66 × 106 m) ⎞ 1 2 1 ⎛ 2πR ⎞ 1 12 mv = m ⎜ ⎟ = 2.59 × 10 J. ⎟ = (86,400 kg) ⎜ 2 2 ⎝ T ⎠ 2 ⎝ (90.1 min) (60 s min) ⎠

(b) (1 2) mv 2 = (1 2) (86,400 kg) ((1.00 m) (3.00 s))2 = 4.80 × 103 J. EVALUATE: The kinetic energy of an object depends on the reference frame in which it is measured. IDENTIFY: W = Fs cos φ . Wtot = K 2 − K1 . SET UP: f k = μ k n . The normal force is n = mg cosθ , with θ = 12.0° . The component of the weight parallel to the incline is mg sin θ . EXECUTE:

(a) φ = 180° and W f = − f k s = −( μk mg cos θ ) s = −(0.31)(5.00 kg)(9.80 m s 2 )(cos 12.0°)(1.50 m) = −22.3 J

(b) (5.00 kg)(9.80 m s 2 )(sin12.0°)(1.50 m) = 15.3 J. (c) The normal force does no work. (d) Wtot = 15.3 J − 22.3 J = −7.0 J.

6.63.

(e) K 2 = K1 + Wtot = (1 2)(5.00 kg)(2.2 m s)2 − 7.0 J = 5.1 J , and so v2 = 2(5.1 J) /(5.00 kg) = 1.4 m / s . EVALUATE: Friction does negative work and gravity does positive work. The net work is negative and the kinetic energy of the object decreases. IDENTIFY: The effective force constant is defined by keff = F / x , where F is the force applied to each end of the spring combination and x is the amount the spring combination is stretched. SET UP: Consider a force F applied to each end of the combination. Then F1 and F2 are the forces applied to each

spring and F = F1 + F2 . Each spring stretches the same amount x. EXECUTE:

(a) F = keff x . F = F1 + F2 = k1 x + k2 x . Equating the two expressions for F gives keff = k1 + k2 .

(b) The same procedure as in part (a) gives keff = k1 + k2 + $ + k N . EVALUATE: The effective force constant of the configuration is greater than any of the force constants of the individual springs. More force is required to stretch the parallel combination that is required to stretch each separate spring the same amount.


6.64.

6-17

IDENTIFY: The effective force constant is defined by keff = F/x , where F is the force applied to each end of the spring combination and x is the amount the spring combination is stretched. SET UP: Consider a force F applied to each end of the combination. The same force F is applied to each spring. Spring 1 stretches a distance x1 and spring 2 stretches a distance x2 , where x1 = F / k1 and x2 = F / k2 . The total

distance the combination stretches is x = x1 + x2 . EXECUTE:

(a) x = x1 + x2 gives

F F F 1 1 1 = + . = = and keff k1 k2 keff k1 k2

(b) The same procedure as in part (a) gives

k1k2 . The effective force constant for k1 + k2 the two springs in series is less than the force constant for each individual spring. It takes less force to stretch the combination an amount x than to stretch either separate spring an amount x. IDENTIFY: Apply Eq.(6.7). dx 1 SET UP: ∫ 2 = − . x x EVALUATE:

6.65.

1 1 1 1 . = + +$+ keff k1 k2 kN

For two springs the result in part (a) can be written as keff =

x

2 ⎛1 1⎞ dx ⎡ 1⎤ = − k ⎢ − ⎥ = k ⎜ − ⎟ . The force is given to be attractive, so Fx < 0 , 2 x1 x1 x x ⎣ ⎦ x1 ⎝ x2 x1 ⎠ 1 1 < , and W < 0 . and k must be positive. If x2 > x1 , x2 x1 (b) Taking “slowly” to be constant speed, the net force on the object is zero. The force applied by the hand is ⎛1 1⎞ opposite Fx , and the work done is negative of that found in part (a), or k ⎜ − ⎟ , which is positive if x2 > x1 . ⎝ x1 x2 ⎠

EXECUTE:

6.66.

x2

(a) W = ∫ Fx dx = −k ∫

x2

(c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero. EVALUATE: Your force is directed away from the origin, so when the object moves away from the origin your force does positive work. IDENTIFY: Apply Eq.(6.6) to the motion of the asteroid. SET UP: Let point 1 be at a great distance and let point 2 be at the surface of the earth. Assume K1 = 0. From the information given about the gravitational force its magnitude as a function of distance r from the center of the earth must be F = mg ( RE / r ) 2 . This force is directed in the − rˆ direction since it is a “pull”. F is not constant so Eq.(6.7) must be used to calculate the work it does. 2 RE ⎛ mgR ⎞ 2 EXECUTE: W = − ∫ F ds = − ∫ ⎜ 2 E ⎟ dr = − mgRE2 −(1/ r ) ∞RE = mgRE 1 ∞ ⎝ r ⎠

(

)

Wtot = K 2 − K1 , K1 = 0 This gives K 2 = mgRE = 1.25 × 1012 J K 2 = 12 mv22 so v2 = 2 K 2 / m = 11,000 m/s EVALUATE: 6.67.

Note that v2 = 2 gRE , the impact speed is independent of the mass of the asteroid.

IDENTIFY: Calculate the work done by friction and apply Wtot = K 2 − K1 . Since the friction force is not constant, use Eq.(6.7) to calculate the work. SET UP: Let x be the distance past P. Since μ k increases linearly with x, μ k = 0.100 + Ax . When x = 12.5 m ,

μ k = 0.600 , so A = 0.500 /(12.5 m) = 0.0400 /m 1 (a) Wtot = ΔK = K 2 − K1 gives − ∫ μk mgdx = 0 − mv12 . Using the above expression for μ k , 2 2 x2 ⎡ ⎡ x ⎤ 1 x2 ⎤ 1 1 g ∫ (0.100 + Ax)dx = v12 and g ⎢(0.100) x2 + A 2 ⎥ = v12 . (9.80 m/s 2 ) ⎢ (0.100) xf + (0.0400/m) f ⎥ = (4.50 m/s)2 . 0 2 2⎦ 2 2⎦ 2 ⎣ ⎣

EXECUTE:

Solving for x2 gives x2 = 5.11 m . (b) μ k = 0.100 + (0.0400/m)(5.11 m) = 0.304

6-18

6.68.

Chapter 6

v2 (4.50 m/s) 2 1 (c) Wtot = K 2 − K1 gives − μk mgx2 = 0 − mv12 . x2 = 1 = = 10.3 m . 2 2 μk g 2(0.100)(9.80 m/s 2 ) EVALUATE: The box goes farther when the friction coefficient doesn’t increase. IDENTIFY: Use Eq.(6.7) to calculate W. SET UP: x1 = 0 . In part (a), x2 = 0.050 m . In part (b), x2 = −0.050 m . k 2 b 3 c 4 x2 − x2 + x2 . 2 3 4 2 2 3 3 4 W = (50.0 N / m) x2 − (233 N / m ) x2 + (3000 N / m ) x2 . When x2 = 0.050 m , W = 0.12 J . x2

x2

(a) W = ∫ Fdx = ∫ ( kx − bx 2 + cx 3 )dx =

EXECUTE:

0

0

(b) When x2 = −0.050 m, W = 0.17 J . (c) It’s easier to stretch the spring; the quadratic −bx 2 term is always in the − x -direction, and so the needed force, and hence the needed work, will be less when x2 > 0 .

When x = 0.050 m , Fx = 4.75 N . When x = −0.050 m , Fx = 8.25 N . " " IDENTIFY and SET UP: Use ∑ F = ma to find the tension force T. The block moves in uniform circular motion and " " a = arad . (a) The free-body diagram for the block is given in Figure 6.69. EVALUATE:

6.69.

EXECUTE:

T =m

∑F

x

= max

v2 R

T = (0.120 kg)

(0.70 m/s) 2 = 0.15 N 0.40 m

Figure 6.69 (b) T = m

v2 (2.80 m/s) 2 = (0.120 kg) = 9.4 N 0.10 m R

(c) SET UP:

x2

The tension changes as the distance of the block from the hole changes. We could use W = ∫ Fx dx to x1

calculate the work. But a much simpler approach is to use Wtot = K 2 − K1. EXECUTE:

The only force doing work on the block is the tension in the cord, so Wtot = WT . K1 = 12 mv12 = 12 (0.120 kg)(0.70 m/s) 2 = 0.0294 J K 2 = 12 mv22 = 12 (0.120 kg)(2.80 m/s) 2 = 0.470 J Wtot = K 2 − K1 = 0.470 J − 0.029 J = 0.44 J

6.70.

This is the amount of work done by the person who pulled the cord. EVALUATE: The block moves inward, in the direction of the tension, so T does positive work and the kinetic energy increases. IDENTIFY: Use Eq.(6.7) to find the work done by F. Then apply Wtot = K 2 − K1 . SET UP:

dx

∫x

2

1 =− . x

⎛1 1⎞ dx = α ⎜ − ⎟ . W = (2.12 × 10−26 N ⋅ m 2 )((0.200 m −1 ) − (1.25 × 109 m −1 )) = −2.65 × 10−17 J . ⎝ x1 x2 ⎠ Note that x1 is so large compared to x2 that the term 1/ x1 is negligible. Then, using Eq. (6.13)) and solving for v2 ,

EXECUTE:

W =∫

x2

α

x1

x2

v2 = v12 +

2W 2(−2.65 × 10−17 J) = (3.00 × 105 m/s) 2 + = 2.41 × 105 m/s. (1.67 × 10−27 kg) m

(b) With K 2 = 0, W = − K1 . Using W = −

x2 =

α K1

=

α x2

,

2α 2(2.12 × 10−26 N ⋅ m 2 ) = = 2.82 × 10−10 m. 2 mv1 (1.67 × 10−27 kg)(3.00 × 105 m/s) 2


6.71.

6-19

(c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00 × 105 m/s . EVALUATE: As the proton moves toward the uranium nucleus the repulsive force does negative work and the kinetic energy of the proton decreases. As the proton moves away from the uranium nucleus the repulsive force does positive work and the kinetic energy of the proton increases. " " " " IDENTIFY and SET UP: Use vx = dx/dt and ax = dvx /dt. Use ∑ F = ma to calculate F from a.

dx = 2α t + 3β t 2 dt t = 4.00 s: vx = 2(0.200 m/s 2 )(4.00 s) + 3(0.0200 m/s3 )(4.00 s) 2 = 2.56 m/s.

EXECUTE:

(a) x(t ) = α t 2 + β t 3 , vx (t ) =

dvx = 2α + 6 β t dt Fx = max = m(2α + 6 β t )

(b) ax (t ) =

t = 4.00 s: Fx = 6.00 kg(2(0.200 m/s 2 ) + 6(0.0200 m/s3 )(4.00 s)) = 5.28 N (c) IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work. EXECUTE: Wtot = K 2 − K1 At t1 = 0, v1 = 0 so K1 = 0. Wtot = WF K 2 = 12 mv22 = 12 (6.00 kg)(2.56 m/s) 2 = 19.7 J

6.72.

Then Wtot = K 2 − K1 gives that WF = 19.7 J EVALUATE: v increases with t so the kinetic energy increases and the work done is positive. We can also calculate WF directly from Eq.(6.7), by writing dx as vx dt and performing the integral. IDENTIFY: Since the capsule comes to rest, the amount of work the capsule does on the ground equals its original kinetic energy. Use constant acceleration kinematic equations to calculate the stopping time t; Δt = t . SET UP: 311 km/h = 86.4 m/s . Let + y be the direction the capsule is traveling before the crash. EXECUTE:

6.73.

ΔW = K1 = 12 mv12 = 12 (210 kg)(86.4 m/s) 2 = 7.84 × 105 J . y − y0 = 0.810 m , v0 y = 86.4 m/s and v y = 0 .

⎛ v + vy ⎞ 2( y − y0 ) 2(0.810 m) ΔW 7.84 × 105 J y − y0 = ⎜ 0 y = = 0.01875 s . = = 4.18 × 107 W ⎟ t gives t = v0 y 86.4 m/s 0.01875 s Δt ⎝ 2 ⎠ EVALUATE: A large amount of work is done in a very small amount of time. IDENTIFY and SET UP: Use Eq.(6.6). You do positive work and gravity does negative work. Let point 1 be at the base of the bridge and point 2 be at the top of the bridge. EXECUTE: (a) Wtot = K 2 − K1 K1 = 12 mv12 = 12 (80.0 kg)(5.00 m/s) 2 = 1000 J K 2 = 12 mv22 = 12 (80.0 kg)(1.50 m/s)2 = 90 J Wtot = 90 J − 1000 J = −910 J (b) Neglecting friction, work is done by you (with the force you apply to the pedals) and by gravity: Wtot = Wyou + Wgravity . The gravity force is w = mg = (80.0 kg)(9.80 m/s 2 ) = 784 N, downward. The displacement is 5.20 m, upward. Thus φ = 180° and Wgravity = ( F cos φ ) s = (784 N)(5.20 m)cos180° = −4077 J Then Wtot = Wyou + Wgravity gives Wyou = Wtot − Wgravity = −910 J − (−4077 J) = +3170 J

6.74.

EVALUATE: The total work done is negative and you lose kinetic energy. IDENTIFY: Use Eq.(6.7) to calculate W. 1 − ( n −1) SET UP: ∫ x − n dx = − x n −1 EXECUTE:

(a) W = ∫

∞ x0

b b dx = ( n − 1)) x n −1 xn

∞

= x0

b . Note that for this part, for n > 1, x1− n → 0 as x → ∞ . (n − 1) x0n −1

6-20

Chapter 6

⎡ b ⎤ (b) When 0 < n < 1 , the improper integral must be used, W = lim ⎢ ( x2n −1 − x0n −1 ) ⎥ , and because the exponent on x2 →∞ ( n − 1) ⎣ ⎦ the x2n −1 is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an

6.75.

infinite amount of work, even though F → 0 as x2 → ∞. EVALUATE: The work-energy theorem says that an object gains an infinite amount of kinetic energy when an infinite amount of work is done on it. IDENTIFY: The negative work done by the spring equals the change in kinetic energy of the car. SET UP: The work done by a spring when it is compressed a distance x from equilibrium is − 12 kx 2 . K 2 = 0 . EXECUTE:

6.76.

− 12 kx 2 = K 2 − K1 gives

1 2

kx 2 = 12 mv12 and

2 2 k = ( mv12 ) x 2 = ⎡⎣(1200 kg ) ( 0.65 m/s ) ⎤⎦ ( 0.070 m ) = 1.0 × 105 N/m . EVALUATE: When the spring is compressed, the spring force is directed opposite to the displacement of the object and the work done by the spring is negative. IDENTIFY: Apply Wtot = K 2 − K1 .

SET UP: Let x0 be the initial distance the spring is compressed. The work done by the spring is 12 kx02 − 12 kx 2 , where x is the final distance the spring is compressed. EXECUTE: (a) Equating the work done by the spring to the gain in kinetic energy, 12 kx02 = 12 mv 2 , so

v=

400 N / m k (0.060 m) = 6.93 m/s. x0 = 0.0300 kg m

(b) Wtot must now include friction, so

v=

1 2

mv 2 = Wtot = 12 kx02 − fx0 , where f is the magnitude of the friction force. Then,

400 N/m 2(6.00 N) k 2 2f (0.060 m) 2 − (0.060 m) = 4.90 m/s. x0 − x0 = 0.0300 kg (0.0300 kg) m m

(c) The greatest speed occurs when the acceleration (and the net force) are zero. Let x be the amount the spring is still f 6.00 N compressed, so the distance the ball has moved is x0 − x . kx = f , x = = = 0.0150 m . To find the speed, k 400 N/m

the net work is Wtot = 12 k ( x02 − x 2 ) − f ( x0 − x) , so the maximum speed is vmax = vmax =

6.77.

k 2 2f ( x0 − x 2 ) − ( x0 − x) . m m

400 N / m 2(6.00 N) ((0.060 m) 2 − (0.0150 m) 2 ) − (0.060 m − 0.0150 m) = 5.20 m/s (0.0300 kg) (0.0300 kg)

EVALUATE: The maximum speed with friction present (part (c)) is larger than the result of part (b) but smaller than the result of part (a). IDENTIFY and SET UP: Use Eq.(6.6). Work is done by the spring and by gravity. Let point 1 be where the textbook is released and point 2 be where it stops sliding. x2 = 0 since at point 2 the spring is neither stretched nor compressed. The situation is sketched in Figure 6.77. EXECUTE:

Wtot = K 2 − K1 K1 = 0, K 2 = 0 Wtot = Wfric + Wspr Figure 6.77

Wspr = 12 kx12 , where x1 = 0.250 m (Spring force is in direction of motion of block so it does positive work.) Wfric = − μ k mgd Then Wtot = K 2 − K1 gives

1 2

kx12 − μ k mgd = 0

(250 N/m)(0.250 m) 2 kx12 = = 1.1 m, measured from the point where the block was released. 2μ k mg 2(0.30)(2.50 kg)(9.80 m/s 2 ) EVALUATE: The positive work done by the spring equals the magnitude of the negative work done by friction. The total work done during the motion between points 1 and 2 is zero and the textbook starts and ends with zero kinetic energy. d=


6.78.

6-21

IDENTIFY: Apply Wtot = K 2 − K1 to the cat. SET UP: Let point 1 be at the bottom of the ramp and point 2 be at the top of the ramp. EXECUTE: The work done by gravity is Wg = −mgL sin θ (negative since the cat is moving up), and the work done

by the applied force is FL, where F is the magnitude of the applied force. The total work is Wtot = (100 N)(2.00 m) − (7.00 kg)(9.80 m/s 2 )(2.00 m)sin 30° = 131.4 J . The cat’s initial kinetic energy is

1 2

mv12 = 12 (7.00 kg)(2.40 m/s) 2 = 20.2 J , and

v2 =

2( K1 + W ) 2(20.2 J + 131.4 J) = = 6.58 m/s. (7.00 kg) m

EVALUATE: The net work done on the cat is positive and the cat gains speed. Without your push, Wtot = Wgrav = −68.6 J and the cat wouldn’t have enough initial kinetic energy to reach the top of the ramp. 6.79.

Apply Wtot = K 2 − K1 to the vehicle.

IDENTIFY: SET UP:

Call the bumper compression x and the initial speed v0 . The work done by the spring is − 12 kx 2 and

K2 = 0 . 1 2 1 2 kx = mv0 , kx < 5 mg . Combining to eliminate k and then x, the two 2 2 (20.0 m/s) 2 v2 mg 2 inequalities are x > and k < 25 2 . Using the given numerical values, x > = 8.16 m and 5g v 5(9.80 m/s 2 ) (a) The necessary relations are

EXECUTE:

(1700 kg)(9.80 m/s 2 ) 2 = 1.02 × 104 N/m. (20.0 m/s) 2 (b) A distance of 8 m is not commonly available as space in which to stop a car. Also, the car stops only momentarily and then returns to its original speed when the spring returns to its original length. EVALUATE: If k were doubled, to 2.04 × 104 N/m , then x = 5.77 m . The stopping distance is reduced by a factor of k < 25

1/ 2 , but the maximum acceleration would then be kx/m = 69.2 m/s 2 , which is 7.07g . 6.80.

IDENTIFY: Apply Wtot = K 2 − K1 . W = Fs cos φ . SET UP: The students do positive work, and the force that they exert makes an angle of 30.0° with the direction of motion. Gravity does negative work, and is at an angle of 120.0° with the chair’s motion, EXECUTE: The total work done is Wtot = ((600 N)cos30.0° + (85.0 kg)(9.80 m/s 2 )cos120.0°)(2.50 m) = 257.8 J ,

and so the speed at the top of the ramp is v2 = v12 + EVALUATE:

6.81.

2Wtot 2(257.8 J) = (2.00 m/s) 2 + = 3.17 m/s. (85.0 kg) m

The component of gravity down the incline is mg sin 30° = 417 N and the component of the push up

the incline is (600 N)cos30° = 520 N . The force component up the incline is greater than the force component down the incline, the net work done is positive and the speed increases. IDENTIFY: Apply Wtot = K 2 − K1 to the blocks. SET UP:

If X is the distance the spring is compressed, the work done by the spring is − 12 kX 2 . At maximum

compression, the spring (and hence the block) is not moving, so the block has no kinetic energy and x2 = 0 . EXECUTE: (a) The work done by the block is equal to its initial kinetic energy, and the maximum compression is 5.00 kg m found from 12 kX 2 = 12 mv02 and X = (6.00 m/s) = 0.600 m. v= 500 N/m k (b) Solving for v0 in terms of a known X, v0 = 6.82.

500 N/m k (0.150 m) = 1.50 m/s. X= 5.00 kg m

EVALUATE: The negative work done by the spring removes the kinetic energy of the block. IDENTIFY: Apply Wtot = K 2 − K1 to the system of the two blocks. The total work done is the sum of that done by gravity (on the hanging block) and that done by friction (on the block on the table). SET UP: Let h be the distance the 6.00 kg block descends. The work done by gravity is (6.00 kg)gh and the work done by friction is − μk (8.00 kg)gh .

6-22

Chapter 6

EXECUTE:

6.83.

Wtot = (6.00 kg − (0.25)(8.00 kg)) (9.80 m/s 2 ) (1.50 m) = 58.8 J. This work increases the kinetic energy

2(58.8 J) 1 of both blocks: Wtot = ( m1 + m2 )v 2 , so v = = 2.90 m/s. 2 (14.00 kg) EVALUATE: Since the two blocks are connected by the rope, they move the same distance h and have the same speed v. IDENTIFY and SET UP: Apply Wtot = K 2 − K1 to the system consisting of both blocks. Since they are connected by the cord, both blocks have the same speed at every point in the motion. Also, when the 6.00-kg block has moved downward 1.50 m, the 8.00-kg block has moved 1.50 m to the right. The target variable, μ k , will be a factor in the work done by friction. The forces on each block are shown in Figure 6.83.

EXECUTE:

K1 = 12 mAv12 + 12 mB v12 = 12 (mA + mB )v12

K2 = 0

Figure 6.83

The tension T in the rope does positive work on block B and the same magnitude of negative work on block A, so T does no net work on the system. Gravity does work Wmg = mA gd on block A, where d = 2.00 m. (Block B moves horizontally, so no work is done on it by gravity.) Friction does work Wfric = − μ k mB gd on block B. Thus Wtot = Wmg + Wfric = mA gd − μ k mB gd . Then Wtot = K 2 − K1 gives mA gd − μ k mB gd = − 12 (mA + mB )v12 and

μk =

mA 12 (mA + mB )v12 6.00 kg (6.00 kg + 8.00 kg)(0.900 m/s) 2 + = + = 0.786 mB mB gd 8.00 kg 2(8.00 kg)(9.80 m/s 2 )(2.00 m)

EVALUATE: The weight of block A does positive work and the friction force on block B does negative work, so the net work is positive and the kinetic energy of the blocks increases as block A descends. Note that K1 includes the

6.84.

kinetic energy of both blocks. We could have applied the work-energy theorem to block A alone, but then Wtot includes the work done on block A by the tension force. IDENTIFY: Apply Wtot = K 2 − K1 . The work done by the force from the bow is the area under the graph of Fx versus the draw length. SET UP: One possible way of estimating the work is to approximate the F versus x curve as a parabola which goes to zero at x = 0 and x = x0 , and has a maximum of F0 at x = x0 / 2 , so that F ( x) = (4 F0 / x02 ) x( x0 − x). This may seem like a crude approximation to the figure, but it has the advantage of being easy to integrate. x0 4 F x0 4 F ⎛ x 2 x3 ⎞ 2 EXECUTE: ∫ Fdx = 20 ∫ ( x0 x − x 2 ) dx = 20 ⎜ x0 0 − 0 ⎟ = F0 x0 . With F0 = 200 N and x0 = 0.75 m, 0 x0 0 x0 ⎝ 2 3 ⎠ 3 W = 100 J. The speed of the arrow is then

6.85.

2W 2(100 J) = = 89 m/s . m (0.025 kg)

EVALUATE: We could alternatively represent the area as that of a rectangle 180 N by 0.55 m. This gives W = 99 J , in close agreement with our more elaborate estimate. IDENTIFY: Apply Eq.(6.6) to the skater. SET UP: Let point 1 be just before she reaches the rough patch and let point 2 be where she exits from the patch. Work is done by friction. We don’t know the skater’s mass so can’t calculate either friction or the initial kinetic energy. Leave her mass m as a variable and expect that it will divide out of the final equation. EXECUTE: f k = 0.25mg so W f = Wtot = −(0.25mg ) s, where s is the length of the rough patch.

Wtot = K 2 − K1

K1 = 12 mv02 , K 2 = 12 mv22 = 12 m(0.45v0 ) 2 = 0.2025 ( 12 mv02 )

The work-energy relation gives −(0.25mg ) s = ( 0.2025 − 1) 12 mv02 The mass divides out, and solving gives s = 1.5 m. EVALUATE: Friction does negative work and this reduces her kinetic energy.


6.86.

Pav = F!vav . Use F = ma to calculate the force.

IDENTIFY: SET UP:

6-23

vav =

0 + 6.00 m/s = 3.00 m/s 2

Your friend’s average acceleration is a =

EXECUTE:

v − v0 6.00 m/s = = 2.00 m/s 2 . Since there are no other 3.00 s t

horizontal forces acting, the force you exert on her is given by Fnet = ma = (65.0 kg)(2.00 m/s 2 ) = 130 N . Pav = (130 N)(3.00 m/s) = 390 W . EVALUATE:

We could also use the work-energy theorem: W = K 2 − K1 = 12 (65.0 kg)(6.00 m/s)2 = 1170 J .

W 1170 J = = 390 W , the same as obtained by our other approach. t 3.00 s IDENTIFY: To lift a mass m a height h requires work W = mgh . To accelerate mass m from rest to speed v requires Pav =

6.87.

W = K 2 − K1 = 12 mv 2 . Pav =

ΔW . Δt

SET UP: t = 60 s EXECUTE: (a) (800 kg)(9.80 m/s 2 )(14.0 m) = 1.10 × 105 J (b) (1/ 2)(800 kg)(18.0 m/s 2 ) = 1.30 × 105 J.

1.10 × 105 J + 1.30 × 105 J = 3.99 kW. 60 s EVALUATE: Approximately the same amount of work is required to lift the water against gravity as to accelerate it to its final speed. IDENTIFY: P = F!v and F! = ma . (c)

6.88.

SET UP: EXECUTE:

6.89.

From Problem 6.71, v = 2α t + 3β t 2 and a = 2α + 6β t . P = F!v = mav = m(2α + 6 βt )(2α t + 3 βt 2 ) = m(4α 2t + 18α βt 2 + 18 β 2t 3 ) .

P = (0.96 N/s)t + (0.43 N/s 2 )t 2 + (0.043 N/s3 )t 3 . At t = 4.00 s, the power output is 13.5 W. EVALUATE: P increases in time because v increase and because a increases. IDENTIFY and SET UP: Energy is Pavt. The total energy expended in one day is the sum of the energy expended in each type of activity. EXECUTE: 1 day = 8.64 × 104 s Let twalk be the time she spends walking and tother be the time she spends in other activities; tother = 8.64 × 104 s − t walk . The energy expended in each activity is the power output times the time, so E = Pt = (280 W)t walk + (100 W)tother = 1.1 × 107 J (280 W)twalk + (100 W)(8.64 × 104 s − twalk ) = 1.1 × 107 J (180 W)twalk = 2.36 × 106 J t walk = 1.31 × 104 s = 218 min = 3.6 h.

6.90.

6.91.

EVALUATE: Her average power for one day is (1.1 × 107 J)/([24][3600 s]) = 127 W. This is much closer to her 100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate. IDENTIFY and SET UP: W = Pt EXECUTE: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat. (b) The steady output of the athlete is (500 W)/(70 kg) = 7 W/kg, which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could stay aloft for very long. EVALUATE: Movies of early attempts at human-powered flight bear out our results. IDENTIFY and SET UP: Use Eq.(6.15). The work done on the water by gravity is mgh, where h = 170 m. Solve for the mass m of water for 1.00 s and then calculate the volume of water that has this mass.

6-24

Chapter 6

ΔW and 92% of the work done on the water Δt by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is P Δt (2.00 × 109 W)(1.00 s) W = av = = 2.174 × 109 J 0.92 0.92 W = mgh, so the mass of water flowing over the dam in 1.00 s must be EXECUTE:

m=

2.174 × 109 J W = = 1.30 × 106 kg gh (9.80 m/s 2 )(170 m)

density =

6.92.

The power output is Pav = 2000 MW = 2.00 × 109 W. Pav =

1.30 × 106 kg m m so V = = = 1.30 × 103 m3 . V density 1.00 × 103 kg/m 3

EVALUATE: The dam is 1270 m long, so this volume corresponds to about a m3 flowing over each 1 m length of the dam, a reasonable amount. W dv and W = 12 mv 2 , if the object starts from rest. a = and x − x0 = ∫ vdt . IDENTIFY: P = t dt d 1/ 2 1 −1/ 2 t = 2t SET UP: . ∫ t1/ 2 dt = 32 t 3 / 2 . dt EXECUTE: (b) a =

(a) The power P is related to the speed by Pt = K = 12 mv 2 , so v =

2Pt . m

2P d 2P 1 dv d 2 Pt P . = = t= = 2mt dt dt m m dt m 2 t

(c) x − x0 = ∫ v dt = EVALUATE:

2 P 12 2 P 2 32 8P 23 t dt = t = t . ∫ m m 3 9m

v, a, and x − x0 at a particular time are all proportional to P1/ 2 . The result in part (b) could also be

P . vm IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by the heart in one day. For part (b) use W calculated in part (a) in Eq.(6.15). EXECUTE: (a) W = mgh, as in Example 6.11. We need the mass of blood lifted; we are given the volume

obtained from P = Fv and a = F / m , so a = 6.93.

⎛ 1 × 10−3 m3 ⎞ 3 V = (7500 L) ⎜ ⎟ = 7.50 m . 1L ⎝ ⎠ m = density × volume = (1.05 × 103 kg/m3 )(7.50 m3 ) = 7.875 × 103 kg Then W = mgh = (7.875 × 103 kg)(9.80 m/s 2 )(1.63 m) = 1.26 × 105 J. 1.26 × 105 J ΔW = = 1.46 W. (24 h)(3600 s/h) Δt EVALUATE: Compared to light bulbs or common electrical devices, the power output of the heart is rather small. IDENTIFY: P = F!v = Mav . To overcome gravity on a slope that is at an angle α above the horizontal, P = (Mg sin α )v. (b) Pav = 6.94.

SET UP: 1 MW = 106 W . 1 kN = 103 N . When α is small, tan α ≈ sin α . EXECUTE: (a) The number of cars is the total power available divided by the power needed per car, 13.4 × 106 W = 177, rounding down to the nearest integer. (2.8 × 103 N)(27 m/s) (b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav. To climb a hill of angle α , the extra power needed is (Mg sin α )v. This will be nearly the same if a ~ g sin α ; if

g sin α ~ g tan α ~ 0.10 m/s 2 , the power is about the same as that needed to accelerate at 0.10 m/s 2 . (c) P = ( Mg sin α )v , where M is the total mass of the diesel units. P = (1.10 × 106 kg)(9.80 m/s 2 )(0.010)(27 m/s) = 2.9 MW. (d) The power available to the cars is 13.4 MW, minus the 2.9 MW needed to maintain the speed of the diesel units 13.4 × 106 W − 2.9 × 106 W on the incline. The total number of cars is then = 36, (2.8 × 103 N + (8.2 × 104 kg)(9.80 m/s 2 )(0.010))(27 m/s) rounding to the nearest integer.


6.95.

6-25

EVALUATE: For a single car, Mg sin α = (8.2 × 104 kg)(9.80 m/s 2 )(0.010) = 8.0 × 103 N , which is over twice the 2.8 kN required to pull the car at 27 m/s on level tracks. Even a slope as gradual as 1.0% greatly increases the power requirements, or for constant power greatly decreases the number of cars that can be pulled. IDENTIFY: P = F!v . The force required to give mass m an acceleration a is F = ma . For an incline at an angle

α above the horizontal, the component of mg down the incline is mg sin α . SET UP: For small α , sin α ≈ tan α . EXECUTE: (a) P0 = Fv = (53 × 103 N)(45 m/s) = 2.4 MW. (b) P1 = mav = (9.1 × 105 kg)(1.5 m/s 2 )(45 m/s) = 61 MW. (c) Approximating sinα , by tanα , and using the component of gravity down the incline as mgsinα ,

P2 = (mgsinα )v = (9.1 × 105 kg)(9.80 m/s 2 )(0.015)(45 m/s) = 6.0 MW. EVALUATE: From Problem 6.94, we would expect that a 0.15 m/s 2 acceleration and a 1.5% slope would require the same power. We found that a 1.5 m/s 2 acceleration requires ten times more power than a 1.5% slope, which is consistent. 6.96.


x2

W = ∫ Fx dx , and Fx depends on both x and y. x1

In each case, use the value of y that applies to the specified path. (a) Along this path, y is constant, with the value y = 3.00 m .

∫ xdx =

1 2

x2 .

∫ x dx = 2

1 3

x3

x2 ⎛ 2.00 m ⎞ W = αy ∫ xdx = (2.50 N/m 2 )(3.00 m) ⎜ ⎟ = 15.0 J , since x1 = 0 and x2 = 2.00 m . x1 ⎝ 2 ⎠ (b) Since the force has no y-component, no work is done moving in the y-direction. (c) Along this path, y varies with position along the path, given by y = 1.5 x, so Fx = α (1.5 x) x = 1.5α x 2 , and

(2.00 m)3 = 10.0 J. x1 x1 3 EVALUATE: The force depends on the position of the object along its path. IDENTIFY and SET UP: Use Eq.(6.18) to relate the forces to the power required. The air resistance force is Fair = 12 CAρ v 2 , where C is the drag coefficient. x2

x2

W = ∫ Fdx = 1.5α ∫ x 2 dx = 1.5(2.50 N/m 2 )

6.97.

EXECUTE:

(a) P = Ftot v, with Ftot = Froll + Fair

Fair = CAρ v 2 = 12 (1.0)(0.463 m3 )(1.2 kg/m3 )(12.0 m/s) 2 = 40.0 N 1 2

Froll = μr n = μr w = (0.0045)(490 N + 118 N) = 2.74 N P = ( Froll + Fair )v = (2.74 N + 40.0 N)(12.0 s) = 513 W (b) Fair = 12 CAρ v 2 = 12 (0.88)(0.366 m3 )(1.2 kg/m3 )(12.0 m/s) 2 = 27.8 N

Froll = μr n = μr w = (0.0030)(490 N + 88 N) = 1.73 N P = ( Froll + Fair )v = (1.73 N + 27.8 N)(12.0 s) = 354 W (c) Fair = 12 CAρ v 2 = 12 (0.88)(0.366 m3 )(1.2 kg/m3 )(6.0 m/s) 2 = 6.96 N

Froll = μr n = 1.73 N (unchanged) P = ( Froll + Fair )v = (1.73 N + 6.96 N)(6.0 s) = 52.1 W

6.98.

EVALUATE: Since Fair is proportional to v 2 and P = Fv, reducing the speed greatly reduces the power required. IDENTIFY: P = F!v

1 m/s = 3.6 km/h 28.0 × 103 W P EXECUTE: (a) F = = = 1.68 × 103 N. v (60.0 km/h)((1 m/s)/(3.6 km/h)) (b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35/ 4), and so the power at the lower speed is (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp. (c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 + 0.35 × 4) = 114.8 kW = 154 hp. EVALUATE: At low speeds rolling friction dominates the power requirement but at high speeds air resistance dominates. SET UP:

6-26

6.99.

Chapter 6

IDENTIFY and SET UP: Use Eq.(6.18) to relate F and P. In part (a), F is the retarding force. In parts (b) and (c), F includes gravity. EXECUTE: (a) P = Fv, so F = P / v.

⎛ 746 W ⎞ P = (8.00 hp) ⎜ ⎟ = 5968 W ⎝ 1 hp ⎠ ⎛ 1000 m ⎞⎛ 1 h ⎞ v = (60.0 km/h) ⎜ ⎟⎜ ⎟ = 16.67 m/s ⎝ 1 km ⎠⎝ 3600 s ⎠ P 5968 W F= = = 358 N. v 16.67 m/s (b) The power required is the 8.00 hp of part (a) plus the power Pg required to lift the car against gravity. The situation is sketched in Figure 6.99. 10 m = 0.10 100 m α = 5.71° tan α =

Figure 6.99 The vertical component of the velocity of the car is v sin α = (16.67 m/s)sin 5.71° = 1.658 m/s.

Then Pg = F (v sin a) = mgv sin α = (1800 kg)(9.80 m/s 2 )(1.658 m/s) = 2.92 × 104 W ⎛ 1 hp ⎞ Pg = 2.92 × 104 W ⎜ ⎟ = 39.1 hp ⎝ 746 W ⎠ The total power required is 8.00 hp + 39.1 hp = 47.1 hp. (c) The power required from the engine is reduced by the rate at which gravity does positive work. The road incline angle α is given by tan α = 0.0100, so α = 0.5729°. Pg = mg (v sin α ) = (1800 kg)(9.80 m/s 2 )(16.67 m/s)sin 0.5729° = 2.94 × 103 W = 3.94 hp. The power required from the engine is then 8.00 hp − 3.94 hp = 4.06 hp. (d) No power is needed from the engine if gravity does work at the rate of Pg = 8.00 hp = 5968 W

Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in parts (a), (b), and (c). IDENTIFY: Apply Wtot = K 2 − K1 to relate the initial speed v0 to the distance x along the plank that the box moves before coming to rest. SET UP: The component of weight down the incline is mg sin α , the normal force is mg cos α and the friction force is f = μ mg cos α . Pg = mgv sin α , so sin α =

6.100.

x

EXECUTE:

1 ΔK = 0 − mv02 and W = ∫ ( −mg sin α − μmg cos α )dx. Then, 2 0 x ⎡ ⎤ Ax 2 cos α ⎥ . W = − mg ∫ (sin α + Ax cos α )dx, W = −mg ⎢sin α x + 2 ⎣ ⎦ 0

⎡ ⎤ 1 Ax 2 Set W = ΔK : − mv02 = −mg ⎢sin α x + cos α ⎥ . To eliminate x, note that the box comes to a rest when the force of 2 2 ⎣ ⎦ static friction balances the component of the weight directed down the plane. So, mg sin α = Ax mg cos α . Solve this for x and substitute into the previous equation: x =

2 ⎡ ⎤ sin α 1 sin α 1 ⎛ sin α ⎞ . Then, v02 = + g ⎢sin α + A⎜ ⎟ cos α ⎥ , and A cos α 2 A cos α 2 ⎝ A cos α ⎠ ⎣⎢ ⎦⎥

upon canceling factors and collecting terms, v02 =

3g sin 2 α 3g sin 2 α . The box will remain stationary whenever v02 ≥ . A cos α A cos α


EVALUATE: 6.101.

6-27

If v0 is too small the box stops at a point where the friction force is too small to hold the box in place.

sin α increases and cos α decreases as α increases, so the v0 required increases as α increases. IDENTIFY: In part (a) follow the steps outlined in the problem. For parts (b), (c) and (d) apply the work-energy theorem. SET UP: ∫ x 2 dx = 13 x3

EXECUTE: (a) Denote the position of a piece of the spring by l; l = 0 is the fixed point and l = L is the moving end of the spring. Then the velocity of the point corresponding to l, denoted u, is u (l ) = v(I/L) (when the spring is moving, l will be a function of time, and so u is an implicit function of time). The mass of a piece of length dl is dm = ( M/L)dl ,

1 1 Mv 2 2 Mv 2 L Mv 2 . and so dK = (dm)u 2 = l dl , and K = ∫ dK = 3 ∫ l 2 dl = 3 0 2L 6 2 2 L (b) 12 kx 2 = 12 mv 2 , so v = (k/m) x = (3200 N/m)/(0.053 kg)(2.50 × 10−2 m) = 6.1 m/s. (c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so 12 kx 2 = 12 mv 2 + 16 Mv 2 . Solving for v,

v= (d) Algebraically,

1 2 (1/2)kx 2 1 (1/2) kx 2 mv = = 0.40 J and Mv 2 = = 0.60 J. 2 (1 + M/3m) 6 (1 + 3m/M )

⎛ 0.053 kg ⎞ K ball 3m = = 3⎜ ⎟ = 0.65 . The percentage of the final kinetic energy K spring M ⎝ 0.243 kg ⎠ that ends up with each object depends on the ratio of the masses of the two objects. As expected, when the mass of the spring is a small fraction of the mass of the ball, the fraction of the kinetic energy that ends up in the spring is small. IDENTIFY: In both cases, a given amount of fuel represents a given amount of work W0 that the engine does in EVALUATE:

6.102.

k (3200 N/m) (2.50 × 10−2 m) = 3.9 m/s. x= (0.053 kg) + (0.243 kg)/3 m + M/3

For this ball and spring,

moving the plane forward against the resisting force. Write W0 in terms of the range R and speed v and in terms of the time of flight T and v. SET UP: In both cases assume v is constant, so W0 = RF and R = vT . EXECUTE:

β⎞ ⎛ In terms of the range R and the constant speed v, W0 = RF = R ⎜ α v 2 + 2 ⎟ . v ⎝ ⎠

β⎞ ⎛ In terms of the time of flight T , R = vt , so W0 = vTF = T ⎜ α v 3 + ⎟ . v⎠ ⎝ (a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, setting dW0 dR dF dR dF = 0 to obtain F+R = 0. For the maximum range, = 0, so = 0. Performing the differentiation, dv dv dv dv dv dF = 2α v − 2 β/v3 = 0, which is solved for dv 14

⎛β ⎞ v=⎜ ⎟ ⎝α ⎠

14

⎛ 3.5 × 105 N ⋅ m 2 /s 2 ⎞ =⎜ ⎟ = 32.9 m/s = 118 km/h. 2 2 ⎝ 0.30 N ⋅ s /m ⎠ d (b) Similarly, the maximum time is found by setting ( Fv) = 0; performing the differentiation, 3α v 2 − β /v 2 = 0 . dv 1/ 4

⎛ β ⎞ v=⎜ ⎟ ⎝ 3α ⎠

EVALUATE:

1/ 4

⎛ 3.5 × 105 N ⋅ m 2 /s 2 ⎞ =⎜ 2 2 ⎟ ⎝ 3(0.30 N ⋅ s /m ) ⎠

= 25 m/s = 90 km/h.

When v = ( β /α )1/ 4 , Fair has its minimum value Fair = 2 αβ . For this v, R1 = (0.50)

T1 = (0.50)α −1/ 4 β −3/ 4 . When v = ( β /3α )1/ 4 , Fair = 2.3 αβ . For this v, R2 = (0.43)

6.103.

W0

αβ

W0

αβ

and

and T2 = (0.57)α −1/ 4 β −3/ 4 .

R1 > R2 and T2 > T1 , as they should be. IDENTIFY: For each speed, calculate the time. Then use the graph to find the oxygen consumption and from that the energy consumption. SET UP: t = d/v

6-28

Chapter 6

EXECUTE: (a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate appears to be about 12 cm3 /kg ⋅ min, and so the total energy is

(12 cm3/kg ⋅ min) (70 kg) (12 min) (20 J/cm3 ) = 2.0 × 105 J. (b) The run will take 6 min. Using an estimation of the rate from the graph of about 33 cm3 /kg ⋅ min gives an energy

6.104.

consumption of about 2.8 × 105 J. (c) The run takes 4 min, and with an estimated rate of about 50 cm3 /kg ⋅ min, the energy used is about 2.8 × 105 J. (d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about 5 km/h. EVALUATE: In an exercise program, for a fixed distance, running burns more energy than walking. IDENTIFY: Write equations similar to (6.11) for each component. Eq.(6.12) will now involve the sum of three integrals, one for each component. SET UP: v 2 = vx2 + v y2 + vz2 " " EXECUTE: From F = ma , Fx = max , Fy = ma y and Fz = maz . The generalization of Eq. (6.11) is then a x = vx

dv dvx dv , a y = v y y , az = vz z . The total work is then dx dy dz

y2 z2 dv ⎛ x2 dv ⎞ dv Fx dx + Fy dy + Fz dz = m ⎜ ∫ vx x dx + ∫ v y y dy + ∫ vz z dz ⎟ . x1 y1 z1 dx dy dz ⎝ ⎠ vx 2 vy 2 vz 2 1 1 1 Wtot = m ⎛⎜ ∫ vx dvx + ∫ v y dv y + ∫ vz dvz ⎞⎟ = m(vx22 − vx21 + v y2 2 − v y21 + vz22 − vz21 = mv22 − mv12 . v v v x 1 y 1 z 1 ⎝ ⎠ 2 2 2 " " EVALUATE: F and dl are vectors and have components. W and K are scalars and we never speak of their components.

Wtot = ∫

( x2 , y2 , z2 )

( x1 , y1 , z1 )

POTENTIAL ENERGY AND ENERGY CONSERVATION

7.1.

U grav = mgy so ΔU grav = mg ( y2 − y1 )

IDENTIFY: SET UP:

7

+ y is upward.

EXECUTE:

(a) ΔU = (75 kg)(9.80 m/s 2 )(2400 m − 1500 m) = +6.6 × 105 J

(b) ΔU = (75 kg)(9.80 m/s 2 )(1350 m − 2400 m) = −7.7 × 105 J EVALUATE: 7.2.

7.3.

IDENTIFY:

U grav increases when the altitude of the object increases. ! ! Apply ∑ F = ma to the sack to find the force. W = Fs cos φ .

SET UP: The lifting force acts in the same direction as the sack’s motion, so φ = 0° EXECUTE: (a) For constant speed, the net force is zero, so the required force is the sack’s weight, (5.00 kg)(9.80 m/s 2 ) = 49.0 N. (b) W = (49.0 N) (15.0 m) = 735 J . This work becomes potential energy. EVALUATE: The results are independent of the speed. IDENTIFY: Use the free-body diagram for the bag and Newton's first law to find the force the worker applies. Since the bag starts and ends at rest, K 2 − K1 = 0 and Wtot = 0 .

2.0 m A sketch showing the initial and final positions of the bag is given in Figure 7.3a. sin φ = and 3.5 m ! φ = 34.85° . The free-body diagram is given in Figure 7.3b. F is the horizontal force applied by the worker. In the calculation of U grav take + y upward and y = 0 at the initial position of the bag. SET UP:

EXECUTE:

(a)

∑F

y

= 0 gives T cos φ = mg and

∑F

x

= 0 gives F = T sin φ . Combining these equations to

eliminate T gives F = mg tan φ = (120 kg)(9.80 m/s ) tan 34.85° = 820 N . (b) (i) The tension in the rope is radial and the displacement is tangential so there is no component of T in the direction of the displacement during the motion and the tension in the rope does no work. (ii) Wtot = 0 so 2

Wworker = −Wgrav = U grav,2 − U grav,1 = mg ( y2 − y1 ) = (120 kg)(9.80 m/s 2 )(0.6277 m) = 740 J . EVALUATE: The force applied by the worker varies during the motion of the bag and it would be difficult to calculate Wworker directly.

Figure 7.3 7.4.

IDENTIFY: Only gravity does work on him from the point where he has just left the board until just before he enters the water, so Eq.(7.4) applies. SET UP: Let point 1 be just after he leaves the board and point 2 be just before he enters the water. + y is upward and y = 0 at the water. 7-1

7-2

Chapter 7

EXECUTE:

(a) K1 = 0 . y2 = 0 . y1 = 3.25 m . K1 + U grav,1 = K 2 + U grav,2 gives U grav,1 = K 2 and mgy1 = 12 mv22 .

v2 = 2 gy1 = 2(9.80 m/s 2 )(3.25 m) = 7.98 m/s . (b) v1 = 2.50 m/s , y2 = 0 , y1 = 3.25 m . K1 + U grav,1 = K 2 and

1 2

mv12 + mgy1 = 12 mv22 .

v2 = v12 + 2 gy1 = (2.50 m/s) 2 + 2(9.80 m/s 2 )(3.25 m) = 8.36 m/s .

7.5.

(c) v1 = 2.5 m/s and v2 = 8.36 m/s , the same as in part (b). EVALUATE: Kinetic energy depends only on the speed, not on the direction of the velocity. IDENTIFY and SET UP: Use energy methods. (a) K1 + U1 + Wother = K 2 + U 2 . Solve for K 2 and then use K 2 = 12 mv22 to obtain v2 .

Wother = 0 (The only force on the ball while it is in the air is gravity.) K1 = 12 mv12 ; K 2 = 12 mv22 U1 = mgy1 , y1 = 22.0 m U 2 = mgy2 = 0, since y2 = 0 for our choice of coordinates. Figure 7.5 EXECUTE:

1 2

mv + mgy1 = 12 mv22 2 1

v2 = v12 + 2 gy1 = (12.0 m/s) 2 + 2(9.80 m/s 2 )(22.0 m) = 24.0 m/s EVALUATE: The projection angle of 53.1° doesn’t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity. (b) Nothing changes in the calculation. The expression derived in part (a) for v2 is independent of the angle, so

7.6.

v2 = 24.0 m/s, the same as in part (a). (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect. IDENTIFY: The normal force does no work, so only gravity does work and Eq.(7.4) applies. SET UP: K1 = 0 . The crate’s initial point is at a vertical height of d sin α above the bottom of the ramp. EXECUTE:

(a) y2 = 0, y1 = d sin α . K1 + U grav,1 = K 2 + U grav,2 gives U grav,1 = K 2 . mgd sin α = 12 mv22 and

v2 = 2 gd sin α . (b) y1 = 0 , y2 = − d sin α . K1 + U grav,1 = K 2 + U grav,2 gives 0 = K 2 + U grav,2 . 0 = 12 mv22 + (− mgd sin α ) and

v2 = 2 gd sin α , the same as in part (a). (c) The normal force is perpendicular to the displacement and does no work. EVALUATE: When we use U grav = mgy we can take any point as y = 0 but we must take + y to be upward. 7.7.

IDENTIFY: SET UP:

As in Example 7.6, K 2 = 0, U 2 = 94 J, and U 3 = 0.

EXECUTE:

7.8.

Apply Eq.(7.7) to points 2 and 3. Take results from Example 7.6. Wother = − fs, the work done by friction.

The work done by friction is −(35 N) (1.6 m) = −56 J . K 3 = 38 J, and v3 =

2(38 J) = 2.5 m/s. 12 kg

EVALUATE: The value of v3 we obtained is the same as calculated in Example 7.6. For the motion from point 2 to point 3, gravity does positive work, friction does negative work and the net work is positive. IDENTIFY and SET UP: Apply Eq.(7.7) and consider how each term depends on the mass. EXECUTE: The speed is v and the kinetic energy is 4K. The work done by friction is proportional to the normal force, and hence to the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four times the mass, the kinetic energy is quadrupled. ! ! EVALUATE: The same result is obtained if we apply ∑ F = ma to the motion. Each force is proportional to m

and m divides out, so a is independent of m.

Potential Energy and Energy Conservation

7.9.

IDENTIFY: SET UP:

7-3

Wtot = K B − K A . The forces on the rock are gravity, the normal force and friction.

Let y = 0 at point B and let + y be upward. y A = R = 0.50 m . The work done by friction is negative;

W f = −0.22 J . K A = 0 . The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of the rock at this point is arad = v 2 / R , upward. EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work. (ii) Wgrav = U grav, A − U grav, B = mgy A = (0.20 kg)(9.80 m/s 2 )(0.50 m) = 0.98 J . (b) Wtot = Wn + W f + Wgrav = 0 + (−0.22 J) + 0.98 J = 0.76 J . Wtot = K B − K A gives

vB =

1 2

mvB2 = Wtot .

2Wtot 2(0.76 J) = = 2.8 m/s . m 0.20 kg

(c) Gravity is constant and equal to mg. n is not constant; it is zero at A and not zero at B. Therefore, f k = μ k n is also not constant. (d) ∑ Fy = ma y applied to Figure 7.9 gives n − mg = marad .

⎛ ⎛ [2.8 m/s]2 ⎞ v2 ⎞ n = m ⎜ g + ⎟ = (0.20 kg) ⎜ 9.80 m/s 2 + ⎟ = 5.1 N . 0.50 m ⎠ R⎠ ⎝ ⎝ EVALUATE: In the absence of friction, the speed of the rock at point B would be 2 gR = 3.1 m/s . As the rock slides through point B, the normal force is greater than the weight mg = 2.0 N of the rock.

Figure 7.9 7.10.

IDENTIFY: Only gravity does work, so Eq.(7.4) applies. SET UP: Let point 1 be just after the rock leaves the thrower and point 2 be at the maximum height. Let y1 = 0 and + y be upward. v1 = v0 . At the highest point, v2 = v0 cosθ . sin 2 θ + cos 2 θ = 1 . EXECUTE:

7.11.

K1 + U grav,1 = K 2 + U grav,2 gives

1 2

mv02 = 12 m(v0 cosθ ) 2 + mgy2 . y2 =

v02 v 2 sin 2 θ (1 − cos 2 θ ) = 0 , was to 2g 2g

be shown. EVALUATE: The initial kinetic energy is independent of the angle θ but the kinetic energy at the maximum height depends on θ , so the maximum height depends on θ . IDENTIFY: Apply Eq.(7.7) to the motion of the car. SET UP: Take y = 0 at point A. Let point 1 be A and point 2 be B. K1 + U1 + Wother = K 2 + U 2 EXECUTE:

U1 = 0, U 2 = mg (2 R ) = 28,224 J, Wother = W f

K1 = 12 mv = 37,500 J, K 2 = 12 mv22 = 3840 J 2 1

The work-energy relation then gives W f = K 2 + U 2 − K1 = −5400 J. EVALUATE: 7.12.

Friction does negative work. The final mechanical energy ( K 2 + U 2 = 32,064 J) is less than the

initial mechanical energy ( K1 + U1 = 37,500 J) because of the energy removed by friction work. IDENTIFY: Only gravity does work, so apply Eq.(7.5). SET UP: v1 = 0 , so 12 mv22 = mg ( y1 − y2 ) . EXECUTE:

Tarzan is lower than his original height by a distance y1 − y2 = l (cos30° − cos 45°) so his speed is

v = 2 gl (cos30°− cos 45°) = 7.9 m/s, a bit quick for conversation. EVALUATE: The result is independent of Tarzan’s mass.

7-4

Chapter 7

7.13.

y1 = 0 y2 = (8.00 m)sin 36.9° y2 = 4.80 m Figure 7.13a ! (a) IDENTIFY and SET UP: F is constant so Eq.(6.2) can be used. The situation is sketched in Figure 7.13a. EXECUTE: WF = ( F cos φ ) s = (110 N)(cos0°)(8.00 m) = 880 J ! EVALUATE: F is in the direction of the displacement and does positive work. (b) IDENTIFY and SET UP: Calculate W using Eq.(6.2) but first must calculate the friction force. Use the freebody diagram for the oven sketched in Figure 7.13b to calculate the normal force n; then the friction force can be calculated from f k = μ k n. For this calculation use coordinates parallel and perpendicular to the incline.

EXECUTE:

∑F

y

= ma y

n − mg cos36.9° = 0 n = mg cos36.9° f k = μ k n = μ k mg cos36.9° f k = (0.25)(10.0 kg)(9.80 m/s 2 )cos36.9° = 19.6 N Figure 7.13b

W f = ( f k cos φ ) s = (19.6 N)(cos180°)(8.00 m) = −157 J EVALUATE: Friction does negative work. (c) IDENTIFY and SET UP: U = mgy; take y = 0 at the bottom of the ramp. EXECUTE: ΔU = U 2 − U1 = mg ( y2 − y1 ) = (10.0 kg)(9.80 m/s 2 )(4.80 m − 0) = 470 J EVALUATE: The object moves upward and U increases. (d) IDENTIFY and SET UP: Use Eq.(7.7). Solve for ΔK . EXECUTE: K1 + U1 + Wother = K 2 + U 2

ΔK = K 2 − K1 = U1 − U 2 + Wother ΔK = Wother − ΔU Wother = WF + W f = 880 J − 157 J = 723 J ΔU = 470 J Thus ΔK = 723 J − 470 J = 253 J. EVALUATE: Wother is positive. Some of Wother goes to increasing U and the rest goes to increasing K. ! ! ! (e) IDENTIFY: Apply ∑ F = ma to the oven. Solve for a and then use a constant acceleration equation to calculate v2 . SET UP: We can use the free-body diagram that is in part (b): ∑ Fx = max F − f k − mg sin 36.9° = ma EXECUTE: SET UP:

a=

F − f k − mg sin 36.9° 110 N − 19.6 N − (10 kg)(9.80 m/s 2 )sin 36.9° = 3.16 m/s 2 = 10.0 kg m

v1x = 0, ax = 3.16 m/s 2 , x − x0 = 8.00 m, v2 x = ?

v22x = v12x + 2ax ( x − x0 ) EXECUTE:

v2 x = 2ax ( x − x0 ) = 2(3.16 m/s 2 )(8.00 m) = 7.11 m/s 2

Then ΔK = K 2 − K1 = 12 mv22 = 12 (10.0 kg)(7.11 m/s) 2 = 253 J. EVALUATE: This agrees with the result calculated in part (d) using energy methods.


7.14.

IDENTIFY: SET UP:

7-5

! ! Only gravity does work, so apply Eq.(7.4). Use ∑ F = ma to calculate the tension.

Let y = 0 at the bottom of the arc. Let point 1 be when the string makes a 45° angle with the vertical and

point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration arad = v 2 / r EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is mgl (1 − cos θ ), where l is the length of the string and θ is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so mgl (1 − cos θ ) = 12 mv 2 , or v = 2 gl (1 − cos θ ) = 2(9.80 m/s 2 ) (0.80 m) (1 − cos 45°) = 2.1 m/s . (b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or mg cosθ = (0.12 kg) (9.80 m/s 2 ) cos 45° = 0.83 N. (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, mg + mv22 l = mg (1 + 2(1 − cos 45°)) = 1.9 N

7.15.

EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward. IDENTIFY: Apply U el = 12 kx 2 . SET UP: kx = F , so U = 12 Fx ,where F is the magnitude of force required to stretch or compress the spring a distance x. EXECUTE: (a) (1 2)(800 N)(0.200 m) = 80.0 J. (b) The potential energy is proportional to the square of the compression or extension; (80.0 J) (0.050 m 0.200 m ) 2 = 5.0 J.

F 800 N = = 4000 N/m and then used U el = 12 kx 2 directly. x 0.200 m IDENTIFY: Use the information given in the problem with F = kx to find k. Then U el = 12 kx 2 . SET UP: x is the amount the spring is stretched. When the weight is hung from the spring, F = mg . EVALUATE:

7.16.

EXECUTE:

We could have calculated k =

k=

F mg (3.15 kg)(9.80 m/s 2 ) = = = 2205 N/m . x x 0.1340 m − 0.1200 m

2U el 2(10.0 J) =± = ±0.0952 m = ±9.52 cm . The spring could be either stretched 9.52 cm or k 2205 N/m compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm + 9.52 cm = 21.52 cm . If it were compressed, the total length of the spring would be 12.00 cm − 9.52 cm = 2.48 cm . EVALUATE: To stretch or compress the spring 9.52 cm requires a force F = kx = 210 N . IDENTIFY: Apply U el = 12 kx 2 . x=±

7.17.

SET UP: EXECUTE:

U 0 = 12 kx02 . x is the distance the spring is stretched or compressed. (a) (i) x = 2 x0 gives U el = 12 k (2 x0 ) 2 = 4( 12 kx02 ) = 4U 0 . (ii) x = x0 / 2 gives

U el = 12 k ( x0 / 2) 2 = 14 ( 12 kx02 ) = U 0 / 4 . (b) (i) U = 2U 0 gives 7.18.

1 2

kx 2 = 2( 12 kx02 ) and x = x0 2 . (ii) U = U 0 / 2 gives

1 2

kx 2 = 12 ( 12 kx02 ) and x = x0 / 2 .

EVALUATE: U is proportional to x 2 and x is proportional to U . IDENTIFY: Apply Eq.(7.13). SET UP: Initially and at the highest point, v = 0 , so K1 = K 2 = 0 . Wother = 0 . EXECUTE: (a) In going from rest in the slingshot’s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; U = mgy = (10 × 10−3 kg)(9.80 m/s 2 ) (22.0 m) = 2.16 J. (b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m. (c) The lack of air resistance and no deformation of the rubber band are two possible assumptions. EVALUATE: The potential energy stored in the rubber band depends on k for the rubber band and the maximum distance it is stretched.

7-6

7.19.

Chapter 7

IDENTIFY and SET UP: Use energy methods. There are changes in both elastic and gravitational potential energy; elastic; U = 12 kx 2 , gravitational: U = mgy.

2U 2(3.20 J) = = 0.0632 m = 6.32 cm k 1600 N/m (b) Points 1 and 2 in the motion are sketched in Figure 7.19. K1 + U1 + Wother = K 2 + U 2 EXECUTE:

(a) U = 12 kx 2 so x =

Wother = 0 (Only work is that done by gravity and spring force) K1 = 0, K 2 = 0 y = 0 at final position of book U1 = mg ( h + d ) , U 2 = 12 kd 2 Figure 7.19

0 + mg (h + d ) + 0 = 12 kd 2 The original gravitational potential energy of the system is converted into potential energy of the compressed spring. 1 kd 2 − mgd − mgh = 0 2

⎞ 1⎛ ⎛1 ⎞ d = ⎜ mg ± (mg ) 2 + 4 ⎜ k ⎟ (mgh) ⎟ ⎜ ⎟ k⎝ ⎝2 ⎠ ⎠ 1 d must be positive, so d = mg + (mg ) 2 + 2kmgh k 1 d= ((1.20 kg)(9.80 m/s 2 ) + 1600 N/m

(

)

((1.20 kg)(9.80 m/s 2 )) 2 + 2(1600 N/m)(1.20 kg)(9.80 m/s 2 )(0.80 m) d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm

7.20.

EVALUATE: It was important to recognize that the total displacement was h + d ; gravity continues to do work as the book moves against the spring. Also note that with the spring compressed 0.12 m it exerts an upward force (192 N) greater than the weight of the book (11.8 N). The book will be accelerated upward from this position. IDENTIFY: Use energy methods. There are changes in both elastic and gravitational potential energy. SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.20.

The spring force and gravity are the only forces doing work on the cheese, so Wother = 0 and U = U grav + U el .

Figure 7.20 EXECUTE:

Cheese released from rest implies K1 = 0.

At the maximum height v2 = 0 so K 2 = 0. U1 = U1,el + U1, grav y1 = 0 implies U1,grav = 0 U1,el = 12 kx12 = 12 (1800 N/m)(0.15 m) 2 = 20.25 J (Here x1 refers to the amount the spring is stretched or compressed when the cheese is at position 1; it is not the x-coordinate of the cheese in the coordinate system shown in the sketch.) U 2 = U 2,el + U 2,grav


7-7

U 2,grav = mgy2 , where y2 is the height we are solving for. U 2, el = 0 since now the spring is no longer compressed. Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el = U 2,grav

y2 =

7.21.

20.25 J 20.25 J = = 1.72 m mg (1.20 kg)(9.80 m/s 2 )

EVALUATE: The description in terms of energy is very simple; the elastic potential energy originally stored in the spring is converted into gravitational potential energy of the system. IDENTIFY: Apply Eq.(7.13). SET UP: Wother = 0 . As in Example 7.7, K1 = 0 and U1 = 0.0250 J. EXECUTE:

For v2 = 0.20 m s, K 2 = 0.0040 J . U 2 = 0.0210 J = 12 kx 2 , and x = ±

2(0.0210 J) = ±0.092 m. The 5.00 N m

glider has this speed when the spring is stretched 0.092 m or compressed 0.092 m. EVALUATE: Example 7.7 showed that vx = 0.30 m/s when x = 0.0800 m . As x increases, vx decreases, so our result of vx = 0.20 m/s at x = 0.092 m is consistent with the result in the example. 7.22.


Use energy methods. The elastic potential energy changes. In part (a) solve for K 2 and

from this obtain v2 . In part (b) solve for U1 and from this obtain x1. (a) K1 + U1 + Wother = K 2 + U 2

point 1: the glider is at its initial position, where x1 = 0.100 m and v1 = 0 point 2: the glider is at x = 0 EXECUTE: K1 = 0 (released from rest), K 2 = 12 mv22 U1 = 12 kx12 , U 2 = 0, Wother = 0 (only the spring force does work) Thus 12 kx12 = 12 mv22 . (The initial potential energy of the stretched spring is converted entirely into kinetic energy of the glider.) v2 = x1

k 5.00 N/m = (0.100 m) = 0.500 m/s m 0.200 kg

(b) The maximum speed occurs at x = 0, so the same equation applies. 1 2

x1 = v2 EVALUATE: 7.23.

IDENTIFY:

kx12 = 12 mv22

m 0.200 kg = 2.50 m/s = 0.500 m k 5.00 N/m

Elastic potential energy is converted into kinetic energy. A larger x1 gives a larger v2 . Only the spring does work and Eq.(7.11) applies. a =

F − kx = , where F is the force the spring exerts m m

on the mass. SET UP: Let point 1 be the initial position of the mass against the compressed spring, so K1 = 0 and U1 = 11.5 J . Let point 2 be where the mass leaves the spring, so U el,2 = 0 . EXECUTE:

(a) K1 + U el,1 = K 2 + U el,2 gives U el,1 = K 2 .

1 2

mv22 = U el,1 and v2 =

2U el,1 m

=

2(11.5 J) = 3.03 m/s . 2.50 kg

K is largest when U el is least and this is when the mass leaves the spring. The mass achieves its maximum speed of 3.03 m/s as it leaves the spring and then slides along the surface with constant speed. (b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has its 2U el 2(11.5 J) maximum compression. U el = 12 kx 2 so x = − =− = −0.0959 m . The minus sign indicates k 2500 N/m kx (2500 N/m)(−0.0959 m) compression. F = − kx = max and ax = − = − = 95.9 m/s 2 . m 2.50 kg EVALUATE: If the end of the spring is displaced to the left when the spring is compressed, then ax in part (b) is to the right, and vice versa.

7-8

7.24.

Chapter 7

(a) IDENTIFY and SET UP: Use energy methods. Both elastic and gravitational potential energy changes. Work is done by friction. Choose point 1 as in Example 7.9 and let that be the origin, so y1 = 0. Let point 2 be 1.00 m below point 1, so

y2 = −1.00 m. K1 + U1 + Wother = K 2 + U 2

EXECUTE:

K1 = 12 mv = 12 (2000 kg)(25 m/s) 2 = 625,000 J, U1 = 0 2 1

Wother = − f y2 = −(17,000 N)(1.00 m) = −17,000 J K 2 = 12 mg 22 U 2 = U 2,grav + U 2,el = mgy2 + 12 ky22 U 2 = (2000 kg)(9.80 m/s 2 )(−1.00 m) + 12 (1.41 × 105 N/m)(1.00 m)2 U 2 = −19,600 J + 70,500 J = +50,900 J Thus 625,000 J − 17,000 J = 12 mv22 + 50,900 J 1 2

mv22 = 557,100 J

v2 =

2(557,100 J) = 23.6 m/s 2000 kg

EVALUATE: The elevator stops after descending 3.00 m. After descending 1.00 m it is still moving but has slowed down. ! ! ! (b) IDENTIFY: Apply ∑ F = ma to the elevator. We know the forces and can solve for a .

The free-body diagram for the elevator is given in Figure 7.24.

SET UP:

EXECUTE:

Fspr = kd , where d is the

distance the spring is compressed ∑ Fy = ma y f k + Fspr − mg = ma f k + kd − mg = ma Figure 7.24

a=

7.25.

f k + kd − mg 17,000 N + (1.41 × 105 N/m)(1.00 m) − (2000 kg)(9.80 m/s 2 ) = = 69.2 m/s 2 m 2000 kg

We calculate that a is positive, so the acceleration is upward. EVALUATE: The velocity is downward and the acceleration is upward, so the elevator is slowing down at this point. Note that a = 7.1g ; this is unacceptably high for an elevator. IDENTIFY: Apply Eq.(7.13) and F = ma . SET UP: Wother = 0 . There is no change in U grav . K1 = 0 , U 2 = 0 . EXECUTE:

1 2

kx 2 = 12 mvx2 . The relations for m, vx , k and x are kx 2 = mvx2 and kx = 5mg .

Dividing the first equation by the second gives x = (a) k = 25 (b) x =

vx2 mg 2 , and substituting this into the second gives k = 25 2 . vx 5g

(1160 kg)(9.80 m/s 2 ) 2 = 4.46 × 105 N/m (2.50 m/s) 2

(2.50 m/s) 2 = 0.128 m 5(9.80 m/s 2 )

EVALUATE:

Our results for k and x do give the required values for ax and vx : ax =

kx (4.46 × 105 N/m)(0.128 m) k = = 49.2 m/s 2 = 5.0 g and vx = x = 2.5 m/s . m 1160 kg m


7.26.

IDENTIFY:

7-9

Wgrav = mg cos φ .

SET UP: When he moves upward, φ = 180° and when he moves downward, φ = 0° . When he moves parallel to the ground, φ = 90° . EXECUTE:

(a) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos180° = −5100 J .

(b) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos0° = +5100 J . (c) φ = 90° in each case and Wgrav = 0 in each case.

7.27.

(d) The total work done on him by gravity during the round trip is −5100 J + 5100 J = 0 . (e) Gravity is a conservative force since the total work done for a round trip is zero. EVALUATE: The gravity force is independent of the position and motion of the object. When the object moves upward gravity does negative work and when the object moves downward gravity does positive work. IDENTIFY: Apply W fk = f k s cos φ . f k = μ k n . SET UP: For a circular trip the distance traveled is d = 2π r . At each point in the motion the friction force and the displacement are in opposite directions and φ = 180° . Therefore, W fk = − f k d = − f k (2π r ) . n = mg so f k = μ k mg . EXECUTE:

7.28.

(a) W fk = − μ k mg 2π r = −(0.250)(10.0 kg)(9.80 m/s 2 )(2π )(2.00 m) = −308 J .

(b) The distance along the path doubles so the work done doubles and becomes −616 J . (c) The work done for a round trip displacement is not zero and friction is a nonconservative force. EVALUATE: The direction of the friction force depends on the direction of motion of the object and that is why friction is a nonconservative force. IDENTIFY and SET UP: The force is not constant so we must use Eq.(6.14) to calculate W. The properties of work done by a conservative force are described in Section 7.3. ! ! 2 ! W = ∫ F ⋅ dl , F = −α x 2 iˆ 1

! (a) dl = dyˆj (x is constant; the displacement is in the + y -direction )

EXECUTE: ! ! F ⋅ dl = 0 (since iˆ ⋅ ˆj = 0) and thus W = 0. ! (b) dl = dxiˆ ! ! F ⋅ dl = (−α x 2 iˆ) ⋅ ( dxiˆ) = −α x 2 dx x2

W = ∫ (−α x 2 )dx = − 13 ax3 x1

x2 x1

= − 13 α ( x23 − x13 ) = −

12 N/m 2 ((0.300 m)3 − (0.10 m)3 ) = −0.10 J 3

! (c) dl = dxiˆ as in part (b), but now x1 = 0.30 m and x2 = 0.10 m

W = − 13 α ( x23 − x13 ) = +0.10 J (d) EVALUATE: The total work for the displacement along the x-axis from 0.10 m to 0.30 m and then back to 0.10 m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when the starting and ending points are the same, so the force is conservative. EXECUTE: Wx 1 → x2 = − 13 α ( x23 − x13 ) = 13 α x13 − 13 α x23

The definition of the potential energy function is Wx1 → x2 = U1 − U 2 . Comparison of the two expressions for W gives

7.29.

U = 13 α x3 . This does correspond to U = 0 when x = 0. EVALUATE: In part (a) the work done is zero because the force and displacement are perpendicular. In part (b) the force is directed opposite to the displacement and the work done is negative. In part (c) the force and displacement are in the same direction and the work done is positive. IDENTIFY: Since the force is constant, use W = Fs cos φ . SET UP: For both displacements, the direction of the friction force is opposite to the displacement and φ = 180° . EXECUTE: (a) When the book moves to the left, the friction force is to the right, and the work is −(1.2 N)(3.0 m) = −3.6 J. (b) The friction force is now to the left, and the work is again −3.6 J. (c) −7.2 J. (d) The net work done by friction for the round trip is not zero, and friction is not a conservative force. EVALUATE: The direction of the friction force depends on the motion of the object. For the gravity force, which is conservative, the force does not depend on the motion of the object.

7-10

7.30.

Chapter 7

IDENTIFY and SET UP: The friction force is constant during each displacement and Eq.(6.2) can be used to calculate work, but the direction of the friction force can be different for different displacements. ! f = μ k mg = (0.25)(1.5 kg)(9.80 m/s 2 ) = 3.675 N; direction of f is opposite to the motion. EXECUTE: (a) The path of the book is sketched in Figure 7.30a.

Figure 7.30a

! For the motion from you to Beth the friction force is directed opposite to the displacement s and W1 = − fs = −(3.675 N)(8.0 m) = −29.4 J. For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and W2 = −29.4 J. Wtot = W1 + W2 = −29.4 J − 29.4 J = −59 J (b) The path of the book is sketched in Figure 7.30b.

s = 2(8.0 m) 2 = 11.3 m

Figure 7.30b ! ! f is opposite to s , so W = − fs = −(3.675 N)(11.3 m) = −42 J (c)

For the motion from you to Kim W = − fs W = −(3.675 N)(8.0 m) = −29.4 J Figure 7.30c

For the motion from Kim to you W = − fs = −29.4 J Figure 7.30d

7.31.

The total work for the round trip is −29.4 J − 29.4 J = −59 J. (d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero. Both these results show that the friction force is nonconservative. IDENTIFY: The work done by a spring on an object attached to its end when the object moves from xi to xf is W = 12 kxi2 − 12 kxf2 . This result holds for any xi and xf . SET UP:

Assume for simplicity that x1 , x2 and x3 are all positive, corresponding to the spring being stretched.

EXECUTE:

(a)

1 2

k ( x12 − x22 )

(b) − 12 k ( x12 − x22 ). The total work is zero; the spring force is conservative. (c) From x1 to x3 , W = − 12 k ( x32 − x12 ). From x3 to x2 , W = − 12 k ( x22 − x32 ). The net work is − 12 k ( x22 − x12 ). This is the same as the result of part (a). EVALUATE: The results of part (c) illustrate that the work done by a conservative force is path independent.


7.32.

7.33.

7-11

IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U ( x ). Use coordinates where the origin is at one atom. The other atom then has coordinate x. EXECUTE: dU d ⎛ C ⎞ d ⎛ 1⎞ 6C Fx = − = − ⎜ − 66 ⎟ = +C6 ⎜ 6 ⎟ = − 76 dx dx ⎝ x ⎠ dx ⎝ x ⎠ x

The minus sign mean that Fx is directed in the − x -direction, toward the origin. The force has magnitude 6C6 / x 7 and is attractive. ! EVALUATE: U depends only on x so F is along the x-axis; it has no y or z components. IDENTIFY: Apply Eq.(7.16). SET UP: The sign of Fx indicates its direction. Fx = −

EXECUTE:

dU 4 4 = −4α x3 = −(4.8 J m ) x3 . Fx ( −0.800 m) = −(4.8 J m )(−0.80 m)3 = 2.46 N. The force is dx

in the + x -direction. EVALUATE: Fx > 0 when x < 0 and Fx < 0 when x > 0 , so the force is always directed towards the origin. 7.34.

Apply F ( x) = −

IDENTIFY:

dU ( x ) . dx

d (1/ x) 1 =− 2 dx x d (−Gm1m2 / x ) Gm1m2 ⎡ d (1/ x) ⎤ = Gm1m2 ⎢ EXECUTE: Fx ( x) = − ⎥ = − x 2 . The force on m2 is in the − x-direction . This dx ⎣ dx ⎦ SET UP:

is toward m1 , so the force is attractive.

7.35.

EVALUATE: By Newton's 3rd law the force on m1 due to m2 is Gm1m2 / x 2 , in the + x -direction (toward m2 ). The gravitational potential energy belongs to the system of the two masses. ∂U ∂U and Fy = − IDENTIFY: Apply Fx = − . ∂x ∂y SET UP:

r = ( x 2 + y 2 )1/ 2 .

EXECUTE:

Fy = −

(a) U (r ) = −

∂ (1/ r ) x ∂ (1/ r ) y =− 2 and =− 2 . 2 3/ 2 ∂x (x + y ) ∂y ( x + y 2 )3 / 2

∂U Gm m x Gm1m2 ⎡ ∂ (1/ r ) ⎤ = +Gm1m2 ⎢ = − 2 1 22 3 / 2 and . Fx = − ⎥ r (x + y ) ∂x ⎣ ∂x ⎦

⎡ ∂ (1/ r ) ⎤ ∂U Gm1m2 y = +Gm1m2 ⎢ . ⎥=− 2 y ( x ∂y ∂ + y 2 )3 / 2 ⎣ ⎦

Gm1m2 x Gm1m2 y Gm1m2 2 Gm1m2 and Fy = − . F = Fx2 + Fy2 = x + y2 = . 3 3 3 r r r r2 ! ! (c) Fx and Fy are negative. Fx = α x and Fy = α y , where α is a constant, so F and the vector r from m1 to m2 are ! ! in the same direction. Therefore, F is directed toward m1 at the origin and F is attractive. (b) ( x 2 + y 2 )3 / 2 = r 3 so Fx = −

! x y If θ is the angle between the vector r that points from m1 to m2 , then = cosθ and = sin θ . This r r gives Fx = − F cosθ and Fy = − F sin θ , our more usual way of writing the components of a vector. EVALUATE:

7.36.

Apply Eq.(7.18). d ⎛ 1 ⎞ 2 d ⎛ 1 ⎞ 2 SET UP: ⎜ ⎟=− 3 . ⎜ ⎟ = − 3 and dx ⎝ x 2 ⎠ x dy ⎝ y 2 ⎠ y ! ∂U ˆ ∂U ˆ EXECUTE: F = − i− j since U has no z-dependence. ∂U = −23α and ∂U = −23α , so ∂x ∂y x y ∂x ∂y ! ! ! ⎛ i ⎛ −2 −2 ⎞ j ⎞ F = −α ⎜ 3 iˆ + 3 ˆj ⎟ = 2α ⎜ 3 + 3 ⎟ . y ⎠ y ⎠ ⎝x ⎝x EVALUATE: Fx and x have the same sign and Fy and y have the same sign. When x > 0 , Fx is in the IDENTIFY:

+ x-direction, and so forth.

7-12

7.37.

Chapter 7

IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U. At equilibrium F = 0. (a) EXECUTE: The graphs are sketched in Figure 7.37.

a b − r12 r 6 dU 12a 6b F =− = + 13 − 7 dr r r

U=

Figure 7.37 (b) At equilibrium F = 0, so

dU =0 dr F = 0 implies

+12a 6b − 7 =0 r13 r

6br 6 = 12a; solution is the equilibrium distance r0 = (2a / b)1/ 6 U is a minimum at this r; the equilibrium is stable. (c) At r = (2a / b)1/ 6 , U = a / r12 − b / r 6 = a (b / 2a ) 2 − b(b / 2a ) = −b 2 / 4a. At r → ∞, U = 0. The energy that must be added is −ΔU = b 2 / 4a. (d) r0 = (2a / b)1/ 6 = 1.13 × 10−10 m gives that

2a / b = 2.082 × 10−60 m 6 and b / 4a = 2.402 × 1059 m −6 b 2 / 4a = b(b / 4a ) = 1.54 × 10−18 J b(2.402 × 1059 m −6 ) = 1.54 × 10−18 J and b = 6.41 × 10−78 J ⋅ m 6 . Then 2a / b = 2.082 × 10−60 m 6 gives a = (b / 2)(2.082 × 10−60 m 6 ) = (6.41 × 10−78 J ⋅ m 6 )(2.082 × 10−60 m 6 ) = 6.67 × 10−138 J ⋅ m12 EVALUATE: As the graphs in part (a) show, F (r ) is the slope of U (r ) at each r. U (r ) has a minimum where F = 0. IDENTIFY: Apply Eq.(7.16). dU SET UP: is the slope of the U versus x graph. dx dU EXECUTE: (a) Considering only forces in the x-direction, Fx = − and so the force is zero when the slope of dx the U vs x graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. EVALUATE: At point b, Fx is negative when the marble is displaced slightly to the right and Fx is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable. ! ! IDENTIFY: Apply ∑ F = ma to the bag and to the box. Apply Eq.(7.7) to the motion of the system of the box 1 2

7.38.

7.39.

and bucket after the bag is removed. SET UP: Let y = 0 at the final height of the bucket, so y1 = 2.00 m and y2 = 0 . K1 = 0 . The box and the bucket move with the same speed v, so K 2 = 12 ( mbox + mbucket )v 2 . Wother = − f k d , with d = 2.00 m and f k = μk mbox g . Before the bag is removed, the maximum possible friction force the roof can exert on the box is (0.700)(80.0 kg + 50.0 kg)(9.80 m/s 2 ) = 892 N . This is larger than the weight of the bucket (637 N), so before the bag is removed the system is at rest. EXECUTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag for friction to oppose. The static friction force on the box equals the weight of the bucket, 637 N.


(b) Eq.(7.7) gives mbucket gy1 − f k d = 12 mtot v 2 , with mtot = 145.0 kg . v =

v=

2 ( mbucket gy1 − μ k mbox gd ) . mtot

2 ⎡(65.0 kg)(9.80 m/s 2 )(2.00 m) − (0.400)(80.0 kg)(9.80 m/s 2 )(2.00 m) ⎤⎦ . 145.0 kg ⎣

v = 2.99 m/s . EVALUATE: If we apply

7.40.

7-13

!

!

∑ F = ma to the box and to the bucket we can calculate their common acceleration a.

Then a constant acceleration equation applied to either object gives v = 2.99 m/s , in agreement with our result obtained using energy methods. IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq.(7.5). SET UP: Call the blocks A and B, where A is the more massive one. v A1 = vB1 = 0 . Let y = 0 for each block to be at the initial height of that block, so y A1 = yB1 = 0 . y A 2 = −1.20 m and yB 2 = +1.20 m . v A 2 = vB 2 = v2 = 3.00 m/s . Eq.(7.5) gives 0 = 12 (m A + mB )v22 + g (1.20 m)( mB − mA ) . m A + mB = 15.0 kg .

EXECUTE: 1 2

7.41.

(15.0 kg)(3.00 m/s) 2 + (9.80 m/s 2 )(1.20 m)(15.0 kg − 2m A ) . Solving for mA gives mA = 10.4 kg . And then

mB = 4.6 kg . EVALUATE: The final kinetic energy of the two blocks is 68 J. The potential energy of block A decreases by 122 J. The potential energy of block B increases by 54 J. The total decrease in potential energy is 122 J − 54 J = 68 J, and this equals the increase in kinetic energy of the system. IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 SET UP:

U1 = U 2 = K 2 = 0 . Wother = W f = − μ k mgs, with s = 280 ft = 85.3 m (a) The work-energy expression gives

EXECUTE:

1 2

mv12 − μ k mgs = 0 .

v1 = 2 μk gs = 22.4 m/s = 50 mph; the driver was speeding.

7.42.

(b) 15 mph over speed limit so $150 ticket. EVALUATE: The negative work done by friction removes the kinetic energy of the object. IDENTIFY: Apply Eq.(7.14). SET UP: Only the spring force and gravity do work, so Wother = 0 . Let y = 0 at the horizontal surface. (a) Equating the potential energy stored in the spring to the block's kinetic energy,

EXECUTE:

v=

7.43.

1 2

kx 2 = 12 mv 2 , or

k 400 N/m x= (0.220 m) = 3.11 m/s. m 2.00 kg

(b) Using energy methods directly, the initial potential energy of the spring equals the final gravitational potential 1 1 kx 2 (400 N/m)(0.220 m) 2 2 energy, 12 kx 2 = mgL sin θ , or L = 2 = = 0.821 m. mg sin θ (2.00 kg)(9.80 m/s 2 )sin 37.0° EVALUATE: The total energy of the system is constant. Initially it is all elastic potential energy stored in the spring, then it is all kinetic energy and finally it is all gravitational potential energy. IDENTIFY: Use the work-energy theorem, Eq(7.7). The target variable μ k will be a factor in the work done by friction. SET UP: Let point 1 be where the block is released and let point 2 be where the block stops, as shown in Figure 7.43. K1 + U1 + Wother = K 2 + U 2 Work is done on the block by the spring and by friction, so Wother = W f

and U = U el . Figure 7.43

K1 = K 2 = 0

EXECUTE:

U1 = U1,el = kx12 = 12 (100 N/m)(0.200 m)2 = 2.00 J 1 2

U 2 = U 2,el = 0, since after the block leaves the spring has given up all its stored energy Wother = W f = ( f k cos φ ) s = μ k mg (cos φ ) s = − μ k mgs, since φ = 180° (The friction force is directed opposite to the displacement and does negative work.)

7-14

Chapter 7

Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el + W f = 0

μ k mgs = U1,el μk =

U1,el mgs

=

EVALUATE: 7.44.

200 J = 0.41. (0.50 kg)(9.80 m/s 2 )(1.00 m) U1,el + W f = 0 says that the potential energy originally stored in the spring is taken out of the system

by the negative work done by friction. IDENTIFY: Apply Eq.(7.14). Calculate f k from the fact that the crate slides a distance x = 5.60 m before coming to rest. Then apply Eq.(7.14) again, with x = 2.00 m . SET UP: U1 = U el = 360 J . U 2 = 0 . K1 = 0 . Wother = − f k x . EXECUTE:

Work done by friction against the crate brings it to a halt: U1 = −Wother .

360 J = 64.29 N . 5.60 m The friction force working over a 2.00-m distance does work equal to − f k x = −(64.29 N)(2.00 m) = −128.6 J. The kinetic energy of the crate at this point is thus 360 J − 128.6 J = 231.4 J, and its speed is found from f k x = potential energy of compressed spring , and f k =

mv 2 / 2 = 231.4 J , so v =

7.45.

7.46.

2(231.4 J) = 3.04 m/s . 50.0 kg

EVALUATE: The energy of the compressed spring goes partly into kinetic energy of the crate and is partly removed by the negative work done by friction. After the crate leaves the spring the crate slows down as friction does negative work on it. IDENTIFY: At its highest point between bounces all the mechanical energy of the ball is in the form of gravitational potential energy. SET UP: E = U = mgh , where h is the height at the highest point of the motion. EXECUTE: (a) mgh = (0.650 kg)(9.80 m/s 2 )(2.50 m) = 15.9 J (b) The second height is 0.75(2.50 m) = 1.875 m, so the second mgh = 11.9 J ; it loses 15.9 J − 11.9 J = 4.0 J on first bounce. This energy is converted to thermal energy. (c) The third height is 0.75(1.875 m) = 1.40 m, , so third mgh = 8.9 J ; it loses 11.9 J − 8.9 J = 3.0 J on second bounce. EVALUATE: In each bounce the ball loses 25% of its mechanical energy. ! ! IDENTIFY: Apply Eq.(7.14) to relate h and vB . Apply ∑ F = ma at point B to find the minimum speed required

at B for the car not to fall off the track. SET UP: At B, a = vB2 / R , downward. The minimum speed is when n → 0 and mg = mvB2 / R . The minimum speed required is vB = gR . K1 = 0 and Wother = 0 . EXECUTE:

(a) Eq.(7.14) applied to points A and B gives U A − U B = 12 mvB2 . The speed at the top must be at least

1 5 gR . Thus, mg (h − 2 R) > mgR, or h > R. 2 2 (b) Apply Eq.(7.14) to points A and C. U A − U C = (2.50) Rmg = K C , so vC = (5.00) gR = (5.00)(9.80 m/s 2 )(20.0 m) = 31.3 m/s. vC2 = 49.0 m/s 2 . The tangential direction is down, the normal force at point C is R horizontal, there is no friction, so the only downward force is gravity, and atan = g = 9.80 m/s 2 .

The radial acceleration is arad =

EVALUATE: 7.47.

If h > 52 R , then the downward acceleration at B due to the circular motion is greater than g and the

track must exert a downward normal force n. n increases as h increases and hence vB increases. (a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom. SET UP: Let point 1 be where the piece of wood is released and point 2 be just before it enters the rough bottom. Let y = 0 be at point 2. EXECUTE: U1 = K 2 gives K 2 = mgy1 = 78.4 J. IDENTIFY: Now apply work-energy relation to the motion along the rough bottom.


SET UP:

7-15

Let point 1 be where it enters the rough bottom and point 2 be where it stops. K1 + U1 + Wother = K 2 + U 2

EXECUTE:

7.48.

Wother = W f = − μ k mgs, K 2 = U1 = U 2 = 0; K1 = 78.4 J

78.4 J − μ k mgs = 0; solving for s gives s = 20.0 m. The wood stops after traveling 20.0 m along the rough bottom. (b) Friction does −78.4 J of work. EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system. IDENTIFY: Apply Eq.(7.14) to the rock. Wother = W fk . SET UP: Let y = 0 at the foot of the hill, so U1 = 0 and U 2 = mgh , where h is the vertical height of the rock above the foot of the hill when it stops. EXECUTE: (a) At the maximum height, K 2 = 0 . Eq.(7.14) gives K Bottom + W fk = U Top .

1 2 1 h = gh . mv0 − μ k mg cos θ d = mgh . d = h sin θ , so v02 − μ k g cosθ 2 2 sin θ 1 cos 40° (15 m/s) 2 − (0.20)(9.8 m/s 2 ) h = (9.8 m/s 2 )h and h = 9.3 m . 2 sin 40° (b) Compare maximum static friction force to the weight component down the plane. f s = μs mg cosθ = (0.75)(28 kg)(9.8 m/s 2 )cos 40° = 158 N . mg sinθ = (28 kg)(9.8 m/s 2 )(sin 40°) = 176 N > f s , so the rock will slide down. (c) Use same procedure as in part (a), with h = 9.3 m and vB being the speed at the bottom of the hill. U Top + W fk = K B . mgh − μk mg cosθ

h 1 = mvB2 and sin θ 2

vB = 2 gh − 2 μ k gh cosθ sin θ = 11.8 m/s .

7.49.

EVALUATE: For the round trip up the hill and back down, there is negative work done by friction and the speed of the rock when it returns to the bottom of the hill is less than the speed it had when it started up the hill. IDENTIFY: Apply Eq.(7.7) to the motion of the stone. SET UP: K1 + U1 + Wother = K 2 + U 2 Let point 1 be point A and point 2 be point B. Take y = 0 at point B. EXECUTE:

mgy1 + 12 mv12 = 12 mv22 , with h = 20.0 m and v1 = 10.0 m/s

v2 = v12 + 2 gh = 22.2 m/s EVALUATE: The loss of gravitational potential energy equals the gain of kinetic energy. (b) IDENTIFY: Apply Eq.(7.8) to the motion of the stone from point B to where it comes to rest against the spring. SET UP: Use K1 + U1 + Wother = K 2 + U 2 , with point 1 at B and point 2 where the spring has its maximum compression x. EXECUTE: U1 = U 2 = K 2 = 0; K1 = 12 mv12 with v1 = 22.2 m/s

Wother = W f + Wel = − μ k mgs − 12 kx 2 , with s = 100 m + x The work-energy relation gives K1 + Wother = 0. 1 2

mv12 − μ k mgs − 12 kx 2 = 0

Putting in the numerical values gives x 2 + 29.4 x − 750 = 0. The positive root to this equation is x = 16.4 m. EVALUATE: Part of the initial mechanical (kinetic) energy is removed by friction work and the rest goes into the potential energy stored in the spring. (c) IDENTIFY and SET UP: Consider the forces. EXECUTE: When the spring is compressed x = 16.4 m the force it exerts on the stone is Fel = kx = 32.8 N. The maximum possible static friction force is max f s = μs mg = (0.80)(15.0 kg)(9.80 m/s 2 ) = 118 N. EVALUATE: The spring force is less than the maximum possible static friction force so the stone remains at rest.

7-16

7.50.

Chapter 7

IDENTIFY: Once the block leaves the top of the hill it moves in projectile motion. Use Eq.(7.14) to relate the speed vB at the bottom of the hill to the speed vTop at the top and the 70 m height of the hill. SET UP:

For the projectile motion, take + y to be downward. ax = 0 , a y = g . v0 x = vTop , v0 y = 0 . For the motion

up the hill only gravity does work. Take y = 0 at the base of the hill. EXECUTE:

First get speed at the top of the hill for the block to clear the pit. y =

1 2 1 gt . 20 m = (9.8 m/s 2 )t 2 . 2 2

40 m = 20 m/s . 2.0 s Energy conservation applied to the motion up the hill: K Bottom = U Top + K Top gives t = 2.0 s . Then vTopt = 40 m gives vTop =

7.51.

7.52.

1 2 1 2 2 + 2 gh = (20 m/s) 2 + 2(9.8 m/s 2 )(70 m) = 42 m/s . mvB = mgh + mvTop . vB = vTop 2 2 EVALUATE: The result does not depend on the mass of the block. IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the person. SET UP: Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at

point 2. y1 = 41.0 m. Wother = − 12 kx 2 , where x = 11.0 m is the amount the cord is stretched at point 2. The cord does negative work. EXECUTE: K1 = K 2 = U 2 = 0, so mgy1 − 12 kx 2 = 0 and k = 631 N/m. Now apply F = kx to the test pulls: F = kx so x = F/k = 0.602 m. EVALUATE: All his initial gravitational potential energy is taken away by the negative work done by the force exerted by the cord, and this amount of energy is stored as elastic potential energy in the stretched cord. IDENTIFY: Apply Eq.(7.14) to the motion of the skier from the gate to the bottom of the ramp. SET UP: Wother = −4000 J . Let y = 0 at the bottom of the ramp. EXECUTE: For the skier to be moving at no more than 30.0 m/s ; his kinetic energy at the bottom of the ramp can be mv 2 (85.0 kg)(30.0 m/s) 2 no bigger than = = 38,250 J . Friction does −4000 J of work on him during his run, which 2 2 means his combined U and K at the top of the ramp must be no more than 38,250 J + 4000 J = 42,250 J. His K at the mv 2 (85.0 kg)(2.0 m/s) 2 = = 170 J . His U at the top should thus be no more than 42,250 J − 170 J = 42,080 J, 2 2 42,080 J 42,080 J = = 50.5 m. which gives a height above the bottom of the ramp of h = mg (85.0 kg)(9.80 m/s 2 ) EVALUATE: In the absence of air resistance, for this h his speed at the bottom of the ramp would be 31.5 m/s. The work done by air resistance is small compared to the kinetic and potential energies that enter into the calculation. IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for K 2 and then for v2 . SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel, as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by the 4400 N force. K1 + U1 + Wother = K 2 + U 2 top is

7.53.

Take y = 0 at his initial position. EXECUTE:

K1 = 0, K 2 = 12 mv22

Wother = Wfric = − fs Wother = −(40 N)(4.0 m) = −160 J Figure 7.53

U1,grav = 0, U1,el = 12 kd 2 , where d is the distance the spring is initially compressed. F 4400 N = = 4.00 m k 1100 N/m = 12 (1100 N/m)(4.00 m)2 = 8800 J

F = kd so d = and U1,el

U 2,grav = mgy2 = (60 kg)(9.80 m/s 2 )(2.5 m) = 1470 J, U 2, el = 0


7-17

Then K1 + U1 + Wother = K 2 + U 2 gives 8800 J − 160 J = 12 mv22 + 1470 J 2(7170 J) = 15.5 m/s 60 kg EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy. IDENTIFY: To be at equilibrium at the bottom, with the spring compressed a distance x0 , the spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or kx0 = w sin θ + f . Apply Eq.(7.14) to the motion down the ramp. 1 2

7.54.

mv22 = 7170 J and v2 =

SET UP: K 2 = 0 , K1 = 12 mv 2 , where v is the speed at the top of the ramp. Let U 2 = 0 , so U1 = wL sin θ , where L is the total length traveled down the ramp. 1 1 EXECUTE: Eq.(7.14) gives kx02 = ( w sin θ − f ) L + mv 2 . With the given parameters, 12 kx02 = 248 J and 2 2 kx0 = 1.10 × 103 N. Solving for k gives k = 2440 N/m. EVALUATE:

7.55.

x0 = 0.451 m . w sin θ = 551 N . The decrease in gravitational potential energy is only slightly larger

than the amount of mechanical energy removed by the negative work done by friction. 12 mv 2 = 243 J . The energy stored in the spring is only slightly larger than the initial kinetic energy of the crate at the top of the ramp. IDENTIFY: Apply Eq.(7.7) to the system consisting of the two buckets. If we ignore the inertia of the pulley we ignore the kinetic energy it has. SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.55.

Figure 7.55

The tension force does positive work on the 4.0 kg bucket and an equal amount of negative work on the 12.0 kg bucket, so the net work done by the tension is zero. Work is done on the system only by gravity, so Wother = 0 and U = U grav EXECUTE:

K1 = 0

K2 = m v

+ 12 mB vB2 , 2 But since the two buckets are connected by a rope they move together and have the same

1 2

2 A A, 2

speed: v A, 2 = vB , 2 = v2 . Thus K 2 = 12 ( mA + mB )v22 = (8.00 kg)v22 . U1 = mA gy A,1 = (12.0 kg)(9.80 m/s 2 )(2.00 m) = 235.2 J. U 2 = mB gyB , 2 = (4.0 kg)(9.80 m/s 2 )(2.00 m) = 78.4 J. Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1 = K 2 + U 2 235.2 J = (8.00 kg)v22 + 78.4 J 235.2 J − 78.4 J = 4.4 m/s 8.00 kg EVALUATE: The gravitational potential energy decreases and the kinetic energy increases by the same amount. We could apply Eq.(7.7) to one bucket, but then we would have to include in Wother the work done on the bucket by the tension T. v2 =

7-18

7.56.

Chapter 7

IDENTIFY:

Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the rocket from the starting point to the base of the

ramp. Wother is the work done by the thrust and by friction. Let point 1 be at the starting point and let point 2 be at the base of the ramp. v1 = 0 , v2 = 50.0 m/s . Let

SET UP:

y = 0 at the base and take + y upward. Then y2 = 0 and y1 = d sin 53° , where d is the distance along the ramp from the base to the starting point. Friction does negative work. EXECUTE: K1 = 0 , U 2 = 0 . U1 + Wother = K 2 . Wother = (2000 N)d − (500 N)d = (1500 N)d . mgd sin 53° + (1500 N) d = 12 mv22 . d=

mv22 (1500 kg)(50.0 m/s)2 = = 142 m . 2[mg sin 53° + 1500 N] 2[(1500 kg)(9.80 m/s 2 )sin 53° + 1500 N]

EVALUATE:

The initial height is y1 = (142 m)sin 53° = 113 m . An object free-falling from this distance attains a

speed v = 2 gy1 = 47.1 m/s . The rocket attains a greater speed than this because the forward thrust is greater than 7.57.

the friction force. IDENTIFY: The force exerted by a spring is Fx = − kx . The acceleration of the object is given by Fx = max . Apply Eq.(7.14) to relate position and speed. SET UP: Let + x be when the spring is stretched. EXECUTE: (a) U = 12 kx 2 . Let point 1 be when the spring is initially compressed a distance x0 , so x1 = − x0 .

K1 = 0 . Wother = 0 .

1 2

kx02 = U 2 + K 2 . The speed is maximum when x = 0 , so U 2 = 0 . Then

1 2

kx02 = 12 mv22 and

v2 = x0 k / m is this maximum speed. k k x . a is maximum when x is maximum, so a = x0 . m m (c) The speed is maximum when x = 0 , when the spring has returned to its natural length, and the acceleration is maximum when x = − x0 , at the initial compression of the spring. (b) Fx = − kx and Fx = max give ax = −

7.58.

(d) When the spring has maximum extension, v2 = 0 . 12 kx02 = 12 kx 2 and x = x0 .The magnitude of the maximum extension equals the magnitude of the maximum compression. (e) The machine part oscillates between x = − x0 and x = + x0 and never stops permanently. EVALUATE: In any real system there are mechanical energy losses, for example due to negative work done by friction, and the object eventually comes to rest. IDENTIFY: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy. SET UP: Since the two animals are equidistant from the axis, they each have the same speed v. EXECUTE: One mass rises while the other falls, so the net loss of potential energy is (0.500 kg − 0.200 kg)(9.80 m/s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the animals and is

2(1.176 J) = 1.83 m/s. (0.700 kg) EVALUATE: The mouse gains both gravitational potential energy and kinetic energy. The rat’s gain in kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse. (a) IDENTIFY and SET UP: Apply Eq.(7.7) to the motion of the potato. Let point 1 be where the potato is released and point 2 be at the lowest point in its motion, as shown in Figure 7.59a. K1 + U1 + Wother = K 2 + U 2 equal to

7.59.

1 2

mtot v 2 , and v =

y1 = 2.50 m y2 = 0 The tension in the string is at all points in the motion perpendicular to the displacement, so WT = 0 The only force that does work on the potato is gravity, so Wother = 0. Figure 7.59a


EXECUTE:

7-19

K1 = 0, K 2 = 12 mv22 , U1 = mgy1 , U 2 = 0

Thus U1 = K 2 .

mgy1 = 12 mv22 v2 = 2 gy1 = 2(9.80 m/s 2 )(2.50 m) = 7.00 m/s v2 is the same as if the potato fell through 2.50 m. ! ! ! (b) IDENTIFY: Apply ∑ F = ma to the potato. The potato moves in an arc of a circle so its acceleration is arad , EVALUATE:

where arad = v 2 / R and is directed toward the center of the circle. Solve for one of the forces, the tension T in the string. SET UP: The free-body diagram for the potato as it swings through its lowest point is given in Figure 7.59b.

! The acceleration arad is directed in toward the center of the circular path, so at this point it is upward.


∑F

y

= ma y

T − mg = marad ⎛ v2 ⎞ T = m( g + arad ) = m ⎜ g + 2 ⎟ , where the radius R for the circular motion is the length L of the string. R⎠ ⎝ It is instructive to use the algebraic expression for v2 from part (a) rather than just putting in the numerical value: v2 = 2 gy1 = 2 gL , so v22 = 2 gL ⎛ v2 ⎞ 2 gL ⎞ ⎛ Then T = m ⎜ g + 2 ⎟ = m ⎜ g + ⎟ = 3mg ; the tension at this point is three times the weight of the potato. L⎠ L ⎠ ⎝ ⎝ T = 3mg = 3(0.100 kg)(9.80 m/s 2 ) = 2.94 N 7.60.

EVALUATE: The tension is greater than the weight; the acceleration is upward so the net force must be upward. IDENTIFY: Eq.(7.14) says Wother = K 2 + U 2 − ( K1 + U1 ) . Wother is the work done on the baseball by the force exerted by the air. SET UP: U = mgy . K = 12 mv 2 , where v 2 = vx2 + v y2 . (a) The change in total energy is the work done by the air, ⎛1 ⎞ = ( K 2 + U 2 ) − ( K1 + U1 ) = m ⎜ (v22 − v12 ) + gy2 ⎟ . 2 ⎝ ⎠

EXECUTE:

Wother

(

)

Wother = (0.145 kg) (1/ 2 ⎡⎣(18.6 m/s) 2 − (30.0 m/s) 2 − (40.0 m/s) 2 ⎤⎦ + (9.80 m/s 2 )(53.6 m) . Wother = −80.0 J . (b) Similarly, Wother = ( K 3 + U 3 ) − ( K 2 + U 2 ) .

(

)

Wother = (0.145 kg) (1/ 2) ⎡⎣(11.9 m/s) 2 + (−28.7 m/s) 2 − (18.6 m/s) 2 ⎤⎦ − (9.80 m/s 2 )(53.6 m) . Wother = −31.3 J. (c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by the air is smaller in magnitude. EVALUATE: The initial kinetic energy of the baseball is 12 (0.145 kg)(50.0 m/s) 2 = 181 J . For the total motion from the ground, up to the maximum height, and back down the total work done by the air is 111 J. The ball returns to the ground with 181 J − 111 J = 70 J of kinetic energy and a speed of 31 m/s, less than its initial speed of 50 m/s.

7-20

7.61.

Chapter 7

IDENTIFY and SET UP: There are two situations to compare: stepping off a platform and sliding down a pole. Apply the work-energy theorem to each. (a) EXECUTE: Speed at ground if steps off platform at height h: K1 + U1 + Wother = K 2 + U 2

mgh = 12 mv22 , so v22 = 2 gh Motion from top to bottom of pole: (take y = 0 at bottom) K1 + U1 + Wother = K 2 + U 2 mgd − fd = 12 mv22 Use v22 = 2 gh and get mgd − fd = mgh fd = mg ( d − h) f = mg (d − h) / d = mg (1 − h / d ) EVALUATE: For h = d this gives f = 0 as it should (friction has no effect). For h = 0, v2 = 0 (no motion). The equation for f gives f = mg in this special case. When f = mg the forces on him cancel and he doesn’t accelerate down the pole, which agrees with v2 = 0. (b) EXECUTE:

f = mg (1 − h / d ) = (75 kg)(9.80 m/s 2 )(1 − 1.0 m/2.5 m) = 441 N.

(c) Take y = 0 at bottom of pole, so y1 = d and y2 = y.

K1 + U1 + Wother = K 2 + U 2 0 + mgd − f ( d − y ) = 12 mv 2 + mgy 1 2

mv 2 = mg ( d − y ) − f ( d − y )

Using f = mg (1 − h / d ) gives

1 2

mv 2 = mg (d − y ) − mg (1 − h / d )(d − y )

mv 2 = mg ( h / d )(d − y ) and v = 2 gh(1 − y / d ) EVALUATE: This gives the correct results for y = 0 and for y = d . IDENTIFY: Apply Eq.(7.14) to each stage of the motion. SET UP: Let y = 0 at the bottom of the slope. In part (a), Wother is the work done by friction. In part (b), Wother is 1 2

7.62.

the work done by friction and the air resistance force. In part (c), Wother is the work done by the force exerted by the snowdrift. EXECUTE: (a) The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, K1 = mgh − W f = (60.0 kg)(9.8 N/kg)(65.0 m) − 10,500 J, or

K1 = 38,200 J − 10,500 J = 27,720 J . Then v1 =

2 K1 2(27,720 J) = = 30.4 m/s . 60 kg m

(b) K 2 = K1 − (W f + Wair ) = 27,720 J − ( μk mgd + f air d ). K 2 = 27,720 J − [(0.2)(588 N)(82 m) + (160 N)(82 m)] or

K 2 = 27,720 J − 22,763 J = 4957 J . Then, v2 =

2K 2(4957 J) = = 12.9 m/s 60 kg m

(c) Use the Work-Energy Theorem to find the force. W = ΔK , F = K/d = (4957 J) (2.5 m) = 2000 N .

In each case, Wother is negative and removes mechanical energy from the system. ! ! IDENTIFY and SET UP: First apply ∑ F = ma to the skier. EVALUATE:

7.63.

Find the angle α where the normal force becomes zero, in terms of the speed v2 at this point. Then apply the work-energy theorem to the motion of the skier to obtain another equation that relates v2 and α . Solve these two equations for α .

Let point 2 be where the skier loses contact with the snowball, as sketched in Figure 7.63a Loses contact implies n → 0. y1 = R, y2 = R cos α

Figure 7.63a


7-21

First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her acceleration is arad = v 2 / R, directed in towards the center of the snowball. EXECUTE:

∑F

y

= ma y

mg cos α − n = mv / R 2 2

But n = 0 so mg cos α = mv22 / R v22 = Rg cos α Figure 7.63b

Now use conservation of energy to get another equation relating v2 to α : K1 + U1 + Wother = K 2 + U 2 The only force that does work on the skier is gravity, so Wother = 0. K1 = 0, K 2 = 12 mv22 U1 = mgy1 = mgR, U 2 = mgy2 = mgR cos α Then mgR = 12 mv22 + mgR cos α v22 = 2 gR(1 − cos α ) Combine this with the

7.64.

∑F

y

= ma y equation:

Rg cos α = 2 gR (1 − cos α ) cos α = 2 − 2cos α 3cos α = 2 so cos α = 2 / 3 and α = 48.2° EVALUATE: She speeds up and her arad increases as she loses gravitational potential energy. She loses contact when she is going so fast that the radially inward component of her weight isn’t large enough to keep her in the circular path. Note that α where she loses contact does not depend on her mass or on the radius of the snowball. IDENTIFY: Use conservation of energy to relate the speed at the lowest point to the speed at the highest point. ! ! Use ∑ F = ma to calculate the tension. SET UP:

The rock has acceleration arad = v 2 / R , directed toward the center of the circle.

EXECUTE:

If the speed of the rock at the top is vt , then conservation of energy gives the speed vb at the bottom

from mv = 12 mvt2 + mg (2 R ) , R being the radius of the circle, and so vb2 = vt2 + 4 gR . The tension at the top and 1 2

2 b

mvt2 mv 2 m and Tb − mg = b , so Tb − Tt = (vb2 − vt2 ) + 2mg = 6mg = 6w . R R R EVALUATE: The tensions Tt and Tb depend on the speed of the rock and on R, but the difference Tb − Tt is independent of the speed of the rock and the radius of the circle. IDENTIFY and SET UP: bottom are found from Tt + mg =

7.65.

yA = R yB = yC = 0

Figure 7.65 (a) Apply conservation of energy to the motion from B to C: K B + U B + Wother = K C + U C . The motion is described in Figure 7.65. EXECUTE: The only force that does work on the package during this part of the motion is friction, so Wother = W f = f k (cos φ ) s = μ k mg (cos180°) s = − μ k mgs

K B = 12 mvB2 , K C = 0 U B = 0, U C = 0

7-22

Chapter 7

Thus K B + W f = 0 1 2

mvB2 − μ k mgs = 0

μ B2

(4.80 m/s) 2 = 0.392 2 gs 2(9.80 m/s 2 )(3.00 m) EVALUATE: The negative friction work takes away all the kinetic energy. (b) IDENTIFY and SET UP: Apply conservation of energy to the motion from A to B:

μk =

=

K A + U A + Wother = K B + U B Work is done by gravity and by friction, so Wother = W f .

EXECUTE:

K A = 0, K B = 12 mvB2 = 12 (0.200 kg)(4.80 m/s) 2 = 2.304 J U A = mgy A = mgR = (0.200 kg)(9.80 m/s 2 )(1.60 m) = 3.136 J, U B = 0 Thus U A + W f = K B W f = K B − U A = 2.304 J − 3.136 J = −0.83 J EVALUATE: 7.66.

W f is negative as expected; the friction force does negative work since it is directed opposite to the

displacement. IDENTIFY: Apply Eq.(7.14) to the initial and final positions of the truck. SET UP: Let y = 0 at the lowest point of the path of the truck. Wother is the work done by friction. f r = μr n = μr mg cos β . Denote the distance the truck moves up the ramp by x. K1 = 12 mv02 , U1 = mgL sin α , K 2 = 0 ,

EXECUTE:

U 2 = mgx sin β and Wother = − μ r mgx cos β . From Wother = ( K 2 + U 2 ) − ( K1 + U1 ) , and solving for x, x= EVALUATE: 7.67.

(v 2 /2 g ) + L sin α K1 + mgL sin α = 0 . mg (sin β + μ r cos β ) sin β + μ r cos β

x increases when v0 increases and decreases when μ r increases.

Fx = −α x − β x 2 , α = 60.0 N/m and β = 18.0 N/m 2 (a) IDENTIFY: Use Eq.(6.7) to calculate W and then use W = −ΔU to identify the potential energy function U ( x ). SET UP:

x2

WFx = U1 − U 2 = ∫ Fx ( x) dx x1

Let x1 = 0 and U1 = 0. Let x2 be some arbitrary point x, so U 2 = U ( x). EXECUTE:

x

x

0

0

x

U ( x) = − ∫ Fx ( x) dx = − ∫ ( −α x − β x 2 ) dx = ∫ (α x + β x 2 ) dx = 12 α x 2 + 13 β x3 . 0

EVALUATE: If β = 0, the spring does obey Hooke’s law, with k = α , and our result reduces to (b) IDENTIFY: Apply Eq.(7.15) to the motion of the object. SET UP: The system at points 1 and 2 is sketched in Figure 7.67.

1 2

kx 2 .

K1 + U1 + Wother = K 2 + U 2 The only force that does work on the object is the spring force, so Wother = 0.


K1 = 0, K 2 = 12 mv22

U1 = U ( x1 ) = 12 α x12 + 13 β x13 = 12 (60.0 N/m)(1.00 m) 2 + 13 (18.0 N/m 2 )(1.00 m)3 = 36.0 J U 2 = U ( x2 ) = 12 α x22 + 13 β x23 = 12 (60.0 N/m)(0.500 m) 2 + 13 (18.0 N/m 2 )(0.500 m)3 = 8.25 J Thus 36.0 J = 12 mv22 + 8.25 J 2(36.0 J − 8.25 J) = 7.85 m/s 0.900 kg EVALUATE: The elastic potential energy stored in the spring decreases and the kinetic energy of the object increases. v2 =


7.68.

7.69.

IDENTIFY:

7-23

Apply Eq.(7.14). Wother is the work done by F.

SET UP: Wother = ΔK + ΔU . The distance the spring stretches is aθ . y2 − y1 = a sin θ . EXECUTE: The force increases both the gravitational potential energy of the block and the potential energy of the spring. If the block is moved slowly, the kinetic energy can be taken as constant, so the work done by the force is the increase in potential energy, ΔU = mga sin θ + 12 k (aθ ) 2 . EVALUATE: The force is kept tangent to the surface so the block will stay in contact with the surface. IDENTIFY: Apply Eq.(7.14) to the motion of the block. SET UP: Let y = 0 at the floor. Let point 1 be the initial position of the block against the compressed spring and let point 2 be just before the block strikes the floor. EXECUTE: With U 2 = 0, K1 = 0 , K 2 = U1 . 12 mv22 = 12 kx 2 + mgh . Solving for v2 ,

(1900 N/m)(0.045 m) 2 kx 2 + 2 gh = + 2(9.80 m/s 2 )(1.20 m) = 7.01 m/s . (0.150 kg) m EVALUATE: The potential energy stored in the spring and the initial gravitational potential energy all go into the final kinetic energy of the block. IDENTIFY: Apply Eq.(7.14). U is the total elastic potential energy of the two springs. SET UP: Call the two points in the motion where Eq.(7.14) is applied A and B to avoid confusion with springs 1 and 2, that have force constants k1 and k2 . At any point in the motion the distance one spring is stretched equals the distance the other spring is compressed. Let + x be to the right. Let point A be the initial position of the block, where it is released from rest, so x1 A = +0.150 m and x2 A = −0.150 m . v2 =

7.70.

EXECUTE:

(a) With no friction, Wother = 0 . K A = 0 and U A = K B + U B . The maximum speed is when U B = 0 and

this is at x1B = x2 B = 0 , when both springs are at their natural length. 12 k1 x12A + 12 k2 x22A = 12 mvB2 . x12A = x22A = (0.150 m) 2 , so vB =

k1 + k2 2500 N/m + 2000 N/m (0.150 m) = (0.150 m) = 5.81 m/s . m 3.00 kg

(b) At maximum compression of spring 1, spring 2 has its maximum extension and vB = 0 . Therefore, at this point

U A = U B . The distance spring 1 is compressed equals the distance spring 2 is stretched, and vice versa:

7.71.

x1 A = − x2 A and x1B = − x2 B . Then U A = U B gives 12 (k1 + k2 ) x12A = 12 (k1 + k2 ) x12B and x1B = − x1 A = −0.150 m . The maximum compression of spring 1 is 15.0 cm. EVALUATE: When friction is not present mechanical energy is conserved and is continually transformed between kinetic energy of the block and potential energy in the springs. If friction is present, its work removes mechanical energy from the system. ! ! IDENTIFY: Apply conservation of energy to relate x and h. Apply ∑ F = ma to relate a and x. SET UP: The first condition, that the maximum height above the release point is h, is expressed as 12 kx 2 = mgh . The magnitude of the acceleration is largest when the spring is compressed to a distance x; at this point the net upward force is kx − mg = ma , so the second condition is expressed as x = (m/k )( g + a ) . EXECUTE: (a) Substituting the second expression into the first gives 2 1 ⎛m⎞ m( g + a ) 2 . k ⎜ ⎟ ( g + a ) 2 = mgh, or k = 2 ⎝k⎠ 2 gh (b) Substituting this into the expression for x gives x =

mg and x = 2h . The initial spring force is kx = mg and the 2h net upward force approaches zero. But 12 kx 2 = mgh and sufficient potential energy is stored in the spring to move the mass to height h. IDENTIFY: At equilibrium the upward spring force equals the weight mg of the object. Apply conservation of energy to the motion of the fish. SET UP: The distance that the mass descends equals the distance the spring is stretched. K1 = K 2 = 0 , so EVALUATE:

7.72.

2gh . g+a

When a → 0 , our results become k =

U1 (gravitational) = U 2 (spring) EXECUTE: Following the hint, the force constant k is found from mg = kd , or k = mg / d . When the fish falls from rest, its gravitational potential energy decreases by mgy; this becomes the potential energy of the spring, 1 mg 2 y = mgy, or y = 2d . which is 12 ky 2 = 12 (mg / d ) y 2 . Equating these, 2 d

7-24

7.73.

Chapter 7

EVALUATE: At its lowest point the fish is not in equilibrium. The upward spring force at this point is ky = 2kd , and this is equal to twice the weight. At this point the net force is mg, upward, and the fish has an upward acceleration equal to g. IDENTIFY: Apply Eq.(7.15) to the motion of the block. SET UP: The motion from A to B is described in Figure 7.73.

Figure 7.73

The normal force is n = mg cosθ , so f k = μ k n = μ k mg cosθ . y A = 0; yB = (60.0 m)sin 30.0° = 3.00 m K A + U A + Wother = K B + U B EXECUTE:

Work is done by gravity, by the spring force, and by friction, so Wother = W f and U = U el + U grav

K A = 0, K B = 12 mvB2 = 12 (1.50 kg)(7.00 m/s) 2 = 36.75 J U A = U el, A + U grav, A = U el, A , since U grav, A = 0 U B = U el, B + U grav, B = 0 + mgyB = (1.50 kg)(9.80 m/s 2 )(3.00 m) = 44.1 J Wother = W f = ( f k cos φ ) s = μ k mg cosθ (cos180°) s = − μ k mg cosθ s Wother = −(0.50)(1.50 kg)(9.80 m/s 2 )(cos30.0°)(6.00 m) = −38.19 J Thus U el, A − 38.19 J = 36.75 J + 44.10 J U el, A = 38.19 J + 36.75 J + 44.10 J = 119 J

7.74.

EVALUATE: U el must always be positive. Part of the energy initially stored in the spring was taken away by friction work; the rest went partly into kinetic energy and partly into an increase in gravitational potential energy. IDENTIFY: Apply Eq.(7.14) to the motion of the package. Wother = W fk , the work done by the kinetic friction

force. SET UP: f k = μ k n = μ k mg cosθ , with θ = 53.1° . Let L = 4.00 m , the distance the package moves before reaching the spring and let d be the maximum compression of the spring. Let point 1 be the initial position of the package, point 2 be just as it contacts the spring, point 3 be at the maximum compression of the spring, and point 4 be the final position of the package after it rebounds. EXECUTE: (a) K1 = 0 , U 2 = 0 , Wother = − f k L = − μ k L cosθ . U1 = mgL sin θ . K 2 = 12 mv 2 , where v is the speed before the block hits the spring. Eq.(7.14) applied to points 1 and 2, with y2 = 0 , gives U1 + Wother = K 2 . Solving for v, v = 2 gL(sin θ − μ k cosθ ) = 2(9.80 m/s 2 )(4.00 m)(sin 53.1° − (0.20)cos53.1°) = 7.30 m/s. (b) Apply Eq.(7.14) to points 1 and 3. Let y3 = 0 . K1 = K3 = 0 . U1 = mg ( L + d )sin θ . U 2 = 12 kd 2 .

Wother = − f k ( L + d ) . Eq.(7.14) gives mg ( L + d )sin θ − μk mg cosθ ( L + d ) = 12 kd 2 . This can be written as k − d − L = 0. The factor multiplying d 2 is 4.504 m −1 , and use of the quadratic formula 2mg (sin θ − μk cosθ ) gives d = 1.06 m . (c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has moved a distance L + d down the incline and L + d − y up the incline. The magnitude of the friction force is the d2

same in both directions, μ k mg cos θ , and so the work done by friction is − μk (2 L + 2d − y ) mg cos θ . This must be equal to the change in gravitational potential energy, which is − mgy sin θ . Equating these and solving for y gives 2μk cosθ 2μk = (L + d ) . Using the value of d found in part (b) and the given values for μ k sin θ + μ k cosθ tan θ + μk and θ gives y = 1.32 m . y = (L + d )


7.75.

7-25

EVALUATE: Our expression for y gives the reasonable results that y = 0 when μ k = 0 ; in the absence of friction the package returns to its starting point. (a) IDENTIFY and SET UP: Apply K A + U A + Wother = K B + U B to the motion from A to B. EXECUTE:

K A = 0, K B = 12 mvB2

U A = 0, U B = U el, B = 12 kxB2 , where xB = 0.25 m Wother = WF = FxB Thus FxB = 12 mvB2 + 12 kxB2 . (The work done by F goes partly to the potential energy of the stretched spring and partly to the kinetic energy of the block.) FxB = (20.0 N)(0.25 m) = 5.0 J and 12 kxB2 = 12 (40.0 N/m)(0.25 m) 2 = 1.25 J 2(3.75 J) = 3.87 m/s 0.500 kg (b) IDENTIFY: Apply Eq.(7.15) to the motion of the block. Let point C be where the block is closest to the wall. When the block is at point C the spring is compressed an amount xC , so the block is 0.60 m − xC from the wall,

Thus 5.0 J = 12 mvB2 + 1.25 J and vB =

and the distance between B and C is xB + xC . SET UP:

The motion from A to B to C is described in Figure 7.75. K B + U B + Wother = K C + U C EXECUTE:

Wother = 0

K B = 12 mv = 5.0 J − 1.25 J = 3.75 J (from part (a)) U B = 12 kxB2 = 1.25 J 2 B

K C = 0 (instantaneously at rest at point closest to wall) 2 1 U C = 2 k xC Figure 7.75

Thus 3.75 J + 1.25 J = 12 k xC

2

2(5.0 J) = 0.50 m 40.0 N/m The distance of the block from the wall is 0.60 m − 0.50 m = 0.10 m. EVALUATE: The work (20.0 N)(0.25 m) = 5.0 J done by F puts 5.0 J of mechanical energy into the system. No mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J. IDENTIFY: Apply Eq.(7.14) to the motion of the student. SET UP: Let x0 = 0.18 m , x1 = 0.71 m . The spring constants (assumed identical) are then known in terms of the xC =

7.76.

unknown weight w, 4kx0 = w . Let y = 0 at the initial position of the student. EXECUTE: (a) The speed of the brother at a given height h above the point of maximum compression is then ⎛ x2 ⎞ 1 1⎛ w⎞ (4k ) g 2 found from (4k ) x12 = ⎜ ⎟ v 2 + mgh, or v 2 = x1 − 2 gh = g ⎜ 1 − 2h ⎟ . Therefore, 2 2⎝ g ⎠ w ⎝ x0 ⎠ v = (9.80 m/s 2 )((0.71 m) 2 (0.18 m) − 2(0.90 m)) = 3.13 m/s , or 3.1 m/s to two figures. (b) Setting v = 0 and solving for h, h =

2kx12 x2 = 1 = 1.40 m, or 1.4 m to two figures. mg 2 x0 2

(c) No; the distance x0 will be different, and the ratio

⎛ 0.53 m ⎞ x12 ( x1 + 0.53 m) 2 = = x1 ⎜1 + ⎟ will be different. x0 x1 x1 ⎠ ⎝

Note that on a planet with lower g, x1 will be smaller and h will be larger. EVALUATE: We are able to solve the problem without knowing either the mass of the student or the force constant of the spring.

7-26

7.77.

Chapter 7

ax = d 2 x/dt 2 , a y = d 2 y/dt 2 . Fx = max , Fy = ma y . U = ∫ Fx dx + ∫ Fy dy .

IDENTIFY: SET UP:

1 1 d d (cos ω0t ) = −ω0 sin ω0t . (sin ω0t ) = ω0 cos ω0t . ∫ cos ω0t dt = sin ω0t , ∫ sin ω0t dt = − cos ω0t . ω0 ω0 dt dt

vx = dx / dt , v y = dy / dt . E = K + U . (a) ax = d 2 x/dt 2 = −ω02 x, Fx = max = − mω02 x. a y = d 2 y/dt 2 = −ω02 y, Fy = ma y = − mω02 y

EXECUTE:

1 (b) U = − ⎡ ∫ Fx dx + ∫ Fy dy ⎤ = mω02 ⎡ ∫ xdx + ∫ ydy ⎤ = mω02 ( x 2 + y 2 ) ⎣ ⎦ ⎣ ⎦ 2 (c) vx = dx/dt = − x0ω0 sin ω0t = − x0ω0 ( y/y0 ). v y = dy/dt = + y0ω0 cos ω0t = + y0ω0 ( x/x0 ).

(i) When x = x0 and y = 0, vx = 0 and v y = y0ω0 , 1 1 1 1 K = m(vx2 + v y2 ) = my02ω02 , U = ω02 mx02 and E = K + U = mω02 ( x02 + y02 ) 2 2 2 2 (ii) When x = 0 and y = y0 , vx = − x0ω0 and v y = 0 ,

7.78.

1 1 1 K = ω02 mx02 , U = mω02 y02 and E = K + U = mω02 ( x02 + y02 ) 2 2 2 EVALUATE: The total energy is the same at the two points in part (c); the total energy of the system is constant. IDENTIFY: Calculate the increase in kinetic energy for the car. SET UP: The car gets (0.15)(1.3 × 108 J) of energy from one gallon of gasoline. (a) The mechanical energy increase of the car is K 2 − K1 = 12 (1500 kg)(37 m/s) 2 = 1.027 × 106 J. Let

EXECUTE:

α be the number of gallons of gasoline consumed. α (1.3 × 108 J)(0.15) = 1.027 × 106 J and α = 0.053gallons . (b) (1.00 gallons) α = 19 accelerations 7.79.

EVALUATE: The time over which the increase in velocity occurs doesn't enter into the calculation. IDENTIFY: U = mgh . Use h = 150 m for all the water that passes through the dam. SET UP: m = ρV and V = AΔh is the volume of water in a height Δh of water in the lake. EXECUTE: (a) Stored energy = mgh = ( ρ V ) gh = ρ A(1 m) gh .

stored energy = (1000 kg/m3 )(3.0 × 106 m 2 )(1 m)(9.8 m/s 2 )(150 m) = 4.4 × 1012 J. (b) 90% of the stored energy is converted to electrical energy, so (0.90)(mgh) = 1000 kWh . (0.90) ρVgh = 1000 kWh . V =

(1000 kWh)((3600 s) (1 h)) = 2.7 × 103 m3 . (0.90)(1000 kg/m3 )(150 m)(9.8 m/s 2 )

V 2.7 × 103 m3 = = 9.0 × 10−4 m . A 3.0 × 106 m 2 EVALUATE: Δh is much less than 150 m, so using h = 150 m for all the water that passed through the dam was a very good approximation. IDENTIFY and SET UP: The potential energy of a horizontal layer of thickness dy, area A, and height y is dU = (dm) gy. Let ρ be the density of water. EXECUTE: dm = ρ dV = ρ A dy, so dU = ρ Agy dy. The total potential energy U is Change in level of the lake: AΔh = Vwater . Δh =

7.80.

h

h

0

0

U = ∫ dU = ρ Ag ∫ y dy = 12 ρ Agh 2 . A = 3.0 × 106 m 2 and h = 150 m, so U = 3.3 × 1014 J = 9.2 × 107 kWh EVALUATE: The volume is Ah and the mass of water is ρV = ρ Ah. The average depth is hav = h/2, so U = mghav . 7.81.

IDENTIFY: SET UP:

Apply Fx = −

∂U ∂U ∂U , Fy = − and Fz = − . ∂x ∂y ∂z

r = ( x 2 + y 2 + z 2 )1/ 2 .

∂ (1/r ) x ∂ (1/r ) y ∂ (1/r ) z =− 2 , =− 2 and =− 2 . ( x + y 2 )3/ 2 ( x + y 2 )3/ 2 ( x + y 2 )3/ 2 ∂x ∂y ∂z


EXECUTE:

Fy = −

(a) U (r ) = −

7-27

Gm1m2 x ∂U Gm1m2 ⎡ ∂ (1/r ) ⎤ . Fx = − = +Gm1m2 ⎢ ⎥ = − ( x 2 + y 2 + z 2 )3/ 2 . Similarly, r ∂x ⎣ ∂x ⎦

Gm1m2 y Gm1m2 z and Fz = − 2 . ( x 2 + y 2 + z 2 )3/ 2 ( x + y 2 + z 2 )3/ 2

Gm1m2 x Gm1m2 y Gm1m2 z , Fy = − and Fz = − . r3 r3 r3 Gm1m2 2 Gm1m2 F = Fx2 + Fy2 + Fz2 = x + y2 + z2 = . 3 r r2 ! ! (c) Fx , Fy and Fz are negative. Fx = α x , Fy = α y and Fz = α z , where α is a constant, so F and the vector r from ! ! m1 to m2 are in the same direction. Therefore, F is directed toward m1 at the origin and F is attractive. (b) ( x 2 + y 2 + z 2 )3/ 2 = r 3 so Fx = −

7.82.

EVALUATE: When m2 moves to larger r, the work done on it by the attractive gravity force is negative. Since W = −ΔU , negative work done by gravity means the gravitational potential energy increases. Gm1m2 U (r ) = − does increase (becomes less negative) as r increases. For an object near the surface of the earth, r Gm1m2 U (r ) = − will be shown in Chapter 12 to be equivalent to U grav = mgy . r IDENTIFY: Calculate the work W done by this force. If the force is conservative, the work is path independent. ! P2 ! SET UP: W = ∫ F ⋅ dl . P1

EXECUTE:

P2

P2

P1

P1

(a) W = ∫ Fy dy = C ∫

y 2 dy . W doesn't depend on x, so it is the same for all paths between P1 and

P2 . The force is conservative. P2

P2

P1

P1

(b) W = ∫ Fx dx = C ∫

y 2 dx . W will be different for paths between points P1 and P2 for which y has different

values. For example, if y has the constant value y0 along the path, then W = Cy0 ( x2 − x1 ) . W depends on the value

7.83.

of y0 . The force is not conservative. ! Cy 3 EVALUATE: F = Cy 2 ˆj has the potential energy function U ( y ) = − . We cannot find a potential energy 3 ! function for F = Cy 2 iˆ . ! F = −α xy 2 ˆj , α = 2.50 N/m3 ! ! IDENTIFY: F is not constant so use Eq.(6.14) to calculate W. F must be evaluated along the path. (a) SET UP: The path is sketched in Figure 7.83a. ! dl = dxiˆ + dyˆj ! ! F ⋅ dl = −α xy 2 dy ! ! On the path, x = y so F ⋅ dl = −α y 3 dy

EXECUTE:

Figure 7.83a ! 2 ! y2 y2 W = ∫ F ⋅ dl = ∫ (−α y 3 ) dy = −(α / 4) y 4| = −(α / 4)( y24 − y14 ) 1

( ) y1

y1

y1 = 0, y2 = 3.00 m, so W = − (2.50 N/m )(3.00 m) = −50.6 J (b) SET UP: The path is sketched in Figure 7.83b. 1 4

3

4

Figure 7.83b ! ! ! For the displacement from point 1 to point 2, dl = dxiˆ, so F ⋅ dl = 0 and W = 0. (The force is perpendicular to the displacement at each point along the path, so W = 0.)

7-28

Chapter 7

! ! ! For the displacement from point 2 to point 3, dl = dyˆj, so F ⋅ dl = −α xy 2 dy. On this path, x = 3.00 m, so ! ! F ⋅ dl = −(2.50 N/m3 )(3.00 m)y 2 dy = −(7.50 N/m 2 ) y 2 dy. ! 3 ! y3 EXECUTE: W = ∫ F ⋅ dl = − (7.50 N/m 2 ) ∫ y 2 dy = −(7.50 N/m 2 ) 13 ( y33 − y23 ) 2

W = −(7.50 N/m ) ( ) (3.00 m) = −67.5 J 2

7.84.

y2

3

1 3

(c) EVALUATE: For these two paths between the same starting and ending points the work is different, so the force is nonconservative. ! P2 ! IDENTIFY: Use W = ∫ F ⋅ dl to calculate W for each segment of the path. P1 ! ! SET UP: F ⋅ dl = Fx dx = α xy dx EXECUTE: (a) The path is sketched in Figure 7.84. ! ! ! (b) (1): x = 0 along this leg, so F = 0 and W = 0 . (2): Along this leg, y = 1.50 m , so F ⋅ dl = (3.00 N m) xdx , ! ! ! and W = (1.50 N m)((1.50 m) 2 − 0) = 3.38 J (3) F ⋅ dl = 0 , so W = 0 (4) y = 0 , so F = 0 and W = 0 . The work done in moving around the closed path is 3.38 J. (c) The work done in moving around a closed path is not zero, and the force is not conservative. EVALUATE: There is no potential energy function for this force.

Figure 7.84 7.85.

IDENTIFY: Use Eq.(7.16) to relate Fx and U ( x) . The equilibrium is stable where U ( x ) is a local minimum and the equilibrium is unstable where U ( x) is a local maximum. SET UP: The maximum and minimum values of x are those for which U ( x) = E . K = E − U , so the maximum speed is where U is a minimum. dU EXECUTE: (a) For the given proposed potential U ( x), − = −kx + F , so this is a possible potential function. dx For this potential, U (0) = − F 2 2k , not zero. Setting the zero of potential is equivalent to adding a constant to the potential; any additive constant will not change the derivative, and will correspond to the same force. (b) At equilibrium, the force is zero; solving − kx + F = 0 for x gives x0 = F/k . U ( x0 ) = − F 2 /k , and this is a minimum of U, and hence a stable point. (c) The graph is given in Figure 7.85. (d) No; Ftot = 0 at only one point, and this is a stable point. (e) The extreme values of x correspond to zero velocity, hence zero kinetic energy, so U ( x± ) = E , where x± are

the extreme points of the motion. Rather than solve a quadratic, note that

1 2

k ( x − F/k ) 2 − F 2 /k , so U ( x± ) = E

2

becomes

1 ⎛ F⎞ F2 F F F F x− = − . . x± − = ±2 , so x+ = 3 k ⎜ x± − ⎟ − F 2 /k = k k k k 2 ⎝ k⎠ k

(f) The maximum kinetic energy occurs when U ( x ) is a minimum, the point x0 = F/k found in part (b). At this

point K = E − U = ( F 2 /k ) − (− F 2 /k ) = 2 F 2 /k , so v = 2 F

mk .


7-29

EVALUATE: As E increases, the magnitudes of x+ and x− increase. The particle cannot reach values of x for which E < U ( x) because K cannot be negative.

Figure 7.85 7.86.

7.87.

IDENTIFY: Use Eq.(7.16) to relate Fx and U ( x) . The equilibrium is stable where U ( x ) is a local minimum and the equilibrium is unstable where U ( x) is a local maximum. SET UP: dU/dx is the slope of the graph of U versus x. K = E − U , so K is a maximum when U is a minimum. The maximum x is where E = U . EXECUTE: (a) The slope of the U vs. x curve is negative at point A, so Fx is positive (Eq. (7.16)). (b) The slope of the curve at point B is positive, so the force is negative. (c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m. (d) The curve at point C looks pretty close to flat, so the force is zero. (e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than U ( A) , the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at any point where the potential energy is larger than U ( A) . On the graph, that looks to be at about 2.2 m. (f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m. (g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C. EVALUATE: If E is less than U at point C, the particle is trapped in one or the other of the potential "wells" and cannot move from one allowed region of x to the other. IDENTIFY: K = E − U determines v( x) . SET UP: v is a maximum when U is a minimum and v is a minimum when U is a maximum. Fx = − dU/dx . The extreme values of x are where E = U ( x ) . EXECUTE:

(a) Eliminating β in favor of α and x0 ( β = α /x0 ) ,

U ( x) =

α x

2

−

β x

=

α x02 2 0

x x

2

−

α x0 x

=

2 α ⎡⎛ x0 ⎞ ⎛ x0 ⎞ ⎤

⎢⎜ ⎟ − ⎜ ⎟ ⎥ . x02 ⎣⎢⎝ x ⎠ ⎝ x ⎠ ⎦⎥

⎛α ⎞ U ( x0 ) = ⎜ 2 ⎟ (1 − 1) = 0 . U ( x) is positive for x < x0 and negative for x > x0 ( α and β must be taken as ⎝ x0 ⎠ positive). The graph of U ( x) is sketched in Figure 7.87a. 2 ⎛ 2α ⎞ ⎛ ⎛ x ⎞ ⎛ x ⎞ ⎞ 2 U = ⎜ 2 ⎟ ⎜ ⎜ 0 ⎟ − ⎜ 0 ⎟ ⎟ . The proton moves in the positive x-direction, speeding up until it m ⎝ mx0 ⎠ ⎝⎜ ⎝ x ⎠ ⎝ x ⎠ ⎠⎟ reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the square root in the expression for v ( x ) indicates that the particle will be found only in the region where U < 0 , that

(b) v( x) = −

is, x > x0 . The graph of v( x) is sketched in Figure 7.87b. (c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy. 3 2 dU α ⎡ ⎛ x0 ⎞ ⎛ x0 ⎞ ⎤ This minimum occurs when dU = 0 , or = 3 ⎢ −2 ⎜ ⎟ + ⎜ ⎟ ⎥ = 0, dx dx x0 ⎣⎢ ⎝ x ⎠ ⎝ x ⎠ ⎦⎥ which has the solution x = 2 x0 . U (2 x0 ) = −

α 4 x02

, so v =

α 2mx02

.

(d) The maximum speed occurs at a point where dU = 0 , and from Eq. (7.15), the force at this point is zero. dx

7-30

Chapter 7

(e) x1 = 3 x0 , and U (3x0 ) = −

2α . 9 x02

2 ⎛ ⎛ x ⎞ ⎛ x ⎞2 ⎞ 2 2 ⎡⎛ −2 α ⎞ α ⎛ ⎛ x0 ⎞ x0 ⎞ ⎤ − 2 ⎜ ⎜ ⎟ − ⎟ ⎥ = 2α2 ⎜ ⎜ 0 ⎟ − ⎜ 0 ⎟ − 2 9 ⎟ . (U ( x1 ) − U ( x)) = ⎢⎜ 2 ⎟ ⎟ mx0 ⎜⎝ ⎝ x ⎠ ⎝ x ⎠ m m ⎢⎣⎝ 9 x0 ⎠ x0 ⎜⎝ ⎝ x ⎠ x ⎟⎠ ⎦⎥ ⎠ The particle is confined to the region where U ( x) < U ( x1 ) . The maximum speed still occurs at x = 2 x0 , but now

v( x) =

the particle will oscillate between x1 and some minimum value (see part (f)). (f) Note that U ( x) − U ( x1 ) can be written as 2 α ⎡⎛ x0 ⎞ ⎛ x0 ⎞ ⎛ 2 ⎞ ⎤

α ⎡⎛ x0 ⎞ 1 ⎤ ⎡⎛ x0 ⎞ 2 ⎤ , ⎢⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ ⎥ = ⎜ ⎟− ⎜ ⎟− x02 ⎣⎢⎝ x ⎠ ⎝ x ⎠ ⎝ 9 ⎠ ⎥⎦ x02 ⎢⎣⎝ x ⎠ 3 ⎥⎦ ⎢⎣⎝ x ⎠ 3 ⎥⎦ which is zero (and hence the kinetic energy is zero) at x = 3 x0 = x1 and x = 32 x0 . Thus, when the particle is released from x0 , it goes on to infinity, and doesn’t reach any maximum distance. When released from x1 , it oscillates between EVALUATE:

3 2

x0 and 3x0 .

In each case the proton is released from rest and E = U ( xi ) , where xi is the point where it is

released. When xi = x0 the total energy is zero. When xi = x1 the total energy is negative. U ( x ) → 0 as x → ∞ , so for this case the proton can't reach x → ∞ and the maximum x it can have is limited.

Figure 7.87

8

MOMENTUM, IMPULSE, AND COLLISIONS

8.1.


(a) p = (10,000 kg)(12.0 m/s) = 1.20 × 105 kg ⋅ m/s

EXECUTE: (b) (i) v =

p = mv. K = 12 mv 2 .

p 1.20 × 105 kg ⋅ m/s = = 60.0 m/s . (ii) m 2000 kg

vSUV =

8.2.

1 2

2 mT vT2 = 12 mSUV vSUV , so

10,000 kg mT (12.0 m/s) = 26.8 m/s vT = 2000 kg mSUV

EVALUATE: The SUV must have less speed to have the same kinetic energy as the truck than to have the same momentum as the truck. IDENTIFY: Example 8.1 shows that the two iceboats have the same kinetic energy at the finish line. K = 12 mv 2 . p = mv . SET UP:

Let A be the iceboat with mass m and let B be the iceboat with mass 2m, so mB = 2mA .

EXECUTE:

K A = K B gives

1 2

mv A2 = 12 mvB2 . v A =

mB vB = 2vB . mA

(

)

p A = mAv A . pB = mB vB = (2mA ) v A / 2 = 2mAv A = 2 p A .

8.3.

EVALUATE: The more massive boat must have less speed but greater momentum than the other boat in order to have the same kinetic energy. IDENTIFY and SET UP: p = mv . K = 12 mv 2 . 2

EXECUTE:

(a) v =

p2 p ⎛ p⎞ and K = 12 m ⎜ ⎟ = . m ⎝ m ⎠ 2m

(b) K c = K b and the result from part (a) gives

pc2 p2 mb 0.145 kg = b . pb = pc = pc = 1.90 pc . The baseball mc 2mc 2mb 0.040 kg

has the greater magnitude of momentum. pc / pb = 0.526 . (c) p 2 = 2mK so pm = pw gives 2mm K m = 2mw K w . w = mg , so wm K m = ww K w .

8.4.

⎛w ⎞ ⎛ 700 N ⎞ Kw = ⎜ m ⎟ Km = ⎜ ⎟ K m = 1.56 K m . ⎝ 450 N ⎠ ⎝ ww ⎠ The woman has greater kinetic energy. K m / K w = 0.641 . EVALUATE: For equal kinetic energy, the more massive object has the greater momentum. For equal momenta, the less massive object has the greater kinetic energy. IDENTIFY: Each momentum component is the mass times the corresponding velocity component. SET UP: Let +x be along the horizontal motion of the shotput. Let +y be vertically upward. vx = v cosθ , v y = v sin θ . EXECUTE:

The horizontal component of the initial momentum is px = mvx = mv cosθ = (7.30 kg)(15.0 m/s)cos 40.0° = 83.9 kg ⋅ m/s .

The vertical component of the initial momentum is p y = mv y = mv sin θ = (7.30 kg)(15.0 m/s)sin40.0° = 70.4 kg ⋅ m/s EVALUATE:

The initial momentum is directed at 40.0° above the horizontal.

8-1

8-2

8.5.

Chapter 8

! ! IDENTIFY: For each object, p = mv and K = 12 mv 2 . The total momentum is the vector sum of the momenta of each object. The total kinetic energy is the scalar sum of the kinetic energies of each object. SET UP: Let object A be the 110 kg lineman and object B the 125 kg lineman. Let +x be the object to the right, so v Ax = +2.75 m/s and vBx = −2.60 m/s . EXECUTE: (a) Px = mAv Ax + mB vBx = (110 kg)(2.75 m/s) + (125 kg)(−2.60 m/s) = −22.5 kg ⋅ m/s . The net momentum has magnitude 22.5 kg ⋅ m/s and is directed to the left.

8.6.

(b) K = 12 m Av A2 + 12 mB vB2 = 12 (110 kg)(2.75 m/s) 2 + 12 (125 kg)(2.60 m/s) 2 = 838 J EVALUATE: The kinetic energy of an object is a scalar and is never negative. It depends only on the magnitude of the velocity of the object, not on its direction. The momentum of an object is a vector and has both magnitude and direction. When two objects are in motion, their total kinetic energy is greater than the kinetic energy of either one. But if they are moving in opposite directions, the net momentum of the system has a smaller magnitude than the magnitude of the momentum of either object. ! ! ! ! ! IDENTIFY: For each object p = mv and the net momentum of the system is P = pA + pB . The momentum vectors are added by adding components. The magnitude and direction of the net momentum is calculated from its x and y components. SET UP: Let object A be the pickup and object B be the sedan. v Ax = −14.0 m/s , v Ay = 0 . vBx = 0 , vBy = +23.0 m/s . (a) Px = p Ax + pBx = mAv Ax + mB vBx = (2500 kg)(−14.0 m/s) + 0 = −3.50 × 104 kg ⋅ m/s

EXECUTE:

Py = p Ay + pBy = mAv Ay + mB vBy = (1500 kg)(+23.0 m/s) = +3.45 × 104 kg ⋅ m/s (b) P = Px2 + Py2 = 4.91 × 104 kg ⋅ m/s . From Figure 8.6, tan θ =

Px Py

=

3.50 × 104 kg ⋅ m/s and θ = 45.4° . The net 3.45 × 104 kg ⋅ m/s

momentum has magnitude 4.91 × 104 kg ⋅ m/s and is directed at 45.4° west of north. EVALUATE: The momenta of the two objects must be added as vectors. The momentum of one object is west and the other is north. The momenta of the two objects are nearly equal in magnitude, so the net momentum is directed approximately midway between west and north.

Figure 8.6 8.7.

IDENTIFY: The average force on an object and the object’s change in momentum are related by Eq. 8.9. The weight of the ball is w = mg . SET UP:

Let +x be in the direction of the final velocity of the ball, so v1x = 0 and v2 x = 25.0 m/s .

EXECUTE:

8.8.

( Fav ) x (t2 − t1 ) = mv2 x − mv1x gives ( Fav ) x =

mv2 x − mv1x (0.0450 kg)(25.0 m/s) = = 562 N . t2 − t1 2.00 × 10−3 s

w = (0.0450 kg)(9.80 m/s 2 ) = 0.441 N . The force exerted by the club is much greater than the weight of the ball, so the effect of the weight of the ball during the time of contact is not significant. EVALUATE: Forces exerted during collisions typically are very large but act for a short time. IDENTIFY: The change in momentum, the impulse and the average force are related by Eq. 8.9. SET UP: Let the direction in which the batted ball is traveling be the +x direction, so v1x = −45.0 m/s and v2 x = 55.0 m/s . EXECUTE:

(a) Δpx = p2 x − p1x = m(v2 x − v1x ) = (0.145 kg)(55.0 m/s − [ −45.0 m/s]) = 14.5 kg ⋅ m/s . J x = Δpx , so

J x = 14.5 kg ⋅ m/s . Both the change in momentum and the impulse have magnitude 14.5 kg ⋅ m/s . J x 14.5 kg ⋅ m/s = = 7250 N . Δt 2.00 × 10−3 s EVALUATE: The force is in the direction of the momentum change. IDENTIFY: Use Eq. 8.9. We know the intial momentum and the impluse so can solve for the final momentum and then the final velocity. (b) ( Fav ) x = 8.9.

Momentum, Impulse, and Collisions

8-3

SET UP: Take the x-axis to be toward the right, so v1x = +3.00 m / s. Use Eq. 8.5 to calculate the impulse, since the force is constant. EXECUTE: (a) J x = p2 x − p1x

J x = Fx (t2 − t1 ) = (+25.0 N)(0.050 s) = +1.25 kg ⋅ m/s Thus p2 x = J x + p1x = +1.25 kg ⋅ m/s + (0.160 kg)( +3.00 m/s) = +1.73 kg ⋅ m/s

v2 x =

p2 x 1.73 kg ⋅ m/s = = +10.8 kg ⋅ m/s (to the right) 0.160 kg m

(b) J x = Fx (t2 − t1 ) = (−12.0 N)(0.050 s) = −0.600 kg ⋅ m/s (negative since force is to left)

p2 x = J x + p1x = −0.600 kg ⋅ m/s + (0.160 kg)(+3.00 m/s) = −0.120 kg ⋅ m/s v2 x =

8.10.

p2 x −0.120 kg ⋅ m/s = = −0.75 m/s (to the left) 0.160 kg m

EVALUATE: In part (a) the impulse and initial momentum are in the same direction and vx increases. In part (b) the impulse and initial momentum are in opposite directions and the velocity decreases. IDENTIFY: The impulse, change in momentum and change in velocity are related by Eq. 8.9. SET UP: Fy = 26,700 N and Fx = 0 . The force is constant, so ( Fav ) y = Fy . (a) J y = Fy Δt = (26,700 N)(3.90 s) = 1.04 × 105 N ⋅ s .

EXECUTE:

(b) Δp y = J y = 1.04 × 105 kg ⋅ m/s . (c) Δp y = mΔv y . Δv y =

8.11.

Δp y m

=

1.04 × 105 kg ⋅ m/s = 1.09 m/s . 95,000 kg

(d) The initial velocity of the shuttle isn’t known. The change in kinetic energy is ΔK = K 2 − K1 = 12 m(v22 − v12 ) . It depends on the initial and final speeds and isn’t determined solely by the change in speed. EVALUATE: The force in the +y direction produces an increase of the velocity in the +y direction. ! t2 ! IDENTIFY: The force is not constant so J = ∫ F dt . The impulse is related to the change in velocity by Eq. 8.9. t1

! t2 Only the x component of the force is nonzero, so J x = ∫ Fx dt is the only nonzero component of J .

SET UP:

t1

J x = m(v2 x − v1x ) . t1 = 2.00 s , t2 = 3.50 s . EXECUTE:

(a) A =

Fx 781.25 N = = 500 N/s 2 . (1.25 s) 2 t2

t2

(b) J x = ∫ At 2 dt = 13 A(t23 − t13 ) = 13 (500 N/s 2 )([3.50 s]3 − [2.00 s]3 ) = 5.81×103 N ⋅ s . t1

(c) Δvx = v2 x − v1x =

8.12.

J x 5.81× 103 N ⋅ s = = 2.70 m/s . The x component of the velocity of the rocket increases by m 2150 kg

2.70 m/s. EVALUATE: The change in velocity is in the same direction as the impulse, which in turn is in the direction of the net force. In this problem the net force equals the force applied by the engine, since that is the only force on the rocket. IDENTIFY: Apply Eq. 8.9 to relate the change in momentum of the momentum to the components of the average force on it. SET UP: Let +x be to the right and +y be upward. EXECUTE: (a) J x = Δpx = mv2 x − mv1x = (0.145 kg)(−[65.0 m/s]cos30° − 50.0 m/s) = −15.4 kg ⋅ m/s . J y = Δp y = mv2 y − mv1 y = (0.145 kg)([65.0 m/s]sin 30° − 0) = 4.71 kg ⋅ m/s The horizontal component is 15.4 kg ⋅ m/s , to the left and the vertical component is 4.71 kg ⋅ m/s , upward. J 4.71 kg ⋅ m/s J x −15.4 kg ⋅ m/s = = −8800 N . Fav-y = y = = 2690 N . −3 Δt 1.75 × 10 s Δt 1.75 × 10−3 s The horizontal component is 8800 N, to the left, and the vertical component is 2690 N, upward. EVALUATE: The ball gains momentum to the left and upward and the force components are in these directions. ! ! ! ! ! ! ! IDENTIFY: The force is constant during the 1.0 ms interval that it acts, so J = F Δt . J 5 p2 2 p1 5 m(v 2 2 v1 ) . ! SET UP: Let +x be to the right, so v1x = +5.00 m/s . Only the x component of J is nonzero, and (b) Fav-x =

8.13.

J x = m(v2 x − v1x ) .

8-4

Chapter 8

EXECUTE: (a) The magnitude of the impulse is J = F Δt = (2.50 × 103 N)(1.00 × 10−3 s) = 2.50 N ⋅ s . The direction of the impulse is the direction of the force. +2.50 N ⋅ s J (b) (i) v2 x = x + v1x . J x = +2.50 N ⋅ s . v2 x = + 5.00 m/s = 6.25 m/s . The stone’s velocity has magnitude 2.00 kg m

6.25 m/s and is directed to the right. (ii) Now J x = −2.50 N ⋅ s and v2 x =

8.14.

stone’s velocity has magnitude 3.75 m/s and is directed to the right. EVALUATE: When the force and initial velocity are in the same direction the speed increases and when they are in opposite directions the speed decreases. IDENTIFY: Apply conservation of momentum to the system of the astronaut and tool. SET UP: Let A be the astronaut and B be the tool. Let +x be the direction in which she throws the tool, so vB 2 x = +3.20 m/s . Assume she is initially at rest, so v A1x = vB1x = 0 . Solve for v A 2 x . EXECUTE:

8.15.

−2.50 N ⋅ s + 5.00 m/s = 3.75 m/s . The 2.00 kg

P1x = P2 x . P1x = mAv A1x + mB vB1x = 0 . P2 x = mAv A 2 x + mB vB 2 x = 0 and

mv (2.25 kg)(3.20 m/s) vA2 x = − B A2 x = − = −0.105 m/s . Her speed is 0.105 m/s and she moves opposite to the mA 68.5 kg direction in which she throws the tool. EVALUATE: Her mass is much larger than that of the tool so to have the same magnitude of momentum as the tool her speed is much less. IDENTIFY: Since drag effects are neglected there is no net external force on the system of squid plus expelled water and the total momentum of the system is conserved. Since the squid is initially at rest, with the water in its cavity, the initial momentum of the system is zero. For each object, K = 12 mv 2 . SET UP:

Let A be the squid and B be the water it expels, so mA = 6.50 kg − 1.75 kg = 4.75 kg . Let +x be the

direction in which the water is expelled. v A 2 x = −2.50 m/s . Solve for vB 2 x . EXECUTE:

(a) P1x = 0 . P2 x = P1x , so 0 = mAv A 2 x + mB vB 2 x . vB 2 x = −

m Av A 2 x (4.75 kg)(−2.50 m/s) =− = +6.79 m/s . mB 1.75 kg

(b) K 2 = K A 2 + K B 2 = 12 mAv A2 2 + 12 mB vB2 2 = 12 (4.75 kg)(2.50 m/s) 2 + 12 (1.75 kg)(6.79 m/s) 2 = 55.2 J The initial kinetic

8.16.

energy is zero, so the kinetic energy produced is K 2 = 55.2 J . EVALUATE: The two objects end up with momenta that are equal in magnitude and opposite in direction, so the total momentum of the system remains zero. The kinetic energy is created by the work done by the squid as it expels the water. IDENTIFY: Apply conservation of momentum to the system of you and the ball. In part (a) both objects have the same final velocity. SET UP: Let +x be in the direction the ball is traveling initially. m A = 0.400 kg (ball). mB = 70.0 kg (you). EXECUTE:

8.17.

(a) P1x = P2 x gives (0.400 kg)(10.0 m/s) = (0.400 kg + 70.0 kg)v2 and v2 = 0.0568 m/s .

(b) P1x = P2 x gives (0.400 kg)(10.0 m/s) = (0.400 kg)( − 8.00 m/s) + (70.0 kg)vB 2 and vB 2 = 0.103 m/s . EVALUATE: When the ball bounces off it has a greater change in momentum and you acquire a greater final speed. IDENTIFY: Apply conservation of momentum to the system of the two pucks. SET UP: Let +x be to the right. EXECUTE: (a) P1x = P2 x says (0.250)v A1 = (0.250 kg)(−0.120 m/s) + (0.350 kg)(0.650 m/s) and v A1 = 0.790 m/s . (b) K1 = 12 (0.250 kg)(0.790 m/s) 2 = 0.0780 J .

8.18.

K 2 = 12 (0.250 kg)(0.120 m/s) 2 + 12 (0.350 kg)(0.650 m/s) 2 = 0.0757 J and ΔK = K 2 − K1 = −0.0023 J . EVALUATE: The total momentum of the system is conserved but the total kinetic energy decreases. IDENTIFY: Since road friction is neglected, there is no net external force on the system of the two cars and the total momentum of the system is conserved. For each object, K = 12 mv 2 . SET UP:

Let A be the 1750 kg car and B be the 1450 kg car. Let +x be to the right, so v A1x = +1.50 m/s ,

vB1x = −1.10 m/s , and v A 2 x = +0.250 m/s . Solve for vB 2 x . EXECUTE:

(a) P1x = P2 x . mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x . vB 2 x =

vB 2 x =

mAv A1x + mB vB1x − mAv A 2 x . mB

(1750 kg)(1.50 m/s) + (1450 kg)(−1.10 m/s) − (1750 kg)(0.250 m/s) = 0.409 m/s . 1450 kg

After the collision the lighter car is moving to the right with a speed of 0.409 m/s.


8-5

(b) K1 = 12 mAv A21 + 12 mB vB21 = 12 (1750 kg)(1.50 m/s)2 + 12 (1450 kg)(1.10 m/s) 2 = 2846 J .

K 2 = 12 mAv A2 2 + 12 mB vB2 2 = 12 (1750 kg)(0.250 m/s) 2 + 12 (1450 kg)(0.409 m/s) 2 = 176 J .

8.19.

The change in kinetic energy is ΔK = K 2 − K1 = 176 J − 2846 J = −2670 J . EVALUATE: The total momentum of the system is constant because there is no net external force during the collision. The kinetic energy of the system decreases because of negative work done by the forces the cars exert on each other during the collision. IDENTIFY: Since the rifle is loosely held there is no net external force on the system consisting of the rifle, bullet and propellant gases and the momentum of this system is conserved. Before the rifle is fired everything in the system is at rest and the initial momentum of the system is zero. SET UP: Let +x be in the direction of the bullet’s motion. The bullet has speed 601 m/s − 1.85 m/s = 599 m/s relative to the earth. P2 x = prx + pbx + pgx , the momenta of the rifle, bullet and gases. vrx = −1.85 m/s and vbx = +599 m/s . EXECUTE:

P2 x = P1x = 0 . prx + pbx + pgx = 0 . pgx = − prx − pbx = −(2.80 kg)( −1.85 m/s) − (0.00720 kg)(599 m/s)

and pgx = +5.18 kg ⋅ m/s − 4.31 kg ⋅ m/s = 0.87 kg ⋅ m/s . The propellant gases have momentum 0.87 kg ⋅ m/s , in the

8.20.

same direction as the bullet is traveling. EVALUATE: The magnitude of the momentum of the recoiling rifle equals the magnitude of the momentum of the bullet plus that of the gases as both exit the muzzle. IDENTIFY: In part (a) no horizontal force implies Px is constant. In part (b) use the energy expression, Eq. 7.14, to find the potential energy intially in the spring. SET UP: Initially both blocks are at rest.


(a) mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x

0 = mAv A 2 x + mB vB 2 x ⎛m ⎞ ⎛ 3.00 kg ⎞ v A 2 x = − ⎜ B ⎟ vB 2 x = − ⎜ ⎟ (+1.20 m/s) = −3.60 m/s ⎝ 1.00 kg ⎠ ⎝ mA ⎠ Block A has a final speed of 3.60 m/s, and moves off in the opposite direction to B. (b) Use energy conservation: K1 + U1 + Wother = K 2 + U 2 . Only the spring force does work so Wother = 0 and U = U el . K1 = 0 (the blocks initially are at rest) U 2 = 0 (no potential energy is left in the spring) K 2 = 12 mAv A2 2 + 12 mB vB2 2 = 12 (1.00 kg)(3.60 m/s) 2 + 12 (3.00 kg)(1.20 m/s) 2 = 8.64 J U1 = U1,el the potential energy stored in the compressed spring. Thus U1,el = K 2 = 8.64 J

8.21.

EVALUATE: The blocks have equal and opposite momenta as they move apart, since the total momentum is zero. The kinetic energy of each block is positive and doesn’t depend on the direction of the block’s velocity, just on its magnitude. IDENTIFY: Since friction at the pond surface is neglected, there is no net external horizontal force and the horizontal component of the momentum of the system of hunter plus bullet is conserved. Both objects are initially at rest, so the initial momentum of the system is zero. Gravity and the normal force exerted by the ice together produce a net vertical force while the rifle is firing, so the vertical component of momentum is not conserved. SET UP: Let object A be the hunter and object B be the bullet. Let +x be the direction of the horizontal component of velocity of the bullet. Solve for v A 2 x .

8-6

Chapter 8

(a) vB 2 x = +965 m/s . P1x = P2 x = 0 . 0 = mAv A 2 x + mB vB 2 x and

EXECUTE:

vA2 x = −

8.22.

⎛ 4.20 × 10−3 kg ⎞ mB vB 2 x = − ⎜ ⎟ (965 m/s) = −0.0559 m/s . mA ⎝ 72.5 kg ⎠

⎛ 4.20 × 10−3 kg ⎞ (b) vB 2 x = vB 2 cosθ = (965 m/s)cos56.0° = 540 m/s . v A 2 x = − ⎜ ⎟ (540 m/s) = −0.0313 m/s . ⎝ 72.5 kg ⎠ EVALUATE: The mass of the bullet is much less than the mass of the hunter, so the final mass of the hunter plus gun is still 72.5 kg, to three significant figures. Since the hunter has much larger mass, her final speed is much less than the speed of the bullet. IDENTIFY: Assume the nucleus is initially at rest. K = 12 mv 2 . SET UP:

Let +x be to the right. v A 2 x = −v A and vB 2 x = +vB .

⎛m ⎞ (a) P2 x = P1x = 0 gives m Av A 2 x + mB vB 2 x = 0 . vB = ⎜ A ⎟ v A . ⎝ mB ⎠ 2 2 1 mv K m Av A m (b) A = 12 A A2 = = B . K B 2 mB vB mB ( mAv A / mB )2 mA EXECUTE:

8.23.

EVALUATE: The lighter fragment has the greater kinetic energy. IDENTIFY: Apply conservation of momentum to the nucleus and its fragments. The initial momentum is zero. The 214 Po nucleus has mass 214(1.67 × 10−27 kg) = 3.57 × 10−25 kg , where 1.67 × 10−27 kg is the mass of a nucleon

(proton or neutron). K = 12 mv 2 . SET UP: Let +x be the direction in which the alpha particle is emitted. The nucleus that is left after the decay has mass mn = 3.75 × 10−25 kg − mα = 3.57 × 10−25 kg − 6.65 × 10−27 kg = 3.50 × 10−25 kg . EXECUTE:

8.24.

P2 x = P1x = 0 gives mα vα + mn vn = 0 . vn =

⎛ 6.65 × 10−27 kg ⎞ 7 5 vn = ⎜ ⎟ (1.92 × 10 m/s) = 3.65 × 10 m/s . −25 ⎝ 3.50 × 10 kg ⎠ EVALUATE: The recoil velocity of the more massive nucleus is much less than the speed of the emitted alpha particle. IDENTIFY and SET UP: Let the +x-direction be horizontal, along the direction the rock is thrown. There is no net horizontal force, so Px is constant. Let object A be you and object B be the rock. EXECUTE:

0 = −mAv A + mB vB cos35.0° vA =

EVALUATE: 8.25.

mα 2 Kα 2(1.23 × 10−12 J) vα . vα = = = 1.92 × 107 m/s . mn 6.65 × 10−27 kg mα

mB vB cos35.0° = 2.11 m/s mA

Py is not conserved because there is a net external force in the vertical direction; as you throw the

rock the normal force exerted on you by the ice is larger than the total weight of the system. IDENTIFY: Each horizontal component of momentum is conserved. K = 12 mv 2 . SET UP: Let +x be the direction of Rebecca’s initial velocity and let the +y axis make an angle of 36.9° with respect to the direction of her final velocity. vD1x = vD1 y = 0 . vR1x = 13.0 m/s ; vR1 y = 0 .

vR 2 x = (8.00 m/s)cos53.1° = 4.80 m/s ; vR 2 y = (8.00 m/s)sin 53.1° = 6.40 m/s . Solve for vD2 x and vD2 y . EXECUTE:

vD2 x =

(a) P1x = P2 x gives mR vR1x = mR vR 2 x + mD vD2 x .

mR (vR1x − vR 2 x ) (45.0 kg)(13.0 m/s − 4.80 m/s) = = 5.68 m/s . mD 65.0 kg ⎛ 45.0 kg ⎞ mR vR 2 y = − ⎜ ⎟ (6.40 m/s) = −4.43 m/s . mD ⎝ 65.0 kg ⎠ v 4.43 m/s are sketched in Figure 8.25. tan θ = D2 y = and θ = 38.0° . vD2 x 5.68 m/s

P1 y = P2 y gives 0 = mR vR 2 y + mDvD2 y . vD2 y = −

! ! ! The directions of vR1 , vR2 and vD2 2 2 vD = vD2 x + vD2 y = 7.20 m/s .


8-7

2 (b) K1 = 12 mR vR1 = 12 (45.0 kg)(13.0 m/s) 2 = 3.80 × 103 J . 2 2 K 2 = 12 mR vR2 + 12 mD vD2 = 12 (45.0 kg)(8.00 m/s) 2 + 12 (65.0 kg)(7.20 m/s) 2 = 3.12 × 103 J .

ΔK = K 2 − K1 = −680 J . EVALUATE: Each component of momentum is separately conserved. The kinetic energy of the system increases. y

vR2

vR1 u

x

vD2

Figure 8.25 8.26.

IDENTIFY: There is no net external force on the system of astronaut plus canister, so the momentum of the system is conserved. SET UP: Let object A be the astronaut and object B be the canister. Assume the astronaut is initially at rest. After the collision she must be moving in the same direction as the canister. Let +x be the direction in which the canister is traveling initially, so v A1x = 0 , v A 2 x = +2.40 m/s , vB1x = +3.50 m/s , and vB 2 x = +1.20 m/s . Solve for mB .

mA (v A 2 x − v A1x ) (78.4 kg)(2.40 m/s − 0) = = 81.8 kg . vB1x − vB 2 x 3.50 m/s − 1.20 m/s EVALUATE: She must exert a force on the canister in the − x direction to reduce its velocity component in the +x direction. By Newton’s third law, the canister exerts a force on her that is in the +x direction and she gains velocity in that direction. IDENTIFY: The horizontal component of the momentum of the system of the rain and freight car is conserved. SET UP: Let +x be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the x axis. EXECUTE: (a) P1x = P2 x says (24,000 kg)(4.00 m/s) = (27,000 kg)v2 x and v2 x = 3.56 m/s . (b) After it lands in the car the water must gain horizontal momentum, so the car loses horizontal momentum. EVALUATE: The vertical component of the momentum is not conserved, because of the vertical external force exerted by the track. IDENTIFY: The x and y components of the momentum of the system of the two asteroids are separately conserved. SET UP: The before and after diagrams are given in Figure 8.28 and the choice of coordinates is indicated. Each asteroid has mass m. EXECUTE: (a) P1x = P2 x gives mv A1 = mv A 2 cos30.0° + mvB 2 cos 45.0° . 40.0 m/s = 0.866v A 2 + 0.707vB 2 and EXECUTE:

8.27.

8.28.

P1x = P2 x . mAv A1x + mB vB1x = m Av A 2 x + mB vB 2 x . mB =

0.707vB 2 = 40.0 m/s − 0.866v A 2 .

P2 y = P2 y gives 0 = mv A 2 sin 30.0° − mvB 2 sin 45.0° and 0.500v A 2 = 0.707vB 2 . Combining these two equations gives 0.500v A 2 = 40.0 m/s − 0.866v A 2 and v A 2 = 29.3 m/s . Then

⎛ 0.500 ⎞ vB 2 = ⎜ ⎟ (29.3 m/s) = 20.7 m/s . ⎝ 0.707 ⎠ (b) K1 = 12 mv A21 . K 2 = 12 mv A2 2 + 12 mvB2 2 .

K 2 v A2 2 + vB2 2 (29.3 m/s) 2 + (20.7 m/s) 2 = = = 0.804 . K1 v A21 (40.0 m/s) 2 ΔK K 2 − K1 K 2 = = − 1 = −0.196 . K1 K1 K1

19.6% of the original kinetic energy is dissipated during the collision. EVALUATE: We could use any directions we wish for the x and y coordinate directions, but the particular choice we have made is especially convenient.

8-8

Chapter 8

Figure 8.28 8.29.

IDENTIFY: Since drag effects are neglected there is no net external force on the system of two fish and the momentum of the system is conserved. The mechanical energy equals the kinetic energy, which is K = 12 mv 2 for each object. SET UP: Let object A be the 15.0 kg fish and B be the 4.50 kg fish. Let +x be the direction the large fish is moving initially, so v A1x = 1.10 m/s and vB1x = 0 . After the collision the two objects are combined and move with ! velocity v2 . Solve for v2 x . EXECUTE:

(a) P1x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x .

v2 x =

mAv A1x + mB vB1x (15.0 kg)(1.10 m/s) + 0 = = 0.846 m/s . 15.0 kg + 4.50 kg mA + mB

(b) K1 = 12 mAv A21 + 12 mB vB21 = 12 (15.0 kg)(1.10 m/s) 2 = 9.08 J . K 2 = 12 ( mA + mB )v22 = 12 (19.5 kg)(0.846 m/s) 2 = 6.98 J .

8.30.

ΔK = K 2 − K1 = −2.10 J . 2.10 J of mechanical energy is dissipated. EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined. IDENTIFY: There is no net external force on the system of the two otters and the momentum of the system is conserved. The mechanical energy equals the kinetic energy, which is K = 12 mv 2 for each object. ! SET UP: Let A be the 7.50 kg otter and B be the 5.75 kg otter. After the collision their combined velocity is v2 . Let +x be to the right, so v A1x = −5.00 m/s and vB1x = +6.00 m/s . Solve for v2 x . EXECUTE:

(a) P1x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x .

v2 x =

mAv A1x + mB vB1x (7.50 kg)(−5.00 m/s) + (5.75)(+6.00 m/s) = = −0.226 m/s . 7.50 kg + 5.75 kg mA + mB

(b) K1 = 12 mAv A21 + 12 mB vB21 = 12 (7.50 kg)(5.00 m/s)2 + 12 (5.75 kg)(6.00 m/s) 2 = 197.2 J .

K 2 = 12 ( mA + mB )v22 = 12 (13.25 kg)(0.226 m/s) 2 = 0.338 J .

8.31.

ΔK = K 2 − K1 = −197 J . 197 J of mechanical energy is dissipated. EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined. IDENTIFY: Treat the comet and probe as an isolated system for which momentum is conserved. SET UP: In part (a) let object A be the probe and object B be the comet. Let − x be the direction the probe is traveling just before the collision. After the collision the combined object moves with speed v2 . The change in velocity is Δv = v2 x − vB1x . In part (a) the impact speed of 37,000 km/h is the speed of the probe relative to the comet just before impact: v A1x − vB1x = −37,000 km/h . In part (b) let object A be the comet and object B be the earth. Let − x be the direction the comet is traveling just before the collision. The impact speed is 40,000 km/h, so v A1x − vB1x = −40,000 km/h . EXECUTE:

(a) P1x = P2 x . v2 x =

mAv A1x + mB vB1x . mA + mB

⎛ mA ⎞ ⎛ mB − mA − mB ⎞ ⎛ mA ⎞ Δv = v2 x − vB1x = ⎜ ⎟ v A1x + ⎜ ⎟ vB1x = ⎜ ⎟ ( v A1x − vB1x ) . m + m m + m ⎝ A ⎝ ⎠ ⎝ mA + mB ⎠ B ⎠ A B ⎛ ⎞ 372 kg −6 Δv = ⎜ ⎟ (−37,000 km/h) = −1.4 × 10 km/h . 14 + × 372 kg 0.10 10 kg ⎝ ⎠ The speed of the comet decreased by 1.4 × 10−6 km/h . This change is not noticeable.


8.32.

8-9

⎛ ⎞ 0.10 × 1014 kg −8 (b) Δv = ⎜ ⎟ (−40,000 km/h) = −6.7 × 10 km/h . The speed of the earth would change 14 24 × + × 0.10 10 kg 5.97 10 kg ⎝ ⎠ −8 by 6.7 × 10 km/h . This change is not noticeable. EVALUATE: v A1x − vB1x is the velocity of the projectile (probe or comet) relative to the target (comet or earth). The expression for Δv can be derived directly by applying momentum conservation in coordinates in which the target is initially at rest. IDENTIFY: The forces the two vehicles exert on each other during the collision are much larger than the horizontal forces exerted by the road, and it is a good approximation to assume momentum conservation. ! SET UP: Let +x be eastward. After the collision two vehicles move with a common velocity v2 . (a) P1x = P2 x gives mSCvSCx + mT vTx = (mSC + mT )v2 x .

EXECUTE:

v2 x =

mSCvSCx + mT vTx (1050 kg)(−15.0 m/s) + (6320 kg)(+10.0 m/s) = = 6.44 m/s . 1050 kg + 6320 kg mSC + mT

The final velocity is 6.44 m/s, eastward.

⎛m ⎞ ⎛ 1050 kg ⎞ (b) P1x = P2 x = 0 gives mSCvSCx + mT vTx = 0 . vTx = − ⎜ SC ⎟ vSCx = − ⎜ ⎟ (−15.0 m/s) = 2.50 m/s . The truck m ⎝ 6320 kg ⎠ ⎝ T ⎠ would need to have initial speed 2.50 m/s. (c) part (a): ΔK = 12 (7370 kg)(6.44 m/s) 2 − 12 (1050 kg)(15.0 m/s) 2 − 12 (6320 kg)(10.0 m/s) 2 = −2.81 × 105 J

8.33.

part (b): ΔK = 0 − 12 (1050 kg)(15.0 m/s) 2 − 12 (6320 kg)(2.50 m/s) 2 = −1.38 × 105 J . The change in kinetic energy has the greater magnitude in part (a). EVALUATE: In part (a) the eastward momentum of the truck has a greater magnitude than the westward momentum of the car and the wreckage moves eastward after the collision. In part (b) the two vehicles have equal magnitudes of momentum, the total momentum of the system is zero, and the wreckage is at rest after the collision. IDENTIFY: The forces the two players exert on each other during the collision are much larger than the horizontal forces exerted by the slippery ground and it is a good approximation to assume momentum conservation. Each component of momentum is separately conserved. ! SET UP: Let +x be east and +y be north. After the collision the two players have velocity v2 . Let the linebacker be object A and the halfback be object B, so v A1x = 0 , v A1 y = 8.8 m/s , vB1x = 7.2 m/s and vB1 y = 0 . Solve for

v2 x and v2 y . P1x = P2 x gives mAv A1x + mB vB1x = (mA + mB )v2 x .

EXECUTE:

v2 x =

mAv A1x + mB vB1x (85 kg)(7.2 m/s) = = 3.14 m/s . 110 kg + 85 kg mA + mB

P1 y = P2 y gives mAv A1 y + mB vB1 y = (mA + mB )v2 y . v2 y = v= v +v 2 2x

2 2y

mAv A1 y + mB vB1 y mA + mB

(110 kg)(8.8 m/s) = 4.96 m/s . 110 kg + 85 kg

= 5.9 m/s . tan θ =

8.34.

=

v2 y v2 x

=

4.96 m/s and θ = 58° . 3.14 m/s

The players move with a speed of 5.9 m/s and in a direction 58° north of east. EVALUATE: Each component of momentum is separately conserved. IDENTIFY: There is no net external force on the system of the two skaters and the momentum of the system is conserved. SET UP: Let object A be the skater with mass 70.0 kg and object B be the skater with mass 65.0 kg. Let +x be to the right, so v A1x = +2.00 m/s and vB1x = −2.50 m/s . After the collision the two objects are combined and move with ! velocity v2 . Solve for v2 x . EXECUTE:

P1x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x . v2 x =

mAv A1x + mB vB1x (70.0 kg)(2.00 m/s) + (65.0)(−2.50 m/s) = = −0.167 m/s . 70.0 kg + 65.0 kg mA + mB

The two skaters move to the left at 0.167 m/s. EVALUATE: There is a large decrease in kinetic energy.

8-10

8.35.

8.36.

Chapter 8

IDENTIFY: Neglect external forces during the collision. Then the momentum of the system of the two cars is conserved. SET UP: mS = 1200 kg , mL = 3000 kg . The small car has velocity vS and the large car has velocity vL . EXECUTE: (a) The total momentum of the system is conserved, so the momentum lost by one car equals the ! ! momentum gained by the other car. They have the same magnitude of change in momentum. Since p = mv and ! Δp is the same, the car with the smaller mass has a greater change in velocity.

⎛m ⎞ ⎛ 3000 kg ⎞ mSΔvS = mL ΔvL and ΔvS = ⎜ L ⎟ ΔvL = ⎜ ⎟ Δv = 2.50Δv . ⎝ 1200 kg ⎠ ⎝ mS ⎠ (b) The acceleration of the small car is greater, since it has a greater change in velocity during the collision. The large acceleration means a large force on the occupants of the small car and they would sustain greater injuries. EVALUATE: Each car exerts the same magnitude of force on the other car but the force on the compact has a greater effect on its velocity since its mass is less. IDENTIFY: The collision forces are large so gravity can be neglected during the collision. Therefore, the horizontal and vertical components of the momentum of the system of the two birds are conserved. SET UP: The system before and after the collision is sketched in Figure 8.36. Use the coordinates shown.


There is no external force on the system so P1x = P2 x and P1 y = P2 y .

P1x = P2 x gives (1.5 kg)(9.0 m/s) = (1.5 kg)vraven-2 cos φ and vraven-2 cos φ = 9.0 m/s . P1 y = P2 y gives (0.600 kg)(20.0 m/s) = (0.600 kg)( −5.0 m/s) + (1.5 kg)vraven-2 sin φ and vraven-2 sin φ = 10.0 m/s . 10.0 m/s and φ = 48° . 9.0 m/s EVALUATE: Due to its large initial speed the lighter falcon was able to produce a large change in the raven’s direction of motion. IDENTIFY: Since friction forces from the road are ignored, the x and y components of momentum are conserved. SET UP: Let object A be the subcompact and object B be the truck. After the collision the two objects move ! together with velocity v2 . Use the x and y coordinates given in the problem. v A1 y = vB1 y = 0 .

Combining these two equations gives tan φ =

8.37.

v2 x = (16.0 m/s)sin 24.0° = 6.5 m/s ; v2 y = (16.0 m/s)cos 24.0° = 14.6 m/s . EXECUTE:

P1x = P2 x gives mAv A1x = (mA + mB )v2 x . ⎛ m + mB ⎞ ⎛ 950 kg + 1900 kg ⎞ v A1x = ⎜ A ⎟ v2 x = ⎜ ⎟ (6.5 m/s) = 19.5 m/s . 950 kg m ⎝ ⎠ A ⎝ ⎠

P1 y = P2 y gives mAvB1 y = (mA + mB )v2 y . ⎛ m + mB ⎞ ⎛ 950 kg + 1900 kg ⎞ vB1 y = ⎜ A ⎟ v2 y = ⎜ ⎟ (14.6 m/s) = 21.9 m/s . 1900 kg m ⎝ ⎠ A ⎝ ⎠

8.38.

Before the collision the subcompact car has speed 19.5 m/s and the truck has speed 21.9 m/s. EVALUATE: Each component of momentum is independently conserved. IDENTIFY: Apply conservation of momentum to the collision. Apply conservation of energy to the motion of the block after the collision.


8-11

SET UP: Conservation of momentum applied to the collision between the bullet and the block: Let object A be the bullet and object B be the block. Let v A be the speed of the bullet before the collision and let V be the speed of the block with the bullet inside just after the collision.

Figure 8.38a

Px is constant gives mAv A = (mA + mB )V . Conservation of energy applied to the motion of the block after the collision: V

v50

#2

#1

y

A1B

x

0.230 m

Figure 8.38b

K1 + U1 + Wother = K 2 + U 2 EXECUTE:

Work is done by friction so Wother = W f = ( f k cos φ ) s = − f k s = − μ k mgs

U1 = U 2 = 0 (no work done by gravity) K1 = 12 mV 2 ; K 2 = 0 (block has come to rest) Thus

1 2

mV 2 − μ k mgs = 0

V = 2μ k gs = 2(0.20)(9.80 m/s 2 )(0.230 m) = 0.9495 m/s Use this in the conservation of momentum equation ⎛ m + mB ⎞ ⎛ 5.00 × 10−3 kg + 1.20 kg ⎞ = vA = ⎜ A V ⎟ ⎜ ⎟ (0.9495 m/s) = 229 m/s 5.00 × 10−3 kg ⎝ ⎠ ⎝ mA ⎠

8.39.

EVALUATE: When we apply conservation of momentum to the collision we are ignoring the impulse of the friction force exerted by the surface during the collision. This is reasonable since this force is much smaller than the forces the bullet and block exert on each other during the collision. This force does work as the block moves after the collision, and takes away all the kinetic energy. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. After the collision the kinetic energy of the combined object is converted to gravitational potential energy. SET UP: Immediately after the collision the combined object has speed V. Let h be the vertical height through which the pendulum rises. EXECUTE: (a) Conservation of momentum applied to the collision gives (12.0 ×10−3 kg)(380 m/s) = (6.00 kg + 12.0 × 10−3 kg)V and V = 0.758 m/s .

Conservation of energy applied to the motion after the collision gives h=

1 2

mtotV 2 = mtot gh and

V 2 (0.758 m/s) 2 = = 0.0293 m = 2.93 cm . 2 g 2(9.80 m/s 2 )

(b) K = 12 mbvb2 = 12 (12.0 × 10−3 kg)(380 m/s) 2 = 866 J .

8.40.

(c) K = 12 mtotV 2 = 12 (6.00 kg + 12.0 × 10−3 kg)(0.758 m/s) 2 = 1.73 J . EVALUATE: Most of the initial kinetic energy of the bullet is dissipated in the collision. IDENTIFY: Each component of horizontal momentum is conserved. SET UP: Let +x be east and +y be north. vS1 y = vA1x = 0 . vS2 x = (6.00 m/s)cos37.0° = 4.79 m/s ,

vS2 y = (6.00 m/s)sin 37.0° = 3.61 m/s , v A 2 x = (9.00 m/s)cos 23.0° = 8.28 m/s and v A 2 y = −(9.00 m/s)sin 23.0° = −3.52 m/s . EXECUTE:

P1x = P2 x gives mSvS1x = mSvS2x + mA vA 2 x . vS1x =

mSvS2x + mA vA 2 x (80.0 kg)(4.79 m/s) + (50.0 kg)(8.28 m/s) = = 9.97 m/s . mS 80.0 kg

Sam’s speed before the collision was 9.97 m/s.

8-12

Chapter 8

P1 y = P2 y gives mA vA1 y = mSvS2y + mA vA2 y . vA1 y =

8.41.

mSvS2y + mA vA2 y mS

=

(80.0 kg)(3.61 m/s) + (50.0 kg)(−3.52 m/s) = 2.26 m/s . 50.0 kg

Abigail’s speed before the collision was 2.26 m/s. (b) ΔK = 12 (80.0 kg)(6.00 m/s)2 + 12 (50.0 kg)(9.00 m/s)2 − 12 (80.0 kg)(9.97 m/s)2 − 12 (50.0 kg)(2.26 m/s)2 . ΔK = −639 J . EVALUATE: The total momentum is conserved because there is no net external horizontal force. The kinetic energy decreases because the forces between the objects do negative work during the collision. IDENTIFY: When the spring is compressed the maximum amount the two blocks aren’t moving relative to each ! other and have the same velocity V relative to the surface. Apply conservation of momentum to find V and conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give the final velocity of each block after the collision. SET UP: Let +x be the direction of the initial motion of A. EXECUTE: (a) Momentum conservation gives (2.00 kg)(2.00 m/s) = (12.0 kg)V and V = 0.333 m/s . Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy U b stored in the bumpers: 12 (2.00 kg)(2.00 m/s) 2 = U b + 12 (12.0 kg)(0.333 m/s) 2 and U b = 3.33 J .

8.42.

⎛ m − mB ⎞ ⎛ 2.00 kg − 10.0 kg ⎞ (b) v A 2 x = ⎜ A ⎟ v A1x = ⎜ ⎟ (2.00 m/s) = −1.33 m/s . Block A is moving in the − x direction at m + m 12.0 kg ⎝ ⎠ B ⎠ ⎝ A 1.33 m/s. ⎛ 2mA ⎞ 2(2.00 kg) vB 2 x = ⎜ (2.00 m/s) = +0.667 m/s . Block B is moving in the +x direction at 0.667 m/s. ⎟ v A1x = + m m 12.0 kg ⎝ A B ⎠ EVALUATE: When the spring is compressed the maximum amount the system must still be moving in order to conserve momentum. IDENTIFY: No net external horizontal force so Px is conserved. Elastic collision so K1 = K 2 and can use Eq. 8.27. SET UP:


From conservation of x-component of momentum: mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x mAv A1 − mB vB1 = mAv A 2 x + mB vB 2 x (0.150 kg)(0.80 m/s) − (0.300 kg)(2.20 m/s) = (0.150 kg)v A 2 x + (0.300 kg)vB 2 x −3.60 m/s = vA2x + 2vB 2 x

From the relative velocity equation for an elastic collision Eq. 8.27: vB 2 x − v A 2 x = −(vB1x − v A1x ) = −(−2.20 m/s − 0.80 m/s) = +3.00 m/s 3.00 m/s = −vA2x + vB 2 x

8.43.

Adding the two equations gives −0.60 m/s = 3vB 2 x and vB 2 x = −0.20 m/s. Then v A 2 x = vB 2 x − 3.00 m/s = −3.20 m/s. The 0.150 kg glider (A) is moving to the left at 3.20 m/s and the 0.300 kg glider (B) is moving to the left at 0.20 m/s. EVALUATE: We can use our v A 2 x and vB 2 x to show that Px is constant and K1 = K 2 IDENTIFY: Since the collision is elastic, both momentum conservation and Eq. 8.27 apply. SET UP: Let object A be the 30.0 kg marble and let object B be the 10.0 g marble. Let +x be to the right. EXECUTE: (a) Conservation of momentum gives (0.0300 kg)(0.200 m/s) + (0.0100 kg)(−0.400 m/s) = (0.0300 kg)v A 2 x + (0.0100 kg)vB 2 x . 3v A 2 x + vB 2 x = 0.200 m/s . Eq. 8.27 says vB 2 x − v A 2 x = −(−0.400 m/s − 0.200 m/s) = +0.600 m/s . Solving this pair of equations gives v A 2 x = −0.100 m/s and vB 2 x = +0.500 m/s . The 30.0 g marble is moving to the left at 0.100 m/s and the 10.0 g marble is moving to the right at 0.500 m/s.


8-13

(b) For marble A, ΔPAx = mAv A 2 x − mAv A1x = (0.0300 kg)(−0.100 m/s − 0.200 m/s) = −0.00900 kg ⋅ m/s .

For marble B, ΔPBx = mB vB 2 x − mB vB1x = (0.0100 kg)(0.500 m/s − [−0.400 m/s]) = +0.00900 kg ⋅ m/s . The changes in momentum have the same magnitude and opposite sign. (c) For marble A, ΔK A = 12 mAv A2 2 − 12 mAv A21 = 12 (0.0300 kg)([0.100 m/s]2 − [0.200 m/s]2 ) = −4.5 × 10−4 J .

8.44.

For marble B, ΔK B = 12 mB vB2 2 − 12 mB vB21 = 12 (0.0100 kg)([0.500 m/s]2 − [0.400 m/s]2 ) = +4.5 × 10−4 J . The changes in kinetic energy have the same magnitude and opposite sign. EVALUATE: The results of parts (b) and (c) show that momentum and kinetic energy are conserved in the collision. IDENTIFY and SET UP: Without rounding, the calculation in Example 8.12 gives vB 2 = 20 m/s . EXECUTE: The two equations in Example 8.12 for α and β are (0.500 kg)(4.00 m/s) = (0.500 kg)(2.00 m/s)(cosα ) + (0.300 kg)( 20 m/s)(cos β ) Eq. 1 and 0 = (0.500 kg)(2.00 m/s)(sin α ) − (0.300 kg)( 20 m/s)sin β Eq. 2. Dividing each equation by (0.500 kg)(1.00 m/s) gives 4.00 = 2.00cos α + 0.6 20 cos β Eq. 3 and 0 = 2.00sin α − 0.6 20 sin β Eq. 4. 4.00 − 2.00cos α and cos 2 β = 2.222 − 2.222cos α + 0.5556cos 2 α . 0.6 20 Eq. 4 gives sin β = 0.7454sin α and sin 2 β = 0.5556sin 2 α = 0.5556 − 0.5556cos 2 α . Eq. 3 gives cos β =

8.45.

Adding the two equations and using sin 2 β + cos 2 β = 1 gives 1 = 2.778 − 2.222cos α and cos α = 0.8002 . α = 36.9° . Then sin β = 0.7454sin α gives β = 26.6° . EVALUATE: For these values of α and β , the x component of momentum, the y component of momentum and the kinetic energy are all conserved in the collision. IDENTIFY: Eqs. 8.24 and 8.25 apply, with object A being the neutron. SET UP: Let +x be the direction of the initial momentum of the neutron. The mass of a neutron is mn = 1.0 u . ⎛ m − mB ⎞ 1.0 u − 2.0 u v A1x = −v A1x / 3.0 . The speed of the neutron after the collision (a) v A 2 x = ⎜ A ⎟ v A1x = 1.0 u + 2.0 u ⎝ mA + mB ⎠ is one-third its initial speed. 1 (b) K 2 = 12 mn vn2 = 12 mn (v A1 / 3.0) 2 = K1 . 9.0 EXECUTE:

n

8.46.

n

1 ⎛ 1 ⎞ ⎛ 1 ⎞ (c) After n collisions, v A 2 = ⎜ , so 3.0n = 59,000 . n log3.0 = log59,000 and n = 10 . ⎟ = ⎟ v A1 . ⎜ ⎝ 3.0 ⎠ 59,000 ⎝ 3.0 ⎠ EVALUATE: Since the collision is elastic, in each collision the kinetic energy lost by the neutron equals the kinetic energy gained by the deuteron. IDENTIFY: Elastic collision. Solve for mass and speed of target nucleus. SET UP: (a) Let A be the proton and B be the target nucleus. The collision is elastic, all velocities lie along a line, and B is at rest before the collision. Hence the results of Eqs. 8.24 and 8.25 apply. EXECUTE: Eq. 8.24: mB (vx + v Ax ) = mA (vx − v Ax ), where vx is the velocity component of A before the collision and v Ax is the velocity component of A after the collision. Here, vx = 1.50 × 107 m/s (take direction of incident beam to be positive) and v Ax = −1.20 × 107 m/s (negative since traveling in direction opposite to incident beam). ⎛v −v ⎞ ⎛ 1.50 × 107 m/s + 1.20 × 107 m/s ⎞ ⎛ 2.70 ⎞ mB = mA ⎜ x Ax ⎟ = m ⎜ ⎟ = m⎜ ⎟ = 9.00m. 7 7 + × − × v v 1.50 10 m/s 1.20 10 m/s 0.30 ⎠ ⎝ ⎝ ⎠ Ax ⎠ ⎝ x ⎛ 2mA ⎞ 2m ⎛ ⎞ 7 6 (b) Eq. 8.25: vBx = ⎜ ⎟v = ⎜ ⎟ (1.50 × 10 m/s) = 3.00 × 10 m/s. + + m m m 9.00 m ⎝ ⎠ ⎝ A B ⎠ EVALUATE: Can use our calculated vBx and mB to show that Px is constant and that K1 = K 2 .

8-14

8.47.

Chapter 8

IDENTIFY: Apply Eq. 8.28. SET UP: mA = 0.300 kg , mB = 0.400 kg , mC = 0.200 kg . EXECUTE:

xcm =

mA x A + mB xB + mC xC . mA + mB + mC

xcm =

(0.300 kg)(0.200 m) + (0.400 kg)(0.100 m) + (0.200 kg)(−0.300 m) = 0.0444 m . 0.300 kg + 0.400 kg + 0.200 kg ycm =

ycm =

8.48.

mA y A + mB yB + mC yC . mA + mB + mC

(0.300 kg)(0.300 m) + (0.400 kg)(−0.400 m) + (0.200 kg)(0.600 m) = 0.0556 m . 0.300 kg + 0.400 kg + 0.200 kg

EVALUATE: There is mass at both positive and negative x and at positive and negative y and therefore the center of mass is close to the origin. IDENTIFY: Calculate xcm . SET UP: Apply Eq. 8.28 with the sun as mass 1 and Jupiter as mass 2. Take the origin at the sun and let Jupiter lie on the positive x-axis.

Figure 8.48

xcm = EXECUTE:

x1 = 0 and x2 = 7.78 × 1011 m xcm =

8.49.

m1 x1 + m2 x2 m1 + m2

(1.90 ×10

27

kg )( 7.78 × 1011 m )

1.99 × 1030 kg + 1.90 × 1027 kg

= 7.42 × 108 m

The center of mass is 7.42 × 108 m from the center of the sun and is on the line connecting the centers of the sun and Jupiter. The sun’s radius is 6.96 × 108 m so the center of mass lies just outside the sun. EVALUATE: The mass of the sun is much greater than the mass of Jupiter so the center of mass is much closer to the sun. For each object we have considered all the mass as being at the center of mass (geometrical center) of the object. IDENTIFY: The location of the center of mass is given by Eq. 8.48. The mass can be expressed in terms of the diameter. Each object can be replaced by a point mass at its center. SET UP: Use coordinates with the origin at the center of Pluto and the +x direction toward Charon, so xP = 0 xC = 19,700 km . m = ρV = ρ 43 π r 3 = 16 ρπ d 3 . EXECUTE:

xcm =

1 ⎛ ⎞ ⎛ dC3 ⎞ ρπ d C3 mP xP + mC xC ⎛ mC ⎞ 6 x x . =⎜ = ⎟ ⎟ xC = ⎜ 1 ⎜ C 3 3 3 3 ⎟ C 1 mP + mC ⎝ mP + mC ⎠ ⎝ dP + dC ⎠ ⎝ 6 ρπ d P + 6 ρπ d C ⎠

⎛ ⎞ [1250 km]3 xcm = ⎜ (19,700 km) = 2.52 × 103 km . 3 3 ⎟ + [2370 km] [1250 km] ⎝ ⎠

8.50.

The center of mass of the system is 2.52 × 103 km from the center of Pluto. EVALUATE: The center of mass is closer to Pluto because Pluto has more mass than Charon. IDENTIFY: Apply Eqs. 8.28, 8.30 and 8.32. There is only one component of position and velocity. SET UP: mA = 1200 kg , mB = 1800 kg . M = mA + mB = 3000 kg . Let +x be to the right and let the origin be at the center of mass of the station wagon. m x + mB xB 0 + (1800 kg)(40.0 m) = = 24.0 m. EXECUTE: (a) xcm = A A mA + mB 1200 kg + 1800 kg The center of mass is between the two cars, 24.0 m to the right of the station wagon and 16.0 m behind the lead car.


8-15

(b) Px = mAv A1 + mB vB1 = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 5.04 × 104 kg ⋅ m/s. (c) vcm, x =

8.51.

8.52.

mAv A, x + mB vB , x mA + mB

=

(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 16.8 m/s. 1200 kg + 1800 kg

(d) Px = Mvcm-x = (3000 kg)(16.8 m/s) = 5.04 × 104 kg ⋅ m/s , the same as in part (b). EVALUATE: The total momentum can be calculated either as the vector sum of the momenta of the individual objects in the system, or as the total mass of the system times the velocity of the center of mass. IDENTIFY: Use Eq. 8.28 to find the x and y coordinates of the center of mass of the machine part for each configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be represented by a point mass at its geometrical center. SET UP: Use coordinates with the axis at the hinge and the +x and +y axes along the horizontal and vertical bars in the figure in the problem. Let ( xi , yi ) and ( xf , yf ) be the coordinates of the bar before and after the vertical bar is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar and 3 be the ball. m x + m2 x2 + m3 x3 (4.00 kg)(0.750 m) + 0 + 0 = = 0.333 m . EXECUTE: xi = 1 1 m1 + m2 + m3 4.00 kg + 3.00 kg + 2.00 kg

yi =

m1 y1 + m2 y2 + m3 y3 0 + (3.00 kg)(0.900 m) + (2.00 kg)(1.80 m) = = 0.700 m . m1 + m2 + m3 9.00 kg

xf =

(4.00 kg)(0.750 m) + (3.00 kg)( −0.900 m) + (2.00 kg)( −1.80 m) = −0.366 m . 9.00 kg

yf = 0 . xf − xi = −0.700 m and yf − yi = −0.700 m . The center of mass moves 0.700 m to the right and 0.700 m upward. EVALUATE: The vertical bar moves upward and to the right so it is sensible for the center of mass of the machine part to move in these directions. (a) IDENTIFY: Use Eq. 8.28. SET UP: The target variable is m1. EXECUTE:

xcm = 2.0 m, ycm = 0 xcm =

m1 x1 + m2 x2 m1 ( 0 ) + ( 0.10 kg )( 8.0 m ) 0.80 kg ⋅ m = = . m1 + m2 m1 + ( 0.10 kg ) m1 + 0.10 kg xcm = 2.0 m gives 2.0 m = m1 + 0.10 kg =

0.80 kg ⋅ m . m1 + 0.10 kg

0.80 kg ⋅ m = 0.40 kg. 2.0 m

m1 = 0.30 kg. The cm is closer to m1 so its mass is larger then m2 . ! (b) IDENTIFY: Use Eq. 8.32 to calculate P . ! SET UP: v 5 ( 5.0 m/s ) ˆj. EVALUATE:

cm

! ! P 5 Mvcm 5 ( 0.10 kg + 0.30 kg )( 5.0 m/s ) iˆ 5 ( 2.0 kg ⋅ m/s ) iˆ. (c) IDENTIFY: Use Eq. 8.31. ! ! ! ! m v + m2v2 SET UP: vcm 5 1 1 . The target variable is v1. Particle 2 at rest says v2 = 0. m1 + m2

⎛ 0.30 kg + 0.10 kg ⎞ ! ⎛ m + m2 ⎞ ! ˆ ˆ v1 5 ⎜ 1 ⎟ vcm 5 ⎜ ⎟ ( 5.00 m/s ) i 5 ( 6.7 m/s ) i . m 0.30 kg ⎝ ⎠ ⎝ ⎠ 1 ! ! ! EVALUATE: Using the result of part (c) we can calculate p1 and p2 and show that P as calculated in part (b) ! ! does equal p1 1 p2 . IDENTIFY: There is no net external force on the system of James, Ramon and the rope and the momentum of the system is conserved and the velocity of its center of mass is constant. Initially there is no motion, and the velocity of the center of mass remains zero after Ramon has started to move. EXECUTE:

8.53.

8-16

Chapter 8

SET UP:

Let +x be in the direction of Ramon’s motion. Ramon has mass mR = 60.0 kg and James has mass

mJ = 90.0 kg . vcm-x =

EXECUTE:

8.54.

mR vRx + mJ vJx =0. mR + mJ

⎛m ⎞ ⎛ 60.0 kg ⎞ vJx = − ⎜ R ⎟ vRx = − ⎜ ⎟ (0.700 m/s) = −0.47 m/s . James’ speed is 0.47 m/s. ⎝ 90.0 kg ⎠ ⎝ mJ ⎠ EVALUATE: As they move, the two men have momenta that are equal in magnitude and opposite in direction, and the total momentum of the system is zero. Also, Example 8.14 shows that Ramon moves farther than James in the same time interval. This is consistent with Ramon having a greater speed. (a) IDENTIFY and SET UP: Apply Eq. 8.28 and solve for m1 and m2 . ycm =

EXECUTE:

m1 + m2 =

m1 y1 + m2 y2 m1 + m2

m1 y1 + m2 y2 m1 (0) + (0.50 kg)(6.0 m) = = 1.25 kg and m1 = 0.75 kg. ycm 2.4 m

EVALUATE: ycm is closer to m1 since m1 > m2 . ! ! (b) IDENTIFY and SET UP: Apply a 5 d v / dt for the cm motion. ! ! dv EXECUTE: acm 5 cm 5 (1.5 m/s3 ) tiˆ. dt (c) IDENTIFY and SET UP: Apply Eq. 8.34. ! ! EXECUTE: ∑ Fext 5 Macm 5 (1.25 kg ) (1.5 m/s3 ) tiˆ. ! At t = 3.0 s, ∑ Fext 5 (1.25 kg ) (1.5 m/s3 ) ( 3.0 s ) iˆ 5 ( 5.6 N ) iˆ.

8.55.

8.56.

8.57.

! EVALUATE: vcm- x is positive and increasing so acm − x is positive and Fext is in the + x -direction. There is no motion and no force component in the y-direction. ! ! dP to the airplane. IDENTIFY: Apply ∑ F = dt d n SET UP: ( t ) = nt n −1 . 1 N = 1 kg ⋅ m/s2 . dt ! ! ! dP 5 [ −(1.50 kg ⋅ m/s3 )t ]i 1 (0.25 kg ⋅ m/s 2 ) j . Fx = −(1.50 N/s)t , Fy = 0.25 N , Fz = 0 . EXECUTE: dt EVALUATE: There is no momentum or change in momentum in the z direction and there is no force component in this direction. IDENTIFY: Use Eq. 8.38, applied to a finite time interval. SET UP: vex = 1600 m/s Δm −0.0500 kg = −(1600 m/s) = +80.0 N . Δt 1.00 s (b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by ejecting the gas in a direction with a component perpendicular to the rocket’s velocity and braked by ejecting it in a direction parallel (as opposed to antiparallel) to the rocket’s velocity. EVALUATE: The thrust depends on the speed of the ejected gas relative to the rocket and on the mass of gas ejected per second. v dm IDENTIFY: a = − ex . Assume that dm / dt is constant over the 5.0 s interval, since m doesn’t change much m dt dm . during that interval. The thrust is F = −vex dt SET UP: Take m to have the constant value 110 kg + 70 kg = 180 kg . dm / dt is negative since the mass of the MMU decreases as gas is ejected. dm m ⎛ 180 kg ⎞ 2 = − a = −⎜ EXECUTE: (a) ⎟ (0.029 m/s ) = −0.0106 kg/s . In 5.0 s the mass that is ejected is dt vex ⎝ 490 m/s ⎠ (0.0106 kg/s)(5.0 s) = 0.053 kg . EXECUTE:

(a) F = −vex

(b) F = −vex

dm = −(490 m/s)( −0.0106 kg/s) = 5.19 N . dt


8.58.

8-17

EVALUATE: The mass change in the 5.0 s is a very small fraction of the total mass m, so it is accurate to take m to be constant. v dm IDENTIFY and SET UP: Apply Eq. 8.39: a = − ex . Solve for dm / dt. m dt EXECUTE:

( 6000 kg ) ( 25.0 m/s 2 ) dm ma =− =− = −75.0 kg/s . dt vex 2000 m/s

8.59.

So in 1 s the rocket must eject 75.0 kg of gas. EVALUATE: We have approximated dm / dt by Δm / Δt. We have assumed that 25.0 m/s 2 is the average acceleration for the first second. IDENTIFY: Use Eq. 8.39, applied to a finite time interval. Solve for vex .

8.60.

Δm m =− . 160 Δt a −15.0 m/s 2 v Δm = = 2.40 × 103 m/s = 2.40 km/s EXECUTE: a = − ex . vex = − ⎛ Δm ⎞ ⎛ m ⎞ m Δt /m⎟ ⎜− ⎜ ⎟/m ⎝ Δt ⎠ ⎝ 160 ⎠ EVALUATE: The acceleration is proportional to the speed of the exhaust gas and to the rate at which mass is ejected. IDENTIFY and SET UP: ( Fav ) Δt = J relates the impulse J to the average thrust Fav . Eq. 8.38 applied to a finite time SET UP:

interval gives Fav = −vex EXECUTE: (b) vex = −

8.61.

(a) F =

Δm ⎛m ⎞ . v − v0 = vex ln ⎜ 0 ⎟ . The remaining mass m after 1.70 s is 0.0133 kg. Δt ⎝m⎠

J 10.0 N ⋅ s = = 5.88 N . Fav / Fmax = 0.442 . Δt 1.70 s

Fav Δt = 800 m/s . −0.0125 kg

⎛ 0.0258 kg ⎞ ⎛m ⎞ (c) v0 = 0 and v = vex ln ⎜ 0 ⎟ = (800 m/s)ln ⎜ ⎟ = 530 m/s . m ⎝ ⎠ ⎝ 0.0133 kg ⎠ EVALUATE: The acceleration of the rocket is not constant. It increases as the mass remaining decreases. ⎛m ⎞ IDENTIFY: v − v0 = vex ln ⎜ 0 ⎟ . ⎝m⎠ SET UP: v0 = 0 . 3 m ⎛ m ⎞ v 8.00 × 10 m/s ln ⎜ 0 ⎟ = = = 3.81 and 0 = e3.81 = 45.2 . 2100 m/s m ⎝ m ⎠ vex EVALUATE: Note that the final speed of the rocket is greater than the relative speed of the exhaust gas. IDENTIFY and SET UP: Use Eq. 8.40: v − v0 = vex ln ( m0 / m ) .

EXECUTE: 8.62.

v0 = 0 (“fired from rest”), so v / vex = ln ( m0 / m ) . Thus m0 / m = ev / vex , or m / m0 = e − v / vex . If v is the final speed then m is the mass left when all the fuel has been expended; m / m0 is the fraction of the initial mass that is not fuel. (a) EXECUTE: v = 1.00 × 10−3 c = 3.00 × 105 m/s gives 5

m / m0 = e − (3.00×10

8.63.

m/s) /(2000 m/s)

= 7.2 × 10−66 .

EVALUATE: This is clearly not feasible, for so little of the initial mass to not be fuel. (b) EXECUTE: v = 3000 m/s gives m / m0 = e − (3000 m/s)/(2000 m/s) = 0.223 . EVALUATE: 22.3% of the total initial mass not fuel, so 77.7% is fuel; this is possible. IDENTIFY: Use the heights to find v1 y and v2 y , the velocity of the ball just before and just after it strikes the slab.

Then apply J y = Fy Δt = Δp y . SET UP:

Let +y be downward.

8-18

Chapter 8

EXECUTE:

(a)

1 2

mv 2 = mgh so v = ± 2 gh .

v1 y = + 2(9.80 m/s 2 )(2.00 m) = 6.26 m/s . v2 y = − 2(9.80 m/s 2 )(1.60 m) = −5.60 m/s . J y = Δp y = m(v2 y − v1 y ) = (40.0 × 10−3 kg)(−5.60 m/s − 6.26 m/s) = −0.474 kg ⋅ m/s . The impulse is 0.474 kg ⋅ m/s , upward. −0.474 kg ⋅ m/s = = −237 N . The average force on the ball is 237 N, upward. Δt 2.00 × 10−3 s EVALUATE: The upward force on the ball changes the direction of its momentum. IDENTIFY: Momentum is conserved in the explosion. At the highest point the velocity of the boulder is zero. Since one fragment moves horizontally the other fragment also moves horizontally. Use projectile motion to relate the initial horizontal velocity of each fragment to its horizontal displacement. SET UP: Use coordinates where +x is north. Since both fragments start at the same height with zero vertical component of velocity, the time in the air, t, is the same for both. Call the fragments A and B, with A being the one that lands to the north. Therefore, mB = 3mA . (b) Fy = 8.64.

Jy

EXECUTE:

Apply P1x = P2 x to the collision: 0 = mAv Ax + mB vBx . vBx = −

to the motion after the collision: x − x0 = v0 xt . Since t is the same,

8.65.

mA v Ax = −v Ax / 3 . Apply projectile motion mB

( x − x0 ) A ( x − x0 ) B = and v Ax vBx

⎛v ⎞ ⎛ −v / 3 ⎞ ( x − x0 ) B = ⎜ Bx ⎟ ( x − x0 ) A = ⎜ Ax ⎟ ( x − x0 ) A = −(274 m) / 3 = −91.3 m . The other fragment lands 91.3 m v ⎝ Ax ⎠ ⎝ v Ax ⎠ directly south of the point of explosion. EVALUATE: The fragment that has three times the mass travels one-third as far. IDENTIFY: The impulse, force and change in velocity are related by Eq. 8.9 ! ! SET UP: m = w / g = 0.0571 kg . Since the force is constant, F = Fav . EXECUTE:

(a) J x = Fx Δt = ( −380 N)(3.00 × 10−3 s) = −1.14 N ⋅ s . J y = Fy Δt = (110 N)(3.00 ×10−3 s) = 0.330 N ⋅ s .

J Jx −1.14 N ⋅ s 0.330 N ⋅ s + v1x = + 20.0 m/s = 0.04 m/s . v2 y = y − v1 y = + (−4.0 m/s) = +1.8 m/s . m 0.0571 kg 0.0571 kg m ! ! EVALUATE: The change in velocity Δv is in the same direction as the force, so Δv has a negative x component and a positive y component. IDENTIFY: The horizontal component of the momentum of the system of cars is conserved. SET UP: Let +x be the direction the cars are traveling. Each car has mass m. Let v1 be the initial speed of the (b) v2 x =

8.66.

three cars. v2 = 15 v1 . Let N be the number of cars in the final collection. 3v1 v = 3 1 = 15 . v2 v1 / 5 EVALUATE: In the complete absence of friction or other external horizontal forces this process of adding cars and slowing down continues forever. IDENTIFY: Px = p Ax + pBx and Py = p Ay + pBy . EXECUTE:

8.67.

SET UP:

P1x = P2 x . (3m)v1 = ( Nm)v2 . N =

Let object A be the convertible and object B be the SUV. Let +x be west and +y be south, p Ax = 0 and

pBy = 0 . EXECUTE:

vBx =

Px = (8000 kg ⋅ m/s)sin 60.0° = 6928 kg ⋅ m/s , so pBx = 6928 kg ⋅ m/s and

6928 kg ⋅ m/s = 3.46 m/s . 2000 kg

Py = (8000 kg ⋅ m/s)cos 60.0° = 4000 kg ⋅ m/s , so pBx = 4000 kg ⋅ m/s and v Ay =

8.68.

4000 kg ⋅ m/s = 2.67 m/s . 1500 kg

The convertible has speed 2.67 m/s and the SUV has speed 3.46 m/s. EVALUATE: Each component of the total momentum arises from a single vehicle. IDENTIFY: The total momentum of the system is conserved and is equal to zero, since the pucks are released from rest. SET UP: Each puck has the same mass m. Let +x be east and +y be north. Let object A be the puck that moves west. All three pucks have the same speed v.


EXECUTE:

8-19

P1x = P2 x gives 0 = −mv + mvBx + mvCx and v = vBx + vCx . P1 y = P2 y gives 0 = mvBy + mvCy and

vBy = −vCy . Since vB = vC and the y components are equal in magnitude, the x components must also be equal: vBx = vCx and v = vBx + vCx says vBx = vCx = v / 2 . If vBy is positive then vCy is negative. The angle θ that puck B v/2 and θ = 60° . One puck moves in a direction 60° north of east and v the other puck moves in a direction 60° south of east. EVALUATE: Each component of momentum is separately conserved. IDENTIFY: The x and y components of the momentum of the system are conserved. ! Set Up: After the collision the combined object with mass mtot = 0.100 kg moves with velocity v2 . Solve for makes with the x axis is given by cosθ =

8.69.

vCx and vCy . EXECUTE:

(a) P1x = P2 x gives m Av Ax + mB vBx + mC vCx = mtot v2 x .

vCx = − vCx = −

mAv Ax + mB vBx − mtot v2 x mC

(0.020 kg)( −1.50 m/s) + (0.030 kg)( −0.50 m/s)cos60° − (0.100 kg)(0.50 m/s) . 0.050 kg vCx = 1.75 m/s .

P1 y = P2 y gives mAv Ay + mB vBy + mC vCy = mtot v2 y . vCy = −

mAv Ay + mB vBy − mtot v2 y mC

=−

(0.030 kg)( −0.50 m/s)sin 60° = +0.260 m/s . 0.050 kg

2 2 (b) vC = vCx + vCy = 1.77 m/s . ΔK = K 2 − K1 .

8.70.

ΔK = 12 (0.100 kg)(0.50 m/s) 2 − [ 12 (0.020 kg)(1.50 m/s) 2 + 12 (0.030)(0.50 m/s) 2 + 12 (0.050 kg)(1.77 m/s) 2 ] ΔK = −0.092 J . EVALUATE: Since there is no horizontal external force the vector momentum of the system is conserved. The forces the spheres exert on each other do negative work during the collision and this reduces the kinetic energy of the system. IDENTIFY: Use a coordinate system attached to the ground. Take the x-axis to be east (along the tracks) and the y-axis to be north (parallel to the ground and perpendicular to the tracks). Then Px is conserved and Py is not conserved, due to the sideways force exerted by the tracks, the force that keeps the handcar on the tracks. (a) SET UP: Let A be the 25.0 kg mass and B be the car (mass 175 kg). After the mass is thrown sideways relative to the car it still has the same eastward component of velocity, 5.00 m/s, as it had before it was thrown.

Figure 8.70a

Px is conserved so ( mA + mB ) v1 = mAv A 2 x + mB vB 2 x EXECUTE:

( 200 kg )( 5.00 m/s ) = ( 25.0 kg )( 5.00 m/s ) + (175 kg ) vB 2 x . vB 2 x =

1000 kg ⋅ m/s − 125 kg ⋅ m/s = 5.00 m/s. 175 kg

The final velocity of the car is 5.00 m/s, east (unchanged). EVALUATE: The thrower exerts a force on the mass in the y-direction and by Newton’s 3rd law the mass exerts an equal and opposite force in the − y -direction on the thrower and car. (b) SET UP: We are applying Px = constant in coordinates attached to the ground, so we need the final velocity of A relative to the ground. Use the relative velocity addition equation. Then use Px = constant to find the final velocity of the car.

8-20

Chapter 8

EXECUTE:

! ! ! v A / E = v A / B + vB / E

vB / E = +5.00 m/s v A / B = −5.00 m/s (minus since the mass is moving west relative to the car). This gives v A / E = 0; the mass is at rest relative to the earth after it is thrown backwards from the car. As in part (a), ( mA + mB ) v1 = mAv A 2 x + mB vB 2 x . Now v A 2 x = 0, so ( mA + mB ) v1 = mB vB 2 x . ⎛ m + mB ⎞ ⎛ 200 kg ⎞ vB 2 x = ⎜ A ⎟ v1 = ⎜ ⎟ ( 5.00 m/s ) = 5.71 m/s. m ⎝ 175 kg ⎠ ⎝ ⎠ B The final velocity of the car is 5.71 m/s, east. EVALUATE: The thrower exerts a force in the − x-direction so the mass exerts a force on him in the + x -direction and he and the car speed up. (c) SET UP: Let A be the 25.0 kg mass and B be the car (mass mB = 200 kg).

Figure 8.70b Px is conserved so mAv A1x + mB vB1x = ( m A + mB ) v2 x .

EXECUTE:

− mAv A1 + mB vB1 = ( mA + mB ) v2 x . v2 x =

8.71.

8.72.

mB vB1 − mAv A1 ( 200 kg )( 5.00 m/s ) − ( 25.0 kg )( 6.00 m/s ) = = 3.78 m/s. mA + mB 200 kg + 25.0 kg

The final velocity of the car is 3.78 m/s, east. EVALUATE: The mass has negative px so reduces the total Px of the system and the car slows down. IDENTIFY: The horizontal component of the momentum of the sand plus railroad system is conserved. SET UP: As the sand leaks out it retains its horizontal velocity of 15.0 m/s. EXECUTE: The horizontal component of the momentum of the sand doesn’t change when it leaks out so the speed of the railroad car doesn’t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car. Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down. EVALUATE: The horizontal momentum of the sand does change when it strikes the ground, due to the force that is external to the system of sand plus railroad car. IDENTIFY: Kinetic energy is K = 12 mv 2 and the magnitude of the momentum is p = mv . The force and the time t it acts are related to the change in momentum whereas the force and distance d it acts are related to the change in kinetic energy. SET UP: Assume the net forces are constant and let the forces and the motion be along the x axis. The impulsemomentum theorem then says Ft = Δp and the work-energy theorem says Fd = ΔK .

EXECUTE: (a) K N = 12 (840 kg)(9.0 m/s) 2 = 3.40 × 104 J . K P = 12 (1620 kg)(5.0 m/s) 2 = 2.02 × 104 J . The Nash has the greater kinetic energy and

KN = 1.68 . KP

(b) pN = (840 kg)(9.0 m/s) = 7.56 × 103 kg ⋅ m/s . pP = (1620 kg)(5.0 m/s) = 8.10 × 103 kg ⋅ m/s . The Packard has the greater magnitude of momentum and

pN = 0.933 . pP

(c) Since the cars stop the magnitude of the change in momentum equals the initial momentum. Since pP > pN , FN pN = = 0.933 . FP pP (d) Since the cars stop the magnitude of the change in kinetic energy equals the initial kinetic energy. Since F K K N > K P , FN > FP and N = N = 1.68 . FP K P EVALUATE: If the stopping forces were the same, the Packard would have a larger stopping time but would travel a shorter distance while stopping. This consistent with it having a smaller initial speed. FP > FN and


8.73.

8.74.

8-21

IDENTIFY: Use the impulse-momentum theorem to relate the average force on the bullets to their rate of change in momentum. By Newton’s third law, the average force the weapon exerts on the bullets is equal in magnitude and opposite in direction to the recoil force the bullets exert on the weapon. SET UP: Consider a time interval of 1.00 minute. Let +x be the direction of motion of the bullets and use coordinated fixed to the ground. The bullets start from rest. (1000)(7.45 × 10−3 kg)(293 m/s) = 36.4 N . The recoil force is 36.4 N. EXECUTE: Fav Δt = Δp gives Fav = 60.0 s EVALUATE: The change in momentum for each bullet is small since the mass is small, but over 16 bullets are fired per second. IDENTIFY: Find k for the spring from the forces when the frame hangs at rest, use constant acceleration equations to find the speed of the putty just before it strikes the frame, apply conservation of momentum to the collision between the putty and the frame and then apply conservation of energy to the motion of the frame after the collision. SET UP: Use the free-body diagram for the frame when it hangs at rest on the end of the spring to find the force constant k of the spring. Let s be the amount the spring is stretched.

EXECUTE:

Figure 8.74a

∑ Fy = ma y .

− mg + ks = 0 . k=

mg ( 0.150 kg ) ( 9.80 m/s = s 0.050 m

2

) = 29.4 N/m .

SET UP: Next find the speed of the putty when it reaches the frame. The putty falls with acceleration a = g , downward.

Figure 8.74b v0 = 0 y − y0 = 0.300 m a = +9.80 m/s 2 v=? v 2 = v02 + 2a ( y − y0 )

EXECUTE: SET UP:

v = 2a ( y − y0 ) = 2 ( 9.80 m/s 2 ) ( 0.300 m ) = 2.425 m/s . Apply conservation of momentum to the collision between the putty (A) and the frame (B):

Figure 8.74c

Py is conserved, so − mAv A1 = − ( mA + mB ) v2 . EXECUTE:

⎛ mA ⎞ ⎛ 0.200 kg ⎞ v2 = ⎜ ⎟ v A1 = ⎜ ⎟ ( 2.425 m/s ) = 1.386 m/s . ⎝ 0.350 kg ⎠ ⎝ mA + mB ⎠

8-22

Chapter 8

SET UP: Apply conservation of energy to the motion of the frame on the end of the spring after the collision. Let point 1 be just after the putty strikes and point 2 be when the frame has its maximum downward displacement. Let d be the amount the frame moves downward.

Figure 8.74d

When the frame is at position 1 the spring is stretched a distance x1 = 0.050 m. When the frame is at position 2 the spring is stretched a distance x2 = 0.050 m + d . Use coordinates with the y-direction upward and y = 0 at the lowest point reached by the frame, so that y1 = d and y2 = 0. Work is done on the frame by gravity and by the spring force, so Wother = 0, and U = U el + U gravity . EXECUTE:

K1 + U1 + Wother = K 2 + U 2 .

Wother = 0 . K1 = 12 mv12 = 12 ( 0.350 kg )(1.386 m/s ) = 0.3362 J . 2

U1 = U1, el + U1, grav = 12 kx12 + mgy1 = 12 ( 29.4 N/m )( 0.050 m ) + ( 0.350 kg ) ( 9.80 m/s 2 ) d . 2

U1 = 0.03675 J + ( 3.43 N ) d .

( 29.4 N/m )( 0.050 m + d ) . U 2 = 0.03675 J + (1.47 N ) d + (14.7 N/m ) d 2 . Thus 0.3362 J + 0.03675 J + ( 3.43 N ) d = 0.03675 J + (1.47 N ) d + (14.7 N/m ) d 2 . (14.7 N/m ) d 2 − (1.96 N ) d − 0.3362 J = 0 . U 2 = U 2, el + U 2, grav = 12 kx22 + mgy2 =

8.75.

2

1 2

2 d = (1/ 29.4 ) ⎡1.96 ± (1.96 ) − 4 (14.7 )( −0.3362 ) ⎤ m = 0.0667 m ± 0.1653 m. ⎣⎢ ⎦⎥ The solution we want is a positive (downward) distance, so d = 0.0667 m + 0.1653 m = 0.232 m. EVALUATE: The collision is inelastic and mechanical energy is lost. Thus the decrease in gravitational potential energy is not equal to the increase in potential energy stored in the spring. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. SET UP: Let +x be to the right. The total mass is m = mbullet + mblock = 1.00 kg . The spring has force constant

k=

F 0.750 N = = 300 N/m . Let V be the velocity of the block just after impact. x 0.250 × 10−2 m

EXECUTE: (a) Conservation of energy for the motion after the collision gives K1 = U el2 . V =x

1 2

mV 2 = 12 kx 2 and

k 300 N/m = (0.150 m) = 2.60 m/s . m 1.00 kg

(b) Conservation of momentum applied to the collision gives mbullet v1 = mV . v1 =

8.76.

mV (1.00 kg)(2.60 m/s) = = 325 m/s . mbullet 8.00 × 10−3 kg

EVALUATE: The initial kinetic energy of the bullet is 422 J. The energy stored in the spring at maximum compression is 3.38 J. Most of the initial mechanical energy of the bullet is dissipated in the collision. IDENTIFY: The horizontal components of momentum of the system of bullet plus stone are conserved. The collision is elastic if K1 = K 2 .


8-23

SET UP: Let A be the bullet and B be the stone. (a)


Px is conserved so mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x .

m Av A1 = mB vB 2 x . ⎛m ⎞ ⎛ 6.00 × 10−3 kg ⎞ vB 2 x = ⎜ A ⎟ v A1 = ⎜ ⎟ ( 350 m/s ) = 21.0 m/s ⎝ 0.100 kg ⎠ ⎝ mB ⎠ Py is conserved so m Av A1 y + mB vB1 y = m Av A 2 y + mB vB 2 y . 0 = −mAv A 2 + mB vB 2 y . ⎛m ⎞ ⎛ 6.00 × 10−3 kg ⎞ vB 2 y = ⎜ A ⎟ v A 2 = ⎜ ⎟ ( 250 m/s ) = 15.0 m/s . ⎝ 0.100 kg ⎠ ⎝ mB ⎠ vB 2 = vB2 2 x + vB2 2 y = tan θ =

vB 2 y vB 2 x

=

( 21.0 m/s )

2

+ (15.0 m/s ) = 25.8 m/s . 2

15.0 m/s = 0.7143; θ = 35.5° (defined in the sketch). 21.0 m/s

(b) To answer this question compare K1 and K 2 for the system: K1 = 12 mAv A21 + 12 mB vB21 =

1 2

( 6.00 × 10

−3

kg ) ( 350 m/s ) = 368 J . 2

K 2 = 12 mAv A2 2 + 12 mB vB2 2 = 12 ( 6.00 × 10−3 kg ) ( 250 m/s ) + 12 ( 0.100 kg )( 25.8 m/s ) = 221 J . 2

2

ΔK = K 2 − K1 = 221 J − 368 J = −147 J .

EVALUATE: The kinetic energy of the system decreases by 147 J as a result of the collision; the collision is not elastic. Momentum is conserved because ∑ Fext, x = 0 and ∑ Fext, y = 0. But there are internal forces between the 8.77.

bullet and the stone. These forces do negative work that reduces K. IDENTIFY: Apply conservation of momentum to the collision between the two people. Apply conservation of energy to the motion of the stuntman before the collision and to the entwined people after the collision. SET UP: For the motion of the stuntman, y1 − y2 = 5.0 m . Let vS be the magnitude of his horizontal velocity just before the collision. Let V be the speed of the entwined people just after the collision. Let d be the distance they slide along the floor. EXECUTE: (a) Motion before the collision: K1 + U1 = K 2 + U 2 . K1 = 0 and 12 mvS2 = mg ( y1 − y2 ) . vS = 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(5.0 m) = 9.90 m/s . Collision: mSvS = mtotV . V =

⎛ 80.0 kg ⎞ mS vS = ⎜ ⎟ (9.90 m/s) = 5.28 m/s . mtot ⎝ 150.0 kg ⎠

(b) Motion after the collision: K1 + U1 + Wother = K 2 + U 2 gives

mtotV 2 − μ k mtot gd = 0 .

V2 (5.28 m/s) 2 = = 5.7 m . 2 μ k g 2(0.250)(9.80 m/s 2 ) EVALUATE: Mechanical energy is dissipated in the inelastic collision, so the kinetic energy just after the collision is less than the initial potential energy of the stuntman. IDENTIFY: Apply conservation of energy to the motion before and after the collision and apply conservation of momentum to the collision. SET UP: Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object have mass m. EXECUTE: Conservation of energy says 12 mv 2 = mgR; v = 2 gR . d=

8.78.

1 2

8-24

Chapter 8

SET UP: This is speed v1 for the collision. Let v2 be the speed of the combined object just after the collision. EXECUTE: Conservation of momentum applied to the collision gives mv1 = 2mv2 so v2 = v1 / 2 = gR / 2 SET UP: Apply conservation of energy to the motion of the combined object after the collision. Let y3 be the final height above the bottom of the bowl. EXECUTE: 12 ( 2m ) v22 = ( 2m ) gy3 . v22 1 ⎛ gR ⎞ = ⎜ ⎟ = R/4. 2g 2g ⎝ 2 ⎠ EVALUATE: Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. Apply conservation of energy to the motion of the block after the collision. SET UP: Object B is the block, initially at rest. If L is the length of the wire and θ is the angle it makes with the vertical, the height of the block is y = L (1 − cosθ ) . Initially, y1 = 0 . y3 =

8.79.

8.80.

⎛ 2m A ⎞ ⎛ 2M ⎞ EXECUTE: Eq. 8.25 gives vB = ⎜ ⎟ vA = ⎜ ⎟ (5.00 m/s) = 2.50 m/s . Conservation of energy gives m m + ⎝ M + 3M ⎠ B ⎠ ⎝ A v2 (2.50 m/s) 2 1 mB vB2 = mB gL(1 − cosθ ) . cosθ = 1 − B = 1 − = 0.362 and θ = 68.8° . 2 2 gL 2(9.80 m/s 2 )(0.500 m) EVALUATE: Only a portion of the initial kinetic energy of the ball is transferred to the block in the collision. IDENTIFY: Apply conservation of energy to the motion before and after the collision. Apply conservation of momentum to the collision. SET UP: First consider the motion after the collision. The combined object has mass mtot = 25.0 kg. Apply ! ! ∑ F 5 ma to the object at the top of the circular loop, where the object has speed v3. The acceleration is arad = v32 / R, downward.

EXECUTE:

T + mg = m

v32 . R

The minimum speed v3 for the object not to fall out of the circle is given by setting T = 0. This gives v3 = Rg , where R = 3.50 m. SET UP: Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the loop. Take y = 0 at point 2. Only gravity does work, so K 2 + U 2 = K 3 + U 3

EXECUTE:

1 2

mtot v22 = 12 mtot v32 + mtot g ( 2 R ) .

Use v3 = Rg and solve for v2 : v2 = 5 gR = 13.1 m/s .

SET UP: Now apply conservation of momentum to the collision between the dart and the sphere. Let v1 be the speed of the dart before the collision. EXECUTE: ( 5.00 kg ) v1 = ( 25.0 kg )(13.1 m/s ) .

8.81.

v1 = 65.5 m/s . EVALUATE: The collision is inelastic and mechanical energy is removed from the system by the negative work done by the forces between the dart and the sphere. ! ! IDENTIFY: Use Eq. 8.25 to find the speed of the hanging ball just after the collision. Apply ∑ F = ma to find the tension in the wire. After the collision the hanging ball moves in an arc of a circle with radius R = 1.35 m and acceleration arad = v 2 / R . ! ! SET UP: Let A be the 2.00 kg ball and B be the 8.00 kg ball. For applying ∑ F = ma to the hanging ball, let +y ! be upward, since arad is upward. The free-body force diagram for the 8.00 kg ball is given in Figure 8.81. ⎛ 2m A ⎞ ⎛ ⎞ 2[2.00kg] vB 2 x = ⎜ ⎟ v A1x = ⎜ ⎟ (5.00 m/s) = 2.00 m/s . Just after the collision the 8.00 kg m + m 2.00 kg + 8.00 kg ⎝ ⎠ ⎝ A B ⎠ ball has speed v = 2.00 m/s . Using the free-body diagram, ∑ Fy = ma y gives T − mg = marad .

EXECUTE:

⎛ ⎛ v2 ⎞ [2.00 m/s]2 ⎞ T = m ⎜ g + ⎟ = (8.00 kg) ⎜ 9.80 m/s 2 + ⎟ = 102 N . 1.35 m ⎠ R⎠ ⎝ ⎝


8-25

EVALUATE: The tension before the collision is the weight of the ball, 78.4 N. Just after the collision, when the ball has started to move, the tension is greater than this. y T

arad x

mg

8.82.

Figure 8.81 IDENTIFY: The impulse applied to the ball equals its change in momentum. The height of the ball and its speed are related by conservation of energy. SET UP: Let +y be upward. EXECUTE: Applying conservation of energy to the motion of the ball from its height h to the floor gives 1 mv12 = mgh , where v1 is its speed just before it hits the floor. Just before it hits, it is traveling downward, so the 2 velocity of the ball just before it hits the floor is v1 y = − 2 gh . Applying conservation of energy to the motion of the ball from just after it bounces off the floor with speed v2 to its maximum height of 0.90h gives

1 2

mv22 = mg (0.90h) .

It is moving upward, so v2 y = + 2 g (0.90h) . The impulse applied to the ball is J y = p2 y − p1 y = m(v2 y − v1 y ) =

8.83.

m 2 g (0.90h) + m 2 gh = 2.76m gh . The floor exerts an upward impulse of 2.76m gh to the ball. EVALUATE: The impulse increases when m increases and when h increases. The ball does not return to its initial height because some mechanical energy is dissipated during the collision with the floor. IDENTIFY: Apply conservation of momentum to the collision between the bullet and the block and apply conservation of energy to the motion of the block after the collision. (a) SET UP: Collision between the bullet and the block: Let object A be the bullet and object B be the block. Apply momentum conservation to find the speed vB 2 of the block just after the collision.



m Av A1 = mAv A 2 + mB vB 2 x . mA ( v A1 − v A 2 ) 4.00 × 10−3 kg ( 400 m/s − 120 m/s ) = = 1.40 m/s . mB 0.800 kg SET UP: Motion of the block after the collision. Let point 1 in the motion be just after the collision, where the block has the speed 1.40 m/s calculated above, and let point 2 be where the block has come to rest. vB 2 x =

Figure 8.83b K1 + U1 + Wother = K 2 + U 2 .

EXECUTE: Work is done on the block by friction, so Wother = W f . Wother = W f = ( f k cos φ ) s = − f k s = − μ k mgs, where s = 0.450 m U1 = 0,

U2 = 0

K1 = mv , 1 2

Thus

1 2

2 1

K 2 = 0 (block has come to rest)

mv − μ k mgs = 0. 2 1

v12 (1.40 m/s ) = = 0.222 . 2 gs 2 ( 9.80 m/s 2 ) ( 0.450 m ) 2

μk =

8-26

Chapter 8

(b) For the bullet,

K1 = 12 mv12 = 12 ( 4.00 × 10−3 kg ) ( 400 m/s ) = 320 J . 2

K 2 = 12 mv22 = 12 ( 4.00 × 10−3 kg ) (120 m/s ) = 28.8 J . 2

ΔK = K 2 − K1 = 28.8 J − 320 J = −291 J . The kinetic energy of the bullet decreases by 291 J. (c) Immediately after the collision the speed of the block is 1.40 m/s so its kinetic energy is K = 12 mv 2 =

8.84.

1 2

( 0.800 kg )(1.40 m/s )

2

= 0.784 J.

EVALUATE: The collision is highly inelastic. The bullet loses 291 J of kinetic energy but only 0.784 J is gained by the block. But momentum is conserved in the collision. All the momentum lost by the bullet is gained by the block. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion of the block after the collision. SET UP: Let +x be to the right. Let the bullet be A and the block be B. Let V be the velocity of the block just after the collision. EXECUTE: Motion of block after the collision: K1 = U grav2 . 12 mBV 2 = mB gh .

V = 2 gh = 2(9.80 m/s 2 )(0.450 × 10−2 m) = 0.297 m/s . Collision: vB 2 = 0.297 m/s . P1x = P2 x gives mAv A1 = mAv A 2 + mB vB 2 . vA2 =

8.85.

mAv A1 − mB vB 2 (5.00 × 10−3 kg)(450 m/s) − (1.00 kg)(0.297 m/s) = = 391 m/s . mA 5.00 × 10−3 kg

EVALUATE: We assume the block moves very little during the time it takes the bullet to pass through it. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. The value of M where the kinetic energy loss K loss of the neutron is a maximum satisfies dK loss / dM = 0 . SET UP: Let object A be the neutron and object B be the nucleus. Let the initial speed of the neutron be v A1 . All

motion is along the x-axis. K 0 = 12 mv A21 . EXECUTE: (a) v A 2 =

⎛ ⎡ m − M ⎤2 ⎞ 2 2m 2 M 2 4 K 0 mM m−M v A1 . K loss = 12 mv A21 − 12 mv A2 2 = 12 m ⎜1 − ⎢ v = v = , as ⎟ ⎜ ⎣ m + M ⎥⎦ ⎟ A1 ( M + m) 2 A1 ( M + m) 2 m+M ⎝ ⎠

was to be shown. ⎡ 1 2M ⎤ dK loss 2M (b) = 1 and M = m . The incident neutron loses the most = 4K0m ⎢ − =0. 2 3⎥ M +m dM ⎣ ( M + m ) ( M + m) ⎦ kinetic energy when the target has the same mass as the neutron. (c) When mA = mB , Eq. 8.24 says v A 2 = 0 . The final speed of the neutron is zero and the neutron loses all of its kinetic energy. EVALUATE: When M >> m , v A 2 x ≈ −v A1x and the neutron rebounds with speed almost equal to its initial speed. 8.86.

In this case very little kinetic energy is lost; K loss = 4 K 0 m / M , which is very small. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. SET UP: Let all the motion be along the x axis. v A1x = v0 . ⎛ m − mB ⎞ ⎛ 2mA ⎞ 2 1 EXECUTE: (a) v A 2 x = ⎜ A ⎟ v0 and vB 2 x = ⎜ ⎟ v0 . K1 = 2 mAv0 . ⎝ mA + mB ⎠ ⎝ mA + mB ⎠ 2

2

2

⎛ m − mB ⎞ 2 ⎛ mA − mB ⎞ K A 2 ⎛ mA − mB ⎞ =⎜ K A 2 = 12 mAv A2 2 x = 12 mA ⎜ A ⎟ . ⎟ v0 = ⎜ ⎟ K1 and K1 ⎝ mA + mB ⎠ ⎝ mA + mB ⎠ ⎝ m A + mB ⎠ 2

⎛ 2m A ⎞ 2 4mA mB K 4mAmB . K B 2 = 12 mB vB2 2 x = 12 mB ⎜ K1 and B 2 = ⎟ v0 = 2 2 K m + m m m + m ( A B) ( A + mB ) B ⎠ 1 ⎝ A K K K 4 K 5 (b) (i) For mA = mB , A 2 = 0 and B 2 = 1 . (ii) For mA = 5mB , A 2 = and B 2 = . K1 K1 K1 9 K1 9


8-27

2

(c) Equal sharing of the kinetic energy means

K A 2 K B 2 1 ⎛ mA − mB ⎞ 1 = = . ⎜ ⎟ = . K1 K1 2 ⎝ mA + mB ⎠ 2

2mA2 + 2mB2 − 4mAmB = mA2 + 2mA mB + mB2 . mA2 − 6mAmB + mB2 = 0 . The quadratic formula gives

mA = 5.83 or mB

mA K 1 = 0.172 . We can also verify that these values give B 2 = . mB K1 2 8.87.

EVALUATE: When mA > mB , object A retains almost all of the original kinetic energy. IDENTIFY: Apply conservation of energy to the motion of the package before the collision and apply conservation of the horizontal component of momentum to the collision. (a) SET UP: Apply conservation of energy to the motion of the package from point 1 as it leaves the chute to point 2 just before it lands in the cart. Take y = 0 at point 2, so y1 = 4.00 m. Only gravity does work, so

K1 + U1 = K 2 + U 2 . EXECUTE:

1 2

mv12 + mgy1 = 12 mv22 .

v2 = v12 + 2 gy1 = 9.35 m/s . (b) SET UP: In the collision between the package and the cart momentum is conserved in the horizontal direction. (But not in the vertical direction, due to the vertical force the floor exerts on the cart.) Take + x to be to the right. Let A be the package and B be the cart. EXECUTE: Px is constant gives mAv A1x + mB vB1x = ( mA + mB ) v2 x .

vB1x = −5.00 m/s .

v A1x = ( 3.00 m/s ) cos37.0° . (The horizontal velocity of the package is constant during its free-fall.)

8.88.

Solving for v2 x gives v2 x = −3.29 m/s. The cart is moving to the left at 3.29 m/s after the package lands in it. EVALUATE: The cart is slowed by its collision with the package, whose horizontal component of momentum is in the opposite direction to the motion of the cart. IDENTIFY: Eqs. 8.24, 8.25, and 8.27 give the outcome of the elastic collision. SET UP: The blue puck is object A and the red puck is object B. Let +x be the direction of the initial motion of A. v A1x = 0.200 m/s , v A 2 x = 0.050 m/s and vB1x = 0 EXECUTE: (a) Eq. 8.27 gives vB 2 x = v A 2 x − vB1x + v A1x = 0.250 m/s .

⎛ v ⎞ ⎛ ⎡ 0.200 m/s ⎤ ⎞ (b) Eq. 8.25 gives mB = mA ⎜ 2 A1x − 1⎟ = (0.0400 kg) ⎜ 2 ⎢ ⎥ − 1⎟ = 0.024 kg . v ⎝ ⎣ 0.250 m/s ⎦ ⎠ ⎝ B2x ⎠ EVALUATE: We can verify that our results give K1 = K 2 and P1x = P2 x , as required in an elastic collision. 8.89.

(a) IDENTIFY and SET UP: K = 12 mAv A2 + 12 mB vB2 . ! ! ! ! ! ! Use v A = v′A + vcm and v B = v′B + vcm to replace v A and vB in this equation. Note v′A and v′B as defined in the problem are the velocities of A and B in coordinates moving with the center of mass. Note also that ! ! ! ! ′ is the velocity of the car in these coordinates. But that’s zero, so ′ where vcm mAv′A + mB v′B = Mvcm ! ! mAv′A 1 mB v′B 5 0; we can use this in the proof. ! In part (b), use that P is conserved in a collision. ! ! ! ! 2 EXECUTE: v A = v′A + vcm , so v A2 = v′A2 + vcm + 2v′A ⋅ vcm . ! ! ! ! 2 v B = v′B + vcm , so vB2 = v′B2 + vcm + 2v′B ⋅ vcm . ! ! ! (We have used that for a vector A, A2 = A ⋅ A.) ! ! ! ! 2 2 Thus K = 12 mAv′A2 + 12 mAvcm + mAv′A ⋅ vcm + 12 mB v′B2 + 12 mB vcm + mB v′B ⋅ vcm .

K=

1 2

!

!

!

( mA + mB ) vcm2 + 12 ( mAv′A2 + mBv′B2 ) + ( mAv′A + mBv′B ) ⋅ vcm .

! ! 2 + 12 ( mAv′A2 + mB v′B2 ) . This is the result the But mA + mB = M and as noted earlier mAv′A + mB v′B = 0, so K = 12 Mvcm

problem asked us to derive. ! ! (b) EVALUATE: In the collision P = Mvcm is constant, so relative kinetic energy but the

1 2

2 Mvcm must remain.

1 2

2 Mvcm stays constant. The asteroids can lose all their

8-28

8.90.

Chapter 8

IDENTIFY: Eq. 8.27 describes the elastic collision, with x replaced by y. Speed and height are related by conservation of energy. SET UP: Let +y be upward. Let A be the large ball and B be the small ball, so vB1 y = −v and v A1 y = + v . If the large ball has much greater mass than the small ball its speed is changed very little in the collision and v A 2 y = +v .

EXECUTE: (a) vB 2 y − v A 2 y = −(vB1 y − v A1 y ) gives vB 2 y = + v A 2 y − vB1 y + v A1 y = v − ( −v) + v = +3v . The small ball moves upward with speed 3v after the collision. (b) Let h1 be the height the small ball fell before the collision. Conservation of energy applied to the motion from the release point to the floor gives U1 = K 2 and mgh1 = 12 mv 2 . h1 =

v2 . Conservation of energy applied to the 2g

motion of the small ball from immediately after the collision to its maximum height h2 (rebound distance) gives 9v 2 = 9h1 . The ball’s rebound distance is nine times the distance it fell. 2g EVALUATE: The mechanical energy gained by the small ball comes from the energy of the large ball. But since the large ball’s mass is much larger it can give up this energy with very little decrease in speed. IDENTIFY: Apply conservation of momentum to the system consisting of Jack, Jill and the crate. The speed of Jack or Jill relative to the ground will be different from 4.00 m/s. SET UP: Use an inertial coordinate system attached to the ground. Let +x be the direction in which the people jump. Let Jack be object A, Jill be B, and the crate be C. EXECUTE: (a) If the final speed of the crate is v, vC 2 x = −v , and v A 2 x = vB 2 x = 4.00 m/s − v . P2 x = P1x gives K1 = U 2 and

8.91.

1 2

m(3v) 2 = mgh2 . h2 =

mAv A 2 x + mB vBx 2 + mC vCx 2 = 0 . (75.0 kg)(4.00 m/s − v) + (45.0 kg)(4.00 m/s − v) + (15.0 kg)( −v) = 0 and v=

(75.0 kg + 45.0 kg)(4.00 m/s) = 3.56 m/s . 75.0 kg + 45.0 kg + 15.0 kg

(b) Let v′ be the speed of the crate after Jack jumps. Apply momentum conservation to Jack jumping: (75.0 kg)(4.00 m/s) (75.0 kg)(4.00 m/s − v′) + (60.0 kg)( −v′) = 0 and v′ = = 2.22 m/s . Then apply momentum 135.0 kg conservation to Jill jumping, with v being the final speed of the crate: P1x = P2 x gives (60.0 kg)(−v′) = (45.0 kg)(4.00 m/s − v) + (15.0 kg)(−v ) . v=

(45.0 kg)(4.00 m/s) + (60.0 kg)(2.22 m/s) = 5.22 m/s . 60.0 kg

(c) Repeat the calculation in (b), but now with Jill jumping first. Jill jumps: (45.0 kg)(4.00 m/s − v′) + (90.0 kg)(−v′) = 0 and v′ = 1.33 m/s . Jack jumps: (90.0 kg)(−v′) = (75.0 kg)(4.00 m/s − v ) + (15.0 kg)(−v) . v=

8.92.

(75.0 kg)(4.00 m/s) + (90.0 kg)(1.33 m/s) = 4.66 m/s . 90.0 kg

EVALUATE: The final speed of the crate is greater when Jack jumps first, then Jill. In this case Jack leaves with a speed of 1.78 m/s relative to the ground, whereas when they both jump simultaneously Jack and Jill each leave with a speed of only 0.44 m/s relative to the ground. IDENTIFY: Momentum is conserved in the explosion. The total kinetic energy of the two fragments is Q. SET UP: Let the final speed of the two fragments be v A and vB . They must move in opposite directions after the explosion. EXECUTE: (a) Since the initial momentum of the system is zero, conservation of momentum says mAv A = mB vB ⎛m ⎞ and vB = ⎜ A ⎟ v A . K A + K B = Q gives ⎝ mB ⎠

2

1 2

⎛m ⎞ mAv A2 + 12 mB ⎜ A ⎟ v A2 = Q . ⎝ mB ⎠

1 2

⎛ m ⎞ mAv A2 ⎜1 + A ⎟ = Q . ⎝ mB ⎠

⎛ mB ⎞ ⎛ Q mB ⎞ ⎛ mA ⎞ =⎜ ⎟ Q . K B = Q − K A = Q ⎜1 − ⎟=⎜ ⎟Q . 1 + mA / mB ⎝ mA + mB ⎠ ⎝ mA + mB ⎠ ⎝ mA + mB ⎠ 4 1 (b) If mB = 4mA , then K A = Q and K B = Q . The lighter fragment gets 80% of the energy that is released. 5 5 EVALUATE: If mA = mB the fragments share the energy equally. In the limit that mB >> mA , the lighter fragment gets almost all of the released energy. KA =


8.93.

8-29

IDENTIFY: Apply conservation of momentum to the system of the neutron and its decay products. SET UP: Let the proton be moving in the +x direction with speed vp after the decay. The initial momentum of the neutron is zero, so to conserve momentum the electron must be moving in the − x direction after the decay. Let the speed of the electron be ve .

EXECUTE:

⎛m ⎞ P1x = P2 x gives 0 = mp vp − meve and ve = ⎜ p ⎟ vp . The total kinetic energy after the decay is ⎝ me ⎠ 2

⎛m ⎞ ⎛ m ⎞ K tot = 12 meve2 + 12 mp vp2 = 12 me ⎜ p ⎟ vp2 + 12 mp vp2 = 12 mpvp2 ⎜1 + p ⎟ . ⎝ me ⎠ ⎝ me ⎠ K 1 1 Thus, p = = = 5.44 × 10−4 = 0.0544% . K tot 1 + mp / me 1 + 1836

8.94.

EVALUATE: Most of the released energy goes to the electron, since it is much lighter than the proton. IDENTIFY: Momentum is conserved in the decay. The results of Problem 8.92 give the kinetic energy of each fragment. SET UP: Let A be the alpha particle and let B be the radium nucleus, so mA / mB = 0.0176 . Q = 6.54 × 10−13 J . Q 6.54 × 10−13 J = = 6.43 × 10−13 J and K B = 0.11 × 10−13 J . 1 + mA / mB 1 + 0.0176 EVALUATE: The lighter particle receives most of the released energy. IDENTIFY: The momentum of the system is conserved. SET UP: Let +x be to the right. P1x = 0 . pex , pnx and panx are the momenta of the electron, polonium nucleus and antineutrino, respectively. EXECUTE: P1x = P2 x gives pex + pnx + panx = 0 . panx = −( pex + pnx ) .

EXECUTE:

8.95.

KA =

panx = −(5.60 × 10−22 kg ⋅ m/s + [3.50 × 10−25 kg][−1.14 × 103 m/s]) = −1.66 × 10−22 kg ⋅ m/s .

8.96.

The antineutrino has momentum to the left with magnitude 1.66 × 10−22 kg ⋅ m/s . EVALUATE: The antineutrino interacts very weakly with matter and most easily shows its presence by the momentum it carries away. IDENTIFY: Momentum components in the x and y directions are separately conserved. For an elastic collision K1 = K 2 .

SET UP:

v A1x = + v A1 , vB1x = 0 . v A 2 x = v A 2 cos α , v A 2 y = v A 2 sin α . vB 2 x = vB 2 cos α , vB 2 y = −vB 2 sin α .

sin θ + cos 2 θ = 1 , for any angle θ . cos(α + β ) = cos α cos β − sin α sin β . 2

EXECUTE: (a) P1x = P2 x gives mAv A1 = mAv A 2 cos α + mB vB 2 cos β . P1 y = P2 y gives 0 = mAv A 2 sin α − mB vB 2 sin β .

(b) mA2 v A21 = mA2 v A2 2 cos 2 α + mB2 vB2 2 cos 2 β + 2mA mB v A 2vB 2 cosα cos β and 0 = mA2 v A2 2 sin 2 α + mB2 vB2 2 sin 2 β − 2mAmB v A 2vB 2 sin α sin β . Adding these two equations and using the trig identities in the SET UP step gives mA2 v A21 = mA2 v A2 2 + mB2 vB2 2 + 2mA mB v A 2vB 2 cos(α + β ) .

(c) K1 = K 2 says

1 2

mAv A21 = 12 mAv A2 2 + 12 mB vB2 2 . The result in part (b) agrees with this expression only if

cos(α + β ) = 0 . This requires that α + β = 90° =

8.97.

π

rad . 2 EVALUATE: The result of part (c) says that the two protons move in perpendicular directions after the collision. IDENTIFY and SET UP:

Figure 8.97 Px and Py are conserved in the collision since there is no external horizontal force.

8-30

Chapter 8

The result of Problem 8.96 part (d) applies here since the collision is elastic This says that 25.0° + θ B = 90°, so that

θ B = 65.0°. (A and B move off in perpendicular directions.) EXECUTE:


But mA = mB so v A1 = v A 2 cos 25.0° + vB 2 cos65.0° . Py is conserved so mAv A1 y + mB vB1 y = m Av A 2 y + mB vB 2 y . 0 = v A 2 y + vB 2 y . 0 = v A 2 sin 25.0° − vB 2 sin 65.0° . vB 2 = ( sin 25.0° / sin 65.0° ) v A 2 .

⎛ sin 25.0° cos65.0° ⎞ This result in the first equation gives v A1 = v A 2 cos 25.0° + ⎜ ⎟ v A2 . sin 65.0° ⎝ ⎠ v A1 = 1.103v A 2 . v A 2 = v A1 /1.103 = (15.0 m/s)/1.103 = 13.6 m/s .

And then vB 2 = ( sin 25.0° / sin 65.0° )(13.6 m/s ) = 6.34 m/s.

EVALUATE: We can use our numerical results to show that K1 = K 2 and that P1x = P2 x and P1 y = P2 y . 8.98.

IDENTIFY: Since there is no friction, the horizontal component of momentum of the system of Jonathan, Jane and the sleigh is conserved. SET UP: Let +x be to the right. wA = 800 N , wB = 600 N and wC = 1000 N . EXECUTE:

P1x = P2 x gives 0 = mAv A 2 x + mB vB 2 x + mC vC 2 x . vC 2 x =

mAv A 2 x + mB vB 2 x wAv A 2 x + wB vB 2 x = . mC wC

(800 N)(−[5.00 m/s]cos30.0°) + (600 N)(+[7.00 m/s]cos36.9°) = −0.105 m/s . 1000 N The sleigh’s velocity is 0.105 m/s, to the left. EVALUATE: The vertical component of the momentum of the system consisting of the two people and the sleigh is not conserved, because of the net force exerted on the sleigh by the ice while they jump. IDENTIFY: In Eq. 8.28 treat each straight piece as an object in the system. SET UP: The center of mass of each piece of length L is at its center. EXECUTE: (a) From symmetry, the center of mass is on the vertical axis, a distance ( L / 2)cos(α / 2) below the apex. (b) The center of mass is on the vertical axis of symmetry, a distance 2( L / 2) / 3 = L / 3 above the center of the horizontal segment. (c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal and each is equal to ( L / 2) / 2 = L / 4 . The center of mass is along the bisector of the angle, a distance L / 8 from the corner. vC 2 x =

8.99.

8.100.

(d) By symmetry, the center of mass is at the center of the equilateral triangle, a distance ( L / 3)sin 60° = L / 12 above the center of the horizontal segment. EVALUATE: The center of mass need not lie on any point of the object, it can be in empty space. IDENTIFY: There is no net horizontal external force so vcm is constant. SET UP: Let +x be to the right, with the origin at the initial position of the left-hand end of the canoe. mA = 45.0 kg , mB = 60.0 kg . The center of mass of the canoe is at its center. EXECUTE: Initially, vcm = 0 , so the center of mass doesn’t move. Initially, xcm1 =

mA x A1 + mB xB1 . After she mA + mB

mA x A 2 + mB xB 2 . xcm1 = xcm2 gives mA x A1 + mB xB1 = mA x A 2 + mB xB 2 . She walks to a point 1.00 m from mA + mB the right-hand end of the canoe, so she is 1.50 m to the right of the center of mass of the canoe and x A 2 = xB 2 + 1.50 m .

walks, xcm2 =

(45.0 kg)(1.00 m) + (60.0 kg)(2.50 m) = (45.0 kg)( xB 2 + 1.50 m) + (60.0 kg) xB 2 . (105.0 kg) xB 2 = 127.5 kg ⋅ m and xB 2 = 1.21 m . xB 2 − xB1 = 1.21 m − 2.50 m = −1.29 m . The canoe moves 1.29 m to the left.


8.101.

8-31

EVALUATE: When the woman walks to the right, the canoe moves to the left. The woman walks 3.00 m to the right relative to the canoe and the canoe moves 1.29 m to the left, so she moves 3.00 m − 1.29 m = 1.71 m to the right relative to the water. Note that this distance is (60.0 kg / 45.0 kg)(1.29 m) . IDENTIFY: Take as the system you and the slab. There is no horizontal force, so horizontal momentum is ! ! conserved. By Eq. 8.32, P is constant vcm is constant (for a system of constant mass). Use coordinates fixed to ! ! the ice, with the direction you walk as the x-direction. vcm is constant and initially vcm = 0.

Figure 8.101 ! vcm 5

! ! mp vp 1 ms vs

mp + ms ! ! mpvcm 1 ms vs = 0 .

50 .

mpvpx + ms vsx = 0 . vsx = − ( mp / ms ) vpx = − ( mp / 5mp ) 2.00 m/s = −0.400 m/s .

8.102.

The slab moves at 0.400 m/s, in the direction opposite to the direction you are walking. EVALUATE: The initial momentum of the system is zero. You gain momentum in the + x -direction so the slab gains momentum in the − x-direction. The slab exerts a force on you in the + x -direction so you exert a force on the slab in the − x-direction. IDENTIFY: Conservation of x and y components of momentum applies to the collision. At the highest point of the trajectory the vertical component of the velocity of the projectile is zero. SET UP: Let +y be upward and +x be horizontal and to the right. Let the two fragments be A and B, each with mass m. For the projectile before the explosion and the fragments after the explosion. ax = 0 , a y = −9.80 m/s 2 .

EXECUTE: (a) v y2 = v02y + 2a y ( y − y0 ) with v y = 0 gives that the maximum height of the projectile is h=−

v02y 2a y

=−

([80.0 m/s]sin 60.0°) 2 = 244.9 m . Just before the explosion the projectile is moving to the right with 2( −9.80 m/s 2 )

horizontal velocity vx = v0 x = v0 cos60.0° = 40.0 m/s . After the explosion v Ax = 0 since fragment A falls vertically. Conservation of momentum applied to the explosion gives (2m)(40.0 m/s) = mvBx and vBx = 80.0 m/s . Fragment B has zero initial vertical velocity so y − y0 = v0 yt + 12 a yt 2 gives a time of fall of t= −

2h 2(244.9 m) = − = 7.07 s . During this time the fragment travels horizontally a distance ay −9.80 m/s 2

(80.0 m/s)(7.07 s) = 566 m . It also took the projectile 7.07 s to travel from launch to maximum height and during this time it travels a horizontal distance of ([80.0 m/s]cos60.0°)(7.07 s) = 283 m . The second fragment lands 283 m + 566 m = 849 m from the firing point. (b) For the explosion, K1 = 12 (20.0 kg)(40.0 m/s) 2 = 1.60 × 104 J . K 2 = 12 (10.0 kg)(80.0 m/s) 2 = 3.20 × 104 J . The energy released in the explosion is 1.60 × 104 J . EVALUATE: The kinetic energy of the projectile just after it is launched is 6.40 × 104 J . We can calculate the speed of each fragment just before it strikes the ground and verify that the total kinetic energy of the fragments just before they strike the ground is 6.40 × 104 J + 1.60 × 104 J = 8.00 × 104 J . Fragment A has speed 69.3 m/s just before it strikes the ground, and hence has kinetic energy 2.40 × 104 J . Fragment B has speed (80.0 m/s) 2 + (69.3 m/s) 2 = 105.8 m/s just before it strikes the ground, and hence has kinetic energy 5.60 × 104 J . v02 sin(2α 0 ) = 565 m that the projectile g would have had if no explosion had occurred. One fragment lands at R / 2 so the other, equal mass fragment lands at a distance 3R / 2 from the launch point. IDENTIFY: The rocket moves in projectile motion before the explosion and its fragments move in projectile motion after the explosion. Apply conservation of energy and conservation of momentum to the explosion. Also, the center of mass of the system has the same horizontal range R =

8.103.

8-32

Chapter 8

SET UP: Apply conservation of energy to the explosion. Just before the explosion the shell is at its maximum height and has zero kinetic energy. Let A be the piece with mass 1.40 kg and B be the piece with mass 0.28 kg. Let v A and vB be the speeds of the two pieces immediately after the collision. EXECUTE: 12 mAv A2 + 12 mB vB2 = 860 J SET UP: Since the two fragments reach the ground at the same time, their velocities just after the explosion must be horizontal. The initial momentum of the shell before the explosion is zero, so after the explosion the pieces must be moving in opposite horizontal directions and have equal magnitude of momentum: mAv A = mB vB . EXECUTE: Use this to eliminate v A in the first equation and solve for vB : 1 2

mB vB2 (1 + mB / mA ) = 860 J and vB = 71.6 m/s.

Then v A = ( mB / mA ) vB = 14.3 m/s.

(b) SET UP: Use the vertical motion from the maximum height to the ground to find the time it takes the pieces to fall to the ground after the explosion. Take + y downward. v0 y = 0, a y = +9.80 m/s 2 , y − y0 = 80.0 m, t = ?

EXECUTE:

y − y0 = v0 yt + 12 a yt 2 gives t = 4.04 s.

During this time the horizontal distance each piece moves is x A = v At = 57.8 m and xB = vBt = 289.1 m. They move

8.104.

in opposite directions, so they are x A + xB = 347 m apart when they land. EVALUATE: Fragment A has more mass so it is moving slower right after the collision, and it travels horizontally a smaller distance as it falls to the ground. IDENTIFY: Apply conservation of momentum to the collision. At the highest point of its trajectory the shell is moving horizontally. If one fragment received some upward momentum in the collision, the other fragment would have had to receive a downward component. Since they each the ground at the same time, each must have zero vertical velocity immediately after the explosion. SET UP: Let +x be horizontal, along the initial direction of motion of the projectile and let +y be upward. At its maximum height the projectile has vx = v0 cos55.0° = 86.0 m/s . Let the heavier fragment be A and the lighter fragment be B. mA = 9.00 kg and mB = 3.00 kg .

EXECUTE: Since fragment A returns to the launch point, immediately after the explosion it has v Ax = −86.0 m/s . Conservation of momentum applied to the explosion gives (12.0 kg)(86.0 m/s) = (9.00 kg)(−86.0 m/s) + (3.00 kg)vBx and vBx = 602 m/s . The horizontal range of the v02 sin(2α 0 ) = 2157 m . The horizontal distance each fragment g travels is proportional to its initial speed and the heavier fragment travels a horizontal distance R / 2 = 1078 m after ⎛ 602 m ⎞ the explosion, so the lighter fragment travels a horizontal distance ⎜ ⎟ (1078 m) = 7546 m from the point of ⎝ 86 m ⎠ explosion and 1078 m + 7546 m = 8624 m from the launch point. The energy released in the explosion is K 2 − K1 = 12 (9.00 kg)(86.0 m/s) 2 + 12 (3.00 kg)(602 m/s) 2 − 12 (12.0 kg)(86.0 m/s) 2 = 5.33 × 105 J . EVALUATE: The center of mass of the system has the same horizontal range R = 2157 m as if the explosion didn’t occur. This gives (12.0 kg)(2157 m) = (9.00 kg)(0) + (3.00 kg)d and d = 8630 m , where d is the distance from the launch point to where the lighter fragment lands. This agrees with our calculation. ! IDENTIFY: No external force, so P is conserved in the collision. SET UP: Apply momentum conservation in the x and y directions: projectile, if no explosion occurred, would be R =

8.105.

Figure 8.105 Solve for v1 and v2 .


EXECUTE:

8-33

Px is conserved so mv0 = m ( v1 cos 45° + vf cos10° + v2 cos30° ) .

v0 − vf cos10° = v1 cos 45° + v2 cos30° . 1030.4 m/s = v1 cos 45° + v2 cos30° .

Px is conserved so 0 = m ( v1 sin 45° − v2 sin 30° + vf sin10° ) .

v1 sin 45° = v2 sin 30° − 347.3 m/s . sin 45° = cos 45° so 1030.4 m/s = v2 sin 30° − 347.3 m/s + v2 cos30° . v2 =

1030.4 m/s + 347.3 m/s = 1010 m/s . sin 30° + cos30.0°

v2 sin 30° − 347.3 m/s = 223 m/s. Then two emitted neutrons have speeds of 223 m/s and 1010 m/s. sin 45° The speeds of the Ba and Kr nuclei are related by Pz conservation.

And then v1 =

Pz is constant implies that 0 = mBa vBa − mKr vKr ⎛m ⎞ ⎛ 2.3 × 10−25 kg ⎞ vKr = ⎜ Ba ⎟ vBa = ⎜ ⎟ vBa = 1.5vBa . −25 ⎝ 1.5 × 10 kg ⎠ ⎝ mKr ⎠ We can’t say what these speeds are but they must satisfy this relation. The value of vBa depends on energy considerations.

EVALUATE:

K1 = 12 mn ( 3.0 × 103 m/s ) = ( 4.5 × 106 J/kg ) mn . 2

K 2 = 12 mn ( 2.0 × 103 m/s ) + 12 mn ( 223 m/s ) + 12 mn (1010 m/s ) + K Ba + K Kr = ( 2.5 × 106 J/kg ) mn + K Ba + K Kr . 2

2

2

We don’t know what K Ba and K Kr are, but they are positive. We will study such nuclear reactions further in Chapter 43 and will find that energy is released in this process; K 2 > K1. Some of the potential energy stored in the 235

8.106.

U nucleus is released as kinetic energy and shared by the collision fragments. IDENTIFY: The velocity of the center of mass of the system of the two blocks is given by Eq. 8.30. Conservation of momentum says the center of mass moves at constant speed. ! ! SET UP: v A1x = v A1 , vB1x = 0 . The velocity u in the center of mass frame is related to the velocity v in the ! ! ! p2 stationary frame by u = v − vcm . We can express kinetic energy as K = . 2m mAv A1 EXECUTE: (a) vcm-x = . mA + mB (b) The center of mass moves with constant speed so this coordinate system is an inertial frame. mB v A1 mv (c) u A1x = v A1x − vcm-x = . u B1x = vB1x − vcm-x = − A A1 . In this frame P1x = mAu A1x + mBuB1x = 0 . mA + mB mA + mB

(d) P2 x = P1x = 0 gives p A1x + pB1x = 0 and p A 2 x + pB 2 x = 0 , so pB1x = − p A1x and pB 2 x = − p A 2 x . Conservation of kinetic energy gives

p A2 2 x pB2 2 x p A21x pB21x + = + . Using pB 2 x = − p A 2 x and pB1x = − p A1x gives p A2 2 x = p A21x and 2mA 2mB 2mA 2mB

p A 2 x = ± p A1x . If a collision occurs p Ax changes and p A 2 x = − p A1x . But pB 2 x = − p A 2 x and pB1x = − p A1x , so pB 2 x = − pB1x . In the center of mass frame the momentum and hence the velocity of each puck keeps the same magnitude and reverses direction. ⎛ 0.400 kg ⎞ (e) vcm-x = ⎜ ⎟ (6.00 m/s) = 4.00 m/s . u A1x = 6.00 m/s − 4.00 m/s = 2.00 m/s . ⎝ 0.600 kg ⎠ u B1x = 0 − 4.00 m/s = −4.00 m/s . u A 2 x = −2.00 m/s and u B 2 x = +4.00 m/s . v A 2 x = u A 2 x + vcm-x = −2.00 m/s + 4.00 m/s = 2.00 m/s . vB 2 x = u B 2 x + vcm-x = 4.00 m/s + 4.00 m/s = 8.00 m/s . ⎛ 0.400 kg − 0.200 kg ⎞ Eq. 8.24 says v A 2 x = ⎜ ⎟ (6.00 m/s) = 2.00 m/s . Eq. 8.25 says ⎝ 0.400 kg + 0.200 kg ⎠ ⎛ ⎞ 2[0.400 kg] vA2 x = ⎜ ⎟ (6.00 m/s) = 8.00 m/s . Our result agrees with Eqs. 8.24 and 8.25. + 0.400 kg 0.200 kg ⎝ ⎠

8-34

8.107.

Chapter 8

EVALUATE: Eqs. 8.24 and 8.25 apply only when vB1 = 0 . The result that the velocity of each puck in the center of mass frame reverses direction and retains the same magnitude applies to all elastic collisions, even when both are moving initially. IDENTIFY and SET UP: Apply conservation of energy to find the total energy before and after the collision with the floor from the initial and final maximum heights. EXECUTE: (a) Objects stick together says that the relative speed after the collision is zero, so P = 0. (b) In an elastic collision the relative velocity of the two bodies has the same magnitude before and after the collision, so P = 1. (c) Speed of ball just before collision: mgh = 12 mv12 . v1 = 2 gh Speed of ball just after collision: mgH1 = 12 mv22 . v2 = 2 gH1 The second object (the surface) is stationary, so P = v2 / v1 = H1 / h .

(d) P = H1 / h implies H1 = hP2 = (1.2 m )( 0.85 ) = 0.87 m . 2

(e) H1 = hP2 . H 2 = H1P2 = hP4 .

H 3 = H 2P2 = ( hP4 ) P2 = hP6 . Generalize to H n = H n −1P2 = hP2( n −1)P2 = hP2 n . (f) 8th bounce implies n = 8 . H 8 = hP16 = 1.2 m ( 0.85 ) = 0.089 m . 16

8.108.

EVALUATE: P is a measure of the kinetic energy lost in the collision. The collision here is between a ball and the earth. Momentum lost by the ball is gained by the earth, but the velocity gained by the earth is very small and can be taken to be zero. IDENTIFY: Momentum is conserved in the collision. Conservation of energy says K 2 = K1 + Δ . ! ! SET UP: For part (b) let v0 be the common speed of each atom before the collision and let V and v3 be the velocities after the collision of the molecule and the atom that remains. m = 1.67 × 10−27 kg is the mass of one hydrogen atom. EXECUTE: (a) In the center of mass frame P1x = 0 so P2 x = 0 and vcm2 = 0 . But in this frame the potential energy 2 decreases and the kinetic energy increases. This is inconsistent with K 2cm = 12 mtot vcm2 =0.

(b) Before the collision vcm = 0 . After the collision the molecule and remaining atom move in opposite directions and (2m)V = mv3 ; v3 = 2V . Conservation of energy gives 12 (2m)V 2 + 12 mv32 = 3( 12 mv02 ) 2 + Δ . With v3 = 2V this becomes V 2 = 12 v02 +

EVALUATE:

8.109.

Δ .V= 3m

1 2

(1.00 × 103 m/s) 2 +

7.23 × 10−19 J = 1.20 × 104 m/s and v3 = 2V = 2.40 × 104 m/s . 3(1.67 × 10−27 )

K = 3 ( 12 mv02 ) = 2.50 × 10−21 J , which is much less than the binding energy of the molecule. Other

initial conditions also lead to molecule formation; the one of zero initial momentum is just particularly simple to analyze. IDENTIFY: Apply conservation of energy to the motion of the wagon before the collision. After the collision the combined object moves with constant speed on the level ground. In the collision the horizontal component of momentum is conserved. SET UP: Let the wagon be object A and treat the two people together as object B. Let +x be horizontal and to the right. Let V be the speed of the combined object after the collision. EXECUTE: (a) The speed v A1 of the wagon just before the collision is given by conservation of energy applied to the motion of the wagon prior to the collision. U1 = K 2 says mA g ([50 m][sin 6.0°]) = 12 mAv A21 . v A1 = 10.12 m/s . ⎛ ⎞ 300 kg P1x = P2 x for the collision says mAv A1 = (mA + mB )V and V = ⎜ ⎟ (10.12 m/s) = 6.98 m/s . ⎝ 300 kg + 75.0 kg + 60.0 kg ⎠ In 5.0 s the wagon travels (6.98 m/s)(5.0 s) = 34.9 m , and the people will have time to jump out of the wagon before it reaches the edge of the cliff.


8-35

(b) For the wagon, K1 = 12 (300 kg)(10.12 m/s)2 = 1.54 × 104 J . Assume that the two heroes drop from a small height, so their kinetic energy just before the wagon can be neglected compared to K1 of the wagon.

8.110.

K 2 = 12 (435 kg)(6.98 m/s) 2 = 1.06 × 104 J . The kinetic energy of the system decreases by K1 − K 2 = 4.8 × 103 J . EVALUATE: The wagon slows down when the two heroes drop into it. The mass that is moving horizontally increases, so the speed decreases to maintain the same horizontal momentum. In the collision the vertical momentum is not conserved, because of the net external force due to the ground. IDENTIFY: Gravity gives a downward external force of magnitude mg. The impulse of this force equals the change in momentum of the rocket. SET UP: Let +y be upward. Consider an infinitesimal time interval dt. In Example 8.15, vex = 2400 m/s and dm m = − 0 . In Example 8.16, m = m0 / 4 after t = 90 s . dt 120 s EXECUTE: (a) The impulse-momentum theorem gives − mgdt = (m + dm)(v + dv) + (dm)(v − vex ) − mv . This simplifies to − mgdt = mdv + vex dm and m

dv dm = −vex − mg . dt dt

dv v dm = − ex −g. dt m dt 1 ⎞ v dm ⎛ 2 2 (c) At t = 0 , a = − ex − g = −(2400 m/s) ⎜ − ⎟ − 9.80 m/s = 10.2 m/s . m0 dt ⎝ 120 s ⎠

(b) a =

vex m dm − gdt . Integrating gives v − v0 = + vex ln 0 − gt . v0 = 0 and m m v = + (2400 m/s)ln 4 − (9.80 m/s 2 )(90 s) = 2445 m/s . EVALUATE: Both the initial acceleration in Example 8.15 and the final speed of the rocket in Example 8.16 are reduced by the presence of gravity. IDENTIFY and SET UP: Apply Eq. 8.40 to the single-stage rocket and to each stage of the two-stage rocket. (a) EXECUTE: v − v0 = vex ln ( m0 / m ) ; v0 = 0 so v = vex ln ( m0 / m )

(d) dv = −

8.111.

The total initial mass of the rocket is m0 = 12,000 kg + 1000 kg = 13,000 kg. Of this, 9000 kg + 700 kg = 9700 kg is fuel, so the mass m left after all the fuel is burned is 13,000 kg − 9700 kg = 3300 kg. v = vex ln (13,000 kg/3300 kg ) = 1.37vex .

(b) First stage: v = vex ln ( m0 / m ) m0 = 13,000 kg The first stage has 9000 kg of fuel, so the mass left after the first stage fuel has burned is 13,000 kg − 9000 kg = 4000 kg. v = vex ln (13,000 kg/4000 kg ) = 1.18vex .

(c) Second stage: m0 = 1000 kg, m = 1000 kg − 700 kg = 300 kg .

v = v0 + vex ln ( m0 / m ) = 1.18vex + vex ln (1000 kg/300 kg ) = 2.38vex .

(d) v = 7.00 km/s

8.112.

vex = v / 2.38 = ( 7.00 km/s ) / 2.38 = 2.94 km/s . q

EVALUATE: The two-stage rocket achieves a greater final speed because it jetisons the left-over mass of the first stage before the second-state fires and this reduces the final m and increases m0 / m. IDENTIFY: During an interval where the mass is constant the speed of the rocket is constant. During an interval where the mass is changing at a constant rate, the equations of Section 8.6 apply. dm m SET UP: For 0 ≤ t ≤ 90 s , = − 0 . From Example 8.15, vex = 2400 m/s . dt 120 s EXECUTE: (a) For t ≤ 0 , v = 0 . For 0 ≤ t ≤ 90 s , Eq. 8.40 says v = (2400 m/s)ln 4 = 3327 m/s . For t > 90 s , v has the constant value 3327 m/s. The graph of v(t ) is given in Fig. 8.112a. (b) For 0 ≤ t ≤ 90 s , Eq. 8.39 gives a = −

vex dm 2400 m/s 20 m/s 2 ⎛ m0 ⎞ =− − = . a = 20 m/s 2 at t = 0 ⎜ ⎟ m dt m0 (1 − t /[120 s]) ⎝ 120 s ⎠ 1 − t /[120 s]

(as in Example 8.15) and a = 80 m/s 2 at t = 90 s . For t > 90 s , a = 0 . The graph of a (t ) is given in Fig. 8.112b. (c) The astronaut has the same acceleration as the rocket. This is maximum at t = 90 s and Fmax = mastronaut amax = (75 kg)(80 m/s 2 ) = 6.0 × 103 N . This is 8.2 times her weight on earth, since amax is 8.2 times g.

8-36

Chapter 8

EVALUATE: The acceleration increases because the mass decreases while the thrust F = −vex

dm remains constant. dt

a (m/s2) 100

v (m/s) 4000

80

3000

60

2000

40

1000

20

0

0

20

40

60

80

100 120

t (s)

0

0

20

(a)

40

60

80

100 120

t (s)

(b)

Figure 8.112 8.113.

dm = ρ dV . dV = Adx . Since the thin rod lies along the x axis, ycm = 0 . The mass of the


rod is given by M = ∫ dm .

EXECUTE: (a) xcm = xcm =

1 M

∫

L 0

xdm =

ρ M

L

ρ A L2

0

M 2

A∫ xdx =

. The volume of the rod is AL and M = ρ AL .

ρ AL2 L = . The center of mass of the uniform rod is at its geometrical center, midway between its ends. 2 ρ AL 2

(b) xcm =

1 M

∫

L 0

xdm =

1 M

∫

L 0

xρ Adx =

Aα M

∫

L 0

x 2 dx =

L L Aα L3 α AL2 . Therefore, . M = ∫ dm = ∫ ρ Adx = α A∫ xdx = 0 0 2 3M

⎛ Aα L xcm = ⎜ ⎝ 3 EVALUATE:

⎞ ⎛ 2 ⎞ 2L = . ⎟⎜ 2 ⎟ ⎠ ⎝ α AL ⎠ 3 When the density increases with x, the center of mass is to the right of the center of the rod. 1 1 ydm. At the upper surface of the plate, y 2 + x 2 = a 2 . IDENTIFY: xcm = xdm and ycm = ∫ M∫ M SET UP: To find xcm , divide the plate into thin strips parallel to the y-axis, as shown in Fig. 8.114a. To find ycm , divide the plate into thin strips parallel to the x-axis as shown in Fig. 8.114b. The plate has volume one-half that of a circular disk, so V = 12 π a 2t and M = 12 ρπ a 2t. 3

8.114.

EXECUTE: In Fig.114a each strip has length y = a 2 − x 2 . xcm = xcm =

ρt

M∫

a −a

1 xdm, where dm = ρ tydx = ρ t a 2 − x 2 dx. M∫

x a 2 − x 2 dx = 0, since the integrand is an odd function of x. xcm = 0 because of symmetry. In

Fig.114b each strip has length 2 x = 2 a 2 − y 2 . ycm =

1 ydm, where dm = 2 ρ txdy = 2 ρ t a 2 − y 2 dy. M∫

2ρt a y a 2 − y 2 dy . The integral can be evaluated using u = a 2 − y 2 , du = −2 ydy . This substitution gives M ∫0 2 ρ t ⎛ 1 ⎞ 0 1/ 2 2 ρ ta 3 ⎛ 2 ρ ta 3 ⎞⎛ 2 ⎞ 4a = =⎜ = . ⎟⎜ ⎜ − ⎟ ∫ a 2 u du = 2 ⎟ 3M M ⎝ 2⎠ ⎝ 3 ⎠ ⎝ ρπ a t ⎠ 3π

ycm = ycm

EVALUATE:

4 = 0.424. ycm is less than a/2, as expected, since the plate becomes wider as y decreases. 3π y

y

dy

y y x

x

x

dx

2x (b)

(a)

Figure 8.114


8.115.

8-37

x2

IDENTIFY: The work is related to the force by W = ∫ Fdx . The force the person must apply equals the weight of x1

the hanging portion. Since the rope is uniform, the center of mass of the hanging portion is at its geometrical center. SET UP: Let y be the length of the rope hanging over the edge and use coordinates where the origin is at the edge of the table and +y is downward. When the rope is pulled onto the table, y goes from l / 4 to zero. A length y of the rope has mass λ y . EXECUTE: (a) When a length y hangs over the edge, the person must apply an upward force 0 0 λ gl 2 Fy = − m( y ) g = −λ yg . W = ∫ Fy ( y )dy = −λ g ∫ ydy = . l/4 l/4 32 (b) Initially, ycm = l /8 . The work done to raise an object of mass M a distance ycm is W = Mgycm .

8.116.

2 ⎛ λl ⎞ ⎛ l ⎞ λ gl . W = ⎜ ⎟g⎜ ⎟ = 32 ⎝ 4 ⎠ ⎝8⎠ EVALUATE: The answers from methods (a) and (b) agree. The change in gravitational potential energy of the rope can be calculated by considering all its mass acting at its center of mass, and the work done by the person equals the increase in gravitational potential energy of the rope. IDENTIFY: From our analysis of motion with constant acceleration, if v = at and a is constant, then x − x0 = v0t + 12 at 2 .

SET UP: Take v0 = 0 , x0 = 0 and let +x downward. dv dv = a , v = at and x = 12 at 2 . Substituting into xg = x + v 2 gives dt dt 2 2 2 2 3 2 2 1 1 at g = at a + a t = a t . The nonzero solution is a = g / 3 . 2 2 2

EXECUTE: (a)

(b) x = 12 at 2 = 16 gt 2 = 16 (9.80 m/s 2 )(3.00 s)2 = 14.7 m . (c) m = kx = (2.00 g/m)(14.7 m) = 29.4 g . EVALUATE: The acceleration is less than g because the small water droplets are initially at rest, before they adhere to the falling drop. The small droplets are suspended by buoyant forces that we ignore for the raindrops.

9

ROTATION OF RIGID BODIES

9.1.

9.2.

IDENTIFY: s = rθ , with θ in radians. SET UP: π rad = 180° . s 1.50 m EXECUTE: (a) θ = = = 0.600 rad = 34.4° r 2.50 m s 14.0 cm (b) r = = = 6.27 cm θ (128°)(π rad /180°) (c) s = rθ = (1.50 m)(0.700 rad) = 1.05 m EVALUATE: An angle is the ratio of two lengths and is dimensionless. But, when s = rθ is used, θ must be in radians. Or, if θ = s / r is used to calculate θ , the calculation gives θ in radians. IDENTIFY: θ − θ 0 = ωt , since the angular velocity is constant. SET UP: 1 rpm = (2π / 60) rad/s . EXECUTE: (a) ω = (1900)(2π rad / 60 s) = 199 rad/s (b) 35° = (35°)(π /180°) = 0.611 rad . t =

In t =

EVALUATE:

θ − θ 0 0.611 rad = = 3.1× 10−3 s ω 199 rad/s

θ − θ0 we must use the same angular measure (radians, degrees or revolutions) for both ω

θ − θ 0 and ω . 9.3.

t2 dω z . Writing Eq.(2.16) in terms of angular quantities gives θ − θ = ∫ ω z dt . t 1 dt d n 1 n +1 n −1 SET UP: t t = nt and ∫ t n dt = n +1 dt EXECUTE: (a) A must have units of rad/s and B must have units of rad/s3 . (b) α z (t ) = 2 Bt = (3.00 rad/s3 )t . (i) For t = 0 , α z = 0 . (ii) For t = 5.00 s , α z = 15.0 rad/s 2 .

IDENTIFY:

α z (t ) =

t2

(c) θ 2 − θ1 = ∫ ( A + Bt 2 )dt = A(t2 − t1 ) + 13 B (t23 − t13 ) . For t1 = 0 and t2 = 2.00 s , t1

9.4.

θ 2 − θ1 = (2.75 rad/s)(2.00 s) + 13 (1.50 rad/s3 )(2.00 s)3 = 9.50 rad . EVALUATE: Both α z and ω z are positive and the angular speed is increasing. Δω z IDENTIFY: α z = d ω z / dt . α av-z = . Δt

SET UP:

d 2 (t ) = 2t dt

dωz = −2 βt = ( − 1.60 rad s3 )t. dt (b) αz (3.0 s) = ( − 1.60 rad s3 )(3.0 s) = −4.80 rad s 2 . EXECUTE:

(a) αz (t ) =

ωz (3.0 s) − ωz (0) −2.20 rad s − 5.00 rad s = = −2.40 rad s 2. , 3.0 s 3.0 s which is half as large (in magnitude) as the acceleration at t = 3.0 s. α (0) + α z (3.0 s) EVALUATE: α z (t ) increases linearly with time, so α av-z = z . α z (0) = 0 . 2 αav-z =

9-1

9-2

9.5.

Chapter 9

IDENTIFY and SET UP: Use Eq.(9.3) to calculate the angular velocity and Eq.(9.2) to calculate the average angular velocity for the specified time interval. EXECUTE: θ = γ t + β t 3 ; γ = 0.400 rad/s, β = 0.0120 rad/s3 dθ (a) ω z = = γ + 3β t 2 dt (b) At t = 0, ω z = γ = 0.400 rad/s (c) At t = 5.00 s, ω z = 0.400 rad/s + 3(0.0120 rad/s3 )(5.00 s) 2 = 1.30 rad/s

ωav-z =

Δθ θ 2 − θ1 = t2 − t1 Δt

For t1 = 0, θ1 = 0. For θ 2 = 5.00 s, θ 2 = (0.400 rad/s)(5.00 s) + (0.012 rad/s3 )(5.00 s)3 = 3.50 rad 3.50 rad − 0 = 0.700 rad/s. 5.00 s − 0 EVALUATE: The average of the instantaneous angular velocities at the beginning and end of the time interval is 1 2 (0.400 rad/s + 1.30 rad/s) = 0.850 rad/s. This is larger than ωav-z , because ω z (t ) is increasing faster than linearly. So ωav-z =

9.6.

IDENTIFY: SET UP:

dθ dω z Δθ . α z (t ) = . ωav-z = . dt dt Δt ωz = (250 rad s) − (40.0 rad s 2 )t − (4.50 rad s3 )t 2 . αz = −(40.0 rad s 2 ) − (9.00 rad s3 )t .

EXECUTE:

ω z (t ) =

(a) Setting ωz = 0 results in a quadratic in t. The only positive root is t = 4.23 s .

(b) At t = 4.23 s , αz = −78.1 rad s 2 . (c) At t = 4.23 s , θ = 586 rad = 93.3 rev . (d) At t = 0 , ωz = 250 rad/s . (e) ωav-z = 586 rad = 138 rad s. 4.23 s EVALUATE: Between t = 0 and t = 4.23 s , ω z decreases from 250 rad/s to zero. ω z is not linear in t, so ωav-z is

not midway between the values of ω z at the beginning and end of the interval. 9.7.

IDENTIFY:

ω z (t ) =

dθ dω z . α z (t ) = . Use the values of θ and ω z at t = 0 and α z at 1.50 s to calculate a, b, dt dt

and c.

d n t = nt n −1 dt EXECUTE: (a) ω z (t ) = b − 3ct 2 . α z (t ) = −6ct . At t = 0 , θ = a = π / 4 rad and ω z = b = 2.00 rad/s . At t = 1.50 s , SET UP:

α z = −6c(1.50 s) = 1.25 rad/s 2 and c = −0.139 rad/s3 . (b) θ = π / 4 rad and α z = 0 at t = 0 . (c) α z = 3.50 rad/s 2 at t = −

θ=

αz 6c

=−

3.50 rad/s 2 = 4.20 s . At t = 4.20 s , 6(−0.139 rad/s3 )

π

rad + (2.00 rad/s)(4.20 s) − ( −0.139 rad/s 3 )(4.20 s) 3 = 19.5 rad . 4 ω z = 2.00 rad/s − 3( −0.139 rad/s3 )(4.20 s) 2 = 9.36 rad/s .

EVALUATE: 9.8.

IDENTIFY:

ωav-z =

θ , ω z and α z all increase as t increases. dω α z = z . θ − θ 0 = ωav-zt . When ω z is linear in t, ωav-z for the time interval t1 to t2 is

ω z1 + ω z 2 t2 − t1

dt

.

SET UP: From the information given, ω z (t ) = −6.00 rad/s + (2.00 rad/s 2 )t EXECUTE: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value. (b) It takes 3.00 seconds for the wheel to stop (ωz = 0) . During this time its speed is decreasing. For the next

4.00 s its speed is increasing from 0 rad s to + 8.00 rad s .

Rotation of Rigid Bodies

9-3

−6.00 rad s + 8.00 rad s = 1.00 rad s . θ − θ 0 = ωav-zt then leads to 2 displacement of 7.00 rad after 7.00 s. EVALUATE: When α z and ω z have the same sign, the angular speed is increasing; this is the case for t = 3.00 s to (c) The average angular velocity is

9.9.

t = 7.00 s . When α z and ω z have opposite signs, the angular speed is decreasing; this is the case between t = 0 and t = 3.00 s . IDENTIFY: Apply the constant angular acceleration equations. SET UP: Let the direction the wheel is rotating be positive. EXECUTE: (a) ω z = ω0 z + α z t = 1.50 rad s + (0.300 rad s 2 )(2.50 s) = 2.25 rad s. (b) θ − θ 0 = ω0 z t + 12 α zt 2 = (1.50 rad/s)(2.50 s) + 12 (0.300 rad/s 2 )(2.50 s) 2 = 4.69 rad .

⎛ ω0 z + ω z ⎞ ⎛ 1.50 rad/s + 2.25 rad/s ⎞ ⎟t = ⎜ ⎟ (2.50 s) = 4.69 rad , the same as calculated with 2 2 ⎝ ⎠ ⎝ ⎠ another equation in part (b). IDENTIFY: Apply the constant angular acceleration equations to the motion of the fan. (a) SET UP: ω0 z = (500 rev/min)(1 min/60 s) = 8.333 rev/s, ω z = (200 rev/min)(1 min/60 s) = 3.333 rev/s,

θ − θ0 = ⎜

EVALUATE:

9.10.

t = 4.00 s, α z = ?

ω z = ω0 z + α z t αz =

EXECUTE:

ω z − ω0 z t

=

3.333 rev/s − 8.333 rev/s = −1.25 rev/s 2 4.00 s

θ − θ0 = ? θ − θ 0 = ω0 z t + 12 α zt 2 = (8.333 rev/s)(4.00 s) + 12 ( −1.25 rev/s 2 )(4.00 s) 2 = 23.3 rev (b) SET UP: ω z = 0 (comes to rest); ω0 z = 3.333 rev/s; α z = −1.25 rev/s 2 ; t =? ω z = ω0 z + α z t

t=

EXECUTE:

ω z − ω0 z 0 − 3.333 rev/s = = 2.67 s αz −1.25 rev/s 2

EVALUATE: The angular acceleration is negative because the angular velocity is decreasing. The average angular velocity during the 4.00 s time interval is 350 rev/min and θ − θ 0 = ωav-z t gives θ − θ 0 = 23.3 rev, which checks. 9.11.

IDENTIFY: SET UP:

Apply the constant angular acceleration equations to the motion. The target variables are t and θ − θ 0 . (a) α z = 1.50 rad/s 2 ; ω0 z = 0 (starts from rest); ω z = 36.0 rad/s; t = ?

ω z = ω0 z + α z t EXECUTE:

t=

ω z − ω0 z 36.0 rad/s − 0 = = 24.0 s αz 1.50 rad/s 2

(b) θ − θ 0 = ?

θ − θ 0 = ω0 z t + 12 α zt 2 = 0 + 12 (1.50 rad/s 2 )(2.40 s) 2 = 432 rad θ − θ 0 = 432 rad(1 rev/2π rad) = 68.8 rev EVALUATE: We could use θ − θ 0 = 12 (ω z + ω0 z )t to calculate θ − θ 0 = 12 (0 + 36.0 rad/s)(24.0 s) = 432 rad, which 9.12.

checks. IDENTIFY: In part (b) apply the equation derived in part (a). SET UP: Let the direction the propeller is rotating be positive. ω − ω0 z . Rewriting Eq. (9.11) as θ − θ 0 = t (ω0 z + 12 ω z t ) and EXECUTE: (a) Solving Eq. (9.7) for t gives t = z

αz

substituting for t gives ⎛ ω − ω0 z ⎞ ⎛ 1 1 ⎞ 1 ⎛ ω z + ω0 z ⎞ θ − θ0 = ⎜ z (ω z2 − ω02z ), ⎟ ⎜ ω0 z + (ω z − ω0 z ) ⎟ = (ω z − ω0 z ) ⎜ ⎟= α 2 α 2 2 α ⎠ ⎝ ⎠ ⎝ ⎠⎝ z z z which when rearranged gives Eq. (9.12). ⎛ 1 ⎞ 2 1 2 2 ⎞ 2 2 1⎛ (b) α z = 12 ⎜ ⎟ ( ω z − ω0 z ) = 2 ⎜ ⎟ (16.0 rad s ) − (12.0 rad s ) = 8.00 rad/s θ θ 7.00 rad − ⎝ ⎠ 0 ⎠ ⎝

(

)

9-4

Chapter 9

EVALUATE:

⎛ ω + ωz ⎞ We could also use θ − θ 0 = ⎜ 0 z ⎟ t to calculate t = 0.500 s . Then ω z = ω0 z + α zt gives 2 ⎝ ⎠

α z = 8.00 rad/s 2 , which agrees with our results in part (b). 9.13.

IDENTIFY: Use a constant angular acceleration equation and solve for ω0 z . SET UP: Let the direction of rotation of the flywheel be positive. θ − θ0 1 60.0 rad 1 − 2αz = − 2 (2.25 rad/s 2 )(4.00 s) = 10.5 rad/s . EXECUTE: θ − θ 0 = ω0 z t + 12 α z2 gives ω0 z = 4.00 s t EVALUATE: At the end of the 4.00 s interval, ω z = ω0 z + α z t = 19.5 rad/s .

⎛ ω0 z + ω z ⎞ ⎛ 10.5 rad/s + 19.5 rad/s ⎞ ⎟t = ⎜ ⎟ (4.00 s) = 60.0 rad , which checks. 2 2 ⎝ ⎠ ⎝ ⎠ IDENTIFY: Apply the constant angular acceleration equations. SET UP: Let the direction of the rotation of the blade be positive. ω0 z = 0 .

θ − θ0 = ⎜

9.14.

EXECUTE:

ω z = ω0 z + α z gives α z =

ω z − ω0 z t

=

140 rad/s − 0 = 23.3 rad/s 2 . 6.00 s

⎛ ω + ω z ⎞ ⎛ 0 + 140 rad/s ⎞ (θ − θ 0 ) = ⎜ 0 z ⎟t = ⎜ ⎟ (6.00 s) = 420 rad 2 2 ⎝ ⎠ ⎝ ⎠ EVALUATE:

We could also use θ − θ 0 = ω0 z t + 12 α z t 2 . This equation gives

θ − θ 0 = 12 (23.3 rad/s 2 )(6.00 s) 2 = 419 rad , in agreement with the result obtained above. 9.15.

IDENTIFY: Apply constant angular acceleration equations. SET UP: Let the direction the flywheel is rotating be positive. θ − θ 0 = 200 rev, ω0 z = 500 rev min = 8.333 rev s, t = 30.0 s . EXECUTE:

⎛ ω + ωz ⎞ (a) θ − θ 0 = ⎜ 0 z ⎟ t gives ω z = 5.00 rev s = 300 rpm 2 ⎝ ⎠

(b) Use the information in part (a) to find α z : ω z = ω0 z + α zt gives α z = −0.1111 rev s 2 . Then ω z = 0,

⎛ ω0 z + ω z ⎞ ⎟ t gives 2 ⎝ ⎠

α z = −0.1111 rev s 2 , ω0 z = 8.333 rev s in ω z = ω0 z + α zt gives t = 75.0 s and θ − θ 0 = ⎜

9.16.

θ − θ 0 = 312 rev . EVALUATE: The mass and diameter of the flywheel are not used in the calculation. IDENTIFY: Use the constant angular acceleration equations, applied to the first revolution and to the first two revolutions. SET UP: Let the direction the disk is rotating be positive. 1 rev = 2π rad . Let t be the time for the first revolution. The time for the first two revolutions is t + 0.750 s . EXECUTE: (a) θ − θ 0 = ω0 zt + 12 α z t 2 applied to the first and to the first two revolutions gives 2π rad = 12 α zt 2 and 4π rad = 12 α z (t + 0.750 s) 2 . Eliminating α z between these equations gives 4π rad = 2t 2 = (t + 0.750 s) 2 .

2t = ± (t + 0.750 s) . The positive root is t =

2π rad (t + 0.750 s) 2 . t2

0.750 s = 1.81 s . 2 −1

(b) 2π rad = 12 α zt 2 and t = 1.81 s gives α z = 3.84 rad/s 2 EVALUATE:

9.17.

At the start of the second revolution, ω0 z = (3.84 rad/s 2 )(1.81 s) = 6.95 rad/s . The distance the disk

rotates in the next 0.750 s is θ − θ 0 = ω0 z t + 12 α zt 2 = (6.95 rad/s)(0.750 s) + 12 (3.84 rad/s 2 )(0.750 s) 2 = 6.29 rad , which is two revolutions. IDENTIFY: Apply Eq.(9.12) to relate ω z to θ − θ 0 . SET UP: Establish a proportionality. EXECUTE: From Eq.(9.12), with ω0 z = 0, the number of revolutions is proportional to the square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.00 rev. EVALUATE: We don't have enough information to calculate α z ; all we need to know is that it is constant.


9.18.

IDENTIFY:

In each case we apply constant acceleration equations to determine θ (t ) and ω z (t ) .

SET UP: Let θ 0 = 0 . The following table gives the revolutions and the angle θ (in degrees) through which the wheel has rotated for each instant in time (in seconds) and each of the three situations:

t

0.05 0.10 0.15 0.20 EXECUTE: EVALUATE:

(a) rev 0.50 1.00 1.50 2.00

(b)

θ 180 360 540 720

rev 0.03 0.13 0.28 0.50

(c)

θ 11.3 45 101 180

rev 0.44 0.75 0.94 1.00

θ 158 270 338 360

The θ and ω z graphs for each case are given in Figures 9.18 a–c. The slope of the θ (t ) graph is ω z (t ) and the slope of the ω z (t ) graph is α z (t ) .

Figure 9.18

9-5

9-6

9.19.

Chapter 9

IDENTIFY: Apply the constant angular acceleration equations separately to the time intervals 0 to 2.00 s and 2.00 s until the wheel stops. (a) SET UP: Consider the motion from t = 0 to t = 2.00 s: θ − θ 0 = ?; ω0 z = 24.0 rad/s; α z = 30.0 rad/s 2 ; t = 2.00 s EXECUTE:

θ − θ 0 = ω0 z t + 12 α zt 2 = (24.0 rad/s)(2.00 s) + 12 (30.0 rad/s 2 )(2.00 s) 2

θ − θ 0 = 48.0 rad + 60.0 rad = 108 rad Total angular displacement from t = 0 until stops: 108 rad + 432 rad = 540 rad Note: At t = 2.00 s, ω z = ω0 z + α z t = 24.0 rad/s + (30.0 rad/s 2 )(2.00 s) = 84.0 rad/s; angular speed when breaker trips. (b) SET UP: Consider the motion from when the circuit breaker trips until the wheel stops. For this calculation let t = 0 when the breaker trips. t = ?; θ − θ 0 = 432 rad; ω z = 0; ω0 z = 84.0 rad/s (from part (a)) ⎛ ω0 z + ω z ⎞ ⎟t 2 ⎝ ⎠ 2(θ − θ 0 ) 2(432 rad) = = 10.3 s EXECUTE: t = ω0 z + ω z 84.0 rad/s + 0 The wheel stops 10.3 s after the breaker trips so 2.00 s + 10.3 s = 12.3 s from the beginning. (c) SET UP: α z = ?; consider the same motion as in part (b):

θ − θ0 = ⎜

ω z = ω0 z + α z t

9.20.

ω z − ω0 z

0 − 84.0 rad/s = = −8.16 rad/s 2 10.3 s t EVALUATE: The angular acceleration is positive while the wheel is speeding up and negative while it is slowing ω 2 − ω02z 0 − (84.0 rad/s) 2 down. We could also use ω z2 = ω02z + 2α z (θ − θ 0 ) to calculate α z = z = = −8.16 rad/s 2 for 2(θ − θ 0 ) 2(432 rad) the acceleration after the breaker trips. IDENTIFY: The linear distance the elevator travels, its speed and the magnitude of its acceleration are equal to the tangential displacement, speed and acceleration of a point on the rim of the disk. s = rθ , v = rω and a = rα . In these equations the angular quantities must be in radians. SET UP: 1 rev = 2π rad . 1 rpm = 0.1047 rad/s . π rad = 180° . For the disk, r = 1.25 m . EXECUTE:

αz =

v 0.250 m/s = = 0.200 rad/s = 1.91 rpm . 1.25 m r a 1.225 m/s 2 (b) a = 18 g = 1.225 m/s 2 . α = = = 0.980 rad/s 2 . 1.25 m r s 3.25 m (c) s = 3.25 m . θ = = = 2.60 rad = 149° . r 1.25 m EVALUATE: When we use s = rθ , v = rω and atan = rα to solve for θ , ω and α , the results are in rad, rad/s EXECUTE:

9.21.

(a) v = 0.250 m/s so ω =

and rad/s 2 . IDENTIFY: When the angular speed is constant, ω = θ / t . vtan = rω , atan = rα and arad = rω 2 . In these equations radians must be used for the angular quantities. SET UP: The radius of the earth is RE = 6.38×106 m and the earth rotates once in 1 day = 86,400 s . The orbit radius of the earth is 1.50 ×1011 m and the earth completes one orbit in 1 y = 3.156 ×107 s . When ω is constant, ω = θ / t . EXECUTE:

(a) θ = 1 rev = 2π rad in t = 3.156 × 107 s . ω =

(b) θ = 1 rev = 2π rad in t = 86,400 s . ω =

2π rad = 1.99 ×10 −7 rad/s . 3.156 ×107 s

2π rad = 7.27 ×10−5 rad/s 86,400 s

(c) v = rω = (1.50 ×1011 m)(1.99 ×10−7 rad/s) = 2.98 ×104 m/s . (d) v = rω = (6.38 × 106 m)(7.27 ×10−5 rad/s) = 464 m/s . (e) arad = rω 2 = (6.38 ×106 m)(7.27 ×10−5 rad/s) 2 = 0.0337 m/s 2 . atan = rα = 0 . α = 0 since the angular velocity is constant. EVALUATE: The tangential speeds associated with these motions are large even though the angular speeds are very small, because the radius for the circular path in each case is quite large.


9.22.

9.23.

9-7

IDENTIFY: Linear and angular velocities are related by v = rω . Use ω z = ω0 z + α zt to calculate α z . SET UP: ω = v / r gives ω in rad/s. 1.25 m/s 1.25 m/s EXECUTE: (a) = 50.0 rad/s, = 21.6 rad/s. −3 25.0 × 10 m 58.0 × 10−3 m (b) (1.25 m/s) (74.0 min) (60 s/min) = 5.55 km. (c) α z = 21.55 rad/s − 50.0 rad/s = −6.41 × 10−3 rad/s 2 . (74.0 min) (60 s/min) EVALUATE: The width of the tracks is very small, so the total track length on the disc is huge. IDENTIFY: Use constant acceleration equations to calculate the angular velocity at the end of two revolutions. v = rω . SET UP: 2 rev = 4π rad. r = 0.200 m. EXECUTE:

(a) ω z2 = ω02z + 2α z (θ − θ 0 ). ω z = 2α z (θ − θ 0 ) = 2(3.00 rad/s 2 )(4π rad) = 8.68 rad/s.

arad = rω 2 = (0.200 m)(8.68 rad/s) 2 = 15.1 m/s 2 . v 2 (1.74 m/s) 2 = = 15.1 m/s 2 . 0.200 m r rω 2 and v 2 / r are completely equivalent expressions for arad .

(b) v = rω = (0.200 m)(8.68 rad/s) = 1.74 m/s. arad = EVALUATE: 9.24.

IDENTIFY: arad = rω 2 , with ω in rad/s. Solve for ω . SET UP: 1 rpm = (2π / 60) rad/s EXECUTE:

9.25.

ω=

arad (400,000)(9.80 m/s 2 ) = = 1.25 ×104 rad/s = 1.20 ×105 rpm 0.0250 m r

EVALUATE: In arad = rω 2 , ω must be in rad/s. IDENTIFY and SET UP: Use constant acceleration equations to find ω and α after each displacement. The use Eqs.(9.14) and (9.15) to find the components of the linear acceleration. EXECUTE: (a) at the start t = 0 flywheel starts from rest so ω z = ω0 z = 0

atan = rα = (0.300 m)(0.600 rad/s 2 ) = 0.180 m/s 2 arad = rω 2 = 0 2 2 a = arad + atan = 0.180 m/s 2

(b) θ − θ 0 = 60°

atan = rα = 0.180 m/s 2 Calculate ω: θ − θ 0 = 60°(π rad/180°) = 1.047 rad; ω0 z = 0; α z = 0.600 rad/s 2 ; ω z = ?

ω z2 = ω02z + 2α z (θ − θ 0 ) ω z = 2α z (θ − θ 0 ) = 2(0.600 rad/s 2 )(1.047 rad) = 1.121 rad/s and ω = ω z . Then arad = rω 2 = (0.300 m)(1.121 rad/s) 2 = 0.377 m/s 2 . 2 2 a = arad + atan = (0.377 m/s 2 ) 2 + (0.180 m/s 2 ) 2 = 0.418 m/s 2

(c) θ − θ 0 = 120°

atan = rα = 0.180 m/s 2 Calculate ω: θ − θ 0 = 120°(π rad/180°) = 2.094 rad; ω0 z = 0; α z = 0.600 rad/s 2 ; ω z = ?

ω z2 = ω02z + 2α z (θ − θ 0 ) ω z = 2α z (θ − θ 0 ) = 2(0.600 rad/s 2 )(2.094 rad) = 1.585 rad/s and ω = ω z . Then arad = rω 2 = (0.300 m)(1.585 rad/s) 2 = 0.754 m/s 2 . 2 2 a = arad + atan = (0.754 m/s 2 )2 + (0.180 m/s 2 ) 2 = 0.775 m/s 2

EVALUATE:

α is constant so α tan is constant. ω increases so arad increases.

9-8

9.26.

Chapter 9

IDENTIFY: Apply constant angular acceleration equations. v = rω . A point on the rim has both tangential and radial components of acceleration. SET UP: atan = rα and arad = rω 2 . (a) ω z = ω0 z + α z t = 0.250 rev/s + (0.900 rev/s 2 )(0.200 s) = 0.430 rev/s

EXECUTE:

(Note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to convert to radians). (b) ωav-z Δt = (0.340 rev s)(0.2 s) = 0.068 rev . (c) Here, the conversion to radians must be made to use Eq. (9.13), and

⎛ 0.750 m ⎞ v = rω = ⎜ ⎟ ( 0.430 rev/s )( 2π rad rev ) = 1.01 m s. 2 ⎝ ⎠ (d) Combining equations (9.14) and (9.15), 2 2 a = arad + atan = (ω 2 r ) 2 + (α r ) 2 . 2

2

a = ⎡⎣((0.430 rev/s)(2π rad/rev)) 2 (0.375 m) ⎤⎦ + ⎡⎣(0.900 rev/s 2 )(2π rad/rev)(0.375 m) ⎤⎦ . a = 3.46 m s 2 . 9.27.

EVALUATE: If the angular acceleration is constant, atan is constant but arad increases as ω increases. IDENTIFY: Use Eq.(9.15) and solve for r. SET UP: arad = rω 2 so r = arad / ω 2 , where ω must be in rad/s

arad = 3000 g = 3000(9.80 m/s 2 ) = 29,400 m/s 2

EXECUTE:

⎛ 1 min ⎞⎛ 2π rad ⎞ ⎟⎜ ⎟ = 523.6 rad/s ⎝ 60 s ⎠⎝ 1 rev ⎠

ω = (5000 rev/min) ⎜

29,400 m/s 2 = 0.107 m. ω (523.6 rad/s) 2 EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic. IDENTIFY: In part (b) apply the result derived in part (a). SET UP: arad = rω 2 and v = rω ; combine to eliminate r.

Then r = 9.28.

arad 2

=

⎛v⎞ (a) arad = ω 2 r = ω 2 ⎜ ⎟ = ωv. ⎝ω ⎠

EXECUTE:

(b) From the result of part (a), ω = EVALUATE: 9.29.

arad 0.500 m s 2 = = 0.250 rad s. 2.00 m s v

arad = rω 2 and v = rω both require that ω be in rad/s, so in arad = ωv , ω is in rad/s.

IDENTIFY: v = rω and arad = rω 2 = v 2 / r . SET UP: 2π rad = 1 rev , so π rad/s = 30 rev/min . EXECUTE:

(a) ω r = (1250 rev min )

rad/s ⎛ 12.7 × 10 )⎜⎝ 2 ( 30πrev/min

−3

m⎞ ⎟ = 0.831 m s. ⎠

v2 (0.831 m s) 2 = = 109 m s 2 . r (12.7 ×10−3 m) 2 EVALUATE: In v = rω , ω must be in rad/s. IDENTIFY: atan = rα , v = rω and arad = v 2 / r . θ − θ 0 = ωav-zt . (b)

9.30.

SET UP:

When α z is constant, ωav-z =

ω0 z + ω z 2

. Let the direction the wheel is rotating be positive.

atan −10.0 m s 2 = = −50.0 rad s 2 r 0.200 m v 50.0 m s (b) At t = 3.00 s , v = 50.0 m s and ω = = = 250 rad s and at t = 0, r 0.200 m EXECUTE:

(a) α =

v = 50.0 m s + ( − 10.0 m s 2 )(0 − 3.00 s) = 80.0 m s , ω = 400 rad s. (c) ωav-z t = (325 rad s)(3.00 s) = 975 rad = 155 rev .


9-9

50.0 m/s − 1.40 m/s = 4.86 s 10.0 m/s after t = 3.00 s , or at t = 7.86 s . (There are many equivalent ways to do this calculation.) EVALUATE: At t = 0 , arad = rω 2 = 3.20 ×104 m/s 2 . At t = 3.00 s , arad = 1.25 ×104 m/s 2 . For arad = g the wheel must be rotating more slowly than at 3.00 s so it occurs some time after 3.00 s. ! ! IDENTIFY and SET UP: Use Eq.(9.15) to relate ω to arad and ∑ F = ma to relate arad to Frad . Use Eq.(9.13) to (d) v = arad r = (9.80 m/s 2 )(0.200 m) = 1.40 m/s. This speed will be reached at time

9.31.

relate ω and v, where v is the tangential speed. EXECUTE: (a) arad = rω 2 and Frad = marad = mrω 2 2

2

⎛ ω ⎞ ⎛ 640 rev/min ⎞ =⎜ 2 ⎟ =⎜ ⎟ = 2.29 Frad,1 ⎝ ω1 ⎠ ⎝ 423 rev/min ⎠ (b) v = rω v2 ω2 640 rev/min = = = 1.51 v1 ω1 423 rev/min (c) v = rω ⎛ 1 min ⎞⎛ 2π rad ⎞ ω = (640 rev/min) ⎜ ⎟⎜ ⎟ = 67.0 rad/s ⎝ 60 s ⎠⎝ 1 rev ⎠ Then v = rω = (0.235 m)(67.0 rad/s) = 15.7 m/s. Frad, 2

arad = rω 2 = (0.235 m)(67.0 rad/s) 2 = 1060 m/s 2 arad 1060 m/s 2 = = 108; a = 108 g g 9.80 m/s 2 EVALUATE: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert ω to rad/s. In part (c), v and arad are calculated from ω , and ω must be in rad/s. 9.32.

IDENTIFY: SET UP:

v = rω and atan = rα . The linear acceleration of the bucket equals atan for a point on the rim of the axle.

⎛ 7.5 rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ (a) v = Rω . 2.00 cm s = R ⎜ ⎟⎜ ⎟⎜ ⎟ gives R = 2.55 cm . ⎝ min ⎠⎝ 60 s ⎠⎝ 1 rev ⎠ D = 2 R = 5.09 cm . a 0.400 m s 2 (b) atan = Rα . α = tan = = 15.7 rad s 2 . 0.0255 m R EVALUATE: In v = Rω and atan = Rα , ω and α must be in radians. IDENTIFY: Apply v = rω . SET UP: Points on the chain all move at the same speed, so rrωr = rf ωf . EXECUTE:

9.33.

vr 5.00 m s = = 15.15 rad s. 0.330 m r The angular velocity of the front wheel is ωf = 0.600 rev s = 3.77 rad s . rr = rf (ωf ωr ) = 2.99 cm . EXECUTE:

9.34.

The angular velocity of the rear wheel is ωr =

EVALUATE: The rear sprocket and wheel have the same angular velocity and the front sprocket and wheel have the same angular velocity. rω is the same for both, so the rear sprocket has a smaller radius since it has a larger angular velocity. The speed of a point on the chain is v = rrωr = (2.99 ×10−2 m)(15.15 rad/s) = 0.453 m/s . The linear speed of the bicycle is 5.50 m/s. IDENTIFY and SET UP: Use Eq.(9.16). Treat the spheres as point masses and ignore I of the light rods. EXECUTE: The object is shown in Figure 9.34a. (a)

r = (0.200 m) 2 + (0.200 m) 2 = 0.2828 m I = ∑ mi ri 2 = 4(0.200 kg)(0.2828 m) 2 I = 0.0640 kg ⋅ m 2

Figure 9.34a

9-10

Chapter 9

(b) The object is shown in Figure 9.34b.

r = 0.200 m I = ∑ mi ri 2 = 4(0.200 kg)(0.200 m) 2 I = 0.0320 kg ⋅ m 2 Figure 9.34b (c) The object is shown in Figure 9.34c.

r = 0.2828 m I = ∑ mi ri 2 = 2(0.200 kg)(0.2828 m) 2 I = 0.0320 kg ⋅ m 2

Figure 9.34c

9.35.

EVALUATE: In general I depends on the axis and our answer for part (a) is larger than for parts (b) and (c). It just happens that I is the same in parts (b) and (c). IDENTIFY: Use Table 9.2. The correct expression to use in each case depends on the shape of the object and the location of the axis. SET UP: In each case express the mass in kg and the length in m, so the moment of inertia will be in kg ⋅ m 2 . EXECUTE:

(a) (i) I = 13 ML2 = 13 (2.50 kg)(0.750 m) 2 = 0.469 kg ⋅ m 2 .

(ii) I = 121 ML2 = 14 (0.469 kg ⋅ m 2 ) = 0.117 kg ⋅ m 2 . (iii) For a very thin rod, all of the mass is at the axis and I = 0 . (b) (i) I = 52 MR 2 = 52 (3.00 kg)(0.190 m) 2 = 0.0433 kg ⋅ m 2 .

(ii) I = 32 MR 2 = 53 (0.0433 kg ⋅ m 2 ) = 0.0722 kg ⋅ m 2 . (c) (i) I = MR 2 = (8.00 kg)(0.0600 m) 2 = 0.0288 kg ⋅ m 2 .

9.36.

(ii) I = 12 MR 2 = 12 (8.00 kg)(0.0600 m) 2 = 0.0144 kg ⋅ m 2 . EVALUATE: I depends on how the mass of the object is distributed relative to the axis. IDENTIFY: Treat each block as a point mass, so for each block I = mr 2 , where r is the distance of the block from the axis. The total I for the object is the sum of the I for each of its pieces. SET UP: In part (a) two blocks are a distance L / 2 from the axis and the third block is on the axis. In part (b) two blocks are a distance L / 4 from the axis and one is a distance 3L / 4 from the axis. EXECUTE: (a) I = 2m( L / 2) 2 = 12 mL2 . 1 11 mL2 (2 + 9) = mL2 . 16 16 EVALUATE: For the same object I is in general different for different axes. IDENTIFY: I for the object is the sum of the values of I for each part. SET UP: For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2. For a point mass, I = mr 2 , where r is the distance of the mass from the axis. (b) I = 2m( L / 4) 2 + m(3L / 4) 2 =

9.37.

2

EXECUTE:

(a) I = I bar + I balls =

I=

1 ⎛L⎞ M bar L2 + 2mballs ⎜ ⎟ . 12 ⎝2⎠

1 2 2 ( 4.00 kg )( 2.00 m ) + 2 ( 0.500 kg )(1.00 m ) = 2.33 kg ⋅ m2 12

1 1 2 2 (b) I = mbar L2 + mball L2 = ( 4.00 kg )( 2.00 m ) + ( 0.500 kg ) ( 2.00 m ) = 7.33 kg ⋅ m 2 3 3 (c) I = 0 because all masses are on the axis. (d) All the mass is a distance d = 0.500 m from the axis and I = mbar d 2 + 2mball d 2 = M Total d 2 = (5.00 kg)(0.500 m) 2 = 1.25 kg ⋅ m 2 . EVALUATE: I for an object depends on the location and direction of the axis.


9.38.

9-11

IDENTIFY and SET UP: According to Eq.(9.16), I for the entire object equals the sum of I for each piece, the rod plus the end caps. The object is shown in Figure 9.38. EXECUTE:

I = I rod + 2 I cap

I = 121 ML + 2(m)( L / 2) 2 = ( 121 M + 12 m ) L2 2

Figure 9.38 EVALUATE: 9.39.

Table 9.2 was used for I rod and I = mr 2 for the end caps, since they are treated as point particles.


I = ∑ mi ri 2 implies I = I rim + I spokes

EXECUTE: I rim = MR 2 = (1.40 kg)(0.300 m) 2 = 0.126 kg ⋅ m 2 Each spoke can be treated as a slender rod with the axis through one end, so I spokes = 8 ( 13 ML2 ) = 83 (0.280 kg)(0.300 m) 2 = 0.0672 kg ⋅ m 2

I = I rim + I spokes = 0.126 kg ⋅ m 2 + 0.0672 kg ⋅ m 2 = 0.193 kg ⋅ m 2

9.40.

9.41.

EVALUATE: Our result is smaller than mtot R 2 = (3.64 kg)(0.300 m) 2 = 0.328 kg ⋅ m 2 , since the mass of each spoke is distributed between r = 0 and r = R. IDENTIFY: Compare this object to a uniform disk of radius R and mass 2M. SET UP: With an axis perpendicular to the round face of the object at its center, I for a uniform disk is the same as for a solid cylinder. EXECUTE: (a) The total I for a disk of mass 2M and radius R, I = 12 (2 M ) R 2 = MR 2 . Each half of the disk has the

same I, so for the half-disk, I = 12 MR 2 . (b) The same mass M is distributed the same way as a function of distance from the axis. (c) The same method as in part (a) says that I for a quarter-disk of radius R and mass M is half that of a half-disk of radius R and mass 2M, so I = 12 ( 12 [2 M ]R 2 ) = 12 MR 2 . EVALUATE: I depends on how the mass of the object is distributed relative to the axis, and this is the same for any segment of a disk. IDENTIFY: I for the compound disk is the sum of I of the solid disk and of the ring. SET UP: For the solid disk, I = 12 md rd2 . For the ring, I r = 12 mr (r12 + r22 ) , where r1 = 50.0 cm, r2 = 70.0 cm . The mass of the disk and ring is their area times their area density. EXECUTE: I = I d + I r . 1 2 Disk: md = (3.00 g cm )π rd2 = 23.56 kg . I d = md rd2 = 2.945 kg ⋅ m 2 . 2 1 2 2 2 Ring: mr = (2.00 g cm )π (r2 − r1 ) = 15.08 kg . I r = mr (r12 + r22 ) = 5.580 kg ⋅ m 2 . 2 2 I = I d + I r = 8.52 kg ⋅ m . EVALUATE:

9.42.

IDENTIFY:

Even though mr < md , I r > I d since the mass of the ring is farther from the axis.

K = 12 I ω 2 . Use Table 9.2b to calculate I. I = 121 ML2 . 1 rpm = 0.1047 rad/s

SET UP: EXECUTE:

⎛ 0.1047 rad/s ⎞ (a) I = 121 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 . ω = (2400 rev/min) ⎜ ⎟ = 251 rad/s . ⎝ 1 rev/min ⎠

K = 12 I ω 2 = 12 (42.2 kg ⋅ m 2 )(251 rad/s) 2 = 1.33 ×106 J . (b) K1 = 121 M 1L12ω12 , K 2 = 121 M 2 L22ω22 . L1 = L2 and K1 = K 2 , so M 1ω12 = M 2ω22 .

ω2 = ω1

9.43.

M1 M1 = (2400 rpm) = 2770 rpm M2 0.750 M 1

EVALUATE: The rotational kinetic energy is proportional to the square of the angular speed and directly proportional to the mass of the object. IDENTIFY: K = 12 I ω 2 . Use Table 9.2 to calculate I. SET UP:

I = 52 MR 2 . For the moon, M = 7.35 ×1022 kg and R = 1.74 ×106 m . The moon moves through

1 rev = 2π rad in 27.3 d. 1 d = 8.64 ×104 s .

9-12

Chapter 9

EXECUTE:

ω=

(a) I = 52 (7.35 ×1022 kg)(1.74 ×106 m) 2 = 8.90 ×1034 kg ⋅ m 2 .

2π rad = 2.66 ×10−6 rad/s . (27.3 d)(8.64 ×10 4 s/d)

K = 12 I ω 2 = 12 (8.90 ×1034 kg ⋅ m 2 )(2.66 × 10−6 rad/s) 2 = 3.15 ×1023 J . 3.15 ×1023 J = 158 years . Considering the expense involved in tapping the moon’s rotational energy, this 5(4.0 ×1020 J) does not seem like a worthwhile scheme for only 158 years worth of energy. EVALUATE: The moon has a very large amount of kinetic energy due to its motion. The earth has even more, but changing the rotation rate of the earth would change the length of a day. IDENTIFY: K = 12 I ω 2 . Use Table 9.2 to relate I to the mass M of the disk. (b)

9.44.

SET UP: EXECUTE:

45.0 rpm = 4.71 rad/s . For a uniform solid disk, I = 12 MR 2 . (a) I =

2K

ω2

=

2(0.250 J) = 0.0225 kg ⋅ m 2 . (4.71 rad/s) 2

2 I 2(0.0225 kg ⋅ m 2 ) = = 0.500 kg . (0.300 m) 2 R2 EVALUATE: No matter what the shape is, the rotational kinetic energy is proportional to the mass of the object. IDENTIFY: K = 12 I ω 2 , with ω in rad/s. Solve for I. SET UP: 1 rev/min = (2π / 60) rad/s . ΔK = −500 J (b) I = 12 MR 2 and M =

9.45.

EXECUTE:

I=

9.46.

ωi = 650 rev/min = 68.1 rad/s . ωf = 520 rev/min = 54.5 rad/s . ΔK = K f − K i = 12 I (ωf2 − ωi2 ) and

2( ΔK ) 2(−500 J) = = 0.600 kg ⋅ m 2 . 2 2 ωf − ωi (54.5 rad/s) 2 − (68.1 rad/s) 2

EVALUATE: In K = 12 I ω 2 , ω must be in rad/s. IDENTIFY: The work done on the cylinder equals its gain in kinetic energy. SET UP: The work done on the cylinder is PL, where L is the length of the rope. K1 = 0 . K 2 = 12 I ω 2 .

⎛ w⎞ I = 12 mr 2 = 12 ⎜ ⎟ r 2 . ⎝g⎠ 1w 2 1 w v2 (40.0 N)(6.00 m s) 2 v , or P = = = 14.7 N. 2g 2 g L 2(9.80 m s 2 )(5.00 m) EVALUATE: The linear speed v of the end of the rope equals the tangential speed of a point on the rim of the cylinder. When K is expressed in terms of v, the radius r of the cylinder doesn't appear. IDENTIFY and SET UP: Combine Eqs.(9.17) and (9.15) to solve for K. Use Table 9.2 to get I. EXECUTE: K = 12 I ω 2 EXECUTE:

9.47.

PL =

arad = Rω 2 , so ω = arad / R = (3500 m/s 2 ) /1.20 m = 54.0 rad/s For a disk, I = 12 MR 2 = 12 (70.0 kg)(1.20 m) 2 = 50.4 kg ⋅ m 2 Thus K = 12 I ω 2 = 12 (50.4 kg ⋅ m 2 )(54.0 rad/s) 2 = 7.35 ×104 J 9.48.

EVALUATE: The limit on arad limits ω which in turn limits K. IDENTIFY: Repeat the calculation in Example 9.9, but with a different expression for I. SET UP: For the solid cylinder in Example 9.9, I = 12 MR 2 . For the thin-walled, hollow cylinder, I = MR 2 . EXECUTE:

(a) With I = MR 2 , the expression for v is v =

2 gh . 1+ M m

(b) This expression is smaller than that for the solid cylinder; more of the cylinder’s mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is converted to the kinetic energy of the cylinder, and so less is available for the falling mass. EVALUATE: When M is much larger than m, v is very small. When M is much less than m, v becomes v = 2 gh , the same as for a mass that falls freely from a height h.


9.49.

9-13

IDENTIFY: Apply conservation of energy to the system of stone plus pulley. v = rω relates the motion of the stone to the rotation of the pulley. SET UP: For a uniform solid disk, I = 12 MR 2 . Let point 1 be when the stone is at its initial position and point 2 be when it has descended the desired distance. Let + y be upward and take y = 0 at the initial position of the stone, so

y1 = 0 and y2 = − h , where h is the distance the stone descends. (a) K p = 12 I pω 2 . I p = 12 M p R 2 = 12 (2.50 kg)(0.200 m) 2 = 0.0500 kg ⋅ m 2 .

EXECUTE:

ω=

2Kp

Ip

2(4.50 J) = 13.4 rad/s . The stone has speed v = Rω = (0.200 m)(13.4 rad/s) = 2.68 m/s . The 0.0500 kg ⋅ m 2

=

stone has kinetic energy K s = 12 mv 2 = 12 (1.50 kg)(2.68 m/s) 2 = 5.39 J . K1 + U1 = K 2 + U 2 gives 0 = K 2 + U 2 . 0 = 4.50 J + 5.39 J + mg (− h) . h =

4.50 J = 45.5% . K tot 9.89 J EVALUATE: The gravitational potential energy of the pulley doesn’t change as it rotates. The tension in the wire does positive work on the pulley and negative work of the same magnitude on the stone, so no net work on the system. IDENTIFY: K p = 12 I ω 2 for the pulley and K b = 12 mv 2 for the bucket. The speed of the bucket and the rotational (b) K tot = K p + K s = 9.89 J .

9.50.

Kp

9.89 J = 0.673 m . (1.50 kg)(9.80 m/s 2 )

=

speed of the pulley are related by v = Rω . SET UP: K p = 12 K b

9.51.

EXECUTE: 12 I ω 2 = 12 ( 12 mv 2 ) = 14 mR 2ω 2 . I = 12 mR 2 . EVALUATE: The result is independent of the rotational speed of the pulley and the linear speed of the mass. IDENTIFY: The general expression for I is Eq.(9.16). K = 12 I ω 2 . SET UP: R will be multiplied by f. EXECUTE: (a) In the expression of Eq. (9.16), each term will have the mass multiplied by f 3 and the distance

multiplied by f , and so the moment of inertia is multiplied by f 3 ( f ) 2 = f 5 .

9.52.

(b) (2.5 J)(48)5 = 6.37 × 108 J. EVALUATE: Mass and volume are proportional to each other so both scale by the same factor. IDENTIFY: The work the person does is the negative of the work done by gravity. Wgrav = U grav,1 − U grav,2 .

U grav = Mgycm . SET UP: The center of mass of the ladder is at its center, 1.00 m from each end. ycm,1 = (1.00 m)sin 53.0° = 0.799 m . ycm,2 = 1.00 m . EXECUTE:

9.53.

Wgrav = (9.00 kg)(9.80 m/s 2 )(0.799 m − 1.00 m) = −17.7 J . The work done by the person is 17.7 J.

EVALUATE: The gravity force is downward and the center of mass of the ladder moves upward, so gravity does negative work. The person pushes upward and does positive work. IDENTIFY: U = Mgycm . ΔU = U 2 − U1 . SET UP: Half the rope has mass 1.50 kg and length 12.0 m. Let y = 0 at the top of the cliff and take + y to be

upward. The center of mass of the hanging section of rope is at its center and ycm,2 = −6.00 m . EXECUTE: 9.54.

9.55.

ΔU = U 2 − U1 = mg ( ycm,2 − ycm,1 ) = (1.50 kg)(9.80 m/s 2 )(−6.00 m − 0) = −88.2 J .

EVALUATE: The potential energy of the rope decreases when part of the rope moves downward. IDENTIFY: Apply Eq.(9.19), the parallel-axis theorem. SET UP: The center of mass of the hoop is at its geometrical center. EXECUTE: In Eq. (9.19), I cm = MR 2 and d = R 2 , so I P = 2 MR 2 . EVALUATE: I is larger for an axis at the edge than for an axis at the center. Some mass is closer than distance R from the axis but some is also farther away. Since I for each piece of the hoop is proportional to the square of the distance from the axis, the increase in distance has a larger effect. IDENTIFY: Use Eq.(9.19) to relate I for the wood sphere about the desired axis to I for an axis along a diameter. SET UP: For a thin-walled hollow sphere, axis along a diameter, I = 23 MR 2 .

For a solid sphere with mass M and radius R, I cm = 52 MR 2 , for an axis along a diameter.

9-14

Chapter 9

EXECUTE: 2 3

Find d such that I P = I cm + Md 2 with I P = 23 MR 2 :

MR 2 = 52 MR 2 + Md 2

The factors of M divide out and the equation becomes

( 23 − 52 ) R 2 = d 2

d = (10 − 6) /15R = 2 R / 15 = 0.516 R. The axis is parallel to a diameter and is 0.516R from the center. EVALUATE: I cm (lead) > I cm (wood) even though M and R are the same since for a hollow sphere all the mass is a 9.56.

distance R from the axis. Eq.(9.19) says I P > I cm , so there must be a d where I P (wood) = I cm (lead). IDENTIFY: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod. SET UP: The center of mass of the rod is at its center, and I cm = 121 ML2 . 2

M 2 M 2 ⎛L⎞ L +M⎜ ⎟ = L. 12 3 ⎝2⎠ EVALUATE: I is larger when the axis is not at the center of mass. IDENTIFY and SET UP: Use Eq.(9.19). The cm of the sheet is at its geometrical center. The object is sketched in Figure 9.57. EXECUTE: I P = I cm + Md 2 . EXECUTE:

9.57.

I p = I cm + Md 2 =

From part (c) of Table 9.2, I cm = 121 M (a 2 + b 2 ). The distance d of P from the cm is

d = (a / 2) 2 + (b / 2) 2 . Figure 9.57

Thus I P = I cm + Md 2 = 121 M (a 2 + b 2 ) + M ( 14 a 2 + 14 b 2 ) = ( 121 + 14 ) M (a 2 + b 2 ) = 1 3

9.58.

9.59.

M (a 2 + b 2 )

EVALUATE: I P = 4 I cm . For an axis through P mass is farther from the axis. IDENTIFY: Consider the plate as made of slender rods placed side-by-side. SET UP: The expression in Table 9.2(a) gives I for a rod and an axis through the center of the rod. EXECUTE: (a) I is the same as for a rod with length a: I = 121 Ma 2 . (b) I is the same as for a rod with length b: I = 121 Mb 2 . EVALUATE: I is smaller when the axis is through the center of the plate than when it is along one edge. IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis theorem says I p = I cm + Md 2 . SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length L / 2 and mass M / 2 . EXECUTE: (a) For each segment the moment of inertia is for a rod with mass M / 2 , length L / 2 and the axis 2

1 ⎛ M ⎞⎛ L ⎞ 1 1 ML2 . For the rod, I a = 2 I s = ML2 . through one end. For one segment, I s = ⎜ ⎟⎜ ⎟ = 3 ⎝ 2 ⎠⎝ 2 ⎠ 24 12 (b) The center of mass of each segment is at the center of the segment, a distance of L / 4 from each end. For each 2

segment, I cm =

1 ⎛ M ⎞⎛ L ⎞ 1 2 ⎜ ⎟⎜ ⎟ = ML . Axis b is a distance L / 4 from the cm of each segment, so for each 12 ⎝ 2 ⎠⎝ 2 ⎠ 96

segment the parallel axis theorem gives I for axis b to be I s =

1 M ML2 + 96 2

2

1 1 ⎛L⎞ 2 2 ⎜ ⎟ = ML and I b = 2 I s = ML . 4 24 12 ⎝ ⎠


EVALUATE:

9-15

I for these two axes are the same.

Figure 9.59 9.60.

9.61.

9.62.

Apply the parallel-axis theorem. M 2 L and d = ( L 2 − h ) . SET UP: In Eq .( 9.19 ) , I cm = 12 2 ⎡1 1 ⎛L ⎞ ⎤ ⎡1 ⎤ ⎡1 ⎤ EXECUTE: I P = M ⎢ L2 + ⎜ − h ⎟ ⎥ = M ⎢ L2 + L2 − Lh + h 2 ⎥ = M ⎢ L2 − Lh + h 2 ⎥ , 4 ⎝2 ⎠ ⎥⎦ ⎣12 ⎦ ⎣3 ⎦ ⎢⎣12 which is the same as found in Example 9.11. EVALUATE: Example 9.11 shows that this result gives the expected result for h = 0 , h = L and h = L / 2 . IDENTIFY: Apply Eq.(9.20). SET UP: dm = ρ dV = ρ (2π rL dr ) , where L is the thickness of the disk. M = π Lρ R 2 . EXECUTE: The analysis is identical to that of Example 9.12, with the lower limit in the integral being zero and the upper limit being R. The result is I = 12 MR 2 . EVALUATE: Our result agrees with Table 9.2(f). IDENTIFY: Eq.(9.20), I = ∫ r 2 dm IDENTIFY:

SET UP:

Figure 9.62

Take the x-axis to lie along the rod, with the origin at the left end. Consider a thin slice at coordinate x and width dx, as shown in Figure 9.62. The mass per unit length for this rod is M / L, so the mass of this slice is dm = ( M / L) dx. EXECUTE: 9.63.

L

L

0

0

I = ∫ x 2 ( M / L) dx = ( M / L) ∫ x 2 dx = ( M / L)( L3 / 3) = 13 ML2

EVALUATE: This result agrees with Table 9.2. IDENTIFY: Apply Eq.(9.20). SET UP: For this case, dm = γ dx. L

EXECUTE:

(a) M = ∫ dm = ∫ γ x dx = γ 0

L

(b) I = ∫ x 2 (γ x)dx = γ 0

4 L

x 4

0

=

γ L4 4

x2 2

L

= 0

γ L2 2

= M L2 . This is larger than the moment of inertia of a uniform rod of the same 2

mass and length, since the mass density is greater further away from the axis than nearer the axis. L

9.64.

L L ⎛ x2 x3 x 4 ⎞ L4 M 2 = L . (c) I = ∫ (L − x) 2γ xdx = γ ∫ ( L2 x − 2 Lx 2 + x 3 )dx = γ ⎜ L2 − 2 L + ⎟ = γ 3 4 ⎠0 12 6 ⎝ 2 0 0 This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end. EVALUATE: For a uniform rod with an axis at one end, I = 13 ML2 . The result in (b) is larger than this and the result in (a) is smaller than this. ! ! IDENTIFY: We know that v = rω and v is tangential. We know that arad = rω 2 and arad is in toward the center of the wheel. See if the vector product expressions give these results. ! ! SET UP: A × B = AB sin φ , where φ is the angle between A and B . ! EXECUTE: (a) For a counterclockwise rotation, ω will be out of the page.

9-16

9.65.

Chapter 9

! ! (b) The upward direction crossed into the radial direction is, by the right-hand rule, counterclockwise. ω and r are ! ! perpendicular, so the magnitude of ω × r is ω r = v . ! ! ! ! (c) ω is perpendicular to v and so ω × v has magnitude ωv = arad, and from the right-hand rule, the upward ! direction crossed into the counterclockwise direction is inward, the direction of arad . ! ! ! EVALUATE: If the wheel rotates clockwise, the directions of ω and v are reversed, but arad is still inward. IDENTIFY: Apply θ = ωt . SET UP: For alignment, the earth must move through 60° more than Mars, in the same time t. ωe = 360° / yr .

ωM = 360° /(1.9 yr) . EXECUTE: θ e = θ M + 60° . ωet = ωMt + 60° . 60° 60° 60° (1/[0.9yr /1.9 yr 2 ]) = 0.352 yr = 128 days . = = 360 360 ° ° ωe − ω M 360 ° − 1 yr 1.9 yr EVALUATE: Earth has a larger angular velocity than Mars, and completes one orbit in less time. IDENTIFY and SET UP: Use Eqs.(9.3) and (9.5). As long as α z > 0, ω z increases. At the t when α z = 0, ω z is at t=

9.66.

its maximum positive value and then starts to decrease when α z becomes negative.

θ (t ) = γ t 2 − β t 3 ; γ = 3.20 rad/s 2 , β = 0.500 rad/s3 dθ d (γ t 2 − β t 3 ) = = 2γ t − 3β t 2 dt dt d ω z d (2γ t − 3β t 2 ) (b) α z (t ) = = = 2γ − 6 β t dt dt (c) The maximum angular velocity occurs when α z = 0. EXECUTE:

(a) ω z (t ) =

2γ − 6 β t = 0 implies t =

9.67.

2γ γ 3.20 rad/s 2 = = = 2.133 s 6 β 3β 3(0.500 rad/s3 )

At this t, ω z = 2γ t − 3β t 2 = 2(3.20 rad/s 2 )(2.133 s) − 3(0.500 rad/s3 )(2.133 s) 2 = 6.83 rad/s The maximum positive angular velocity is 6.83 rad/s and it occurs at 2.13 s. EVALUATE: For large t both ω z and α z are negative and ω z increases in magnitude. In fact, ω z → −∞ at t → ∞. So the answer in (c) is not the largest angular speed, just the largest positive angular velocity. IDENTIFY: The angular acceleration α of the disk is related to the linear acceleration a of the ball by a = Rα . t

t

0

0

Since the acceleration is not constant, use ω z − ω0 z = ∫ α z dt and θ − θ 0 = ∫ ω z dt to relate θ , ω z , α z and t for the disk. ω0 z = 0 . 1

∫ t dt = n + 1 t n

SET UP:

n +1

. In a = Rα , α is in rad/s 2 .

a 1.80 m/s 2 = = 0.600 m/s3 t 3.00 s a (0.600 m/s3 )t (b) α = = = (2.40 rad/s3 )t R 0.250 m EXECUTE:

(a) A =

t

(c) ω z = ∫ (2.40 rad/s3 )tdt = (1.20 rad/s3 )t 2 . ω z = 15.0 rad/s for t = 0

t

t

0

0

15.0 rad/s = 3.54 s . 1.20 rad/s3

(d) θ − θ 0 = ∫ ω z dt = ∫ (1.20 rad/s3 )t 2 dt = (0.400 rad/s3 )t 3 . For t = 3.54 s , θ − θ 0 = 17.7 rad .

9.68.

EVALUATE: If the disk had turned at a constant angular velocity of 15.0 rad/s for 3.54 s it would have turned through an angle of 53.1 rad in 3.54 s. It actually turns through less than half this because the angular velocity is increasing in time and is less than 15.0 rad/s at all but the end of the interval. IDENTIFY and SET UP: The translational kinetic energy is K = 12 mv 2 and the kinetic energy of the rotating

flywheel is K = 12 I ω 2 . Use the scale speed to calculate the actual speed v. From that calculate K for the car and then solve for ω that gives this K for the flywheel.


EXECUTE:

(a)

vtoy vscale

=

9-17

Ltoy Lreal

⎛L ⎞ ⎛ 0.150 m ⎞ vtoy = vscale ⎜ toy ⎟ = (700 km/h) ⎜ ⎟ = 35.0 km/h ⎝ 3.0 m ⎠ ⎝ Lreal ⎠ vtoy = (35.0 km/h)(1000 m/1 km)(1 h/3600 s) = 9.72 m/s (b) K = 12 mv 2 = 12 (0.180 kg)(9.72 m/s)2 = 8.50 J (c) K = 12 I ω 2 gives that ω =

9.69.

2K 2(8.50 J) = = 652 rad/s I 4.00 ×10−5 kg ⋅ m 2

EVALUATE: K = 12 I ω 2 gives ω in rad/s. 652 rad/s = 6200 rev/min so the rotation rate of the flywheel is very large. ! ! IDENTIFY: atan = rα , arad = rω 2 . Apply the constant acceleration equations and ∑ F = ma . ! 2 2 SET UP: atan and arad are perpendicular components of a , so a = arad + atan .

atan 3.00 m s 2 = = 0.050 rad s 2 r 60.0 m (b) α t = (0.05 rad s 2 )(6.00 s) = 0.300 rad s. EXECUTE:

(a) α =

(c) arad = ω 2 r = (0.300 rad s) 2 (60.0 m) = 5.40 m s 2 . (d) The sketch is given in Figure 9.69. 2

(e) a = a 2 rad + a 2 tan = (5.40 m s ) 2 + (3.00 m s 2 ) 2 = 6.18 m s 2 , and the magnitude of the force is

F = ma = (1240 kg)(6.18 m s 2 ) = 7.66 kN. ⎛a ⎞ ⎛ 5.40 ⎞ (f) arctan ⎜ rad ⎟ = arctan ⎜ ⎟ = 60.9°. a ⎝ 3.00 ⎠ ⎝ tan ⎠ ! ! EVALUATE: atan is constant and arad increases as ω increases. At t = 0 , a is parallel to v . As t increases, ! ! a moves toward the radial direction and the angle between a increases toward 90° .

Figure 9.69 9.70.

IDENTIFY: Apply conservation of energy to the system of drum plus falling mass, and compare the results for earth and for Mars. SET UP: K drum = 12 I ω 2 . K mass = 12 mv 2 . v = Rω so if K drum is the same, ω is the same and v is the same on both

planets. Therefore, K mass is the same. Let y = 0 at the initial height of the mass and take + y upward. Configuration 1 is when the mass is at its initial position and 2 is when the mass has descended 5.00 m, so y1 = 0 and y2 = − h , where h is the height the mass descends. EXECUTE:

(a) K1 + U1 = K 2 + U 2 gives 0 = K drum + K mass − mgh . K drum + K mass are the same on both planets, so

⎛g ⎞ ⎛ 9.80 m/s 2 ⎞ mg E hE = mg M hM . hM = hE ⎜ E ⎟ = (5.00 m) ⎜ = 13.2 m . 2 ⎟ ⎝ 3.71 m/s ⎠ ⎝ gM ⎠

9-18

Chapter 9

(b) mg M hM = K drum + K mass .

v = 2 g M hM −

9.71.

1 2

mv 2 = mg M hM − K drum and

2 K drum 2(250.0 J) = 2(3.71 m/s 2 )(13.2 m) − = 8.04 m/s m 15.0 kg

EVALUATE: We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius of the drum. IDENTIFY and SET UP: All points on the belt move with the same speed. Since the belt doesn’t slip, the speed of the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are related to the angular speed of each circular object by v = rω . EXECUTE:

Figure 9.71 (a) v1 = r1ω1

ω1 = (60.0 rev/s)(2π rad/1 rev) = 377 rad/s v1 = r1ω1 = (0.45 ×10−2 m)(377 rad/s) = 1.70 m/s (b) v1 = v2

r1ω1 = r2ω2 ω2 = ( r1 / r2 )ω1 = (0.45 cm/2.00 cm)(377 rad/s) = 84.8 rad/s

9.72.

EVALUATE: The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for points on the rim. IDENTIFY: The speed of all points on the belt is the same, so r1ω1 = r2ω2 applies to the two pulleys. SET UP: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade. π rad/s = 30 rev/min .

⎛ π rad s ⎞ ⎛ 0.208 m ⎞ (a) v2 = (2(3450 rev min)) ⎜ ⎟⎜ ⎟ = 75.1 m s. 2 ⎠ ⎝ 30 rev min ⎠ ⎝

EXECUTE:

2

⎛ ⎛π rad s ⎞ ⎞ ⎛ 0.208 m ⎞ 4 2 (b) arad = ω 2 r = ⎜⎜ 2(3450 rev min) ⎜ ⎟ ⎟⎟ ⎜ ⎟ = 5.43 ×10 m s , 30 rev min 2 ⎠ ⎝ ⎠⎠ ⎝ ⎝ so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity. EVALUATE: In v = rω and arad = rω 2 , ω must be in rad/s. 9.73.

IDENTIFY and SET UP: Use Eq.(9.15) to relate arad to ω and then use a constant acceleration equation to replace ω. EXECUTE: (a) arad = rω 2 , arad,1 = rω12 , arad, 2 = rω22

Δarad = arad, 2 − arad,1 = r (ω22 − ω12 ) One of the constant acceleration equations can be written ω22z = ω12z + 2α (θ 2 − θ1 ), or ω22z − ω12z = 2α z (θ 2 − θ1 ) Thus Δarad = r 2α z (θ 2 − θ1 ) = 2rα z (θ 2 − θ1 ), as was to be shown. (b) α z =

Δarad 85.0 m/s 2 − 25.0 m/s 2 = = 8.00 rad/s 2 2r (θ 2 − θ1 ) 2(0.250 m)(15.0 rad)

Then atan = rα = (0.250 m)(8.00 rad/s 2 ) = 2.00 m/s 2

ω 2 is proportional to α z and (θ − θ 0 ) so arad is also proportional to these quantities. arad increases while r stays fixed, ω z increases, and α z is positive. IDENTIFY and SET UP: Use Eq.(9.17) to relate K and ω and then use a constant acceleration equation to replace ω . EXECUTE: (c) K = 12 I ω 2 ; K 2 = 12 I ω22 , K1 = 12 I ω12 EVALUATE:

ΔK = K 2 − K1 = 12 I (ω22 − ω12 ) = 12 I (2α z (θ 2 − θ1 )) = Iα z (θ 2 − θ1 ), as was to be shown. (d) I =

ΔK

=

45.0 J − 20.0 J

= 0.208 kg ⋅ m 2

α z (θ 2 − θ1 ) (8.00 rad/s 2 )(15.0 rad) EVALUATE: α z is positive, ω increases, and K increases.


9.74.

IDENTIFY: SET UP:

9-19

I = I wood + I lead . m = ρV , where ρ is the volume density and m = σ A , where σ is the area density. For a solid sphere, I = 52 mR 2 . For the hollow sphere (foil), I = 32 mR 2 . For a sphere, V = 34 π R 3 and

4 A = 4π R 2 . mw = ρ wVw = ρ w π R 3 . mL = σ L AL = σ L 4π R 2 . 3 2 2 2⎛ 4 2 8 ⎞ ⎛ρ R ⎞ 2 EXECUTE: I = mw R + mL R 2 = ⎜ ρ w π R 3 ⎟ R 2 + (σ L 4π R 2 )R 2 = π R 4 ⎜ w + σ L ⎟ . 5 3 5⎝ 3 3 3 ⎠ ⎝ 5 ⎠ I=

9.75.

9.76.

⎡ (800 kg m3 )(0.20 m) ⎤ 8π (0.20 m) 4 ⎢ + 20 kg m 2 ⎥ = 0.70 kg ⋅ m 2 . 3 5 ⎣ ⎦

EVALUATE: mW = 26.8 kg and I W = 0.429 kg ⋅ m 2 . mL = 10.1 kg and I L = 0.268 kg ⋅ m 2 . Even though the foil is only 27% of the total mass its contribution to I is about 38% of the total. IDENTIFY: Estimate the shape and dimensions of your body and apply the approximate expression from Table 9.2. SET UP: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and diameter 0.30 m (radius 0.15 m) 1 1 EXECUTE: I = mR 2 = (80 kg) (0.15 m) 2 = 0.9 kg ⋅ m 2 2 2 EVALUATE: I depends on your mass and width but not on your height. IDENTIFY: Treat the V like two thin 0.160 kg bars, each 25 cm long. SET UP: For a slender bar with the axis at one end, I = 13 mL2 .

⎛1 ⎞ ⎛1⎞ I = 2 ⎜ mL2 ⎟ = 2 ⎜ ⎟ (0.160 kg)(0.250 m) 2 = 6.67 ×10−3 kg ⋅ m 2 3 ⎝ ⎠ ⎝3⎠ EVALUATE: The value of I is independent of the angle between the two sides of the V; the angle 70.0° didn't enter into the calculation. IDENTIFY: K = 12 I ω 2 . arad = rω 2 . m = ρV . EXECUTE:

9.77.

SET UP:

For a disk with the axis at the center, I = 12 mR 2 . V = tπ R 2 , where t = 0.100 m is the thickness of the

flywheel. ρ = 7800 kg m 3 is the density of the iron. EXECUTE:

(a) ω = 90.0 rpm = 9.425 rad s . I =

2K

ω2

=

2(10.0 × 106 J) = 2.252 × 105 kg ⋅ m 2 . (9.425 rad s) 2

1 1 m = ρV = ρπ R 2t . I = mR 2 = ρπ tR 4 . This gives R = (2 I ρπ t )1 4 = 3.68 m and the diameter is 7.36 m. 2 2 (b) arad = Rω 2 = 327 m s 2

9.78.

EVALUATE: In K = 12 I ω 2 , ω must be in rad/s. arad is about 33g; the flywheel material must have large cohesive strength to prevent the flywheel from flying apart. IDENTIFY: K = 12 I ω 2 . To have the same K for any ω the two parts must have the same I. Use Table 9.2 for I. SET UP:

9.79.

For a solid sphere, I solid = 52 M solid R 2 . For a hollow sphere, I hollow = 32 M hollow R 2 .

EXECUTE: I solid = I hollow gives 52 M solid R 2 = 32 M hollow R 2 and M hollow = 35 M solid = 53 M . EVALUATE: The hollow sphere has less mass since all its mass is distributed farther from the rotation axis. 2π rad IDENTIFY: K = 12 I ω 2 . ω = , where T is the period of the motion. For the earth's orbital motion it can be T treated as a point mass and I = MR 2 . SET UP: The earth's rotational period is 24 h = 86,164 s . Its orbital period is 1 yr = 3.156 ×107 s .

M = 5.97 ×1024 kg . R = 6.38 ×106 m . EXECUTE:

(a) K = 2

2π 2 I 2π 2 (0.3308)(5.97 ×1024 kg)(6.38 × 106 m) 2 = = 2.14 ×1029 J. T2 (86,164 s) 2

1 ⎛ 2π R ⎞ 2π 2 (5.97 ×10 24 kg)(1.50 ×1011 m) 2 = 2.66 ×1033 J. M⎜ ⎟ = 2 ⎝ T ⎠ (3.156 ×107 s) 2 (c) Since the Earth’s moment of inertia is less than that of a uniform sphere, more of the Earth’s mass must be concentrated near its center. EVALUATE: These kinetic energies are very large, because the mass of the earth is very large. (b)

9-20

9.80.

Chapter 9

IDENTIFY: SET UP:

Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR. 1 1 The kinetic energy is K = I ω 2 + mv 2 , with I = 12 mR 2 for the disk. v = Rω . 2 2

EXECUTE: EVALUATE:

K=

4g 1 2 1 1 4g . I ω + m(ω R ) 2 = ( I + mR 2 ) . Using Ι = 12 mR 2 and solving for ω, ω 2 = and ω = 3R 2 2 2 3R

The small object has speed v =

height h, it would attain a speed 9.81.

2 2 gR . If it was not attached to the disk and was dropped from a 3 2 . 3

2gR . Being attached to the disk reduces its final speed by a factor of

IDENTIFY: Use Eq.(9.20) to calculate I. Then use K = 12 I ω 2 to calculate K. (a) SET UP: The object is sketched in Figure 9.81.

Consider a small strip of width dy and a distance y below the top of the triangle. The length of the strip is x = ( y / h)b.


The strip has area x dy and the area of the sign is 12 bh, so the mass of the strip is

⎛ x dy ⎞ ⎛ yb ⎞⎛ 2 dy ⎞ ⎛ 2 M ⎞ dm = M ⎜ 1 ⎟ = M ⎜ ⎟⎜ ⎟ = ⎜ 2 ⎟ y dy ⎝ h ⎠⎝ bh ⎠ ⎝ h ⎠ ⎝ 2 bh ⎠ ⎛ 2 Mb 2 ⎞ 3 dI = 13 (dm) x 2 = ⎜ y dy 4 ⎟ ⎝ 3h ⎠ h 2 Mb 2 h 3 2 Mb 2 ⎛ 1 4 h ⎞ 1 2 = I = ∫ dI = y dy ⎜ y 0 ⎟ = Mb 0 3h 4 ∫ 0 3h 4 ⎝ h ⎠ 6 (b) I = 16 Mb 2 = 2.304 kg ⋅ m 2 ω = 2.00 rev/s = 4.00π rad/s K = 12 I ω 2 = 182 J

9.82.

EVALUATE: From Table (9.2), if the sign were rectangular, with length b, then I = 13 Mb 2 . Our result is one-half this, since mass is closer to the axis for the triangular than for the rectangular shape. IDENTIFY: Apply conservation of energy to the system. SET UP: For the falling mass K = 12 mv 2 . For the wheel K = 12 I ω 2 . EXECUTE:

( 8.00 kg )( 5.00 m/s ) = 100 J. mgh = ( 8.00 kg ) ( 9.8 m/s 2 ) ( 2.00 m ) = 156.8 J . The wheel must

(a) The kinetic energy of the falling mass after 2.00 m is K = 12 mv 2 =

The change in its potential energy while falling is

1 2

2

have the “missing” 56.8 J in the form of rotational kinetic energy. Since its outer rim is moving at the same speed v 5.00 m/s 1 as the falling mass, 5.00 m/s , v = rω gives ω = = = 13.51 rad/s . K = I ω 2 ; therefore r 0.370 m 2 2 56.8 J 2K ( ) = 0.622 kg ⋅ m 2 . I= 2 = 2 ω (13.51 rad s ) (b) The wheel’s mass is (280 N) (9.8 m s 2 ) = 28.6 kg . The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim. Its moment of inertia would be 2 I = MR 2 = ( 28.6 kg )( 0.370 m ) = 3.92 kg ⋅ m 2 . The boss’s wheel is physically impossible.


9.83.

9-21

EVALUATE: If the mass falls from rest in free-fall its speed after it has descended 2.00 m is v = 2 g (2.00 m) = 6.26 m/s . Its actual speed is less because some of the energy of the system is in the form of rotational kinetic energy of the wheel. IDENTIFY: Use conservation of energy. The stick rotates about a fixed axis so K = 12 I ω 2 . Once we have ω use v = rω to calculate v for the end of the stick. SET UP: The object is sketched in Figure 9.83.

Take the origin of coordinates at the lowest point reached by the stick and take the positive y-direction to be upward.


(a) Use Eq.(9.18): U = Mgycm

ΔU = U 2 − U1 = Mg ( ycm2 − ycm1 ) The center of mass of the meter stick is at its geometrical center, so ycm1 = 1.00 m and ycm2 = 0.50 m Then ΔU = (0.160 kg)(9.80 m/s 2 )(0.50 m − 1.00 m) = −0.784 J (b) Use conservation of energy: K1 + U1 + Wother = K 2 + U 2

Gravity is the only force that does work on the meter stick, so Wother = 0. K1 = 0. Thus K 2 = U1 − U 2 = −ΔU , where ΔU was calculated in part (a). K 2 = 12 I ω22 so

1 2

I ω22 = −ΔU and ω2 = 2( −ΔU ) / I

For stick pivoted about one end, I = 13 ML2 where L = 1.00 m, so

ω2 =

6( −ΔU ) 6(0.784 J) = = 5.42 rad/s 2 ML (0.160 kg)(1.00 m) 2

(c) v = rω = (1.00 m)(5.42 rad/s) = 5.42 m/s (d) For a particle in free-fall, with + y upward,

v0 y = 0; y − y0 = −1.00 m; a y = −9.80 m/s 2 ; v y = ? v y2 = v02y + 2a y ( y − y0 ) v y = − 2a y ( y − y0 ) = − 2(−9.80 m/s 2 )( −1.00 m) = −4.43 m/s EVALUATE:

The magnitude of the answer in part (c) is larger. U1,grav is the same for the stick as for a particle

falling from a height of 1.00 m. For the stick K = 12 I ω22 =

9.84.

1 2

(

1 3

ML2 ) (v / L) 2 = 16 Mv 2 . For the stick and for the

particle, K 2 is the same but the same K gives a larger v for the end of the stick than for the particle. The reason is that all the other points along the stick are moving slower than the end opposite the axis. IDENTIFY: Apply conservation of energy to the system of cylinder and rope. SET UP: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance π R below the axle, since the length of the rope is 2π R and half this distance is the position of the center of the mass. Initially, every part of the rope is moving with speed ω0 R, and when the rope has unwound, and the cylinder has angular speed ω , the speed of the rope is ω R (the upper end of the rope has the same tangential speed at the edge of the cylinder). I = (1 2) MR 2 for a uniform cylinder,

9-22

Chapter 9

⎛M m⎞ ⎛M m⎞ K1 = K 2 + U 2 . ⎜ + ⎟ R 2ω02 = ⎜ + ⎟ R 2ω 2 − mgπ R. Solving for ω gives ⎝ 4 2⎠ ⎝ 4 2⎠

EXECUTE:

ω = ω02 +

( 4π mg R ) , and the speed of any part of the rope is ( M + 2m )

EVALUATE:

9.85.

v = ω R.

When m → 0 , ω → ω0 . When m >> M , ω = ω02 +

2π g and v = v02 + 2π gR . This is the final R

speed when an object with initial speed v0 descends a distance π R . IDENTIFY: Apply conservation of energy to the system consisting of blocks A and B and the pulley. SET UP: The system at points 1 and 2 of its motion is sketched in Figure 9.85.

Figure 9.85

Use the work-energy relation K1 + U1 + Wother = K 2 + U 2 . Use coordinates where + y is upward and where the origin is at the position of block B after it has descended. The tension in the rope does positive work on block A and negative work of the same magnitude on block B, so the net work done by the tension in the rope is zero. Both blocks have the same speed. EXECUTE: Gravity does work on block B and kinetic friction does work on block A. Therefore Wother = W f = − μ k mA gd . K1 = 0 (system is released from rest) U1 = mB gyB1 = mB gd ; U 2 = mB gyB 2 = 0 K 2 = 12 mAv22 + 12 mB v22 + 12 I ω22 . But v(blocks) = Rω (pulley), so ω2 = v2 / R and K 2 = 12 ( mA + mB )v22 + 12 I (v2 / R) 2 = 12 (m A + mB + I / R 2 )v22 Putting all this into the work-energy relation gives mB gd − μ k mA gd = 12 (mA + mB + I / R 2 )v22 (mA + mB + I / R 2 )v22 = 2 gd (mB − μk mA ) v2 =

2 gd ( mB − μ k m A ) mA + mB + I / R 2

EVALUATE:

9.86.

If mB >> mA and I / R 2 , then v2 = 2 gd ; block B falls freely. If I is very large, v2 is very small.

Must have mB > μ k mA for motion, so the weight of B will be larger than the friction force on A. I / R 2 has units of mass and is in a sense the “effective mass” of the pulley. IDENTIFY: Apply conservation of energy to the system of two blocks and the pulley. SET UP: Let the potential energy of each block be zero at its initial position. The kinetic energy of the system is the sum of the kinetic energies of each object. v = Rω , where v is the common speed of the blocks and ω is the angular velocity of the pulley. EXECUTE: The amount of gravitational potential energy which has become kinetic energy is K = ( 4.00 kg − 2.00 kg ) ( 9.80 m s 2 ) ( 5.00 m ) = 98.0 J. In terms of the common speed v of the blocks, the kinetic 2

1 1 ⎛v⎞ energy of the system is K = (m1 + m2 )v 2 + I ⎜ ⎟ . 2 2 ⎝R⎠ 1⎛ (0.480 kg ⋅ m 2 ) ⎞ 2 98.0 J K = v 2 ⎜ 4.00 kg + 2.00 kg + = 2.81 m s. ⎟ = v (12.4 kg). Solving for v gives v = 2⎝ (0.160 m) 2 ⎠ 12.4 kg


9.87.

9-23

EVALUATE: If the pulley is massless, 98.0 J = 12 (4.00 kg + 2.00 kg)v 2 and v = 5.72 m/s . The moment of inertia of the pulley reduces the final speed of the blocks. IDENTIFY and SET UP: Apply conservation of energy to the motion of the hoop. Use Eq.(9.18) to calculate U grav .

Use K = 12 I ω 2 for the kinetic energy of the hoop. Solve for ω . The center of mass of the hoop is at its geometrical center.

Take the origin to be at the original location of the center of the hoop, before it is rotated to one side, as shown in Figure 9.87.

Figure 9.87

ycm1 = R − R cos β = R (1 − cos β ) ycm2 = 0 (at equilibrium position hoop is at original position) K1 + U1 + Wother = K 2 + U 2

EXECUTE:

Wother = 0 (only gravity does work) K1 = 0 (released from rest), K 2 = 12 I ω22 For a hoop, I cm = MR 2 , so I = Md 2 + MR 2 with d = R and I = 2 MR 2 , for an axis at the edge. Thus K 2 = 12 (2MR 2 )ω22 = MR 2ω22 . U1 = Mgycm1 = MgR(1 − cos β ), U 2 = mgycm2 = 0 Thus K1 + U1 + Wother = K 2 + U 2 gives MgR (1 − cos β ) = MR 2ω22 and ω = g (1 − cos β ) / R EVALUATE: 9.88.


If β = 0, then ω2 = 0. As β increases, ω2 increases.

energy t 2 1 For a solid cylinder, I = 2 MR . 1 rev/min = (2π / 60) rad/s K = 12 I ω 2 , with ω in rad/s. P =

(a) ω = 3000 rev/min = 314 rad/s . I = 12 (1000 kg)(0.900 m) 2 = 405 kg ⋅ m 2

K = 12 (405 kg ⋅ m 2 )(314 rad/s) 2 = 2.00 × 107 J . K 2.00 ×107 J = = 1.08 ×103 s = 17.9 min . P 1.86 ×104 W EVALUATE: In K = 12 I ω 2 , we must use ω in rad/s. (b) t =

9.89.

IDENTIFY: SET UP:

I = I1 + I 2 . Apply conservation of energy to the system. The calculation is similar to Example 9.9.

ω=

v v for part (b) and ω = for part (c). R1 R2

1 1 1 (a) I = M 1R12 + M 2 R22 = ((0.80 kg)(2.50 ×10−2 m) 2 + (1.60 kg)(5.00 × 10−2 m) 2 ) 2 2 2 −3 2 I = 2.25 ×10 kg ⋅ m .

EXECUTE:

(b) The method of Example 9.9 yields v =

v=

2 gh . 1 + ( I mR12 )

2(9.80 m s 2 )(2.00 m) = 3.40 m s. (1 + ((2.25 ×10−3 kg ⋅ m 2 ) (1.50 kg)(0.025 m) 2 ))

The same calculation, with R2 instead of R1 gives v = 4.95 m s. EVALUATE: The final speed of the block is greater when the string is wrapped around the larger disk. v = Rω , so when R = R2 the factor that relates v to ω is larger. For R = R2 a larger fraction of the total kinetic energy resides

9-24

9.90.

Chapter 9

with the block. The total kinetic energy is the same in both cases (equal to mgh), so when R = R2 the kinetic energy and speed of the block are greater. IDENTIFY: Apply conservation of energy to the motion of the mass after it hits the ground. 2 gh SET UP: From Example 9.9, the speed of the mass just before it hits the ground is v = . 1 + M / 2m (a) In the case that no energy is lost, the rebound height h′ is related to the speed v by h′ =

EXECUTE:

v2 , and 2g

h . 1 + M 2m (b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved. EVALUATE: If m >> M , h′ = h and the mass does rebound to its initial height. IDENTIFY: Apply conservation of energy to relate the height of the mass to the kinetic energy of the cylinder. SET UP: First use K (cylinder) = 250 J to find ω for the cylinder and v for the mass.

with the form for v given in Example 9.9, h′ =

9.91.

I = 12 MR 2 = 12 (10.0 kg)(0.150 m) 2 = 0.1125 kg ⋅ m 2

EXECUTE:

K = 12 I ω 2 so ω = 2 K / I = 66.67 rad/s v = Rω = 10.0 m/s SET UP: Use conservation of energy K1 + U1 = K 2 + U 2 to solve for the distance the mass descends. Take y = 0 at lowest point of the mass, so y2 = 0 and y1 = h, the distance the mass descends. K1 = U 2 = 0 so U1 = K 2 .

EXECUTE:

mgh = 12 mv + 12 I ω 2 , where m = 12.0 kg 2

For the cylinder, I = 12 MR 2 and ω = v / R, so

1 2

I ω 2 = 14 Mv 2 .

mgh = 12 mv 2 + 14 Mv 2 h=

v2 ⎛ M ⎞ ⎜1 + ⎟ = 7.23 m 2 g ⎝ 2m ⎠

EVALUATE:

For the cylinder K cyl = 12 I ω 2 = 12 ( 12 MR 2 ) (v / R ) 2 = 14 Mv 2 .

K mass = 12 mv 2 , so K mass = (2m / M ) K cyl = [2(12.0 kg)/10.0 kg](250 J) = 600 J. The mass has 600 J of kinetic energy when the cylinder has 250 J of kinetic energy and at this point the system has total energy 850 J since U 2 = 0. 9.92.

Initially the total energy of the system is U1 = mgy1 = mgh = 850 J, so the total energy is shown to be conserved. IDENTIFY: Energy conservation: Loss of U of box equals gain in K of system. Both the cylinder and pulley have 1 1 1 2 2 2 + I pulleyωpulley + I cylinderωcylinder . kinetic energy of the form K = 12 I ω 2 . mbox gh = mbox vbox 2 2 2 v v SET UP: ωpulley = Box and ωcylinder = Box . rp rcylinder EXECUTE:

vB =

9.93.

1 1 ⎛1 ⎞ mB gh = mBvB2 + ⎜ mP rp2 ⎟ 2 2 ⎝2 ⎠

2

2

⎛ vB ⎞ 1 ⎛ 1 1 1 1 2⎞ ⎛v ⎞ 2 2 2 ⎜⎜ ⎟⎟ + ⎜ mC rC ⎟ ⎜ B ⎟ . mB gh = mBvB + mP vB + mCvB and r 2 2 r 2 4 4 ⎝ ⎠ ⎝ C⎠ ⎝ p⎠

mB gh (3.00 kg)(9.80 m s 2 )(1.50 m) = = 3.68 m s . 1 1 1 1.50 kg + 14 (7.00 kg) 2 mB + 4 m p + 4 mC

EVALUATE: If the box was disconnected from the rope and dropped from rest, after falling 1.50 m its speed would be v = 2 g (1.50 m) = 5.42 m/s . Since in the problem some of the energy of the system goes into kinetic energy of the cylinder and of the pulley, the final speed of the box is less than this. IDENTIFY: I = I disk − I hole , where I hole is I for the piece punched from the disk. Apply the parallel-axis theorem to calculate the required moments of inertia. SET UP: For a uniform disk, I = 12 MR 2 .


EXECUTE:

(a) The initial moment of inertia is I 0 = 12 MR 2 . The piece punched has a mass of

9-25

M and a moment 16

of inertia with respect to the axis of the original disk of 2 2 M ⎡1 ⎛ R ⎞ ⎛ R ⎞ ⎤ 9 MR 2 . ⎢ ⎜ ⎟ +⎜ ⎟ ⎥ = 16 ⎣⎢ 2 ⎝ 4 ⎠ ⎝ 2 ⎠ ⎦⎥ 512

9.94.

1 9 247 MR 2 = MR 2 . The moment of inertia of the remaining piece is then I = MR 2 − 2 512 512 383 (b) I = 12 MR 2 + M ( R / 2) 2 − 12 ( M /16)( R / 4) 2 = 512 MR 2 . EVALUATE: For a solid disk and an axis at a distance R / 2 from the disk's center, the parallel-axis theorem gives 2 I = 12 MR 2 = 34 MR 2 = 384 512 MR . For both choices of axes the presence of the hole reduces I, but the effect of the hole is greater in part (a), when it is farther from the axis. IDENTIFY: In part (a) use the parallel-axis theorem to relate the moment of inertia I cm for an axis through the center of the sphere to I P , the moment of inertia for an axis at the pivot. SET UP:

I for a uniform solid sphere and the axis through its center is

2 5

MR 2 . I for a slender rod and an axis at

one end is 13 mL2 , where m is the mass of the rod and L is its length. EXECUTE:

(a) From the parallel-axis theorem, the moment of inertia is I P = (2 5) MR 2 + ML2 , and

⎛ ⎛ 2 ⎞⎛ R ⎞ 2 ⎞ IP 2 = ⎜1 + ⎜ ⎟⎜ ⎟ ⎟ . If R = (0.05) L, the difference is (2 5)(0.05) = 0.001 = 0.1%. ML2 ⎜⎝ ⎝ 5 ⎠⎝ L ⎠ ⎟⎠ (b) ( I rod ML2 ) = (mrod 3M ), which is 0.33% when mrod = (0.01) M . 9.95.

EVALUATE: In both these cases the correction to I = ML2 is very small. IDENTIFY: Follow the instructions in the problem to derive the perpendicular-axis theorem. Then apply that result in part (b). SET UP: I = ∑ mi ri 2 . The moment of inertia for the washer and an axis perpendicular to the plane of the washer i

at its center is 1 12

1 2

M ( R12 + R22 ) . In part (b), I for an axis perpendicular to the plane of the square at its center is

M ( L2 + L2 ) = 16 ML2 .

EXECUTE:

(a) With respect to O, ri 2 = xi 2 + yi 2 , and so

I O = ∑ mi ri 2 = ∑ mi ( xi 2 + yi 2 ) = ∑ mi xi 2 + ∑ mi yi 2 = I x + I y . i

i

i

i

(b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts that they will sum to the moment of inertia about the washer axis, which is I = 12 M ( R12 + R2 2 ), and so I x = I y = 14 M ( R12 + R2 2 ). (c) I 0 = 16 mL2 . Since I 0 = I x + I y , and I x = I y , both I x and I y must be

9.96.

1 12

mL2 .

EVALUATE: The result in part (c) says that I is the same for an axis that bisects opposite sides of the square as for an axis along the diagonal of the square, even though the distribution of mass relative to the two axes is quite different in these two cases. IDENTIFY: Apply the parallel-axis theorem to each side of the square. SET UP: Each side has length a and mass M / 4, and the moment of inertia of each side about an axis

perpendicular to the side and through its center is 121 14 Ma 2 = 481 Ma 2 . EXECUTE: The moment of inertia of each side about the axis through the center of the square is, from the

()

2

2 2 perpendicular axis theorem, Ma + M a = Ma . The total moment of inertia is the sum of the contributions 48 4 2 12 2 2 from the four sides, or 4 × Ma = Ma . 12 3 EVALUATE: If all the mass of a side were at its center, a distance a / 2 from the axis, we would have 2

⎛ M ⎞⎛ a ⎞ 1 I = 4 ⎜ ⎟⎜ ⎟ = Ma 2 . If all the mass was divided equally among the four corners of the square, a distance ⎝ 4 ⎠⎝ 2 ⎠ 4 2

⎛ M ⎞⎛ a ⎞ 1 2 a / 2 from the axis, we would have I = 4 ⎜ ⎟⎜ ⎟ = Ma . The actual I is between these two values. ⎝ 4 ⎠⎝ 2 ⎠ 2

9-26

9.97.

Chapter 9

IDENTIFY: Use Eq.(9.20) to calculate I. (a) SET UP: Let L be the length of the cylinder. Divide the cylinder into thin cylindrical shells of inner radius r and outer radius r + dr. An end view is shown in Figure 9.97.

ρ = αr The mass of the thin cylindrical shell is dm = ρ dV = ρ (2π r dr ) L = 2πα Lr 2 dr Figure 9.97 EXECUTE:

I = ∫ r 2 dm = 2πα L ∫ r 4 dr = 2πα L ( 15 R 5 ) = 52 πα LR 5 R

0

Relate M to α : M = ∫ dm = 2πα L ∫ r 2 dr = 2πα L ( 13 R 3 ) = 23 πα LR 3 , so πα LR 3 = 3M / 2. R

0

Using this in the above result for I gives I = 52 (3M / 2) R 2 = 35 MR 2 .

9.98.

(b) EVALUATE: For a cylinder of uniform density I = 12 MR 2 . The answer in (a) is larger than this. Since the density increases with distance from the axis the cylinder in (a) has more mass farther from the axis than for a cylinder of uniform density. IDENTIFY: Write K in terms of the period T and take derivatives of both sides of this equation to relate dK / dt to dT / dt . 2π SET UP: ω = and K = 12 I ω 2 . The speed of light is c = 3.00 ×108 m/s . T 2π 2 I dK 4π 2 I dT 4π 2 I dT EXECUTE: (a) K = 2 . =− 3 . The rate of energy loss is . Solving for the moment of T dt T dt T 3 dt inertia I in terms of the power P,

I= (b) R = (c) v =

PT 3 1 (5 ×1031 W)(0.0331 s)3 1s = = 1.09 × 1038 kg ⋅ m 2 4π dT dt 4π 2 4.22 ×10−13 s

5I 5(1.08 ×1038 kg ⋅ m 2 ) = = 9.9 × 103 m, about 10 km. 2M 2(1.4)(1.99 ×1030 kg)

2π R 2π (9.9 × 103 m) = = 1.9 × 106 m s = 6.3 × 10−3 c. T (0.0331 s)

M M = = 6.9 ×1017 kg m3 , which is much higher than the density of ordinary rock by 14 orders of V (4π 3) R 3 magnitude, and is comparable to nuclear mass densities. EVALUATE: I is huge because M is huge. A small rate of change in the period corresponds to a large release of energy. IDENTIFY: In part (a), do the calculations as specified in the hint. In part (b) calculate the mass of each shell of inner radius R1 and outer radius R2 and sum to get the total mass. In part (c) use the expression in part (a) to calculate I for each shell and sum to get the total I. SET UP: m = ρV . For a solid sphere, V = 34 π R 3 . EXECUTE: (a) Following the hint, the moment of inertia of a uniform sphere in terms of the mass density is I = 52 MR 2 = 158 πρ R 5 , and so the difference in the moments of inertia of two spheres with the same density ρ but (d) ρ =

9.99.

different radii R2 and R1 is I = ρ (8π 15)( R25 − R15 ). (b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii that bound the regions and multiplying by 4π 3, gives M = 5.97 ×1024 kg. (c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii that bound the regions and multiplying by 8π 15, gives I = 8.02 ×1022 kg ⋅ m 2 = 0.334 MR 2 . EVALUATE: The calculated value of I = 0.334 MR 2 agrees closely with the measured value of 0.3308MR 2 . This simple model is fairly accurate.


9.100.

IDENTIFY: SET UP:

9-27

Apply Eq.(9.20) Let z be the coordinate along the vertical axis. r ( z ) =

πρ R 4

R2 z 2 πρ R 4 4 zR z dz . . dm = πρ 2 and dI = h h 2 h4

πρ R 4

h 1 ⎡⎣ z 5 ⎤⎦ = πρ R 4 h . The volume of a right circular cone is 0 2 h 10 h 10 2 ⎛ R h⎞ 2 3 3 πρ 2 V = 13 π R 2 h, the mass is 13 πρ R 2 h and so I = ⎜ ⎟ R = MR . 10 ⎝ 3 ⎠ 10

EXECUTE:

9.101.

I = ∫ dI =

∫

h

0

z 4 dz =

4

EVALUATE: For a uniform cylinder of radius R and for an axis through its center, I = 12 MR 2 . I for the cone is less, as expected, since the cone is constructed from a series of parallel discs whose radii decrease from R to zero along the vertical axis of the cone. IDENTIFY: Follow the steps outlined in the problem. SET UP: ω z = dθ / dt . α z = d 2ω z / dt 2 . EXECUTE:

(a) ds = r dθ = r0 dθ + βθ dθ so s (θ ) = r0θ +

(b) Setting s = vt = r0θ +

θ (t ) =

β 2

β 2

θ 2 . θ must be in radians.

θ 2 gives a quadratic in θ . The positive solution is

1⎡ 2 r0 + 2 β vt − r0 ⎤ . ⎦ β⎣

(The negative solution would be going backwards, to values of r smaller than r0 .) (c) Differentiating, ω z (t ) =

dω z β v2 dθ v = , αz = =− . The angular acceleration α z is not 32 dt dt r02 + 2 β vt ( r02 + 2β vt )

constant. (d) r0 = 25.0 mm. θ must be measured in radians, so β = (1.55 μ m rev )(1 rev 2π rad ) = 0.247 μ m rad. Using

θ (t ) from part (b), the total angle turned in 74.0 min = 4440 s is θ=

2 1 ⎛ ⎞ −7 −3 −3 ⎜ 2 ( 2.47 × 10 m/rad ) (1.25 m/s )( 4440 s ) + ( 25.0 × 10 m ) − 25.0 × 10 m ⎟ −7 2.47 × 10 m/rad ⎝ ⎠

θ = 1.337 ×105 rad , which is 2.13 ×10 4 rev . (e) The graphs are sketched in Figure 9.101. EVALUATE: ω z must decrease as r increases, to keep v = rω constant. For ω z to decrease in time, α z must be negative.

Figure 9.101

10

DYNAMICS OF ROTATIONAL MOTION

10.1.

IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Fig.(10.4) to calculate the torque direction. (a) SET UP: Consider Figure 10.1a. EXECUTE: τ = Fl l = r sin φ = (4.00 m)sin 90° l = 4.00 m τ = (10.0 N)(4.00 m) = 40.0 N ⋅ m Figure 10.1a

! This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is directed out of the plane of the figure. (b) SET UP: Consider Figure 10.1b. EXECUTE: τ = Fl l = r sin φ = (4.00 m)sin120° l = 3.464 m τ = (10.0 N)(3.464 m) = 34.6 N ⋅ m Figure 10.1b

! This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is directed out of the plane of the figure. (c) SET UP: Consider Figure 10.1c. EXECUTE: τ = Fl l = r sin φ = (4.00 m)sin 30° l = 2.00 m τ = (10.0 N)(2.00 m) = 20.0 N ⋅ m Figure 10.1c ! This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is directed out of the plane of the figure. (d) SET UP: Consider Figure 10.1d. EXECUTE: τ = Fl l = r sin φ = (2.00 m)sin 60° = 1.732 m τ = (10.0 N)(1.732 m) = 17.3 N ⋅ m Figure 10.1d

! This force tends to produce a clockwise rotation about the axis; by the right-hand rule the vector τ is directed into the plane of the figure. (e) SET UP: Consider Figure 10.1e. EXECUTE: τ = Fl r = 0 so l = 0 and τ = 0 Figure 10.1e

10-1

10-2

Chapter 10

(f) SET UP:

Consider Figure 10.1f.

EXECUTE: τ = Fl l = r sin φ , φ = 180°, so l = 0 and τ = 0

Figure 10.1f

10.2.

EVALUATE: The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action of the force passes through the axis. IDENTIFY: τ = Fl with l = r sin φ . Add the two torques to calculate the net torque. SET UP: Let counterclockwise torques be positive. EXECUTE: τ 1 = − F1l1 = −(8.00 N)(5.00 m) = −40.0 N ⋅ m . τ 2 = + F2l2 = (12.0 N)(2.00 m)sin 30.0° = +12.0 N ⋅ m .

∑τ = τ 10.3.

1

+ τ 2 = −28.0 N ⋅ m . The net torque is 28.0 N ⋅ m , clockwise.

EVALUATE: Even though F1 < F2 , the magnitude of τ 1 is greater than the magnitude of τ 2 , because F1 has a larger moment arm. IDENTIFY and SET UP: Use Eq.(10.2) to calculate the magnitude of each torque and use the right-hand rule (Fig.10.4) to determine the direction. Consider Figure 10.3

Figure 10.3

Let counterclockwise be the positive sense of rotation. EXECUTE:

r1 = r2 = r3 = (0.090 m) 2 + (0.090 m) 2 = 0.1273 m

τ 1 = − F1l1 l1 = r1 sin φ1 = (0.1273 m)sin135° = 0.0900 m τ 1 = −(18.0 N)(0.0900 m) = −1.62 N ⋅ m ! τ 1 is directed into paper τ 2 = + F2l2 l2 = r2 sin φ2 = (0.1273 m)sin135° = 0.0900 m τ 2 = + (26.0 N)(0.0900 m) = +2.34 N ⋅ m ! τ 2 is directed out of paper τ 3 = + F3l3 l3 = r3 sin φ3 = (0.1273 m)sin 90° = 0.1273 m τ 3 = +(14.0 N)(0.1273 m) = +1.78 N ⋅ m ! τ 3 is directed out of paper

∑τ = τ 10.4.

1

+ τ 2 + τ 3 = −1.62 N ⋅ m + 2.34 N ⋅ m + 1.78 N ⋅ m = 2.50 N ⋅ m

EVALUATE: The net torque is positive, which means it tends to produce a counterclockwise rotation; the vector torque is directed out of the plane of the paper. In summing the torques it is important to include + or − signs to show direction. IDENTIFY: Use τ = Fl = rF sin φ to calculate the magnitude of each torque and use the right-hand rule to determine the direction of each torque. Add the torques to find the net torque.

Dynamics of Rotational Motion

SET UP:

10-3

Let counterclockwise torques be positive. For the 11.9 N force ( F1 ), r = 0 . For the 14.6 N force ( F2 ),

r = 0.350 m and φ = 40.0° . For the 8.50 N force ( F3 ), r = 0.350 m and φ = 90.0° EXECUTE:

τ 1 = 0 . τ 2 = −(14.6 N)(0.350 m)sin 40.0° = −3.285 N ⋅ m .

τ 3 = + (8.50 N)(0.350 m)sin 90.0° = +2.975 N ⋅ m .

10.5.

∑τ = −3.285 N ⋅ m + 2.975 N ⋅ m = −0.31 N ⋅ m .The net torque

is 0.31 N ⋅ m and is clockwise. ! ! EVALUATE: If we treat the torques as vectors, τ 2 is into the page and τ 3 is out of the page. IDENTIFY and SET UP: Calculate the torque using Eq.(10.3) and also determine the direction of the torque using the right-hand rule. ! ! (a) r = (−0.450 m)iˆ + (0.150 m) ˆj; F = (−5.00 N)iˆ + (4.00 N) ˆj. The sketch is given in Figure 10.5.

Figure 10.5

! ! EXECUTE: (b) When the fingers of your right hand curl from the direction of r into the direction of F (through ! the smaller of the two angles, angle φ ) your thumb points into the page (the direction of τ , the − z -direction). ! ! ! (c) τ = r × F = ⎡⎣( −0.450 m)iˆ +(0.150 m) ˆj ⎤⎦ × ⎡⎣( −5.00 N)iˆ + (4.00 N) ˆj ⎤⎦ ! τ = + (2.25 N ⋅ m)iˆ × iˆ − (1.80 N ⋅ m)iˆ × ˆj − (0.750 N ⋅ m) ˆj × iˆ + (0.600 N ⋅ m) ˆj × ˆj iˆ × iˆ = ˆj × ˆj = 0

10.6.

10.7.

iˆ × ˆj = kˆ , ˆj × iˆ = − kˆ ! Thus τ = −(1.80 N ⋅ m)kˆ − (0.750 N ⋅ m)( − kˆ ) = (−1.05 N ⋅ m)kˆ. ! EVALUATE: The calculation gives that τ is in the − z -direction. This agrees with what we got from the righthand rule. IDENTIFY: Use τ = Fl = rF sin φ for the magnitude of the torque and the right-hand rule for the direction. SET UP: In part (a), r = 0.250 m and φ = 37° EXECUTE: (a) τ = (17.0 N)(0.250 m)sin 37° = 2.56 N ⋅ m . The torque is counterclockwise. (b) The torque is maximum when φ = 90° and the force is perpendicular to the wrench. This maximum torque is (17.0 N)(0.250 m) = 4.25 N ⋅ m . EVALUATE: If the force is directed along the handle then the torque is zero. The torque increases as the angle between the force and the handle increases. IDENTIFY: Apply ∑τ z = Iα z .

⎛ 2π rad/rev ⎞ ⎟ = 41.9 rad/s ⎝ 60 s/min ⎠ ω − ω0 z 41.9 rad/s = ( 2.50 kg ⋅ m 2 ) = 13.1 N ⋅ m. EXECUTE: τ z = Iαz = I z 8.00 s t EVALUATE: In τ z = Iα z , α z must be in rad/s 2 . SET UP:

10.8.

IDENTIFY: SET UP:

ω0 z = 0 . ω z = (400 rev/min) ⎜

Use a constant acceleration equation to calculate α z and then apply

∑τ

z

= Iα z .

I = MR + 2mR , where M = 8.40 kg, m = 2.00 kg , so I = 0.600 kg ⋅ m . 2 3

2

2

2

ω0 z = 75.0 rpm = 7.854 rad s; ωz = 50.0 rpm = 5.236 rad s; t = 30.0 s . EXECUTE:

10.9.

2

ωz = ω0z + αzt gives αz = −0.08726 rad s . τ z = Iαz = −0.0524 N ⋅ m

EVALUATE: The torque is negative because its direction is opposite to the direction of rotation, which must be the case for the speed to decrease. IDENTIFY: Use ∑τ z = Iα z to calculate α . Use a constant angular acceleration kinematic equation to relate α z ,

ω z and t. SET UP: For a solid uniform sphere and an axis through its center, I = 52 MR 2 . Let the direction the sphere is spinning be the positive sense of rotation. The moment arm for the friction force is l = 0.0150 m and the torque due to this force is negative.

10-4

Chapter 10

(a) α z =

EXECUTE:

τz I

=

−(0.0200 N)(0.0150 m) = −14.8 rad/s 2 2 2 (0.225 kg)(0.0150 m) 5

(b) ω z − ω0 z = −22.5 rad/s . ω z = ω0 z + α z t gives t =

ω z − ω0 z −22.5 rad/s = = 1.52 s . αz −14.8 rad/s 2

The fact that α z is negative means its direction is opposite to the direction of spin. The negative

EVALUATE:

α z causes ω z to decrease. 10.10.

IDENTIFY:

Apply

∑τ

z

= Iα z to the wheel. The acceleration a of a point on the cord and the angular acceleration

α of the wheel are related by a = Rα . SET UP: Let the direction of rotation of the wheel be positive. The wheel has the shape of a disk and I = 12 MR 2 . The free-body diagram for the wheel is sketched in Figure 10.10a for a horizontal pull and in Figure 10.10b for a vertical pull. P is the pull on the cord and F is the force exerted on the wheel by the axle. τ (40.0 N)(0.250 m) EXECUTE: (a) α z = z = 1 = 34.8 rad/s 2 . a = Rα = (0.250 m)(34.8 rad/s 2 ) = 8.70 m/s 2 . 2 (9.20 kg)(0.250 m) I 2 (b) Fx = − P , Fy = − Mg . F = P 2 + ( Mg ) 2 = (40.0 N) 2 + ([9.20 kg][9.80 m/s 2 ]) 2 = 98.6 N .

tan φ =

Fy Fx

=

Mg (9.20 kg)(9.80 m/s 2 ) and φ = 66.1° . The force exerted by the axle has magnitude 98.6 N and = 40.0 N P

is directed at 66.1° above the horizontal, away from the direction of the pull on the cord. (c) The pull exerts the same torque as in part (a), so the answers to part (a) don’t change. In part (b), F + P = Mg and F = Mg − P = (9.20 kg)(9.80 m/s 2 ) − 40.0 N = 50.2 N . The force exerted by the axle has magnitude 50.2 N and is upward. EVALUATE: The weight of the wheel and the force exerted by the axle produce no torque because they act at the axle.

Figure 10.10 10.11.

IDENTIFY:

Use a constant angular acceleration equation to calculate α z and then apply

∑τ

z

= Iα z to the motion

of the cylinder. f k = μ k n . SET UP:

I = 12 mR 2 =

1 2

( 8.25 kg )( 0.0750 m )

2

= 0.02320 kg ⋅ m 2 . Let the direction the cylinder is rotating be

positive. ω0 z = 220 rpm = 23.04 rad/s; ωz = 0; θ − θ0 = 5.25 rev = 33.0 rad . EXECUTE:

ωz2 = ω02z + 2αz ( θ − θ0 ) gives αz = −8.046 rad/s 2 . ∑ τ z = τ f = − f k R = − μk nR . Then

∑τ

z

Iαz = 7.47 N . μk R EVALUATE: The friction torque is directed opposite to the direction of rotation and therefore produces an angular acceleration that slows the rotation. ! ! IDENTIFY: Apply ∑ F = ma to the stone and ∑τ z = Iα z to the pulley. Use a constant acceleration equation to − μk nR = Iα z and n =

10.12.

= Iα z gives

find a for the stone. SET UP: For the motion of the stone take + y to be downward. The pulley has I = 12 MR 2 . a = Rα .


EXECUTE:

∑F

(a) y − y0 = v0 yt + 12 a yt 2 gives 12.6 m = 12 a y ( 3.00 s ) and a y = 2.80 m s 2 . Then 2

to the stone gives mg − T = ma .

∑τ

z

y

10-5

= ma y applied

= Iα z applied to the pulley gives TR = MR α = MR ( a / R ) . T = 12 Ma . 1 2

2

1 2

2

Combining these two equations to eliminate T gives

M=

10.13.

M 2

⎞ ⎛ a ⎞ ⎛ 10.0 kg ⎞ ⎛ 2.80 m/s 2 = 2.00 kg . ⎜ ⎟=⎜ ⎟⎜ 2 2 ⎟ − − 2 9.80 m/s 2.80 m/s g a ⎠⎝ ⎝ ⎠ ⎝ ⎠

1 1 (b) T = Ma = (10.0 kg ) ( 2.80 m/s 2 ) = 14.0 N 2 2 EVALUATE: The tension in the wire is less than the weight mg = 19.6 N of the stone, because the stone has a downward acceleration. IDENTIFY: Use the kinematic information to solve for the angular acceleration of the grindstone. Assume that the grindstone is rotating counterclockwise and let that be the positive sense of rotation. Then apply Eq.(10.7) to calculate the friction force and use f k = μ k n to calculate μ k .

ω0 z = 850 rev/min(2π rad/1 rev)(1 min/60 s) = 89.0 rad/s t = 7.50 s; ω z = 0 (comes to rest); α z = ? EXECUTE: ω z = ω0 z + α z t SET UP:

0 − 89.0 rad/s = −11.9 rad/s 2 7.50 s SET UP: Apply ∑τ z = Iα z to the grindstone. The free-body diagram is given in Figure 10.13.

αz =

Figure 10.13

The normal force has zero moment arm for rotation about an axis at the center of the grindstone, and therefore zero torque. The only torque on the grindstone is that due to the friction force f k exerted by the ax; for this force the moment arm is l = R and the torque is negative. EXECUTE: ∑τ z = − f k R = − μ k nR I = 12 MR 2 (solid disk, axis through center) Thus

∑τ

z

= Iα z gives − μk nR = ( 12 MR 2 )α z

MRα z (50.0 kg)(0.260 m)( − 11.9 rad/s 2 ) =− = 0.483 2n 2(160 N) EVALUATE: The friction torque is clockwise and slows down the counterclockwise rotation of the grindstone. IDENTIFY: Apply ∑ Fy = ma y to the bucket, with + y downward. Apply ∑τ z = Iα z to the cylinder, with the

μk = −

10.14.

direction the cylinder rotates positive. SET UP: The free-body diagram for the bucket is given in Fig.10.14a and the free-body diagram for the cylinder is given in Fig.10.14b. I = 12 MR 2 . a (bucket) = Rα (cylinder) EXECUTE:

(a) For the bucket, mg − T = ma . For the cylinder,

∑τ

z

= Iα z gives TR = 12 MR 2α . α = a / R then

gives T = 12 Ma . Combining these two equations gives mg − 12 Ma = ma and

a=

⎛ ⎞ mg 15.0 kg 2 2 =⎜ ⎟ (9.80 m/s ) = 7.00 m/s . m + M / 2 ⎝ 15.0 kg + 6.0 kg ⎠

T = m( g − a) = (15.0 kg)(9.80 m/s 2 − 7.00 m/s 2 ) = 42.0 N . (b) v y2 = v02y + 2a y ( y − y0 ) gives v y = 2(7.00 m/s 2 )(10.0 m) = 11.8 m/s . (c) a y = 7.00 m/s 2 , v0 y = 0 , y − y0 = 10.0 m . y − y0 = v0 y t + 12 α yt 2 gives t =

2( y − y0 ) 2(10.0 m) = = 1.69 s ay 7.00 m/s 2

10-6

Chapter 10

(d)

∑F

y

= ma y applied to the cylinder gives n − T − Mg = 0 and

n = T + mg = 42.0 N + (12.0 kg)(9.80 m/s 2 ) = 160 N . EVALUATE: The tension in the rope is less than the weight of the bucket, because the bucket has a downward acceleration. If the rope were cut, so the bucket would be in free-fall, the bucket would strike the water in 2(10.0 m) t= = 1.43 s and would have a final speed of 14.0 m/s. The presence of the cylinder slows the fall of 9.80 m/s 2 the bucket.

10.15.

Figure 10.14 ! ! IDENTIFY: Apply ∑ F = ma to each book and apply ∑τ z = Iα z to the pulley. Use a constant acceleration

equation to find the common acceleration of the books. SET UP: m1 = 2.00 kg , m2 = 3.00 kg . Let T1 be the tension in the part of the cord attached to m1 and T2 be the tension in the part of the cord attached to m2 . Let the + x -direction be in the direction of the acceleration of each book. a = Rα . 2( x − x0 ) 2(1.20 m) = = 3.75 m/s 2 . a1 = 3.75 m/s 2 so EXECUTE: (a) x − x0 = v0 xt + 12 axt 2 gives ax = t2 (0.800 s) 2 T1 = m1a1 = 7.50 N and T2 = m2 ( g − a1 ) = 18.2 N . (b) The torque on the pulley is (T2 − T1 ) R = 0.803 N ⋅ m, and the angular acceleration is

α = a1 R = 50 rad/s 2 , so I = τ α = 0.016 kg ⋅ m 2 .

10.16.

EVALUATE: The tensions in the two parts of the cord must be different, so there will be a net torque on the pulley. ! ! IDENTIFY: Apply ∑ F = ma to each box and ∑τ z = Iα z to the pulley. The magnitude a of the acceleration of

each box is related to the magnitude of the angular acceleration α of the pulley by a = Rα . SET UP: The free-body diagrams for each object are shown in Figure 10.16a-c. For the pulley, R = 0.250 m and I = 12 MR 2 . T1 and T2 are the tensions in the wire on either side of the pulley. m1 = 12.0 kg , m2 = 5.00 kg and ! M = 2.00 kg . F is the force that the axle exerts on the pulley. For the pulley, let clockwise rotation be positive.

∑F = m a . ∑τ

EXECUTE:

m2 g − T2

∑F

(a)

x

= max for the 12.0 kg box gives T1 = m1a .

2

z

= Iα z for the pulley gives (T2 − T1 ) R = ( 12 MR 2 )α . a = Rα and T2 − T1 = 12 Ma . Adding these

y

= ma y for the 5.00 kg weight gives

three equations gives m2 g = (m1 + m2 + 12 M )a and ⎛ ⎞ ⎛ ⎞ m2 5.00 kg 2 2 a =⎜ ⎟g =⎜ ⎟ (9.80 m/s ) = 2.72 m/s . Then 1 m + m + M 12.0 kg + 5.00 kg + 1.00 kg ⎝ ⎠ 2 2 ⎝ 1 ⎠ T1 = m1a = (12.0 kg)(2.72 m/s 2 ) = 32.6 N . m2 g − T2 = m2 a gives T2 = m2 ( g − a ) = (5.00 kg)(9.80 m/s 2 − 2.72 m/s 2 ) = 35.4 N . The tension to the left of the pulley is 32.6 N and below the pulley it is 35.4 N. (b) a = 2.72 m/s 2 (c) For the pulley, ∑ Fx = max gives Fx = T1 = 32.6 N and ∑ Fy = ma y gives Fy = Mg + T2 = (2.00 kg)(9.80 m/s 2 ) + 35.4 N = 55.0 N .


EVALUATE: objects.

10.17.

IDENTIFY:

10-7

The equation m2 g = (m1 + m2 + 12 M )a says that the external force m2 g must accelerate all three

Figure 10.16 ! ! ! Apply ∑τ z = Iα z to the post and ∑ F = ma to the hanging mass. The acceleration a of the mass has

the same magnitude as the tangential acceleration atan = rα of the point on the post where the string is attached; r = 1.75 m − 0.500 m = 1.25 m . SET UP: The free-body diagrams for the post and mass are given in Figures 10.17a and b. The post has I = 13 ML2 , with M = 15.0 kg and L = 1.75 m . EXECUTE:

(a)

∑τ

z

= Iα z for the post gives Tr = ( 13 ML2 )α . a = rα so α =

⎛ ML2 ⎞ a and T = ⎜ 2 ⎟ a . r ⎝ 3r ⎠

∑F

y

= ma y for

the mass gives mg − T = ma . These two equations give mg = (m + ML2 /[3r 2 ])a and

⎛ ⎞ ⎛ ⎞ m 5.00 kg (9.80 m/s 2 ) = 3.31 m/s 2 . a =⎜ g =⎜ 2 2 ⎟ 2 2 ⎟ /[3 ] 5.00 kg [15.0 kg][1.75 m] / 3[1.25 m] m + ML r + ⎝ ⎠ ⎝ ⎠ a 3.31 m/s 2 = = 2.65 rad/s 2 . r 1.25 m (b) No. As the post rotates and the point where the string is attached moves in an arc of a circle, the string is no longer perpendicular to the post. The torque due to this tension changes and the acceleration due to this torque is not constant. (c) From part (a), a = 3.31 m/s 2 . The acceleration of the mass is not constant. It changes as α for the post changes. EVALUATE: At the instant the cable breaks the tension in the string is less than the weight of the mass because the mass accelerates downward and there is a net downward force on it.

α=

Figure 10.17 10.18.


10.19.

Apply

∑τ

z

= Iα z to the rod.

For the rod and axis at one end, I = 13 Ml 2 .

α=

τ I

=

1 3

Fl 3F . = Ml 2 Ml

EVALUATE: Note that α decreases with the length of the rod, even though the torque increases. IDENTIFY: Since there is rolling without slipping, vcm = Rω . The kinetic energy is given by Eq.(10.8). The velocities of points on the rim of the hoop are as described in Figure 10.13 in chapter 10. SET UP: ω = 3.00 rad/s and R = 0.600 m . For a hoop rotating about an axis at its center, I = MR 2 .

10-8

Chapter 10

(a) vcm = Rω = (0.600 m)(3.00 rad/s) = 1.80 m/s .

EXECUTE:

2 2 2 (b) K = Mvcm + 12 I ω 2 = 12 Mvcm + 12 ( MR 2 )(vcm / R 2 ) = Mvcm = (2.20 kg)(1.80 m/s) 2 = 7.13 J ! (c) (i) v = 2vcm = 3.60 m/s . v is to the right. (ii) v = 0 ! 2 2 2 + vtan = vcm + ( Rω ) 2 = 2vcm = 2.55 m/s . v at this point is at 45° below the horizontal. (iii) v = vcm 1 2

10.20.

(d) To someone moving to the right at v = vcm , the hoop appears to rotate about a stationary axis at its center. (i) v = Rω = 1.80 m/s , to the right. (ii) v = 1.80 m/s , to the left. (iii) v = 1.80 m/s , downward. EVALUATE: For the special case of a hoop, the total kinetic energy is equally divided between the motion of the center of mass and the rotation about the axis through the center of mass. In the rest frame of the ground, different points on the hoop have different speed. IDENTIFY: Only gravity does work, so Wother = 0 and conservation of energy gives K i + U i = K f + U f . 2 K f = 12 Mvcm + 12 I cmω 2 .

SET UP:

Let yf = 0 , so U f = 0 and yi = 0.750 m . The hoop is released from rest so K i = 0 . vcm = Rω . For a

hoop with an axis at its center, I cm = MR 2 . EXECUTE:

(a) Conservation of energy gives U i = K f . K f = 12 MR 2ω 2 + 12 ( MR 2 )ω 2 = MR 2ω 2 , so MR 2ω 2 = Mgyi .

gyi (9.80 m/s 2 )(0.750 m) = = 33.9 rad/s . R 0.0800 m (b) v = Rω = (0.0800 m)(33.9 rad/s) = 2.71 m/s EVALUATE: An object released from rest and falling in free-fall for 0.750 m attains a speed of 2 g (0.750 m) = 3.83 m/s . The final speed of the hoop is less than this because some of its energy is in kinetic energy of rotation. Or, equivalently, the upward tension causes the magnitude of the net force of the hoop to be less than its weight. IDENTIFY: Apply Eq.(10.8). SET UP: For an object that is rolling without slipping, vcm = Rω . EXECUTE: The fraction of the total kinetic energy that is rotational is

ω=

10.21.

1 (1 2 ) I cmω 2 = (1 2 ) Mvcm2 + (1 2 ) I cmω 2 1( M/I cm )vcm2 /ω 2

=

1 1 + ( MR 2 / I cm )

(a) I cm = (1 2) MR 2 , so the above ratio is 1 3. (b) I cm = (2 5) MR 2 so the above ratio is 2 7 . (c) I cm = (2 3) MR 2 so the ratio is 2 5 . (d) I cm = (5 8)MR 2 so the ratio is 5 13.

The moment of inertia of each object takes the form I = β MR 2 . The ratio of rotational kinetic 1 β energy to total kinetic energy can be written as = . The ratio increases as β increases. 1 + 1/ β 1 + β ! ! IDENTIFY: Apply ∑ F = ma to the translational motion of the center of mass and ∑τ z = Iα z to the rotation EVALUATE:

10.22.

about the center of mass. SET UP: Let + x be down the incline and let the shell be turning in the positive direction. The free-body diagram for the shell is given in Fig.10.22. From Table 9.2, I cm = 32 mR 2 . EXECUTE:

∑F

x

= max gives mg sin β − f = macm .

∑τ

z

= Iα z gives fR = ( 23 mR 2 )α . With α = acm / R this

becomes f = 23 macm . Combining the equations gives mg sin β − 32 macm = macm and 3g sin β 3(9.80 m/s 2 )(sin 38.0°) = = 3.62 m/s 2 . f = 32 macm = 32 (2.00 kg)(3.62 m/s 2 ) = 4.83 N . The friction is 5 5 f 4.83 N = 0.313 . static since there is no slipping at the point of contact. n = mg cos β = 15.45 N . μs = = n 15.45 N (b) The acceleration is independent of m and doesn’t change. The friction force is proportional to m so will double; f = 9.66 N . The normal force will also double, so the minimum μs required for no slipping wouldn’t change.

acm =


10-9

EVALUATE: If there is no friction and the object slides without rolling, the acceleration is g sin β . Friction and rolling without slipping reduce a to 0.60 times this value.

Figure 10.22 10.23.

! ! IDENTIFY: Apply ∑ Fext = macm and (a) SET UP:

∑τ

z

= I cmα z to the motion of the ball.

The free-body diagram is given in Figure 10.23a. EXECUTE:

∑F

y

= ma y

n = mg cosθ and f s = μs mg cosθ

∑F

x

= max

mg sin θ − μs mg cosθ = ma g (sin θ − μs cosθ ) = a

(eq. 1)

Figure 10.23a SET UP:

Consider Figure 10.23b.

n and mg act at the center of the ball and provide no torque


∑τ

z

∑τ = τ

f

= μs mg cosθ R; I = 52 mR 2

= I cmα z gives μs mg cosθ R = 52 mR 2α

No slipping means α = a / R, so μs g cosθ = 52 a

(eq.2)

We have two equations in the two unknowns a and μs . Solving gives a = 75 g sin θ and

μs = 72 tan θ = 72 tan 65.0° = 0.613 (b) Repeat the calculation of part (a), but now I = 32 mR 2 . a = 35 g sin θ and μs = 52 tan θ = 52 tan 65.0° = 0.858

10.24.

The value of μs calculated in part (a) is not large enough to prevent slipping for the hollow ball. (c) EVALUATE: There is no slipping at the point of contact. More friction is required for a hollow ball since for a given m and R it has a larger I and more torque is needed to provide the same α . Note that the required μs is independent of the mass or radius of the ball and only depends on how that mass is distributed. IDENTIFY: Apply conservation of energy to the motion of the marble. SET UP: K = 12 mv 2 + 12 I ω 2 , with I = 52 MR 2 . vcm = Rω for no slipping . Let y = 0 at the bottom of the bowl. The marble at its initial and final locations is sketched in Figure 10.24.

10-10

Chapter 10

EXECUTE:

1 1 (a) Motion from the release point to the bottom of the bowl: mgh = mv 2 + I ω 2 . 2 2 2

1 1⎛ 2 10 ⎞⎛ v ⎞ gh . mgh = mv 2 + ⎜ mR 2 ⎟⎜ ⎟ and v = 2 2⎝ 5 7 ⎠⎝ R ⎠ Motion along the smooth side: The rotational kinetic energy does not change, since there is no friction torque on v 2 107 gh 5 1 = = h the marble, mv 2 + K rot = mgh′ + K rot . h′ = 2g 2g 7 2 (b) mgh = mgh′ so h′ = h . EVALUATE: (c) With friction on both halves, all the initial potential energy gets converted back to potential energy. Without friction on the right half some of the energy is still in rotational kinetic energy when the marble is at its maximum height.

10.25.

Figure 10.24 IDENTIFY: Apply conservation of energy to the motion of the wheel. SET UP: The wheel at points 1 and 2 of its motion is shown in Figure 10.25.

Take y = 0 at the center of the wheel when it is at the bottom of the hill. Figure 10.25 2 . The wheel has both translational and rotational motion so its kinetic energy is K = 12 I cmω 2 + 12 Mvcm

EXECUTE:

K1 + U1 + Wother = K 2 + U 2

Wother = Wfric = −3500 J (the friction work is negative) K1 = 12 I ω12 + 12 Mv12 ; v = Rω and I = 0.800 MR 2 so K1 = 12 (0.800) MR 2ω12 + 12 MR 2ω12 = 0.900 MR 2ω12 K 2 = 0, U1 = 0, U 2 = Mgh Thus 0.900MR 2ω12 + Wfric = Mgh M = w / g = 392 N/(9.80 m/s 2 ) = 40.0 kg h=

0.900MR 2ω12 + Wfric Mg

(0.900)(40.0 kg)(0.600 m) 2 (25.0 rad/s) 2 − 3500 J = 11.7 m (40.0 kg)(9.80 m/s 2 ) EVALUATE: Friction does negative work and reduces h. ! ! IDENTIFY: Apply ∑τ z = Iα z and ∑ F = ma to the motion of the bowling ball. h=

10.26.

SET UP: acm = Rα . f s = μs n . Let + x be directed down the incline. EXECUTE: (a) The free-body diagram is sketched in Figure 10.26. The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill. ! ! (b) The friction force results in an angular acceleration, given by Iα = fR. ∑ F = ma applied to the motion of the

center of mass gives mg sin β − f = macm, and the acceleration and angular acceleration are related by acm = Rα . I ⎞ ⎛ Combining, mg sin β = ma ⎜1 + = ma ( 7 5) . acm = ( 5 7 ) g sin β . 2 ⎟ ⎝ mR ⎠


10-11

2 2 (c) From either of the above relations between f and acm , f = macm = mg sin β ≤ μs n = μs mg cos β . 5 7 μs ≥ ( 2 7 ) tanβ . EVALUATE: If μs = 0 , acm = mg sin β . acm is less when friction is present. The ball rolls farther uphill when friction is present, because the friction removes the rotational kinetic energy and converts it to gravitational potential energy. In the absence of friction the ball retains the rotational kinetic energy that is has initially.

Figure 10.26 10.27.

(a) IDENTIFY: Use Eq.(10.7) to find α z and then use a constant angular acceleration equation to find ω z . SET UP: The free-body diagram is given in Figure 10.27. EXECUTE: Apply ∑τ z = Iα z to find the angular

acceleration: FR = Iα z

αz =

FR (18.0 N)(2.40 m) = = 0.02057 rad/s 2 I 2100 kg ⋅ m 2

Figure 10.27 SET UP:

Use the constant α z kinematic equations to find ω z .

ω z = ?; ω0 z (initially at rest); α z = 0.02057 rad/s 2 ; t = 15.0 s EXECUTE: ω z = ω0 z + α z t = 0 + (0.02057 rad/s 2 )(15.0 s) = 0.309 rad/s (b) IDENTIFY and SET UP: Calculate the work from Eq.(10.21), using a constant angular acceleration equation to calculate θ − θ 0 , or use the work-energy theorem. We will do it both ways. EXECUTE:

(1) W = τ z Δθ (Eq.(10.21))

Δθ = θ − θ 0 = ω0 zt + 12 α zt 2 = 0 + 12 (0.02057 rad/s 2 )(15.0 s) 2 = 2.314 rad t z = FR = (18.0 N)(2.40 m) = 43.2 N ⋅ m Then W = τ z Δθ = (43.2 N ⋅ m)(2.314 rad) = 100 J. or (2) Wtot = K 2 − K1 (the work-energy relation from chapter 6) Wtot = W , the work done by the child K1 = 0; K 2 = 12 I ω 2 = 12 (2100 kg ⋅ m 2 )(0.309 rad/s)2 = 100 J Thus W = 100 J, the same as before. EVALUATE: Either method yields the same result for W. (c) IDENTIFY and SET UP: Use Eq.(6.15) to calculate Pav ΔW 100 J = = 6.67 W Δt 15.0 s EVALUATE: Work is in joules, power is in watts. IDENTIFY: Apply P = τω and W = τΔθ . SET UP: P must be in watts, Δθ must be in radians, and ω must be in rad/s. 1 rev = 2π rad . 1 hp = 746 W . π rad/s = 30 rev/min . P (175 hp )( 746 W / hp ) = 519 N ⋅ m. EXECUTE: (a) τ = = π rad/s ⎞ ω ( 2400 rev/min ) ⎛⎜ ⎟ ⎝ 30 rev/min ⎠ (b) W = τΔθ = ( 519 N ⋅ m )( 2π rad ) = 3260 J EXECUTE: 10.28.

Pav =

10-12

Chapter 10

10.29.

EVALUATE: ω = 40 rev/s , so the time for one revolution is 0.025 s. P = 1.306 × 105 W , so in one revolution, W = Pt = 3260 J , which agrees with our previous result. IDENTIFY: Apply ∑τ z = Iα z and constant angular acceleration equations to the motion of the wheel.

1 rev = 2π rad . π rad/s = 30 rev/min . ω − ω0 z EXECUTE: (a) τ z = Iα z = I z . t SET UP:

((1 2)(1.50 kg )( 0.100 m ) ) (1200 rev min ) ⎛⎜⎝ 30π revradmins ⎞⎟⎠ 2

τz = (b) ωav Δt =

2.5 s

= 0.377 N ⋅ m

( 600 rev/min )( 2.5 s ) = 25.0 rev = 157 rad.

60 s/min (c) W = τΔθ = (0.377 N ⋅ m)(157 rad) = 59.2 J .

2

(d) K =

10.30.

1 2 1 ⎛ ⎛ π rad/s ⎞ ⎞ I ω = ( (1/ 2)(1.5 kg)(0.100 m) 2 ) ⎜ (1200 rev/min) ⎜ ⎟ ⎟ = 59.2 J . 2 2 ⎝ 30 rev/min ⎠ ⎠ ⎝

the same as in part (c). EVALUATE: The agreement between the results of parts (c) and (d) illustrates the work-energy theorem IDENTIFY: The power output of the motor is related to the torque it produces and to its angular velocity by P = τ zω z , where ω z must be in rad/s. The work output of the motor in 60.0 s is

SET UP:

2 6.00 kJ (9.00 kJ) = 6.00 kJ , so P = = 100 W . 3 60.0 s

ω z = 2500 rev/min = 262 rad/s . 100 W = 0.382 N ⋅ m 262 rad/s For a constant power output, the torque developed decreases and the rotation speed of the motor

τz =

EXECUTE:

10.31.

10.32.

P

ωz

=

EVALUATE: increases. IDENTIFY: Apply τ = FR and P = τω . SET UP: 1 hp = 746 W . π rad/s = 30 rev/min EXECUTE: (a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction), and the bearing radius is small compared to the blade radius, so any frictional torque can be neglected. τ P /ω (1.9 hp)(746 W/hp) (b) F = = = = 65.6 N. ⎛ π rad/s ⎞ R R (2400 rev/min) ⎜ ⎟ (0.086 m) ⎝ 30 rev/min ⎠ EVALUATE: In P = I ω , τ must be in watts and ω must be in rad/s. IDENTIFY: Apply ∑τ z = Iα z to the motion of the propeller and then use constant acceleration equations to

analyze the motion. W = τΔθ . SET UP: I = 12 mL2 = 12 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 . (a) α =

EXECUTE:

τ I

=

1950 N ⋅ m = 46.2 rad/s 2 . 42.2 kg ⋅ m 2

(b) ω = ω + 2α z (θ − θ 0 ) gives ω = 2αθ = 2(46.2 rad/s 2 )(5.0 rev)(2π rad/rev) = 53.9 rad/s. 2 z

2 0z

(c) W = τθ = (1950 N ⋅ m)(5.00 rev)(2π rad/rev) = 6.13 × 104 J.

ω z − ω0 z 53.9 rad/s W 6.13 × 104 J = = 1.17 s . Pav = = = 52.5 kW . 2 αz 46.2 rad/s Δt 1.17 s EVALUATE: P = τω . τ is constant and ω is linear in t, so Pav is half the instantaneous power at the end of the (d) t =

5.00 revolutions. We could also calculate W from W = ΔK = 12 I ω 2 = 12 (42.2 kg ⋅ m 2 )(53.9 rad/s) 2 = 6.13 × 104 J . 10.33.


Use Eq.(10.23) and solve for τ z .

P = τ zω z , where ω z must be in rad/s EXECUTE:

τz =

P

ωz

=

ω z = (4000 rev/min)(2π rad/1 rev)(1 min/60 s) = 418.9 rad/s

1.50 × 105 W = 358 N ⋅ m 418.9 rad/s


10-13

! ! (b) IDENTIFY and SET UP: Apply ∑ F = ma to the drum. Find the tension T in the rope using τ z from part (a). The system is sketched in Figure 10.33. EXECUTE: v constant implies a = 0 and T = w τ z = TR implies

T = τ z / R = 358 N ⋅ m/0.200 m = 1790 N Thus a weight w = 1790 N can be lifted. Figure 10.33

10.34.

(c) IDENTIFY and SET UP: Use v = Rω. EXECUTE: The drum has ω = 418.9 rad/s, so v = (0.200 m)(418.9 rad/s) = 83.8 m/s EVALUATE: The rate at which T is doing work on the drum is P = Tv = (1790 N)(83.8 m/s) = 150 kW. This agrees with the work output of the motor. IDENTIFY: L = I ω and I = I disk + I woman . SET UP:

10.35.

ω = 0.50 rev/s = 3.14 rad/s . I disk = 12 mdisk R 2 and I woman = mwoman R 2 .

EXECUTE: I = (55 kg + 50.0 kg)(4.0 m) 2 = 1680 kg ⋅ m 2 . L = (1680 kg ⋅ m 2 )(3.14 rad/s) = 5.28 × 103 kg ⋅ m 2 /s EVALUATE: The disk and the woman have similar values of I, even though the disk has twice the mass. (a) IDENTIFY: Use L = mvr sin φ (Eq.(10.25)): SET UP: Consider Figure 10.35. EXECUTE: L = mvr sin φ = (2.00 kg)(12.0 m/s)(8.00 m)sin143.1°

L = 115 kg ⋅ m 2 / s

10.36.

Figure 10.35 ! ! ! To find the direction of L apply the right-hand rule by turning r into the direction of v by pushing on it with the ! fingers of your right hand. Your thumb points into the page, in the direction of L. (b) IDENTIFY and SET UP: By Eq.(10.26) the rate of change of the angular momentum of the rock equals the torque of the net force acting on it. EXECUTE: τ = mg (8.00 m)cos36.9° = 125 kg ⋅ m 2 / s 2 ! ! ! To find the direction of τ and hence of dL / dt , apply the right-hand rule by turning r into the direction of the gravity force by pushing on it with the fingers of your right hand. Your thumb points out of the page, in the ! direction of dL / dt. ! ! EVALUATE: L and dL / dt are in opposite directions, so L is decreasing. The gravity force is accelerating the rock downward, toward the axis. Its horizontal velocity is constant but the distance l is decreasing and hence L is decreasing. IDENTIFY: Lz = I ω z SET UP: For a particle of mass m moving in a circular path at a distance r from the axis, I = mr 2 and v = rω . For a uniform sphere of mass M and radius R and an axis through its center, I = 52 MR 2 . The earth has mass

mE = 5.97 × 1024 kg , radius RE = 6.38 × 106 m and orbit radius r = 1.50 × 1011 m . The earth completes one rotation on its axis in 24 h = 86,400 s and one orbit in 1 y = 3.156 × 107 s . ⎛ 2π rad ⎞ 40 2 (a) Lz = I ω z = mr 2ω z = (5.97 × 1024 kg)(1.50 × 1011 m) 2 ⎜ ⎟ = 2.67 × 10 kg ⋅ m /s . 7 ⎝ 3.156 × 10 s ⎠ The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. ⎛ 2π rad ⎞ 33 2 (b) Lz = I ω z = ( 52 MR 2 ) ω = 52 (5.97 × 1024 kg)(6.38 × 106 m) 2 ⎜ ⎟ = 7.07 × 10 kg ⋅ m /s 86,400 s ⎝ ⎠ EVALUATE: The angular momentum associated with each of these motions is very large. EXECUTE:

10-14

Chapter 10

10.37.

IDENTIFY and SET UP: Use L = I ω EXECUTE: The second hand makes 1 revolution in 1 minute, so ω = (1.00 rev/min)(2π rad/1 rev)(1 min/60 s) = 0.1047 rad/s For a slender rod, with the axis about one end, I = 13 ML2 = 13 (6.00 × 10−3 kg)(0.150 m)2 = 4.50 × 10−5 kg ⋅ m 2

10.38.

Then L = I ω = (4.50 × 10−5 kg ⋅ m 2 )(0.1047 rad/s) = 4.71 × 10−6 kg ⋅ m 2 / s. ! EVALUATE: L is clockwise. IDENTIFY: ω z = dθ / dt . Lz = I ω z and τ z = dLz dt . For a hollow, thin-walled sphere rolling about an axis through its center, I = 32 MR 2 . R = 0.240 m .

SET UP: EXECUTE:

(a) A = 1.50 rad/s 2 and B = 1.10 rad/s 4 , so that θ (t ) will have units of radians.

dθ = 2 At + 4 Bt 3 . At t = 3.00 s , ω z = 2(1.50 rad/s 2 )(3.00 s) + 4(1.10 rad/s 4 )(3.00 s)3 = 128 rad/s . dt Lz = ( 32 MR 2 )ω z = 32 (12.0 kg)(0.240 m) 2 (128 rad/s) = 59.0 kg ⋅ m 2 /s .

(b) (i) ω z =

dLz dω z =I = I (2 A + 12 Bt 2 ) and dt dt τ z = 32 (12.0 kg)(0.240 m) 2 (2[1.50 rad/s 2 ] + 12[1.10 rad/s 4 ][3.00 s]2 ) = 56.1 N ⋅ m .

(ii) τ z =

EVALUATE:

10.39.

The angular speed of rotation is increasing. This increase is due to an acceleration α z that is

produced by the torque on the sphere. When I is constant, as it is here, τ z = dLz dt = Idω z / dt = Iα z and Equations (10.29) and (10.7) are identical. IDENTIFY: Apply conservation of angular momentum. SET UP: For a uniform sphere and an axis through its center, I = 52 MR 2 . EXECUTE: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is 2

2

⎛ R1 ⎞ ⎛ ⎞ ⎛ 7.0 × 105 km ⎞ 2π rad 3 ⎟ =⎜ ⎟ = 4.6 × 10 rad s. ⎟⎜ ⎝ R2 ⎠ ⎝ (30 d)(86,400 s d) ⎠⎝ 16 km ⎠

ω2 = ω1 ⎜

10.40.

EVALUATE: K = 12 I ω 2 = 12 Lω . L is constant and ω increases by a large factor, so there is a large increase in the rotational kinetic energy of the star. This energy comes from potential energy associated with the gravity force within the star. ! IDENTIFY and SET UP: L is conserved if there is no net external torque. Use conservation of angular momentum to find ω at the new radius and use K = 12 I ω 2 to find the change in kinetic energy, which is equal to the work done on the block. EXECUTE: (a) Yes, angular momentum is conserved. The moment arm for the tension in the cord is zero so this force exerts no torque and there is no net torque on the block. (b) L1 = L2 so I1ω1 = I 2ω2. Block treated as a point mass, so I = mr 2, where r is the distance of the block from the hole. mr12ω1 = mr22ω2 2

2

⎛r ⎞ ⎛ 0.300 m ⎞ ω2 = ⎜ 1 ⎟ ω1 = ⎜ ⎟ (1.75 rad/s) = 7.00 rad/s r ⎝ 0.150 m ⎠ ⎝ 2⎠ (c) K1 = 12 I1ω12 = 12 mr12ω12 = 12 mv12 v1 = r1ω1 = (0.300 m)(1.75 rad/s) = 0.525 m/s K1 = 12 mv12 = 12 (0.0250 kg)(0.525 m/s)2 = 0.00345 J K 2 = 12 mv22 v2 = r2ω2 = (0.150 m)(7.00 rad/s) = 1.05 m/s K 2 = 12 mv22 = 12 (0.0250 kg)(1.05 m/s) 2 = 0.01378 J ΔK = K 2 − K1 = 0.01378 J − 0.00345 J = 0.0103 J


10-15

(d) Wtot = ΔK

10.41.

But Wtot = W , the work done by the tension in the cord, so W = 0.0103 J EVALUATE: Smaller r means smaller I. L = I ω is constant so ω increases and K increases. The work done by the tension is positive since it is directed inward and the block moves inward, toward the hole. IDENTIFY: Apply conservation of angular momentum to the motion of the skater. SET UP: For a thin-walled hollow cylinder I = mR 2 . For a slender rod rotating about an axis through its center, I = 121 ml 2 . EXECUTE:

Li = Lf so I iωi = I f ωf .

I i = 0.40 kg ⋅ m 2 + 121 (8.0 kg)(1.8 m)2 = 2.56 kg ⋅ m 2 . I f = 0.40 kg ⋅ m 2 + (8.0 kg)(0.25 m) 2 = 0.90 kg ⋅ m 2 . ⎛ Ii ⎞ ⎛ 2.56 kg ⋅ m 2 ⎞ (0.40 rev/s)=1.14 rev/s . ⎟ ωi = ⎜ 2 ⎟ ⎝ 0.90 kg ⋅ m ⎠ ⎝ If ⎠ EVALUATE: K = 12 I ω 2 = 12 Lω . ω increases and L is constant, so K increases. The increase in kinetic energy comes from the work done by the skater when he pulls in his hands. IDENTIFY: Apply conservation of angular momentum to the diver. SET UP: The number of revolutions she makes in a certain time is proportional to her angular velocity. The ratio of her untucked to tucked angular velocity is (3.6 kg ⋅ m 2 ) /(18 kg ⋅ m 2 ) .

ωf = ⎜

10.42.

EXECUTE:

10.43.

If she had tucked, she would have made (2 rev)(3.6 kg ⋅ m 2 ) (18 kg ⋅ m 2 ) = 0.40 rev in the last 1.0 s,

so she would have made (0.40 rev)(1.5 1.0) = 0.60 rev in the total 1.5 s. EVALUATE: Untucked she rotates slower and completes fewer revolutions. IDENTIFY and SET UP: There is no net external torque about the rotation axis so the angular momentum L = I ω is conserved. EXECUTE: (a) L1 = L2 gives I1ω1 = I 2ω2 , so ω2 = ( I1 / I 2 )ω1 I1 = I tt = 12 MR 2 = 12 (120 kg)(2.00 m) 2 = 240 kg ⋅ m 2 I 2 = I tt + I p = 240 kg ⋅ m 2 + mR 2 = 240 kg ⋅ m 2 + (70 kg)(2.00 m) 2 = 520 kg ⋅ m 2

ω2 = ( I1 / I 2 )ω1 = (240 kg ⋅ m 2 / 520 kg ⋅ m 2 )(3.00 rad/s) = 1.38 rad/s (b) K1 = 12 I1ω12 = 12 (240 kg ⋅ m 2 )(3.00 rad/s) 2 = 1080 J

10.44.

K 2 = 12 I 2ω22 = 12 (520 kg ⋅ m 2 )(1.38 rad/s) 2 = 495 J EVALUATE: The kinetic energy decreases because of the negative work done on the turntable and the parachutist by the friction force between these two objects. The angular speed decreases because I increases when the parachutist is added to the system. IDENTIFY: Apply conservation of angular momentum to the collision. SET UP: Let the width of the door be l. The initial angular momentum of the mud is mv (l / 2) , since it strikes the door at its center. For the axis at the hinge, I door = 13 Ml 2 and I mud = m(l / 2) 2 . EXECUTE:

ω=

10.45.

ω=

mv ( l 2 ) L = . I (1 3) Ml 2 + m ( l 2 )2

( 0.500 kg )(12.0 m s )( 0.500 m ) 2 2 (1 3)( 40.0 kg )(1.00 m ) + ( 0.500 kg )( 0.500 m )

= 0.223 rad s.

Ignoring the mass of the mud in the denominator of the above expression gives ω = 0.225 rad s, so the mass of the mud in the moment of inertia does affect the third significant figure. EVALUATE: Angular momentum is conserved but there is a large decrease in the kinetic energy of the system. ! (a) IDENTIFY and SET UP: Apply conservation of angular momentum L, with the axis at the nail. Let object A be the bug and object B be the bar. Initially, all objects are at rest and L1 = 0. Just after the bug jumps, it has angular momentum in one direction of rotation and the bar is rotating with angular velocity ωB in the opposite direction. EXECUTE: L2 = mAv Ar − I BωB where r = 1.00 m and I B = 13 mB r 2 L1 = L2 gives mAv A r = 13 mB r 2ωB

ωB =

3mAv A = 0.120 rad/s mB r

10-16

Chapter 10

(b) K1 = 0; K 2 = 12 mAv A2 + 12 I BωB2 = 1 2

10.46.

(0.0100 kg)(0.200 m/s)2 + 12 ( 13 [0.0500 kg][1.00 m]2 ) (0.120 rad/s) 2 = 3.2 × 10−4 J.

The increase in kinetic energy comes from work done by the bug when it pushes against the bar in order to jump. EVALUATE: There is no external torque applied to the system and the total angular momentum of the system is constant. There are internal forces, forces the bug and bar exert on each other. The forces exert torques and change the angular momentum of the bug and the bar, but these changes are equal in magnitude and opposite in direction. These internal forces do positive work on the two objects and the kinetic energy of each object and of the system increases. IDENTIFY: Apply conservation of angular momentum to the system of earth plus asteroid. SET UP: Take the axis to be the earth’s rotation axis. The asteroid may be treated as a point mass and it has zero angular momentum before the collision, since it is headed toward the center of the earth. For the earth, 2π rad Lz = I ω z and I = 52 MR 2 ,where M is the mass of the earth and R is its radius. The length of a day is T = ,

ω

where ω is the earth’s angular rotation rate. EXECUTE: Conservation of angular momentum applied to the collision between the earth and asteroid gives ⎛ ω − ω2 ⎞ 1 1.250 2 MR 2ω1 = (mR 2 + 52 MR 2 )ω2 and m = 52 M ⎜ 1 = and ω1 = 1.250ω2 . ⎟ . T2 = 1.250T1 gives 5 ω ω1 ω ⎝ ⎠ 2 2 ω1 − ω2 = 0.250 . m = 52 (0.250) M = 0.100 M .

ω2

10.47.

EVALUATE: If the asteroid hit the surface of the earth tangentially it could have some angular momentum with respect to the earth’s rotation axis, and could either speed up or slow down the earth’s rotation rate. IDENTIFY: Apply conservation of angular momentum to the collision. SET UP: The system before and after the collision is sketched in Figure 10.47. Let counterclockwise rotation be positive. The bar has I = 13 m2 L2 . EXECUTE:

(a) Conservation of angular momentum: m1v0 d = − m1vd + 13 m2 L2ω .

1 ⎛ 90.0 N ⎞ 2 (3.00 kg)(10.0 m s)(1.50 m) = −(3.00 kg)(6.00 m s)(1.50 m) + ⎜ ⎟ (2.00 m) ω 3 ⎝ 9.80 m s 2 ⎠

ω = 5.88 rad s . (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. EVALUATE: Kinetic energy is not conserved in the collision.

10.48.

Figure 10.47 ! ! ! ! IDENTIFY: dL = τ dt , so dL is in the direction of τ . ! SET UP: The direction of ω is given by the right-hand rule, as described in Figure 10.26 in the textbook. EXECUTE: The sketches are given in Figures 10.48a–d.


10-17

EVALUATE: In figures (a) and (c) the precession is counterclockwise and in figures (b) and (d) it is clockwise. ! ! When the direction of either ω or τ reverses, the direction of precession reverses.

Figure 10.48 10.49.

IDENTIFY:

The precession angular velocity is Ω =

wr , where ω is in rad/s. Also apply Iω

!

!

∑ F = ma to the

gyroscope. SET UP: The total mass of the gyroscope is mr + mf = 0.140 kg + 0.0250 kg = 0.165 kg . 2π rad 2π rad = = 2.856 rad/s . T 2.20 s EXECUTE: (a) Fp = wtot = (0.165 kg)(9.80 m/s 2 ) = 1.62 N Ω=

wr (0.165 kg)(9.80 m/s 2 )(0.0400 m) = = 189 rad/s = 1.80 × 103 rev/min I Ω (1.20 × 10−4 kg ⋅ m 2 )(2.856 rad/s) ! ! (c) If the figure in the problem is viewed from above, τ is in the direction of the precession and L is along the axis of the rotor, away from the pivot. EVALUATE: There is no vertical component of acceleration associated with the motion, so the force from the pivot equals the weight of the gyroscope. The larger ω is, the slower the rate of precession. IDENTIFY: The precession angular speed is related to the acceleration due to gravity by Eq.(10.33), with w = mg . (b) ω =

10.50.

SET UP: Ω E = 0.50 rad/s , g E = g and g M = 0.165 g . For the gyroscope, m, r, I, and ω are the same on the moon as on the earth. Ω Ω mgr Ω mr EXECUTE: Ω = . = = constant , so E = M . Iω g Iω gE gM

10.51.

10.52.

⎛g ⎞ Ω M = Ω E ⎜ M ⎟ = 0.165Ω E = (0.165)(0.50 rad/s) = 0.0825 rad/s . ⎝ gE ⎠ EVALUATE: In the limit that g → 0 the precession rate → 0 . IDENTIFY and SET UP: Apply Eq.(10.33). EXECUTE: (a) halved (b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed) (c) halved (assuming that w and r are not changed) (d) doubled (e) unchanged. EVALUATE: Ω is directly proportional to w and r and is inversely proportional to I and ω . IDENTIFY: Apply Eq.(10.33), where τ = wr . SET UP: 1 day = 86,400 s . 1 yr = 3.156 × 107 s . The earth has mass M = 5.97 × 1024 kg and radius R = 6.38 × 106 m . For a uniform sphere and an axis through its center, I = 52 MR 2 . 2π rad (a) τ = I ωΩ = (2 / 5) MR 2ωΩ. Using ω = 2π rad and Ω = , and the mass 86,400 s (26,000 y)(3.156 × 107 s/y) and radius of the earth from Appendix F, τ = 5.4 N ⋅ m . EVALUATE: If the torque is applied by the sun, r = 1.5 × 1011 m and F⊥ = 3.6 × 1011 N .

EXECUTE:

10-18

Chapter 10

10.53.

IDENTIFY:

Apply

∑τ

z

= Iα z and constant acceleration equations to the motion of the grindstone.

Let the direction of rotation of the grindstone be positive. The friction force is f = μ k n and produces

SET UP:

⎛ 2π rad ⎞⎛ 1 min ⎞ 2 2 1 torque fR . ω = ⎜ ⎟⎜ ⎟ = 4π rad . I = 2 MR = 1.69 kg ⋅ m . ⎝ 1 rev ⎠⎝ 60 s ⎠ EXECUTE: (a) The net torque must be 4π rad/s = (1.69 kg ⋅ m 2 ) = 2.36 N ⋅ m. t 9.00 s This torque must be the sum of the applied force FR and the opposing frictional torques 1 τ f at the axle and fR = μ k nR due to the knife. F = (τ + τ f + μ k nR ) . R 1 F= ( (2.36 N ⋅ m) + (6.50 N ⋅ m) + (0.60)(160 N)(0.260 m) ) = 67.6 N. 0.500 m (b) To maintain a constant angular velocity, the net torque τ is zero, and the force F ′ is 1 F′ = (6.50 N ⋅ m + 24.96 N ⋅ m) = 62.9 N. 0.500 m (c) The time t needed to come to a stop is found by taking the magnitudes in Eq.(10.27), with τ = τ f constant;

τ = Iα = I

ω z − ω0 z

2 ω I (4π rad/s) (1.69 kg ⋅ m ) = = 3.27 s. τf τf 6.50 N ⋅ m EVALUATE: The time for a given change in ω is proportional to α , which is in turn proportional to the net torque,

t=

10.54.

L

=

so the time in part (c) can also be found as t = ( 9.00 s ) 2.36 N ⋅ m . 6.50 N ⋅ m IDENTIFY: Apply ∑τ z = Iα z and use the constant acceleration equations to relate α to the motion. Let the direction the wheel is rotating be positive. 100 rev/min = 10.47 rad/s ω − ω0 z 10.47 rad/s − 0 = = 5.23 rad/s 2 . EXECUTE: (a) ω z = ω0 z + α z t gives α z = z t 2.00 s SET UP:

I=

∑τ αz

z

=

5.00 N ⋅ m = 0.956 kg ⋅ m 2 5.23 rad/s 2

(b) ω0 z = 10.47 rad/s , ω z = 0 , t = 125 s . ω z = ω0 z + α z t gives α z =

∑τ

10.55.

z

ω z − ω0 z t

=

0 − 10.47 rad/s = −0.0838 rad/s 2 125 s

= Iα z = (0.956 kg ⋅ m 2 )(−0.0838 rad/s 2 ) = −0.0801 N ⋅ m

⎛ ω + ω z ⎞ ⎛ 10.47 rad/s + 0 ⎞ (c) θ = ⎜ 0 z ⎟t = ⎜ ⎟ (125 s) = 654 rad = 104 rev 2 2 ⎝ ⎠ ⎝ ⎠ EVALUATE: The applied net torque ( 5.00 N ⋅ m ) is much larger than the magnitude of the friction torque ( 0.0801 N ⋅ m ), so the time of 2.00 s that it takes the wheel to reach an angular speed of 100 rev/min is much less than the 125 s it takes the wheel to be brought to rest by friction. IDENTIFY and SET UP: Apply v = rω. v is the tangential speed of a point on the rim of the wheel and equals the linear speed of the car. EXECUTE: (a) v = 60 mph = 26.82 m/s r = 12 in. = 0.3048 m v ω = = 88.0 rad/s = 14.0 rev/s = 840 rpm r (b) Same ω as in part (a) since speedometer reads same. r = 15 in. = 0.381 m v = rω = (0.381 m)(88.0 rad/s) = 33.5 m/s = 75 mph (c) v = 50 mph = 22.35 m/s r = 10 in. = 0.254 m v ω = = 88.0 rad/s. This is the same as for 60 mph with correct tires, so speedometer read 60 mph. r EVALUATE: For a given ω , v increases when r increases.


10.56.

10-19

2 IDENTIFY: The kinetic energy of the disk is K = 12 Mvcm + 12 I ω 2 . As it falls its gravitational potential energy decreases and its kinetic energy increases. The only work done on the disk is the work done by gravity, so K1 + U1 = K 2 + U 2 .

SET UP:

I cm = 12 M ( R22 + R12 ) , where R1 = 0.300 m and R2 = 0.500 m . vcm = R2ω . Take y1 = 0 , so

y2 = −1.20 m . K1 + U1 = K 2 + U 2 . K1 = 0 , U1 = 0 . K 2 = −U 2 .

EXECUTE: 1 2

1 2

2 Mvcm + 12 I cmω 2 = − Mgy2 .

2 2 2 I cmω 2 = 14 M (1 + [ R1 / R2 ]2 ) vcm = 0.340 Mvcm . Then 0.840 Mvcm = − Mgy2 and

− gy2 −(9.80 m/s 2 )( −1.20 m) = = 3.74 m/s 0.840 0.840 EVALUATE: A point mass in free-fall acquires a speed of 4.85 m/s after falling 1.20 m. The disk has a value of vcm that is less than this, because some of the original gravitational potential energy has been converted to rotational kinetic energy. IDENTIFY: Use ∑τ z = Iα z to find the angular acceleration just after the ball falls off and use conservation of vcm =

10.57.

energy to find the angular velocity of the bar as it swings through the vertical position. SET UP: The axis of rotation is at the axle. For this axis the bar has I = 121 mbar L2 , where mbar = 3.80 kg and

L = 0.800 m . Energy conservation gives K1 + U1 = K 2 + U 2 . The gravitational potential energy of the bar doesn’t change. Let y1 = 0 , so y2 = − L / 2 .

(a) τ z = mball g ( L / 2) and I = I ball + I bar = 121 mbar L2 + mball ( L / 2) 2 .

EXECUTE:

∑τ

z

= Iα z gives

⎞ ⎞ mball g ( L / 2) 2g ⎛ mball 2(9.80 m/s ) ⎛ 2.50 kg 2 = ⎜ ⎟ and α z = ⎜ ⎟ = 16.3 rad/s . 2 2 mbar L + mball ( L / 2) L ⎝ mball + mbar / 3 ⎠ 0.800 m ⎝ 2.50 kg + [3.80 kg]/ 3 ⎠ (b) As the bar rotates, the moment arm for the weight of the ball decreases and the angular acceleration of the bar decreases. (c) K1 + U1 = K 2 + U 2 . 0 = K 2 + U 2 . 12 ( I bar + I ball )ω 2 = − mball g ( − L / 2) .

αz =

1 12

mball gL = mball L / 4 + mbar L2 /12

ω=

10.58.

2

2

⎞ ⎞ g⎛ 4mball 9.80 m/s 2 ⎛ 4[2.50 kg] ⎜ ⎟= ⎜ ⎟ L ⎝ mball + mbar / 3 ⎠ 0.800 m ⎝ 2.50 kg + [3.80 kg]/ 3 ⎠

ω = 5.70 rad/s . EVALUATE: As the bar swings through the vertical, the linear speed of the ball that is still attached to the bar is v = (0.400 m)(5.70 rad/s) = 2.28 m/s . A point mass in free-fall acquires a speed of 2.80 m/s after falling 0.400 m; the ball on the bar acquires a speed less than this. IDENTIFY: Use ∑τ z = Iα z to find α z , and then use the constant α z kinematic equations to solve for t. SET UP:

The door is sketched in Figure 10.58. EXECUTE:

∑τ

z

= Fl = (220 N)(1.25 m) = 275 N ⋅ m

From Table 9.2(d), I = 13 Ml 2 I = 13 (750 N/9.80 m/s 2 )(1.25 m) 2 = 39.9 kg ⋅ m 2 Figure 10.58

∑τ

z

= Iα z so α z =

SET UP:

∑τ I

z

=

275 N ⋅ m = 6.89 rad/s 2 39.9 kg ⋅ m 2

α z = 6.89 rad/s 2 ; θ − θ 0 = 90°(π rad/180°) = π /2 rad; ω0 z = 0 (door initially at rest); t = ?

θ − θ 0 = ω0 z t + 12 α z t 2 EXECUTE: 10.59.

t=

2(θ − θ 0 )

αz

−

2(π / 2 rad) = 0.675 s 6.89 rad/s 2

EVALUATE: The forces and the motion are connected through the angular acceleration. IDENTIFY: τ = rF sin φ SET UP: Let x be the distance from the left end of the rod where the string is attached. For the value of x where f ( x) is a maximum, df / dx = 0 .

10-20

Chapter 10

""! ! EXECUTE: (a) From geometric consideration, the lever arm and the sine of the angle between F and r are both maximum if the string is attached at the end of the rod.

(b) In terms of the distance x where the string is attached, the magnitude of the torque is Fxh x 2 + h 2 . This function attains its maximum at the boundary, where x = h, so the string should be attached at the right end of the rod. xh (c) As a function of x, l and h, the torque has magnitude τ = F . Differentiating τ with respect to x ( x − l 2) 2 + h 2

10.60.

and setting equal to zero gives xmax = (l 2)(1 + (2 h l ) 2 ). This will be the point at which to attach the string unless 2h > l , in which case the string should be attached at the furthest point to the right, x = l . EVALUATE: In part (a) the maximum torque is independent of h. In part (b) the maximum torque is independent of l. In part (c) the maximum torque depends on both h and l. IDENTIFY: Apply ∑τ z = Iα z , where τ z is due to the gravity force on the object. SET UP:

I = I rod + I clay . I rod = 13 ML2 . In part (b), I clay = ML2 . In part (c), I clay = 0 . (a) A distance L 4 from the end with the clay.

EXECUTE:

(b) In this case I = (4 3) ML2 and the gravitational torque is (3L 4)(2Mg )sin θ = (3Mg L 2)sin θ , so

α = (9 g 8L)sin θ .

10.61.

(c) In this case I = (1 3) ML2 and the gravitational torque is ( L 4)(2Mg )sin θ = ( Mg L 2)sin θ , so α = (3 g 2 L)sin θ . This is greater than in part (b). (d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome deviations from the vertical. EVALUATE: In part (b), I is 4 times larger than in part (c) and τ is 3 times larger. α = τ / I , so the net effect is that α is smaller in (b) than in (c). IDENTIFY: Calculate W using the procedure specified in the problem. In part (c) apply the work-energy theorem. In part (d), atan = Rα and ∑τ z = Iα z . arad = Rω 2 . SET UP:

Let θ be the angle the disk has turned through. The moment arm for F is R cosθ .

EXECUTE:

(a) The torque is τ = FR cosθ . W = ∫

π 2 0

FR cosθ dθ = FR .

(b) In Eq.(6.14), dl is the horizontal distance the point moves, and so W = F ∫ dl = FR , the same as part (a). (c) From K 2 = W = (MR 2 4)ω 2 , ω = 4 F MR . (d) The torque, and hence the angular acceleration, is greatest when θ = 0, at which point α = (τ I ) = 2 F MR , and

so the maximum tangential acceleration is 2 F M . (e) Using the value for ω found in part (c), arad = ω 2 R = 4 F M . EVALUATE:

10.62.

atan = ω 2 R is maximum initially, when the moment arm for F is a maximum, and it is zero after the

disk has rotated one-quarter of a revolution. arad is zero initially and is a maximum at the end of the motion, after the disk has rotated one-quarter of a revolution. ! ! IDENTIFY: Apply ∑ F = ma to the crate and ∑τ z = Iα z to the cylinder. The motions are connected by a (crate) = Rα (cylinder). SET UP: The force diagram for the crate is given in Figure 10.62a. EXECUTE:

∑F

y

= ma y

T − mg = ma T = m( g + a ) = 50 kg(9.80 m/s 2 + 0.80 m/s 2 ) = 530 N Figure 10.62a


SET UP:

10-21

The force diagram for the cylinder is given in Figure 10.62b.

EXECUTE:

∑τ

z

= Iα z

Fl − TR = Iα z , where l = 0.12 m and R = 0.25 m a = Rα so α z = a / R Fl = TR + Ia / R

Figure 10.62b 2 2 ⎛ R ⎞ Ia ⎛ 0.25 m ⎞ (2.9 kg ⋅ m )(0.80 m/s ) = 530 N ⎜ + = 1200 N F =T⎜ ⎟+ ⎟ (0.25 m)(0.12 m) ⎝ l ⎠ Rl ⎝ 0.12 m ⎠

10.63.

EVALUATE: The tension in the rope is greater than the weight of the crate since the crate accelerates upward. If F were applied to the rim of the cylinder (l = 0.25 m), it would have the value F = 567 N. This is greater than T because it must accelerate the cylinder as well as the crate. And F is larger than this because it is applied closer to the axis than R so has a smaller moment arm and must be larger to give the same torque. ! ! IDENTIFY: Apply ∑ Fext = macm and ∑τ z = I cmα z to the roll. SET UP: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero. EXECUTE: (a) Balancing vertical forces, Frod cosθ = f + w + F , and balancing horizontal forces

Frod sin θ = n. With f = μk n, these equations become Frod cosθ = μk n + F + w, Frod sin θ = n. Eliminating n and

solving for Frod gives Frod =

w+ F (16.0 kg) (9.80 m/s 2 ) + (40.0 N) = = 266 N. cosθ − μ k sin θ cos 30° − (0.25)sin 30°

(b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is ( F − f ) R, and f = μk n may be found by insertion of the value found for Frod into either of the above

relations; i.e., f = μk Frod sin θ = 33.2 N. Then,

α=

10.64.

τ I

=

(40.0 N − 31.54 N)(18.0 × 10−2 m) = 4.71 rad/s 2 . (0.260 kg ⋅ m 2 )

EVALUATE: If the applied force F is increased, Frod increases and this causes n and f to increase. The angle φ changes as the amount of paper unrolls and this affects α for a given F. ! ! IDENTIFY: Apply ∑τ z = Iα z to the flywheel and ∑ F = ma to the block. The target variables are the tension in

the string and the acceleration of the block. (a) SET UP: Apply ∑τ z = Iα z to the rotation of the flywheel about the axis. The free-body diagram for the flywheel is given in Figure 10.64a. EXECUTE: The forces n and Mg act at the axis so have zero torque. ∑τ z = TR

TR = Iα z Figure 10.64a

10-22

Chapter 10

Apply

SET UP:

!

!

∑ F = ma

to the translational motion of the block. The free-body diagram for the block is given

in Figure 10.64b.

EXECUTE:

∑F

y

= ma y

n − mg cos36.9° = 0 n = mg cos36.9° f k = μ k n = μ k mg cos36.9°

Figure 10.64b

∑F

x

= max

mg sin 36.9° − T − μ k mg cos36.9° = ma mg (sin 36.9° − μ k cos36.9°) − T = ma

But we also know that ablock = Rα wheel , so α = a / R. Using this in the T = ( I / R ) a. Use this to replace T in the 2

∑F

x

∑τ

z

= Iα z equation gives TR = Ia / R and

= max equation:

mg (sin 36.9° − μ k cos36.9°) − ( I / R )a = ma 2

mg (sin 36.9° − μ k cos36.9°) m + I / R2 (5.00 kg)(9.80 m/s 2 )(sin 36.9° − (0.25)cos36.9°) a= = 1.12 m/s 2 5.00 kg + 0.500 kg ⋅ m 2 /(0.200 m) 2

a=

0.500 kg ⋅ m 2 (1.12 m/s 2 ) = 14.0 N (0.200 m)2 EVALUATE: If the string is cut the block will slide down the incline with a = g sin 36.9° − μ k g cos36.9° = 3.92 m/s 2 . The actual acceleration is less than this because mg sin 36.9° must also (b) T =

10.65.

accelerate the flywheel. mg sin 36.9° − f k = 19.6 N. T is less than this; there must be more force on the block directed down the incline than up then incline since the block accelerates down the incline. ! ! IDENTIFY: Apply ∑ F = ma to the block and ∑τ z = Iα z to the combined disks. SET UP:

For a disk, I disk = 12 MR 2 , so I for the disk combination is I = 2.25 × 10−3 kg ⋅ m 2 .

For a tension T in the string, mg − T = ma and TR = Iα = I a . Eliminating T and solving for a gives R m g a=g = , where m is the mass of the hanging block and R is the radius of the disk to which the m + I / R 2 1 + I / mR 2 string is attached. (a) With m = 1.50 kg and R = 2.50 × 10−2 m, a = 2.88 m/s 2 . EXECUTE:

10.66.

(b) With m = 1.50 kg and R = 5.00 × 10−2 m, a = 6.13 m/s 2 . The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable of applying a larger torque. EVALUATE: ω = v / R , where v is the speed of the block and ω is the angular speed of the disks. When R is larger, in part (b), a smaller fraction of the kinetic energy resides with the disks. The block gains more speed as it falls a certain distance and therefore has a larger acceleration. ! ! IDENTIFY: Apply both ∑ F = ma and ∑τ z = Iα z to the motion of the roller. Rolling without slipping means

acm = Rα . Target variables are acm and f.


SET UP:

10-23

The free-body diagram for the roller is given in Figure 10.66.

EXECUTE:

Apply

!

!

∑ F = ma

to the

translational motion of the center of mass: ∑ Fx = max F − f = Macm

Figure 10.66

Apply

∑τ

z

∑τ

z

= Iα z to the rotation about the center of mass:

= fR

thin-walled hollow cylinder: I = MR 2 Then ∑τ z = Iα z implies fR = MR 2α . But α cm = Rα , so f = Macm . Using this in the

10.67.

∑F

x

= max equation gives F − Macm = Macm

acm = F / 2 M , and then f = Macm = M ( F / 2M ) = F / 2. EVALUATE: If the surface were frictionless the object would slide without rolling and the acceleration would be acm = F / M . The acceleration is less when the object rolls. ! ! IDENTIFY: Apply ∑ F = ma to each object and apply ∑τ z = Iα z to the pulley. SET UP:

Call the 75.0 N weight A and the 125 N weight B. Let TA and TB be the tensions in the cord to the left

and to the right of the pulley. For the pulley, I = 12 MR 2 , where Mg = 50.0 N and R = 0.300 m . The 125 N weight accelerates downward with acceleration a, the 75.0 N weight accelerates upward with acceleration a and the pulley rotates clockwise with angular acceleration α , where a = Rα . ! ! ! ! EXECUTE: ∑ F = ma applied to the 75.0 N weight gives TA − wA = mAa . ∑ F = ma applied to the 125.0 N weight gives wB − TB = mB a .

∑τ

z

= Iα z applied to the pulley gives (TB − TA ) R = ( 12 MR 2 )α z and TB − TA = 12 M .

Combining these three equations gives wB − wA = (m A + mB + M / 2)a and ⎛ ⎞ wB − wA 125 N − 75.0 N ⎛ ⎞ a =⎜ g =⎜ ⎟ g = 0.222 g . TA = wA (1 + a / g ) = 1.222 wA = 91.65 N . ⎜ wA + wB + wpulley / 2 ⎟⎟ + + 75.0 N 125 N 25.0 N ⎝ ⎠ ⎝ ⎠ ! ! TB = wB (1 − a / g ) = 0.778wB = 97.25 N . ∑ F = ma applied to the pulley gives that the force F applied by the hook to the pulley is F = TA + TB + wpulley = 239 N . The force the ceiling applies to the hook is 239 N.

10.68.

EVALUATE: The force the hook exerts on the pulley is less than the total weight of the system, since the net effect of the motion of the system is a downward acceleration of mass. ! ! IDENTIFY: This problem can be done either with conservation of energy or with ∑ Fext = ma. We will do it both

ways. (a) SET UP: (1) Conservation of energy: K1 + U1 + Wother = K 2 + U 2 . Take position 1 to be the location of the disk at the base of the ramp and 2 to be where the disk momentarily stops before rolling back down, as shown in Figure 10.68a. Figure 10.68a

Take the origin of coordinates at the center of the disk at position 1 and take + y to be upward. Then y1 = 0 and y2 = d sin 30°, where d is the distance that the disk rolls up the ramp. “Rolls without slipping” and neglect rolling

friction says W f = 0; only gravity does work on the disk, so Wother = 0

10-24

Chapter 10

EXECUTE:

U1 = Mgy1 = 0

K1 = Mv + 12 I cmω12 (Eq.10.11). But ω1 = v1 / R and I cm = 12 MR 2 , so 1 2

2 1

1 2

I cmω12 = 12 ( 12 MR 2 ) (v1 / R) 2 = 14 Mv12 . Thus

K1 = 12 Mv12 + 14 Mv12 = 34 Mv12 . U 2 = Mgy2 = Mgd sin 30° K 2 = 0 (disk is at rest at point 2).

Thus

3 4

Mv12 = Mgd sin 30°

3v12 3(2.50 m/s) 2 = = 0.957 m 4 g sin 30° 4(9.80 m/s 2 )sin 30° SET UP: (2) force and acceleration The free-body diagram is given in Figure 10.68b. d=

Apply

EXECUTE:

∑F

x

= max to the

translational motion of the center of mass: Mg sin θ − f = Macm Apply

∑τ

z

= Iα z to the rotation about the

center of mass: f R = ( 12 MR 2 )α z f = 12 MRα z Figure 10.68b

But acm = Rα in this equation gives f = 12 Macm . Use this in the

∑F

x

= max equation to eliminate f.

Mg sin θ − 12 Macm = Macm M divides out and 32 acm = g sin θ . acm = 23 g sin θ = 32 (9.80 m/s 2 )sin 30° = 3.267 m/s 2 SET UP: Apply the constant acceleration equations to the motion of the center of mass. Note that in our coordinates the positive x-direction is down the incline. v0 x = −2.50 m/s (directed up the incline); ax = +3.267 m/s 2 ; vx = 0 (momentarily comes to rest); x − x0 = ? vx2 = v02x + 2ax ( x − x0 ) v02x ( −2.50 m/s) 2 =− = −0.957 m 2a x 2(3.267 m/s 2 ) (b) EVALUATE: The results from the two methods agree; the disk rolls 0.957 m up the ramp before it stops. The mass M enters both in the linear inertia and in the gravity force so divides out. The mass M and radius R enter in both the rotational inertia and the gravitational torque so divide out. ! ! IDENTIFY: Apply ∑ Fext = macm to the motion of the center of mass and apply ∑τ z = I cmα z to the rotation about EXECUTE:

10.69.

x − x0 = −

the center of mass. SET UP: I = 2 ( 12 MR 2 ) = MR 2 . The moment arm for T is b. EXECUTE:

The tension is related to the acceleration of the yo-yo by (2m) g − T = (2m)a, and to the angular

acceleration by Tb = Iα = I a . Dividing the second equation by b and adding to the first to eliminate T yields b 2m 2 2 =g . The tension is found by substitution into either of the two , α=g a=g (2m + I b 2 ) 2 + ( R b) 2 2b + R 2 b equations: ⎛ ⎞ 2 ( R b) 2 2mg = = 2 . T = (2m)( g − a ) = (2mg ) ⎜1 − mg ⎟ 2 2 2 + ( R b) (2(b R) 2 + 1) ⎝ 2 + ( R b) ⎠ EVALUATE:

a → 0 when b → 0 . As b → R , a → 2 g / 3 .


10.70.

10-25

IDENTIFY: Apply conservation of energy to the motion of the shell, to find its linear speed v at points A and B. ! ! Apply ∑ F = ma to the circular motion of the shell in the circular part of the track to find the normal force exerted

by the track at each point. Since r 0 has been chosen for a wave traveling from the center to = = = dt dt m μ 2 the edge. Separating variables and integrating, the time t is v(r ) =

2 L dr . ω ∫ 0 L2 − r 2 The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting dr r 2 π arcsin(1) = . r = L sin θ , dr = L cos θ dθ , L2 − r 2 = L cos θ , so that ∫ = θ = arcsin , and t = 2 2 L ω ω 2 L −r EVALUATE: An equivalent method for obtaining T (r ) is to consider the net force on a piece of the rope with t = ∫ dt =

length dr and mass dm = dr m L . The tension must vary in such a way that dT = −(mω2 L) r dr. This is integrated to obtained T ( r ) = −(mω2 2 L)r 2 + C , where dr C is a constant of integration. The tension must vanish at r = L, from which C = (mω2 L 2) and the previous result is obtained. IDENTIFY: Carry out the calculation specified in part (a). ∂y ∂y SET UP: = kASW cos kx sin ωt, = −ω ASW sin kx cos ωt . sin θ cosθ = 12 sin 2θ . ∂x ∂t EXECUTE: The instantaneous power is T ( r ) − T ( r + dr ) = −ω2 r dm, or

15.84.

1 2 FASWω k sin(2kx)sin(2ωt ). 4 (b) The average value of P is proportional to the average value of sin(2ωt ), and the average of the sine function is 2 ω k (sin kx cos kx)(sin ωt cos ωt ) = P = FASW

zero; Pav = 0. (c) The graphs are given in Figure 15.84. The waveform is the solid line, and the power is the dashed line. At time t = 0 , y = 0 and P = 0 and the graphs coincide. (d) When the standing wave is at its maximum displacement at all points, all of the energy is potential, and is concentrated at the places where the slope is steepest (the nodes). When the standing wave has zero displacement, all of the energy is kinetic, concentrated where the particles are moving the fastest (the antinodes). Thus, the energy must be transferred from the nodes to the antinodes, and back again, twice in each cycle. Note that P is greatest midway between adjacent nodes and antinodes, and that P vanishes at the nodes and antinodes.

Mechanical Waves

EVALUATE: string.

15-29

There is energy flow back and forth between the nodes, but there is no net flow of energy along the

Figure 15.84 15.85.

IDENTIFY: SET UP:

For a string, f n = n F . 2L μ For the fundamental, n = 1. Solving for F gives F = μ 4 L2 f 2 . Note that μ = π r 2 ρ , so

μ = π (0.203 × 10−3 m)2 (7800 kg/m3 ) = 1.01 × 10−3 kg/m . EXECUTE:

(a) F = (1.01 × 10−3 kg/m)4(0.635 m) 2 (247.0 Hz) 2 = 99.4 N

(b) To find the fractional change in the frequency we must take the ratio of Δf to f : f =

1 F and 2L μ

⎛ 1 ⎛ 1 F⎞ 1 ⎞ 1 1 1 1 ΔF F 2 ⎟⎟ = Δf = Δ ⎜⎜ Δ F2 = ⎟ = Δ ⎜⎜ ⎟ 2 L μ 2 2 2 L L L μ μ μ2 F ⎝ ⎠ ⎝ ⎠

( )

1 Δf 2 L μ Now divide both sides by the original equation for f and cancel terms: = f 1 2L

1 ΔF 2 F 1 ΔF = . 2 F F

μ

(c) The coefficient of thermal expansion α is defined by Δl = l0αΔT . Combining this with Y =

F/A gives Δl/l0

ΔF = − Y α AΔT . ΔF = − ( 2.00 × 1011 Pa )(1.20 × 10−5 /C° ) π (0.203 × 10−3 m)2 (11°C) = 3.4 N . Then ΔF/F − 0.034 ,

Δf / f = −0.017 and Δf = −4.2 Hz . The pitch falls. This also explains the constant tuning in the string sections of symphonic orchestras. EVALUATE: An increase in temperature causes a decrease in tension of the string, and this lowers the frequency of each standing wave.

16

SOUND AND HEARING

16.1.

IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) λ = v / f = (344 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk is constant gives pmax1 / A1 = pmax2 / A2

⎛p ⎞ 30 Pa ⎛ ⎞ −5 A2 = A1 ⎜ max2 ⎟ = 1.2 × 10−8 m ⎜ ⎟ = 1.2 × 10 m −2 × 3.0 10 Pa p ⎝ ⎠ ⎝ max1 ⎠ (c) pmax = BkA = 2π BA / λ

16.2.

⎛p ⎞ ⎛ 3.0 × 10−2 Pa ⎞ pmax λ = 2π BA = constant so pmax1λ1 = pmax2λ2 and λ2 = λ1 ⎜ max1 ⎟ = (0.344 m) ⎜ ⎟ = 6.9 m −3 ⎝ 1.5 × 10 Pa ⎠ ⎝ pmax2 ⎠ f = v / λ = (344 m/s)/6.9 m = 50 Hz EVALUATE: The pressure amplitude and displacement amplitude are directly proportional. For the same displacement amplitude, the pressure amplitude decreases when the frequency decreases and the wavelength increases. IDENTIFY: Apply pmax = BkA and solve for A. 2π f 2π f BA . and p = v v p v (3.0 × 10−2 Pa) (1480 m s) = 3.21 × 10−12 m. EXECUTE: A = max = 2πBf 2π (2.2 × 109 Pa) (1000 Hz) EVALUATE: Both v and B are larger, but B is larger by a much greater factor, so v / B is a lot smaller and therefore A is a lot smaller. IDENTIFY: Use Eq.(16.5) to relate the pressure and displacement amplitudes. SET UP: As stated in Example 16.1 the adiabatic bulk modulus for air is B = 1.42 × 105 Pa. Use Eq.(15.1) to calculate λ from f, and then k = 2π / λ. EXECUTE: (a) f = 150 Hz Need to calculate k: λ = v / f and k = 2π / λ so k = 2π f / v = (2π rad)(150 Hz)/344 m/s = 2.74 rad/m. Then SET UP:

16.3.

k=

2π

λ

and v = f λ , so k =

pmax = BkA = (1.42 × 105 Pa)(2.74 rad/m)(0.0200 × 10−3 m) = 7.78 Pa. This is below the pain threshold of 30 Pa. (b) f is larger by a factor of 10 so k = 2π f / v is larger by a factor of 10, and pmax = BkA is larger by a factor of

10. pmax = 77.8 Pa, above the pain threshold.

16.4.

16.5.

(c) There is again an increase in f, k, and pmax of a factor of 10, so pmax = 778 Pa, far above the pain threshold. EVALUATE: When f increases, λ decreases so k increases and the pressure amplitude increases. 2π 2π f 2π f BA IDENTIFY: Apply pmax = BkA. k = = , so pmax = . v λ v SET UP: v = 344 m/s vp (344 m/s)(10.0 Pa) EXECUTE: f = max = = 3.86 × 103 Hz 2π BA 2π (1.42 × 105 Pa)(1.00 × 10−6 m) EVALUATE: Audible frequencies range from about 20 Hz to about 20,000 Hz, so this frequency is audible. IDENTIFY: v = f λ . Apply Eq.(16.7) for the waves in the liquid and Eq.(16.8) for the waves in the metal bar. SET UP:

In part (b) the wave speed is v =

d 1.50 m = t 3.90 × 10−4 s

16-1

16-2

Chapter 16

(a) Using Eq.(16.7), B = v 2 ρ = (λf ) 2 ρ , so B = [ (8 m)(400 Hz) ] (1300 kg m3 ) = 1.33 × 1010 Pa. 2

EXECUTE:

2

16.6.

(b) Using Eq.(16.8), Y = v 2 ρ = ( L t ) 2 ρ = ⎡⎣(1.50 m) (3.90 × 10−4 s) ⎤⎦ (6400 kg m3 ) = 9.47 × 1010 Pa. EVALUATE: In the liquid, v = 3200 m/s and in the metal, v = 3850 m/s. Both these speeds are much greater than the speed of sound in air. IDENTIFY: v = d / t. Apply Eq.(16.7) to calculate B. SET UP: ρ = 3.3 × 103 kg/m3 EXECUTE: (a) The time for the wave to travel to Caracas was 9 min 39 s = 579 s and the speed was 1.08 × 104 m/s. Similarly, the time for the wave to travel to Kevo was 680 s for a speed of 1.28 × 104 m/s, and the

time to travel to Vienna was 767 s for a speed of 1.26 × 104 m/s. The average for these three measurements is 1.21× 104 m/s. Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds. (b) From Eq.(16.7), B = v 2 ρ , and using the given value of ρ = 3.3 × 103 kg/m3 and the speeds found in part (a),

16.7.

the values for the bulk modulus are, respectively, 3.9 × 1011 Pa, 5.4 × 1011 Pa and 5.2 × 1011 Pa. EVALUATE: These are larger, by a factor of 2 or 3, than the largest values in Table 11.1. IDENTIFY: d = vt for the sound waves in air an in water. SET UP: Use vwater = 1482 m/s at 20°C, as given in Table 16.1. In air, v = 344 m/s. EXECUTE: Since along the path to the diver the sound travels 1.2 m in air, the sound wave travels in water for the same time as the wave travels a distance 22.0 m − 1.20 m = 20.8 m in air. The depth of the diver is

(20.8 m)

vwater 1482 m/s = (20.8 m) = 89.6 m. This is the depth of the diver; the distance from the horn is 90.8 m. vair 344 m/s

EVALUATE:

16.8.

The time it takes the sound to travel from the horn to the person on shore is t1 =

22.0 m = 0.0640 s. 344 m/s

The time it takes the sound to travel from the horn to the diver is 1.2 m 89.6 m t2 = + = 0.0035 s + 0.0605 s = 0.0640 s. These times are indeed the same. For three figure 344 m/s 1482 m/s accuracy the distance of the horn above the water can’t be neglected. IDENTIFY: Apply Eq.(16.10) to each gas. SET UP: In each case, express M in units of kg/ mol. For H 2 , γ = 1.41. For He and Ar, γ = 1.67. (a) vH2 =

EXECUTE: (b) vHe = (c) vAr =

(1.41)(8.3145 J

(1.67 )( 8.3145 J

( 2.02 ×10

−3

kg mol )

mol ⋅ K )( 300.15 K )

( 4.00 × 10

(1.67 )( 8.3145 J

mol ⋅ K )( 300.15 K )

−3

kg mol )

mol ⋅ K )( 300.15 K )

( 39.9 ×10

−3

kg mol )

= 1.32 × 103 m s

= 1.02 × 103 m s = 323 m s.

(d) Repeating the calculation of Example 16.5 at T = 300.15 K gives vair = 348m s, and so

vH2 = 3.80vair , vHe = 2.94vair and vAr = 0.928vair . EVALUATE: 16.9.

v = f λ . The relation of v to gas temperature is given by v =

IDENTIFY: SET UP:

v is larger for gases with smaller M.

γ RT M

.

Let T = 22.0°C = 295.15 K.

EXECUTE:

At 22.0°C, λ =

v 325 m/s v 1 γ RT . = = 0.260 m = 26.0 cm. λ = = f 1250 Hz f f M 2

2

λ T

=

1 f

γR M

, which is

⎛λ ⎞ ⎛ 28.5 cm ⎞ . T2 = T1 ⎜ 2 ⎟ = (295.15 K) ⎜ ⎟ = 354.6 K = 81.4°C. λ T1 T2 ⎝ 26.0 cm ⎠ ⎝ 1⎠ EVALUATE: When T increases v increases and for fixed f, λ increases. Note that we did not need to know either γ or M for the gas.

constant, so

λ1

=

λ2

Sound and Hearing

16.10.

16.11.

IDENTIFY:

v=

EXECUTE:

(a)

16-3

γ RT

. Take the derivative of v with respect to T. In part (b) replace dv by Δv and dT by ΔT in M the expression derived in part (a). d ( x1/ 2 ) 1 −1/ 2 SET UP: = 2 x . In Eq.(16.10), T must be in kelvins. 20°C = 293 K. ΔT = 1 C° = 1 K. dx dv γ R dT 1/ 2 γ R 1 −1/ 2 1 T = = = 2T dT M dT M 2

γ RT M

=

v dv 1 dT . Rearranging gives , the desired = 2T v 2 T

result. v ΔT ⎛ 344 m/s ⎞⎛ 1 K ⎞ Δv 1 ΔT =⎜ (b) . Δv = = ⎟⎜ ⎟ = 0.59 m/s. 2 T 2 v 2 T ⎝ ⎠⎝ 293 K ⎠ ΔT Δv EVALUATE: Since is one-half this, replacing dT by ΔT and dv by Δv is accurate. Using = 3.4 × 10−3 and T v the result from part (a) is much simpler than calculating v for 20°C and for 21°C and subtracting, and is not subject to round-off errors. IDENTIFY and SET UP: Use t = distance/speed. Calculate the time it takes each sound wave to travel the L = 80.0 m length of the pipe. Use Eq.(16.8) to calculate the speed of sound in the brass rod. EXECUTE: wave in air: t = 80.0 m/(344 m/s) = 0.2326 s wave in the metal: v =

Y

ρ

=

9.0 × 1010 Pa = 3235 m/s 8600 kg/m3

80.0 m = 0.0247 s 3235 m/s The time interval between the two sounds is Δt = 0.2326 s − 0.0247 s = 0.208 s EVALUATE: The restoring forces that propagate the sound waves are much greater in solid brass than in air, so v is much larger in brass. IDENTIFY: Repeat the calculation of Example 16.5 at each temperature. SET UP: 27.0°C = 300.15 K and −13.0°C = 260.15 K t=

16.12.

(1.40)(8.3145 J/mol ⋅ K)(300.15 K) (1.40)(8.3145 J/mol ⋅ K)(260.15 K) − = 24 m/s −3 (28.8 × 10 kg/mol) (28.8 × 10−3 kg/mol)


IDENTIFY:

The speed is greater at the higher temperature. The difference in speeds corresponds to a 7% increase. F Y For transverse waves, vtrans = . For longitudinal waves, vlong = .

μ

ρ

The mass per unit length μ is related to the density (assumed uniform) and the cross-section area A by

SET UP: μ = Aρ.

vlong = 30vtrans gives

EXECUTE: EVALUATE:

Y

ρ

= 30

F

μ

and

Υ F Υ = 900 . Therefore, F A = . ρ Αρ 900

Typical values of Y are on the order of 1011 Pa, so the stress must be about 108 Pa. If A is on the

order of 1 mm = 10−6 m 2 , this requires a force of about 100 N. IDENTIFY: The intensity I is given in terms of the displacement amplitude by Eq.(16.12) and in terms of the pressure amplitude by Eq.(16.14). ω = 2π f . Intensity is energy per second per unit area. 2

16.14.

For part (a), I = 10−12 W/m 2 . For part (b), I = 3.2 × 10−3 W/m 2 .

SET UP: EXECUTE:

A=

1

ω

(a) I =

1 2

ρ Bω 2 A2 .

2I 1 = ρ B 2π (1000 Hz)

2(1 × 10−12 W/m 2 ) (1.20 kg/m )(1.42 × 10 Pa) 3

5

= 1.1 × 10−11 m. I =

2 pmax . 2 ρB

pmax = 2 I ρ B = 2(1 × 10−12 W/m 2 ) (1.20 kg/m3 )(1.42 × 105 Pa) = 2.9 × 10−5 Pa = 2.8 × 10−10 atm (b) A is proportional to

I , so A = (1.1 × 10−11 m)

I , so pmax = (2.9 × 10−5 Pa)

3.2 × 10−3 W/m 2 = 6.2 × 10−7 m. pmax is also proportional to 1 × 10−12 W/m 2

3.2 × 10−3 W/m 2 = 1.6 Pa = 1.6 × 10−5 atm. 1 × 10−12 W/m 2

16-4

Chapter 16

(c) area = (5.00 mm)2 = 2.5 × 10−5 m 2 . Part (a): (1 × 10−12 W/m 2 )(2.5 × 10−5 m 2 ) = 2.5 × 10−17 J/s.

16.15.

Part (b): (3.2 × 10−3 W/m 2 )(2.5 × 10−5 m 2 ) = 8.0 × 10−8 J/s. EVALUATE: For faint sounds the displacement and pressure variation amplitudes are very small. Intensities for audible sounds vary over a very wide range. IDENTIFY: Apply Eq.(16.12) and solve for A. λ = v / f , with v = B / ρ . SET UP: ω = 2π f . For air, B = 1.42 × 105 Pa. EXECUTE: (a) The amplitude is

2Ι = ρΒω2

A=

2(3.00 × 10−6 W m 2 ) (1000 kg m )(2.18 × 10 Pa)(2π (3400 Hz))

The wavelength is λ =

16.16.

3

v = f

9

= 9.44 × 10−11 m.

B ρ (2.18 × 109 Pa) (1000 kg m3 ) = = 0.434 m. f 3400 Hz

(b) Repeating the above with B = 1.42 × 105 Pa and the density of air gives A = 5.66 × 10−9 m and λ = 0.100 m. EVALUATE: (c) The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense air molecules must have a larger amplitude to transfer the same amount of energy. vp 2 IDENTIFY and SET UP: Use Eq.(16.7) to eliminate either v or B in I = max . 2B EXECUTE:

From Eq. (19.21), v 2 = B ρ . Using Eq.(16.7) to eliminate v, I =

(

)

2 2 B ρ pmax 2 B = pmax 2 ρB .

2(v ρ) = p 2 ρv. Using Eq. (16.7) to eliminate B, I = vp EVALUATE: The equation in this form shows the dependence of I on the density of the material in which the wave propagates. IDENTIFY and SET UP: Apply Eqs.(16.5), (16.11) and (16.15). EXECUTE: (a) ω = 2π f = (2π rad)(150 Hz) = 942.5 rad/s 2 max

16.17.

2

2 max

2π f ω 942.5 rad/s = = = 2.74 rad/m λ v v 344 m/s B = 1.42 × 105 Pa (Example 16.1) Then pmax = BkA = (1.42 × 105 Pa)(2.74 rad/m)(5.00 × 10−6 m) = 1.95 Pa. k=

2π

2

=

(b) Eq.(16.11): I = 12 ω BkA2

I = 12 (942.5 rad/s)(1.42 × 105 Pa)(27.4 rad/m)(5.00 × 10 −6 m) 2 = 4.58 × 10−3 W/m 2 . (c) Eq.(16.15): β = (10 dB)log( I / I 0 ), with I 0 = 1 × 10−12 W/m 2 .

β = (10 dB)log ( (4.58 × 10−3 W/m 2 ) /(1 × 10−12 W/m 2 ) ) = 96.6 dB.

16.18.

EVALUATE: Even though the displacement amplitude is very small, this is a very intense sound. Compare the sound intensity level to the values in Table 16.2. IDENTIFY: Apply β = (10 dB)log (I / I 0 ). In part (b), use Eq.(16.14) to calculate I from the information that is given. SET UP: I 0 = 10−12 W/m 2 . From Table 16.1 the speed of sound in air at 20.0°C is 344 m/s. The density of air at

that temperature is 1.20kg m3 . ⎛ 0.500 μ W/m 2 ⎞ = 57 dB. (a) β = (10 dB) log ⎜ −12 2 ⎟ ⎝ 10 W/m ⎠ p2 (0.150 N m 2 ) 2 (b) I = max = = 2.73 × 10−5 W m 2 . Using this in Equation (16.15), 2ρv 2(1.20 kg m3 )(344 m s) EXECUTE:

2.73 × 10−5 W m 2 = 74.4 dB. 10−12 W m 2 EVALUATE: As expected, the sound intensity is larger for the jack hammer. IDENTIFY: Use Eq.(16.13) to relate I and pmax . β = (10 dB)log( I / I 0 ). Eq.(16.4) says the pressure amplitude and β = (10 dB) log

16.19.

⎛ 2π f displacement amplitude are related by pmax = BkA = B ⎜ ⎝ v SET UP:

⎞ ⎟ A. ⎠

At 20°C the bulk modulus for air is 1.42 × 105 Pa and v = 344 m/s. I 0 = 1 × 10−12 W/m 2 .

Sound and Hearing

(a) I =

EXECUTE:

16.20.

2 vpmax (344 m/s)(6.0 × 10−5 Pa) 2 = = 4.4 × 10−12 W/m 2 2B 2(1.42 × 105 Pa)

⎛ 4.4 × 10−12 W/m 2 ⎞ = 6.4 dB (b) β = (10 dB)log ⎜ −12 2 ⎟ ⎝ 1 × 10 W/m ⎠ vp (344 m/s)(6.0 × 10−5 Pa) = 5.8 × 10−11 m (c) A = max = 2π fB 2π (400 Hz)(1.42 × 105 Pa) EVALUATE: This is a very faint sound and the displacement and pressure amplitudes are very small. Note that the displacement amplitude depends on the frequency but the pressure amplitude does not. IDENTIFY and SET UP: Apply the relation β 2 − β1 = (10 dB)log (I 2 / I1 ) that is derived in Example 16.10.

( )

(a) Δβ = (10 dB)log 4I = 6.0 dB I (b) The total number of crying babies must be multiplied by four, for an increase of 12 kids. EVALUATE: For I 2 = α I1 , where α is some factor, the increase in sound intensity level is Δβ = (10 dB)log α . For α = 4, Δβ = 6.0 dB. IDENTIFY and SET UP: Let 1 refer to the mother and 2 to the father. Use the result derived in Example 16.11 for the difference in sound intensity level for the two sounds. Relate intensity to distance from the source using Eq.(15.26). EXECUTE: From Example 16.11, β 2 − β1 = (10 dB)log( I 2 / I1 ) EXECUTE:

16.21.

16-5

Eq.(15.26): I1 / I 2 = r22 / r12 or I 2 / I1 = r12 / r22

16.22.

Δβ = β 2 − β1 = (10 dB)log( I 2 / I1 ) = (10 dB)log( r1 / r2 ) 2 = (20 dB)log( r1 / r2 ) Δβ = (20 dB)log (1.50 m/0.30 m) = 14.0 dB. EVALUATE: The father is 5 times closer so the intensity at his location is 25 times greater. I I I IDENTIFY: β = (10 dB)log . β 2 − β1 = (10 dB)log 2 . Solve for 2 . I0 I1 I1 If log y = x then y = 10 x. Let β 2 = 70 dB and β1 = 95 dB.

SET UP:

70.0 dB − 95.0 dB = −25.0 dB = (10 dB)log


I2 I I . log 2 = −2.5 and 2 = 10−2.5 = 3.2 × 10−3. I1 I1 I1

I 2 < I1 when β 2 < β1.


From Example 16.11 Δβ = (10 dB)log( I 2 / I1 )

Set Δβ = 13.0 dB and solve for I 2 / I1. EXECUTE: 13.0 dB = 10 dBlog( I 2 / I1 ) so 1.3 = log( I 2 / I1 ) and I 2 / I1 = 20.0. (b) EVALUATE: According to the equation in part (a) the difference in two sound intensity levels is determined by the ratio of the sound intensities. So you don’t need to know I1 , just the ratio I 2 / I1. 16.24.

IDENTIFY: SET UP: open end.

v v . For a stopped pipe, f1 = . v = f λ. 2L 4L v = 344 m/s. For a pipe, there must be a displacement node at a closed end and an antinode at the For an open pipe, f1 =

v 344 m/s = = 0.290 m. 2 f1 2(594 Hz) (b) There is a node at one end, an antinode at the other end and no other nodes or antinodes in between, so EXECUTE:

λ1 4

= L and λ1 = 4 L = 4(0.290 m) = 1.16 m.

(c) f1 =

v 1⎛ v ⎞ 1 = ⎜ ⎟ = (594 Hz) = 297 Hz. 4L 2 ⎝ 2L ⎠ 2

EVALUATE: 16.25.

(a) L =

We could also calculate f1 for the stopped pipe as f1 =

v

λ1

=

344 m/s = 297 Hz, which agrees with 1.16 m

our result in part (a). IDENTIFY and SET UP: An open end is a displacement antinode and a closed end is a displacement node. Sketch the standing wave pattern and use the sketch to relate the node-to-antinode distance to the length of the pipe. A displacement node is a pressure antinode and a displacement antinode is a pressure node.

16-6

Chapter 16

EXECUTE: (a) The placement of the displacement nodes and antinodes along the pipe is as sketched in Figure 16.25a. The open ends are displacement antinodes.

Figure 16.25a

Location of the displacement nodes (N) measured from the left end: fundamental 0.60 m 1st overtone 0.30 m, 0.90 m 2nd overtone 0.20 m, 0.60 m, 1.00 m Location of the pressure nodes (displacement antinodes (A)) measured from the left end: fundamental 0, 1.20 m 1st overtone 0, 0.60 m, 1.20 m 2nd overtone 0, 0.40 m, 0.80 m, 1.20 m (b) The open end is a displacement antinode and the closed end is a displacement node. The placement of the displacement nodes and antinodes along the pipe is sketched in Figure 16.25b.

Figure 16.25b

Location of the displacement nodes (N) measured from the closed end: fundamental 0 1st overtone 0, 0.80 m 2nd overtone 0, 0.48 m, 0.96 m

16.26.

Location of the pressure nodes (displacement antinodes (A)) measured from the closed end: fundamental 1.20 m 1st overtone 0.40 m, 1.20 m 2nd overtone 0.24 m, 0.72 m, 1.20 m EVALUATE: The node-to-node or antinode-to-antinode distance is λ / 2. For the higher overtones the frequency is higher and the wavelength is smaller. IDENTIFY: A pipe closed at one end is a stopped pipe. Apply Eqs.(16.18) and (16.22) to find the frequencies and Eqs.(16.19) and (16.23) to find the highest audible harmonic in each case. SET UP: For the open pipe n = 1, 2, and 3 for the first three harmonics and for the stopped pipe n = 1, 3, and 5. v EXECUTE: (a) f1 = and f n = nf1. 2L 344 m/s f1 = = 382 Hz. f 2 = 764 Hz, f 3 = 1146 Hz, f 4 = 1528 Hz 2(0.450 m) v and f n = nf1 , n = 1, 3, 5, …. 4L 344 m/s f1 = = 191 Hz. f 3 = 573 Hz, f 5 = 955 Hz, f 7 = 1337 Hz 4(0.450 m)

(b) f1 =

(c) open pipe: n =

f 20,000 Hz f 20,000 Hz = = 52. closed pipe: = = 104. But only odd n are present, so f1 382 Hz f1 191 Hz

n = 103. EVALUATE: For an open pipe all harmonics are present. For a stopped pipe only odd harmonics are present. For pipes of a given length, f1 for a stopped pipe is half what it is for an open pipe.

Sound and Hearing

16.27.

16.28.

16-7

IDENTIFY: For a stopped pipe, the standing wave frequencies are given by Eq.(16.22). SET UP: The first three standing wave frequencies correspond to n = 1, 3 and 5. (344 m/s) EXECUTE: f1 = = 506 Hz, f3 = 3 f1 = 1517 Hz, f5 = 5 f1 = 2529 Hz. 4(0.17 m) EVALUATE: All three of these frequencies are in the audible range, which is about 20 Hz to 20,000 Hz. IDENTIFY: Model the auditory canal as a stopped pipe of length L = 2.40 cm. For a stopped pipe, λ1 = 4 L,

v and f n = nf1 , n = 1, 3, 5, …. 4L SET UP: Take the highest audible frequency to be 20,000 Hz. v = 344 m/s. v 344 m/s EXECUTE: (a) f1 = = = 3.58 × 103 Hz. λ1 = 4 L = 4(0.0240 m) = 0.0960 m. This frequency is 4 L 4(0.0240 m) audible. f 20,000 Hz (b) For f = 20,000 Hz, = = 5.6; the highest harmonic which is audible is for n = 5 (fifth harmonic). f1 3580 Hz f1 =

16.29.

f 5 = 5 f1 = 1.79 × 104 Hz. EVALUATE: For a stopped pipe there are no even harmonics. IDENTIFY: For either type of pipe, stopped or open, the fundamental frequency is proportional to the wave speed v. The wave speed is given in turn by Eq.(16.10). SET UP: For He, γ = 5/ 3 and for air, γ = 7 / 5. EXECUTE:

f He = f air

16.30.

γ M

, so

γ He M air (5 3) 28.8 ⋅ = (262 Hz) ⋅ = 767 Hz. γ air M He (7 5) 4.00

(b) No. In either case the frequency is proportional to the speed of sound in the gas. EVALUATE: The frequency is much higher for helium, since the rms speed is greater for helium. IDENTIFY: There must be a node at each end of the pipe. For the fundamental there are no additional nodes and each successive overtone has one additional node. v = f λ . SET UP: v = 344 m/s. The node to node distance is λ / 2. EXECUTE:

16.31.

(a) The fundamental frequency is proportional to the square root of the ratio

(a)

λ1 2

= L so λ1 = 2 L. Each successive overtone adds an additional λ / 2 along the pipe, so

v nv 2L ⎛λ ⎞ , where n = 1, 2, 3, …. f n = n ⎜ n ⎟ = L and λn = = . 2L 2 n λ ⎝ ⎠ n v 344 m/s (b) f1 = = = 68.8 Hz. f 2 = 2 f1 = 138 Hz. f 3 = 3 f1 = 206 Hz. All three of these frequencies are 2 L 2(2.50 m) audible. EVALUATE: A pipe of length L closed at both ends has the same standing wave wavelengths, frequencies and nodal patterns as for a string of length L that is fixed at both ends. IDENTIFY and SET UP: Use the standing wave pattern to relate the wavelength of the standing wave to the length of the air column and then use Eq.(15.1) to calculate f. There is a displacement antinode at the top (open) end of the air column and a node at the bottom (closed) end, as shown in Figure 16.31 EXECUTE: (a)

λ/4 = L λ = 4 L = 4(0.140 m) = 0.560 m f =

v

λ

=

344 m/s = 614 Hz 0.560 m

Figure 16.31 (b) Now the length L of the air column becomes 12 (0.140 m) = 0.070 m and λ = 4 L = 0.280 m.

344 m/s = 1230 Hz λ 0.280 m EVALUATE: Smaller L means smaller λ which in turn corresponds to larger f. f =

v

=

16-8

Chapter 16

16.32.

IDENTIFY:

The wire will vibrate in its second overtone with frequency f 3wire when f 3wire = f1pipe . For a stopped

pipe, f1pipe =

⎛ v ⎞ v . The first overtone standing wave frequency for a wire fixed at both ends is f 3wire = 3⎜ wire ⎟ . 4 Lpipe 2 L ⎝ wire ⎠

vwire = F / μ . SET UP: EXECUTE:

The wire has μ = vwire =

m 7.25 × 10−3 kg = = 8.53 × 10−3 kg/m. The speed of sound in air is v = 344 m/s. Lwire 0.850 m

v v 4110 N = 694 m/s. f 3wire = f1pipe gives 3 wire = . −3 2 Lwire 4 Lpipe 8.53 × 10 kg/m

2 Lwirev 2(0.850 m)(344 m/s) = = 0.0702 m = 7.02 cm. 12vwire 12(694 m/s) EVALUATE: The fundamental for the pipe has the same frequency as the third harmonic of the wire. But the wave speeds for the two objects are different and the two standing waves have different wavelengths. Lpipe =

16.33.

Figure 16.33 (a) IDENTIFY and SET UP: Path difference from points A and B to point Q is 3.00 m − 1.00 m = 2.00 m, as shown in Figure 16.33. Constructive interference implies path difference = nλ , n = 1, 2, 3, … EXECUTE: 2.00 m = nλ so λ = 2.00 m / n v nv n(344 m/s) f = = = = n(172 Hz,) n = 1, 2, 3, … 2.00 m λ 2.00 m The lowest frequency for which constructive interference occurs is 172 Hz. (b) IDENTIFY and SET UP: Destructive interference implies path difference = ( n / 2)λ , n = 1, 3, 5, … EXECUTE: 2.00 m = (n / 2)λ so λ = 4.00 m / n

nv n(344 m/s) = = n(86 Hz), n = 1, 3, 5, …. λ 4.00 m (4.00 m) The lowest frequency for which destructive interference occurs is 86 Hz. EVALUATE: As the frequency is slowly increased, the intensity at Q will fluctuate, as the interference changes between destructive and constructive. IDENTIFY: Constructive interference occurs when the difference of the distances of each source from point P is an integer number of wavelengths. The interference is destructive when this difference of path lengths is a half integer number of wavelengths. SET UP: The wavelength is λ = v f = ( 344 m s ) (206 Hz) = 1.67 m. Since P is between the speakers, x must be f =

16.34.

v

=

in the range 0 to L, where L = 2.00 m is the distance between the speakers. EXECUTE: The difference in path length is Δl = ( L − x) − x = L − 2 x, or x = ( L − Δl ) / 2. For destructive

16.35.

interference, Δl = ( n + (1 2))λ, and for constructive interference, Δl = nλ . (a) Destructive interference: n = 0 gives Δl = 0.835 m and x = 0.58 m. n = 1 gives Δl = −0.835 m and x = 1.42 m. No other values of n place P between the speakers. (b) Constructive interference: n = 0 gives Δl = 0 and x = 1.00 m. n = 1 gives Δl = 1.67 m and x = 0.17 m. n = −1 gives Δl = −1.67 m and x = 1.83 m. No other values of n place P between the speakers. (c) Treating the speakers as point sources is a poor approximation for these dimensions, and sound reaches these points after reflecting from the walls, ceiling, and floor. EVALUATE: Points of constructive interference are a distance λ / 2 apart, and the same is true for the points of destructive interference. IDENTIFY: For constructive interference the path difference is an integer number of wavelengths and for destructive interference the path difference is a half-integer number of wavelengths. SET UP: λ = v / f = (344 m/s) /(688 Hz) = 0.500 m EXECUTE: To move from constructive interference to destructive interference, the path difference must change by λ/2. If you move a distance x toward speaker B, the distance to B gets shorter by x and the distance to A gets longer by x so the path difference changes by 2x. 2 x = λ/2 and x = λ / 4 = 0.125 m. EVALUATE: If you walk an additional distance of 0.125 m farther, the interference again becomes constructive.

Sound and Hearing

16.36.

16-9

IDENTIFY: Destructive interference occurs when the path difference is a half integer number of wavelengths. SET UP: v = 344 m s, so λ = v f = ( 344 m/s ) (172 Hz ) = 2.00 m. If rA = 8.00 m and rB are the distances of the

person from each speaker, the condition for destructive interference is rB − rA = ( n + 12 ) λ, where n is any integer. Requiring rB = rA + ( n + 12 ) λ > 0 gives n + 12 > − rA λ = − ( 8.00 m ) ( 2.00 m ) = −4, so the smallest

EXECUTE:

value of rB occurs when n = −4, and the closest distance to B is rB = 8.00 m + ( −4 + 12 ) ( 2.00 m ) = 1.00 m. 16.37.

EVALUATE: For rB = 1.00 m, the path difference is rA − rB = 7.00 m. This is 3.5λ. IDENTIFY: Compare the path difference to the wavelength. SET UP: λ = v f = ( 344 m s ) ( 860 Hz ) = 0.400 m

The path difference is 13.4 m − 12.0 m = 1.4 m.

EXECUTE:

16.38.

= 3.5. The path difference is a half-

v = 344 m/s, Let λ1 = 6.50 cm and λ2 = 6.52 cm. λ2 > λ1 so f1 > f 2 . ⎛ 1 1 ⎞ v(λ2 − λ1 ) (344 m/s)(0.02 × 10−2 m) = = 16 Hz. There are 16 beats per f1 − f 2 = v ⎜ − ⎟ = λ1λ2 (6.50 × 10−2 m)(6.52 × 10−2 m) ⎝ λ1 λ2 ⎠

EXECUTE:

16.40.

λ

integer number of wavelengths, so the interference is destructive. EVALUATE: The interference is destructive at any point where the path difference is a half-integer number of wavelengths. IDENTIFY: f beat = f1 − f 2 . v = f λ . SET UP:

16.39.

path difference

second. EVALUATE: We could have calculated f1 and f 2 and subtracted, but doing it this way we would have to be careful to retain enough figures in intermediate calculations to avoid round-off errors. v IDENTIFY: f beat = f a − fb . For a stopped pipe, f1 = . 4L SET UP: v = 344 m/s. Let La = 1.14 m and Lb = 1.16 m. Lb > La so f1a > f1b . EXECUTE:

v ⎛ 1 1 ⎞ v( Lb − La ) (344 m/s)(2.00 × 10−2 m) f1a − f1b = ⎜ − ⎟ = = = 1.3 Hz. There are 1.3 beats per 4 ⎝ La Lb ⎠ 4 La Lb 4(1.14 m)(1.16 m)

second. EVALUATE: frequency.

Increasing the length of the pipe increases the wavelength of the fundamental and decreases the

IDENTIFY:

f beat = f − f 0 . f =

v . Changing the tension changes the wave speed and this alters the frequency. 2L

FL 1 F 1 F0 so f = , where F = F0 + ΔF . Let f 0 = . We can assume that ΔF / F0 is very 2 mL 2 mL m small. Increasing the tension increases the frequency, so f beat = f − f 0 .

SET UP:

EXECUTE:

16.41.

v=

(a) f beat = f − f 0 =

1 2 mL

(

)

F0 + ΔF − F0 =

1/ 2 ⎞ 1 F0 ⎛ ⎡ ΔF ⎤ ⎜ ⎢1 + − 1⎟ . ⎥ ⎟ 2 mL ⎜ ⎣ F0 ⎦ ⎝ ⎠

1/ 2

⎡ ΔF ⎤ ⎢1 + ⎥ F0 ⎦ ⎣

=1+

ΔF when 2 F0

⎛ ΔF ⎞ ΔF / F0 is small. This gives that f beat = f 0 ⎜ ⎟. ⎝ 2 F0 ⎠ ΔF 2 f beat 2(1.5 Hz) (b) = = = 0.68%. F0 f0 440 Hz EVALUATE: The fractional change in frequency is one-half the fractional change in tension. ⎛ v + vL ⎞ IDENTIFY: Apply the Doppler shift equation f L = ⎜ ⎟ fS . ⎝ v + vS ⎠ SET UP:

The positive direction is from listener to source. fS = 1200 Hz. f L = 1240 Hz.

⎛ v ⎞ vS f L (−25 m/s)(1240 Hz) vL = 0. vS = −25.0 m/s. f L = ⎜ = = 780 m/s. ⎟ fS gives v = fS − f L 1200 Hz − 1240 Hz ⎝ v + vS ⎠ EVALUATE: f L > fS since the source is approaching the listener. EXECUTE:

16-10

Chapter 16

16.42.

IDENTIFY: Follow the steps of Example 16.19. SET UP: In the first step, vS = +20.0 m/s instead of −30.0 m/s. In the second step, vL = −20.0 m/s instead of +30.0 m/s.

⎛ v ⎞ 340 m/s ⎛ ⎞ fW = ⎜ ⎟ fS = ⎜ ⎟ (300 Hz) = 283 Hz. Then 340 m/s 20.0 m/s v v + + ⎝ ⎠ S ⎠ ⎝ v + v 340 m/s − 20.0 m/s ⎛ ⎛ ⎞ L ⎞ fL = ⎜ ⎟ fW = ⎜ ⎟ (283 Hz) = 266 Hz. 340 m/s ⎝ v ⎠ ⎝ ⎠

EXECUTE:

EVALUATE: When the car is moving toward the reflecting surface, the received frequency back at the source is higher than the emitted frequency. When the car is moving away from the reflecting surface, as is the case here, the received frequency back at the source is lower than the emitted frequency. 16.43.

⎛ v + vL ⎞ Apply the Doppler shift equation f L = ⎜ ⎟ fS . ⎝ v + vS ⎠

IDENTIFY:

The positive direction is from listener to source. fS = 392 Hz.

SET UP:

⎛ v + vL ⎞ ⎛ 344 m/s − 15.0 m/s ⎞ (a) vS = 0. vL = −15.0 m/s. f L = ⎜ ⎟ fS = ⎜ ⎟ (392 Hz) = 375 Hz 344 m/s v + v ⎝ ⎠ S ⎠ ⎝ ⎛ v + vL ⎞ ⎛ 344 m/s + 15.0 m/s ⎞ (b) vS = +35.0 m/s. vL = +15.0 m/s. f L = ⎜ ⎟ fS = ⎜ ⎟ (392 Hz) = 371 Hz v + v ⎝ 344 m/s + 35.0 m/s ⎠ S ⎠ ⎝ (c) f beat = f1 − f 2 = 4 Hz EVALUATE:

The distance between whistle A and the listener is increasing, and for whistle A f L < fS . The

distance between whistle B and the listener is also increasing, and for whistle B f L < fS . 16.44.


Apply Eqs.(16.27) and (16.28) for the wavelengths in front of and behind the source. v 344 m/s = 0.860 m. Then f = v / λ . When the source is at rest λ = = fS 400 Hz EXECUTE:

(a) Eq.(16.27): λ =

(b) Eq.(16.28): λ =

v − vS 344 m/s − 25.0 m/s = = 0.798 m fS 400 Hz

v + vS 344 m/s + 25.0 m/s = = 0.922 m fS 400 Hz

(c) f L = v / λ (since vL = 0), so f L = (344 m/s)/0.798 m = 431 Hz (d) f L = v / λ = (344 m/s)/0.922 m = 373 Hz

16.45.

EVALUATE: In front of the source (source moving toward listener) the wavelength is decreased and the frequency is increased. Behind the source (source moving away from listener) the wavelength is increased and the frequency is decreased. v + vS v − vS IDENTIFY: The distance between crests is λ . In front of the source λ = and behind the source λ = . fS fS

fS = 1/ T . T = 1.6 s. v = 0.32 m/s. The crest to crest distance is the wavelength, so λ = 0.12 m.

SET UP: EXECUTE: (b) λ =

(a) fS = 1/ T = 0.625 Hz. λ =

v − vS gives vS = v − λ fS = 0.32 m/s − (0.12 m)(0.625 Hz) = 0.25 m/s. fS

v + vS 0.32 m/s + 0.25 m/s = = 0.91 m fS 0.625 Hz

EVALUATE: If the duck was held at rest but still paddled its feet, it would produce waves of wavelength 0.32 m/s λ= = 0.51 m. In front of the duck the wavelength is decreased and behind the duck the wavelength is 0.625 Hz increased. The speed of the duck is 78% of the wave speed, so the Doppler effects are large.

Sound and Hearing

16.46.

IDENTIFY: SET UP:

16-11

⎛ v + vL ⎞ Apply f L = ⎜ ⎟ fS . ⎝ v + vS ⎠ fS = 1000 Hz. The positive direction is from the listener to the source. v = 344 m/s.

⎛ v + vL ⎞ 344 m/s ⎛ ⎞ (a) vS = −(344 m/s) / 2 = −172 m/s, vL = 0. f L = ⎜ ⎟ fS = ⎜ ⎟ (1000 Hz) = 2000 Hz 344 m/s 172 m/s v + v − ⎝ ⎠ S ⎝ ⎠ ⎛ v + vL ⎞ ⎛ 344 m/s + 172 m/s ⎞ (b) vS = 0, vL = +172 m/s. f L = ⎜ ⎟ fS = ⎜ ⎟ (1000 Hz) = 1500 Hz 344 m/s ⎝ ⎠ ⎝ v + vS ⎠

16.47.

EVALUATE: (c) The answer in (b) is much less than the answer in (a). It is the velocity of the source and listener relative to the air that determines the effect, not the relative velocity of the source and listener relative to each other. ⎛ v + vL ⎞ IDENTIFY: Apply f L = ⎜ ⎟ fS . ⎝ v + vS ⎠ SET UP:

(

)

⎛ f ⎞ v + vL fS = (1 + vL v ) fS , which gives vL = v ⎜ L − 1⎟ = ( 344 m s ) 490 Hz − 1 = −19.8 m/s. 520 Hz v + vS f ⎝ S ⎠ You must be traveling at 19.8 m/s. EVALUATE: vL < 0 means that the listener is moving away from the source. IDENTIFY: Apply the Doppler effect formula, Eq.(16.29). (a) SET UP: The positive direction is from the listener toward the source, as shown in Figure 16.48a. EXECUTE:

16.48.

The positive direction is from the motorcycle toward the car. The car is stationary, so vS = 0. fL =

Figure 16.48a

⎛ v + vL ⎞ ⎛ 344 m/s + 18.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (262 Hz) = 302 Hz v + v ⎝ 344 m/s − 30.0 m/s ⎠ S ⎠ ⎝ EVALUATE: Listener and source are approaching and f L > fS . (b) SET UP: See Figure 16.48b. EXECUTE:

Figure 16.48b

⎛ v + vL ⎞ ⎛ 344 m/s − 18.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (262 Hz) = 228 Hz v + v ⎝ 344 m/s + 30.0 m/s ⎠ S ⎠ ⎝ EVALUATE: Listener and source are moving away from each other and f L < fS . IDENTIFY: The radar beam consists of electromagnetic waves and Eq.(16.30) applies. Apply the Doppler formula twice, once with the storm as a receiver and then again with the storm as a source. SET UP: c = 3.00 × 108 m/s. When the source and receiver are moving toward each other, as is the case here, then v is negative. EXECUTE:

16.49.

EXECUTE:

Let f ′ be the frequency received by the storm; f ′ =

frequency when the waves are reflected and f R = ⎛c+ v ⎞ ⎛ 2v Δf = f R − f S = ⎜ − 1 ⎟ fS = ⎜ ⎜c− v ⎟ ⎜c− v ⎝ ⎠ ⎝

c+ v c− v

fS . Then f ′ serves as the source

c+ v ⎛ c+ v ⎞ ⎛c+ v ⎞ ⎜ fS ⎟ = ⎜ ⎟ fS . c − v ⎜⎝ c − v ⎟⎠ ⎜⎝ c − v ⎟⎠

⎞ 2(20.1 m/s) ⎡ ⎤ 6 ⎟⎟ fS = ⎢ 8 ⎥ (200.0 × 10 Hz) = 26.8 Hz 3.00 10 m/s 20.1 m/s × − ⎣ ⎦ ⎠

16-12

Chapter 16

⎛ 2v ⎞ Since v ! c, in the expression Δf = ⎜ f it is a very good approximation to replace c − v ⎜ c − v ⎟⎟ S ⎝ ⎠ Δf v v Δf by c and then is very small. Since the storm is approaching the station the final = 2 . is very small, so fS c c fS received frequency is larger than the original transmitted frequency. IDENTIFY: Apply Eq.(16.30). The source is moving away, so v is positive. SET UP: c = 3.00 × 108 m/s. v = +50.0 × 103 m/s. EVALUATE:

16.50.

EXECUTE:

16.51.

c−v 3.00 × 108 m/s − 50.0 × 103 m/s fS = (3.330 × 1014 Hz) = 3.329 × 1014 Hz 3.00 × 108 m/s + 50.0 × 103 m/s c+v f R < fS since the source is moving away. The difference between f R and fS is very small since

fR =

EVALUATE: v ! c. IDENTIFY: Apply Eq.(16.30). SET UP: Require f R = 1.100 fS . Since f R > fS the star would be moving toward us and v < 0, so v = − v .

c = 3.00 × 108 m/s. EXECUTE:

fR =

c+ v c− v

fS . f R = 1.100 fS gives

c+ v = (1.100) 2 . Solving for v gives c− v

⎡(1.100) 2 − 1⎤⎦ c v =⎣ = 0.0950c = 2.85 × 107 m/s. 1 + (1.100) 2 Δf f − fS Δf v v = 9.5%. and are approximately equal. = R = 10.0%. c c fS fS fS IDENTIFY: Apply Eq.(16.31). SET UP: The Mach number is the value of vS / v, where vS is the speed of the shuttle and v is the speed of sound at the altitude of the shuttle. v v 1 = 1.18. EXECUTE: (a) = sin α = sin 58.0° = 0.848. The Mach number is S = v 0.848 vS EVALUATE: 16.52.

v 331 m/s = = 390 m/s sin α sin 58.0° v 344 m/s v 390 m/s = 1.13. The Mach number would be 1.13. sin α = = (c) S = and α = 61.9° v 344 m/s vS 390 m/s EVALUATE: The smaller the Mach number, the larger the angle of the shock-wave cone. IDENTIFY: Apply Eq.(16.31) to calculate α . Use the method of Example 16.20 to calculate t. SET UP: Mach 1.70 means vS / v = 1.70. (b) vS =

16.53.

EXECUTE:

(a) In Eq.(16.31), v vS = 1 1.70 = 0.588 and α = arcsin(0.588) = 36.0°.

(b) As in Example 16.20, t =

16.54.

( 950 m )

(1.70)(344 m s)( tan(36.0°))

= 2.23 s.

EVALUATE: The angle α decreases when the speed vS of the plane increases. IDENTIFY: The displacement y ( x, t ) is given in Eq.(16.1) and the pressure variation is given in Eq.(16.4). The pressure variation is related to the displacement by Eq.(16.3). SET UP: k = 2π / λ EXECUTE: (a) Mathematically, the waves given by Eq.(16.1) and Eq.(16.4) are out of phase. Physically, at a displacement node, the air is most compressed or rarefied on either side of the node, and the pressure gradient is zero. Thus, displacement nodes are pressure antinodes. (b) The graphs have the same form as in Figure 16.3 in the textbook. ∂y ( x, t ) . When y ( x, t ) versus x is a straight line with positive slope, p ( x, t ) is constant and (c) p ( x, t ) = − B ∂x negative. When y ( x, t ) versus x is a straight line with negative slope, p ( x, t ) is constant and positive. The graph of

p( x,0) is given in Figure 16.54. The slope of the straightline segments for y ( x,0) is 1.6 × 10−4 , so for the wave in

Figure 16.42 in the textbook, pmax-non = (1.6 × 10−4 ) B. The sinusoidal wave has amplitude pmax = BkA = (2.5 × 10−4 ) B. The difference in the pressure amplitudes is because the two y ( x,0) functions have different slopes.

Sound and Hearing

16-13

EVALUATE: (d) p ( x, t ) has its largest magnitude where y ( x, t ) has the greatest slope. This is where y ( x, t ) = 0 for a sinusoidal wave but it is not true in general.

Figure 16.54 16.55.

IDENTIFY: The sound intensity level is β = (10 dB)log (I / I 0 ), so the same sound intensity level β means the same intensity I. The intensity is related to pressure amplitude by Eq.(16.13) and to the displacement amplitude by Eq.(16.12). SET UP: v = 344 m/s. ω = 2π f . Each octave higher corresponds to a doubling of frequency, so the note sung by the bass has frequency (932 Hz) / 8 = 116.5 Hz. Let 1 refer to the note sung by the soprano and 2 refer to the note

sung by the bass. I 0 = 1 × 10−12 W/m 2 . 2 vpmax and I1 = I 2 gives pmax,1 = pmax,2 ; the ratio is 1.00. 2B A f (b) I = 12 ρ Bω 2 A2 = 12 ρ B 4π 2 f 2 A2 . I1 = I 2 gives f1 A1 = f 2 A2 . 2 = 1 = 8.00. A1 f 2 I = 107.2 and I = 1.585 × 10−5 W/m 2 . I = 12 ρ B 4π 2 f 2 A2 . (c) β = 72.0 dB gives log( I / I 0 ) = 7.2. I0

EXECUTE:

A=

16.56.

1 2π f

(a) I =

2I 1 = ρ B 2π (932 Hz)

2(1.585 × 10−5 W/m 2 ) (1.20 kg/m3 )(1.42 × 105 Pa)

= 4.73 × 10−8 m = 47.3 nm.

EVALUATE: Even for this loud note the displacement amplitude is very small. For a given intensity, the displacement amplitude depends on the frequency of the sound wave but the pressure amplitude does not. IDENTIFY: Use the equations that relate intensity level and intensity, intensity and pressure amplitude, pressure amplitude and displacement amplitude, and intensity and distance. (a) SET UP: Use the intensity level β to calculate I at this distance. β = (10 dB)log( I / I 0 ) EXECUTE:

52.0 dB = (10 dB)log( I /(10−12 W/m 2 ))

log( I /(10−12 W/m 2 )) = 5.20 implies I = 1.585 × 10−7 W/m 2 SET UP: Then use Eq.(16.14) to calculate pmax : 2 pmax so pmax = 2 ρ vI 2ρv From Example 16.6, ρ = 1.20 kg/m3 for air at 20°C.

I=

EXECUTE:

pmax = 2 ρ vI = 2(1.20 kg/m 3 )(344 m/s)(1.585 × 10 −7 W/m 2 ) = 0.0114 Pa

(b) SET UP:

Eq.(16.5): pmax = BkA so A =

pmax Bk

For air B = 1.42 × 105 Pa (Example 16.1). 2π 2π f (2π rad)(587 Hz) EXECUTE: k = = = = 10.72 rad/m λ v 344 m/s p 0.0114 Pa = 7.49 × 10−9 m A = max = Bk (1.42 × 105 Pa)(10.72 rad/m) (c) SET UP:

β 2 − β1 = (10 dB)log( I 2 / I1 ) (Example 16.11).

Eq.(15.26): I1 / I 2 = r22 / r12 so I 2 / I1 = r12 / r22

β 2 − β1 = (10 dB)log( r1 / r2 ) 2 = (20 dB)log( r1 / r2 ). β 2 = 52.0 dB and r2 = 5.00 m. Then β1 = 30.0 dB and we need to calculate r1.

EXECUTE:

52.0 dB − 30.0 dB = (20 dB)log(r1 / r2 ) 22.0 dB = (20 dB)log(r1 / r2 ) log(r1 / r2 ) = 1.10 so r1 = 12.6r2 = 63.0 m.

16-14

Chapter 16

16.57.

EVALUATE: The decrease in intensity level corresponds to a decrease in intensity, and this means an increase in distance. The intensity level uses a logarithmic scale, so simple proportionality between r and β doesn’t apply. IDENTIFY: The sound is first loud when the frequency f 0 of the speaker equals the frequency f1 of the

γ RT v . v= . The sound M 4L is next loud when the speaker frequency equals the first overtone frequency for the tube. SET UP: A stopped pipe has only odd harmonics, so the frequency of the first overtone is f 3 = 3 f1.

fundamental standing wave for the gas in the tube. The tube is a stopped pipe, and f1 =

EXECUTE:

(a) f 0 = f1 =

1 γ RT 16 L2 Mf 02 v = . This gives T = . γR 4L 4L M

(b) 3 f 0 .

16.58.

EVALUATE:

(c) Measure f 0 and L. Then f 0 =

IDENTIFY:

f beat = f A − f B . f1 =

v and v = 2L

v gives v = 4 Lf 0 . 4L FL 1 F gives f1 = . Apply m 2 mL

∑τ

z

= 0 to the bar to find the

tension in each wire. SET UP: For ∑τ z = 0 take the pivot at wire A and let counterclockwise torques be positive. The free-body diagram for the bar is given in Figure 16.58. Let L be the length of the bar. EXECUTE: ∑τ z = 0 gives FB L − wlead (3L / 4) − wbar ( L / 2) = 0. FB = 3wlead / 4 + wbar / 2 = 3(185 N) / 4 + (165 N) / 2 = 221 N. FA + FB = wbar + wlead so FA = wbar + wlead − FB = 165 N + 185 N − 221 N = 129 N. f1 A =

1 129 N = 88.4 Hz. 2 (5.50 × 10−3 kg)(0.750 m)

221 N = 115.7 Hz. f beat = f1B − f1 A = 27.3 Hz. 129 N EVALUATE: The frequency increases when the tension in the wire increases. f1B = f1 A

Figure 16.58 16.59.

IDENTIFY: The flute acts as a stopped pipe and its harmonic frequencies are given by Eq.(16.23). The resonant frequencies of the string are f n = nf1 , n = 1, 2, 3,... The string resonates when the string frequency equals the flute frequency. SET UP: For the string f1s = 600.0 Hz. For the flute, the fundamental frequency is

344.0 m s v = = 800.0 Hz. Let nf label the harmonics of the flute and let ns label the harmonics of the 4 L 4(0.1075 m) string. EXECUTE: For the flute and string to be in resonance, nf f1f = ns f1s , where f1s = 600.0 Hz is the fundamental f1f =

frequency for the string. ns = nf ( f1f f1s ) = 43 nf . ns is an integer when nf = 3N , N = 1, 3, 5, … (the flute has only odd harmonics). nf = 3N gives ns = 4 N Flute harmonic 3N resonates with string harmonic 4 N , N = 1, 3, 5, …. EVALUATE: We can check our results for some specific values of N. For N = 1, nf = 3 and f 3f = 2400 Hz. For this N, ns = 4 and f 4s = 2400 Hz. For N = 3, nf = 9 and f 9f = 7200 Hz, and ns = 12, f12s = 7200 Hz. Our general results do give equal frequencies for the two objects.

Sound and Hearing

16.60.

IDENTIFY:

16-15

⎛ v ⎞ The harmonics of the string are f n = nf1 = n ⎜ ⎟ , where l is the length of the string. The tube is a ⎝ 2l ⎠

stopped pipe and its standing wave frequencies are given by Eq.(16.22). For the string, v = F / μ , where F is the tension in the string. SET UP: The length of the string is d = L 10, so its third harmonic has frequency f3string = 3 1 F μ . The 2d vs pipe stopped pipe has length L, so its first harmonic has frequency f1 = . 4L 1 EXECUTE: (a) Equating f1string and f1pipe and using d = L 10 gives F = μ vs2 . 3600 (b) If the tension is doubled, all the frequencies of the string will increase by a factor of 2. In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound produced by the instrument will be diminished. (c) The string will be in resonance with a standing wave in the pipe when their frequencies are equal. Using f1pipe = 3 f1string , the frequencies of the pipe are nf1pipe = 3nf1string , (where n = 1, 3, 5, …). Setting this equal to the

16.61.

frequencies of the string n′f1string , the harmonics of the string are n′ = 3n = 3, 9, 15, … The nth harmonic of the pipe is in resonance with the 3nth harmonic of the string. EVALUATE: Each standing wave for the air column is in resonance with a standing wave on the string. But the reverse is not true; not all standing waves of the string are in resonance with a harmonic of the pipe. IDENTIFY and SET UP: The frequency of any harmonic is an integer multiple of the fundamental. For a stopped pipe only odd harmonics are present. For an open pipe, all harmonics are present. See which pattern of harmonics fits to the observed values in order to determine which type of pipe it is. Then solve for the fundamental frequency and relate that to the length of the pipe. EXECUTE: (a) For an open pipe the successive harmonics are f n = nf1 , n = 1, 2, 3, …. For a stopped pipe the successive harmonics are f n = nf1 , n = 1, 3, 5, …. If the pipe is open and these harmonics are successive, then

f n = nf1 = 1372 Hz and f n +1 = (n + 1) f1 = 1764 Hz. Subtract the first equation from the second: 1372 Hz = 3.5. But n must be an integer, so 392 Hz the pipe can’t be open. If the pipe is stopped and these harmonics are successive, then f n = nf1 = 1372 Hz and

(n + 1) f1 − nf1 = 1764 Hz − 1372 Hz. This gives f1 = 392 Hz. Then n =

f n + 2 = (n + 2) f1 = 1764 Hz (in this case successive harmonics differ in n by 2). Subtracting one equation from the other gives 2 f1 = 392 Hz and f1 = 196 Hz. Then n = 1372 Hz / f1 = 7 so 1372 Hz = 7 f1 and 1764 Hz = 9 f1. The solution gives integer n as it should; the pipe is stopped. (b) From part (a) these are the 7th and 9th harmonics. (c) From part (a) f1 = 196 Hz. 344 m/s v v and L = = = 0.439 m. 4L 4 f1 4(196 Hz) EVALUATE: It is essential to know that these are successive harmonics and to realize that 1372 Hz is not the fundamental. There are other lower frequency standing waves; these are just two successive ones. IDENTIFY: The steel rod has standing waves much like a pipe open at both ends, since the ends are both displacement nv antinodes. An integral number of half wavelengths must fit on the rod, that is, f n = , with n = 1, 2, 3, …. 2L SET UP: Table 16.1 gives v = 5941 m/s for longitudinal waves in steel. EXECUTE: (a) The ends of the rod are antinodes because the ends of the rod are free to oscillate. (b) The fundamental can be produced when the rod is held at the middle because a node is located there. (1)( 5941 m s ) = 1980 Hz (c) f1 = 2 (1.50 m )

For a stopped pipe f1 =

16.62.

(d) The next harmonic is n = 2, or f 2 = 3961 Hz. We would need to hold the rod at an n = 2 node, which is located

at L 4 = 0.375 m from either end. EVALUATE: For the 1.50 m long rod the wavelength of the fundamental is x = 2 L = 3.00 m. The node to antinode distance is λ / 4 = 0.75 m. For the second harmonic λ = L = 1.50 m and the node to antinode distance is 0.375 m. There is a node at the middle of the rod, but forcing a node at 0.375 m from one end, by holding the rod there, prevents rod from vibrating in the fundamental.

16-16

Chapter 16

16.63.

IDENTIFY: The shower stall can be modeled as a pipe closed at both ends, and hence there are nodes at the two end walls. Figure 15.23 in the textbook shows standing waves on a string fixed at both ends but the sequence of harmonics is the same, namely that an integral number of half wavelengths must fit in the stall. SET UP: The first three normal modes correspond to one half, two halves or three halves of a wavelength along the length of the air column. EXECUTE: (a) The condition for standing waves is f n = nv , so the first three harmonics are for n = 1, 2, 3. 2L (b) A particular physics professor’s shower has a length of L = 1.48 m. Using f n = nv and v = 344 m s gives 2L resonant frequencies 116 Hz, 232 Hz and 349 Hz. Note that the fundamental and second harmonic, which would have the greatest amplitude, are frequencies typically in the normal range of male singers. Hence, men do sing better in the shower! (For a further discussion of resonance and the human voice, see Thomas D. Rossing, The Science of Sound, Second Edition, Addison-Wesley, 1990, especially Chapters 4 and 17.) EVALUATE: The standing wave frequencies for a pipe closed at both ends are the same as for an open pipe of the same length, even though the nodal patterns are different. IDENTIFY: Stress is F / A, where F is the tension in the string and A is its cross sectional area.

16.64.

SET UP:

A = π r 2 . For a string fixed at each end, f1 =

v 1 F 1 F = = 2 L 2 L μ 2 mL

EXECUTE: (a) The cross-section area of the string would be A = (900 N) (7.0 × 108 Pa) = 1.29 × 10−6 m 2 , corresponding to a radius of 0.640 mm. The length is the volume divided by the area, and the volume is V = m / ρ , so L=

V m ρ (4.00 × 10 −3 kg) = = = 0.40 m. A A (7.8 × 103 kg m 3 )(1.29 × 10 −6 m 2 )

(b) For the maximum tension of 900 N, f1 =

16.65.

1 900 N = 375 Hz, or 380 Hz to two figures. 2 (4.00 × 10−3 kg)(0.40 m)

EVALUATE: The string could be shorter and thicker. A shorter string of the same mass would have a higher fundamental frequency. IDENTIFY and SET UP: There is a node at the piston, so the distance the piston moves is the node to node distance, λ / 2. Use Eq.(15.1) to calculate v and Eq.(16.10) to calculate γ from v. EXECUTE: (a) λ / 2 = 37.5 cm, so λ = 2(37.5 cm) = 75.0 cm = 0.750 m. v = f λ = (500 Hz)(0.750 m) = 375 m/s (b) v = γ RT / M (Eq.16.10) Mv 2 (28.8 × 10 −3 kg/mol)(375 m/s) 2 = = 1.39. RT (8.3145 J/mol ⋅ K)(350 K) (c) EVALUATE: There is a node at the piston so when the piston is 18.0 cm from the open end the node is inside the pipe, 18.0 cm from the open end. The node to antinode distance is λ / 4 = 18.8 cm, so the antinode is 0.8 cm beyond the open end of the pipe. The value of γ we calculated agrees with the value given for air in Example 16.5. IDENTIFY: Model the auditory canal as a stopped pipe with length 2.5 cm. SET UP: The frequencies of a stopped pipe are given by Eq.(16.22). EXECUTE: (a) The frequency of the fundamental is f1 = v/4 L = (344 m/s) /[4(0.025 m)] = 3440 Hz. 3500 Hz is near the resonant frequency, and the ear will be sensitive to this frequency. (b) The next resonant frequency would be 3 f1 = 10,500 Hz and the ear would be sensitive to sounds with frequencies close to this value. But 7000 Hz is not a resonant frequency for a stopped pipe and the ear is not sensitive at this frequency. EVALUATE: For a stopped pipe only odd harmonics are present. IDENTIFY: The tuning fork frequencies that will cause this to happen are the standing wave frequencies of the v F wire. For a wire of mass m, length L and with tension F the fundamental frequency is f1 = . The = 2L 4mL standing wave frequencies are f n = nf1 , n = 1, 2, 3, …

γ=

16.66.

16.67.

SET UP: EXECUTE:

F = Mg , where M = 0.420 kg. The mass of the wire is m = ρV = ρ Lπ d 2 / 4, where d is the diameter.

(a) f1 =

F Mg (420.0 × 10 −3 kg)(9.80 m s 2 ) = = = 77.3 Hz. 4mL π d 2 L2 ρ π (225 × 10−6 m) 2 (0.45 m) 2 (21.4 × 103 kg m 3 )

The tuning fork frequencies for which the fork would vibrate are integer multiples of 77.3 Hz.

Sound and Hearing

16.68.

16.69.

EVALUATE: (b) The ratio m M ≈ 9 × 10−4 , so the tension does not vary appreciably along the string due to the mass of the wire. Also, the suspended mass has a large inertia compared to the mass of the wire and assuming that it is stationary is an excellent approximation. v IDENTIFY: For a stopped pipe the frequency of the fundamental is f1 = . The speed of sound in air depends on 4L temperature, as shown by Eq.(16.10). SET UP: Example 16.5 shows that the speed of sound in air at 20°C is 344 m/s. v 344 m/s = 0.246 m = EXECUTE: (a) L = 4 f 4(349 Hz) (b) The frequency will be proportional to the speed, and hence to the square root of the Kelvin temperature. The temperature necessary to have the frequency be higher is (293.15 K)([370 Hz]/[349 Hz]) 2 = 329.5 K, which is 56.3°C. EVALUATE: 56.3°C = 133°F, so this extreme rise in pitch won't occur in practical situations. But changes in temperature can have noticeable effects on the pitch of the organ notes. γ RT IDENTIFY: v = f λ . v = . Solve for γ . M SET UP: The wavelength is twice the separation of the nodes, so λ = 2 L, where L = 0.200 m. EXECUTE:

v = λ f = 2 Lf =

γ=

16.70.

16-17

γ RT M

. Solving for γ ,

M (16.0 × 10−3 kg/mol) 2 (2 Lf ) 2 = ( 2(0.200 m)(1100 Hz) ) = 1.27. RT (8.3145 J mol ⋅ K) (293.15 K)

EVALUATE: This value of γ is smaller than that of air. We will see in Chapter 19 that this value of γ is a typical value for polyatomic gases. IDENTIFY: Destructive interference occurs when the path difference is a half-integer number of wavelengths. Constructive interference occurs when the path difference is an integer number of wavelengths. v 344 m/s SET UP: λ = = = 0.439 m f 784 Hz EXECUTE: (a) If the separation of the speakers is denoted h, the condition for destructive interference is

x 2 + h 2 − x = βλ , where β is an odd multiple of one-half. Adding x to both sides, squaring, canceling the x 2 term from both sides and solving for x gives x =

h2 β − λ . Using λ = 0.439 m and h = 2.00 m yields 9.01 m 2βλ 2

for β = 12 , 2.71 m for β = 32 , 1.27 m for β = 52 , 0.53 m for β = 72 , and 0.026 m for β = 92 . These are the only allowable values of β that give positive solutions for x. (b) Repeating the above for integral values of β , constructive interference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m. Note that these are between, but not midway between, the answers to part (a). (c) If h = λ 2, there will be destructive interference at speaker B. If λ 2 > h, the path difference can never be as large as λ 2. (This is also obtained from the above expression for x, with x = 0 and β = 12 .) The minimum

16.71.

frequency is then v 2h = (344 m s) (4.0 m) = 86 Hz. EVALUATE: When f increases, λ is smaller and there are more occurrences of points of constructive and destructive interference. ⎛ v + vL ⎞ IDENTIFY: Apply f L = ⎜ ⎟ fS . ⎝ v + vS ⎠ SET UP:

The positive direction is from the listener to the source. (a) The wall is the listener. vS = −30 m/s.

vL = 0. f L = 600 Hz. (b) The wall is the source and the car is the listener. vS = 0. vL = +30 m/s. fS = 600 Hz. ⎛ v + vL ⎞ ⎛ v + vS ⎞ ⎛ 344 m/s − 30 m/s ⎞ (a) f L = ⎜ ⎟ fS . fS = ⎜ ⎟ fL = ⎜ ⎟ (600 Hz) = 548 Hz + v v v + v 344 m/s ⎝ ⎠ ⎝ S ⎠ L ⎠ ⎝ ⎛ v + vL ⎞ ⎛ 344 m/s + 30 m/s ⎞ (b) f L = ⎜ ⎟ fS = ⎜ ⎟ (600 Hz) = 652 Hz 344 m/s ⎝ ⎠ ⎝ v + vS ⎠ EXECUTE:

EVALUATE: Since the singer and wall are moving toward each other the frequency received by the wall is greater than the frequency sung by the soprano, and the frequency she hears from the reflected sound is larger still.

16-18

Chapter 16

16.72.

IDENTIFY:

⎛ v + vL ⎞ Apply f L = ⎜ ⎟ fS . The wall first acts as a listener and then as a source. ⎝ v + vS ⎠ SET UP: The positive direction is from listener to source. The bat is moving toward the wall so the Doppler effect increases the frequency and the final frequency received, f L2 , is greater than the original source frequency, fS1. fS1 = 2000 Hz. f L2 − fS1 = 10.0 Hz. ⎛ v + vL ⎞ The wall receives the sound: fS = fS1. f L = f L1. vS = −vbat and vL = 0. f L = ⎜ ⎟ fS gives ⎝ v + vS ⎠ ⎛ v ⎞ f L1 = ⎜ ⎟ fS1. The wall receives the sound: fS2 = f L1. vS = 0 and vL = + vbat . ⎝ v − vbat ⎠ ⎛ v + vbat ⎞ ⎛ v + vbat ⎞ ⎛ v + vbat ⎞ ⎛ v ⎞ f L2 = ⎜ ⎟ fS1 = ⎜ ⎟ fS1. ⎟ fS2 = ⎜ ⎟⎜ ⎝ v ⎠ ⎝ v ⎠ ⎝ v − vbat ⎠ ⎝ v − vbat ⎠

EXECUTE:

⎛ v + vbat ⎞ ⎛ 2vbat ⎞ vΔf (344 m/s)(10.0 Hz) = = 0.858 m/s. − 1⎟ fS1 = ⎜ f L2 − fS1 = Δf = ⎜ ⎟ fS1. vbat = 2 fS1 + Δf 2(2000 Hz) + 10.0 Hz ⎝ v − vbat ⎠ ⎝ v − vbat ⎠ ⎛ 2v ⎞ fS1 < Δf , so we can write our result as the approximate but accurate expression Δf = ⎜ bat ⎟ f . ⎝ v ⎠ IDENTIFY and SET UP: Use Eq.(16.12) for the intensity and Eq.(16.14) to relate the intensity and pressure amplitude. EXECUTE: (a) The amplitude of the oscillations is ΔR. EVALUATE: 16.73.

I=

ρ B (2π f ) 2 A2 = 2 ρ Bπ 2 f 2 (ΔR ) 2

1 2

(b) P = I (4π R 2 ) = 8π 3 ρ B f 2 R 2 (ΔR) 2 (c) I R / I d = d 2 / R 2

I d = ( R / d ) 2 I R = 2π 2 ρ B ( Rf / d ) 2 (ΔR) 2 2 I = pmax / 2 ρ B so

pmax = A=

(2

)

ρ BI = 2π ρ B ( Rf / d ) ΔR

pmax pmax λ p v = = max = v ρ / B ( R / d ) ΔR Bk B 2π B 2π f

But v = B / ρ so v ρ / B = 1 so A = ( R / d ) ΔR. EVALUATE: 16.74.

IDENTIFY:

The pressure amplitude and displacement amplitude fall off like 1/ d and the intensity like 1/ d 2 . ⎛ v + vL ⎞ Apply f L = ⎜ ⎟ fS . The heart wall first acts as the listener and then as the source. ⎝ v + vS ⎠

SET UP: The positive direction is from listener to source. The heart wall is moving toward the receiver so the Doppler effect increases the frequency and the final frequency received, f L2 , is greater than the source frequency,

fS1. f L2 − fS1 = 85 Hz. EXECUTE:

⎛ v + vL ⎞ Heart wall receives the sound: fS = fS1. f L = f L1. vS = 0. vL = −vwall . f L = ⎜ ⎟ fS gives ⎝ v + vS ⎠

⎛ v − vwall ⎞ f L1 = ⎜ ⎟ fS1. Heart wall emits the sound: fS2 = f L1. vS = +vwall . vL = 0. ⎝ v ⎠ ⎛ v ⎞ ⎛ v ⎞ ⎛ v − vwall ⎞ ⎛ v − vwall ⎞ ⎛ v − vwall ⎞ ⎛ 2vwall ⎞ f L2 = ⎜ ⎟ fS2 = ⎜ ⎟⎜ ⎟ fS1. f L2 − fS1 = ⎜1 − ⎟ fS1 = ⎜ ⎟ fS1. ⎟ fS1 = ⎜ ⎝ v + vwall ⎠ ⎝ v + vwall ⎠ ⎝ v ⎠ ⎝ v + vwall ⎠ ⎝ v + vwall ⎠ ⎝ v + vwall ⎠ ( f L2 − fS1 )v ( f − f )v (85 Hz)(1500 m/s) = 0.0319 m/s = 3.19 cm/s. vwall = . fS1 " f L2 − fS1 and vwall = L2 S1 = 2 fS1 − ( f L2 − fS1 ) 2 fS1 2(2.00 × 106 Hz) EVALUATE: fS1 = 2.00 × 106 Hz and f L2 − fS1 = 85 Hz, so the approximation we made is very accurate. Within this approximation, the frequency difference between the reflected and transmitted waves is directly proportional to the speed of the heart wall.

Sound and Hearing

16.75.

16-19

(a) IDENTIFY and SET UP: Use Eq.(15.1) to calculate λ. v 1482 m/s EXECUTE: λ = = = 0.0674 m f 22.0 × 103 Hz (b) IDENTIFY: Apply the Doppler effect equation, Eq.(16.29). The Problem-Solving Strategy in the text (Section 16.8) describes how to do this problem. The frequency of the directly radiated waves is fS = 22,000 Hz. The moving whale first plays the role of a moving listener, receiving waves with frequency f L′. The whale then acts as

a moving source, emitting waves with the same frequency, fS′ = f L′ with which they are received. Let the speed of the whale be vW . SET UP: whale receives waves (Figure 16.75a) EXECUTE:

vL = + vW

⎛ v + vL ⎞ ⎛ v + vW ⎞ f L′ = fS ⎜ ⎟ = fS ⎜ ⎟ v v + ⎝ v ⎠ S ⎠ ⎝ Figure 16.75a SET UP:

whale re-emits the waves (Figure 16.75b) EXECUTE:

vS = −vW

⎛ v + vL ⎞ ⎛ v ⎞ f L = fS ⎜ ⎟ = fS′ ⎜ ⎟ ⎝ v + vS ⎠ ⎝ v − vW ⎠ Figure 16.75b

⎛ v + vW ⎞ ⎛ v + vW ⎞ ⎛ v ⎞ But fS′ = f L′ so f L = fS ⎜ ⎟ = fS ⎜ ⎟. ⎟⎜ ⎝ v ⎠ ⎝ v − vW ⎠ ⎝ v − vW ⎠ ⎛ v + vW ⎞ ⎛ v − vW − v − vW ⎞ −2 fSvW Then Δf = fS − f L = fS ⎜1 − . ⎟ = fS ⎜ ⎟= v − vW ⎝ v − vW ⎠ ⎝ ⎠ v − vW −2(2.20 × 104 Hz)(4.95 m/s) = 147 Hz. 1482 m/s − 4.95 m/s EVALUATE: Listener and source are moving toward each other so frequency is raised. ⎛ v + vL ⎞ IDENTIFY: Apply the Doppler effect formula f L = ⎜ ⎟ fS . In the SHM the source moves toward and away ⎝ v + vS ⎠ Δf =

16.76.

from the listener, with maximum speed ωp Ap . SET UP: The direction from listener to source is positive. EXECUTE: (a) The maximum velocity of the siren is ωP AP = 2ω f P AP . You hear a sound with frequency

f L = fsiren v ( v + vS ) , where vS varies between +2π f P AP and −2π f P AP . f L − max = f siren v ( v − 2π f P AP ) and f L − min = fsiren v ( v + 2π f P AP ) .

(b) The maximum (minimum) frequency is heard when the platform is passing through equilibrium and moving up (down). EVALUATE: When the platform is moving upward the frequency you hear is greater than f siren and when it is

moving downward the frequency you hear is less than fsiren . When the platform is at its maximum displacement 16.77.

from equilibrium its speed is zero and the frequency you hear is fsiren . IDENTIFY: Follow the method of Example 16.19 and apply the Doppler shift formula twice, once with the insect as the listener and again with the insect as the source. SET UP: Let vbat be the speed of the bat, vinsect be the speed of the insect, and f i be the frequency with which the sound waves both strike and are reflected from the insect. The positive direction in each application of the Doppler shift formula is from the listener to the source.

16-20

Chapter 16

EXECUTE: The frequencies at which the bat sends and receives the signals are related by ⎛ v + vbat ⎞ ⎛ v + vinsect ⎞⎛ v + vbat ⎞ f L = fi ⎜ ⎟ = fS ⎜ ⎟⎜ ⎟ . Solving for vinsect , v v − insect ⎠ ⎝ ⎝ v − vbat ⎠⎝ v − vinsect ⎠

vinsect

⎡ ⎢1 − = v⎢ ⎢ ⎢1 + ⎢⎣

fS ⎛ v + vbat ⎞ ⎤ ⎜ ⎟⎥ ⎡ f ( v − vbat ) − fS ( v + vbat ) ⎤ f L ⎝ v − vbat ⎠ ⎥ = v⎢ L ⎥. fS ⎛ v + vbat ⎞ ⎥ ⎢⎣ f L ( v − vbat ) + fS ( v + vbat ) ⎦⎥ ⎜ ⎟⎥ f L ⎝ v − vbat ⎠ ⎥⎦

Letting f L = f refl and fS = f bat gives the result. (b) If f bat = 80.7 kHz, f refl = 83.5 kHz, and vbat = 3.9 m s, then vinsect = 2.0 m s. 16.78.

EVALUATE: f refl > f bat because the bat and insect are approaching each other. IDENTIFY: Follow the steps specified in the problem. v is positive when the source is moving away from the receiver and v is negative when the source is moving toward the receiver. f L − f R is the beat frequency. SET UP:

The source and receiver are approaching, so f R > fS and f R − fS = 46.0 Hz. 12

EXECUTE:

(a) f R = f L

c−v 1− v / c ⎛ v⎞ ⎛ v⎞ = fS = fS ⎜ 1 − ⎟ ⎜ 1 + ⎟ c+v 1+ v / c ⎝ c⎠ ⎝ c⎠

−1 2

.

(b) For small x, the binomial theorem (see Appendix B) gives (1 − x )

12

≈ 1 − x 2, (1 + x )

−1 2

≈ 1 − x 2. Therefore

2

16.79.

v ⎞ 2 ⎛ ⎛ v⎞ f L ≈ fS ⎜1 − ⎟ ≈ fS ⎜1 − ⎟ , where the binomial theorem has been used to approximate (1 − x 2 ) ≈ 1 − x. ⎝ 2c ⎠ ⎝ c⎠ (c) For an airplane, the approximation v ! c is certainly valid. Solving the expression found in part (b) for v , f − fR f −46.0 Hz v=c S = c beat = (3.00 × 108 m s) = −56.8 m s. The speed of the aircraft is 56.8 m/s. fS fS 2.43 × 108 Hz EVALUATE: The approximation v ! c is seen to be valid. v is negative because the source and receiver are Δf , is very small. approaching. Since v ! c, the fractional shift in frequency, f IDENTIFY: Apply the result derived in part (b) of Problem 16.78. The radius of the nebula is R = vt , where t is the time since the supernova explosion. SET UP: When the source and receiver are moving toward each other, v is negative and f R > fS . The light from the explosion reached earth 952 years ago, so that is the amount of time the nebula has expanded. 1 ly = 9.46 × 1015 m. EXECUTE:

(a) v = c

fS − f R −0.018 × 1014 Hz = (3.00 × 108 m s) = −1.2 × 106 m s, with the minus sign indicating fS 4.568 × 108 Hz

that the gas is approaching the earth, as is expected since f R > fS .

16.80.

(b) The radius is (952 yr)(3.156 × 107 s yr)(1.2 × 106 m s) = 3.6 × 1016 m = 3.8 ly. (c) The ratio of the width of the nebula to 2π times the distance from the earth is the ratio of the angular width (taken as 5 arc minutes) to an entire circle, which is 60 × 360 arc minutes. The distance to the nebula is then (60)(360) 2(3.75 ly) = 5.2 × 103 ly. The time it takes light to travel this distance is 5200 yr, so the explosion 5 actually took place 5200 yr before 1054 C.E., or about 4100 B.C.E. Δf v is accurate. EVALUATE: = 4.0 × 10−3 , so even though v is very large the approximation required for v = c c f IDENTIFY and SET UP: Use Eq.(16.30) that describes the Doppler effect for electromagnetic waves. v ! c, so the simplified form derived in Problem 16.78b can be used. (a) EXECUTE: From Problem 16.78b, f R = fS (1 − v / c). v is negative since the source is approaching: v = −(42.0 km/h)(1000 m/1 km)(1 h/3600 s) = −11.67 m/s Approaching means that the frequency is increased. ⎛ v⎞ ⎛ −11.67 m/s ⎞ Δf = fS ⎜ − ⎟ = 2800 × 106 Hz ⎜ − ⎟ = 109 Hz 8 ⎝ c⎠ ⎝ 3.00 × 10 m/s ⎠

Sound and Hearing

16-21

(b) EVALUATE: Approaching, so the frequency is increased. v ! c and therefore Δf / fS ! 1. The frequency of the waves received and reflected by the water is very close to 2880 MHz, so get an additional shift of 109 Hz and the total shift in frequency is 2(109 Hz) = 218 Hz. 16.81.

IDENTIFY: Follow the method of Example 16.19 and apply the Doppler shift formula twice, once for the wall as a listener and then again with the wall as a source. SET UP: In each application of the Doppler formula, the positive direction is from the listener to the source v EXECUTE: (a) The wall will receive and reflect pulses at a frequency f 0 , and the woman will hear this v − vw

reflected wave at a frequency

⎛ v + vw ⎞ ⎛ 2vw v + vw v v + vw −1 ⎟ = f0 ⎜ f0 = f 0 . The beat frequency is f beat = f 0 ⎜ v v − vw v − vw ⎝ v − vw ⎠ ⎝ v − vw

⎞ ⎟. ⎠

(b) In this case, the sound reflected from the wall will have a lower frequency, and using f 0 (v − vw )/(v + vw ) as the

16.82.

⎛ v − vw ⎞ ⎛ 2vw ⎞ detected frequency. vw is replaced by −vw in the calculation of part (a) and f beat = f 0 ⎜1 − ⎟ = f0 ⎜ ⎟. ⎝ v + vw ⎠ ⎝ v + vw ⎠ EVALUATE: The beat frequency is larger when she runs toward the wall, even though her speed is the same in both cases. IDENTIFY and SET UP: Use Fig.(16.37) to relate α and T. Use this in Eq.(16.31) to eliminate sin α . sin α sin α EXECUTE: Eq.(16.31): sin α = v / vS From Fig.16.37 tan α = h / vST . And tan α = . = cos α 1 − sin 2 α Combining these equations we get 1 − (v / vS ) 2 = vS =

h v h v / vS = and = . T vST 1 − (v / vS ) 2 1 − (v / vS ) 2

v 2T 2 v2 and vS2 = 2 2 2 h 1 − v T / h2

hv

as was to be shown. h − v 2T 2 EVALUATE: For a given h, the faster the speed vS of the plane, the greater is the delay time T. The maximum 16.83.

2

delay time is h / v, and T approaches this value as vS → ∞. T → 0 as v → vS . IDENTIFY: The phase of the wave is determined by the value of x − vt , so t increasing is equivalent to x decreasing with t constant. The pressure fluctuation and displacement are related by Eq.(16.3). 1 SET UP: y ( x, t ) = − ∫ p ( x, t ) dx. If p ( x, t ) versus x is a straight line, then y ( x, t ) versus x is a parabola. For air, B B = 1.42 × 105 Pa. EXECUTE: (a) The graph is sketched in Figure 16.83a. (b) From Eq.(16.4), the function that has the given p ( x, 0) at t = 0 is given graphically in Figure 16.83b. Each section is a parabola, not a portion of a sine curve. The period is λ v = (0.200 m) (344 m s) = 5.81 × 10−4 s and the amplitude is equal to the area under the p versus x curve between x = 0 and x = 0.0500 m divided by B, or 7.04 × 10−6 m. (c) Assuming a wave moving in the + x -direction, y (0, t ) is as shown in Figure 16.83c. (d) The maximum velocity of a particle occurs when a particle is moving through the origin, and the particle speed ∂y pv is v y = − v = . The maximum velocity is found from the maximum pressure, and ∂x B v ymax = (40 Pa)(344 m/s) /(1.42 × 105 Pa) = 9.69 cm/s. The maximum acceleration is the maximum pressure gradient divided by the density, amax =

(80.0 Pa) (0.100 m) = 6.67 × 102 m s 2 . (1.20 kg m3 )

(e) The speaker cone moves with the displacement as found in part (c ); the speaker cone alternates between moving forward and backward with constant magnitude of acceleration (but changing sign). The acceleration as a function of time is a square wave with amplitude 667 m/s 2 and frequency f = v/λ = (344 m/s) / (0.200 m) = 1.72 kHz.

16-22

Chapter 16

EVALUATE: We can verify that p ( x, t ) versus x has a shape proportional to the slope of the graph of y ( x, t ) versus x. The same is also true of the graphs versus t.

Figure 16.83 16.84.

IDENTIFY:

At a distance r from a point source with power output P, I =

P . β = (10 dB)log( I / I 0 ). For two 4π r 2

sources the amplitudes are combined according to the phase difference. SET UP: The amplitude is proportional to the square root of the intensity. Taking the speed of sound to be 344 m s, the wavelength of the waves emitted by each speaker is 2.00 m. EXECUTE: (a) Point C is two wavelengths from speaker A and one and one-half from speaker B, and so the phase difference is 180° = π rad. P 8.00 × 10−4 W (b) I = = = 3.98 × 10−6 W m 2 and the sound intensity level is 2 4π r 4π (4.00 m) 2 (10 dB)log (3.98 × 106 ) = 66.0 dB. Repeating with P = 6.00 × 10−5 W and r = 3.00 m gives I = 5.31 × 10−7 W/m 2 and β = 57.2 dB. (c) With the result of part (a), the amplitudes, either displacement or pressure, must be subtracted. That is, the intensity is found by taking the square roots of the intensities found in part (b), subtracting, and squaring the difference. The result is that I = 1.60 × 10−6 W/m 2 and β = 62.1 dB. EVALUATE: Subtracting the intensities of A and B gives 3.98 × 10−6 W/m 2 − 5.31 × 10−7 W/m 2 = 3.45 × 10−6 W/m 2 . This is very different from the correct intensity at C.

TEMPERATURE AND HEAT

17.1.

IDENTIFY and SET UP: EXECUTE:

17

TF = 95 TC + 32°.

(a) TF = (9/5)(−62.8) + 32 = −81.0°F

(b) TF = (9/5)(56.7) + 32 = 134.1°F (c) TF = (9/5)(31.1) + 32 = 88.0°F

17.2.

EVALUATE: Fahrenheit degrees are smaller than Celsius degrees, so it takes more F° than C° to express the difference of a temperature from the ice point. IDENTIFY and SET UP: TC = 59 (TF − 32°) EXECUTE:

(a) TC = (5/9)(41.0 − 32) = 5.0°C

(b) TC = (5/9)(107 − 32) = 41.7°C (c) TC = (5/9)(−18 − 32) = −27.8°C

17.3.

EVALUATE: Fahrenheit degrees are smaller than Celsius degrees, so it takes more F° than C° to express the difference of a temperature from the ice point. IDENTIFY: Convert each temperature from °C to °F. SET UP: TF = 95 TC + 32°C EXECUTE:

18°C equals 95 (18°) + 32° = 64°F and 39°C equals 95 (39°) + 32° = 102°F. The temperature increase is

102°F − 64°F = 38 F°. ⎛ 9 F° ⎞ The temperature increase is 21 C°, and this corresponds to (21 C°) ⎜ 5 ⎟ = 38 F°. ⎝ 1 C° ⎠ IDENTIFY: Convert ΔT = 10 K to F°. SET UP: 1 K = 1 C° = 95 F°. EVALUATE: 17.4.

17.5.

EXECUTE: A temperature increase of 10 K corresponds to an increase of 18 F°. Beaker B has the higher temperature. EVALUATE: Kelvin and Celsius degrees are the same size. Fahrenheit degrees are smaller, so it takes more of them to express a given ΔT value. IDENTIFY: Convert ΔT in kelvins to C° and to F°. SET UP: 1 K = 1 C° = 95 F° EXECUTE:

(a) ΔTF = 95 ΔTC = 95 ( −10.0 C° ) = −18.0 F°

(b) ΔTC = ΔTK = −10.0 C°

17.6.

EVALUATE: Kelvin and Celsius degrees are the same size. Fahrenheit degrees are smaller, so it takes more of them to express a given ΔT value. IDENTIFY: Convert ΔT between different scales. SET UP: ΔT is the same on the Celsius and Kelvin scales. 180 F° = 100 C°, so 1 C° = 95 F°. EXECUTE:

⎛ 1 C° ⎞ (a) ΔT = 49.0 F°. ΔT = (49.0 F°) ⎜ 9 ⎟ = 27.2 C°. ⎝ 5 F° ⎠

⎛ 1 C° ⎞ (b) ΔT = −100 F°. ΔT = (−100.0 F°) ⎜ 9 ⎟ = −55.6 C° ⎝ 5 F° ⎠ EVALUATE: The magnitude of the temperature change is larger in F° than in C°.

17-1

17-2

17.7.

Chapter 17

IDENTIFY: Convert T in °C to °F. SET UP: TF = 95 (TC + 32°) EXECUTE:

17.8.

(a) TF = 95 (40.2°) + 32° = 104.4°F. Yes, you should be concerned.

(b) TF = 95 (TC + 32°) = 95 (12°C) + 32° = 54°F. EVALUATE: In doing the temperature conversion we account for both the size of the degrees and the different zero points on the two temperature scales. IDENTIFY: Set TC = TF and TF = TK . SET UP: EXECUTE:

TF = 95 TC + 32°C and TK = TC + 273.15 = 59 (TF − 32°) + 273.15 (a) TF = TC = T gives T = 95 T + 32° and T = −40°; −40°C = −40°F.

(b) TF = TK = T gives T = 95 (T − 32°) + 273.15 and T = 17.9.

17.10.

9 4

( − ( ) (32°) + 273.15) = 575°; 5 9

575°F = 575 K.

EVALUATE: Since TK = TC + 273.15 there is no temperature at which Celsius and Kelvin thermometers agree. IDENTIFY: Convert to the Celsius scale and then to the Kelvin scale. SET UP: Combining Eq.(17.2) and Eq.(17.3), TK = 59 (TF − 32° ) + 273.15, EXECUTE: Substitution of the given Fahrenheit temperatures gives (a) 216.5 K (b) 325.9 K (c) 205.4 K EVALUATE: All temperatures on the Kelvin scale are positive. IDENTIFY: Convert TK to TC and then convert TC to TF . SET UP: EXECUTE:

TK = TC + 273.15 and TF = 95 TC + 32°. (a) TC = 400 − 273.15 = 127°C, TF = (9/5)(126.85) + 32 = 260°F

(b) TC = 95 − 273.15 = −178°C, TF = (9/5)(−178.15) + 32 = −289°F (c) TC = 1.55 × 107 − 273.15 = 1.55 × 107°C, TF = (9/5)(1.55 × 107 ) + 32 = 2.79 × 107°F

17.11.

EVALUATE: All temperatures on the Kelvin scale are positive. TC is negative if the temperature is below the freezing point of water. IDENTIFY: Convert TF to TC and then convert TC to TK . SET UP: EXECUTE:

17.12.

TC = 59 (TF − 32°). TK = TC + 273.15. (a) TC = 59 (−346° − 32°) = −210°C

(b) TK = −210° + 273.15 = 63 K EVALUATE: The temperature is negative on the Celsius and Fahrenheit scales but all temperatures are positive on the Kelvin scale. IDENTIFY: Apply Eq.(17.5) and solve for p. SET UP: ptriple = 325 mm of mercury

(

17.13.

)

p = (325.0 mm of mercury) 373.15 K = 444 mm of mercury 273.16 K EVALUATE: mm of mercury is a unit of pressure. Since Eq.(17.5) involves a ratio of pressures, it is not necessary to convert the pressure to units of Pa. T p IDENTIFY: When the volume is constant, 2 = 2 , for T in kelvins. T1 p1 EXECUTE:

SET UP:

Ttriple = 273.16 K. Figure 17.7 in the textbook gives that the temperature at which CO 2 solidifies is

TCO2 = 195 K. ⎛T ⎞ ⎛ 195 K ⎞ p2 = p1 ⎜ 2 ⎟ = (1.35 atm) ⎜ ⎟ = 0.964 atm ⎝ 273.16 K ⎠ ⎝ T1 ⎠ EVALUATE: The pressure decreases when T decreases. IDENTIFY: 1 K = 1 C° and 1 C° = 95 F°, so 1 K = 95 R °. SET UP: On the Kelvin scale, the triple point is 273.16 K. EXECUTE: Ttriple = (9/5)273.16 K = 491.69°R. EXECUTE: 17.14.

EVALUATE: One could also look at Figure 17.7 in the textbook and note that the Fahrenheit scale extends from −460°F to + 32°F and conclude that the triple point is about 492°R.

Temperature and Heat

17.15.

Fit the data to a straight line for p (T ) and use this equation to find T when p = 0.


(a) If the pressure varies linearly with temperature, then p2 = p1 + γ (T2 − T1 ).

EXECUTE:

γ=

17-3

p2 − p1 6.50 × 104 Pa − 4.80 × 104 Pa = = 170.0 Pa/C° 100°C − 0.01°C T2 − T1

Apply p = p1 + γ (T − T1 ) with T1 = 0.01°C and p = 0 to solve for T. 0 = p1 + γ (T − T1 ) T = T1 −

p1

γ

= 0.01°C −

4.80 × 104 Pa = −282°C. 170 Pa/C°

(b) Let T1 = 100°C and T2 = 0.01°C; use Eq.(17.4) to calculate p2 . Eq.(17.4) says T2 /T1 = p2 /p1 , where T is in kelvins.

17.16.

⎛T ⎞ ⎛ 0.01 + 273.15 ⎞ 4 4 p2 = p1 ⎜ 2 ⎟ = 6.50 × 104 Pa ⎜ ⎟ = 4.76 × 10 Pa; this differs from the 4.80 × 10 Pa that was measured + 100 273.15 T ⎝ ⎠ ⎝ 1⎠ so Eq.(17.4) is not precisely obeyed. EVALUATE: The answer to part (a) is in reasonable agreement with the accepted value of −273°C IDENTIFY: Apply ΔL = α L0 ΔT and calculate ΔT . Then T2 = T1 + ΔT , with T1 = 15.5°C. SET UP:

Table 17.1 gives α = 1.2 × 10−5 (C°) −1 for steel. ΔT =


17.18.

=

0.471 ft = 23.5 C°. T2 = 15.5°C + 23.5 C° = 39.0°C. [1.2 × 10−5 (C°) −1 ][1671 ft]

Since then the lengths enter in the ratio ΔL / L0 , we can leave the lengths in ft. ΔL = L0αΔT

IDENTIFY: SET UP:

ΔL

α L0

For steel, α = 1.2 × 10−5 (C°) −1

EXECUTE: ΔL = (1.2 × 10−5 (C°) −1 )(1410 m)(18.0°C − ( − 5.0°C)) = +0.39 m EVALUATE: The length increases when the temperature increases. The fractional increase is very small, since αΔT is small. IDENTIFY: Apply L = L0 (1 + αΔT ) to the diameter d of the rivet. SET UP:

For aluminum, α = 2.4 ×10−5 (C°) −1. Let d 0 be the diameter at –78.0°C and d be the diameter at 23.0°C.

EXECUTE: d = d 0 + Δd = d 0 (1 + αΔT ) = (0.4500 cm)(1 + (2.4 × 10 −5 (C°) −1 )(23.0°C − [ −78.0°C])). d = 0.4511 cm = 4.511 mm. EVALUATE: We could have let d 0 be the diameter at 23.0°C and d be the diameter at −78.0°C. Then 17.19.

ΔT = −78.0°C − 23.0°C. IDENTIFY: Apply L = L0 (1 + αΔT ) to the diameter D of the penny. SET UP:

1 K = 1 C°, so we can use temperatures in °C.

EXECUTE:

17.20.

Death Valley: α D0 ΔT = (2.6 × 10−5 (C°) −1 )(1.90 cm)(28.0 C°) = 1.4 × 10−3 cm, so the diameter is

1.9014 cm. Greenland: α D0 ΔT = −3.6 × 10−3 cm, so the diameter is 1.8964 cm. EVALUATE: When T increases the diameter increases and when T decreases the diameter decreases. IDENTIFY: ΔV = βV0 ΔT . Use the diameter at −15°C to calculate the value of V0 at that temperature. SET UP: For a hemisphere of radius R, the volume is V = 23 π R 3. Table 17.2 gives β = 7.2 ×10−5 (C°) −1 for aluminum. EXECUTE: V0 = 23 π R 3 = 23 π (27.5 m)3 = 4.356 × 104 m3 . ΔV = (7.2 × 10−5 (C°) −1 )(4.356 × 104 m3 )(35°C − [−15°C]) = 160 m3 EVALUATE:

17.21.

We could also calculate R = R0 (1 + αΔT ) and calculate the new V from R. The increase in volume is

V − V0 , but we would have to be careful to avoid round-off errors when two large volumes of nearly the same size are subtracted. IDENTIFY: Linear expansion; apply Eq.(17.6) and solve for α . SET UP: Let L0 = 40.125 cm; T0 = 20.0°C. ΔT = 45.0°C − 20.0°C = 25.0 C° gives ΔL = 0.023 cm EXECUTE: EVALUATE:

ΔL 0.023 cm = = 2.3 × 10−5 (C°) −1. L0 ΔT (40.125 cm)(25.0 C°) The value we calculated is the same order of magnitude as the values for metals in Table 17.1.

ΔL = α L0 ΔT implies α =

17-4

Chapter 17

17.22.

IDENTIFY: SET UP:

17.23.

17.24.

Apply ΔV = V0 βΔT . For copper, β = 5.1 × 10−5 (C°)−1. ΔV / V0 = 0.150 × 10−2.

0.150 × 10−2 = 29.4 C°. Tf = Ti + ΔT = 49.4°C. β 5.1 × 10−5 (C°)−1 EVALUATE: The volume increases when the temperature increases. IDENTIFY: Volume expansion; apply Eq.(17.8) to calculate ΔV for the ethanol. SET UP: From Table 17.2, β for ethanol is 75 × 10−5 K −1 ΔV / V0

EXECUTE:

ΔT =

=

EXECUTE:

ΔT = 10.0°C − 19.0°C = −9.0 K. Then ΔV = βV0 ΔT = (75 × 10−5 K −1 )(1700 L)(−9.0 K) = −11 L. The

volume of the air space will be 11 L = 0.011 m3 . EVALUATE: The temperature decreases, so the volume of the liquid decreases. The volume change is small, less than 1% of the original volume. IDENTIFY: Apply ΔV = V0 βΔT to the tank and to the ethanol. SET UP: For ethanol, β e = 75 × 10−5 (C°)−1. For steel, β s = 3.6 × 10−5 (C°)−1. EXECUTE: The volume change for the tank is ΔVs = V0 βs ΔT = (2.80 m3 )(3.6 × 10−5 (C°)−1 )(−14.0 C°) = −1.41 × 10−3 m3 = −1.41 L. The volume change for the ethanol is ΔVe = V0 β e ΔT = (2.80 m3 )(75 × 10−5 (C°)−1 )(−14.0 C°) = −2.94 × 10−2 m3 = −29.4 L.

17.25.

The empty volume in the tank is ΔVe − ΔVs = −29.4 L − ( −1.4 L) = −28.0 L. 28.0 L of ethanol can be added to the tank. EVALUATE: Both volumes decrease. But β e > βs , so the magnitude of the volume decrease for the ethanol is less than it is for the tank. IDENTIFY: Apply ΔV = V0 βΔT to the volume of the flask and to the mercury. When heated, both the volume of the flask and the volume of the mercury increase. SET UP: For mercury, β Hg = 18 × 10−5 (C°)−1. EXECUTE:

8.95 cm3 of mercury overflows, so ΔVHg − ΔVglass = 8.95 cm 3.

EXECUTE:

ΔVHg = V0 β Hg ΔT = (1000.00 cm3 )(18 × 10−5 (C°)−1 )(55.0 C°) = 9.9 cm3 .

17.26.

ΔVglass

0.95 cm3 = 1.7 × 10−5 (C°)−1. V0 ΔT (1000.00 cm3 )(55.0 C°) EVALUATE: The coefficient of volume expansion for the mercury is larger than for glass. When they are heated, both the volume of the mercury and the inside volume of the flask increase. But the increase for the mercury is greater and it no longer all fits inside the flask. IDENTIFY: Apply ΔL = L0αΔT to each linear dimension of the surface. ΔVglass = ΔVHg − 8.95 cm3 = 0.95 cm3 . β glass =

SET UP:

=

The area can be written as A = aL1L2 , where a is a constant that depends on the shape of the surface. For

example, if the object is a sphere, a = 4π and L1 = L2 = r. If the object is a cube, a = 6 and L1 = L2 = L, the length of one side of the cube. For aluminum, α = 2.4 × 10−5 (C°) −1. EXECUTE:

(a) A0 = aL01L02 . L1 = L01 (1 + αΔT ). L2 = L02 (1 + αΔT ).

A = aL1L2 = aL01L02 (1 + αΔT ) 2 = A0 (1 + 2αΔT + [αΔT ]2 ). αΔT is very small, so [αΔT ]2 can be neglected and A = A0 (1 + 2αΔT ). ΔA = A − A0 = (2α ) A0 ΔT

17.27.

(b) ΔA = (2α ) A0 ΔT = (2)(2.4 × 10−5 (C°) −1 )(π (0.275 m) 2 )(12.5 C°) = 1.4 × 10−4 m 2 EVALUATE: The derivation assumes the object expands uniformly in all directions. IDENTIFY and SET UP: Apply the result of Exercise 17.26a to calculate ΔA for the plate, and then A = A0 + ΔA. EXECUTE:

(a) A0 = π r02 = π (1.350 cm/2) 2 = 1.431 cm 2

(b) Exercise 17.26 says ΔA = 2α A0 ΔT , so ΔA = 2(1.2 × 10−5 C°−1 )(1.431 cm 2 )(175°C − 25°C) = 5.15 × 10 −3 cm 2

17.28.

A = A0 + ΔA = 1.436 cm 2 EVALUATE: A hole in a flat metal plate expands when the metal is heated just as a piece of metal the same size as the hole would expand. IDENTIFY: Apply ΔL = L0αΔT to the diameter DST of the steel cylinder and the diameter DBR of the brass piston. SET UP:

For brass, α BR = 2.0 × 10−5 (C°) −1. For steel, α ST = 1.2 × 10−5 (C°) −1.


17-5

EXECUTE: (a) No, the brass expands more than the steel. (b) Call D0 the inside diameter of the steel cylinder at 20°C. At 150°C, DST = DBR . D0 + ΔDST = 25.000 cm + ΔDBR . This gives D0 + α ST D0 ΔT = 25.000 cm + α BR (25.000 cm) ΔT . −5 −1 25.000 cm(1 + α BR ΔT ) (25.000 cm) ⎣⎡1 + (2.0 × 10 (C°) )(130 C°) ⎦⎤ = = 25.026 cm. 1 + α ST ΔT 1 + (1.2 × 10−5 (C°) −1 )(130 C°) EVALUATE: The space inside the steel cylinder expands just like a solid piece of steel of the same size. IDENTIFY: Find the change ΔL in the diameter of the lid. The diameter of the lid expands according to Eq.(17.6). SET UP: Assume iron has the same α as steel, so α = 1.2 × 10−5 (C°) −1.

D0 =

17.29.

17.30.

17.31.

EXECUTE: ΔL = α L0 ΔT = (1.2 × 10−5 (C°) −1 )(725 mm)(30.0 C°) = 0.26 mm EVALUATE: In Eq.(17.6), ΔL has the same units as L. IDENTIFY: Apply Eq.(17.12) and solve for F. SET UP: For brass, Y = 0.9 × 1011 Pa and α = 2.0 × 10−5 (C°) −1. EXECUTE: F = −Y αΔTA = −(0.9 × 1011 Pa)(2.0 × 10 −5 (C°) −1 )( −110 C°)(2.01 × 10 −4 m 2 ) = 4.0 × 104 N EVALUATE: A large force is required. ΔT is negative and a positive tensile force is required. IDENTIFY and SET UP: For part (a), apply Eq.(17.6) to the linear expansion of the wire. For part (b), apply Eq.(17.12) and calculate F/A. EXECUTE: (a) ΔL = α L0 ΔT ΔL 1.9 × 10−2 m = = 3.2 × 10−5 (C°) −1 L0 ΔT (1.50 m)(420°C − 20°C) (b) Eq.(17.12): stress F/A = −YαΔT ΔT = 20°C − 420°C = −400 C° (ΔT always means final temperature minus initial temperature)

α=

17.32.

F/A = −(2.0 × 1011 Pa)(3.2 × 10−5 (C°) −1 )( −400 C°) = +2.6 × 109 Pa EVALUATE: F/A is positive means that the stress is a tensile (stretching) stress. The answer to part (a) is consistent with the values of α for metals in Table 17.1. The tensile stress for this modest temperature decrease is huge. IDENTIFY: Apply ΔL = L0αΔT and stress = F / A = −YαΔT .

SET UP: EXECUTE:

17.33.

For steel, α = 1.2 × 10 −5 (C°)−1 and Y = 2.0 × 1011 Pa. (a) ΔL = L0αΔT = (12.0 m)(1.2 × 10−5 (C°)−1 )(35.0 C°) = 5.0 mm

(b) stress = −YαΔT = −(2.0 × 1011 Pa)(1.2 × 10−5 (C°) −1 )(35.0 C°) = −8.4 × 107 Pa. The minus sign means the stress is compressive. EVALUATE: Commonly occurring temperature changes result in very small fractional changes in length but very large stresses if the length change is prevented from occurring. IDENTIFY and SET UP: Apply Eq.(17.13) to the kettle and water. EXECUTE: kettle Q = mcΔT , c = 910 J/kg ⋅ K (from Table 17.3) Q = (1.50 kg)(910 J/kg ⋅ K)(85.0°C − 20.0°C) = 8.873 × 104 J water Q = mcΔT , c = 4190 J/kg ⋅ K (from Table 17.3) Q = (1.80 kg)(4190 J/kg ⋅ K)(85.0°C − 20.0°C) = 4.902 × 105 J

17.34.

Total Q = 8.873 × 104 J + 4.902 × 105 J = 5.79 × 105 J EVALUATE: Water has a much larger specific heat capacity than aluminum, so most of the heat goes into raising the temperature of the water. IDENTIFY: The heat required is Q = mcΔT . P = 200 W = 200 J/s, which is energy divided by time. SET UP: EXECUTE:

(a) Q = mcΔT = (0.320 kg)(4.19 × 103 J/kg ⋅ K)(60.0 C°) = 8.04 × 104 J

8.04 × 104 J = 402 s = 6.7 min 200.0 J/s EVALUATE: 0.320 kg of water has volume 0.320 L. The time we calculated in part (b) is consistent with our everyday experience. IDENTIFY: Apply Q = mcΔT . m = w/g . SET UP: The temperature change is ΔT = 18.0 K. Q gQ (9.80 m/s 2 )(1.25 × 104 J) EXECUTE: c = = = = 240 J/kg ⋅ K. mΔT wΔT (28.4 N)(18.0 K) EVALUATE: The value for c is similar to that for silver in Table 17.3, so it is a reasonable result. (b) t =

17.35.

For water, c = 4.19 × 103 J/kg ⋅ K.

17-6

Chapter 17

17.36.

IDENTIFY and SET UP: Use Eq.(17.13) EXECUTE: (a) Q = mcΔT

m = 12 (1.3 × 10−3 kg) = 0.65 × 10−3 kg Q = (0.65 × 10−3 kg)(1020 J/kg ⋅ K)(37°C − ( −20°C)) = 38 J (b) 20 breaths/min (60 min/1 h) = 1200 breaths/h So Q = (1200)(38 J) = 4.6 × 104 J. EVALUATE: The heat loss rate is Q/t = 13 W. 17.37.

IDENTIFY: Apply Q = mcΔT to find the heat that would raise the temperature of the student's body 7 C°. SET UP: 1 W = 1 J/s EXECUTE: Find Q to raise the body temperature from 37°C to 44°C. Q = mcΔT = (70 kg)(3480 J/kg ⋅ K)(7 C°) = 1.7 × 106 J.

1.7 × 106 J = 1400 s = 23 min. 1200 J/s EVALUATE: Heat removal mechanisms are essential to the well-being of a person. IDENTIFY and SET UP: Set the change in gravitational potential energy equal to the quantity of heat added to the water. EXECUTE: The change in mechanical energy equals the decrease in gravitational potential energy, ΔU = − mgh; |ΔU | = mgh. Q = |ΔU | = mgh implies mcΔT = mgh t=

17.38.

17.39.

ΔT = gh/c = (9.80 m/s 2 )(225 m)/(4190 J/kg ⋅ K) = 0.526 K = 0.526 C° EVALUATE: Note that the answer is independent of the mass of the object. Note also the small change in temperature that corresponds to this large change in height! IDENTIFY: The work done by friction is the loss of mechanical energy. The heat input for a temperature change is Q = mcΔT SET UP: EXECUTE:

The crate loses potential energy mgh, with h = (8.00 m)sin 36.9°, and gains kinetic energy

1 2

mv22 .

(a) W f = mgh − 12 mv22 = (35.0 kg) ( (9.80 m/s2 )(8.00 m)sin 36.9° − 12 (2.50 m/s) 2 ) = 1.54 × 103 J.

(b) Using the results of part (a) for Q gives ΔT = (1.54 × 103 J)/ ( (35.0 kg)(3650 J/kg ⋅ K) ) = 1.21 × 10−2 C°. 17.40.

EVALUATE: The temperature rise is very small. IDENTIFY: The work done by the brakes equals the initial kinetic energy of the train. Use the volume of the air to calculate its mass. Use Q = mcΔT applied to the air to calculate ΔT for the air. SET UP: EXECUTE:

K = 12 mv 2 . m = ρV . The initial kinetic energy of the train is K = 12 (25,000 kg)(15.5 m/s) 2 = 3.00 × 106 J. Therefore, Q for

the air is 3.00 × 106 J. m = ρV = (1.20 kg/m3 )(65.0 m)(20.0 m)(12.0 m) = 1.87 × 104 kg. Q = mcΔT gives 3.00 × 106 J Q = = 0.157 C°. mc (1.87 × 104 kg)(1020 J/kg ⋅ K) EVALUATE: The mass of air in the station is comparable to the mass of the train and the temperature rise is small. IDENTIFY: Set K = 12 mv 2 equal to Q = mcΔT for the nail and solve for ΔT . ΔT =

17.41.

SET UP: For aluminum, c = 0.91 × 103 J/kg ⋅ K. EXECUTE: The kinetic energy of the hammer before it strikes the nail is K = 12 mv 2 = 12 (1.80 kg)(7.80 m/s) 2 = 54.8 J. Each strike of the hammer transfers 0.60(54.8 J) = 32.9 J, and with

329 J Q = = 45.2 C° mc (8.00 × 10−3 kg)(0.91 × 103 J/kg ⋅ K) EVALUATE: This agrees with our experience that hammered nails get noticeably warmer. IDENTIFY and SET UP: Use the power and time to calculate the heat input Q and then use Eq.(17.13) to calculate c. (a) EXECUTE: P = Q/t , so the total heat transferred to the liquid is Q = Pt = (65.0 W)(120 s) = 7800 J 10 strikes Q = 329 J. Q = mcΔT and ΔT =

17.42.

7800 K Q = = 2.51 × 103 J/kg ⋅ K mΔT 0.780 kg(22.54°C − 18.55°C) (b) EVALUATE: Then the actual Q transferred to the liquid is less than 7800 J so the actual c is less than our calculated value; our result in part (a) is an overestimate.

Then Q = mcΔT gives c =


17.43.


17-7

Q = mcΔT . The mass of n moles is m = nM . For iron, M = 55.845 × 10−3 kg/mol and c = 470 J/kg ⋅ K. (a) The mass of 3.00 mol is m = nM = (3.00 mol)(55.845 × 10−3 kg/mol) = 0.1675 kg.

ΔT = Q/mc = (8950 J) / [ (0.1675 kg)(470 J/kg ⋅ K) ] = 114 K = 114 C°.

17.44.

(b) For m = 3.00 kg, ΔT = Q/mc = 6.35 C°. EVALUATE: (c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. IDENTIFY: The latent heat of fusion Lf is defined by Q = mLf for the solid → liquid phase transition. For a temperature change, Q = mcΔT . SET UP: At t = 1 min the sample is at its melting point and at t = 2.5 min all the sample has melted. EXECUTE: (a) It takes 1.5 min for all the sample to melt once its melting point is reached and the heat input during this time interval is (1.5 min)(10.0 × 103 J/min) = 1.50 × 104 J. Q = mLf .

Lf =

Q 1.50 × 104 J = = 3.00 × 104 J/kg. 0.500 kg m

(b) The liquid's temperature rises 30 C° in 1.5 min. Q = mcΔT .

cliquid =

1.50 × 104 J Q = = 1.00 × 103 J/kg ⋅ K. mΔT (0.500 kg)(30 C°)

The solid's temperature rises 15 C° in 1.0 min. csolid =

17.45.

1.00 × 104 J Q = = 1.33 × 103 J/kg ⋅ K. mΔT (0.500 kg)(15 C°)

EVALUATE: The specific heat capacities for the liquid and solid states are different. The values of c and Lf that we calculated are within the range of values in Tables 17.3 and 17.4. IDENTIFY and SET UP: Heat comes out of the metal and into the water. The final temperature is in the range 0 < T < 100°C, so there are no phase changes. Qsystem = 0. (a) EXECUTE:

Qwater + Qmetal = 0

mwater cwater ΔTwater + mmetalcmetal ΔTmetal = 0 (1.00 kg)(4190 J/kg ⋅ K)(2.0 C°) + (0.500 kg)(cmetal )(−78.0 C°) = 0

17.46.

cmetal = 215 J/kg ⋅ K (b) EVALUATE: Water has a larger specific heat capacity so stores more heat per degree of temperature change. (c) If some heat went into the styrofoam then Qmetal should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE:

(a) Qsystem = 0 gives mmCm ΔTm + mw Cw ΔTw = 0. mmCm (T − Tm ) + mw Cw (T − Tw ) = 0.

mmCm (Tm − T ) = mw Cw (T − Tw ). Solving for T, T = T=

17.47.

mmCmTm + mw CwTw . mmCm + mw Cw

(70.0 kg) (3480 J/kg ⋅ K) (37.0°C) + (0.355 kg) (4190 J/kg ⋅ C°) (12.0°C) = 36.85°C (70.0 kg)(3480 J/kg ⋅ C°) + (0.355 kg) (4190 J/kg ⋅ C°)

(b) It is possible a sensitive digital thermometer could measure this change since they can read to 0.1°C. It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. EVALUATE: Heat comes out of the body and its temperature falls. Heat goes into the soft drink and its temperature rises. IDENTIFY: For the man's body, Q = mcΔT . SET UP: From Exercise 17.46, ΔT = 0.15 C° when the body returns to 37.0°C. Q mC ΔT EXECUTE: The rate of heat loss is Q/t. = and t = mC ΔT . (Q/t ) t t

(70.355 kg)(3480 J/kg ⋅ C°)(0.15 C°) = 0.00525 d = 7.6 minutes. 7.00 × 106 J/day EVALUATE: Even if all the BMR energy stays in the body, it takes the body several minutes to return to its normal temperature. t=

17-8

Chapter 17

17.48.


For a temperature change Q = mcΔT and for the liquid to solid phase change Q = −mLf . For water, c = 4.19 × 103 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. Q = mcΔT − mLf = (0.350 kg)([4.19 × 103 J/kg ⋅ K][ − 18.0 C°] − 3.34 × 105 J/kg) = −1.43 × 105 J. The

⎛ 1 cal ⎞ 4 minus sign says 1.43 × 105 J must be removed from the water. (1.43 ×105 J) ⎜ ⎟ = 3.42 ×10 cal = 34.2 kcal. ⎝ 4.186 J ⎠ EVALUATE: Q < 0 when heat comes out of an object the equation Q = mcΔT puts in the correct sign 17.49.

automatically, from the sign of ΔT = Tf − Ti . But in Q = ± L we must select the correct sign. IDENTIFY and SET UP: Use Eq.(17.13) for the temperature changes and Eq.(17.20) for the phase changes. EXECUTE: Heat must be added to do the following ice at −10.0°C → ice at 0°C Qice = mcice ΔT = (12.0 × 10−3 kg)(2100 J/kg ⋅ K)(0°C − ( −10.0°C)) = 252 J phase transition ice (0°C) → liquid water (0°C) (melting) Qmelt = + mLf = (12.0 × 10−3 kg)(334 × 103 J/kg) = 4.008 × 103 J water at 0°C (from melted ice → water at 100°C Qwater = mcwater ΔT = (12.0 × 10−3 kg)(4190 J/kg ⋅ K)(100°C − 0°C) = 5.028 × 103 J phase transition water (100°C) → steam (100°C) (boiling) Qboil = + mLv = (12.0 × 10−3 kg)(2256 × 103 J/kg) = 2.707 × 104 J

The total Q is Q = 252 J + 4.008 × 103 J + 5.028 × 103 J + 2.707 × 104 J = 3.64 × 104 J (3.64 × 104 J)(1 cal/4.186 J) = 8.70 × 103 cal

17.50.

(3.64 × 104 J)(1 Btu/1055 J) = 34.5 Btu EVALUATE: Q is positive and heat must be added to the material. Note that more heat is needed for the liquid to gas phase change than for the temperature changes. IDENTIFY: Q = mcΔT for a temperature change and Q = + mLf for the solid to liquid phase transition. The ice starts to melt when its temperature reaches 0.0°C. The system stays at 0.00°C until all the ice has melted. SET UP: For ice, c = 2.01 × 103 J/kg ⋅ K. For water, Lf = 3.34 × 105 J/kg. EXECUTE:

(a) Q to raise the temperature of ice to 0.00°C:

1.66 × 104 J = 20.8 min. 800.0 J/min 1.84 × 105 J (b) To melt all the ice requires Q = mLf = (0.550 kg)(3.34 × 105 J/kg) = 1.84 × 105 J. t = = 230 min. 800.0 J/min The total time after the start of the heating is 251 min. (c) A graph of T versus t is sketched in Figure 17.50. EVALUATE: It takes much longer for the ice to melt than it takes the ice to reach the melting point. Q = mcΔT = (0.550 kg)(2.01 × 103 J/kg ⋅ K)(15.0 C°) = 1.66 × 10 4 J. t =

Figure 17.50 17.51.

IDENTIFY and SET UP: Use Eq.(17.20) to calculate Q and then P = Q/t. Must convert the quantity of ice from lb to kg. EXECUTE: “two-ton air conditioner” means 2 tons (4000 lbs) of ice can be frozen from water at 0°C in 24 h. Find the mass m that corresponds to 4000 lb (weight of water): m = (4000 lb)(1 kg/2.205 lb) = 1814 kg (The kg to lb equivalence from Appendix E has been used.) The heat that must be removed from the water to freeze it is Q = −mLf = −(1814 kg)(334 × 103 J/kg) = −6.06 × 108 J. The power required if this is to be done in 24 hours is

|Q | 6.06 × 108 J = = 7010 W or P = (7010 W)((1 Btu/h)/(0.293 W)) = 2.39 × 104 Btu/h. t (24 h)(3600 s/1 h) EVALUATE: The calculated power, the rate at which heat energy is removed by the unit, is equivalent to seventy 100-W light bulbs. P=


17.52.

17-9

For a temperature change, Q = mcΔT . For the vapor → liquid phase transition, Q = − mLv .

IDENTIFY:

For water, Lv = 2.256 × 106 J/kg and c = 4.19 × 103 J/kg ⋅ K.

SET UP:

(a) Q = + m( − Lv + cΔT )

EXECUTE:

Q = +(25.0 × 10−3 kg)(−2.256 × 106 J/kg + [4.19 × 103 J/kg ⋅ K][ − 66.0 C°]) = −6.33 × 10 4 J

17.53.

(b) Q = mcΔT = (25.0 × 10−3 kg)(4.19 × 103 J/kg ⋅ K)( − 66.0 C°) = −6.91 × 103 J. (c) The total heat released by the water that starts as steam is nearly a factor of ten larger than the heat released by water that starts at 100°C. Steam burns are much more severe than hot-water burns. EVALUATE: For a given amount of material, the heat for a phase change is typically much more than the heat for a temperature change. IDENTIFY and SET UP: The heat that must be added to a lead bullet of mass m to melt it is Q = mcΔt + mLf

( mcΔT is the heat required to raise the temperature from 25°C to the melting point of 327.3°C; mLf is the heat required to make the solid → liquid phase change.) The kinetic energy of the bullet if its speed is v is K = 12 mv 2 . K = Q says

EXECUTE:

1 2

mv 2 = mcΔT + mLf

v = 2(cΔT + Lf )

17.54.

v = 2[(130 J/kg ⋅ K)(327.3°C − 25°C) + 24.5 × 103 J/kg] = 357 m/s EVALUATE: This is a typical speed for a rifle bullet. A bullet fired into a block of wood does partially melt, but in practice not all of the initial kinetic energy is converted to heat that remains in the bullet. IDENTIFY: For a temperature change, Q = mcΔT . For the liquid → vapor phase change, Q = + mLv .

The density of water is 1000 kg/m3 .

SET UP:

(a) The heat that goes into mass m of water to evaporate it is Q = + mLv . The heat flow for the man is

EXECUTE:

Q = mman cΔT , where ΔT = −1.00 C°. m=−

v

+ mman cΔT and

mman cΔT (70.0 kg)(3480 J/kg ⋅ K)(−1.00 C°) =− = 0.101 kg = 101 g. Lv 2.42 × 106 J/kg

0.101 kg = 1.01× 10−4 m3 = 101 cm3 . This is about 28% of the volume of a soft-drink can. ρ 1000 kg/m3 EVALUATE: Fluid loss by evaporation from the skin can be significant. IDENTIFY: Use Q = McΔT to find Q for a temperature rise from 34.0°C to 40.0°C. Set this equal to (b) V =

17.55.

∑ Q = 0 so mL

m

=

Q = mLv and solve for m, where m is the mass of water the camel would have to drink.

17.56.

SET UP: c = 3480 J/kg ⋅ K and Lv = 2.42 × 106 J/kg. For water, 1.00 kg has a volume 1.00 L. M = 400 kg is the mass of the camel. McΔT (400 kg)(3480 J/kg ⋅ K)(6.0 K) EXECUTE: The mass of water that the camel saves is m = = = 3.45 kg Lv (2.42 × 106 J/kg) which is a volume of 3.45 L. EVALUATE: This is nearly a gallon of water, so it is an appreciable savings. IDENTIFY: The asteroid's kinetic energy is K = 12 mv 2 . To boil the water, its temperature must be raised to

100.0°C and the heat needed for the phase change must be added to the water. SET UP: For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg. EXECUTE:

K = 12 (2.60 × 1015 kg)(32.0 × 103 m/s) 2 = 1.33 × 1024 J. Q = mcΔT + mLv .

Q 1.33 × 1022 J = = 5.05 × 1015 kg. cΔT + Lv (4190 J/kg ⋅ K)(90.0 K) + 2256 × 103 J/kg EVALUATE: The mass of water boiled is 2.5 times the mass of water in Lake Superior. IDENTIFY: Apply Q = mcΔT to the air in the refrigerator and to the turkey. m=

17.57.

SET UP: EXECUTE:

17.58.

For the air mair = ρV mair = (1.20 kg/m3 )(1.50 m3 ) = 1.80 kg. Q = mair cair ΔT + mt ct ΔT .

Q = ([1.80 kg][1020 J/kg ⋅ K] + [10.0 kg][3480 J/kg ⋅ K])(−15.0 C°) = −5.50 × 105 J EVALUATE: Q is negative because heat is removed. 5% of the heat removed comes from the air. IDENTIFY: Q = mcΔT for a temperature change. The net Q for the system (sample, can and water) is zero. SET UP:

For water, cw = 4.19 × 103 J/kg ⋅ K. For copper, cc = 390 J/kg ⋅ K.

17-10

Chapter 17

EXECUTE:

For the water, Qw = mw cw ΔTw = (0.200 kg)(4.19 × 103 J/kg ⋅ K)(7.1 C°) = 5.95 × 103 J.

For the copper can, Qc = mccc ΔTc = (0.150 kg)(390 J/kg ⋅ K)(7.1 C°) = 415 J. For the sample, Qs = ms cs ΔTs = (0.085 kg)cs ( −73.9 C°).

∑Q = 0 17.59.

gives (0.085 kg)(−73.9 C°)cs + 415 J + 5.95 × 103 J = 0. cs = 1.01 × 103 J/kg ⋅ K.

EVALUATE: Heat comes out of the sample and goes into the water and the can. The value of cs we calculated is consistent with the values in Table 17.3. IDENTIFY and SET UP: Heat flows out of the water and into the ice. The net heat flow for the system is zero. The ice warms 0°C, melts, and then the water from the melted ice warms from 0°C to the final temperature. EXECUTE: Qsystem = 0; calculate Q for each component of the system: (Beaker has small mass says that

Q = mcΔT for beaker can be neglected.) 0.250 kg of water (cools from 75.0°C to 30.0°C) Qwater = mcΔT = (0.250 kg)(4190 J/kg ⋅ K)(30.0°C − 75.0°C) = −4.714 × 104 J ice (warms to 0°C; melts; water from melted ice warms to 30.0°C) Qice = mcice ΔT + mLf + mcwater ΔT Qice = m[(2100 J/kg ⋅ K)(0°C − ( −20.0°C)) + 334 × 103 J/kg + (4190 J/kg ⋅ K)(30.0°C − 0°C)] Qice = (5.017 × 105 J/kg)m Qsystem = 0 says Qwater + Qice = 0

−4.714 × 104 J + (5.017 × 105 J/kg)m = 0

4.714 × 104 J = 0.0940 kg 5.017 × 105 J/kg EVALUATE: Since the final temperature is 30.0°C we know that all the ice melts and the final system is all liquid water. The mass of ice added is much less than the mass of the 75°C water; the ice requires a large heat input for the phase change. IDENTIFY: For a temperature change Q = mcΔT . For a melting phase transition Q = mLf . The net Q for the system (sample, vial and ice) is zero. SET UP: Ice remains, so the final temperature is 0.0°C. For water, Lf = 3.34 × 105 J/kg. m=

17.60.

EXECUTE:

For the sample, Qs = ms cs ΔTs = (16.0 × 10−3 kg)(2250 J/kg ⋅ K)(−19.5 C°) = −702 J. For the vial,

Qv = mv cv ΔTv = (6.0 × 10−3 kg)(2800 J/kg ⋅ K)(−19.5 C°) = −328 J. For the ice that melts, Qi = mLf .

∑Q = 0

gives

−3

mLf − 702 J − 328 J = 0 and m = 3.08 × 10 kg = 3.08 g.

17.61.

EVALUATE: Only a small fraction of the ice melts. The water for the melted ice remains at 0°C and has no heat flow. IDENTIFY and SET UP: Large block of ice implies that ice is left, so T2 = 0°C (final temperature). Heat comes out of the ingot and into the ice. The net heat flow is zero. The ingot has a temperature change and the ice has a phase change. EXECUTE: Qsystem = 0; calculate Q for each component of the system:

ingot Qingot = mcΔT = (4.00 kg)(234 J/kg ⋅ K)(0°C − 750°C) = −7.02 × 105 J ice Qice = + mLf , where m is the mass of the ice that changes phase (melts) Qsystem = 0 says Qingot + Qice = 0

−7.02 × 105 J + m(334 × 103 J/kg) = 0

7.02 × 105 J = 2.10 kg 334 × 103 J/kg EVALUATE: The liquid produced by the phase change remains at 0°C since it is in contact with ice. IDENTIFY: The initial temperature of the ice and water mixture is 0.0°C. Assume all the ice melts. We will know that assumption is incorrect if the final temperature we calculate is less than 0.0°C. The net Q for the system (can, water, ice and lead) is zero. SET UP: For copper, cc = 390 J/kg ⋅ K. For lead, cl = 130 J/kg ⋅ K. For water, cw = 4.19 × 103 J/kg ⋅ K and m=

17.62.

Lf = 3.34 × 105 J/kg.


EXECUTE:

17-11

For the copper can, Qc = mccc ΔTc = (0.100 kg)(390 J/kg ⋅ K)(T − 0.0°C) = (39.0 J/K)T .

For the water, Qw = mw cw ΔTw = (0.160 kg)(4.19 × 103 J/kg ⋅ K)(T − 0.0°C) = (670.4 J/K)T . For the ice, Qi = mi Lf + mi cw ΔTw Qi = (0.018 kg)(3.34 × 105 J/kg ⋅ K) + (0.018 kg)(4.19 × 103 J/kg ⋅ K)(T − 0.0°C) = 6012 J + (75.4 J/K)T

For the lead, Ql = mlcl ΔTl = (0.750 kg)(130 J/kg ⋅ K)(T − 255°C) = (97.5 J/K)T − 2.486 × 104 J

∑Q = 0

gives (39.0 J/K)T + (670.4 J/K)T + 6012 J + (75.4 J/K) + (97.5 J/K)T − 2.486 × 104 J = 0.

1.885 × 104 J = 21.4°C. 882.3 J/K EVALUATE: T > 0.0°C, which confirms that all the ice melts. IDENTIFY: Set Qsystem = 0, for the system of water, ice and steam. Q = mcΔT for a temperature change and T=

17.63.

Q = ± mL for a phase transition. SET UP: For water, c = 4190 J/kg ⋅ K, Lf = 334 × 103 J/kg and Lv = 2256 × 103 J/kg. EXECUTE: The steam both condenses and cools, and the ice melts and heats up along with the original water. mi Lf + mi c(28.0 C°)+ mw c(28.0 C°)− msteam Lv + msteamc (−72.0 C°) = 0. The mass of steam needed is

msteam =

(0.450 kg)(334 × 103 J/kg) + (2.85 kg)(4190 J/kg ⋅ K)(28.0 C°) = 0.190 kg. 2256 × 103 J/kg + (4190 J/kg ⋅ K)(72.0 C°)

Since the final temperature is greater than 0.0°C, we know that all the ice melts. HL IDENTIFY: H = kAΔT/L and k = . AΔT SET UP: The SI units of H are watts, the units of area are m 2 , the temperature difference is in K, the length is in [W][m] W . meters, so the SI units for thermal conductivity are = [m 2 ][K] m ⋅ K EVALUATE: An equivalent way to express the units of k is J/(s ⋅ m ⋅ K). EVALUATE:

17.64.

17.65.

IDENTIFY and SET UP: The temperature gradient is (TH − TC )/L and can be calculated directly. Use Eq.(17.21) to calculate the heat current H. In part (c) use H from part (b) and apply Eq.(17.21) to the 12.0-cm section of the left end of the rod. T2 = TH and T1 = T , the target variable. EXECUTE: (a) temperature gradient = (TH − TC )/L = (100.0°C − 0.0°C)/0.450 m = 222 C°/m = 222 K/m (b) H = kA(TH − TC )/L. From Table 17.5, k = 385 W/m ⋅ K, so

H = (385 W/m ⋅ K)(1.25 × 10−4 m 2 )(222 K/m) = 10.7 W (c) H = 10.7 W for all sections of the rod.

Figure 17.65

Apply H = kaΔT/L to the 12.0 cm section (Figure 17.65): TH − T = LH/kA and (0.120 m)(10.7 W) = 73.3°C (1.25 × 10−4 m 2 )(385 W/m ⋅ K) EVALUATE: H is the same at all points along the rod, so ΔT/Δx is the same for any section of the rod with length Δx. Thus (TH − T )/(12.0 cm) = (TH − TC )/(45.0 cm) gives that TH = T = 26.7 C° and T = 73.3°C, as we already calculated. Q kA(TH − TC ) = . IDENTIFY: For a melting phase transition, Q = mLf . The rate of heat conduction is t L SET UP: For water, Lf = 3.34 × 105 J/kg. EXECUTE: The heat conducted by the rod in 10.0 min is Q 2.84 × 103 J = = 4.73 W. Q = mLf = (8.50 × 10−3 kg)(3.34 × 105 J/kg) = 2.84 × 103 J. t 600 s (Q / t ) L (4.73 W)(0.600 m) = = 227 W/m ⋅ K. k= A(TH − TC ) (1.25 × 10−4 m 2 )(100 C°) EVALUATE: The heat conducted by the rod is the heat that enters the ice and produces the phase change. T = TH − LH/Ak = 100.0°C −

17.66.

17-12

Chapter 17

17.67.

IDENTIFY and SET UP: Call the temperature at the interface between the wood and the styrofoam T. The heat current in each material is given by H = kA(TH − TC )/L.

See Figure 17.67 Heat current through the wood: H w = kw A(T − T1 ) Lw Heat current through the styrofoam: H s = ks A(T2 − T )/Ls

Figure 17.67

In steady-state heat does not accumulate in either material. The same heat has to pass through both materials in succession, so H w = H s . EXECUTE:

(a) This implies kw A(T − T1 )/Lw = ks A(T2 − T )/Ls

kw Ls (T − T1 ) = ks Lw (T2 − T ) T=

17.68.

kw LsT1 + ks LwT2 −0.0176 W ⋅ °C/K + 00057 W ⋅ °C/K = = −5.8°C kw Ls + ks Lw 0.00206 W/K

EVALUATE: The temperature at the junction is much closer in value to T1 than to T2 . The styrofoam has a very large k, so a larger temperature gradient is required for than for wood to establish the same heat current. H ⎛T −T ⎞ (b) IDENTIFY and SET UP: Heat flow per square meter is = k ⎜ H C ⎟ . We can calculate this either for the A ⎝ L ⎠ wood or for the styrofoam; the results must be the same. EXECUTE: wood (−5.8°C − ( −10.0°C)) Hw T − T1 = kw = (0.080 W/m ⋅ K) = 11 W/m 2 . A Lw 0.030 m styrofoam (19.0°C − ( −5.8°C)) Hs T −T = ks 2 = (0.010 W/m ⋅ K) = 11 W/m 2 . A Ls 0.022 m EVALUATE: H must be the same for both materials and our numerical results show this. Both materials are good insulators and the heat flow is very small. Q kA(TH − TC ) = IDENTIFY: t L SET UP: TH − TC = 175°C − 35°C. 1 K = 1 C°, so there is no need to convert the temperatures to kelvins.

Q (0.040 W/m ⋅ K)(1.40 m 2 )(175°C − 35°C) = = 196 W. t 4.0 × 10−2 m (b) The power input must be 196 W, to replace the heat conducted through the walls. EVALUATE: The heat current is small because k is small for fiberglass. IDENTIFY: Apply Eq.(17.23). Q = Ht. SET UP: 1 Btu = 1055 J AΔT (125 ft 2 )(34 F°) EXECUTE: The energy that flows in time t is Q = Ht = t= (5.0 h) = 708 Btu = 7.5 × 105 J. R (30 ft 2 ⋅ F° ⋅ h/Btu) EXECUTE:

17.69.

EVALUATE: Q in Btu. 17.70.

(a)

With the given units of R, we can use A in ft 2 , ΔT in F° and t in h, and the calculation then gives

Q kAΔT = . Q/t is the same for both sections of the rod. t L SET UP: For copper, kc = 385 W/m ⋅ K. For steel, ks = 50.2 W/m ⋅ K. IDENTIFY:

Q (385 W/m ⋅ K)(4.00 × 10−4 m 2 )(100°C − 65.0°C) = = 5.39 J/s. t 1.00 m kAΔT (50.2 W/m ⋅ K)(4.00 × 10−4 m 2 )(65.0°C − 0°C) (b) For the steel section, L = = = 0.242 m. (Q/t ) 5.39 J/s EVALUATE: The thermal conductivity for steel is much less than that for copper, so for the same ΔT and A a smaller L for steel would be needed for the same heat current as in copper. EXECUTE:

(a) For the copper section,


17.71.

17-13

IDENTIFY and SET UP: The heat conducted through the bottom of the pot goes into the water at 100°C to convert it to steam at 100°C. We can calculate the amount of heat flow from the mass of material that changes phase. Then use Eq.(17.21) to calculate TH , the temperature of the lower surface of the pan. EXECUTE:

Q = mLv = (0.390 kg)(2256 × 103 J/kg) = 8.798 × 105 J

H = Q/t = 8.798 × 105 J/180 s = 4.888 × 103 J/s Then H = kA(TH − TC )/L says that TH − TC =

HL (4.888 × 103 J/s)(8.50 × 10−3 m) = = 5.52 C° kA (50.2 W/m ⋅ K)(0.150 m 2 )

TH = TC + 5.52 C° = 100°C + 5.52 C° = 105.5°C 17.72.

EVALUATE: The larger TH − TC is the larger H is and the faster the water boils. IDENTIFY: Apply Eq.(17.21) and solve for A. SET UP: The area of each circular end of a cylinder is related to the diameter D by A = π R 2 = π ( D / 2) 2 . For

steel, k = 50.2 W/m ⋅ K. The boiling water has T = 100°C, so ΔT = 300 K. Q ΔT ⎛ 300 K ⎞ −3 2 = kA and 150 J/s = ( 50.2 W/m ⋅ K ) A ⎜ EXECUTE: ⎟ . This gives A = 4.98 × 10 m , and t L ⎝ 0.500 m ⎠

17.73.

17.74.

17.75.

D = 4 A/π = 4(4.98 × 10−3 m 2 )/π = 8.0 × 10−2 m = 8.0 cm. EVALUATE: H increases when A increases. IDENTIFY: Assume the temperatures of the surfaces of the window are the outside and inside temperatures. Use the concept of thermal resistance. For part (b) use the fact that when insulating materials are in layers, the R values are additive. SET UP: From Table 17.5, k = 0.8 W/m ⋅ K for glass. R = L/k . 5.20 × 10−3 m = 6.50 × 10−3 m 2 ⋅ K/W. EXECUTE: (a) For the glass, Rglass = 0.8 W/m ⋅ K A(TH − TC ) (1.40 m)(2.50 m)(39.5 K) H= = = 2.1 × 104 W R 6.50 × 10−3 m 2 ⋅ K/W 0.750 × 10−3 m = 0.015 m 2 ⋅ K/W. The total R is R = Rglass + Rpaper = 0.0215 m 2 ⋅ K/W. (b) For the paper, Rpaper = 0.05 W/m ⋅ K A(TH − TC ) (1.40 m)(2.50 m)(39.5 K) H= = = 6.4 × 103 W. R 0.0215 m 2 ⋅ K/W EVALUATE: The layer of paper decreases the rate of heat loss by a factor of about 3. H = eσ T 4 . IDENTIFY: The rate of energy radiated per unit area is A SET UP: A blackbody has e = 1. H EXECUTE: (a) = (1)(5.67 × 10−8 W/m 2 ⋅ K 4 )(273 K) 4 = 315 W/m 2 A H (b) = (1)(5.67 × 10−8 W/m 2 ⋅ K 4 )(2730 K)4 = 3.15 × 106 W/m 2 A EVALUATE: When the Kelvin temperature increases by a factor of 10 the rate of energy radiation increases by a factor of 104. IDENTIFY: Use Eq.(17.26) to calculate H net . SET UP:

H net = Aeσ (T 4 − Ts4 ) (Eq.(17.26); T must be in kelvins)

Example 17.16 gives A = 1.2 m 2 , e = 1.0, and T = 30°C = 303 K (body surface temperature) Ts = 5.0°C = 278 K EXECUTE:

17.76.

H net = 573.5 W − 406.4 W = 167 W

EVALUATE: Note that this is larger than H net calculated in Example 17.16. The lower temperature of the surroundings increases the rate of heat loss by radiation. IDENTIFY: The net heat current is H = Aeσ (T 4 − Ts4 ). A power input equal to H is required to maintain constant temperature of the sphere. SET UP: The surface area of a sphere is 4π r 2 . EXECUTE: H = 4π (0.0150 m) 2 (0.35)(5.67 × 10−8 W/m 2 ⋅ K 4 )([3000 K]4 − [290 K]4 ) = 4.54 × 104 W EVALUATE: Since 3000 K > 290 K and H is proportional to T 4 , the rate of emission of heat energy is much greater than the rate of absorption of heat energy from the surroundings.

17-14

Chapter 17

17.77.

IDENTIFY: Use Eq.(17.26) to calculate A. SET UP: H = Aeσ T 4 so A = H/eσ T 4 150-W and all electrical energy consumed is radiated says H = 150 W 150 W EXECUTE: A = = 2.1 × 10−4 m 2 (1 × 104 cm 2 /1 m 2 ) = 2.1 cm 2 (0.35)(5.67 × 10−8 W/m 2 ⋅ K 4 )(2450 K) 4 EVALUATE: Light bulb filaments are often in the shape of a tightly wound coil to increase the surface area; larger A means a larger radiated power H. IDENTIFY: Apply H = Aeσ T 4 and calculate A. SET UP: For a sphere of radius R, A = 4π R 2 . σ = 5.67 × 10−8 W/m 2 ⋅ K 4 . The radius of the earth is RE = 6.38 × 106 m, the radius of the sun is Rsun = 6.96 × 108 m, and the distance between the earth and the sun is

17.78.

r = 1.50 × 1011 m. The radius is found from R =

EXECUTE: (a) Ra =

A = 4π

H/(σ T 4 ) H 1 = . 4π 4πσ T 2

(2.7 × 1032 W) 1 = 1.61 × 1011 m 4π (5.67 × 10−8 W/m 2 ⋅ K 4 ) (11,000 K) 2

(2.10 × 1023 W) 1 = 5.43 × 106 m 4π (5.67 × 10−8 W/m 2 ⋅ K 4 ) (10,000 K) 2 EVALUATE: (c) The radius of Procyon B is comparable to that of the earth, and the radius of Rigel is comparable to the earth-sun distance. IDENTIFY and SET UP: Use the temperature difference in M° and in C° between the melting and boiling points of mercury to relate M° to C°. Also adjust for the different zero points on the two scales to get an equation for TM in terms of TC. (a) EXECUTE: normal melting point of mercury: −39°C = 0.0°M normal boiling point of mercury: 357°C = 100.0°M 100.0°M = 396 C° so 1 M° = 3.96 C° Zero on the M scale is −39 on the C scale, so to obtain TC multiply TM by 3.96 and then subtract 39°: TC = 3.96TM − 39° (b) Rb =

17.79.

1 (TC + 39°) Solving for TM gives TM = 3.96

17.80.

1 (100° + 39°) = 35.1°M The normal boiling point of water is 100°C; TM = 3.96 (b) 10.0 M° = 39.6 C° EVALUATE: A M° is larger than a C° since it takes fewer of them to express the difference between the boiling and melting points for mercury. IDENTIFY: Apply ΔL = L0αΔT to the radius of the hoop. The thickness of the space equals the increase in radius of the hoop. SET UP: The earth has radius RE = 6.38 × 106 m and this is the initial radius R0 of the hoop. For steel,

α = 1.2 × 10−5 K −1. 1 K = 1 C°. EXECUTE: The increase in the radius of the hoop would be ΔR = RαΔT = (6.38 × 106 m)(1.2 × 10−5 K −1 )(0.5 K) = 38 m. EVALUATE: 17.81.

IDENTIFY: SET UP:

Even though ΔR is large, the fractional change in radius, ΔR/R0 , is very small. The volume increases by ΔV = V0 βΔT and the mass is constant. ρ = m/V .

Copper has density ρ 0 = 8.9 × 103 kg/m3 and coefficient of volume expansion β = 5.1 × 10−5 (C°) −1. The

tube is initially at temperature T0 , has sides of length L0 , volume V0 , density ρ 0 , and coefficient of volume expansion β . EXECUTE: (a) When the temperature increase to T0 + ΔT , the volume changes by an amount ΔV , where m m . Divide the top and bottom by V0 and ΔV = βV0 ΔT . Then, ρ = , or eliminating ΔV , ρ = V0 + βV0ΔT V0 + ΔV ρ0 . This can be rewritten as ρ = ρ0 (1 + βΔT ) −1 . Then using the expression substitute ρ 0 = m/V0 . Then ρ = 1 + β ΔT

(1 + x )

n

≈ 1 + nx, where n = −1, ρ = ρ 0 (1 − βΔT ) . This is accurate when βΔT is small, which is the case if

ΔT ! 1/β . 1/β is on the order of 104 C° and ΔT is typically about 102 C° or less, so this approximation is accurate.


17-15

(b) The copper cube has sides of length 1.25 cm = 0.0125 m and ΔT = 70.0°C − 20.0°C = 50.0 C°. ΔV = βV0 ΔT = (5.1 × 10−5 (C°)−1 )(0.0125 m)3 (50.0 C°) = 5 × 10−9 m3 . Similarly,

17.82.

ρ = (8.9 × 103 kg/m3 )(1 − (5.1 × 10−5 (C°) −1 )(50.0 C°)) = 8.877 × 103 kg/m3 . Therefore, Δρ = −23 kg/m3 . EVALUATE: When the temperature increases, the volume decreases and the density increases. v 1 F . F, v and λ change when T IDENTIFY: v = F/μ = FL/m . For the fundamental, λ = 2L and f = = λ 2 mL changes because L changes. ΔL = LαΔT , where L is the original length. SET UP: For copper, α = 1.7 × 10−5 (C°) −1. EXECUTE: (a) We can use differentials to find the frequency change because all length changes are small percents. Δf ≈

⎛ F ⎞ ΔL 1 Δ L ( F/mL ) −1/ 2 ( F/m)( −1/L2 ) ΔL = 12 ⎜⎜ 12 . =2f ⎟⎟ L ⎝ mL ⎠ L (C°)−1 )(40 C°)(440 Hz) = −0.15 Hz. The frequency decreases since the length

∂f ΔL (only L changes due to heating). Δf = ∂L

1 1 2 2

Δf = − 12 (αΔT ) f = − 12 (1.7 × 10−5 increases. Δv 12 ( FL m) −1 2 ( F m)ΔL ΔL αΔT 1 ∂v = = = = (1.7 × 10−5 (C°) −1 )(40 C°) = 3.4 × 10−4 = 0.034%. (b) Δv = ΔL. v 2L 2 2 ∂L FL m

17.83.

Δλ (d) λ = 2L so Δλ = 2ΔL → Δλ = 2ΔL = ΔL = αΔT . = (1.7 × 10 −5 (C°) −1 )(40 C°) = 6.8 × 10−4 = 0.068%. 2L λ L λ λ increases. EVALUATE: The wave speed and wavelength increase when the length increases and the frequency decreases. The percentage change in the frequency is −0.034%. The fractional change in all these quantities is very small. IDENTIFY and SET UP: Use Eq.(17.8) for the volume expansion of the oil and of the cup. Both the volume of the cup and the volume of the olive oil increase when the temperature increases, but β is larger for the oil so it expands more. When the oil starts to overflow, ΔVoil = ΔVglass + (1.00 × 10 −3 m)A, where A is the cross-sectional

area of the cup. EXECUTE: ΔVoil = V0,oil β oil ΔT = (9.9 cm)Aβ oil ΔT

ΔVglass = V0,glass β glass ΔT = (10.0 cm)Aβ glass ΔT (9.9 cm)Aβ oil ΔT − (10.0 cm)Aβ glass ΔT + (1.00 × 10 −3 m)A

The A divides out. Solving for ΔT gives ΔT = 15.5 C° T2 = T1 + ΔT = 37.5°C EVALUATE: If the expansion of the cup is neglected, the olive oil will have expanded to fill the cup when (0.100 cm)A = (9.9 cm)Aβ oil ΔT , so ΔT = 15.0 C° and T2 = 37.0°C. Our result is slightly higher than this. The cup also expands but not very much since β glass ! βoil . 17.84.

17.85.

dV/dT . V SET UP: dV/dT is the slope of the graph of V versus T, the graph given in Figure 17.12 in the textbook. Slope of graph EXECUTE: β = . Construct the tangent to the graph at 2°C and 8°C and measure the slope of this line. V 0.10 cm3 0.10 cm3 /3 C° ≈ −3 ×10−5 (C°) −1. The slope in negative, as the At 22°C: Slope ≈ − and V ≈ 1000 cm3 . β ≈ − 3 C° 1000 cm3 0.24 cm3 0.24 cm3 /4 C° ≈ 6 × 10−5 (C°) −1. and V ≈ 1000 cm3 . β ≈ water contracts or it is heated. At 8°C: slope ≈ 4 C° 1000 cm3 The water now expands when heated. EVALUATE: β > 0 when the material expands when heated and β < 0 when the material contracts when it is heated. The minimum volume is at about 4°C and β has opposite signs above and below this temperature. IDENTIFY: Use Eq.(17.6) to find the change in diameter of the sphere and the change in length of the cable. Set the sum of these two increases in length equal to 2.00 mm. SET UP: α brass = 2.0 × 10−5 K −1 and α steel = 1.2 × 10−5 K −1. EXECUTE: ΔL = (α brass L0,brass + α steel L0,steel ) ΔT .

IDENTIFY:

Volume expansion: dV = βV dT . β =

2.00 × 10−3 m = 15.0 C°. T2 = T1 + ΔT = 35.0°C. (2.0 × 10−5 K −1 )(0.350 m) + (1.2 × 10−5 K −1 )(10.5 m) EVALUATE: The change in diameter of the brass sphere is 0.10 mm. This is small, but should not be neglected. ΔT =

17-16

Chapter 17

17.86.

IDENTIFY: Conservation of energy says Qe + Qc = 0, where Qe and Qc are the heat changes for the ethanol and cylinder. To find the volume of ethanol that overflows calculate ΔV for the ethanol and for the cylinder. SET UP: For ethanol, ce = 2428 J/kg ⋅ K and β e = 75 × 10−5 K −1. EXECUTE: (a) Qe + Qc = 0 gives mece (Tf − [−10.0°C]) + mccc (Tf − 20.0°C]) = 0. (20.0°C)mccc − (10.0°C)mece (20.0°C)(0.110 kg)(840 J/kg ⋅ K) − (10.0°C)(0.0873 kg)(2428 J/kg ⋅ K) Tf = . Tf = . mece + mccc (0.0873 kg)(2428 J/kg ⋅ K) + (0.110 kg)(840 J/kg ⋅ K) −271.6°C Tf = = −0.892°C. 304.4 (b) ΔVe = β eVe ΔT = (75 × 10−5 K −1 )(108 cm3 )( −0.892°C − [ −10.0°C]) = +0.738 cm 3. ΔVc = β cVc ΔT = (1.2 × 10−5 K −1 )(108 cm3 )(−0.892°C − 20.0°C) = −0.0271 cm3 . The volume that overflows is

17.87.

0.738 cm 3 − (−0.0271 cm3 ) = 0.765 cm 3 . EVALUATE: The cylinder cools so its volume decreases. The ethanol warms, so its volume increases. The sum of the magnitudes of the two volume changes gives the volume that overflows. IDENTIFY and SET UP: Call the metals A and B. Use the data given to calculate α for each metal. EXECUTE: ΔL = L0αΔT so α = ΔL/( L0 ΔT )

ΔL 0.0650 cm = = 2.167 × 10−5 (C°) −1 L0 ΔT (30.0 cm)(100 C°) ΔL 0.0350 cm = = 1.167 × 10−5 (C°) −1 metal B: α B = L0 ΔT (30.0 cm)(100 C°) EVALUATE: L0 and ΔT are the same, so the rod that expands the most has the larger α . IDENTIFY and SET UP: Now consider the composite rod (Figure 17.87). Apply Eq.(17.6). The target variables are LA and LB, the lengths of the metals A and B in the composite rod. metal A: α A =

ΔT = 100 C° ΔL = 0.058 cm Figure 17.87 EXECUTE: ΔL = ΔLA + ΔLB = (α A LA + α B LB )ΔT ΔL/ΔT = α A LA + α B (0.300 m − LA )

ΔL/ΔT − (0.300 m)α B (0.058 × 10−2 m/100 C°) − (0.300 m)(1.167 × 10−5 (C°) −1 ) = α A − αB 1.00 × 10−5 (C°) −1 LB = 30.0 cm − LA = 30.0 cm − 23.0 m = 7.0 cm EVALUATE: The expansion of the composite rod is similar to that of rod A, so the composite rod is mostly metal A. IDENTIFY: Apply ΔV = V0 β ΔT to the gasoline and to the volume of the tank. LA =

17.88.

SET UP: For aluminum, β = 7.2 × 10−5 K −1. 1 L = 10−3 m3 . EXECUTE: (a) The lost volume, 2.6 L, is the difference between the expanded volume of the fuel and the tanks, and the maximum temperature difference is ΔV (2.6 × 10−3 m3 ) ΔT = = = 28 C°. −4 −1 (β fuel − β A1 )V0 (9.5 × 10 (C°) − 7.2 × 10−5 (C°) −1 )(106.0 × 10−3 m3 )

17.89.

The maximum temperature was 32°C. (b) No fuel can spill if the tanks are filled just before takeoff. EVALUATE: Both the volume of the gasoline and the capacity of the tanks increased when T increased. But β is larger for gasoline than for aluminum so the volume of the gasoline increased more. When the tanks have returned to 4.0°C on Sunday morning there is 2.6 L of air space in the tanks. IDENTIFY: The change in length due to heating is ΔLT = L0αΔT and this need not equal ΔL. The change in FL length due to the tension is ΔLF = 0 . Set ΔL = ΔLF + ΔLT . AY SET UP: α brass = 2.0 × 10 −5 (C°) −1. α steel = 1.5 × 10 −5 (C°) −1. Ysteel = 20 × 1010 Pa. EXECUTE: (a) The change in length is due to the tension and heating . ΔL = F + αΔT . Solving for F/A, L0 AY ⎛ ΔL ⎞ F =Y⎜ − αΔT ⎟ . A ⎝ L0 ⎠


17-17

(b) The brass bar is given as “heavy” and the wires are given as “fine,” so it may be assumed that the stress in the bar due to the fine wires does not affect the amount by which the bar expands due to the temperature increase. This means that ΔL is not zero, but is the amount α brass L0 ΔT that the brass expands, and so

17.90.

F = Ysteel (α brass − α steel )ΔT = (20 × 1010 Pa)(2.0 × 10−5 (C°) −1 − 1.2 × 10−5 (C°) −1 )(120 C°) = 1.92 × 108 Pa. A EVALUATE: The length of the brass bar increases more than the length of the steel wires. The wires remain taut and are under tension when the temperature of the system is raised above 20°C. IDENTIFY: Apply the equation derived in part (a) of Problem 17.89 to the steel and aluminum sections. The sum of the ΔL values of the two sections must be zero. SET UP: For steel, Y = 20 × 1010 Pa and α = 1.2 × 10 −5 (C°) −1. For aluminum, Y = 7.0 × 1010 Pa and

α = 2.4 × 10−5 (C°) −1. EXECUTE:

In deriving Eq.(17.12), it was assumed that ΔL = 0; if this is not the case when there are both thermal

F ⎞ ⎛ and tensile stresses, Eq. (17.12) becomes ΔL = L0 ⎜ αΔT + ⎟ . (See Problem 17.89.) For the situation in this AY ⎠ ⎝ problem, there are two length changes which must sum to zero, and so Eq.(17.12) may be extended to two ⎛ ⎛ F ⎞ F ⎞ materials a and b in the form L0a ⎜ α a ΔT + ⎟ + L0b ⎜ α b ΔT + ⎟ = 0. Note that in the above, ΔT , F and A are AY AY a ⎠ b ⎠ ⎝ ⎝ the same for the two rods. Solving for the stress F/A,

α a L0a + α b L0b F =− ΔT . (( Loa Ya ) + ( L0b Yb )) A

F (1.2 × 10−5 (C°) −1 )(0.350 m) + (2.4 × 10−5 (C°) −1 )(0.250 m) = (60.0 C°) = −1.2 × 108 Pa. A ((0.350 m 20 × 1010 Pa) + (0.250 m 7 × 1010 Pa)) EVALUATE: F / A is negative and the stress is compressive. If the steel rod was considered alone and its length was held fixed, the stress would be −Ysteelα steel ΔT = −1.4 × 108 Pa. For the aluminum rod alone the stress would be 17.91.

−Yaluminumα aluminum ΔT = −1.0 × 108 Pa. The stress for the combined rod is the average of these two values. (a) IDENTIFY and SET UP: The diameter of the ring undergoes linear expansion (increases with T ) just like a solid steel disk of the same diameter as the hole in the ring. Heat the ring to make its diameter equal to 2.5020 in. ΔL 0.0020 in. = = 66.7 C° EXECUTE: ΔL = α L0 ΔT so ΔT = L0α (2.5000 in.)(1.2 × 10−5 (C°) −1 )

T = T0 + ΔT = 20.0°C + 66.7 C° = 87°C (b) IDENTIFY and SET UP: Apply the linear expansion equation to the diameter of the brass shaft and to the diameter of the hole in the steel ring. EXECUTE: L = L0 (1 + αΔT ) Want Ls (steel) = Lb (brass) for the same ΔT for both materials: L0s (1 + α s ΔT ) = L0b (1 + α b ΔT ) so L0s + L0sα s ΔT = L0b + L0bα b ΔT L0b − L0s 2.5020 in. − 2.5000 in. = L0sα s − L0bα b (2.5000 in.)(1.2 × 10−5 (C°) −1 ) − (2.5050 in.)(2.0 × 10−5 (C°) −1 ) 0.0020 ΔT = C° = −100 C° 3.00 × 10−5 − 5.00 × 10−5 T = T0 + ΔT = 20.0°C − 100 C° = −80°C EVALUATE: Both diameters decrease when the temperature is lowered but the diameter of the brass shaft decreases more since α b > α s ; |ΔLb | − |ΔLs | = 0.0020 in. IDENTIFY: Follow the derivation of Eq.(17.12). SET UP: For steel, the bulk modulus is B = 1.6 × 1011 Pa and the volume expansion coefficient is β = 3.0 × 10−5 K −1. EXECUTE: (a) The change in volume due to the temperature increase is βV ΔT , and the change in volume due to Δp , or Δp = BβΔV . the pressure increase is − V Δp. Setting the net change equal to zero, βV ΔT = V B B (b) From the above, Δp = (1.6 × 1011 Pa)(3.0 × 10−5 K −1 )(15.0 K) = 8.6 × 107 Pa. EVALUATE: Δp in part (b) is about 850 atm. A small temperature increase corresponds to a very large pressure increase. ΔT =

17.92.

17-18

Chapter 17

17.93.

IDENTIFY: Apply Eq.(11.14) to the volume increase of the liquid due to the pressure decrease. Eq.(17.8) gives the volume decrease of the cylinder and liquid when they are cooled. Can think of the liquid expanding when the pressure is reduced and then contracting to the new volume of the cylinder when the temperature is reduced. SET UP: Let β1 and β m be the coefficients of volume expansion for the liquid and for the metal. Let ΔT be the (negative) change in temperature when the system is cooled to the new temperature. EXECUTE: Change in volume of cylinder when cool: ΔVm = β mV0 ΔT (negative)

Change in volume of liquid when cool: ΔV1 = β1V0 ΔT (negative) The difference ΔV1 = ΔVm must be equal to the negative volume change due to the increase in pressure, which is −ΔpV0 /B = −k ΔpV0 . Thus ΔV1 − ΔVm = −k ΔpV0 . ΔT = −

k Δp β1 − β m

(8.50 × 10−10 Pa −1 )(50.0 atm)(1.013 × 105 Pa/1 atm) = −9.8 C° 4.80 × 10−4 K −1 − 3.90 × 10−5 K −1 T = T0 + ΔT = 30.0°C − 9.8 C° = 20.2°C. EVALUATE: A modest temperature change produces the same volume change as a large change in pressure; B " β for the liquid. IDENTIFY: Qsystem = 0. Assume that the normal melting point of iron is above 745°C, so the iron initially is solid. ΔT = −

17.94.

SET UP:

For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg. For solid iron, c = 470 J/kg ⋅ K.

EXECUTE: The heat released when the iron slug cools to 100°C is Q = mcΔT = (0.1000 kg)(470 J/kg ⋅ K)(645 K) = 3.03 × 104 J. The heat absorbed when the temperature of the water

is raised to 100°C is Q = mcΔT = (0.0750 kg)(4190 J/kg ⋅ K)(80.0 K) = 2.51× 104 J. This is less than the heat released from the iron and 3.03 × 104 J − 2.51 × 104 J = 5.20 × 103 J of heat is available for converting some of the 5.20 × 103 J liquid water at 100°C to vapor. The mass m of water that boils is m = = 2.30 × 10 −3 kg = 2.3 g 2256 × 103 J/kg

17.95.

(a) The final temperature is 100°C. (b) There is 75.0 g − 2.3 g = 72.7 g of liquid water remaining, so the final mass of the iron and remaining water is 172.7 g. EVALUATE: If we ignore the phase change of the water and write miron ciron (T − 745°C) + mwater cwater (T − 200°C) = 0, when we solve for T we will get a value larger than 100°C. That result is unphysical and tells us that some of the water changes phase. (a) IDENTIFY: Calculate K / Q. We don't know the mass m of the spacecraft, but it divides out of the ratio. SET UP: The kinetic energy is K = 12 mv 2 . The heat required to raise its temperature by 600 C° (but not to melt it) is Q = mcΔT .

(7700 m/s) 2 K 12 mv 2 v2 = = = = 54.3. Q mcΔT 2cΔT 2(910 J/kg ⋅ K)(600 C°) (b) EVALUATE: The heat generated when friction work (due to friction force exerted by the air) removes the kinetic energy of the spacecraft during reentry is very large, and could melt the spacecraft. Manned space vehicles must have heat shields made of very high melting temperature materials, and reentry must be made slowly. IDENTIFY: The rate at which thermal energy is being generated equals the rate at which the net torque due to the rope is doing work. The energy input associated with a temperature change is Q = mcΔT . EXECUTE:

17.96.

The ratio is

SET UP: The rate at which work is being done is P = τω . For iron, c = 470 J/kg ⋅ K. 1 C° = 1 K EXECUTE: (a) The net torque that the rope exerts on the capstan, and hence the net torque that the capstan exerts on the rope, is the difference between the forces of the ends of the rope times the radius of the capstan. The capstan 2π rad 2π rad is doing work on the rope at a rate P = τω = Fnet r = ( 520 N ) ( 5.0 × 10−2 m ) = 182 W, or 180 W to T ( 0.90 s )

two figures. A larger number of turns might increase the force, but for given forces, the torque is independent of the number of turns. (182 W) ΔT Q / t P (b) = = = = 0.064 C° s. t mc mc (6.00 kg)(470 J kg⋅ K) EVALUATE: The rate of temperature rise is proportional to the difference in tension between the ends of the rope and to the rate at which the capstan is rotating.


17.97.

17-19

IDENTIFY and SET UP: To calculate Q, use Eq.(17.18) in the form dQ = nC dT and integrate, using C (T ) given in the problem. Cav is obtained from Eq.(17.19) using the finite temperature range instead of an infinitesimal dT. EXECUTE: (a) dQ = mcdT T2

T2

T2

T1

T1

T1

Q = n ∫ C dT = n ∫ k (T 3 /Θ3 ) dT = ( nk/Θ3 ) ∫ T 3 dt = ( nk/Θ3 )

(

1 4

T4

T2 T1

)

nk (1.50 mol)(1940 J/mol ⋅ K) (T24 − T14 ) = ((40.0 K) 4 − (10.0 K) 4 ) = 83.6 J 4Θ3 4(281 K)3 1 ΔQ 1 83.6 J ⎛ ⎞ = (b) Cav = ⎜ ⎟ = 1.86 J/mol ⋅ K n ΔT 1.50 mol ⎝ 40.0 K − 10.0 K ⎠ Q=

17.98.

(c) C = k (t/Θ)3 = (1940 J/mol ⋅ K)(40.0 K/281 K)3 = 5.60 J/mol ⋅ K EVALUATE: C is increasing with T, so C at the upper end of the temperature integral is larger than its average value over the interval. IDENTIFY: For a temperature change, Q = mcΔT , and for the liquid → solid phase change, Q = −mLf . SET UP:

The volume Vw of the water determines its mass. mw = ρ wVw . For water, ρ w = 1000 kg/m3 ,

c = 4190 J/kg ⋅ K and Lf = 334 × 103 J/kg. EXECUTE: Set the heat energy that flows into the water equal to the final gravitational potential energy. Lf ρ wVw + cw ρ wVw ΔT = mgh. Solving for h, and inserting numbers:

h=

17.99.

(1000 kg m3 )(1.9 × 0.8 × 0.1 m 3 ) ⎡⎣334 × 103 J kg + (4190 J kg ⋅ K)(37 C°) ⎤⎦ (70 kg)(9.8 m s 2 )

.

h = 1.08 × 105 m = 108 km. EVALUATE: The heat associated with temperature and phase changes corresponds to a large amount of mechanical energy. IDENTIFY: Apply Q = mcΔT to the air in the room. SET UP: EXECUTE:

The mass if air in the room is m = ρV = (1.20 kg/m3 )(3200 m3 ) = 3840 kg. 1 W = 1 J/s. (a) Q = (3000 s)(90 students)(100 J/s ⋅ student) = 2.70 × 107 J.

Q 2.70 × 107 J = = 6.89 C° mc (3840 kg)(1020 J/kg ⋅ K) ⎛ 280 W ⎞ (c) ΔT = (6.89 C°) ⎜ ⎟ = 19.3 C°. ⎝ 100 W ⎠ EVALUATE: In the absence of a cooling mechanism for the air, the air temperature would rise significantly.

(b) Q = mcΔT . ΔT =

17.100. IDENTIFY: SET UP:

T2

dQ = nCdT so for the temperature change T1 → T2 , Q = n ∫ C (T )dT . T1

∫ dT = T and ∫ TdT =

1 2

T 2 . Express T1 and T2 in kelvins: T1 = 300 K, T2 = 500 K.

EXECUTE: Denoting C by C = a + bT , a and b independent of temperature, integration gives b ⎛ ⎞ Q = n ⎜ a (T2 − T1 ) + (T22 − T12 ) ⎟ . 2 ⎝ ⎠

Q = (3.00 mol)(29.5 J mol ⋅ K)(500 K − 300 K) + (4.10 × 10−3 J mol ⋅ K 2 )((500 K) 2 − (300 K) 2 )). Q = 1.97 × 104 J. EVALUATE: If C is assumed to have the constant value 29.5 J/mol ⋅ K, then Q = 1.77 × 104 J for this temperature change. At T1 = 300 K, C = 32.0 J/mol ⋅ K and at T2 = 500 K, C = 33.6 J/mol ⋅ K. The average value of C is

32.8 J/mol ⋅ K, If C is assumed to be constant and to have this average value, then Q = 2.02 × 104 J, which is within 3% of the correct value. 17.101. IDENTIFY: Use Q = mLf to find the heat that goes into the ice to melt it. This amount of heat must be conducted through the walls of the box; Q = Ht. Assume the surfaces of the styrofoam have temperatures of 5.00°C and 21.0°C. SET UP:

For water Lf = 334 × 103 J/kg. For Styrofoam k = 0.01 W/m ⋅ K. One week is 6.048 × 105 s. The surface

area of the box is 4(0.500 m)(0.800 m) + 2(0.500 m) 2 = 2.10 m 2 . TH − TC . Q = Ht gives L tkA(TH − TC ) (6.048 × 105 s)(0.01 W/m ⋅ K)(2.10 m 2 )(21.0°C − 5.00°C) L= = = 2.5 cm Q 8.016 × 106 J

EXECUTE:

Q = mLf = (25.0 kg)(334 × 103 J/kg) = 8.016 × 106 J. H = kA

17-20

Chapter 17

EVALUATE: We have assumed that the liquid water that is produced by melting the ice remains in thermal equilibrium with the ice so has a temperature of 0°C. The interior of the box and the ice are not in thermal equilibrium, since they have different temperatures. 17.102. IDENTIFY: For a temperature change Q = mcΔT . For the vapor → liquid phase transition, Q = −mLv . SET UP: For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg. EXECUTE: The requirement that the heat supplied in each case is the same gives mw cw ΔTw = ms (cw ΔTs + Lv ), where ΔTw = 42.0 K and ΔTs = 65.0 K. The ratio of the masses is ms cw ΔTw (4190 J kg ⋅ K)(42.0 K) = = = 0.0696, mw cw ΔTs + Lv (4190 J kg ⋅ K)(65.0 K) + 2256 × 103 J kg so 0.0696 kg of steam supplies the same heat as 1.00 kg of water. EVALUATE: Note the heat capacity of water is used to find the heat lost by the condensed steam, since the phase transition produces liquid water at an initial temperature of 100°C. 17.103. (a) IDENTIFY and SET UP: Assume that all the ice melts and that all the steam condenses. If we calculate a final temperature T that is outside the range 0°C to 100°C then we know that this assumption is incorrect. Calculate Q for each piece of the system and then set the total Qsystem = 0. EXECUTE: copper can (changes temperature form 0.0° to T; no phase change) Qcan = mcΔT = (0.446 kg)(390 J/kg ⋅ K)(T − 0.0°C) = (173.9 J/K)T ice (melting phase change and then the water produced warms to T ) Qice = + mLf + mcΔT = (0.0950 kg)(334 × 103J/kg) + (0.0950 kg)(4190 J/kg ⋅ K)(T − 0.0°C) Qice = 3.173 × 104 J + (398.0 J/K)T . steam (condenses to liquid and then water produced cools to T ) Qsteam = −mLv + mcΔT = −(0.0350 kg)(2256 × 103 J/kg) + (0.0350 kg)(4190 J/kg ⋅ K)(T − 100.0°C) Qsteam = −7.896 × 104 J + (146.6 J/K)T − 1.466 × 104 J = −9.362 × 10 4 J + (146.6J/K)T Qsystem = 0 implies Qcan + Qice + Qsteam = 0. (173.9 J/K)T + 3.173 × 10 4 J + (398.0 J/K)T − 9.362 × 10 4 J + (146.6 J/K)T = 0 (718.5 J/K)T = 6.189 × 104 J 6.189 × 104 J = 86.1°C 718.5 J/K EVALUATE: This is between 0°C and 100°C so our assumptions about the phase changes being complete were correct. (b) No ice, no steam 0.0950 kg + 0.0350 kg = 0.130 kg of liquid water. 17.104. IDENTIFY: The final amount of ice is less than the initial mass of water, so water remains and the final temperature is 0°C. The ice added warms to 0°C and heat comes out of water to convert it to ice. Conservation of energy says Qi + Qw = 0, where Qi and Qw are the heat flows for the ice that is added and for the water that freezes. SET UP: Let mi be the mass of ice that is added and mw is the mass of water that freezes. The mass of ice T=

increases by 0.328 kg, so mi + mw = 0.328 kg. For water, Lf = 334 × 103 J/kg and for ice ci = 2100 J/kg ⋅ K. Heat comes out of the water when it freezes, so Qw = −mLf EXECUTE:

Qi + Qw = 0 gives mici (15.0 C°) + (−mw Lf ) = 0, mw = 0.328 kg − mi , so

mici (15.0 C°) + (−0.328 + mi ) Lf = 0. mi =

(0.328 kg) Lf (0.328 kg)(334 × 103 J/kg) = = 0.300 kg. ci (15.0 C°) + Lf (2100 J/kg ⋅ K)(15.0 K) + 334 × 103 J/kg

0.300 kg of ice was added. EVALUATE: The mass of water that froze when the ice at −15.0°C was added was 0.778 kg − 0.450 kg − 0.300 kg = 0.028 kg. 17.105. IDENTIFY and SET UP: Heat comes out of the steam when it changes phase and heat goes into the water and causes its temperature to rise. Qsystem = 0. First determine what phases are present after the system has come to a uniform final temperature. (a) EXECUTE: Heat that must be removed from steam if all of it condenses is Q = −mLv = −(0.0400 kg)(2256 × 103 J/kg) = −9.02 × 104 J Heat absorbed by the water if it heats all the way to the boiling point of 100°C: Q = mcΔT = (0.200 kg)(4190 J/kg ⋅ K)(50.0 C°) = 4.19 × 104 J EVALUATE: The water can’t absorb enough heat for all the steam to condense. Steam is left and the final temperature then must be 100°C.


17-21

(b) EXECUTE: Mass of steam that condenses is m = Q/Lv = 4.19 × 104 J/2256 × 103 J/kg = 0.0186 kg Thus there is 0.0400 kg − 0.0186 kg = 0.0214 kg of steam left. The amount of liquid water is 0.0186 kg + 0.200 kg = 0.219 kg. 17.106. IDENTIFY:

Qsystem = 0.

SET UP: The mass of the system increases by 0.525 kg − 0.490 kg = 0.035 kg, so the mass of the steam that condensed is 0.035 kg. EVALUATE: The heat lost by the steam as it condenses and cools is (0.035 kg) Lv + (0.035 kg)(4190 J kg ⋅ K)(29.0 K), and the heat gained by the original water and calorimeter

is ((0.150 kg)(420 J kg ⋅ K) + (0.340 kg)(4190 J kg ⋅ K ))(56.0 K) = 8.33 × 104 J. Setting the heat lost equal to the heat gained and solving for Lv gives 2.26 × 106 J kg, or 2.3 × 106 J kg to two figures (the mass of steam condensed is known to only two figures). EVALUATE: Qsystem = 0 means the magnitude of the heat that flows out of the 0.035 kg of steam as it condenses and cools equals the heat that flows into the calorimeter and 0.340 kg of water as their temperature increases. To the accuracy of the calculation, our result agrees with the value of Lv given in Table 17.4. 17.107. IDENTIFY:

Heat Ql comes out of the lead when it solidifies and the solid lead cools to Tf . If mass ms of steam is

produced, the final temperature is Tf = 100°C and the heat that goes into the water is

Qw = mw Cw (25.0 C°) + ms Lv,w , where mw = 0.5000 kg. Conservation of energy says Ql + Qw = 0. Solve for ms . The mass that remains is 1.250 kg + 0.5000 kg − ms . SET UP:

For lead, Lf, l = 24.5 × 103 J/kg, cl = 130 J/kg ⋅ K and the normal melting point of lead is 327.3°C. For

water, cw = 4190 J/kg ⋅ K and Lv,w = 2256 × 103 J/kg. EXECUTE:

ms = ms =

Ql + Qw = 0. −ml Lf,l + mlcl (−227.3 C°) + mw cw (25.0 C°) + ms Lv,w = 0.

ml Lf,l + mlcl (+227.3 C°) − mw cw (25.0 C°) Lv,w

.

+ (1.250 kg)(24.5 × 103 J/kg) + (1.250 kg)(130 J/kg ⋅ K)(227.3 K) − (0.5000 kg)(4190 J/kg ⋅ K)(25.0 K) 2256 × 103 J/kg

1.519 × 104 J = 0.0067 kg. The mass of water and lead that remains is 1.743 kg. 2256 × 103 J/kg EVALUATE: The magnitude of heat that comes out of the lead when it goes from liquid at 327.3°C to solid at 100.0°C is 6.76 × 104 J. The heat that goes into the water to warm it to 100°C is 5.24 × 10 4 J. The additional heat that goes into the water, 6.76 × 10 4 J − 5.24 × 104 J = 1.52 × 10 4 J converts 0.0067 kg of water at 100°C to steam. ΔT 17.108. IDENTIFY: Apply H = kA and solve for k. L SET UP: H equals the power input required to maintain a constant interior temperature L (3.9 × 10−2 m) = (180 W) = 5.0 × 10−2 W m ⋅ K. EXECUTE: k = H AΔT (2.18 m 2 )(65.0 K) EVALUATE: Our result is consistent with the values for insulating solids in Table 17.5. ΔT 17.109. IDENTIFY: Apply H = kA . L SET UP: For the glass use L = 12.45 cm, to account for the thermal resistance of the air films on either side of the glass. 28.0 C° ⎛ ⎞ EXECUTE: (a) H = (0.120 J mol.K) (2.00 × 0.95 m 2 ) ⎜ ⎟ = 93.9 W. −2 −2 ⎝ 5.0 × 10 m + 1.8 × 10 m ⎠ (0.50) 2 = 0.868, so it (b) The heat flow through the wood part of the door is reduced by a factor of 1 − (2.00 × 0.95) ms =

28.0 C° ⎞ ⎛ becomes 81.5 W. The heat flow through the glass is H glass = (0.80 J/mol ⋅ K)(0.50 m) 2 ⎜ ⎟ = 45.0 W, −2 ⎝ 12.45 × 10 m ⎠ and so the ratio is 81.5 + 45.0 = 1.35. 93.9 EVALUATE: The single-pane window produces a significant increase in heat loss through the door. (See Problem 17.111).

17-22

Chapter 17

Apply Eq.(17.23). HR1 HR2 be the temperature difference across the wood and let ΔT2 = be the temperature SET UP: Let ΔT1 = A A difference across the insulation. The temperature difference across the combination is ΔT = ΔT1 + ΔT2 . The

17.110. IDENTIFY:

effective thermal resistance R of the combination is defined by ΔT =

HR . A

H H ( R1 + R2 ) = R, and R = R1 + R2 . A A EVALUATE: A good insulator has a large value of R. R for the combination is larger than the R for any one of the layers. 17.111. IDENTIFY and SET UP: Use H written in terms of the thermal resistance R: H = AΔT/R, where R = L/k and R = R1 + R2 + … (additive). EXECUTE:

ΔT = ΔT1 + ΔT2 gives

EXECUTE:

single pane Rs = Rglass + Rfilm , where Rfilm = 0.15 m 2 ⋅ K / W is the combined thermal resistance of the

air films on the room and outdoor surfaces of the window. Rglass = L/k = (4.2 × 10−3 m)/(0.80 W/m ⋅ K) = 0.00525 m 2 ⋅ K / W Thus Rs = 0.00525 m 2 ⋅ K / W + .15 m 2 ⋅ K / W = 0.1553 m 2 ⋅ K / W. double pane Rd = 2 Rglass + Rair + Rfilm , where Rair is the thermal resistance of the air space between the panes. Rair = L /k = (7.0 × 10−3 m)/(0.024 W/m ⋅ K) = 0.2917 m 2 ⋅ K / W Thus Rd = 2(0.00525 m 2 ⋅ K / W) + 0.2917 m 2 ⋅ K / W + 0.15 m 2 ⋅ K / W = 0.4522 m 2 ⋅ K / W H s = AΔT / Rs , H d = AΔT / Rd , so H s / H d = Rd / Rs (since A and ΔT are same for both) H s / H d = (0.4522 m 2 ⋅ K / W)/(0.1553 m 2 ⋅ K / W) = 2.9 EVALUATE: The heat loss is about a factor of 3 less for the double-pane window. The increase in R for a doublepane is due mostly to the thermal resistance of the air space between the panes. kAΔT to each rod. Conservation of energy requires that the heat current through the copper 17.112. IDENTIFY: H = L equals the sum of the heat currents through the brass and the steel. SET UP: Denote the quantities for copper, brass and steel by 1, 2 and 3, respectively, and denote the temperature at the junction by T0 . EXECUTE:

(a) H1 = H 2 + H 3 . Using Eq.(17.21) and dividing by the common area gives,

k1 k k ( k1 L1 ) (100°C − T0 ) = 2 T0 + 3 T0 . Solving for T0 gives T0 = (100°C ). Substitution of L1 L2 L3 k L + ( 1 1 ) ( k2 L2 ) + ( k3 L3 ) numerical values gives T0 = 78.4°C. (b) Using H = kA ΔT for each rod, with ΔT1 = 21.6 C°, ΔT2 = ΔT3 = 78.4 C° gives H1 = 12.8 W, H 2 = 9.50 W L and H 3 = 3.30 W. EVALUATE:

In part (b), H1 is seen to be the sum of H 2 and H 3 .

17.113. (a) EXECUTE: Heat must be conducted from the water to cool it to 0°C and to cause the phase transition. The entire volume of water is not at the phase transition temperature, just the upper surface that is in contact with the ice sheet. (b) IDENTIFY: The heat that must leave the water in order for it to freeze must be conducted through the layer of ice that has already been formed. SET UP: Consider a section of ice that has area A. At time t let the thickness be h. Consider a short time interval t to t + dt. Let the thickness that freezes in this time be dh. The mass of the section that freezes in the time interval dt is dm = ρ dV = ρ A dh. The heat that must be conducted away from this mass of water to freeze it is dQ = dmLf = ( ρ ALf )dh. H = dQ/dt = kA(ΔT/h), so the heat dQ conducted in time dt throughout the thickness h

⎛T −T ⎞ that is already there is dQ = kA ⎜ H C ⎟ dt. Solve for dh in terms of dt and integrate to get an expression relating ⎝ h ⎠ h and t.


17-23

Equate these expressions for dQ. ⎛T −T ⎞ ρ ALf dh = kA ⎜ H C ⎟ dt ⎝ h ⎠

EXECUTE:

⎛ k (TH − TC ) ⎞ h dh = ⎜ ⎟ dt ⎝ ρ Lf ⎠ Integrate from t = 0 to time t. At t = 0 the thickness h is zero.

∫

h 0

t

h dh = [k (TH − TC ) ρ Lf ]∫ dt 0

k (TH − TC ) 2k (TH − TC ) 1 2 h = t and h = t 2 ρ Lf ρ Lf The thickness after time t is proportional to t . h 2 ρ Lf (0.25 m) 2 (920 kg/m3 )(334 × 103 J/kg) = = 6.0 × 105 s (c) The expression in part (b) gives t = 2k (TH − TC ) 2(1.6 W/m ⋅ K)(0°C − (−10°C)) t = 170 h. (d) Find t for h = 40 m. t is proportional to h 2 , so t = (40 m/0.25 m) 2 (6.00 × 105 s) = 1.5 × 1010 s. This is about 500 years. With our current climate this will not happen. EVALUATE: As the ice sheet gets thicker, the rate of heat conduction through it decreases. Part (d) shows that it takes a very long time for a moderately deep lake to totally freeze. 17.114. IDENTIFY: Apply Eq.(17.22) at each end of the short element. In part (b) use the fact that the net heat current into the element provides the Q for the temperature increase, according to Q = mcΔT . SET UP: dT / dx is the temperature gradient. EXECUTE: (a) H = (380 W m ⋅ K)(2.50 × 10−4 m 2 )(140 C° m) = 13.3 W. ΔT Q ΔT = mc , where = 0.250 C°/s. t t t ρ cΔx ⎛ ΔT ⎞ dT dT = kA + The mass m is ρ AΔx, so kA ⎜ ⎟. dx 2 dx 1 k ⎝ t ⎠

(b) Denoting the two ends of the element as 1 and 2, H 2 − H1 =

kA

dT dT ⎛ ΔT − kA = mc ⎜ dx 2 dx 1 ⎝ t

kA

dT dx

= 140 C°/m + 2

⎞ ⎟. ⎠

(1.00 × 10 4 kg/m 3 )(520 J/kg ⋅ K)(1.00 × 10−2 m)(0.250 C°/s) = 174 C°/m. 380 W/m ⋅ K

EVALUATE: At steady-state temperature of the short element is no longer changing and H1 = H 2 . 17.115. IDENTIFY: The rate of heat conduction through the walls is 1.25 kW. Use the concept of thermal resistance and the fact that when insulating materials are in layers, the R values are additive. SET UP: The total area of the four walls is 2(3.50 m)(2.50 m) + 2(3.00 m)(2.50 m) = 32.5 m 2

TH − TC A(TH − TC ) (32.5 m 2 )(17.0 K) gives R = = = 0.442 m 2 ⋅ K / W. For the wood, H 1.25 × 10−3 W R L 1.80 × 10−2 m Rw = = = 0.300 m 2 ⋅ K / W. For the insulating material, Rin = R − Rw = 0.142 m 2 ⋅ K / W. k 0.060 W/m ⋅ K L 1.50 × 10−2 m L = 0.106 W/m ⋅ K. Rin = in and kin = in = Rin 0.142 m 2 ⋅ K / W kin EVALUATE: The thermal conductivity of the insulating material is larger than that of the wood, the thickness of the insulating material is less than that of the wood, and the thermal resistance of the wood is about three times that of the insulating material. 17.116. IDENTIFY: I1r12 = I 2 r22 . Apply H = Aeσ T 4 (Eq.17.25) to the sun. EXECUTE:

H=A

SET UP: I1 = 1.50 × 103 W/m 2 when r = 1.50 × 1011 m. EXECUTE: (a) The energy flux at the surface of the sun is 2

⎛ 1.50 × 1011 m ⎞ 7 2 I 2 = (1.50 × 103 W/m 2 ) ⎜ ⎟ = 6.97 × 10 W/m . 8 × 6.96 10 m ⎝ ⎠ 1

1

7 2 ⎤4 ⎡ H 1 ⎤ 4 ⎡ 6.97 × 10 W m = = 5920 K. (b) Solving Eq.(17.25) with e = 1, T = ⎢ ⎢ ⎥ − 8 2 4 ⎥ ⎣ A σ ⎦ ⎣ 5.67 × 10 W m ⋅ K ⎦ EVALUATE: The total power output of the sun is P = 4π r22 I 2 = 2.0 × 1031 W.

17-24

Chapter 17

17.117. IDENTIFY and SET UP: Use Eq.(17.26) to find the net heat current into the can due to radiation. Use Q = Ht to find the heat that goes into the liquid helium, set this equal to mL and solve for the mass m of helium that changes phase. EXECUTE: Calculate the net rate of radiation of heat from the can. H net = Aeσ (T 4 − Ts4 ).

The surface area of the cylindrical can is A = 2π rh + 2π r 2 . (See Figure 17.117.) Figure 17.117

A = 2π r ( h + r ) = 2π (0.045 m)(0.250 m + 0.045 m) = 0.08341 m 2 . H net = (0.08341 m 2 )(0.200)(5.67 × 10−8 W/m 2 ⋅ K 4 )((4.22 K) 4 − (77.3 K)4 ) H net = −0.0338 W (the minus sign says that the net heat current is into the can). The heat that is put into the can by

radiation in one hour is Q = −( H net )t = (0.0338 W)(3600 s) = 121.7 J. This heat boils a mass m of helium according

Q 121.7 J = = 5.82 × 10−3 kg = 5.82 g. Lf 2.09 × 104 J/kg EVALUATE: In the expression for the net heat current into the can the temperature of the surroundings is raised to the fourth power. The rate at which the helium boils away increases by about a factor of (293/77) 4 = 210 if the walls surrounding the can are at room temperature rather than at the temperature of the liquid nitrogen. 17.118. IDENTIFY: The coefficient of volume expansion β is defined by ΔV = V0 βΔT . to the equation Q = mLf , so m =

SET UP:

For copper, β = 5.1 × 10−5 K −1. (a) With Δp = 0, pΔV = nRΔT =

EXECUTE: (b)

ΔV ΔT 1 pV = , and β = . ΔT , or V T T T

β air 1 = = 67. β copper (293 K)(5.1 × 10−5 K −1 )

EVALUATE: The coefficient of volume expansion for air is much greater than that for copper. For a given ΔT , gases expand much more than solids do. 17.119. IDENTIFY: For the water, Q = mcΔT . SET UP: For water, c = 4190 J/kg ⋅ K. (a) At steady state, the input power all goes into heating the water, so P =

EXECUTE:

Q mcΔT = and t t

Pt (1800 W)(60 s/min) = = 51.6 K, and the output temperature is 18.0°C + 51.6°C = 69.6°C. cm (4190 J kg ⋅ K)(0.500 kg/min) EVALUATE: (b) At steady state, the temperature of the apparatus is constant and the apparatus will neither remove heat from nor add heat to the water. 17.120. IDENTIFY: For the air the heat input is related to the temperature change by Q = mcΔT . SET UP: The rate P at which heat energy is generated is related to the rate P0 at which food energy is consumed ΔT =

by the hamster by P = 0.10 P0 . EXECUTE: (a) The heat generated by the hamster is the heat added to the box; Q ΔT P = = mc = (1.20 kg m3 )(0.0500 m3 )(1020 J kg ⋅ K)(1.60 C° h ) = 97.9 J h. t t (b) Taking the efficiency into account, the mass M of seed that must be eaten in time t is M P0 P (10%) 979 J h = = = = 40.8 g h. 24 J g t Lc Lc EVALUATE: This is about 1.5 ounces of seed consumed in one hour. 17.121. IDENTIFY: Heat Qi goes into the ice when it warms to 0°C, melts, and the resulting water warms to the final temperature Tf . Heat Qow comes out of the ocean water when it cools to Tf . Conservation of energy gives Qi + Qow = 0.

SET UP:

For ice, ci = 2100 J/kg ⋅ K. For water, Lf = 334 × 103 J/kg and cw = 4190 J/kg ⋅ K. Let m be the total

mass of the water on the earth's surface. So mi = 0.0175m and mow = 0.975m.


EXECUTE:

17-25

Qi + Qow = 0 gives mici (30 C°) + mi Lf + micwTf + mow cw (Tf − 5.00°C) = 0.

Tf =

−mici (30 C°) − mi Lf + mow cw (5.00 C°) . ( mi + mow )cw

Tf =

−(0.0175m)(2100 J/kg ⋅ K)(30 K) − (0.0175m)(334 × 103 J/kg) + (0.975m)(4190 J/kg ⋅ K)(5.00 K) (0.0175m + 0.975m)(4190 J/kg ⋅ K)

Tf =

1.348 × 104 J/kg = 3.24°C. The temperature decrease is 1.76 C°. 4.159 × 103 J/kg ⋅ K

EVALUATE: The mass of ice in the icecaps is much less than the mass of the water in the oceans, but much more heat is required to change the phase of 1 kg of ice than to change the temperature of 1 kg of water 1 C°, so the lowering of the temperature of the oceans would be appreciable. 17.122. IDENTIFY: Apply Eq.(17.21). For a spherical or cylindrical surface, the area A in Eq.(17.21) is not constant, and the material must be considered to consist of shells with thickness dr and a temperature difference between the inside and outside of the shell dT . The heat current will be a constant, and must be found by integrating a differential equation. SET UP: The surface area of a sphere is 4π r 2 . The surface area of the curved side of a cylinder is 2π rl . ln(1 + ε ) ≈ ε when ε ! 1. (a) Equation (17.21) becomes H = k (4π r 2 )

dT H dr = k dT . Integrating both sides between the appropriate or 4π r 2 dr

H ⎛1 1⎞ ⎜ − ⎟ = k (T2 − T1 ). In this case the “appropriate limits” have been chosen so that if the inner 4π ⎝ a b ⎠ temperature T2 is at the higher temperature T1 , the heat flows outward; that is, dT < 0. Solving for the heat dr k 4π ab(T2 − T1 ) . current, H = b−a (b) The rate of change of temperature with radius is of the form dT = B2 , with B a constant. Integrating from dr r ⎛1 1⎞ ⎛1 1⎞ r = a to r and from r = a to r = b gives T ( r ) − T2 = B ⎜ − ⎟ and T1 − R2 = B ⎜ − ⎟ . Using the second of these ⎝a r⎠ ⎝a b⎠

limits,

⎛ r − a ⎞⎛ b ⎞ to eliminate B and solving for T(r) gives T ( r ) = T2 − (T2 − T1 ) ⎜ ⎟⎜ ⎟ . There are, of course, many equivalent ⎝ b − a ⎠⎝ r ⎠ forms. As a check, note that at r = a, T = T2 and at r = b, T = T1.

dT H or = kLdT , which integrates, with πr 2 dr 2π kL(T2 − T1 ) H . ln(b a ) = kL(T2 − T1 ), or H = the same condition on the limits, to ln(b a ) 2π (c) As in part (a), the expression for the heat current is H = k (2π rL)

(d) A method similar to that used in part (b) gives T ( r ) = T2 + (T1 − T2 ) EVALUATE:

ln( r a) . ln(b a )

(e) For the sphere: Let b − a = l , and approximate b ~a, with a the common radius. Then the surface

area of the sphere is A = 4π a 2 , and the expression for H is that of Eq. (17.21) (with l instead of L, which has

l⎞ l ⎛b⎞ ⎛ another use in this problem). For the cylinder: with the same notation, consider ln ⎜ ⎟ = ln ⎜1 + ⎟ ~ , a a ⎝ ⎠ ⎝ ⎠ a where the approximation for ln(1 + ε ) for small ε has been used. The expression for H then reduces to

k ( 2π La )( ΔT l ) , which is Eq. (17.21) with A = 2π La. 17.123. IDENTIFY: From the result of Problem 17.122, the heat current through each of the jackets is related to the temperature difference by H = 2π lk ΔT , where l is the length of the cylinder and b and a are the inner and ln ( b a ) outer radii of the cylinder.

17-26

Chapter 17

SET UP: Let the temperature across the cork be ΔT1 and the temperature across the styrofoam be ΔT2 , with similar notation for the thermal conductivities and heat currents. EXECUTE: (a) ΔT1 + ΔT2 = ΔT = 125 C°. Setting H1 = H 2 = H and canceling the common factors, −1

⎛ ΔT1k1 ΔT2k2 k ln 1.5 ⎞ = . Eliminating ΔT2 and solving for ΔT1 gives ΔT1 = ΔT ⎜1 + 1 ⎟ . Substitution of numerical ln 2 ln 1.5 k 2 ln 2 ⎠ ⎝ values gives ΔT1 = 37 C°, and the temperature at the radius where the layers meet is 140°C − 37°C = 103°C. (b) Substitution of this value for ΔT1 into the above expression for H1 = H gives 2π ( 2.00 m )( 0.0400 W m ⋅ K ) H= ( 37 C° ) = 27 W. ln 2 2π ( 2.00 m )( 0.0100 W m ⋅ K ) EVALUATE: ΔT = 103°C − 15°C = 88 C°. H 2 = ( 88 C° ) = 27 W. This is the same ln(6.00/4.00)

as H1 , as it should be. 17.124. IDENTIFY:

Apply Eq.(17.22) to different points along the rod, where

dT is the temperature gradient at each point. dx

SET UP: For copper, k = 385 W/m ⋅ K. EXECUTE: (a) The initial temperature distribution, T = (100°C)sin π x/L, is shown in Figure 17.124a. (b) After a very long time, no heat will flow, and the entire rod will be at a uniform temperature which must be that of the ends, 0°C. (c) The temperature distribution at successively greater times T1 < T2 < T3 is sketched in Figure 17.124b. dT = (100°C )(π L ) cosπ x L. At the ends, x = 0 and x = L, the cosine is ±1 and the temperature gradient is (d) dx ± (100°C )(π 0.100 m ) = ±3.14 × 103 C° m. (e) Taking the phrase “into the rod” to mean an absolute value, the heat current will be kA dT = (385.0 W m ⋅ K) (1.00 × 10−4 m 2 )(3.14 × 103 C° m) = 121 W. dx (f) Either by evaluating dT at the center of the rod, where π x L = π 2 and cos (π 2 ) = 0, or by checking the dx figure in part (a), the temperature gradient is zero, and no heat flows through the center; this is consistent with the symmetry of the situation. There will not be any heat current at the center of the rod at any later time. k (385 W m ⋅ K) = = 1.1 × 10−4 m 2 s. (g) ρ c (8.9 × 103 kg m3 )(390 J kg ⋅ K) (h) Although there is no net heat current, the temperature of the center of the rod is decreasing; by considering the heat current at points just to either side of the center, where there is a non-zero temperature gradient, there must be a net flow of heat out of the region around the center. Specifically, ⎛ ∂T ⎞ ∂T ∂T ∂ 2T H (( L/2) + Δ x ) − H (( L/2) − Δ x ) = ρ A(Δ x)c = kA ⎜ − = kA 2 Δ x, from which the Heat ⎟ ⎜ ∂x ∂t ∂x ( L/ 2) −Δ x ⎟⎠ ∂x ( L/ 2) + Δ x ⎝

Equation,

2 ∂T k ∂ 2T = is obtained. At the center of the rod, ∂ T2 = −(100 C°)( π L)2 , and so 2 ∂x ∂t ρ c ∂x 2

∂T ⎛ π ⎞ = −(1.11 × 10−4 m 2 s)(100 C°) ⎜ ⎟ = −10.9 C° s, or −11C° s to two figures. ∂t ⎝ 0.100 m ⎠ 100 C° = 9.17 s (i) 10.9 C° s 2 (j) Decrease (that is, become less negative), since as T decreases, ∂ T2 decreases. This is consistent with the ∂x graphs, which correspond to equal time intervals. 2 (k) At the point halfway between the end and the center, at any given time ∂ T2 is a factor of sin (π 4 ) = 1 2 less ∂x than at the center, and so the initial rate of change of temperature is −7.71C° s.


EVALUATE: A plot of temperature as a function of both position and time for 0 ≤ t ≤ 50 s is shown in Figure 17.124c.

Figure 17.124

17-27

17-28

Chapter 17

17.125. IDENTIFY: Apply the concept of thermal expansion. In part (b) the object can be treated as a simple pendulum. SET UP: For steel α = 1.2 × 10−5 (C°) −1. 1 yr = 86,400 s. EXECUTE: (a) In hot weather, the moment of inertia I and the length d in Eq.(13.39) will both increase by the same factor, and so the period will be longer and the clock will run slow (lose time). Similarly, the clock will run fast (gain time) in cold weather. (b) ΔL = αΔT = (1.2 × 10−5 (C°) −1 )(10.0 C°) = 1.2 × 10−4 . L0 (c) See Problem 13.98. To avoid possible confusion, denote the pendulum period by τ . For this problem, Δτ = 1 ΔL = 6.0 × 10−5 so in one day the clock will gain (86,400 s)(6.0 × 10−5 ) = 5.2 s. τ 2 L Δτ 1 Δτ 1.0 s = 2 αΔT . = gives ΔT = 2[(1.2 × 10−5 (C°) −1 )(86,400)]−1 = 1.9 C°. T must be controlled to (d) τ τ 86,400 s

within 1.9 C°. EVALUATE: In part (d) the answer does not depend on the period of the pendulum. It depends only on the fractional change in the period. 17.126. IDENTIFY: The rate at which heat is absorbed at the blackened end is the heat current in the rod, kA Aeσ (TS4 − T24 ) = (T2 − T1 ) where T1 = 20.00 K and T2 is the temperature of the blackened end of the rod. L SET UP: Since the end is blackened, e = 1. Ts = 500.0 K. If the equation were to be solved exactly for T2 , the equation would be a quartic, very likely not

EXECUTE:

worth the trouble. Following the hint, approximate T2 on the left side of the above expression as T1 to obtain

σL

(T 2 − T14 ) = T1 + (6.79 × 10−12 K −3 )(Ts4 − T14 ) = T1 + 0.424 K = 20.42 K. k S EVALUATE: This approximation for T2 is indeed only slightly than T1 , and is a good estimate of the temperature.

T2 = T1 +

Using this for T2 in the original expression to find a better value of ΔT gives the same ΔT to eight figures, and further iterations are not worthwhile. 17.127. IDENTIFY: The rate in (iv) is given by Eq.(17.26), with T = 309 K and Ts = 320 K. The heat absorbed in the evaporation of water is Q = mL.

m = ρ. V EXECUTE: (a) The rates are: (i) 280 W, (ii) (54 J/h ⋅ C° ⋅ m 2 )(1.5 m 2 )(11 C°)/(3600 s/h) = 0.248 W, SET UP:

m = ρV , so

(iii) (1400 W/m 2 )(1.5 m 2 ) = 2.10 × 103 W, (iv) (5.67 × 10 −8 W/m 2 ⋅ K 4 )(1.5 m 2 )((320 K) 4 − (309 K) 4 ) = 116 W. The total is 2.50 kW, with the largest portion due to radiation from the sun.

P 2.50 × 103 W = = 1.03 × 10−6 m 3/s. This is equal to = 3.72 L/h. ρ Lv (1000 kg m3 )(2.42 × 106 J kg ⋅ K) (c) Redoing the above calculations with e = 0 and the decreased area gives a power of 945 W and a corresponding evaporation rate of 1.4 L/h. Wearing reflective clothing helps a good deal. Large areas of loose weave clothing also facilitate evaporation. EVALUATE: The radiant energy from the sun absorbed by the area covered by clothing is assumed to be zero, since e ≈ 0 for the clothing and the clothing reflects almost all the radiant energy incident on it. For the same reason, the exposed skin area is the area used in Eq.(17.26). (b)

THERMAL PROPERTIES OF MATTER

18.1.

18

(a) IDENTIFY: We are asked about a single state of the system. SET UP: Use Eq.(18.2) to calculate the number of moles and then apply the ideal-gas equation. m 0.225 kg = 56.2 mol EXECUTE: n = tot = M 4.00 × 10−3 kg/mol (b) pV = nRT implies p = nRT / V T must be in kelvins; T = (18 + 273) K = 291 K

(56.2 mol)(8.3145 J/mol ⋅ K)(291 K) = 6.80 × 106 Pa 20.0 × 10−3 m3 p = (6.80 × 106 Pa)(1.00 atm/1.013 × 105 Pa) = 67.1 atm EVALUATE: Example 18.1 shows that 1.0 mol of an ideal gas is about this volume at STP. Since there are 56.2 moles the pressure is about 60 times greater than 1 atm. IDENTIFY: pV = nRT . p=

18.2.

SET UP: EXECUTE:

18.3.

18.4.

T1 = 41.0°C = 314 K . R = 0.08206 L ⋅ atm/mol ⋅ K . n, R constant so

pV pV pV = nR is constant. 1 1 = 2 2 . T1 T2 T

⎛ p ⎞⎛ V ⎞ T2 = T1 ⎜ 2 ⎟⎜ 2 ⎟ = (314 K)(2)(2) = 1.256 × 103 K = 983°C . ⎝ p1 ⎠⎝ V1 ⎠ (1.30 atm)(2.60 L) pV (b) n = = = 0.131 mol . mtot = nM = (0.131 mol)(4.00 g/mol) = 0.524 g . RT (0.08206 L ⋅ atm/mol ⋅ K)(314 K) EVALUATE: T is directly proportional to p and to V, so when p and V are each doubled the Kelvin temperature increases by a factor of 4. IDENTIFY: pV = nRT . SET UP: T is constant. ⎛V ⎞ ⎛ 0.110 m3 ⎞ EXECUTE: nRT is constant so p1V1 = p2V2 . p2 = p1 ⎜ 1 ⎟ = (3.40 atm) ⎜ = 0.959 atm . 3 ⎟ ⎝ 0.390 m ⎠ ⎝ V2 ⎠ EVALUATE: For T constant, p decreases. IDENTIFY: pV = nRT . SET UP: EXECUTE:

T1 = 20.0°C = 293 K . (a) n, R, and V are constant.

p p p nR = = constant . 1 = 2 . T1 T2 T V

⎛p ⎞ ⎛ 1.00 atm ⎞ T2 = T1 ⎜ 2 ⎟ = (293 K) ⎜ ⎟ = 97.7 K = −175°C . p ⎝ 3.00 atm ⎠ ⎝ 1⎠ (b) p2 = 1.00 atm , V2 = 3.00 L . p3 = 3.00 atm . n, R, and T are constant so pV = nRT = constant . p2V2 = p3V3 . ⎛p ⎞ ⎛ 1.00 atm ⎞ V3 = V2 ⎜ 2 ⎟ = (3.00 L) ⎜ ⎟ = 1.00 L . ⎝ 3.00 atm ⎠ ⎝ p3 ⎠ EVALUATE: The final volume is one-third the initial volume. The initial and final pressures are the same, but the final temperature is one-third the initial temperature.

18-1

18-2

18.5.

Chapter 18

pV = nRT

IDENTIFY:

SET UP: Assume a room size of 20 ft × 20 ft × 10 ft . V = 4000 ft 3 = 113 m3 . Assume a temperature of T = 20°C = 293 K and a pressure of p = 1.01 × 105 Pa . 1 m3 = 106 cm3 . (a) n =

EXECUTE:

pV (1.01 × 105 Pa)(113 m3 ) = = 4.68 × 103 mol . RT (8.315 J/mol ⋅ K)(293 K)

N = nN A = (4.68 × 103 mol)(6.022 × 1023 molecules/mol) = 3 × 1027 molecules . N 3 × 1027 molecules = = 3 × 1025 molecules/m3 = 3 × 1019 molecules/cm3 V 113 m3 EVALUATE: The solution doesn't rely on the assumption that air is all N 2 . (b)

18.6.

pV = nRT and the mass of the gas is mtot = nM .

IDENTIFY: SET UP:

The temperature is T = 22.0°C = 295.15K. The average molar mass of air is M = 28.8 × 10−3 kg mol .

For helium M = 4.00 × 10−3 kg mol . (a) mtot = nM =

EXECUTE:

(b) mtot = nM =

pV (1.00 atm)(0.900 L)(4.00 × 10−3 kg/mol) M= = 1.49 × 10−4 kg. RT (0.08206 L ⋅ atm/mol ⋅ K)(295.15 K)

N pV = says that in each case the balloon contains the same number of molecules. The mass N A RT is greater for air since the mass of one molecule is greater than for helium. IDENTIFY: We are asked to compare two states. Use the ideal gas law to obtain T2 in terms of T1 and ratios of pressures and volumes of the gas in the two states. SET UP: pV = nRT and n, R constant implies pV / T = nR = constant and p1V1 / T1 = p2V2 / T2 n=

EVALUATE:

18.7.

pV (1.00 atm)(0.900 L)(28.8 × 10−3 kg/mol) M= = 1.07 × 10−3 kg. RT (0.08206 L ⋅ atm/mol ⋅ K)(295.15 K)

T1 = (27 + 273) K = 300 K

EXECUTE:

p1 = 1.01 × 10 Pa 5

p2 = 2.72 × 106 Pa + 1.01 × 105 Pa = 2.82 × 106 Pa (in the ideal gas equation the pressures must be absolute, not gauge, pressures) ⎛ p ⎞⎛ V ⎞ ⎛ 2.82 × 106 Pa ⎞⎛ 46.2 cm3 ⎞ = 776 K T2 = T1 ⎜ 2 ⎟⎜ 2 ⎟ = 300 K ⎜ ⎟⎜ 5 3 ⎟ ⎝ 1.01 × 10 Pa ⎠⎝ 499 cm ⎠ ⎝ p1 ⎠⎝ V1 ⎠ T2 = (776 − 273)°C = 503°C EVALUATE: 18.8.

The units cancel in the V2 / V1 volume ratio, so it was not necessary to convert the volumes in cm3

to m3 . It was essential, however, to use T in kelvins. IDENTIFY: pV = nRT and m = nM . SET UP:

We must use absolute pressure in pV = nRT . p1 = 4.01 × 105 Pa , p2 = 2.81 × 105 Pa . T1 = 310 K ,

T2 = 295 K . EXECUTE:

(a) n1 =

p1V1 (4.01 × 105 Pa)(0.075 m3 ) = = 11.7 mol . m = nM = (11.7 mol)(32.0 g/mol) = 374 g . RT1 (8.315 J/mol ⋅ K)(310 K)

p2V2 (2.81 × 105 Pa)(0.075 m 3 ) = = 8.59 mol . m = 275 g . RT2 (8.315 J/mol ⋅ K)(295 K) The mass that has leaked out is 374 g − 275 g = 99 g . EVALUATE: In the ideal gas law we must use absolute pressure, expressed in Pa, and T must be in kelvins. IDENTIFY: pV = nRT . (b) n2 =

18.9.

SET UP: EXECUTE:

T1 = 300 K , T2 = 430 K . (a) n, R are constant so

pV pV pV = nR = constant . 1 1 = 2 2 . T1 T2 T

⎛ V ⎞⎛ T ⎞ ⎛ 0.750 m3 ⎞ ⎛ 430 K ⎞ 5 p2 = p1 ⎜ 1 ⎟⎜ 2 ⎟ = (1.50 × 105 Pa) ⎜ ⎟ = 3.36 × 10 Pa . 3 ⎟⎜ V T 0.480 m 300 K ⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠⎝ 1 ⎠ EVALUATE: In pV = nRT , T must be in kelvins, even if we use a ratio of temperatures.

Thermal Properties of Matter

18.10.

18-3

Use the ideal-gas equation to calculate the number of moles, n. The mass mtotal of the gas is

IDENTIFY:

mtotal = nM . SET UP:

The volume of the cylinder is V = π r 2l , where r = 0.450 m and l = 1.50 m. T = 22.0°C = 293.15 K.

1 atm = 1.013 × 105 Pa. M = 32.0 × 10−3 kg/mol. R = 8.314 J/mol ⋅ K. (a) pV = nRT gives n =

EXECUTE:

18.11.

(b) mtotal = (827 mol)(32.0 × 10−3 kg/mol) = 26.5 kg EVALUATE: In the ideal-gas law, T must be in kelvins. Since we used R in units of J/mol ⋅ K we had to express p in units of Pa and V in units of m3 . IDENTIFY: We are asked to compare two states. Use the ideal-gas law to obtain V1 in terms of V2 and the ratio of the temperatures in the two states. SET UP: pV = nRT and n, R, p are constant so V / T = nR / p = constant and V1 / T1 = V2 / T2

T1 = (19 + 273) K = 292 K (T must be in kelvins)

EXECUTE:

18.12.

pV (21.0 atm)(1.013 × 105 Pa/atm)π (0.450 m) 2 (1.50 m) = = 827 mol. RT (8.314 J/mol ⋅ K)(295.15 K)

V2 = V1 (T2 / T1 ) = (0.600 L)(77.3 K/292 K) = 0.159 L EVALUATE: p is constant so the ideal-gas equation says that a decrease in T means a decrease in V. IDENTIFY: Apply pV = nRT and the van der Waals equation (Eq.18.7) to calculate p. SET UP: 400 cm3 = 400 × 10−6 m3 . R = 8.314 J/mol ⋅ K. EXECUTE: (a) The ideal gas law gives p = nRT V = 7.28 × 106 Pa while Eq.(18.7) gives 5.87 × 106 Pa. (b) The van der Waals equation, which accounts for the attraction between molecules, gives a pressure that is 20% lower. (c) The ideal gas law gives p = 7.28 × 105 Pa. Eq.(18.7) gives p = 7.13 × 105 Pa, for a 2.1% difference.

18.13.

18.14.

EVALUATE: (d) As n V decreases, the formulas and the numerical values for the two equations approach each other. IDENTIFY: pV = nRT . SET UP: T is constant. EXECUTE: n, R, T are constant, so pV = nRT = constant. p1V1 = p2V2 .

⎛V ⎞ ⎛ 6.00 L ⎞ p2 = p1 ⎜ 1 ⎟ = (1.00 atm) ⎜ ⎟ = 1.05 atm. ⎝ 5.70 L ⎠ ⎝ V2 ⎠ EVALUATE: For constant T, when V decreases, p increases. Since the volumes enter as a ratio we don't have to convert from L to m3 . IDENTIFY: pV = nRT . SET UP:

T1 = 277 K. T2 = 296 K. Assume the number of moles of gas in the bubble remains constant. (a) n, R are constant so

EXECUTE:

18.15.

pV pV pV = nR = constant. 1 1 = 2 2 and T1 T2 T

V2 ⎛ p1 ⎞⎛ T2 ⎞ ⎛ 3.50 atm ⎞⎛ 296 K ⎞ = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = 3.74. V1 ⎝ p2 ⎠⎝ T1 ⎠ ⎝ 1.00 atm ⎠⎝ 277 K ⎠ (b) This increase in volume of air in the lungs would be dangerous. EVALUATE: The large decrease in pressure results in a large increase in volume. IDENTIFY: We are asked to compare two states. First use pV = nRT to calculate p1. Then use it to obtain T2 in terms of T1 and the ratio of pressures in the two states. (a) SET UP: EXECUTE: SET UP:

pV = nRT . Find the initial pressure p1:

nRT1 (11.0 mol)(8.3145 J/mol ⋅ K)((23.0 + 273.13)K) = = 8.737 × 106 Pa V 3.10 × 10−3 m3 p2 = 100 atm(1.013 × 105 Pa/1 atm) = 1.013 × 107 Pa p1 =

p / T = nR / V = constant, so p1 / T1 = p2 / T2 ⎛p ⎞ ⎛ 1.013 × 107 Pa ⎞ T2 = T1 ⎜ 2 ⎟ = (296.15 K) ⎜ ⎟ = 343.4 K = 70.2°C 6 ⎝ 8.737 × 10 Pa ⎠ ⎝ p1 ⎠ (b) EVALUATE: The coefficient of volume expansion for a gas is much larger than for a solid, so the expansion of the tank is negligible. EXECUTE:

18-4

Chapter 18

18.16.

IDENTIFY: F = pA and pV = nRT SET UP: For a cube, V / A = L. EXECUTE: (a) The force of any side of the cube is F = pA = ( nRT V ) A = (nRT ) L, since the ratio of area to

volume is A / V = 1/ L. For T = 20.0°C = 293.15 K, nRT (3 mol) (8.3145 J mol ⋅ K) (293.15 K) = = 3.66 × 104 N. L 0.200 m (b) For T = 100.00°C = 373.15 K, F=

F=

18.17.

nRT (3 mol)(8.3145 J mol ⋅ K)(373.15 K) = = 4.65 × 104 N. L 0.200 m

EVALUATE: When the temperature increases while the volume is kept constant, the pressure increases and therefore the force increases. The force increases by the factor T2 / T1. IDENTIFY: Example 18.4 assumes a temperature of 0°C at all altitudes and neglects the variation of g with

elevation. With these approximations, p = p0e − Mgy / RT . SET UP:

RT ln(0.90) = 850 m. Mg EVALUATE: This is a commonly occurring elevation, so our calculation shows that 10% variations in atmospheric pressure occur at many locations. IDENTIFY: From Example 18.4, the pressure at elevation y above sea level is p = p0e − Mgy / RT . EXECUTE:

18.18.

ln(e − x ) = − x. For air, M = 28.8 × 10−3 kg/mol.

SET UP: EXECUTE:

We want y for p = 0.90 p0 so 0.90 = e − Mgy / RT and y = −

The average molar mass of air is M = 28.8 × 10−3 kg/mol. At an altitude of 100 m,

Mgy1 (28.8 × 10−3 kg mol)(9.80 m s 2 )(100 m) = = 0.01243, and the percent RT (8.3145 J mol ⋅ K)(273.15 K)

decrease in pressure is 1 − p p0 = 1 − e −0.01243 = 0.0124 = 1.24%. At an altitude of 1000 m, Mgy2 RT = 0.1243 and

18.19.

the percent decrease in pressure is 1 − e −0.1243 = 0.117 = 11.7%. EVALUATE: These answers differ by a factor of (11.7%) (1.24%) = 9.44, which is less than 10 because the variation of pressure with altitude is exponential rather than linear. IDENTIFY: p = p0e − Myg RT from Example 18.4. Eq.(18.5) says p = ( ρ M )RT. Example 18.4 assumes a constant T = 273 K, so p and ρ are directly proportional and we can write ρ = ρ0e − Mgy RT . Mgy = 1.10 when y = 8863 m. RT Mgy EXECUTE: For y = 100 m, = 0.0124, so ρ = ρ 0e−0.0124 = 0.988 ρ 0 . The density at sea level is 1.2% larger RT than the density at 100 m. m EVALUATE: The pressure decreases with altitude. pV = tot RT , so when the pressure decreases and T is M constant the volume of a given mass of gas increases and the density decreases. IDENTIFY: p = p0e − Mgy / RT from Example 18.4 gives the variation of air pressure with altitude. The density ρ SET UP:

18.20.

From Example 18.4,

of the air is ρ =

pM , so ρ is proportional to the pressure p. Let ρ0 be the density at the surface, where the RT

pressure is p0 . SET UP:

From Example 18.4,

Mg (28.8 × 10−3 kg/mol)(9.80 m/s 2 ) = = 1.244 × 10−4 m −1. RT (8.314 J/mol ⋅ K)(273 K)

⎛ p⎞ M ρ ρ = constant, so = 0 and ρ = ρ 0 ⎜ ⎟ = 0.883ρ 0 . p RT p p0 ⎝ p0 ⎠ The density at an altitude of 1.00 km is 88.3% of its value at the surface. EVALUATE: If the temperature is assumed to be constant, then the decrease in pressure with increase in altitude corresponds to a decrease in density. EXECUTE:

p = p0e − (1.244×10

−4

m −1 )(1.00×103 m)

= 0.883 p0 .

ρ

=


18.21.

18-5

IDENTIFY: Use Eq.(18.5) and solve for p. SET UP: ρ = pM / RT and p = RT ρ / M T = ( −56.5 + 273.15) K = 216.6 K

For air M = 28.8 × 10−3 kg/mol (Example 18.3) (8.3145 J/mol ⋅ K)(216.6 K)(0.364 kg/m3 ) = 2.28 × 104 Pa 28.8 × 10−3 kg/mol EVALUATE: The pressure is about one-fifth the pressure at sea-level. IDENTIFY: The molar mass is M = N A m , where m is the mass of one molecule. p=

EXECUTE: 18.22.

N A = 6.02 × 1023 molecules/mol .

SET UP:

M = N A m = (6.02 × 1023 molecules mol)(1.41 × 10−21 kg molecule) = 849 kg/mol.

EXECUTE: EVALUATE:

For a carbon atom, M = 12 × 10−3 kg/mol . If this molecule is mostly carbon, so the average mass of

its atoms is the mass of carbon, the molecule would contain 18.23.

849 kg/mol = 71,000 atoms . 12 × 10−3 kg/mol

The mass mtot is related to the number of moles n by mtot = nM . Mass is related to volume by

IDENTIFY: ρ = m /V .

For gold, M = 196.97 g/mol and ρ = 19.3 × 103 kg/m3 . The volume of a sphere of radius r is V = 34 π r 3 .

SET UP:

EXECUTE: (a) mtot = nM = (3.00 mol)(196.97 g/mol) = 590.9 g. The value of this mass of gold is (590.9 g)($14.75/ g) = $8720 . (b) V =

m

ρ

=

0.5909 kg = 3.06 × 10−5 m3 . V = 34 π r 3 gives 19.3 × 103 kg/m 3 1/ 3

1/ 3

18.24.

⎛ 3[3.06 × 10−5 m3 ] ⎞ ⎛ 3V ⎞ r =⎜ ⎟ = 0.0194 m = 1.94 cm . The diameter is 2r = 3.88 cm . ⎟ =⎜ 4π ⎝ 4π ⎠ ⎝ ⎠ EVALUATE: The mass and volume are directly proportional to the number of moles. IDENTIFY: Use pV = nRT to calculate the number of moles and then the number of molecules would be N = nN A . 1 atm = 1.013 × 105 Pa . 1.00 cm3 = 1.00 × 10−6 m3 . N A = 6.022 × 1023 molecules/mol .

SET UP: EXECUTE:

(a) n =

pV (9.00 × 10−14 atm)(1.013 × 105 Pa/atm)(1.00 × 10−6 m3 ) = = 3.655 × 10−18 mol . RT (8.314 J/mol ⋅ K)(300.0 K)

N = nN A = (3.655 × 10−18 mol)(6.022 × 1023 molecules/mol) = 2.20 × 106 molecules . (b) N =

18.25.

N VN A N N pVN A = = constant and 1 = 2 . so p RT RT p1 p2

⎛p ⎞ 1.00 atm ⎛ ⎞ 19 N 2 = N1 ⎜ 2 ⎟ = (2.20 × 106 molecules) ⎜ ⎟ = 2.44 × 10 molecules . −14 ⎝ 9.00 × 10 atm ⎠ ⎝ p1 ⎠ EVALUATE: The number of molecules in a given volume is directly proportional to the pressure. Even at the very low pressure in part (a) the number of molecules in 1.00 cm3 is very large. IDENTIFY: We are asked about a single state of the system. SET UP: Use the ideal-gas law. Write n in terms of the number of molecules N. (a) EXECUTE: pV = nRT , n = N / N A so pV = ( N / N A ) RT ⎛ N ⎞⎛ R ⎞ p = ⎜ ⎟⎜ ⎟T ⎝ V ⎠ ⎝ NA ⎠ 8.3145 J/mol ⋅ K ⎛ 80 molecules ⎞⎛ ⎞ −12 p=⎜ ⎟ (7500 K) = 8.28 × 10 Pa 3 ⎟⎜ 23 −6 ⎝ 1 × 10 m ⎠⎝ 6.022 × 10 molecules/mol ⎠ p = 8.2 × 10−17 atm. This is much lower than the laboratory pressure of 1 × 10−13 atm in Exercise 18.24. (b) EVALUATE: The Lagoon Nebula is a very rarefied low pressure gas. The gas would exert very little force on an object passing through it.

18-6

Chapter 18

18.26.

IDENTIFY:

pV = nRT = NkT At STP, T = 273 K , p = 1.01 × 105 Pa . N = 6 × 109 molecules .

SET UP:

EXECUTE:. V =

18.27.

L3 = V so L = V 1/ 3 = 6.1 × 10−6 m . EVALUATE: This is a small cube. m N IDENTIFY: n = = M NA N A = 6.022 × 1023 molecules/mol . For water, M = 18 × 10−3 kg/mol .

SET UP:

EXECUTE:. n =

18.28.

NkT (6 × 109 molecules)(1.381 × 10−23 J/molecule ⋅ K)(273 K) = = 2.24 × 10−16 m3 . p 1.01 × 105 Pa

m 1.00 kg = = 55.6 mol . M 18 × 10−3 kg/mol

N = nN A = (55.6 mol)(6.022 × 1023 molecules/mol) = 3.35 × 1025 molecules . EVALUATE: Note that we converted M to kg/mol. N IDENTIFY: Use pV = nRT and n = with N = 1 to calculate the volume V occupied by 1 molecule. The length NA l of the side of the cube with volume V is given by V = l 3 . SET UP: T = 27°C = 300 K. p = 1.00 atm = 1.013 ×105 Pa. R = 8.314 J/mol ⋅ K. N A = 6.022 ×1023 molecules/mol. The diameter of a typical molecule is about 10−10 m. 0.3 nm = 0.3 × 10−9 m. N EXECUTE: (a) pV = nRT and n = gives NA NRT (1.00)(8.314 J/mol ⋅ K)(300 K) = = 4.09 × 10−26 m3 . l = V 1/ 3 = 3.45 × 10−9 m . N A p (6.022 × 1023 molecules/mol)(1.013 × 105 Pa) (b) The distance in part (a) is about 10 times the diameter of a typical molecule. (c) The spacing is about 10 times the spacing of atoms in solids. EVALUATE: There is space between molecules in a gas whereas in a solid the atoms are closely packed together. (a) IDENTIFY and SET UP: Use the density and the mass of 5.00 mol to calculate the volume. ρ = m / V implies V = m / ρ , where m = mtot , the mass of 5.00 mol of water. V=

18.29.

mtot = nM = (5.00 mol)(18.0 × 10−3 kg/mol) = 0.0900 kg

EXECUTE:

Then V =

m

ρ

=

0.0900 kg = 9.00 × 10−5 m 3 1000 kg/m3

(b) One mole contains N A = 6.022 × 1023 molecules, so the volume occupied by one molecule is

9.00 × 10−5 m3 / mol = 2.989 × 10−29 m3 / molecule (5.00 mol)(6.022 × 1023 molecules/mol) V = a 3 , where a is the length of each side of the cube occupied by a molecule. a 3 = 2.989 × 10−29 m3 , so

18.30.

a = 3.1 × 10−10 m. (c) EVALUATE: Atoms and molecules are on the order of 10−10 m in diameter, in agreement with the above estimates. 3RT IDENTIFY: K av = 32 kT . vrms = . M SET UP: M Ne = 20.180 g/mol , M Kr = 83.80 g/mol and M Rn = 222 g/mol . EXECUTE: (a) K av = 32 kT depends only on the temperature so it is the same for each species of atom in the mixture. v v M Kr 83.80 g/mol M Rn 222 g/mol (b) rms,Ne = = = 2.04 . rms,Ne = = = 3.32 . vrms,Kr M Ne 20.18 g/mol vrms,Rn M Ne 20.18 g/mol

vrms,Kr vrms,Rn

=

M Rn 222 g/mol = = 1.63 . M Kr 83.80 g/mol

EVALUATE:

The average kinetic energies are the same. The gas atoms with smaller mass have larger vrms .


18.31.

3RT . M (a) vrms is different for the two different isotopes, so the 235 isotope diffuses more rapidly.

IDENTIFY and SET UP: EXECUTE: (b)

vrms,235 vrms,238

=

EVALUATE: 18.32.

18-7

vrms =

M 238 0.352 kg/mol = = 1.004 . M 235 0.349 kg/mol The vrms values each depend on T but their ratio is independent of T.


⎛ 1 ⎞ With the multiplicity of each score denoted by ni , the average score is ⎜ ⎟ ∑ ni xi and ⎝ 150 ⎠ 1/ 2

18.33.

⎡⎛ 1 ⎞ 2⎤ the rms score is ⎢⎜ ⎟ ∑ ni xi ⎥ . 150 ⎠ ⎣⎝ ⎦ EXECUTE: (a) 54.6 (b) 61.1 EVALUATE: The rms score is higher than the average score since the rms calculation gives more weight to the higher scores. N m IDENTIFY: pV = nRT = RT = tot RT . NA M SET UP: We known that VA = VB and that TA > TB . EXECUTE: (a) p = nRT / V ; we don’t know n for each box, so either pressure could be higher.

⎛ N ⎞ pVN A , where N A is Avogadro’s number. We don’t know how the pressures compare, (b) pV = ⎜ ⎟ RT so N = RT N ⎝ A⎠ so either N could be larger. (c) pV = ( mtot M ) RT . We don’t know the mass of the gas in each box, so they could contain the same gas or different gases. (d) 12 m ( v 2 ) = 32 kT . TA > TB and the average kinetic energy per molecule depends only on T, so the statement av

must be true. (e) vrms = 3kT m . We don’t know anything about the masses of the atoms of the gas in each box, so either set of

18.34.

18.35.

molecules could have a larger vrms . EVALUATE: Only statement (d) must be true. We need more information in order to determine whether the other statements are true or false. IDENTIFY: Use pV = nRT to solve for V. SET UP: Use R = 0.08206 L ⋅ atm/mol ⋅ K . T = 273.15 K . nRT (1.00 mol)(0.08206 L ⋅ atm/mol ⋅ K)(273.15 K) = = 22.4 L EXECUTE: (a) V = 1.00 atm p

⎛p ⎞ ⎛ 1.00 atm ⎞ (b) pV = nRT = constant , so p1V1 = p2V2 . V2 = ⎜ 1 ⎟V1 = ⎜ ⎟ (22.4 L) = 0.243 L . p ⎝ 92 atm ⎠ ⎝ 2⎠ EVALUATE: For constant T, the volume of 1.00 mol is inversely proportional to the pressure. 3kT IDENTIFY: vrms = m SET UP:

The mass of a deuteron is m = mp + mn = 1.673 × 10−27 kg + 1.675 × 10−27 kg = 3.35 × 10−27 kg .

c = 3.00 × 108 m/s . k = 1.381 × 10−23 J/molecule ⋅ K . EXECUTE:

(a) vrms =

3(1.381 × 10−23 J/molecule ⋅ K)(300 × 106 K) v = 1.93 × 106 m/s . rms = 6.43 × 10−3 . −27 3.35 × 10 kg c

⎛ ⎞ 3.35 × 10−27 kg ⎛m⎞ 7 2 10 (b) T = ⎜ ⎟ (vrms ) 2 = ⎜ ⎟ (3.0 × 10 m/s) = 7.3 × 10 K . −23 × ⋅ 3 3(1.381 10 J/molecule K) k ⎝ ⎠ ⎝ ⎠ EVALUATE: Even at very high temperatures and for this light nucleus, vrms is a small fraction of the speed of light.

18-8

Chapter 18

18.36.

IDENTIFY:

vrms =

3RT n p , where T is in kelvins. pV = nRT gives = . M V RT

R = 8.314 J/mol ⋅ K . M = 44.0 × 10−3 kg/mol .

SET UP: EXECUTE:

(a) For T = 0.0°C = 273.15 K , vrms =

3(8.314 J/mol ⋅ K)(273.15 K) = 393 m/s . For 44.0 × 10−3 kg/mol

T = −100.0°C = 173 K , vrms = 313 m/s . The range of speeds is 393 m/s to 313 m/s. (b) For T = 273.15 K ,

18.37.

n 650 Pa n = = 0.286 mol/m3 . For T = 173.15 K , = 0.452 mol/m 3 . V (8.314 J/mol ⋅ K)(273.15 K) V

The range of densities is 0.286 mol/m3 to 0.452 mol/m 3 . EVALUATE: When the temperature decreases the rms speed decreases and the density increases. IDENTIFY and SET UP: Apply the analysis of Section 18.3. EXECUTE: (a) 12 m(v 2 )av = 32 kT = 32 (1.38 × 10−23 J/molecule ⋅ K)(300 K) = 6.21 × 10−21 J (b) We need the mass m of one atom: m =

Then

1 2

M 32.0 × 10−3 kg/mol = = 5.314 × 10−26 kg/molecule N A 6.022 × 1023 molecules/mol

m(v 2 )av = 6.21 × 10−21 J (from part (a)) gives (v 2 )av =

2(6.21 × 10−21 J) 2(6.21 × 10−21 J) = = 2.34 × 105 m 2 / s 2 5.314 × 10−26 kg m

(c) vrms = (v 2 ) rms = 2.34 × 104 m 2 / s 2 = 484 m/s (d) p = mvrms = (5.314 × 10−26 kg)(484 m/s) = 2.57 × 10−23 kg ⋅ m/s (e) Time between collisions with one wall is t =

0.20 m 0.20 m = = 4.13 × 10−4 s vrms 484 m/s

! In a collision v changes direction, so Δp = 2mvrms = 2(2.57 × 10−23 kg ⋅ m/s) = 5.14 × 10−23 kg ⋅ m/s

Δp 5.14 × 10−23 kg ⋅ m/s dp = = 1.24 × 10−19 N so Fav = Δt 4.13 × 10−4 s dt (f ) pressure = F / A = 1.24 × 10−19 N/(0.10 m)2 = 1.24 × 10−17 Pa (due to one atom) F=

(g) pressure = 1 atm = 1.013 × 105 Pa

Number of atoms needed is 1.013 × 105 Pa/(1.24 × 10−17 Pa/atom) = 8.17 × 1021 atoms (h) pV = NkT (Eq.18.18), so N =

18.38.

pV (1.013 × 105 Pa)(0.10 m)3 = = 2.45 × 1022 atoms kT (1.381 × 10−23 J/molecule ⋅ K)(300 K)

(i) From the factor of 13 in (vx2 )av = 13 (v 2 )av . EVALUATE: This Exercise shows that the pressure exerted by a gas arises from collisions of the molecules of the gas with the walls. IDENTIFY: Apply Eq.(18.22) and calculate λ SET UP: 1 atm = 1.013 × 105 Pa , so p = 3.55 × 10−8 Pa . r = 2.0 × 10−10 m and k = 1.38 × 10−23 J/K .

kT (1.38 × 10−23 J/K)(300 K) = = 1.5 × 105 m 4π 2r 2 ρ 4π 2(2.0 × 10−10 m) 2 (3.55 × 10−8 Pa) EVALUATE: At this very low pressure the mean free path is very large. If v = 484 m/s , as in Example 18.8, then EXECUTE:

tmean = 18.39.

λ v

λ=

= 330 s . Collisions are infrequent.

IDENTIFY and SET UP: 2 rms

v

Use equal vrms to relate T and M for the two gases. vrms = 3RT / M (Eq.18.19), so

/ 3R = T / M , where T must be in kelvins. Same vrms so same T / M for the two gases and

TN2 / M N 2 = TH2 / M H2 . ⎛ M N2 ⎞ ⎛ 28.014 g/mol ⎞ 3 TN2 = TH2 ⎜ ⎟ = ((20 + 273) K) ⎜ ⎟ = 4.071 × 10 K ⎜ MH ⎟ 2.016 g/mol ⎝ ⎠ 2 ⎠ ⎝ = (4071 − 273)°C = 3800°C

EXECUTE:

TN2

EVALUATE:

A N 2 molecule has more mass so N 2 gas must be at a higher temperature to have the same vrms .


18.40.

IDENTIFY: SET UP:

3kT . m k = 1.381 × 10−23 J/molecule ⋅ K. vrms =

(a) vrms =

EXECUTE:

18.41.

18-9

3(1.381 × 10−23 J/molecule ⋅ K)(300 K) = 6.44 × 10 −3 m/s = 6.44 mm/s 3.00 × 10 −16 kg

EVALUATE: (b) No. The rms speed depends on the average kinetic energy of the particles. At this T, H2 molecules would have larger vrms than the typical air molecules but would have the same average kinetic energy and the average kinetic energy of the smoke particles would be the same. IDENTIFY: Use Eq.(18.24), applied to a finite temperature change. SET UP: CV = 5R/2 for a diatomic ideal gas and CV = 3R/2 for a monatomic ideal gas. (a) Q = nCV ΔT = n ( 52 R ) ΔT

EXECUTE:

Q = (2.5 mol) ( 52 ) (8.3145 J/mol ⋅ K)(30.0 K) = 1560 J

(b) Q = nCV ΔT = n ( 32 R ) ΔT

Q = (2.5 mol) ( 32 ) (8.3145 J/mol ⋅ K)(30.0 K) = 935 J

18.42.

18.43.

EVALUATE: More heat is required for the diatomic gas; not all the heat that goes into the gas appears as translational kinetic energy, some goes into energy of the internal motion of the molecules (rotations). IDENTIFY: The heat Q added is related to the temperature increase ΔT by Q = nCV ΔT . SET UP: For H 2 , CV ,H 2 = 20.42 J/mol ⋅ K and for Ne (a monatomic gas), CV , Ne = 12.47 J/mol ⋅ K. Q EXECUTE: CV ΔT = = constant , so CV ,H2 ΔTH2 = CV ,Ne ΔTNe . n ⎛ CV ,H2 ⎞ ⎛ 20.42 J/mol ⋅ K ⎞ ΔTNe = ⎜ ⎟⎟ ΔTH 2 = ⎜ ⎟ (2.50 C°) = 4.09 C°. ⎜C ⎝ 12.47 J/mol ⋅ K ⎠ ⎝ V , Ne ⎠ EVALUATE: The same amount of heat causes a smaller temperature increase for H 2 since some of the energy input goes into the internal degrees of freedom. m RT . IDENTIFY: C = Mc , where C is the molar heat capacity and c is the specific heat capacity. pV = nRT = M SET UP: M N2 = 2(14.007 g/mol) = 28.014 × 10−3 kg/mol . For water, cw = 4190 J/kg ⋅ K . For N 2 ,

CV = 20.76 J/mol ⋅ K . (a) cN2 =

EXECUTE:

c 20.76 J/mol ⋅ K C = = 741 J/kg ⋅ K . w = 5.65 ; cw is over five time larger. −3 cN 2 M 28.014 × 10 kg/mol

(b) To warm the water, Q = mcw ΔT = (1.00 kg)(4190 J/mol ⋅ K)(10.0 K) = 4.19 × 104 J . For air,

m=

Q 4.19 × 104 J (5.65 kg)(8.314 J/mol ⋅ K)(293 K) mRT = = 5.65 kg . V = = = 4.85 m3 . cN2 ΔT (741 J/kg ⋅ K)(10.0 K) Mp (28.014 × 10−3 kg/mol)(1.013 × 105 Pa)

EVALUATE: 18.44.

c is smaller for N 2 , so less heat is needed for 1.0 kg of N 2 than for 1.0 kg of water.

(a) IDENTIFY and SET UP: 12 R contribution to CV for each degree of freedom. The molar heat capacity C is related to the specific heat capacity c by C = Mc. EXECUTE: CV = 6 ( 12 R ) = 3R = 3(8.3145 J/mol ⋅ K) = 24.9 J/mol ⋅ K. The specific heat capacity is

cV = CV / M = (24.9 J/mol ⋅ K)/(18.0 × 10−3 kg/mol) = 1380 J/kg ⋅ K. (b) For water vapor the specific heat capacity is c = 2000 J/kg ⋅ K. The molar heat capacity is C = Mc = (18.0 × 10−3 kg/mol)(2000 J/kg ⋅ K) = 36.0 J/mol ⋅ K. EVALUATE: 18.45.

The difference is 36.0 J/mol ⋅ K − 24.9 J/mol ⋅ K = 11.1 J/mol ⋅ K, which is about 2.7 ( 12 R ) ; the

vibrational degrees of freedom make a significant contribution. IDENTIFY: CV = 3R gives CV in units of J/mol ⋅ K . The atomic mass M gives the mass of one mole. SET UP:

For aluminum, M = 26.982 × 10−3 kg/mol.

24.9 J/mol ⋅ K = 923 J/kg ⋅ K . 26.982 × 10−3 kg/mol (b) Table 17.3 gives 910 J/kg ⋅ K. The value from Eq.(18.28) is too large by about 1.4%. EVALUATE: As shown in Figure 18.21 in the textbook, CV approaches the value 3R as the temperature increases. The values in Table 17.3 are at room temperature and therefore are somewhat smaller than 3R. EXECUTE:

(a) CV = 3R = 24.9 J/mol ⋅ K . cV =

18-10

Chapter 18

18.46.

IDENTIFY:

Table 18.2 gives the value of v / vrms for which 94.7% of the molecules have a smaller value of v / vrms .

3RT . vrms = M SET UP: For N 2 , M = 28.0 ×10−3 kg/mol . v / vrms = 1.60 . v 3RT = , so the temperature is 1.60 M Mv 2 (28.0 × 10 −3 kg/mol) T= = v 2 = (4.385 × 10 −4 K ⋅ s 2 /m 2 )v 2 . 3(1.60) 2 R 3(1.60) 2 (8.3145 J/mol ⋅ K)

EXECUTE:

vrms =

(a) T = (4.385 × 10−4 K ⋅ s 2 /m 2 )(1500 m/s) 2 = 987 K (b) T = (4.385 × 10−4 K ⋅ s 2 /m 2 )(1000 m/s) 2 = 438 K

18.47.

(c) T = (4.385 × 10−4 K ⋅ s 2 /m 2 )(500 m/s) 2 = 110 K. EVALUATE: As T decreases the distribution of molecular speeds shifts to lower values. IDENTIFY and SET UP: Make the substitution P = 12 mv 2 in Eq.(18.32). 3/2

3/2

⎛ m ⎞ 2P − P/kT 8π ⎛ m ⎞ − P/kT f ( v ) = 4π ⎜ e . = ⎜ ⎟ ⎟ Pe m ⎝ 2πkT ⎠ ⎝ 2πkT ⎠ m EVALUATE: The shape of the distribution of molecular speeds versus the temperature is a function only of the kinetic energy of the molecules. 3/2 8π ⎛ m ⎞ − P/kT IDENTIFY and SET UP: Eq.(18.33): f (v ) = ⎜ ⎟ Pe m ⎝ 2π kT ⎠ df At the maximum of f (P), = 0. dP 3/2 df 8π ⎛ m ⎞ d EXECUTE: (Pe − P/kT ) = 0 = ⎜ ⎟ d P m ⎝ 2π kT ⎠ d P d This requires that (Pe − P/kT ) = 0. dP e−P/kT − (P/kT )e −P/kT = 0

EXECUTE:

18.48.

(1 − P/kT )e− P/kT = 0 This requires that 1 − P/kT = 0 so P = kT , as was to be shown. And then since P = 12 mv 2 , this gives

1 2

2 mvmp = kT

and vmp = 2kT/m , which is Eq.(18.34). EVALUATE: 18.49.

vrms =

v . The average of v 2 weights larger v.

3 2 mp

Apply Eqs.(18.34) (18.35) and (18.36). k R/N A R = = SET UP: Note that . M = 44.0 × 10−3 kg/mol . m M/N A M IDENTIFY:

EXECUTE:

(a) vmp = 2(8.3145 J/mol ⋅ K)(300 K)/(44.0 × 10−3 kg/mol) = 3.37 × 102 m/s.

(b) vav = 8(8.3145 J mol ⋅ K)(300 K) (π (44.0 × 10−3 kg mol)) = 3.80 × 102 m s.

18.50.

(c) vrms = 3(8.3145 J mol ⋅ K)(300 K) (44.0 × 10−3 kg mol) = 4.12 × 102 m s. EVALUATE: The average speed is greater than the most probable speed and the rms speed is greater than the average speed. IDENTIFY and SET UP: If the temperature at altitude y is below the freezing point only cirrus clouds can form. Use T = T0 − α y to find the y that gives T = 0.0°C. T − T 15.0°C − 0.0°C EXECUTE: y = 0 = = 2.5 km 6.0 C° /km α EVALUATE: The solid-liquid phase transition occurs at 0°C only for p = 1.01 × 105 Pa. Use the results of Example 18.4 to estimate the pressure at an altitude of 2.5 km. p2 = p1e Mg ( y2 − y1 ) / RT Mg ( y2 − y1 ) / RT = 1.10(2500 m/8863 m) = 0.310 (using the calculation in Example 18.4)

Then p2 = (1.01× 105 Pa)e −0.31 = 0.74 × 105 Pa. This pressure is well above the triple point pressure for water. Figure 18.21 shows that the fusion curve has large slope and it takes a large change in pressure to change the phase transition temperature very much. Using 0.0°C introduces little error.


18.51.

18.52.

18.53.

18.54.

18.55.

18-11

IDENTIFY: Refer to the phase diagram in Figure 18.24 in the textbook. SET UP: For water the triple-point pressure is 610 Pa and the critical-point pressure is 2.212 × 107 Pa . EXECUTE: (a) To observe a solid to liquid (melting) phase transition the pressure must be greater than the triplepoint pressure, so p1 = 610 Pa . For p < p1 the solid to vapor (sublimation) phase transition is observed. (b) No liquid to vapor (boiling) phase transition is observed if the pressure is greater than the critical-point pressure. p2 = 2.212 × 107 Pa . For p1 < p < p2 the sequence of phase transitions are solid to liquid and then liquid to vapor. EVALUATE: Normal atmospheric pressure is approximately 1.0 × 105 Pa , so the solid to liquid to vapor sequence of phase transitions is normally observed when the material is water. IDENTIFY: Refer to Figure 18.24 in the textbook. SET UP: The triple-point temperature for water is 273.16 K = 0.01°C . EXECUTE: The temperature is less than the triple-point temperature so the solid and vapor phases are in equilibrium. The box contains ice and water vapor but no liquid water. EVALUATE: The fusion curve terminates at the triple point. IDENTIFY: Figure 18.24 in the textbook shows that there is no liquid phase below the triple point pressure. SET UP: Table 18.3 gives the triple point pressure to be 610 Pa for water and 5.17 × 105 Pa for CO2. EXECUTE: The atmospheric pressure is below the triple point pressure of water, and there can be no liquid water on Mars. The same holds true for CO2. EVALUATE: On earth patm = 1× 105 Pa , so on the surface of the earth there can be liquid water but not liquid CO2. IDENTIFY: ΔV = βV0 ΔT − V0 k Δp SET UP: For steel, β = 3.6 × 10−5 K −1 and k = 6.25 × 10−12 Pa −1 . EXECUTE: βV0 ΔT = (3.6 × 10−5 K −1 )(11.0 L)(21 C°) = 0.0083 L . −kVo Δp = (6.25 × 10−12 Pa)(11 L) (2.1 × 107 Pa) = −0.0014 L . The total change in volume is ΔV = 0.0083 L − 0.0014 L = 0.0069 L. (b) Yes; ΔV is much less than the original volume of 11.0 L. EVALUATE: Even for a large pressure increase and a modest temperature increase, the magnitude of the volume change due to the temperature increase is much larger than that due to the pressure increase. IDENTIFY: We are asked to compare two states. Use the ideal-gas law to obtain m2 in terms of m1 and the ratio of pressures in the two states. Apply Eq.(18.4) to the initial state to calculate m1. SET UP: pV = nRT can be written pV = (m / M ) RT T, V, M, R are all constant, so p / m = RT / MV = constant. So p1 / m1 = p2 / m2 , where m is the mass of the gas in the tank. EXECUTE:

p1 = 1.30 × 106 Pa + 1.01 × 105 Pa = 1.40 × 106 Pa

p2 = 2.50 × 105 Pa + 1.01 × 105 Pa = 3.51 × 105 Pa m1 = p1VM / RT ; V = hA = hπ r 2 = (1.00 m)π (0.060 m) 2 = 0.01131 m3 m1 =

(1.40 × 106 Pa)(0.01131 m3 )(44.1 × 10−3 kg/mol) = 0.2845 kg (8.3145 J/mol ⋅ K)((22.0 + 273.15) K)

⎛p ⎞ ⎛ 3.51 × 105 Pa ⎞ Then m2 = m1 ⎜ 2 ⎟ = (0.2845 kg) ⎜ ⎟ = 0.0713 kg. 6 ⎝ 1.40 × 10 Pa) ⎠ ⎝ p1 ⎠ m2 is the mass that remains in the tank. The mass that has been used is

18.56.

m1 − m2 = 0.2848 kg − 0.0713 kg = 0.213 kg. EVALUATE: Note that we have to use absolute pressures. The absolute pressure decreases by a factor of four and the mass of gas in the tank decreases by a factor of four. IDENTIFY: Apply pV = nRT to the air inside the diving bell. The pressure p at depth y below the surface of the water is p = patm + ρ gy . SET UP: p = 1.013 × 105 Pa . T = 300.15 K at the surface and T ′ = 280.15 K at the depth of 13.0 m. EXECUTE: (a) The height h′ of the air column in the diving bell at this depth will be proportional to the volume, and hence inversely proportional to the pressure and proportional to the Kelvin temperature: p T′ patm T′ =h h′ = h . p′ T patm + ρgy T ⎛ 280.15 K ⎞ (1.013 × 105 Pa) h′ = (2.30 m) ⎜ ⎟ = 0.26 m . 5 3 2 (1.013 × 10 Pa) + (1030 kg m )(9.80 m s )(73.0 m) ⎝ 300.15 K ⎠ The height of the water inside the diving bell is h − h′ = 2.04 m .

18-12

Chapter 18

(b) The necessary gauge pressure is the term ρgy from the above calculation, pgauge = 7.37 × 105 Pa. .

The gauge pressure required in part (b) is about 7 atm. N p IDENTIFY: pV = NkT gives = . V kT SET UP: 1 atm = 1.013 × 105 Pa . TK = TC + 273.15 . k = 1.381 × 10−23 J/molecule ⋅ K . EVALUATE:

18.57.

EXECUTE: (b)

(a) TC = TK − 273.15 = 94 K − 273.15 = − 179°C

(1.5 atm)(1.013 × 105 Pa/atm) N p = = = 1.2 × 1026 molecules/m3 V kT (1.381 × 10−23 J/molecule ⋅ K)(94 K)

(c) For the earth, p = 1.0 atm = 1.013 × 105 Pa and T = 22°C = 295 K .

18.58.

(1.0 atm)(1.013 × 105 Pa/atm) N = = 2.5 × 1025 molecules/m3 . The atmosphere of Titan is about five times V (1.381 × 10−23 J/molecule ⋅ K)(295 K) denser than earth's atmosphere. EVALUATE: Though it is smaller than Earth and has weaker gravity at its surface, Titan can maintain a dense atmosphere because of the very low temperature of that atmosphere. IDENTIFY: For constant temperature, the variation of pressure with altitude is calculated in Example 18.4 to be 3RT p = p0e− Mgy / RT . vrms = . M SET UP: EXECUTE:

g Earth = 9.80 m/s 2 . T = 460°C = 733 K . M = 44.0 g/mol = 44.0 × 10−3 kg/mol . (a)

Mgy (44.0 × 10−3 kg/mol)(0.894)(9.80 m/s 2 )(1.00 × 103 m) = = 0.06326 . RT (8.314 J/mol ⋅ K)(733 K)

p = p0e− Mgy / RT = (92 atm)e−0.06326 = 86 atm . The pressure is 86 Earth-atmospheres, or 0.94 Venus-atmospheres. (b) vrms =

18.59.

of 1.00 km. EVALUATE: vrms depends only on T and the molar mass of the gas. For Venus compared to earth, the surface temperature, in kelvins, is nearly a factor of three larger and the molecular mass of the gas in the atmosphere is only about 50% larger, so vrms for the Venus atmosphere is larger than it is for the Earth's atmosphere. IDENTIFY: pV = nRT SET UP: EXECUTE:

18.60.

3RT 3(8.314 J/mol ⋅ K)(733 K) = = 645 m/s . vrms has this value both at the surface and at an altitude M 44.0 × 10−3 kg/mol

In pV = nRT we must use the absolute pressure. T1 = 278 K . p1 = 2.72 atm . T2 = 318 K . n, R constant, so

pV pV pV = nR = constant . 1 1 = 2 2 and T T1 T2

⎛ V ⎞⎛ T ⎞ ⎛ 0.0150 m3 ⎞ ⎛ 318 K ⎞ p2 = p1 ⎜ 1 ⎟⎜ 2 ⎟ = (2.72 atm) ⎜ ⎟ = 2.94 atm . The final gauge pressure is 3 ⎟⎜ ⎝ 0.0159 m ⎠ ⎝ 278 K ⎠ ⎝ V2 ⎠⎝ T1 ⎠ 2.94 atm − 1.02 atm = 1.92 atm . EVALUATE: Since a ratio is used, pressure can be expressed in atm. But absolute pressures must be used. The ratio of gauge pressures is not equal to the ratio of absolute pressures. IDENTIFY: In part (a), apply pV = nRT to the ethane in the flask. The volume is constant once the stopcock is in mtot RT to the ethane at its final temperature and pressure. M 1.50 L = 1.50 × 10−3 m3 . M = 30.1 × 10−3 kg/mol . Neglect the thermal expansion of the flask.

place. In part (b) apply pV = SET UP: EXECUTE:

(a) p2 = p1 (T2 T1 ) = (1.013 × 105 Pa)(300 K 380 K) = 8.00 × 104 Pa.

⎛ pV ⎞ ⎛ (8.00 × 104 Pa)(1.50 × 10−3 m3 ) ⎞ −3 (b) mtot = ⎜ 2 ⎟ M = ⎜ ⎟ (30.1 × 10 kg mol) = 1.45 g. ⎝ (8.3145 J mol ⋅ K)(300 K) ⎠ ⎝ RT2 ⎠ EVALUATE: We could also calculate mtot with p = 1.013 × 105 Pa and T = 380 K , and we would obtain the same result. Originally, before the system was warmed, the mass of ethane in the flask was ⎛ 1.013 × 105 Pa ⎞ m = (1.45 g) ⎜ ⎟ = 1.84 g . 4 ⎝ 8.00 × 10 Pa ⎠


18.61.

18-13

(a) IDENTIFY: Consider the gas in one cylinder. Calculate the volume to which this volume of gas expands when the pressure is decreased from (1.20 × 106 Pa + 1.01 × 105 Pa) = 1.30 × 106 Pa to 1.01 × 105 Pa. Apply the ideal-gas

law to the two states of the system to obtain an expression for V2 in terms of V1 and the ratio of the pressures in the two states. SET UP: pV = nRT n, R, T constant implies pV = nRT = constant, so p1V1 = p2V2 . ⎛ 1.30 × 106 Pa ⎞ 3 V2 = V1 ( p1 / p2 ) = (1.90 m3 ) ⎜ ⎟ = 24.46 m 5 ⎝ 1.01 × 10 Pa ⎠ The number of cylinders required to fill a 750 m3 balloon is 750 m3 / 24.46 m3 = 30.7 cylinders.

EXECUTE:

EVALUATE: The ratio of the volume of the balloon to the volume of a cylinder is about 400. Fewer cylinders than this are required because of the large factor by which the gas is compressed in the cylinders. (b) IDENTIFY: The upward force on the balloon is given by Archimedes’ principle (Chapter 14): B = weight of

air displaced by balloon = ρ airVg. Apply Newton’s 2nd law to the balloon and solve for the weight of the load that can be supported. Use the ideal-gas equation to find the mass of the gas in the balloon. SET UP: The free-body diagram for the balloon is given in Figure 18.61. mgas is the mass of the gas that is inside the balloon; mL is the mass of the load that is supported by the balloon EXECUTE:

∑F

y

= ma y

B − mL g − mgas g = 0 Figure 18.61

ρairVg − mL g − mgas g = 0 mL = ρairV − mgas Calculate mgas , the mass of hydrogen that occupies 750 m3 at 15°C and p = 1.01 × 105 Pa. pV = nRT = ( mgas / M ) RT gives mgas = pVM / RT =

(1.01 × 105 Pa)(750 m3 )(2.02 × 10−3 kg/mol) = 63.9 kg (8.3145 J/mol ⋅ K)(288 K)

Then mL = (1.23 kg/m3 )(750 m3 ) − 63.9 kg = 859 kg, and the weight that can be supported is wL = mL g = (859 kg)(9.80 m/s 2 ) = 8420 N. (c) mL = ρairV − mgas

mgas = pVM / RT = (63.9 kg)((4.00 g/mol)/(2.02 g/mol)) = 126.5 kg (using the results of part (b)). Then mL = (1.23 kg/m3 )(750 m3 ) − 126.5 kg = 796 kg. wL = mL g = (796 kg)(9.80 m/s 2 ) = 7800 N. 18.62.

EVALUATE: A greater weight can be supported when hydrogen is used because its density is less. IDENTIFY: The upward force exerted by the gas on the piston must equal the piston's weight. Use pV = nRT to calculate the volume of the gas, and from this the height of the column of gas in the cylinder. SET UP: F = pA = pπ r 2 , with r = 0.100 m and p = 1.00 atm = 1.013 × 105 Pa . For the cylinder, V = π r 2 h . EXECUTE: (b) V =

(a) pπ r 2 = mg and m =

pπ r 2 (1.013 × 105 Pa)π (0.100 m) 2 = = 325 kg . g 9.80 m/s 2

4.33 × 10−2 m3 nRT (1.80 mol)(8.31 J/mol ⋅ K)(293.15 K) V − = 4.33 × 10−2 m3 . h = 2 = = 1.38 m . 5 πr π (0.100 m) 2 p 1.013 × 10 Pa

EVALUATE:

The calculation assumes a vacuum ( p = 0) in the tank above the piston.

18-14

Chapter 18

18.63.

IDENTIFY: Apply Bernoulli’s equation to relate the efflux speed of water out the hose to the height of water in the tank and the pressure of the air above the water in the tank. Use the ideal-gas equation to relate the volume of the air in the tank to the pressure of the air. (a) SET UP: Points 1 and 2 are shown in Figure 18.63.

p1 = 4.20 × 105 Pa p2 = pair = 1.00 × 105 Pa large tank implies v1 ≈ 0

Figure 18.63 EXECUTE: 1 2

p1 + ρ gy1 + 12 ρ v12 = p2 + ρ gy2 + 12 ρ v22

ρ v22 = p1 − p2 + ρ g ( y1 − y2 )

v2 = (2 / ρ )( p1 − p2 ) + 2 g ( y1 − y2 ) v2 = 26.2 m/s (b) h = 3.00 m

The volume of the air in the tank increases so its pressure decreases. pV = nRT = constant, so pV = p0V0 ( p0 is the pressure for h0 = 3.50 m and p is the pressure for h = 3.00 m) p(4.00 m − h) A = p0 (4.00 m − h0 ) A ⎛ 4.00 m − h0 ⎞ ⎛ 4.00 m − 3.50 m ⎞ 5 5 p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ = 2.10 × 10 Pa ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − 3.00 m ⎠ Repeat the calculation of part (a), but now p1 = 2.10 × 105 Pa and y1 = 3.00 m. v2 =

( 2 / ρ ) ( p1 − p2 ) + 2 g ( y1 − y2 )

v2 = 16.1 m/s h = 2.00 m ⎛ 4.00 m − h0 ⎞ ⎛ 4.00 m − 3.50 m ⎞ 5 5 p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ = 1.05 × 10 Pa ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − 2.00 m ⎠ v2 = (2 / ρ )( p1 − p2 ) + 2 g ( y1 − y2 ) v2 = 5.44 m/s (c) v2 = 0 means (2 / ρ )( p1 − p2 ) + 2 g ( y1 − y2 ) = 0

p1 − p2 = − ρ g ( y1 − y2 ) y1 − y2 = h − 1.00 m ⎛ 0.50 m ⎞ ⎛ 0.50 m ⎞ 5 p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ . This is p1 , so ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − h ⎠ ⎛ 0.50 m ⎞ 5 2 3 (4.20 × 105 Pa) ⎜ ⎟ − 1.00 × 10 Pa = (9.80 m/s )(1000 kg/m )(1.00 m − h) 4.00 m h − ⎝ ⎠ (210 /(4.00 − h)) − 100 = 9.80 − 9.80h, with h in meters. 210 = (4.00 − h)(109.8 − 9.80h) 9.80h 2 − 149h + 229.2 = 0 and h 2 − 15.20h + 23.39 = 0

(

)

quadratic formula: h = 12 15.20 ± (15.20) 2 − 4(23.39) = (7.60 ± 5.86) m h must be less than 4.00 m, so the only acceptable value is h = 7.60 m − 5.86 m = 1.74 m EVALUATE:

The flow stops when p + ρ g ( y1 − y2 ) equals air pressure. For h = 1.74 m, p = 9.3 × 104 Pa and

ρ g ( y1 − y2 ) = 0.7 × 104 Pa, so p + ρ g ( y1 − y2 ) = 1.0 × 105 Pa, which is air pressure.


18.64.

18-15

IDENTIFY: Use the ideal gas law to find the number of moles of air taken in with each breath and from this calculate the number of oxygen molecules taken in. Then find the pressure at an elevation of 2000 m and repeat the calculation. SET UP: The number of molecules in a mole is N A = 6.022 × 1023 molecules/mol . R = 0.08206 L ⋅ atm/mol ⋅ K . Example 18.4 shows that the pressure variation with altitude y, when constant temperature is assumed, is p = p0e − Mgy / RT . For air, M = 28.8 × 10−3 kg/mol . EXECUTE:

(a) pV = nRT gives n =

pV (1.00 atm)(0.50 L) = = 0.0208 mol . RT (0.08206 L ⋅ atm/mol ⋅ K)(293.15 K)

N = (0.210)nN A = (0.210)(0.0208 mol)(6.022 × 1023 molecules/mol) = 2.63 × 1021 molecules . Mgy (28.8 × 10−3 kg/mol)(9.80 m/s 2 )(2000 m) = = 0.2316 . p = p0e − Mgy / RT = (1.00 atm)e −0.2316 = 0.793 atm . RT (8.314 J/mol ⋅ K)(293.15 K) N is proportional to n, which is in turn proportional to p, so ⎛ 0.793 atm ⎞ 21 21 N =⎜ ⎟ (2.63 × 10 molecules) = 2.09 × 10 molecules . ⎝ 1.00 atm ⎠ (c) Less O 2 is taken in with each breath at the higher altitude, so the person must take more breaths per minute. EVALUATE: A given volume of gas contains fewer molecules when the pressure is lowered and the temperature is kept constant. IDENTIFY and SET UP: Apply Eq.(18.2) to find n and then use Avogadro’s number to find the number of molecules. EXECUTE: Calculate the number of water molecules N. m 50 kg Number of moles: n = tot = = 2.778 × 103 mol M 18.0 × 10−3 kg/mol (b)

18.65.

N = nN A = (2.778 × 103 mol)(6.022 × 1023 molecules/mol) = 1.7 × 1027 molecules Each water molecule has three atoms, so the number of atoms is 3(1.7 × 1027 ) = 5.1 × 1027 atoms We could also use the masses in Example 18.5 to find the mass m of one H 2O molecule:

EVALUATE:

m = 2.99 × 10

18.66.

18.67.

−26

kg. Then N = mtot / m = 1.7 × 1027 molecules, which checks. N RT . Deviations will be noticeable when the volume V of a molecule is on the order IDENTIFY: pV = nRT = NA of 1% of the volume of gas that contains one molecule. 4 SET UP: The volume of a sphere of radius r is V = π r 3 . 3 RT EXECUTE: The volume of gas per molecule is , and the volume of a molecule is about NA p 4 V0 = π (2.0 × 10−10 m)3 = 3.4 × 10−29 m3 . Denoting the ratio of these volumes as f, 3 RT (8.3145 J mol ⋅ K)(300 K) p= f = f = (1.2 × 108 Pa) f . N AV0 (6.023 × 1023 molecules mol)(3.4 × 10−29 m3 ) “Noticeable deviations” is a subjective term, but f on the order of 1.0% gives a pressure of 106 Pa. EVALUATE: The forces between molecules also cause deviations from ideal-gas behavior. IDENTIFY: Eq.(18.16) says that the average translational kinetic energy of each molecule is equal to 32 kT . vrms =

3kT . m

SET UP: k = 1.381 × 10−23 J/molecule ⋅ K . EXECUTE: (a) 12 m(v 2 )av depends only on T and both gases have the same T, so both molecules have the same

average translational kinetic energy. vrms is proportional to m −1/ 2 , so the lighter molecules, A, have the greater vrms . (b) The temperature of gas B would need to be raised. ⎛m ⎞ ⎛ 5.34 × 10−26 kg ⎞ T T T v 3 (c) = rms = constant , so A = B . TB = ⎜ B ⎟ TA = ⎜ ⎟ (283.15 K) = 4.53 × 10 K = 4250°C . −27 mA mB m 3.34 10 kg m × 3k ⎝ ⎠ ⎝ A⎠ (d) TB > TA so the B molecules have greater translational kinetic energy per molecule. EVALUATE:

In

1 2

m(v 2 )av = 32 kT and vrms =

3kT the temperature T must be in kelvins. m

18-16

Chapter 18

18.68.

IDENTIFY: The equations derived in the subsection Collisions between Molecules in Section 18.3 can be applied to the bees. The average distance a bee travels between collisions is the mean free path, λ . The average time dN 1 between collisions is the mean free time, tmean . The number of collisions per second is = . dt tmean SET UP:

V = (1.25 m)3 = 1.95 m3 . r = 0.750 × 10−2 m . v = 1.10 m/s . N = 2500 .

EXECUTE:

(a) λ =

V 1.95 m3 = = 0.780 m = 78.0 cm 4π 2r 2 N 4π 2(0.750 × 10−2 m) 2 (2500)

(b) λ = vtmean , so tmean =

λ v

=

0.780 m = 0.709 s . 1.10 m/s

dN 1 1 = = = 1.41 collisions/s dt tmean 0.709 s EVALUATE: The calculation is valid only if the motion of each bee is random. IDENTIFY: Apply the iteration procedure that is described in the problem. SET UP: Let x = n / V . T = 400.15 K . EXECUTE: (a) Dividing both sides of Eq.(18.7) by the product RTV gives the result. (b) The algorithm described is best implemented on a programmable calculator or computer; for a calculator, the numerical procedure is an iteration of (c)

18.69.

⎡ (9.80 × 105 ) ⎤ (0.448) + x=⎢ x 2 ⎥ ⎡⎣1 − (4.29 × 10−5 ) x ⎤⎦ . (8.3145)(400.15) (8.3145)(400.15) ⎣ ⎦ Starting at x = 0 gives a fixed point at x = 3.03 × 102 after four iterations. The number density is 3.03 × 102 mol m3 .

18.70.

(c) The ideal-gas equation is the result after the first iteration, 295mol m3 . EVALUATE: The van der Waals density is larger. The term corresponding to a represents the attraction of the molecules, and hence more molecules will be in a given volume for a given pressure. IDENTIFY: Calculate vrms and use conservation of energy to relate the initial speed of the molecules (vrms ) to the maximum height they reach. SET UP: T = 298.15 K . M = 28.0 × 10−3 kg/mol . EXECUTE:

3RT 3(8.314 J/mol ⋅ K)(298.15 K) = = 515 m/s . Conservation of energy gives 28.0 × 10−3 kg/mol M

2 vrms (515 m/s) 2 = = 1.02 × 105 m = 102 km 2 g 2(1.30 m/s 2 ) EVALUATE: The result does not depend on the amount of gas in the canister. IDENTIFY: The mass of one molecule is the molar mass, M, divided by the number of molecules in a mole, N A . 1 2

18.71.

vrms =

2 mvrms = mgy and y =

The average translational kinetic energy of a single molecule is 12 m(v 2 )av = 32 kT . Use pV = NkT to calculate N, the number of molecules. SET UP: k = 1.381 × 10−23 J/molecule ⋅ K . M = 28.0 × 10−3 kg/mol . T = 295.15 K . The volume of the balloon is V = 34 π (0.250 m)3 = 0.0654 m3 . p = 1.25 atm = 1.27 × 105 Pa . EXECUTE: (b)

1 2

(a) m =

M 28.0 × 10−3 kg/mol = = 4.65 × 10−26 kg N A 6.022 × 1023 molecules/mol

m(v 2 )av = 32 kT = 23 (1.381 × 10−23 J/molecule ⋅ K)(295.15 K) = 6.11 × 10−21 J

pV (1.27 × 105 Pa)(0.0654 m3 ) = = 2.04 × 1024 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(295.15 K) (d) The total average translational kinetic energy is N ( 12 m(v 2 )av ) = (2.04 × 1024 molecules)(6.11× 10−21 J/molecule) = 1.25 × 104 J . (c) N =

EVALUATE:

The number of moles is n =

N 2.04 × 1024 molecules = = 3.39 mol . N A 6.022 × 1023 molecules/mol

K tr = 32 nRT = 32 (3.39 mol)(8.314 J/mol ⋅ K)(295.15 K) = 1.25 × 104 J , which agrees with our results in part (d).


18.72.

IDENTIFY:

18-17

U = mgy . The mass of one molecule is m = M/N A . K av = kT . 3 2

Let y = 0 at the surface of the earth and h = 400 m . N A = 6.023 × 1023 molecules/mol and k = 1.38 × 10 −23 J/K . 15.0°C = 288 K . ⎛ ⎞ M 28.0 × 10−3 kg/mol 2 −22 EXECUTE: (a) U = mgh = gh = ⎜ ⎟ (9.80 m/s )(400 m) = 1.82 × 10 J. 23 × NA 6.023 10 molecules/mol ⎝ ⎠ SET UP:

18.73.

3 2 ⎛ 1.82 × 10−22 J ⎞ (b) Setting U = kT , T = ⎜ ⎟ = 8.80 K. 2 3 ⎝ 1.38 × 10 −23 J/K ⎠ EVALUATE: (c) The average kinetic energy at 15.0°C is much larger than the increase in gravitational potential energy, so it is energetically possible for a molecule to rise to this height. But Example 18.8 shows that the mean free path will be very much less than this and a molecule will undergo many collisions as it rises. These numerous collisions transfer kinetic energy between molecules and make it highly unlikely that a given molecule can have very much of its translational kinetic energy converted to gravitational potential energy. IDENTIFY and SET UP: At equilibrium F ( r ) = 0. The work done to increase the separation from r2 to ∞ is U (∞ ) − U ( r2 ).

U ( r ) = U 0 ⎡⎣ ( R0 / r )12 − 2( R0 / r )6 ⎤⎦

(a) EXECUTE:

Eq.(13.26): F ( r ) = 12(U 0 / R0 ) ⎡⎣( R0 / r )13 − ( R0 / r ) 7 ⎤⎦ . The graphs are given in Figure 18.73.

Figure 18.73 (b) equilibrium requires F = 0; occurs at point r2 . r2 is where U is a minimum (stable equilibrium). (c) U = 0 implies ⎡⎣( R0 / r )12 − 2( R0 / r )6 ⎤⎦ = 0

(r1 / R0 )6 = 1/ 2 and r1 = R0 /(2)1/ 6 F = 0 implies ⎡⎣( R0 / r )13 − ( R0 / r )7 ⎤⎦ = 0 (r2 / R0 )6 = 1 and r2 = R0 Then r1 / r2 = ( R0 / 21/ 6 ) / R 0 = 2−1/ 6 (d) Wother = ΔU

At r → ∞, U = 0, so W = −U ( R0 ) = −U 0 ⎡⎣( R0 / R0 )12 − 2( R0 / R0 )6 ⎤⎦ = +U 0 EVALUATE: 18.74.

IDENTIFY:

The answer to part (d), U 0 , is the depth of the potential well shown in the graph of U (r ). Use pV = nRT to calculate the number of moles, n. Then K tr = 32 nRT . The mass of the gas, mtot , is

given by mtot = nM . 5.00 L = 5.00 × 10−3 m 3 pV (1.01 × 105 Pa)(5.00 × 10 −3 m 3 ) EXECUTE: (a) n = = = 0.2025 moles . (8.314 J/mol ⋅ K)(300 K) RT SET UP:

K tr = 32 (0.2025 mol)(8.314 J/mol ⋅ K)(300 K) = 758 J . (b) mtot = nM = (0.2025 mol)(2.016 × 10−3 kg/mol) = 4.08 × 10−4 kg . The kinetic energy due to the speed of the jet

is K = 12 mv 2 = 12 (4.08 × 10−4 kg)(300.0 m/s) 2 = 18.4 J . The total kinetic energy is

K 18.4 J × 100% = × 100% = 2.37% . 776 J K tot (c) No. The temperature is associated with the random translational motion, and that hasn't changed. EVALUATE: Eq.(18.13) gives K tr = 32 pV = 32 (1.01× 105 Pa)(5.00 × 10−3 m3 ) = 758 J , which agrees with our result K tot = K + K tr = 18.4 J + 758 J = 776 J . The percentage increase is

3RT = 1.93 × 103 m/s . vrms is a lot larger than the speed of the jet, so the percentage increase in M the total kinetic energy, calculated in part (b), is small.

in part (a). vrms =

18-18

Chapter 18

18.75.


(a) vrms = 3RT / M

EXECUTE:

3(8.3145 J/mol ⋅ K)(300 K) = 517 m/s 28.0 × 10−3 kg/mol

vrms =

(b) (vx2 )av = 13 (v 2 )av so

18.76.

Apply Eq.(18.19) for vrms . The equation preceeding Eq.(18.12) relates vrms and (vx ) rms .

(

(vx2 )av = 1/ 3

)

(

)

(

)

(v 2 )av = 1/ 3 vrms = 1/ 3 (517 m/s) = 298 m/s

EVALUATE: The speed of sound is approximately equal to (vx ) rms since it is the motion along the direction of propagation of the wave that transmits the wave. 3kT IDENTIFY: vrms = m SET UP:

M = 1.99 × 1030 kg , R = 6.96 × 108 m and G = 6.673 × 10−11 N ⋅ m 2 /kg 2 .

EXECUTE:

(a) vrms =

3kT 3(1.38 × 10−23 J K) (5800 K) = = 1.20 × 104 m s. m (1.67 × 10−27 kg)

2GM 2(6.673 × 10−11 N ⋅ m 2 kg 2 ) (1.99 × 1030 kg) = = 6.18 × 105 m s. (6.96 × 108 m) R EVALUATE: (c) The escape speed is about 50 times the rms speed, and any of Figure 18.23 in the textbook, Eq.(18.32) or Table (18.2) will indicate that there is a negligibly small fraction of molecules with the escape speed. (a) IDENTIFY and SET UP: Apply conservation of energy K1 + U1 + Wother = K 2 + U 2 , where U = −Gmmp / r. Let (b) vescape =

18.77.

point 1 be at the surface of the planet, where the projectile is launched, and let point 2 be far from the earth. Just barely escapes says v2 = 0. EXECUTE:

Only gravity does work says Wother = 0.

U1 = −Gmmp / Rp ; r2 → ∞ so U 2 = 0; v2 = 0 so K 2 = 0. The conservation of energy equation becomes K1 − Gmmp / Rp = 0 and K1 = Gmmp / Rp . But g = Gmp / Rp2 so Gmp / Rp = Rp g and K1 = mgRp , as was to be shown. The greater gRp is the more initial kinetic energy is required for escape.

EVALUATE:


Eq.(18.16). EXECUTE:

Set K1 from part (a) equal to the average kinetic energy of a molecule as given by

m(v )av = mgRp (from part (a)). But also, 2

1 2

T=

1 2

m(v 2 )av = 32 kT , so mgRp = 32 kT

2mgRp

3k nitrogen mN2 = (28.0 × 10−3 kg/mol)/(6.022 × 1023 molecules/mol) = 4.65 × 10 −26 kg/molecule

T=

2mgRp

=

2(4.65 × 10−26 kg/molecule)(9.80 m/s 2 )(6.38 × 106 m) = 1.40 × 105 K 3(1.381 × 10−23 J/molecule ⋅ K)

=

2(3.354 × 10−27 kg/molecule)(9.80 m/s 2 )(6.38 × 106 m) = 1.01 × 104 K 3(1.381 × 10−23 J/molecule ⋅ K)

3k hydrogen mH2 = (2.02 × 10−3 kg/mol)/(6.022 × 1023 molecules/mol) = 3.354 × 10−27 kg/molecule

T=

2mgRp 3k

(c) T =

2mgRp

3k nitrogen 2(4.65 × 10−26 kg/molecule)(1.63 m/s 2 )(1.74 × 106 m) T= = 6730 K 3(1.381 × 10−23 J/molecule ⋅ K) hydrogen 2(3.354 × 10−27 kg/molecule)(1.63 m/s 2 )(1.74 × 106 m) T= = 459 K 3(1.381 × 10−23 J/molecule ⋅ K) (d) EVALUATE: The “escape temperatures” are much less for the moon than for the earth. For the moon a larger fraction of the molecules at a given temperature will have speeds in the Maxwell-Boltzmann distribution larger than the escape speed. After the long time most of the molecules will have escaped from the moon.


18.78.

IDENTIFY: SET UP:

vrms =

18-19

3RT . M

M H2 = 2.02 × 10−3 kg/mol . M O2 = 32.0 × 10−3 kg/mol . For Earth, M = 5.97 × 1024 kg and

4 R = 6.38 × 106 m . For Jupiter, M = 1.90 × 1027 kg and R = 6.91 × 107 m . For a sphere, M = ρV = ρ π r 3 . The 3 escape speed is vescape = EXECUTE:

2GM . R

(a) Jupiter: vrms = 3(8.3145J mol ⋅ K)(140K) (2.02 × 10−3 kg mol) = 1.31 × 103 m s .

vescape = 6.06 × 104 m/s . vrms = 0.022vescape . Earth: vrms = 3(8.3145J mol ⋅ K)(220K) (2.02 × 10−3 kg mol) = 1.65 × 103 m s . vescape = 1.12 × 104 m/s .

vrms = 0.15vescape . (b) Escape from Jupiter is not likely for any molecule, while escape from earth is much more probable. (c) vrms = 3(8.3145J mol ⋅ K)(200K) (32.0 × 10−3 kg mol) = 395m s. The radius of the asteroid is

R = (3M 4πρ )1/ 3 = 4.68 × 105 m, and the escape speed is vescape = 2GM R = 542m s . Over time the O 2

18.79.

molecules would essentially all escape and there can be no such atmosphere. EVALUATE: As Figure 18.23 in the textbook shows, there are some molecules in the velocity distribution that have speeds greater than vrms . But as the speed increases above vrms the number with speeds in that range decreases. 3kT m IDENTIFY: vrms = . The number of molecules in an object of mass m is N = nN A = NA . M m 4 SET UP: The volume of a sphere of radius r is V = π r 3 . 3 3kT 3(1.381 × 10−23 J K)(300K) EXECUTE: (a) m = 2 = = 1.24 × 10−14 kg. (0.0010m s) 2 vrms (b) N = mN A M = (1.24 × 10−14 kg)(6.023 × 1023 molecules mol) (18.0 × 10−3 kg mol)

N = 4.16 × 1011 molecules. 1/3

1/3

⎛ 3V ⎞ ⎛ 3m/ρ ⎞ (c) The diameter is D = 2r = 2 ⎜ ⎟ = 2⎜ ⎟ 4 π ⎝ ⎠ ⎝ 4π ⎠ to see. EVALUATE: vrms decreases as m increases. 18.80.


18.81.

1/3

⎛ 3(1.24 × 10−14 kg) ⎞ = 2⎜ ⎟ 3 ⎝ 4π (920 kg/m ) ⎠

= 2.95 × 10−6 m which is too small

For a simple harmonic oscillator, x = A cos ωt and vx = −ω A sin ωt , with ω = k / m . The average value of cos(2ωt ) over one period is zero, so (sin 2 ωt )av = (cos 2 ωt )av = 12 .

x = A cos ωt , vx = −ω A sin ωt , U av = 12 kA2 (cos 2 ωt )av , K av = 12 mω 2 A2 (sin 2 ωt )av . Using

(sin 2 ωt )av = (cos 2 ωt )av = 12 and mω 2 = k shows that K av = U av . EVALUATE: In general, at any given instant of time U ≠ K . It is only the values averaged over one period that are equal. IDENTIFY: The equipartition principle says that each atom has an average kinetic energy of 12 kT for each degree of freedom. There is an equal average potential energy. SET UP: The atoms in a three-dimensional solid have three degrees of freedom and the atoms in a twodimensional solid have two degrees of freedom. EXECUTE: (a) In the same manner that Eq.(18.28) was obtained, the heat capacity of the two-dimensional solid would be 2 R = 16.6 J/mol ⋅ K . (b) The heat capacity would behave qualitatively like those in Figure 18.21 in the textbook, and the heat capacity would decrease with decreasing temperature. EVALUATE: At very low temperatures the equipartition theorem doesn't apply. Most of the atoms remain in their lowest energy states because the next higher energy level is not accessible.

18-20

Chapter 18

18.82.

IDENTIFY:

The equipartition principle says that each molecule has average kinetic energy of 12 kT for each degree

of freedom. I = 2m( L / 2) 2 , where L is the distance between the two atoms in the molecule. K rot = 12 I ω 2 .

ωrms = (ω 2 )av . SET UP: The mass of one atom is m = M/N A = (16.0 ×10−3 kg/mol) /(6.02 ×1023 molecules/mol) = 2.66 ×10−26 kg. EXECUTE: (a) The two degrees of freedom associated with the rotation for a diatomic molecule account for twofifths of the total kinetic energy, so K rot = nRT = (1.00 mol)(8.3145 J mol ⋅ K)(300 K) = 2.49 × 103 J .

⎛ ⎞ 16.0 × 10−3 kg mol -11 2 −46 2 (b) I = 2m( L 2) 2 = 2 ⎜ ⎟ (6.05 × 10 m) = 1.94 × 10 kg ⋅ m 23 ⎝ 6.023 × 10 molecules mol ⎠ (c) Since the result in part (b) is for one mole, the rotational kinetic energy for one atom is K rot / N A and

ωrms =

2 K rot N A 2(2.49 × 103 J) = = 6.52 × 1012 rad s . This is much larger −46 I (1.94 × 10 kg ⋅ m 2 )(6.023 × 1023 molecules/mol)

than the typical value for a piece of rotating machinery. 2π rad EVALUATE: The average rotational period, T = , for molecules is very short.

ωrms

18.83.

IDENTIFY:

CV = N ( R ) , where N is the number of degrees of freedom. 1 2

SET UP: There are three translational degrees of freedom. EXECUTE: For CO 2 , N = 5 and the contribution to CV other than from vibration is

5 2

R = 20.79 J/mol ⋅ K and

CV − R = 0.270 CV . So 27% of CV is due to vibration. For both SO2 and H2S, N = 6 and the contribution to CV 5 2

other than from vibration is

The vibrational contribution is much less for H 2S . In H 2S the vibrational energy steps are larger

EVALUATE: 18.84.

R = 24.94 J/mol ⋅ K . The respective fractions of CV from vibration are 21% and 3.9%.

6 2

because the two hydrogen atoms have small mass and ω = k / m . IDENTIFY: Evaluate the integral, as specified in the problem. SET UP: Use the integral formula given in Problem 18.85, with α = m / 2kT . 32

∞

18.85.

(a)

EXECUTE:

∫

∞ 0

∞

v 2 f (v) dv = 4π (m / 2π kT )3/ 2 ∫ v 4e − mv / 2 kT dv

Apply with a = m / 2kT ,

2

0

The integral formula with n = 2 gives

18.86.

32

∞ ⎞ 1 π ⎛ m ⎞ ⎛ m ⎞ ⎛ 2 − mv 2 / 2 kT =1 dv = 4π ⎜ ⎟ ⎟ ⎜ ∫0 f (v) dv = 4π ⎜⎝ 2πkT ⎟⎠ ∫0 v e 2 4( 2 ) 2kT πkT m kT m ⎝ ⎠ ⎝ ⎠ EVALUATE: (b) f (v )dv is the probability that a particle has speed between v and v + dv; the probability that the particle has some speed is unity, so the sum (integral) of f (v )dv must be 1. IDENTIFY and SET UP: Evaluate the integral in Eq.(18.31) as specified in the problem.

EXECUTE:

∫

∞ 0

∫

∞ 0

2

v 4e − av dv = (3/8a 2 ) π /a

v 2 f (v) dv = 4π (m/2π kT )3 / 2 (3/8)(2kT/m) 2 2π kT/m = (3/2)(2kT/m) = 3kT/m

EVALUATE: Equation (18.16) says 12 m(v 2 )av = 3kT / 2, so (v 2 )av = 3kT / m, in agreement with our calculation. IDENTIFY: Follow the procedure specified in the problem. SET UP: If v 2 = x , then dx = 2vdv . ∞

EXECUTE:

⎛ m ⎞

32∞

∫ vf (v)dv = 4π ⎜⎝ 2πkT ⎟⎠ ∫ v e 0

3 − mv 2 2 kT

dv. Making the suggested change of variable, v 2 = x. 2vdv = dx,

0

v3dv = (1/2) x dx, and the integral becomes ∞

⎛ m ⎞ ∫0 vf (v)dv = 2π ⎜⎝ 2πkT ⎟⎠

3/2 ∞

∫ 0

which is Eq. (18.35). ∞

EVALUATE:

The integral ∫ vf (v)dv is the definition of vav . 0

3/2

2

2 ⎛ m ⎞ ⎛ 2kT ⎞ xe− mx/ 2 kT dx = 2π ⎜ ⎟ ⎜ ⎟ = π ⎝ 2πkT ⎠ ⎝ m ⎠

2 KT 8KT = m πm


18.87.

IDENTIFY:

18-21

f (v)dv is the probability that a particle has a speed between v and v + dv . Eq.(18.32) gives f (v ) .

vmp is given by Eq.(18.34). SET UP: For O 2 , the mass of one molecule is m = M / N A = 5.32 × 10−26 kg . EXECUTE: (a) f (v)dv is the fraction of the particles that have speed in the range from v to v + dv . The number v + Δv

of particles with speeds between v and v + dv is therefore dN = Nf (v)dv and ΔN = N ∫ v 3/2

⎛ m ⎞ ⎛ 2kT (b) Setting v = vmp = 2kT in f (v ) gives f (vmp ) = 4π ⎜ ⎟ ⎜ m ⎝ 2πkT ⎠ ⎝ m

f (v )dv.

4 ⎞ −1 . For oxygen gas at 300 K, ⎟e = e πvmp ⎠

vmp = 3.95 × 102 m/s and f (v) Δv = 0.0421. (c) Increasing v by a factor of 7 changes f by a factor of 7 2e −48 , and f (v )Δv = 2.94 × 10−21. (d) Multiplying the temperature by a factor of 2 increases the most probable speed by a factor of

2, and the

−21

18.88.

answers are decreased by 2: 0.0297 and 2.08 × 10 . (e) Similarly, when the temperature is one-half what it was parts (b) and (c), the fractions increase by 2 to 0.0595 and 4.15 × 10−21. EVALUATE: (f ) At lower temperatures, the distribution is more sharply peaked about the maximum (the most probable speed), as is shown in Figure 18.23a in the textbook. m IDENTIFY: Apply the definition of relative humidity given in the problem. pV = nRT = tot RT . M SET UP: M = 18.0 × 10−3 kg/mol . EXECUTE:

(a) The pressure due to water vapor is (0.60)(2.34 × 103 Pa) = 1.40 × 103 Pa.

MpV (18.0 × 10−3 kg mol)(1.40 × 103 Pa)(1.00 m3 ) = = 10 g (8.3145 J mol ⋅ K)(293.15 K) RT EVALUATE: The vapor pressure of water vapor at this temperature is much less than the total atmospheric pressure of 1.0 × 105 Pa . IDENTIFY: The measurement gives the dew point. Relative humidity is defined in Problem 18.88. partial pressure of water vapor at temperature T SET UP: relative humidity = vapor pressure of water at temperature T EXECUTE: The experiment shows that the dew point is 16.0°C, so the partial pressure of water vapor at 30.0°C (b) mtot =

18.89.

is equal to the vapor pressure at 16.0°C, which is 1.81 × 103 Pa. 1.81 × 103 Pa = 0.426 = 42.6%. 4.25 × 103 Pa EVALUATE: The lower the dew point is compared to the air temperature, the smaller the relative humidity. IDENTIFY: Use the definition of relative humidity in Problem 18.88 and the vapor pressure table in Problem 18.89. SET UP: At 28.0°C the vapor pressure of water is 3.78 × 103 Pa . EXECUTE: For a relative humidity of 35%, the partial pressure of water vapor is (0.35)(3.78 × 103 Pa) = 1.323 × 103 Pa. This is close to the vapor pressure at 12°C, which would be at an altitude

Thus the relative humidity = 18.90.

18.91.

(30°C − 12°C) (0.6° C 100 m) = 3 km above the ground. For a relative humidity of 80%, the vapor pressure will be the same as the water pressure at around 24°C, corresponding to an altitude of about 1 km. EVALUATE: Clouds form at a lower height when the relative humidity at the surface is larger. 3RT IDENTIFY: Eq.(18.21) gives the mean free path λ . In Eq.(18.20) use vrms = in place of v. M

pV = nRT = NkT . The escape speed is vescape = SET UP: EXECUTE:

2GM . R

For atomic hydrogen, M = 1.008 × 10−3 kg/mol . (a) From Eq.(18.21), λ = (4π 2r 2 ( N V )) −1 = (4π 2(5.0 × 10−11 m) 2 (50 × 106 m −3 )) −1 = 4.5 × 1011 m .

(b) vrms = 3RT / M = 3(8.3145 J mol ⋅ K)(20 K) (1.008 × 10−3 kg mol) = 703 m s, and the time between

collisions is then (4.5 × 1011 m) (703 m s) = 6.4 × 108 s, about 20 yr. Collisions are not very important. (c) p = ( N V )kT = (50 /1.0 × 10−6 m3 )(1.381 × 10 −23 J K)(20 K) = 1.4 × 10−14 Pa.

18-22

Chapter 18

(d) vescape =

2GM 2G ( Nm V )(4πR 3 3) = = (8π 3)G ( N V )mR 2 R R

vescape = (8π/3)(6.673 × 10−11 N ⋅ m 2 /kg 2 )(50 × 106 m −3 )(1.67 × 10−27 kg)(10 × 9.46 × 1015 m) 2 vescape = 650 m s. This is lower than vrms and the cloud would tend to evaporate. (e) In equilibrium (clearly not thermal equilibrium), the pressures will be the same; from pV = NkT ,

kTISM ( N V ) ISM = kTnebula ( N V ) nebula and the result follows. (f ) With the result of part (e),

18.92.

⎛ (V N ) nebula ⎞ ⎛ 50 × 106 m3 ⎞ = 2 × 105 K, TISM = Tnebula ⎜ ⎟ = (20 K) ⎜ −6 3 −1 ⎟ × V N ( ) (200 10 m ) ⎝ ⎠ ISM ⎠ ⎝ more than three times the temperature of the sun. This indicates a high average kinetic energy, but the thinness of the ISM means that a ship would not burn up. EVALUATE: The temperature of a gas is determined by the average kinetic energy per atom of the gas. The energy density for the gas also depends on the number of atoms per unit volume, and this is very small for the ISM. IDENTIFY: Follow the procedure of Example 18.4, but use T = T0 − α y . SET UP: ln(1 + x) ≈ x when x is very small. EXECUTE:

(a)

dp Mg dy dp pM =− , which in this case becomes =− . This integrates to dy RT p R T0 − αy Mg / Rα

⎛ p ⎞ Mg ⎛ αy ⎞ ⎛ αy ⎞ ln ⎜ ⎟ = ln ⎜1 − ⎟ , or p = p0 ⎜1 − ⎟ . p Rα T 0 ⎠ ⎝ 0⎠ ⎝ ⎝ T0 ⎠ αy αy (b) For sufficiently small α, ln(1 − ) ≈ − , and this gives the expression derived in Example 18.4. T0 T0 ⎛ (0.6 × 10−2 C°/m)(8863 m) ⎞ Mg (28.8 × 10−3 )(9.80 m/s 2 ) = = 5.6576 and (c) ⎜1 − ⎟ = 0.8154, Rα (8.3145 J/mol ⋅ K)(0.6 × 10−2 C°/m) (288 K) ⎝ ⎠

18.93.

p0 (0.8154)5.6576 = 0.315 atm, which is 0.95 of the result found in Example 18.4. EVALUATE: The pressure is calculated to decrease more rapidly with altitude when we assume that T also decreases with altitude. IDENTIFY and SET UP: The behavior of isotherms for a real gas above and below the critical point are shown in Figure 18.7 in the textbook. EXECUTE: (a) A positive slope ∂P would mean that an increase in pressure causes an increase in volume, or ∂V that decreasing volume results in a decrease in pressure, which cannot be the case for any real gas. (b) See Figure 18.7 in the textbook. From part (a), p cannot have a positive slope along an isotherm, and so can have no extremes (maxima or minima) along an isotherm. When

∂p vanishes along an isotherm, the point on the ∂V

∂2 p =0. ∂V 2 nRT an 2 ∂p nRT 2an 2 ∂ 2 p 2nRT 6an 2 =− + 3 . = − 4 . Setting the last two of these equal to (c) p = − 2 . 2 2 3 V V ∂V (V − nb) ∂V (V − nb) V − nb V curve in a p -V diagram must be an inflection point, and

zero gives V 3nRT = 2an 2 (V − nb) 2 and V 4 nRT = 3an 2 (V − nb)3 . (d) Following the hint, V = (3 2)(V − nb), which is solved for (V n)c = 3b. Substituting this into either of the last

two expressions in part (c) gives Tc = 8a 27 Rb. (e) pc = (f )

RT a R (8a / 27 Rb) a a . − = − 2= (V n)c − b (V n)c 2b 9b 27b 2

RTc (8a / 27b) 8 = = . 2 pc (V n)c (a / 27b )3b 3

(g) H 2 : 3.28. N 2 : 3.44. H 2O : 4.35. EVALUATE: (h) While all are close to 8/3, the agreement is not good enough to be useful in predicting critical point data. The van der Waals equation models certain gases, and is not accurate for substances near critical points.


18.94.


For N particles, vav =

∑v

i

N

and vrms =

∑v

2 i

N

18-23

.

1 v12 + v22 and 2 1 1 1 1 2 vrms − vav2 = (v12 + v22 ) − (v12 + v22 + 2v1v2 ) = (v12 + v22 − 2v1v2 ) = (v1 − v2 ) 2 2 4 4 4 This shows that vrms ≥ vav , with equality holding if and only if the particles have the same speeds.

EXECUTE:

(a) vav = 12 (v1 + v2 ) , vrms =

1 ( Nv 2 + u 2 ), v′ = 1 ( Nv + u ), and the given forms follow immediately. rms av av N +1 N +1 (c) The algebra is similar to that in part (a); it helps somewhat to express 1 ′2 = ( N ((N + 1) − 1)vav2 + 2Nvavu + (( N + 1) − N )u 2 ) . vav (N + 1) 2 ′2 = (b) vrms

′2 = vav

N 2 N 1 (−vav2 + 2vavu − u 2 ) + vav + u2 2 (N + 1) N +1 N +1

Then,

N N N N 2 2 (vrms (vav2 − 2vavu + u 2 ) = (vrms (vav − u ) 2 . If vrms > vav , then − vav2 ) + − vav2 ) + 2 (N + 1) (N + 1) (N + 1) 2 N +1 ′. this difference is necessarily positive, and v′rms > vav ′2 − vav ′2 = vrms

(d) The result has been shown for N = 1, and it has been shown that validity for N implies validity for N + 1; by induction, the result is true for all N. EVALUATE: vrms > vav because vrms gives more weight to particles that have greater speed.

19

THE FIRST LAW OF THERMODYNAMICS

19.1.


The pressure is constant and the volume increases. The pV-diagram is sketched in Figure 19.1 Figure 19.1

(b) W = ∫

V2 V1

p dV V2

Since p is constant, W = p ∫ dV = p (V2 − V1 ) V1

The problem gives T rather than p and V, so use the ideal gas law to rewrite the expression for W. EXECUTE: pV = nRT so p1V1 = nRT1 , p2V2 = nRT2 ; subtracting the two equations gives

p (V2 − V1 ) = nR (T2 − T1 ) Thus W = nR (T2 − T1 ) is an alternative expression for the work in a constant pressure process for an ideal gas.

19.2.

Then W = nR (T2 − T1 ) = (2.00 mol)(8.3145 J/mol ⋅ K)(107°C − 27°C) = +1330 J EVALUATE: The gas expands when heated and does positive work. IDENTIFY: At constant pressure, W = pΔV = nRΔT . R = 8.3145 J/mol ⋅ K. ΔT has the same numerical value in kelvins and in C°. W 1.75 × 103 J = = 35.1 K. ΔTK = ΔTC and T2 = 27.0°C + 35.1°C = 62.1°C. EXECUTE: ΔT = nR (6 mol) (8.3145 J/mol ⋅ K) EVALUATE: When W > 0 the gas expands. When p is constant and V increases, T increases. IDENTIFY: Example 19.1 shows that for an isothermal process W = nRT ln( p1 / p2 ). pV = nRT says V decreases when p increases and T is constant. SET UP: T = 358.15 K. p2 = 3 p1. EXECUTE: (a) The pV-diagram is sketched in Figure 19.3. ⎛ p ⎞ (b) W = (2.00 mol)(8.314 J/mol ⋅ K)(358.15 K)ln ⎜ 1 ⎟ = −6540 J. ⎝ 3 p1 ⎠ SET UP:

19.3.

EVALUATE:

Since V decreases, W is negative.

Figure 19.3 19.4.

IDENTIFY: Use the expression for W that is appropriate to this type of process. SET UP: The volume is constant. EXECUTE: (a) The pV diagram is given in Figure 19.4. (b) Since ΔV = 0, W = 0. 19-1

19-2

Chapter 19

EVALUATE:

For any constant volume process the work done is zero.

Figure 19.4 19.5.

IDENTIFY: Example 19.1 shows that for an isothermal process W = nRT ln( p1 / p2 ). Solve for p1. SET UP: For a compression (V decreases) W is negative, so W = −518 J. T = 295.15 K.

⎛p ⎞ p W W −518 J = = −0.692. = ln ⎜ 1 ⎟ . 1 = eW / nRT . p nRT (0.305 mol)(8.314 J/mol ⋅ K)(295.15 K) nRT p 2 ⎝ 2⎠ p1 = p2eW / nRT = (1.76 atm)e −0.692 = 0.881 atm. (b) In the process the pressure increases and the volume decreases. The pV-diagram is sketched in Figure 19.5. EVALUATE: W is the work done by the gas, so when the surroundings do work on the gas, W is negative. EXECUTE:

(a)

Figure 19.5 19.6.


The pV-diagram is sketched in Figure 19.6.

Figure 19.6

19.7.

(b) Calculate W for each process, using the expression for W that applies to the specific type of process. EXECUTE: 1 → 2, ΔV = 0, so W = 0 2→3 p is constant; so W = p ΔV = (5.00 × 105 Pa)(0.120 m3 − 0.200 m3 ) = −4.00 × 104 J (W is negative since the volume decreases in the process.) Wtot = W1→ 2 + W2 →3 = −4.00 × 104 J EVALUATE: The volume decreases so the total work done is negative. IDENTIFY: Calculate W for each step using the appropriate expression for each type of process. SET UP: When p is constant, W = pΔV . When ΔV = 0, W = 0. EXECUTE:

19.8.

(a) W13 = p1 (V2 − V1 ), W32 = 0, W24 = p2 (V1 − V2 ) and W41 = 0. The total work done by the system is

W13 + W32 + W24 + W41 = ( p1 − p2 )(V2 − V1 ), which is the area in the pV plane enclosed by the loop. (b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a). EVALUATE: When ΔV > 0, W > 0 and when ΔV < 0, W < 0. IDENTIFY: Apply ΔU = Q − W . SET UP: For an ideal gas, U depends only on T.

The First Law of Thermodynamics

19.9.

EXECUTE: (a) V decreases and W is negative. (b) Since T is constant, ΔU = 0 and Q = W . Since W is negative, Q is negative. (c) Q = W , the magnitudes are the same. EVALUATE: Q < 0 means heat flows out of the gas. The plunger does positive work on the gas. The energy added by the positive work done on the gas leaves as heat flow out of the gas and the internal energy of the gas is constant. IDENTIFY: ΔU = Q − W . For a constant pressure process, W = pΔV . SET UP: EXECUTE:

19.10.

19.12.

Q = +1.15 × 105 J, since heat enters the gas. (a) W = pΔV = (1.80 × 105 Pa)(0.320 m3 − 0.110 m3 ) = 3.78 × 104 J.

(b) ΔU = Q − W = 1.15 × 105 J − 3.78 × 104 J = 7.72 × 104 J. EVALUATE: (c) W = pΔV for a constant pressure process and ΔU = Q − W both apply to any material. The ideal gas law wasn’t used and it doesn’t matter if the gas is ideal or not. IDENTIFY: The type of process is not specified. We can use ΔU = Q − W because this applies to all processes. Calculate ΔU and then from it calculate ΔT . SET UP: Q is positive since heat goes into the gas; Q = +1200 J W positive since gas expands; W = +2100 J EXECUTE: ΔU = 1200 J − 2100 J = −900 J We can also use ΔU = n ( 32 R ) ΔT since this is true for any process for an ideal gas.

ΔT =

19.11.

19-3

2 ΔU 2(−900 J) = = −14.4C° 3nR 3(5.00 mol)(8.3145 J/mol ⋅ K)

T2 = T1 + ΔT = 127°C − 14.4C° = 113°C EVALUATE: More energy leaves the gas in the expansion work than enters as heat. The internal energy therefore decreases, and for an ideal gas this means the temperature decreases. We didn’t have to convert ΔT to kelvins since ΔT is the same on the Kelvin and Celsius scales. IDENTIFY: Apply ΔU = Q − W to the air inside the ball. SET UP: Since the volume decreases, W is negative. Since the compression is sudden, Q = 0. EXECUTE: ΔU = Q − W with Q = 0 gives ΔU = −W . W < 0 so ΔU > 0. ΔU = +410 J. (b) Since ΔU > 0, the temperature increases. EVALUATE: When the air is compressed, work is done on the air by the force on the air. The work done on the air increases its energy. No energy leaves the gas as a flow of heat, so the internal energy increases. IDENTIFY and SET UP: Calculate W using the equation for a constant pressure process. Then use ΔU = Q − W to calculate Q. (a) EXECUTE:

W =∫

V2 V1

p dV = p (V2 − V1 ) for this constant pressure process.

W = (2.3 × 10 Pa)(1.20 m3 − 1.70 m3 ) = −1.15 × 105 J (The volume decreases in the process, so W is negative.) (b) ΔU = Q − W 5

Q = ΔU + W = −1.40 × 105 J + (−1.15 × 105 J) = −2.55 × 105 J Q negative means heat flows out of the gas. (c) EVALUATE:

19.13.

W =∫

V2 V1

p dV = p (V2 − V1 ) (constant pressure) and ΔU = Q − W apply to any system, not just to

an ideal gas. We did not use the ideal gas equation, either directly or indirectly, in any of the calculations, so the results are the same whether the gas is ideal or not. IDENTIFY: Calculate the total food energy value for one doughnut. K = 12 mv 2 . SET UP: 1 cal = 4.186 J EXECUTE: (a) The energy is (2.0 g)(4.0 kcal g) + (17.0 g)(4.0 kcal g) + (7.0 g)(9.0 kcal g) = 139 kcal. The time required is (139 kcal) (510 kcal h) = 0.273 h = 16.4 min.

19.14.

(b) v = 2K m = 2(139 × 103 cal)(4.186 J cal) (60 kg) = 139 m s = 501 km h. EVALUATE: When we set K = Q, we must express Q in J, so we can solve for v in m/s. IDENTIFY: Apply ΔU = Q − W . SET UP: W > 0 when the system does work.

19-4

19.15.

19.16.

19.17.

19.18.

19.19.

Chapter 19

EXECUTE: (a) The container is said to be well-insulated, so there is no heat transfer. (b) Stirring requires work. The stirring needs to be irregular so that the stirring mechanism moves against the water, not with the water. (c) The work mentioned in part (b) is work done on the system, so W < 0, and since no heat has been transferred, ΔU = −W > 0. EVALUATE: The stirring adds energy to the liquid and this energy stays in the liquid as an increase in internal energy. IDENTIFY: Apply ΔU = Q − W to the gas. SET UP: For the process, ΔV = 0. Q = +400 J since heat goes into the gas. EXECUTE: (a) Since ΔV = 0, W = 0. p nR (b) pV = nRT says = = constant. Since p doubles, T doubles. Tb = 2Ta . T V (c) Since W = 0, ΔU = Q = +400 J. U b = U a + 400 J. EVALUATE: For an ideal gas, when T increases, U increases. IDENTIFY: Apply ΔU = Q − W . W is the area under the path in the pV-plane. SET UP: W > 0 when V increases. EXECUTE: (a) The greatest work is done along the path that bounds the largest area above the V-axis in the p-V plane, which is path 1. The least work is done along path 3. (b) W > 0 in all three cases; Q = ΔU + W , so Q > 0 for all three, with the greatest Q for the greatest work, that along path 1. When Q > 0, heat is absorbed. EVALUATE: ΔU is path independent and depends only on the initial and final states. W and Q are path independent and can have different values for different paths between the same initial and final states. IDENTIFY: ΔU = Q − W . W is the area under the path in the pV-diagram. When the volume increases, W > 0. SET UP: For a complete cycle, ΔU = 0. EXECUTE: (a) and (b) The clockwise loop (I) encloses a larger area in the p-V plane than the counterclockwise loop (II). Clockwise loops represent positive work and counterclockwise loops negative work, so WI > 0 and WII < 0. Over one complete cycle, the net work WI + WII > 0, and the net work done by the system is positive. (c) For the complete cycle, ΔU = 0 and so W = Q. From part (a), W > 0, so Q > 0, and heat flows into the system. (d) Consider each loop as beginning and ending at the intersection point of the loops. Around each loop, ΔU = 0, so Q = W ; then, QI = WI > 0 and QII = WII < 0. Heat flows into the system for loop I and out of the system for loop II. EVALUATE: W and Q are path dependent and are in general not zero for a cycle. IDENTIFY and SET UP: Deduce information about Q and W from the problem statement and then apply the first law, ΔU = Q − W , to infer whether Q is positive or negative. EXECUTE: (a) For the water ΔT > 0, so by Q = mc ΔT heat has been added to the water. Thus heat energy comes from the burning fuel-oxygen mixture, and Q for the system (fuel and oxygen) is negative. (b) Constant volume implies W = 0. (c) The 1st law (Eq.19.4) says ΔU = Q − W . Q < 0, W = 0 so by the 1st law ΔU < 0. The internal energy of the fuel-oxygen mixture decreased. EVALUATE: In this process internal energy from the fuel-oxygen mixture was transferred to the water, raising its temperature. IDENTIFY: ΔU = Q − W . For a constant pressure process, W = pΔV . SET UP: Q = +2.20 × 106 J; Q > 0 since this amount of heat goes into the water. p = 2.00 atm = 2.03 × 105 Pa. EXECUTE:

(a) W = pΔV = (2.03 × 105 Pa)(0.824 m3 − 1.00 × 10−3 m3 ) = 1.67 × 105 J

(b) ΔU = Q − W = 2.20 × 106 J − 1.67 × 105 J = 2.03 × 106 J.

19.20.

EVALUATE: 2.20 × 106 J of energy enters the water. 1.67 × 105 J of energy leaves the materials through expansion work and the remainder stays in the material as an increase in internal energy. IDENTIFY: ΔU = Q − W SET UP: Q < 0 when heat leaves the gas. EXECUTE: For an isothermal process, ΔU = 0, so W = Q = −335 J. EVALUATE: In a compression the volume decreases and W < 0.


19.21.

19-5

For a constant pressure process, W = pΔV , Q = nC p ΔT and ΔU = nCV ΔT . ΔU = Q − W and

IDENTIFY:

C p = CV + R. For an ideal gas, pΔV = nRΔT . SET UP: From Table 19.1, CV = 28.46 J/mol ⋅ K. EXECUTE: (a) The pV diagram is given in Figure 19.21. (b) W = pV2 − pV1 = nR (T2 − T1 ) = (0.250 mol)(8.3145 J mol ⋅ K)(100.0 K) = 208 J. (c) The work is done on the piston. (d) Since Eq. (19.13) holds for any process, ΔU = nCV ΔT = (0.250 mol)(28.46 J mol ⋅ K)(100.0 K) = 712 J. (e) Either Q = nC p ΔT or Q = ΔU + W gives Q = 920 J to three significant figures. ( f ) The lower pressure would mean a correspondingly larger volume, and the net result would be that the work done would be the same as that found in part (b). EVALUATE: W = nRΔT , so W, Q and ΔU all depend only on ΔT . When T increases at constant pressure, V increases and W > 0. ΔU and Q are also positive when T increases.

Figure 19.21 19.22.

For constant volume Q = nCV ΔT . For constant pressure, Q = nC p ΔT . For any process of an ideal

IDENTIFY:

gas, ΔU = nCV ΔT . SET UP:

R = 8.315 J/mol ⋅ K. For helium, CV = 12.47 J/mol ⋅ K and C p = 20.78 J/mol ⋅ K.

EXECUTE: (a) Q = nCV ΔT = (0.0100 mol)(12.47 J mol ⋅ K)(40.0 C°) = 4.99 J. The pV-diagram is sketched in Figure 19.22a. (b) Q = nC p ΔT = (0.0100 mol)(20.78 J/mol ⋅ K)(40.0 C°) = 8.31 J. The pV-diagram is sketched in Figure 19.22b. (c) More heat is required for the constant pressure process. ΔU is the same in both cases. For constant volume W = 0 and for constant pressure W > 0. The additional heat energy required for constant pressure goes into expansion work. (d) ΔU = nCV ΔT = 4.99 J for both processes. ΔU is path independent and for an ideal gas depends only on ΔT . EVALUATE:

C p = CV + R, so C p > CV .

Figure 19.22 19.23.


For constant volume, Q = nCV ΔT . For constant pressure, Q = nC p ΔT . From Table 19.1, CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K (a) Using Equation (19.12), ΔT =

The pV-diagram is sketched in Figure 19.23a.

Q 645 J = = 167.9 K and T = 948 K. nCV (0.185 mol)(20.76 J mol ⋅ K)

19-6

Chapter 19

(b) Using Equation (19.14), ΔT =

645 J Q = = 119.9 K and T = 900 K. nC p (0.185 mol)(29.07 J mol ⋅ K)

The pV-diagram is sketched in Figure 19.23b. EVALUATE: At constant pressure some of the heat energy added to the gas leaves the gas as expansion work and the internal energy change is less than if the same amount of heat energy is added at constant volume. ΔT is proportional to ΔU .

Figure 19.23 19.24.

IDENTIFY and SET UP: Use information about the pressure and volume in the ideal gas law to determine the sign of ΔT , and from that the sign of Q. EXECUTE: For constant p, Q = nC p ΔT

Since the gas is ideal, pV = nRT and for constant p, p ΔV = nR ΔT . ⎛ p ΔV Q = nC p ⎜ ⎝ nR

⎞ ⎛ Cp ⎞ ⎟ p ΔV ⎟=⎜ ⎠ ⎝ R ⎠

Since the gas expands, ΔV > 0 and therefore Q > 0. Q > 0 means heat goes into gas. EVALUATE: Heat flows into the gas, W is positive and the internal energy increases. It must be that Q > W . 19.25.

ΔU = Q − W . For an ideal gas, ΔU = CV ΔT , and at constant pressure, W = p ΔV = nR ΔT .

IDENTIFY:

CV = 32 R for a monatomic gas.

SET UP:

ΔU = n( 32 R )ΔT = 32 p ΔV = 32 W . Then Q = ΔU + W = 52 W , so W Q = 52 .

EXECUTE:

19.26.

EVALUATE: For diatomic or polyatomic gases, CV is a different multiple of R and the fraction of Q that is used for expansion work is different. IDENTIFY: For an ideal gas, ΔU = CV ΔT , and at constant pressure, pΔV = nRΔT .

CV = 32 R for a monatomic gas.

SET UP:

ΔU = n ( 32 R ) ΔT = 32 pΔV = 32 (4.00 × 104 Pa)(8.00 × 10−3 m3 − 2.00 × 10−3 m3 ) = 360 J.


W = nRΔT = 32 ΔU = 240 J. Q = nC p ΔT = n( 52 R )ΔT = 53 ΔU = 600 J. 600 J of heat energy flows into

the gas. 240 J leaves as expansion work and 360 J remains in the gas as an increase in internal energy. IDENTIFY: For a constant volume process, Q = nCV ΔT . For a constant pressure process, Q = nC p ΔT . For any process of an ideal gas, ΔU = nCV ΔT . From Table 19.1, for N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. Heat is added, so Q is

SET UP:

positive and Q = +1557 J. EXECUTE: (b) ΔT =

(a) ΔT =

Q 1557 J = = +25.0 K nCV (3.00 mol)(20.76 J/mol ⋅ K)

Q 1557 J = = +17.9 K nC p (3.00 mol)(29.07 J/mol ⋅ K)

(c) ΔU = nCV ΔT for either process, so ΔU is larger when ΔT is larger. The final internal energy is larger for the constant volume process in (a). EVALUATE: For constant volume W = 0 and all the energy added as heat stays in the gas as internal energy. For the constant pressure process the gas expands and W > 0. Part of the energy added as heat leaves the gas as expansion work done by the gas.


19.28.

19-7

Apply pV = nRT to calculate T. For this constant pressure process, W = pΔV . Q = nC p ΔT . Use

IDENTIFY:

ΔU = Q − W to relate Q, W and ΔU . 2.50 atm = 2.53 × 105 Pa. For a monatomic ideal gas, CV = 12.47 J/mol ⋅ K and C p = 20.78 J/mol ⋅ K.

SET UP:

T2 =

pV1 (2.53 × 105 Pa)(3.20 × 10−2 m 2 ) = = 325 K. nR (3.00 mol)(8.314 J/mol ⋅ K)

(a) T1 =

EXECUTE:

pV2 (2.53 × 105 Pa)(4.50 × 10−2 m 2 ) = = 456 K. nR (3.00 mol)(8.314 J/mol ⋅ K)

(b) W = pΔV = (2.53 × 105 Pa)(4.50 × 10−2 m3 − 3.20 × 10−2 m3 ) = 3.29 × 103 J (c) Q = nC p ΔT = (3.00 mol)(20.78 J/mol ⋅ K)(456 K − 325 K) = 8.17 × 103 J (d) ΔU = Q − W = 4.88 × 103 J

19.29.

EVALUATE: We could also calculate ΔU as ΔU = nCV ΔT = (3.00 mol)(12.47 J/mol ⋅ K)(456 K − 325 K) = 4.90 × 103 J, which agrees with the value we calculated in part (d). IDENTIFY: Calculate W and ΔU and then use the first law to calculate Q. (a) SET UP:

W =∫

V2 V1

p dV

pV = nRT so p = nRT / V V2

V2

V1

V1

W = ∫ (nRT / V ) dV = nRT ∫ dV / V = nRT ln(V2 / V1 ) (work done during an isothermal process). EXECUTE: W = (0.150 mol)(8.3145 J/mol ⋅ K)(350 K)ln(0.25V1 / V1 ) = (436.5 J)ln(0.25) = −605 J. EVALUATE: W for the gas is negative, since the volume decreases. (b) EXECUTE: ΔU = nCV ΔT for any ideal gas process. ΔT = 0 (isothermal) so ΔU = 0. EVALUATE: ΔU = 0 for any ideal gas process in which T doesn’t change. (c) EXECUTE: ΔU = Q − W ΔU = 0 so Q = W = −605 J. (Q is negative; the gas liberates 605 J of heat to the surroundings.) EVALUATE:

Q = nCV ΔT is only for a constant volume process so doesn’t apply here.

Q = nC p ΔT is only for a constant pressure process so doesn’t apply here. 19.30.

IDENTIFY:

C p = CV + R and γ =

Cp CV

.

R = 8.315 J/mol ⋅ K

SET UP: EXECUTE:

C p = CV + R. γ =

Cp CV

=1+

R R 8.315 J/mol ⋅ K . CV = = = 65.5 J/mol ⋅ K. Then 0.127 CV γ −1

C p = CV + R = 73.8 J/mol ⋅ K.

19.31.

EVALUATE: The value of CV is about twice the values for the polyatomic gases in Table 19.1. A propane molecule has more atoms and hence more internal degrees of freedom than the polyatomic gases in the table. IDENTIFY: ΔU = Q − W . Apply Q = nC p ΔT to calculate C p . Apply ΔU = nCV ΔT to calculate CV . γ = C p / CV . SET UP: ΔT = 15.0 C° = 15.0 K. Since heat is added, Q = +970 J. EXECUTE: (a) ΔU = Q − W = +970 J − 223 J = 747 J (b) C p =

37.0 J/mol ⋅ K = 1.30 CV 28.5 J/mol ⋅ K EVALUATE: The value of γ we calculated is similar to the values given in Tables 19.1 for polyatomic gases. IDENTIFY and SET UP: For an ideal gas ΔU = nCV ΔT . The sign of ΔU is the same as the sign of ΔT . Combine Eq.(19.22) and the ideal gas law to obtain an equation relating T and p, and use it to determine the sign of ΔT . γ −1 EXECUTE: TV = T2V2γ −1 and V = nRT / p so, T1γ p11−γ = T2γ p12−γ and T2γ = T1γ ( p2 / p1 )γ −1 1 1

γ=

19.32.

Q 970 J ΔU 747 J = = 37.0 J/mol ⋅ K. CV = = = 28.5 J/mol ⋅ K. nΔT (1.75 mol)(15.0 K) nΔT (1.75 mol)(15.0 K)

Cp

=

p2 < p1 and γ − 1 is positive so T2 < T1. ΔT is negative so ΔU is negative; the energy of the gas decreases. EVALUATE: Eq.(19.24) shows that the volume increases for this process, so it is an adiabatic expansion. In an adiabatic expansion the temperature decreases.

19-8

Chapter 19

19.33.

IDENTIFY: SET UP:

For an adiabatic process of an ideal gas, p1V1γ = p2V2γ , W =

1 γ −1 = T2V2γ −1. ( p1V1 − p2V2 ) and TV 1 1 γ −1

For a monatomic ideal gas γ = 5 / 3. γ

5/3

⎛V ⎞ ⎛ 0.0800 m3 ⎞ (a) p2 = p1 ⎜ 1 ⎟ = (1.50 × 105 Pa ) ⎜ = 4.76 × 105 Pa. 3 ⎟ V 0.0400 m ⎝ ⎠ ⎝ 2⎠ (b) This result may be substituted into Eq.(19.26), or, substituting the above form for p2 , EXECUTE:

W=

⎛ ⎛ 0.0800 ⎞ 2 / 3 ⎞ 1 3 γ −1 p1V1 1 − (V1 / V2 ) = (1.50 × 105 Pa )( 0.0800 m 3 ) ⎜1 − ⎜ = −1.06 × 10 4 J. ⎜ ⎝ 0.0400 ⎠⎟ ⎟⎟ 2 γ −1 ⎝ ⎠

(

)

(c) From Eq.(19.22), (T2 T1 ) = (V2 V1 )

γ −1

19.34.

= ( 0.0800 0.0400 )

23

= 1.59, and since the final temperature is higher than

the initial temperature, the gas is heated. EVALUATE: In an adiabatic compression W < 0 since ΔV < 0. Q = 0 so ΔU = −W . ΔU > 0 and the temperature increases. IDENTIFY and SET UP: (a) In the process the pressure increases and the volume decreases. The pV-diagram is sketched in Figure 19.34.

Figure 19.34 (b) For an adiabatic process for an ideal gas γ −1 TV = T2V2γ −1 , p1V1γ = p2V2γ , and pV = nRT 1 1

From the first equation, T2 = T1 (V1 / V2 )γ −1 = (293 K)(V1 / 0.0900V1 )1.4 −1

EXECUTE:

T2 = (293 K)(11.11)0.4 = 768 K = 495°C γ −1 (Note: In the equation TV = T2V2γ −1 the temperature must be in kelvins.) 1 1

p1V1γ = p2V2γ implies p2 = p1 (V1 / V2 )γ = (1.00 atm)(V1 / 0.0900V1 )1.4 p2 = (1.00 atm)(11.11)1.4 = 29.1 atm EVALUATE:

Alternatively, we can use pV = nRT to calculate p2 : n, R constant implies pV / T = nR = constant

so p1V1 / T1 = p2V2 / T2 p2 = p1 (V1 / V2 )(T2 / T1 ) = (1.00 atm)(V1 / 0.0900V1 )(768 K/293 K) = 29.1 atm, which checks. 19.35.


For an adiabatic process of an ideal gas, W =

1 ( p1V1 − p2V2 ) and p1V1γ = p2V2γ . γ −1

γ = 1.40 for an ideal diatomic gas. 1 atm = 1.013 × 105 Pa and 1 L = 10−3 m3 . Q = ΔU + W = 0 for an adiabatic process, so ΔU = −W =

1

γ −1

( p2V2 − p1V1 ).

p1 = 1.22 × 105 Pa.

p2 = p1 (V1 / V2 ) = (1.22 × 105 Pa)(3)1.4 = 5.68 × 105 Pa. γ

1 ([5.68 ×105 Pa][10 ×10−3 m −3 ] − [1.22 ×105 Pa][30 ×10−3 m −3 ]) = 5.05 ×103 J. The internal energy 0.40 increases because work is done on the gas (ΔU > 0) and Q = 0. The temperature increases because the internal energy has increased. EVALUATE: In an adiabatic compression W < 0 since ΔV < 0. Q = 0 so ΔU = −W . ΔU > 0 and the temperature increases. γ −1 IDENTIFY: Assume the expansion is adiabatic. TV = T2V2γ −1 relates V and T. Assume the air behaves as an ideal 1 1 W=

19.36.

gas, so ΔU = nCV ΔT . Use pV = nRT to calculate n. SET UP:

For air, CV = 29.76 J/mol ⋅ K and γ = 1.40. V2 = 0.800V1. T1 = 293.15 K. p1 = 2.026 × 105 Pa. For a

sphere, V = 34 π r 3 .

The First Law of Thermodynamics γ −1

19-9

0.40

⎛V ⎞ ⎛ V1 ⎞ (a) T2 = T1 ⎜ 1 ⎟ = (293.15 K) ⎜ ⎟ = 320.5 K = 47.4°C. ⎝ V2 ⎠ ⎝ 0.800V1 ⎠ p V (2.026 × 105 Pa)(7.15 × 10−3 m3 ) 4π = 0.594 mol. (0.1195 m)3 = 7.15 × 10−3 m3 . n = 1 1 = (b) V1 = 34 π r 3 = RT1 (8.314 J/mol ⋅ K)(293.15 K) 3 EXECUTE:

19.37.

ΔU = nCV ΔT = (0.594 mol)(20.76 J/mol ⋅ K)(321 K − 293 K) = 345 J. 1 EVALUATE: We could also use ΔU = W = ( p1V1 − p2V2 ) to calculate ΔU , if we first found p2 from pV = nRT . γ −1 (a) IDENTIFY and SET UP: In the expansion the pressure decreases and the volume increases. The pV-diagram is sketched in Figure 19.37.

Figure 19.37

19.38.

(b) Adiabatic means Q = 0. Then ΔU = Q − W gives W = −ΔU = − nCV ΔT = nCV (T1 − T2 ) (Eq.19.25). CV = 12.47 J/mol ⋅ K (Table 19.1) EXECUTE: W = (0.450 mol)(12.47 J/mol ⋅ K)(50.0°C − 10.0°C) = +224 J W positive for ΔV > 0 (expansion) (c) ΔU = −W = −224 J. EVALUATE: There is no heat energy input. The energy for doing the expansion work comes from the internal energy of the gas, which therefore decreases. For an ideal gas, when T decreases, U decreases. γ −1 IDENTIFY: pV = nRT . For an adiabatic process, TV = T2V2γ −1. 1 1 SET UP: For an ideal monatomic gas, γ = 5/ 3. 5 −3 3 pV (1.00 × 10 Pa) (2.50 × 10 m ) EXECUTE: (a) T = = = 301 K. nR (0.1 mol) (8.3145 J mol ⋅ K ) (b) (i) Isothermal: If the expansion is isothermal, the process occurs at constant temperature and the final temperature is the same as the initial temperature, namely 301 K. p2 = p1 (V1 / V2 ) = 12 p1 = 5.00 × 104 Pa.

(ii) Isobaric: Δp = 0 so p2 = 1.00 × 105 Pa. T2 = T1 (V2 / V1 ) = 2T1 = 602 K. γ −1 0.67 TV (301 K)(V1 ) 0.67 1 1 = = (301 K) ( 12 ) = 189 K. γ −1 V2 (2V1 )0.67 EVALUATE: In an isobaric expansion, T increases. In an adiabatic expansion, T decreases. γ −1 IDENTIFY: Combine TV = T2V2γ −1 with pV = nRT to obtain an expression relating T and p for an adiabatic 1 1 process of an ideal gas. SET UP: T1 = 299.15 K

(iii) Adiabatic: Using Equation (19.22), T2 = 19.39.

γ −1

EXECUTE:

⎛ nRT1 ⎞ nRT V= so T1 ⎜ ⎟ p ⎝ p1 ⎠ ( γ −1) / γ

19.40.

γ −1

⎛ nRT2 ⎞ = T2 ⎜ ⎟ ⎝ p2 ⎠

and

T1γ

γ −1

p1

=

T2γ

γ −1

p2

.

0.4 /1.4

⎛p ⎞ ⎛ 0.850 × 105 Pa ⎞ T2 = T1 ⎜ 2 ⎟ = (299.15 K) ⎜ = 284.8 K = 11.6 °C ⎟ 5 ⎝ 1.01 × 10 Pa ⎠ ⎝ p1 ⎠ EVALUATE: For an adiabatic process of an ideal gas, when the pressure decreases the temperature decreases. IDENTIFY: Apply ΔU = Q − W . For any process of an ideal gas, ΔU = nCV ΔT . For an isothermal expansion, ⎛V ⎞ ⎛p ⎞ W = nRT ln ⎜ 2 ⎟ = nRT ln ⎜ 1 ⎟ . ⎝ V1 ⎠ ⎝ p2 ⎠ p V SET UP: T = 288.15 K. 1 = 2 = 2.00. p2 V1 EXECUTE: (a) ΔU = 0 since ΔT = 0. (b) W = (1.50 mol)(8.314 J/mol ⋅ K)(288.15 K)ln(2.00) = 2.49 × 103 J. W > 0 and work is done by the gas. Since ΔU = 0, Q = W = +2.49 × 103 J. Q > 0 so heat flows into the gas. EVALUATE: When the volume increases, W is positive.

19-10

Chapter 19

19.41.

IDENTIFY and SET UP: For an ideal gas, pV = nRT . The work done is the area under the path in the pV-diagram. EXECUTE: (a) The product pV increases and this indicates a temperature increase. (b) The work is the area in the pV plane bounded by the blue line representing the process and the verticals at

Va and Vb . The area of this trapezoid is 12 ( pb + pa )(Vb − Va ) = 12 (2.40 × 105 Pa)(0.0400 m3 ) = 4800 J. 19.42.

EVALUATE: The work done is the average pressure, 12 ( p1 + p2 ), times the volume increase. IDENTIFY: Use pV = nRT to calculate T. W is the area under the process in the pV-diagram. Use

ΔU = nCV ΔT and ΔU = Q − W to calculate Q. SET UP: EXECUTE:

19.43.

In state c, pc = 2.0 × 105 Pa and Vc = 0.0040 m3 . In state a, pa = 4.0 × 105 Pa and Va = 0.0020 m3 . (a) Tc =

pcVc (2.0 × 105 Pa)(0.0040 m3 ) = = 192 K nR (0.500 mol)(8.314 J/mol ⋅ K)

(b) W = 12 (4.0 × 105 Pa + 2.0 × 105 Pa)(0.0030 m3 − 0.0020 m3 ) + (2.0 × 105 Pa)(0.0040 m3 − 0.0030 m3 ) W = +500 J. 500 J of work is done by the gas. pV (4.0 × 105 Pa)(0.0020 m3 ) = 192 K. For the process, ΔT = 0, so ΔU = 0 and Q = W = +500 J. (c) Ta = a a = nR (0.500 mol)(8.314 J/mol ⋅ K) 500 J of heat enters the system. EVALUATE: The work done by the gas is positive since the volume increases. IDENTIFY: Use ΔU = Q − W and the fact that ΔU is path independent. W > 0 when the volume increases, W < 0 when the volume decreases, and W = 0 when the volume is constant. Q > 0 if heat flows into the system. SET UP: The paths are sketched in Figure 19.43.

Qacb = +90.0 J (positive since heat flows in) Wacb = +60.0 J (positive since ΔV > 0)

Figure 19.43 EXECUTE: (a) ΔU = Q − W ΔU is path independent; Q and W depend on the path. ΔU = U b − U a

This can be calculated for any path from a to b, in particular for path acb: ΔU a → b = Qacb − Wacb = 90.0 J − 60.0 J = 30.0 J. Now apply ΔU = Q − W to path adb; ΔU = 30.0 J for this path also. Wadb = +15.0 J (positive since ΔV > 0) ΔU a →b = Qadb − Wadb so Qacb = ΔU a →b + Wadb = 30.0 J + 15.0 J = +45.0 J (b) Apply ΔU = Q − W to path ba: ΔU b → a = Qba − Wba

Wba = −35.0 J (negative since ΔV < 0) ΔU b → a = U a − U b = −(U b − U a ) = −ΔU a →b = −30.0 J Then Qba = ΔU b → a + Wba = −30.0 J − 35.0 J = −65.0 J. (Qba < 0; the system liberates heat.) (c) U a = 0, U d = 8.0 J

ΔU a → b = U b − U a = +30.0 J, so U b = +30.0 J. process a → d ΔU a → d = Qad − Wad ΔU a → d = U d − U a = +8.0 J Wadb = +15.0 J and Wadb = Wad + Wdb . But the work Wdb for the process d → b is zero since ΔV = 0 for that process. Therefore Wad = Wadb = +15.0 J. Then Qad = ΔU a → d + Wad = +8.0 J + 15.0 J = +23.0 J (positive implies heat absorbed).


19-11

process d → b ΔU d →b = Qdb − Wdb Wdb = 0, as already noted. ΔU d →b = U b − U d = 30.0 J − 8.0 J = +22.0 J. Then Qdb = ΔU d →b + Wdb = +22.0 J (positive; heat absorbed).

19.44.

EVALUATE: The signs of our calculated Qad and Qdb agree with the problem statement that heat is absorbed in these processes. IDENTIFY: ΔU = Q − W . SET UP: W = 0 when ΔV = 0. EXECUTE: For each process, Q = ΔU + W . No work is done in the processes ab and dc, and so Wbc = Wabc = 450 J

and Wad = Wadc = 120 J. The heat flow for each process is: for ab, Q = 90 J. For bc, Q = 440 J + 450 J = 890 J. For ad , Q = 180 J + 120 J = 300 J. For dc, Q = 350 J. Heat is absorbed in each process. Note that the arrows representing the processes all point in the direction of increasing temperature (increasing U ). EVALUATE: ΔU is path independent so is the same for paths adc and abc. Qadc = 300 J + 350 J = 650 J. Qabc = 90 J + 890 J = 980 J. Q and W are path dependent and are different for these two paths. 19.45.

IDENTIFY: Use pV = nRT to calculate Tc / Ta . Calculate ΔU and W and use ΔU = Q − W to obtain Q. SET UP: For path ac, the work done is the area under the line representing the process in the pV-diagram. T pV (1.0 × 105 J)(0.060 m3 ) = 1.00. Tc = Ta . EXECUTE: (a) c = c c = Ta paVa (3.0 × 105 J)(0.020 m3 ) (b) Since Tc = Ta , ΔU = 0 for process abc. For ab, ΔV = 0 and Wab = 0. For bc, p is constant and

Wbc = pΔV = (1.0 × 105 Pa)(0.040 m3 ) = 4.0 × 103 J. Therefore, Wabc = +4.0 × 103 J. Since ΔU = 0, Q = W = +4.0 × 103 J. 4.0 × 103 J of heat flows into the gas during process abc.

19.46.

(c) W = 12 (3.0 × 105 Pa + 1.0 × 105 Pa)(0.040 m3 ) = +8.0 × 103 J. Qac = Wac = +8.0 × 103 J. EVALUATE: The work done is path dependent and is greater for process ac than for process abc, even though the initial and final states are the same. IDENTIFY: For a cycle, ΔU = 0 and Q = W . Calculate W. SET UP: The magnitude of the work done by the gas during the cycle equals the area enclosed by the cycle in the pV-diagram. EXECUTE: (a) The cycle is sketched in Figure 19.46. (b) W = (3.50 × 104 Pa − 1.50 × 104 Pa)(0.0435 m3 − 0.0280 m3 ) = +310 J. More negative work is done for cd than

positive work for ab and the net work is negative. W = −310 J. (c) Q = W = −310 J. Since Q < 0, the net heat flow is out of the gas. EVALUATE: During each constant pressure process W = pΔV and during the constant volume process W = 0.

Figure 19.46 19.47.

IDENTIFY:

Use the 1st law to relate Qtot to Wtot for the cycle.

Calculate Wab and Wbc and use what we know about Wtot to deduce Wca

19-12

Chapter 19

(a) SET UP: We aren’t told whether the pressure increases or decreases in process bc. The two possibilities for the cycle are sketched in Figure 19.47.

Figure 19.47

In cycle I, the total work is negative and in cycle II the total work is positive. For a cycle, ΔU = 0, so Qtot = Wtot The net heat flow for the cycle is out of the gas, so heat Qtot < 0 and Wtot < 0. Sketch I is correct. (b) EXECUTE:

Wtot = Qtot = −800 J

Wtot = Wab + Wbc + Wca Wbc = 0 since ΔV = 0. Wab = pΔV since p is constant. But since it is an ideal gas, pΔV = nRΔT Wab = nR(Tb − Ta ) = 1660 J

19.48.

Wca = Wtot − Wab = −800 J − 1660 J = −2460 J EVALUATE: In process ca the volume decreases and the work W is negative. IDENTIFY: Apply the appropriate expression for W for each type of process. pV = nRT and C p = CV + R. SET UP: R = 8.315 J/mol ⋅ K EXECUTE: Path ac has constant pressure, so Wac = pΔV = nRΔT , and

Wac = nR (Tc − Ta ) = (3 mol)(8.3145 J mol ⋅ K)(492 K − 300 K) = 4.789 × 103 J.

Path cb is adiabatic (Q = 0), so Wcb = Q − ΔU = −ΔU = −nCV ΔT , and using CV = C p − R, Wcb = − n(C p − R )(Tb − Tc ) = −(3 mol)(29.1 J mol ⋅ K − 8.3145 J mol ⋅ K)(600 K − 492 K) = −6.735 × 103 J.

Path ba has constant volume, so Wba = 0. So the total work done is

19.49.

W = Wac + Wcb + Wba = 4.789 × 103 J − 6.735 × 103 J + 0 = −1.95 × 103 J. EVALUATE: W > 0 when ΔV > 0, W < 0 when ΔV < 0 and W = 0 when ΔV = 0. IDENTIFY: Use Q = nCV ΔT to calculate the temperature change in the constant volume process and use pV = nRT to calculate the temperature change in the constant pressure process. The work done in the constant

volume process is zero and the work done in the constant pressure process is W = pΔV . Use Q = nC p ΔT to calculate the heat flow in the constant pressure process. ΔU = nCV ΔT , or ΔU = Q − W . SET UP: EXECUTE:

For N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. (a) For process ab, ΔT =

Q 1.52 × 104 J = = 293 K. Ta = 293 K, so Tb = 586 K. nCV (2.50 mol)(20.76 J/mol ⋅ K)

pV = nRT says T doubles when V doubles and p is constant, so Tc = 2(586 K) = 1172 K = 899°C. (b) For process ab, Wab = 0. For process bc,

Wbc = pΔV = nRΔT = (2.50 mol)(8.314 J/mol ⋅ K)(1172 K − 586 K) = 1.22 × 104 J. W = Wab + Wbc = 1.22 × 104 J. (c) For process bc, Q = nC p ΔT = (2.50 mol)(29.07 J/mol ⋅ K)(1172 K − 586 K) = 4.26 × 104 J. (d) ΔU = nCV ΔT = (2.50 mol)(20.76 J/mol ⋅ K)(1172 K − 293 K) = 4.56 × 104 J. EVALUATE: The total Q is 1.52 × 104 J + 4.26 × 104 J = 5.78 × 104 J. ΔU = Q − W = 5.78 × 104 J − 1.22 × 104 J = 4.56 × 104 J, which agrees with our results in part (d). 19.50.


For a constant pressure process, Q = nC p ΔT . ΔU = Q − W . ΔU = nCV ΔT for any ideal gas process. For N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. Q < 0 if heat comes out of the gas. (a) n =

Q +2.5×104 J = = 21.5 mol. C p ΔT (29.07 J mol ⋅ K)(40.0 K)

(b) ΔU = nCV ΔT = Q(CV / C p ) = (−2.5 × 104 J)(20.76/29.07) = −1.79 × 104 J.


19-13

(c) W = Q − ΔU = −7.15 × 103 J.

19.51.

(d) ΔU is the same for both processes, and if ΔV = 0, W = 0 and Q = ΔU = −1.79 × 104 J. EVALUATE: For a given ΔT , Q is larger when the pressure is constant than when the volume is constant. IDENTIFY and SET UP: Use the first law to calculate W and then use W = pΔV for the constant pressure process to calculate ΔV . EXECUTE: ΔU = Q − W

Q = −2.15 × 105 J (negative since heat energy goes out of the system)

ΔU = 0 so W = Q = −2.15 × 105 J Constant pressure, so W = ∫

V2 V1

pdV = p(V2 − V1 ) = pΔV .

W −2.15 × 105 J = = −0.226 m3 . p 9.50 × 105 Pa EVALUATE: Positive work is done on the system by its surroundings; this inputs to the system the energy that then leaves the system as heat. Both Eq.(19.4) and (19.2) apply to all processes for any system, not just to an ideal gas. IDENTIFY: pV = nRT . For an isothermal process W = nRT ln(V2 / V1 ). For a constant pressure process, W = pΔV .

Then ΔV =

19.52.

SET UP: 1 L = 10−3 m3 . EXECUTE: (a) The pV-diagram is sketched in Figure 19.52.

⎛ 1.00 × 105 Pa ⎞ (b) At constant temperature, the product pV is constant, so V2 = V1 ( p1 / p2 ) = (1.5 L) ⎜ ⎟ = 6.00 L. The 4 ⎝ 2.50 × 10 Pa ⎠ final pressure is given as being the same as p3 = p2 = 2.5 × 104 Pa. The final volume is the same as the initial volume, so T3 = T1 ( p3 p1 ) = 75.0 K. (c) Treating the gas as ideal, the work done in the first process is W = nRT ln(V2 V1 ) = p1V1 ln( p1 p2 ).

⎛ 1.00 × 105 Pa ⎞ W = (1.00 × 105 Pa)(1.5 × 10−3 m3 )ln ⎜ ⎟ = 208 J. 4 ⎝ 2.50 × 10 Pa ⎠ For the second process, W = p2 (V3 − V2 ) = p2 (V1 − V2 ) = p2V1 (1 − ( p1 p2 )). ⎛ 1.00 × 105 Pa ⎞ W = (2.50 × 104 Pa)(1.5 × 10−3 m3 ) ⎜1 − ⎟ = −113 J. 4 ⎝ 2.50 × 10 Pa ⎠ The total work done is 208 J − 113 J = 95 J. (d) Heat at constant volume. No work would be done by the gas or on the gas during this process. EVALUATE: When the volume increases, W > 0. When the volume decreases, W < 0.

Figure 19.52 19.53.

IDENTIFY:

ΔV = V0 βΔT . W = pΔV since the force applied to the piston is constant. Q = mc p ΔT . ΔU = Q − W .

SET UP: m = ρV EXECUTE: (a) The fractional change in volume is ΔV = V0 β ΔT = (1.20 × 10−2 m3 )(1.20 × 10−3 K −1 )(30.0 K) = 4.32 × 10−4 m3 . (b) W = pΔV = ( F A)ΔV = ((3.00 × 104 N) (0.0200 m 2 ))(4.32 × 10−4 m3 ) = 648 J. (c) Q = mc p ΔT = V0 ρ c p ΔT = (1.20 × 10−2 m3 )(791 kg m3 )(2.51 × 103 J kg ⋅ K)(30.0 K).

Q = 7.15 × 105 J.

19-14

Chapter 19

(d) ΔU = Q − W = 7.15 × 105 J to three figures. (e) Under these conditions W is much less than Q and there is no substantial difference between cV and c p .

ΔU = Q − W is valid for any material. For liquids the expansion work is much less than Q.

EVALUATE: 19.54.

ΔV = βV0 ΔT . W = pΔV since the applied pressure (air pressure) is constant. Q = mc p ΔT .

IDENTIFY:

ΔU = Q − W . SET UP: EXECUTE:

For copper, β = 5.1 × 10−3 (C°)−1 , c p = 390 J/kg ⋅ K and ρ = 8.90 × 103 kg/m3 . (a) ΔV = βΔTV0 = (5.1 × 10−5 (C°) −1 )(70.0 C°)(2.00 × 10−2 m)3 = 2.86 × 10−8 m3 .

(b) W = pΔV = 2.88 × 10−3 J. (c) Q = mc p ΔT = ρV0c p ΔT = (8.9 × 103 kg m3 )(8.00 × 10−6 m3 )(390 J kg ⋅ K)(70.0 C°) = 1944 J. (d) To three figures, ΔU = Q = 1940 J. (e) Under these conditions, the difference is not substantial, since W is much less than Q. EVALUATE: ΔU = Q − W applies to any material. For solids the expansion work is much less than Q. 19.55.

IDENTIFY and SET UP: The heat produced from the reaction is Qreaction = mLreaction , where Lreaction is the heat of reaction of the chemicals. Qreaction = W + ΔU spray EXECUTE:

For a mass m of spray, W = 12 mv 2 = 12 m(19 m/s) 2 = (180.5 J/kg)m and

ΔU spray = Qspray = mcΔT = m(4190 J/kg ⋅ K)(100°C − 20°C) = (335,200 J/kg)m. Then Qreaction = (180 J/kg + 335,200 J/kg)m = (335,380 J/kg)m and Qreaction = mLreaction implies mLreaction = (335,380 J/kg)m.

19.56.

The mass m divides out and Lreaction = 3.4 × 105 J/kg EVALUATE: The amount of energy converted to work is negligible for the two significant figures to which the answer should be expressed. Almost all of the energy produced in the reaction goes into heating the compound. IDENTIFY: The process is adiabatic. Apply p1V1γ = p2V2γ and pV = nRT . Q = 0 so ΔU = −W = − SET UP:

1 ( p1V1 − p2V2 ). γ −1

For helium, γ = 1.67. p1 = 1.00 atm = 1.013 × 105 Pa. V1 = 2.00 × 103 m3 .

p2 = 0.900 atm = 9.117 × 104 Pa. T1 = 288.15 K. 1/ γ

⎛p ⎞ ⎛p ⎞ EXECUTE: (a) V2 = V1 ⎜ 1 ⎟ . V2 = V1 ⎜ 1 ⎟ ⎝ p2 ⎠ ⎝ p2 ⎠ T1 T2 (b) pV = nRT gives = . p1V1 p2V2 γ

19.57.

γ

1/1.67

⎛ 1.00 atm ⎞ = (2.00 × 103 m3 ) ⎜ ⎟ ⎝ 0.900 atm ⎠

= 2.13 × 103 m3 .

3 3 ⎛ p ⎞⎛ V ⎞ ⎛ 0.900 atm ⎞ ⎛ 2.13 × 10 m ⎞ = 276.2 K = 3.0°C. T2 = T1 ⎜ 2 ⎟⎜ 2 ⎟ = (288.15 K) ⎜ ⎟⎜ 3 3 ⎟ ⎝ 1.00 atm ⎠ ⎝ 2.00 × 10 m ⎠ ⎝ p1 ⎠⎝ V1 ⎠ 1 (c) ΔU = − ([1.013 × 105 Pa)(2.00 × 103 m3 )] − [9.117 × 104 Pa)(2.13 × 103 m3 )] = −1.25 × 107 J. 0.67 EVALUATE: The internal energy decreases when the temperature decreases. γ −1 IDENTIFY: For an adiabatic process of an ideal gas, TV = T2V2γ −1. pV = nRT . 1 1

SET UP: For air, γ = 1.40 = 75 . EXECUTE: (a) As the air moves to lower altitude its density increases; under an adiabatic compression, the temperature rises. If the wind is fast-moving, Q is not as likely to be significant, and modeling the process as adiabatic (no heat loss to the surroundings) is more accurate. nRT γ −1 (b) V = , so TV = T2V2γ −1 gives T1γ p11−γ = T2γ p12−γ . The temperature at the higher pressure is 1 1 p

T2 = T1 ( p1 / p2 )

( γ −1) / γ

= (258.15 K) ([8.12 × 104 Pa]/[5.60 × 104 Pa])

2/7

= 287.1 K = 13.0°C so the temperature would

rise by 11.9 C°. EVALUATE: In an adiabatic compression, Q = 0 but the temperature rises because of the work done on the gas.


19.58.

IDENTIFY:

For constant pressure , W = pΔV . For an adiabatic process of an ideal gas, W =

19-15

CV ( p1V1 − p2V2 ) and R

p1V1γ = p2V2γ . Cp

C p + CV

R CV CV CV EXECUTE: (a) The pV-diagram is sketched in Figure 19.58. C (b) The work done is W = p0 (2V0 − V0 ) + V ( p0 (2V0 ) − p3 (4V0 )). p3 = p0 (2V0 4V0 )γ and so R ⎡ C ⎤ W = p0V0 ⎢1 + V (2 − 22 −γ ) ⎥ . Note that p0 is the absolute pressure. R ⎣ ⎦ (c) The most direct way to find the temperature is to find the ratio of the final pressure and volume to the original SET UP:

γ=

=

=1+

γ

γ

⎛V ⎞ ⎛V ⎞ and treat the air as an ideal gas. p3 = p2 ⎜ 2 ⎟ = p1 ⎜ 2 ⎟ , since p1 = p2 . Then ⎝ V3 ⎠ ⎝ V3 ⎠ γ

γ

⎛V ⎞ ⎛V ⎞ p3V3 2 −γ ⎛1⎞ = T0 ⎜ 2 ⎟ ⎜ 3 ⎟ = T0 ⎜ ⎟ 4 = T0 ( 2 ) . 2 p1V1 V V ⎝ ⎠ ⎝ 3⎠ ⎝ 1⎠ p0V0 pV ⎛C ⎞ (d) Since n = , Q = 0 0 ( CV + R )( 2T0 − T0 ) = p0V0 ⎜ V + 1⎟ . This amount of heat flows into the gas, since RT0 RT0 ⎝ R ⎠ Q > 0. EVALUATE: In the isobaric expansion the temperature doubles and in the adiabatic expansion the temperature decreases. If the gas is diatomic, with γ = 75 , 2 − γ = 53 and T3 = 3.03T0 , W = 2.21 p0V0 and Q = 3.50 p0V0 . T3 = T0

ΔU = 1.29 p0V0 . ΔU > 0 and this is consistent with an increase in temperature.

Figure 19.58 19.59.

IDENTIFY: Assume that the gas is ideal and that the process is adiabatic. Apply Eqs.(19.22) and (19.24) to relate pressure and volume and temperature and volume. The distance the piston moves is related to the volume of the gas. Use Eq.(19.25) to calculate W. (a) SET UP: γ = C p / CV = (CV + R) / CV = 1 + R / CV = 1.40. The two positions of the piston are shown in

Figure 19.59. p1 = 1.01 × 105 Pa p2 = 4.20 × 105 Pa + pair = 5.21 × 106 Pa V1 = h1 A V2 = h2 A Figure 19.59 EXECUTE:

adiabatic process: p1V1γ = p2V2γ

p1h1γ Aγ = p2 h2γ Aγ 1/ γ

1/1.40

⎛p ⎞ ⎛ 1.01 × 105 Pa ⎞ = 0.0774 m h2 = h1 ⎜ 1 ⎟ = (0.250 m) ⎜ ⎟ 5 ⎝ 5.21 × 10 Pa ⎠ ⎝ p2 ⎠ The piston has moved a distance h1 − h2 = 0.250 m − 0.0774 m = 0.173 m.

19-16

Chapter 19 γ −1 (b) TV = T2V2γ −1 1 1

T1h1γ −1 Aγ −1 = T2 h2γ −1 Aγ −1 γ −1

⎛h ⎞ ⎛ 0.250 m ⎞ T2 = T1 ⎜ 1 ⎟ = 300.1 K ⎜ ⎟ ⎝ 0.0774 m ⎠ ⎝ h2 ⎠ (c) W = nCV (T1 − T2 ) (Eq.19.25)

19.60.

0.40

= 479.7 K = 207°C

W = (20.0 mol)(20.8 J/mol ⋅ K)(300.1 K − 479.7 K) = −7.47 × 104 J EVALUATE: In an adiabatic compression of an ideal gas the temperature increases. In any compression the work W is negative. pM γ −1 IDENTIFY: m = ρV . The density of air is given by ρ = . For an adiabatic process, TV = T2V2γ −1. 1 1 RT pV = nRT nRT γ −1 in TV = T2V2γ −1 gives T1 p11−γ = T2 p12−γ . 1 1 p EXECUTE: (a) The pV-diagram is sketched in Figure 19.60. (b) The final temperature is the same as the initial temperature, and the density is proportional to the absolute pressure. The mass needed to fill the cylinder is then SET UP:

Using V =

m = p0V

p 1.45 × 105 Pa = (1.23 kg/m3 )(575 × 10−6 m3 ) = 1.02 × 10−3 kg. 1.01 × 105 Pa pair

Without the turbocharger or intercooler the mass of air at T = 15.0°C and p = 1.01 × 105 Pa in a cylinder is m = ρ 0V = 7.07 × 10−4 kg. The increase in power is proportional to the increase in mass of air in the cylinder; the

percentage increase is

1.02 × 10−3 kg − 1 = 0.44 = 44%. 7.07 × 10−4 kg

⎛p ⎞ (c) The temperature after the adiabatic process is T2 = T1 ⎜ 2 ⎟ ⎝ p1 ⎠ ⎛ T ⎞⎛ p ⎞ ⎛p ⎞ ρ = ρ 0 ⎜ 1 ⎟⎜ 2 ⎟ = ρ 0 ⎜ 2 ⎟ ⎝ T2 ⎠⎝ p1 ⎠ ⎝ p1 ⎠

(1− γ ) / γ

( γ −1) / γ

. The density becomes

1/ γ

⎛ p2 ⎞ ⎛ p2 ⎞ ⎜ ⎟ = ρ0 ⎜ ⎟ ⎝ p1 ⎠ ⎝ p1 ⎠

. The mass of air in the cylinder is 1 1.40

⎛ 1.45 × 105 Pa ⎞ m = (1.23 kg/m3 )(575 × 10−6 m3 ) ⎜ ⎟ 5 ⎝ 1.01× 10 Pa ⎠

= 9.16 × 10−4 kg,

9.16 × 10−4 kg − 1 = 0.30 = 30%. 7.07 × 10−4 kg EVALUATE: The turbocharger and intercooler each have an appreciable effect on the engine power.

The percentage increase in power is

Figure 19.60 19.61.

IDENTIFY: In each case calculate either ΔU or Q for the specific type of process and then apply the first law. (a) SET UP: isothermal (ΔT = 0) ΔU = Q − W ; W = +300 J

For any process of an ideal gas, ΔU = nCV ΔT . EXECUTE: Therefore, for an ideal gas, if ΔT = 0 then ΔU = 0 and Q = W = +300 J. (b) SET UP: adiabatic (Q = 0) ΔU = Q − W ; W = +300 J EXECUTE: Q = 0 says ΔU = −W = −300 J


19-17

(c) SET UP: isobaric Δp = 0 Use W to calculate ΔT and then calculate Q. EXECUTE: W = pΔT = nRΔT ; ΔT = W / nR

Q = nC p ΔT and for a monatomic ideal gas C p = 52 R

Thus Q = n 52 RΔT = (5Rn/2)(W/nR ) = 5W/2 = +750 J. ΔU = nCV ΔT for any ideal gas process and CV = C p − R = 32 R.

19.62.

Thus ΔU = 3W / 2 = +450 J EVALUATE: 300 J of energy leaves the gas when it performs expansion work. In the isothermal process this energy is replaced by heat flow into the gas and the internal energy remains the same. In the adiabatic process the energy used in doing the work decreases the internal energy. In the isobaric process 750 J of heat energy enters the gas, 300 J leaves as the work done and 450 J remains in the gas as increased internal energy. IDENTIFY: pV = nRT . For the isobaric process, W = pΔV = nRΔT . For the isothermal process, ⎛V ⎞ W = nRT ln ⎜ f ⎟ . ⎝ Vi ⎠ SET UP: R = 8.315 J/mol ⋅ K EXECUTE: (a) The pV diagram for these processes is sketched in Figure 19.62. T T T p (b) Find T2 . For process 1 → 2, n, R, and p are constant so = = constant. 1 = 2 and V1 V2 V nR ⎛V ⎞ T2 = T1 ⎜ 2 ⎟ = (355 K)(2) = 710 K. ⎝ V1 ⎠ (c) The maximum pressure is for state 3. For process 2 → 3, n, R, and T are constant. p2V2 = p3V3 and ⎛V ⎞ p3 = p2 ⎜ 2 ⎟ = (2.40 × 105 Pa)(2) = 4.80 × 105 Pa. ⎝ V3 ⎠ (d) process 1 → 2: W = pΔV = nRΔT = (0.250 mol)(8.315 J/mol ⋅ K)(710 K − 355 K) = 738 K. ⎛V ⎞ ⎛1⎞ process 2 → 3: W = nRT ln ⎜ 3 ⎟ = (0.250 mol)(8.315 J/mol ⋅ K)(710 K)ln ⎜ ⎟ = −1023 J. ⎝2⎠ ⎝ V2 ⎠ process 3 → 1: ΔV = 0 and W = 0. The total work done is 738 J + (−1023 J) = −285 J. This is the work done by the gas. The work done on the gas is 285 J. EVALUATE: The final pressure and volume are the same as the initial pressure and volume, so the final state is the same as the initial state. For the cycle, ΔU = 0 and Q = W = −285 J. During the cycle, 285 J of heat energy must leave the gas.

Figure 19.62 19.63.

IDENTIFY and SET UP: Use the ideal gas law, the first law and expressions for Q and W for specific types of processes. EXECUTE: (a) initial expansion (state 1 → state 2)

p1 = 2.40 × 105 Pa, T1 = 355 K, p2 = 2.40 × 105 Pa, V2 = 2V1 pV = nRT ; T / V = p / nR = constant, so T1 / V1 = T2 / V2 and T2 = T1 (V2 / V1 ) = 355 K(2V1 / V1 ) = 710 K Δp = 0 so W = pΔV = nRΔT = (0.250 mol)(8.3145 J/mol ⋅ K)(710 K − 355 K) = +738 J Q = nC p ΔT = (0.250 mol)(29.17 J/mol ⋅ K)(710 K − 355 K) = +2590 J

ΔU = Q − W = 2590 J − 738 J = 1850 J

19-18

Chapter 19

(b) At the beginning of the final cooling process (cooling at constant volume), T = 710 K. The gas returns to its original volume and pressure, so also to its original temperature of 355 K. ΔV = 0 so W = 0 Q = nCV ΔT = (0.250 mol)(20.85 J/mol ⋅ K)(355 K − 710 K) = −1850 J ΔU = Q − W = −1850 J. (c) For any ideal gas process ΔU = nCV ΔT . For an isothermal process ΔT = 0, so ΔU = 0. EVALUATE: 19.64.

IDENTIFY:

The three processes return the gas to its initial state, so ΔU total = 0; our results agree with this. γ −1 pV = nRT . For an adiabatic process of an ideal gas, TV = T2V2γ −1. 1 1

SET UP: For N 2 , γ = 1.40. EXECUTE: (a) The pV-diagram is sketched in Figure 19.64. (b) At constant pressure, halving the volume halves the Kelvin temperature, and the temperature at the beginning of the adiabatic expansion is 150 K. The volume doubles during the adiabatic expansion, and from Eq. (19.22), the

temperature at the end of the expansion is (150 K)(1 2)0.40 = 114 K. (c) The minimum pressure occurs at the end of the adiabatic expansion (state 3). During the final heating the volume is held constant, so the minimum pressure is proportional to the Kelvin temperature, pmin = (1.80 × 105 Pa)(114K 300 K) = 6.82 × 104 Pa. EVALUATE: In the adiabatic expansion the temperature decreases.

Figure 19.64 19.65.

IDENTIFY: Use the appropriate expressions for Q, W and ΔU for each type of process. ΔU = Q − W can also be used. SET UP: For N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. EXECUTE:

(a) W = pΔV = nRΔT = (0.150 mol)(8.3145 J mol ⋅ K)( − 150 K) = −187 J,

Q = nC p ΔT = (0.150 mol)(29.07 mol ⋅ K)( − 150 K) = −654 J, ΔU = Q − W = −467 J. (b) From Eq. (19.24), using the expression for the temperature found in Problem 19.64, 1 W= (0.150 mol)(8.3145 J/mol ⋅ K)(150 K)(1 − (1/20.40 ) = 113 J. Q = 0 for an adiabatic process, and 0.40 ΔU = Q − W = −W = −113 J. (c) ΔV = 0, so W = 0. Using the temperature change as found in Problem 19.64 and part (b), Q = nCV ΔT = (0.150 mol)(20.76 J mol ⋅ K)(300 K − 113.7 K) = 580 J and ΔU = Q − W = Q = 580 J. 19.66.

EVALUATE: For each process we could also use ΔU = nCV ΔT to calculate ΔU . IDENTIFY: Use the appropriate expression for W for each type of process. SET UP: For a monatomic ideal gas, γ = 5/ 3 and CV = 3R / 2. EXECUTE:

(a) W = nRT ln (V2 / V1 ) = nRT ln (3) = 3.29 × 103 J.

γ −1 (b) Q = 0 so W = −ΔU = −nCV ΔT . TV = T2V2γ −1 gives T2 = T1 (1/ 3) 2 / 3 . Then 1 1

W = nCV T1 (1 − (1 3) 2 / 3 ) = 2.33 × 103 J. (c) V2 = 3V1 , so W = pΔV = 2pV1 = 2nRT1 = 6.00 × 103 J. (d) Each process is shown in Figure 19.66. The most work done is in the isobaric process, as the pressure is maintained at its original value. The least work is done in the adiabatic process. (e) The isobaric process involves the most work and the largest temperature increase, and so requires the most heat. Adiabatic processes involve no heat transfer, and so the magnitude is zero. ( f ) The isobaric process doubles the Kelvin temperature, and so has the largest change in internal energy. The isothermal process necessarily involves no change in internal energy.


19-19

EVALUATE: The work done is the area under the path for the process in the pV-diagram. Figure 19.66 shows that the work done is greatest in the isobaric process and least in the adiabatic process.

Figure 19.66 19.67.

IDENTIFY: SET UP:

γ −1 Assume the compression is adiabatic. Apply TV = T2V2γ −1 and pV = nRT . 1 1

For N 2 , γ = 1.40. V1 = 3.00 L, p = 1.00 atm = 1.013 × 105 Pa, T = 273.15 K. V2 = V1 / 2 = 1.50 L. γ −1

EXECUTE:

⎛V ⎞ (a) T2 = T1 ⎜ 1 ⎟ ⎝ V2 ⎠

⎛ V ⎞ = (273.15 K) ⎜ 1 ⎟ ⎝ V1 / 2 ⎠

0.40

= (273.15 K)(2)0.4 = 360.4 K = 87.3°C.

p1V1 p2V2 = . T1 T2

⎛ V ⎞⎛ T ⎞ ⎛ V ⎞ ⎛ 360.4 K ⎞ p2 = p1 ⎜ 1 ⎟⎜ 2 ⎟ = (1.00 atm) ⎜ 1 ⎟ ⎜ ⎟ = 2.64 atm. V T ⎝ 2 ⎠⎝ 1 ⎠ ⎝ V1 / 2 ⎠ ⎝ 273.15 K ⎠ (b) p is constant, so 19.68.

⎛T ⎞ V V V nR ⎛ 273.15 K ⎞ = = constant and 2 = 3 . V3 = V2 ⎜ 3 ⎟ = (1.50 L) ⎜ ⎟ = 1.14 L. T2 T3 T T ⎝ 360.4 K ⎠ ⎝ T2 ⎠

EVALUATE: In an adiabatic compression the temperature increases. IDENTIFY: At equilibrium the net upward force of the gas on the piston equals the weight of the piston. When the piston moves upward the gas expands, the pressure of the gas drops and there is a net downward force on the piston. For simple harmonic motion the net force has the form Fy = − ky , for a displacement y from equilibrium,

1 k . 2π m SET UP: pV = nRT . T is constant. (a) The difference between the pressure, inside and outside the cylinder, multiplied by the area of the piston, must mg mg be the weight of the piston. The pressure in the trapped gas is p0 + = p0 + 2 . A πr

and f =

mg ⎞ ⎛ h ⎞ ⎛ (b) When the piston is a distance h + y above the cylinder, the pressure in the trapped gas is ⎜ p0 + 2 ⎟ ⎜ ⎟ π r ⎠⎝ h + y ⎠ ⎝ −1

and for values of y small compared to h,

h y⎞ y ⎛ = ⎜1 + ⎟ ~ 1 − . The net force, taking the positive direction to h h+ y ⎝ h⎠

⎡⎛ mg ⎞⎛ y⎞ ⎤ ⎛ y⎞ be upward, is the then Fy = ⎢⎜ p0 + 2 ⎟⎜1 − ⎟ − p0 ⎥ (π r 2 ) − mg = − ⎜ ⎟ ( p0π r 2 + mg ) . π r ⎠⎝ h ⎠ ⎝h⎠ ⎣⎝ ⎦ This form shows that for positive h, the net force is down; the trapped gas is at a lower pressure than the equilibrium pressure, and so the net force tends to restore the piston to equilibrium. ( p0π r 2 + mg ) h = g ⎛1 + p0π r 2 ⎞. (c) The angular frequency of small oscillations would be given by ω 2 = ⎜ ⎟ m h⎝ mg ⎠

ω 1 = f = 2π 2π

1/ 2

g⎛ p0π r 2 ⎞ ⎜1 + ⎟ . h⎝ mg ⎠

19-20

Chapter 19

If the displacements are not small, the motion is not simple harmonic. This can be seen be considering what happens if y ~ −h; the gas is compressed to a very small volume, and the force due to the pressure of the gas would become unboundedly large for a finite displacement, which is not characteristic of simple harmonic motion. If y >> h (but not so large that the piston leaves the cylinder), the force due to the pressure of the gas becomes small, and the restoring force due to the atmosphere and the weight would tend toward a constant, and this is not characteristic of simple harmonic motion. h EVALUATE: The assumption of small oscillations was made when was replaced by 1 − y / h; this is h+ y accurate only when y / h is small. 19.69.

IDENTIFY: SET UP:

W =∫

V2 V1

pdV .

For an isothermal process of an ideal gas, W = nRT ln (V2 V1 ) .

EXECUTE: (a) Solving for p as a function of V and T and integrating with respect to V, V2 ⎡V − nb ⎤ nRT an 2 1⎤ 2⎡ 1 p= − 2 and W = ∫ pdV = nRT ln ⎢ 2 ⎥ + an ⎢ − ⎥ . V1 − V − nb V V nb V V 1⎦ ⎣ 1 ⎦ ⎣ 2

When a = b = 0, W = nRT ln (V2 V1 ) , as expected. (b) (i) Using the expression found in part (a), W = (1.80 mol )( 8.3145 J/mol ⋅ K )( 300 K )

⎡ ( 4.00 × 10−3 m3 ) − (1.80 mol ) ( 6.38 × 10−5 m 2 / mol ) ⎤ ⎥ × ln ⎢ −3 3 −5 2 ⎣⎢ ( 2.00 × 10 m ) − (1.80 mol ) ( 6.38 × 10 m / mol ) ⎦⎥ 1 1 2⎡ ⎤ + ( 0.554 J ⋅ m3 mol2 ) (1.80 mol ) ⎢ − −3 3 −3 3⎥ ⎣ 4.00 × 10 m 2.00 × 10 m ⎦ W = 2.80 × 103 J. (ii) W = nRT ln(2) = 3.11× 103 J. (c) The work for the ideal gas is larger by about 300 J. For this case, the difference due to nonzero a is more than that due to nonzero b. The presence of a nonzero a indicates that the molecules are attracted to each other and so do not do as much work in the expansion. EVALUATE: The difference in the two results for W is about 10%, which can be considered to be important.

THE SECOND LAW OF THERMODYNAMICS

20.1.

For a heat engine, W = QH − QC . e =

IDENTIFY: SET UP:

20

W . QH > 0, QC < 0. QH

W = 2200 J. QC = 4300 J. (a) QH = W + QC = 6500 J.

EXECUTE:

2200 J = 0.34 = 34%. 6500 J EVALUATE: Since the engine operates on a cycle, the net Q equal the net W. But to calculate the efficiency we use the heat energy input, QH . (b) e =

20.2.

For a heat engine, W = QH − QC . e =

IDENTIFY: SET UP:

W . QH > 0, QC < 0. QH

QH = 9000 J. QC = 6400 J.

EXECUTE: (a) W = 9000 J − 6400 J = 2600 J. W 2600 J (b) e = = = 0.29 = 29%. QH 9000 J EVALUATE: Since the engine operates on a cycle, the net Q equal the net W. But to calculate the efficiency we use the heat energy input, QH . 20.3.


The problem deals with a heat engine. W = +3700 W and QH = +16,100 J. Use

Eq.(20.4) to calculate the efficiency e and Eq.(20.2) to calculate QC . Power = W / t. EXECUTE:

(a) e =

work output W 3700 J = = = 0.23 = 23%. heat energy input QH 16,100 J

(b) W = Q = QH − QC

Heat discarded is QC = QH − W = 16,100 J − 3700 J = 12,400 J. (c) QH is supplied by burning fuel; QH = mLc where Lc is the heat of combustion. QH 16,100 J = = 0.350 g. Lc 4.60 × 104 J/g (d) W = 3700 J per cycle In t = 1.00 s the engine goes through 60.0 cycles. P = W / t = 60.0(3700 J)/1.00 s = 222 kW m=

P = (2.22 × 105 W)(1 hp/746 W) = 298 hp EVALUATE: 20.4.

IDENTIFY:

QC = −12,400 J. In one cycle Qtot = QC + QH = 3700 J. This equals Wtot for one cycle. W = QH − QC . e =

W . QH > 0, QC < 0. QH

For 1.00 s, W = 180 × 103 J. W 180 × 103 J EXECUTE: (a) QH = = = 6.43 × 105 J. 0.280 e (b) QC = QH − W = 6.43 × 105 J − 1.80 × 105 J = 4.63 × 105 J. SET UP:

EVALUATE: Of the 6.43 × 105 J of heat energy supplied to the engine each second, 1.80 × 105 J is converted to mechanical work and the remaining 4.63 × 105 J is discarded into the low temperature reservoir. 20-1

20-2

20.5.

Chapter 20

W = QH − QC . e =

IDENTIFY:

W . QH > 0, QC < 0. Dividing by t gives equivalent equations for the rate of QH

heat flows and power output. SET UP: W / t = 330 MW. QH / t = 1300 MW. (a) e =

EXECUTE:

W W /t 330 MW = = = 0.25 = 25%. QH QH / t 1300 MW

(b) QC = QH − W so QC / t = QH / t − W / t = 1300 MW − 330 MW = 970 MW.

The equations for e and W have the same form when written in terms of power output and rate of

EVALUATE: heat flow. 20.6.

Apply e = 1 −

IDENTIFY: SET UP:

Q 1 . e =1− C . r γ −1 QH

In part (b), QH = 10,000 J. The heat discarded is QC .

1 = 0.594 = 59.4%. 9.500.40 (b) QC = QH (1 − e) = (10,000 J)(1 − 0.594) = 4060 J. (a) e = 1 −

EXECUTE:

The work output of the engine is W = QH − QC = 10,000 J − 4060 J = 5940 J

EVALUATE: 20.7.

20.8.

20.9.

1 . r γ −1 SET UP: γ = 1.40 and e = 0.650. 1 1 EXECUTE: = 1 − e = 0.350. r 0.40 = and r = 13.8. γ −1 r 0.350 EVALUATE: e increases when r increases. IDENTIFY: e = 1 − r1− γ SET UP: r is the compression ratio. EXECUTE: (a) e = 1 − (8.8) −0.40 = 0.581, which rounds to 58%. e =1−

IDENTIFY:

(b) e = 1 − (9.6) −0.40 = 0.595 an increase of 1.4%. EVALUATE: An increase in r gives an increase in e. IDENTIFY and SET UP: For the refrigerator K = 2.10 and QC = +3.4 × 104 J. Use Eq.(20.9) to calculate W and

then Eq.(20.2) to calculate QH . (a) EXECUTE:

Performance coefficient K = QC / W (Eq.20.9)

W = QC / K = 3.40 × 104 J/2.10 = 1.62 × 104 J (b) SET UP:

The operation of the device is illustrated in Figure 20.9 EXECUTE: W = QC + QH QH = W − QC QH = −1.62 × 104 J − 3.40 × 104 J = −5.02 × 104 J (negative because heat goes out of the system) Figure 20.9

EVALUATE 20.10.

QH = W + QC . The heat QH delivered to the high temperature reservoir is greater than the heat

taken in from the low temperature reservoir. Q IDENTIFY: K = C and QH = QC + W . W SET UP:

The heat removed from the room is QC and the heat delivered to the hot outside is QH .

W = (850 J/s)(60.0 s) = 5.10 × 10 4 J. EXECUTE:

(a) QC = K W = (2.9)(5.10 × 104 J) = 1.48 × 105 J

(b) QH = QC + W = 1.48 × 105 J + 5.10 × 104 J = 1.99 × 105 J. EVALUATE:

(c) QH = QC + W , so QH > QC .

The Second Law of Thermodynamics

20.11.


20-3

Apply Eq.(20.2) to the cycle and calculate W and then P = W / t. Section 20.4 shows

that EER = (3.413) K . (a) The operation of the device is illustrated in Figure 20.11. EXECUTE: QC = +9.80 × 104 J

QH = −1.44 × 105 J Figure 20.11

W = QC + QH = +9.80 × 104 J − 1.44 × 105 J = −4.60 × 104 J P = W / t = −4.60 × 104 J/60.0 s = −767 W (b) EER = (3.413) K K = QC / W = 9.80 × 104 J/4.60 × 104 J = 2.13

EER = (3.413)(2.13) = 7.27 EVALUATE: W negative means power is consumed, not produced, by the device. QH = W + QC . 20.12.


QC . W For water, cw = 4190 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. For ice, cice = 2010 J/kg ⋅ K. QH = QC + W . K =

(a) Q = mcice ΔTice − mLf + mcw ΔTw .

Q = (1.80 kg)([2010 J/kg ⋅ K][−5.0 C°] − 3.34 × 105 J/kg + [4190 J/kg ⋅ K][−25.0 C°]) = −8.08 × 105 J Q = −8.08 × 105 J. Q is negative for the water since heat is removed from it. QC 8.08 × 105 J = = 3.37 × 105 J. K 2.40 (c) QH = 8.08 × 105 J + 3.37 × 105 J = 1.14 × 106 J. (b) QC = 8.08 × 105 J. W =

20.13.

EVALUATE: For this device, QC > 0 and QH < 0. More heat is rejected to the room than is removed from the water. IDENTIFY: Use Eq.(20.2) to calculate W . Since it is a Carnot device we can use Eq.(20.13) to relate the heat

flows out of the reservoirs. The reservoir temperatures can be used in Eq.(20.14) to calculate e. (a) SET UP: The operation of the device is sketched in Figure 20.13.

EXECUTE: W = QC + QH W = −335 J + 550 J = 215 J

Figure 20.13 (b) For a Carnot cycle,

TC = TH

QC TC = (Eq.20.13) QH TH

QC ⎛ 335 J ⎞ = 620 K ⎜ ⎟ = 378 K QH ⎝ 550 J ⎠

(c) e(Carnot) = 1 − TC / TH = 1 − 378 K/620 K = 0.390 = 39.0% EVALUATE: We could use the underlying definition of e (Eq.20.4): e = W / QH = (215 J)/(550 J) = 39%, which checks.

20-4

Chapter 20

20.14.

IDENTIFY:

W = QH − QC . QC < 0, QH > 0. e =

W Q T . For a Carnot cycle, C = − C . QH QH TH

TC = 300 K, TH = 520 K. QH = 6.45 × 103 J.

SET UP:

⎛T ⎞ ⎛ 300 K ⎞ 3 (a) QC = −QH ⎜ C ⎟ = −(6.45 × 103 J) ⎜ ⎟ = −3.72 × 10 J. ⎝ 520 K ⎠ ⎝ TH ⎠ (b) W = QH − QC = 6.45 × 103 J − 3.72 × 103 J = 2.73 × 103 J EXECUTE:

(c) e =

W 2.73 × 103 J = = 0.423 = 42.3%. QH 6.45 × 103 J

EVALUATE: 20.15.

We can verify that e = 1 − TC / TH also gives e = 42.3%. e=

IDENTIFY:

W Q T for any engine. For the Carnot cycle, C = − C . QH QH TH

TC = 20.0°C + 273.15 K = 293.15 K

SET UP:

W 2.5 × 104 J = = 4.24 × 104 J e 0.59 (b) W = QH + QC so QC = W − QH = 2.5 × 104 J − 4.24 × 104 J = −1.74 × 104 J. (a) QH =

EXECUTE:

TH = −TC

20.16.

⎛ 4.24 × 104 J ⎞ QH = −(293.15 K) ⎜ ⎟ = 714 K = 441°C. 4 QC ⎝ −1.74 × 10 J ⎠

EVALUATE: For a heat engine, W > 0, QH > 0 and QC < 0. IDENTIFY and SET UP: The device is a Carnot refrigerator. We can use Eqs.(20.2) and (20.13). (a) The operation of the device is sketched in Figure 20.16.

TH = 24.0°C = 297 K TC = 0.0°C = 273 K Figure 20.16

The amount of heat taken out of the water to make the liquid → solid phase change is

Q = −mLf = −(85.0 kg)(334 × 103 J/kg) = −2.84 × 107 J. This amount of heat must go into the working substance of the refrigerator, so QC = +2.84 × 107 J. For Carnot cycle QC / QH = TC / TH

QH = QC (TH / TC ) = 2.84 × 107 J(297 K/273 K) = 3.09 × 107 J

EXECUTE:

20.17.

(b) W = QC + QH = +2.84 × 107 J − 3.09 × 107 J = −2.5 × 106 J EVALUATE: W is negative because this much energy must be supplied to the refrigerator rather than obtained from it. Note that in Eq.(20.13) we must use Kelvin temperatures. Q Q T IDENTIFY: QH = W + QC . QH < 0, QC > 0. K = C . For a Carnot cycle, C = − C . W QH TH

TC = 270 K, TH = 320 K. QC = 415 J.

SET UP: EXECUTE:

⎛T ⎞ ⎛ 320 K ⎞ (a) QH = − ⎜ H ⎟ QC = − ⎜ ⎟ (415 J) = −492 J. T ⎝ 270 K ⎠ ⎝ C⎠

(b) For one cycle, W = QH − QC = 492 J − 415 J = 77 J. P = (c) K =

(165)(77 J) = 212 W. 60 s

QC 415 J = = 5.4. W 77 J

EVALUATE:

The amount of heat energy QH delivered to the high-temperature reservoir is greater than the

amount of heat energy QC removed from the low-temperature reservoir.


20.18.

W = QH − QC . For a Carnot cycle,

IDENTIFY:

20-5

QC T = − C , where the temperatures must be in kelvins. QH TH

SET UP: −10.0°C = 263.15 K, 25.0°C = 298.15 K, 0.0°C = 273.15 K and −25.0°C = 248.15 K. EXECUTE: (a) The heat is discarded at a higher temperature, and a refrigerator is required. QH = QC (TH / TC )

and | W | = | QC | ((TH TC ) − 1) = (5.00 × 103 J)(( 298.15 K 263.15 K ) − 1) = 665 J.

20.19.

(b) Again, the device is a refrigerator, and | W | = (5.00 × 103 J)((273.15 K / 263.15 K) − 1) = 190 J. (c) The device is an engine; the heat is taken from the hot reservoir, and the work done by the engine is | W | = (5.00 × 103 J)(1 − ( 248.15 K 263.15 K)) = 285 J. EVALUATE: For a refrigerator work must be supplied to the device. For a heat engine, there is mechanical work output from the device. Q TC IDENTIFY: The theoretical maximum performance coefficient is K Carnot = . K = C . QC is the heat W TH − TC

removed from the water to convert it to ice. For the water, Q = mcw ΔT + mLf . SET UP: TC = −5.0°C = 268 K. TH = 20.0°C = 293 K. cw = 4190 J/kg ⋅ K and Lf = 334 × 103 J/kg. EXECUTE: (a) In one year the freezer operates (5 h/day)(365 days) = 1825 h.

730 kWh = 0.400 kW = 400 W. 1825 h 268 K =10.7 (b) K Carnot = 293 K − 268 K (c) W = Pt = (400 W)(3600 s) = 1.44 × 106 J. QC = K W = 1.54 × 107 J. Q = mcw ΔT + mLf gives

P=

m=

QC 1.54 × 107 J = = 36.9 kg. cw ΔT + Lf (4190 J/kg ⋅ K)(20.0 K) + 334 × 103 J/kg For any actual device, K < K Carnot , QC is less than we calculated and the freezer makes less ice in

EVALUATE: 20.20.

one hour than the mass we calculated in part (c). IDENTIFY: The total work that must be done is Wtot = mg Δy. W = QH − QC . QH > 0, W > 0 and QC < 0. For a Carnot cycle, SET UP:

QC T =− C, QH TH

TC = 373 K, TH = 773 K. QH = 250 J.

⎛T ⎞ ⎛ 373 K ⎞ QC = −QH ⎜ C ⎟ = −(250 J) ⎜ ⎟ = −121 J. W = 250 J − 121 J = 129 J. This is the work done in T ⎝ 773 K ⎠ ⎝ H⎠ one cycle. Wtot = (500 kg)(9.80 m/s 2 )(100 m) = 4.90 × 105 J. The number of cycles required is

EXECUTE:

Wtot 4.90 × 105 J = = 3.80 × 103 cycles. 129 J/cycle W

20.21.

EVALUATE:

In

IDENTIFY:

e=

SET UP:

QC T = − C , the temperatures must be in kelvins. QH TH W Q Q T T = 1 − C . For a Carnot cycle, C = − C and e = 1 − C . QH QH QH TH TH

TH = 800 K. QC = −3000 J.

For a heat engine, QH = −QC / (1 − e ) = −(−3000 J) (1 − 0.600 ) = 7500 J, and then

EXECUTE:

W = eQH = (0.600)(7500 J) = 4500 J. EVALUATE:

This does not make use of the given value of TH . If TH is used,

then TC = TH (1 − e ) = ( 800 K )(1 − 0.600 ) = 320 K and QH = −QCTH / TC , which gives the same result. 20.22.

IDENTIFY: SET UP:

W = QC + QH . For a Carnot cycle, For water, Lf = 334 × 103 J/kg

QC T = − C . For the ice to liquid water phase transition, Q = mLf . QH TH

20-6

Chapter 20

QC = −mLf = − ( 0.0400 kg ) ( 334 × 103 J/kg ) = −1.336 × 104 J.

EXECUTE:

QC T = − C gives QH TH

QH = − (TH TC ) QC = − ( −1.336 × 104 J ) ⎡⎣( 373.15 K ) ( 273.15 K ) ⎤⎦ = +1.825 × 104 J. W = QC + QH = 4.89 × 103 J. EVALUATE: For a heat engine, QC is negative and QH is positive. The heat that comes out of the engine (Q < 0) goes into the ice ( Q > 0 ). 20.23.

The power output is P =

IDENTIFY: SET UP:

T W W . The theoretical maximum efficiency is eCarnot = 1 − C . e = . t TH QH

QH = 1.50 × 104 J. TC = 350 K. TH = 650 K. 1 hp = 746 W. eCarnot = 1 −

EXECUTE:

TC 350 K =1− = 0.4615. W = eQH = (0.4615)(1.50 × 104 J) = 6.923 × 103 J; this is the TH 650 K

W (240)(6.923 × 103 J) = = 2.77 × 104 W = 37.1 hp. t 60.0 s ⎛T ⎞ Q T ⎛ 350 K ⎞ 4 3 EVALUATE: We could also use C = − C to calculate QC = − ⎜ C ⎟ QH = − ⎜ ⎟ (1.50 × 10 J) = −8.08 × 10 J. QH TH T 650 K ⎝ ⎠ ⎝ H⎠ work output in one cycle. P =

Then W = QC + QH = 6.92 × 103 J, the same as previously calculated. 20.24.


eCarnot = 1 −

TC TC . K Carnot = . TH TH − TC

TH (1 − e) 1− e = . TH − TH (1 − e) e EVALUATE: (b) When e → 1, K → 0. When e → 0, K → ∞. e → 1 when QC 0 and the process is irreversible. IDENTIFY: Apply Qsystem = 0 to calculate the final temperature. Q = mcΔT . Example 20.6 shows that occurs at T = 273 K, so ΔS =

20.26.

ΔS = mc ln(T2 / T1 ) when an object undergoes a temperature change. SET UP:

For water c = 4190 J/kg ⋅ K. Boiling water has T = 100.0°C = 373 K.

EXECUTE: (a) The heat transfer between 100°C water and 30°C water occurs over a finite temperature difference and the process is irreversible. (b) (270 kg)c(T2 − 30.0°C) + (5.00 kg)c (T2 − 100°C) = 0. T2 = 31.27 °C = 304.42 K.

⎛ 304.42 K ⎞ ⎛ 304.42 K ⎞ (c) ΔS = (270 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ + (5.00 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟. ⎝ 303.15 K ⎠ ⎝ 373.15 K ⎠ ΔS = 4730 J/K + (−4265 J/K) = +470 J/K. EVALUATE: 20.27.

ΔSsystem > 0, as it should for an irreversible process.

Q . For the melting phase transition, T Q = mLf . Conservation of energy requires that the quantity of heat that goes into the ice is the amount of heat that comes out of the room. IDENTIFY:

Both the ice and the room are at a constant temperature, so ΔS =


20.28.

20-7

SET UP: For ice, Lf = 334 × 103 J/kg. When heat flows into an object, Q > 0, and when heat flows out of an object, Q < 0. EXECUTE: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water. mLf −mLf (15.0 kg)(334 × 103 J/kg) −(15.0 kg)(334 × 103 J/kg) + = + . ΔS = +1250 J/K. (b) ΔS = ΔSice + ΔS room = Tice Troom 273 K 293 K EVALUATE: This result is consistent with the answer in (a) because ΔS > 0 for irreversible processes. IDENTIFY: Q = mcΔT for the water. Example 20.6 shows that ΔS = mc ln(T2 /T1 ) when an object undergoes a temperature change. ΔS = Q / T for an isothermal process. SET UP:

For water, c = 4190 J/kg ⋅ K. 85.0°C = 358.2 K. 20.0°C = 293.2 K.

⎛T ⎞ ⎛ 293.2 K ⎞ (a) ΔS = mc ln ⎜ 2 ⎟ = (0.250 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ = −210 J/K. Heat comes out of the ⎝ 358.2 K ⎠ ⎝ T1 ⎠ water and its entropy decreases. (b) Q = mcΔT = (0.250 kg)(4190 J/kg ⋅ K)(−65.0 K) = −6.81 × 104 J. The amount of heat that goes into the air is EXECUTE:

Q +6.81 × 104 J = = +232 J/K. ΔSsystem = −210 J/K + 232 J/K = +22 J/K. T 293.1 K > 0 and the process is irreversible.

+6.81× 104 J. For the air, ΔS = EVALUATE:

ΔSsystem

Q . ΔU = Q − W . T SET UP: For an isothermal process of an ideal gas, ΔU = 0 and Q = W . For a compression, ΔV < 0 and W < 0. The process is at constant temperature, so ΔS =

20.29.

IDENTIFY:

20.30.

−1850 J = −6.31 J/K. 293 K EVALUATE: The entropy change of the gas is negative. Heat must be removed from the gas during the compression to keep its temperature constant and therefore the gas is not an isolated system. IDENTIFY and SET UP: The initial and final states are at the same temperature, at the normal boiling point of 4.216 K. Calculate the entropy change for the irreversible process by considering a reversible isothermal process that connects the same two states, since ΔS is path independent and depends only on the initial and final states. For the reversible isothermal process we can use Eq.(20.18). The heat flow for the helium is Q = − mLv , negative since in condensation heat flows out of the helium. The heat of EXECUTE:

Q = W = −1850 J. ΔS =

vaporization Lv is given in Table 17.4 and is Lv = 20.9 × 103 J/kg.

20.31.

EXECUTE: Q = −mLv = −(0.130 kg)(20.9 × 103 J/kg) = −2717 J ΔS = Q / T = −2717 J/4.216 K = −644 J/K. EVALUATE: The system we considered is the 0.130 kg of helium; ΔS is the entropy change of the helium. This is not an isolated system since heat must flow out of it into some other material. Our result that ΔS < 0 doesn’t violate the 2nd law since it is not an isolated system. The material that receives the heat that flows out of the helium would have a positive entropy change and the total entropy change would be positive. Q IDENTIFY: Each phase transition occurs at constant temperature and ΔS = . Q = mLv . T 3 SET UP: For vaporization of water, Lv = 2256 × 10 J/kg.

Q mLv (1.00 kg)(2256 × 103 J/kg) = = = 6.05 × 103 J/K. Note that this is the change of T T (373.15 K) entropy of the water as it changes to steam. (b) The magnitude of the entropy change is roughly five times the value found in Example 20.5. EVALUATE: Water is less ordered (more random) than ice, but water is far less random than steam; a consideration of the density changes indicates why this should be so. Q IDENTIFY: The phase transition occurs at constant temperature and ΔS = . Q = mLv . The mass of one mole is T the molecular mass M. SET UP: For water, Lv = 2256 × 103 J/kg. For N 2 , M = 28.0 × 10−3 kg/mol, the boiling point is 77.34 K and EXECUTE:

20.32.

(a) ΔS =

Lv = 201 × 103 J/kg. For silver (Ag), M = 107.9 × 10−3 kg/mol, the boiling point is 2466 K and Lv = 2336 × 103 J/kg. For mercury (Hg), M = 200.6 × 10−3 kg/mol, the boiling point is 630 K and Lv = 272 × 103 J/kg.

20-8

Chapter 20

EXECUTE: (b) N 2 :

(a) ΔS =

Q mLv (18.0 × 10−3 kg)(2256 × 103 J kg) = = = 109 J K. T T (373.15 K)

(28.0 × 10−3 kg)(201 × 103 J kg) (107.9 × 10−3 kg)(2336 × 103 J/kg) = 72.8 J K. Ag: = 102.2 J/K. (77.34 K) (2466 K)

(200.6 × 10−3 kg)(272 × 103 J/kg) = 86.6 J/K (630 K) (c) The results are the same order or magnitude, all around 100 J/K. EVALUATE: The entropy change is a measure of the increase in randomness when a certain number (one mole) goes from the liquid to the vapor state. The entropy per particle for any substance in a vapor state is expected to be roughly the same, and since the randomness is much higher in the vapor state (see Exercise 20.31), the entropy change per molecule is roughly the same for these substances. IDENTIFY: During the phase transition the gallium is at a constant temperature equal to the melting point of gallium. Your hand is at a constant temperature of 98.6°F = 37.0°C = 310.1 K. Heat Q = mLf flows out of your Q hand and into the gallium. For heat flow at constant temperature, ΔS = . T SET UP: For gallium, Lf = 8.04 × 104 J/kg and the melting point is 29.8°C = 303.0 K. Hg:

20.33.

EXECUTE:

Q = mLf = (25.0 × 10−3 kg)(8.04 × 104 J/kg) = 2.01 × 103 J. For your hand,

Q −2.01 × 103 J = = −6.48 J/K. Heat flows out of your hand, Q is negative, and ΔS is negative. For the T 310.1 K Q gallium, ΔS = . The temperature of the gallium is less than that of your hand and Q is the same, so the 303.0 K magnitude of the entropy change of the gallium is greater than the magnitude of the entropy change of your hand. EVALUATE: For the gallium, ΔS > 0, so ΔSsystem > 0 and the process is irreversible. ΔS =

20.34.

20.35.

IDENTIFY: Apply Eq.(20.23) and follow the procedure used in Example 20.11. SET UP: After the partition is punctured each molecule has equal probability of being on each side of the box. The probability of two independent events occurring simultaneously is the product of the probabilities of each separate event. EXECUTE: (a) On the average, each half of the box will contain half of each type of molecule, 250 of nitrogen and 50 of oxygen. (b) See Example 20.11. The total change in entropy is ΔS = kN1ln(2) + kN 2ln(2) = (N1 + N 2 )k ln(2) = (600)(1.381 × 10−23 J K) ln(2) = 5.74 × 10−21 J K. (c) The probability is (1/2)500 × (1/2)100 = (1/2)600 = 2.4 × 10−181 , and is not likely to happen. The numerical result for part (c) above may not be obtained directly on some standard calculators. For such calculators, the result may be found by taking the log base ten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of the sum. The result is then 10−181 × 100.87 = 2.4 × 10−181. EVALUATE: The contents of the box constitutes an isolated system. ΔS > 0 and the process is irreversible. (a) IDENTIFY and SET UP: The velocity distribution of Eq.(18.32) depends only on T, so in an isothermal process it does not change. (b) EXECUTE: Calculate the change in the number of available microscopic states and apply Eq.(20.23). Following the reasoning of Example 20.11, the number of possible positions available to each molecule is altered by a factor of 3 (becomes larger). Hence the number of microscopic states the gas occupies at volume 3V is w2 = (3) N w1 , where N is the number of molecules and w1 is the number of possible microscopic states at the start of the process, where the volume is V. Then, by Eq.(20.23), ΔS = k ln( w2 / w1 ) = k ln(3) N = Nk ln(3) = nN A k ln(3) = nR ln(3) ΔS = (2.00 mol)(8.3145 J/mol ⋅ K)ln(3) = +18.3 J/K (c) IDENTIFY and SET UP: For an isothermal reversible process ΔS = Q / T . EXECUTE: Calculate W and then use the first law to calculate Q. ΔT = 0 implies ΔU = 0, since system is an ideal gas. Then by ΔU = Q − W , Q = W .

For an isothermal process, W = ∫

V2 V1

V2

p dV = ∫ (nRT / V ) dV = nRT ln(V2 / V1 ) V1

Thus Q = nRT ln(V2 / V1 ) and ΔS = Q / T = nR ln(V2 / V1 ) ΔS = (2.00 mol)(8.3145 J/mol ⋅ K)ln(3V1 / V1 ) = +18.3 J/K EVALUATE: This is the same result as obtained in part (b).


20.36.

IDENTIFY: SET UP:

20-9

Example 20.8 shows that for a free expansion, ΔS = nR ln(V2 / V1 ). V1 = 2.40 L = 2.40 × 10−3 m3

⎛ ⎞ 425 m3 ΔS = (0.100 mol)(8.314 J/mol ⋅ K)ln ⎜ = 10.0 J/K −3 3 ⎟ × 2.40 10 m ⎝ ⎠ EVALUATE: ΔSsystem > 0 and the free expansion is irreversible. EXECUTE:

20.37.


TC W . W = QH + QC . e = . TH QH pV = nRT ; the 300 K isotherm lies below the 400 K isotherm in the pV-diagram. eCarnot = 1 −

(a) eCarnot = 1 −

400 K = 0.200 = 20.0%. 500 K

W 2000 J = = 10,000 J. QC = QH − W = 10,000 J − 2000 J = 8000 J. 0.200 e (c) The 500 K and 400 K isotherms and the Carnot cycle operating between those isotherms are sketched in Figure 20.37. (d) The 300 K isotherm and the Carnot cycle operating between the 500 K and 300 K isotherms are also sketched in Figure 20.37. (e) The cycle with TC = 300 K encloses more area than the cycle with TC = 400 K. (f) Less work is done on the gas during the compression at lower temperature, so less heat is ejected to keep the internal energy and temperature constant. EVALUATE: For TC = 300 K, eCarnot = 0.400. W = eQH = (0.400)(10,000 J) = 4000 J. QC = 6000 J. (b) QH =

Figure 20.37 20.38.


W = QC + QH . Since it is a Carnot cycle,

QC T = − C . The heat required to melt the ice is Q = mLf . QH TH

For water, Lf = 334 × 103 J/kg. QH > 0, QC < 0. QC = − mLf . TH = 527°C = 800.15 K. (a) QH = +400 J, W = +300 J. QC = W − QH = −100 J.

TC = −TH (QC QH ) = −(800.15 K) [ ( − 100 J) (400 J)] = +200 K = −73°C (b) The total QC required is − mLf = − (10.0 kg ) ( 334 × 103 J kg ) = −3.34 × 106 J. QC for one cycle is −100 J, so

−3.34 × 106 J = 3.34 × 104 cycles. −100 J cycle EVALUATE: The results depend only on the maximum temperature of the gas, not on the number of moles or the maximum pressure. T IDENTIFY: eCarnot = 1 − C , where TC and TH must be in kelvins. TH the number of cycles required is

20.39.

SET UP:

TC = −90.0°C = 183 K.

TC 183 K 183 K . For e = 0.400, TH = = 305 K. For e = 0.450, TH = = 333 K. TH 1− e 1 − 0.400 1 − 0.450 must be increased 28 K = 28 C°. (b) TC = (1 − e)TH = (1 − 0.450)(305 K) = 168 K. TC must be decreased 15 K = 15 C°. EVALUATE: A Kelvin degree is the same size as a Celsius degree, so a temperature change ΔT has the same numerical value whether it is expressed in K or in C°. EXECUTE:

(a) TH =

20-10

Chapter 20

20.40.

IDENTIFY: Use the ideal gas law to calculate p and V for each state. Use the first law and specific expressions for Q, W, and ΔU for each process. Use Eq.(20.4) to calculate e. QH is the net heat flow into the gas. SET UP: γ = 1.40 CV = R /(γ − 1) = 20.79 J/mol ⋅ K; C p = CV + R = 29.10 J/mol ⋅ K. The cycle is sketched in Figure 20.40.

T1 = 300 K T2 = 600 K T3 = 492 K Figure 20.40 EXECUTE: (a) point 1 p1 = 1.00 atm = 1.013 × 105 Pa (given); pV = nRT ;

V1 =

nRT1 (0.350 mol)(8.3145 J/mol ⋅ K)(300 K) = = 8.62 × 10−3 m3 1.013 × 105 Pa p1

point 2 process 1 → 2 at constant volume so V2 = V1 = 8.62 × 10−3 m3 pV = nRT and n, R, V constant implies p1 / T1 = p2 / T2 p2 = p1 (T2 / T1 ) = (1.00 atm)(600 K/300 K) = 2.00 atm = 2.03 × 105 Pa

point 3 Consider the process 3 → 1, since it is simpler than 2 → 3. Process 3 → 1 is at constant pressure so p3 = p1 = 1.00 atm = 1.013 × 105 Pa pV = nRT and n, R, p constant implies V1 / T1 = V3 / T3 V3 = V1 (T3 / T1 ) = (8.62 × 10−3 m3 )(492 K/300 K) = 14.1 × 10−3 m3 (b) process 1 → 2

constant volume (ΔV = 0) Q = nCV ΔT = (0.350 mol)(20.79 J/mol ⋅ K)(600 K − 300 K) = 2180 J ΔV = 0 and W = 0. Then ΔU = Q − W = 2180 J

process 2 → 3 Adiabatic means Q = 0. ΔU = nCV ΔT (any process), so ΔU = (0.350 mol)(20.79 J/mol ⋅ K)(492 K − 600 K) = −780 J Then ΔU = Q − W gives W = Q − ΔU = +780 J. (It is correct for W to be positive since ΔV is positive.)

process 3 → 1 For constant pressure W = pΔV = (1.013 × 105 Pa)(8.62 × 10−3 m3 − 14.1 × 10−3 m3 ) = −560 J or W = nRΔT = (0.350 mol)(8.3145 J/mol ⋅ K)(300 K − 492 K) = −560 J, which checks. (It is correct for W to be negative, since ΔV is negative for this process.) Q = nC p ΔT = (0.350 mol)(29.10 J/mol ⋅ K)(300 K − 492 K) = −1960 J ΔU = Q − W = −1960 J − (−560 K) = −1400 J or ΔU = nCV ΔT = (0.350 mol)(20.79 J/mol ⋅ K)(300 K − 492 K) = −1400 J, which checks (c) Wnet = W1→ 2 + W2 →3 + W3→1 = 0 + 780 J − 560 J = +220 J (d) Qnet = Q1→ 2 + Q2 →3 + Q3→1 = 2180 J + 0 − 1960 J = +220 J (e) e =

work output W 220 J = = = 0.101 = 10.1%. heat energy input QH 2180 J

e(Carnot) = 1 − TC / TH = 1 − 300 K/600 K = 0.500.


EVALUATE:

20.41.

20-11

For a cycle ΔU = 0, so by ΔU = Q − W it must be that Qnet = Wnet for a cycle. We can also check

that ΔU net = 0: ΔU net = ΔU1→ 2 + ΔU 2 →3 + ΔU 3→1 = 2180 J − 1050 J − 1130 J = 0 e < e(Carnot), as it must. IDENTIFY: pV = nRT , so pV is constant when T is constant. Use the appropriate expression to calculate Q and W for each process in the cycle. e = SET UP:

W . QH

For an ideal diatomic gas, CV = 52 R and C p = 72 R.

EXECUTE:

(a) paVa = 2.0 × 103 J. pbVb = 2.0 × 103 J. pV = nRT so paVa = pbVb says Ta = Tb .

(b) For an isothermal process, Q = W = nRT ln(V2 / V1 ). ab is a compression, with Vb < Va , so Q < 0 and heat is

rejected. bc is at constant pressure, so Q = nC p ΔT =

Cp R

pΔV . ΔV is positive, so Q > 0 and heat is absorbed. cd is

CV V Δp. Δp is negative, so Q < 0 and heat is rejected. R pV 2.0 × 103 J pV = 241 K. Tb = b b = Ta = 241 K. (c) Ta = a a = (1.00)(8.314 J/mol ⋅ K) nR nR

at constant volume, so Q = nCV ΔT =

Tc =

pcVc 4.0 × 103 J = = 481 K. nR (1.00)(8.314 J/mol ⋅ K)

⎛V ⎞ ⎛ 0.0050 m3 ⎞ = −1.39 × 103 J. (d) Qab = nRT ln ⎜ b ⎟ = (1.00 mol)(8.314 J/mol ⋅ K)(241 K)ln ⎜ 3 ⎟ 0.010 m V ⎝ ⎠ ⎝ a⎠ Qbc = nC p ΔT = (1.00)( 72 )(8.314 J/mol ⋅ K)(241 K) = 7.01 × 103 J. Qca = nCV ΔT = (1.00)( 52 )(8.314 J/mol ⋅ K)( − 241 K) = −5.01 × 103 J. Qnet = Qab + Qbc + Qca = 610 J. Wnet = Qnet = 610 J. (e) e =

W 610 J = = 0.087 = 8.7% QH 7.01 × 103 J

EVALUATE:

We can calculate W for each process in the cycle. Wab = Qab = −1.39 × 103 J.

Wbc = pΔV = (4.0 × 105 Pa)(0.0050 m3 ) = 2.00 × 103 J. Wca = 0. Wnet = Wab + Wbc + Wca = 610 J, which does equal Qnet . 20.42.


Combine Eqs.(20.13) and (20.2) to eliminate QC and obtain an expression for QH in

terms of W, TC , and TH . W = 1.00 J, TC = 268.15 K, TH = 290.15 K

For the heat pump QC > 0 and QH < 0 QC T W 1.00 J = = 13.2 J = − C gives QH = 1 − TC / TH 1 − (268.15/ 290.15) QH TH (b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required. W (c) EVALUATE: From part (a), QH = . QH decreases as TC decreases. The heat pump is less efficient as 1 − TC / TH EXECUTE:

W = QC + QH ; combining this with

the temperature difference through which the heat has to be “pumped” increases. In an engine, heat flows from TH

20.43.

to TC and work is extracted. The engine is more efficient the larger the temperature difference through which the heat flows. IDENTIFY: Tb = Tc and is equal to the maximum temperature. Use the ideal gas law to calculate Ta . Apply the appropriate expression to calculate Q for each process. e =

W . ΔU = 0 for a complete cycle and for an QH

isothermal process of an ideal gas. SET UP: For helium, CV = 3R / 2 and C p = 5R / 2. The maximum efficiency is for a Carnot cycle, and eCarnot = 1 − TC / TH .

20-12

Chapter 20

(a) Qin = Qab + Qbc . Qout = Qca . Tmax = Tb = Tc = 327°C = 600 K.

EXECUTE:

paVa pbVb p 1 = → Ta = a Tb = (600 K) = 200 K. 3 Ta Tb pb pbVb = nRTb → Vb =

nRTb (2 moles)(8.31 J/mol ⋅ K)(600 K) = = 0.0332 m3 . 3.0 × 105 Pa pb

pbVb pcVc p ⎛3⎞ = → Vc = Vb b = (0.0332 m3 ) ⎜ ⎟ = 0.0997 m3 = Va . Tb Tc pc ⎝1⎠ ⎛3⎞ Qab = nCV ΔTab = (2 mol) ⎜ ⎟ ( 8.31 J/mol ⋅ K ) (400 K) = 9.97 × 103 J ⎝ 2⎠ c c nRT V b Qbc = Wbc = ∫ pdV = ∫ dV = nRTb ln c = nRTb ln 3. b b V Vb Qbc = (2.00 mol) ( 8.31 J/mol ⋅ K ) (600 K)ln 3 = 1.10 × 104 J. Qin = Qab + Qbc = 2.10 × 104 J. ⎛5⎞ Qout = Qca = nC p ΔTca = (2.00 mol) ⎜ ⎟ ( 8.31 J/mol ⋅ K ) (400 K) = 1.66 × 104 J. ⎝2⎠ (b) Q = ΔU + W = 0 + W → W = Qin − Qout = 2.10 × 104 J − 1.66 × 104 J = 4.4 × 103 J.

4.4 × 103 J = 0.21 = 21%. 2.10 × 104 J T (c) emax = eCarnot = 1 − C = 1 − 200 K = 0.67 = 67% 600 K TH EVALUATE: The thermal efficiency of this cycle is about one-third of the efficiency of a Carnot cycle that operates between the same two temperatures. Q T T IDENTIFY: For a Carnot engine, C = − C . eCarnot = 1 − C . W = QH − QC . QH > 0, QC < 0. pV = nRT . QH TH TH e = W Qin =

20.44.

SET UP:

The work done by the engine each cycle is mg Δy , with m = 15.0 kg and Δy = 2.00 m. TH = 773 K.

QH = 500 J. EXECUTE: (a) The pV diagram is sketched in Figure 20.44. (b) W = mg Δy = (15.0 kg)(9.80 m/s 2 )(2.00 m) = 294 J. QC = QH − W = 500 J − 294 J = 206 J, and QC = −206 J. ⎛Q ⎞ ⎛ −206 J ⎞ TC = −TH ⎜ C ⎟ = −(773 K) ⎜ ⎟ = 318 K = 45°C. ⎝ 500 J ⎠ ⎝ QH ⎠ T 318 K = 0.589 = 58.9%. (c) e = 1 − C = 1 − 773 K TH (d) QC = 206 J. (e) The maximum pressure is for state a. This is also where the volume is a minimum, so nRTa (2.00 mol)(8.315 J/mol ⋅ K)(773 K) = = 2.57 × 106 Pa. Va = 5.00 L = 5.00 × 10−3 m3 . Ta = TH = 773 K. pa = 5.00 × 10−3 m3 Va EVALUATE:

We can verify that e =

W gives the same value for e as calculated in part (c). QH

Figure 20.44


20.45.

emax = eCarnot = 1 − TC / TH . e =

IDENTIFY:

20-13

W W /t W QC QH . W = QH + QC so = . For a temperature change = + QH QH / t t t t

Q = mcΔT . SET UP: TH = 300.15 K, TC = 279.15 K. For water, ρ = 1000 kg/m3 , so a mass of 1 kg has a volume of 1 L. For water, c = 4190 J/kg ⋅ K. (a) e = 1 − 279.15 K = 7.0%. 300.15 K Q Q W QH Pout 210 kW (b) = = = 3.0 MW. C = H − = 3.0 MW − 210 kW = 2.8 MW. 0.070 t t t t e 6 m QC / t (2.8 × 10 W) (3600 s h) = = = 6 × 105 kg h = 6 × 105 L h. (c) t cΔT (4190 J kg ⋅ K) (4 K) EXECUTE:

20.46.

EVALUATE: The efficiency is small since TC and TH don’t differ greatly. IDENTIFY: Use Eq.(20.4) to calculate e. SET UP: The cycle is sketched in Figure 20.46.

CV = 5 R / 2 for an ideal gas C p = CV + R = 7 R / 2

Figure 20.46 SET UP:

Calculate Q and W for each process.

process 1 → 2 ΔV = 0 implies W = 0 ΔV = 0 implies Q = nCV ΔT = nCV (T2 − T1 ) But pV = nRT and V constant says p1V = nRT1 and p2V = nRT2 . Thus ( p2 − p1 )V = nR (T2 − T1 ); V Δp = nRΔT (true when V is constant). Then Q = nCV ΔT = nCV (V Δp / nR ) = (CV /R)V Δp = (CV /R)V0 (2 p0 − p0 ) = (CV /R ) p0V0 . Q > 0; heat is absorbed by the gas.) process 2 → 3 Δp = 0 so W = pΔV = p(V3 − V2 ) = 2 p0 (2V0 − V0 ) = 2 p0V0 (W is positive since V increases.) Δp = 0 implies Q = nC p ΔT = nC p (T2 − T1 ) But pV = nRT and p constant says pV1 = nRT1 and pV2 = nRT2 . Thus p (V2 − V1 ) = nR (T2 − T1 ); pΔV = nRΔT (true when p is constant). Then Q = nC p ΔT = nC p ( pΔV/nR ) = (C p /R ) pΔV = (C p /R)2 p0 (2V0 − V0 ) = (C p /R)2 p0V0 . (Q > 0; heat is absorbed by the gas.) process 3 → 4 ΔV = 0 implies W = 0 ΔV = 0 so Q = nCV ΔT = nCV (V Δp/nR) = (CV /R)(2V0 )( p0 − 2 p0 ) = −2(CV /R) p0V0 (Q < 0 so heat is rejected by the gas.) process 4 → 1 Δp = 0 so W = pΔV = p(V1 − V4 ) = p0 (V0 − 2V0 ) = − p0V0 (W is negative since V decreases) Δp = 0 so Q = nC p ΔT = nC p ( pΔV/nR ) = (C p /R ) pΔV = (C p /R) p0 (V0 − 2V0 ) = −(C p /R) p0V0 (Q < 0 so heat is rejected by the gas.)

20-14

Chapter 20

total work performed by the gas during the cycle: Wtot = W1→ 2 + W2 →3 + W3→ 4 + W4 →1 = 0 + 2 p0V0 + 0 − p0V0 = p0V0 (Note that Wtot equals the area enclosed by the cycle in the pV-diagram.) total heat absorbed by the gas during the cycle (QH ): Heat is absorbed in processes 1 → 2 and 2 → 3. C ⎛ C + 2C p ⎞ C QH = Q1→ 2 + Q2 →3 = V p0V0 + 2 p p0V0 = ⎜ V ⎟ p0V0 R R R ⎝ ⎠ CV + 2(CV + R ) ⎛ 3C + 2 R ⎞ p0V0 = ⎜ V ⎟ p0V0 . R R ⎝ ⎠ total heat rejected by the gas during the cycle (QC ): Heat is rejected in processes 3 → 4 and 4 → 1. C ⎛ 2C + C p ⎞ C QC = Q3→ 4 + Q4 →1 = −2 V p0V0 − p p0V0 = − ⎜ V ⎟ p0V0 R R R ⎝ ⎠ But C p = CV + R so QH =

But C p = CV + R so QC = −

2CV + (CV + R ) ⎛ 3C + R ⎞ p0V0 = − ⎜ V ⎟ p0V0 . R ⎝ R ⎠

efficiency W p0V0 R R 2 e= = = = = . QH ([3CV + 2 R ]/ R ) ( p0V0 ) 3CV + 2 R 3(5R/2) + 2 R 19 e = 0.105 = 10.5% EVALUATE:

⎛ 3C + R ⎞ ⎛ 3CV + 2 R ⎞ As a check on the calculations note that QC + QH = − ⎜ V ⎟ p0V0 + ⎜ ⎟ p0V0 = p0V0 = W , R R ⎝ ⎠ ⎝ ⎠

as it should. 20.47.

Use pV = nRT . Apply the expressions for Q and W that apply to each type of process. e =

IDENTIFY: SET UP:

W . QH

For O 2 , CV = 20.85 J/mol ⋅ K and C p = 29.17 J/mol ⋅ K. (a) p1 = 2.00 atm, V1 = 4.00 L, T1 = 300 K.

EXECUTE:

p2 = 2.00 atm. V3 = 6.00 L.

⎛T ⎞ V1 V2 ⎛ 450 K ⎞ = . V2 = ⎜ 2 ⎟V1 = ⎜ ⎟ (4.00 L) = 6.00 L. T T1 T2 ⎝ 300 K ⎠ ⎝ 1⎠

⎛T ⎞ p2 p3 ⎛ 250 K ⎞ = . p3 = ⎜ 3 ⎟ p2 = ⎜ ⎟ (2.00 atm) = 1.11 atm T2 T3 T ⎝ 450 K ⎠ ⎝ 2⎠

⎛V ⎞ ⎛ 6.00 L ⎞ V4 = 4.00 L. p3V3 = p4V4 . p4 = p3 ⎜ 3 ⎟ = (1.11 atm) ⎜ ⎟ = 1.67 atm. ⎝ 4.00 L ⎠ ⎝ V4 ⎠ These processes are shown in Figure 20.47. pV (2.00 atm)(4.00 L) = 0.325 mol (b) n = 1 1 = RT1 (0.08206 L ⋅ atm/mol ⋅ K)(300 K) process 1 → 2: W = pΔV = nRΔT = (0.325 mol)(8.315 J/mol ⋅ K)(150 K) = 405 J. Q = nC p ΔT = (0.325 mol)(29.17 J/mol ⋅ K)(150 K) = 1422 J. process 2 → 3: W = 0. Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)( −200 K) = −1355 J. ⎛V ⎞ ⎛ 4.00 L ⎞ process 3 → 4: ΔU = 0 and Q = W = nRT3 ln ⎜ 4 ⎟ = (0.325 mol)(8.315 J/mol ⋅ K)(250 K)ln ⎜ ⎟ = −274 J. ⎝ 6.00 L ⎠ ⎝ V3 ⎠ process 4 → 1: W = 0. Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)(50 K) = 339 J. (c) W = 405 J − 274 J = 131 J W 131 J = = 0.0744 = 7.44%. (d) e = QH 1422 J + 339 J eCarnot = 1 −

TC 250 K =1− = 0.444 = 44.4%; eCarnot is much larger. TH 450 K


EVALUATE:

20-15

Qtot = +1422 J + (−1355 J) + (−274 J) + 339 J = 132 J. This is equal to Wtot , apart from a slight

difference due to rounding. For a cycle, Wtot = Qtot , since ΔU = 0.

Figure 20.47 20.48.

IDENTIFY and SET UP: For the constant pressure processes ab and cd calculate W and use the first law to calculate Q. Calculate Qtot and use that Wtot = Qtot for a cycle. The coefficient of performance is given by

Eq.(20.9); QC is the net heat that goes into the system. The cycle is sketched in Figure 20.48.


(a) process c → d

ΔU = U d − U c = 1657 × 103 J − 1005 × 103 J = 6.52 × 105 J W =∫

Vd Vc

p dV = pΔV (since is a constant pressure process)

W = (363 × 103 Pa)(0.4513 m3 − 0.2202 m3 ) = +8.39 × 104 J (positive since process is an expansion) ΔU = Q − W so Q = ΔU + W = 6.52 × 105 J + 8.39 × 104 J = 7.36 × 105 J. (Q positive so heat goes into the coolant) (b) process a → b

ΔU = U b − U a = 1171 × 103 J − 1969 × 103 J = −7.98 × 105 J W = pΔV = (2305 × 103 Pa)(0.00946 m3 − 0.0682 m3 ) = −1.35 × 105 J (negative since ΔV < 0 for the process) Q = ΔU + W = −7.98 × 105 J − 1.35 × 105 J = −9.33 × 105 J (negative so heat comes out of coolant). (c) The coolant cannot be treated as an ideal gas, so we can’t calculate W for the adiabatic processes. But ΔU = 0 (for cycle) so Wnet = Qnet . Q = 0 for the two adiabatic processes, so Qnet = Qcd + Qab = 7.36 × 105 J − 9.33 × 105 J = −1.97 × 105 J

Thus Wnet = −1.97 × 105 J (negative since work is done on the coolant, the working substance). (d) K = QC / W = (+7.36 × 105 J) /( +1.97 × 105 J) = 3.74. EVALUATE:

Wnet < 0 when the cycle is taken in the counterclockwise direction, as is the case here.

20-16

Chapter 20

20.49.

IDENTIFY:

Use ΔU = Q − W and the appropriate expressions for Q, W and ΔU for each type of process. W pV = nRT relates ΔT to p and V values. e = , where QH is the heat that enters the gas during the cycle. QH

SET UP: For a monatomic ideal gas, CP = 52 R and CV = 32 R. (a) ab: The temperature changes by the same factor as the volume, and so C Q = nCP ΔT = P pa (Va − Vb ) = (2.5)(3.00 × 105 Pa)(0.300 m3 ) = 2.25 × 105 J. R The work pΔV is the same except for the factor of 52 , so W = 0.90 × 105 J.

ΔU = Q − W = 1.35 × 105 J. bc: The temperature now changes in proportion to the pressure change, and 3 Q = 2 ( pc − pb )Vb = (1.5)( −2.00 × 105 Pa)(0.800 m3 ) = −2.40 × 105 J, and the work is zero

(ΔV = 0). ΔU = Q − W = −2.40 × 105 J. ca: The easiest way to do this is to find the work done first; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is 1 (3.00 × 105 Pa + 1.00 × 105 Pa)(0.800 m3 − 0.500 m3 ) = 6.00 × 104 J and so the work is −0.60 × 105 J. ΔU must 2 be 1.05 × 105 J (since ΔU = 0 for the cycle, anticipating part (b)), and so Q must be ΔU + W = 0.45 × 105 J. (b) See above; Q = W = 0.30 × 105 J, ΔU = 0. (c) The heat added, during process ab and ca, is 2.25 ×105 J + 0.45 × 105 J = 2.70 × 105 J and the efficiency is W 0.30 × 105 e= = = 0.111 = 11.1%. QH 2.70 × 105 EVALUATE: For any cycle, ΔU = 0 and Q = W . 20.50.

IDENTIFY:

Use the appropriate expressions for Q, W and ΔU for each process. e = W / QH and eCarnot = 1 − TC / TH .

SET UP: For this cycle, TH = T2 and TC = T1 EXECUTE: (a) ab: For the isothermal process, ΔT = 0 and ΔU = 0. W = nRT1 ln(Vb Va ) = nRT1ln(1/r ) = −nRT1ln(r ) and Q = W = − nRT1 ln(r ).

bc: For the isochoric process, ΔV = 0 and W = 0. Q = ΔU = nCV ΔT = nCV (T2 − T1 ). cd: As in the process ab, ΔU = 0 and W = Q = nRT2ln(r ). da: As in process bc, ΔV = 0 and W = 0; ΔU = Q = nCV (T1 − T2 ). (b) The values of Q for the processes are the negatives of each other. (c) The net work for one cycle is Wnet = nR (T2 − T1 )ln(r ), and the heat added (neglecting the heat exchanged during

the isochoric expansion and compression, as mentioned in part (b)) is Qcd = nRT2 ln(r ), and the efficiency is Wnet = 1 − (T1 T2 ). This is the same as the efficiency of a Carnot-cycle engine operating between the two Qcd temperatures. EVALUATE: For a Carnot cycle two steps in the cycle are isothermal and two are adiabatic and all the heat flow occurs in the isothermal processes. For the Stirling cycle all the heat flow is also in the isothermal steps, since the net heat flow in the two constant volume steps is zero. W + W2 IDENTIFY: The efficiency of the composite engine is e12 = 1 , where QH1 is the heat input to the first engine QH1 e=

20.51.

and W1 and W2 are the work outputs of the two engines. For any heat engine, W = QC + QH , and for a Carnot engine, Qlow T = − low , where Qlow and Qhigh are the heat flows at the two reservoirs that have temperatures Tlow and Thigh . Qhigh Thigh SET UP: EXECUTE:

Qhigh,2 = −Qlow,1. Tlow,1 = T ′, Thigh,1 = TH , Tlow,2 = TC and Thigh,2 = T ′. e12 =

Qlow,2 = −Qhigh,2

Q W1 + W2 Qhigh,1 + Qlow,1 + Qhigh,2 + Qlow,2 = . Since Qhigh,2 = −Qlow,1 , this reduces to e12 = 1 + low,2 . QH1 Qhigh,1 Qhigh,1

Tlow,2 Thigh,2

= Qlow,1

⎛T ⎞T ⎛ T′ ⎞T TC T = −Qhigh,1 ⎜ low,1 ⎟ C = −Qhigh,1 ⎜ ⎟ C . This gives e12 = 1 − C . The efficiency of ⎜ ⎟ TH T′ ⎝ TH ⎠ T ′ ⎝ Thigh,1 ⎠ T ′

the composite system is the same as that of the original engine. EVALUATE: The overall efficiency is independent of the value of the intermediate temperature T ′.


20.52.

IDENTIFY:

e=

20-17

W . 1 day = 8.64 × 10 4 s. For the river water, Q = mcΔT , where the heat that goes into the water QH

is the heat QC rejected by the engine. The density of water is 1000 kg/m 3 . When an object undergoes a temperature change, ΔS = mc ln(T2 / T1 ). SET UP: 18.0°C = 291.1 K. 18.5°C = 291.6 K. W P 1000 MW = 2.50 × 103 MW. EXECUTE: (a) QH = so PH = W = e 0.40 e (b) The heat input in one day is (2.50 × 109 W)(8.64 × 10 4 s) = 2.16 × 1014 J. The mass of coal used per day is 2.16 × 1014 J = 8.15 × 106 kg. 2.65 × 107 J/kg

(c) QH = W + QC . QC = QH − W . PC = PH − PW = 2.50 × 103 MW − 1000 MW = 1.50 × 103 MW. (d) The heat input to the river is 1.50 × 109 J/s. Q = mcΔT and ΔT = 0.5 C° gives m=

Q 1.50 × 109 J m = = 7.16 × 105 kg. V = = 716 m3 . The river flow rate must be 716 m 3 / s. ρ cΔT (4190 J/kg ⋅ K)(0.5 K)

(e) In one second, 7.16 × 105 kg of water goes from 291.1 K to 291.6 K.

20.53.

⎛T ⎞ ⎛ 291.6 K ⎞ 6 ΔS = mc ln ⎜ 2 ⎟ = (7.16 × 105 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ = 5.1 × 10 J/K. 291.1 K T ⎝ ⎠ ⎝ 1⎠ EVALUATE: The entropy of the river increases because heat flows into it. The mass of coal used per second is huge. (a) IDENTIFY and SET UP: Calcualte e from Eq.(20.6), QC from Eq.(20.4) and then W from Eq.(20.2).

EXECUTE:

e = 1 − 1/(r γ −1 ) = 1 − 1/(10.60.4 ) = 0.6111

e = (QH + QC ) / QH and we are given QH = 200 J; calculate QC. QC = (e − 1)QH = (0.6111 − 1)(200 J) = −78 J (negative since corresponds to heat leaving)

Then W = QC + QH = −78 J + 200 J = 122 J. (Positive, in agreement with Fig. 20.6.) EVALUATE: QH , W > 0, and QC < 0 for an engine cycle. (b) IDENTIFY and SET UP: The stoke times the bore equals the change in volume. The initial volume is the final volume V times the compression ratio r. Combining these two expressions gives an equation for V. For each cylinder of area A = π (d / 2) 2 the piston moves 0.864 m and the volume changes from rV to V, as shown in Figure 20.53a.

l1 A = rV l2 A = V and l1 − l2 = 86.4 × 10−3 m Figure 20.53a EXECUTE:

l1 A − l2 A = rV − V and (l1 − l2 ) A = (r − 1)V

(l1 − l2 ) A (86.4 × 10−3 m)π (41.25 × 10−3 m) 2 = = 4.811 × 10−5 m3 10.6 − 1 r −1 At point a the volume is rV = 10.6(4.811 × 10−5 m 3 ) = 5.10 × 10−4 m 3 . V=

(c) IDENTIFY and SET UP: The processes in the Otto cycle are either constant volume or adiabatic. Use the QH that is given to calculate ΔT for process bc. Use Eq.(19.22) and pV = nRT to relate p, V and T for the adiabatic processes ab and cd. EXECUTE: point a: Ta = 300 K, pa = 8.50 × 104 Pa, and Va = 5.10 × 10−4 m 3

point b: Vb = Va / r = 4.81 × 10−5 m3 . Process a → b is adiabatic, so TaVaγ −1 = TbVbγ −1. Ta (rV )γ −1 = TbV γ −1 Tb = Ta r γ −1 = 300 K(10.6)0.4 = 771 K pV = nRT so pV / T = nR = constant, so paVa / Ta = pbVb / Tb pb = pa (Va / Vb )(Tb / Ta ) = (8.50 × 104 Pa)(rV / V )(771 K/300 K) = 2.32 × 106 Pa

20-18

Chapter 20

point c: Process b → c is at constant volume, so Vc = Vb = 4.81 × 10−5 m3 QH = nCV ΔT = nCV (Tc − Tb ). The problem specifies QH = 200 J; use to calculate Tc . First use the p, V, T values at point a to calculate the number of moles n. pV (8.50 × 104 Pa)(5.10 × 10−4 m3 ) n= = = 0.01738 mol (8.3145 J/mol ⋅ K)(300 K) RT

Then Tc − Tb =

QH 200 J = = 561.3 K, and Tc = Tb + 561.3 K = 771 K + 561 K = 1332 K nCV (0.01738 mol)(20.5 J/mol ⋅ K)

p / T = nR / V = constant so pb / Tb = pc / Tc pc = pb (Tc / Tb ) = (2.32 × 106 Pa)(1332 K/771 K) = 4.01 × 106 Pa

point d: Vd = Va = 5.10 × 10−4 m3 process c → d is adiabatic, so TdVdγ −1 = TcVcγ −1 Td ( rV )γ −1 = TcV γ −1 Td = Tc / r γ −1 = 1332 K/10.60.4 = 518 K pcVc / Tc = pdVd / Td pd = pc (Vc / Vd )(Td / Tc ) = (4.01 × 106 Pa)(V / rV )(518 K/1332 K) = 1.47 × 105 Pa EVALUATE: Can look at process d → a as a check. QC = nCV (Ta − Td ) = (0.01738 mol)(20.5 J/mol ⋅ K)(300 K − 518 K) = −78 J, which agrees with part (a). The cycle is sketched in Figure 20.53b.

Figure 20.53b (d) IDENTIFY and SET UP:

The Carnot efficiency is given by Eq.(20.14). TH is the highest temperature reached

in the cycle and TC is the lowest. EXECUTE: From part (a) the efficiency of this Otto cycle is e = 0.611 = 61.1%. The efficiency of a Carnot cycle operating between 1332 K and 300 K is e(Carnot) = 1 − TC / TH = 1 − 300 K /1332 K = 0.775 = 77.5%, which is larger. EVALUATE: The 2nd law requires that e ≤ e(Carnot), and our result obeys this law. 20.54.

IDENTIFY:

K=

QC . QH = QC + W . The heat flows for the inside and outside air occur at constant T, so W

ΔS = Q / T . SET UP: 21.0°C = 294.1 K. 35.0°C = 308.1 K. EXECUTE: (a) QC = K W . PC = KPW = (2.80)(800 W) = 2.24 × 103 W. (b) PH = PC + PW = 2.24 × 103 W + 800 W = 3.04 × 103 W. (c) In 1 h = 3600 s, QH = PHt = 1.094 × 107 J. ΔSout =

QH 1.094 × 107 J = = 3.55 × 104 J/K. 308.1 K TH


20-19

(d) QC = PCt = 8.064 × 106 J. Heat QC is removed from the inside air.

−QC −8.064 × 106 J = = −2.74 × 104 J/K. ΔS net = ΔSout + ΔSin = 8.1 × 103 J/K. 294.1 K TC EVALUATE: The increase in the entropy of the outside air is greater than the entropy decrease of the air in the room. IDENTIFY and SET UP: Use Eq.(20.13) for an infinitesimal heat flow dQH from the hot reservoir and use that ΔSin =

20.55.

expression with Eq.(20.19) to relate ΔS H , the entropy change of the hot reservoir, to QC (a) EXECUTE:

Consider an infinitesimal heat flow dQH that occurs when the temperature of the hot reservoir is T ′:

dQC = −(TC / T ′)dQH dQH T′ dQH QC = TC ∫ = TC ΔS H T′ (b) The 1.00 kg of water (the high-temperature reservoir) goes from 373 K to 273 K. QH = mcΔT = (1.00 kg)(4190 J/kg ⋅ K)(100 K) = 4.19 × 105 J

∫ dQ

C

= −TC ∫

ΔS H = mc ln(T2 / T1 ) = (1.00 kg)(4190 J/kg ⋅ K)ln(273/ 373) = −1308 J/K

The result of part (a) gives QC = (273 K)(1308 J/K) = 3.57 × 105 J QC comes out of the engine, so QC = −3.57 × 105 J

Then W = QC + QH = −3.57 × 105 J + 4.19 × 105 J = 6.2 × 104 J. (c) 2.00 kg of water goes from 323 K to 273 K QH = − mcΔT = (2.00 kg)(4190 J/kg ⋅ K)(50 K) = 4.19 × 105 J ΔS H = mc ln(T2 / T1 ) = (2.00 kg)(4190 J/kg ⋅ K)ln(272 / 323) = −1.41 × 103 J/K QC = −TC ΔS H = −3.85 × 105 J

20.56.

W = QC + QH = 3.4 × 104 J (d) EVALUATE: More work can be extracted from 1.00 kg of water at 373 K than from 2.00 kg of water at 323 K even though the energy that comes out of the water as it cools to 273 K is the same in both cases. The energy in the 323 K water is less available for conversion into mechanical work. IDENTIFY: The maximum power that can be extracted is the total kinetic energy K of the mass of air that passes over the turbine blades in time t. SET UP: The volume of a cylinder of diameter d and length L is (π d 2 / 4) L. Kinetic energy is 12 mv 2 . EXECUTE:

(a) The cylinder described contains a mass of air m = ρ(πd 2 4 )L, and so the total kinetic energy is

K = ρ(π 8 )d 2 Lv 2 . This mass of air will pass by the turbine in a time t = L v, and so the maximum power is P=

K = ρ(π 8) d 2v3 . Numerically, the product ρair (π 8 ) ≈ 0.5 kg m3 = 0.5 W ⋅ s3 m5 . This completes the proof. t 1/ 3

20.57.

1/ 3

⎛ (3.2 × 106 W) (0.25) ⎞ ⎛P e⎞ = 14 m s = 50 km h. (b) v = ⎜ 2 ⎟ = ⎜ 3 5 2 ⎟ ⎝ kd ⎠ ⎝ (0.5 W ⋅ s m )(97 m) ⎠ (c) Wind speeds tend to be higher in mountain passes. EVALUATE: The maximum power is proportional to v 3 , so increases rapidly with increase in wind speed. T Q IDENTIFY: For a Carnot device, C = − C . W = QH + QC . TH QH SET UP:

QC = 1000 J. 10.0°C = 283.1 K. 35.0°C = 308.1 K. 15.0°C = 288.1 K.

⎛T ⎞ ⎛ 308.1 K ⎞ 3 3 (a) QH = − ⎜ H ⎟ QC = − ⎜ ⎟ (1000 J) = −1.088 × 10 J. W = 1000 J + (−1.088 × 10 J) = −88 J. T ⎝ 283.1 K ⎠ ⎝ C⎠ ⎛ 288.1 K ⎞ 3 3 (b) Now QH = − ⎜ ⎟ (1000 J) = −1.018 × 10 J. W = 1000 J + (−1.018 × 10 J) = −18 J. ⎝ 283.1 K ⎠ (c) The pV-diagrams for the two Carnot cycles are sketched in Figure 20.57. EXECUTE:

20-20

Chapter 20

EVALUATE:

More work must be done to move the heat energy through a greater temperature difference.

Figure 20.57 20.58.

IDENTIFY and SET UP: First use the methods of Chapter 17 to calculate the final temperature T of the system. EXECUTE: 0.600 kg of water (cools from 45.0°C to T ) Q = mcΔT = (0.600 kg)(4190 J/kg ⋅ K)(T − 45.0°C) = (2514 J/K)T − 1.1313 × 105 J 0.0500 kg of ice (warms to 0°C, melts, and water warms from 0°C to T ) Q = mcice (0°C − ( −15.0°C)) + mLf + mcwater (T − 0°C) Q = 0.0500 kg ⎡⎣ (2100 J/kg ⋅ K)(15.0°C) + 334 × 103 J/kg + (4190 J/kg ⋅ K)(T − 0°C) ⎤⎦

Q = 1575 J + 1.67 × 104 J + (209.5 J/K)T = 1.828 × 10 4 J + (209.5 J/K)T Qsystem = 0 gives (2514 J/K)T − 1.1313 × 105 J + 1.828 × 104 J + (209.5 J/K)T = 0

(2.724 × 103 J/K)T = 9.485 × 10 4 J T = (9.485 × 104 J)/(2.724 × 103 J/K) = 34.83°C = 308 K EVALUATE: The final temperature must lie between –15.0°C and 45.0°C. A final temperature of 34.8°C is consistent with only liquid water being present at equilibrium. IDENTIFY and SET UP: Now we can calculate the entropy changes. Use ΔS = Q / T for phase changes and the method of Example 20.6 to calculate ΔS for temperature changes. EXECUTE: ice: The process takes ice at –15°C and produces water at 34.8°C. Calculate ΔS for a reversible process between these two states, in which heat is added very slowly. ΔS is path independent, so ΔS for a reversible process is the same as ΔS for the actual (irreversible) process as long as the initial and final states are the same. 2

ΔS = ∫ dQ / T , where T must be in kelvins 1

T2

For a temperature change dQ = mc dT so ΔS = ∫ (mc / T ) dT = mc ln(T2 / T1 ). T1

For a phase change, since it occurs at constant T, 2

ΔS = ∫ dQ / T = Q / T = ± mL / T . 1

Therefore ΔSice = mcice ln(273 K/258 K) + mLf / 273 K + mcwater ln(308 K/273 K)

20.59.

ΔSice = (0.0500 kg)[(2100 J/kg ⋅ K)ln(273 K/258 K) + (334 × 103 J/kg)/273 K + (4190 J/kg ⋅ K)ln(308 K/273 K)] ΔSice = 5.93 J/K + 61.17 J/K + 25.27 J/K = 92.4 J/K water: ΔS water = mc ln(T2 / T1 ) = (0.600 kg)(4190 J/kg ⋅ K)ln(308 K/318 K) = −80.3 J/K For the system, ΔS = ΔSice + ΔS water = 92.4 J/K − 80.3 J/K = +12 J/K EVALUATE: Our calculation gives ΔS > 0, as it must for an irreversible process of an isolated system. IDENTIFY: Apply Eq.(20.19). From the derivation of Eq. (20.6), Tb = r γ −1Ta and Tc = r γ −1Td . SET UP: For a constant volume process, dQ = nCV dT . EXECUTE: (a) For a constant-volume process for an ideal gas, where the temperature changes from T1 to T2, T2 dT ⎛T ⎞ ΔS = nCV ∫ = nCV ln ⎜ 2 ⎟ . The entropy changes are nCV ln( Tc Tb ) and nCV ln( Ta Td ). T1 T ⎝ T1 ⎠ (b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the other processes in the cycle are adiabatic, with Q = 0 and ΔS = 0. The total is then

⎛TT ⎞ TT Tc T r γ −1T T + nCV ln a = nCV ln ⎜ c a ⎟ . c a = γ −1 d a = 1. ln(1) = 0, so ΔS = 0. Tb Td ⎝ TbTd ⎠ TbTd r Td Ta (c) The system is not isolated, and a zero change of entropy for an irreversible system is certainly possible. EVALUATE: In an irreversible process for an isolated system, ΔS > 0. But the entropy change for some of the components of the system can be negative or zero. ΔS = nCV ln


20.60.

20-21

Q . For a reversible adiabatic process, Q = 0 and ΔS = 0. T The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes. SET UP: Use the results for the Stirling cycle from Problem 20.50. EXECUTE: (a) The graph is given in Figure 20.60. dQ , and so dQ = T dS , and Q = ∫ dQ = ∫ T dS , which is the area under the curve (b) For a reversible process, dS = T in the TS plane. (c) QH is the area under the rectangle bounded by the horizontal part of the rectangle at TH and the verticals. | QC |

IDENTIFY:

For a reversible isothermal process, ΔS =

is the area bounded by the horizontal part of the rectangle at TC and the verticals. The net work is then QH − | QC |, the area bounded by the rectangle that represents the process. The ratio of the areas is the ratio of the lengths of the W TH − TC = . vertical sides of the respective rectangles, and the efficiency is e = QH TH (d) As explained in Problem 20.50, the substance that mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a). As found in that problem, the ideal efficiency is the same as for a Carnot-cycle engine. EVALUATE: The derivation of eCarnot using the concept of entropy is much simpler than the derivation in Section 20.6, but yields the same result.

Figure 20.60 20.61.

IDENTIFY:

The temperatures of the ice-water mixture and of the boiling water are constant, so ΔS =

flow for the melting phase transition of the ice is Q = + mLf .

20.62.

Q . The heat T

SET UP: For water, Lf = 3.34 × 105 J/kg. EXECUTE: (a) The heat that goes into the ice-water mixture is Q = mLf = (0.160 kg)(3.34 × 105 J/kg) = 5.34 × 104 J. This is same amount of heat leaves the boiling water, so Q −5.34 × 10 4 J ΔS = = = −143 J/K. T 373 K Q 5.34 × 104 J = +196 J/K (b) ΔS = = 273 K T (c) For any segment of the rod, the net heat flow is zero, so ΔS = 0. (d) ΔStot = −143 J/K + 196 J/K = +53 J/K. EVALUATE: The heat flow is irreversible, since the system is isolated and the total entropy change is positive. IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change. SET UP: For water, c = 4190 J/kg ⋅ K. 20°C = 293.15 K, 65°C = 338.15 K and 120°C = 393.15 K. EXECUTE:

(a) ΔS = mcln( T2 T1 ) = (250 × 10−3 kg)(4190 J kg ⋅ K)ln(338.15 K 293.15 K) = 150 J K.

− mcΔT −(250 × 10−3 kg)(4190 J kg ⋅ K)(338.15 K − 293.15 K) = = −120 J/K. Telement 393.15 K (c) The sum of the result of parts (a) and (b) is ΔSsystem = 30 J/K.

(b) ΔS =

20-22

20.63.

20.64.

Chapter 20

EVALUATE: (d) Heating a liquid is not reversible. Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water. The net entropy change is positive. IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change. For the value of T for which ΔS is a maximum, d (ΔS ) / dT = 0. SET UP: The heat flow for a temperature change is Q = mcΔT EXECUTE: (a) As in Example 20.10, the entropy change of the first object is m1c1ln( T T 1 ) and that of the second is m2c2ln( T ′ T 2 ), and so the net entropy change is as given. Neglecting heat transfer to the surroundings, Q1 + Q2 = 0, m1c1 (T − T1 ) + m2c2 (T ′ − T2 ) = 0, which is the given expression. (b) Solving the energy-conservation relation for T ′ and substituting into the expression for ΔS gives ⎛ ⎛T ⎞ m c ⎛ T T ⎞⎞ ΔS = m1c1ln ⎜ ⎟ + m2c21n ⎜⎜1 − 1 1 ⎜ − 1 ⎟ ⎟⎟ . Differentiating with respect to T and setting the derivative equal to ⎝ T1 ⎠ ⎝ m2c2 ⎝ T2 T2 ⎠ ⎠ m c T + m2c2T2 m1c1 ( m2c2 )(m1c1 m2c2 )( −1 T2 ) . Using this value for T + . This may be solved for T = 1 1 1 0 gives 0 = m1c1 + m2c2 T ⎛ ⎛ T T1 ⎞ ⎞ ⎜⎜1 − (m1c1 m2c2 ) ⎜ − ⎟ ⎟⎟ ⎝ T2 T2 ⎠ ⎠ ⎝ m c T + m2c2T2 . Therefore, in the conservation of energy expression in part (a) and solving for T ′ gives T ′ = 1 1 1 m1c1 + m2c2 T = T ′ when ΔS is a maximum. EVALUATE: (c) The final state of the system will be that for which no further entropy change is possible. If T < T ′, it is possible for the temperatures to approach each other while increasing the total entropy, but when T = T ′, no further spontaneous heat exchange is possible. IDENTIFY: Calculate QC and QH in terms of p and V at each point. Use the ideal gas law and the pressure-volume Q relation for adiabatic processes for an ideal gas. e = 1 − C . QH SET UP: For an ideal gas, C p = CV + R, and taking air to be diatomic, C p = 72 R, CV = 52 R and γ = 75 . EXECUTE: Referring to Figure 20.7 in the textbook, QH = n 72 R(Tc − Tb ) = 72 ( pcVc − pbVb ). Similarly, QC = n 52 R ( paVa − pdVd ). What needs to be done is to find the relations between the product of the pressure and the

volume at the four points. For an ideal gas,

⎛T ⎞ pcVc pbVb = so pcVc = paVa ⎜ c ⎟ . For a compression ratio r, and given Tc Tb ⎝ Ta ⎠ γ −1

γ −1

⎛V ⎞ ⎛V ⎞ that for the Diesel cycle the process ab is adiabatic, pbVb = paVa ⎜ a ⎟ = paVa r γ −1. Similarly, pdVd = pcVc ⎜ c ⎟ . ⎝ Vb ⎠ ⎝ Va ⎠ Note that the last result uses the fact that process da is isochoric, and Vd = Va ; also, pc = pb (process bc is isobaric),

⎛T ⎞ and so Vc = Vb ⎜ c ⎟ . Then, ⎝ Ta ⎠ −γ

Vc Tc Vb Tb Ta Va Tc ⎛ TaVaγ −1 ⎞⎛ Va ⎞ Tc γ = ⋅ = ⋅ ⋅ = ⋅⎜ ⎟ = r γ −1 ⎟⎜ Va Tb Va Ta Tb Vb Ta ⎝ TbVb ⎠⎝ Vb ⎠ Ta γ

⎛T ⎞ 2 Combining the above results, pdVd = paVa ⎜ c ⎟ r γ − γ . Substitution of the above results into Eq. (20.4) gives ⎝ Ta ⎠ ⎡⎛ T ⎞ γ ⎤ 2 ⎢ ⎜ c ⎟ r γ −γ − 1 ⎥ 5⎢ T ⎥ . e =1− ⎢⎝ a ⎠ 7 ⎛ Tc ⎞ γ −1 ⎥ ⎢ ⎜ ⎟−r ⎥ ⎢ ⎝ Ta ⎠ ⎥ ⎣ ⎦ −0.56 T 1 ⎡ (5.002)r − 1⎤ , where c = 3.167 and γ = 1.40 have been used. Substitution of r = 21.0 yields (b) e = 1 − ⎢ 0.40 ⎥ Ta 1.4 ⎣ (3.167) − r ⎦ e = 0.708 = 70.8%. EVALUATE: The efficiency for an Otto cycle with r = 21.0 and γ = 1.40 is e = 1 − r1−γ = 1 − (21.0) −0.40 = 70.4%. This is very close to the value for the Diesel cycle.

ELECTRIC CHARGE AND ELECTRIC FIELD

21.1.

21

(a) IDENTIFY and SET UP: Use the charge of one electron ( −1.602 × 10 −19 C) to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce net charge q is q −3.20 × 10−9 C = = 2.00 × 1010 electrons. −e −1.602 × 10−19 C/electron (b) IDENTIFY and SET UP: Use the atomic mass of lead to find the number of lead atoms in 8.00 × 10 −3 kg of lead. From this and the total number of excess electrons, find the number of excess electrons per lead atom. EXECUTE: The atomic mass of lead is 207 × 10−3 kg/mol, so the number of moles in 8.00 × 10 −3 kg is

mtot 8.00 × 10−3 kg = = 0.03865 mol. N A (Avogadro’s number) is the number of atoms in 1 mole, so the M 207 × 10−3 kg/mol number of lead atoms is N = nN A = (0.03865 mol)(6.022 × 1023 atoms/mol) = 2.328 × 1022 atoms. The number of n=

2.00 × 1010 electrons = 8.59 × 10−13. 2.328 × 1022 atoms EVALUATE: Even this small net charge corresponds to a large number of excess electrons. But the number of atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small. IDENTIFY: The charge that flows is the rate of charge flow times the duration of the time interval. SET UP: The charge of one electron has magnitude e = 1.60 × 10−19 C. EXECUTE: The rate of charge flow is 20,000 C/s and t = 100 μs = 1.00 × 10−4 s. Q = 1.25 × 1019. Q = (20,000 C/s)(1.00 × 10 −4 s) = 2.00 C. The number of electrons is ne = 1.60 × 10−19 C EVALUATE: This is a very large amount of charge and a large number of electrons. IDENTIFY: From your mass estimate the number of protons in your body. You have an equal number of electrons. SET UP: Assume a body mass of 70 kg. The charge of one electron is −1.60 × 10−19 C. EXECUTE: The mass is primarily protons and neutrons of m = 1.67 × 10−27 kg. The total number of protons and 70 kg neutrons is np and n = = 4.2 × 10 28. About one-half are protons, so np = 2.1 × 1028 = ne . The number of 1.67 × 10−27 kg electrons is about 2.1× 1028. The total charge of these electrons is Q = ( −1.60 × 10 −19 C/electron)(2.10 × 1028 electrons) = −3.35 × 109 C. EVALUATE: This is a huge amount of negative charge. But your body contains an equal number of protons and your net charge is zero. If you carry a net charge, the number of excess or missing electrons is a very small fraction of the total number of electrons in your body. IDENTIFY: Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms. Each atom has 79 protons and an equal number of electrons. SET UP: N A = 6.02 × 1023 atoms/mol . A proton has charge +e. excess electrons per lead atom is

21.2.

21.3.

21.4.

The mass of gold is 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms ⎛ 17.7 g ⎞ 22 is N A n = (6.02 × 1023 atoms/mol) ⎜ ⎟ = 5.41 × 10 atoms . The number of protons is ⎝ 197 g mol ⎠ np = (79 protons/atom)(5.41× 1022 atoms) = 4.27 × 1024 protons . Q = (np )(1.60 × 10−19 C/proton) = 6.83 × 105 C . EXECUTE:

(b) The number of electrons is ne = np = 4.27 × 1024. EVALUATE: charge.

The total amount of positive charge in the ring is very large, but there is an equal amount of negative

21-1

21-2

21.5.

Chapter 21

Apply F =

IDENTIFY: SET UP:

r2

and solve for r.

F = 650 N .

k q1q2 (8.99 × 109 N ⋅ m 2 /C2 )(1.0 C) 2 = = 3.7 × 103 m = 3.7 km 650 N F EVALUATE: Charged objects typically have net charges much less than 1 C. IDENTIFY: Apply Coulomb's law and calculate the net charge q on each sphere. r=

EXECUTE:

21.6.

k q1q2

SET UP:

The magnitude of the charge of an electron is e = 1.60 × 10−19 C .

1 q2 . This gives q = 4π P0 Fr 2 = 4π P0 (4.57 × 10−21 N)(0.200 m) 2 = 1.43 × 10 −16 C. And 4π P0 r 2 therefore, the total number of electrons required is n = q /e = (1.43 ×10−16 C)/(1.60 ×10−19 C/electron) = 890 electrons. F=

EXECUTE:

21.7.

EVALUATE: Each sphere has 890 excess electrons and each sphere has a net negative charge. The two like charges repel. IDENTIFY: Apply Coulomb’s law. SET UP: Consider the force on one of the spheres. (a) EXECUTE: q1 = q2 = q

F=

1 q1q2 q2 F 0.220 N = so q = r = 0.150 m = 7.42 × 10−7 C (on each) 4π P0 r 2 4π P0 r 2 (1/4π P0 ) 8.988 × 109 N ⋅ m 2 /C 2

(b) q2 = 4q1

F=

21.8.

1 q1q2 4q12 F F = so q1 = r = 1r = 1 (7.42 × 10−7 C) = 3.71 × 10 −7 C. 2 4π P0 r 4π P0 r 2 4(1/4π P0 ) 2 (1/4π P0 ) 2

And then q2 = 4q1 = 1.48 × 10 −6 C. EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the sphere have equal charges or not. IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in one sphere. Each atom has 13 electrons. Apply Coulomb's law and calculate the magnitude of charge q on each sphere. SET UP:

N A = 6.02 × 10 23 atoms/mol . q = ne′e , where ne′ is the number of electrons removed from one sphere

and added to the other. EXECUTE: (a) The total number of electrons on each sphere equals the number of protons. ⎛ ⎞ 0.0250 kg 24 ne = np = (13)( N A ) ⎜ ⎟ = 7.25 × 10 electrons . ⎝ 0.026982 kg mol ⎠ (b) For a force of 1.00 × 104 N to act between the spheres, F = 1.00 × 104 N =

1 q2 . This gives 4π P0 r 2

q = 4π P0 (1.00 × 10 4 N)(0.0800 m) 2 = 8.43 × 10 −4 C . The number of electrons removed from one sphere and

added to the other is ne′ = q / e = 5.27 ×1015 electrons.

21.9.

(c) ne′ / ne = 7.27 × 10 −10 . EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of electrons in the object is very small . qq IDENTIFY: Apply F = ma , with F = k 1 2 2 . r SET UP: a = 25.0 g = 245 m/s 2 . An electron has charge −e = −1.60 × 10−19 C. EXECUTE:

F = ma = (8.55 × 10−3 kg)(245 m/s 2 ) = 2.09 N . The spheres have equal charges q, so F = k

q2 and r2

q 2.29 × 10−6 C F 2.09 N −6 2.29 10 C N = (0.150 m) = × . = = = 1.43 × 1013 electrons . The 8.99 × 109 N ⋅ m 2 /C 2 e 1.60 × 10−19 C k charges on the spheres have the same sign so the electrical force is repulsive and the spheres accelerate away from each other. q =r

Electric Charge and Electric Field

21.10.

21-3

EVALUATE: As the spheres move apart the repulsive force they exert on each other decreases and their acceleration decreases. (a) IDENTIFY: The electrical attraction of the proton gives the electron an acceleration equal to the acceleration due to gravity on earth. SET UP: Coulomb’s law gives the force and Newton’s second law gives the acceleration this force produces.

ma =

1 q1q2 e2 . and r = 4π P0 r 2 4π P0 ma

( 9.00 ×10 N ⋅ m /C )(1.60 ×10 C ) ( 9.11× 10 kg )( 9.80 m/s ) 9

EXECUTE:

r=

2

−31

2

−19

2

2

= 5.08 m

EVALUATE: The electron needs to be about 5 m from a single proton to have the same acceleration as it receives from the gravity of the entire earth. (b) IDENTIFY: The force on the electron comes from the electrical attraction of all the protons in the earth. SET UP: First find the number n of protons in the earth, and then find the acceleration of the electron using Newton’s second law, as in part (a).

n = mE/mp = (5.97 × 1024 kg)/(1.67 × 10−27 kg) = 3.57 × 1051 protons. 1 qp qe 1 ne2 4π P0 RE2 4π P0 a = F/m = = . me me RE2

21.11.

EXECUTE: a = (9.00 × 109 N ⋅ m2/C2)(3.57 × 1051)(1.60 × 10−19 C)2/[(9.11 × 10−31 kg)(6.38 × 106 m)2] = 2.22 × 1040 m/s2. One can ignore the gravitation force since it produces an acceleration of only 9.8 m/s2 and hence is much much less than the electrical force. EVALUATE: With the electrical force, the acceleration of the electron would nearly 1040 times greater than with gravity, which shows how strong the electrical force is. IDENTIFY: In a space satellite, the only force accelerating the free proton is the electrical repulsion of the other proton. SET UP: Coulomb’s law gives the force, and Newton’s second law gives the acceleration: a = F/m = (1/ 4π P0 ) (e2/r2)/m. EXECUTE: (a) a = (9.00 × 109 N ⋅ m2/C2)(1.60 × 10-19 C)2/[(0.00250 m)2(1.67 × 10-27 kg)] = 2.21 × 104 m/s2. (b) The graphs are sketched in Figure 21.11. EVALUATE: The electrical force of a single stationary proton gives the moving proton an initial acceleration about 20,000 times as great as the acceleration caused by the gravity of the entire earth. As the protons move farther apart, the electrical force gets weaker, so the acceleration decreases. Since the protons continue to repel, the velocity keeps increasing, but at a decreasing rate.

Figure 21.11 21.12.

IDENTIFY: Apply Coulomb’s law. SET UP: Like charges repel and unlike charges attract. EXECUTE:

(a) F =

−6 1 q1q2 1 (0.550 × 10 C) q2 . This gives 0.200 N = and q2 = +3.64 × 10−6 C . The 2 4π P0 r 4π P0 (0.30 m) 2

force is attractive and q1 < 0 , so q2 = +3.64 × 10−6 C . (b) F = 0.200 N. The force is attractive, so is downward. EVALUATE:

The forces between the two charges obey Newton's third law.

21-4

Chapter 21

21.13.

IDENTIFY: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions. SET UP: Like charges repel and unlike charges attract. qq EXECUTE: The force F2 that q2 exerts on q3 has magnitude F2 = k 2 2 3 and is in the +x direction. F1 must be in r2

the − x direction, so q1 must be positive. F1 = F2 gives k 2

21.14.

If q1 is −2.0 μ C and q2 is +2.0 μ C , then the net force is in the +y-direction.

Apply Coulomb’s law and find the vector sum of the two forces on q1 .

IDENTIFY:

Like charges repel and unlike charges attract, so F2 and F3 are both in the +x-direction.

SET UP:

F2 = k

EXECUTE:

21.16.

2

⎛r ⎞ ⎛ 2.00 cm ⎞ q1 = q2 ⎜ 1 ⎟ = (3.00 nC) ⎜ ⎟ = 0.750 nC . ⎝ 4.00 cm ⎠ ⎝ r2 ⎠ EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 . IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q. SET UP: The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that q2 exerts is attractive. EXECUTE: The x-components cancel. We only need the y-components, and each charge contributes equally. 1 (2.0 × 10−6 C) (4.0 × 10−6 C) sin α = −0.173 N (since sinα = 0.600). Therefore, the total force is F1 y = F2 y = − 4π P0 (0.500 m) 2 2 F = 0.35 N, in the − y -direction . EVALUATE:

21.15.

q1 q3 q q =k 2 2 3 . 2 r1 r2

q1q2 qq = 6.749 × 10−5 N, F3 = k 1 2 3 = 1.124 × 10−4 N . F = F2 + F3 = 1.8 × 10−4 N . r122 r13

F = 1.8 × 10−4 N and is in the +x-direction. EVALUATE: Comparing our results to those in Example 21.3, we see that F1 on 3 = − F3 on 1 , as required by Newton’s third law. IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q2 .

F2 on 1 is in the +y-direction.

SET UP: EXECUTE:

F2on 1 =

(9.0 × 109 N ⋅ m 2 C2 ) (2.0 × 10−6 C) (2.0 × 10−6 C)

( F2 on 1 ) y = +0.100 N .

(F

)

Q on 1 y

( 0.60 m )

2

= 0.100 N . ( F2 on 1 ) x = 0 and

FQ on 1 is equal and opposite to F1 on Q (Example 21.4), so ( FQ on 1 ) = −0.23N and x

= 0.17 N . Fx = ( F2 on 1 ) x + ( FQ on 1 ) = −0.23 N . Fy = ( F2 on 1 ) y + ( FQ on 1 ) = 0.100 N + 0.17 N = 0.27 N . x

y

0.23 = 40° , so F is 0.27 40° counterclockwise from the +y axis, or 130° counterclockwise from the +x axis. EVALUATE: Both forces on q1 are repulsive and are directed away from the charges that exert them. The magnitude of the total force is F =

21.17.


( 0.23 N )

2

+ ( 0.27 N ) = 0.35 N. tan −1 2

Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1. Add these

forces as vectors to get the net force. The target variable is the x-coordinate of q3 . EXECUTE:

F2 is in the x-direction.

qq F2 = k 1 2 2 = 3.37 N, so F2 x = +3.37 N r12 Fx = F2 x + F3 x and Fx = −7.00 N F3 x = Fx − F2 x = −7.00 N − 3.37 N = −10.37 N

For F3 x to be negative, q3 must be on the − x -axis.

F3 = k

q1q3 k q1q3 , so x = = 0.144 m, so x = −0.144 m 2 x F3

EVALUATE:

q2 attracts q1 in the + x -direction so q3 must attract q1 in the − x -direction, and q3 is at negative x.


21.18.

21-5

IDENTIFY: Apply Coulomb’s law. SET UP: Like charges repel and unlike charges attract. Let F21 be the force that q2 exerts on q1 and let F31 be

the force that q3 exerts on q1 . EXECUTE: The charge q3 must be to the right of the origin; otherwise both q2 and q3 would exert forces in the + x direction. Calculating the two forces: 1 q1q2 (9 × 109 N ⋅ m 2 C2 )(3.00 × 10−6 C)(5.00 × 10−6 C) F21 = = = 3.375 N , in the +x direction. 4π P0 r122 (0.200 m) 2

F31 =

(9 × 109 N ⋅ m 2 C2 ) (3.00 × 10−6 C) (8.00 × 10−6 C) 0.216 N ⋅ m 2 = , in the − x direction. r132 r132

We need Fx = F21 − F31 = −7.00 N , so 3.375 N −

0.216 N ⋅ m 2 0.216 N ⋅ m 2 = −7.00 N . r13 = = 0.144 m . q3 2 r13 3.375 N + 7.00 N

is at x = 0.144 m . EVALUATE: F31 = 10.4 N. F31 is larger than F21 , because q3 is larger than q2 and also because r13 is less than r12 . 21.19.

IDENTIFY: Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge. Add these forces as vectors. SET UP: The three charges are placed as shown in Figure 21.19a.


Like charges repel and unlike attract, so the free-body diagram for q3 is as shown in Figure 21.19b. F1 =

1 q1q3 4π P0 r132

F2 =

1 q2 q3 4π P0 r232

Figure 21.19b F1 = (8.988 × 109 N ⋅ m 2 /C2 )

(1.50 × 10−9 C)(5.00 × 10−9 C) = 1.685 × 10−6 N (0.200 m) 2

F2 = (8.988 × 109 N ⋅ m 2 /C2 )

(3.20 × 10−9 C)(5.00 × 10−9 C) = 8.988 × 10−7 N (0.400 m) 2

The resultant force is R = F1 + F2 . Rx = 0. Ry = F1 + F2 = 1.685 × 10−6 N +8.988 × 10−7 N = 2.58 × 10−6 N. The resultant force has magnitude 2.58 × 10−6 N and is in the –y-direction. EVALUATE: The force between q1 and q3 is attractive and the force between q2 and q3 is replusive. 21.20.

IDENTIFY: SET UP:

Apply F = k

qq′

to each pair of charges. The net force is the vector sum of the forces due to q1 and q2 . r2 Like charges repel and unlike charges attract. The charges and their forces on q3 are shown in Figure 21.20.

21-6

Chapter 21

F1 = k

EXECUTE: F2 = k

q2 q3 2 2

r

q1q3 (4.00 × 10−9 C)(0.600 × 10−9 C) = (8.99 × 109 N ⋅ m 2 /C 2 ) = 5.394 × 10−7 N . 2 r1 (0.200 m) 2

= (8.99 × 109 N ⋅ m 2 /C 2 )

(5.00 × 10−9 C)(0.600 × 10−9 C) = 2.997 × 10 −7 N . (0.300 m) 2

Fx = F1x + F2 x = + F1 − F2 = 2.40 × 10−7 N . The net force has magnitude 2.40 × 10−7 N and is in the + x direction. EVALUATE: Each force is attractive, but the forces are in opposite directions because of the placement of the charges. Since the forces are in opposite directions, the net force is obtained by subtracting their magnitudes.

Figure 21.20 21.21.

Apply Coulomb’s law to calculate each force on −Q .

IDENTIFY: SET UP:

Let F1 be the force exerted by the charge at y = a and let F2 be the force exerted by the charge at y = − a. (a) The two forces on −Q are shown in Figure 21.21a. sin θ =

EXECUTE:

a and r = ( a 2 + x 2 )1/ 2 is the ( a 2 + x 2 )1/ 2

distance between q and −Q and between − q and −Q . (b) Fx = F1x + F2 x = 0 . Fy = F1 y + F2 y = 2 (c) At x = 0, Fy =

1 qQ 1 2qQa sin θ = . 2 2 2 4π P0 (a + x ) 4π P0 (a + x 2 )3 2

1 2qQ , in the +y direction. 4π P0 a 2

(d) The graph of Fy versus x is given in Figure 21.21b. EVALUATE:

Fx = 0 for all values of x and Fy > 0 for all x.

Figure 21.21 21.22.

IDENTIFY: SET UP:

Apply Coulomb’s law to calculate each force on −Q . Let F1 be the force exerted by the charge at y = a and let F2 be the force exerted by the charge at

y = − a . The distance between each charge q and Q is r = ( a 2 + x 2 )

1/ 2

EXECUTE:

. cosθ =

x

(a

2

+ x2 )

1/ 2

.

(a) The two forces on −Q are shown in Figure 21.22a.

(b) When x > 0 , F1x and F2 x are negative. Fx = F1x + F2 x = −2

1 qQ 1 −2qQx cosθ = . When 4π P0 ( a 2 + x 2 ) 4π P0 (a 2 + x 2 )3 / 2

x < 0 , F1x and F2 x are positive and the same expression for Fx applies. Fy = F1 y + F2 y = 0 . (c) At x = 0 , Fx = 0 . (d) The graph of Fx versus x is sketched in Figure 21.22b.


EVALUATE:

21-7

The direction of the net force on −Q is always toward the origin.

Figure 21.22 21.23.

IDENTIFY: Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other three and then add these forces as vectors. (a) SET UP: The charges are placed as shown in Figure 21.23a.

q1 = q2 = q3 = q4 = q

Figure 21.23a

Consider forces on q4 . The free-body diagram is given in Figure 21.23b. Take the y-axis to be parallel to the

diagonal between q2 and q4 and let + y be in the direction away from q2 . Then F2 is in the + y -direction. EXECUTE:

F2 =

F3 = F1 =

1 q2 4π P0 L2

1 q2 4π P0 2 L2

F1x = − F1 sin 45° = − F1/ 2

F1 y = + F1 cos 45° = + F1/ 2 F3 x = + F3 sin 45° = + F3 / 2

F3 y = + F3 cos 45° = + F3/ 2 F2 x = 0, F2 y = F2

Figure 21.23b (b) Rx = F1x + F2 x + F3 x = 0

Ry = F1 y + F2 y + F3 y = (2/ 2) R=

1 q2 1 q2 q2 (1 + 2 2) + = 4π P0 L2 4π P0 2 L2 8π P0 L2

q2 (1 + 2 2). Same for all four charges. 8π P0 L2

21-8

21.24.

Chapter 21

EVALUATE: In general the resultant force on one of the charges is directed away from the opposite corner. The forces are all repulsive since the charges are all the same. By symmetry the net force on one charge can have no component perpendicular to the diagonal of the square. k qq′ IDENTIFY: Apply F = 2 to find the force of each charge on + q . The net force is the vector sum of the r individual forces. SET UP: Let q1 = +2.50 μ C and q2 = −3.50 μ C . The charge + q must be to the left of q1 or to the right of q2 in order for the two forces to be in opposite directions. But for the two forces to have equal magnitudes, + q must be closer to the charge q1 , since this charge has the smaller magnitude. Therefore, the two forces can combine to give zero net force only in the region to the left of q1 . Let + q be a distance d to the left of q1 , so it is a distance d + 0.600 m from q2 . EXECUTE:

F1 = F2 gives

kq q1 d

2

=

kq q2 ( d + 0.600 m)

2

. d =±

q1 q2

(d + 0.600 m) = ± (0.8452)( d + 0.600 m) . d must

(0.8452)(0.600 m) = 3.27 m . The net force would be zero when + q is at x = −3.27 m . 1 − 0.8452 When + q is at x = −3.27 m , F1 is in the − x direction and F2 is in the +x direction.

be positive, so d = EVALUATE: 21.25.

IDENTIFY:

F = q E . Since the field is uniform, the force and acceleration are constant and we can use a constant

acceleration equation to find the final speed. SET UP: A proton has charge +e and mass 1.67 × 10−27 kg . EXECUTE: (b) a =

21.26.

F 4.40 × 10−16 N = = 2.63 × 1011 m/s 2 m 1.67 × 10−27 kg

(c) vx = v0 x + axt gives v = (2.63 × 1011 m/s 2 )(1.00 × 10−6 s) = 2.63 × 105 m/s EVALUATE: The acceleration is very large and the gravity force on the proton can be ignored. q IDENTIFY: For a point charge, E = k 2 . r SET UP: E is toward a negative charge and away from a positive charge. EXECUTE: (a) The field is toward the negative charge so is downward. 3.00 × 10−9 C E = (8.99 × 109 N ⋅ m 2 /C 2 ) = 432 N/C . (0.250 m) 2

(8.99 × 109 N ⋅ m 2 /C 2 )(3.00 × 10−9 C) = 1.50 m E 12.0 N/C EVALUATE: At different points the electric field has different directions, but it is always directed toward the negative point charge. IDENTIFY: The acceleration that stops the charge is produced by the force that the electric field exerts on it. Since the field and the acceleration are constant, we can use the standard kinematics formulas to find acceleration and time. (a) SET UP: First use kinematics to find the proton’s acceleration. vx = 0 when it stops. Then find the electric field needed to cause this acceleration using the fact that F = qE. EXECUTE: vx2 = v02x + 2ax ( x − x0 ) . 0 = (4.50 × 106 m/s)2 + 2a(0.0320 m) and a = 3.16 × 1014 m/s2. Now find the (b) r =

21.27.

(a) F = (1.60 × 10−19 C)(2.75 × 103 N/C) = 4.40 × 10−16 N

kq

=

electric field, with q = e. eE = ma and E = ma/e = (1.67 × 10−27 kg)(3.16 × 1014 m/s2)/(1.60 × 10−19 C) = 3.30 × 106 N/C, to the left. (b) SET UP: Kinematics gives v = v0 + at, and v = 0 when the electron stops, so t = v0/a. EXECUTE: t = v0/a = (4.50 × 106 m/s)/(3.16 × 1014 m/s2) = 1.42 × 10−8 s = 14.2 ns (c) SET UP: In part (a) we saw that the electric field is proportional to m, so we can use the ratio of the electric fields. Ee / Ep = me / mp and Ee = ( me / mp ) Ep . EXECUTE: Ee = [(9.11 × 10−31 kg)/(1.67 × 10−27 kg)](3.30 × 106 N/C) = 1.80 × 103 N/C, to the right EVALUATE: Even a modest electric field, such as the ones in this situation, can produce enormous accelerations for electrons and protons.


21.28.

21-9

IDENTIFY: Use constant acceleration equations to calculate the upward acceleration a and then apply F = qE to calculate the electric field. SET UP: Let +y be upward. An electron has charge q = −e . (a) v0 y = 0 and a y = a , so y − y0 = v0 yt + 12 a yt 2 gives y − y0 = 12 at 2 . Then

EXECUTE:

2

2( y − y0 ) 2(4.50 m) F ma (9.11 × 10−31 kg) (1.00 × 1012 m s ) 2 12 = = 1.00 × 10 m s . E = = = = 5.69 N C q q 1.60 × 10−19 C t2 (3.00 × 10−6 s) 2 The force is up, so the electric field must be downward since the electron has negative charge. (b) The electron’s acceleration is ~ 1011 g , so gravity must be negligibly small compared to the electrical force. EVALUATE: Since the electric field is uniform, the force it exerts is constant and the electron moves with constant acceleration. (a) IDENTIFY: Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If the particle is to remain stationary the net force on it must be zero. SET UP: The free-body diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For the net force to be zero the force exerted by the electric field must be upward. The electric field is downward. Since the electric field and the electric force are in opposite directions the charge of the particle is negative. a=

21.29.

mg = q E

Figure 21.29

mg (1.45 × 10−3 kg)(9.80 m/s 2 ) = = 2.19 × 10−5 C and q = −21.9 μ C 650 N/C E (b) SET UP: The electrical force has magnitude FE = q E = eE. The weight of a proton is w = mg . FE = w so q =

EXECUTE:

eE = mg mg (1.673 ×10−27 kg)(9.80 m/s 2 ) = = 1.02 ×10−7 N/C. e 1.602 ×10−19 C This is a very small electric field. EVALUATE: In both cases q E = mg and E = ( m / q ) g . In part (b) the m / q ratio is much smaller (∼ 10−8 ) than E=

EXECUTE:

21.30.

in part (a) (∼ 10−2 ) so E is much smaller in (b). For subatomic particles gravity can usually be ignored compared to electric forces. 1 q IDENTIFY: Apply E = . 4π P0 r 2 SET UP: The iron nucleus has charge +26e. A proton has charge +e . 1 (26)(1.60 × 10−19 C) = 1.04 × 1011 N/C. EXECUTE: (a) E = 4π P0 (6.00 × 10−10 m) 2 1 (1.60 × 10−19 C) = 5.15 × 1011 N/C. 4π P0 (5.29 × 10−11 m) 2 EVALUATE: These electric fields are very large. In each case the charge is positive and the electric fields are directed away from the nucleus or proton. q IDENTIFY: For a point charge, E = k 2 . The net field is the vector sum of the fields produced by each charge. A r charge q in an electric field E experiences a force F = qE . SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q2 and 0.150 m from q1. Point B is 0.100 m from q1 and 0.350 m from q2. EXECUTE: (a) The electric fields due to the charges at point A are shown in Figure 21.31a. q 6.25 × 10−9 C = 2.50 × 103 N/C E1 = k 21 = (8.99 × 109 N ⋅ m 2 /C2 ) (0.150 m) 2 rA1 (b) Eproton =

21.31.

E2 = k

q2 2 A2

r

= (8.99 × 109 N ⋅ m 2 /C 2 )

12.5 × 10−9 C = 1.124 × 104 N/C (0.100 m) 2

21-10

Chapter 21

Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. E = E2 − E1 = 8.74 × 103 N/C, to the right. (b) The electric fields at points B are shown in Figure 21.31b. q 6.25 × 10−9 C = 5.619 × 103 N/C E1 = k 21 = (8.99 × 109 N ⋅ m 2 /C2 ) (0.100 m) 2 rB1 E2 = k

q2 rB22

= (8.99 × 109 N ⋅ m 2 /C 2 )

12.5 × 10−9 C = 9.17 × 102 N/C (0.350 m) 2

Since the fields are in the same direction, we add their magnitudes to find the net field. E = E1 + E2 = 6.54 ×103 N/C, to the right. (c) At A, E = 8.74 × 103 N/C , to the right. The force on a proton placed at this point would be F = qE = (1.60 × 10−19 C)(8.74 × 103 N/C) = 1.40 × 10−15 N, to the right. EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same direction as the field.

Figure 21.31 21.32.

IDENTIFY: The electric force is F = qE . SET UP: The gravity force (weight) has magnitude w = mg and is downward. EXECUTE: (a) To balance the weight the electric force must be upward. The electric field is downward, so for an upward force the charge q of the person must be negative. w = F gives mg = q E and

mg (60 kg)(9.80 m/s 2 ) = = 3.9 C . 150 N/C E qq′ (3.9 C) 2 = 1.4 × 107 N . The repulsive force is immense and this is not a (b) F = k 2 = (8.99 × 109 N ⋅ m 2 /C 2 ) r (100 m) 2 feasible means of flight. EVALUATE: The net charge of charged objects is typically much less than 1 C. IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion; use constant acceleration equations for the horizontal and vertical components of the motion. (a) SET UP: The motion is sketched in Figure 21.33a. q =

21.33.

For an electron q = −e. Figure 21.33a

F = qE and q negative gives that F and E are in opposite directions, so F is upward. The free-body diagram for the electron is given in Figure 21.33b. EXECUTE:

∑F

y

= ma y

eE = ma Figure 21.33b

Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x − x0 = 2.00 cm when y − y0 = +0.500 cm. x-component v0 x = v0 = 1.60 × 106 m/s, ax = 0, x − x0 = 0.0200 m, t = ?

x − x0 = v0 xt + 12 axt 2


21-11

x − x0 0.0200 m = = 1.25 × 10−8 s v0 x 1.60 × 106 m/s In this same time t the electron travels 0.0050 m vertically: y-component t = 1.25 × 10−8 s, v0 y = 0, y − y0 = +0.0050 m, a y = ? t=

y − y0 = v0 yt + 12 a yt 2 ay =

2( y − y0 ) 2(0.0050 m) = = 6.40 × 1013 m/s 2 t2 (1.25 × 10−8 s) 2

(This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration is upward rather than downward.) This acceleration must be produced by the electric-field force: eE = ma E=

ma (9.109 × 10−31 kg)(6.40 × 1013 m/s 2 ) = = 364 N/C e 1.602 × 10−19 C

Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem. eE (1.602 × 10−19 C)(364 N/C) = = 3.49 × 1010 m/s 2 (b) a = mp 1.673 × 10−27 kg This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time t = 1.25 × 10−8 s to travel horizontally the length of the plates. The force on the proton is downward (in the same direction as E , since q is positive), so the acceleration is downward and a y = −3.49 × 1010 m/s 2 . y − y0 = v0 y t + 12 a y t 2 = 12 ( −3.49 × 1010 m/s 2 )(1.25 × 10−8 s) 2 = −2.73 × 10−6 m. The displacement is 2.73 × 10−6 m,

21.34.

downward. (c) EVALUATE: The displacements are in opposite directions because the electron has negative charge and the proton has positive charge. The electron and proton have the same magnitude of charge, so the force the electric field exerts has the same magnitude for each charge. But the proton has a mass larger by a factor of 1836 so its acceleration and its vertical displacement are smaller by this factor. IDENTIFY: Apply Eq.(21.7) to calculate the electric field due to each charge and add the two field vectors to find the resultant field. SET UP: For q1 , rˆ = ˆj . For q2 , rˆ = cosθ iˆ + sin θ ˆj , where θ is the angle between E2 and the +x-axis. (a) E1 =

EXECUTE:

E2 =

q1 ˆ (9.0 × 109 N ⋅ m 2 /C2 )(−5.00 × 10−9 C) = ( −2.813 × 104 N/C) ˆj . j= 2 4π P0 r12 ( 0.0400 m )

q2 (9.0 × 109 N ⋅ m 2 /C2 )(3.00 × 10−9 C) = = 1.080 × 104 N/C . The angle of E2 , measured from the 2 2 2 4π P0 r2 0.0300 m + (0.0400 m) ( )

)

(

x-axis, is 180° − tan −1 4.00 cm = 126.9° Thus 3.00 cm 4 ˆ E2 = (1.080 × 10 N/C)(i cos126.9° + ˆj sin126.9°) = ( − 6.485 × 103 N/C) iˆ + (8.64 × 103 N/C) ˆj (b) The resultant field is E + E = (−6.485 × 103 N/C) iˆ + ( −2.813 × 104 N/C + 8.64 × 103 N/C) ˆj . 1

2

E1 + E 2 = (−6.485 × 103 N/C) iˆ − (1.95 × 104 N/C) ˆj . 21.35.

EVALUATE: E1 is toward q1 since q1 is negative. E2 is directed away from q2 , since q2 is positive. IDENTIFY: Apply constant acceleration equations to the motion of the electron. SET UP: Let +x be to the right and let + y be downward. The electron moves 2.00 cm to the right and 0.50 cm downward. EXECUTE: Use the horizontal motion to find the time when the electron emerges from the field. x − x0 = 0.0200 m, ax = 0, v0 x = 1.60 × 106 m s . x − x0 = v0 xt + 12 axt 2 gives t = 1.25 × 10−8 s . Since ax = 0 ,

⎛ v + vy vx = 1.60 × 106 m s . y − y0 = 0.0050 m, v0y = 0, t = 1.25 × 10−8 s . y − y0 = ⎜ 0 y ⎝ 2

⎞ 5 ⎟ t gives v y = 8.00 × 10 m s . ⎠

Then v = vx2 + v y2 = 1.79 × 106 m s . EVALUATE:

E=

ma y e

=

v y = v0 y + a yt gives a y = 6.4 × 1013 m/s 2 . The electric field between the plates is

(9.11× 10−31 kg)(6.4 × 1013 m/s 2 ) = 364 V/m . This is not a very large field. 1.60 × 10−19 C

21-12

Chapter 21

21.36.

IDENTIFY: Use the components of E from Example 21.6 to calculate the magnitude and direction of E . Use F = qE to calculate the force on the −2.5 nC charge and use Newton's third law for the force on the −8.0 nC charge. SET UP: From Example 21.6, E = (−11 N/C) iˆ + (14 N/C) ˆj .

⎛ Ey ⎞ ⎟ = tan −1 (14 11) = 51.8° , so (a) E = Ex2 + E y2 = (−11 N/C) 2 + (14 N/C) 2 = 17.8 N/C . tan −1 ⎜ ⎜ Ex ⎟ ⎝ ⎠ θ = 128° counterclockwise from the +x-axis. (b) (i) F = Eq so F = (17.8 N C)(2.5 × 10−9 C) = 4.45 × 10−8 N , at 52° below the +x-axis. EXECUTE:

21.37.

(ii) 4.45 × 10−8 N at 128° counterclockwise from the +x-axis. EVALUATE: The forces in part (b) are repulsive so they are along the line connecting the two charges and in each case the force is directed away from the charge that exerts it. IDENTIFY and SET UP: The electric force is given by Eq. (21.3). The gravitational force is we = me g . Compare these forces. (a) EXECUTE: we = (9.109 × 10−31 kg)(9.80 m/s 2 ) = 8.93 × 10−30 N In Examples 21.7 and 21.8, E = 1.00 × 104 N/C, so the electric force on the electron has magnitude FE = q E = eE = (1.602 ×10−19 C)(1.00 ×104 N/C) = 1.602 × 10−15 N. we 8.93 × 10−30 N = = 5.57 × 10−15 FE 1.602 × 10−15 N The gravitational force is much smaller than the electric force and can be neglected. (b) mg = q E m = q E / g = (1.602 × 10−19 C)(1.00 × 104 N/C)/(9.80 m/s 2 ) = 1.63 × 10−16 kg m 1.63 × 10−16 kg = = 1.79 × 1014 ; m = 1.79 × 1014 me . me 9.109 × 10−31 kg EVALUATE:

m is much larger than me . We found in part (a) that if m = me the gravitational force is much smaller

than the electric force. q is the same so the electric force remains the same. To get w large enough to equal FE ,

21.38.

the mass must be made much larger. (c) The electric field in the region between the plates is uniform so the force it exerts on the charged object is independent of where between the plates the object is placed. IDENTIFY: Apply constant acceleration equations to the motion of the proton. E = F / q . SET UP: EXECUTE:

E=

A proton has mass mp = 1.67 × 10−27 kg and charge +e . Let +x be in the direction of motion of the proton. (a) v0 x = 0 . a =

2(0.0160 m)(1.67 × 10−27 kg) = 148 N C. (1.60 × 10−19 C)(1.50 × 10−6 s) 2

(b) vx = v0 x + axt =

21.39.

eE 1 1 eE 2 t . Solving for E gives . x − x0 = v0 xt + 12 axt 2 gives x − x0 = axt 2 = mp 2 2 mp

eE t = 2.13 × 104 m s. mp

EVALUATE: The electric field is directed from the positively charged plate toward the negatively charged plate and the force on the proton is also in this direction. IDENTIFY: Find the angle θ that rˆ makes with the +x-axis. Then rˆ = (cosθ ) iˆ + (sin θ ) ˆj . SET UP: EXECUTE:

tan θ = y / x

π ⎛ −1.35 ⎞ ˆ (a) tan −1 ⎜ ⎟ = − rad . rˆ = − j . 0 2 ⎝ ⎠

2ˆ 2ˆ ⎛ 12 ⎞ π (b) tan −1 ⎜ ⎟ = rad . rˆ = i+ j. 2 2 ⎝ 12 ⎠ 4 ⎛ 2.6 ⎞ ˆ ˆ (c) tan −1 ⎜ ⎟ = 1.97 rad = 112.9° . rˆ = −0.39i + 0.92 j (Second quadrant). ⎝ +1.10 ⎠ EVALUATE: In each case we can verify that rˆ is a unit vector, because rˆ ⋅ rˆ = 1 .


21.40.

21-13

IDENTIFY: The net force on each charge must be zero. SET UP: The force diagram for the −6.50 μ C charge is given in Figure 21.40. FE is the force exerted on the charge by the uniform electric field. The charge is negative and the field is to the right, so the force exerted by the field is to the left. Fq is the force exerted by the other point charge. The two charges have opposite signs, so the force is attractive. Take the +x axis to be to the right, as shown in the figure. EXECUTE: (a) F = q E = (6.50 × 10−6 C)(1.85 × 108 N/C) = 1.20 × 103 N Fq = k

∑F

x

q1q2 r2

= (8.99 × 109 N ⋅ m 2 /C 2 )

(6.50 × 10−6 C)(8.75 × 10−6 C) = 8.18 × 102 N (0.0250 m) 2

= 0 gives T + Fq − FE = 0 and T = FE − Fq = 382 N .

(b) Now Fq is to the left, since like charges repel. ∑ Fx = 0 gives T − Fq − FE = 0 and T = FE + Fq = 2.02 × 103 N . EVALUATE: The tension is much larger when both charges have the same sign, so the force one charge exerts on the other is repulsive.

Figure 21.40 21.41.

IDENTIFY and SET UP: Use E in Eq. (21.3) to calculate F , F = ma to calculate a , and a constant acceleration equation to calculate the final velocity. Let +x be east. (a) EXECUTE: Fx = q E = (1.602 × 10 −19 C)(1.50 N/C) = 2.403 × 10−19 N

ax = Fx /m = (2.403 × 10−19 N)/(9.109 × 10−31 kg) = +2.638 × 1011 m/s 2 v0 x = +4.50 × 105 m/s, ax = +2.638 × 1011 m/s 2 , x − x0 = 0.375 m, vx = ?

vx2 = v02x + 2ax ( x − x0 ) gives vx = 6.33 × 105 m/s EVALUATE: E is west and q is negative, so F is east and the electron speeds up. (b) EXECUTE: Fx = − q E = −(1.602 × 10−19 C)(1.50 N/C) = − 2.403 × 10−19 N

ax = Fx / m = (−2.403 × 10−19 N)/(1.673 × 10−27 kg) = −1.436 × 108 m/s 2 v0 x = +1.90 × 104 m/s, ax = −1.436 × 108 m/s 2 , x − x0 = 0.375 m, vx = ?

vx2 = v02x + 2ax ( x − x0 ) gives vx = 1.59 × 104 m/s 21.42.

21.43.

EVALUATE: q > 0 so F is west and the proton slows down. IDENTIFY: Coulomb’s law for a single point-charge gives the electric field. (a) SET UP: Coulomb’s law for a point-charge is E = (1/ 4π P0 )q / r 2 . EXECUTE: E = (9.00 × 109 N ⋅ m2/C2)(1.60 × 10−19 C)/(1.50 × 10−15 m)2 = 6.40 × 1020 N/C (b) Taking the ratio of the electric fields gives E/Eplates = (6.40 × 1020 N/C)/(1.00 × 104 N/C) = 6.40 × 1016 times as strong EVALUATE: The electric field within the nucleus is huge compared to typical laboratory fields! IDENTIFY: Calculate the electric field due to each charge and find the vector sum of these two fields. SET UP: At points on the x-axis only the x component of each field is nonzero. The electric field of a point charge points away from the charge if it is positive and toward it if it is negative. EXECUTE: (a) Halfway between the two charges, E = 0. (b) For | x | < a , Ex =

For x > a , Ex =

1 4πP0

1 4πP0

⎛ q q ⎞ 4q ax − =− . ⎜ 2 2 ⎟ 2 ( a x ) ( a x ) 4 ( x a 2 )2 π P + − − ⎝ ⎠ 0

⎛ q q ⎞ 2q x 2 + a 2 . + ⎜ ⎟= 2 (a − x) 2 ⎠ 4π P0 ( x 2 − a 2 ) 2 ⎝ (a + x)

⎛ q q ⎞ 2q x 2 + a 2 + =− . ⎜ 2 2 ⎟ (a − x) ⎠ 4π P0 ( x 2 − a 2 ) 2 ⎝ (a + x) The graph of Ex versus x is sketched in Figure 21.43.

For x < − a , Ex =

−1 4πP0

21-14

Chapter 21

EVALUATE:

The magnitude of the field approaches infinity at the location of one of the point charges.

Figure 21.43 21.44.

IDENTIFY:

For a point charge, E = k

q r2

. For the net electric field to be zero, E1 and E2 must have equal

magnitudes and opposite directions. SET UP: Let q1 = +0.500 nC and q2 = +8.00 nC. E is toward a negative charge and away from a positive charge. EXECUTE: The two charges and the directions of their electric fields in three regions are shown in Figure 21.44. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from q1 so a 0.500 nC 8.00 nC =k . 16 x 2 = (1.20 m − x) 2 . 4 x = ± (1.20 m − x ) (1.20 − x) 2 x2 and x = 0.24 m is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge and 0.96 m from the 8.00 nC charge. EVALUATE: There is only one point along the line connecting the two charges where the net electric field is zero. This point is closer to the charge that has the smaller magnitude. distance 1.20 m − x from q2 . E1 = E2 gives k

Figure 21.44 21.45.

IDENTIFY: Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a). (a) SET UP: The placement of charges is sketched in Figure 21.45a.

Figure 21.45a

The electric field of a point charge is directed away from the point charge if the charge is positive and toward the 1 q point charge if the charge is negative. The magnitude of the electric field is E = , where r is the distance 4π P0 r 2 between the point where the field is calculated and the point charge. (i) At point a the fields E1 of q1 and E 2 of q2 are directed as shown in Figure 21.45b.

Figure 21.45b


EXECUTE: E2 =

E1 =

21-15

−9

1 q1 2.00 × 10 C = (8.988 × 109 N ⋅ m 2 /C 2 ) = 449.4 N/C 4π P0 r12 (0.200 m) 2

−9 1 q2 C 9 2 2 5.00 × 10 = (8.988 × 10 N ⋅ m /C ) = 124.8 N/C 2 2 4π P0 r2 (0.600 m)

E1x = 449.4 N/C, E1 y = 0 E2 x = 124.8 N/C, E2 y = 0 Ex = E1x + E2 x = +449.4 N/C + 124.8 N/C = +574.2 N/C

E y = E1 y + E2 y = 0 The resultant field at point a has magnitude 574 N/C and is in the +x-direction. (ii) SET UP: At point b the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45c.

Figure 21.45c EXECUTE: E2 =

E1 =

−9 1 q1 C 9 2 2 2.00 × 10 = (8.988 × 10 N ⋅ m /C ) = 12.5 N/C 2 2 4π P0 r1 (1.20 m )

1 q2 5.00 × 10−9 C = ( 8.988 × 109 N ⋅ m 2 /C2 ) = 280.9 N/C 2 2 4π P0 r2 ( 0.400 m )

E1x = 12.5 N/C, E1 y = 0 E2 x = −280.9 N/C, E2 y = 0 Ex = E1x + E2 x = +12.5 N/C − 280.9 N/C = −268.4 N/C

E y = E1 y + E2 y = 0 The resultant field at point b has magnitude 268 N/C and is in the − x -direction. (iii) SET UP: At point c the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45d.

Figure 21.45d EXECUTE: E2 =

E1 =

1 q1 2.00 × 10−9 C = (8.988 × 109 N ⋅ m 2 /C2 ) = 449.4 N/C 2 2 4π P0 r1 ( 0.200 m )

1 q2 5.00 × 10−9 C = ( 8.988 × 109 N ⋅ m 2 /C 2 ) = 44.9 N/C 2 4π P0 r2 (1.00 m) 2

E1x = −449.4 N/C, E1 y = 0 E2 x = +44.9 N/C, E2 y = 0 Ex = E1x + E2 x = −449.4 N/C + 44.9 N/C = −404.5 N/C E y = E1 y + E2 y = 0 The resultant field at point b has magnitude 404 N/C and is in the − x -direction. (b) SET UP: Since we have calculated E at each point the simplest way to get the force is to use F = −eE . EXECUTE: (i) F = (1.602 × 10−19 C)(574.2 N/C) = 9.20 × 10−17 N, − x-direction (ii) F = (1.602 × 10−19 C)(268.4 N/C) = 4.30 × 10−17 N, + x-direction (iii) F = (1.602 × 10−19 C)(404.5 N/C) = 6.48 × 10−17 N, + x-direction EVALUATE: The general rule for electric field direction is away from positive charge and toward negative charge. Whether the field is in the + x - or − x -direction depends on where the field point is relative to the charge that produces the field. In part (a) the field magnitudes were added because the fields were in the same direction and in (b) and (c) the field magnitudes were subtracted because the two fields were in opposite directions. In part (b) we could have used Coulomb's law to find the forces on the electron due to the two charges and then added these force vectors, but using the resultant electric field is much easier.

21-16

Chapter 21

21.46.

IDENTIFY: Apply Eq.(21.7) to calculate the field due to each charge and then require that the vector sum of the two fields to be zero. SET UP: The field of each charge is directed toward the charge if it is negative and away from the charge if it is positive . EXECUTE: The point where the two fields cancel each other will have to be closer to the negative charge, because it is smaller. Also, it can’t be between the two charges, since the two fields would then act in the same direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the −4.00 nC charge. The zero point will therefore have to be a factor of 2 farther from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the –4.00 nC

21.47.

charge: 1.20 + x = 2 x and x = 2.90 m . EVALUATE: This point is 4.10 m from the 8.00 nC charge. The two fields at this point are in opposite directions and have equal magnitudes. q IDENTIFY: E = k 2 . The net field is the vector sum of the fields due to each charge. r SET UP: The electric field of a negative charge is directed toward the charge. Label the charges q1, q2 and q3, as shown in Figure 21.47a. This figure also shows additional distances and angles. The electric fields at point P are shown in Figure 21.47b. This figure also shows the xy coordinates we will use and the x and y components of the fields E1 , E2 and E3 . EXECUTE:

E1 = E3 = (8.99 × 109 N ⋅ m 2 /C 2 )

E2 = (8.99 × 109 N ⋅ m 2 /C 2 )

5.00 × 10−6 C = 4.49 × 106 N/C (0.100 m) 2

2.00 × 10−6 C = 4.99 × 106 N/C (0.0600 m) 2

E y = E1 y + E2 y + E3 y = 0 and Ex = E1x + E2 x + E3 x = E2 + 2 E1 cos53.1° = 1.04 × 107 N/C E = 1.04 × 107 N/C , toward the −2.00 μ C charge. EVALUATE:

The x-components of the fields of all three charges are in the same direction.

Figure 21.47 21.48.

IDENTIFY: A positive and negative charge, of equal magnitude q, are on the x-axis, a distance a from the origin. Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of these fields. SET UP: E due to a point charge is directed away from the charge if it is positive and directed toward the charge if it is negative. 1 2q EXECUTE: (a) Halfway between the charges, both fields are in the − x -direction and E = , in the 4π P0 a 2 − x-direction . 1 ⎛ −q q ⎞ 1 ⎛ −q q ⎞ (b) E x = − + ⎜ ⎟ for | x | < a . E x = ⎜ ⎟ for x > a . 4π P0 ⎝ (a + x) 2 (a − x) 2 ⎠ 4π P0 ⎝ (a + x) 2 ( a − x ) 2 ⎠ Ex =

1 ⎛ −q q ⎞ − ⎜ ⎟ for x < − a . Ex is graphed in Figure 21.48. 2 4π P0 ⎝ (a + x) (a − x) 2 ⎠


21-17

EVALUATE: At points on the x axis and between the charges, Ex is in the − x-direction because the fields from both charges are in this direction. For x < −a and x > + a , the fields from the two charges are in opposite directions and the field from the closer charge is larger in magnitude.

Figure 21.48 21.49.

IDENTIFY: The electric field of a positive charge is directed radially outward from the charge and has magnitude 1 q E= . The resultant electric field is the vector sum of the fields of the individual charges. 4π P0 r 2 SET UP: The placement of the charges is shown in Figure 21.49a.


(a) The directions of the two fields are shown in Figure 21.49b. E1 = E2 =

1 q with r = 0.150 m. 4π P0 r 2

E = E2 − E1 = 0; Ex = 0, E y = 0 Figure 21.49b (b) The two fields have the directions shown in Figure 21.49c.

E = E1 + E2 , in the + x-direction

Figure 21.49c E1 =

−9 1 q1 C 9 2 2 6.00 × 10 = (8.988 × 10 N ⋅ m /C ) = 2396.8 N/C 2 2 4π P0 r1 (0.150 m)

E2 =

1 q2 6.00 × 10−9 C = (8.988 × 109 N ⋅ m 2 /C 2 ) = 266.3 N/C 2 4π P0 r2 (0.450 m) 2

E = E1 + E2 = 2396.8 N/C + 266.3 N/C = 2660 N/C; Ex = +2260 N/C, E y = 0

21-18

Chapter 21

(c) The two fields have the directions shown in Figure 21.49d.

0.400 m = 0.800 0.500 m 0.300 m cosθ = = 0.600 0.500 m

sin θ =

Figure 21.49d E1 =

1 q1 4π P0 r12

E1 = (8.988 × 109 N ⋅ m 2 /C 2 ) E2 =

6.00 × 10−9 C = 337.1 N/C (0.400 m) 2

1 q2 4π P0 r22

E2 = (8.988 × 109 N ⋅ m 2 /C 2 )

6.00 × 10−9 C = 215.7 N/C (0.500 m) 2

E1x = 0, E1 y = − E1 = −337.1 N/C E2x = + E2 cosθ = + (215.7 N/C)(0.600) = +129.4 N/C E2y = − E2 sin θ = −(215.7 N/C)(0.800) = −172.6 N/C Ex = E1x + E2 x = +129 N/C Ey = E1 y + E2 y = −337.1 N/C − 172.6 N/C = −510 N/C E = Ex2 + E y2 = (129 N/C) 2 + (−510 N/C) 2 = 526 N/C E and its components are shown in Figure 21.49e.

tan α =

Ey Ex

−510 N/C = −3.953 +129 N/C α = 284°C, counterclockwise from + x-axis tan α =

Figure 21.49e (d) The two fields have the directions shown in Figure 21.49f.

sin θ =

Figure 21.49f

0.200 m = 0.800 0.250 m


21-19

The components of the two fields are shown in Figure 21.49g. E1 = E2 =

1 q 4π P0 r 2

E1 = (8.988 ×109 N ⋅ m 2 /C2 )

6.00 ×10−9 C (0.250 m) 2

E1 = E2 = 862.8 N/C Figure 21.49g

E1x = − E1 cosθ , E2 x = + E2 cosθ Ex = E1x + E2 x = 0 E1 y = + E1 sin θ , E2 y = + E2 sin θ

E y = E1 y + E2 y = 2 E1 y = 2 E1 sin θ = 2 ( 862.8 N/C )( 0.800 ) = 1380 N/C E = 1380 N/C, in the + y -direction. EVALUATE: Point a is symmetrically placed between identical charges, so symmetry tells us the electric field must be zero. Point b is to the right of both charges and both electric fields are in the +x-direction and the resultant field is in this direction. At point c both fields have a downward component and the field of q2 has a component to

21.50.

the right, so the net E is in the 4th quadrant. At point d both fields have an upward component but by symmetry they have equal and opposite x-components so the net field is in the +y-direction. We can use this sort of reasoning to deduce the general direction of the net field before doing any calculations. IDENTIFY: Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields. SET UP: The fields due to q1 and to q2 are sketched in Figure 21.50. EXECUTE:

E2 =

1 (6.00 × 10−9 C) ˆ ( −i ) = −150iˆ N/C . 4π P0 (0.6 m) 2

⎛ ⎞ 1 1 1 (4.00 × 10−9 C) ⎜ (0.600) iˆ + (0.800) ˆj ⎟ = (21.6 iˆ + 28.8 ˆj ) N C . 2 2 4π P0 (1.00 m) (1.00 m) ⎝ ⎠ E = E1 + E2 = (−128.4 N/C) iˆ + (28.8 N/C) ˆj . E = (128.4 N/C) 2 + (28.8 N/C) 2 = 131.6 N/C at ⎛ 28.8 ⎞ θ = tan −1 ⎜ ⎟ = 12.6° above the − x axis and therefore 196.2° counterclockwise from the +x axis. ⎝ 128.4 ⎠ EVALUATE: E1 is directed toward q1 because q1 is negative and E 2 is directed away from q2 because q2 is positive. E1 =

Figure 21.50 21.51.

IDENTIFY: The resultant electric field is the vector sum of the field E1 of q1 and E 2 of q2 . SET UP: The placement of the charges is shown in Figure 21.51a.

Figure 21.51a

21-20

Chapter 21

EXECUTE:

(a) The directions of the two fields are shown in Figure 21.51b. 1 q1 E1 = E2 = 4π P0 r12

E1 = (8.988 × 109 N ⋅ m 2 /C2 )

6.00 × 10−9 C (0.150 m) 2

E1 = E2 = 2397 N/C

Figure 21.51b

E1x = −2397 N/C, E1 y = 0 E2x = −2397 N/C, E2 y = 0 E x = E1x + E2 x = 2(−2397 N/C) = −4790 N/C E y = E1 y + E2 y = 0

The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the − x -direction. (b) The directions of the two fields are shown in Figure 21.51c.

Figure 21.51c

1 q1 6.00 × 10−9 C = (8.988 × 109 N ⋅ m 2 /C2 ) = 2397 N/C 2 4π P0 r1 (0.150 m) 2 −9 1 q2 C 9 2 2 6.00 × 10 = × ⋅ = 266 N/C E2 = (8.988 10 N m /C ) 4π P0 r22 (0.450 m) 2 E1x = +2397 N/C, E1 y = 0 E2 x = −266 N/C, E2 y = 0 E1 =

E x = E1x + E2 x = +2397 N/C − 266 N/C = +2130 N/C E y = E1 y + E2 y = 0

The resultant electric field at point b in the sketch has magnitude 2130 N/C and is in the + x -direction. (c) The placement of the charges is shown in Figure 21.51d. 0.300 m = 0.600 0.500 m 0.400 m = 0.800 cosθ = 0.500 m

sin θ =

Figure 21.51d

The directions of the two fields are shown in Figure 21.51e.

E1 =

1 q1 4π P0 r12

E1 = (8.988 × 109 N ⋅ m 2 /C2 )

6.00 × 10−9 C (0.400 m) 2

E1 = 337.0 N/C 1 q2 E2 = 4π P0 r22

E2 = (8.988 × 109 N ⋅ m 2 /C 2 ) E2 = 215.7 N/C

Figure 21.51e

E1x = 0, E1 y = − E1 = −337.0 N/C E2 x = − E2 sin θ = −(215.7 N/C)(0.600) = −129.4 N/C E2 y = + E2 cosθ = +(215.7 N/C)(0.800) = +172.6 N/C

6.00 × 10−9 C (0.500 m) 2


21-21

Ex = E1x + E2 x = −129 N/C E y = E1 y + E2 y = −337.0 N/C + 172.6 N/C = −164 N/C E = Ex2 + E y2 = 209 N/C The field E and its components are shown in Figure 21.51f. tan α =

Ey Ex

−164 N/C = +1.271 −129 N/C α = 232°, counterclockwise from + x-axis tan α =

Figure 21.51f (d) The placement of the charges is shown in Figure 21.51g.

0.200 m = 0.800 0.250 m 0.150 m = 0.600 cosθ = 0.250 m

sin θ =

Figure 21.51g

The directions of the two fields are shown in Figure 21.51h.

E1 = E2 =

1 q 4π P0 r 2

E1 = (8.988 × 109 N ⋅ m 2 /C 2 )

6.00 × 10−9 C (0.250 m) 2

E1 = 862.8 N/C E 2 = E1 = 862.8 N/C

Figure 21.51h

E1x = − E1 cosθ , E2 x = − E2 cosθ

Ex = E1x + E2 x = −2 ( 862.8 N/C )( 0.600 ) = −1040 N/C E1 y = + E1 sin θ , E2 y = − E2 sin θ

E y = E1 y + E2 y = 0 E = 1040 N/C, in the − x-direction. EVALUATE: The electric field produced by a charge is toward a negative charge and away from a positive charge. As in Exercise 21.45, we can use this rule to deduce the direction of the resultant field at each point before doing any calculations. 21.52.

For a long straight wire, E =

IDENTIFY: SET UP:

λ 2π P0 r

.

1 = 4.49 × 109 N ⋅ m 2 /C2 . 2π P0

EXECUTE: EVALUATE:

r=

1.5 × 10−10 C m = 1.08 m 2π P0 (2.50 N C)

For a point charge, E is proportional to 1/r 2 . For a long straight line of charge, E is proportional to 1/r.

21-22

Chapter 21

21.53.

IDENTIFY:

Apply Eq.(21.10) for the finite line of charge and E =

λ for the infinite line of charge. 2π P0

SET UP: For the infinite line of positive charge, E is in the +x direction. EXECUTE: (a) For a line of charge of length 2a centered at the origin and lying along the y-axis, the electric field 1 λ is given by Eq.(21.10): E = iˆ . 2π P0 x x 2 a 2 + 1 (b) For an infinite line of charge: E =

λ ˆ i . Graphs of electric field versus position for both distributions of 2π P0 x

charge are shown in Figure 21.53. EVALUATE: For small x, close to the line of charge, the field due to the finite line approaches that of the infinite line of charge. As x increases, the field due to the infinite line falls off more slowly and is larger than the field of the finite line.

Figure 21.53 21.54.

(a) IDENTIFY: The field is caused by a finite uniformly charged wire. SET UP: The field for such a wire a distance x from its midpoint is ⎛ 1 ⎞ 1 λ λ E= = 2⎜ . ⎟ 2 2π P0 x ( x / a ) + 1 4 π P x ( x / a) 2 + 1 0 ⎠ ⎝

(18.0 ×10

9

E=

EXECUTE:

(b) IDENTIFY: SET UP:

N ⋅ m 2 / C2 )(175 × 10−9 C/m ) 2

= 3.03 × 104 N/C, directed upward.

⎛ 6.00 cm ⎞ (0.0600 m) ⎜ ⎟ +1 ⎝ 4.25 cm ⎠ The field is caused by a uniformly charged circular wire.

The field for such a wire a distance x from its midpoint is E =

1 Qx . We first find the radius 4π P0 ( x 2 + a 2 )3/ 2

of the circle using 2πr = l. EXECUTE: Solving for r gives r = l/2π = (8.50 cm)/2π = 1.353 cm The charge on this circle is Q = λl = (175 nC/m)(0.0850 m) = 14.88 nC The electric field is

E=

21.55.

(9.00 ×109 N ⋅ m2/C2 )(14.88 × 10−9 C/m ) ( 0.0600 m ) 1 Qx = 3 / 2 3/ 2 4π P0 ( x 2 + a 2 ) ⎡⎣(0.0600 m) 2 + (0.01353 m) 2 ⎤⎦

E = 3.45 × 104 N/C, upward. EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different because the charge has been bent into different shapes. IDENTIFY: For a ring of charge, the electric field is given by Eq. (21.8). F = qE . In part (b) use Newton's third law to relate the force on the ring to the force exerted by the ring. SET UP: Q = 0.125 × 10−9 C, a = 0.025 m and x = 0.400 m . 1 Qx EXECUTE: (a) E = iˆ = (7.0 N/C) iˆ . 2 4π P0 ( x + a 2 )3/ 2 = − F = −qE = −(−2.50 × 10−6 C) (7.0 N/C) iˆ = (1.75 × 10−5 N) iˆ (b) F on ring

EVALUATE:

on q

Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive.


21.56.

21-23

IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because all the charge is along the rim of the disk, and a point-charge in (c). (a) SET UP: First find the surface charge density (Q/A), then use the formula for the field due to a disk of charge, ⎤ σ ⎡ 1 Ex = ⎢1 − ⎥. 2 2P0 ⎢⎣ ( R / x) + 1 ⎥⎦ EXECUTE:

The surface charge density is σ =

Q Q 6.50 × 10−9 C = 2= = 1.324 × 10−5 C/m2. π (0.0125 m) 2 A πr

The electric field is Ex =

σ ⎡

⎤ 1 1.324 × 10−5 C/m 2 1 ⎡ ⎤ 1− ⎢1 − ⎥ = ⎥ 2 2 2P0 ⎢⎣ 2(8.85 × 10−12 C2 /N ⋅ m 2 ) ⎢ ( R / x) + 1 ⎥⎦ ⎛ 1.25 cm ⎞ ⎢ ⎥ + 1 ⎜ ⎟ ⎢ ⎥ 2.00 cm ⎝ ⎠ ⎣ ⎦

Ex = 1.14 × 105 N/C, toward the center of the disk. (b) SET UP: EXECUTE:

For a ring of charge, the field is E =

1 Qx . 2 4π P0 ( x + a 2 )3 / 2

Substituting into the electric field formula gives

E=

(9.00 × 109 N ⋅ m 2 /C2 )(6.50 × 10−9 C)(0.0200 m) 1 Qx = 3/ 2 3 / 2 4π P0 ( x 2 + a 2 ) ⎡⎣(0.0200 m) 2 + (0.0125 m) 2 ⎤⎦

E = 8.92 × 104 N/C, toward the center of the disk. (c) SET UP:

21.57.

For a point charge, E = (1/ 4π P0 ) q / r 2 .

EXECUTE: E = (9.00 × 109 N ⋅ m2/C2)(6.50 × 10−9 C)/(0.0200 m)2 = 1.46 × 105 N/C (d) EVALUATE: With the ring, more of the charge is farther from P than with the disk. Also with the ring the component of the electric field parallel to the plane of the ring is greater than with the disk, and this component cancels. With the point charge in (c), all the field vectors add with no cancellation, and all the charge is closer to point P than in the other two cases. IDENTIFY: By superposition we can add the electric fields from two parallel sheets of charge. SET UP: The field due to each sheet of charge has magnitude σ / 2P0 and is directed toward a sheet of negative charge and away from a sheet of positive charge. (a) The two fields are in opposite directions and E = 0. (b) The two fields are in opposite directions and E = 0. (c) The fields of both sheets are downward and E = 2

21.58.

σ

=

σ

, directed downward. 2P0 P0 EVALUATE: The field produced by an infinite sheet of charge is uniform, independent of distance from the sheet. IDENTIFY and SET UP: The electric field produced by an infinite sheet of charge with charge density σ has magnitude E =

σ

. The field is directed toward the sheet if it has negative charge and is away from the sheet if it 2P0 has positive charge. EXECUTE: (a) The field lines are sketched in Figure 21.58a. (b) The field lines are sketched in Figure 21.58b. EVALUATE: The spacing of the field lines indicates the strength of the field. In part (a) the two fields add between the sheets and subtract in the regions to the left of A and to the right of B. In part (b) the opposite is true.

Figure 21.58

21-24

Chapter 21

21.59.

IDENTIFY: The force on the particle at any point is always tangent to the electric field line at that point. SET UP: The instantaneous velocity determines the path of the particle. EXECUTE: In Fig.21.29a the field lines are straight lines so the force is always in a straight line and velocity and acceleration are always in the same direction. The particle moves in a straight line along a field line, with increasing speed. In Fig.21.29b the field lines are curved. As the particle moves its velocity and acceleration are not in the same direction and the trajectory does not follow a field line. EVALUATE: In two-dimensional motion the velocity is always tangent to the trajectory but the velocity is not always in the direction of the net force on the particle. IDENTIFY: The field appears like that of a point charge a long way from the disk and an infinite sheet close to the disk’s center. The field is symmetrical on the right and left. SET UP: For a positive point charge, E is proportional to 1/r 2 and is directed radially outward. For an infinite sheet of positive charge, the field is uniform and is directed away from the sheet. EXECUTE: The field is sketched in Figure 21.60. EVALUATE: Near the disk the field lines are parallel and equally spaced, which corresponds to a uniform field. Far from the disk the field lines are getting farther apart, corresponding to the 1/r 2 dependence for a point charge.

21.60.

Figure 21.60 21.61.

IDENTIFY: Use symmetry to deduce the nature of the field lines. (a) SET UP: The only distinguishable direction is toward the line or away from the line, so the electric field lines are perpendicular to the line of charge, as shown in Figure 21.61a.

Figure 21.61a (b) EXECUTE and EVALUATE: The magnitude of the electric field is inversely proportional to the spacing of the field lines. Consider a circle of radius r with the line of charge passing through the center, as shown in Figure 21.61b.

Figure 21.61b

21.62.

The spacing of field lines is the same all around the circle, and in the direction perpendicular to the plane of the circle the lines are equally spaced, so E depends only on the distance r. The number of field lines passing out through the circle is independent of the radius of the circle, so the spacing of the field lines is proportional to the reciprocal of the circumference 2π r of the circle. Hence E is proportional to 1/r. IDENTIFY: Field lines are directed away from a positive charge and toward a negative charge. The density of field lines is proportional to the magnitude of the electric field. SET UP: The field lines represent the resultant field at each point, the net field that is the vector sum of the fields due to each of the three charges. EXECUTE: (a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. (b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the y-components of the field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges. EVALUATE: Far from all three charges the field is the same as the field of a point charge equal to the algebraic sum of the three charges.


21.63.

21-25

(a) IDENTIFY and SET UP: Use Eq.(21.14) to relate the dipole moment to the charge magnitude and the separation d of the two charges. The direction is from the negative charge toward the positive charge. EXECUTE: p = qd = (4.5 × 10−9 C)(3.1 × 10−3 m) = 1.4 × 10−11 C ⋅ m; The direction of p is from q1 toward q2 . (b) IDENTIFY and SET UP: Use Eq. (21.15) to relate the magnitudes of the torque and field. EXECUTE: τ = pE sin φ , with φ as defined in Figure 21.63, so

E= E=

τ p sin φ 7.2 × 10−9 N ⋅ m = 860 N/C (1.4 × 10−11 C ⋅ m)sin 36.9°

Figure 21.63 EVALUATE: Eq.(21.15) gives the torque about an axis through the center of the dipole. But the forces on the two charges form a couple (Problem 11.53) and the torque is the same for any axis parallel to this one. The force on each charge is q E and the maximum moment arm for an axis at the center is d/2, so the maximum torque is

2( q E )(d/2) = 1.2 × 10−8 N ⋅ m. The torque for the orientation of the dipole in the problem is less than this

21.64.

maximum. (a) IDENTIFY: The potential energy is given by Eq.(21.17). SET UP: U (φ ) = − p ⋅ E = − pE cos φ , where φ is the angle between p and E . parallel: φ = 0 and U ( 0° ) = − pE

EXECUTE:

perpendicular: φ = 90° and U ( 90° ) = 0

ΔU = U ( 90° ) − U ( 0° ) = pE = ( 5.0 × 10−30 C ⋅ m )(1.6 × 106 N/C ) = 8.0 × 10−24 J. (b)

21.65.

3 2

kT = ΔU so T =

2 ( 8.0 × 10−24 J ) 2ΔU = = 0.39 K 3k 3 (1.381 × 10−23 J/K )

EVALUATE: Only at very low temperatures are the dipoles of the molecules aligned by a field of this strength. A much larger field would be required for alignment at room temperature. IDENTIFY: Follow the procedure specified in part (a) of the problem. SET UP: Use that y >> d . EXECUTE:

Ey =

(a)

1 1 ( y + d 2) 2 − ( y − d 2) 2 2 yd − = = 2 . This gives 2 2 2 2 2 ( y − d 2) ( y + d 2) ( y − d 4) ( y − d 2 4) 2

q 2 yd qd y p = . Since y 2 >> d 2 / 4 , E y ≈ . 4π P0 ( y 2 − d 2 4) 2 2π P0 ( y 2 − d 2 4) 2 2π P0 y 3

(b) For points on the − y -axis , E− is in the +y direction and E+ is in the − y direction. The field point is closer to

− q , so the net field is upward. A similar derivation gives E y ≈

p . E y has the same magnitude and direction 2π P0 y 3

at points where y >> d as where y > ( d 2) 2 , so the expression in part (a) reduces to the approximation Edipole ≈ . 4π P0 x 3 qd EVALUATE: Example 21.15 shows that at points on the +y axis far from the dipole, Edipole ≈ . The 2π P0 y 3 expression in part (b) for points on the x axis has a similar form. IDENTIFY: The torque on a dipole in an electric field is given by τ = p × E . SET UP:

τ = pE sin φ , where φ is the angle between the direction of p and the direction of E .

EXECUTE: (a) The torque is zero when p is aligned either in the same direction as E or in the opposite direction, as shown in Figure 21.69a. (b) The stable orientation is when p is aligned in the same direction as E . In this case a small rotation of the

dipole results in a torque directed so as to bring p back into alignment with E . When p is directed opposite to E , a small displacement results in a torque that takes p farther from alignment with E . (c) Field lines for Edipole in the stable orientation are sketched in Figure 21.69b.

EVALUATE:

The field of the dipole is directed from the + charge toward the − charge.

Figure 21.69 21.70.

IDENTIFY: The plates produce a uniform electric field in the space between them. This field exerts torque on a dipole and gives it potential energy. SET UP: The electric field between the plates is given by E = σ / P0 , and the dipole moment is p = ed. The

potential energy of the dipole due to the field is U = − p ⋅ E = − pE cos φ , and the torque the field exerts on it is τ =

pE sin φ.


EXECUTE:

21-27

(a) The potential energy, U = − p ⋅ E = − pE cos φ , is a maximum when φ = 180°. The field between

the plates is E = σ / P0 , giving

Umax = (1.60 × 10–19 C)(220 × 10–9 m)(125 × 10–6 C/m2)/(8.85 × 10–12 C2/N ⋅ m2) = 4.97 × 10–19 J The orientation is parallel to the electric field (perpendicular to the plates) with the positive charge of the dipole toward the positive plate. (b) The torque, τ = pE sin φ, is a maximum when φ = 90° or 270°. In this case

τ max = pE = pσ / P0 = edσ / P0

τ max = (1.60 × 10−19 C )( 220 × 10−9 m )(125 × 10−6 C/m 2 ) (8.85 × 10−12 C2 / N ⋅ m 2 ) τ max = 4.97 × 10−19 N ⋅ m

21.71.

The dipole is oriented perpendicular to the electric field (parallel to the plates). (c) F = 0. EVALUATE: When the potential energy is a maximum, the torque is zero. In both cases, the net force on the dipole is zero because the forces on the charges are equal but opposite (which would not be true in a nonuniform electric field). (a) IDENTIFY: Use Coulomb's law to calculate each force and then add them as vectors to obtain the net force. Torque is force times moment arm. SET UP: The two forces on each charge in the dipole are shown in Figure 21.71a.

sin θ = 1.50 / 2.00 so θ = 48.6° Opposite charges attract and like charges repel. Fx = F1x + F2 x = 0


F1 = k

qq′ (5.00 ×10−6 C)(10.0 ×10−6 C) =k = 1.124 × 103 N 2 r (0.0200 m) 2

F1 y = − F1 sin θ = −842.6 N F2 y = −842.6 N so Fy = F1 y + F2 y = −1680 N (in the direction from the +5.00-μ C charge toward the −5.00-μ C charge). EVALUATE: The x-components cancel and the y-components add. (b) SET UP: Refer to Figure 21.71b.

The y-components have zero moment arm and therefore zero torque. F1x and F2 x both produce clockwise torques.


F1 X = F1 cosθ = 743.1 N

τ = 2( F1x )(0.0150 m) = 22.3 N ⋅ m, clockwise EVALUATE:

The electric field produced by the −10.00μ C charge is not uniform so Eq. (21.15) does not apply.

21-28

21.72.

Chapter 21

qq′ for each pair of charges and find the vector sum of the forces that q1 and q2 exert on q3 . r2 SET UP: Like charges repel and unlike charges attract. The three charges and the forces on q3 are shown in Figure 21.72. IDENTIFY:

Apply F = k

Figure 21.72 q1q3 (5.00 × 10−9 C)(6.00 × 10−9 C) = (8.99 × 109 N ⋅ m 2 /C2 ) = 1.079 × 10−4 C . 2 r1 (0.0500 m) 2 θ = 36.9° . F1x = + F1 cosθ = 8.63 × 10−5 N . F1 y = + F1 sin θ = 6.48 × 10−5 N .

EXECUTE:

(a) F1 = k

q2 q3 (2.00 × 10−9 C)(6.00 × 10−9 C) = (8.99 × 109 N ⋅ m 2 /C2 ) = 1.20 × 10−4 C . 2 r2 (0.0300 m) 2 F2 x = 0 , F2 y = − F2 = −1.20 × 10−4 N . Fx = F1x + F2 x = 8.63 × 10−5 N .

F2 = k

Fy = F1 y + F2 y = 6.48 × 10−5 N + (−1.20 × 10−4 N) = −5.52 × 10−5 N .

(b) F = Fx2 + Fy2 = 1.02 × 10−4 N . tan φ =

21.73.

Fy Fx

= 0.640 . φ = 32.6° , below the + x axis.

EVALUATE: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using components. (a) IDENTIFY: Use Coulomb's law to calculate the force exerted by each Q on q and add these forces as vectors to find the resultant force. Make the approximation x >> a and compare the net force to F = − kx to deduce k and

then f = (1/ 2π ) k / m . SET UP: The placement of the charges is shown in Figure 21.73a.


Find the net force on q.

Fx = F1x + F2 x and F1x = + F1 , F2 x = − F2 Figure 21.73b F1 =

1 qQ 1 qQ , F2 = 4π P0 ( a + x )2 4π P0 ( a − x )2

Fx = F1 − F2 =

Fx =

qQ ⎡ 1 1 ⎤ − 4π P0 ⎢⎣ ( a + x )2 ( a − x )2 ⎥⎦

−2 −2 qQ ⎡ ⎛ x⎞ ⎛ x⎞ ⎤ 1 1 + + − − ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ 4π P0 a 2 ⎢⎣ ⎝ a ⎠ ⎝ a ⎠ ⎥⎦


21-29

Since x 0 and in the − y -direction when y < 0. The charge moves away from the origin on the y-axis and never returns. EVALUATE: The directions of the forces and of the net force depend on where q is located relative to the other two charges. In part (a), F = 0 at x = 0 and when the charge q is displaced in the +x- or –x-direction the net force is a restoring force, directed to return q to x = 0. The charge oscillates back and forth, similar to a mass on a spring. 21.74.

IDENTIFY:

Apply

∑F

x

= 0 and

∑F

y

= 0 to one of the spheres.

SET UP: The free-body diagram is sketched in Figure 21.74. Fe is the repulsive Coulomb force between the spheres. For small θ , sin θ ≈ tan θ . EXECUTE:

∑ Fx = T sin θ − Fe = 0 and ∑ Fy = T cosθ − mg = 0 . So

mg sin θ kq 2 = Fe = 2 . But tan θ ≈ sin θ = d , cosθ 2L d

1/3

⎛ q2L ⎞ 2kq 2 L and d = ⎜ ⎟ . mg ⎝ 2π P0 mg ⎠ EVALUATE: d increases when q increases.

so d 3 =

Figure 21.74 21.75.

IDENTIFY: Use Coulomb's law for the force that one sphere exerts on the other and apply the 1st condition of equilibrium to one of the spheres. (a) SET UP: The placement of the spheres is sketched in Figure 21.75a.

Figure 21.75a

21-30

Chapter 21

The free-body diagrams for each sphere are given in Figure 21.75b.

Figure 21.75b Fc is the repulsive Coulomb force exerted by one sphere on the other. (b) EXECUTE: From either force diagram in part (a): ∑ Fy = ma y mg T cos 25.0° − mg = 0 and T = cos 25.0° ∑ Fx = max

T sin 25.0° − Fc = 0 and Fc = T sin 25.0°

Use the first equation to eliminate T in the second: Fc = ( mg/ cos 25.0° )( sin 25.0° ) = mg tan 25.0° Fc =

1 q1q2 1 q2 1 q2 = = 2 2 4π P0 r 4π P0 r 4π P0 [2(1.20 m)sin 25.0°]2

Combine this with Fc = mg tan 25.0° and get mg tan 25.0° = q = ( 2.40 m ) sin 25.0°

q = ( 2.40 m ) sin 25.0°

1 q2 4π P0 [2(1.20 m)sin 25.0°]2

mg tan 25.0° (1/ 4π P0 )

(15.0 ×10

−3

kg )( 9.80 m/s 2 ) tan 25.0°

= 2.80 × 10−6 C 8.988 × 109 N ⋅ m 2 /C 2 (c) The separation between the two spheres is given by 2 L sin θ . q = 2.80μ C as found in part (b). Fc = (1/ 4π P0 ) q 2 / ( 2 L sin θ ) and Fc = mg tan θ . Thus (1/ 4π P0 ) q 2 / ( 2 L sin θ ) = mg tan θ . 2

( sin θ )

2

( 2.80 ×10 C ) 1 q2 tan θ = = ( 8.988 × 109 N ⋅ m 2 / C2 ) = 0.3328. 2 2 4π P0 4 L mg 4 ( 0.600 m ) (15.0 × 10−3 kg )( 9.80 m/s 2 ) −6

2

2

Solve this equation by trial and error. This will go quicker if we can make a good estimate of the value of θ that solves the equation. For θ small, tan θ ≈ sin θ . With this approximation the equation becomes sin 3 θ = 0.3328 and sin θ = 0.6930, so θ = 43.9°. Now refine this guess:

θ 45.0° 40.0° 39.6° 39.5° 39.4°

21.76.

sin 2 θ tan θ 0.5000 0.3467 0.3361 0.3335 0.3309

so θ = 39.5°

EVALUATE: The expression in part (c) says θ → 0 as L → ∞ and θ → 90° as L → 0. When L is decreased from the value in part (a), θ increases. IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to each sphere. SET UP: (a) Free body diagrams are given in Figure 21.76. Fe is the repulsive electric force that one sphere exerts on the other. kq q EXECUTE: (b) T = mg cos 20° = 0.0834 N , so Fe = T sin 20° = 0.0285 N = 12 2 . (Note: r1

r1 = 2(0.500 m)sin 20° = 0.342 m.) (c) From part (b), q1q2 = 3.71 × 10−13 C2 .


21-31

(d) The charges on the spheres are made equal by connecting them with a wire, but we still have q +q Q2 , where Q = 1 2 . But the separation r2 is known: Fe = mg tan θ = 0.0453 N = 1 2 4π P0 r22 q1 + q2 = 4π P0 Fe r22 = 1.12 × 10 −6 C. This equation, along 2 with that from part (c), gives us two equations in q1 and q2 : q1 + q2 = 2.24 × 10 −6 C and q1q2 = 3.71 × 10 −13 C 2 .

r2 = 2(0.500 m)sin 30° = 0.500 m. Hence: Q =

By elimination, substitution and after solving the resulting quadratic equation, we find: q1 = 2.06 × 10 −6 C and q2 = 1.80 × 10 −7 C . EVALUATE: After the spheres are connected by the wire, the charge on sphere 1 decreases and the charge on sphere 2 increases. The product of the charges on the sphere increases and the thread makes a larger angle with the vertical.

Figure 21.76 21.77.

IDENTIFY and SET UP: Use Avogadro's number to find the number of Na + and Cl − ions and the total positive and negative charge. Use Coulomb's law to calculate the electric force and F = ma to calculate the acceleration. (a) EXECUTE: The number of Na + ions in 0.100 mol of NaCl is N = nN A . The charge of one ion is +e, so the

total charge is q1 = nN A e = (0.100 mol)(6.022 × 1023 ions/mol)(1.602 × 10−19 C/ion) = 9.647 × 103 C There are the same number of Cl − ions and each has charge −e, so q2 = −9.647 × 103 C.

F=

1 q1q2 (9.647 × 103 C) 2 = (8.988 × 109 N ⋅ m 2 /C2 ) = 2.09 × 1021 N 2 4π P0 r (0.0200 m)2

(b) a = F/m. Need the mass of 0.100 mol of Cl − ions. For Cl, M = 35.453 × 10 −3 kg/mol, so F 2.09 × 1021 N m = ( 0.100 mol ) 35.453 × 10−3 kg/mol = 35.45 × 10−4 kg. Then a = = = 5.90 × 1023 m/s 2 . m 35.45 × 10−4 kg (c) EVALUATE: Is is not reasonable to have such a huge force. The net charges of objects are rarely larger than 1 μC; a charge of 104 C is immense. A small amount of material contains huge amounts of positive and negative charges. IDENTIFY: For the acceleration (and hence the force) on Q to be upward, as indicated, the forces due to q1 and q2 must have equal strengths, so q1 and q2 must have equal magnitudes. Furthermore, for the force to be upward, q1 must be positive and q2 must be negative. SET UP: Since we know the acceleration of Q, Newton’s second law gives us the magnitude of the force on it. We can then add the force components using F = FQq1 cosθ + FQq2 cosθ = 2 FQq1 cosθ . The electrical force on Q is

(

21.78.

)

1 Qq1 (for q1) and likewise for q2. 4π P0 r 2 EXECUTE: First find the net force: F = ma = (0.00500 kg)(324 m/s2) = 1.62 N. Now add the force components, calling θ the angle between the line connecting q1 and q2 and the line connecting q1 and Q. F 1.62 N = F = FQq1 cosθ + FQq2 cosθ = 2 FQq1 cosθ and FQq1 = = 1.08 N. Now find the charges by ⎛ 2.25 cm ⎞ 2cosθ 2⎜ ⎟ ⎝ 3.00 cm ⎠ solving for q1 in Coulomb’s law and use the fact that q1 and q2 have equal magnitudes but opposite signs. r 2 FQq1 (0.0300 m) 2 (1.08 N) 1 Qq1 q = = FQq1 = and = 6.17 × 10−8 C. 1 2 −6 9 2 2 1 4π P0 r 9.00 10 N m /C 1.75 10 C × ⋅ × ( )( ) Q 4π P0

given by Coulomb’s law, FQq1 =

q2 = − q1 = −6.17 × 10−8 C.

21-32

21.79.

Chapter 21

EVALUATE: Simple reasoning allows us first to conclude that q1 and q2 must have equal magnitudes but opposite signs, which makes the equations much easier to set up than if we had tried to solve the problem in the general case. As Q accelerates and hence moves upward, the magnitude of the acceleration vector will change in a complicated way. IDENTIFY: Use Coulomb's law to calculate the forces between pairs of charges and sum these forces as vectors to find the net charge. (a) SET UP: The forces are sketched in Figure 21.79a.

F1 + F3 = 0, so the net force is F = F2 .

EXECUTE:

F=

1 q(3q) 6q 2 = , away from the vacant corner. 4π P0 ( L/ 2) 2 4π P0 L2

Figure 21.79a (b) SET UP: The forces are sketched in Figure 21.79b.

EXECUTE:

F1 = F3 =

F2 =

1 q ( 3q ) 3q 2 = 4π P0 2 L 2 4π P0 ( 2 L2 )

(

)

1 q ( 3q ) 3q = 4π P0 L2 4π P0 L2 2

The vector sum of F1 and F3 is F13 = F12 + F32 .

Figure 21.79b

F13 = 2 F1 =

3 2q 2 ; F13 and F2 are in the same direction. 4π P0 L2

3q 2 ⎛ 1⎞ ⎜ 2 + ⎟ , and is directed toward the center of the square. 4π P0 L2 ⎝ 2⎠ EVALUATE: By symmetry the net force is along the diagonal of the square. The net force is only slightly larger when the −3q charge is at the center. Here it is closer to the charge at point 2 but the other two forces cancel. IDENTIFY: Use Eq.(21.7) for the electric field produced by each point charge. Apply the principle of superposition and add the fields as vectors to find the net field. (a) SET UP: The fields due to each charge are shown in Figure 21.80a. F = F13 + F2 =

21.80.

cosθ =

Figure 21.80a

x x + a2 2


21-33

EXECUTE: The components of the fields are given in Figure 21.80b.

E1 = E2 =

E3 =

1 ⎛ q ⎞ ⎜ ⎟ 4π P0 ⎝ a 2 + x 2 ⎠

1 ⎛ 2q ⎞ ⎜ ⎟ 4π P0 ⎝ x 2 ⎠

Figure 21.80b

E1 y = − E1 sin θ , E2 y = + E2 sin θ so E y = E1 y + E2 y = 0. ⎞ 1 ⎛ q ⎞⎛ x ⎟ , E3 x = − E3 ⎜ 2 2 ⎟⎜ 2 2 4π P0 ⎝ a + x ⎠ ⎝ x + a ⎠ ⎛ 1 ⎛ q ⎞⎛ ⎞⎞ x 2q E x = E1x + E2 x + E3 x = 2 ⎜ − ⎜ ⎜ ⎟ 2 2 ⎜ 4π P ⎝ a + x ⎠ x 2 + a 2 ⎟ ⎟⎟ 4π P x 2 0 0 ⎝ ⎠⎠ ⎝ 2q ⎛ 1 x 2q ⎛ 1 ⎞ ⎞ Ex = − 1− − =− 3/ 2 ⎟ 3/ 2 ⎟ ⎜ 2 2 ⎜ 2 2 2 2 4π P0 ⎜ x ( a + x ) ⎟ 4π P0 x ⎜ (1 + a / x ) ⎟ ⎝ ⎠ ⎝ ⎠ 2q ⎛ 1 ⎞ Thus E = 1− , in the − x-direction. 4π P0 x 2 ⎜⎜ (1 + a 2 / x 2 )3 / 2 ⎟⎟ ⎝ ⎠ E1x = E2 x = + E1 cosθ =

(b) x >> a implies a 2 / x 2 c, as shown in Figure 22.37b. EXECUTE:

Φ E = E ( 2π rl )

Qencl = λl (the charge on the length l of the inner conductor that is inside the Gaussian surface; the outer conductor carries no net charge). Figure 22.37b

Qencl λl gives E ( 2π rl ) = P0 P0 G λ E= . The enclosed charge is positive so the direction of E is radially outward. 2π P0 r (c) E = 0 within a conductor. Thus E = 0 for r < a; ΦE =

λ for a < r < b; E = 0 for b < r < c; 2π P0 r λ E= for r > c. The graph of E versus r is sketched in Figure 22.37c. 2π P0 r E=

Figure 22.37c

22.38.

EVALUATE: Inside either conductor E = 0. Between the conductors and outside both conductors the electric field is the same as for a line of charge with linear charge density λ lying along the axis of the inner conductor. (d) IDENTIFY and SET UP: inner surface: Apply Gauss’s law to a Gaussian cylinder with radius r, where b < r < c. We know E on this surface; calculate Qencl . EXECUTE: This surface lies within the conductor of the outer cylinder, where E = 0, so Φ E = 0. Thus by Gauss’s law Qencl = 0. The surface encloses charge λ l on the inner conductor, so it must enclose charge − λ l on the inner surface of the outer conductor. The charge per unit length on the inner surface of the outer cylinder is − λ . outer surface: The outer cylinder carries no net charge. So if there is charge per unit length − λ on its inner surface there must be charge per until length + λ on the outer surface. EVALUATE: The electric field lines between the conductors originate on the surface charge on the outer surface of the inner conductor and terminate on the surface charges on the inner surface of the outer conductor. These surface charges are equal in magnitude (per unit length) and opposite in sign. The electric field lines outside the outer conductor originate from the surface charge on the outer surface of the outer conductor. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r, length l and that has the line of charge along its axis. The charge on a length l of the line of charge or of the tube is q = α l . α Q αl EXECUTE: (a) (i) For r < a , Gauss’s law gives E (2πrl ) = encl = and E = . P0 P0 2πP0 r (ii) The electric field is zero because these points are within the conducting material. Q 2αl α (iii) For r > b , Gauss’s law gives E (2πrl ) = encl = and E = . P0 P0 πP0 r The graph of E versus r is sketched in Figure 22.38.

22-12

Chapter 22

(b) (i) The Gaussian cylinder with radius r, for a < r < b , must enclose zero net charge, so the charge per unit length on the inner surface is −α . (ii) Since the net charge per length for the tube is +α and there is −α on the inner surface, the charge per unit length on the outer surface must be +2α . EVALUATE: For r > b the electric field is due to the charge on the outer surface of the tube.

Figure 22.38 22.39.

(a) IDENTIFY: Use Gauss’s law to calculate E(r). (i) SET UP: r < a: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r < a, as sketched in Figure 22.39a. EXECUTE:

Φ E = E ( 2π rl )

Qencl = α l (the charge on the length l of the line of charge) Figure 22.39a ΦE =

Qencl αl gives E ( 2π rl ) = P0 P0

G α . The enclosed charge is positive so the direction of E is radially outward. 2π P0 r (ii) a < r < b: Points in this region are within the conducting tube, so E = 0. (iii) SET UP: r > b: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r > b, as sketched in Figure 22.39b.

E=

EXECUTE:

Φ E = E ( 2π rl )

Qencl = α l (the charge on length l of the line of charge) −α l (the charge on length l of the tube) Thus Qencl = 0. Figure 22.39b ΦE =

Qencl gives E ( 2π rl ) = 0 and E = 0. The graph of E versus r is sketched in Figure 22.39c. P0

Figure 22.39c (b) IDENTIFY: Apply Gauss’s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the tube. We know E so can calculate Qencl . (i) SET UP: inner surface Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where a < r < b.

Gauss’s Law

EXECUTE:

22.40.

22-13

This surface lies within the conductor of the tube, where E = 0, so Φ E = 0. Then by Gauss’s law

Qencl = 0. The surface encloses charge α l on the line of charge so must enclose charge −α l on the inner surface of the tube. The charge per unit length on the inner surface of the tube is −α . (ii) outer surface The net charge per unit length on the tube is −α . We have shown in part (i) that this must all reside on the inner surface, so there is no net charge on the outer surface of the tube. EVALUATE: For r < a the electric field is due only to the line of charge. For r > b the electric field of the tube is the same as for a line of charge along its axis. The fields of the line of charge and of the tube are equal in magnitude and opposite in direction and sum to zero. For r < a the electric field lines originate on the line of charge and terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no surface charge on the outer surface of the tube. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r and length l, and that is coaxial with the cylindrical charge distributions. The volume of the Gaussian cylinder is π r 2l and the area of its curved surface is 2π rl . The charge on a length l of the charge distribution is q = λl , where λ = ρπ R 2 .

EXECUTE:

(a) For r < R , Qencl = ρπ r 2l and Gauss’s law gives E (2πrl ) =

Qencl ρπr 2l ρr and E = , radially = 2P0 P0 P0

outward. (b) For r > R , Qencl =λl = ρπR2l and Gauss’s law gives E (2πrl ) =

q ρπR 2l ρR 2 and E = = = λ , radially 2P0 r 2πP0 r P0 P0

outward. (c) At r = R, the electric field for BOTH regions is E =

ρR , so they are consistent. 2P0

(d) The graph of E versus r is sketched in Figure 22.40. EVALUATE: For r > R the field is the same as for a line of charge along the axis of the cylinder.

Figure 22.40 22.41.

IDENTIFY: First make a free-body diagram of the sphere. The electric force acts to the left on it since the electric field due to the sheet is horizontal. Since it hangs at rest, the sphere is in equilibrium so the forces on it add to zero, by Newton’s first law. Balance horizontal and vertical force components separately. SET UP: Call T the tension in the thread and E the electric field. Balancing horizontal forces gives T sin θ = qE. Balancing vertical forces we get T cos θ = mg. Combining these equations gives tan θ = qE/mg, which means that θ = arctan(qE/mg). The electric field for a sheet of charge is E = σ 2P0 . EXECUTE:

Substituting the numbers gives us E =

σ 2P0

=

2.50 × 10−7 C/m 2 = 1.41 × 104 N/C . Then 2 ( 8.85 × 10−12 C2 /N ⋅ m 2 )

⎡ ( 5.00 × 10−8 C )(1.41× 104 N/C ) ⎤ ⎥ = 19.8° −2 2 ⎢⎣ ( 2.00 × 10 kg )( 9.80 m/s ) ⎥⎦

θ = arctan ⎢ EVALUATE: angle.

Increasing the field, or decreasing the mass of the sphere, would cause the sphere to hang at a larger

22-14

Chapter 22

22.42.

IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the conducting spheres. EXECUTE: (a) For r < a, E = 0, since these points are within the conducting material.

q For a < r < b, E = 1 2 , since there is +q inside a radius r. 4π P0 r For b < r < c, E = 0, since since these points are within the conducting material 1 q , since again the total charge enclosed is +q. For r > c, E = 4π P0 r 2 (b) The graph of E versus r is sketched in Figure 22.42a. (c) Since the Gaussian sphere of radius r, for b < r < c , must enclose zero net charge, the charge on inner shell surface is –q. (d) Since the hollow sphere has no net charge and has charge −q on its inner surface, the charge on outer shell surface is +q. (e) The field lines are sketched in Figure 22.42b. Where the field is nonzero, it is radially outward. EVALUATE: The net charge on the inner solid conducting sphere is on the surface of that sphere. The presence of the hollow sphere does not affect the electric field in the region r < b .

Figure 22.42 22.43.

IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charge distributions. EXECUTE: (a) For r < R, E = 0, since these points are within the conducting material. For R < r < 2 R,

1 2Q Q since the charge enclosed is 2Q. E = 1 2 , since the charge enclosed is Q. For r > 2 R , E = 4πP0 r 4π P0 r 2 (b) The graph of E versus r is sketched in Figure 22.43. EVALUATE: For r < 2 R the electric field is unaffected by the presence of the charged shell.

Figure 22.43 22.44.

IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: Use a Gaussian surface that is a sphere of radius r and that has the point charge at its center. 1 Q , radially outward, since the charge enclosed is Q, the charge of the point EXECUTE: (a) For r < a , E = 4π P0 r 2 2Q charge. For a < r < b , E = 0 since these points are within the conducting material. For r > b , E = 1 , 4πP0 r 2 radially inward, since the total enclosed charge is –2Q.

Gauss’s Law

22-15

(b) Since a Gaussian surface with radius r, for a < r < b , must enclose zero net charge, the total charge on the inner Q surface is −Q and the surface charge density on inner surface is σ = − . 4πa 2 (c) Since the net charge on the shell is −3Q and there is −Q on the inner surface, there must be −2Q on the outer 2Q . 4π b 2 (d) The field lines and the locations of the charges are sketched in Figure 22.44a. (e) The graph of E versus r is sketched in Figure 22.44b.

surface.

The surface charge density on the outer surface is σ = −

Figure 22.44

22.45.

EVALUATE: For r < a the electric field is due solely to the point charge Q. For r > b the electric field is due to the charge −2Q that is on the outer surface of the shell. IDENTIFY: Apply Gauss’s law to a spherical Gaussian surface with radius r. Calculate the electric field at the surface of the Gaussian sphere. (a) SET UP: (i) r < a : The Gaussian surface is sketched in Figure 22.45a.

EXECUTE:

Φ E = EA = E (4π r 2 )

Qencl = 0; no charge is enclosed ΦE =

Qencl says E (4π r 2 ) = 0 and E = 0. P0

Figure 22.45a

(ii) a < r < b: Points in this region are in the conductor of the small shell, so E = 0. (iii) SET UP: b < r < c: The Gaussian surface is sketched in Figure 22.45b. Apply Gauss’s law to a spherical Gaussian surface with radius b < r < c.

EXECUTE: Φ E = EA = E (4π r 2 ) The Gaussian surface encloses all of the small shell and none of the large shell, so Qencl = +2q.

Figure 22.45b

Qencl 2q 2q gives E (4π r 2 ) = so E = . Since the enclosed charge is positive the electric field is radially 4π P0 r 2 P0 P0 outward. (iv) c < r < d : Points in this region are in the conductor of the large shell, so E = 0. ΦE =

22-16

Chapter 22

(v) SET UP:

r > d : Apply Gauss’s law to a spherical Gaussian surface with radius r > d, as shown in Figure 22.45c.

EXECUTE:

Φ E = EA = E ( 4π r 2 )

The Gaussian surface encloses all of the small shell and all of the large shell, so Qencl = +2q + 4q = 6q.

Figure 22.45c

Qencl 6q gives E ( 4π r 2 ) = P0 P0 6q E= . Since the enclosed charge is positive the electric field is radially outward. 4π P0 r 2 The graph of E versus r is sketched in Figure 22.45d. ΦE =

Figure 22.45d (b) IDENTIFY and SET UP: Apply Gauss’s law to a sphere that lies outside the surface of the shell for which we want to find the surface charge. EXECUTE: (i) charge on inner surface of the small shell: Apply Gauss’s law to a spherical Gaussian surface with radius a < r < b. This surface lies within the conductor of the small shell, where E = 0, so Φ E = 0. Thus by Gauss’s

22.46.

law Qencl = 0, so there is zero charge on the inner surface of the small shell. (ii) charge on outer surface of the small shell: The total charge on the small shell is +2q. We found in part (i) that there is zero charge on the inner surface of the shell, so all +2q must reside on the outer surface. (iii) charge on inner surface of large shell: Apply Gauss’s law to a spherical Gaussian surface with radius c < r < d . The surface lies within the conductor of the large shell, where E = 0, so Φ E = 0. Thus by Gauss’s law Qencl = 0. The surface encloses the +2q on the small shell so there must be charge −2q on the inner surface of the large shell to make the total enclosed charge zero. (iv) charge on outer surface of large shell: The total charge on the large shell is +4q. We showed in part (iii) that the charge on the inner surface is −2q, so there must be +6q on the outer surface. EVALUATE: The electric field lines for b < r < c originate from the surface charge on the outer surface of the inner shell and all terminate on the surface charge on the inner surface of the outer shell. These surface charges have equal magnitude and opposite sign. The electric field lines for r > d originate from the surface charge on the outer surface of the outer sphere. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. EXECUTE: (a) (i) For r < a, E = 0, since the charge enclosed is zero. (ii) For a < r < b, E = 0, since the points 1 2q , outward, since charge enclosed is +2q. 4π P0 r 2 (iv) For c < r < d , E = 0, since the points are within the conducting material. (v) For r > d , E = 0, since the net charge enclosed is zero. The graph of E versus r is sketched in Figure 22.46.

are within the conducting material. (iii) For b < r < c, E =

Gauss’s Law

22-17

(b) (i) small shell inner surface: Since a Gaussian surface with radius r, for a < r < b , must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: +2q . (iii) large shell inner surface: Since a Gaussian surface with radius r, for c < r < d , must enclose zero net charge, the charge on this surface is −2q . (iv) large shell outer surface: Since there is −2q on the inner surface and the total charge on this conductor is −2q , the charge on this surface is zero. EVALUATE: The outer shell has no effect on the electric field for r < c . For r > d the electric field is due only to the charge on the outer surface of the larger shell.

Figure 22.46 22.47.

IDENTIFY: Apply Gauss’s law SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. EXECUTE: (a) (i) For r < a, E = 0, since charge enclosed is zero. (ii) a < r < b, E = 0, since the points are 2q within the conducting material. (iii) For b < r < c, E = 1 2 , outward, since charge enclosed is +2q. 4πP0 r 2q (iv) For c < r < d , E = 0, since the points are within the conducting material. (v) For r > d , E = 1 2 , inward, 4πP0 r since charge enclosed is –2q. The graph of the radial component of the electric field versus r is sketched in Figure 22.47, where we use the convention that outward field is positive and inward field is negative. (b) (i) small shell inner surface: Since a Gaussian surface with radius r, for a < r < b, must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: +2q. (iii) large shell inner surface: Since a Gaussian surface with radius r, for c < r < d, must enclose zero net charge, the charge on this surface is –2q. (iv) large shell outer surface: Since there is –2q on the inner surface and the total charge on this conductor is –4q, the charge on this surface is –2q. EVALUATE: The outer shell has no effect on the electric field for r < c . For r > d the electric field is due only to the charge on the outer surface of the larger shell.

Figure 22.47 22.48.

IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the sphere and shell. The 4 28π 3 R . volume of the insulating shell is V = π [2 R]3 − R3 = 3 3 3Q 28π ρR 3 EXECUTE: (a) Zero net charge requires that −Q = , so ρ = − . 28πR 3 3 (b) For r < R, E = 0 since this region is within the conducting sphere. For r > 2 R, E = 0, since the net charge enclosed Q 4π ρ 3 (r − R3 ) and by the Gaussian surface with this radius is zero. For R < r < 2R , Gauss’s law gives E (4πr 2 ) = + P0 3P0 Q ρ Q Qr E= + (r 3 − R 3 ) . Substituting ρ from part (a) gives E = 2 2 − . The net enclosed charge for 7πP0 r 28πP0 R 3 4πP0 r 2 3P0r 2 each r in this range is positive and the electric field is outward.

(

)

22-18

Chapter 22

(c) The graph is sketched in Figure 22.48. We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere. But we see a smooth transition from the uniform insulator to the surrounding space. EVALUATE: The expression for E within the insulator gives E = 0 at r = 2 R .

22.49.

Figure 22.48 G IDENTIFY: Use Gauss’s law to find the electric field E produced by the shell for r < R and r > R and then use G G F = qE to find the force the shell exerts on the point charge. (a) SET UP: Apply Gauss’s law to a spherical Gaussian surface that has radius r > R and that is concentric with the shell, as sketched in Figure 22.49a.

EXECUTE:

Φ E = E ( 4π r 2 )

Qencl = −Q

Figure 22.49a

ΦE =

Qencl −Q gives E ( 4π r 2 ) = P0 P0

Q qQ and it is directed toward the center of the shell. Then F = qE = , 4π P0 r 2 4π P0 r 2 G G directed toward the center of the shell. (Since q is positive, E and F are in the same direction.) (b) SET UP: Apply Gauss’s law to a spherical Gaussian surface that has radius r < R and that is concentric with the shell, as sketched in Figure 22.49b.

The magnitude of the field is E =

EXECUTE:

Φ E = E ( 4π r 2 )

Qencl = 0

Figure 22.49b

Qencl gives E ( 4π r 2 ) = 0 P0 Then E = 0 so F = 0. EVALUATE: Outside the shell the electric field and the force it exerts is the same as for a point charge −Q located at the center of the shell. Inside the shell E = 0 and there is no force. IDENTIFY: The method of Example 22.9 shows that the electric field outside the sphere is the same as for a point charge of the same charge located at the center of the sphere. SET UP: The charge of an electron has magnitude e = 1.60 × 10−19 C . ΦE =

22.50.

Gauss’s Law

22-19

q Er 2 (1150 N/C)(0.150 m) 2 . For r = R = 0.150 m , E = 1150 N/C so q = = = 2.88 × 10−9 C . 2 k r 8.99 × 109 N ⋅ m 2 /C2 2.88 × 10−9 C = 1.80 × 1010 electrons . The number of excess electrons is 1.60 × 10−19 C/electron EXECUTE:

(a) E = k

(b) r = R + 0.100 m = 0.250 m . E = k

22.51.

q r

2

= (8.99 × 109 N ⋅ m 2 /C 2 )

2.88 × 10−9 C = 414 N/C . (0.250 m) 2

EVALUATE: The magnitude of the electric field decreases according to the square of the distance from the center of the sphere. IDENTIFY: The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the plate. SET UP: The electric field due to the sheet of charge on each surface is E = σ / 2P0 and is directed away from the surface. EXECUTE: (a) For the conductor the charge sheet on each surface produces fields of magnitude σ / 2P0 and in the

same direction, so the total field is twice this, or σ / P0 . (b) At points inside the plate the fields of the sheets of charge on each surface are equal in magnitude and opposite in direction, so their vector sum is zero. At points outside the plate, on either side, the fields of the two sheets of charge are in the same direction so their magnitudes add, giving E = σ / P0 . 22.52.

EVALUATE: Gauss’s law can also be used directly to determine the fields in these regions. IDENTIFY: Example 22.9 gives the expression for the electric field both inside and outside a uniformly charged G G sphere. Use F = −eE to calculate the force on the electron. SET UP: The sphere has charge Q = + e . EXECUTE:

(a) Only at r = 0 is E = 0 for the uniformly charged sphere.

(b) At points inside the sphere, Er =

er 1 e2r . The field is radially outward. Fr = −eE = − . The minus sign 3 4π P0 R 3 4πP0 R

denotes that Fr is radially inward. For simple harmonic motion, Fr = − kr = − mω2 r , where ω = k / m = 2π f . Fr = − mω2 r = −

1 e2r 1 e2 1 1 e2 so ω = and f = . 3 3 4πP0 R 4πP0 mR 2π 4πP0 mR 3

(c) If f = 4.57 × 1014 Hz =

1 2π

1 e2 1 (1.60 × 10−19 C) 2 then R = 3 = 3.13 × 10−10 m. 3 2 4πP0 mR 4πP0 4π (9.11 × 10−31 kg)(4.57 × 1014 Hz) 2

The atom radius in this model is the correct order of magnitude. e e2 (d) If r > R , Er = and . The electron would still oscillate because the force is directed toward = − F r 4π P0 r 2 4π P0 r 2

22.53.

the equilibrium position at r = 0 . But the motion would not be simple harmonic, since Fr is proportional to 1/ r 2 and simple harmonic motion requires that the restoring force be proportional to the displacement from equilibrium. EVALUATE: As long as the initial displacement is less than R the frequency of the motion is independent of the initial displacement. IDENTIFY: There is a force on each electron due to the other electron and a force due to the sphere of charge. Use Coulomb’s law for the force between the electrons. Example 22.9 gives E inside a uniform sphere and Eq.(21.3) gives the force. SET UP: The positions of the electrons are sketched in Figure 22.53a.

If the electrons are in equilibrium the net force on each one is zero.

Figure 22.53a

22-20

Chapter 22

EXECUTE:

Consider the forces on electron 2. There is a repulsive force F1 due to the other electron, electron 1. F1 =

1 e2 4π P0 ( 2d )2

The electric field inside the uniform distribution of positive charge is E =

Qr (Example 22.9), where Q = +2e. 4π P0 R 3

At the position of electron 2, r = d. The force Fcd exerted by the positive charge distribution is Fcd = eE =

e ( 2e ) d 4π P0 R 3

and is attractive. The force diagram for electron 2 is given in Figure 22.53b.

Figure 22.53b 1 e2 2e 2 d Net force equals zero implies F1 = Fcd and = 2 4π P0 4d 4π P0 R 3

Thus (1/ 4d 2 ) = 2d / R 3 , so d 3 = R 3 /8 and d = R / 2.

22.54.

EVALUATE: The electric field of the sphere is radially outward; it is zero at the center of the sphere and increases with distance from the center. The force this field exerts on one of the electrons is radially inward and increases as the electron is farther from the center. The force from the other electron is radially outward, is infinite when d = 0 and decreases as d increases. It is reasonable therefore for there to be a value of d for which these forces balance. IDENTIFY: Use Gauss’s law to find the electric field both inside and outside the slab. SET UP: Use a Gaussian surface that has one face of area A in the y z plane at x = 0, and the other face at a general value x. The volume enclosed by such a Gaussian surface is Ax. EXECUTE: (a) The electric field of the slab must be zero by symmetry. There is no preferred direction in the y z plane, so the electric field can only point in the x-direction. But at the origin, neither the positive nor negative x-directions should be singled out as special, and so the field must be zero. ρA x ρx x ˆ Q (b) For x ≤ d , Gauss’s law gives EA = encl = and E = , with direction given by i (away from the P0 P0 P0 |x| center of the slab). Note that this expression does give E = 0 at x = 0. Outside the slab, the enclosed charge does Q ρAd ρd not depend on x and is equal to ρ Ad . For x ≥ d , Gauss’s law gives EA = encl = and E = , again with P0 P0 P0

x ˆ i. | x| EVALUATE: At the surfaces of the slab, x = ± d . For these values of x the two expressions for E (for inside and outside the slab) give the same result. The charge per unit area σ of the slab is given by σ A = ρ A(2d ) and

direction given by

22.55.

ρ d = σ / 2 . The result for E outside the slab can therefore be written as E = σ / 2P0 and is the same as for a thin sheet of charge. (a) IDENTIFY and SET UP: Consider the direction of the field for x slightly greater than and slightly less than zero. The slab is sketched in Figure 22.55a.

ρ ( x ) = ρ0 ( x / d )

2


The charge distribution is symmetric about x = 0, so by symmetry E ( x ) = E ( − x ) . But for x > 0 the

field is in the + x direction and for x < 0 the field is in the − x direction. At x = 0 the field can’t be both in the + x and − x directions so must be zero. That is, Ex ( x ) = − Ex ( − x ) . At point x = 0 this gives Ex ( 0 ) = − Ex ( 0 ) and this equation is satisfied only for Ex ( 0 ) = 0.

Gauss’s Law


22-21

x > d (outside the slab)

Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A and are the same distance x > d from x = 0, as shown in Figure 22.55b.

EXECUTE:

Φ E = 2 EA

Figure 22.55b

To find Qencl consider a thin disk at coordinate x and with thickness dx, as shown in Figure 22.55c. The charge within this disk is dq = ρ dV = ρ Adx = ( ρ0 A / d 2 ) x 2 dx. Figure 22.55c

The total charge enclosed by the Gaussian cylinder is Qencl = 2 ∫ dq = ( 2 ρ0 A / d 2 ) ∫ x 2 dx = ( 2 ρ 0 A / d 2 )( d 3 / 3) = 32 ρ0 Ad . Then Φ E =

d

d

0

0

Qencl gives 2 EA = 2 ρ 0 Ad / 3P0 . P0

E = ρ 0 d / 3P0 G G E is directed away from x = 0, so E = ( ρ 0 d / 3P0 ) ( x / x ) iˆ. IDENTIFY and SET UP:

x < d (inside the slab)

Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A and are the same distance x < d from x = 0, as shown in Figure 22.55d.

EXECUTE:

Φ E = 2 EA

Figure 22.55d

Qencl is found as above, but now the integral on dx is only from 0 to x instead of 0 to d. Qencl = 2 ∫ dq = ( 2 ρ0 A / d 2 ) ∫ x 2 dx = ( 2 ρ 0 A / d 2 )( x 3 / 3) .

Then Φ E =

x

x

0

0

Qencl gives 2 EA = 2 ρ 0 Ax3 / 3P0 d 2 . P0 E = ρ 0 x 3 / 3P0 d 2

G G E is directed away from x = 0, so E = ( ρ0 x3 / 3P0 d 2 ) iˆ. EVALUATE: 22.56.

Note that E = 0 at x = 0 as stated in part (a). Note also that the expressions for x > d and x < d

agree for x = d . G G IDENTIFY: Apply F = qE to relate the force on q to the electric field at the location of q. SET UP: Flux is negative if the electric field is directed into the enclosed volume.

22-22

Chapter 22

EXECUTE: (a) We could place two charges +Q on either side of the charge + q , as shown in Figure 22.56. (b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the equilibrium position. (c) If q is moved to infinity and we require there to be an electric field always pointing in to the region where q had been, we could draw a small Gaussian surface there. We would find that we need a negative flux into the surface. That is, there has to be a negative charge in that region. However, there is none, and so we cannot get such a stable equilibrium. (d) For a negative charge to be in stable equilibrium, we need the electric field to always point away from the charge position. The argument in (c) carries through again, this time implying that a positive charge must be in the space where the negative charge was if stable equilibrium is to be attained. EVALUATE: If q is displaced to the left or right in Figure 22.56, the net force is directed back toward the equilibrium position. But if q is displaced slightly in a direction perpendicular to the line connecting the two charges Q, then the net force on q is directed away from the equilibrium position and the equilibrium is not stable for such a displacement.

22.57.

Figure 22.56 ρ ( r ) = ρ 0 (1 − r / R ) for r ≤ R where ρ 0 = 3Q / π R . ρ ( r ) = 0 for r ≥ R 3

(a) IDENTIFY: The charge density varies with r inside the spherical volume. Divide the volume up into thin concentric shells, of radius r and thickness dr. Find the charge dq in each shell and integrate to find the total charge. SET UP: The thin shell is sketched in Figure 22.57a. EXECUTE: The volume of such a shell is dV = 4π r 2 dr The charge contained within the shell is dq = ρ ( r ) dV = 4π r 2 ρ 0 (1 − r / R ) dr Figure 22.57a

The total charge Q in the charge distribution is obtained by integrating dq over all such shells into which the sphere can be subdivided: R

R

0

0

Q = ∫ dq = ∫ 4π r 2 ρ0 (1 − r / R ) dr = 4πρ 0 ∫

(r

2

− r 3 / R ) dr

R

⎡ r3 r 4 ⎤ ⎛ R3 R 4 ⎞ 3 3 3 − Q = 4πρ 0 ⎢ − ⎟ = 4πρ 0 ( R /12 ) = 4π ( 3Q / π R )( R /12 ) = Q, as was to be shown. ⎥ = 4πρ 0 ⎜ 3 4 3 4 R R ⎣ ⎦0 ⎝ ⎠ (b) IDENTIFY: Apply Gauss’s law to a spherical surface of radius r, where r > R. SET UP: The Gaussian surface is shown in Figure 22.57b. EXECUTE:

E ( 4π r 2 ) =

ΦE =

Qencl P0

Q P0

Figure 22.57b

Q ; same as for point charge of charge Q. 4π P0 r 2 (c) IDENTIFY: Apply Gauss’s law to a spherical surface of radius r, where r < R: SET UP: The Gaussian surface is shown in Figure 22.57c. E=

EXECUTE:

ΦE =

Φ E = E ( 4π r 2 ) Figure 22.57c

Qencl P0

Gauss’s Law

22-23

The calculate the enclosed charge Qencl use the same technique as in part (a), except integrate dq out to r rather than R. (We want the charge that is inside radius r.) r r⎛ r′3 ⎞ ⎛ r′ ⎞ Qencl = ∫ 4π r ′2 ρ0 ⎜1 − ⎟ dr ′ = 4πρ 0 ∫ ⎜ r′2 − ⎟dr ′ 0 0 R⎠ ⎝ R⎠ ⎝ r

⎡ r ′3 r ′4 ⎤ ⎛ r3 r4 ⎞ r ⎞ 3⎛1 Qencl = 4πρ0 ⎢ − ⎟ = 4πρ 0 r ⎜ − ⎥ = 4πρ 0 ⎜ − ⎟ 3 4 3 4 3 4 R R R⎠ ⎝ ⎣ ⎦0 ⎝ ⎠

ρ0 =

⎛ r3 ⎞⎛ 3Q r3 ⎛ 1 r ⎞ r⎞ = − = so 12 Q Q Q ⎜ 3 ⎟ ⎜ 4 − 3 ⎟. encl ⎟ 3 3⎜ πR R ⎝ 3 4R ⎠ R⎠ ⎝ R ⎠⎝

Thus Gauss’s law gives E ( 4π r 2 ) =

Q ⎛ r3 ⎞⎛ r⎞ ⎜ 3 ⎟⎜ 4 − 3 ⎟ P0 ⎝ R ⎠ ⎝ R⎠

3r ⎞ Qr ⎛ ⎜ 4 − ⎟, r ≤ R R⎠ 4π P0 R 3 ⎝ (d) The graph of E versus r is sketched in Figure 22.57d. E=

Figure 22.57d dE d ⎛ 3r 2 ⎞ = 0. Thus (e) Where the electric field is a maximum, ⎜ 4r − ⎟ = 0 so 4 − 6r / R = 0 and r = 2 R / 3. dr dr ⎝ R ⎠

At this value of r, E = EVALUATE:

3 2R ⎞ Q ⎛ 2 R ⎞⎛ Q ⎜ ⎟⎜ 4 − ⎟= R 3 ⎠ 3π P0 R 2 4π P0 R 3 ⎝ 3 ⎠⎝

Our expressions for E ( r ) for r < R and for r > R agree at r = R. The results of part (e) for the

value of r where E ( r ) is a maximum agrees with the graph in part (d). 22.58.

IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the spherical distribution of charge. The volume of a thin spherical shell of radius r and thickness dr is dV = 4π r 2 dr . ∞ R R ⎡R 2 4r ⎞ 2 4 3 ⎤ ⎛ EXECUTE: (a) Q = ∫ ρ(r )dV = 4π ∫ ρ(r )r 2 dr = 4πρ0 ∫ ⎜ 1 − r dr ⎥ ⎟ r dr = 4πρ0 ⎢ ∫ r dr − 3R ⎠ 3R ∫0 0 0⎝ ⎣0 ⎦

⎡ R3 4 R 4 ⎤ Q = 4πρ0 ⎢ − ⎥ = 0 . The total charge is zero. ⎣ 3 3R 4 ⎦ G G Q (b) For r ≥ R , ú E ⋅ dA = encl = 0 , so E = 0 . P0 G G Qencl 4π r 4πρ0 ⎡ r 2 4 r 3 ⎤ = (c) For r ≤ R , ú E ⋅ dA = ρ( r′)r ′2 dr ′ . E 4πr 2 = r′ dr ′ − r ′ dr′ ⎥ and ∫ ∫ ⎢ 0 0 P0 P0 P0 ⎣ 3R ∫ 0 ⎦ ρ0 1 ⎡ r 3 r 4 ⎤ ρ0 ⎡ r⎤ − r ⎢1 − ⎥ . ⎥= 2 ⎢ P0 r ⎣ 3 3R ⎦ 3P0 ⎣ R ⎦ (d) The graph of E versus r is sketched in Figure 22.58. 2ρ r ρ R ⎡ 1⎤ ρ R ρ R dE (e) Where E is a maximum, = 0 . This gives 0 − 0 max = 0 and rmax = . At this r, E = 0 ⎢1 − ⎥ = 0 . 3P0 2 ⎣ 2 ⎦ 12P0 2 dr 3P0 3P0 R E=

22-24

Chapter 22

EVALUATE: The result in part (b) for r ≤ R gives E = 0 at r = R ; the field is continuous at the surface of the charge distribution.

Figure 22.58 22.59.

IDENTIFY: SET UP:

Follow the steps specified in the problem. G In spherical polar coordinates dA = r 2 sin θ dθ dφ rˆ .

ú sin θ dθ dφ = 4π .

r sinθ dθ dφ G G (a) Φ g = ú g ⋅ dA = −Gm ú = −4πGm. r2 (b) For any closed surface, mass OUTSIDE the surface contributes zero to the flux passing through the surface. Thus the formula above holds for any situation where m is the mass enclosed by the Gaussian surface. G G That is, Φ g = ú g ⋅ dA = −4πGM encl . 2

EXECUTE:

EVALUATE: 22.60.

IDENTIFY: SET UP: G G ú g ⋅ dA

The minus sign in the expression for the flux signifies that the flux is directed inward. G G Apply ú g ⋅ dA = −4πGM encl .

Use a Gaussian surface that is a sphere of radius r, concentric with the mass distribution. Let Φ g denote

EXECUTE: (a) Use a Gaussian sphere with radius r > R , where R is the radius of the mass distribution. g is constant on this surface and the flux is inward. The enclosed mass is M. Therefore, Φ g = − g 4πr 2 = −4πGM and

GM , which is the same as for a point mass. r2 (b) For a Gaussian sphere of radius r < R , where R is the radius of the shell, M encl = 0, so g = 0. g=

⎛4 ⎞ (c) Use a Gaussian sphere of radius r < R , where R is the radius of the planet. Then M encl = ρ ⎜ π r 3 ⎟ = Mr 3 / R 3 . ⎝3 ⎠

22.61.

⎛ r3 ⎞ GMr This gives Φ g = − g 4πr 2 = − 4πGM encl = − 4πG ⎜ M 3 ⎟ and g = 3 , which is linear in r. R ⎝ R ⎠ G EVALUATE: The spherically synmetric distribution of mass results in an acceleration due to gravity g that is radical and that depends only on r, the distance from the center of the mass distribution. G G (a) IDENTIFY: Use E ( r ) from Example (22.9) (inside the sphere) and relate the position vector of a point inside the sphere measured from the origin to that measured from the center of the sphere. SET UP: For an insulating sphere of uniform charge density ρ and centered at the origin, the electric field inside G the sphere is given by E = Qr ′ / 4π P0 R3 (Example 22.9), where r ′ is the vector from the center of the sphere to the point where E is calculated. G But ρ = 3Q / 4π R 3 so this may be written as E = ρ r / 3P0 . And E is radially outward, in the direction of G G G r ′, so E = ρ r ′ / 3P0 .

Gauss’s Law

22-25

G G For a sphere whose center is located by vector b , a point inside the sphere and located by r is located by the vector G G G r ′ = r − b relative to the center of the sphere, as shown in Figure 22.61.

EXECUTE:

G G G ρ r −b Thus E = 3P0

(

)

Figure 22.61

G G EVALUATE: When b = 0 this reduces to the result of Example 22.9. When r = b , this gives E = 0, which is correct since we know that E = 0 at the center of the sphere. (b) IDENTIFY: The charge distribution can be represented as a uniform sphere with charge density ρ and centered G G at the origin added to a uniform sphere with charge density − ρ and centered at r = b. G G G G SET UP: E = Euniform + Ehole , where Euniform is the field of a uniformly charged sphere with charge density ρ and G Ehole is the field of a sphere located at the hole and with charge density − ρ . (Within the spherical hole the net charge density is + ρ − ρ = 0. ) G G ρr G EXECUTE: E uniform = , where r is a vector from the center of the sphere. 3P0 G G −ρ r − b G , at points inside the hole. Ehole = 3P0 G G G G ρ rG ⎛ − ρ r − b ⎞ ρ b ⎟= Then E = . +⎜ ⎟ 3P0 3P0 ⎜ 3P0 ⎝ ⎠ G G G EVALUATE: E is independent of r so is uniform inside the hole. The direction of E inside the hole is in the G direction of the vector b , the direction from the center of the insulating sphere to the center of the hole. IDENTIFY: We first find the field of a cylinder off-axis, then the electric field in a hole in a cylinder is the difference between two electric fields: that of a solid cylinder on-axis, and one off-axis, at the location of the hole. G G SET UP: Let r locate a point within the hole, relative to the axis of the cylinder and let r ′ locate this point relative G to the axis of the hole. Let b locate the axis of the hole relative to the axis of the cylinder. As shown in Figure 22.62, G ρ rG G G G r ′ = r − b . Problem 23.48 shows that at points within a long insulating cylinder, E = . 2P0 G G G G G G G G G G ρr ′ ρ ( r − b ) G ρ r ρ( r − b ) ρ b EXECUTE: Eoff-axis = . Ehole = Ecylinder − Eoff-axis = . = − = 2P0 2P0 2P0 2P0 2P0 G Note that E is uniform. EVALUATE: If the hole is coaxial with the cylinder, b = 0 and Ehole = 0 .

(

)

(

22.62.

)

Figure 22.62

22-26

Chapter 22

22.63.

IDENTIFY: SET UP:

ρ=

4 3

The electric field at each point is the vector sum of the fields of the two charge distributions. ρr . Inside a sphere of uniform positive charge, E = 3P0

Q 3Q Qr = so E = , directed away from the center of the sphere. Outside a sphere of uniform 3 3 4π P0 R 3 π R 4π R

Q , directed away from the center of the sphere. 4π P0 r 2 EXECUTE: (a) x = 0. This point is inside sphere 1 and outside sphere 2. The fields are shown in Figure 22.63a.

positive charge, E =

E1 =

Qr = 0, since r = 0. 4π P0 R 3

Figure 22.63a

Q Q with r = 2 R so E2 = , in the − x-direction. 2 4π P0 r 16π P0 R 2 G G G Q Thus E = E1 + E2 = iˆ. 16π P0 R 2 (b) x = R / 2. This point is inside sphere 1 and outside sphere 2. Each field is directed away from the center of the sphere that produces it. The fields are shown in Figure 22.63b. E2 =

E1 =

Qr with r = R / 2 so 4π P0 R 3

E1 =

Q 8π P0 R 2

Figure 22.63b

E2 =

Q Q with r = 3R / 2 so E2 = 4π P0 r 2 9π P0 R 2

G Q Q , in the + x-direction and E = iˆ 2 72π P0 R 72π P0 R 2 (c) x = R. This point is at the surface of each sphere. The fields have equal magnitudes and opposite directions, so E = 0. (d) x = 3R. This point is outside both spheres. Each field is directed away from the center of the sphere that produces it. The fields are shown in Figure 22.63c.

E = E1 − E2 =

E1 =

Q with r = 3R so 4π P0 r 2

E1 =

Q 36π P0 R 2

Figure 22.63c

E2 =

Q Q with r = R so E2 = 4π P0 r 2 4π P0 R 2

G 5Q 5Q ˆ i , in the + x-direction and E = 2 18π P0 R 18π P0 R 2 EVALUATE: The field of each sphere is radially outward from the center of the sphere. We must use the correct expression for E(r) for each sphere, depending on whether the field point is inside or outside that sphere. IDENTIFY: The net electric field at any point is the vector sum of the fields at each sphere. SET UP: Example 22.9 gives the electric field inside and outside a uniformly charged sphere. For the positively charged sphere the field is radially outward and for the negatively charged sphere the electric field is radially inward.

E = E1 + E2 =

22.64.

Gauss’s Law

22-27

EXECUTE: (a) At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is toward the center of that sphere, in the +x-direction. This point is outside the right-hand sphere, a distance r = 2R from its G 1 Q ˆ i. center. E = + 4πP0 4 R 2 (b) This point is inside the left-hand sphere, at r = R / 2 , and is outside the right-hand sphere, at r = 3R / 2 . Both fields are in the +x-direction.

G ⎤ˆ 1 ⎡ Q ( R / 2) Q 1 ⎡ Q 4Q ⎤ ˆ 1 17Q ˆ + + E= i= i. ⎢ ⎥i = 4πP0 ⎣ R 3 (3 R 2) 2 ⎦ 4πP0 ⎢⎣ 2 R 2 9 R 2 ⎥⎦ 4πP0 18R 2 (c) This point is outside both spheres, at a distance r = R from their centers. Both fields are in the +x-direction. G 1 ⎡Q Q ⎤ˆ Q ˆ E= i= i. + 4πP0 ⎢⎣ R 2 R 2 ⎥⎦ 2πP0 R 2 (d) This point is outside both spheres, a distance r = 3R from the center of the left-hand sphere and a distance r = R from the center of the right-hand sphere. The field of the left-hand sphere is in the +x-direction and the field G 1 ⎡ Q Q⎤ 1 ⎡ Q Q ⎤ˆ −1 8Q ˆ i= i. of the right-hand sphere is in the − x -direction . E = − ⎥ iˆ = − ⎢ 4πP0 ⎣ (3R ) 2 R 2 ⎦ 4πP0 ⎢⎣ 9 R 2 R 2 ⎥⎦ 4πP0 9 R 2 22.65.

EVALUATE: At all points on the x-axis the net field is parallel to the x-axis. IDENTIFY: Let − dQ be the electron charge contained within a spherical shell of radius r ′ and thickness dr ′ . Integrate r ′ from 0 to r to find the electron charge within a sphere of radius r. Apply Gauss’s law to a sphere of radius r to find the electric field E ( r ) .

The volume of the spherical shell is dV = 4π r ′2 dr ′ . Q 4π −2 r / a0 2 4Q r EXECUTE: (a) Q (r ) = Q − ∫ ρdV = Q − e r dr = Q − 3 ∫ r ′2e −2 r ′ / a0 dr′ . 3 ∫ πa0 a0 0 SET UP:

Q(r ) = Q −

4Qe −αr (2eαr − α 2 r 2 − 2αr − 2) = Qe −2 r / a0 [2(r / a0 ) 2 + 2(r / a0 ) + 1]. a03α 3

Note if r → ∞, Q (r ) → 0 ; the total net charge of the atom is zero. (b) The electric field is radially outward. Gauss’s law gives E (4π r 2 ) =

Q(r ) and P0

kQe −2 r / a0 (2(r a0 ) 2 + 2(r a0 ) + 1) . r2 (c) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E, equal to E / Ept charge , versus E=

kQ is the field of a point charge. r2 EVALUATE: As r → 0 , the field approaches that of the positive point charge (the proton). For increasing r the electric field falls to zero more rapidly than the 1/ r 2 dependence for a point charge. scaled r, equal to r / a0 . Ept charge =

22.66.

IDENTIFY: SET UP:

Figure 22.65 The charge in a spherical shell of radius r and thickness dr is dQ = ρ ( r )4π r 2 dr . Apply Gauss’s law.

Use a Gaussian surface that is a sphere of radius r. Let Qi be the charge in the region r ≤ R / 2 and let

Q0 be the charge in the region where R / 2 ≤ r ≤ R .

22-28

Chapter 22

4π ( R 2)3 απR 3 = and 3 6 R ⎛ ( R 3 − R 3 8) ( R 4 − R 4 16) ⎞ 11απR 3 15απR 3 8Q . Therefore, Q = . Q0 = 4π (2α ) ∫ (r 2 − r 3 / R )dr = 8απ ⎜ and α = − = ⎟ R/2 5 πR 3 24 3 4 R 24 ⎝ ⎠ αr 8Qr α 4πr 3 (b) For r ≤ R 2 , Gauss’s law gives E 4πr 2 = and E = . For R 2 ≤ r ≤ R , = 3P0 15πP0 R 3 3P0 EXECUTE:

E 4πr 2 =

E=

(a) The total charge is Q = Qi + Q0 , where Qi = α

⎛ (r 3 − R 3 8) (r 4 − R 4 16) ⎞ ⎞ Qi 1 ⎛ + ⎜⎜ 8απ ⎜ − ⎟ ⎟⎟ and P0 P0 ⎝ 3 4R ⎝ ⎠⎠

απR 3 kQ (64(r R)3 − 48( r R ) 4 − 1) = (64(r R)3 − 48( r R ) 4 − 1). 2 24P0 (4πr ) 15r 2

For r ≥ R , E (4π r 2 ) = (c)

Q Q and E = . P0 4πP0 r 2

Qi (4Q /15) 4 = = = 0.267. Q Q 15

8eQ r , so the restoring force depends upon displacement to the first power, and 15πP0 R 3 we have simple harmonic motion.

(d) For r ≤ R / 2 , Fr = − eE = −

(e) Comparing to F = − kr , k =

22.67.

8eQ k 8eQ 2π 15πP0 R 3me . Then ω = and T = = = 2π . 3 3 15πP0 R me ω 8eQ 15πP0 R me

EVALUATE: (f ) If the amplitude of oscillation is greater than R / 2, the force is no longer linear in r , and is thus no longer simple harmonic. IDENTIFY: The charge in a spherical shell of radius r and thickness dr is dQ = ρ ( r )4π r 2 dr . Apply Gauss’s law.

Use a Gaussian surface that is a sphere of radius r. Let Qi be the charge in the region r ≤ R / 2 and let

SET UP:

Q0 be the charge in the region where R / 2 ≤ r ≤ R . 3αr 3 6πα 1 R 4 3 dr = = παR 3 and 2R R 4 16 32 R 31 ⎞ 47 47 ⎞ 233 ⎛ 7 ⎛ 3 3 Q0 = 4πα ∫ (1 − (r/R ) 2 ) r 2 dr = 4παR 3 ⎜ − παR 3 . Therefore, Q = ⎜ + παR 3 and ⎟= ⎟ παR = R/2 480 ⎝ 24 160 ⎠ 120 ⎝ 32 120 ⎠ 480Q α= . 233πR 3 4π r 3αr ′3 3παr 4 6αr 2 180Qr 2 ′ dr E = = (b) For r ≤ R 2 , Gauss’s law gives E 4πr 2 = and . For R 2 ≤ r ≤ R , = 2P0 R 16P0 R 233πP0 R 4 P0 ∫ 0 2 R EXECUTE:

(a) The total charge is Q = Qi + Q0 , where Qi = 4π ∫

R/2

0

E 4πr 2 =

Qi 4πα r Q 4πα ⎛ r 3 R 3 r5 R3 ⎞ + (1 − ( r′ / R ) 2 )r ′2 dr ′ = i + − 2+ ⎜ − ⎟. ∫ R / 2 P0 P0 P0 P0 ⎝ 3 24 5 R 160 ⎠

E 4πr 2 =

3 4παR 3 4παR 3 + P0 128 P0

E=

3 5 ⎛ 1 ⎛ r ⎞3 1 ⎛ r ⎞5 17 ⎞ 480Q ⎛ 1 ⎛ r ⎞ 1 ⎛ r ⎞ 23 ⎞ − ⎜ ⎟ . For r ≥ R , ⎜⎜ ⎜ ⎟ − ⎜ ⎟ − ⎟⎟ and E = ⎜ ⎟ ⎜ ⎟ − 2 ⎜ 233πP0 r ⎝ 3 ⎝ R ⎠ 5 ⎝ R ⎠ 1920 ⎟⎠ ⎝ 3 ⎝ R ⎠ 5 ⎝ R ⎠ 480 ⎠

Q , since all the charge is enclosed. 4πP0 r 2

(c) The fraction of Q between R 2 ≤ r ≤ R is

Q0 47 480 = = 0.807. Q 120 233

Q (d) E = 180 using either of the electric field expressions above, evaluated at r = R / 2. 233 4πP0 R 2 EVALUATE: (e) The force an electron would feel never is proportional to − r which is necessary for simple harmonic oscillations. It is oscillatory since the force is always attractive, but it has the wrong power of r to be simple harmonic motion.

23

ELECTRIC POTENTIAL

23.1.

IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position of q2 be point a and the final position be point b, as shown in Figure 23.1.

ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m


Wa →b = U a − U b

Ua =

1 q1q2 (+2.40 × 10−6 C)(−4.30 × 10−6 C) = (8.988 × 109 N ⋅ m 2 / C 2 ) 4π P0 ra 0.150 m

U a = −0.6184 J Ub =

1 q1q2 ( +2.40 × 10−6 C)(−4.30 × 10−6 C) = (8.988 × 109 N ⋅ m 2 / C 2 ) 4π P0 rb 0.3536 m

U b = −0.2623 J Wa →b = U a − U b = −0.6184 J − (−0.2623 J) = −0.356 J

23.2.

EVALUATE: The attractive force on q2 is toward the origin, so it does negative work on q2 when q2 moves to larger r. IDENTIFY: Apply Wa → b = U a − U b . SET UP:

23.3.

U a = +5.4 × 10−8 J. Solve for U b .

EXECUTE: Wa →b = −1.9 × 10−8 J = U a − U b . U b = U a − Wa →b = 1.9 × 10−8 J − (−5.4 × 10−8 J) = 7.3 × 10−8 J. EVALUATE: When the electric force does negative work the electrical potential energy increases. IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity. SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is U = (1/ 4π P0 )( qq0 / r ). Each charge is e and the charges are equidistant from each other, so the total

potential energy is U = EXECUTE:

1 ⎛ e2 e2 e2 ⎞ 3e2 . + + = ⎜ ⎟ 4π P0 ⎝ r r r ⎠ 4π P0 r

Adding the potential energies gives U=

3e 2 3(1.60 × 10−19 C) 2 (9.00 × 109 N ⋅ m 2 /C2 ) = = 3.46 × 10−13 J = 2.16 MeV 4π P0 r 2.00 × 10−15 m

EVALUATE: This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a lot of energy. 23-1

23-2

23.4.

Chapter 23

IDENTIFY: The work required is the change in electrical potential energy. The protons gain speed after being released because their potential energy is converted into kinetic energy. (a) SET UP: Using the potential energy of a pair of point charges relative to infinity, U = (1/ 4π P0 )( qq0 / r ). we have

W = ΔU = U 2 − U 1 = EXECUTE:

1 ⎛ e2 e2 ⎞ ⎜ − ⎟. 4π P0 ⎝ r2 r1 ⎠

Factoring out the e2 and substituting numbers gives 2⎛ 1 1 ⎞ −14 − W = ( 9.00 × 109 N ⋅ m 2 /C 2 )(1.60 × 10−19 C ) ⎜ ⎟ = 7.68 × 10 J −15 −15 × × 3.00 10 m 2.00 10 m ⎝ ⎠

The protons have equal momentum, and since they have equal masses, they will have equal speeds ⎛1 ⎞ and hence equal kinetic energy. ΔU = K1 + K 2 = 2 K = 2 ⎜ mv 2 ⎟ = mv 2 . ⎝2 ⎠

(b) SET UP:

EXECUTE:

23.5.

Solving for v gives v =

ΔU 7.68 × 10−14 J = 6.78 × 106 m/s = 1.67 × 10−27 kg m

EVALUATE: The potential energy may seem small (compared to macroscopic energies), but it is enough to give each proton a speed of nearly 7 million m/s. (a) IDENTIFY: Use conservation of energy:

K a + U a + Wother = K b + U b U for the pair of point charges is given by Eq.(23.9). SET UP: Let point a be where q2 is 0.800 m from q1 and point b be where q2 is 0.400 m from q1, as shown in Figure 23.5a.


Only the electric force does work, so Wother = 0 and U =

1 q1q2 . 4π P0 r

K a = 12 mva2 = 12 (1.50 × 10−3 kg)(22.0 m/s) 2 = 0.3630 J Ua =

1 q1q2 (−2.80 × 10−6 C)(−7.80 × 10−6 C) = (8.988 × 109 N ⋅ m 2 /C 2 ) = +0.2454 J 4π P0 ra 0.800 m

K b = 12 mvb2 Ub =

1 q1q2 (−2.80 × 10−6 C)(−7.80 × 10−6 C) = (8.988 × 109 N ⋅ m 2 /C 2 ) = +0.4907 J 4π P0 rb 0.400 m

The conservation of energy equation then gives K b = K a + (U a − U b ) 1 2

mvb2 = +0.3630 J + (0.2454 J − 0.4907 J) = 0.1177 J vb =

2(0.1177 J) = 12.5 m/s 1.50 × 10−3 kg

EVALUATE: The potential energy increases when the two positively charged spheres get closer together, so the kinetic energy and speed decrease. (b) IDENTIFY: Let point c be where q2 has its speed momentarily reduced to zero. Apply conservation of energy to points a and c: K a + U a + Wother = K c + U c .

Electric Potential

SET UP:

23-3

Points a and c are shown in Figure 23.5b.

EXECUTE:

K a = +0.3630 J (from part (a))

U a = +0.2454 J (from part (a))

Figure 23.5b

K c = 0 (at distance of closest approach the speed is zero)

Uc =

1 q1q2 = +0.3630 J + 0.2454 J = 0.6084 J 4π P0 rc

Thus conservation of energy K a + U a = U c gives rc =

1 q1q2 4π P0 rc

1 q1q2 (−2.80 × 10−6 C)( −7.80 × 10−6 C) = (8.988 × 109 N ⋅ m 2 /C2 ) = 0.323 m. 4π P0 0.6084 J +0.6084 J

U → ∞ as r → 0 so q2 will stop no matter what its initial speed is. qq IDENTIFY: Apply U = k 1 2 and solve for r. r −6 SET UP: q1 = −7.2 × 10 C , q2 = +2.3 × 10−6 C EVALUATE:

23.6.

kq1q2 (8.99 × 109 N ⋅ m 2 /C2 )(−7.20 × 10−6 C)(+2.30 × 10−6 C) = = 0.372 m −0.400 J U EVALUATE: The potential energy U is a scalar and can take positive and negative values. (a) IDENTIFY and SET UP: U is given by Eq.(23.9). 1 qq′ EXECUTE: U = 4επ 0 r EXECUTE:

23.7.

r=

(+4.60 × 10−6 C)(+1.20 × 10−6 C) = +0.198 J 0.250 m EVALUATE: The two charges are both of the same sign so their electric potential energy is positive. (b) IDENTIFY: Use conservation of energy: K a + U a + Wother = K b + U b SET UP: Let point a be where q is released and point b be at its final position, as shown in Figure 23.7. U = (8.988 × 109 N ⋅ m 2 /C 2 )

EXECUTE:

K a = 0 (released from rest)

U a = +0.198 J (from part (a)) K b = 12 mvb2 Figure 23.7

Only the electric force does work, so Wother = 0 and U =

1 qQ . 4π P0 r

(i) rb = 0.500 m Ub =

1 qQ (+4.60 × 10−6 C)(+1.20 × 10−6 C) = (8.988 × 109 N ⋅ m 2 /C 2 ) = +0.0992 J 4π P0 r 0.500 m

Then K a + U a + Wother = K b + U b gives K b = U a − U b and vb =

1 2

mvb2 = U a − U b and

2(U a − U b ) 2(+0.198 J − 0.0992 J) = = 26.6 m/s. m 2.80 × 10−4 kg

(ii) rb = 5.00 m rb is now ten times larger than in (i) so U b is ten times smaller: U b = +0.0992 J /10 = +0.00992 J. vb =

2(U a − U b ) 2(+0.198 J − 0.00992 J) = = 36.7 m/s. m 2.80 × 10−4 kg

23-4

Chapter 23

(iii) rb = 50.0 m rb is now ten times larger than in (ii) so Ub is ten times smaller: U b = +0.00992 J/10 = +0.000992 J. vb =

23.8.

EVALUATE: The force between the two charges is repulsive and provides an acceleration to q. This causes the speed of q to increase as it moves away from Q. IDENTIFY: Call the three charges 1, 2 and 3. U = U12 + U13 + U 23 SET UP:

U12 = U 23 = U13 because the charges are equal and each pair of charges has the same separation, 0.500 m.

3kq 2 3k (1.2 × 10−6 C) 2 = = 0.078 J. 0.500 m 0.500 m EVALUATE: When the three charges are brought in from infinity to the corners of the triangle, the repulsive electrical forces between each pair of charges do negative work and electrical potential energy is stored. ⎛qq qq qq ⎞ IDENTIFY: U = k ⎜ 1 2 + 1 3 + 2 3 ⎟ r13 r23 ⎠ ⎝ r12 EXECUTE:

23.9.

2(U a − U b ) 2( +0.198 J − 0.000992 J) = = 37.5 m/s. m 2.80 × 10−4 kg

SET UP:

U=

In part (a), r12 = 0.200 m , r23 = 0.100 m and r13 = 0.100 m. In part (b) let particle 3 have coordinate x, so

r12 = 0.200 m , r13 = x and r23 = 0.200 − x. EXECUTE:

⎛ (4.00 nC)(−3.00 nC) (4.00 nC)(2.00 nC) (−3.00 nC)(2.00 nC) ⎞ −7 (a) U = k ⎜ + + ⎟ = 3.60 × 10 J (0.200 m) (0.100 m) (0.100 m) ⎝ ⎠

⎛qq qq qq ⎞ (b) If U = 0 , then 0 = k ⎜ 1 2 + 1 3 + 2 3 ⎟ . Solving for x we find: r x r ⎝ 12 12 − x ⎠ 8 6 0 = − 60 + − ⇒ 60 x 2 − 26 x + 1.6 = 0 ⇒ x = 0.074 m, 0.360 m. Therefore, x = 0.074 m since it is the only x 0.2 − x value between the two charges. EVALUATE: U13 is positive and both U 23 and U12 are negative. If U = 0 , then U13 = U 23 + U12 . For 23.10.

x = 0.074 m , U13 = +9.7 × 10−7 J , U 23 = −4.3 × 10−7 J and U12 = −5.4 × 10−7 J. It is true that U = 0 at this x. IDENTIFY: The work done on the alpha particle is equal to the difference in its potential energy when it is moved from the midpoint of the square to the midpoint of one of the sides. SET UP: We apply the formula Wa →b = U a − U b . In this case, a is the center of the square and b is the midpoint of one of the sides. Therefore Wcenter →side = U center − U side . There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single alpha-electron pair. At the center of the square, the alpha particle is a distance r1 = 50 nm from each electron. At the midpoint of the side, the alpha is a distance r2 = 5.00 nm from the two nearest electrons and a distance r2 = 125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of two point charges, U = (1/ 4π P0 )(qq0 / r ), the total work is Wcenter →side = U center − U side = 4

1 qα qe ⎛ 1 qα qe 1 qα qe ⎞ −⎜2 +2 ⎟ 4π P0 r1 4π P0 r3 ⎠ ⎝ 4π P0 r2

Substituting qe = e and qα = 2e and simplifying gives Wcenter →side = −4e2 EXECUTE:

1 ⎡ 2 ⎛ 1 1 ⎞⎤ ⎢ − ⎜ + ⎟⎥ 4π P0 ⎢⎣ r1 ⎝ r2 r3 ⎠ ⎥⎦

Substituting the numerical values into the equation for the work gives 2⎡ 2 1 1 ⎛ ⎞⎤ −21 W = −4 (1.60 × 10−19 C ) ⎢ −⎜ + ⎟ ⎥ = 6.08 × 10 J ??? 5.00 nm 50 m 125 nm ⎝ ⎠ ⎣ ⎦

EVALUATE: Since the work is positive, the system has more potential energy with the alpha particle at the center of the square than it does with it at the midpoint of a side.

Electric Potential

23.11.

23-5

IDENTIFY: Apply Eq.(23.2). The net work to bring the charges in from infinity is equal to the change in potential energy. The total potential energy is the sum of the potential energies of each pair of charges, calculated from Eq.(23.9). SET UP: Let 1 be where all the charges are infinitely far apart. Let 2 be where the charges are at the corners of the triangle, as shown in Figure 23.11.

Let qc be the third, unknown charge.

Figure 23.11 EXECUTE: W = −ΔU = −(U 2 − U1 )

U1 = 0 U 2 = U ab + U ac + U bc =

1 (q 2 + 2qqc ) 4π P0 d

Want W = 0, so W = −(U 2 − U1 ) gives 0 = −U 2 0=

1 (q 2 + 2qqc ) 4π P0 d

q 2 + 2qqc = 0 and qc = −q/2. EVALUATE: The potential energy for the two charges q is positive and for each q with qc it is negative. There are two of the q, qc terms so must have qc < q. 23.12.

IDENTIFY:

Use conservation of energy U a + K a = U b + K b to find the distance of closest approach rb . The

maximum force is at the distance of closest approach, F = k SET UP:

q1q2 . rb2

K b = 0. Initially the two protons are far apart, so U a = 0. A proton has mass 1.67 × 10−27 kg and charge

q = + e = +1.60 × 10−19 C. EXECUTE:

rb =

K a = U b . 2( 12 mva2 ) = k

q1q2 e2 . mva2 = k and rb rb

ke 2 (8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 = = 1.38 × 10−13 m. mva2 (1.67 × 10−27 kg)(1.00 × 106 m/s) 2

−19 e2 C) 2 9 2 2 (1.60 × 10 = (8.99 × 10 N ⋅ m /C ) = 0.012 N. rb2 (1.38 × 10−13 C) 2 EVALUATE: The acceleration a = F/m of each proton produced by this force is extremely large. W IDENTIFY: E points from high potential to low potential. a →b = Va − Vb . q0

F =k

23.13.

23.14.

SET UP: The force on a positive test charge is in the direction of E . EXECUTE: V decreases in the eastward direction. A is east of B, so VB > VA . C is east of A, so VC < VA . The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular), and VD = VA . EVALUATE: The electric potential is constant in a direction perpendicular to the electric field. Wa →b kq IDENTIFY: = Va − Vb . For a point charge, V = . r q0 SET UP: Each vacant corner is the same distance, 0.200 m, from each point charge. EXECUTE: Taking the origin at the center of the square, the symmetry means that the potential is the same at the two corners not occupied by the +5.00 μ C charges. This means that no net work is done is moving from one corner to the other. EVALUATE: If the charge q0 moves along a diagonal of the square, the electrical force does positive work for part of the path and negative work for another part of the path, but the net work done is zero.

23-6

Chapter 23

23.15.

IDENTIFY and SET UP: Apply conservation of energy to points A and B. EXECUTE: K A + U A = K B + U B U = qV , so K A + qVA = K B + qVB

K B = K A + q(VA − VB ) = 0.00250 J + (−5.00 × 10−6 C)(200 V − 800 V) = 0.00550 J vB = 2 K B /m = 7.42 m/s It is faster at B; a negative charge gains speed when it moves to higher potential. W IDENTIFY: The work-energy theorem says Wa →b = K b − K a . a →b = Va − Vb . q SET UP: Point a is the starting and point b is the ending point. Since the field is uniform, Wa →b = Fs cos φ = E q s cos φ . The field is to the left so the force on the positive charge is to the left. The particle EVALUATE: 23.16.

moves to the left so φ = 0° and the work Wa →b is positive. EXECUTE:

(a) Wa →b = K b − K a = 1.50 × 10−6 J − 0 = 1.50 × 10−6 J

(b) Va − Vb =

Wa →b 1.50 × 10−6 J = = 357 V. Point a is at higher potential than point b. q 4.20 × 10−9 C

(c) E q s = Wa →b , so E = EVALUATE: 23.17.

IDENTIFY:

Wa →b Va − Vb 357 V = = = 5.95 × 103 V/m. 6.00 × 10−2 m qs s

A positive charge gains kinetic energy when it moves to lower potential; Vb < Va . b

Apply the equation that precedes Eq.(23.17): Wa →b = q′∫ E ⋅ dl . a

SET UP: Use coordinates where + y is upward and + x is to the right. Then E = Eˆj with E = 4.00 × 104 N/C. (a) The path is sketched in Figure 23.17a.


b E ⋅ dl = ( Eˆj ) ⋅ (dxiˆ) = 0 so Wa →b = q′∫ E ⋅ dl = 0. a

EVALUATE: The electric force on the positive charge is upward (in the direction of the electric field) and does no work for a horizontal displacement of the charge. (b) SET UP: The path is sketched in Figure 23.17b.

dl = dyˆj


E ⋅ dl = ( Eˆj ) ⋅ (dyˆj ) = E dy b

b

a

a

Wa →b = q′∫ E ⋅ dl = q′E ∫ dy = q′E ( yb − ya ) yb − ya = +0.670 m, positive since the displacement is upward and we have taken + y to be upward.

Wa →b = q′E ( yb − ya ) = (+28.0 × 10−9 C)(4.00 × 104 N/C)(+0.670 m) = +7.50 × 10−4 J. EVALUATE: The electric force on the positive charge is upward so it does positive work for an upward displacement of the charge.

Electric Potential

(c) SET UP:

23-7

The path is sketched in Figure 23.17c. ya = 0 yb = − r sinθ = −(2.60 m)sin 45° = −1.838 m The vertical component of the 2.60 m displacement is 1.838 m downward. Figure 23.17c

dl = dxiˆ + dyˆj (The displacement has both horizontal and vertical components.) E ⋅ dl = ( Eˆj ) ⋅ (dxiˆ + dyˆj ) = E dy (Only the vertical component of the displacement contributes to the work.)

EXECUTE:

b

b

a

a

Wa →b = q′∫ E ⋅ dl = q′E ∫ dy = q′E ( yb − ya ) Wa →b = q′E ( yb − ya ) = (+28.0 × 10−9 C)(4.00 × 104 N/C)(−1.838 m) = −2.06 × 10−3 J.

23.18.

EVALUATE: The electric force on the positive charge is upward so it does negative work for a displacement of the charge that has a downward component. IDENTIFY: Apply K a + U a = K b + U b . SET UP:

Let q1 = +3.00 nC and q2 = +2.00 nC. At point a, r1a = r2 a = 0.250 m . At point b, r1b = 0.100 m and

r2b = 0.400 m . The electron has q = −e and me = 9.11 × 10−31 kg . K a = 0 since the electron is released from rest. EXECUTE:

−

keq1 keq2 keq1 keq2 1 − =− − + mevb2 . r1a r2 a r1b r2b 2

⎛ (3.00 × 10−9 C) (2.00 × 10−9 C) ⎞ −17 Ea = K a + U a = k (− 1.60 × 10−19 C) ⎜ + ⎟ = − 2.88 × 10 J . 0.250 m ⎠ ⎝ 0.250 m ⎛ (3.00 × 10−9 C) (2.00 × 10−9 C ⎞ 1 1 2 2 −17 + Eb = K b + U b = k (− 1.60 × 10−19 C) ⎜ ⎟ + mevb = − 5.04 × 10 J + mevb 0.400 m ⎠ 2 2 ⎝ 0.100 m Setting Ea = Eb gives vb =

23.19.

23.20.

2 (5.04 × 10−17 J − 2.88 × 10−17 J) = 6.89 × 106 m s. 9.11 × 10−31 kg

EVALUATE: Va = V1a + V2 a = 180 V. Vb = V1b + V2b = 315 V. Vb > Va . The negatively charged electron gains kinetic energy when it moves to higher potential. kq IDENTIFY and SET UP: For a point charge V = . Solve for r. r kq (8.99 × 109 N ⋅ m 2 /C 2 )(2.50 × 10−11 C) EXECUTE: (a) r = = = 2.50 × 10−3 m = 2.50 mm V 90.0 V ⎛V ⎞ ⎛ 90.0 V ⎞ (b) Vr = kq = constant so V1r1 = V2 r2 . r2 = r1 ⎜ 1 ⎟ = (2.50 mm) ⎜ ⎟ = 7.50 mm . ⎝ 30.0 V ⎠ ⎝ V2 ⎠ EVALUATE: The potential of a positive charge is positive and decreases as the distance from the point charge increases. IDENTIFY: The total potential is the scalar sum of the individual potentials, but the net electric field is the vector sum of the two fields. SET UP: The net potential can only be zero if one charge is positive and the other is negative, since it is a scalar. The electric field can only be zero if the two fields point in opposite directions. EXECUTE: (a) (i) Since both charges have the same sign, there are no points for which the potential is zero. (ii) The two electric fields are in opposite directions only between the two charges, and midway between them the fields have equal magnitudes. So E = 0 midway between the charges, but V is never zero. (b) (i) The two potentials have equal magnitude but opposite sign midway between the charges, so V = 0 midway between the charges, but E ≠ 0 there since the fields point in the same direction. (ii) Between the two charges, the fields point in the same direction, so E cannot be zero there. In the other two regions, the field due to the nearer charge is always greater than the field due to the more distant charge, so they cannot cancel. Hence E is not zero anywhere. EVALUATE: It does not follow that the electric field is zero where the potential is zero, or that the potential is zero where the electric field is zero.

23-8

23.21.

Chapter 23

1 qi ∑ 4π P0 i ri SET UP: The locations of the changes and points A and B are sketched in Figure 23.21. V=

IDENTIFY:

Figure 23.21 (a) VA =

EXECUTE:

1 ⎛ q1 q2 ⎞ ⎜ + ⎟ 4π P0 ⎝ rA1 rA2 ⎠ ⎛ +2.40 × 10 −9 C −6.50 × 10 −9 C ⎞ + VA = (8.988 × 109 N ⋅ m 2 /C 2 ) ⎜ ⎟ = −737 V 0.050 m ⎠ ⎝ 0.050 m

(b) VB =

1 ⎛ q1 q2 ⎞ + ⎜ ⎟ 4π P0 ⎝ rB1 rB 2 ⎠

⎛ +2.40 × 10−9 C −6.50 × 10−9 C ⎞ + VB = (8.988 × 109 N ⋅ m2 /C2 ) ⎜ ⎟ = −704 V 0.060 m ⎠ ⎝ 0.080 m

23.22.

(c) IDENTIFY and SET UP: Use Eq.(23.13) and the results of parts (a) and (b) to calculate W. −9 −8 EXECUTE: WB → A = q′(VB − VA ) = (2.50 ×10 C)(−704 V − (−737 V)) = +8.2 ×10 J EVALUATE: The electric force does positive work on the positive charge when it moves from higher potential (point B) to lower potential (point A). kq IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the potentials of the r two charges. SET UP:

(a) The positions of the two charges are shown in Figure 23.22a. r = a 2 + x 2 .

Figure 23.22a

1 q . 4π P0 a

EXECUTE:

(b) V0 = 2

(c) V ( x) = 2

1 q 1 =2 4π P0 r 4π P0

q a + x2 2

Electric Potential

23-9

(d) The graph of V versus x is sketched in Figure 23.22b.

Figure 23.22b 1 2q , just like a point charge of charge +2q. At distances from the charges 4π P0 x much greater than their separation, the two charges act like a single point charge. kq IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the potentials of the r two charges. SET UP: (a) The positions of the two charges are shown in Figure 23.23. kq k ( − q) EXECUTE: (b) V = + = 0. r r (c) The potential along the x-axis is always zero, so a graph would be flat. (d) If the two charges are interchanged, then the results of (b) and (c) still hold. The potential is zero. EVALUATE: The potential is zero at any point on the x-axis because any point on the x-axis is equidistant from the two charges.

EVALUATE:

23.23.

(e) When x >> a, V =

Figure 23.23 23.24.

IDENTIFY:

For a point charge, V =

kq . The total potential at any point is the algebraic sum of the potentials of the r

two charges. SET UP: Consider the distances from the point on the y-axis to each charge for the three regions −a ≤ y ≤ a (between the two charges), y > a (above both charges) and y < −a (below both charges). 2kqy kq kq kq kq − 2kqa . . y > a :V = − = 2 − = 2 2 (a + y ) (a − y ) y − a (a + y) y − a y − a 2 2kqa kq − kq . y < − a :V = − = (a + y ) (− y + a) y 2 − a 2

EXECUTE:

(a) | y | < a : V =

⎛ −q q ⎞ A general expression valid for any y is V = k ⎜ + ⎟. ⎝| y − a | | y + a |⎠ (b) The graph of V versus y is sketched in Figure 23.24. −2kqa −2kqa (c) y >> a : V = 2 . ≈ y − a2 y2 (d) If the charges are interchanged, then the potential is of the opposite sign.

23-10

Chapter 23

V = 0 at y = 0 . V → +∞ as the positive charge is approached and V → −∞ as the negative charge is

EVALUATE: approached.

Figure 23.24 23.25.

IDENTIFY:



two charges. SET UP: (a) The positions of the two charges are shown in Figure 23.25a.

Figure 23.25a

kq 2kq − kq( x + a) kq 2kq kq(3x − a) . 0 < x < a :V = . − = − = x x−a x( x − a) x a−x x( x − a ) kq( x + a ) − kq 2kq q 2q ⎞ . A general expression valid for any y is V = k ⎛⎜ x < 0 :V = + = − ⎟. x x−a x( x − a) ⎝| x| | x − a |⎠ (c) The potential is zero at x = − a and a /3. (d) The graph of V versus x is sketched in Figure 23.25b. (b) x > a : V =

Figure 23.25b − kqx − kq = , which is the same as the potential of a point charge –q. Far from x2 x the two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges.

EVALUATE:

(e) For x >> a : V ≈

Electric Potential

23.26.


IDENTIFY:

23-11


two charges. SET UP: The distance of a point with coordinate y from the positive charge is y and the distance from the negative charge is r = a 2 + y 2 . ⎞ ⎟. ⎟ ⎠ 2 2 a + y a (b) V = 0, when y 2 = ⇒ 3 y2 = a2 ⇒ y = ± . 4 3 (c) The graph of V versus y is sketched in Figure 23.26. V → ∞ as the positive charge at the origin is approached. (a) V =

EXECUTE:

⎛ 1 kq 2kq 2 − = kq ⎜ − 2 ⎜ | y| r a + y2 ⎝| y|

⎛1 2⎞ kq (d) y >> a : V ≈ kq ⎜ − ⎟ = − , which is the potential of a point charge − q . Far from the two y ⎝ y y⎠ charges they appear to be a point charge with a charge that is the algebraic sum of their two charges. EVALUATE:

Figure 23.26 23.27.

K a + qVa = K b + qVb .

IDENTIFY: SET UP:

Let point a be at the cathode and let point b be at the anode. K a = 0 . Vb − Va = 295 V . An electron has

q = −e and m = 9.11 × 10−31 kg . EXECUTE:

vb =

1 K b = q(Va − Vb ) = −(1.60 × 10−19 C)( − 295 V) = 4.72 × 10−17 J . K b = mvb2 , so 2

2(4.72 × 10−17 J) = 1.01 × 107 m s. 9.11 × 10−31 kg

The negatively charged electron gains kinetic energy when it moves to higher potential. kq kq IDENTIFY: For a point charge, E = 2 and V = . r r SET UP: The electric field is directed toward a negative charge and away from a positive charge. 2 V kq/r 4.98 V ⎛ kq ⎞ ⎛ r ⎞ EXECUTE: (a) V > 0 so q > 0 . = = = 0.415 m . ⎜ ⎟=r. r= ⎜ ⎟ 2 E k q /r 12.0 V/m ⎝ r ⎠ ⎝ kq ⎠ EVALUATE:

23.28.

(0.415 m)(4.98 V) rV = = 2.30 × 10−10 C 8.99 × 109 N ⋅ m 2 /C2 k (c) q > 0 , so the electric field is directed away from the charge. EVALUATE: The ratio of V to E due to a point charge increases as the distance r from the charge increases, because E falls off as 1/r 2 and V falls off as 1/r . (a) IDENTIFY and SET UP: The direction of E is always from high potential to low potential so point b is at higher potential. (b) Apply Eq.(23.17) to relate Vb − Va to E. (b) q =

23.29.

EXECUTE:

b

b

a

a

Vb − Va = − ∫ E ⋅ dl = ∫ E dx = E ( xb − xa ). E=

Vb − Va +240 V = = 800 V/m xb − xa 0.90 m − 0.60 m

23-12

23.30.

Chapter 23

(c) Wb → a = q (Vb − Va ) = ( −0.200 × 10−6 C)( +240 V) = −4.80 × 10−5 J. EVALUATE: The electric force does negative work on a negative charge when the negative charge moves from high potential (point b) to low potential (point a). kq IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the potentials of the r kq two charges. For a point charge, E = 2 . The net electric field is the vector sum of the electric fields of the two r charges. SET UP: E produced by a point charge is directed away from the point charge if it is positive and toward the charge if it is negative. EXECUTE: (a) V = VQ + V2Q > 0, so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges, because only there are the fields of the two charges in opposite directions. Consider a kQ k (2Q) point a distance x from Q and d − x from 2Q, as shown in Figure 23.30a. EQ = E2Q → 2 = → (d − x)2 = 2 x 2 . x (d − x)2 x = d . The other root, x = d , does not lie between the charges. 1+ 2 1− 2 (b) V can be zero in 2 places, A and B, as shown in Figure 23.30b. Point A is a distance x from −Q and d − x from

2Q. B is a distance y from −Q and d + y from 2Q. At A :

k ( − Q) k (2Q) + =0→ x = d 3. x d−x

k ( − Q) k (2Q) + =0→ y=d . y d+y The two electric fields are in opposite directions to the left of −Q or to the right of 2Q in Figure 23.30c. But for the magnitudes to be equal, the point must be closer to the charge with smaller magnitude of charge. This can be the kQ k (2Q) d . and x = case only in the region to the left of −Q . EQ = E2Q gives 2 = 2 x (d + x) 2 −1 At B :

EVALUATE: (d) E and V are not zero at the same places. E is a vector and V is a scalar. E is proportional to 1/r 2 and V is proportional to 1/r . E is related to the force on a test charge and ΔV is related to the work done on a test charge when it moves from one point to another.

Figure 23.30 23.31.

IDENTIFY and SET UP: Apply conservation of energy, Eq.(23.3). Use Eq.(23.12) to express U in terms of V. (a) EXECUTE: K1 + qV1 = K 2 + qV2

q (V1 − V2 ) = K 2 − K1; K1 = 12 mev12 = 4.099 × 10−18 J;

q = −1.602 × 10−19 C K 2 = 12 mev22 = 2.915 × 10−17 J

K 2 − K1 = −156 V q EVALUATE: The electron gains kinetic energy when it moves to higher potential. (b) EXECUTE: Now K1 = 2.915 × 10−17 J, K 2 = 0 V1 − V2 =

V1 − V2 = 23.32.

K 2 − K1 = +182 V q

EVALUATE: The electron loses kinetic energy when it moves to lower potential. IDENTIFY and SET UP: Expressions for the electric potential inside and outside a solid conducting sphere are derived in Example 23.8. kq k (3.50 × 10−9 C) EXECUTE: (a) This is outside the sphere, so V = = = 65.6 V. r 0.480 m

k (3.50 × 10−9 C) = 131 V . 0.240 m (c) This is inside the sphere. The potential has the same value as at the surface, 131 V. EVALUATE: All points of a conductor are at the same potential. (b) This is at the surface of the sphere, so V =

Electric Potential

23.33.

23-13

(a) IDENTIFY and SET UP: The electric field on the ring’s axis is calculated in Example 21.10. The force on the electron exerted by this field is given by Eq.(21.3). EXECUTE: When the electron is on either side of the center of the ring, the ring exerts an attractive force directed toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the ring, with amplitude 30.0 cm. The force on the electron is not of the form F = −kx so the oscillatory motion is not simple harmonic motion. (b) IDENTIFY: Apply conservation of energy to the motion of the electron. SET UP: K a + U a = K b + U b with a at the initial position of the electron and b at the center of the ring. From

Example 23.11, V = EXECUTE:

1 4π P0

Q

, where R is the radius of the ring.

x + R2 xa = 30.0 cm, xb = 0. 2

K a = 0 (released from rest), K b = 12 mv 2 Thus

1 2

mv 2 = U a − U b 2e(Vb − Va ) . m

And U = qV = −eV so v = Va =

1 4π P0

Q xa2 + R 2

= (8.988 × 109 N ⋅ m 2 / C2 )

24.0 × 10−9 C (0.300 m)2 + (0.150 m) 2

Va = 643 V Vb =

v=

1 4π P0

Q xb2 + R 2

= (8.988 × 109 N ⋅ m 2 / C 2 )

24.0 × 10−9 C = 1438 V 0.150 m

2e(Vb − Va ) 2(1.602 × 10−19 C)(1438 V − 643 V) = = 1.67 × 107 m/s m 9.109 × 10−31 kg

EVALUATE: The positively charged ring attracts the negatively charged electron and accelerates it. The electron has its maximum speed at this point. When the electron moves past the center of the ring the force on it is opposite to its motion and it slows down. 23.34.

IDENTIFY:

Example 23.10 shows that for a line of charge, Va − Vb =

λ ln( rb / ra ) . Apply conservation of energy 2π P0

to the motion of the proton. SET UP: Let point a be 18.0 cm from the line and let point b be at the distance of closest approach, where K b = 0 . EXECUTE:

(a) K a = 12 mv 2 = 12 (1.67 × 10−27 kg)(1.50 × 103 m/s) 2 = 1.88 × 10−21 J .

(b) K a + qVa = Kb + qVb . Va − Vb =

23.35.

K b − K a −1.88 × 10−21 J ⎛ 2π P0 ⎞ = = −0.01175 V . ln(rb / ra ) = ⎜ ⎟ ( −0.01175 V) . −19 q 1.60 × 10 C ⎝ λ ⎠

⎛ 2π P0 (0.01175 V) ⎞ ⎛ 2π P0 ( −0.01175 V) ⎞ rb = ra exp ⎜ ⎟ = 0.158 m . ⎟ = (0.180 m)exp ⎜ − −12 λ ⎝ ⎠ ⎝ 5.00 × 10 C/m ⎠ EVALUATE: The potential increases with decreasing distance from the line of charge. As the positively charged proton approaches the line of charge it gains electrical potential energy and loses kinetic energy. IDENTIFY: The voltmeter measures the potential difference between the two points. We must relate this quantity to the linear charge density on the wire. SET UP: EXECUTE:

For a very long (infinite) wire, the potential difference between two points is ΔV =

λ

2π P0

ln ( rb / ra ) .

(a) Solving for λ gives

λ=

( ΔV )2π P0 = ln ( rb / ra )

575 V

= 9.49 × 10-8 C/m ⎛ 3.50 cm ⎞ (18 ×10 N ⋅ m /C ) ln ⎜⎝ 2.50 cm ⎟⎠ (b) The meter will read less than 575 V because the electric field is weaker over this 1.00-cm distance than it was over the 1.00-cm distance in part (a). (c) The potential difference is zero because both probes are at the same distance from the wire, and hence at the same potential. EVALUATE: Since a voltmeter measures potential difference, we are actually given ΔV, even though that is not stated explicitly in the problem. We must also be careful when using the formula for the potential difference because each r is the distance from the center of the cylinder, not from the surface. 9

2

2

23-14

Chapter 23

23.36.

IDENTIFY: The voltmeter reads the potential difference between the two points where the probes are placed. Therefore we must relate the potential difference to the distances of these points from the center of the cylinder. For points outside the cylinder, its electric field behaves like that of a line of charge. SET UP:

23.37.

Using ΔV =

ln ( rb / ra ) and solving for rb, we have rb = ra e2π P0 ΔV / λ .

1 ⎛ ⎞ ⎜ ⎟ (175 V) 2 × 9.00 × 109 N ⋅ m 2 /C2 ⎠ ⎝ = 0.648 , which gives EXECUTE: The exponent is 15.0 × 10−9 C/m rb = (2.50 cm) e0.648 = 4.78 cm. The distance above the surface is 4.78 cm – 2.50 cm = 2.28 cm. EVALUATE: Since a voltmeter measures potential difference, we are actually given ΔV, even though that is not stated explicitly in the problem. We must also be careful when using the formula for the potential difference because each r is the distance from the center of the cylinder, not from the surface. IDENTIFY: For points outside the cylinder, its electric field behaves like that of a line of charge. Since a voltmeter reads potential difference, that is what we need to calculate. SET UP: EXECUTE:

The potential difference is ΔV =

λ

2π P0

ln ( rb / ra ) .

(a) Substituting numbers gives

ΔV =

23.38.

λ

2π P0

λ 2π P0

⎛ 10.0 cm ⎞ ln ( rb / ra ) = ( 8.50 × 10−6 C/m )( 2 × 9.00 × 109 N ⋅ m 2 /C2 ) ln ⎜ ⎟ ⎝ 6.00 cm ⎠ ΔV = 7.82 × 104 V = 78,200 V = 78.2 kV

(b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero. EVALUATE: Caution! The fact that the voltmeter reads zero in part (b) does not mean that V = 0 inside the cylinder. The electric field is zero, but the potential is constant and equal to the potential at the surface. IDENTIFY: The work required is equal to the change in the electrical potential energy of the charge-ring system. We need only look at the beginning and ending points, since the potential difference is independent of path for a conservative field. ⎛ 1 Q ⎞ − 0⎟ SET UP: (a) W = ΔU = qΔV = q (Vcenter − V∞ ) = q ⎜ ⎝ 4πε 0 a ⎠

Substituting numbers gives ΔU = (3.00 × 10-6 C)(9.00 × 109 N ⋅ m2/C2)(5.00 × 10–6 C)/(0.0400 m) = 3.38 J (b) We can take any path since the potential is independent of path. (c) SET UP: The net force is away from the ring, so the ball will accelerate away. Energy conservation gives U 0 = K max = 12 mv 2 . EXECUTE: Solving for v gives 2U 0 2(3.38 J) = 67.1 m/s v= = m 0.00150 kg EVALUATE: Direct calculation of the work from the electric field would be extremely difficult, and we would need to know the path followed by the charge. But, since the electric field is conservative, we can bypass all this calculation just by looking at the end points (infinity and the center of the ring) using the potential. IDENTIFY: The electric field is zero everywhere except between the plates, and in this region it is uniform and points from the positive to the negative plate (to the left in Figure 23.32). SET UP: Since the field is uniform between the plates, the potential increases linearly as we go from left to right starting at b. EXECUTE: Since the potential is taken to be zero at the left surface of the negative plate (a in Figure 23.32), it is zero everywhere to the left of b. It increases linearly from b to c, and remains constant (since E = 0) past c. The graph is sketched in Figure 23.39. EVALUATE: When the electric field is zero, the potential remains constant but not necessarily zero (as to the right of c). When the electric field is constant, the potential is linear. EXECUTE:

23.39.

Figure 23.39

Electric Potential

23.40.

23.41.

23-15

IDENTIFY and SET UP: For oppositely charged parallel plates, E = σ / P0 between the plates and the potential difference between the plates is V = Ed . σ 47.0 × 10−9 C m 2 EXECUTE: (a) E = = = 5310 N C. P0 P0 (b) V = Ed = (5310 N/C)(0.0220 m) = 117 V. (c) The electric field stays the same if the separation of the plates doubles. The potential difference between the plates doubles. EVALUATE: The electric field of an infinite sheet of charge is uniform, independent of distance from the sheet. The force on a test charge between the two plates is constant because the electric field is constant. The potential difference is the work per unit charge on a test charge when it moves from one plate to the other. When the distance doubles the work, which is force times distance, doubles and the potential difference doubles. IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to the potential difference between them and their separation. The force this field exerts on the particle is given by Eq.(21.3). Use the equation that precedes Eq.(23.17) to calculate the work. V 360 V = 8000 V/m EXECUTE: (a) From Example 23.9, E = ab = d 0.0450 m (b) F = q E = (2.40 × 10−9 C)(8000 V/m) = +1.92 × 10−5 N (c) The electric field between the plates is shown in Figure 23.41.

Figure 23.41

The plate with positive charge (plate a) is at higher potential. The electric field is directed from high potential toward low potential (or, E is from + charge toward − charge), so E points from a to b. Hence the force that E exerts on the positive charge is from a to b, so it does positive work. b

W = ∫ F ⋅ dl = Fd , where d is the separation between the plates. a

W = Fd = (1.92 × 10−5 N)(0.0450 m) = +8.64 × 10−7 J (d) Va − Vb = +360 V (plate a is at higher potential)

ΔU = U b − U a = q (Vb − Va ) = (2.40 × 10−9 C)(−360 V) = −8.64 × 10−7 J. 23.42.

EVALUATE: We see that Wa →b = −(U b − U a ) = U a − U b . IDENTIFY: The electric field is zero inside the sphere, so the potential is constant there. Thus the potential at the center must be the same as at the surface, where it is equivalent to that of a point-charge. SET UP: At the surface, and hence also at the center of the sphere, the field is that of a point-charge, E = Q /(4π P0 R ). EXECUTE: (a) Solving for Q and substituting the numbers gives

Q = 4π P0 RV = (0.125 m)(1500 V)/(9.00 × 109 N ⋅ m2/C2) = 2.08 × 10-8 C = 20.8 nC

23.43.

(b) Since the potential is constant inside the sphere, its value at the surface must be the same as at the center, 1.50 kV. EVALUATE: The electric field inside the sphere is zero, so the potential is constant but is not zero. IDENTIFY and SET UP: Consider the electric field outside and inside the shell and use that to deduce the potential. EXECUTE: (a) The electric field outside the shell is the same as for a point charge at the center of the shell, so the potential outside the shell is the same as for a point charge:

V=

q for r > R. 4π P0 r

The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell and all q points inside the shell are at the same potential as the surface of the shell: V = for r ≤ R. 4π P0 R kq RV (0.15 m)( −1200 V) so q = = = −20 nC R k k (c) EVALUATE: No, the amount of charge on the sphere is very small. Since U = qV the total amount of electric (b) V =

energy stored on the balloon is only (20 nC)(1200 V) = 2.4 × 10−5 J.

23-16

Chapter 23

23.44.

IDENTIFY: Example 23.8 shows that the potential of a solid conducting sphere is the same at every point inside the sphere and is equal to its value V = q / 2π P0 R at the surface. Use the given value of E to find q. SET UP: For negative charge the electric field is directed toward the charge. For points outside this spherical charge distribution the field is the same as if all the charge were concentrated at the center. q (3800 N/C)(0.200 m) 2 2 EXECUTE: E = and q = π P Er = = 1.69 × 10−8 C . 4 0 4π P0 r 2 8.99 × 109 N ⋅ m 2 /C 2 Since the field is directed inward, the charge must be negative. The potential of a point charge, taking ∞ as zero, is q (8.99 × 109 N ⋅ m 2 /C2 )( −1.69 × 10−8 C) V= = = −760 V at the surface of the sphere. Since the charge all resides 4π P0 r 0.200 m on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero. No work is therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must also be − 760 V. EVALUATE: Inside the sphere the electric field is zero and the potential is constant. IDENTIFY: Example 23.9 shows that V ( y ) = Ey , where y is the distance from the negatively charged plate, whose potential is zero. The electric field between the plates is uniform and perpendicular to the plates. SET UP: V increases toward the positively charged plate. E is directed from the positively charged plated toward the negatively charged plate. V 480 V V EXECUTE: (a) E = = = 2.82 × 104 V/m and y = . V = 0 at y = 0 , V = 120 V at y = 0.43 cm , E d 0.0170 m V = 240 V at y = 0.85 cm , V = 360 V at y = 1.28 cm and V = 480 V at y = 1.70 cm . The equipotential surfaces are sketched in Figure 23.45. The surfaces are planes parallel to the plates. (b) The electric field lines are also shown in Figure 23.45. The field lines are perpendicular to the plates and the equipotential lines are parallel to the plates, so the electric field lines and the equipotential lines are mutually perpendicular. EVALUATE: Only differences in potential have physical significance. Letting V = 0 at the negative plate is a choice we are free to make.

23.45.

Figure 23.45 23.46.

IDENTIFY: By the definition of electric potential, if a positive charge gains potential along a path, then the potential along that path must have increased. The electric field produced by a very large sheet of charge is uniform and is independent of the distance from the sheet. (a) SET UP: No matter what the reference point, we must do work on a positive charge to move it away from the negative sheet. EXECUTE: Since we must do work on the positive charge, it gains potential energy, so the potential increases. (b) SET UP:

Since the electric field is uniform and is equal to σ /2ε0, we have ΔV = Ed =

σ

2P0

d.

EXECUTE: Solving for d gives

d=

2P0 ΔV

=

2 ( 8.85 × 10−12 C 2 /N ⋅ m 2 ) (1.00 V)

= 0.00295 m = 2.95 mm 6.00 × 10−9 C/m 2 EVALUATE: Since the spacing of the equipotential surfaces (d = 2.95 mm) is independent of the distance from the sheet, the equipotential surfaces are planes parallel to the sheet and spaced 2.95 mm apart.

σ

Electric Potential

23.47

23-17

IDENTIFY and SET UP: Use Eq.(23.19) to calculate the components of E . EXECUTE: V = Axy − Bx 2 + Cy (a) Ex = −

∂V = − Ay + 2 Bx ∂x

∂V = − Ax − C ∂y ∂V Ez = − =0 ∂z (b) E = 0 requires that Ex = E y = Ez = 0. Ey = −

E z = 0 everywhere. E y = 0 at x = −C/A.

23.48.

And Ex is also equal zero for this x, any value of z, and y = 2Bx /A = (2 B/A)(−C/A) = −2 BC/A2 . EVALUATE: V doesn’t depend on z so E z = 0 everywhere. IDENTIFY: Apply Eq.(21.19). 1 q SET UP: Eq.(21.7) says E = rˆ is the electric field due to a point charge q. 4π P0 r 2

⎞ ∂V ∂ ⎛ kQ kQx kQx ⎟= =− ⎜ = 3 . ∂x ∂x ⎜ x 2 + y 2 + z 2 ⎟ ( x 2 + y 2 + z 2 )3 2 r ⎝ ⎠ kQy kQz Similarly, E y = 3 and Ez = 3 . r r ˆ ⎛ kQ xi yˆj zkˆ ⎞ kQ (b) From part (a), E = 2 ⎜ + + rˆ, which agrees with Equation (21.7). ⎟= r ⎜⎝ r r r ⎟⎠ r 2 EVALUATE: V is a scalar. E is a vector and has components. kq kq IDENTIFY and SET UP: For a solid metal sphere or for a spherical shell, V = outside the sphere and V = at R r ∂V . all points inside the sphere, where R is the radius of the sphere. When the electric field is radial, E = − ∂r ⎛1 1⎞ kq kq EXECUTE: (a) (i) r < ra : This region is inside both spheres. V = − = kq ⎜ − ⎟ . ra rb ⎝ ra rb ⎠ EXECUTE:

23.49.

(a) Ex = −

(ii) ra < r < rb : This region is outside the inner shell and inside the outer shell. V =

⎛1 1 ⎞ kq kq − = kq ⎜ − ⎟ . r rb ⎝ r rb ⎠

(iii) r > rb : This region is outside both spheres and V = 0 since outside a sphere the potential is the same as for point charge. Therefore the potential is the same as for two oppositely charged point charges at the same location. These potentials cancel. 1 ⎛q q⎞ 1 ⎛1 1⎞ (b) Va = q⎜ − ⎟ . ⎜ − ⎟ and Vb = 0 , so Vab = 4π P0 ⎝ ra rb ⎠ 4π P0 ⎝ ra rb ⎠ ⎛1 1 ⎞ ∂V q ∂ ⎛1 1 ⎞ 1 q Vab 1 (c) Between the spheres ra < r < rb and V = kq ⎜ − ⎟ . E = − =− = . ⎜ − ⎟=+ 2 ∂r ⎛ 1 1 ⎞ r2 4π P0 ∂r ⎝ r rb ⎠ 4π P0 r ⎝ r rb ⎠ − ⎜ ⎟ ⎝ ra rb ⎠ (d) From Equation (23.23): E = 0, since V is constant (zero) outside the spheres. (e) If the outer charge is different, then outside the outer sphere the potential is no longer zero but is 1 q 1 Q 1 (q − Q) V= − = . All potentials inside the outer shell are just shifted by an amount 4π P0 r 4π P0 r 4π P0 r 1 Q V =− . Therefore relative potentials within the shells are not affected. Thus (b) and (c) do not change. 4π P0 rb However, now that the potential does vary outside the spheres, there is an electric field there: ∂V ∂ ⎛ kq −kQ ⎞ kq ⎛ Q ⎞ k E=− =− ⎜ + ⎟ = ⎜1 − ⎟ = 2 (q − Q) . ∂r ∂r ⎝ r r ⎠ r2 ⎝ q⎠ r EVALUATE: In part (a) the potential is greater than zero for all r < rb .

23-18

23.50.

Chapter 23

⎛1 1⎞ ⎛1 1 ⎞ Exercise 23.49 shows that V = kq ⎜ − ⎟ for r < ra , V = kq ⎜ − ⎟ for ra < r < rb and r r ⎝ a b⎠ ⎝ r rb ⎠ ⎛1 1⎞ Vab = kq ⎜ − ⎟ . ⎝ ra rb ⎠ kq SET UP: E = 2 , radially outward, for ra ≤ r ≤ rb r ⎛1 1⎞ 500 V EXECUTE: (a) Vab = kq ⎜ − ⎟ = 500 V gives q = = 7.62 × 10−10 C . ⎛ ⎞ r r 1 1 b ⎠ ⎝ a − k⎜ ⎟ ⎝ 0.012 m 0.096 m ⎠ IDENTIFY:

1 1 V = + . V = 400 V at r ra kq r = 1.45 cm , V = 300 V at r = 1.85 cm , V = 200 V at r = 2.53 cm , V = 100 V at r = 4.00 cm , V = 0 at r = 9.60 cm . The equipotential surfaces are sketched in Figure 23.50. EVALUATE: (c) The equipotential surfaces are concentric spheres and the electric field lines are radial, so the field lines and equipotential surfaces are mutually perpendicular. The equipotentials are closest at smaller r, where the electric field is largest. (b) Vb = 0 so Va = 500 V . The inner metal sphere is an equipotential with V = 500 V .

Figure 23.50 23.51.

IDENTIFY: Outside the cylinder it is equivalent to a line of charge at its center. SET UP: The difference in potential between the surface of the cylinder (a distance R from the central axis) and a

general point a distance r from the central axis is given by ΔV =

λ ln( r / R ) . 2π P0

EXECUTE: (a) The potential difference depends only on r, and not direction. Therefore all points at the same value of r will be at the same potential. Thus the equipotential surfaces are cylinders coaxial with the given cylinder.

23.52.

λ

ln( r / R ) for r, gives r = R e 2π P0 ΔV/λ . 2π P0 For 10 V, the exponent is (10 V)/[(2 × 9.00 × 109 N · m2/C2)(1.50 × 10–9 C/m)] = 0.370, which gives r = (2.00 cm) e0.370 = 2.90 cm. Likewise, the other radii are 4.20 cm (for 20 V) and 6.08 cm (for 30 V). (c) Δr1 = 2.90 cm – 2.00 cm = 0.90 cm; Δr2 = 4.20 cm – 2.90 cm = 1.30 cm; Δr3 = 6.08 cm – 4.20 cm = 1.88 cm EVALUATE: As we can see, Δr increases, so the surfaces get farther apart. This is very different from a sheet of charge, where the surfaces are equally spaced planes. IDENTIFY: The electric field is the negative gradient of the potential. ∂V SET UP: Ex = − , so Ex is the negative slope of the graph of V as a function of x. ∂x EXECUTE: The graph is sketched in Figure 23.52. Up to a, V is constant, so Ex = 0. From a to b, V is linear with a positive slope, so Ex is a negative constant. Past b, the V-x graph has a decreasing positive slope which approaches zero, so Ex is negative and approaches zero. (b) Solving ΔV =

Electric Potential

23-19

Notice that an increasing potential does not necessarily mean that the electric field is increasing.

EVALUATE:

Figure 23.52 23.53.

(a) IDENTIFY: Apply the work-energy theorem, Eq.(6.6). SET UP: Points a and b are shown in Figure 23.53a.


−5

Wtot = ΔK = K b − K a = K b = 4.35 × 10 J

The electric force FE and the additional force F both do work, so that Wtot = WFE + WF . WFE = Wtot − WF = 4.35 × 10 −5 J − 6.50 × 10−5 J = −2.15 × 10−5 J

The forces on the charged particle are shown in Figure 23.53b.

EVALUATE:

Figure 23.53b

The electric force is to the left (in the direction of the electric field since the particle has positive charge). The displacement is to the right, so the electric force does negative work. The additional force F is in the direction of the displacement, so it does positive work. (b) IDENTIFY and SET UP: For the work done by the electric force, Wa →b = q (Va − Vb ) EXECUTE:

Va − Vb =

Wa →b −2.15 × 10−5 J = = −2.83 × 103 V. 7.60 × 10−9 C q

EVALUATE: The starting point (point a) is at 2.83 × 103 V lower potential than the ending point (point b). We know that Vb > Va because the electric field always points from high potential toward low potential. (c) IDENTIFY:

Calculate E from Va − Vb and the separation d between the two points.

SET UP: Since the electric field is uniform and directed opposite to the displacement Wa →b = − FE d = − qEd , where d = 8.00 cm is the displacement of the particle. W V − V −2.83 × 103 V = 3.54 × 104 V/m. EXECUTE: E = − a → b = − a b = 0.0800 m qd d

23.54.

EVALUATE: In part (a), Wtot is the total work done by both forces. In parts (b) and (c) Wa →b is the work done just by the electric force. ke 2 IDENTIFY: The electric force between the electron and proton is attractive and has magnitude F = 2 . For r 2 e circular motion the acceleration is arad = v 2 /r . U = − k . r SET UP: e = 1.60 × 10−19 C . 1 eV = 1.60 × 10−19 J . EXECUTE:

(a)

mv 2 ke 2 ke 2 = 2 and v = . r r mr

1 1 ke 2 1 (b) K = mv 2 = =− U 2 2 r 2

23-20

23.55.

Chapter 23

1 1 ke 2 1 k (1.60 × 10−19 C) 2 (c) E = K + U = U = − =− = −2.17 × 10−18 J = −13.6 eV . 2 2 r 2 5.29 × 10−11 m EVALUATE: The total energy is negative, so the electron is bound to the proton. Work must be done on the electron to take it far from the proton. IDENTIFY and SET UP: Calculate the components of E from Eq.(23.19). Eq.(21.3) gives F from E. EXECUTE: (a) V = Cx 4 / 3 C = V / x 4 / 3 = 240 V /(13.0 × 10−3 m) 4 / 3 = 7.85 × 104 V/m 4 / 3 ∂V 4 = − Cx1/ 3 = −(1.05 × 105 V/m 4 / 3 ) x1/ 3 ∂x 3 The minus sign means that Ex is in the − x -direction, which says that E points from the positive anode toward the negative cathode. (c) F = qE so Fx = −eEx = 43 eCx1/3

(b) E x = −

Halfway between the electrodes means x = 6.50 × 10−3 m.

Fx = 34 (1.602 × 10−19 C)(7.85 × 104 V/m 4 / 3 )(6.50 × 10−3 m)1/ 3 = 3.13 × 10−15 N Fx is positive, so the force is directed toward the positive anode. G EVALUATE: V depends only on x, so E y = Ez = 0. E is directed from high potential (anode) to low potential 23.56.

(cathode). The electron has negative charge, so the force on it is directed opposite to the electric field. IDENTIFY: At each point (a and b), the potential is the sum of the potentials due to both spheres. The voltmeter reads the difference between these two potentials. The spheres behave like a point-charges since the meter is connected to the surface of each one. SET UP: (a) Call a the point on the surface of one sphere and b the point on the surface of the other sphere, call r the radius of each sphere, and call d the center-to-center distance between the spheres. The potential difference Vab between points a and b is then Vb – Va = Vab = EXECUTE:

1 ⎡ −q q −q ⎞ ⎤ 2q ⎛ 1 1⎞ ⎛q − ⎟ + −⎜ + ⎜ ⎟ = 4πε 0 ⎝ d − r r ⎠ 4π P0 ⎢⎣ r d − r ⎝ r d − r ⎠ ⎥⎦

Substituting the numbers gives 1 1 ⎛ ⎞ 6 Vb – Va = 2(175 µC) ( 9.00 × 109 N ⋅ m 2 /C 2 ) ⎜ − ⎟ = –8.40 × 10 V 0.750 m 0.250 m ⎝ ⎠

23.57.

The meter reads 8.40 MV. (b) Since Vb – Va is negative, Va > Vb, so point a is at the higher potential. EVALUATE: An easy way to see that the potential at a is higher than the potential at b is that it would take positive work to move a positive test charge from b to a since this charge would be attracted by the negative sphere and repelled by the positive sphere. kq q IDENTIFY: U = 1 2 r SET UP: Eight charges means there are 8(8 − 1) / 2 = 28 pairs. There are 12 pairs of q and −q separated by d, 12 pairs of equal charges separated by

2d and 4 pairs of q and − q separated by

3d .

4 ⎞ 12kq ⎛ 1 1 ⎞ ⎛ 12 12 2 (a) U = kq 2 ⎜ − + − + ⎟ = − d ⎜1 − ⎟ = −1.46q /π P0 d d d d 2 3 2 3 3 ⎝ ⎠ ⎝ ⎠ EVALUATE: (b) The fact that the electric potential energy is less than zero means that it is energetically favorable for the crystal ions to be together. kq q IDENTIFY: For two small spheres, U = 1 2 . For part (b) apply conservation of energy. r SET UP: Let q1 = 2.00 μ C and q2 = −3.50 μ C . Let ra = 0.250 m and rb → ∞ . 2

EXECUTE:

23.58.

(8.99 × 109 N ⋅ m 2 /C 2 )(2.00 × 10−6 C)( −3.50 × 10−6 C) = −0.252 J 0.250 m (b) K b = 0 . U b = 0 . U a = −0.252 J . K a + U a = K b + U b gives K a = 0.252 J . K a = 12 mva2 , so EXECUTE:

va =

(a) U =

2Ka 2(0.252 J) = = 18.3 m/s 1.50 × 10−3 kg m

Electric Potential

23.59.

23-21

EVALUATE: As the sphere moves away, the attractive electrical force exerted by the other sphere does negative work and removes all the kinetic energy it initially had. Note that it doesn’t matter which sphere is held fixed and which is shot away; the answer to part (b) is unaffected. (a) IDENTIFY: Use Eq.(23.10) for the electron and each proton. SET UP: The positions of the particles are shown in Figure 23.59a.

r = (1.07 × 10−10 m) / 2 = 0.535 × 10−10 m Figure 23.59a EXECUTE: The potential energy of interaction of the electron with each proton is 1 ( −e 2 ) , so the total potential energy is U= 4π P0 r

U =−

2e 2 2(8.988 × 109 N ⋅ m 2 /C 2 )(1.60 × 10−19 C) 2 =− = −8.60 × 10−18 J 4π P0 r 0.535 × 10−10 m

U = −8.60 × 10−18 J(1 eV /1.602 × 10−19 J) = −53.7 eV EVALUATE: The electron and proton have charges of opposite signs, so the potential energy of the system is negative. (b) IDENTIFY and SET UP: The positions of the protons and points a and b are shown in Figure 23.59b.

rb = ra2 + d 2

ra = r = 0.535 × 10−10 m Figure 23.59b

Apply K a + U a + Wother = K b + U b with point a midway between the protons and point b where the electron instantaneously has v = 0 (at its maximum displacement d from point a). EXECUTE: Only the Coulomb force does work, so Wother = 0.

U a = −8.60 × 10−18 J (from part (a))

K a = 12 mv 2 = 12 (9.109 × 10−31 kg)(1.50 × 106 m/s) 2 = 1.025 × 10−18 J Kb = 0 U b = −2ke2 /rb Then U b = K a + U a − K b = 1.025 × 10−18 J − 8.60 × 10−18 J = −7.575 × 10−18 J.

rb = −

2ke2 2(8.988 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 =− = 6.075 × 10−11 m Ub −7.575 × 10−18 J

Then d = rb2 − ra2 = (6.075 × 10−11 m) 2 − (5.35 × 10−11 m) 2 = 2.88 × 10−11 m.

23.60.

EVALUATE: The force on the electron pulls it back toward the midpoint. The transverse distance the electron moves is about 0.27 times the separation of the protons. IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to the sphere. The electric force on the sphere is Fe = qE . The

potential difference between the plates is V = Ed . SET UP: The free-body diagram for the sphere is given in Figure 23.56. 2 EXECUTE: T cosθ = mg and T sin θ = Fe gives Fe = mg tanθ = (1.50 × 10 −3 kg)(9.80 m s )tan(30°) = 0.0085 N . Fe = Eq =

Fd (0.0085 N)(0.0500 m) Vq = = 47.8 V. and V = d q 8.90 × 10−6 C

23-22

Chapter 23

EVALUATE:

E = V/d = 956 V/m . E = σ /P0 and σ = EP0 = 8.46 × 10−9 C/m 2 .

Figure 23.60 23.61.

(a) IDENTIFY: The potential at any point is the sum of the potentials due to each of the two charged conductors. SET UP: From Example 23.10, for a conducting cylinder with charge per unit length λ the potential outside the cylinder is given by V = (λ /2π P0 )ln( r0 /r ) where r is the distance from the cylinder axis and r0 is the distance from

the axis for which we take V = 0. Inside the cylinder the potential has the same value as on the cylinder surface. The electric field is the same for a solid conducting cylinder or for a hollow conducting tube so this expression for V applies to both. This problem says to take r0 = b. EXECUTE: For the hollow tube of radius b and charge per unit length −λ : outside V = −(λ /2π P0 )ln(b /r ); inside V = 0 since V = 0 at r = b. For the metal cylinder of radius a and charge per unit length λ : outside V = (λ /2π P0 )ln(b /r ), inside V = (λ /2π P0 )ln(b /a ), the value at r = a.

(i) r < a; inside both V = (λ /2π P0 )ln(b /a )

(ii) a < r < b; outside cylinder, inside tube V = (λ /2π P0 )ln(b /r ) (iii) r > b; outside both the potentials are equal in magnitude and opposite in sign so V = 0. (b) For r = a, Va = (λ /2π P0 )ln(b /a ). For r = b, Vb = 0. Thus Vab = Va − Vb = (λ /2π P0 )ln(b /a ). (c) IDENTIFY and SET UP: Use Eq.(23.23) to calculate E. ∂V λ ∂ ⎛b⎞ λ ⎛ r ⎞⎛ b ⎞ Vab 1 ln ⎜ ⎟ = − . =− EXECUTE: E = − ⎜ ⎟⎜ − 2 ⎟ = ∂r 2π P0 ∂r ⎝ r ⎠ 2π P0 ⎝ b ⎠⎝ r ⎠ ln(b /a) r (d) The electric field between the cylinders is due only to the inner cylinder, so Vab is not changed,

Vab = (λ /2π P0 )ln(b /a ). EVALUATE: The electric field is not uniform between the cylinders, so Vab ≠ E (b − a ). 23.62.

IDENTIFY:

The wire and hollow cylinder form coaxial cylinders. Problem 23.61 gives E ( r ) =

Vab 1 . ln(b /a ) r

a = 145 × 10−6 m , b = 0.0180 m . Vab 1 EXECUTE: E = and Vab = E ln (b/a)r = (2.00 × 104 N C)(ln (0.018 m 145 × 10−6 m))0.012 m = 1157 V. ln( b a ) r EVALUATE: The electric field at any r is directly proportional to the potential difference between the wire and the cylinder. IDENTIFY and SET UP: Use Eq.(21.3) to calculate F and then F = ma gives a. EXECUTE: (a) FE = qE. Since q = −e is negative FE and E are in opposite directions; E is upward so FE is SET UP:

23.63.

downward. The magnitude of FE is FE = q E = eE = (1.602 × 10−19 C)(1.10 × 103 N/C) = 1.76 × 10−16 N. (b) Calculate the acceleration of the electron produced by the electric force:

a=

F 1.76 × 10−16 N = = 1.93 × 1014 m/s 2 m 9.109 × 10−31 kg

EVALUATE: This is much larger than g = 9.80 m/s 2 , so the gravity force on the electron can be neglected. FE is downward, so a is downward. (c) IDENTIFY and SET UP: The acceleration is constant and downward, so the motion is like that of a projectile. Use the horizontal motion to find the time and then use the time to find the vertical displacement.

Electric Potential

23-23

EXECUTE: x-component v0 x = 6.50 × 106 m/s; a x = 0; x − x0 = 0.060 m; t = ?

x − x0 = v0 xt + 12 axt 2 and the a x term is zero, so x − x0 0.060 m t= = = 9.231× 10−9 s v0 x 6.50 × 106 m/s y-component v0 y = 0; a y = 1.93 × 1014 m/s 2 ; t = 9.231 × 10 −9 m/s; y − y0 = ?

y − y0 = v0 y t + 12 a y t 2 y − y0 = 12 (1.93 × 1014 m/s 2 )(9.231× 10−9 s)2 = 0.00822 m = 0.822 cm (d) The velocity and its components as the electron leaves the plates are sketched in Figure 23.63. vx = v0 x = 6.50 × 106 m/s (since ax = 0 ) v y = v0 y + a yt

v y = 0 + (1.93 × 1014 m/s 2 )(9.231× 10−9 s) v y = 1.782 × 106 m/s Figure 23.63 1.782 × 10 m/s = 0.2742 so α = 15.3°. vx 6.50 × 106 m/s EVALUATE: The greater the electric field or the smaller the initial speed the greater the downward deflection. (e) IDENTIFY and SET UP: Consider the motion of the electron after it leaves the region between the plates. Outside the plates there is no electric field, so a = 0. (Gravity can still be neglected since the electron is traveling at such high speed and the times are small.) Use the horizontal motion to find the time it takes the electron to travel 0.120 m horizontally to the screen. From this time find the distance downward that the electron travels. EXECUTE: x-component v0 x = 6.50 × 106 m/s; a x = 0; x − x0 = 0.120 m; t = ? tan α =

vy

=

6

x − x0 = v0 xt + 12 axt 2 and the a x term is term is zero, so x − x0 0.120 m t= = = 1.846 × 10−8 s v0 x 6.50 × 106 m/s y-component v0 y = 1.782 × 106 m/s (from part (b)); a y = 0; t = 1.846 × 10 −8 m/s; y − y0 = ?

y − y0 = v0 yt + 12 a yt 2 = (1.782 × 106 m/s)(1.846 × 10−8 s) = 0.0329 m = 3.29 cm

23.64.

EVALUATE: The electron travels downward a distance 0.822 cm while it is between the plates and a distance 3.29 cm while traveling from the edge of the plates to the screen. The total downward deflection is 0.822 cm + 3.29 cm = 4.11 cm. The horizontal distance between the plates is half the horizontal distance the electron travels after it leaves the plates. And the vertical velocity of the electron increases as it travels between the plates, so it makes sense for it to have greater downward displacement during the motion after it leaves the plates. IDENTIFY: The charge on the plates and the electric field between them depend on the potential difference across the plates. Since we do not know the numerical potential, we shall call this potential V and find the answers in terms of V. σ Qd (a) SET UP: For two parallel plates, the potential difference between them is V = Ed = d = . P0 P0 A EXECUTE: Solving for Q gives

Q = P0 AV / d = (8.85 × 10–12 C2/N ⋅ m2)(0.030 m)2V/(0.0050 m) Q = 1.59V × 10–12 C = 1.59V pC, when V is in volts. (b) E = V/d = V/(0.0050 m) = 200V V/m, with V in volts. (c) SET UP: Energy conservation gives 12 mv 2 = eV . EXECUTE: Solving for v gives v=

2 (1.60 × 10−19 C )V 2eV = = 5.93 × 105V 1/ 2 m/s , with V in volts m 9.11 × 10−31 kg

EVALUATE: Typical voltages in student laboratory work run up to around 25 V, so the charge on the plates is typically about around 40 pC, the electric field is about 5000 V/m, and the electron speed would be about 3 million m/s.

23-24

Chapter 23

23.65.


Problem 23.61 derived that E =

Vab 1 , where a is the radius of the inner cylinder ln(b / a ) r

(wire) and b is the radius of the outer hollow cylinder. The potential difference between the two cylinders is Vab . Use this expression to calculate E at the specified r. EXECUTE: Midway between the wire and the cylinder wall is at a radius of r = ( a + b)/2 = (90.0 × 10−6 m + 0.140 m)/2 = 0.07004 m. E=

Vab 1 50.0 × 103 V = = 9.71 × 104 V/m ln(b / a ) r ln(0.140 m / 90.0 × 10−6 m)(0.07004 m)

(b) IDENTIFY and SET UP: The electric force is given by Eq.(21.3). Set this equal to ten times the weight of the particle and solve for q , the magnitude of the charge on the particle.

FE = 10mg

EXECUTE:

10mg 10(30.0 × 10−9 kg)(9.80 m/s 2 ) = = 3.03 × 10−11 C E 9.71 × 104 V/m EVALUATE: It requires only this modest net charge for the electric force to be much larger than the weight. (a) IDENTIFY: Calculate the potential due to each thin ring and integrate over the disk to find the potential. V is a scalar so no components are involved. SET UP: Consider a thin ring of radius y and width dy. The ring has area 2π y dy so the charge on the ring is dq = σ (2π y dy ). EXECUTE: The result of Example 23.11 then says that the potential due to this thin ring at the point on the axis at a distance x from the ring is 1 dq 2πσ y dy dV = = 2 2 4π P0 x + y 4π P0 x 2 + y 2 q E = 10mg and q =

23.66.

V = ∫ dV =

For x

EVALUATE:

σ 2P0

∫

R 0

y dy x +y 2

2

=

R σ ⎡ 2 σ x + y2 ⎤ = ( x 2 + R 2 − x)

2P0 ⎣

⎦0

R this result should reduce to the potential of a point charge with Q = σπ R 2 . x 2 + R 2 = x(1 + R 2 /x 2 )1/ 2 ≈ x(1 + R 2 /2 x 2 ) so

Then V ≈

σ R2 2P0 2 x

=

23.67.

(a) IDENTIFY: SET UP:

E (r ) =

Use Eq.(23.19) to calculate Ex .

⎞ σx⎛1 ∂V σ ⎛ x 1 =− − 1⎟ = ⎜ ⎜ − 2 ∂x x 2P0 ⎝ x 2 + R 2 2 P x + R2 0 ⎝ ⎠ Our result agrees with Eq.(21.11) in Example 21.12.

Ex = −

EVALUATE:

x 2 + R 2 − x ≈ R 2 /2 x

σπ R 2 Q = , as expected. 4π P0 x 4π P0 x

(b) IDENTIFY and SET UP: EXECUTE:

2P0

⎞ ⎟. ⎠

b

Use Va − Vb = ∫ E ⋅ dl. a

From Problem 22.48, E ( r ) =

λr for r ≤ R (inside the cylindrical charge distribution) and 2π P0 R 2

λr for r ≥ R. Let V = 0 at r = R (at the surface of the cylinder). 2π P0 r

EXECUTE: r > R Take point a to be at R and point b to be at r, where r > R. Let dl = dr. E and dr are both radially outward, so r

r

R

R

E ⋅ d r = E dr. Thus VR − Vr = ∫ E dr. Then VR = 0 gives Vr = − ∫ E dr. In this interval (r > R ), E (r ) = λ /2π P0 r , so Vr = − ∫

r R

λ λ r dr λ ⎛r⎞ =− ln ⎜ ⎟ . dr = − ∫ R 2π P0 r 2π P0 2π P0 ⎝ R ⎠ r

EVALUATE: This expression gives Vr = 0 when r = R and the potential decreases (becomes a negative number of larger magnitude) with increasing distance from the cylinder.

Electric Potential

EXECUTE:

23-25

r R : E =

r

R r r kQ kQ kQ kQ 1 2 kQ kQ kQr 2 kQ ⎡ r2 ⎤ kQr ′ ′ ′ ′ ′ = − ⋅ − ⋅ = − = − = + − = − V d d r dr r 3 and E r E r ⎢ ⎥. ∫ ∫R R3 R R 3 ∫R R R3 2 R R 2 R 2 R3 2 R ⎣ R 2 ⎦ ∞ (b) The graphs of V and E versus r are sketched in Figure 23.72. EVALUATE: For r < R the potential depends on the electric field in the region r to ∞ .

For r < R : E =

Figure 23.72 23.73.

IDENTIFY: SET UP:

Problem 23.70 shows that Vr = V0 =

Q Q (3 − r 2 R 2 ) for r ≤ R and Vr = for r ≥ R . 8π P0 R 4π P0 r

3Q Q , VR = 8π P0 R 4π P0 R

Q 8π P0 R (b) If Q > 0 , V is higher at the center. If Q < 0 , V is higher at the surface. EXECUTE:

(a) V0 − VR =

EVALUATE: For Q > 0 the electric field is radially outward, E is directed toward lower potential, so V is higher at the center. If Q < 0 , the electric field is directed radially inward and V is higher at the surface. 23.74.


For r < c , E = 0 and the potential is constant. For r > c , E is the same as for a point charge and V = V∞ = 0 (a) Points a, b, and c are all at the same potential, so Va − Vb = Vb − Vc = Va − Vc = 0 . 2

kq (8.99 × 109 N ⋅ m 2 C )(150 × 10−6 C) = = 2.25 × 106 V R 0.60 m (b) They are all at the same potential. (c) Only Vc − V∞ would change; it would be −2.25 × 106 V. Vc − V∞ =

kq . r

Electric Potential

23.75.

23.76.

23-27

EVALUATE: The voltmeter reads the potential difference between the two points to which it is connected. IDENTIFY and SET UP: Apply Fr = − dU / dr and Newton's third law. EXECUTE: (a) The electrical potential energy for a spherical shell with uniform surface charge density and a point charge q outside the shell is the same as if the shell is replaced by a point charge at its center. Since Fr = − dU dr , this means the force the shell exerts on the point charge is the same as if the shell were replaced by a point charge at its center. But by Newton’s 3rd law, the force q exerts on the shell is the same as if the shell were a point charge. But q can be replaced by a spherical shell with uniform surface charge and the force is the same, so the force between the shells is the same as if they were both replaced by point charges at their centers. And since the force is the same as for point charges, the electrical potential energy for the pair of spheres is the same as for a pair of point charges. (b) The potential for solid insulating spheres with uniform charge density is the same outside of the sphere as for a spherical shell, so the same result holds. (c) The result doesn’t hold for conducting spheres or shells because when two charged conductors are brought close together, the forces between them causes the charges to redistribute and the charges are no longer distributed uniformly over the surfaces. qq kq q EVALUATE: For the insulating shells or spheres, F = k 1 2 2 and U = 1 2 , where q1 and q2 are the charges of r r the objects and r is the distance between their centers. IDENTIFY: Apply Newton's second law to calculate the acceleration. Apply conservation of energy and conservation of momentum to the motions of the spheres. qq kq q SET UP: Problem 23.75 shows that F = k 1 2 2 and U = 1 2 , where q1 and q2 are the charges of the objects and r r r is the distance between their centers. EXECUTE: Maximum speed occurs when the spheres are very far apart. Energy conservation gives kq1q2 1 1 2 2 . Momentum conservation gives m50v50 = m150v150 and v50 = 3v150 . r = 0.50 m. Solve for v50 = m50v50 + m150v150 r 2 2 and v150 : v50 = 12.7 m s, v150 = 4.24 m s . Maximum acceleration occurs just after spheres are released. ∑ F = ma

gives

(9 × 109 N ⋅ m 2 C 2 )(10−5 C)(3 × 10−5 C) kq1q2 2 = (0.15 kg)a150 . a150 = 72.0 m s and = m150 a150 . 2 (0.50 m) 2 r 2

23.77.

a50 = 3a150 = 216 m s . EVALUATE: The more massive sphere has a smaller acceleration and a smaller final speed. IDENTIFY: Use Eq.(23.17) to calculate Vab . SET UP: From Problem 22.43, for R ≤ r ≤ 2 R (between the sphere and the shell) E = Q / 4π P0 r 2 Take a at R and b at 2R. 2R 2R Q 2 R dr Q ⎡ 1⎤ Q ⎛1 1 ⎞ = − = EXECUTE: Vab = Va − Vb = ∫ E dr = ⎜ − ⎟ R 4π P0 ∫ R r 2 4π P0 ⎣⎢ r ⎥⎦ R 4π P0 ⎝ R 2 R ⎠ Q Vab = 8π P0 R EVALUATE: The electric field is radially outward and points in the direction of decreasing potential, so the sphere is at higher potential than the shell.

23.78.

IDENTIFY: SET UP:

b

Va − Vb = ∫ E ⋅ dl a

E is radially outward, so E ⋅ dl = E dr . Problem 22.42 shows that E ( r ) = 0 for r ≤ a , E (r ) = kq / r 2 for

a < r < b , E ( r ) = 0 for b < r < c and E (r ) = kq / r 2 for r > c . c

EXECUTE:

kq kq dr = . 2 c ∞ r

(a) At r = c : Vc = − ∫ c

b

(b) At r = b : Vb = − ∫ E ⋅ dr − ∫ E ⋅ dr = ∞

c

kq kq −0= . c c

c

b

a

∞

c

b

(c) At r = a : Va = − ∫ E ⋅ dr − ∫ E ⋅ dr − ∫ E ⋅ dr =

a

kq dr ⎡1 1 1 ⎤ − kq ∫ 2 = kq ⎢ − + ⎥ c r ⎣c b a ⎦ b

⎡1 1 1 ⎤ (d) At r = 0 : V0 = kq ⎢ − + ⎥ since it is inside a metal sphere, and thus at the same potential as its surface. ⎣c b a ⎦ ⎡1 1⎤ EVALUATE: The potential difference between the two conductors is Va − Vb = kq ⎢ − ⎥ . ⎣a b⎦

23-28

Chapter 23

23.79.

IDENTIFY: Slice the rod into thin slices and use Eq.(23.14) to calculate the potential due to each slice. Integrate over the length of the rod to find the total potential at each point. (a) SET UP: An infinitesimal slice of the rod and its distance from point P are shown in Figure 23.79a.

Figure 23.79a

Use coordinates with the origin at the left-hand end of the rod and one axis along the rod. Call the axes x′ and y′ so as not to confuse them with the distance x given in the problem. EXECUTE: Slice the charged rod up into thin slices of width dx′. Each slice has charge dQ = Q (dx′/a ) and a distance r = x + a − x′ from point P. The potential at P due to the small slice dQ is dV =

1 ⎛ dQ ⎞ 1 Q ⎛ dx′ ⎞ ⎜ ⎟= ⎜ ⎟. 4π P0 ⎝ r ⎠ 4π P0 a ⎝ x + a − x′ ⎠

Compute the total V at P due to the entire rod by integrating dV over the length of the rod ( x′ = 0 to x′ = a ) : a Q dx′ Q Q ⎛ x+a⎞ = [− ln( x + a − x′)]0a = ln ⎜ ⎟. ∫ 0 4π P0 a ( x + a − x′) 4π P0 a 4π P0 a ⎝ x ⎠

V = ∫ dV = EVALUATE: (b) SET UP:

Q ⎛ x⎞ ln ⎜ ⎟ = 0. 4π P0 a ⎝ x ⎠ An infinitesimal slice of the rod and its distance from point R are shown in Figure 23.79b. As x → ∞, V →

Figure 23.79b

dQ = (Q / a )dx′ as in part (a)

Each slice dQ is a distance r = y 2 + (a − x′) 2 from point R. EXECUTE: The potential dV at R due to the small slice dQ is dV =

dx′

1 ⎛ dQ ⎞ 1 Q ⎜ ⎟= 4π P0 ⎝ r ⎠ 4π P0 a

V = ∫ dV =

Q 4π P0 a

∫

y + (a − x′) 2 2

dx′

a 0

y + (a − x′) 2 2

.

.

In the integral make the change of variable u = a − x′; du = − dx′ V =−

V =−

0 0 Q du Q ⎡ 2 2 ⎤ = − + + ln( ) u y u ⎦a 4π P0 a ∫ a y 2 + u 2 4π P0 a ⎣

Q Q ⎡ ⎛ a + a2 + y2 ⎢ln ⎜ [ln y − ln(a + y 2 + a 2 )] = 4π P0 a 4π P0 a ⎢ ⎜⎝ y ⎣

(The expression for the integral was found in appendix B.) ⎛ y⎞ Q EVALUATE: As y → ∞, V → ln ⎜ ⎟ = 0. 4π P0 a ⎝ y ⎠

⎞⎤ ⎟⎥ . ⎟⎥ ⎠⎦

Electric Potential

part (a): V =

(c) SET UP:

23-29

Q Q ⎛ x+a⎞ ⎛ a⎞ ln ⎜ ln ⎜ 1 + ⎟ . ⎟= 4π P0 a ⎝ x ⎠ 4π P0 a ⎝ x⎠

From Appendix B, ln(1 + u ) = u − u 2 / 2 . . . , so ln(1 + a / x) = a / x − a 2 / 2 x 2 and this becomes a / x when x is large. Thus V →

EXECUTE:

part (b): V =

Q ⎛a⎞ Q . For large x, V becomes the potential of a point charge. ⎜ ⎟= 4π P0 a ⎝ x ⎠ 4π P0 a

Q ⎡ ⎛ a + a2 + y2 ⎢ln ⎜ 4π P0 a ⎢ ⎜⎝ y ⎣

From Appendix B,

⎞⎤ ⎛a Q a2 ⎞ ⎟⎥ = ln ⎜ + 1 + 2 ⎟ . ⎟ ⎥ 4π P0 a ⎜ y y ⎟⎠ ⎝ ⎠⎦

1 + a 2 / y 2 = (1 + a 2 / y 2 )1/ 2 = 1 + a 2 / 2 y 2 + …

Thus a / y + 1 + a 2 / y 2 → 1 + a / y + a 2 / 2 y 2 + … → 1 + a / y. And then using ln(1 + u ) ≈ u gives V→ EVALUATE: 23.80.

IDENTIFY:

Q Q ⎛a⎞ Q ln(1 + a / y ) → . ⎜ ⎟= 4π P0 a 4π P0 a ⎝ y ⎠ 4π P0 y

For large y, V becomes the potential of a point charge. The potential at the surface of a uniformly charged sphere is V =

kQ . R

4 For a sphere, V = π R 3 . When the raindrops merge, the total charge and volume is conserved. 3 kQ k (−1.20 × 10−12 C) = = −16.6 V . EXECUTE: (a) V = R 6.50 × 10−4 m SET UP:

(b) The volume doubles, so the radius increases by the cube root of two: Rnew = 3 2 R = 8.19 × 10−4 m and the new

kQnew k (−2.40 × 10−12 C) = = −26.4 V . 8.19 × 10−4 m Rnew The charge doubles but the radius also increases and the potential at the surface increases by only a

charge is Qnew = 2Q = − 2.40 × 10−12 C. The new potential is Vnew =

23.81.

EVALUATE: 2 factor of 1/ 3 = 22 / 3 . 2 (a) IDENTIFY and SET UP: The potential at the surface of a charged conducting sphere is given by Example 23.8: 1 q . For spheres A and B this gives V= 4π P0 R

VA = EXECUTE:

QA QB and VB = . 4π P0 RA 4π P0 RB

VA = VB gives QA / 4π P0 RA = QB / 4π P0 RB and QB / QA = RB / RA . And then RA = 3RB implies

QB / QA = 1/ 3. (b) IDENTIFY and SET UP: Example 22.5:

The electric field at the surface of a charged conducting sphere is given in E=

EXECUTE:

1 q . 4π P0 R 2

For spheres A and B this gives EA =

QA QB and EB = . 4π P0 RA2 4π P0 RB2

EB ⎛ QB ⎞ ⎛ 4π P0 RA2 ⎞ 2 2 =⎜ ⎟ = QB /QA ( RA /RB ) = (1/3)(3) = 3. ⎟⎜ E A ⎝ 4π P0 R 2 B ⎠ ⎜⎝ QA ⎟⎠

23.82.

EVALUATE: The sphere with the larger radius needs more net charge to produce the same potential. We can write E = V / R for a sphere, so with equal potentials the sphere with the smaller R has the larger V. IDENTIFY: Apply conservation of energy, K a + U a = K b + U b . SET UP:

Assume the particles initially are far apart, so U a = 0 , The alpha particle has zero speed at the distance of

closest approach, so K b = 0 . 1 eV = 1.60 × 10−19 J . The alpha particle has charge +2e and the lead nucleus has charge +82e .

23-30

Chapter 23

Set the alpha particle’s kinetic energy equal to its potential energy: K a = U b gives

EXECUTE:

23.83.

k (164)(1.60 × 10−19 C) 2 k (2e)(82e) 11.0 MeV = and r = = 2.15 × 10−14 m . r (11.0 × 106 eV)(1.60 × 10−19 J eV) EVALUATE: The calculation assumes that at the distance of closest approach the alpha particle is outside the radius of the lead nucleus. IDENTIFY and SET UP: The potential at the surface is given by Example 23.8 and the electric field at the surface is given by Example 22.5. The charge initially on sphere 1 spreads between the two spheres such as to bring them to the same potential. 1 Q1 1 Q1 EXECUTE: (a) E1 = = R1E1 , V1 = 2 4π P0 R1 4π P0 R1 (b) Two conditions must be met: 1) Let q1and q2 be the final potentials of each sphere. Then q1 + q2 = Q1 (charge conservation) 2) Let V1 and V2 be the final potentials of each sphere. All points of a conductor are at the same potential, so V1 = V2 . V1 = V2 requires that

1 q1 1 q2 = and then q1 / R1 = q2 / R2 4π P0 R1 4π P0 R2 q1R2 = q2 R1 = (Q1 − q1 ) R1

This gives q1 = ( R1 /[ R1 + R2 ])Q1 and q2 = Q1 − q1 = Q1 (1 − R1 /[ R1 + R2 ]) = Q1 ( R2 /[ R1 + R2 ]) (c) V1 =

1 q1 Q1 1 q2 Q1 = = , which equals V1 as it should. and V2 = 4π P0 R1 4π P0 ( R1 + R2 ) 4π P0 R2 4π P0 ( R1 + R2 )

(d) E1 =

V1 Q1 V Q1 = . E2 = 2 = . R1 4π P0 R1 ( R1 + R2 ) R2 4π P0 R2 ( R1 + R2 )

EVALUATE:

Part (a) says q2 = q1 ( R2 / R1 ). The sphere with the larger radius needs more charge to produce the

same potential at its surface. When R1 = R2 , q1 = q2 = Q1 / 2. The sphere with the larger radius has the smaller electric field at its surface. 23.84.

b

Apply Va − Vb = ∫ E ⋅ dl

IDENTIFY:

a

From Problem 22.57, for r ≥ R , E =

SET UP:

kQ ⎡ r 3 r4 ⎤ kQ . For r ≤ R , E = 2 ⎢ 4 3 − 3 4 ⎥ . 2 r r ⎣ R R ⎦

r

EXECUTE:

(a) r ≥ R : E =

kQ kQ kQ ⇒ V = − ∫ 2 dr ′ = , which is the potential of a point charge. 2 r r′ r ∞

R r ⎤ kQ ⎡ r 3 r4 ⎤ kQ ⎡ r2 R 2 r 3 R 3 ⎤ kQ ⎡ r 3 r2 4 3 − 3 4 ⎥ and V = − ∫ Edr′ − ∫ Edr′ = ⎢1 − 2 2 + 2 2 + 3 − 3 ⎥ = ⎢ 3 − 2 2 + 2⎥ . 2 ⎢ r ⎣ R R ⎦ R R R R R R R R ⎣ ⎦ ⎣ ⎦ R ∞ kQ 2kQ At r = R , V = . At r = 0 , V = . The electric field is radially outward and V increases as r R R

(b) r ≤ R : E = EVALUATE: 23.85.

decreases. IDENTIFY: Apply conservation of energy: Ei = Ef . SET UP: In the collision the initial kinetic energy of the two particles is converted into potential energy at the distance of closest approach. EXECUTE: (a) The two protons must approach to a distance of 2rp , where rp is the radius of a proton. 2 k (1.60 × 10−19 C) 2 ⎡1 ⎤ ke Ei = Ef gives 2 ⎢ mpv 2 ⎥ = and v = = 7.58 × 106 m s . 2(1.2 × 10−15 m)(1.67 × 10−27 kg) ⎣2 ⎦ 2rp (b) For a helium-helium collision, the charges and masses change from (a) and

v=

k (2(1.60 × 10−19 C)) 2 = 7.26 × 106 m s. (3.5 × 10−15 m)(2.99)(1.67 × 10−27 kg)

(c) K =

m v 2 (1.67 × 10−27 kg)(7.58 × 106 m s) 2 3kT mv 2 = . Tp = p = = 2.3 × 109 K . 2 2 3k 3(1.38 × 10−23 J K) THe =

mHev 2 (2.99)(1.67 × 10−27 kg)(7.26 × 106 m s) 2 = = 6.4 × 109 K . 3k 3(1.38 × 10−23 J K)

Electric Potential

23.86.

23-31

(d) These calculations were based on the particles’ average speed. The distribution of speeds ensures that there are always a certain percentage with a speed greater than the average speed, and these particles can undergo the necessary reactions in the sun’s core. EVALUATE: The kinetic energies required for fusion correspond to very high temperatures. b W IDENTIFY and SET UP: Apply Eq.(23.20). a → b = Va − Vb and Va − Vb = ∫ E ⋅ dl . a q0 EXECUTE:

(a) E = −

∂V ˆ ∂V ˆ ∂V ˆ i− j− k = −2 Axiˆ + 6 Ayˆj − 2 Azkˆ ∂x ∂y ∂z 0

0

z0

z0

(b) A charge is moved in along the z -axis. The work done is given by W = q ∫ E ⋅ kˆdz = q ∫ (−2 Az ) dz = + ( Aq ) z02 . −5

Wa → b 6.00 × 10 J = = 640 V m 2 . qz02 (1.5 × 10−6 C)(0.250 m) 2 (c) E (0,0,0.250) = −2(640 V m 2 )(0.250 m)kˆ = −(320 V m)kˆ .

Therefore, A =

(d) In every plane parallel to the xz -plane, y is constant, so V ( x, y, z ) = Ax 2 + Az 2 − C , where C = 3Ay 2 .

V +C = R 2 , which is the equation for a circle since R is constant as long as we have constant potential on A those planes. 2 1280 V + 3(640 V m )(2.00 m) 2 (e) V = 1280 V and y = 2.00 m , so x 2 + z 2 = = 14.0 m 2 and the radius of the circle 2 640 V m is 3.74 m. x2 + z 2 =

23.87.

EVALUATE: In any plane parallel to the xz-plane, E projected onto the plane is radial and hence perpendicular to the equipotential circles. IDENTIFY: Apply conservation of energy to the motion of the daughter nuclei. SET UP: Problem 23.73 shows that the electrical potential energy of the two nuclei is the same as if all their charge was concentrated at their centers. EXECUTE: (a) The two daughter nuclei have half the volume of the original uranium nucleus, so their radii are 7.4 × 10−15 m smaller by a factor of the cube root of 2: r = = 5.9 × 10−15 m. 3 2 k (46e) 2 k (46) 2 (1.60 × 10−19 C) 2 (b) U = = = 4.14 × 10−11 J . U = 2 K , where K is the final kinetic energy of each 2r 1.18 × 10−14 m

nucleus. K = U 2 = (4.14 × 10−11 J) 2 = 2.07 × 10−11 J . (c) If we have 10.0 kg of uranium, then the number of nuclei is n =

23.88.

10.0 kg = 2.55 × 1025 nuclei . (236 u)(1.66 × 10−27 kg u)

And each releases energy U, so E = nU = (2.55 × 1025 )(4.14 × 10−11 J) = 1.06 × 1015 J = 253 kilotons of TNT . (d) We could call an atomic bomb an “electric” bomb since the electric potential energy provides the kinetic energy of the particles. EVALUATE: This simple model considers only the electrical force between the daughter nuclei and neglects the nuclear force. ∂V IDENTIFY and SET UP: In part (a) apply E = − . In part (b) apply Gauss's law. ∂r ∂V ρ a2 ⎡ r ∂V r2 ⎤ ρ a ⎡ r r2 ⎤ EXECUTE: (a) For r ≤ a , E = − = 0 . E has = − 0 ⎢ −6 2 + 6 3 ⎥ = 0 ⎢ − 2 ⎥ . For r ≥ a , E = − ∂r ∂r 18P0 ⎣ a a ⎦ 3P0 ⎣ a a ⎦ only a radial component because V depends only on r. Q ρ a ⎡ r r2 ⎤ (b) For r ≤ a , Gauss's law gives Er 4π r 2 = r = 0 ⎢ − 2 ⎥ 4π r 2 and P0 3P0 ⎣ a a ⎦ Er + dr 4π (r 2 + 2rdr ) =

Qr + dr ρ 0 a ⎡ r + dr (r 2 + 2rdr ) ⎤ 2 = − ⎢ ⎥ 4π (r + 2rdr ) . Therefore, P0 a2 3P0 ⎣ a ⎦

Qr + dr − Qr ρ (r )4π r 2 dr ρ0 a 4π r 2 dr ⎡ 2r 2 2r 1 ⎤ ρ ⎡ 4r ⎤ ⎡ 4r ⎤ = ≈ − 2 + − 2 + ⎥ and ρ (r ) = 0 ⎢3 − ⎥ = ρ0 ⎢1 − ⎥ . ⎢ 3P0 a a a⎦ 3 ⎣ a⎦ P0 P0 ⎣ a ⎣ 3a ⎦

23-32

Chapter 23

(c) For r ≥ a , ρ (r ) = 0 , so the total charge enclosed will be given by a

23.89.

a a⎡ ⎡1 4r 3 ⎤ r4 ⎤ Q = 4π ∫ ρ (r )r 2 dr = 4πρ 0 ∫ ⎢ r 2 − dr = 4πρ 0 ⎢ r 3 − ⎥ = 0 . ⎥ 0 3a ⎦ 3a ⎦ 0 ⎣ ⎣3 0 EVALUATE: Apply Gauss's law to a sphere of radius r > R . The result of part (c) says that Qencl = 0 , so E = 0 . This agrees with the result we calculated in part (a). IDENTIFY: Angular momentum and energy must be conserved. SET UP: At the distance of closest approach the speed is not zero. E = K + U . q1 = 2e , q2 = 82e .

EXECUTE:

1 kq q mv1b = mv2 r2 . E1 = E2 gives E1 = mv2 2 + 1 2 . E1 = 11 MeV = 1.76 × 10−12 J . r2 is the distance of 2 r2

⎛b⎞ b 2 kq q closest approach. Substituting in for v2 = v1 ⎜ ⎟ we find E1 = E1 2 + 1 2 . r2 r2 ⎝ r2 ⎠ 2 2 −12 −12 ( E1 )r2 − (kq1q2 )r2 − E1b = 0 . For b = 10 m , r2 = 1.01 × 10 m . For b = 10−13 m , r2 = 1.11 × 10−13 m . And for

23.90.

b = 10−14 m , r2 = 2.54 × 10−14 m . EVALUATE: As b decreases the collision is closer to being head-on and the distance of closest approach decreases. Problem 23.82 shows that the distance of closest approach is 2.15 × 10−14 m when b = 0 . IDENTIFY: Consider the potential due to an infinitesimal slice of the cylinder and integrate over the length of the ∂V . cylinder to find the total potential. The electric field is along the axis of the tube and is given by E = − ∂x SET UP: Use the expression from Example 23.11 for the potential due to each infinitesimal slice. Let the slice be at coordinate z along the x-axis, relative to the center of the tube. EXECUTE: (a) For an infinitesimal slice of the finite cylinder, we have the potential k dQ kQ dz dV = = . Integrating gives L ( x − z )2 + R 2 ( x − z )2 + R 2 L 2− x

L2

V=

kQ dz kQ du = where u = x − z . Therefore, L − L∫ 2 ( x − z ) 2 + R 2 L − L ∫2 − x u 2 + R 2

V=

kQ ⎡ ( L 2 − x) 2 + R 2 + ( L 2 − x) ⎤ ⎥ on the cylinder axis. ln ⎢ L ⎢⎣ ( L 2 + x) 2 + R 2 − L 2 − x ⎥⎦

(b) For L mg and FI − mg is the net force that accelerates the bar upward. Use Newton's 2nd law to find the acceleration.

27-12

Chapter 27

mg ( 0.750 kg ) ( 9.80 m/s ) = = 32.67 A lB ( 0.500 m )( 0.450 T ) 2

(a) EXECUTE:

IlB = mg , I =

E = IR = ( 32.67 A )( 25.0 Ω ) = 817 V (b) R = 2.0 Ω, I = E / R = ( 816.7 V ) / ( 2.0 Ω ) = 408 A

FI = IlB = 92 N a = ( FI − mg ) / m = 113 m/s 2

27.40.

EVALUATE: I increases by over an order of magnitude when R changes to FI >> mg and a is an order of magnitude larger than g. ! ! IDENTIFY: The magnetic force FB must be upward and equal to mg. The direction of FB is determined by the direction of I in the circuit. V SET UP: FB = IlB sin φ , with φ = 90° . I = , where V is the battery voltage. R EXECUTE: (a) The forces are shown in Figure 27.40. The current I in the bar must be to the right to produce ! FB upward. To produce current in this direction, point a must be the positive terminal of the battery.

IlB VlB (175 V)(0.600 m)(1.50 T) = = = 3.21 kg . g Rg (5.00 Ω)(9.80 m/s 2 ) EVALUATE: If the battery had opposite polarity, with point a as the negative terminal, then the current would be clockwise and the magnetic force would be downward. (b) FB = mg . IlB = mg . m =

27.41.

Figure 27.40 ! ! ! IDENTIFY: Apply F = Il × B to each segment of the conductor: the straight section parallel to the x axis, the semicircular section and the straight section that is perpendicular to the plane of the figure in Example 27.8. ! ! SET UP: B = B iˆ . The force is zero when the current is along the direction of B . x

27.42.

EXECUTE: (a) The force on the straight section along the –x-axis is zero. For the half of the semicircle at negative x the force is out of the page. For the half of the semicircle at positive x the force is into the page. The net force on the semicircular section is zero. The force on the straight section that is perpendicular to the plane of the figure is in the –y-direction and has magnitude F = ILB. The total magnetic force on the conductor is ILB, in the –y-direction. EVALUATE: (b) If the semicircular section is replaced by a straight section along the x -axis, then the magnetic force on that straight section would be zero, the same as it is for the semicircle. IDENTIFY: τ = IAB sin φ . The magnetic moment of the loop is μ = IA . SET UP: Since the plane of the loop is parallel to the field, the field is perpendicular to the normal to the loop and φ = 90° . EXECUTE:

(a) τ = IAB = (6.2 A)(0.050 m)(0.080 m)(0.19 T) = 4.7 × 10−3 N ⋅ m

(b) μ = IA = (6.2 A)(0.050 m)(0.080 m) = 0.025 A ⋅ m 2 27.43.

EVALUATE: The torque is a maximum when the field is in the plane of the loop and φ = 90° . IDENTIFY: The period is T = 2π r / v , the current is Q / t and the magnetic moment is μ = IA SET UP: The electron has charge −e . The area enclosed by the orbit is π r 2 . EXECUTE: (a) T = 2πr v = 1.5 × 10−16 s (b) Charge −e passes a point on the orbit once during each period, so I = Q t = e t = 1.1 mA . (c) μ = IA = I πr 2 = 9.3 × 10−24 A ⋅ m 2 EVALUATE: Since the electron has negative charge, the direction of the current is opposite to the direction of motion of the electron.

Magnetic Field and Magnetic Forces

27.44.

27-13

! IDENTIFY: τ = IAB sin φ , where φ is the angle between B and the normal to the loop. SET UP: The coil as viewed along the axis of rotation is shown in Figure 27.44a for its original position and in Figure 27.44b after it has rotated 30.0° . ! ! ! ! EXECUTE: (a) The forces on each side of the coil are shown in Figure 27.44a. F1 + F2 = 0 and F3 + F4 = 0 . The net force on the coil is zero. φ = 0° and sin φ = 0 , so τ = 0 . The forces on the coil produce no torque. (b) The net force is still zero. φ = 30.0° and the net torque is τ = (1)(1.40 A)(0.220 m)(0.350 m)(1.50 T)sin30.0° = 0.0808 N ⋅ m . The net torque is clockwise in Figure 27.44b and is directed so as to increase the angle φ . EVALUATE: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop depends on the orientation of the plane of the loop relative to the magnetic field direction.

Figure 27.44 27.45.

IDENTIFY: The magnetic field exerts a torque on the current-carrying coil, which causes it to turn. We can use the rotational form of Newton’s second law to find the angular acceleration of the coil. ! ! ! SET UP: The magnetic torque is given by τ = μ × B , and the rotational form of Newton’s second law is

∑τ = Iα . The magnetic field is parallel to the plane of the loop. EXECUTE: (a) The coil rotates about axis A2 because the only torque is along top and bottom sides of the coil. (b) To find the moment of inertia of the coil, treat the two 1.00-m segments as point-masses (since all the points in them are 0.250 m from the rotation axis) and the two 0.500-m segments as thin uniform bars rotated about their centers. Since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of the coil. Each 1.00-m segment is 1/3 of the total perimeter, so its mass is (1/3)(210 g) = 70 g = 0.070 kg. The mass of each 0.500-m segment is half this amount, or 0.035 kg. The result is

I = 2(0.070 kg)(0.250 m) 2 + 2 121 (0.035 kg)(0.500 m) 2 = 0.0102 kg ⋅ m 2 The torque is

!

!

!

τ = μ × B = IAB sin 90° = (2.00 A)(0.500 m)(1.00 m)(3.00 T) = 3.00 N ⋅ m

27.46.

27.47.

Using the above values, the rotational form of Newton’s second law gives τ α = = 290 rad/s 2 I EVALUATE: This angular acceleration will not continue because the torque changes as the coil turns. ! ! ! IDENTIFY: τ = μ × B and U = − μ B cosφ , where μ = NIB . τ = μ B sin φ . ! SET UP: φ is the angle between B and the normal to the plane of the loop. EXECUTE: (a) φ = 90°. τ = NIABsin(90°) = NIAB, direction kˆ × ˆj = − iˆ. U = − μ B cos φ = 0. (b) φ = 0. τ = NIABsin(0) = 0, no direction. U = − μ Bcosφ = − NIAB. (c) φ = 90°. τ = NIAB sin(90°) = NIAB, direction − kˆ × ˆj = iˆ. U = − μ B cos φ = 0. (d) φ = 180° : τ = NIAB sin(180°) = 0, no direction, U = − μ B cos(180°) = NIAB. EVALUATE: When τ is maximum, U = 0 . When U is maximum, τ = 0 . ! ! IDENTIFY and SET UP: The potential energy is given by Eq.(27.27): U = μ ⋅ B. The scalar product depends on ! ! the angle between μ and B. ! ! ! ! ! ! EXECUTE: For μ and B parallel, φ = 0° and μ ⋅ B = μ B cosφ = μ B. For μ and B antiparallel, ! ! φ = 180° and μ ⋅ B = μ B cos φ = − μ B. U1 = + μ B, U 2 = − μ B ΔU = U 2 − U1 = −2μ B = −2(1.45 A ⋅ m 2 )(0.835 T) = −2.42 J ! ! EVALUATE: U is maximum when μ and B are antiparallel and minimum when they are parallel. When the coil is rotated as specified its magnetic potential energy decreases.

27-14

Chapter 27

27.48.

IDENTIFY:

Apply Eq.(27.29) in order to calculate I. The power drawn from the line is Psupplied = IVab . The

mechanical power is the power supplied minus the I 2 r electrical power loss in the internal resistance of the motor. SET UP: Vab = 120V , E = 105 V , and r = 3.2 Ω . V − E 120 V − 105 V = = 4.7 A. EXECUTE: (a) Vab = E + Ir ⇒ I = ab r 3.2 Ω (b) Psupplied = IVab = (4.7 A)(120 V) = 564 W.

27.49.

(c) Pmech = IVab − I 2r = 564 W − (4.7 A)2 (3.2 Ω) = 493 W. EVALUATE: If the rotor isn’t turning, when the motor is first turned on or if the rotor bearings fail, then E = 0 120V = 37.5 A . This large current causes large I 2 r heating and can trip the circuit breaker. and I = 3.2 Ω IDENTIFY: The circuit consists of two parallel branches with the potential difference of 120 V applied across each. One branch is the rotor, represented by a resistance Rr and an induced emf that opposes the applied potential. Apply the loop rule to each parallel branch and use the junction rule to relate the currents through the field coil and through the rotor to the 4.82 A supplied to the motor. SET UP: The circuit is sketched in Figure 27.49.

E is the induced emf developed by the motor. It is directed so as to oppose the current through the rotor.

Figure 27.49 EXECUTE: (a) The field coils and the rotor are in parallel with the applied potential difference V , so V = I f Rf . V 120 V If = = = 1.13 A. Rf 106 Ω (b) Applying the junction rule to point a in the circuit diagram gives I − I f − I r = 0. I r = I − I f = 4.82 A − 1.13 A = 3.69 A. (c) The potential drop across the rotor, I r Rr + E , must equal the applied potential difference V : V = I r Rr + E E = V − I r Rr = 120 V − ( 3.69 A )( 5.9 Ω ) = 98.2 V (d) The mechanical power output is the electrical power input minus the rate of dissipation of electrical energy in the resistance of the motor: electrical power input to the motor Pin = IV = ( 4.82 A )(120 V ) = 578 W

electrical power loss in the two resistances 2 2 Ploss = I f2 Rf + I f2 R = (1.13 A ) (106 Ω ) + ( 3.69 A ) ( 5.9 Ω ) = 216 W mechanical power output Pout = Pin − Ploss = 578 W − 216 W = 362 W The mechanical power output is the power associated with the induced emf E Pout = PE = EI r = ( 98.2 V )( 3.69 A ) = 362 W, which agrees with the above calculation.

27.50.

EVALUATE: The induced emf reduces the amount of current that flows through the rotor. This motor differs from the one described in Example 27.12. In that example the rotor and field coils are connected in series and in this problem they are in parallel. IDENTIFY: The field and rotor coils are in parallel, so Vab = I f Rf = E + I r Rr and I = I f + I r , where I is the current

drawn from the line. The power input to the motor is P = Vab I . The power output of the motor is the power input minus the electrical power losses in the resistances and friction losses. SET UP: Vab = 120 V. I = 4.82 A. EXECUTE:

(a) Field current I f =

120 V = 0.550 A. 218 Ω

(b) Rotor current I r = I total − I f = 4.82 A − 0.550 A = 4.27 A.


27-15

(c) V = E + I r Rr and E = V − I r Rr = 120 V − (4.27 A)(5.9 Ω) = 94.8 V. (d) Pf = I f2 Rf = (0.550 A)2 (218 Ω) = 65.9 W. (e) Pr = I r2 Rr = (4.27 A) 2 (5.9 Ω) = 108 W. (f ) Power input = (120 V) (4.82 A) = 578 W. Poutput (578 W − 65.9 W − 108 W − 45 W) 359 W = = = 0.621. (g) Efficiency = Pinput 578 W 578 W

27.51.

EVALUATE: I 2 R losses in the resistance of the rotor and field coils are larger than the friction losses for this motor. IDENTIFY: The drift velocity is related to the current density by Eq.(25.4). The electric field is determined by the requirement that the electric and magnetic forces on the current-carrying charges are equal in magnitude and opposite in direction. (a) SET UP: The section of the silver ribbon is sketched in Figure 27.51a.

J x = n q vd so vd =

Jx nq

Figure 27.51a I I 120 A = = = 4.42 × 107 A/m 2 A y1 z1 (0.23 × 10−3 m)(0.0118 m) J 4.42 ×107 A/m 2 vd = x = = 4.7 ×10−3 m/s = 4.7 mm/s n q ( 5.85 ×1028 / m3 )(1.602 ×10−19 C ) ! (b) magnitude of E

EXECUTE:

Jx =

q Ez = q vd By Ez = vd By = (4.7 ×10−3 m/s)(0.95 T) = 4.5 ×10−3 V/m ! direction of E The drift velocity of the electrons is in the opposite direction to the current, as shown in Figure 27.51b. ! ! v×B ↑ ! ! ! ! ! FB = qv × B = −ev × B ↓ Figure 27.51b

The directions of the electric and magnetic forces on an electron in the ribbon are shown in Figure 27.51c.

! ! ! FE must oppose FB so FE is in the − z -direction Figure 27.51c ! ! ! ! ! ! FE = qE = −eE so E is opposite to the direction of FE and thus E is in the + z -direction. (c) The Hall emf is the potential difference between the two edges of the strip (at z = 0 and z = z1 ) that results from

27.52.

the electric field calculated in part (b). EHall = Ez1 = (4.5 × 10−3 V/m)(0.0118 m) = 53 μ V EVALUATE: Even though the current is quite large the Hall emf is very small. Our calculated Hall emf is more than an order of magnitude larger than in Example 27.13. In this problem the magnetic field and current density are larger than in the example, and this leads to a larger Hall emf. IDENTIFY: Apply Eq.(27.30). SET UP: A = y1 z1. E = E/z1. q = e. EXECUTE:

n=

J x By

=

IBy

=

IBy z1

=

IBy

q Ez A q Ez A q E y1 q E (78.0 A)(2.29 T) n= = 3.7 × 1028 electrons / m3 (2.3 × 10−4 m)(1.6 × 10−19 C)(1.31 × 10−4 V) EVALUATE: The value of n for this metal is about one-third the value of n calculated in Example 27.12 for copper.

27-16

Chapter 27

27.53.

! ! ! Use Eq.(27.2) to relate v , B, and F . ! ! SET UP: The directions of v1 and F1 are shown in Figure 27.53a. (a) IDENTIFY:

! ! ! ! F = qv × B says that F is perpendicular ! ! to v and B. The information given here ! means that B can have no z-component. Figure 27.53a ! ! The directions of v2 and F2 are shown in Figure 27.53b.

! ! ! ! F is perpendicular to v and B, so B can have no x-component. Figure 27.53b

! ! Both pieces of information taken together say that B is in the y-direction; B = B y ˆj. ! ! ! ! EXECUTE: Use the information given about F2 to calculate Fy : F2 = F2 iˆ, v 2 = v2 kˆ , B = By ˆj. ! ! ! F2 = qv2 × B says F2 iˆ = qv2 By kˆ × ˆj = qv2 By ( − iˆ) and F2 = − qv2 By ! By = − F2 /(qv2 ) = − F2 /(qv1 ). B has the maginitude F2 /(qv1 ) and is in the − y -direction. (b) F1 = qvB sin φ = qv1 By / 2 = F2 / 2 ! ! ! ! EVALUATE: v1 = v2 . v2 is perpendicular to B whereas only the component of v1 perpendicular to B contributes 27.54.

to the force, so it is expected that F2 > F1 , as we found. ! ! ! IDENTIFY: Apply F = qv × B. SET UP:

Bx = 0.450 T, By = 0 and Bz = 0.

EXECUTE:

Fx = q(v y Bz − vz By ) = 0.

Fy = q(vz Bx − vx Bz ) = (9.45 × 10−8 C)(5.85 × 104 m/s)(0.450 T) = 2.49 × 10−3 N.

27.55.

Fz = q(vx By − v y Bx ) = − (9.45 × 10−8 C)(−3.11 × 104 m/s)(0.450 T) = 1.32 × 10−3 N. ! ! ! ! ! ! EVALUATE: F is perpendicular to both v and B. We can verify that F ⋅ v = 0. Since B is along the x-axis, vx does not affect the force components. IDENTIFY: The sum of the magnetic, electrical, and gravitational forces must be zero to aim at and hit the target. SET UP: The magnetic field must point to the left when viewed in the direction of the target for no net force. The net force is zero, so ∑ F = FB − FE − mg = 0 and qvB – qE – mg = 0. EXECUTE:

Solving for B gives

B=

27.56.

qE + mg (2500 ×10−6 C)(27.5 N/C) + (0.0050 kg)(9.80 m/s 2 ) = = 3.7 T qv (2500 ×10−6 C)(12.8 m/s)

The direction should be perpendicular to the initial velocity of the coin. EVALUATE: This is a very strong magnetic field, but achievable in some labs. IDENTIFY: Apply R = mv / q B . ω = v / R SET UP: 1 eV = 1.60 × 10−19 J EXECUTE: (a) K = 2.7 MeV = (2.7 × 106eV) (1.6 × 10−19 J/eV) = 4.32 × 10−13 J. v=

R=

2K = m

2(4.32 × 10 −13 J) = 2.27 × 107 m/s . 1.67 × 10 −27 kg

mv (1.67 × 10 −27 kg) (2.27 × 107 m/s) v 2.27 × 107 m/s = = 0.068 m. Also, ω = = = 3.34 × 108 rad/s. −19 qB R (1.6 × 10 C) (3.5 T) 0.068 m

(b) If the energy reaches the final value of 5.4 MeV, the velocity increases by The angular frequency is unchanged from part (a) so is 3.34 × 108 rad / s. EVALUATE:

2 , as does the radius, to 0.096 m.

ω = q B / m , so ω is independent of the energy of the protons. The orbit radius increases when the

energy of the proton increases.


27.57.

27-17

(a) IDENTIFY and SET UP: The maximum radius of the orbit determines the maximum speed v of the protons. Use Newton's 2nd law and ac = v 2 /R for circular motion to relate the variables. The energy of the particle is the

kinetic energy K = 12 mv 2 . ! ! EXECUTE: ∑ F = ma gives q vB = m(v 2 /R )

v=

q BR m

=

(1.60 × 10−19 C)(0.85 T)(0.40 m) = 3.257 × 107 m/s. The kinetic energy of a proton moving with this 1.67 × 10−27 kg

speed is K = 12 mv 2 = 12 (1.67 × 10−27 kg)(3.257 × 107 m/s) 2 = 8.9 × 10−13 J = 5.6 MeV (b) The time for one revolution is the period T = 2

⎛ q BR ⎞ (c) K = 12 mv 2 = 12 m ⎜ ⎟ = ⎝ m ⎠

2π R 2π (0.40 m) = = 7.7 × 10 −8 s 3.257 × 107 m/s v

2

1 2

q B2R2 . Or, B = m

factor of 2 then B must be increased by a factor of (d) v =

2 Km . B is proportional to qR

K , so if K is increased by a

2. B = 2(0.85 T) = 1.2 T.

q BR (3.20 × 10−19 C)(0.85 T)(0.40 m) = = 1.636 × 107 m/s m 6.65 × 10−27 kg

K = 12 mv 2 = 12 (6.65 × 10 −27 kg)(1.636 × 10 7 m/s) 2 = 8.9 × 10 −13 J = 5.5 MeV, the same as the maximum energy for protons. EVALUATE: We can see that the maximum energy must be approximately the same as follows: From part (c), 2

⎛ q BR ⎞ K = 12 m ⎜ ⎟ . For alpha particles q is larger by a factor of 2 and m is larger by a factor of 4 (approximately). ⎝ m ⎠ 2

27.58.

27.59.

Thus q / m is unchanged and K is the same. ! ! ! IDENTIFY: Apply F = qv × B. ! SET UP: v = −vˆj ! EXECUTE: (a) F = −qv[ Bx ( ˆj × iˆ) + By ( ˆj × ˆj ) + Bz ( ˆj × kˆ )] = qvBx kˆ − qvBz iˆ (b) Bx > 0, Bz < 0, sign of By doesn't matter. ! ! (c) F = q vBx iˆ − q vBx kˆ and F = 2 q vBx . ! ! ! EVALUATE: F is perpendicular to v , so F has no y-component. IDENTIFY: The contact at a will break if the bar rotates about b. The magnetic field is directed out of the page, so the magnetic torque is counterclockwise, whereas the gravity torque is clockwise in the figure in the problem. The maximum current corresponds to zero net torque, in which case the torque due to gravity is just equal to the torque due to the magnetic field. SET UP: The magnetic force is perpendicular to the bar and has moment arm l / 2 , where l = 0.750 m is the ⎛l ⎞ length of the bar. The gravity torque is mg ⎜ cos60.0° ⎟ 2 ⎝ ⎠ l l EXECUTE: τ gravity = τ B and mg cos 60.0° = IlB sin 90° . This gives 2 2

I=

2 mg cos60.0° (0.458 kg) ( 9.80 m/s ) (cos 60.0°) = = 1.93 A lB sin 90° (0.750 m)(1.55 T)(1)

Once contact is broken, the magnetic torque ceases. The 90.0° angle in the expression for τ B is the ! angle between the direction of I and the direction of B. EVALUATE:

27.60.

IDENTIFY: SET UP:

Apply R =

mv . qB

Assume D > D , d ≈ R ⎢1 − ⎜ 1 − ⎟⎥ = 2 2 R R ⎦ 2 R2 ⎠⎦ 2R ⎣ ⎣ ⎝

particle moving in a magnetic field, R =

mv . But qB

d≈

1 2

mv 2 = qV , so R =

1 2mV . Thus, the deflection B q

D2B q D2B e . = 2 2mV 2 2mV

(c) d =

(0.50 m) 2 (5.0 × 10 −5 T) (1.6 × 10 −19 C) = 0.067 m = 6.7 cm. d ≈ 13% of D, which is fairly 2 2(9.11 × 10 −31 kg)(750 V)

significant. 2

EVALUATE:

In part (c), R =

1 2mV D 2 ⎛ D ⎞ ⎛R⎞ = =⎜ ⎟ D = 3.7 D and ⎜ ⎟ = 14 , so the approximation made in 2d ⎝ 2d ⎠ B e ⎝D⎠

part (b) is valid.

27.61.

Figure 27.60 ! ! ! ! ! IDENTIFY and SET UP: Use Eq.(27.2) to relate q, v , B and F . The force F and a are related by Newton's 2nd law. ! ! B = −(0.120 T)kˆ , v = (1.05 ×106 m/s)( − 3iˆ + 4 ˆj + 12kˆ ), F = 1.25 N ! ! ! (a) EXECUTE: F = qv × B ! F = q( −0.120 T)(1.05 ×106 m/s)( − 3iˆ × kˆ + 4 ˆj × kˆ + 12kˆ × kˆ )

iˆ × kˆ = − ˆj , ˆj × kˆ = iˆ, kˆ × kˆ = 0 ! F = −q (1.26 × 105 N/C)(+3 ˆj + 4iˆ) = −q(1.26 × 105 N/C)( +4iˆ + 3 ˆj ) The magnitude of the vector +4iˆ + 3 ˆj is 32 + 42 = 5. Thus F = − q(1.26 × 105 N/C)(5). F 1.25 N =− = −1.98 × 10−6 C 5(1.26 × 105 N/C) 5(1.26 × 105 N/C) ! ! ! ! (b) ∑ F = ma so a = F / m ! F = −q (1.26 × 105 N/C)(+4iˆ + 3 ˆj ) = −(−1.98 × 10−6 C)(1.26 × 105 N/C)( +4iˆ + 3 ˆj ) = +0.250 N(+4iˆ + 3 ˆj ) q=−

⎛ 0.250 N ⎞ ! ! 13 2 ˆ ˆ ˆ ˆ Then a = F / m = ⎜ ⎟ (+4i + 3 j ) = (9.69 × 10 m/s )(+4i + 3 j ) −15 × 2.58 10 kg ⎝ ⎠ ! (c) IDENTIFY and SET UP: F is in the xy-plane, so in the z-direction the particle moves with constant speed ! ! 12.6 × 106 m/s. In the xy-plane the force F causes the particle to move in a circle, with F directed in towards the center of the circle. ! ! EXECUTE: ∑ F = ma gives F = m(v 2 / R ) and R = mv 2 / F v 2 = vx2 + v y2 = (−3.15 × 106 m/s) 2 + (+4.20 × 106 m/s) 2 = 2.756 × 1013 m 2 /s 2 F = Fx2 + Fy2 = (0.250 N) 42 + 32 = 1.25 N R=

mv 2 (2.58 × 10−15 kg)(2.756 × 1013 m 2 /s 2 ) = = 0.0569 m = 5.69 cm F 1.25 N


By Eq.(27.12) the cyclotron frequency is f = ω / 2π = v / 2π R.

(d) IDENTIFY and SET UP:

The circular motion is in the xy-plane, so v = vx2 + v y2 = 5.25 × 106 m/s.

EXECUTE:

5.25 × 106 m/s = 1.47 × 107 Hz, and ω = 2π f = 9.23 × 107 rad/s 2π R 2π (0.0569 m) (e) IDENTIFY and SET UP Compare t to the period T of the circular motion in the xy-plane to find the x and y coordinates at this t. In the z-direction the particle moves with constant speed, so z = z0 + vz t. 1 1 EXECUTE: The period of the motion in the xy-plane is given by T = = = 6.80 × 10 −8 s f 1.47 × 107 Hz In t = 2T the particle has returned to the same x and y coordinates. The z-component of the motion is motion with a constant velocity of vz = +12.6 × 106 m/s. Thus z = z0 + vz t = 0 + (12.6 ×106 m/s)(2)(6.80 ×10−8 s) = +1.71 m. The coordinates at t = 2T are x = R, y = 0, z = +1.71 m. ! EVALUATE: The circular motion is in the plane perpendicular to B. The radius of this motion ! gets smaller when B increases and it gets larger when v increases. There is no magnetic force in the direction of B so the particle moves with constant velocity in that direction. The superposition of circular motion in the xy-plane and constant speed motion in the z-direction is a helical path. IDENTIFY: The net magnetic force on the wire is the vector sum of the force on the straight segment plus the force on the curved section. We must integrate to get the force on the curved section. f =

27.62.

v

=

SET UP:

27.63.

∑F = F

straight, top

π

+ Fcurved + Fstraight, bottom and Fstraight, top = Fstraight, bottom = iLstraight B. Fcurved, x = ∫ iRB sin θ dθ = 2iRB 0

(the same as if it were a straight segment 2R long) and Fy = 0 due to symmetry. Therefore, F = 2iLstraightB + 2iRB EXECUTE: Using Lstraight = 0.55 m, R = 0.95 m, i = 3.40 A, and B = 2.20 T gives F = 22 N, to right. EVALUATE: Notice that the curve has no effect on the force. In other words, the force is the same as if the wire were simply a straight wire 3.00 m long. IDENTIFY: τ = NIAB sin φ . SET UP: The area A is related to the diameter D by A = 14 π D 2 . EXECUTE:

27.64.

τ = NI ( 14 π D2 ) B sin φ . τ is proportional to D 2 . Increasing D by a factor of 3 increases τ by a factor of

32 = 9 . EVALUATE: The larger diameter means larger length of wire in the loop and also larger moment arms because parts of the loop are farther from the axis. ! ! ! IDENTIFY: Apply F = qv × B ! SET UP: v = vkˆ ! ! EXECUTE: (a) F = − qvB iˆ + qvB ˆj. But F = 3F iˆ + 4 F ˆj , so 3F = −qvB and 4 F = qvB y

x

0

0

0

y

0

x

3F 4F Therefore, By = − 0 , Bx = 0 and Bz is undetermined. qv qv 2

2

6 F0 F F 11F0 ⎛ qv ⎞ ⎛ qv ⎞ = Bx 2 + By 2 + Bz 2 = 0 9 + 16 + ⎜ ⎟ Bz 2 = 0 25 + ⎜ ⎟ Bz 2 , so Bz = ± ⋅ qv qv F qv F qv 0 0 ⎝ ⎠ ⎝ ⎠ ! EVALUATE: The force doesn’t depend on Bz , since v is along the z-direction. mv IDENTIFY: For the velocity selector, E = vB . For the circular motion in the field B′ , R = . q B′ SET UP: B = B′ = 0.701 T. E 1.88 × 104 N/C mv EXECUTE: v = = , so = 2.68 × 104 m/s. R = B 0.701 T qB′ (b) B =

27.65.

27-19

82(1.66 × 10−27 kg)(2.68 × 104 m/s) = 0.0325 m. (1.60 × 10−19 C)(0.701 T) 84(1.66 × 10−27 kg)(2.68 × 104 m/s) R84 = = 0.0333 m. (1.60 × 10−19 C)(0.701 T) 86(1.66 × 10−27 kg)(2.68 × 104 m/s) R86 = = 0.0341 m. (1.60 × 10−19 C)(0.701 T) The distance between two adjacent lines is ΔR = 1.6 mm . R82 =

27-20

27.66.

Chapter 27

EVALUATE: The distance between the 82 Kr line and the 84 Kr line is 1.6 mm and the distance between the 84 Kr line and the 86 Kr line is 1.6 mm. Adjacent lines are equally spaced since the 82 Kr versus 84 Kr and 84 Kr versus 86 Kr mass differences are the same. IDENTIFY: Apply conservation of energy to the acceleration of the ions and Newton’s second law to their motion in the magnetic field. SET UP: The singly ionized ions have q = +e . A 12 C ion has mass 12 u and a 14 C ion has mass 14 u, where

1 u = 1.66 × 10−27 kg EXECUTE:

2qV . In the magnetic field, m

(a) During acceleration of the ions, qV = 12 mv 2 and v =

qB 2 R 2 mv m 2qV / m and m = . = 2V qB qB qB 2 R 2 (1.60 × 10−19 C)(0.150 T) 2 (0.500 m) 2 (b) V = = = 2.26 × 104 V 2m 2(12)(1.66 × 10−27 kg) R=

(c) The ions are separated by the differences in the diameters of their paths. D = 2 R = 2 ΔD = D14 − D12 = 2

ΔD = 2

2Vm qB 2

−2 14

=2 12

2(2.26 × 104 V)(1.66 × 10−27 kg) (1.6 × 10−19 C)(0.150 T) 2

EVALUATE: 27.67.

2Vm qB 2

The speed of the

12

(

2V (1 u) qB 2

(

2Vm , so qB 2

)

14 − 12 .

)

14 − 12 = 8.01×10−2 m. This is about 8 cm and is easily distinguishable.

C ion is v =

2(1.60 × 10 −19 C)(2.26 × 10 4 V) = 6.0 × 105 m/s . This is very fast, but 12(1.66 × 10−27 kg)

well below the speed of light, so relativistic mechanics is not needed. IDENTIFY: The force exerted by the magnetic field is given by Eq.(27.19). The net force on the wire must be zero. SET UP: For the wire to remain at rest the force exerted on it by the magnetic field must have a component directed up the incline. To produce a force in this direction, the current in the wire must be directed from right to left in Figure 27.61 in the textbook. Or, viewing the wire from its left-hand end the directions are shown in Figure 27.67a.

Figure 27.67a

The free-body diagram for the wire is given in Figure 27.67b.

EXECUTE:

∑F

y

=0

FI cosθ − Mg sinθ = 0 FI = ILB sin φ

!

φ = 90° since B is perpendicular to the current direction. Figure 27.67b

Mg tan θ LB EVALUATE: The magnetic and gravitational forces are in perpendicular directions so their components parallel to the incline involve different trig functions. As the tilt angle θ increases there is a larger component of Mg down the incline and the component of FI up the incline is smaller; I must increase with θ to compensate. As θ → 0, I → 0 and as θ → 90°, I → ∞. IDENTIFY: The current in the bar is downward, so the magnetic force on it is vertically upwards. The net force on the bar is equal to the magnetic force minus the gravitational force, so Newton’s second law gives the acceleration. The bar is in parallel with the 10.0-Ω resistor, so we must use circuit analysis to find the initial current through it. Thus (ILB) cosθ − Mg sin θ = 0 and I =

27.68.


27-21

SET UP: First find the current. The equivalent resistance across the battery is 30.0 Ω, so the total current is 4.00 A, half of which goes through the bar. Applying Newton’s second law to the bar gives ∑ F = ma = FB − mg = iLB − mg. EXECUTE:

27.69.

Solving for the acceleration gives iLB − mg (2.0 A)(1.50 m)(1.60 T) − 3.00 N a= = = 5.88 m/s 2 . m (3.00 N/9.80 m/s 2 )

The direction is upward. EVALUATE: Once the bar is free of the conducting wires, its acceleration will become 9.8 m/s2 downward since only gravity will be acting on it. IDENTIFY: Calculate the acceleration of the ions when they first enter the field and assume this acceleration is constant. Apply conservation of energy to the acceleration of the ions by the potential difference. ! ! SET UP: Assume v = v iˆ and neglect the y-component of v that is produced by the magnetic force. x

EXECUTE:

(a)

1 2

mv x 2 = qV , so vx = 2

2qV qv B x . Also, a y = x and t = . m m vx 2

1/ 2

1/ 2

⎛ x ⎞ 1 ⎛ qv B ⎞ ⎛ x ⎞ 1 ⎛ qBx 2 ⎞⎛ m ⎞ q ⎞ 2⎛ y = a yt = a y ⎜ ⎟ = ⎜ x ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟ = Bx ⎜ ⎟ . ⎝ 8mV ⎠ ⎝ vx ⎠ 2 ⎝ m ⎠ ⎝ vx ⎠ 2 ⎝ m ⎠ ⎝ 2qV ⎠ (b) This can be used for isotope separation since the mass in the denominator leads to different locations for different isotopes. EVALUATE: For B = 0.1 T, v = 1 × 104 m/s, q = +e and m = 12 u = 2.0 × 10−26 kg, y = (1.0 m −2 ) x 2 . The approximation y 0. ! ! μ = − IAkˆ , B = Bx iˆ + By ˆj + By kˆ ! ! ! τ = μ × B = ( − IA)( Bx kˆ × iˆ + B y kˆ × ˆj + Bz kˆ × kˆ ) = IAB y iˆ − IABx ˆj ! Compare this to the expression given for τ : IABy = 4D so By = 4D / IA and − IABx = −3D so Bx = 3D / IA ! Bz doesn't contribute to the torque since μ is along the z-direction. But B = B0 and Bx2 + By2 + Bz2 = B02 ; with

B0 = 13D / IA. Thus Bz = ± B02 − Bx2 − By2 = ± ( D / IA ) 169 − 9 − 16 = ±12 ( D / IA ) ! ! ! ! That U = − μ ⋅ B is negative determines the sign of Bz : U = − μ ⋅ B = −( − IAkˆ ) ⋅ ( Bx iˆ + By ˆj + Bz kˆ ) = + IABz So U negative says that Bz is negative, and thus Bz = −12 D / IA.

27-28

Chapter 27

EVALUATE:

!

!

μ is along the z-axis so only Bx and By contribute to the torque. Bx produces a y-component of τ

! ! ! and By produces an x-component of τ . Only Bz affects U, and U is negative when μ and Bz are parallel. 27.86.

27.87.

Δq and μ = IA. Δt ! SET UP: The direction of μ is given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. I is in the direction of flow of positive charge and opposite to the direction of flow of negative charge. dq Δq qu v ev EXECUTE: (a) I u = . = = = dt Δt 2π r 3π r ev evr (b) μu = I u A = π r2 = . 3π r 3 evr 2evr (c) Since there are two down quarks, each of half the charge of the up quark, μd = μu = . Therefore, μtotal = . 3 3 3μ 3(9.66 × 10−27 A ⋅ m 2 ) = = 7.55 × 107 m s. (d) v = 2er 2(1.60 × 10−19 C)(1.20 × 10−15 m) EVALUATE: The speed calculated in part (d) is 25% of the speed of light. IDENTIFY: Eq.(27.8) says that the magnetic field through any closed surface is zero. SET UP: The cylindrical Gaussian surface has its top at z = L and its bottom at z = 0 . The rest of the surface is the curved portion of the cylinder and has radius r and length L. B = 0 at the bottom of the surface, since z = 0 there. ! ! EXECUTE: (a) úB ⋅ dA = ∫ Bz dA + ∫ Br dA = ∫ ( βL) dA + ∫ Br dA = 0. This gives 0 = β Lπ r 2 + Br 2π rL , and

IDENTIFY:

I=

top

curved

top

curved

βr . 2 (b) The two diagrams in Figure 27.87 show views of the field lines from the top and side of the Gaussian surface. EVALUATE: Only a portion of each field line is shown; the field lines are closed loops. Br ( r ) = −

27.88.

IDENTIFY:

Figure 27.87 ! ! U = − μ ⋅ B . In part (b) apply conservation of energy.

SET UP: The kinetic energy of the rotating ring is K = 12 I ω 2 . ! ! ! ! ! ! "! EXECUTE: (a) ΔU = −( μf ⋅ B − μi ⋅ B ) = −( μf − μi ) ⋅ B = ⎡⎣ − μ (− kˆ − (−0.8iˆ + 0.6 ˆj )) ⎤⎦ ⋅ ⎡⎣ B0 (12iˆ + 3 ˆj − 4kˆ ) ⎤⎦ .

ΔU = IAB0 [(−0.8)( +12) + (0.6)(+3) + (+1)( −4)] = (12.5 A)(4.45 × 10−4 m 2 )(0.0115 T)( − 11.8). ΔU = −7.55 × 10−4 J .

(b) ΔK =

1 2 2ΔK 2(7.55 × 10−4 J) Iω . ω = = = 42.1 rad/s. 2 8.50 × 10−7 kg ⋅ m 2 I

EVALUATE:

27.89.

The potential energy of the ring decreases and its kinetic energy increases. mv IDENTIFY and SET UP: In the magnetic field, R = . Once the particle exits the field it travels in a straight line. qB Throughout the motion the speed of the particle is constant. mv (3.20 × 10−11 kg)(1.45 × 105 m/s) EXECUTE: (a) R = = = 5.14 m. qB (2.15 × 10−6 C)(0.420 T)


27-29

0.35 m , so 5.14 m d 0.25 m θ = 2.78° = 0.0486 rad. d = Rθ = (5.14 m)(0.0486 rad) = 0.25 m. And t = = = 1.72 × 10−6 s. v 1.45 × 105 m/s (c) Δx1 = d tan(θ / 2) = (0.25 m)tan (2.79° / 2) = 6.08 × 10−3 m.

(b) See Figure 27.89. The distance along the curve, d , is given by d = Rθ . sin θ =

(d) Δx = Δx1 + Δx2 , where Δx2 is the horizontal displacement of the particle from where it exits the field region to where it hits the wall. Δx2 = (0.50 m) tan 2.79° = 0.0244 m. Therefore, Δx = 6.08 × 10−3 m + 0.0244 m = 0.0305 m. EVALUATE: d is much less than R, so the horizontal deflection of the particle is much smaller than the distance it travels in the y-direction.

27.90.

Figure 27.89 ! IDENTIFY: The current direction is perpendicular to B , so F = IlB . If the liquid doesn’t flow, a force (Δp ) A from the pressure difference must oppose F. SET UP: J = I/A, where A = hw. EXECUTE: (a) Δp = F / A = IlB / A = JlB. Δ p (1.00 atm)(1.013 × 105 Pa/atm) = = 1.32 × 106 A/m 2. lB (0.0350 m)(2.20 T) EVALUATE: A current of 1 A in a wire with diameter 1 mm corresponds to a current density of J = 1.36 × 106 A/m 2 , so the current density calculated in part (c) is a typical value for circuits. IDENTIFY: The electric and magnetic fields exert forces on the moving charge. The work done by the electric v2 and this acceleration must correspond to the net field equals the change in kinetic energy. At the top point, a y = R force. SET UP: The electric field is uniform so the work it does for a displacement y in the y-direction is W = Fy = qEy. ! ! At the top point, FB is in the − y -direction and FE is in the +y-direction. EXECUTE: (a) The maximum speed occurs at the top of the cycloidal path, and hence the radius of curvature is greatest there. Once the motion is beyond the top, the particle is being slowed by the electric field. As it returns to y = 0, the speed decreases, leading to a smaller magnetic force, until the particle stops completely. Then the electric field again provides the acceleration in the y-direction of the particle, leading to the repeated motion. 2qEy 1 (b) W = qEy = mv 2 and v = . 2 m

(b) J =

27.91.

mv 2 m 2qEy 2E . =− = −qE. 2qE = qvB and v = R 2y m B EVALUATE: The speed at the top depends on B because B determines the y-displacement and the work done by the electric force depends on the y-displacement.

(c) At the top, Fy = qE − qvB = −

28

SOURCES OF MAGNETIC FIELD

28.1.

! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! μ qv × rˆ μ0 qv × r r B= 0 = , since rˆ = . 4π r 2 4π r 3 r ! ! 6 ˆ v = ( 8.00 × 10 m/s ) j and r is the vector from the charge to the point where the field is calculated.

! EXECUTE: (a) r = ( 0.500 m ) iˆ, r = 0.500 m ! ! v × r = vrˆj × iˆ = −vrkˆ

! ( 6.00 ×10−6 C )(8.00 ×106 m/s ) kˆ μ qv B = − 0 2 kˆ = − (1 × 10−7 T ⋅ m/A ) 2 4π r ( 0.500 m ) ! B = − (1.92 × 10−5 T ) kˆ ! (b) r = − ( 0.500 m ) ˆj , r = 0.500 m ! ! ! v × r = −vrˆj × ˆj = 0 and B = 0. ! (c) r = ( 0.500 m ) kˆ , r = 0.500 m ! ! v × r = vrˆj × kˆ = vriˆ

! ( 6.00 ×10−6 C )(8.00 × 106 m/s ) iˆ = + 1.92 ×10−5 T iˆ B = (1 × 10−7 T ⋅ m/A ) ( ) 2 ( 0.500 m ) ! 2 2 (d) r = − ( 0.500 m ) ˆj + ( 0.500 m ) kˆ , r = ( 0.500 m ) + ( 0.500 m ) = 0.7071 m ! ! v × r = v ( 0.500 m ) − ˆj × ˆj + ˆj × kˆ = ( 4.00 × 106 m 2 /s ) iˆ

(

28.2.

! ( 6.00 ×10−6 C )( 4.00 ×106 m / s ) iˆ = + 6.79 ×10−6 T iˆ B = (1×10−7 T ⋅ m/A ) ( ) 3 ( 0.7071 m ) ! ! ! ! EVALUATE: At each point B is perpendicular to both v and r . B = 0 along the direction of v . IDENTIFY: A moving charge creates a magnetic field as well as an electric field. 1 e μ qv sin φ SET UP: The magnetic field caused by a moving charge is B = 0 , and its electric field is E = 4π P0 r 2 4π r 2 since q = e. EXECUTE: Substitute the appropriate numbers into the above equations.

B=

μ0 qv sin φ 4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.2 × 106 m/s)sin 90° = = 13 T, out of the page. 4π r 2 4π (5.3 × 10−11 m) 2

E=

28.3.

)

1 e (9.00 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) = = 5.1 × 1011 N/C, toward the electron. 4π P0 r 2 (5.3 × 10−11 m) 2

EVALUATE: There are enormous fields within the atom! IDENTIFY: A moving charge creates a magnetic field. SET UP: The magnetic field due to a moving charge is B =

μ0 qv sin φ

4π

r2

.

28-1

28-2

28.4.

Chapter 28

EXECUTE: Substituting numbers into the above equation gives μ qv sin φ 4π ×10−7 T ⋅ m/A (1.6 ×10−19 C)(3.0 ×107 m/s)sin 30° (a) B = 0 = . 4π r 2 4π (2.00 ×10−6 m) 2 B = 6.00 × 10–8 T, out of the paper, and it is the same at point B. (b) B = (1.00 × 10–7 T ⋅ m/A)(1.60 × 10–19 C)(3.00 × 107 m/s)/(2.00 × 10–6 m)2 B = 1.20 × 10–7 T, out of the page. (c) B = 0 T since sin(180°) = 0. EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the Earth’s magnetic field. IDENTIFY: Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields. SET UP: Both fields point out of the paper, so their magnitudes add, giving B = Balpha + Bel =

μ0v ( e sin 40° + 2e sin140° ) 4π r 2

EXECUTE: Factoring out an e and putting in the numbers gives B=

4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.50 × 105 m/s) ( sin 40° + 2sin140° ) 4π (1.75 × 10−9 m) 2

B = 2.52 × 10−3 T = 2.52 mT, out of the page. At distances very close to the charges, the magnetic field is strong enough to be important. ! μ qv! × r! IDENTIFY: Apply B = 0 . 4π r 3 ! SET UP: Since the charge is at the origin, r = xiˆ + yˆj + zkˆ. ! ! ! ! ! EXECUTE: (a) v = vi , r = riˆ; v × r = 0, B = 0 . ! ! ! ! (b) v = viˆ, r = rˆj; v × r = vrkˆ , r = 0.500 m. EVALUATE:

28.5.

−7 2 2 −6 5 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) = 1.31 × 10−6 T. B=⎜ 0 ⎟ 2 = (0.500 m) 2 ⎝ 4π ⎠ r ! q is negative, so B = −(1.31 × 10−6 T)kˆ. ! ! ! ! (c) v = viˆ, r = (0.500 m)(iˆ + ˆj ); v × r = (0.500 m)vkˆ , r = 0.7071 m.

! ! (1.0 × 10−7 N ⋅ s 2 /C 2 )(4.80 × 10−6 C)(0.500 m)(6.80 × 105 m/s) ⎛μ ⎞ B = ⎜ 0 ⎟( q v × r r3 ) = . (0.7071 m)3 ⎝ 4π ⎠ ! ˆ B = 4.62 × 10 −7 T. B = −(4.62 × 10−7 T) k. ! ! ! ! (d) v = viˆ, r = rkˆ; v × r = − vrˆj , r = 0.500 m

28.6.

−7 2 2 −6 5 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) = 1.31 × 10−6 T. B=⎜ 0 ⎟ 2 = 2 (0.500 m) ⎝ 4π ⎠ r ! B = (1.31 × 10−6 T ) ˆj. ! ! ! EVALUATE: In each case, B is perpendicular to both r and v . ! ! ! μ qv × r IDENTIFY: Apply B = 0 . For the magnetic force, apply the results of Example 28.1, except here the two 4π r 3 charges and velocities are different. ! ! v ×r ! ! v SET UP: In part (a), r = d and r is perpendicular to v in each case, so = 2 . For calculating the force 3 r d between the charges, r = 2d . μ ⎛ qv q′v′ ⎞ EXECUTE: (a) Btotal = B + B′ = 0 ⎜ 2 + 2 ⎟ . 4π ⎝ d d ⎠

μ0 ⎛ (8.0 × 10−6 C)(4.5 × 106 m/s) (3.0 × 10−6 C)(9.0 × 106 m/s) ⎞ −4 + ⎜ ⎟ = 4.38 × 10 T. 4π ⎝ (0.120 m) 2 (0.120 m) 2 ⎠ ! The direction of B is into the page. B=

Sources of Magnetic Field

28-3

(b) Following Example 28.1 we can find the magnetic force between the charges: μ qq′vv′ (8.00 × 10−6 C)(3.00 × 10−6 C)(4.50 × 106 m s )(9.00 × 106 m s ) −7 FB = 0 (10 T m A) = ⋅ 4π r 2 (0.240 m) 2 −3 FB = 1.69 × 10 N. The force on the upper charge points up and the force on the lower charge points down. The

Coulomb force between the charges is FC = k

28.7.

28.8.

q1q2 (8.0 × 10 −6 C)(3.0 × 10 −6 C) = (8.99 × 10 9 N ⋅ m 2 C 2 ) = 3.75 N . 2 r (0.240 m) 2

The force on the upper charge points up and the force on the lower charge points down. The ratio of the Coulomb F c2 3.75 N force to the magnetic force is C = = = 2.22 × 103 ; the Coulomb force is much larger. FB v1v2 1.69 × 10 −3 N (b) The magnetic forces are reversed in direction when the direction of only one velocity is reversed but the magnitude of the force is unchanged. EVALUATE: When two charges have the same sign and move in opposite directions, the force between them is repulsive. When two charges of the same sign move in the same direction, the force between them is attractive. ! ! μ qv! × r! ! ! IDENTIFY: Apply B = 0 . For the magnetic force on q′ , use FB = q′v × Bq and for the magnetic force on 3 4π r ! ! ! q use FB = qv × Bq ′ . ! ! v ×r v SET UP: In part (a), r = d and = 2. r3 d μ qv μ qv′ EXECUTE: (a) q′ = − q; Bq = 0 2 , into the page; Bq ′ = 0 2 , out of the page. 4πd 4πd v μ qv μ0 qv (i) v′ = gives B = 0 2 (1 − 12 ) = , into the page. (ii) v′ = v gives B = 0. 2 4πd 4π (2d 2 ) μ qv (iii) v′ = 2v gives B = 0 2 , out of the page. 4π d ! ! ! μ q 2v′v ! (b) The force that q exerts on q′ is given by F = q′v′ × Bq , so F = 0 . Bq is into the page, so the force on 4π (2d ) 2 q′ is toward q. The force that q′ exerts on q is toward q′. The force between the two charges is attractive. F μ q 2vv′ q2 (c) FB = 0 so B = μ0P0vv′ = μ0P0 (3.00 × 105 m/s)2 = 1.00 × 10−6 . , FC = 2 FC 4π (2d ) 4πP0 (2d ) 2 EVALUATE: When charges of opposite sign move in opposite directions, the force between them is attractive. For the values specified in part (c), the magnetic force between the two charges is much smaller in magnitude than the Coulomb force between them. IDENTIFY: Both moving charges create magnetic fields, and the net field is the vector sum of the two. The magnetic force on a moving charge is Fmag = qvB sin φ and the electrical force obeys Coulomb’s law.

μ0 qv sin φ . 4π r 2 EXECUTE: (a) Both fields are into the page, so their magnitudes add, giving SET UP: The magnetic field due to a moving charge is B =

B = Be + Bp = B=

μ0 ⎛ ev ev ⎞ ⎜ + ⎟ sin 90° 4π ⎜⎝ re2 rp2 ⎟⎠

⎡ ⎤ μ0 1 1 1.60 ×10−19 C ) ( 845,000 m/s ) ⎢ + ( −9 −9 2 2⎥ 4π ⎣ (5.00 ×10 m) (4.00 ×10 m) ⎦

B = 1.39 × 10–3 T = 1.39 mT, into the page.

μ0 qv sin φ , where r = 41 nm and φ = 180° − arctan(5/4) = 128.7°, we get 4π r 2 4π ×10−7 T ⋅ m/A (1.6 ×10−19 C)(845,000 m/s)sin128.7° B= = 2.58 ×10−4 T, into the page. 4π ( 41 ×10−9 m) 2

(b) Using B =

(c) Fmag = qvB sin90° = (1.60 × 10−19 C)(845,000 m/s)(2.58 × 10−4 T) = 3.48 × 10−17 N, in the +x direction.

Felec = (1/ 4π P0 )e2 /r 2 = clockwise. EVALUATE:

(9.00 ×109 N ⋅ m 2 /C2 )(1.60 ×10−19 C) 2 = 5.62 ×10−12 N, at 51.3° below the +x-axis measured ( 41 ×10−9 m) 2

The electric force is much stronger than the magnetic force.

28-4

28.9.

28.10.

Chapter 28

A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . r2 4π EXECUTE: Applying the law of Biot and Savart gives 4π × 10−7 T ⋅ m/A (10.0 A)(0.00110 m) sin90° (a) dB = = 4.40 × 10–7 T, out of the paper. 4π (0.0500 m) 2

IDENTIFY:

(b) The same as above, except r = (5.00 cm) 2 + (14.0 cm) 2 and φ = arctan(5/14) = 19.65°, giving dB = 1.67 × 10–8 T, out of the page. (c) dB = 0 since φ = 0°. EVALUATE: This is a very small field, but it comes from a very small segment of current. ! ! ! μ Idl × rˆ μ0 Idl × r! IDENTIFY: Apply dB = 0 . = 4π r 2 4π r 3 μ Idl sin φ SET UP: The magnitude of the field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2

dBb =

μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2

dBc =

μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2

dBd =

μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2

μ0 Idl sin φ μ0 (200 A)(0.00100 m) 2 = = 0.545 × 10−6 T 4π r 2 4π 3(0.100 m) 2 3 The field vectors at each point are shown in Figure 28.10. ! EVALUATE: In each case dB is perpendicular to the current direction. dBe =

Figure 28.10 28.11.

IDENTIFY and SET UP: The magnetic field produced by an infinitesimal current element is given by Eq.(28.6). ! ! μ0 Il × rˆ dB = As in Example 28.2 use this equation for the finite 0.500-mm segment of wire since the 4π r 2 Δl = 0.500 mm length is much smaller than the distances to the field points. ! ! ! μ I Δl × rˆ μ0 I Δl × r! B= 0 = 4π r 2 4π r 3 ! I is in the + z -direction, so Δl = ( 0.500 × 10−3 m ) kˆ ! EXECUTE: (a) Field point is at x = 2.00 m, y = 0, z = 0 so the vector r from the source point (at the origin) to the ! field point is r = ( 2.00 m ) iˆ. ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × iˆ = + (1.00 × 10−3 m 2 ) ˆj ! (1× 10−7 T ⋅ m/A ) ( 4.00 A ) (1.00 ×10−3 m 2 ) ˆj = ( 5.00 ×10−11 T ) ˆj B= 3 ( 2.00 m )


28-5

! (b) r = ( 2.00 m ) ˆj , r = 2.00 m. ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × ˆj = − (1.00 × 10−3 m 2 ) iˆ ! (1×10−7 T ⋅ m/A ) ( 4.00 A ) (1.00 ×10−3 m2 ) iˆ = − 5.00 ×10−11 T iˆ B=− ( ) 3 ( 2.00 m ) ! (c) r = ( 2.00 m ) iˆ + ˆj , r = 2 ( 2.00 m ) . ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × iˆ + ˆj = (1.00 × 10−3 m 2 ) ˆj – iˆ

(

)

(

! (1× 10 B=

28.12.

−7

)

T ⋅ m/A ) ( 4.00 A ) (1.00 × 10 m −3

⎡ 2 ( 2.00 m ) ⎤ ⎣ ⎦

3

)

( ˆj – iˆ ) = ( −1.77 ×10

)

−11

(

T ) iˆ – ˆj

)

! (d) r = (2.00 m)kˆ , r = 2.00 m. ! ! ! Δl × r = (0.500 × 10−3 m)(2.00 m)kˆ × kˆ = 0; B = 0. ! ! ! EVALUATE: At each point B is perpendicular to both r and Δl . B = 0 along the length of the wire. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . r2 4π Both fields are into the page, so their magnitudes add. EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives

4π × 10−7 T ⋅ m/A dB = 4π

28.13.

(

2

⎛ 2.50 cm ⎞ (12.0 A)(0.00150 m) ⎜ ⎟ ⎝ 8.00 cm ⎠ = 8.79 × 10–8 T 2 (0.0800 m)

The field from the 24.0-A segment is twice this value, so the total field is 2.64 × 10–7 T, into the page. EVALUATE: The rest of each wire also produces field at P. We have calculated just the field from the two segments that are indicated in the problem. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . Both fields are into the page, so their magnitudes add. 4π r2

EXECUTE: Applying the Biot and Savart law, where r = 12 (3.00 cm) 2 + (3.00 cm) 2 = 2.121 cm, we have dB = 2

28.14.

28.15.

4π × 10−7 T ⋅ m/A (28.0 A)(0.00200 m)sin 45.0° = 1.76 × 10–5 T, into the paper. 4π (0.02121 m) 2

EVALUATE: Even though the two wire segments are at right angles, the magnetic fields they create are in the same direction. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . All four fields are of equal magnitude and into the 4π r2 page, so their magnitudes add. 4π × 10−7 T ⋅ m/A (15.0 A)(0.00120 m) sin90° EXECUTE: dB = 4 = 2.88 × 10–6 T, into the page. 4π (0.0500 m) 2 EVALUATE: A small current element causes a small magnetic field. IDENTIFY: We can model the lightning bolt and the household current as very long current-carrying wires. μI SET UP: The magnetic field produced by a long wire is B = 0 . 2π r EXECUTE: Substituting the numerical values gives (4π × 10−7 T ⋅ m/A)(20,000 A) = 8 × 10–4 T (a) B = 2π(5.0 m) (4π × 10−7 T ⋅ m/A)(10 A) = 4.0 × 10–5 T. 2π(0.050 m) EVALUATE: The field from the lightning bolt is about 20 times as strong as the field from the household current.

(b) B =

28-6

Chapter 28

28.16.

IDENTIFY: The long current-carrying wire produces a magnetic field. μI SET UP: The magnetic field due to a long wire is B = 0 . 2π r EXECUTE: First find the current: I = (3.50 × 1018 el/s)(1.60 × 10–19 C/el) = 0.560 A (4π × 10−7 T ⋅ m/A)(0.560 A) = 2.80 × 10–6 T Now find the magnetic field: 2π(0.0400 m) Since electrons are negative, the conventional current runs from east to west, so the magnetic field above the wire points toward the north. EVALUATE: This magnetic field is much less than that of the Earth, so any experiments involving such a current would have to be shielded from the Earth’s magnetic field, or at least would have to take it into consideration. IDENTIFY: The long current-carrying wire produces a magnetic field. μI SET UP: The magnetic field due to a long wire is B = 0 . 2π r EXECUTE: First solve for the current, then substitute the numbers using the above equation. (a) Solving for the current gives

28.17.

I = 2π rB/μ0 = 2π (0.0200 m)(1.00 × 10−4 T)/(4π × 10−7 T ⋅ m/A) = 10.0 A

28.18.

(b) The earth’s horizontal field points northward, so at all points directly above the wire the field of the wire would point northward. (c) At all points directly east of the wire, its field would point northward. EVALUATE: Even though the Earth’s magnetic field is rather weak, it requires a fairly large current to cancel this field. ! μI IDENTIFY: For each wire B = 0 (Eq.28.9), and the direction of B is given by the right-hand rule (Fig. 28.6 in 2π r the textbook). Add the field vectors for each wire to calculate the total field. (a) SET UP: The two fields at this point have the directions shown in Figure 28.18a. EXECUTE: At point P midway between ! ! the two wires the fields B1 and B2 due to the two currents are in opposite directions, so B = B2 − B1. Figure 28.18a But B1 = B2 =

(b) SET UP:

μ0 I , so B = 0. 2π a The two fields at this point have the directions shown in Figure 28.18b.

EXECUTE: At point Q above the upper ! ! wire B1 and B2 are both directed out of the page (+ z -direction), so B = B1 + B2 .

Figure 28.18b B1 =

μ0 I μ0 I , B2 = 2π a 2π (3a )

B=

μ0 I 2μ I ! 2μ I 1 + 13 ) = 0 ; B = 0 kˆ ( 2π a 3π a 3π a


28-7

(c) SET UP: The two fields at this point have the directions shown in Figure 28.18c.

EXECUTE: At point R below the lower ! ! wire B1 and B2 are both directed into the page (− z -direction), so B = B1 + B2 .

Figure 28.18c

μ0 I μI , B2 = 0 2π (3a ) 2π a μ0 I 2 μI ! 2μ I B1 = 1 + 13 ) = 0 ; B = − 0 kˆ ( 2π a 3π a 3π a B1 =

28.19.

! EVALUATE: In the figures we have drawn, B due to each wire is out of the page at points above the wire and into the page at points below the wire. If the two field vectors are in opposite directions the magnitudes subtract. IDENTIFY: The total magnetic field is the vector sum of the constant magnetic field and the wire’s magnetic field. μI SET UP: For the wire, Bwire = 0 and the direction of Bwire is given by the right-hand rule that is illustrated in 2π r ! Figure 28.6 in the textbook. B = (1.50 × 10−6 T) iˆ. 0

! ! μI μ (8.00 A) ˆ EXECUTE: (a) At (0, 0, 1 m), B = B0 − 0 iˆ = (1.50 × 10−6 T)iˆ − 0 i = −(1.0 × 10−7 T)iˆ. 2π r 2π (1.00 m) ! ! μI μ (8.00 A) ˆ (b) At (1 m, 0, 0), B = B0 + 0 kˆ = (1.50 × 10−6 T) iˆ + 0 k. 2πr 2π (1.00 m) ! B = (1.50 × 10−6 T)iˆ + (1.6 × 10−6 T)kˆ = 2.19 × 10−6 T, at θ = 46.8° from x to z.

28.20.

! ! μI μ (8.00 A) ˆ (c) At (0, 0, –0.25 m), B = B0 + 0 iˆ = (1.50 × 10−6 T)iˆ + 0 i = (7.9 × 10−6 T)iˆ. 2πr 2π (0.25 m) EVALUATE: At point c the two fields are in the same direction and their magnitudes add. At point a they are in opposite directions and their magnitudes subtract. At point b the two fields are perpendicular. ! IDENTIFY and SET UP: The magnitude of B is given by Eq.(28.9) and the direction is given by the right-hand rule. (a) EXECUTE: Viewed from above, the current is in the direction shown in Figure 28.20. Directly below the wire the direction of the magnetic field due to the current in the wire is east.

Figure 28.20 B=

μ0 I ⎛ 800 A ⎞ −5 = (2 ×10−7 T ⋅ m/A) ⎜ ⎟ = 2.91×10 T 2π r ⎝ 5.50 m ⎠

(b) EVALUATE: problem. 28.21.

B from the current is nearly equal in magnitude to the earth's field, so, yes, the current really is a

! μ0 I . The direction of B is given by the right-hand rule in Section 20.7. 2π r SET UP: Call the wires a and b, as indicated in Figure 28.21. The magnetic fields of each wire at points P1 and P2 are shown in Figure 28.21a. The fields at point 3 are shown in Figure 28.21b. EXECUTE: (a) At P1 , Ba = Bb and the two fields are in opposite directions, so the net field is zero. ! μI ! μI (b) Ba = 0 . Bb = 0 . Ba and Bb are in the same direction so 2π rb 2π ra IDENTIFY:

B=

μ0 I ⎛ 1 1 ⎞ (4π × 10−7 T ⋅ m/A)(4.00 A) ⎡ 1 1 ⎤ −6 ⎜ + ⎟= ⎢ 0.300 m + 0.200 m ⎥ = 6.67 × 10 T 2π ⎝ ra rb ⎠ 2π ⎣ ⎦ ! B has magnitude 6.67 μ T and is directed toward the top of the page. B = Ba + Bb =

28-8

Chapter 28

! ! ! ! 5 cm (c) In Figure 28.21b, Ba is perpendicular to ra and Bb is perpendicular to rb . tan θ = and θ = 14.04° . 20 cm ra = rb = (0.200 m) 2 + (0.050 m) 2 = 0.206 m and Ba = Bb . ⎛ μI ⎞ 2(4π ×10−7 T ⋅ m/A)(4.0 A)cos14.04° = 7.54 μ T B = Ba cosθ + Bb cosθ = 2 Ba cosθ = 2 ⎜ 0 ⎟ cosθ = 2π (0.206 m) ⎝ 2π ra ⎠ B has magnitude 7.53 μ T and is directed to the left. EVALUATE: At points directly to the left of both wires the net field is directed toward the bottom of the page.

Figure 28.21 28.22.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine points where the fields of the two wires cancel. (a) SET UP: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying I 2 = 75.0 A. Btot will be zero

where B1 = B2 . EXECUTE:

μ0 I1

2π (0.400 m − x)

=

μ0 I 2 2π x

I 2 (0.400 m − x) = I1 x; I1 = 25.0 A, I 2 = 75.0 A x = 0.300 m; Btot = 0 along a line 0.300 m from the wire carrying 75.0 A and 0.100 m from the wire carrying current 25.0 A. (b) SET UP: Let the wire with I1 = 25.0 A be 0.400 m above the wire with I 2 = 75.0 A. The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have B1 = B2 must be closer to wire #1 since I1 < I 2 , so can have Btot = 0 only at points above both wires. Consider a point a distance x from the wire carrying I1 = 25.0 A. Btot will be zero where B1 = B2 . EXECUTE:

μ0 I1 μ0 I 2 = 2π x 2π (0.400 m + x)

I 2 x = I1 (0.400 m + x); x = 0.200 m Btot = 0 along a line 0.200 m from the wire carrying current 25.0 A and 0.600 m from the wire carrying current

28.23.

I 2 = 75.0 A. EVALUATE: For parts (a) and (b) the locations of zero field are in different regions. In each case the points of zero field are closer to the wire that has the smaller current. IDENTIFY: The net magnetic field at the center of the square is the vector sum of the fields due to each wire. ! μI SET UP: For each wire, B = 0 and the direction of B is given by the right-hand rule that is illustrated in 2π r Figure 28.6 in the textbook. EXECUTE: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c) The fields due to each wire are sketched in Figure 28.23. ⎛μI⎞ B = Ba cos 45° + Bb cos 45° + Bc cos 45° + Bd cos 45° = 4 Ba cos 45° = 4 ⎜ 0 ⎟ cos 45° . ⎝ 2πr ⎠ r = (10 cm) 2 + (10 cm) 2 = 10 2 cm = 0.10 2 m , so B=4

(4π × 10−7 T ⋅ m A ) (100 A) cos 45° = 4.0 × 10−4 T, to the left. 2π (0.10 2 m)


28-9

EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the square would be to the right.

Figure 28.23 28.24.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine the field due to each wire. Set the sum of the four fields equal to zero and use that equation to solve for the field and the current of the fourth wire. SET UP: The three known currents are shown in Figure 28.24.

! ! ! B1 ⊗, B2 ⊗, B3 " B=

μ0 I ; r = 0.200 m for each wire 2π r

Figure 28.24 EXECUTE: Let " be the positive z-direction. I1 = 10.0 A, I 2 = 8.0 A, I 3 = 20.0 A. Then B1 = 1.00 ×10−5 T,

B2 = 0.80 × 10−5 T, and B3 = 2.00 × 10−5 T. B1z = −1.00 × 10−5 T, B2z = −0.80 × 10−5 T, B3z = +2.00 × 10−5 T B1z + B2 z + B3 z + B4 z = 0

B4 z = −( B1z + B2 z + B3 z ) = −2.0 × 10−6 T ! To give B4 in the ⊗ direction the current in wire 4 must be toward the bottom of the page. B4 =

28.25.

28.26.

rB4 μ0 I (0.200 m)(2.0 × 10−6 T) so I 4 = = = 2.0 A 2π r ( μ0 / 2π ) (2 × 10−7 T ⋅ m/A)

EVALUATE: The fields of wires #2 and #3 are in opposite directions and their net field is the same as due to a current 20.0 A – 8.0 A = 12.0 A in one wire. The field of wire #4 must be in the same direction as that of wire #1, and 10.0 A + I 4 = 12.0 A. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. μ I I L μ (5.00 A)(2.00 A)(1.20 m) EXECUTE: (a) F = 0 1 2 = 0 = 6.00 × 10−6 N, and the force is repulsive since the 2π r 2π (0.400 m) currents are in opposite directions. (b) Doubling the currents makes the force increase by a factor of four to F = 2.40 × 10−5 N. EVALUATE: Doubling the current in a wire doubles the magnetic field of that wire. For fixed magnetic field, doubling the current in a wire doubles the force that the magnetic field exerts on the wire. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. F 2π r 2π (0.0250 m) F μ0 I1I 2 EXECUTE: (a) = gives I 2 = = (4.0 × 10−5 N m) = 8.33 A. L 2π r L μ0 I1 μ0 (0.60 A) (b) The two wires repel so the currents are in opposite directions. EVALUATE: The force between the two wires is proportional to the product of the currents in the wires.

28-10

Chapter 28

28.27.

IDENTIFY: The lamp cord wires are two parallel current-carrying wires, so they must exert a magnetic force on each other. SET UP: First find the current in the cord. Since it is connected to a light bulb, the power consumed by the bulb is μ I ′I P = IV. Then find the force per unit length using F/L = 0 . 2π r EXECUTE: For the light bulb, 100 W = I(120 V) gives I = 0.833 A. The force per unit length is

F/L =

28.28.

4π × 10−7 T ⋅ m/A (0.833 A) 2 = 4.6 × 10−5 N/m 2π 0.003 m

Since the currents are in opposite directions, the force is repulsive. EVALUATE: This force is too small to have an appreciable effect for an ordinary cord. IDENTIFY: Apply Eq.(28.11) for the force from each wire. SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. F μ0 I 2 ⎛ 1 1 ⎞ μ0 I 2 EXECUTE: On the top wire = − , upward. On the middle wire, the magnetic forces cancel ⎜ ⎟= L 2π ⎝ d 2d ⎠ 4π d F μ0 I 2 ⎛ 1 1 ⎞ μ0 I 2 = − + , downward. ⎜ ⎟= L 2π ⎝ d 2d ⎠ 4π d EVALUATE: The net force on the middle wire is zero because at the location of the middle wire the net magnetic field due to the other two wires is zero. IDENTIFY: The wire CD rises until the upward force FI due to the currents balances the downward force of gravity. SET UP: The forces on wire CD are shown in Figure 28.29.

so the net force is zero. On the bottom wire

28.29.

Currents in opposite directions so the force is repulsive and FI is upward, as shown.

Figure 28.29

Eq.(28.11) says FI = EXECUTE:

μ0 I 2 L where L is the length of wire CD and h is the distance between the wires. 2π h

mg = λ Lg

μ0 I 2 L μ I2 = λ Lg and h = 0 . 2π h 2π g λ EVALUATE: The larger I is or the smaller λ is, the larger h will be. Thus FI − mg = 0 says

28.30.

IDENTIFY: The magnetic field at the center of a circular loop is B =

μ0 I 2R

. By symmetry each segment of the loop

that has length Δl contributes equally to the field, so the field at the center of a semicircle is 12 that of a full loop. μI SET UP: Since the straight sections produce no field at P, the field at P is B = 0 . EXECUTE:

B=

μ0 I 4R

4R

! ! . The direction of B is given by the right-hand rule: B is directed into the page.

EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is B = 28.31.

μ0 I

. 8R IDENTIFY: Calculate the magnetic field vector produced by each wire and add these fields to get the total field. SET UP: First consider the field at P produced by the current I1 in the upper semicircle of wire. See Figure 28.31a. Consider the three parts of this wire a: long straight section, b: semicircle c: long, straight section Figure 28.31a

! ! ! μ0 Idl × rˆ μ0 Idl × r! = Apply the Biot-Savart law dB = to each piece. 4π r 2 4π r 3


EXECUTE: part a See Figure 28.31b.

28-11

! ! dl × r = 0, so dB = 0

Figure 28.31b

The same is true for all the infinitesimal segments that make up this piece of the wire, so B = 0 for this piece. part c See Figure 28.31c.

! ! dl × r = 0, so dB = 0 and B = 0 for this piece. Figure 28.31c

part b See Figure 28.31d.

! ! dl × r is directed into the paper for all infinitesimal! segments that make up this semicircular piece, so B is directed into the paper and B = ∫ dB (the vector sum ! of the dB is obtained by adding their magnitudes since they are in the same direction).

Figure 28.31d ! ! ! ! ! ! dl × r = r dl sinθ . The angle θ between dl and r is 90° and r = R, the radius of the semicircle. Thus dl × r = Rdl ! ! μ0 I dl × r μ0 I1 R ⎛ μI ⎞ = dB = dl = ⎜ 0 12 ⎟ dl 3 3 4π 4π R r ⎝ 4π R ⎠ I I μ μ μI ⎛ ⎞ ⎛ ⎞ B = ∫ dB = ⎜ 0 12 ⎟ ∫ dl = ⎜ 0 12 ⎟ (π R ) = 0 1 4R ⎝ 4π R ⎠ ⎝ 4π R ⎠

∫ dl

is equal to π R , the length of wire in the semicircle.) We have shown that the two straight ! sections make zero contribution to B, so B1 = μ0 I1 / 4 R and is directed into the page.

(We used that

For current in the direction shown in Figure 28.31e, a similar analysis gives B2 = μ0 I 2 / 4 R, out of the paper Figure 28.31e

28.32.

! ! μ I −I B1 and B2 are in opposite directions, so the magnitude of the net field at P is B = B1 − B2 = 0 1 2 . 4R EVALUATE: When I1 = I 2 , B = 0. IDENTIFY: Apply Eq.(28.16). SET UP: At the center of the coil, x = 0. a is the radius of the coil, 0.0240 m. 2aBx 2(0.024 m) (0.0580 T) = = 2.77 A EXECUTE: (a) Bx = μ0 NI 2a, so I = μ0 N (4π × 10−7 T ⋅ m A) (800) (b) At the center, Bc = μ0 NI 2a . At a distance x from the center,

⎞ ⎛ ⎞ a3 a3 a3 ⎛ μ NI ⎞ ⎛ = ⎜ 0 ⎟⎜ 2 = Bc ⎜ 2 . Bx = 12 Bc says 2 = 12 , and ( x 2 + a 2 )3 = 4a 6 . 2 3/ 2 ⎟ 2 3/ 2 ⎟ 2 32 x a + 2( x + a ) ( ) ( x + a ) ⎝ 2a ⎠ ⎝ ( x + a ) ⎠ ⎝ ⎠ Since a = 0.024 m, x = 0.0184 m . EVALUATE: As shown in Figure 28.41 in the textbook, the field has its largest magnitude at the center of the coil and decreases with distance along the axis from the center. IDENTIFY: Apply Eq.(28.16). SET UP: At the center of the coil, x = 0 . a is the radius of the coil, 0.020 m. μ NI μ (600) (0.500 A) EXECUTE: (a) Bcenter = 0 = 0 = 9.42 × 10−3 T. 2a 2(0.020 m) Bx =

28.33.

μ0 NIa 2 2

2 32

μ0 NIa 2

μ0 (600)(0.500 A)(0.020 m)2

= 1.34 × 10−4 T. 2( x + a ) 2((0.080 m) 2 + (0.020 m) 2 )3/ 2 EVALUATE: As shown in Figure 28.41 in the textbook, the field has its largest magnitude at the center of the coil and decreases with distance along the axis from the center. (b) B ( x) =

2

2 3/ 2

. B (0.08 m) =

28-12

Chapter 28

28.34.

IDENTIFY and SET UP: The magnetic field at a point on the axis of N circular loops is given by μ0 NIa 2 . Solve for N and set x = 0.0600 m. Bx = 2( x 2 + a 2 )3 / 2 EXECUTE: EVALUATE:

28.35.

N=

2 Bx ( x 2 + a 2 )3 / 2 2(6.39 × 10−4 T)[(0.0600 m) 2 + (0.0600 m)2 ]3 / 2 = = 69. (4π × 10−7 T ⋅ m/A)(2.50 A)(0.0600 m)2 μ0 Ia 2

At the center of the coil the field is Bx =

μ0 NI 2a

= 1.8 × 10−3 T. The field 6.00 cm from the center is a

factor of 1/ 23 / 2 times smaller. IDENTIFY: Apply Ampere’s law. SET UP: μ0 = 4π × 10−7 T ⋅ m/A ! ! EXECUTE: (a) úB ⋅ dl = μ0 I encl = 3.83 × 10−4 T ⋅ m and I encl = 305 A.

! (b) −3.83 × 10−4 T ⋅ m since at each point on the curve the direction of dl is reversed. ! ! EVALUATE: The line integral úB ⋅ dl around a closed path is proportional to the net current that is enclosed by

28.36.

28.37.

28.38.

the path. IDENTIFY: Apply Ampere’s law. SET UP: From the right-hand rule, when going around the path in a counterclockwise direction currents out of the page are positive and currents into the page are negative. ! ! EXECUTE: Path a: I encl = 0 ⇒ úB ⋅ dl = 0. ! ! Path b: I encl = − I1 = −4.0 A ⇒ úB ⋅ dl = − μ0 (4.0 A) = −5.03 × 10−6 T ⋅ m. ! ! Path c: I encl = − I1 + I 2 = −4.0 A + 6.0 A = 2.0 A ⇒ úB ⋅ dl = μ0 (2.0 A) = 2.51 × 10−6 T ⋅ m ! ! Path d: I encl = − I1 + I 2 + I 3 = 4.0 A ⇒ úB ⋅ dl = + μ0 (4.0 A) = 5.03 × 10−6 T ⋅ m. EVALUATE: If we instead went around each path in the clockwise direction, the sign of the line integral would be reversed. IDENTIFY: Apply Ampere’s law. SET UP: To calculate the magnetic field at a distance r from the center of the cable, apply Ampere’s law to a ! ! circular path of radius r. By symmetry, úB ⋅ dl = B (2π r ) for such a path. ! ! μI EXECUTE: (a) For a < r < b, I encl = I ⇒ úB ⋅ dl = μ0 I ⇒ B 2πr = μ0 I ⇒ B = 0 . 2πr (b) For r > c, the enclosed current is zero, so the magnetic field is also zero. EVALUATE: A useful property of coaxial cables for many applications is that the current carried by the cable doesn’t produce a magnetic field outside the cable. ! IDENTIFY: Apply Ampere's law to calculate B. (a) SET UP: For a < r < b the end view is shown in Figure 28.38a. Apply Ampere's law to a circle of radius r, where a < r < b. Take currents I1 and I 2 to be directed into the page. Take this direction to be positive, so go around the integration path in the clockwise direction.

Figure 28.38a ! ! úB ⋅ dl = μ0 I encl

EXECUTE: ! ! úB ⋅ dl = B(2π r ), I encl = I1

Thus B (2π r ) = μ0 I1 and B =

μ0 I1 2π r


28-13

(b) SET UP: r > c: See Figure 28.38b. Apply Ampere's law to a circle of radius r, where r > c. Both currents are in the positive direction.

Figure 28.38b ! ! úB ⋅ dl = μ0 I encl

EXECUTE: ! ! úB ⋅ dl = B(2π r ), I encl = I1 + I 2

Thus B (2π r ) = μ0 ( I1 + I 2 ) and B =

28.39.

μ0 ( I1 + I 2 ) 2π r

EVALUATE: For a < r < b the field is due only to the current in the central conductor. For r > c both currents contribute to the total field. IDENTIFY: The largest value of the field occurs at the surface of the cylinder. Inside the cylinder, the field increases linearly from zero at the center, and outside the field decreases inversely with distance from the central axis of the cylinder. µI µI r SET UP: At the surface of the cylinder, B = 0 , inside the cylinder, Eq. 28.21 gives B = 0 2 , and outside 2π R 2π R µ0 I the field is B = . 2π r EXECUTE: For points inside the cylinder, the field is half its maximum value when

µ0 I r 1⎛ µ I ⎞ = ⎜ 0 ⎟ , which 2π R 2 2 ⎝ 2π R ⎠

µ0 I 1 ⎛ µ0 I ⎞ = ⎜ ⎟ , which gives r = 2R. 2π r 2 ⎝ 2π R ⎠ EVALUATE: The field has half its maximum value at all points on cylinders coaxial with the wire but of radius R/2 and of radius 2R. μ NI IDENTIFY: B = μ0 nI = 0 L SET UP: L = 0.150 m μ (600) (8.00 A) = 0.0402 T EXECUTE: B = 0 (0.150 m) EVALUATE: The field near the center of the solenoid is independent of the radius of the solenoid, as long as the radius is much less than the length. (a) IDENTIFY and SET UP: The magnetic field near the center of a long solenoid is given by Eq.(28.23), B = μ0 nI . gives r = R/2. Outside the cylinder, we have

28.40.

28.41.

EXECUTE: Turns per unit length n =

B

μ0 I

=

0.0270 T = 1790 turns/m (4π × 10−7 T ⋅ m/A)(12.0 A)

(b) N = nL = (1790 turns/m)(0.400 m) = 716 turns Each turn of radius R has a length 2π R of wire. The total length of wire required is N (2π R ) = (716)(2π )(1.40 × 10−2 m) = 63.0 m.

EVALUATE: resistance. 28.42.

A large length of wire is required. Due to the length of wire the solenoid will have appreciable

IDENTIFY and SET UP: At the center of a long solenoid B = μ0 nI = μ0

N I. L

BL (0.150 T)(1.40 m) = = 41.8 A μ0 N (4π × 10−7 T ⋅ m/A)(4000) EVALUATE: The magnetic field inside the solenoid is independent of the radius of the solenoid, if the radius is much less than the length, as is the case here.

EXECUTE:

I=

28-14

Chapter 28

28.43.

IDENTIFY and SET UP: configuration. EXECUTE: (a) B = (b) B =

Use the appropriate expression for the magnetic field produced by each current

2π B 2π (2.00 × 10−2 m)(37.2 T) μ0 I = = 3.72 × 106 A . so I = 4π × 10−7 T ⋅ m/A μ0 2π r

2 RB 2(0.210 m)(37.2 T) N μ0 I = = 1.24 × 105 A . so I = N μ0 (100)(4π × 10−7 T ⋅ m/A) 2R

(37.2 T)(0.320 m) BL N I so I = = = 237 A . L μ0 N (4π × 10−7 T ⋅ m/A)(40,000) EVALUATE: Much less current is needed for the solenoid, because of its large number of turns per unit length. IDENTIFY: Example 28.10 shows that outside a toroidal solenoid there is no magnetic field and inside it the μ NI magnetic field is given by B = 0 . 2πr SET UP: The torus extends from r1 = 15.0 cm to r2 = 18.0 cm. EXECUTE: (a) r = 0.12 m, which is outside the torus, so B = 0. μ NI μ (250)(8.50 A) (b) r = 0.16 m, so B = 0 = 0 = 2.66 × 10−3 T. 2π r 2π (0.160 m) (c) r = 0.20 m, which is outside the torus, so B = 0. EVALUATE: The magnetic field inside the torus is proportional to 1/ r , so it varies somewhat over the crosssection of the torus. μ NI IDENTIFY: Example 28.10 shows that inside a toroidal solenoid, B = 0 . 2π r SET UP: r = 0.070 m μ NI μ (600)(0.650 A) EXECUTE: B = 0 = 0 = 1.11× 10−3 T. 2π r 2π (0.070 m) EVALUATE: If the radial thickness of the torus is small compared to its mean diameter, B is approximately uniform inside its windings. IDENTIFY: Use Eq.(28.24), with μ0 replaced by μ = K m μ0 , with K m = 80. SET UP: The contribution from atomic currents is the difference between B calculated with μ and B calculated with μ0 .

(c) B = μ0

28.44.

28.45.

28.46.

μ NI K m μ0 NI μ0 (80)(400)(0.25 A) = = = 0.0267 T. 2π r 2π r 2π (0.060 m) (b) The amount due to atomic currents is B′ = 79 B = 79 (0.0267 T) = 0.0263 T. 80 80 EVALUATE: The presence of the core greatly enhances the magnetic field produced by the solenoid. K μ NI IDENTIFY and SET UP: B = m 0 (Eq.28.24, with μ0 replaced by K m μ0 ) 2π r EXECUTE: (a) K m = 1400 EXECUTE: (a) B =

28.47.

I=

2π rB (2.90 × 10−2 m)(0.350 T) = = 0.0725 A μ0 K m N (2 × 10−7 T ⋅ m/A)(1400)(500)

(b) K m = 5200

2π rB (2.90 × 10−2 m)(0.350 T) = = 0.0195 A μ0 K m N (2 × 10−7 T ⋅ m/A)(5200)(500) EVALUATE: If the solenoid were air-filled instead, a much larger current would be required to produce the same magnetic field. K μ NI IDENTIFY: Apply B = m 0 . 2πr SET UP: K m is the relative permeability and χ m = K m − 1 is the magnetic susceptibility. I=

28.48.

EXECUTE: (a) K m =

2πrB 2π (0.2500 m)(1.940 T) = = 2021. μ0 NI μ0 (500)(2.400 A)

(b) χ m = K m − 1 = 2020. EVALUATE: Without the magnetic material the magnetic field inside the windings would be B/2021 = 9.6 × 10−4 T. The presence of the magnetic material greatly enhances the magnetic field inside the windings.


28.49.

28-15

IDENTIFY: The magnetic field from the solenoid alone is B0 = μ0 nI . The total magnetic field is B = K m B0 . M is given by Eq.(28.29). SET UP: n = 6000 turns/m EXECUTE: (a) (i) B0 = μ0 nI = μ0 (6000 m −1 ) (0.15 A) = 1.13 × 10−3 T.

(ii) M =

Km − 1

μ0

B0 =

5199

μ0

(1.13 × 10−3 T) = 4.68 × 106 A m.

(iii) B = K m B0 = (5200)(1.13 × 10−3 T) = 5.88 T. ! ! ! ! ! (b) The directions of B , B0 and M are shown in Figure 28.49. Silicon steel is paramagnetic and B0 and M are in the same direction. EVALUATE: The total magnetic field is much larger than the field due to the solenoid current alone.

Figure 28.49 28.50.

IDENTIFY:

Curie’s law (Eq.28.32) says that 1/ M is proportional to T, so 1/χ m is proportional to T.

SET UP: The graph of 1/χ m versus the Kelvin temperature is given in Figure 28.50. EXECUTE: The material does obey Curie’s law because the graph in Figure 28.50 is a straight line. M = C

M=

B − B0

μ0

says that χ m =

B and T

T C μ0 . 1/ χ m = and the slope of 1/ χ m versus T is 1/(C μ0 ) . Therefore, from the T C μ0

1 1 = = 1.55 × 105 K ⋅ A T ⋅ m. μ0 (slope) μ0 (5.13 K −1 ) EVALUATE: For this material Curie’s law is valid over a wide temperature range.

graph the Curie constant is C =

Figure 28.50 28.51.

IDENTIFY: Moving charges create magnetic fields. The net field is the vector sum of the two fields. A charge moving in an external magnetic field feels a force. µ qv sin φ (a) SET UP: The magnetic field due to a moving charge is B = 0 . Both fields are into the paper, so 4π r 2 µ ⎛ qv sin φ q′v′ sin φ ′ ⎞ their magnitudes add, giving Bnet = B + B′ = 0 ⎜ + ⎟. r ′2 ⎠ 4π ⎝ r 2 EXECUTE: Substituting numbers gives µ ⎡ (8.00 µC)(9.00 × 104 m/s)sin 90° (5.00 µC)(6.50 × 104 m/s)sin 90° ⎤ Bnet = 0 ⎢ + ⎥ 4π ⎣ (0.300 m) 2 (0.400 m) 2 ⎦

Bnet = 1.00 × 10−6 T = 1.00μ T, into the paper.

! ! ! (b) SET UP: The magnetic force on a moving charge is F = qv × B , and the magnetic field of charge q′ at the location of charge q is into the page. The force on q is ! ! µ qv′ × rˆ ! ! ⎛ µ qv′ sin φ ⎞ ˆ ⎛ µ0 qq′vv′ sin φ ⎞ ˆ = (qv) iˆ × ⎜ 0 F = qv × B′ = (qv)iˆ × 0 ⎟ (−k ) = ⎜ ⎟j 2 4π r r2 ⎠ r2 ⎝ 4π ⎝ 4π ⎠ ! where φ is the angle between v′ and rˆ′.

28-16

Chapter 28

EXECUTE: Substituting numbers gives ! µ ⎡ ( 8.00 × 10−6 C )( 5.00 × 10−6 C )( 9.00 × 10−6 m/s )( 6.50 × 10−6 m/s ) ⎛ 0.400 ⎞ ⎤ ˆ F= 0⎢ ⎜ ⎟⎥ j 2 4π ⎢ ⎝ 0.500 ⎠ ⎦⎥ ( 0.500 m ) ⎣ ! F = ( 7.49 × 10−8 N ) ˆj .

28.52.

EVALUATE: These are small fields and small forces, but if the charge has small mass, the force can affect its motion. IDENTIFY: The wire creates a magnetic field near it, and the moving electron feels a force due to this field. µI SET UP: The magnetic field due to the wire is B = 0 , and the force on a moving charge is F = qvB sin φ . 2π r EXECUTE: F = qvB sin φ = (evμ0 I sin φ )/2π r. Substituting numbers gives

F = (1.60 × 10−19 C)(6.00 × 104 m/s)(4π × 10−7 T ⋅ m/A)(2.50 A)(sin 90°) /[2π (0.0450 m)]

28.53.

F = 1.07 × 10–19 N ! ! From the right hand rule for the cross product, the direction of v × B is opposite to the current, but since the electron is negative, the force is in the same direction as the current. EVALUATE: This force is small at an everyday level, but it would give the electron an acceleration of about 1011 m/s2. IDENTIFY: Find the force that the magnetic field of the wire exerts on the electron. SET UP: The force on a moving charge has magnitude F = q vB sin φ and direction given by the right-hand rule. ! μI For a long straight wire, B = 0 and the direction of B is given by the right-hand rule. 2π r F q vB sin φ ev ⎛ μ0 I ⎞ EXECUTE: (a) a = = = ⎜ ⎟ m m m ⎝ 2πr ⎠ a=

(1.6 × 10−17 C)(2.50 × 105 m/s)(4π × 10−7 T ⋅ m/A)(25.0 A) = 1.1 × 1013 m/s 2 , (9.11 × 10−31 kg)(2π )(0.0200 m)

away from the wire. (b) The electric force must balance the magnetic force. eE = evB , and μ I (250, 000 m/s)(4π × 10−7 T ⋅ m/A)(25.0 A) E = vB = v 0 = = 62.5 N/C . The magnetic force is directed away from 2π r 2π (0.0200 m) the wire so the force from the electric field must be toward the wire. Since the charge of the electron is negative, the electric field must be directed away from the wire to produce a force in the desired direction. EVALUATE: (c) mg = (9.11 × 10−31 kg)(9.8 m/s 2 ) ≈ 10−29 N . Fel = eE = (1.6 × 10−19 C)(62.5 N/C) ≈ 10−17 N . Fel ≈ 1012 Fgrav , so we can neglect gravity. 28.54.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to calculate the magnetic field due to each wire. Add these field vectors to calculate the net field and then use Eq.(27.2) to calculate the force. SET UP: Let the wire connected to the 25.0 Ω resistor be #2 and the wire connected to the 10.0 Ω resistor be #1. Both I1 and I 2 are directed toward the right in the figure, so at the location of the proton B2 is ⊗ and B1 = ".

B1 =

μ0 I1 μI and B2 = 0 2 , with r = 0.0250 m. I1 = (100.0 V)/(10.0 Ω) = 10.0 A and I 2 = (100.0 V)/(25.0 Ω) = 4.00 A 2π r 2π r

EXECUTE:

B1 = 8.00 × 10−5 T, B2 = 3.20 × 10−5 T and B = B1 − B2 = 4.80 × 10−5 T and in the direction ". Force is to the right.

Figure 28.54

28.55.

F = qvB = (1.602 × 10−19 C)(650 × 103 m/s)(4.80 × 10−5 T) = 5.00 × 10−18 N ! ! EVALUATE: The force is perpendicular to both v and B. The magnetic force is much larger than the gravity force on the proton. IDENTIFY: Find the net magnetic field due to the two loops at the location of the proton and then find the force these fields exert on the proton. μ0 IR 2 SET UP: For a circular loop, the field on the axis, a distance x from the center of the loop is B = . 2( R 2 + x 2 )3 / 2 R = 0.200 m and x = 0.125 m .


28-17

⎡ ⎤ μ0 IR 2 EXECUTE: The fields add, so B = B1 + B2 = 2 B1 = 2 ⎢ . 2 2 3/ 2 ⎥ + 2( R x ) ⎣ ⎦ (4π × 10−7 T ⋅ m/A)(1.50 A)(0.200 m)2 B= = 5.75 × 10−6 T. [(0.200 m) 2 + (0.125 m)2 ]3/ 2 F = q vB sin φ = (1.6 × 10−19 C)(2400 m/s)(5.75 × 10−6 T) sin 90° = 2.21 × 10−21 N , perpendicular to the line ab and

28.56.

28.57.

to the velocity. EVALUATE: The weight of a proton is w = mg = 1.6 × 10−24 N , so the force from the loops is much greater than the gravity force on the proton. IDENTIFY: The net magnetic field is the vector sum of the fields due to each wire. ! μI SET UP: B = 0 . The direction of B is given by the right-hand rule. 2π r EXECUTE: (a) The currents are the same so points where the two fields are equal in magnitude are equidistant from the two wires. The net field is zero along the dashed line shown in Figure 28.56a. (b) For the magnitudes of the two fields to be the same at a point, the point must be 3 times closer to the wire with the smaller current. The net field is zero along the dashed line shown in Figure 28.56b. (c) As in (a), the points are equidistant from both wires. The net field is zero along the dashed line shown in Figure 28.56c. EVALUATE: The lines of zero net field consist of points at which the fields of the two wires have opposite directions and equal magnitudes.

Figure 28.56 ! μ0 qv!0 × rˆ IDENTIFY: B = 4π r 2 ! SET UP: rˆ = iˆ and r = 0.250 m , so v0 × rˆ = v0 z ˆj − v0 y kˆ. ! μ q μ q EXECUTE: B = 0 2 v0 z ˆj − v0 y kˆ = 6.00 × 10−6 T ˆj. v0 y = 0. 0 2 v0 z = 6.00 × 10−6 T and 4π r 4π r 4π (6.00 × 10−6 T)(0.25 m) 2 v0z = = −521 m/s. v0x = ± v02 − v0 y 2 − v0 z 2 = ± (800 m/s) 2 − (−521 m/s) 2 = ±607 m/s. μ0 ( −7.20 × 10−3 C) The sign of v0x isn’t determined. ! μ0 qv!0 × rˆ μ0 q ! ˆ v0 x kˆ − v0 z iˆ . = (b) Now r = j and r = 0.250 m . B = 4π r 2 4π r 2 μ |q| μ |q| μ (7.20 × 10−3 C) B = 0 2 v02x + v02z = 0 2 v0 = 0 800 m/s = 9.20 × 10 −6 T . 4π r 4π r 4π (0.250 m) 2 EVALUATE: The magnetic field in part (b) doesn’t depend on the sign of v0 x . ! IDENTIFY and SET UP: B = B ( x / a )iˆ

(

) (

)

(

28.58.

)

0

Apply Gauss's law for magnetic fields to a cube with side length L, one corner at the origin, and sides parallel to the x, y and z axes, as shown in Figure 28.58.

Figure 28.58

28-18

Chapter 28

! EXECUTE: Since B is parallel to the x-axis the only sides that have nonzero flux are the front side (parallel to the yz-plane at x =!L) and the back side (parallel to the yz-plane at x = 0.) ! front Φ B = ∫ B ⋅ dA = B0 ( x/a ) ∫ dA(iˆ ⋅ iˆ) = B0 ( x/a ) ∫ dA ! ! x = L on this face so B ⋅ dA = B0 ( L/a ) dA

Φ B = B0 ( L/a) ∫ dA = B0 ( L/a) L2 = B0 ( L3 /a )

28.59.

back On the back face x = 0 so B = 0 and Φ B = 0. The total flux through the cubical Gaussian surface is Φ B = B0 ( L3 /a ). ! EVALUATE: This violates Eq.(27.8), which says that Φ B = 0 for any closed surface. The claimed B is impossible because it has been shown to violate Gauss's law for magnetism. IDENTIFY: Use Eq.(28.9) and the right-hand rule to calculate the magnitude and direction of the magnetic field at P produced by each wire. Add these two field vectors to find the net field. (a) SET UP: The directions of the fields at point P due to the two wires are sketched in Figure 28.59a. ! ! EXECUTE: B1 and B2 must be equal and opposite for the resultant field at P to be zero. ! B2 is to the right so I 2 is out of the page.

Figure 28.59a

μ0 I1 μ0 ⎛ 6.00 A ⎞ μI μ ⎛ I2 ⎞ = B2 = 0 2 = 0 ⎜ ⎜ ⎟ ⎟ 2π r1 2π ⎝ 1.50 m ⎠ 2π r2 2π ⎝ 0.50 m ⎠ μ ⎛ 6.00 A ⎞ μ0 ⎛ I 2 ⎞ B1 = B2 says 0 ⎜ ⎟= ⎜ ⎟ 2π ⎝ 1.50 m ⎠ 2π ⎝ 0.50 m ⎠ ⎛ 0.50 m ⎞ I2 = ⎜ ⎟ ( 6.00 A ) = 2.00 A ⎝ 1.50 m ⎠ (b) SET UP: The directions of the fields at point Q are sketched in Figure 28.59b. μI EXECUTE: B1 = 0 1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.40 × 10 T ⎝ 0.50 m ⎠ μI B2 = 0 2 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.67 × 10 T ⎝ 1.50 m ⎠ Figure 28.59b ! ! B1 and B2 are in opposite directions and B1 > B2 so ! B = B1 − B2 = 2.40 × 10−6 T − 2.67 × 10−7 T = 2.13 × 10−6 T, and B is to the right. (c) SET UP: The directions of the fields at point S are sketched in Figure 28.59c. μI EXECUTE: B1 = 0 1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.00 × 10 T ⎝ 0.60 m ⎠ μI B2 = 0 2 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 5.00 × 10 T ⎝ 0.80 m ⎠ Figure 28.59c ! ! B1 and B2 are right angles to each other, so the magnitude of their resultant is given by B1 =

B = B12 + B22 = (2.00 × 10−6 T) 2 + (5.00 × 10−7 T) 2 = 2.06 × 10−6 T


28.60.

28-19

EVALUATE: The magnetic field lines for a long, straight wire!are concentric circles with the wire at the center. The magnetic field at each point is tangent to the field line, so B is perpendicular to the line from the wire to the point where the field is calculated. IDENTIFY: Find the vector sum of the magnetic fields due to each wire. ! μI SET UP: For a long straight wire B = 0 . The direction of B is given by the right-hand rule and is perpendicular 2π r to the line from the wire to the point where then field is calculated. EXECUTE: (a) The magnetic field vectors are shown in Figure 28.60a. μI μ0 I a μ0 Ia (b) At a position on the x-axis Bnet = 2 0 sinθ = = , in the positive x-direction. 2 2 2 2 2 2πr π ( x + a2 ) π x +a x +a (c) The graph of B versus x/a is given in Figure 28.60b. EVALUATE: (d) The magnetic field is a maximum at the origin, x = 0. μ Ia (e) When x >> a, B ≈ 0 2 . πx

Figure 28.60 28.61.

Apply F = IlB sin φ , with the magnetic field at point P that is calculated in problem 28.60. μ0 Ia SET UP: The net field of the first two wires at the location of the third wire is B = , in the +x-direction. π ( x2 + a2 ) EXECUTE: (a) Wire is carrying current into the page, so it feels a force in the − y -direction . IDENTIFY:

⎛ μ0 Ia ⎞ F μ0 (6.00 A) 2 (0.400 m) = IB = I ⎜ = = 1.11 × 10−5 N/m. ⎟ 2 2 2 2 L ⎝ π ( x + a ) ⎠ π ( (0.600 m) + (0.400 m) )

28.62.

28.63.

(b) If the wire carries current out of the page then the force felt will be in the opposite direction as in part (a). Thus the force will be 1.11× 10−5 N m, in the +y-direction. EVALUATE: We could also calculate the force exerted by each of the first two wires and find the vector sum of the two forces. IDENTIFY: The wires repel each other since they carry currents in opposite directions, so the wires will move away from each other until the magnetic force is just balanced by the force due to the spring. µ I 2L SET UP: The force of the spring is kx and the magnetic force on each wire is Fmag = 0 . 2π x EXECUTE: Call x the distance the springs each stretch. On each wire, Fspr = Fmag, and there are two spring forces µ I 2L µ0 I 2 L , which gives x = . on each wire. Therefore 2kx = 0 2π x 2π k EVALUATE: Since μ0 /2π is small, x will likely be much less than the length of the wires. ! ! IDENTIFY: Apply ∑ F = 0 to one of the wires. The force one wire exerts on the other depends on I so ∑ F = 0

gives two equations for the two unknowns T and I. SET UP: The force diagram for one of the wires is given in Figure 28.63. ⎛ μ I2 ⎞ The force one wire exerts on the other is F = ⎜ 0 ⎟ L, ⎝ 2π r ⎠ where r = 2(0.040 m)sin θ = 8.362 × 10−3 m is the distance between the two wires.

Figure 28.63

28-20

Chapter 28

EXECUTE:

∑F

x

∑F

y

= 0 gives T cosθ = mg and T = mg / cosθ

= 0 gives F = T sin θ = (mg / cosθ )sin θ = mg tan θ

And m = λ L, so F = λ Lg tan θ ⎛ μ0 I 2 ⎞ ⎜ ⎟ L = λ Lg tan θ ⎝ 2π r ⎠ λ gr tan θ I= ( μ0 / 2π ) (0.0125 kg/m)(9.80 m/s) 2 (tan 6.00°)(8.362 ×10−3 m) = 23.2 A 2 × 10−7 T ⋅ m/A EVALUATE: Since the currents are in opposite directions the wires repel. When I is increased, the angle θ from the vertical increases; a large current is required even for the small displacement specified in this problem. IDENTIFY: Consider the forces on each side of the loop. SET UP: The forces on the left and right sides cancel. The forces on the top and bottom segments of the loop are in opposite directions, so the magnitudes subtract. ⎛ μ I ⎞ ⎛ Il Il ⎞ μ IlI ⎛ 1 1 ⎞ EXECUTE: F = Ft − Fb = ⎜ 0 wire ⎟ ⎜ − ⎟ = 0 wire ⎜ − ⎟ . 2π ⎝ rr rl ⎠ ⎝ 2π ⎠ ⎝ rt rb ⎠ I=

28.64.

F=

28.65.

28.66.

28.67.

⎞ μ0 (5.00 A)(0.200 m)(14.0 A) ⎛ 1 1 −5 − ⎜ ⎟ = 7.97 × 10 N . The force on the top segment is away 2π ⎝ 0.100 m 0.026 m ⎠

from the wire, so the net force is away from the wire. EVALUATE: The net force on a current loop in a uniform magnetic field is zero, but the magnetic field of the wire is not uniform, it is stronger closer to the wire. ! ! IDENTIFY: Find the magnetic field of the first loop at the location of the second loop and apply τ = μ × B and ! ! U = − μ ⋅ B to find μ and U. SET UP: Since x is much larger than a′ , assume B is uniform over the second loop and equal to its value on the axis of the first loop. Nμ0 Ia 2 Nμ0 Ia 2 EXECUTE: (a) x >> a ⇒ B = . ≈ 2 2 3/ 2 2( x + a ) 2 x3 ⎛ Nμ Ia 2 ⎞ NN ′μ0πI I ′a 2 a′2 sin θ ! ! τ = μ × B = μB sin θ = ( N ′I ′A′) ⎜ 0 3 ⎟ sin θ = 2 x3 ⎝ 2x ⎠ ⎛ Nμ Ia 2 ⎞ NN ′μ0π II ′a 2 a′2 cosθ ! ! . (b) U = − μ ⋅ B = − μB cosθ = −( N ′I ′π a′2 ) ⎜ 0 3 ⎟ cosθ = − 2 x3 ⎝ 2x ⎠ EVALUATE: (c) Having x >> a allows us to simplify the form of the magnetic field, whereas assuming x >> a′ means we can assume that the magnetic field from the first loop is constant over the second loop. ! ! μ Idl × rˆ IDENTIFY: Apply dB = 0 . 4π r 2 SET UP: The two straight segments produce zero field at P. The field at the center of a circular loop of radius R is μI μI B = 0 , so the field at the center of curvature of a semicircular loop is B = 0 . 2R 4R EXECUTE: The semicircular loop of radius a produces field out of the page at P and the semicircular loop of 1⎛ μ I ⎞⎛ 1 1⎞ μ I ⎛ a ⎞ radius b produces field into the page. Therefore, B = Ba − Bb = ⎜ 0 ⎟ ⎜ − ⎟ = 0 ⎜ 1 − ⎟ , out of page. 2 ⎝ 2 ⎠ ⎝ a b ⎠ 4a ⎝ b ⎠ EVALUATE: If a = b , B = 0 . IDENTIFY: Find the vector sum of the fields due to each loop. μ0 Ia 2 SET UP: For a single loop B = . Here we have two loops, each of N turns, and measuring the field 2( x 2 + a 2 )3/ 2 along the x-axis from between them means that the “x” in the formula is different for each case: EXECUTE: Left coil: x → x +

a μ0 NIa 2 . ⇒ Bl = 2 2(( x + a 2) 2 + a 2 )3 2

Right coil: x → x −

a μ0 NIa 2 . ⇒ Br = 2 2(( x − a 2) 2 + a 2 )3 2


28-21

So, the total field at a point a distance x from the point between them is B=

⎞ μ0 NIa 2 ⎛ 1 1 . + ⎜ 2 2 32 2 2 32 ⎟ 2 ⎝ (( x + a 2) + a ) (( x − a 2) + a ) ⎠

(b) B versus x is graphed in Figure 28.67. Figure 28.67a is the total field and Figure 27.67b is the field from the right-hand coil. 32 ⎞ μ NIa 2 ⎛ 1 1 μ0 NIa 2 ⎛ 4 ⎞ μ0 NI (c) At point P, x = 0 and B = 0 + = = ⎜ ⎟ ⎜ ⎟ 2 ⎝ ((a 2) 2 + a 2 )3 2 ((− a 2) 2 + a 2 )3 2 ⎠ (5a 2 4)3 2 ⎝ 5 ⎠ a ⎛ 4⎞ (d) B = ⎜ ⎟ ⎝5⎠

(e) dB dx

32

μ0 NI ⎛ 4 ⎞ =⎜ ⎟ a ⎝5⎠

32

μ0 (300)(6.00 A) = 0.0202 T. (0.080 m)

⎞ dB μ0 NIa 2 ⎛ −3( x + a 2) −3( x − a 2) . At x = 0 , = + ⎜ 2 2 52 2 2 52 ⎟ 2 ⎝ (( x + a 2) + a ) (( x − a 2) + a ) ⎠ dx = x =0

⎞ μ0 NIa 2 ⎛ −3(a 2) −3( − a 2) + =0. ⎜ 2 2 52 2 2 52 ⎟ 2 ⎝ (( a 2) + a ) ((− a 2) + a ) ⎠

d B μ0 NIa 2 ⎛ −3 6( x + a 2) 2 (5 2) −3 6( x − a 2) 2 (5 2) ⎞ = + + + ⎜ ⎟ 2 2 2 52 2 2 72 2 2 5/ 2 2 ⎝ (( x + a 2) + a ) (( x + a 2) + a ) (( x − a 2) + a ) (( x − a 2) 2 + a 2 )7 / 2 ⎠ dx 2

At x = 0 ,

d 2B dx 2

= x =0

−3 −3 μ0 NIa 2 ⎛ 6(a 2) 2 (5 2) 6(− a 2) 2 (5 2) ⎞ + + + ⎜⎜ ⎟ = 0. 2 2 7/2 2 2 5/ 2 2 2 5/ 2 2 ⎝ ((a 2) + a ) ((a 2) + a ) (( a 2) + a ) ((a 2) 2 + a 2 )7 / 2 ⎠⎟

EVALUATE: Since both first and second derivatives are zero, the field can only be changing very slowly.

Figure 28.67 28.68.

IDENTIFY: A current-carrying wire produces a magnetic field, but the strength of the field depends on the shape of the wire. SET UP: The magnetic field at the center of a circular wire of radius a is B = μ0 I/2a, and the field a distance x µ0 I 2a . 4π x x 2 + a 2 EXECUTE: (a) Since the diameter D = 2a, we have B = μ0 I/2a = μ0 I/D. (b) In this case, the length of the wire is equal to the diameter of the circle, so 2a = π D, giving a = π D/2, and

from the center of a straight wire of length 2a is B =

x = D/2. Therefore B =

28.69.

2 (π D / 2 ) µ0 I µ0 I = . 4π ( D / 2) D 2 / 4 + π 2 D 2 / 4 D 1 + π 2

EVALUATE: The field in part (a) is greater by a factor of 1 + π 2 . It is reasonable that the field due to the circular wire is greater than the field due to the straight wire because more of the current is close to point A for the circular wire than it is for the straight wire. ! ! μ0 Idl × rˆ IDENTIFY: Apply dB = . 4π r 2 ! ! SET UP: The contribution from the straight segments is zero since dl × r = 0. The magnetic field from the curved wire is just one quarter of a full loop. ⎛μI⎞ μI EXECUTE: B = 14 ⎜ 0 ⎟ = 0 and is directed out of the page. ⎝ 2 R ⎠ 8R

28-22

28.70.

28.71.

Chapter 28

EVALUATE: It is very simple to calculate B at point P but it would be much more difficult to calculate B at other points. ! ! μ0 Idl × rˆ IDENTIFY: Apply dB = . 4π r 2 ! ! SET UP: The horizontal wire yields zero magnetic field since dl × r = 0. The vertical current provides the magnetic field of half of an infinite wire. (The contributions from all infinitesimal pieces of the wire point in the same direction, so there is no vector addition or components to worry about.) ⎛ μI ⎞ μI EXECUTE: B = 12 ⎜ 0 ⎟ = 0 and is directed out of the page. ⎝ 2πR ⎠ 4πR EVALUATE: In the equation preceding Eq.(28.8) the limits on the integration are 0 to a rather than −a to a and this introduces a factor of 12 into the expression for B. (a) IDENTIFY: Consider current density J for a small concentric ring and integrate to find the total current in terms of α and R. SET UP: We can't say I = JA = J π R 2 , since J varies across the cross section.

To integrate J over the cross section of the wire divide the wire cross section up into thin concentric rings of radius r and width dr, as shown in Figure 28.71. Figure 28.71 EXECUTE: The area of such a ring is dA, and the current through it is dI = J dA; dA = 2π r dr and dI = J dA = α r (2π r dr ) = 2πα r 2 dr R

I = ∫ dI = 2πα ∫ r 2 dr = 2πα ( R 3 / 3) so α = 0

3I 2π R 3

(b) IDENTIFY and SET UP: (i) r ≤ R Apply Ampere's law to a circle of radius r < R. Use the method of part (a) to find the current enclosed by the Ampere's law path. ! ! ! EXECUTE: úB ⋅ dl = úB dl = Búdl = B (2π r ), by the symmetry and direction of B. The current passing through r

the path is I encl = ∫ dl , where the integration is from 0 to r. I encl = 2πα ∫ r 2 dr = 0

!

2πα r 3 2π ⎛ 3I ⎞ 3 Ir 3 − ⎜ ⎟ r = 3 . Thus 3 3 ⎝ 2π R 3 ⎠ R

!

⎛ Ir 3 ⎞ μ Ir 2 gives B (2π r ) = μ0 ⎜ 3 ⎟ and B = 0 3 2π R ⎝R ⎠ (ii) IDENTIFY and SET UP: r ≥ R Apply Ampere’s law to a circle of radius r > R. ! ! EXECUTE: úB ⋅ dl = úB dl = Búdl = B (2π r )

úB ⋅ dl = μ I

0 encl

28.72.

! ! μI I encl = I ; all the current in the wire passes through this path. Thus úB ⋅ dl = μ0 I encl gives B (2π r ) = μ0 I and B = 0 2π r μI EVALUATE: Note that at r = R the expression in (i) (for r ≤ R ) gives B = 0 . At r = R the expression in (ii) 2π R μ0 I (for r ≥ R ) gives B = , which is the same. 2π R IDENTIFY: Apply Ampere’s law to a circle of radius r in each case. SET UP: Assume that the currents are uniform over the cross sections of the conductors. ! ! ⎛A ⎞ ⎛ r2 ⎞ ⎛ r2 ⎞ μ Ir EXECUTE: (a) r < a ⇒ I encl = I ⎜ r ⎟ = I ⎜ 2 ⎟ . úB ⋅ dl = B 2πr = μ0 I encl = μ0 I ⎜ 2 ⎟ and B = 0 2 . When 2πa ⎝a ⎠ ⎝a ⎠ ⎝ Aa ⎠

r = a, B =

μ0 I , which is just what was found in part (a) of Exercise 28.37. 2πa


⎛A ⎞ ⎛ r 2 − b2 ⎞ (b) b < r < c ⇒ I encl = I − I ⎜ b → r ⎟ = I ⎜ 1 − 2 . 2 ⎟ ⎝ c −b ⎠ ⎝ Ab → c ⎠

!

!

⎛

28-23

⎛ c2 − r 2 ⎞ r 2 − b2 ⎞ and = μ0 I ⎜ 2 2 2 ⎟ 2 ⎟ −b ⎠ ⎝c −b ⎠

úB ⋅ dl = B 2πr = μ I ⎜⎝1 − c 0

μ0 I ⎛ c 2 − r 2 ⎞ μ0 I , just as in part (a) of Exercise 28.37 and when r = c, B = 0 , just as in ⎟ . When r = b, B = ⎜ 2πb 2πr ⎝ c 2 − b 2 ⎠ part (b) of Exercise 28.37. EVALUATE: Unlike E, B is not zero within the conductors. B varies across the cross section of each conductor. ! ! IDENTIFY: Apply úB ⋅ dA = 0. B=

28.73.

SET UP: Take the closed gaussian surface to be a cylinder whose axis coincides with the wire. EXECUTE: If there is a magnetic field component in the z-direction, it must be constant because of the symmetry of the wire. Therefore the contribution to a surface integral over a closed cylinder, encompassing a long straight wire will be zero: no flux through the barrel of the cylinder, and equal but opposite flux through the ends. The radial field will have no contribution through the ends, but through the barrel: ! ! ! ! ! ! 0 = úB ⋅ dA = úBr ⋅ dA = ∫ Br ⋅ dA = ∫ Br dA = Br Abarrel = 0 . Therefore, Br = 0. barrel

28.74.

28.75.

barrel

EVALUATE: The magnetic field of a long straight wire is everywhere tangent to a circular area whose plane is perpendicular to the wire, with the wire passing through the center of the circular area. This field produces zero flux through the cylindrical gaussian surface. IDENTIFY: Apply Ampere’s law to a circular path of radius r. SET UP: Assume the current is uniform over the cross section of the conductor. EXECUTE: (a) r < a ⇒ I encl = 0 ⇒ B = 0. ! ! ⎛ A ⎞ ⎛ π (r 2 − a 2 ) ⎞ (r 2 − a 2 ) μ I (r 2 − a 2 ) (r 2 − a 2 ) . (b) a < r < b ⇒ I encl = I ⎜ a → r ⎟ = I ⎜ and B = 0 . úB ⋅ dl = B 2πr = μ0 I 2 =I 2 2 2 2 ⎟ 2 (b − a ) 2πr (b 2 − a 2 ) (b − a ) ⎝ Aa →b ⎠ ⎝ π (b − a ) ⎠ ! ! μI (c) r > b ⇒ I encl = I . úB ⋅ dl = B 2πr = μ0 I and B = 0 . 2πr EVALUATE: The expression in part (b) gives B = 0 at r = a and this agrees with the result of part (a). The μI expression in part (b) gives B = 0 at r = b and this agrees with the result of part (c). 2π b IDENTIFY: Use Ampere's law to find the magnetic field at r = 2a from the axis. The analysis of Example 28.9 shows that the field outside the cylinder is the same as for a long, straight wire along the axis of the cylinder. SET UP: EXECUTE: Apply Ampere's law to a circular path of radius 2a, as shown in Figure 28.75. B (2π ) = μ0 I encl ⎛ (2a ) 2 − a 2 ⎞ I encl = I ⎜ = 3I /8 2 2 ⎟ ⎝ (3a ) − a ⎠

Figure 28.75 3 μ0 I ; this is the magnetic field inside the metal at a distance of 2a from the cylinder axis. Outside the 16 2π a μI cylinder, B = 0 . The value of r where these two fields are equal is given by 1/ r = 3/(16a ) and r = 16a / 3. 2π r EVALUATE: For r < 3a, as r increases the magnetic field increases from zero at r = 0 to μ0 I /(2π (3a )) at r = 3a. For r > 3a the field decreases as r increases so it is reasonable for there to be a r > 3a where the field is the same as at r = 2a. IDENTIFY: The net field is the vector sum of the fields due to the circular loop and to the long straight wire. μI μI SET UP: For the long wire, B = 0 1 , and for the loop, B = 0 2 . 2π D 2R EXECUTE: At the center of the circular loop the current I 2 generates a magnetic field that is into the page, so the

B=

28.76.

current I1 must point to the right. For complete cancellation the two fields must have the same magnitude: μ0 I1 μ0 I 2 = . Thus, I1 = πD I 2 . R 2πD 2 R EVALUATE: If I1 is to the left the two fields add.

28-24

Chapter 28

28.77.

IDENTIFY: Use the current density J to find dI through a concentric ring and integrate over the appropriate cross ! section to find the current through that cross section. Then use Ampere's law to find B at the specified distance from the center of the wire. (a) SET UP: Divide the cross section of the cylinder into thin concentric rings of radius r and width dr, as shown in Figure 28.77a. The current through each ring is dI = J dA = J 2π r dr.

Figure 28.77a dI =

EXECUTE:

2I0 ⎡ 4I 2 2 1 − ( r / a ) ⎤ 2π r dr = 20 ⎡1 − ( r / a ) ⎤ r dr. The total current I is obtained by integrating dI ⎦ ⎦ a ⎣

π a2 ⎣

a

a 1 ⎛ 4I ⎞ a ⎛ 4I ⎞ ⎡ 1 ⎤ over the cross section I = ∫ dI = ⎜ 20 ⎟ ∫ (1 − r 2 / a 2 ) r dr = ⎜ 20 ⎟ ⎢ r 2 − r 4 / a 2 ⎥ = I 0 , as was to be shown. 0 4 ⎝ a ⎠ 0 ⎝ a ⎠⎣2 ⎦0 (b) SET UP: Apply Ampere's law to a path that is a circle of radius r > a, as shown in Figure 28.77b.

!

!

úB ⋅ dl = B(2π r ) I encl = I 0 (the path encloses the entire cylinder)

Figure 28.77b !

!

úB ⋅ dl = μ I

EXECUTE:

0 encl

says B (2π r ) = μ0 I 0 and B =

μ0 I 0 . 2π r

(c) SET UP: Divide the cross section of the cylinder into concentric rings of radius r ′ and width dr ′, as was done in part (a). See Figure 28.77c. The current dI through each ring 2 4 I ⎡ ⎛ r′ ⎞ ⎤ is dI = 20 ⎢1 − ⎜ ⎟ ⎥ r′ dr ′ a ⎢⎣ ⎝ a ⎠ ⎥⎦

Figure 28.77c EXECUTE: The current I is obtained by integrating dI from r ′ = 0 to r ′ = r: 2 r 4I r ⎡ ⎛ r′ ⎞ ⎤ 4I 2 4 I = ∫ dI = 20 ∫ ⎢1 − ⎜ ⎟ ⎥ r ′ dr′ = 20 ⎡ 12 ( r ′ ) − 14 ( r ′ ) /a 2 ⎤ ⎦0 a 0 ⎣⎢ ⎝ a ⎠ ⎦⎥ a ⎣ 4I I r2 ⎛ r2 ⎞ I = 20 ( r 2 / 2 − r 4 / 4a 2 ) = 0 2 ⎜ 2 − 2 ⎟ a a ⎝ a ⎠ (d) SET UP:

Apply Ampere's law to a path that is a circle of radius r < a, as shown in Figure 28.77d. ! ! úB ⋅ dl = B(2π r ) I encl =

I0r 2 ⎛ r2 ⎞ 2 − ⎜ ⎟ (from part (c)) a2 ⎝ a2 ⎠

Figure 28.77d

28.78.

!

!

I0r 2 μI r (2 − r 2 /a 2 ) and B = 0 0 2 (2 − r 2 / a 2 ) 2 2π a a μI EVALUATE: Result in part (b) evaluated at r = a: B = 0 0 . Result in part (d) evaluated at 2π a I μ0 I 0 a μ r = a: B = (2 − a 2 / a 2 ) = 0 0 . The two results, one for r > a and the other for r < a, agree at r = a. 2π a 2 2π a IDENTIFY: Apply Ampere’s law to a circle of radius r. ! ! SET UP: The current within a radius r is I = ∫ J ⋅ dA , where the integration is over a disk of radius r.

EXECUTE:

úB ⋅ dl = μ I

0 encl

says B (2π r ) = μ0


! ! a ⎛b ⎞ EXECUTE: (a) I 0 = ∫ J ⋅ dA = ∫ ⎜ e(r − a )/δ ⎟ rdrdθ = 2π b ∫ e(r − a )/δ dr = 2πbδ e(r − a )/δ 0 ⎝r ⎠

a 0

28-25

= 2π bδ (1 − e − a/δ ) .

I 0 = 2π (600 A/m) (0.025 m) (1 − e(0.050 / 0.025) ) = 81.5 A. ! ! μI (b) For r ≥ a, úB ⋅ d l = B 2πr = μ0 I encl = μ0 I 0 and B = 0 0 . 2π r r ! ! r b ⎛ ⎞ (c) For r ≤ a, I ( r ) = ∫ J ⋅ dA = ∫ ⎜ e(r ′ − a )/δ ⎟ r ′dr ′dθ = 2πb ∫ e(r − a )/δ dr = 2πbδe(r ′ − a )/δ . 0 0 ⎝ r′ ⎠

I ( r ) = 2πbδ (e(r ′− a )/δ − e − a/δ ) = 2πbδe− a/δ (er/δ − 1) and I ( r ) = I 0 (d) For r ≤ a ,

!

!

úB ⋅ dl = B(r )2πr = μ I

0 encl

(e) At r = δ = 0.025 m, B =

= μ0 I 0

(e r/δ − 1) . (e a/δ − 1)

(e r/δ − 1) μ I (e r/δ − 1) . and B = 0 0 a/δ a/ δ (e − 1) 2π r (e − 1)

μ0 I 0 (e − 1) μ (81.5 A) (e − 1) = 0 = 1.75 × 10−4 T . 2πδ (e a / δ − 1) 2π (0.025 m) (e0.050 / 0.025 − 1)

μ0 I 0 (ea / δ − 1) μ0 (81.5 A) = = 3.26 × 10−4 T. 2π a (e a / δ − 1) 2π (0.050 m) μI μ (81.5 A) At r = 2a = 0.100 m, B = 0 0 = 0 = 1.63 × 10−4 T. 2π r 2π (0.100 m)

At r = a = 0.050 m, B =

28.79.

EVALUATE: At points outside the cylinder, the magnetic field is the same as that due to a long wire running along the axis of the cylinder. IDENTIFY: Evaluate the integral as specified in the problem. μ0 Ia 2 SET UP: Eq.(28.15) says Bx = . 2 2( x + a 2 )3 / 2 EXECUTE:

∫

∞ −∞

Bx dx = ∫ B=

μ0 Ia 2 μI dx = 0 −∞ 2( x + a 2 ) 3 / 2 2 ∞

μ0 I 2

2

∫

∫

∞ −∞

1 d ( x/a ). (( x/a ) 2 + 1)3/ 2

∞ dz μI ⇒ ∫ Bx dx = 0 −∞ ( z + 1)3/ 2 −∞ 2 ∞

2

π/2

∫π

− /2

π /2

cos θdθ =

μ0 I (sinθ ) = μ0 I , 2 − π /2

where we used the substitution z = tan θ to go from the first to second line. EVALUATE: This is just what Ampere’s Law tells us to expect if we imagine the loop runs along the x-axis ! ! closing on itself at infinity: úB ⋅ d l = μ0 I .

28.80.

IDENTIFY: Follow the procedure specified in the problem. SET UP: The field and integration path are sketched in Figure 28.80. ! ! ! EXECUTE: úB ⋅ d l = 0 (no currents in the region). Using the figure, let B = B0 iˆ for y < 0 and B = 0 for y > 0. ! ! Then ∫ B ⋅ d l = Bab L − Bcd L = 0. Bcd = 0 , so Bab L = 0. But we have assumed that Bab ≠ 0. This is a contradiction abcde

and violates Ampere’s Law. EVALUATE: It is often convenient to approximate B as confined to a particular region of space, but this result tells us that the boundary of such a region isn’t sharp.

Figure 28.80 28.81.

IDENTIFY: Use what we know about the magnetic field of a long, straight conductor to deduce the symmetry of the magnetic field. Then apply Ampere's law to calculate the magnetic field at a distance a above and below the current sheet.

28-26

Chapter 28

SET UP: Do parts (a) and (b) together.

Consider the individual currents in pairs, where the currents in each pair are equidistant ! on either side of the point where B is being calculated. Figure 28.81a shows that for each pair the z-components cancel, and that above the sheet the field is in the –x-direction and that below the sheet it is in the +x-direction.

Figure 28.81a

! ! Also, by symmetry the magnitude of B a distance a above the sheet must equal the magnitude of B a distance a ! below the sheet. Now that we have deduced the symmetry of B, apply Ampere's law. Use a path that is a rectangle, as shown in Figure 28.81b. !

!

ú B ⋅ dl = μ I

0 encl

Figure 28.81b I is directed out of the page, so for I to be positive the integral around the path is taken in the counterclockwise direction. ! ! ! EXECUTE: Since B is parallel to the sheet, on the sides of the rectangle that have length 2a, úB ⋅ dl = 0. On the ! long sides of length L, B is parallel to the side, in the direction we are integrating around the path, and has the ! ! same magnitude, B, on each side. Thus úB ⋅ dl = 2 BL. n conductors per unit length and current I out of the page in

28.82.

each conductor gives I encl = InL. Ampere's law then gives 2 BL = μ0 InL and B = 12 μ0 In. EVALUATE: Note that B is independent of the distance a from the sheet. Compare this result to the electric field due to an infinite sheet of charge (Example 22.7). IDENTIFY: Find the vector sum of the fields due to each sheet. ! SET UP: Problem 28.81 shows that for an infinite sheet B = 12 μ0 In . If I is out of the page, B is to the left above ! the sheet and to the right below the sheet. If I is into the page, B is to the right above the sheet and to the left below the sheet. B is independent of the distance from the sheet. The directions of the two fields at points P, R and S are shown in Figure 28.82. EXECUTE: (a) Above the two sheets, the fields cancel (since there is no dependence upon the distance from the sheets). (b) In between the sheets the two fields add up to yield B = μ0 nI , to the right. (c) Below the two sheets, their fields again cancel (since there is no dependence upon the distance from the sheets). EVALUATE: The two sheets with currents in opposite directions produce a uniform field between the sheets and zero field outside the two sheets. This is analogous to the electric field produced by large parallel sheets of charge of opposite sign.

Figure 28.82 28.83.

IDENTIFY and SET UP: Use Eq.(28.28) to calculate the total magnetic moment of a volume V of the iron. Use the density and atomic mass of iron to find the number of atoms in this volume and use that to find the magnetic dipole moment per atom.


28-27

μ total

, so μ total = MV The average magnetic moment per atom is μatom = μ total / N = MV / N , V where N is the number of atoms in volume V. The mass of volume V is m = ρV , where ρ is the density.

EXECUTE:

M=

( ρiron = 7.8 × 103 kg/m3 ). The number of moles of iron in volume V is n=

ρV m = , where 55.847 × 10−3 kg/mol is the atomic mass −3 55.847 × 10 kg/mol 55.847 × 10−3 kg/mol

of iron from appendix D. N = nN A , where N A = 6.022 × 1023 atoms/mol is Avogadro's number. Thus N = nN A =

ρVN A 55.847 × 10−3 kg/mol

.

μatom =

⎛ 55.847 × 10−3 kg/mol ⎞ M (55.847 × 10−3 kg/mol) MV . = MV ⎜ ⎟= N ρVN A ρ NA ⎝ ⎠

μatom =

(6.50 × 104 A/m)(55.847 × 10−3 kg/mol) (7.8 × 103 kg/m 3 )(6.022 × 1023 atoms/mol)

μatom = 7.73 × 10−25 A ⋅ m 2 = 7.73 × 10−25 J/T μ B = 9.274 × 10−24 A ⋅ m 2 , so μatom = 0.0834μB . 28.84.

EVALUATE: The magnetic moment per atom is much less than one Bohr magneton. The magnetic moments of each electron in the iron must be in different directions and mostly cancel each other. IDENTIFY: The force on the cube of iron must equal the weight of the iron cube. The weight is proportional to the density and the magnetic force is proportional to μ , which is in turn proportional to K m . SET UP: The densities if iron, aluminum and silver are ρ Fe = 7.8 × 103 kg/m3 , ρ Al = 2.7 × 103 kg/m3 and

ρ Ag = 10.5 × 103 kg/m3. The relative permeabilities of iron, aluminum and silver are K Fe = 1400 , K Al = 1.00022 and K Ag = 1.00 − 2.6 × 10−5 .

EXECUTE: (a) The microscopic magnetic moments of an initially unmagnetized ferromagnetic material experience torques from a magnet that aligns the magnetic domains with the external field, so they are attracted to the magnet. For a paramagnetic material, the same attraction occurs because the magnetic moments align themselves parallel to the external field. For a diamagnetic material, the magnetic moments align antiparallel to the external field so it is like two magnets repelling each other. (b) The magnet can just pick up the iron cube so the force it exerts is FFe = mFe g = ρ Fe a 3 g = (7.8 × 103 kg/m3 )(0.020 m)3 (9.8 m/s 2 ) = 0.612 N. If the magnet tries to lift the aluminum cube of the same dimensions as the iron block, then the upward force felt by the cube is K 1.000022 FAl = Al (0.612 N) = (0.612 N) = 4.37 × 10−4 N. The weight of the aluminum cube is K Fe 1400

WAl = mAl g = ρAl a3 g = (2.7 × 103 kg m3 )(0.020 m)3 (9.8 m s 2 ) = 0.212 N. Therefore, the ratio of the magnetic force 4.37 × 10−4 N = 2.1 × 10−3 and the magnet cannot lift it. 0.212 N (c) If the magnet tries to lift a silver cube of the same dimensions as the iron block, then the downward force felt K (1.00 − 2.6 × 10−5 ) by the cube is FAl = Ag (0.612 N) = (0.612 N) = 4.37 × 10−4 N. But the weight of the silver cube K Fe 1400

on the aluminum cube to the weight of the cube is

is WAg = mAg g = ρ Ag a 3 g = (10.5 × 103 kg/m 3 )(0.020 m)3 (9.8 m/s 2 ) = 0.823 N. So the ratio of the magnetic force on the silver cube to the weight of the cube is

28.85.

4.37 × 10−4 N = 5.3 × 10−4 and the magnet’s effect would not be 0.823 N

noticeable. EVALUATE: Silver is diamagnetic and is repelled by the magnet. Aluminum is paramagnetic and is attracted by the magnet. But for both these materials the force is much less that the force on a similar cube of ferromagnetic iron. IDENTIFY: The current-carrying wires repel each other magnetically, causing them to accelerate horizontally. Since gravity is vertical, it plays no initial role. F µ0 I 2 SET UP: The magnetic force per unit length is = , and the acceleration obeys the equation F/L = m/L a. L 2π d The rms current over a short discharge time is I 0 / 2 .

28-28

Chapter 28

EXECUTE: (a) First get the force per unit length: 2

2

F µ0 I 2 µ ⎛ I ⎞ µ ⎛V ⎞ µ ⎛Q ⎞ = = 0 ⎜ 0 ⎟ = 0 ⎜ ⎟ = 0 ⎜ 0 ⎟ L 2π d 2π d ⎝ 2 ⎠ 4π d ⎝ R ⎠ 4π d ⎝ RC ⎠

2

2

F m µ ⎛Q ⎞ = a = λ a = 0 ⎜ 0 ⎟ . Solving for a gives L L 4π d ⎝ RC ⎠ μ0Q02 μ0Q02 a= . From the kinematics equation vx = v0 x + axt , we have v0 = at = aRC = 2 2 4πλ R C d 4πλ RCd 2 2 ⎛ μ0Q0 ⎞ 2 ⎜ ⎟ 2 4πλ RCd ⎠ 1 ⎛ μ0Q02 ⎞ v (b) Conservation of energy gives 12 mv02 = mgh and h = 0 = ⎝ = ⎜ ⎟. 2g 2g 2 g ⎝ 4πλ RCd ⎠

Now apply Newton’s second law using the result above:

28.86.

EVALUATE: Once the wires have swung apart, we would have to consider gravity in applying Newton’s second law. IDENTIFY: Approximate the moving belt as an infinite current sheet. SET UP: Problem 28.81 shows that B = 12 μ0 In for an infinite current sheet. Let L be the width of the sheet, so n = I/L. ΔQ Δx = L σ = Lvσ . EXECUTE: The amount of charge on a length Δx of the belt is ΔQ = LΔxσ , so I = Δt Δt μ0 I μ0vσ ! Approximating the belt as an infinite sheet B = . B is directed out of the page, as shown in Figure 28.86. = 2L 2 EVALUATE: The field is uniform above the sheet, for points close enough to the sheet for it to be considered infinite.

Figure 28.86 28.87.

28.88.

IDENTIFY: The rotating disk produces concentric rings of current. Calculate the field due to each ring and integrate over the surface of the disk to find the total field. μI SET UP: At the center of a circular ring carrying current I, B = 0 . 2r 2Qrdr . If the disk rotates at n turns per EXECUTE: The charge on a ring of radius r is q = σ A = σ 2π rdr = a2 dq 2Qnrdr μ I μ 2Qnrdr μ0 nQdr = ndq = . Therefore, dB = 0 = 0 second, then the current from that ring is dI = = . dt a2 2r 2r a 2 a2 a a μ nQdr μ nQ We integrate out from the center to the edge of the disk and find B = ∫ dB = ∫ 0 2 = 0 . 0 0 a a EVALUATE: The magnetic field is proportional to the total charge on the disk and to its rotation rate. IDENTIFY: There are two parts to the magnetic field: that from the half loop and that from the straight wire segment running from − a to a. SET UP: Apply Eq.(28.14). Let the φ be the angle that locates dl around the ring. μ0 Ia 2 EXECUTE: Bx (ring ) = 12 Bloop = − . 4( x 2 + a 2 )3/ 2 μI dl x μ Iax sin φ dφ dBy ( ring ) = dB sin θ sin φ = 0 sin φ = 0 2 and 4π ( x 2 + a 2 ) ( x 2 + a 2 )1 2 4π ( x + a 2 )3 2 π

π

0

0

By ( ring ) = ∫ dBy ( ring ) = ∫

By ( rod ) = Bx = −

π

μ0 Iax sinφ dφ μ0 Iax μ0 Iax = cosφ = − . 2 + a 2 )3/ 2 4π ( x 2 + a 2 )3/ 2 4π ( x 2 + a 2 )3/ 2 2 π ( x 0

μ0 Ia , using Eq. (28.8). The total field components are: 2πx(x 2 + a 2 )1/ 2

μ0 Ia 2 4( x + a ) 2

2 3/ 2

and By =

μ0 Ia

2π x( x + a ) 2

2 1/ 2

⎛ x2 ⎞ μ0 Ia 3 = . ⎜1 − 2 2 ⎟ 2 2 3/ 2 ⎝ x + a ⎠ 2π x( x + a )

2a μI μI Bx . By decreases faster than Bx as x increases. For very small x, Bx = − 0 and By = 0 . 4a 2π a π x In this limit Bx is the field at the center of curvature of a semicircle and By is the field of a long straight wire.

EVALUATE:

By = −

ELECTROMAGNETIC INDUCTION

29.1.

29.2.

29

IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns is Φ = NBA cos φ, and by Faraday’s law the magnitude of the induced emf is E = dΦ/dt. EXECUTE: (a) ΔΦ = NBA = (50)(1.20 T)(0.250 m)(0.300 m) = 4.50 Wb (b) E = dΦ/dt = (4.50 Wb)/(0.222 s) = 20.3 V EVALUATE: This induced potential is certainly large enough to be easily detectable. ΔΦ B . Φ B = BA cos φ . Φ B is the flux through each turn of the coil. IDENTIFY: E = Δt SET UP: EXECUTE:

φi = 0°. φf = 90°. (a) Φ B ,i = BA cos0° = (6.0 × 10−5 T)(12 × 10−4 m 2 )(1) = 7.2 × 10−8 Wb. The total flux through the coil is

N Φ B ,i = (200)(7.2 × 10−8 Wb) = 1.44 × 10−5 Wb . Φ B ,f = BA cos90° = 0 . N Φ i − N Φ f 1.44 × 10−5 Wb = = 3.6 × 10−4 V = 0.36 mV . Δt 0.040 s EVALUATE: The average induced emf depends on how rapidly the flux changes. IDENTIFY and SET UP: Use Faraday’s law to calculate the average induced emf and apply Ohm’s law to the coil to calculate the average induced current and charge that flows. ΔΦI B (a) EXECUTE: The magnitude of the average emf induced in the coil is Eav = N . Initially, Δt

(b) E = 29.3.

Φ Bf − Φ Bi NBA = . The average induced current is Δt Δt E NBA NBA ⎛ NBA ⎞ I = av = . The total charge that flows through the coil is Q = I Δt = ⎜ . ⎟ Δt = R R Δt R ⎝ R Δt ⎠ EVALUATE: The charge that flows is proportional to the magnetic field but does not depend on the time Δt . (b) The magnetic stripe consists of a pattern of magnetic fields. The pattern of charges that flow in the reader coil tell the card reader the magnetic field pattern and hence the digital information coded onto the card. (c) According to the result in part (a) the charge that flows depends only on the change in the magnetic flux and it does not depend on the rate at which this flux changes. IDENTIFY and SET UP: Apply the result derived in Exercise 29.3: Q = NBA / R. In the present exercise the flux Φ Bi = BA cos φ = BA. The final flux is zero, so Eav = N

29.4.

29.5.

changes from its maximum value of Φ B = BA to zero, so this equation applies. R is the total resistance so here R = 60.0 Ω + 45.0 Ω = 105.0 Ω. NBA QR (3.56 × 10−5 C)(105.0 Ω) EXECUTE: Q = says B = = = 0.0973 T. R NA 120(3.20 × 10−4 m 2 ) EVALUATE: A field of this magnitude is easily produced. IDENTIFY: Apply Faraday’s law. ! SET UP: Let +z be the positive direction for A . Therefore, the initial flux is positive and the final flux is zero. ! ΔΦ B 0 − (1.5 T)π (0.120 m) 2 =− = +34 V. Since E is positive and A is toward us, EXECUTE: (a) and (b) E = − −3 Δt 2.0 × 10 s the induced current is counterclockwise. EVALUATE: The shorter the removal time, the larger the average induced emf.

29-1

29-2

29.6.

Chapter 29

IDENTIFY: Apply Eq.(29.4). I = E/R. SET UP: d Φ B /dt = AdB/dt. EXECUTE:

(a) E =

Nd Φ B d d = NA ( B) = NA ( (0.012 T/s)t + (3.00 × 10−5 T/s 4 )t 4 ) dt dt dt

E = NA ( (0.012 T/s) + (1.2 × 10−4 T/s 4 )t 3 ) = 0.0302 V + (3.02 × 10−4 V/s3 )t 3.

E 0.0680 V = = 1.13 × 10−4 A. R 600 Ω EVALUATE: The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time. IDENTIFY: Calculate the flux through the loop and apply Faraday’s law. SET UP: To find the total flux integrate dΦ B over the width of the loop. The magnetic field of a long straight ! μI wire, at distance r from the wire, is B = 0 . The direction of B is given by the right-hand rule. 2π r μ 0i EXECUTE: (a) When B = , into the page. 2π r μi (b) d Φ B = BdA = 0 Ldr. 2π r b μ iL b dr μ0iL (c) Φ B = ∫ d Φ B = 0 ∫ = ln(b/a ). a 2π a r 2π d Φ B μ0 L di (d) E = = ln( b a ) . dt dt 2π μ (0.240 m) ln(0.360/0.120)(9.60 A/s) = 5.06 × 10−7 V. (e) E = 0 2π EVALUATE: The induced emf is proportional to the rate at which the current in the long straight wire is changing IDENTIFY: Apply Faraday’s law. ! SET UP: Let A be upward in Figure 29.28 in the textbook. dΦB EXECUTE: (a) Eind = = d ( B⊥ A) dt dt (b) At t = 5.00 s, E = 0.0302 V + (3.02 × 10−4 V/s3 )(5.00 s)3 = 0.0680 V. I =

29.7.

29.8.

Eind = A sin 60°

(

)

−1 −1 dB d = A sin 60° (1.4 T)e − (0.057s )t = (π r 2 )(sin 60°)(1.4 T)(0.057 s −1 )e − (0.057s )t dt dt

Eind = π (0.75 m) 2 (sin 60°)(1.4 T)(0.057 s −1 )e − (0.057s

= (0.12 V) e− (0.057 s

−1 ) t

.

− (0.057 s−1 ) t

(b) E = E0 = (0.12 V). (0.12 V) = (0.12 V) e . ln(1/10) = −(0.057 s −1 )t and t = 40.4 s. ! ! (c) B is in the direction of A so Φ B is positive. B is getting weaker, so the magnitude of the flux is decreasing and d Φ B /dt < 0. Faraday’s law therefore says E > 0. Since E > 0, the induced current must flow counterclockwise as viewed from above. EVALUATE: The flux changes because the magnitude of the magnetic field is changing. IDENTIFY and SET UP: Use Faraday’s law to calculate the emf (magnitude and direction). The direction of the induced current is the same as the direction of the emf. The flux changes because the area of the loop is changing; relate dA/dt to dc/dt, where c is the circumference of the loop. (a) EXECUTE: c = 2π r and A = π r 2 so A = c 2 /4π Φ B = BA = ( B/4π )c 2 1 10

29.9.

−1 ) t

1 10

1 10

d Φ B ⎛ B ⎞ dc =⎜ ⎟c dt ⎝ 2π ⎠ dt At t = 9.0 s, c = 1.650 m − (9.0 s)(0.120 m/s) = 0.570 m E = (0.500 T)(1/2π )(0.570 m)(0.120 m/s) = 5.44 mV E =

(b) SET UP:

The loop and magnetic field are sketched in Figure 29.9. Take into the page to be the ! positive direction for A. Then the magnetic flux is positive. Figure 29.9

Electromagnetic Induction

29-3

The positive flux is decreasing in magnitude; d Φ B / dt is negative and E is positive. By the right! hand rule, for A into the page, positive E is clockwise. EVALUATE: Even though the circumference is changing at a constant rate, dA/dt is not constant and E is not EXECUTE:

29.10.

constant. Flux ⊗ is decreasing so the flux of the induced current is ⊗ and this means that I is clockwise, which checks. IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = d Φ/dt. EXECUTE: (a) and (c) The magnetic flux is constant, so the induced emf is zero. (b) The area inside the field is changing. If we let x be the length (along the 30.0-cm side) in the field, then A = (0.400 m)x. ΦB = BA = (0.400 m)x E = dΦ/dt = B d[(0.400 m) x]/dt = B(0.400 m)dx/dt = B(0.400 m)v E = (1.25 T)(0.400 m)(0.0200 m/s) = 0.0100 V

29.11.

29.12.

29.13.

EVALUATE: It is not a large flux that induces an emf, but rather a large rate of change of the flux. The induced emf in part (b) is small enough to be ignored in many instances. IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = dΦ/dt. EXECUTE: (a) E = dΦ/dt = d[A(B0 + bx)]/dt = bA dx/dt = bAv (b) clockwise (c) Same answers except the current is counterclockwise. EVALUATE: Even though the coil remains within the magnetic field, the flux through it increases because the strength of the field is increasing. IDENTIFY: Use the results of Example 29.5. 2 ⎛ 2π rad/rev ⎞ SET UP: Emax = NBAω. Eav = Emax . ω = (440 rev/min) ⎜ ⎟ = 46.1 rad/s. π ⎝ 60 s/min ⎠ EXECUTE: (a) Emax = NBAω = (150)(0.060 T)π (0.025 m) 2 (46.1 rad/s) = 0.814 V 2 2 (b) Eav = Emax = (0.815 V) = 0.519 V π π EVALUATE: In Emax = NBAω , ω must be in rad/s. IDENTIFY: Apply the results of Example 29.5. SET UP: Emax = NBAω

Emax 2.40 × 10−2 V = = 10.4 rad/s NBA (120)(0.0750 T)(0.016 m) 2 EVALUATE: We may also express ω as 99.3 rev/min or 1.66 rev/s . IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = d Φ/dt. EXECUTE: The flux is constant in each case, so the induced emf is zero in all cases. EVALUATE: Even though the coil is moving within the magnetic field and has flux through it, this flux is not changing, so no emf is induced in the coil. IDENTIFY and SET UP: The field of the induced current is directed to oppose the change in flux. EXECUTE: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise. (c) The field is constant so the flux is constant and there is no induced emf and no induced current. EVALUATE: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing. IDENTIFY: By Lenz’s law, the induced current flows to oppose the flux change that caused it. SET UP and EXECUTE: The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppose this decrease. Therefore the induced current is counterclockwise. EVALUATE: Careful! Lenz’s law does not say that the induced current flows to oppose the magnetic flux. Instead it says that the current flows to oppose the change in flux. IDENTIFY and SET UP: Apply Lenz's law, in the form that states that the flux of the induced current tends to oppose the change in flux. EXECUTE: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B. When the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced in coil B is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the right the current in the circuit with coil B must flow through the resistor in the direction a to b. EXECUTE: ω =

29.14.

29.15.

29.16.

29.17.

29-4

29.18.

29.19.

29.20.

Chapter 29

(b) With the switch closed the magnetic field of coil A is to the right at the location of coil B. This field is stronger at points closer to coil A so when coil B is brought closer the flux through coil B increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a. (c) With the switch closed the magnetic field of coil A is to the right at the location of coil B. The current in the circuit that includes coil A increases when R is decreased and the magnetic field of coil A increases when the current through the coil increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a. EVALUATE: In parts (b) and (c) the change in the circuit causes the flux through circuit B to increase and in part (a) it causes the flux to decrease. Therefore, the direction of the induced current is the same in parts (b) and (c) and opposite in part (a). IDENTIFY: Apply Lenz’s law. SET UP: The field of the induced current is directed to oppose the change in flux in the primary circuit. EXECUTE: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the induced current in B is to the right. To produce magnetic field to the right, the induced current flows through R from right to left. (b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced current in B is to the right. To produce magnetic field to the right the induced current flows through R from right to left. (c) The magnetic field in A is to the right and is increasing. The flux is increasing so the field due to the induced current in B is to the left. To produce magnetic field to the left the induced current flows through R from left to right. EVALUATE: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing. IDENTIFY and SET UP: Lenz's law requires that the flux of the induced current opposes the change in flux. EXECUTE: (a) Φ B is " and increasing so the flux Φ ind of the induced current is ⊗ and the induced current is clockwise. (b) The current reaches a constant value so Φ B is constant. d Φ B / dt = 0 and there is no induced current. (c) Φ B is " and decreasing, so Φ ind is " and current is counterclockwise. EVALUATE: Only a change in flux produces an induced current. The induced current is in one direction when the current in the outer ring is increasing and is in the opposite direction when that current is decreasing. IDENTIFY: Use the results of Example 29.6. Use the three approaches specified in the problem for determining the direction of the induced current. I = E/R . ! SET UP: Let A be directed into the figure, so a clockwise emf is positive. EXECUTE: (a) E = vBl = (5.0 m/s)(0.750 T)(1.50 m) = 5.6 V (b) (i) Let q be a positive charge in the moving bar, as shown in Figure 29.20a. The magnetic force on this charge is ! ! ! F = qv × B , which points upward. This force pushes the current in a counterclockwise direction through the circuit.

(ii) Φ B is positive and is increasing in magnitude, so d Φ B / dt > 0. Then by Faraday’s law E < 0 and the emf and induced current are counterclockwise. (iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure 29.20b. E 5.6 V (c) E = RI . I = = = 0.22 A. R 25 Ω EVALUATE: All three methods agree on the direction of the induced current.

Figure 29.20 29.21.

IDENTIFY: A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving. SET UP: The induced emf is E = vBL sin φ, where φ is the angle between the velocity and the magnetic field.


29.22.

29.23.

29.24.

29-5

EXECUTE: (a) E = vBL sin φ = (5.00 m/s)(0.450 T)(0.300 m)(sin 90°) = 0.675 V (b) The positive charges are moved to end b, so b is at the higher potential. ! (c) E = V/L = (0.675 V)/(0.300 m) = 2.25 V/m. The direction of E is from, b to a. (d) The positive charge are pushed to b, so b has an excess of positive charge. (e) (i) If the rod has no appreciable thickness, L = 0, so the emf is zero. (ii) The emf is zero because no magnetic force acts on the charges in the rod since it moves parallel to the magnetic field. EVALUATE: The motional emf is large enough to have noticeable effects in some cases. IDENTIFY: The moving bar has a motional emf induced across its ends, so it causes a current to flow. SET UP: The induced potential is E = vBL and Ohm’s law is E = IR. EXECUTE: (a) E = vBL = (5.0 m/s)(0.750 T)(1.50 m) = 5.6 V (b) I = E /R = (5.6 V)/(25 Ω) = 0.23 A EVALUATE: Both the induced potential and the current are large enough to have noticeable effects. IDENTIFY: E = vBL SET UP: L = 5.00 × 10−2 m. 1 mph = 0.4470 m/s. E 1.50 V = = 46.2 m/s = 103 mph. EXECUTE: v = BL (0.650 T)(5.00 × 10−2 m) EVALUATE: This is a large speed and not practical. It is also difficult to produce a 5.00 cm wide region of 0.650 T magnetic field. IDENTIFY: E = vBL. SET UP: 1 mph = 0.4470 m/s . 1 G = 10−4 T . EXECUTE:

⎛ 0.4470 m/s ⎞ −4 (a) E = (180 mph) ⎜ ⎟ (0.50 × 10 T)(1.5 m) = 6.0 mV. This is much too small to be 1 mph ⎝ ⎠

noticeable.

29.25.

29.26.

⎛ 0.4470 m/s ⎞ −4 (b) E = (565 mph) ⎜ ⎟ (0.50 × 10 T)(64.4 m) = 0.813 mV. This is too small to be noticeable. 1 mph ⎝ ⎠ EVALUATE: Even though the speeds and values of L are large, the earth’s field is small and motional emfs due to the earth’s field are not important in these situations. IDENTIFY and SET UP: E = vBL. Use Lenz's law to determine the direction of the induced current. The force Fext required to maintain constant speed is equal and opposite to the force FI that the magnetic field exerts on the rod because of the current in the rod. EXECUTE: (a) E = vBL = (7.50 m/s)(0.800 T)(0.500 m) = 3.00 V ! (b) B is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the induced current must be counterclockwise, so from b to a in the rod. ! E 3.00 V (c) I = = = 2.00 A. FI = ILB sin φ = (2.00 A)(0.500 m)(0.800 T)sin 90° = 0.800 N . FI is to the left. To R 1.50 Ω keep the bar moving to the right at constant speed an external force with magnitude Fext = 0.800 N and directed to the right must be applied to the bar. (d) The rate at which work is done by the force Fext is Fext v = (0.800 N)(7.50 m/s) = 6.00 W. The rate at which thermal energy is developed in the circuit is I 2 R = (2.00 A)(1.50 Ω) = 6.00 W. These two rates are equal, as is required by conservation of energy. EVALUATE: The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz’s law. IDENTIFY: Use Faraday’s law to calculate the induced emf. Ohm’s law applied to the loop gives I. Use Eq.(27.19) to calculate the force exerted on each side of the loop. SET UP: The loop before it starts to enter the magnetic field region is sketched in Figure 29.26a. EXECUTE: For x < −3L/2 or x > 3L/2 the loop is completely outside the field dΦB region. Φ B = 0, and = 0. dt Figure 29.26a

Thus E = 0 and I = 0, so there is no force from the magnetic field and the external force F necessary to maintain constant velocity is zero.

29-6

Chapter 29

SET UP:

The loop when it is completely inside the field region is sketched in Figure 29.26b. EXECUTE: For − L/2 < x < L/2 the loop is completely inside the field region and Φ B = BL2 . Figure 29.26b

! ! ! dΦB = 0 so E = 0 and I = 0. There is no force F = Il × B from the magnetic field and the external force F dt necessary to maintain constant velocity is zero. SET UP: The loop as it enters the magnetic field region is sketched in Figure 29.26c.

But

EXECUTE: For −3L/2 < x < − L/2 the loop is entering the field region. Let x′ be the length of the loop that is within the field. Figure 29.26c dΦB dΦB = Blv. The magnitude of the induced emf is E = = BLv and the induced dt dt ! E BLv = current is I = . Direction of I: Let A be directed into the plane of the figure. Then Φ B is positive. The R R dΦB flux is positive and increasing in magnitude, so is positive. Then by Faraday’s law E is negative, and with dt ! our choice for direction of A a negative E is counterclockwise. The current induced in the loop is counterclockwise. SET UP: The induced current and magnetic force on the loop are shown in Figure 29.26d, for the situation where the loop is entering the field. ! ! ! EXECUTE: FI = Il × B gives that the ! force FI exerted on the loop by the magnetic field is to the left and has B 2 L2v ⎛ BLv ⎞ magnitude FI = ILB = ⎜ = . LB ⎟ R ⎝ R ⎠ Figure 29.26d ! The external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to ! FI so is to the right and has this same magnitude. SET UP: The loop as it leaves the magnetic field region is sketched in Figure 29.26e.

Then Φ B = BLx′ and

EXECUTE: For L/2 < x < 3L/2 the loop is leaving the field region. Let x′ be the length of the loop that is outside the field. Figure 29.26e

dΦB dΦB = BLv. The magnitude of the induced emf is E = = BLv and the induced dt dt ! E BLv current is I = = . Direction of I: Again let A be directed into the plane of the figure. Then Φ B is positive R R dΦB and decreasing in magnitude, so is negative. Then by Faraday’s law E is positive, and with our choice for dt ! direction of A a positive E is clockwise. The current induced in the loop is clockwise. Then Φ B = BL( L − x′) and


29-7

SET UP: The induced current and magnetic force on the loop are shown in Figure 29.26f, for the situation where the loop is leaving the field. ! ! ! EXECUTE: FI = Il × B gives that the ! force FI exerted on the loop by the magnetic field is to the left and has B 2 L2v ⎛ BLv ⎞ magnitude FI = ILB = ⎜ . ⎟ LB = R ⎝ R ⎠ Figure 29.26f ! The ! external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to FI so is to the right and has this same magnitude. (a) The graph of F versus x is given in Figure 29.26g.

Figure 29.26g (b) The graph of the induced current I versus x is given in Figure 29.26h.

Figure 29.26h

29.27.

29.28.

EVALUATE: When the loop is either totally outside or totally inside the magnetic field region the flux isn’t changing, there is no induced current, and no external force is needed for the loop to maintain constant speed. When the loop is entering the field the external force required is directed so as to pull the loop in and when the loop is leaving the field the external force required is directed so as to pull the loop out of the field. These directions agree with Lenz’s law: the force on the induced current (opposite in direction to the required external force) is directed so as to oppose the loop entering or leaving the field. IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends. SET UP: The induced potential is E = vBL sin φ. EXECUTE: Note that φ = 90° in all these cases because the bar moved perpendicular to the magnetic field. But the effective length of the bar, L sin θ, is different in each case. (a) E = vBL sin θ = (2.50 m/s)(1.20 T)(1.41 m) sin (37.0°) = 2.55 V, with a at the higher potential because positive charges are pushed toward that end. (b) Same as (a) except θ = 53.0°, giving 3.38 V, with a at the higher potential. (c) Zero, since the velocity is parallel to the magnetic field. (d) The bar must move perpendicular to its length, for which the emf is 4.23 V. For Vb > Va, it must move upward and to the left (toward the second quadrant) perpendicular to its length. EVALUATE: The orientation of the bar affects the potential induced across its ends. IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E at a distance r from the center of the solenoid. Away from the ends of the solenoid, B = μ0nI inside and B = 0 outside. (a) SET UP: The end view of the solenoid is sketched in Figure 29.28.

Let R be the radius of the solenoid.

Figure 29.28 ! ! dΦB Apply úE ⋅ dl = − to an integration path that is a circle of radius r, where r < R. We need to calculate just the dt magnitude of E so we can take absolute values.

29-8

Chapter 29

EXECUTE:

!

Φ B = Bπ r 2 , − !

!

úE ⋅ dl E = 12 r

= −

!

úE ⋅ dl

= E (2π r )

dΦB dB = π r2 dt dt

dΦB dB implies E (2π r ) = π r 2 dt dt

dB dt

dB dI = μ0n dt dt dI Thus E = 12 r μ 0 n = 12 (0.00500 m)(4π × 10 −7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 1.70 × 10 −4 V/m dt (b) r = 0.0100 cm is still inside the solenoid so the expression in part (a) applies. dI E = 12 r μ 0 n = 12 (0.0100 m)(4π × 10−7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 3.39 × 10 −4 V/m dt EVALUATE: Inside the solenoid E is proportional to r, so E doubles when r doubles. IDENTIFY: Apply Eqs.(29.9) and (29.10). SET UP: Evaluate the integral if Eq.(29.10) for a path which is a circle of radius r and concentric with the solenoid. The magnetic field of the solenoid is confined to the region inside the solenoid, so B(r ) = 0 for r > R B = μ 0 nI , so

29.29.

EXECUTE: (b) E =

(a)

dΦB dB dB . =A = π r12 dt dt dt

! 1 d Φ B π r12 dB r1 dB = = . The direction of E is shown in Figure 29.29a. 2π r1 dt 2π r1 dt 2 dt

(c) All the flux is within r < R, so outside the solenoid E =

1 d Φ B π R 2 dB R 2 dB = = . 2π r2 dt 2π r2 dt 2r2 dt

(d) The graph is sketched in Figure 29.29b. dΦB dB π R 2 dB (e) At r = R 2, E = = π ( R/2) 2 = . dt dt 4 dt dΦB dB (f) At r = R , E = . = π R2 dt dt dΦB dB (g) At r = 2 R , E = . = π R2 dt dt EVALUATE: The emf is independent of the distance from the center of the cylinder at all points outside it. Even though the magnetic field is zero for r > R , the induced electric field is nonzero outside the solenoid and a nonzero emf is induced in a circular turn that has r > R.

Figure 29.29 29.30.

IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E and use this E in Eq.(29.9) to calculate E between two points. (a) SET UP: Because of the axial symmetry and the absence of any electric charge, the field lines are concentric circles.


29-9

(b) See Figure 29.30.

! E is tangent to the ring. The direction ! of E (clockwise or counterclockwise) is the direction in which current will be induced in the ring. Figure 29.30

! Use the sign convention for Faraday’s law to deduce this direction. Let A be into the paper. Then dΦB dΦB Φ B is positive. B decreasing then means is negative, so by E = − , E is positive and therefore dt dt ! ! ! dΦB clockwise. Thus E is clockwise around the ring. To calculate E apply úE ⋅ dl = − to a circular path that dt coincides with the ring. ! ! úE ⋅ dl = E (2π r ) EXECUTE:

Φ B = Bπ r 2 ;

dΦB dB = π r2 dt dt

dB dB 1 and E = 12 r = 2 (0.100 m)(0.0350 T/s) = 1.75 × 10−3 V/m dt dt ! ! (c) The induced emf has magnitude E = úE ⋅ dl = E (2π r ) = (1.75 × 10−3 V/m)(2π )(0.100 m) = 1.100 × 10−3 V. Then E (2π r ) = π r 2

E 1.100 × 10−3 V = = 2.75 × 10−4 A. R 4.00 Ω (d) Points a and b are separated by a distance around the ring of π r so E = E (π r ) = (1.75 × 10−3 V/m)(π )(0.100 m) = 5.50 × 10−4 V I=

29.31.

(e) The ends are separated by a distance around the ring of 2π r so E = 1.10 × 10−3 V as calculated in part (c). EVALUATE: The induced emf, calculated from Faraday’s law and used to calculate the induced current, is associated with the induced electric field integrated around the total circumference of the ring. IDENTIFY: Apply Eq.(29.1) with Φ B = μ0 niA .

A = π r 2 , where r = 0.0110 m . In Eq.(29.11), r = 0.0350 m . dΦB d d di di E 2π r EXECUTE: E = and E = E (2π r ). Therefore, = . = ( BA) = ( μ0 niA) = μ0 nA dt μ0 nA dt dt dt dt SET UP:

29.32.

di (8.00 × 10−6 V/m)2π (0.0350 m) = = 9.21 A/s. μ0 (400 m −1 )π (0.0110 m) 2 dt EVALUATE: Outside the solenoid the induced electric field decreases with increasing distance from the axis of the solenoid. IDENTIFY: A changing magnetic flux through a coil induces an emf in that coil, which means that an electric field is induced in the material of the coil. ! ! dΦB SET UP: According to Faraday’s law, the induced electric field obeys the equation úE ⋅ dl = − . dt EXECUTE: (a) For the magnitude of the induced electric field, Faraday’s law gives E2πr = d(Bπr2)/dt = πr2 dB/dt

r dB 0.0225 m = (0.250 T/s) = 2.81 × 10−3 V/m 2 dt 2 (b) The field points toward the south pole of the magnet and is decreasing, so the induced current is counterclockwise. EVALUATE: This is a very small electric field compared to most others found in laboratory equipment. ΔΦ B IDENTIFY: Apply Faraday’s law in the form Eav = N . Δt E=

29.33.

SET UP:

The magnetic field of a large straight solenoid is B = μ0 nI inside the solenoid and zero outside.

Φ B = BA , where A is 8.00 cm 2 , the cross-sectional area of the long straight solenoid.

29-10

Chapter 29

Eav = N

EXECUTE:

ΔΦ B NA( Bf − Bi ) NAμo nI = = . Δt Δt Δt

μ0 (12)(8.00 × 10−4 m 2 )(9000 m −1 )(0.350 A)

29.34.

= 9.50 × 10−4 V. 0.0400 s EVALUATE: An emf is induced in the second winding even though the magnetic field of the solenoid is zero at the location of the second winding. The changing magnetic field induces an electric field outside the solenoid and that induced electric field produces the emf. IDENTIFY: Apply Eq.(29.14). SET UP: P = 3.5 × 10−11 F/m dΦE EXECUTE: iD = P = (3.5 × 10−11 F/m)(24.0 × 103 V ⋅ m/s3 )t 2 . iD = 21 × 10−6 A gives t = 5.0 s. dt EVALUATE: iD depends on the rate at which Φ E is changing.

29.35.

IDENTIFY:

Eav =

SET UP:

Apply Eq.(29.14), where P = K P0 .

d Φ E / dt = 4(8.76 × 103 V ⋅ m/s 4 )t 3 . P0 = 8.854 × 10−12 F/m.

EXECUTE:

P=

iD 12.9 × 10−12 A = = 2.07 × 10−11 F/m. The dielectric constant is 3 ( d Φ E / dt ) 4(8.76 × 10 V ⋅ m/s 4 )(26.1× 10−3 s)3

K = P = 2.34. P0 EVALUATE: 29.36.

The larger the dielectric constant, the larger is the displacement current for a given d Φ E /dt.

IDENTIFY and SET UP: Eqs.(29.13) and (29.14) show that iC = iD and also relate iD to the rate of change of the electric field flux between the plates. Use this to calculate dE / dt and apply the generalized form of Ampere’s law (Eq.29.15) to calculate B. i i 0.280 A 0.280 A (a) EXECUTE: iC = iD , so jD = D = C = = = 55.7 A/m 2 A A π r2 π (0.0400 m)2

dE dE jD 55.7 A/m 2 = = = 6.29 × 1012 V/m ⋅ s so dt dt P0 8.854 × 10−12 C2 / N ⋅ m 2 ! ! (c) SET UP: Apply Ampere’s law úB ⋅ dl = μ0 (iC + iD )encl (Eq.(28.20)) to a circular path with radius r = 0.0200 m. (b) jD = P0

An end view of the solenoid is given in Figure 29.36. By symmetry the magnetic field is tangent to the path and constant around it.

EXECUTE:

Figure 29.36 ! ! Thus úB ⋅ dl = úBdl = B ∫ dl = B (2π r ).

iC = 0 (no conduction current flows through the air space between the plates) The displacement current enclosed by the path is jDπ r 2 . Thus B (2π r ) = μ0 ( jDπ r 2 ) and B = 12 μ0 jD r = 12 (4π × 10−7 T ⋅ m/A)(55.7 A/m 2 )(0.0200 m) = 7.00 × 10−7 T (d) B = 12 μ0 jD r. Now r is

29.37.

1 2

the value in (c), so B is

1 2

also: B = 12 (7.00 × 10−7 T) = 3.50 × 10−7 T

EVALUATE: The definition of displacement current allows the current to be continuous at the capacitor. The magnetic field between the plates is zero on the axis (r = 0) and increases as r increases. PA E IDENTIFY: q = CV . For a parallel-plate capacitor, C = , where P = K P0 . iC = dq/dt. jD = P . dt d SET UP: E = q/PA so dE/dt = iC /PA. −4

EXECUTE:

(4.70)P0 (3.00 × 10 m )(120 V) ⎛ PA ⎞ (a) q = CV = ⎜ ⎟V = = 5.99 × 10−10 C. 2.50 × 10−3 m ⎝ d ⎠ 2


29-11

dq = iC = 6.00 × 10−3 A. dt dE i i (c) jD = P = K P0 C = C = jC , so iD = iC = 6.00 × 10−3 A. dt K P0 A A

(b)

iD = iC , so Kirchhoff’s junction rule is satisfied where the wire connects to each capacitor plate.

EVALUATE: 29.38.

IDENTIFY and SET UP: Use iC = q / t to calculate the charge q that the current has carried to the plates in time t. The two equations preceeding Eq.(24.2) relate q to the electric field E and the potential difference between the plates. The displacement current density is defined by Eq.(29.16). EXECUTE: (a) iC = 1.80 × 10−3 A q = 0 at t = 0 The amount of charge brought to the plates by the charging current in time t is q = iCt = (1.80 × 10−3 A)(0.500 × 10−6 s) = 9.00 × 10 −10 C

E=

σ P0

=

q 9.00 × 10−10 C = = 2.03 × 105 V/m −12 P0 A (8.854 × 10 C 2 / N ⋅ m 2 )(5.00 × 10−4 m 2 )

V = Ed = (2.03 × 105 V/m)(2.00 × 10−3 m) = 406 V (b) E = q / P0 A

dE dq / dt iC 1.80 × 10−3 A = = = = 4.07 × 1011 V/m ⋅ s −12 dt P0 A P0 A (8.854 × 10 C 2 / N ⋅ m 2 )(5.00 × 10−4 m 2 ) Since iC is constant dE/dt does not vary in time. dE (Eq.(29.16)), with P replaced by P0 since there is vacuum between the plates.) dt jD = (8.854 × 10−12 C2 / N ⋅ m 2 )(4.07 × 1011 V/m ⋅ s) = 3.60 A/m 2

(c) jD = P0

iD = jD A = (3.60 A/m 2 )(5.00 × 10−4 m 2 ) = 1.80 × 10−3 A; iD = iC EVALUATE:

iC = iD . The constant conduction current means the charge q on the plates and the electric field

between them both increase linearly with time and iD is constant. 29.39.

IDENTIFY:

Ohm’s law relates the current in the wire to the electric field in the wire. jD = P

dE . Use Eq.(29.15) to dt

calculate the magnetic fields. SET UP: Ohm’s law says E = ρ J . Apply Ohm’s law to a circular path of radius r.

ρI

(2.0 × 10−8 Ω ⋅ m)(16 A) = 0.15 V/m. 2.1 × 10−6 m 2 A dE d ⎛ ρI ⎞ ρ dI 2.0 × 10−8 Ω ⋅ m (4000 A/s) = 38 V/m ⋅ s. = ⎜ ⎟= = (b) 2.1 × 10−6 m 2 dt dt ⎝ A ⎠ A dt dE (c) jD = P0 = P0 (38 V/m ⋅ s) = 3.4 × 10−10 A/m 2 . dt (d) iD = jD A = (3.4 × 10−10 A/m 2 )(2.1 × 10−6 m 2 ) = 7.14 × 10−16 A. Eq.(29.15) applied to a circular path of radius r EXECUTE:

(a) E = ρ J =

=

μ0 I D μ0 (7.14 × 10−16 A) = = 2.38 × 10−21 T, and this is a negligible contribution. 2π r 2π (0.060 m) μI μ (16 A) BC = 0 C = 0 = 5.33 × 10−5 T. 2π r 2π (0.060 m)

gives BD =

29.40.

EVALUATE: In this situation the displacement current is much less than the conduction current. IDENTIFY: Apply Ampere's law to a circular path of radius r < R, where R is the radius of the wire. SET UP: The path is shown in Figure 29.40. !

!

ú B ⋅ dl = μ Figure 29.40

0

dΦE ⎞ ⎛ ⎜ I C + P0 ⎟ dt ⎠ ⎝

29-12

Chapter 29

EXECUTE:

There is no displacement current, so

!

!

úB ⋅ dl = μ I

0 C

The magnetic field inside the superconducting material is zero, so

29.41.

29.42.

!

!

úB ⋅ dl = 0. But then Ampere’s law says that

I C = 0; there can be no conduction current through the path. This same argument applies to any circular path with r < R, so all the current must be at the surface of the wire. EVALUATE: If the current were uniformly spread over the wire’s cross section, the magnetic field would be like that calculated in Example 28.9. IDENTIFY: A superconducting region has zero resistance. SET UP: If the superconducting and normal regions each lie along the length of the cylinder, they provide parallel conducting paths. EXECUTE: Unless some of the regions with resistance completely fill a cross-sectional area of a long type-II superconducting wire, there will still be no total resistance. The regions of no resistance provide the path for the current. EVALUATE: The situation here is like two resistors in parallel, where one has zero resistance and the other is nonzero. The equivalent resistance is!zero. ! ! IDENTIFY: Apply Eq.(28.29): B = B0 + μ0 M . SET UP: For magnetic fields less than the critical field, there is no internal magnetic field. For fields greater than ! ! the critical field, B is very nearly equal to B0 . ! EXECUTE: (a) The external field is less than the critical field, so inside the superconductor B = 0 and ! ! ! ! ! B (0.130 T ) iˆ M=− 0 =− = −(1.03 × 105 A/m) iˆ. Outside the superconductor, B = B0 = (0.130 T )iˆ and M = 0. μ0 μ0 ! ! (b) The field is greater than the critical field and B = B = (0.260 T)iˆ, both inside and outside the superconductor. 0

29.43.

29.44.

EVALUATE: Below!the !critical field the external field is expelled from the superconducting material. ! IDENTIFY: Apply B = B0 + μ0 M . ! SET UP: When the magnetic flux is expelled from the material the magnetic field B in the material is zero. When the material is completely normal,!the magnetization is close to zero. ! EXECUTE: (a) When B0 is just under Bc1 (threshold of superconducting phase), the magnetic field in the ! ! (55 × 10 −3 T)iˆ Bc1 material must be zero, and M = − =− = −(4.38 × 10 4 A/m) iˆ. μ0 μ0 ! ! ! ! (b) When B0 is just over Bc2 (threshold of normal phase), there is zero magnetization, and B = Bc2 = (15.0 T)iˆ. EVALUATE: Between Bc1 and Bc2 there are filaments of normal phase material and there is magnetic field along these filaments. IDENTIFY and SET UP: Use Faraday’s law to calculate the magnitude of the induced emf and Lenz’s law to determine its direction. Apply Ohm’s law to calculate I. Use Eq.(25.10) to calculate the resistance of the coil. ! (a) EXECUTE: The angle φ between the normal to the coil and the direction of B is 30.0°. dΦB E = = ( N π r 2 )( dB / dt ) and I = E / R. dt For t < 0 and t > 1.00 s, dB/dt = 0 E = 0 and I = 0.

For 0 ≤ t ≤ 1.00 s, dB/dt = (0.120 T) π sin π t E = ( Nπ r 2 )π (0.120 T)sin π t = (0.9475 V)sin π t

ρL = ; ρ = 1.72 × 10 −8 Ω ⋅ m, r = 0.0150 × 10 −3 m A π r2 L = Nc = N 2π r = (500)(2π )(0.0400 m) = 125.7 m Rw = 3058 Ω and the total resistance of the circuit is R = 3058 Ω + 600 Ω = 3658 Ω I = E / R = (0.259 mA)sin π t. The graph of I versus t is sketched in Figure 29.44a.

R for wire: Rw =

ρL

Figure 29.44a


29-13

(b) The coil and the magnetic field are shown in Figure 29.44b.

B increasing so Φ B is " and increasing. Φ B is ⊗ so I is clockwise Figure 29.44b

29.45.

EVALUATE: The long length of small diameter wire used to make the coil has a rather large resistance, larger than the resistance of the 600-Ω resistor connected to it in the circuit. The flux has a cosine time dependence so the rate of change of flux and the current have a sine time dependence. There is no induced current for t < 0 or t > 1.00 s. IDENTIFY: Apply Faraday’s law and Lenz’s law. V SET UP: For a discharging RC circuit, i (t ) = 0 e − t / RC , where V0 is the initial voltage across the capacitor. The R resistance of the small loop is (25)(0.600 m)(1.0 Ω/m) = 15.0 Ω . EXECUTE:

(a) The large circuit is an RC circuit with a time constant of τ = RC = (10 Ω)(20 ×10−6 F) = 200 μs. Thus,

the current as a function of time is i = ( (100 V) /(10 Ω) ) e− t / 200 μ s . At t = 200 μs, we obtain i = (10 A)(e−1 ) = 3.7 A. (b) Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the c + a μ ib μ ib ⎛ a ⎞ 0 dr = 0 ln ⎜1 + ⎟. Therefore, small loop and referring to the solution of Exercise 29.7 we obtain Φ B = ∫ c 2π r 2π ⎝ c ⎠

the emf induced in the small loop at t = 200 μs is E = −

dΦ μ b ⎛ a ⎞ di = − 0 ln ⎜1 + ⎟ . 2π ⎝ c ⎠ dt dt

(4π × 10−7 Wb/A ⋅ m 2 )(0.200 m) 3.7 A ⎞ ⎛ ln(3.0) ⎜ − = +0.81 mV. Thus, the induced current in the small −6 ⎟ 2π ⎝ 200 × 10 s ⎠ loop is i′ = E = 0.81 mV = 54μ A. R 15.0 Ω (c) The magnetic field from the large loop is directed out of the page within the small loop.The induced current will act to oppose the decrease in flux from the large loop. Thus, the induced current flows counterclockwise. EVALUATE: (d) Three of the wires in the large loop are too far away to make a significant contribution to the flux in the small loop–as can be seen by comparing the distance c to the dimensions of the large loop. IDENTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. dΦB SET UP: Faraday’s law says that the induced emf is E = − and the magnetic flux through a coil is defined dt as Φ B = BA cos φ . E=−

29.46.

EXECUTE: In this case, Φ B = BA, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure 29.46. EVALUATE: It is the rate at which the magnetic field is changing, not the field’s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero.

Figure 29.46 29.47.

IDENTIFY: Follow the steps specified in the problem. SET UP: Let the flux through the loop due to the current be positive. μi μ iπ a EXECUTE: (a) Φ B = BA = 0 π a 2 = 0 . 2a 2

29-14

Chapter 29

(b) E = −

dΦB d ⎛ μ iπ a ⎞ μ0π a di di 2R = iR ⇒ − ⎜ 0 = iR ⇒ = −i ⎟=− dt dt ⎝ 2 ⎠ dt 2 dt μ0π a

(c) Solving

di 2R = − dt for i (t ) yields i (t ) = i0e −t (2 R/μ0π a ) . i μ0π a

(d) We want i (t ) = i0 (0.010) = i0e− t (2 R/μ0π a ) , so ln(0.010) = −t (2 R/μ0π a ) and

29.48.

μ0π a

μ0π (0.50 m)

ln(0.010) = 4.55 × 10−5 s. 2(0.10 Ω) EVALUATE: (e) We can ignore the self-induced currents because it takes only a very short time for them to die out. IDENTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. dΦB SET UP: Faraday’s law says that the induced emf is E = − and the magnetic flux through a coil is defined dt as Φ B = BA cos φ . t=−

2R

ln(0.010) = −

EXECUTE: In this case, Φ B = BA, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure 29.48. EVALUATE: It is the rate at which the magnetic field is changing, not the field’s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero.

Figure 29.48 29.49.

dΦB (a) IDENTIFY: (i) E = . The flux is changing because the magnitude of the magnetic field of the wire decreases dt

with distance from the wire. Find the flux through a narrow strip of area and integrate over the loop to find the total flux. SET UP:

Consider a narrow strip of width dx and a distance x from the long wire, as shown in Figure 29.49a. The magnetic field of the wire at the strip is B = μ0 I/2π x. The flux through the strip is d Φ B = Bb dx = ( μ0 Ib/2π )(dx/x)

Figure 29.49a

⎛ μ Ib ⎞ r + a dx The total flux through the loop is Φ B = ∫ dΦ B = ⎜ 0 ⎟ ∫ ⎝ 2π ⎠ r x ⎛ μ Ib ⎞ ⎛ r + a ⎞ Φ B = ⎜ 0 ⎟ ln ⎜ ⎟ ⎝ 2π ⎠ ⎝ r ⎠ ⎞ d Φ B d Φ B dr μ0 Ib ⎛ a = = ⎜− ⎟v 2π ⎜⎝ r ( r + a ) ⎟⎠ dt dt dt μ0 Iabv E = 2π r ( r + a )

EXECUTE:

(ii) IDENTIFY: E = Bvl for a bar of length l moving at speed v perpendicular to a magnetic field B. Calculate the induced emf in each side of the loop, and combine the emfs according to their polarity.


SET UP:

29-15

The four segments of the loop are shown in Figure 29.49b.

The emf in each side ⎛μI⎞ of the loop is E1 = ⎜ 0 ⎟ vb, ⎝ 2π r ⎠ EXECUTE:

⎛ ⎞ μ0 I E2 = ⎜ ⎟ vb, E2 = E4 = 0 + 2 r ( r a ) π ⎝ ⎠

Figure 29.49b

Both emfs E1 and E2 are directed toward the top of the loop so oppose each other. The net emf is E = E1 − E2 =

μ0 Ivb ⎛ 1 1 ⎞ μ0 Iabv ⎜ − ⎟= 2π ⎝ r r + a ⎠ 2π r (r + a )

This expression agrees with what was obtained in (i) using Faraday’s law. (b) (i) IDENTIFY and SET UP: The flux of the induced current opposes the change in flux. ! EXECUTE: B is ⊗ . Φ B is ⊗ and decreasing, so the flux Φ ind of the induced current is ⊗ and the current is clockwise. (ii) IDENTIFY and SET UP: Use the right-hand rule to find the force on the positive charges in each side of the loop. The forces on positive charges in segments 1 and 2 of the loop are shown in Figure 29.49c.

Figure 29.49c

29.50.

29.51.

EXECUTE: B is larger at segment 1 since it is closer to the long wire, so FB is larger in segment 1 and the induced current in the loop is clockwise. This agrees with the direction deduced in (i) using Lenz’s law. (c) EVALUATE: When v = 0 the induced emf should be zero; the expression in part (a) gives this. When a → 0 the flux goes to zero and the emf should approach zero; the expression in part (a) gives this. When r → ∞ the magnetic field through the loop goes to zero and the emf should go to zero; the expression in part (a) gives this. IDENTIFY: Apply Faraday’s law. SET UP: For rotation about the y-axis the situation is the same as in Examples 29.4 and 29.5 and we can apply the results from those examples. EXECUTE: (a) Rotating about the y-axis: the flux is given by Φ B = BA cosφ and dΦB Emax = = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V. dt dΦB (b) Rotating about the x-axis: = 0 and E = 0. dt (c) Rotating about the z-axis: the flux is given by Φ B = BA cosφ and dΦB Emax = = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V. dt EVALUATE: The maximum emf is the same if the loop is rotated about an edge parallel to the z-axis as it is when it is rotated about the z-axis. IDENTIFY: Apply the results of Example 29.4, so Emax = Nω BA for N loops. SET UP: For the minimum ω , let the rotating loop have an area equal to the area of the uniform magnetic field, so A = (0.100 m) 2 . EXECUTE: N = 400 , B = 1.5 T , A = (0.100 m) 2 and Emax = 120 V gives ω = Emax /NBA = (20 rad/s)(1 rev/2π rad)(60 s/1 min) = 190 rpm. EVALUATE:

In Emax = ω BA, ω is in rad/s.

29-16

Chapter 29

29.52.

IDENTIFY: SET UP:

Apply the results of Example 29.4, generalized to N loops: Emax = N ω BA. v = rω . In the expression for Emax , ω must be in rad/s. 30 rpm = 3.14 rad/s

E0 9.0 V = = 18 m 2 . ω NB (3.14 rad/s)(2000 turns)(8.0 × 10−5 T) (b) Assuming a point on the coil at maximum distance from the axis of rotation we have EXECUTE:

(a) Solving for A we obtain A =

A 18 m 2 ω= (3.14 rad/s) = 7.5 m s. π π EVALUATE: The device is not very feasible. The coil would need a rigid frame and the effects of air resistance would be appreciable. ΔΦ B IDENTIFY: Apply Faraday’s law in the form Eav = − N to calculate the average emf. Apply Lenz’s law to Δt calculate the direction of the induced current. SET UP: Φ B = BA . The flux changes because the area of the loop changes. v = rω =

29.53.

EXECUTE:

29.54.

(a) Eav =

ΔΦ B ΔA π r2 π (0.0650/2 m) 2 =B =B = (0.950 T) = 0.0126 V. Δt Δt Δt 0.250 s

(b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore the current flows from point a through the resistor to point b . ! EVALUATE: Faraday’s law can be used to find the direction of the induced current. Let A be into the page. Then Φ B is positive and decreasing in magnitude, so d Φ B /dt < 0. Therefore E > 0 and the induced current is clockwise around the loop. IDENTIFY: By Lenz’s law, the induced current flows to oppose the flux change that caused it. SET UP: When the switch is suddenly closed with an uncharged capacitor, the current in the outer circuit immediately increases from zero to its maximum value. As the capacitor gets charged, the current in the outer circuit gradually decreases to zero. EXECUTE: (a) (i) The current in the outer circuit is suddenly increasing and is in a counterclockwise direction. The magnetic field through the inner circuit is out of the paper and increasing. The magnetic flux through the inner circuit is increasing, so the induced current in the inner circuit is clockwise (a to b) to oppose the flux increase. (ii) The current in the outer circuit is still counterclockwise but is now decreasing, so the magnetic field through the inner circuit is out of the page but decreasing. The flux through the inner circuit is now decreasing, so the induced current is counterclockwise (b to a) to oppose the flux decrease. (b) The graph is sketched in Figure 29.54. EVALUATE: Even though the current in the outer circuit does not change direction, the current in the inner circuit does as the flux through it changes from increasing to decreasing.

Figure 29.54 29.55.

IDENTIFY: Use Faraday’s law to calculate the induced emf and Ohm’s law to find the induced current. Use Eq.(27.19) to calculate the magnetic force FI on the induced current. Use the net force F − FI in Newton’s 2nd law to calculate the acceleration of the rod and use that to describe its motion.


(a) SET UP:

29-17

The forces in the rod are shown in Figure 29.55a. EXECUTE:

I=

E =

dΦB = BLv dt

BLv R

Figure 29.55a

! dΦB to find the direction of I: Let A be into the page. Then Φ B > 0. The area of the circuit is dt ! dΦB > 0. Then E < 0 and with our direction for A this means that E and I are counterclockwise, increasing, so dt ! ! ! ! as shown in the sketch. The force FI on the rod due to the induced current is given by FI = Il × B. This gives FI ! to the left with magnitude FI = ILB = ( BLv / R) LB = B 2 L2v / R. Note that FI is directed to oppose the motion of the rod, as required by Lenz’s law. EVALUATE: The net force on the rod is F − FI , so its acceleration is a = ( F − FI ) / m = ( F − B 2 L2v / R) / m. The rod starts with v = 0 and a = F/m. As the speed v increases the acceleration a decreases. When a = 0 the rod has reached its terminal speed vt . The graph of v versus t is sketched in Figure 29.55b.

Use E = −

(Recall that a is the slope of the tangent to the v versus t curve.)

Figure 29.55b (b) EXECUTE: EVALUATE: 29.56.

F − B 2 L2vt / R RF = 0 and vt = 2 2 . m B L A large F produces a large vt . If B is larger, or R is smaller, the induced current is larger at a given v v = vt when a = 0 so

so FI is larger and the terminal speed is less. IDENTIFY: Apply Newton’s 2nd law to the bar. The bar will experience a magnetic force due to the induced current in the loop. Use a = dv / dt to solve for v. At the terminal speed, a = 0 . SET UP: The induced emf in the loop has a magnitude BLv . The induced emf is counterclockwise, so it opposes the voltage of the battery, E . EXECUTE: (a) The net current in the loop is I = E − BLv . The acceleration of the bar is R ILB sin(90 ° ) ( E − BLv ) LB (E − BLv) LB = a= F = . To find v(t ) , set dv = a = and solve for v using the method m m mR dt mR of separation of variables:

∫

v

0

t LB 2 2 dv E = (1 − e− B L t /mR) = (10 m/s)(1 − e −t /3.1 s ) dt → v = (E − BLv) ∫ 0 mR BL

The graph of v versus t is sketched in Figure 29.56. Note that the graph of this function is similar in appearance to that of a charging capacitor. (b) Just after the switch is closed, v = 0 and I = E/R = 2.4 A, F = ILB = 2.88 N and a = F/m = 3.2 m/s 2 . [12 V − (1.5 T)(0.8 m)(2.0 m/s)](0.8 m)(1.5 T) = 2.6 m/s 2 . (0.90 kg)(5.0 Ω) (d) Note that as the speed increases, the acceleration decreases. The speed will asymptotically approach the 12 V = 10 m/s, which makes the acceleration zero. terminal speed E = BL (1.5 T)(0.8 m) (c) When v = 2.0 m/s, a =

29-18

Chapter 29

EVALUATE: The current in the circuit is counterclockwise and the magnetic force on the bar is to the right. The energy that appears as kinetic energy of the moving bar is supplied by the battery.

29.57.

29.58.

29.59.

Figure 29.56 ! ! IDENTIFY: Apply E = BvL. Use ∑ F = ma applied to the satellite motion to find the speed v of the satellite. mm SET UP: The gravitational force on the satellite is Fg = G 2 E , where m is the mass of the satellite and r is the r radius of its orbit. mm v2 GmE EXECUTE: B = 8.0 ×10−5 T, L = 2.0 m. G 2 E = m and r = 400 ×103 m + RE gives v = = 7.665 ×103 m/s. r r r Using this v in E = vBL gives E = (8.0 × 10−5 T)(7.665 × 103 m/s)(2.0 m) = 1.2 V. EVALUATE: The induced emf is large enough to be measured easily. IDENTIFY: The induced emf is E = BvL, where L is measured in a direction that is perpendicular to both the magnetic field and the velocity of the bar. SET UP: The magnetic force pushed positive charge toward the high potential end of the bullet. EXECUTE: (a) E = BLv = (8 × 10−5 T)(0.004 m)(300 m/s) = 96 μ V. Since a positive charge moving to the east would be deflected upward, the top of the!bullet will be at a higher potential. ! (b) For a bullet that travels south, v and B are along the same line, there is no magnetic force and the induced emf is zero. ! (c) If v is horizontal, the magnetic force on positive charges in the bullet is either upward or downward, perpendicular to the line between the front and back of the bullet. There is no emf induced between the front and back of the bullet. EVALUATE: Since the velocity of a bullet is always in the direction from the back to the front of the bullet, and since the magnetic force is perpendicular to the velocity, there is never an induced emf between the front and back of the bullet, no matter what the direction of the magnetic field is. IDENTIFY: Find the magnetic field at a distance r from the center of the wire. Divide the rectangle into narrow strips of width dr, find the flux through each strip and integrate to find the total flux. SET UP: Example 28.8 uses Ampere’s law to show that the magnetic field inside the wire, a distance r from the axis, is B (r ) = μ0 Ir 2π R 2 . EXECUTE: Consider a small strip of length W and width dr that is a distance r from the axis of the wire, as shown μ IW in Figure 29.59. The flux through the strip is d Φ B = B ( r )W dr = 0 2 r dr . The total flux through the rectangle is 2π R μ0 IW ⎛ μ0 IW ⎞ R ΦB = ∫ dΦB = ⎜ r dr = . 2 ⎟∫ 4π ⎝ 2π R ⎠ 0 EVALUATE: Note that the result is independent of the radius R of the wire.

Figure 29.59


29.60.

29-19

IDENTIFY: Apply ! Faraday’s law to calculate the magnitude and direction of the induced emf. SET UP: Let A be directed out of the page in Figure 29.50 in the textbook. This means that counterclockwise emf is positive. EXECUTE: (a) Φ B = BA = B0πr0 2 (1 − 3(t t0 ) 2 + 2(t t0 )3 ). 2 dΦB d B πr 2 6 B0πr0 2 ⎛ ⎛ t ⎞ ⎛ t ⎞ ⎞ = ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ . At = − B0πr0 2 (1 − 3(t/t0 ) 2 + 2(t/t0 )3 ) = − 0 0 ( −6(t/t0 ) + 6(t/t0 ) 2 ). E = − ⎜ ⎝ t0 ⎠ ⎝ t0 ⎠ ⎟ t0 dt dt t0 ⎝ ⎠ 2 −3 −3 2 ⎛ ⎞ 6 B π (0.0420 m) ⎛ 5.0 × 10 s ⎞ ⎛ 5.0 × 10 s ⎞ ⎜ ⎟ = 0.0665 V. E is positive so it is − t = 5.0 × 10−3 s, E = − 0 ⎜ ⎜⎝ 0.010 s ⎟⎠ ⎜⎝ 0.010 s ⎟⎠ ⎟ 0.010 s ⎝ ⎠ counterclockwise. 0.0665 V E E ⇒ Rtotal = r + R = ⇒ r = − 12 Ω = 10.2 Ω. (c) I = 3.0 × 10−3 A Rtotal I (d) Evaluating the emf at t = 1.21 × 10−2 s and using the equations of part (b), E = −0.0676 V, and the current flows clockwise, from b to a through the resistor. ⎛ ⎛ t ⎞2 ⎛ t ⎞ ⎞ t (e) E = 0 when 0 = ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ . 1 = and t = t0 = 0.010 s. ⎜ ⎝ t 0 ⎠ ⎝ t0 ⎠ ⎟ t0 ⎝ ⎠ ! EVALUATE: At t = t0 , B = 0 . At t = 5.00 × 10 −3 s , B is in the + kˆ direction and is decreasing in magnitude. Lenz’s ! law therefore says E is counterclockwise. At t = 0.0121 s , B is in the + kˆ direction and is increasing in magnitude.

(b) E = −

29.61.

Lenz’s law therefore says E is clockwise. These results for the direction of E agree with the results we obtained from Faraday’s law. (a) and (b) IDENTIFY and Set Up: The magnetic field of the wire is given by μI B = 0 and varies along the length of the 2π r ! bar. At every point along the bar B has direction into the page. Divide the bar up into thin slices, as shown in Figure 29.61a. Figure 29.61a ! ! ! ! ! The emf dE induced in each slice is given by d E = v × B ⋅ dl . v × B is directed toward the wire, so ⎛μI⎞ d E = −vB dr = −v ⎜ 0 ⎟ dr. The total emf induced in the bar is ⎝ 2π r ⎠ b d + L ⎛ μ Iv ⎞ μ Iv d + L dr μ Iv d +L Vba = ∫ d E = − ∫ ⎜ 0 ⎟ dr = − 0 ∫ = − 0 [ ln( r ) ]d a d d r 2π 2π ⎝ 2π r ⎠ μ0 Iv μ0 Iv Vba = − (ln( d + L ) − ln(d )) = − ln(1 + L/d ) 2π 2π EVALUATE: The minus sign means that Vba is negative, point a is at higher potential than point b. (The force ! ! ! F = qv × B on positive charge carriers in the bar is towards a, so a is at higher potential.) The potential difference increases when I or v increase, or d decreases. (c) IDENTIFY: Use Faraday’s law to calculate the induced emf. SET UP: The wire and loop are sketched in Figure 29.61b. EXECUTE:

EXECUTE: As the loop moves to the right the magnetic flux through it doesn’t change. Thus dΦB E =− = 0 and I = 0. dt Figure 29.61b EVALUATE: This result can also be understood as follows. The induced emf in section ab puts point a at higher potential; the induced emf in section dc puts point d at higher potential. If you travel around the loop then these two induced emf’s sum to zero. There is no emf in the loop and hence no current.

29-20

Chapter 29

29.62.

IDENTIFY: E = vBL, where v is the component of velocity perpendicular to the field direction and perpendicular to the bar. SET UP: Wires A and C have a length of 0.500 m and wire D has a length of 2(0.500 m) 2 = 0.707 m. ! ! EXECUTE: Wire A: v is parallel to B , so the induced emf is zero. ! ! ! Wire C: v is perpendicular to B. The component of v perpendicular to the bar is v cos 45° . E = (0.350 m/s)(cos 45°)(0.120 T)(0.500 m) = 0.0148 V. ! ! ! Wire D: v is perpendicular to B. The component of v perpendicular to the bar is v cos 45° . E = (0.350 m/s)(cos 45°)(0.120 T)(0.707 m) = 0.0210 V. ! ! ! EVALUATE: The induced emf depends on the angle between v and B and also on the angle between v and the bar. (a) IDENTIFY: Use the expression for motional emf to calculate the emf induced in the rod. SET UP: The rotating rod is shown in Figure 29.63a.

29.63.

The emf induced !in a!thin ! slice is d E = v × B ⋅ dl .

Figure 29.63a ! ! ! EXECUTE: Assume that B is directed out of the page. Then v × B is directed radially outward and ! ! ! dl = dr , so v × B ⋅ dl = vB dr v = rω so d E = ω Br dr. The d E for all the thin slices that make up the rod are in series so they add: L

E = ∫ d E = ∫ ω Br dr = 12 ω BL2 = 12 (8.80 rad/s)(0.650 T)(0.240 m) 2 = 0.165 V 0

EVALUATE: E increases with ω , B or L2 . (b) No current flows so there is no IR drop in potential. Thus the potential difference between the ends equals the emf of 0.165 V calculated in part (a). (c) SET UP: The rotating rod is shown in Figure 29.63b.

Figure 29.63b EXECUTE: The emf between the center of the rod and each end is E = 12 ω B( L / 2) 2 = 14 (0.165 V) = 0.0412 V, with the direction of the emf from the center of the rod toward each end. The emfs in each half of the rod thus oppose each other and there is no net emf between the ends of the rod. EVALUATE: ω and B are the same as in part (a) but L of each half is 12 L for the whole rod. E is proportional to 29.64.

L2 , so is smaller by a factor of 14 . IDENTIFY: The power applied by the person in moving the bar equals the rate at which the electrical energy is dissipated in the resistance. ( BLv) 2 SET UP: From Example 29.7, the power required to keep the bar moving at a constant velocity is P = . R EXECUTE:

(a) R =

( BLv) 2 [(0.25 T)(3.0 m)(2.0 m s)]2 = = 0.090 Ω. P 25 W

(b) For a 50 W power dissipation we would require that the resistance be decreased to half the previous value. (c) Using the resistance from part (a) and a bar length of 0.20 m, ( BLv) 2 [(0.25 T)(0.20 m)(2.0 m s)]2 = = 0.11 W . P= R 0.090 Ω EVALUATE: When the bar is moving to the right the magnetic force on the bar is to the left and an applied force directed to the right is required to maintain constant speed. When the bar is moving to the left the magnetic force on the bar is to the right and an applied force directed to the left is required to maintain constant speed.


29.65.

29-21

(a) IDENTIFY: Use Faraday’s law to calculate the induced emf, Ohm’s law to calculate I, and Eq.(27.19) to calculate the force on the rod due to the induced current. SET UP: The force on the wire is shown in Figure 29.65. EXECUTE: When the wire has speed v the induced emf is E = Bva and the Bva induced current is I = E / R = R Figure 29.65

! ! ! The induced current flows upward in the wire as shown, so the force F = Il × B exerted by the magnetic field on ! the induced current is to the left. F opposes the motion of the wire, as it must by Lenz’s law. The magnitude of the force is F = IaB = B 2 a 2v / R. ! ! (b) Apply ∑ F = ma to the wire. Take +x to be toward the right and let the origin be at the location of the wire at

t = 0, so x0 = 0.

∑F

x

= max says − F = max

F B 2 a 2v =− m mR Use this expression to solve for v(t): dv B 2 a 2v dv B2a 2 =− and =− ax = dt dt mR v mR v dv′ B2a 2 t = − dt ′ ∫v0 v′ mR ∫ 0 B 2 a 2t ln(v) − ln(v0 ) = − mR ⎛v⎞ 2 2 B 2 a 2t ln ⎜ ⎟ = − and v = v0e − B a t / mR mR ⎝ v0 ⎠ ax = −

Note: At t = 0, v = v0 and v → 0 when t → ∞ Now solve for x(t): 2 2 2 2 dx = v0e − B a t / mR so dx = v0e − B a t / mR dt v= dt

∫

x 0

t

dx′ = ∫ v0e − B 0

2 a 2 t /mR

dt ′

(

)

t 2 2 2 2 mRv ⎛ mR ⎞ x = v0 ⎜ − 2 2 ⎟ ⎡e − B a t ′/mR ⎤ = 2 20 1 − e − B a t / mR ⎣ ⎦ 0 Ba ⎝ Ba ⎠ Comes to rest implies v = 0. This happens when t → ∞. mRv t → ∞ gives x = 2 20 . Thus this is the distance the wire travels before coming to rest. Ba EVALUATE: The motion of the slide wire causes an induced emf and current. The magnetic force on the induced current opposes the motion of the wire and eventually brings it to rest. The force and acceleration depend on v and are constant. If the acceleration were constant, not changing from its initial value of ax = − B 2 a 2v0 / mR, then the

29.66.

stopping distance would be x = −v02 / 2ax = mRv0 / 2 B 2 a 2 . The actual stopping distance is twice this. ! ! ! IDENTIFY: Since the bar is straight and the magnetic field is uniform, integrating dε = v × B ⋅ dl along the length ! ! ! of the bar gives E = (v × B) ⋅ L ! ! SET UP: v = (4.20 m/s)iˆ . L = (0.250 m)(cos36.9°iˆ + sin 36.9° ˆj ). ! ! ! ! EXECUTE: (a) E = (v × B ) ⋅ L = (4.20 m/s)iˆ × ((0.120 T)iˆ − ( 0.220 T ) ˆj − (0.0900 T)kˆ ) ⋅ L.

(

)(

)

E = ( 0.378 V/m ) ˆj – ( 0.924 V/m ) kˆ ⋅ (0.250 m)(cos 36.9°iˆ + sin 36.9° ˆj ) . E = (0.378 V/m)(0.250 m)sin 36.9° = 0.0567 V. (b) The higher potential end is the end to which positive charges in the rod are pushed by the magnetic force. ! ! v × B has a positive y-component, so the end of the rod marked + in Figure 29.66 is at higher potential.

29-22

Chapter 29

! ! ! EVALUATE: Since v × B has nonzero ˆj and kˆ components, and L has nonzero iˆ and ˆj components, only the ! kˆ component of B contributes to E . In fact, E = vx Bz Ly = (4.20 m/s)(0.0900 T)(0.250 m)sin 36.9° = 0.0567 V.

Figure 29.66 29.67.

IDENTIFY: SET UP:

! ! Use Eq.(29.10) to calculate the induced electric field at each point and then use F = qE . !

!

dΦB to a concentric circle of dt ! radius r, as shown in Figure 29.67a. Take A to ! be into the page, in the direction of B.

Apply

ú E ⋅ dl = −

Figure 29.67a ! ! dΦB > 0, so úE ⋅ dl is negative. This means that E is tangent to the circle in dt the counterclockwise direction, as shown in Figure 29.67b.

EXECUTE: B increasing then gives

!

!

úE ⋅ dl = − E (2π r ) dΦB dB = π r2 dt dt

Figure 29.67b dB dB so E = 12 r dt dt The induced electric field and the force on q are shown in Figure 29.67c. dB F = qE = 12 qr dt ! F is to the left ! ! ( F is in the same direction as E since q is positive.) Figure 29.67c

− E (2π r ) = −π r 2

point a

point b The induced electric field and the force on q are shown in Figure 29.67d. dB F = qE = 12 qr dt ! F is toward the top of the page.

Figure 29.67d

29.68.

point c r = 0 here, so E = 0 and F = 0. EVALUATE: If there were a concentric conducting ring of radius r in the magnetic field region, Lenz’s law tells us that the increasing magnetic field would induce a counterclockwise current in the ring. This agrees with the direction of the force we calculated for the individual positive point charges. IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends. The propeller acts as such a bar. SET UP: Different parts of the propeller are moving at different speeds, so we must integrate to get the total induced emf. The potential induced across an element of length dx is d E = vBdx, where B is uniform.


29-23

EXECUTE: (a) Call x the distance from the center to an element of length dx, and L the length of the propeller. The speed of dx is xω, giving d E = vBdx = xω Bdx. E = ∫

L/ 2 0

xω Bdx = ω BL2 /8.

(b) The potential difference is zero since the potential is the same at both ends of the propeller. (2.0 m)3 ⎛ 220 rev ⎞ −4 (c) E = 2π ⎜ =5.8 × 10−4 V = 0.58 mV ⎟ ( 0.50 × 10 T ) 60 s 8 ⎝ ⎠ 29.69.

EVALUATE: A potential difference of about 12 mV is not large enough to be concerned about in a propeller. IDENTIFY: Follow the steps specified in the problem. SET UP: The electric field region is sketched in Figure 29.69. ! ! ! ! ! ! ! dΦB dΦB EXECUTE: úE ⋅ dl = − . If B is constant then = 0, so úE ⋅ dl = 0. ∫ E ⋅ dl = Eab L − Ecd L = 0. But abcda dt dt Ecd = 0 , so Eab L = 0. But since we assumed Eab ≠ 0, this contradicts Faraday’s law. Thus, we can’t have a uniform electric field abruptly drop to zero in a region in which the magnetic field is constant. ! ! EVALUATE: If the magnetic field in the region is constant, then the integral úE ⋅ dl must be zero.

Figure 26.69 29.70.

IDENTIFY and SET UP: At the terminal speed vt , the upward force FI exerted on the loop due to the induced current equals the downward force of gravity: FI = mg . Use Eq.(29.6) to find the induced emf in the side of the loop that is totally within the magnetic field. There is no induced emf in the other sides of the loop. EXECUTE: E = Bvs, I = Bvs / R and FI = IsB − B 2 s 2v / R B 2 s 2vt mgR = mg and vt = 2 2 R Bs m = ρ mV = ρ m (4s )π ( d / 2) 2 = ρ mπ d 2 R=

ρL A

=

ρ R 4 s 16 ρ R s = πd2 πd2

1 4

Using these expressions for m and R gives vt = 16 ρ m ρ R g / B 2

EVALUATE: 29.71.

We know ρ m = 8900 kg/m3 (Table 14.1) and ρ R = 1.72 × 10−8 Ω ⋅ m

(Table 25.1). Taking B = 0.5 T gives vt = 9.6 cm/s. IDENTIFY: Follow the steps specified in the problem. SET UP: (a) The magnetic field region is sketched in Figure 29.71. ! ! ! EXECUTE: (b) úB ⋅ dl = 0 (no currents in the region). Using the figure, let B = B0iˆ for y < 0 and B = 0 for y > 0. ! ! ∫ B ⋅ dl = Bab L − Bcd L = 0 but Bcd = 0. Bab L = 0, but Bab ≠ 0. This is a contradiction and violates Ampere’s Law. abcde

EVALUATE: We often describe a magnetic field as being confined to a region, but this result shows that the edges of such a region can't be sharp.

Figure 29.71

29-24

Chapter 29

29.72.

IDENTIFY and SET UP: Apply Ohm’s law to the dielectric to relate the current in the dielectric to the charge on the plates. Use Eq.(25.1) for the current and obtain a differential equation for q(t). Integrate this equation to obtain q(t) and i(t). Use E = q / PA and Eq.(29.16) to calculate jD . EXECUTE: (a) Apply Ohm’s law to the dielectric: The capacitor is sketched in Figure 29.72. v(t ) R q (t ) PA v(t ) = and C = K 0 C d

i (t ) =

Figure 29.72 ⎛ d ⎞ v(t ) = ⎜ ⎟ q (t ) ⎝ K P0 A ⎠ v(t ) ⎛ q (t ) d ⎞ ⎛ A ⎞ q (t ) . But the current i(t) =⎜ ⎟⎜ ⎟= R ⎝ K P0 A ⎠ ⎝ ρ d ⎠ K P0 ρ in the dielectric is related to the rate of change dq/dt of the charge q(t) on the plates by i(t) = –dq/dt (a positive i in the direction from the + to the – plate of the capacitor corresponds to a decrease in the charge). Using this in the above dq ⎛ 1 ⎞ dq dt gives − =− . Integrate both sides of this equation from t = 0, where q = Q0 , to a later =⎜ ⎟ q (t ). q K ρ P0 dt ⎝ K ρ P0 ⎠

The resistance R of the dielectric slab is R = ρ d / A. Thus i (t ) =

⎛ 1 ⎞ t ⎛ q ⎞ dq t and q (t ) = Q0e−t / K ρ P0 . Then = −⎜ ⎟ ∫ 0 dt. ln ⎜ ⎟ = − ρ P ρ P q K Q K 0 0 0 ⎝ ⎠ ⎝ ⎠ dq ⎛ Q0 ⎞ −t / K ρ P0 i (t ) ⎛ Q0 ⎞ −t / K ρ P0 and jC = i (t ) = − . The conduction current flows from the positive to =⎜ =⎜ ⎟e ⎟e dt ⎝ K ρ P0 ⎠ A ⎝ AK ρ P0 ⎠ the negative plate of the capacitor. q (t ) q (t ) = (b) E (t ) = PA K P0 A dE dE dq (t ) / dt i (t ) jD (t ) = P = K P0 = K P0 = − C = − jC (t ) dt dt K P0 A A ! The minus sign means that jD (t ) is directed from the negative to the positive plate. E is from + to – but dE/dt is negative (E decreases) so jD (t ) is from – to +. EVALUATE: There is no conduction current to and from the plates so the concept of displacement current, with ! ! jD = − jC in the dielectric, allows the current to be continuous at the capacitor. IDENTIFY: The conduction current density is related to the electric field by Ohm's law. The displacement current density is related to the rate of change of the electric field by Eq.(29.16). SET UP: dE/dt = ω E0 cos ωt E 0.450 V/m = 1.96 × 10−4 A/m 2 EXECUTE: (a) jC (max) = 0 = ρ 2300 Ω ⋅ m ⎛ dE ⎞ −9 2 (b) jD (max) = P0 ⎜ ⎟ = P0ω E0 = 2π P0 fE0 = 2π P0 (120 Hz)(0.450 V/m) = 3.00 × 10 A/m dt ⎝ ⎠ max E0 1 = ωP0 E0 and ω = = 4.91 × 107 rad/s (c) If jC = jD then ρP0 ρ time t when the charge is q(t).

29.73.

∫

q

Q0

f =

29.74.

ω 4.91× 107 rad s = = 7.82 × 106 Hz. 2π 2π

EVALUATE: (d) The two current densities are out of phase by 90° because one has a sine function and the other has a cosine, so the displacement current leads the conduction current by 90°. IDENTIFY: A current is induced in the loop because of its motion and because of this current the magnetic field exerts a torque on the loop. SET UP: Each side of the loop has mass m / 4 and the center of mass of each side is at the center of each side. The flux through the loop is Φ B = BA cos φ .


29-25

! ! EXECUTE: (a) τ g = ∑ rcm × mg summed over each leg. !

⎛ L ⎞⎛ m ⎞

⎛ L ⎞⎛ m ⎞

⎛m⎞

τ g = ⎜ ⎟⎜ ⎟g sin(90° − φ ) + ⎜ ⎟⎜ ⎟ g sin(90° − φ ) + ( L) ⎜ ⎟ g sin(90° − φ ) ⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 4 ⎠ ⎝4⎠ mgL τg = cos φ (clockwise). 2 ! ! τ B = τ × B = IAB sin φ (counterclockwise). E BA d BA dφ BAω = cos φ = − sin φ = sin φ . The current is going counterclockwise looking to the − kˆ direction. R R dt R dt R mgL B 2 L4ω 2 B 2 A2ω 2 B 2 L4ω 2 Therefore, τ B = sin φ = sin φ . The net torque is τ = cos φ − sin φ , opposite to the 2 R R R direction of the rotation. 5 (b) τ = Iα (I being the moment of inertia). About this axis I = mL2 . Therefore, 12 12 1 ⎡ mgL B 2 L4ω 2 ⎤ 6 g 12 B 2 L2ω 2 α= cos φ − sin φ ⎥ = cos φ − sin φ . ⎢ 5 mL2 ⎣ 2 R 5mR ⎦ 5L I=

29.75.

EVALUATE: (c) The magnetic torque slows down the fall (since it opposes the gravitational torque). (d) Some energy is lost through heat from the resistance of the loop. IDENTIFY: Apply Eq.(29.10). SET UP: Use an integration path that is a circle of radius r. By symmetry the induced electric field is tangent to this path and constant in magnitude at all points on the path. EXECUTE: (a) The induced electric field at these points is shown in Figure 29.75a. (b) To work out the amount of the electric field that is in the direction of the loop at a general position, we will use E E E cosθ = = . Therefore, the geometry shown in Figure 29.75b. Eloop = E cosθ but E = 2π r 2π ( a /cosθ ) 2π a E cos 2 θ π a 2 dB a dB dΦB dB dB π a 2 dB , so Eloop = = . This is exactly the value . But E = =A = π r2 = 2 dt dt dt cos θ dt 2π a dt 2 dt 2π a for a ring, obtained in Exercise 29.30, and has no dependence on the part of the loop we pick. E A dB L2 dB (0.20 m)2 (0.0350 T/s) (c) I = = = = = 7.37 × 10−4 A. R R dt R dt 1.90 Ω 1 1 dB (0.20 m)2 (0.0350 T/s) (d) Eab = E = L2 = = 1.75 × 10−4 V. But there is potential drop V = IR = −1.75 × 10−4 V, 8 8 dt 8 so the potential difference is zero. EVALUATE: The magnitude of the induced emf between any two points equals the magnitudes of the potential drop due to the current through the resistance of that portion of the loop.

Eloop =

Figure 29.75 29.76.

IDENTIFY: Apply Eq.(29.10). SET UP: Use an integration path that is a circle of radius r. By symmetry the induced electric field is tangent to this path and constant in magnitude at all points on the path. EXECUTE: (a) The induced emf at these points is shown in Figure 29.76. (b) The induced emf on the side ac is zero, because the electric field is always perpendicular to the line ac. dΦB dB dB (c) To calculate the total emf in the loop, E = . E = (0.20 m)2 (0.035 T/s) = 1.40 × 10−3 V. =A = L2 dt dt dt

29-26

Chapter 29

E 1.40 ×10−3 V = = 7.37 ×10−4 A R 1.90 Ω (e) Since the loop is uniform, the resistance in length ac is one quarter of the total resistance. Therefore the potential difference between a and c is Vac = IRac = (7.37 ×10−4 A)(1.90 Ω 4) = 3.50 ×10−4 V and the point a is at a higher potential since the current is flowing from a to c. EVALUATE: This loop has the same resistance as the loop in Challenge Problem 29.75 and the induced current is the same.

(d) I =

Figure 29.76 29.77.

IDENTIFY: The motion of the bar produces an induced current and that results in a magnetic force on the bar. ! ! SET UP: FB is perpendicular to B , so is horizontal. The vertical component of the normal force equals mg cosφ , so the horizontal component of the normal force equals mg tan φ . EXECUTE: (a) As the bar starts to slide, the flux is decreasing, so the current flows to increase the flux, which LB LB d Φ B LB dA LB 2 vL2 B 2 E= (vL cos φ ) = cos φ . At the terminal means it flows from a to b. FB = iLB = = B = R R dt R dt R R v L2 B 2 Rmg tan φ speed the horizontal forces balance, so mg tan φ = t cos φ and vt = 2 2 . R L B cos φ E 1 d Φ B 1 dA B vLB cos φ mg tan φ = = B = (vL cos φ ) = = . R R dt R dt R R LB Rm 2 g 2 tan 2 φ (d) P = i 2 R = . L2 B 2 ⎛ Rmg tan φ ⎞ Rm 2 g 2 tan 2 φ (e) Pg = Fv cos(90° − φ ) = mg ⎜ 2 2 . ⎟ sin φ and Pg = L2 B 2 ⎝ L B cos φ ⎠

(c) i =

29.78.

EVALUATE: The power in part (e) equals that in part (d), as is required by conservation of energy. IDENTIFY: Follow the steps indicated in the problem. SET UP: The primary assumption throughout the problem is that the square patch is small enough so that the velocity is constant over its whole area, that is, v = ω r ≈ ω d . E EA ω dBA ! ! EXECUTE: (a) ω → clockwise, B → into page. E = vBL = ω d BL . I = = = . Since v × B points R ρL ρ ω dBLt ! ! flowing radially outward since v × B points outward, A is just the cross-sectional area tL. Therefore, I =

ρ

outward. ! ! ! ω d 2 B 2 L2t ! ! ! (b) τ = d × F and FB = IL × B = ILB pointing counterclockwise. So τ = pointing out of the page (a

ρ

counterclockwise torque opposing the clockwise rotation). ! ! (c) If ω → counterclockwise and B → into page, then I → inward radially since v × B points inward. τ → clockwise (again opposing the motion). If ω → counterclockwise and B → out of the page, then I → radially outward. τ → clockwise (opposing the motion) The magnitudes of I and τ are the same as in part (a). EVALUATE: In each case the magnetic torque due to the induced current opposes the rotation of the disk, as is required by conservation of energy.

30

INDUCTANCE

30.1.


Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 ×10−4 H)(830 A/s) = 0.270 V; yes, it is constant. dt

EXECUTE:

di2 ; M is a property of the pair of coils so is the same as in part (a). Thus E1 = 0.270 V. dt EVALUATE: The induced emf is the same in either case. A constant di/dt produces a constant emf. (b) E1 = M

30.2.

E1 = M

IDENTIFY:

Δi2 Δi NΦ and E2 = M 1 . M = 2 B 2 , where Φ B 2 is the flux through one turn of the second Δt Δt i1

coil. SET UP:

M is the same whether we consider an emf induced in coil 1 or in coil 2. 1.65 ×10−3 V E2 EXECUTE: (a) M = = = 6.82 ×10−3 H = 6.82 mH Δi1 / Δt 0.242 A/s (b) Φ B 2 =

Mi1 (6.82 ×10−3 H)(1.20 A) = = 3.27 × 10−4 Wb N2 25

Δi2 = (6.82 ×10−3 H)(0.360 A/s) = 2.46 ×10−3 V = 2.46 mV Δt EVALUATE: We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil. IDENTIFY: Replace units of Wb, A and Ω by their equivalents. SET UP: 1 Wb = 1 T ⋅ m 2 . 1 T = 1 N/(A ⋅ m). 1 N ⋅ m = 1 J. 1 A = 1 C/s. 1 V = 1 J/C. 1 V/A = 1 Ω. (c) E1 = M

30.3.

30.4.

EXECUTE: EVALUATE: IDENTIFY: (a) SET UP: EXECUTE:

1 H = 1 Wb/A = 1 T ⋅ m 2 /A = 1 N ⋅ m/A 2 = 1 J/A 2 = 1 (J/[A ⋅ C])s = 1 (V/A)s = 1 Ω ⋅ s. We may use whichever equivalent unit is the most convenient in a particular problem. Changing flux from one object induces an emf in another object. The magnetic field due to a solenoid is B = μ 0 nI . The above formula gives B1 =

( 4π ×10

−7

T ⋅ m/A ) (300)(0.120 A)

0.250 m The average flux through each turn of the inner solenoid is therefore

=1.81×10−4 T

Φ B = B1 A = (1.81 × 10−4 T ) π (0.0100 m) 2 = 5.68 × 10−8 Wb (b) SET UP:

The flux is the same through each turn of both solenoids due to the geometry, so M=

(25) ( 5.68 × 10−8 Wb )

N 2Φ B ,2 N 2Φ B ,1 = i1 i1

= 1.18 ×10−5 H

EXECUTE:

M=

(c) SET UP:

The induced emf is E2 = − M

EXECUTE:

E2 = − (1.18 × 10−5 H ) (1750 A/s) = −0.0207 V

EVALUATE:

0.120 A

di1 . dt

A mutual inductance around 10−5 H is not unreasonable. 30-1

30-2

30.5.

Chapter 30

IDENTIFY and SET UP: Apply Eq.(30.5). 400 ( 0.0320 Wb ) NΦ EXECUTE: (a) M = 2 B 2 = = 1.96 H 6.52 A i1

N1Φ B1 Mi (1.96 H)(2.54 A) so Φ B1 = 2 = = 7.11×10−3 Wb i2 N1 700 EVALUATE: M relates the current in one coil to the flux through the other coil. Eq.(30.5) shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux. IDENTIFY: A changing current in an inductor induces an emf in it. μ N2A (a) SET UP: The self-inductance of a toroidal solenoid is L = 0 . 2π r (4π × 10−7 T ⋅ m/A)(500) 2 (6.25 × 10−4 m 2 ) EXECUTE: L = = 7.81 × 10−4 H 2π (0.0400 m) (b) M =

30.6.

(b) SET UP:

The magnitude of the induced emf is E = L

di . dt

⎛ 5.00 A − 2.00 A ⎞ E = ( 7.81 × 10−4 H ) ⎜ ⎟ = 0.781 V −3 ⎝ 3.00 × 10 s ⎠ (c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to b, making b at a higher potential than a. EVALUATE: This is a reasonable value for self-inductance, in the range of a mH. Δi NΦB IDENTIFY: E = L and L = . i Δt EXECUTE:

30.7.

Δi = 0.0640 A/s Δt E 0.0160 V EXECUTE: (a) L = = = 0.250 H Δi / Δt 0.0640 A/s

SET UP:

Li (0.250 H)(0.720 A) = = 4.50 × 10−4 Wb. N 400 EVALUATE: The self-induced emf depends on the rate of change of flux and therefore on the rate of change of the current, not on the value of the current. IDENTIFY: Combine the two expressions for L: L = N Φ B / i and L = E/(di/dt ).

(b) The average flux through each turn is Φ B =

30.8.

SET UP:

30.9.

30.10.

30.11.

Φ B is the average flux through one turn of the solenoid.

(12.6 × 10−3 V)(1.40 A) = 238 turns. (0.00285 Wb)(0.0260 A/s) EVALUATE: The induced emf depends on the time rate of change of the total flux through the solenoid. IDENTIFY and SET UP: Apply E = L di/dt . Apply Lenz’s law to determine the direction of the induced emf in EXECUTE:

Solving for N we have N = Ei/Φ B (di/dt ) =

the coil. EXECUTE:

(a) E = L( di/dt ) = (0.260 H)(0.0180 A/s) = 4.68 × 10−3 V

(b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at a. EVALUATE: The induced emf is directed so as to oppose the decrease in the current. di IDENTIFY: Apply E = − L . dt SET UP: The induced emf points from low potential to high potential across the inductor. EXECUTE: (a) The induced emf points from b to a, in the direction of the current. Therefore, the current is decreasing and the induced emf is directed to oppose this decrease. (b) E = L Δi / Δt , so Δi/Δt = Vab /L = (1.04 V) /(0.260 H) = 4.00 A/s. In 2.00 s the decrease in i is 8.00 A and the

current at 2.00 s is 12.0 A − 8.0 A = 4.0 A. EVALUATE: When the current is decreasing the end of the inductor where the current enters is at the lower potential. This agrees with our result and with Figure 30.6d in the textbook. IDENTIFY and SET UP: Use Eq.(30.6) to relate L to the flux through each turn of the solenoid. Use Eq.(28.23) for the magnetic field through the solenoid.

Inductance

NΦB . If the magnetic field is uniform inside the solenoid Φ B = BA. From Eq.(28.23), i μ NiA N ⎛ μ NiA ⎞ μ 0 N 2 A ⎛N⎞ . Then L = ⎜ 0 . B = μ 0 ni = μ 0 ⎜ ⎟ i so Φ B = 0 ⎟= l i ⎝ l ⎠ l ⎝ l ⎠ EVALUATE: Our result is the same as L for a torodial solenoid calculated in Example 30.3, except that the average circumference 2π r of the toroid is replaced by the length l of the straight solenoid. IDENTIFY and SET UP: The stored energy is U = 12 LI 2 . The rate at which thermal energy is developed is P = I 2 R. EXECUTE: (a) U = 12 LI 2 = 12 (12.0 H)(0.300 A)2 = 0.540 J (b) P = I 2 R = (0.300 A)2 (180 Ω) = 16.2 W = 16.2 J/s EVALUATE: (c) No. If I is constant then the stored energy U is constant. The energy being consumed by the resistance of the inductor comes from the emf source that maintains the current; it does not come from the energy stored in the inductor. IDENTIFY and SET UP: Use Eq.(30.9) to relate the energy stored to the inductance. Example 30.3 gives the μ N 2A inductance of a toroidal solenoid to be L = 0 , so once we know L we can solve for N. 2π r 2U 2(0.390 J) EXECUTE: U = 12 LI 2 so L = 2 = = 5.417 × 10−3 H (12.0 A) 2 I EXECUTE:

30.12.

30.13.

N= 30.14.

L=

2π rL 2π (0.150 m)(5.417 ×10−3 H) = = 2850. (4π ×10−7 T ⋅ m/A)(5.00 × 10−4 m 2 ) μ0 A

EVALUATE: L and hence U increase according to the square of N. IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. μ NI (a) SET UP: The magnetic field inside a toroidal solenoid is B = 0 . 2π r μ0 (300)(5.00 A) −3 EXECUTE: B = = 2.50 × 10 T = 2.50 mT 2π (0.120 m) μ N2A (b) SET UP: The self-inductance of a toroidal solenoid is L = 0 . 2π r (4π × 10−7 T ⋅ m/A)(300) 2 (4.00 × 10−4 m 2 ) EXECUTE: L = = 6.00 × 10−5 H 2π (0.0120 m) (c) SET UP: The energy stored in an inductor is U L = 12 LI 2 . EXECUTE:

U L = 12 (6.00 × 10−5 H)(5.00 A) 2 = 7.50 ×10−4 J

(d) SET UP:

The energy density in a magnetic field is u =

B2 . 2μ 0

(2.50 × 10−3 T) 2 = 2.49 J/m3 2(4π × 10−7 T ⋅ m/A) energy energy 7.50 × 10−4 J (e) u = = = = 2.49 J/m3 volume 2π rA 2π (0.120 m)(4.00 × 10−4 m 2 ) EVALUATE: An inductor stores its energy in the magnetic field inside of it. IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. (a) SET UP: The magnetic field inside a solenoid is B = μ0 nI . (4π × 10−7 T ⋅ m/A)(400)(80.0 A) = 0.161 T EXECUTE: B = 0.250 m B2 (b) SET UP: The energy density in a magnetic field is u = . 2μ0 (0.161 T) 2 EXECUTE: u = = 1.03 ×104 J/m3 2(4π ×10−7 T ⋅ m/A) (c) SET UP: The total stored energy is U = uV. EXECUTE: U = uV = u (lA) = (1.03 ×104 J/m3 )(0.250 m)(0.500 ×10−4 m2 ) = 0.129 J EXECUTE:

30.15.

30-3

u=

(d) SET UP: The energy stored in an inductor is U = 12 LI 2 . EXECUTE: Solving for L and putting in the numbers gives 2U 2(0.129 J) L= 2 = = 4.02 ×10−5 H I (80.0 A) 2 EVALUATE: An inductor stores its energy in the magnetic field inside of it.

30-4

Chapter 30

30.16.

IDENTIFY: Energy = Pt . U = 12 LI 2 . SET UP: P = 200 W = 200 J/s EXECUTE: (a) Energy = (200 W)(24 h)(3600 s/h) = 1.73 ×107 J 2U 2(1.73 × 107 J) = = 5.41×103 H I2 (80.0 A) 2 EVALUATE: A large value of L and a large current would be required, just for one light bulb. Also, the resistance of the inductor would have to be very small, to avoid a large P = I 2 R rate of electrical energy loss. IDENTIFY and SET UP: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability μ is used in place of μ 0 .

(b) L =

30.17.

EXECUTE:

30.18.

B2 μN 2A μ NI and B = gives u = . 2π r 2π r 2μ

EVALUATE: For a given value of B, the energy density is less when μ is larger than μ 0 . IDENTIFY and SET UP: The energy density (energy per unit volume) in a magnetic field (in vacuum) is given by U B2 (Eq.30.10). u= = V 2μ0 EXECUTE: (b) u =

B=

30.19.

Using L =

(a) V =

2μ 0U 2(4π ×10−7 T ⋅ m/A)(3.60 ×106 J) = = 25.1 m3 . B2 (0.600 T) 2

U B2 = V 2μ0

2μ 0U 2(4π ×10−7 T ⋅ m/A)(3.60 ×106 J) = = 11.9 T 3 V ( 0.400 m )

EVALUATE: Large-scale energy storage in a magnetic field is not practical. The volume in part (a) is quite large and the field in part (b) would be very difficult to achieve. IDENTIFY: Apply Kirchhoff’s loop rule to the circuit. i(t) is given by Eq.(30.14). SET UP: The circuit is sketched in Figure 30.19. di is positive as the current dt increases from its initial value of zero.


E − vR − v L = 0

di E E − iR − L = 0 so i = (1 − e− ( R / L )t ) dt R di (a) Initially (t = 0), i = 0 so E − L = 0 dt di E 6.00 V = = = 2.40 A/s dt L 2.50 H di (b) E − iR − L = 0 (Use this equation rather than Eq.(30.15) since i rather than t is given.) dt di E − iR 6.00 V − (0.500 A)(8.00 Ω) Thus = = = 0.800 A/s dt L 2.50 H E ⎛ 6.00 V ⎞ − (8.00 Ω/ 2.50 H)(0.250 s) (c) i = (1 − e − ( R/L )t ) = ⎜ ) = 0.750 A(1 − e−0.800 ) = 0.413 A ⎟ (1 − e R ⎝ 8.00 Ω ⎠ (d) Final steady state means t → ∞ and

di → 0, so E − iR = 0. dt

E 6.00 V = = 0.750 A R 8.00 Ω EVALUATE: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of E/R. The slope of the current in the figure, which is di/dt, decreases with t.

i=

Inductance

30.20.

IDENTIFY: The current decays exponentially. SET UP: After opening the switch, the current is i = I 0e−tR/L , and the time constant is τ = L /R. EXECUTE:

(a) The initial current is I 0 = (6.30 V)/(15.0 Ω) = 0.420 A. Now solve for L and put in the numbers.

L=

30.21.

30-5

−tR −(2.00 ms)(15.0 Ω) = = 43.3 mH ln(i / I 0 ) ⎛ 0.210 A ⎞ ln ⎜ ⎟ ⎝ 0.420 A ⎠

(b) τ = L/R = (43.3 mH)/(15.0 Ω) = 2.89 ms (c) Solve i = I 0e − t/τ for t, giving t = −τ ln(i/I 0 ) = −(2.89 ms)ln(0.0100) = 13.3 ms. EVALUATE: In less than 5 time constants, the current is only 1% of its initial value. IDENTIFY: i = E/R (1 − e − t/τ ), with τ = L/R. The energy stored in the inductor is U = 12 Li 2 . SET UP: The maximum current occurs after a long time and is equal to E/R. EXECUTE: (a) imax = E/R so i = imax /2 when (1 − e −t/τ ) = 12 and e− t/τ = 12 . −t/τ = ln ( 12 ) .

t=

L ln 2 (ln 2)(1.25 × 10−3 H) = = 17.3 μs R 50.0 Ω

(b) U = 12 U max when i = imax

2 . 1 − e − t/τ = 1

2 , so e− t/τ = 1 − 1

2 = 0.2929. t = − L ln (0.2929)/R = 30.7 μs.

EVALUATE: τ = L/R = 2.50 × 10 s = 25.0 μs. The time in part (a) is 0.692τ and the time in part (b) is 1.23τ . IDENTIFY: With S1 closed and S 2 open, i (t ) is given by Eq.(30.14). With S1 open and S 2 closed, i (t ) is given by Eq.(30.18). SET UP: U = 12 Li 2 . After S1 has been closed a long time, i has reached its final value of I = E/R. −5

30.22.

EXECUTE:

(a) U = 12 LI 2 and I =

2U 2(0.260 J) = = 2.13 A. E = IR = (2.13 A)(120 Ω) = 256 V. L 0.115 H

(b) i = Ie − ( R/L )t and U = 12 Li 2 = 12 LI 2e−2( R/L ) t = 12 U 0 =

t=−

30.23.

30.24.

30.25.

1 2

(

1 2

LI 2 ) . e−2( R/L )t = 12 , so

L 0.115 H ln ( 12 ) = − ln ( 12 ) = 3.32 × 10−4 s. 2R 2(120 Ω)

EVALUATE: τ = L/R = 9.58 × 10−4 s. The time in part (b) is τ ln(2) / 2 = 0.347τ . IDENTIFY: L has units of H and R has units of Ω . SET UP: 1 H = 1 Ω ⋅ s EXECUTE: Units of L/R = H / Ω = (Ω ⋅ s) / Ω = s = units of time. EVALUATE: Rt/L = t/τ is dimensionless. IDENTIFY: Apply the loop rule. SET UP: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the inductor is − Ldi / dt. i di′ R t R di R EXECUTE: − Ldi/dt − iR = 0. = − ∫ dt ′ and ln(i/I 0 ) = − t. i = I 0e− t ( R/L ) . = − i gives ∫ I 0 0 i′ dt L L L EVALUATE: di / dt is negative, so there is a potential rise across the inductor; point c is at higher potential than point b. There is a potential drop across the resistor. IDENTIFY: Apply the concepts of current decay in an R-L circuit. Apply the loop rule to the circuit. i(t) is given by Eq.(30.18). The voltage across the resistor depends on i and the voltage across the inductor depends on di/dt. SET UP: The circuit with S1 closed and S 2 open is sketched in Figure 30.25a.

E − iR − L

Figure 30.25a

Constant current established means EXECUTE:

i=

di = 0. dt

E 60.0 V = = 0.250 A R 240 Ω

di =0 dt

30-6

Chapter 30

(a) SET UP:

The circuit with S 2 closed and S1 open is shown in Figure 30.25b. i = I 0e − ( R / L ) t At t = 0, i = I 0 = 0.250 A Figure 30.25b

The inductor prevents an instantaneous change in the current; the current in the inductor just after S 2 is closed and S1 is opened equals the current in the inductor just before this is done. (b) EXECUTE: i = I 0e− ( R / L )t = (0.250 A)e− (240 Ω / 0.160 H)(4.00×10 (c) SET UP: See Figure 30.25c.

−4

s)

= (0.250 A)e−0.600 = 0.137 A

Figure 30.25c EXECUTE: If we trace around the loop in the direction of the current the potential falls as we travel through the resistor so it must rise as we pass through the inductor: vab > 0 and vbc < 0. So point c is at higher potential than point b. vab + vbc = 0 and vbc = −vab

Or, vcb = vab = iR = (0.137 A)(240 Ω) = 32.9 V (d) i = I 0e − ( R/L )t

i = 12 I 0 says 12 I 0 = I 0e− ( R/L ) t and

1 2

= e − ( R/L )t

Taking natural logs of both sides of this equation gives ln( 12 ) = − Rt / L ⎛ 0.160 H ⎞ −4 t =⎜ ⎟ ln 2 = 4.62 × 10 s ⎝ 240 Ω ⎠

30.26.

EVALUATE: The current decays, as shown in Fig. 30.13 in the textbook. The time constant is τ = L/R = 6.67 × 10−4 s. The values of t in the problem are less than one time constant. At any instant the potential drop across the resistor (in the direction of the current) equals the potential rise across the inductor. IDENTIFY: Apply Eq.(30.14). di SET UP: vab = iR. vbc = L . The current is increasing, so di / dt is positive. dt EXECUTE: (a) At t = 0, i = 0. vab = 0 and vbc = 60 V. (b) As t → ∞, i → E/R and di / dt → 0. vab → 60 V and vbc → 0. (c) When i = 0.150 A, vab = iR = 36.0 V and vbc = 60.0 V − 36.0 V = 24.0 V.

30.27.

EVALUATE: At all times, E = vab + vbc , as required by the loop rule. IDENTIFY: i (t ) is given by Eq.(30.14). SET UP: The power input from the battery is Ei. The rate of dissipation of energy in the resistance is i 2 R. The voltage across the inductor has magnitude Ldi/dt , so the rate at which energy is being stored in the inductor is iLdi/dt. E2 (6.00 V) 2 (1 − e− (8.00 Ω / 2.50 H) t ). EXECUTE: (a) P = Ei = EI 0 (1 − e − ( R/L )t ) = (1 − e − ( R/L )t ) = 8.00 Ω R P = (4.50 W)(1 − e − (3.20 s

−1

)t

).

−1 E (6.00 V) 2 (1 − e − ( R/L )t ) 2 = (1 − e − (8.00 Ω/ 2.50 H) t ) 2 = (4.50 W)(1 − e− (3.20 s ) t ) 2 8.00 Ω R 2 di E ⎛E ⎞ E (c) PL = iL = (1 − e − ( R/L )t ) L ⎜ e − ( R/L )t ⎟ = (e − ( R/L )t − e −2( R/L )t ) dt R ⎝L ⎠ R

(b) PR = i 2 R =

2

−1

−1

PL = (4.50 W)(e− (3.20 s ) t − e− (6.40 s ) t ). EVALUATE: (d) Note that if we expand the square in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor. Conservation of energy requires that this be so.

Inductance

30.28.

30-7

An L-C circuit oscillates, with the energy going back and forth between the inductor and capacitor. 1 1 ω (a) SET UP: The frequency is f = and ω = , giving f = . 2π 2π LC LC 1 EXECUTE: f = = 2.13 × 103 Hz = 2.13 kHz −3 2π ( 0.280 × 10 H )( 20.0 ×10−6 F ) IDENTIFY:

(b) SET UP:

The energy stored in a capacitor is U = 12 CV 2 .

EXECUTE:

U = 12 (20.0 ×10−6 F)(150.0 V) 2 = 0.225 J

(c) SET UP:

The current in the circuit is i = −ωQ sin ωt , and the energy stored in the inductor is U = 12 Li 2 .

First find ω and Q. ω = 2π f = 1.336 × 104 rad/s. Q = CV = (20.0 × 10–6 F)(150.0 V) = 3.00 × 10–3 C Now calculate the current: i = − (1.336 × 104 rad/s)(3.00 × 10–3 C) sin[(1.336 × 104 rad/s)(1.30 × 10–3 s)] Notice that the argument of the sine is in radians, so convert it to degrees if necessary. The result is i = − 39.92 A Now find the energy in the inductor: U = 12 Li 2 = 12 (0.280 ×10−3 H)( −39.92 A) 2 = 0.223 J EVALUATE: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the capacitor. IDENTIFY: The energy moves back and forth between the inductor and capacitor. 1 1 2π (a) SET UP: The period is T = = = = 2π LC . f ω / 2π ω EXECUTE: Solving for L gives T2 (8.60 × 10−5 s) 2 L= 2 = 2 = 2.50 × 10−2 H = 25.0 mH 4π C 4π (7.50 × 10−9 C) EXECUTE:

30.29.

(b) SET UP: The charge on a capacitor is Q = CV. EXECUTE: Q = CV = (7.50 × 10–9 F)(12.0 V) = 9.00 × 10–8 C (c) SET UP: The stored energy is U = Q2/2C.

( 9.00 ×10 C ) = 5.40 ×10 U= 2 ( 7.50 ×10 F ) −8

EXECUTE:

−9

2

−7

J

(d) SET UP: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial energy. U L + U C = U Total . 12 LI 2 + 0 = U Total . EXECUTE: Solve for the current: I=

30.30.

2 ( 5.40 × 10−7 J ) 2U Total = = 6.58 × 10 −3 A = 6.58 mA 2.50 × 10 −2 H L

EVALUATE: The energy oscillates back and forth forever. However if there is any resistance in the circuit, no matter how small, all this energy will eventually be dissipated as heat in the resistor. IDENTIFY: The circuit is described in Figure 30.14 of the textbook. SET UP: The energy stored in the inductor is U L = 12 Li 2 and the energy stored in the capacitor is U C = q 2 /2C. Initially,

U C = 12 CV 2, with V = 12.0 V. The period of oscillation is T = 2π LC = 2π (12.0 ×10−3 H)(18.0 ×10−6 F) = 2.92 ms . EXECUTE:

(a) Energy conservation says U L (max) = U C (max), and

imax = V C/L = (22.5 V)

1 2

2 Limax = 12 CV 2 .

18 × 10−6 F = 0.871 A. The charge on the capacitor is zero because all the energy is in 12 × 10−3 H

the inductor. (b) From Figure 30.14 in the textbook, q = 0 at t = T/4 = 0.730 ms and at t = 3T/4 = 2.19 ms. (c) q0 = CV = (18 μ F) (22.5 V) = 405 μ C is the maximum charge on the plates. The graphs are sketched in Figure 30.30. q refers to the charge on one plate and the sign of i indicates the direction of the current. EVALUATE: If the capacitor is fully charged at t = 0 it is fully charged again at t = T/2, but with the opposite polarity.

Figure 30.30

30-8

Chapter 30

30.31.

IDENTIFY and SET UP: The angular frequency is given by Eq.(30.22). q(t) and i(t) are given by Eqs.(30.21) and (30.23). The energy stored in the capacitor is U C = 12 CV 2 = q 2 /2C. The energy stored in the inductor is U L = 12 Li 2 . EXECUTE:

(a) ω =

1 1 = = 105.4 rad/s, which rounds to 105 rad/s. The period is LC (1.50 H)(6.00 ×10−5 F)

2π

2π = = 0.0596 s ω 105.4 rad/s (b) The circuit containing the battery and capacitor is sketched in Figure 30.31.

given by T =

Q =0 C Q = EC = (12.0 V)(6.00 ×10−5 F) = 7.20 ×10−4 C

E−

Figure 30.31 (c) U = CV = 12 (6.00 × 10−5 F)(12.0 V) 2 = 4.32 × 10 −3 J (d) q = Q cos(ω t + φ ) (Eq.30.21) q = Q at t = 0 so φ = 0 1 2

2

q = Q cos ω t = (7.20 ×10−4 C)cos([105.4 rad/s][0.0230 s]) = −5.42 ×10−4 C The minus sign means that the capacitor has discharged fully and then partially charged again by the current maintained by the inductor; the plate that initially had positive charge now has negative charge and the plate that initially had negative charge now has positive charge. (e) i = −ω Q sin(ω t + φ ) (Eq.30.23) i = −(105 rad/s)(7.20 ×10 −4 C)sin([105.4 rad/s][0.0230 s]) = −0.050 A The negative sign means the current is counterclockwise in Figure 30.15 in the textbook. or q2 Q2 1 2 1 gives i = ± = Q 2 − q 2 (Eq.30.26) 2 Li + 2C 2C LC

i = ±(105 rad/s) (7.20 ×10−4 C) 2 − (−5.42 ×10−4 C) 2 = ±0.050 A, which checks. q 2 (−5.42 ×10−4 C) 2 = = 2.45 ×10−3 J 2C 2(6.00 ×10−5 F) U L = 12 Li 2 = 12 (1.50 H)(0.050 A) 2 = 1.87 × 10−3 J

(f ) U C =

EVALUATE:

Note that U C + U L = 2.45 ×10−3 J + 1.87 ×10−3 J = 4.32 ×10−3 J.

Q 2 (7.20 ×10−4 C) 2 = = 4.32 ×10−3 J. 2C 2(6.00 ×10−5 F) Energy is conserved. At some times there is energy stored in both the capacitor and the inductor. When i = 0 all the energy is stored in the capacitor and when q = 0 all the energy is stored in the inductor. But at all times the total energy stored is the same. 1 IDENTIFY: ω = = 2πf LC SET UP: ω is the angular frequency in rad/s and f is the corresponding frequency in Hz. 1 1 EXECUTE: (a) L = 2 2 = 2 = 2.37 × 10−3 H. 4π f C 4π (1.6 ×106 Hz) 2 (4.18 ×10 −12 F) (b) The maximum capacitance corresponds to the minimum frequency. 1 1 Cmax = = = 3.67 × 10−11 F = 36.7 pF 2 2 2 5 4π f min L 4π (5.40 × 10 Hz) 2 (2.37 ×10 −3 H) EVALUATE: To vary f by a factor of three (approximately the range in this problem), C must be varied by a factor of nine. IDENTIFY: Apply energy conservation and Eqs. (30.22) and (30.23). Q2 SET UP: If I is the maximum current, 12 LI 2 = . For the inductor, U L = 12 Li 2 . 2C Q2 EXECUTE: (a) 12 LI 2 = gives Q = i LC = (0.750 A) (0.0800 H)(1.25 ×10−9 F) = 7.50 ×10−6 C . 2C 1 1 ω (b) ω = = = 1.00 ×105 rad/s . f = = 1.59 ×104 Hz . −9 2 π LC (0.0800 H)(1.25 ×10 F) This agrees with the total energy initially stored in the capacitor, U =

30.32.

30.33.

Inductance

(c) q = Q at t = 0 means φ = 0 . i = −ω Q sin(ω t ) , so

i = −(1.00 ×105 rad/s)(7.50 ×10−6 C)sin([1.00 ×105 rad/s][2.50 ×10−3 s]) = −0.7279 A . U L = 12 Li 2 = 12 (0.0800 H)( − 0.7279 A)2 = 0.0212 J .

30.34.

EVALUATE: The total energy of the system is 12 LI 2 = 0.0225 J . At t = 2.50 ms , the current is close to its maximum value and most of the system’s energy is stored in the inductor. IDENTIFY: Apply Eq.(30.25). SET UP: q = Q when i = 0 . i = imax when q = 0 . 1/ LC = 1917 s −1 . EXECUTE:

(a)

1 2

2 Limax =

Q2 . Q = imax LC = (0.850 ×10−3 A) (0.0850 H)(3.20 × 10−6 F) = 4.43 ×10−7 C 2C 2

30.35.

⎛ 5.00 ×10−4 A ⎞ −7 (b) q = Q 2 − LCi 2 = (4.43 ×10−7 C) 2 − ⎜ ⎟ = 3.58 ×10 C . −1 1917s ⎝ ⎠ EVALUATE: The value of q calculated in part (b) is less than the maximum value Q calculated in part (a). IDENTIFY: q = Q cos(ω t + φ ) and i = −ω Q sin(ω + φ ) q2 . U L = 12 Li 2 . 2C q 2 1 Q 2 cos 2 (ωt + φ ) EXECUTE: (a) U C = 12 = . C 2 C 1 Q 2 sin 2 (ωt + φ ) . , since ω2 = U L = 12 Li 2 = 12 Lω2Q 2 sin 2 (ωt + φ ) = 12 C LC 1 Q2 1 (b) U Total = U C + U L = cos 2 (ωt + φ ) + Lω2Q 2 sin 2 (ωt + φ ) 2 C 2 2 Q 1 Q2 Q2 ⎛ ⎞ 2 2 1 U total = 12 cos 2 (ωt + φ ) + 12 L ⎜ (cos 2 (ωt + φ ) + sin 2 (ωt + φ )) = 12 ⎟ Q sin (ωt + φ ) = 2 C C C ⎝ LC ⎠ U Total is a constant. EVALUATE: Eqs.(30.21) and (30.23) are consistent with conservation of energy in the L-C circuit. d 2q and insert into Eq.(20.20). IDENTIFY: Evaluate dt 2 d 2q 1 SET UP: Equation (30.20) is + q = 0. dt 2 LC dq d 2q EXECUTE: q = Q cos(ωt + φ ) ⇒ = −ωQ sin(ωt + φ ) ⇒ 2 = −ω2Q cos(ωt + φ ). dt dt 2 d q 1 Q 1 1 q = −ω2Q cos(ωt + φ ) + cos(ωt + φ ) = 0 ⇒ ω 2 = + ⇒ω= . dt 2 LC LC LC LC EVALUATE: The value of φ depends on the initial conditions, the value of q at t = 0 . IDENTIFY: The unit of L is H and the unit of C is F. SET UP: C = q / Vab says 1 F = 1 C/V . 1 H = 1 V ⋅ s/A = 1 V ⋅ s 2 /C . SET UP:

30.36.

30.37.

EXECUTE: 30.38.

UC =

1 H ⋅ F = (1 V ⋅ s 2 /C)(1 C/V) = 1 s 2 . Therefore, LC has units of s 2 and

LC has units of s.

EVALUATE: Our result shows that ω t is dimensionless, since ω = 1/ LC . IDENTIFY: The presence of resistance in an L-R-C circuit affects the frequency of oscillation and causes the amplitude of the oscillations to decrease over time. 1 R2 − 2 . (a) SET UP: The frequency of damped oscillations is ω ′ = LC 4 L EXECUTE:

ω′ =

1 (75.0 Ω) 2 − = 5.5 ×104 rad/s ( 22 ×10−3 H )(15.0 ×10−9 F) 4 ( 22 ×10−3 H )2

The frequency f is f =

ω 5.50 ×104 rad/s = = 8.76 ×103 Hz = 8.76 kHz . 2π 2π

(b) SET UP: The amplitude decreases as A(t) = A0 e–(R/2L)t. EXECUTE: Solving for t and putting in the numbers gives:

−2 L ln( A / A0 ) −2 ( 22.0 ×10 H ) ln(0.100) = = 1.35 ×10−3 s = 1.35 ms R 75.0 Ω −3

t=

30-9

30-10

30.39.

Chapter 30

(c) SET UP:

At critical damping, R = 4 L / C .

EXECUTE:

R=

4 ( 22.0 ×10−3 H ) 15.0 ×10−9 F

= 2420 Ω

EVALUATE: The frequency with damping is almost the same as the resonance frequency of this circuit (1/ LC ), which is plausible because the 75-Ω resistance is considerably less than the 2420 Ω required for critical damping. IDENTIFY: Follow the procedure specified in the problem. 1 SET UP: Make the substitutions x → q, m → L, b → R, k → . C d 2 x b dx kx d 2 q R dq q EXECUTE: (a) Eq. (13.41): 2 + + = 0 . This becomes + + = 0 , which is Eq.(30.27). dt m dt m dt 2 L dt LC

1 R2 k b2 − . This becomes ω′ = − 2 , which is Eq.(30.29). 2 m 4m LC 4 L − (b / 2 m )t cos(ω′t + φ ) . This becomes q = Ae− ( R / 2 L )t cos(ω′t + φ ) , which is Eq.(30.28). (c) Eq. (13.42): x = Ae EVALUATE: Equations for the L-R-C circuit and for a damped harmonic oscillator have the same form. IDENTIFY: For part (a), evaluate the derivatives as specified in the problem. For part (b) set q = Q in Eq.(30.28) and set dq / dt = 0 in the expression for dq / dt . (b) Eq. (13.43): ω′ =

30.40.

SET UP:

In terms of ω ′ , Eq.(30.28) is q (t ) = Ae − ( R / 2 L )t cos(ω ′t + φ ) .

EXECUTE:

(a) q = Ae− ( R / 2 L ) t cos(ω′t + φ ) .

dq R = − A e − ( R / 2 L ) t cos(ω′t + φ ) − ω′Ae − ( R / 2 L )t sin(ω′t + φ ). 2L dt

2

d 2q R ⎛ R ⎞ = A ⎜ ⎟ e − ( R / 2 L ) t cos(ω′t + φ ) + 2ω′A e − ( R / 2 L ) t sin(ω′t + φ ) − ω′2 Ae − ( R / 2 L )t cos(ω′t + φ ) 2 2L dt ⎝ 2L ⎠ ⎛ ⎛ R ⎞2 d 2 q R dq q R2 1 ⎞ 1 R2 2 2 + + = q − 2. ⎜⎜ ⎜ ⎟ − ω ′ − 2 + ⎟⎟ = 0 , so ω′ = 2 dt L dt LC 2 L LC ⎠ LC 4 L ⎝ ⎝ 2L ⎠ dq dq R Q = 0 , so q = Acos φ = Q and =− Acos φ − ω′Asin φ = 0 . This gives A = and (b) At t = 0, q = Q, i = dt dt 2L cos φ R R . =− 2 Lω′ 2 L 1/ LC − R 2 / 4 L2 EVALUATE: If R = 0 , then A = Q and φ = 0 . IDENTIFY: Evaluate Eq.(30.29). SET UP: The angular frequency of the circuit is ω ′ . 1 1 EXECUTE: (a) When R = 0, ω0 = = = 298 rad s. LC (0.450 H) (2.50 ×10−5 F)

tan φ = −

30.41.

(b) We want

R=

30.42.

ω (1 LC − R 2 4 L2 ) R 2C = 0.95 , so = 1− = (0.95) 2 . This gives ω0 1 LC 4L

4L 4(0.450 H)(0.0975) = 83.8 Ω. (1 − (0.95) 2 ) = C (2.50 ×10−5 F)

EVALUATE: When R increases, the angular frequency decreases and approaches zero as R → 2 L / C . IDENTIFY: L has units of H and C has units of F. SET UP: 1 H = 1 Ω ⋅ s . C = q / V says 1 F = 1 C/V. V = IR says 1 V/A = 1 Ω . EXECUTE:

The units of L / C are

H Ω⋅s Ω⋅ V = = = Ω 2 . Therefore, the unit of F CV A

L / C is Ω.

1 R2 and must have the same units, so R and LC 4 L2 same units, and we have shown that this is indeed the case. EVALUATE:

30.43.

IDENTIFY:

For Eq.(30.28) to be valid,

L / C must have the

The emf E2 in solenoid 2 produced by changing current i1 in solenoid 1 is given by E2 = M

Δi1 . The Δt

mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem μ AN N M = 0 1 2 , where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid. l

Inductance

SET UP:

30-11

μ 0 = 4π ×10−7 T ⋅ m/A . Let the outer solenoid be solenoid 1.

EXECUTE:

(a) M =

(4π ×10−7 T ⋅ m/A)π (6.00 ×10−4 m) 2 (6750)(15) = 2.88 ×10−7 H = 0.288 μ H 0.500 m

Δi1 = (2.88 ×10−7 H)(37.5 A/s) = 1.08 ×10−5 V Δt EVALUATE: If current in the inner solenoid changed at 37.5 A/s, the emf induced in the outer solenoid would be 1.08 ×10−5 V. di IDENTIFY: Apply E = − L and Li = N Φ B . dt SET UP: Φ B is the flux through one turn. (b) E2 =

30.44.

EXECUTE:

(a) E = − L

ε = (3.50 ×10

−3

di d = −(3.50 ×10−3 H) ((0.680 A)cos(π t /[0.0250 s])). dt dt

H)(0.680 A)

π 0.0250 s

Emax = (3.50 ×10−3 H)(0.680 A)

sin(π t /[0.0250 s]) . Therefore,

π 0.0250 s

= 0.299 V.

Limax (3.50 ×10−3 H)(0.680 A) = = 5.95 ×10−6 Wb. N 400 di (c) E (t ) = − L = − (3.50 ×10−3 H)(0.680 A)(π / 0.0250 s)sin(π t / 0.0250 s). dt E (t ) = − (0.299 V)sin((125.6 s −1 )t ) .Therefore, at t = 0.0180 s , (b) Φ B max =

E (0.0180 s) = − (0.299 V)sin((125.6 s −1 )(0.0180 s)) = 0.230 V . The magnitude of the induced emf is 0.230 V. EVALUATE: 30.45.

The maximum emf is when i = 0 and at this instant Φ B = 0.

di . dt SET UP: During an interval in which the graph of i versus t is a straight line, di / dt is constant and equal to the slope of that line. EXECUTE: (a) The pattern on the oscilloscope is sketched in Figure 30.45. EVALUATE: (b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph. IDENTIFY:

E = −L

Figure 30.45 30.46.

IDENTIFY:

Apply E = − L

di . dt

d cos(ω t ) = −ω sin(ω t ) dt di d EXECUTE: (a) E = − L = − L ((0.124 A)cos[(240 π s)t ]. dt dt SET UP:

E = + (0.250 H) (0.124 A) (240 π / s)sin((240π s)t ) = +(23.4 V) sin ((240 π s)t ).

The graphs are given in Figure 30.46. (b) Emax = 23.4 V; i = 0, since the emf and current are 90° out of phase. (c) imax = 0.124 A; E = 0, since the emf and current are 90° out of phase.

30-12

Chapter 30

EVALUATE:

The induced emf depends on the rate at which the current is changing.

Figure 30.46 30.47.

di to the series and parallel combinations. dt SET UP: In series, i1 = i2 and the voltages add. In parallel the voltages are the same and the currents add. di di di di di di EXECUTE: (a) Series: L1 1 + L2 2 = Leq , but i1 = i2 = i for series components so 1 = 2 = and dt dt dt dt dt dt L1 + L2 = Leq . IDENTIFY:

Apply E = − L

(b) Parallel: Now L1

di di1 di2 di1 di di di L di = + = L2 2 = Leq , where i = i1 + i2 . Therefore, . But 1 = eq and dt dt dt dt dt dt dt L1 dt −1

30.48.

⎛1 1⎞ di2 Leq di di Leq di Leq di = . and Leq = ⎜ + ⎟ . = + dt L2 dt dt L1 dt L2 dt ⎝ L1 L2 ⎠ EVALUATE: Inductors in series and parallel combine in the same way as resistors. IDENTIFY: Follow the steps outlined in the problem. SET UP: The energy stored is U = 12 Li 2 . ! ! μi EXECUTE: (a) AB ⋅ dl = μ0 I encl ⇒ B 2πr = μ0i ⇒ B = 0 . 2πr μ0i ldr. (b) d Φ B = BdA = 2πr b b μ il dr μ0il = ln ( b a ). (c) Φ B = ∫ d Φ B = 0 ∫ 2π a r 2π a

NΦB μ = l 0 ln( b a). i 2π 1 2 1 μ0 μ li 2 (e) U = Li = l ln( b a )i 2 = 0 ln( b a ). 2 2 2π 4π EVALUATE: The magnetic field between the conductors is due only to the current in the inner conductor. (a) IDENTIFY and SET UP: An end view is shown in Figure 30.49. (d) L =

30.49.

Apply Ampere’s law to a circular path of radius r. ! ! AB ⋅ dl = μ 0 I encl Figure 30.49 ! ! EXECUTE: AB ⋅ dl = B (2π r ) I encl = i, the current in the inner conductor μi Thus B(2π r ) = μ 0i and B = 0 . 2π r

Inductance

30-13

(b) IDENTIFY and SET UP: Follow the procedure specified in the problem. B2 EXECUTE: u = 2μ 0 dU = u dV , where dV = 2π rl dr 2

dU =

1 ⎛ μ 0i ⎞ μ 0i 2 l π rl dr dr (2 ) = ⎜ ⎟ 2μ0 ⎝ 2π r ⎠ 4π r

μ 0i 2l b dr μ 0i 2l = [ln r ]ba 4π ∫ a r 4π μ i 2l μ i 2l ⎛ b ⎞ U = 0 (ln b − ln a ) = 0 ln ⎜ ⎟ 4π 4π ⎝a⎠

(c) U = ∫ dU =

(d) Eq.(30.9): U = 12 Li 2

Part (c): U = 1 2

Li 2 =

L=

30.50.

μ 0i 2 l ⎛ b ⎞ ln ⎜ ⎟ 4π ⎝a⎠

μ 0i 2l ⎛ b ⎞ ln ⎜ ⎟ 4π ⎝a⎠

μ 0l ⎛ b ⎞ ln ⎜ ⎟ . 2π ⎝ a ⎠

EVALUATE: The value of L we obtain from these energy considerations agrees with L calculated in part (d) of Problem 30.48 by considering flux and Eq.(30.6) NΦ NΦB IDENTIFY: Apply L = to each solenoid, as in Example 30.3. Use M = 2 B 2 to calculate the mutual i1 i inductance M. SET UP: The magnetic field produced by solenoid 1 is confined to the space within its windings and is equal to μ Ni B1 = 0 1 1 . 2π r N1Φ B1 N1 A ⎛ μ0 N1i1 ⎞ μ0 N12 A N 2Φ B2 N 2 A ⎛ μ0 N 2i2 ⎞ μ0 N 2 2 A . EXECUTE: (a) L1 = = , L2 = = ⎜ ⎟= ⎜ ⎟= 2π r i1 i1 ⎝ 2π r ⎠ i2 i2 ⎝ 2π r ⎠ 2π r 2

⎛ μ N N A⎞ μ N 2 A μ0 N 2 2 A N 2 AB1 μ 0 N1 N 2 A = . M2 =⎜ 0 1 2 ⎟ = 0 1 = L1L2 . 2π r 2π r 2π r i1 ⎝ 2π r ⎠ EVALUATE: If the two solenoids are identical, so that N1 = N 2 , then M = L . (b) M =

30.51.

IDENTIFY:

U = 12 LI 2 . The self-inductance of a solenoid is found in Exercise 30.11 to be L =

μ 0 AN 2 l

.

The length l of the solenoid is the number of turns divided by the turns per unit length. 2U 2(10.0 J) EXECUTE: (a) L = 2 = = 8.89 H I (1.50 A) 2 μ AN 2 (b) L = 0 . If α is the number of turns per unit length, then N = α l and L = μ 0 Aα 2l . For this coil l L 8.89 H = = 56.3 m . α = 10 coils/mm = 10 × 103 coils/m . l = 2 −7 (4π ×10 T ⋅ m/A)π (0.0200 m) 2 (10 ×103 coils/m) 2 μ 0 Aα This is not a practical length for laboratory use. EVALUATE: The number of turns is N = (56.3 m)(10 ×103 coils/m) = 5.63 ×105 turns . The length of wire in the SET UP:

solenoid is the circumference C of one turn times the number of turns. C = π d = π (4.00 ×10−2 m) = 0.126 m . The length of wire is (0.126 m)(5.63 × 105 ) = 7.1× 104 m = 71 km . This length of wire will have a large resistance and 30.52.

I 2 R electrical energy loses will be very large. IDENTIFY: This is an R-L circuit and i (t ) is given by Eq.(30.14). SET UP: When t → ∞ , i → if = V / R . V 12.0 V EXECUTE: (a) R = = = 1860 Ω. if 6.45 ×10−3 A Rt − Rt −(1860 Ω)(7.25 ×10−4 s) (b) i = if (1 − e − ( R / L )t ) so = −ln(1 − i / if ) and L = = = 0.963 H . L ln(1 − i / if ) ln(1 − (4.86 / 6.45))

30-14

Chapter 30

30.53.

EVALUATE: The current after a long time depends only on R and is independent of L. The value of R / Ldetermines how rapidly the final value of i is reached. IDENTIFY and SET UP: Follow the procedure specified in the problem. L = 2.50 H, R = 8.00 Ω,

E = 6.00 V. i = (E / R)(1 − e −t / τ ), τ = L / R EXECUTE: (a) Eq.(30.9): U L = 12 Li 2 t = τ so i = (E / R )(1 − e −1 ) = (6.00 V/8.00 Ω)(1 − e −1 ) = 0.474 A

Then U L = 12 Li 2 = 12 (2.50 H)(0.474 A) 2 = 0.281 J dU L di Exercise 30.27 (c): PL = = Li dt dt di ⎛ E ⎞ − ( R / L ) t E −t /τ ⎛E⎞ = ⎜ ⎟e = e i = ⎜ ⎟ (1 − e −t /τ ); R dt L ⎝ ⎠ ⎝ L⎠ 2 ⎛E ⎞⎛ E ⎞ E PL = L ⎜ (1 − e− t / τ ) ⎟⎜ e− t / τ ⎟ = (e− t / τ − e −2 ± / τ ) R L ⎝ ⎠⎝ ⎠ R τ

E 2 τ −t /τ E2 ⎡ τ ⎤ −τ e −t /τ + e −2t /τ ⎥ (e − e −2t /τ ) dt = ∫ 0 R 0 R ⎢⎣ 2 ⎦0 2 τ E2 E τ 2 / 1 2 −t /τ − t − − ⎤⎦ = τ ⎡⎣1 − 12 − e + 12 e ⎤⎦ U L = − τ ⎡⎣e − 12 e 0 R R 2 2 ⎛ E ⎞⎛ L ⎞ E⎞ −1 −2 −1 −2 1⎛ − + = (1 2 ) UL = ⎜ e e ⎟⎜ ⎟ ⎟ L(1 − 2e + e ) 2⎜ 2 R R R ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ τ

U L = ∫ PL dt =

2

⎛ 6.00 V ⎞ U L = 12 ⎜ ⎟ (2.50 H)(0.3996) = 0.281 J, which checks. ⎝ 8.00 Ω ⎠ 2

⎛E ⎞ E (b) Exercise 30.27(a): The rate at which the battery supplies energy is PE = Ei = E ⎜ (1 − e − t /τ ) ⎟ = (1 − e − t /τ ) ⎝R ⎠ R τ

U E = ∫ PE dt = 0

τ ⎛ E2 ⎞ E2 τ E2 ⎡t + τ e −t /τ ⎦⎤ = ⎜ ⎟ (τ + τ e−1 − τ ) (1 − e −t /τ )dt = ∫ ⎣ 0 R 0 R ⎝ R⎠ 2

⎛ E2 ⎞ ⎛ E 2 ⎞⎛ L ⎞ ⎛E⎞ U E = ⎜ ⎟τ e −1 = ⎜ ⎟⎜ ⎟ e −1 = ⎜ ⎟ Le −1 ⎝R⎠ ⎝ R⎠ ⎝ R ⎠⎝ R ⎠ 2

⎛ 6.00 V ⎞ UE = ⎜ ⎟ (2.50 H)(0.3679) = 0.517 J ⎝ 8.00 Ω ⎠ ⎛ E2 ⎞ E2 (c) PR = i 2 R = ⎜ ⎟ (1 − e−t /τ ) 2 = (1 − 2e−t /τ + e−2t /τ ) R ⎝ R⎠ τ

U R = ∫ PR dt = 0

UR =

τ

E2 τ E2 ⎡ τ ⎤ − t /τ −2 t / τ − + = (1 2 e e ) dt t + 2τ e− t /τ − e−2t /τ ⎥ R ∫0 R ⎢⎣ 2 ⎦0

E2 ⎡ τ τ ⎤ E2 ⎡ τ τ ⎤ τ + 2τ e−1 − e−2 − 2τ + ⎥ = ⎢ − + 2τ e −1 − e −2 ⎥ ⎢ R⎣ 2 2⎦ R ⎣ 2 2 ⎦

⎛ E 2 ⎞⎛ L ⎞ −1 −2 UR = ⎜ ⎟⎜ ⎟ ⎡⎣ −1 + 4e − e ⎤⎦ ⎝ 2R ⎠ ⎝ R ⎠ 2

30.54.

2

⎛E⎞ ⎛ 6.00 V ⎞ 1 U R = ⎜ ⎟ ( 12 L) ⎡⎣ −1 + 4e−1 − e −2 ⎤⎦ = ⎜ ⎟ 2 (2.50 H)(0.3362) = 0.236 J R ⎝ ⎠ ⎝ 8.00 Ω ⎠ (d) EVALUATE: U E = U R + U L . (0.517 J = 0.236 J + 0.281 J) The energy supplied by the battery equals the sum of the energy stored in the magnetic field of the inductor and the energy dissipated in the resistance of the inductor. IDENTIFY: This is a decaying R-L circuit with I 0 = E / R . i (t ) = I 0e − ( R / L )t . SET UP: E = 60.0 V , R = 240 Ω and L = 0.160 H . The rate at which energy stored in the inductor is decreasing is iLdi / dt . 2

EXECUTE:

(a) U =

2

⎛ 60 V ⎞ 1 1 ⎛E⎞ 1 −3 LI 0 2 = L ⎜ ⎟ = (0.160 H) ⎜ ⎟ = 5.00 ×10 J. Ω 2 2 ⎝ R⎠ 2 240 ⎝ ⎠

Inductance

(b) i =

30-15

E − ( R / L )t E2 di R dU L di dU L (60 V) 2 −2(240 / 0.160)(4.00×10−4 ) ⇒ =− i⇒ = iL = − Ri 2 = e−2( R / L ) t . =− = −4.52 W. e e R dt L dt dt R dt 240 Ω

(c) In the resistor, PR =

dU R 2 (60 V) 2 −2(240 / 0.160)(4.00×10−4 ) E2 = i R = e −2( R / L ) t = = 4.52 W . e dt R 240 Ω ∞

(d) PR (t ) = i 2 R =

30.55.

E 2 −2( R / L )t E 2 −2( R / L ) t E 2 L (60 V) 2 (0.160 H) e . UR = = = = 5.00 ×10−3 J, which is the same as e R R ∫0 R 2R 2(240 Ω) 2

part (a). EVALUATE: During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor. IDENTIFY and SET UP: Follow the procedure specified in the problem. 12 Li 2 is the energy stored in the inductor di q − = 0. dt C d d d di di qi ⎛ di ⎞ EXECUTE: Multiplying by –i gives i 2 R + Li + = 0. U L = ( 12 Li 2 ) = 12 L ( i 2 ) = 12 L ⎜ 2i ⎟ = Li , the dt dt dt dt dt C ⎝ dt ⎠ and q 2 / 2C is the energy stored in the capacitor. The equation is −iR − L

d d ⎛ q2 ⎞ 1 d 2 1 dq qi UC = ⎜ (q ) = (2q ) = , the third term. i 2 R = PR , the rate at which ⎟= dt dt ⎝ 2C ⎠ 2C dt 2C dt C d electrical energy is dissipated in the resistance. U L = PL , the rate at which the amount of energy stored in the dt d inductor is changing. U C = PC , the rate at which the amount of energy stored in the capacitor is changing. dt EVALUATE: The equation says that PR + PL + PC = 0; the net rate of change of energy in the circuit is zero. Note second term.

that at any given time one of PC or PL is negative. If the current and U L are increasing the charge on the capacitor and U C are decreasing, and vice versa. 30.56.

IDENTIFY: The energy stored in a capacitor is U C = 12 Cv 2 . The energy stored in an inductor is U L = 12 Li 2 . Energy conservation requires that the total stored energy be constant. SET UP: The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero. EXECUTE: (a) Initially v = 16.0 V and i = 0 . U L = 0 and U C = 12 Cv 2 = 12 (5.00 ×10−6 F)(16.0 V)2 = 6.40 ×10−4 J . The total energy stored is 0.640 mJ . (b) The current is maximum when q = 0 and U C = 0 . U C + U L = 6.40 ×10−4 J so U L = 6.40 ×10−4 J .

2(6.40 ×10−4 J) = 0.584 A . 3.75 ×10−3 H EVALUATE: The maximum charge on the capacitor is Q = CV = 80.0 μ C . 1 2

30.57.

2 Limax = 6.40 ×10−4 J and imax =

IDENTIFY and SET UP: Use U C = 12 CVC2 (energy stored in a capacitor) to solve for C. Then use Eq.(30.22) and ω = 2π f to solve for the L that gives the desired current oscillation frequency. EXECUTE:

1 (2 π f )2 C 2π LC f = 3500 Hz gives L = 9.31 μ H EVALUATE: f is in Hz and ω is in rad/s; we must be careful not to confuse the two. IDENTIFY: Apply energy conservation to the circuit. SET UP: For a capacitor V = q / C and U = q 2 / 2C . For an inductor U = 12 Li 2 f =

30.58.

1

VC = 12.0 V; U C = 12 CVC2 so C = 2U C / VC2 = 2(0.0160 J)/(12.0 V) 2 = 222 μ F

EXECUTE: (b)

1 2 Limax 2

(c) U max =

so L =

Q 6.00 ×10−6 C = = 0.0240 V. C 2.50 ×10−4 F Q2 Q 6.00 ×10−6 C = , so imax = = = 1.55 ×10−3 A 2C LC (0.0600 H)(2.50 ×10−4 F) (a) Vmax =

1 2 1 Limax = (0.0600 H)(1.55 ×10−3 A) 2 = 7.21×10−8 J. 2 2

30-16

Chapter 30

30.59.

3 1 1 (d) If i = imax then U L = U max = 1.80 × 10 −8 J and U C = U max = 4 2 4 3 q= Q = 5.20 ×10−6 C. 4 1 1 q2 EVALUATE: U max = Li 2 + for all times. 2 2C IDENTIFY: Set U B = K , where K = 12 mv 2 .

(

(3/ 4)Q 2C

)

2

=

q2 . This gives 2C

SET UP: The energy density in the magnetic field is u B = B 2 / 2μ 0 . Consider volume V = 1 m 3 of sunspot material. EXECUTE: The energy density in the sunspot is uB = B 2 /2 μ0 = 6.366 ×104 J /m3. The total energy stored in

volume V of the sunspot is U B = u BV . The mass of the material in volume V of the sunspot is m = ρV . K = U B so 30.60.

1 2

mv 2 = U B .

1 2

ρVv 2 = uBV . The volume divides out, and v = 2uB / ρ = 2 ×104 m/s .

EVALUATE: The speed we calculated is about 30 times smaller than the escape speed. IDENTIFY: i (t ) is given by Eq.(30.14). SET UP: The graph shows V = 0 at t = 0 and V approaches the constant value of 25 V at large times. EXECUTE: (a) The voltage behaves the same as the current. Since VR is proportional to i, the scope must be across the 150 Ω resistor. (b) From the graph, as t → ∞, VR → 25 V, so there is no voltage drop across the inductor, so its internal resistance ⎛ 1⎞ must be zero. VR = Vmax (1 − e −t / r ) . When t = τ , VR = Vmax ⎜1 − ⎟ ≈ 0.63Vmax . From the graph, V = 0.63Vmax = 16 V at ⎝ e⎠ t ≈ 0.5 ms . Therefore τ = 0.5 ms . L / R = 0.5 ms gives L = (0.5 ms) (150 Ω) = 0.075 H . (c) The graph if the scope is across the inductor is sketched in Figure 30.60. EVALUATE: At all times VR + VL = 25.0 V . At t = 0 all the battery voltage appears across the inductor since i = 0 . At t → ∞ all the battery voltage is across the resistance, since di / dt = 0 .

30.61.

Figure 30.60 IDENTIFY and SET UP: The current grows in the circuit as given by Eq.(30.14). In an R-L circuit the full emf initially is across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is represented as a resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a resistance is given by Ohm’s law. EXECUTE: (a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. (b) At t → ∞ the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance RL . The final voltage across the solenoid is IRL , where I is the final current in the circuit. (c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. (d) As t → ∞, E − IR − IRL = 0

E = 50 V and from the graph IRL = 15 V (the final voltage across the inductor), so IR = 35 V and I = (35 V)/R = 3.5 A (e) IRL = 15 V, so RL = (15 V)/(3.5 A) = 4.3 Ω E − VL − iR = 0, where VL includes the voltage across the resistance of the solenoid. E VL = E − iR, i = (1 − e−t /τ ) , so VL = E ⎡⎢1 − RR (1 − e−t /τ ) ⎤⎥ Rtot tot ⎣ ⎦ E = 50 V, R = 10 Ω, Rtot = 14.3 Ω, so when t = τ , VL = 27.9 V. From the graph, VL has this value when t = 3.0 ms (read approximately from the graph), so τ = L / Rtot = 3.0 ms. Then L = (3.0 ms)(14.3 Ω ) = 43 mH. EVALUATE: At t = 0 there is no current and the 50 V measured by the oscilloscope is the induced emf due to the inductance of the solenoid. As the current grows, there are voltage drops across the two resistances in the circuit.

Inductance

30-17

We derived an equation for VL , the voltage across the solenoid. At t = 0 it gives VL = E and at t → ∞ it gives 30.62.

30.63.

VL = ER / Rtot = iR. IDENTIFY: At t = 0 , i = 0 through each inductor. At t → ∞ , the voltage is zero across each inductor. SET UP: In each case redraw the circuit. At t = 0 replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire. EXECUTE: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is shown in E 50 V = 0.333 A . V1 = (100 Ω)(0.333 A) = 33.3 V . V4 = (50 Ω)(0.333 A) = 16.7 V . V3 = 0 Figure 30.62a. i = = R 150 Ω since no current flows through it. V2 = V4 = 16.7 V , since the inductor is in parallel with the 50 Ω resistor. A1 = A3 = 0.333 A, A2 = 0 . (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. The simplified 50 V = 0.385 A . V1 = (100 Ω)(0.385 A) = 38.5 V ; V2 = 0 ; circuit is sketched in Figure 30.62b. i = E/R = 130 Ω 11.5 V 11.5 V = 0.153 A; i3 = = 0.230 A . V3 = V4 = 50 V − 38.5 V = 11.5 V . i1 = 0.385 A; i2 = 75 Ω 50 Ω EVALUATE: Just after the switch is closed the current through the battery is 0.333 A. After a long time the current through the battery is 0.385 A. After a long time there is an additional current path, the equivalent resistance of the circuit is decreased and the current has increased.

Figure 30.62 IDENTIFY and SET UP: Just after the switch is closed, the current in each branch containing an inductor is zero and the voltage across any capacitor is zero. The inductors can be treated as breaks in the circuit and the capacitors can be replaced by wires. After a long time there is no voltage across each inductor and no current in any branch containing a capacitor. The inductors can be replaced by wires and the capacitors by breaks in the circuit. EXECUTE: (a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor, so V3 = 0. V2 + V3 = V4 = V5 , and since V5 = 0 and V3 = 0, V4 and V2 are also zero.

V4 = 0 means V3 reads zero. V1 then must equal 40.0 V, and this means the current read by A1 is (40.0 V)/(50.0 Ω) = 0.800 A. A2 + A3 + A4 = A1 , but A2 = A3 = 0 so A4 = A1 = 0.800 A. A1 = A4 = 0.800 A; all other

ammeters read zero. V1 = 40.0 V and all other voltmeters read zero. (b) After a long time the capacitor is fully charged so A4 = 0. The current through the inductor isn’t changing, so V2 = 0. The currents can be calculated from the equivalent circuit that replaces the inductor by a short circuit, as shown in Figure 30.63a.

Figure 30.63a I = (40.0 V)/(83.33 Ω) = 0.480 A; A1 reads 0.480 A V1 = I (50.0 Ω) = 24.0 V The voltage across each parallel branch is 40.0 V – 24.0 V = 16.0 V V2 = 0, V3 = V4 = V5 = 16.0 V

V3 = 16.0 V means A2 reads 0.160 A. V4 = 16.0 V means A3 reads 0.320 A. A4 reads zero. Note that A2 + A3 = A1. (c) V5 = 16.0 V so Q = CV = (12.0 μ F)(16.0 V) = 192 μ C

30-18

Chapter 30

(d) At t = 0 and t → ∞, V2 = 0. As the current in this branch increases from zero to 0.160 A the voltage V2 reflects the rate of change of the current. The graph is sketched in Figure 30.63b.

Figure 30.63b

30.64.

30.65.

EVALUATE: This reduction of the circuit to resistor networks only apply at t = 0 and t → ∞. At intermediate times the analysis is complicated. IDENTIFY: At all times v1 + v2 = 25.0 V . The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing. SET UP: Immediately after closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing. EXECUTE: (a) i = 0 so v1 = 0 and v2 = 25.0 V . The ammeter reading is A = 0 . v 25.0 V (b) After a long time, v2 = 0 and v1 = 25.0 V. v1 = iR and i = 1 = = 1.67 A. The ammeter reading is R 15.0 Ω A = 1.67A. (c) None of the answers in (a) and (b) depend on L so none of them would change. EVALUATE: The inductance L of the circuit affects the rate at which current reaches its final value. But after a long time the inductor doesn’t affect the circuit and the final current does not depend on L. IDENTIFY: At t = 0 , i = 0 through each inductor. At t → ∞ , the voltage is zero across each inductor. SET UP: In each case redraw the circuit. At t = 0 replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire. EXECUTE: (a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the 40.0 Ω and 15.0 Ω resistors and is equal to

(25.0 V) (40.0 Ω + 15.0 Ω) = 0.455 A. A1 = A4 = 0.455 A; A2 = A3 = 0. (b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit . The circuit is equivalent to the circuit sketched in Figure 30.65. I = (25.0 V) (42.73 Ω) = 0.585 A . A1 reads 0.585 A. The voltage across each parallel branch is 25.0 V − (0.585 A)(40.0 Ω) = 1.60 V . A2 reads (1.60 V) /(5.0 Ω) = 0.320 A . A3 reads (1.60 V) (10.0 Ω) = 0.160 A . A4 reads (1.60 V) (15.0 Ω) = 0.107 A. EVALUATE: Just after the switch is closed the current through the battery is 0.455 A. After a long time the current through the battery is 0.585 A. After a long time there are additional current paths, the equivalent resistance of the circuit is decreased and the current has increased.

Figure 30.65 30.66.

IDENTIFY: Closing S 2 and simultaneously opening S1 produces an L-C circuit with initial current through the inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is 2 a maximum the current is zero. Conservation of energy says that the maximum energy 12 Limax stored in the inductor 2 qmax stored in the capacitor. C = 3.50 A , the current in the inductor just after the switch is closed.

equals the maximum energy SET UP:

imax

1 2

Inductance

EXECUTE:

30.67.

(a)

1 2

2 Limax = 12

30-19

2 qmax . C

qmax = ( LC )imax = (2.0 ×10−3 H)(5.0 ×10−6 F)(3.50 A) = 3.50 ×10−4 C = 0.350 mC . (b) When q is maximum, i = 0 . EVALUATE: In the final circuit the current will oscillate. IDENTIFY: Apply the loop rule to each parallel branch. The voltage across a resistor is given by iR and the voltage across an inductor is given by L di / dt . The rate of change of current through the inductor is limited. SET UP:

With S closed the circuit is sketched in Figure 30.67a. The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is closed the current in the inductor has not had time to increase from zero, so i2 = 0. Figure 30.67a

EXECUTE: (a) E − vab = 0, so vab = 60.0 V (b) The voltage drops across R, as we travel through the resistor in the direction of the current, so point a is at higher potential. (c) i2 = 0 so vR2 = i2 R2 = 0

E − vR2 − vL = 0 so vL = E = 60.0 V (d) The voltage rises when we go from b to a through the emf, so it must drop when we go from a to b through the inductor. Point c must be at higher potential than point d. di (e) After the switch has been closed a long time, 2 → 0 so vL = 0. Then E − vR2 = 0 and i2 R2 = E dt E 60.0 V so i2 = = = 2.40 A. R2 25.0 Ω SET UP: The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is opened again the current through the inductor hasn’t had time to change and is still i2 = 2.40 A. The circuit is sketched in Figure 30.67b. EXECUTE: The current through R1 is i2 = 2.40 A, in the direction b to a. Thus vab = −i2 R1 = −(2.40 A)(40.0 Ω)

vab = −96.0 V Figure 30.67b (f ) Point where current enters resistor is at higher potential; point b is at higher potential. (g) vL − vR1 − vR2 = 0

vL = vR1 + vR2 vR1 = −vab = 96.0 V; vR2 = i2 R2 = (2.40 A)(25.0 Ω) = 60.0 V

Then vL = vR1 + vR2 = 96.0 V + 60.0 V = 156 V. As you travel counterclockwise around the circuit in the direction of the current, the voltage drops across each resistor, so it must rise across the inductor and point d is at higher potential than point c. The current is decreasing, so the induced emf in the inductor is directed in the direction of the current. Thus, vcd = −156 V. (h) Point d is at higher potential. EVALUATE: The voltage across R1 is constant once the switch is closed. In the branch containing R2 , just after S is closed the voltage drop is all across L and after a long time it is all across R2 . Just after S is opened the same current flows in the single loop as had been flowing through the inductor and the sum of the voltage across the resistors equals the voltage across the inductor. This voltage dies away, as the energy stored in the inductor is dissipated in the resistors.

30-20

Chapter 30

30.68.

Apply the loop rule to the two loops. The current through the inductor doesn't change abruptly. di SET UP: For the inductor E = L and E is directed to oppose the change in current. dt EXECUTE: (a) Switch is closed, then at some later time di di = 50.0 A/s ⇒ vcd = L = (0.300 H) (50.0 A/s) = 15.0 V. dt dt 60.0 V = 1.50 A. The top circuit loop: 60.0 V = i1R1 ⇒ i1 = 40.0 Ω 45.0 V = 1.80 A. The bottom loop: 60 V − i2 R2 − 15.0 V = 0 ⇒ i2 = 25.0 Ω 60.0 V = 2.40 A, and immediately when the switch is opened, the inductor maintains (b) After a long time: i2 = 25.0 Ω this current, so i1 = i2 = 2.40 A. IDENTIFY:

EVALUATE: 30.69.

The current through R1 changes abruptly when the switch is closed.

IDENTIFY and SET UP: The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after S1 is closed, i = 0. After a long time i has reached its final value and di/dt = 0. The voltage across a resistor depends on i and the voltage across an inductor depends on di/dt.

Figure 30.69a EXECUTE: (a) At time t = 0, i0 = 0 so vac = i0 R0 = 0. By the loop rule E − vac − vcb = 0, so vcb = E − vac = E = 36.0 V.

( i0 R = 0 so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the changing current.) di di (b) After a long time 0 → 0 so the potential − L 0 across the inductor becomes zero. The loop rule gives dt dt E − i0 ( R0 + R) = 0. E 36.0 V i0 = = = 0.180 A R0 + R 50.0 Ω + 150 Ω vac = i0 R0 = (0.180 A)(50.0 Ω) = 9.0 V di Thus vcb = i0 R + L 0 = (0.180 A)(150 Ω) + 0 = 27.0 V (Note that vac + vcb = E. ) dt (c) E − vac − vcb = 0 di E − iR0 − iR − L = 0 dt ⎛ L ⎞ di di E L = E − i( R0 + R) and ⎜ ⎟ = −i + dt R R dt R R0 + + 0 ⎠ ⎝

di ⎛ R + R0 ⎞ =⎜ ⎟ dt −i + E /( R + R0 ) ⎝ L ⎠ Integrate from t = 0, when i = 0, to t, when i = i0 :

∫

i0

0

i0

⎡ di R + R0 t E ⎤ ⎛ R + R0 ⎞ = dt = − ln ⎢ −i + ⎥ =⎜ ⎟ t, ∫ 0 −i + E /( R + R0 ) L R + R0 ⎦ 0 ⎝ L ⎠ ⎣

⎛ ⎛ E ⎞ E ⎞ ⎛ R + R0 ⎞ so ln ⎜ −i0 + ⎟ − ln ⎜ ⎟ = −⎜ ⎟t R R R R + + ⎝ L ⎠ 0 ⎠ 0 ⎠ ⎝ ⎝ ⎛ −i + E /( R + R0 ) ⎞ ⎛ R + R0 ⎞ ln ⎜ 0 ⎟ = −⎜ ⎟t ⎝ L ⎠ ⎝ E /( R + R0 ) ⎠ E −i0 + E /( R + R0 ) 1 − e− ( R + R0 ) t / L = e− ( R + R0 ) t / L and i0 = R + R0 E /( R + R0 ) 36.0 V Substituting in the numerical values gives i0 = (1 − e−(200 Ω / 4.00 H)t ) = (0.180 A) (1 − e−t / 0.020 s ) 50 Ω + 150 Ω

Taking exponentials of both sides gives

(

)

Inductance

30-21

At t → 0, i0 = (0.180 A)(1 − 1) = 0 (agrees with part (a)). At t → ∞, i0 = (0.180 A)(1 − 0) = 0.180 A (agrees with part (b)). ER0 vac = i0 R0 = 1 − e− ( R + R0 )t / L = 9.0 V (1 − e−t / 0.020 s ) R + R0

(

)

vcb = E − vac = 36.0 V − 9.0 V(1 − e−t / 0.020 s ) = 9.0 V(3.00 + e−t / 0.020 s ) At t → 0, vac = 0, vcb = 36.0 V (agrees with part (a)). At t → ∞, vac = 9.0 V, vcb = 27.0 V (agrees with part (b)). The graphs are given in Figure 30.69b.

30.70.

Figure 30.69b EVALUATE: The expression for i(t) we derived becomes Eq.(30.14) if the two resistors R0 and R in series are replaced by a single equivalent resistance R0 + R. IDENTIFY: Apply the loop rule. The current through the inductor doesn't change abruptly. SET UP: With S2 closed, vcb must be zero. EXECUTE: (a) Immediately after S 2 is closed, the inductor maintains the current i = 0.180 A through R. The loop rule around the outside of the circuit yields 36 V = 0.720 A . E + EL − iR − i0 R0 = 36.0 V + (0.18 A)(150 Ω ) − (0.18 A)(150 Ω ) − i0 (50 Ω ) = 0 . i0 = 50 Ω vac = (0.72 A)(50 V) = 36.0 V and vcb = 0 . E 36.0 V = = 0.720 A , iR = 0 and is 2 = 0.720 A . (b) After a long time, vac = 36.0 V, and vcb = 0. Thus i0 = R0 50 Ω −1 E − ( R L )t e (c) i0 = 0.720 A , iR (t ) = and iR (t ) = (0.180 A)e − (12.5 s ) t . Rtotal

is 2 (t ) = (0.720 A) − (0.180 A)e − (12.5 s

−1

)t

(

= (0.180 A) 4 − e − (12.5 s

−1

)t

). The graphs of the currents are given in Figure 30.70.

EVALUATE: R0 is in a loop that contains just E and R0, so the current through R0 is constant. After a long time the current through the inductor isn't changing and the voltage across the inductor is zero. Since vcb is zero, the voltage across R must be zero and iR becomes zero.

Figure 30.70

30-22

Chapter 30

30.71.

IDENTIFY: The current through an inductor doesn't change abruptly. After a long time the current isn't changing and the voltage across each inductor is zero. SET UP: Problem 30.47 shows how to find the equivalent inductance of inductors in series and parallel. EXECUTE: (a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V. (b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent to the circuit shown in Figure 30.71a. I = (20.0 V) (75.0 Ω) = 0.267 A . The voltage between points a and b is zero, so the voltmeter reads zero. (c) Use the results of Problem 30.49 to combine the inductor network into its equivalent, as shown in Figure 30.71b. R = 75.0 Ω is the equivalent resistance. Eq.(30.14) says i = (E R)(1 − e −t τ ) with

τ = L / R = (10.8 mH) (75.0 Ω) = 0.144 ms . E = 20.0 V , R = 75.0 Ω , t = 0.115 ms so i = 0.147 A . VR = iR = (0.147 A)(75.0 Ω) = 11.0 V . 20.0 V − VR − VL = 0 and VL = 20.0 V − VR = 9.0 V . The ammeter reads 0.147 A and the voltmeter reads 9.0 V. EVALUATE: The current through the battery increases from zero to a final value of 0.267 A. The voltage across the inductor network drops from 20.0 V to zero.

Figure 30.71 30.72.

IDENTIFY: At steady state with the switch in position 1, no current flows to the capacitors and the inductors can be replaced by wires. Apply conservation of energy to the circuit with the switch in position 2. SET UP: Replace the series combinations of inductors and capacitors by their equivalents. For the inductors use the results of Problem 30.47. E 75.0 V EXECUTE: (a) At steady state i = = = 0.600 A . R 125 Ω (b) The equivalent circuit capacitance of the two capacitors is given by

1 1 1 = + and Cs = 14.6 μ F . Cs 25 μ F 35 μ F

Ls = 15.0 mH + 5.0 mH = 20.0 mH . The equivalent circuit is sketched in Figure 30.72a. q2 1 2 = Li0 . q = i0 LC = (0.600 A) (20 ×10−3 H)(14.6 ×10−6 F) = 3.24 ×10−4 C . As shown 2C 2 in Figure 30.72b, the capacitors have their maximum charge at t = T / 4 .

Energy conservation:

t = 14 T = 14 (2π LC ) =

π

π

LC = (20 ×10−3 H)(14.6 × 10−6 F) = 8.49 ×10−4 s 2 2 EVALUATE: With the switch closed the battery stores energy in the inductors. This then is the energy in the L-C circuit when the switch is in position 2.

Figure 30.72 30.73.

IDENTIFY: SET UP:

Follow the steps specified in the problem. Find the flux through a ring of height h, radius r and thickness dr. Example 28.19 shows that B =

inside the toroid.

μ 0 Ni 2π r

Inductance

μ Nih dr μ0 Nih ⎛ μ Ni ⎞ = (a) Φ B = ∫ B (h dr ) = ∫ ⎜ 0 ⎟ ( h dr ) = 0 ln(b a ). π r 2 2π ∫a r 2π ⎠ a a ⎝ b

EXECUTE: (b) L =

30-23

b

b

N Φ B μ0 N 2h ln(b a ). = i 2π

b − a (b − a ) 2 μ0 N 2h ⎛ b − a ⎞ + + ⋅⋅⋅ ⇒ ≈ L ⎜ ⎟. a 2a 2 2π ⎝ a ⎠ EVALUATE: h(b − a ) is the cross-sectional area A of the toroid and a is approximately the radius r, so this result is approximately the same as the result derived in Example 30.3. IDENTIFY: The direction of the current induced in circuit A is given by Lenz’s law. SET UP: When the switch is closed current flows counterclockwise in the circuit on the left, from the positive plate of the capacitor. The current decreases as a function of time, as the charge and voltage of the capacitor decrease. EXECUTE: At loop A the magnetic field from the wire of the other circuit adjacent to A is into the page. The magnetic field of this current is decreasing, as the current decreases. Therefore, the magnetic field of the induced current in A is directed into the page inside A and to produce a magnetic field in this direction the induced current is clockwise. EVALUATE: The magnitude of the emf induced in circuit A decreases with time after the switch is closed, because the rate of change of the current in the other circuit decreases. (a) IDENTIFY and SET UP: With switch S closed the circuit is shown in Figure 30.75a. (c) ln(b / a ) = ln(1 − (b − a ) / a ) ≈

30.74.

30.75.

Apply the loop rule to loops 1 and 2. EXECUTE: loop 1 E − i1R1 = 0 i1 =

E (independent of t) R1

Figure 30.75a

loop (2) E − i2 R2 − L

di2 =0 dt

E (1 − e− R2t / L ) R2 E E di The expressions derived in part (a) give that as t → ∞, i1 = and i2 = . Since 2 → 0 at dt R1 R2

This is in the form of equation (30.12), so the solution is analogous to Eq.(30.14): i2 = (b) EVALUATE:

steady-state, the inductance then has no effect on the circuit. The current in R1 is constant; the current in R2 starts at zero and rises to E / R2 . (c) IDENTIFY and SET UP:

The circuit now is as shown in Figure 30.75b. Let t = 0 now be when S is opened. E At t = 0, i = . R2

Figure 30.75b Apply the loop rule to the single current loop. di di EXECUTE: −i ( R1 + R2 ) − L = 0. (Now is negative.) dt dt di di ⎛ R + R2 ⎞ = −⎜ 1 L = −i ( R1 + R2 ) gives ⎟ dt dt i ⎝ L ⎠ Integrate from t = 0, when i = I 0 = E / R2 , to t.

∫

i I0

⎛ i ⎞ di ⎛ R1 + R2 ⎞ ⎛ R + R2 ⎞ t = −⎜ 1 ⎟t ⎟ ∫ 0 dt and ln ⎜ ⎟ = − ⎜ I i L ⎝ L ⎠ ⎝ ⎠ ⎝ 0⎠

Taking exponentials of both sides of this equation gives i = I 0e − ( R1 + R2 )t / L =

E − ( R1 + R2 )t / L e R2

30-24

Chapter 30

(d) IDENTIFY and SET UP: Use the equation derived in part (c) and solve for R2 and E . EXECUTE: L = 22.0 H V2 V 2 (120 V) 2 RR1 = = 40.0 W gives R1 = = = 360 Ω. 40.0 W R1 PR1

We are asked to find R2 and E . Use the expression derived in part (c). I 0 = 0.600 A so E / R2 = 0.600 A i = 0.150 A when t = 0.080 s, so i = 1 4

E − ( R1 + R2 )t / L e gives 0.150 A = (0.600 A)e − ( R1 + R2 )t / L R2

= e − ( R1 + R2 ) t / L so ln 4 = ( R1 + R2 )t / L

L ln 4 (22.0 H)ln 4 − R1 = − 360 Ω = 381.2 Ω − 360 Ω = 21.2 Ω t 0.080 s Then E = (0.600 A)R2 = (0.600 A)(21.2 Ω) = 12.7 V. (e) IDENTIFY and SET UP: Use the expressions derived in part (a). R2 =

E 12.7 V = = 0.0353 A R1 360 Ω EVALUATE: When the switch is opened the current through the light bulb jumps from 0.0353 A to 0.600 A. Since the electrical power dissipated in the bulb (brightness) depend on i 2 , the bulb suddenly becomes much brighter. IDENTIFY: Follow the steps specified in the problem. SET UP: The current in an inductor does not change abruptly. EXECUTE: (a) Using Kirchhoff’s loop rule on the left and right branches: di di Left: E − (i1 + i2 ) R − L 1 = 0 ⇒ R (i1 + i2 ) + L 1 = E . dt dt q2 q2 Right: E − (i1 + i2 ) R − = 0 ⇒ R (i1 + i2 ) + = E . C C E (b) Initially, with the switch just closed, i1 = 0, i2 = and q2 = 0. R (c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise but straightforward exercise. We will show that the initial conditions are satisfied: E −βt E At t = 0, q2 = e sin(ω t ) = sin(0) = 0. ωR ωR E E i1 (t ) = (1 − e − β t [(2ω RC ) −1 sin(ω t ) + cos(ω t )] ⇒ i1 (0) = (1 − [cos(0)]) = 0. R R 1 1 − = 625 rad/s . (d) When does i2 first equal zero? ω = LC (2 RC ) 2 E i2 (t ) = 0 = e − β t [ −(2ω RC ) −1 sin(ω t ) + cos(ω t )] ⇒ − (2ω RC ) −1 tan(ω t ) + 1 = 0 and R tan(ω t ) = +2ω RC = +2(625 rad/s)(400 Ω)(2.00 ×10−6 F) = +1.00. 0.785 ω t = arctan( + 1.00) = +0.785 ⇒ t = = 1.256 ×10−3 s. 625 rad/s EVALUATE: As t → ∞ , i1 → E / R , q2 → 0 and i2 → 0 . NΦB IDENTIFY: Apply L = to calculate L. i μ Ni μ Ni SET UP: In the air the magnetic field is BAir = 0 . In the liquid, BL = W W μ 0 Ni K μ 0 Ni EXECUTE: (a) Φ B = BA = BL AL + BAir AAir = (( D − d )W ) + ( dW ) = μ 0 Ni[( D − d ) + Kd ] . W W NΦB d d ⎛ L − L0 ⎞ = μ 0 N 2 [( D − d ) + Kd ] = L0 − L0 + Lf = L0 + ⎜ f L= ⎟d . i D D ⎝ D ⎠ ⎛ L − L0 ⎞ 2 2 d =⎜ ⎟ D, where L0 = μ 0 N D, and L f = K μ 0 N D. L L − 0 ⎠ ⎝ f EXECUTE:

30.76.

30.77.

The current through the light bulb before the switch is opened is i1 =

Inductance

30-25

d⎞ ⎛ (b) Using K = χ m + 1 we can find the inductance for any height L = L0 ⎜ 1 + χ m ⎟ . D⎠ ⎝ _______________________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury d=D 4 0.63024 H 0.63000 H d=D 2

30.78.

30.79.

0.63048 H

0.62999 H

d = 3D 4 0.63072 H 0.62999 H d=D 0.63096 H 0.62998 H ________________________________________________________________________________ The values χ m (O 2 ) = 1.52 ×10−3 and χ m (Hg) = −2.9 ×10−5 have been used. EVALUATE: (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury. IDENTIFY: The induced emf across the two coils is due to both the self-inductance of each and the mutual inductance of the pair of coils. di SET UP: The equivalent inductance is defined by E = Leq , where E and i are the total emf and current across dt the combination. di di di di di EXECUTE: Series: L1 1 + L2 2 + M 21 1 + M 12 2 ≡ Leq . dt dt dt dt dt di di1 di2 di di + and M 12 = M 21 ≡ M , so ( L1 + L2 + 2M ) = Leq and Leq = L1 + L2 + 2M . But i = i1 + i2 ⇒ = dt dt dt dt dt di di di di di di di di di and L2 2 + M 21 1 = Leq , with 1 + 2 = and M 12 = M 21 ≡ M . Parallel: We have L1 1 + M 12 2 = Leq dt dt dt dt dt dt dt dt dt di di di To simplify the algebra let A = 1 , B = 2 , and C = . So L1 A + MB = LeqC , L2 B + MA = LeqC , A + B = C. Now dt dt dt solve for A and B in terms of C. ( L1 − M ) A + ( M − L2 ) B = 0 using A = C − B . ( L1 − M )(C − B) + ( M − L2 ) B = 0. ( M − L1 ) ( L1 − M )C − ( L1 − M ) B + ( M − L2 ) B = 0. (2M − L1 − L2 ) B = ( M − L1 )C and B = C. (2M − L1 − L2 ) M − L2 ( M − L1 )C (2M − L1 − L2 ) − M + L1 = C , or A = C . Substitute A in B back But A = C − B = C − 2M − L1 − L2 (2M − L1 − L2 ) (2M − L1 − L2 ) into original equation: L1 ( M − L2 )C M ( M − L1 ) M 2 − L1L2 LL −M2 + C = LeqC and . C = LeqC . Finally, Leq = 1 2 2M − L1 − L2 (2M − L1 − L2 ) 2 M − L1 − L2 L1 + L2 − 2 M EVALUATE: If the flux of one coil doesn't pass through the other coil, so M = 0 , then the results reduce to those of problem 30.47. IDENTIFY: Apply Kirchhoff’s loop rule to the top and bottom branches of the circuit. SET UP: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is zero. di E q di i E EXECUTE: E − i1R1 − L 1 = 0 ⇒ i1 = (1 − e− ( R1 L ) t ). E − i2 R2 − 2 = 0 ⇒ − 2 R2 − 2 = 0 ⇒ i2 = e − (1 R2C )t ) . dt R1 C dt C R2 t

q2 = ∫ i2 dt ′ = − 0

t

E R2Ce − (1/ R2C )t ′ = EC (1 − e − (1/ R2C ) t ) . R2 0

E E 48.0 V (1 − e0 ) = 0, i2 = e0 = = 9.60 ×10−3 A. R1 R2 5000 Ω E E 48.0 V E = 1.92 A, i2 = e−∞ = 0. A good definition of a “long time” is (c) As t → ∞ : i1 (∞) = (1 − e−∞ ) = = R1 R1 25.0 Ω R2 many time constants later. (b) i1 (0) =

30-26

Chapter 30

(d) i1 = i2 ⇒

E (1 − e− ( R1 R1

L)t

)=

E − (1 R2C )t e ⇒ (1 − e− ( R1 R2

L )t

)=

R1 − (1 R2C )t e . Expanding the exponentials like R2

2

⎞ R R ⎛ t t2 1⎛ R ⎞ x 2 x3 + 2 2 − "⎟ and + + ", we find: 1 t − ⎜ 1 ⎟ t 2 + " = 1 ⎜1 − L R2 ⎝ RC 2 R C 2⎝ L ⎠ 2 3! ⎠ ⎛R R ⎞ R t ⎜ 1 + 21 ⎟ + O(t 2 ) + ⋅⋅⋅ = 1 , if we have assumed that t X L the phase angle is negative and the source voltage lags the current.

(b) I =

Figure 31.16 31.17.

IDENTIFY and SET UP: Calculate the impendance of the circuit and use Eq.(31.22) to find the current amplitude. The voltage amplitudes across each circuit element are given by Eqs.(31.7), (31.13), and (31.19). The phase angle is calculated using Eq.(31.24). The circuit is shown in Figure 31.17a.

No inductor means X L = 0 R = 200 Ω, C = 6.00 ×10−6 F, V = 30.0 V, ω = 250 rad/s Figure 31.17a

31-6

Chapter 31

EXECUTE:

(a) X C =

1 1 = = 666.7 Ω ω C (250 rad/s)(6.00 ×10−6 F)

Z = R 2 + ( X L − X C ) 2 = (200 Ω) 2 + (666.7 Ω) 2 = 696 Ω V 30.0 V = = 0.0431 A = 43.1 mA Z 696 Ω (c) Voltage amplitude across the resistor: VR = IR = (0.0431 A)(200 Ω) = 8.62 V Voltage amplitude across the capacitor: VC = IX C = (0.0431 A)(666.7 Ω) = 28.7 V X − X C 0 − 666.7 Ω (d) tan φ = L = = −3.333 so φ = −73.3° R 200 Ω The phase angle is negative, so the source voltage lags behind the current. (e) The phasor diagram is sketched qualitatively in Figure 31.17b.

(b) I =

31.18.

31.19.

Figure 31.17b EVALUATE: The voltage across the resistor is in phase with the current and the capacitor voltage lags the current by 90°. The presence of the capacitor causes the source voltage to lag behind the current. Note that VR + VC > V . The instantaneous voltages in the circuit obey the loop rule at all times but because of the phase differences the voltage amplitudes do not. IDENTIFY: vR (t ) is given by Eq.(31.8). vC (t )is given by Eq.(31.16). SET UP: From Exercise 31.17, V = 30.0 V, VR = 8.62 V, VC = 28.7 V and φ = −73.3°. EXECUTE: (a) The graph is given in Figure 31.18. (b) The different voltage are: v = (30.0 V)cos(250t − 73.3°), vR = (8.62 V)cos(250t ), vC = (28.7 V)cos(250t − 90°). At t = 20 ms : v = −25.1 V, vR = 2.45 V, vC = −27.5 V. Note that vR + vC = v. (c) At t = 40 ms: v = −22.9 V, vR = −7.23 V, vC = −15.6 V. Note that vR + vC = v. EVALUATE: It is important to be careful with radians vs. degrees!

Figure 31.18 IDENTIFY: Apply the equations in Section 31.3. SET UP: ω = 250 rad/s, R = 200 Ω, L = 0.400 H, C = 6.00 μ F and V = 30.0 V. EXECUTE:

(a) Z = R 2 + (ω L − 1/ ω C ) 2 .

Z = (200 Ω) 2 + ((250 rad/s)(0.0400 H) − 1/((250 rad/s)(6.00 ×10−6 F))) 2 = 601 Ω V 30 V = 0.0499 A. (b) I = = Z 601 Ω

⎛ 100 Ω − 667 Ω ⎞ ⎛ ω L − 1/ ω C ⎞ (c) φ = arctan ⎜ ⎟ = −70.6°, and the voltage lags the current. ⎟ = arctan ⎜ R 200 Ω ⎝ ⎠ ⎝ ⎠

Alternating Current

31-7

(d) VR = IR = (0.0499 A)(200 Ω) = 9.98 V;

VL = Iω L = (0.0499 A)(250 rad s)(0.400 H) = 4.99 V; VC = EVALUATE:

I (0.0499 A) = = 33.3 V. ω C (250 rad/s)(6.00 × 10−6 F)

(e) At any instant, v = vR + vC + vL . But vC and vL are 180° out of phase, so vC can be larger than v

at a value of t, if vL + vR is negative at that t. 31.20.

IDENTIFY:

vR (t ) is given by Eq.(31.8). vC (t ) is given by Eq.(31.16). vL (t ) is given by Eq.(31.10).

SET UP: From Exercise 31.19, V = 30.0 V, VL = 4.99 V, VR = 9.98 V, VC = 33.3 V and φ = −70.6°. EXECUTE: (a) The graph is sketched in Figure 31.20. The different voltages plotted in the graph are: v = (30 V)cos(250t − 70.6°), vR = (9.98 V)cos(250t ), vL = (4.99 V)cos(250t + 90°) and vC = (33.3 V)cos(250t − 90°). (b) At t = 20 ms: v = −24.3 V, vR = 2.83 V, vL = 4.79 V, vC = − 31.9 V. (c) At t = 40 ms: v = −23.8 V, vR = −8.37 V, vL = 2.71 V, vC = −18.1 V. EVALUATE: In both parts (b) and (c), note that the source voltage equals the sum of the other voltages at the given instant. Be careful with degrees versus radians!

Figure 31.20 31.21.

IDENTIFY and SET UP: The current is largest at the resonance frequency. At resonance, X L = X C and Z = R. For part (b), calculate Z and use I = V / Z . 1 EXECUTE: (a) f 0 = = 113 Hz. I = V / R = 15.0 mA. 2π LC (b) X C = 1/ ω C = 500 Ω. X L = ω L = 160 Ω. Z = R 2 + ( X L − X C ) 2 = (200 Ω) 2 + (160 Ω − 500 Ω)2 = 394.5 Ω.

I = V / Z = 7.61 mA. X C > X L so the source voltage lags the current.

31.22.

EVALUATE: ω 0 = 2π f 0 = 710 rad/s. ω = 400 rad/s and is less than ω 0 . When ω < ω 0 , X C > X L . Note that I in part (b) is less than I in part (a). IDENTIFY: The impedance and individual reactances depend on the angular frequency at which the circuit is driven. 2

1 ⎞ ⎛ The impedance is Z = R 2 + ⎜ ω L − ⎟ , the current amplitude is I = V/Z, and the instantaneous C⎠ ω ⎝ values of the potential and current are v = V cos(ωt + φ), where tan φ = (XL – XC)/R, and i = I cos ωt. 1 1 1 , which gives ω = = 3162 rad/s = EXECUTE: (a) Z is a minimum when ω L = = ωC (8.00 mH)(12.5 µF) LC 3160 rad/s and Z = R = 175 Ω. (b) I = V/Z = (25.0 V)/(175 Ω) = 0.143 A (c) i = I cos ωt = I/2, so cosωt = 12 , which gives ωt = 60° = π/3 rad. v = V cos(ωt + φ), where tan φ = (XL – XC)/R = SET UP:

0/R =0. So, v = (25.0 V) cosωt = (25.0 V)(1/2) = 12.5 V. vR = Ri = (175 Ω)(1/2)(0.143 A) = 12.5 V. 0.143 A cos(60° − 90°) = +3.13 V. vC = VC cos(ωt – 90°) = IXC cos(ωt – 90°) = (3162 rad/s)(12.5 µF)

31-8

Chapter 31

31.23.

vL = VL cos(ωt + 90°) = IXL cos(ωt + 90°) = (0.143 A)(3162 rad/s)(8.00 mH) cos(60° + 90°). vL = –3.13 V. (d) vR + vL + vC = 12.5 V + (–3.13 V) + 3.13 V = 12.5 V = vsource EVALUATE: The instantaneous potential differences across all the circuit elements always add up to the value of the source voltage at that instant. In this case (resonance), the potentials across the inductor and capacitor have the same magnitude but are 180° out of phase, so they add to zero, leaving all the potential difference across the resistor. IDENTIFY and SET UP: Use the equation that preceeds Eq.(31.20): V 2 = VR2 + (VL − VC ) 2

31.24.

EXECUTE: V = (30.0 V) 2 + (50.0 V − 90.0 V) 2 = 50.0 V EVALUATE: The equation follows directly from the phasor diagrams of Fig.31.13 (b or c). Note that the voltage amplitudes do not simply add to give 170.0 V for the source voltage. 1 IDENTIFY and SET UP: X L = ω L and X C = . ωC 1 L 1 1 , then X = ω L − EXECUTE: (a) If ω = ω 0 = and X = − = 0. ωC LC LC C LC (b) When ω > ω 0 , X > 0 (c) When ω > ω0 , X < 0 (d) The graph of X versus ω is given in Figure 31.24. EVALUATE:

Z = R 2 + X 2 and tan φ = X / R.

Figure 31.24 31.25.

2 R. IDENTIFY: For a pure resistance, Pav = Vrms I rms = I rms SET UP: 20.0 W is the average power Pav . EXECUTE: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. P 20.0 W = 0.167 A (b) I rms = av = Vrms 120 V P 20.0 W = 720 Ω (c) R = 2av = I rms (0.167 A) 2

31.26.

VR2,rms

2 (120 V) 2 Vrms = = 20.0 W. 750 Ω R R IDENTIFY: The average power supplied by the source is P = Vrms I rms cos φ . The power consumed in the resistance

EVALUATE:

We can also calculate the average power as Pav =

=

2 R. is P = I rms

SET UP:

ω = 2π f = 2π (1.25 × 103 Hz) = 7.854 ×103 rad/s. X L = ω L = 157 Ω. X C =

1

ωC

= 909 Ω.

EXECUTE: (a) First, let us find the phase angle between the voltage and the current: X − X C 157 Ω − 909 Ω and φ = −65.04°. The impedance of the circuit is tan φ = L = R 350 Ω Z = R 2 + ( X L − X C ) 2 = (350 Ω) 2 + ( −752 Ω) 2 = 830 Ω. The average power provided by the generator is then P = Vrms I rms cos(φ ) =

2 Vrms (120 V) 2 cos(φ ) = cos( −65.04°) = 7.32 W Z 830 Ω

Alternating Current

31-9

2

31.27.

⎛ 120 V ⎞ 2 (b) The average power dissipated by the resistor is PR = I rms R=⎜ ⎟ (350 Ω) = 7.32 W. ⎝ 830 Ω ⎠ EVALUATE: Conservation of energy requires that the answers to parts (a) and (b) are equal. IDENTIFY: The power factor is cos φ , where φ is the phase angle in Fig.31.13. The average power is given by Eq.(31.31). Use the result of part (a) to rewrite this expression. (a) SET UP: The phasor diagram is sketched in Figure 31.27.

EXECUTE: From the diagram V IR R cos φ = R = = , V IZ Z as was to be shown.

31.28.

31.29.

Figure 31.27 V ⎛ R ⎞ ⎛V ⎞ 2 (b) Pav = Vrms I rms cos φ = Vrms I rms ⎜ ⎟ = ⎜ rms ⎟ I rms R. But rms = I rms , so Pav = I rms R. Z ⎝Z⎠ ⎝ Z ⎠ EVALUATE: In an L-R-C circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the resistor. V R IDENTIFY and SET UP: Pav = Vrms I rms cosφ . I rms = rms . cos φ = . Z Z 80.0 V 75.0 Ω EXECUTE: I rms = = 0.714. Pav = (80.0 V)(0.762 A)(0.714) = 43.5 W. = 0.762 A. cos φ = 105 Ω 105 Ω EVALUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also 2 R = (0.762 A) 2 (75.0 Ω) = 43.5 W. calculate the average power as Pav = I rms IDENTIFY and SET UP: Use the equations of Section 31.3 to calculate φ , Z and Vrms . The average power 2 R delivered by the source is given by Eq.(31.31) and the average power dissipated in the resistor is I rms EXECUTE: (a) X L = ω L = 2π f L = 2π (400 Hz)(0.120 H) = 301.6 Ω 1 1 1 XC = = = = 54.51 Ω ω C 2π fC 2π (400 Hz)(7.3 ×10−6 Hz) X − X C 301.6 Ω − 54.41 Ω tan φ = L = , so φ = +45.8°. The power factor is cos φ = +0.697. R 240 Ω (b) Z = R 2 + ( X L − X C ) 2 = (240 Ω) 2 + (301.6 Ω − 54.51 Ω) 2 = 344 Ω

(c) Vrms = I rms Z = (0.450 A)(344 Ω) = 155 V (d) Pav = I rmsVrms cos φ = (0.450 A)(155 V)(0.697) = 48.6 W

31.30.

2 R = (0.450 A)2 (240 Ω) = 48.6 W (e) Pav = I rms EVALUATE: The average electrical power delivered by the source equals the average electrical power consumed in the resistor. (f ) All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another part of the cycle. There is no net dissipation of energy in the capacitor. (g) The answer is the same as for the capacitor. Energy is repeatedly being stored and released in the inductor, but no net energy is dissipated there. IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance X C of the capacitor. The angular frequency and the inductance can be used to calculate the reactance X L of the inductor. Calculate the phase angle φ and then the power factor is cos φ . Calculate the impedance of the circuit and then the rms current in the circuit. The average power is Pav = Vrms I rms cos φ . On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor. V 45 V SET UP: The source has rms voltage Vrms = = = 31.8 V. 2 2

31-10

Chapter 31

EXECUTE:

X L = ω L = (360 rad/s)(15 × 10−3 H) = 5.4 Ω. X C =

1 1 = = 794 Ω. ω C (360 rad/s)(3.5 × 10−6 F)

X L − X C 5.4 Ω − 794 Ω = and φ = −72.4°. The power factor is cos φ = 0.302. R 250 Ω V 31.8 V (b) Z = R 2 + ( X L − X C ) 2 = (250 Ω) 2 + (5.4 Ω − 794 Ω) 2 = 827 Ω. I rms = rms = = 0.0385 A. Z 827 Ω Pav = Vrms I rms cos φ = (31.8 V)(0.0385 A)(0.302) = 0.370 W. tan φ =

31.31.

2 (c) The average power delivered to the resistor is Pav = I rms R = (0.0385 A) 2 (250 Ω) = 0.370 W. The average power delivered to the capacitor and to the inductor is zero. EVALUATE: On average the power delivered to the circuit equals the power consumed in the resistor. The capacitor and inductor store electrical energy during part of the current oscillation but each return the energy to the circuit during another part of the current cycle. IDENTIFY and SET UP: At the resonance frequency, Z = R. Use that V = IZ, VR = IR, VL = IX L and VC = IX C . Pav is given by Eq.(31.31). (a) EXECUTE: V = IZ = IR = (0.500 A)(300 Ω) = 150 V

(b) VR = IR = 150 V X L = ω L = L(1/ LC ) = L / C = 2582 Ω; VL = IX L = 1290 V

X C = 1/(ω C ) = L / C = 2582 Ω; VC = IX C = 1290 V (c) Pav = 12 VI cos φ = 12 I 2 R, since V = IR and cos φ = 1 at resonance.

31.32.

31.33.

31.34.

Pav = 12 (0.500 A) 2 (300 Ω) = 37.5 W EVALUATE: At resonance VL = VC . Note that VL + VC > V . However, at any instant vL + vC = 0. IDENTIFY: The current is maximum at the resonance frequency, so choose C such that ω = 50.0 rad/s is the resonance frequency. At the resonance frequency Z = R. SET UP: VL = Iω L V EXECUTE: (a) The amplitude of the current is given by I = . Thus, the current will have a 2 R + (ω L − 1 ) 2 ωC 1 maximum amplitude when ω L = 1 . Therefore, C = 12 = = 44.4 μ F. ωC ω L (50.0 rad/s) 2 (9.00 H) (b) With the capacitance calculated above we find that Z = R, and the amplitude of the current is 120 V I =V = = 0.300 A. Thus, the amplitude of the voltage across the inductor is R 400 Ω VL = I (ω L) = (0.300 A)(50.0 rad/s)(9.00 H) = 135 V. EVALUATE: Note that VL is greater than the source voltage amplitude. IDENTIFY and SET UP: At resonance X L = X C , φ = 0 and Z = R. R = 150 Ω, L = 0.750 H, C = 0.0180 μ F, V = 150 V X − XC EXECUTE: (a) At the resonance frequency X L = X C and from tan φ = L we have that φ = 0° and the R power factor is cos φ = 1.00. (b) Pav = 12 VI cos φ (Eq.31.31) V V At the resonance frequency Z = R, so I = = Z R 2 2 V V (150 V) ⎛ ⎞ =1 = 75.0 W Pav = 12 V ⎜ ⎟ cos φ = 12 R 2 150 Ω ⎝R⎠ (c) EVALUATE: When C and f are changed but the circuit is kept on resonance, nothing changes in Pav = V 2 /(2 R ), so the average power is unchanged: Pav = 75.0 W. The resonance frequency changes but since Z = R at resonance the current doesn’t change. 1 IDENTIFY: ω 0 = . VC = IX C . V = IZ . LC SET UP: At resonance, Z = R.

Alternating Current

(a) ω 0 =

EXECUTE:

31.35.

31-11

1 1 = = 1.54 ×104 rad/s LC (0.350 H)(0.0120 ×10−6 F)

⎛V ⎞ ⎛V ⎞ 1 1 (b) V = IZ = ⎜ C ⎟ Z = ⎜ C ⎟ R. X C = = = 5.41× 103 Ω. −6 4 ω C × × X X (1.54 10 rad/s)(0.0120 10 F) C C ⎝ ⎠ ⎝ ⎠ ⎛ 550 V ⎞ V =⎜ ⎟ (400 Ω) = 40.7 V. 3 ⎝ 5.41 × 10 Ω ⎠ EVALUATE: The voltage amplitude for the capacitor is more than a factor of 10 times greater than the voltage amplitude of the source. 1 1 . X L = ω L. X C = IDENTIFY and SET UP: The resonance angular frequency is ω 0 = and ωC LC Z = R 2 + ( X L − X C ) 2 . At the resonance frequency X L = X C and Z = R. (a) Z = R = 115 Ω 1

EXECUTE: (b) ω 0 =

−3

(4.50 × 10 H)(1.26 ×10

−6

F)

= 1.33 ×104 rad/s . ω = 2ω 0 = 2.66 × 104 rad/s.

X L = ω L = (2.66 × 104 rad/s)(4.50 × 10−3 H) = 120 Ω. X C =

1 1 = = 30 Ω 4 ω C (2.66 ×10 rad/s)(1.25 × 10−6 F)

Z = (115 Ω) 2 + (120 Ω − 30 Ω) 2 = 146 Ω (c) ω = ω 0 / 2 = 6.65 × 103 rad/s. X L = 30 Ω. X C =

31.36.

1

ωC

= 120 Ω. Z = (115 Ω) 2 + (30 Ω − 120 Ω) 2 = 146 Ω, the

same value as in part (b). EVALUATE: For ω = 2ω 0 , X L > X C . For ω = ω 0 / 2, X L < X C . But ( X L − X C ) 2 has the same value at these two frequencies, so Z is the same. IDENTIFY: At resonance Z = R and X L = X C . SET UP:

1 . V = IZ . VR = IR, VL = IX L and VC = VL . LC 1 1 (a) ω 0 = = = 945 rad s. LC ( 0.280 H ) ( 4.00 × 10−6 F )

ω0 =

EXECUTE: (b)

I = 1.20 A at resonance, so R = Z =

V 120 V = = 70.6 Ω I 1.70 A

(c) At resonance, VR = 120 V, VL = VC = Iω L = (1.70 A )( 945 rad s )( 0.280 H ) = 450 V. 31.37.

EVALUATE: At resonance, VR = V and VL − VC = 0. IDENTIFY and SET UP: Eq.(31.35) relates the primary and secondary voltages to the number of turns in each. I = 2 2 V/R and the power consumed in the resistive load is I rms = Vrms / R. EXECUTE:

(a)

V2 N 2 N V 120 V = so 1 = 1 = = 10 V1 N1 N 2 V2 12.0 V

V2 12.0 V = = 2.40 A R 5.00 Ω (c) Pav = I 22 R = (2.40 A) 2 (5.00 Ω) = 28.8 W (d) The power drawn from the line by the transformer is the 28.8 W that is delivered by the load. (b) I 2 =

Pav = 2

V2 V 2 (120 V) 2 so R = = = 500 Ω R Pav 28.8 W

⎛N ⎞ And ⎜ 1 ⎟ (5.00 Ω) = (10) 2 (5.00 Ω) = 500 Ω, as was to be shown. ⎝ N2 ⎠ EVALUATE: The resistance is “transformed”. A load of resistance R connected to the secondary draws the same power as a resistance ( N1 / N 2 )2 R connected directly to the supply line, without using the transformer.

31-12

Chapter 31

31.38.


31.39.

Pav = VI and Pav,1 = Pav,2 .

N1 V1 = . N 2 V2

V1 = 120 V. V2 = 13,000 V. (a)

N 2 V2 13,000 V = = = 108 N1 V1 120 V

(b) Pav = V2 I 2 = (13,000 V)(8.50 ×10−3 A) = 110 W P 110 W = 0.917 A (c) I1 = av = V1 120 V EVALUATE: Since the power supplied to the primary must equal the power delivered by the secondary, in a stepup transformer the current in the primary is greater than the current in the secondary. V N IDENTIFY: A transformer transforms voltages according to 2 = 2 . The effective resistance of a secondary V1 N1 R V2 . Resistance R is related to Pav and V by Pav = . Conservation of energy circuit of resistance R is Reff = 2 ( N 2 / N1 ) R requires Pav,1 = Pav,2 so V1I1 = V2 I 2 . SET UP:

Let V1 = 240 V and V2 = 120 V, so P2,av = 1600 W. These voltages are rms.

(a) V1 = 240 V and we want V2 = 120 V, so use a step-down transformer with N 2 / N1 = 12 . P 1600 W (b) Pav = VI , so I = av = = 6.67 A. V 240 V V 2 (120 V) 2 (c) The resistance R of the blower is R = = = 9.00 Ω. The effective resistance of the blower is P 1600 W 9.00 Ω Reff = = 36.0 Ω. (1/ 2)2 EVALUATE: I 2V2 = (13.3 A)(120 V) = 1600 W. Energy is provided to the primary at the same rate that it is consumed in the secondary. Step-down transformers step up resistance and the current in the primary is less than the current in the secondary. 1 IDENTIFY: Z = R 2 + ( X L − X C ) 2 , with X L = ω L and X C = . ωC SET UP: The woofer has a R and L in series and the tweeter has a R and C in series. EXECUTE: (a) Z tweeter = R 2 + (1 ω C ) 2 EXECUTE:

31.40.

(b) Z woofer = R 2 + (ω L )

2

(c) If Z tweeter = Z woofer , then the current splits evenly through each branch. (d) At the crossover point, where currents are equal, R 2 + (1 ω C 2 ) = R 2 + (ω L ) . ω = 2

f =

31.41.

31.42.

ω 1 . = 2π 2π LC

1 and LC

EVALUATE: The crossover frequency corresponds to the resonance frequency of a R-C-L circuit, since the crossover frequency is where X L = X C . IDENTIFY and SET UP: Use Eq.(31.24) to relate L and R to φ . The voltage across the coil leads the current in it by 52.3°, so φ = +52.3°. X − XC X EXECUTE: tan φ = L . But there is no capacitance in the circuit so X C = 0. Thus tan φ = L and X L = R R XL 62.1 Ω = = 0.124 H. R tan φ = (48.0 Ω) tan 52.3° = 62.1 Ω. X L = ω L = 2π f L so L = 2π f 2π (80.0 Hz) EVALUATE: φ > 45° when ( X L − X C ) > R, which is the case here. IDENTIFY: SET UP:

Z = R 2 + ( X L − X C ) 2 . I rms = Vrms =

V 30.0 V = = 21.2 V. 2 2

Vrms . Vrms = I rms R. VC ,rms = I rms X C . VL ,rms = I rms X L . Z

Alternating Current

31-13

(a) ω = 200 rad/s , so X L = ω L = (200 rad/s)(0.400 H) = 80.0 Ω and

EXECUTE:

1 1 XC = = = 833 Ω. Z = (200 Ω) 2 + (80.0 Ω − 833 Ω) 2 = 779 Ω. ω C (200 rad/s)(6.00 × 10−6 F) I rms = VL ,rms

Vrms 21.2 V = = 0.0272 A. V1 reads VR ,rms = I rms R = (0.0272 A)(200 Ω) = 5.44 V. V2 reads Z 779 Ω = I rms X L = (0.0272 A)(80.0 Ω) = 2.18 V. V3 reads VC ,rms = I rms X C = (0.0272 A)(833 Ω) = 22.7 V. V4 reads

VL − VC = VL ,rms − VC ,rms = 2.18 V − 22.7 V = 20.5 V. V5 reads Vrms = 21.2 V. 2 1 833 Ω (b) ω = 1000 rad/s so X L = ω L = (5)(80.0 Ω) = 400 Ω and X C = = = 167 Ω. 5 ωC V 21.2 V Z = (200 Ω) 2 + (400 Ω − 167 Ω) 2 = 307 Ω. I rms = rms = = 0.0691 A. V1 reads VR ,rms = 13.8 V. V2 reads 307 Ω Z VL ,rms = 27.6 V. V3 reads VC ,rms = 11.5 V. V4 reads VL ,rms − VC ,rms = 27.6 V − 11.5 V = 16.1 V. V5 reads Vrms = 21.2 V. 1 = 645 rad/s. 200 rad/s is less than the LC resonance frequency and X C > X L . 1000 rad/s is greater than the resonance frequency and X L > X C . IDENTIFY and SET UP: The rectified current equals the absolute value of the current i. Evaluate the integral as specified in the problem. EXECUTE: (a) From Fig.31.3b, the rectified current is zero at the same values of t for which the sinusoidal current is zero. At these t, cos ω t = 0 and ω t = ±π / 2, ± 3π / 2,…. The two smallest positive times are t1 = π / 2ω , t2 = 3π / 2ω . EVALUATE:

31.43.

(b) A =

∫

t2 t1

The resonance frequency for this circuit is ω 0 =

t

2 t2 I ⎡1 ⎤ idt = − ∫ I cos ω tdt = − I ⎢ sin ω t ⎥ = − (sin ω t2 − sin ω t1 ) t1 ω ⎣ω ⎦ t1

sin ω t1 = sin[ω (π / 2ω )] = sin(π / 2) = 1 sin ω t2 = sin[ω (3π / 2ω )] = sin(3π / 2) = −1 2I ⎛I⎞ A = ⎜ ⎟ (1 − ( −1)) = ω ⎝ω ⎠ (c ) I rav (t2 − t1 ) = 2 I / ω 2I 2I 2I = = , which is Eq.(31.3). ω (t2 − t1 ) ω (3π / 2ω − π / 2ω ) π EVALUATE: We have shown that Eq.(31.3) is correct. The average rectified current is less than the current amplitude I, since the rectified current varies between 0 and I. The average of the current is zero, since it has both positive and negative values. IDENTIFY: X L = ω L. Pav = Vrms I rms cos φ SET UP: f = 120 Hz; ω = 2π f . I rav =

31.44.

EXECUTE:

(a) X L = ω L ⇒ L =

(b) Z = R 2 + X L 2 =

Vrms = Z

2

ω

=

250 Ω = 0.332 Ω 2π (120 Hz )

+ ( 250 Ω ) = 472 Ω. cos φ = 2

V R V2 R and I rms = rms . Pav = rms , so Z Z Z Z

800 W Pav = ( 472 Ω ) = 668 V. 400 Ω R

EVALUATE: 31.45.

( 400 Ω )

XL

I rms =

Vrms 668 V 2 R = (1.415 A)2 (400 Ω) = 800 W, which = = 1.415 A. We can calculate Pav as I rms 472 Ω Z

checks. (a) IDENTIFY and SET UP: Source voltage lags current so it must be that X C > X L and we must add an inductor in series with the circuit. When X C = X L the power factor has its maximum value of unity, so calculate the additional L needed to raise X L to equal X C .

31-14

Chapter 31

power factor cosφ equals 1 so φ = 0 and X C = X L . Calculate the present value of X C − X L to

(b) EXECUTE:

see how much more X L is needed: R = Z cosφ = (60.0 Ω)(0.720) = 43.2 Ω X L − XC so X L − X C = R tan φ R cos φ = 0.720 gives φ = −43.95° ( φ is negative since the voltage lags the current) tan φ =

Then X L − X C = R tan φ = (43.2 Ω) tan(−43.95°) = −41.64 Ω. Therefore need to add 41.64 Ω of X L . XL 41.64 Ω = = 0.133 H, amount of inductance to add. 2π f 2π (50.0 Hz) EVALUATE: From the information given we can’t calculate the original value of L in the circuit, just how much to add. When this L is added the current in the circuit will increase. 2 IDENTIFY: Use Vrms = I rms Z to calculate Z and then find R. Pav = I rms R X L = ω L = 2π f L and L =

31.46.

SET UP:

X C = 50.0 Ω

EXECUTE: R=

Z=

Vrms 240 V 2 = = 80.0 Ω = R 2 + X C2 = R 2 + ( 50.0 Ω ) . Thus, I rms 3.00 A

( 80.0 Ω ) − ( 50.0 Ω ) 2

2

= 62.4 Ω. The average power supplied to this circuit is equal to the power dissipated

2 by the resistor, which is P = I rms R = ( 3.00 A ) ( 62.4 Ω ) = 562 W. 2

X L − X C −50.0 Ω and φ = −38.7°. = R 62.4 Ω Pav = Vrms I rms cos φ = (240 V)(3.00 A)cos(−38.7°) = 562 W, which checks. IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values. SET UP: The voltage amplitudes are VR = RI, VL = XLI, and VC = XCI, where I = V/Z and EVALUATE:

31.47.

tan φ =

2

1 ⎞ ⎛ Z = R2 + ⎜ω L − . ω C ⎟⎠ ⎝ EXECUTE: (a) ω = 2πf = 2π(1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives XL = ωL = (7854 rad/s)(3.50 mH) = 27.5 Ω; XC = 1/ωC = 1/[(7854 rad/s)(10.0 µF)] = 12.7 Ω; Z = R 2 + ( X L − X C )2 =

(50.0 Ω)2 + (27.5 Ω − 12.7 Ω) 2 = 52.1 Ω;

I = V/Z = (60.0 V)/(52.1 Ω) = 1.15 A; VR = RI = (50.0 Ω)(1.15 A) = 57.5 V; VL = XLI = (27.5 Ω)(1.15 A) = 31.6 V; VC = XCI = (12.7 Ω)(1.15 A) = 14.7 V. The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of time. At any instant, the instantaneous voltages all add to 60.0 V. (b) All of them will change because they all depend on ω. XL = ωL will double to 55.0 Ω, and XC = 1/ωC will

31.48.

decrease by half to 6.35 Ω. Therefore Z = (50.0 Ω) 2 + (55.0 Ω − 6.35 Ω) 2 = 69.8 Ω; I = V/Z = (60.0 V)/(69.8 Ω) = 0.860 A; VR = IR = (0.860 A)(50.0 Ω) = 43.0 V; VL = IXL = (0.860 A)(55.0 Ω) = 47.3 V; VC = IXC = (0.860 A)(6.35 Ω) = 5.47 V. EVALUATE: The new amplitudes in part (b) are not simple multiples of the values in part (a) because the impedance and reactances are not all the same simple multiple of the angular frequency. 1 IDENTIFY and SET UP: X C = . X L = ω L. ωC 1 XL ωL 1 1 EXECUTE: (a) = ω1L and LC = 2 . At angular frequency ω 2 , = 2 = ω 22 LC = (2ω1 ) 2 2 = 4. ω1C ω1 ω1 X C 1/ ω 2C X L > XC. 2

XL ⎛ω ⎞ ⎛ 1 ⎞ 1 = ω32 LC = ⎜ 1 ⎟ ⎜ 2 ⎟ = . X C > X L . XC ⎝ 3 ⎠ ⎝ ω1 ⎠ 9 When ω increases, X L increases and X C decreases. When ω decreases, X L decreases and X C

(b) At angular frequency ω3 ,

EVALUATE: increases. (c) The resonance angular frequency ω 0 is the value of ω for which X C = X L , so ω 0 = ω1.

Alternating Current

31.49.

31-15

Express Z and I in terms of ω , L, C and R. The voltages across the resistor and the


inductor are 90° out of phase, so Vout = VR2 + VL2 . The circuit is sketched in Figure 31.49.

EXECUTE:

X L = ω L, X C =

1

ωC 2

1 ⎞ ⎛ Z = R2 + ⎜ω L − ⎟ ωC ⎠ ⎝ V Vs I= s = 2 Z 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

Figure 31.49

Vout = I R 2 + X L2 = I R 2 + ω 2 L2 = Vs

Vout = Vs

R 2 + ω 2 L2 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

2

R 2 + ω 2 L2 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

2

ω small 2

1 ⎞ 1 ⎛ 2 2 2 2 As ω gets small, R 2 + ⎜ ω L − ⎟ → 2 2 ,R +ω L → R ωC ⎠ ωC ⎝

Therefore

Vout R2 → = ω RC as ω becomes small. Vs (1/ ω 2C 2 )

ω large 2

1 ⎞ ⎛ 2 2 2 2 2 2 2 2 2 2 As ω gets large, R 2 + ⎜ ω L − ⎟ → R +ω L → ω L , R +ω L → ω L ωC ⎠ ⎝ Therefore,

31.50.

Vout ω 2 L2 → = 1 as ω becomes large. Vs ω 2 L2

EVALUATE: Vout / Vs → 0 as ω becomes small, so there is Vout only when the frequency ω of Vs is large. If the source voltage contains a number of frequency components, only the high frequency ones are passed by this filter. IDENTIFY: V = VC = IX C . I = V / Z . SET UP:

X L = ω L, X C =

EXECUTE:

31.51.

Vout = VC =

1

ωC

.

I V 1 ⇒ out = . 2 Vs ω C R + (ω L − 1 ω C )2 ωC

If ω is large:

Vout 1 1 1 = ≈ = . 2 2 Vs ω C R 2 + (ω L − 1 ω C ) ( LC )ω 2 ω C (ω L )

If ω is small:

Vout ≈ Vs ω C

EVALUATE:

When ω is large, X C is small and X L is large so Z is large and the current is small. Both factors in

1

(1 ω C )

2

=

ωC = 1. ωC

VC = IX C are small. When ω is small, X C is large and the voltage amplitude across the capacitor is much larger than the voltage amplitudes across the resistor and the inductor. IDENTIFY: I = V / Z and Pav = 12 I 2 R. Z = R 2 + (ω L − 1/ ω C ) 2 V V EXECUTE: (a) I = = . 2 2 Z R + (ω L − 1 ω C )

SET UP:

31-16

Chapter 31 2

(b) Pav =

1 2 1 ⎛V ⎞ V 2R 2 . I R= ⎜ ⎟ R= 2 2 2 2⎝Z⎠ R + (ω L − 1 ω C )

(c) The average power and the current amplitude are both greatest when the denominator is smallest, which occurs 1 1 . , so ω 0 = for ω 0 L = ω 0C LC

25ω 2 (100 V ) ( 200 Ω ) 2 = . 2 2 2 ( 200 Ω ) + (ω ( 2.00 H ) − 1 [ω (0.500 ×10−6 F)]) 40,000ω 2 + ( 2ω 2 − 2,000,000 ) 2

(d) Pav =

The graph of Pav versus ω is sketched in Figure 31.51. EVALUATE: Note that as the angular frequency goes to zero, the power and current are zero, just as they are when the angular frequency goes to infinity. This graph exhibits the same strongly peaked nature as the light purple curve in Figure 31.19 in the textbook.

Figure 31.51 31.52.

IDENTIFY:

Problem 31.51 shows that I =

SET UP: EXECUTE: (b) VC =

I VL = Iω L and VC = . ωC

(a) VL = I ω L =

V R + (ω L − 1/[ω C ]) 2 2

.

V ωL V ωL = . 2 Z R 2 + (ω L − 1/[ωC ])

I I 1 = = . 2 ωC ω CZ ω C R + (ω L − 1 [ω C ])2

(c) The graphs are given in Figure 31.52. EVALUATE: (d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of 100 V (equal to the total source voltage). At very high frequencies, the capacitor voltage goes to zero, 1 while the inductor’s voltage goes to 100 V. At resonance, ω 0 = = 1000 rad s, the two voltages are equal, and LC are a maximum, 1000 V.

Figure 31.52

Alternating Current

31.53.

U B = 12 Li 2 . U E = 12 Cv 2 .

IDENTIFY:

Let 〈 x〉 denote the average value of the quantity x. 〈i 2 〉 = 12 I 2 and 〈vC2 〉 = 12 VC2 . Problem 31.51 shows

SET UP:

that I =

31-17

V R + (ω L − 1/[ω C ]) 2

2

. Problem 31.52 shows that VC =

V

ω C R + (ω L − 1/[ω C ]) 2 2

.

2

EXECUTE:

⎛ I ⎞ 1 2 2 (a) U B = 12 Li 2 ⇒ U B = 12 L i 2 = 12 LI rms = 12 L ⎜ ⎟ = 4 LI . ⎝ 2⎠ 2

1 ⎛V ⎞ U E = 12 CvC2 ⇒ U E = C vC2 = 12 CVC2,rms = 12 C ⎜ C ⎟ = 14 CVC2 2 ⎝ 2⎠ (b) Using Problem 31.51a UB

⎛ 1 1 V2 = LI 2 = L ⎜ 4 4 ⎜ R 2 + (ω L − 1 ω C ) 2 ⎝

2

⎞ LV 2 ⎟ = . ⎟ 4 R 2 + (ω L − 1 ω C ) 2 ⎠

(

)

2

1 1 V V2 Using Problem (31.47b): U E = CVC 2 = C . = 2 4 4 ω 2C 2 R 2 + (ω L − 1 ω C ) 2 4ω 2C R 2 + (ω L − 1 ω C )

(

)

(

)

(c) The graphs of the magnetic and electric energies are given in Figure 31.53. EVALUATE: (d) When the angular frequency is zero, the magnetic energy stored in the inductor is zero, while the electric energy in the capacitor is U E = CV 2 4. As the frequency goes to infinity, the energy noted in both

inductor and capacitor go to zero. The energies equal each other at the resonant frequency where ω 0 = UB = UE =

1 and LC

LV 2 . 4R2

Figure 31.53 31.54.

IDENTIFY: At any instant of time the same rules apply to the parallel ac circuit as to parallel dc circuit: the voltages are the same and the currents add. SET UP: For a resistor the current and voltage in phase. For an inductor the voltage leads the current by 90° and for a capacitor the voltage lags the current by 90°. EXECUTE: (a) The parallel L-R-C circuit must have equal potential drops over the capacitor, inductor and resistor, so vR = vL = vC = v. Also, the sum of currents entering any junction must equal the current leaving the junction. Therefore, the sum of the currents in the branches must equal the current through the source: i = iR + iL + iC . (b) iR = v is always in phase with the voltage. iL = v lags the voltage by 90°, and iC = vωC leads the voltage R ωL by 90°. The phase diagram is sketched in Figure 31.54. (c) From the diagram, I = I R + ( I C − I L ) 2

2

2

2

2

(d) From part (c):

I =V

2

V ⎞ ⎛V ⎞ ⎛ = ⎜ ⎟ + ⎜ V ωC − ⎟ . R L⎠ ω ⎝ ⎠ ⎝

V 1 ⎛ 1 ⎞ 1 + ⎜ ωC − = ⎟ . But I = , so 2 Z Z R ωL ⎠ ⎝

2

1 ⎛ 1 ⎞ + ⎜ ωC − . 2 R ⎝ ω L ⎟⎠

31-18

Chapter 31

1 . The current in the capacitor branch is much larger than the current in the ωC other branches. For small ω , Z → ω L. The current in the inductive branch is much larger than the current in the other branches. EVALUATE:

For large ω , Z →

Figure 31.54 31.55.

Apply the expression for I from problem 31.54 when ω 0 = 1/ LC .

IDENTIFY:

2

SET UP:

From Problem 31.54, I = V

EXECUTE:

(a) At resonance, ω 0 =

1 ⎛ 1 ⎞ + ⎜ ωC − 2 R ω L ⎟⎠ ⎝ 1 1 V ⇒ ω 0C = ⇒ I C = V ω 0C = = I so I = I R and I is a minimum. ω0 L ω0 L L LC

2 Vrms V2 cos φ = at resonance where R < Z so power is a maximum. Z R (c) At ω = ω 0 , I and V are in phase, so the phase angle is zero, which is the same as a series resonance.

(b) Pav =

EVALUATE:

(d) The parallel circuit is sketched in Figure 31.55. At resonance, iC = iL and at any instant of time

these two currents are in opposite directions. Therefore, the net current between a and b is always zero. (e) If the inductor and capacitor each have some resistance, and these resistances aren’t the same, then it is no longer true that iC + iL = 0 and the statement in part (d) isn’t valid.

Figure 31.55 31.56.

IDENTIFY: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. SET UP: V = 2Vrms = 311 V EXECUTE:

(a) I R =

V 311 V = = 0.778 A. R 400 Ω

(b) I C = V ωC = ( 311 V )( 360 rad s ) ( 6.00 × 10−6 F ) = 0.672 A.

⎛I ⎞ ⎛ 0.672 A ⎞ (c) φ = arctan ⎜ C ⎟ = arctan ⎜ ⎟ = 40.8°. I ⎝ 0.778 A ⎠ ⎝ R⎠ (d) I = I R2 + I C2 =

31.57.

( 0.778 A )

2

+ ( 0.672 A ) = 1.03 A. 2

(e) Leads since φ > 0. EVALUATE: The phasor diagram shows that the current in the capacitor always leads the source voltage. IDENTIFY and SET UP: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. V V EXECUTE: (a) I R = ; I C = V ωC ; I L = . R ωL (b) The graph of each current versus ω is given in Figure 31.57a. (c) ω → 0 : I C → 0; I L → ∞. ω → ∞: I C → ∞; I L → 0. At low frequencies, the current is not changing much so the inductor’s back-emf doesn’t “resist.” This allows the current to pass fairly freely. However, the current in the capacitor goes to zero because it tends to “fill up” over the slow period, making it less effective at passing charge. At high frequency, the induced emf in the inductor resists the violent changes and passes little current. The capacitor never gets a chance to fill up so passes charge freely.

Alternating Current

(d) ω =

31-19

1 1 = = 1000 rad sec and f = 159 Hz. The phasor diagram is sketched in LC (2.0 H)(0.50 ×10−6 F)

Figure 31.57b. 2

2

V ⎞ ⎛V ⎞ ⎛ (e) I = ⎜ ⎟ + ⎜ V ωC − ⎟ . R L⎠ ω ⎝ ⎝ ⎠ 2

2

⎛ 100 V ⎞ ⎛ ⎞ 100 V −1 −6 I= ⎜ ⎟ + ⎜ (100 V)(1000 s )(0.50 ×10 F) − ⎟ = 0.50 A −1 (1000 s )(2.0 H) ⎠ ⎝ 200 Ω ⎠ ⎝ (f ) At resonance I L = I C = V ωC = (100 V)(1000 s −1 )(0.50 ×10−6 F) = 0.0500 A and I R = EVALUATE: resistor.

V 100 V = = 0.50 A. R 200 Ω

At resonance iC = iL = 0 at all times and the current through the source equals the current through the

Figure 31.57 31.58.

IDENTIFY:

The average power depends on the phase angle φ. 2

SET UP: EXECUTE:

1 ⎞ ⎛ The average power is Pav = VrmsIrmscos φ, and the impedance is Z = R 2 + ⎜ ω L − . ω C ⎟⎠ ⎝

(a) Pav = VrmsIrmscos φ =

1 2

(VrmsIrms), which gives cos φ = 12 , so φ = π/3 = 60°. tan φ = (XL – XC)/R,

which gives tan 60° = (ωL – 1/ωC)/R. Using R = 75.0 Ω, L = 5.00 mH, and C = 2.50 µF and solving for ω we get ω = 28760 rad/s = 28,800 rad/s. (b) Z = R 2 + ( X L − X C ) 2 , where XL = ωL = (28,760 rad/s)(5.00 mH) = 144 Ω and

XC = 1/ωC = 1/[(28,760 rad/s)(2.50 µF)] = 13.9 Ω, giving Z = (75 Ω) 2 + (144 Ω − 13.9 Ω) 2 = 150 Ω;

31.59.

I = V/Z = (15.0 V)/(150 Ω) = 0.100 A and Pav = 12 VI cos φ = 12 (15.0 V)(0.100 A)(1/2) = 0.375 W. EVALUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero. 2 R from Exercise 31.27 to calculate IDENTIFY: We know R, X C and φ so Eq.(31.24) tells us X L . Use Pav = I rms I rms . Then calculate Z and use Eq.(31.26) to calculate Vrms for the source. SET UP: Source voltage lags current so φ = −54.0°. X C = 350 Ω, R = 180 Ω, Pav = 140 W X − XC EXECUTE: (a) tan φ = L R X L = R tan φ + X C = (180 Ω) tan(−54.0°) + 350 Ω = −248 Ω + 350 Ω = 102 Ω 2 R (Exercise 31.27). I rms = (b) Pav = Vrms I rms cosφ = I rms

Pav 140 W = = 0.882 A R 180 Ω

(c) Z = R 2 + ( X L − X C ) 2 = (180 Ω) 2 + (102 Ω − 350 Ω) 2 = 306 Ω

Vrms = I rms Z = (0.882 A)(306 Ω) = 270 V. EVALUATE: We could also use Eq.(31.31): Pav = Vrms I rms cos φ Pav 140 W Vrms = = = 270 V, which agrees. The source voltage lags the current when I rms cos φ (0.882 A)cos( −54.0°) X C > X L , and this agrees with what we found.

31-20

Chapter 31

31.60.

IDENTIFY and SET UP: Calculate Z and I = V / Z . EXECUTE: (a) For ω = 800 rad s: Z = R 2 + (ω L − 1 ω C ) 2 = (500 Ω) 2 + ((800 rad/s)(2.0 H) − 1 ((800 rad/s)(5.0 ×10 −7 F))) 2 . Z = 1030 Ω . V 100 V 1 0.0971 A = = 243 V I= = = 0.0971 A. VR = IR = (0.0971 A)(500 Ω) = 48.6 V, VC = ω C (800 rad s)(5.0 ×10−7 F) Z 1030 Ω ⎛ ω L − 1 (ωC ) ⎞ and VL = IωL = (0.0971 A)(800 rad s)(2.00 H) = 155 V. φ = arctan ⎜ ⎟ = −60.9°. The graph of each R ⎝ ⎠ voltage versus time is given in Figure 31.60a. (b) Repeating exactly the same calculations as above for ω = 1000 rad/s: Z = R = 500 Ω; φ = 0; I = 0.200 A; VR = V = 100 V; VC = VL = 400 V. The graph of each voltage versus time is given in Figure 31.60b. (c) Repeating exactly the same calculations as part (a) for ω = 1250 rad/s : Z = R = 1030 Ω; φ = +60.9°; I = 0.0971 A; VR = 48.6 V; VC = 155 V; VL = 243 V. The graph of each voltage versus time is given in Figure 31.60c. 1 1 EVALUATE: The resonance frequency is ω 0 = = = 1000 rad/s. For ω < ω 0 the phase LC (2.00 H)(0.500 μ F)

angle is negative and for ω > ω 0 the phase angle is positive.

Figure 31.60

Alternating Current

31.61.

31-21

Eq.(31.19) allows us to calculate I and then Eq.(31.22) gives Z. Solve Eq.(31.21) for L. V 360 V = 0.750 A (a) VC = IX C so I = C = X C 480 Ω


V 120 V = = 160 Ω I 0.750 A (c) Z 2 = R 2 + ( X L − X C ) 2

(b) V = IZ so Z =

X L − X C = ± Z 2 − R 2 , so X L = X C ± Z 2 − R 2 = 480 Ω ± (160 Ω) 2 − (80.0 Ω) 2 = 480 Ω ± 139 Ω X L = 619 Ω or 341 Ω 1 and X L = ω L. At resonance, X C = X L . As the frequency is lowered below the ωC resonance frequency X C increases and X L decreases. Therefore, for ω < ω 0 , X L < X C . So for X L = 341 Ω the XC =

(d) EVALUATE:

angular frequency is less than the resonance angular frequency. ω is greater than ω 0 when X L = 619 Ω. But at these two values of X L , the magnitude of X L − X C is the same so Z and I are the same. In one case ( X L = 691 Ω) 31.62.

the source voltage leads the current and in the other ( X L = 341 Ω) the source voltage lags the current. IDENTIFY and SET UP: The maximum possible current amplitude occurs at the resonance angular frequency because the impedance is then smallest. EXECUTE: (a) At the resonance angular frequency ω 0 = 1/ LC , the current is a maximum and Z = R, giving Imax = V/R. At the required frequency, I = Imax/3. I = V/Z = Imax/3 = (V/R)/3, which means that Z = 3R. Squaring gives R2 + (ωL – 1/ωC)2 = 9R2 . Solving for ω gives ω = 3.192 × 105 rad/s and ω = 8.35 × 104 rad/s. I V 49.5 V (b) V = 2Vrms = 2(35.0 V) = 49.5 V. I = max = = = 0.132 A. 3 3R 3(125 Ω) For ω = 8.35 × 104 rad/s: R = 125 Ω and VR = IR = 16.5 Ω; X L = ω L = 125 Ω and VL = 16.5 V; 1 = 479 Ω and VC = 63.2 V. ωC For ω = 3.192 × 105 rad/s: R = 125 Ω and VR = IR = 16.5 Ω; X L = ω L = 479 Ω and VL = 63.2 V; XC =

XC =

1

ωC

= 125 Ω and VC = 16.5 V.

For the lower frequency, X C > X L and VC > VL . For the higher frequency, X L > X C and VL > VC .

EVALUATE: 31.63.


this interval i =

2I0

τ

Consider the cycle of the repeating current that lies between t1 = τ / 2 and t2 = 3τ / 2. In

(t − τ ). I av =

t2 t2 1 1 2 i dt and I rms = i 2 dt t2 − t1 ∫ t1 t2 − t1 ∫t1 3τ / 2

I av =

EXECUTE:

t2 1 1 3τ / 2 2 I 0 2I ⎡ 1 ⎤ (t − τ ) dt = 20 ⎢ t 2 − τ t ⎥ i dt = ∫ ∫ t τ / 2 1 t2 − t1 τ τ τ ⎣2 ⎦τ /2

2 3τ 2 τ 2 τ 2 ⎞ I ⎛ 2 I ⎞ ⎛ 9τ − − + ⎟ = (2 I 0 ) 18 (9 − 12 − 1 + 4) = 0 (13 − 13) = 0. I av = ⎜ 20 ⎟ ⎜ τ 8 2 8 2 4 ⎝ ⎠⎝ ⎠ 2 = ( I 2 )av = I rms

2 = I rms

4 I 02

τ

3

t2 1 1 3τ / 2 4 I 02 (t − τ ) 2 dt i 2 dt = ∫ ∫ t τ τ /2 τ 2 t2 − t1 1

3τ / 2

∫τ

/2

(t − τ ) 2 dt =

4 I 02 ⎡⎛ τ ⎞ ⎛ τ ⎞ 3 3τ / 2 1 ⎡ ⎤ ( ) − = t τ 3 ⎦τ / 2 3τ 3 ⎢⎜⎝ 2 ⎟⎠ − ⎜⎝ − 2 ⎟⎠ τ3 ⎣ ⎢⎣

4 I 02

3

3

⎤ ⎥ ⎥⎦

I 02 [1 + 1] = 13 I 02 6 I 2 I rms = I rms = 0 . 3 EVALUATE: In each cycle the current has as much negative value as positive value and its average is zero. i 2 is always positive and its average is not zero. The relation between I rms and the current amplitude for this current is different from that for a sinusoidal current (Eq.31.4). 2 = I rms

31-22

Chapter 31

31.64.

IDENTIFY:

Apply Vrms = I rms Z 1 and Z = R 2 + ( X L − X C ) 2 . LC 1 1 = = 786 rad s. (a) ω 0 = LC (1.80 H)(9.00 ×10−7 F)

ω0 =

SET UP: EXECUTE:

(b) Z = R 2 + (ωL − 1 ωC ) 2 . Z = (300 Ω) 2 + ((786 rad s)(1.80 H) − 1 ((786 rad s)(9.00 ×10−7 F))) 2 = 300 Ω.

I rms-0 =

Vrms 60 V = = 0.200 A. Z 300 Ω

(c) We want I =

ω 2 L2 +

1 V I rms-0 = rms = 2 Z

Vrms R 2 + (ω L − 1 ωC ) 2

. R 2 + (ω L −1 ωC ) 2 =

2 4Vrms . 2 I rms-0

2 ⎛ ⎞ 1 2 L 4Vrms 1 2L 4V 2 − 2 ⎟ + 2 = 0. − + R 2 − 2 rms = 0 and (ω 2 ) 2 L2 + ω 2 ⎜ R 2 − 2 ωC C I rms-0 C I C rms-0 ⎠ ⎝ 2

Substituting in the values for this problem, the equation becomes (ω 2 ) 2 (3.24) + ω 2 (−4.27 × 106 ) + 1.23 × 1012 = 0. Solving this quadratic equation in ω 2 we find ω 2 = 8.90 ×105 rad 2 s 2 or 4.28 ×105 rad 2 /s 2 and

ω = 943 rad s or 654 rad s. (d) (i) R = 300 Ω, I rms-0 = 0.200 A, ω1 − ω2 = 289 rad s. (ii) R = 30 Ω, I rms-0 = 2A, ω1 − ω 2 = 28 rad/s.

(iii) R = 3 Ω, I rms-0 = 20 A, ω1 − ω 2 = 2.88 rad/s. 31.65.

EVALUATE: The width gets smaller as R gets smaller; I rms-0 gets larger as R gets smaller. IDENTIFY: The resonance frequency, the reactances, and the impedance all depend on the values of the circuit elements. SET UP:

The resonance frequency is ω0 = 1/ LC , the reactances are XL = ωL and XC = 1/ωC, and the impedance

is Z = R 2 + ( X L − X C ) 2 . 1 → 1/ 2, so ω0 decreases by 12 . 2 L 2C (b) Since XL = ωL, if L is doubled, XL increases by a factor of 2. (c) Since XC = 1/ωC, doubling C decreases XC by a factor of 12 .

EXECUTE:

(a) ω0 = 1/ LC becomes

(d) Z = R 2 + ( X L − X C ) 2 → Z = (2 R ) 2 + (2 X L − 12 X C ) 2 , so Z does not change by a simple factor of 2 or

1 2

.

The impedance does not change by a simple factor, even though the other quantities do. V N IDENTIFY: A transformer transforms voltages according to 2 = 2 . The effective resistance of a secondary V1 N1 R . circuit of resistance R is Reff = ( N 2 / N1 ) 2 SET UP: N 2 = 275 and V1 = 25.0 V. EXECUTE: (a) V2 = V1 ( N 2 / N1 ) = (25.0 V)(834 / 275) = 75.8 V

EVALUATE: 31.66.

R 125 Ω = = 13.6 Ω ( N 2 / N1 ) 2 (834 / 275) 2 EVALUATE: The voltage across the secondary is greater than the voltage across the primary since N 2 > N1. The effective load resistance of the secondary is less than the resistance R connected across the secondary. 1 1 . IDENTIFY: The resonance angular frequency is ω 0 = and the resonance frequency is f 0 = LC 2π LC SET UP: ω 0 is independent of R. (b) Reff =

31.67.

EXECUTE:

31.68.

(a) ω 0 (or f 0 ) depends only on L and C so change these quantities.

(b) To double ω 0 , decrease L and C by multiplying each of them by 12 . EVALUATE: Increasing L and C decreases the resonance frequency; decreasing L and C increases the resonance frequency. IDENTIFY: At resonance, Z = R. I = V / R. VR = IR, VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities.

Alternating Current

EXECUTE:

(a) I =

(b) VC = IX C = (c) VL = IX L =

31.69.

V Rω 0C

=

V R + (ω L − 1/ ω C ) 2

2

. At resonance ω L =

1

ωC

and I max =

V . R

V L . R C

V V L ω0 L = . R R C

1 1 V2 L 1 V2 (d) U C = CVC2 = C 2 = L 2 . 2 2 R C 2 R 1 1 V2 (e) U L = LI 2 = L 2 . 2 2 R EVALUATE: At resonance VC = VL and the maximum energy stored in the inductor equals the maximum energy stored in the capacitor. IDENTIFY: I = V / R. VR = IR, VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: (a) I =

V = Z

ω0

ω=

2

.

V ⎛ω L ⎞ R 2 + ⎜ 0 − 2 / ω 0C ⎟ 2 ⎝ ⎠

(b) VC = IX C =

2 ω 0C

ω0 L

V R2 +

2

=

9 L 4C

=

V 9 L R + 4C

.

2

L C

2V R2 +

9 L 4C

.

L V 2 = . C L 9 9 L 2 2 R + R + 4C 4C 1 2 LV 2 2 . (d) U C = CVC = 9 L 2 R2 + 4C 1 2 1 LV 2 . (e) U L = LI = 2 2 R2 + 9 L 4C EVALUATE: For ω < ω 0 , VC > VL and the maximum energy stored in the capacitor is greater than the maximum energy stored in the inductor. IDENTIFY: I = V / R . VR = IR , VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: ω = 2ω 0 . V V V (a) I = = = . 2 2 Z 9 L R + (2ω 0 L − 1/ 2ω 0C ) 2 R + 4C (c) VL = IX L =

31.70.

V = Z

31-23

(b) VC = IX C =

V

2

1 2ω 0C

(c) VL = IX L = 2ω 0 L

V R2 +

9 L 4C

V

9 L R2 + 4C 1 LV 2 2 (d) U C = CVC = . 2 9 L 2 8 R + 4C

=

=

L C L C

V 2 . 9 L R2 + 4C 2V 9 L R + 4C 2

.

31-24

Chapter 31

31.71.

LV 2 . 9 L 2 R2 + 4C EVALUATE: For ω > ω 0 , VL > VC and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor. IDENTIFY and SET UP: Assume the angular frequency ω of the source and the resistance R of the resistor are known. V Iω L ω L EXECUTE: Connect the source, capacitor, resistor, and inductor in series. Measure VR and VL . L = = VR IR R and L can be calculated. EVALUATE: There are a number of other approaches. The frequency could be varied until VC = VL , and then this

31.72.

frequency is equal to 1/ LC . If C is known, then L can be calculated. V IDENTIFY: Pav = Vrms I rms cos φ and I rms = rms . Calculate Z. R = Z cos φ . Z SET UP: f = 50.0 Hz and ω = 2π f . The power factor is cosφ .

(e) U L =

1 2 LI = 2

2 Vrms V 2 cos φ (120 V) 2 (0.560) cos φ . Z = rms = = 36.7 Ω. Z Pav (220 W) R = Z cos φ = (36.7 Ω)(0.560) = 20.6 Ω.

(a) Pav =

EXECUTE:

(b) Z = R 2 + X L 2 ⋅ X L = Z 2 − R 2 = (36.7 Ω) 2 − (20.6 Ω) 2 = 30.4 Ω. But φ = 0 is at resonance, so the inductive

and capacitive reactances equal each other. Therefore we need to add X C = 30.4 Ω. X C = C=

1

ωX C

=

31.73.

IDENTIFY: SET UP:

therefore gives

1 1 = = 1.05 ×10−4 F. 2π f X C 2π (50.0 Hz)(30.4 Ω)

(c) At resonance, Pav = EVALUATE: resonance.

1

ωC

V 2 (120 V) 2 = = 699 W. R 20.6 Ω

2 Pav = I rms R and I rms is maximum at resonance, so the power drawn from the line is maximum at

pR = i 2 R. pL = iL i = I cos ω t

di q . pC = i. C dt

1 (a) pR = i 2 R = I 2 cos 2 (ωt ) R = VR I cos 2 (ωt ) = VR I (1 + cos(2ωt )). 2 1 T VR I T VR I T 1 Pav ( R ) = ∫ pR dt = (1 + cos(2ω t )) dt = [t ]0 = 2 VR I . T 0 2T ∫ 0 2T T di (b) pL = Li = −ω LI 2 cos(ω t )sin(ω t ) = − 12 VL I sin(2ω t ). But ∫ sin(2ω t ) dt = 0 ⇒ Pav ( L) = 0. 0 dt T q (c) pC = i = vC i = VC I sin(ω t )cos(ω t ) = 12 VC I sin(2ω t ). But ∫ sin(2ω t )dt = 0 ⇒ Pav (C ) = 0. 0 C (d) p = pR + pL + pc = VR I cos 2 (ω t ) − 12 VL I sin(2ω t ) + 12 VC I sin(2ω t ) and

EXECUTE:

V −V VR and sin φ = L C , so V V p = VI cos(ω t )(cos φ cos(ω t ) − sin φ sin(ω t )), at any instant of time. EVALUATE: At an instant of time the energy stored in the capacitor and inductor can be changing, but there is no net consumption of electrical energy in these components. dVL dVC IDENTIFY: VL = IX L . = 0 at the ω where VL is a maximum. VC = IX C . = 0 at the ω where VC is a dω dω maximum. V . SET UP: Problem 31.51 shows that I = 2 R + (ω L − 1/ ω C ) 2 p = I cos(ω t )(VR cos(ω t ) − VL sin(ω t ) + VC sin(ω t )). But cos φ =

31.74.

EXECUTE:

(a) VR =maximum when VC = VL ⇒ ω = ω 0 =

1 . LC

Alternating Current

31-25

⎞ dVL d ⎛ V ωL dVL ⎜ ⎟. =0= = 0.Therefore: dω dω ⎜⎝ R 2 + (ω L − 1 ωC ) 2 ⎟⎠ dω VL V ω 2 L( L − 1 ω 2C )( L + 1 ω 2C ) − . R 2 + (ω L − 1 ω C ) 2 = ω 2 ( L2 − 1 ω 4C 2 ) . 2 2 3 2 2 2 ( R + ( ω L − 1 ω C ) ) R + (ω L − 1 ωC )

(b) VL =maximum when 0=

R2 +

1 R 2C 2 1 2L 1 LC = − and ω = . − = − 2 ω 2C 2 C ω 2C 2 ω 2

1 LC − R 2C 2 2

.

⎞ dVC d ⎛ V dVC ⎜ ⎟. =0= = 0. Therefore: dω dω ⎜ ω C R 2 + (ω L − 1 ω C ) 2 ⎟ dω ⎝ ⎠ V V ( L − 1 ω 2C )( L + 1 ω 2C ) 0=− − . R 2 + (ω L − 1 ω C ) 2 = −ω 2 ( L2 − 1 ω 4C 2 ). 2 2 3 2 ω 2C R 2 + (ω L − 1 ω C ) 2 C ( R + (ω L − 1 ω C ) )

(c) VC = maximum when

1 R2 2L − 2. = −ω 2 L2 and ω = C LC 2 L 2L 2 2 2 2 2 = −ω L . R +ω L − C EVALUATE: VL is maximum at a frequency greater than the resonance frequency and X C is a maximum at a frequency less than the resonance frequency. These frequencies depend on R , as well as on L and on C. IDENTIFY: Follow the steps specified in the problem. SET UP: In part (a) use Eq.(31.23) to calculate Z and then I = V / Z . φ is given by Eq.(31.24). In part (b) let Z = R + iX . R 2 + ω 2 L2 −

31.75.

EXECUTE:

(a) From the current phasors we know that Z = R 2 + (ω L − 1 ω C ) 2 . 2

⎛ ⎞ 1 Z = (400 Ω) + ⎜ (1000 rad s)(0.50 H) − ⎟ = 500 Ω. −6 (1000 rad s)(1.25 10 F) × ⎝ ⎠ V 200 V I= = = 0.400 A. Z 500 Ω 2

⎛ (1000 rad s)(0.500 H) − 1 (1000 rad s)(1.25 × 10−6 F) ⎞ ⎛ ω L − 1 (ω C ) ⎞ arctan (b) φ = arctan ⎜ φ = . ⎜ ⎟ = +36.9° ⎟ 400 Ω R ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ 1 1 ⎞ ⎛ (c) Z cpx = R + i ⎜ ω L − ⎟ = 400 Ω − 300 Ωi. ⎟ . Z cpx = 400 Ω − i ⎜ (1000 rad s)(0.50 H) − −6 (1000 rad s)(1.25 10 F) × ω C ⎝ ⎠ ⎝ ⎠ Z = (400 Ω) 2 + ( −300 Ω) 2 = 500 Ω.

200 V V ⎛ 8 + 6i ⎞ ⎛ 8 + 6i ⎞ ⎛ 8 − 6i ⎞ = =⎜ ⎟ ⎜ ⎟ = 0.400 A. ⎟ A = (0.320 A) + (0.240 A)i. I = ⎜ Z cpx (400 − 300i ) Ω ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ Im( I cpx ) 6 25 = = 0.75 ⇒ φ = +36.9°. (e) tan φ = Re( I cpx ) 8 25 (d) I cpx =

⎛ 8 + 6i ⎞ (f ) VRcpx = I cpx R = ⎜ ⎟ (400 Ω) = (128 + 96i)V. ⎝ 25 ⎠ ⎛ 8 + 6i ⎞ VLcpx = iI cpxω L = i ⎜ ⎟ (1000 rad s)(0.500 H) = ( − 120 + 160i ) V. ⎝ 25 ⎠ VCcpx = i (g) Vcpx

1 ⎛ 8 + 6i ⎞ = i⎜ = (+192 − 256i ) V. ⎟ ω C ⎝ 25 ⎠ (1000 rad s)(1.25 × 10−6 F) = VRcpx + VLcpx + VCcpx = (128 + 96i ) V + (−120 + 160i)V + (192 − 256i ) V = 200 V. I cpx

EVALUATE:

Both approaches yield the same value for I and for φ .

ELECTROMAGNETIC WAVES

32.1.

IDENTIFY: Since the speed is constant, distance x = ct. SET UP: The speed of light is c = 3.00 ×108 m/s . 1 yr = 3.156 ×107 s.

x 3.84 ×108 m = = 1.28 s c 3.00 ×108 m/s (b) x = ct = (3.00 ×108 m/s)(8.61 yr)(3.156 ×107 s/yr) = 8.15 × 1016 m = 8.15 × 1013 km EVALUATE: The speed of light is very great. The distance between stars is very large compared to terrestrial distances. IDENTIFY: Since the speed is constant the difference in distance is cΔt. SET UP: The speed of electromagnetic waves in air is c = 3.00 × 108 m/s. EXECUTE: A total time difference of 0.60 μ s corresponds to a difference in distance of (a) t =

EXECUTE:

32.2.

32.3.

32

cΔt = (3.00 ×108 m/s)(0.60 ×10−6 s) = 180 m. EVALUATE: The time delay doesn’t depend on the distance from the transmitter to the receiver, it just depends on the difference in the length of the two paths. IDENTIFY: Apply c = f λ. c = 3.00 ×108 m/s c 3.0 ×108 m s EXECUTE: (a) f = = = 6.0 ×104 Hz. 5000 m λ

SET UP:

(b) f = (c) f =

λ c

λ

=

3.0 ×108 m s = 6.0 ×107 Hz. 5.0 m

=

3.0 ×108 m s = 6.0 ×1013 Hz. 5.0 ×10−6 m

3.0 ×108 m s = 6.0 ×1016 Hz. λ 5.0 × 10−9 m EVALUATE: f increases when λ decreases. 2π IDENTIFY: c = f λ and k = . (d) f =

32.4.

c

c

=

λ

c = 3.00 ×108 m/s . c EXECUTE: (a) f = . UVA: 7.50 ×1014 Hz to 9.38 ×1014 Hz . UVB: 9.38 ×1014 Hz to 1.07 ×1015 Hz . SET UP:

λ

(b) k = 32.5.

2π

λ

. UVA: 1.57 ×107 rad/m to 1.96 × 107 rad/m . UVB: 1.96 × 107 rad/m to 2.24 ×107 rad/m .

EVALUATE: Larger λ corresponds to smaller f and k. IDENTIFY: c = f λ . Emax = cBmax . k = 2π / λ . ω = 2π f .

Since the wave is traveling in empty space, its wave speed is c = 3.00 ×108 m/s . c 3.00 ×108 m/s = 6.94 ×1014 Hz EXECUTE: (a) f = = λ 432 ×10−9 m (b) Emax = cBmax = (3.00 ×108 m/s)(1.25 ×10−6 T) = 375 V/m SET UP:

32-1

32-2

Chapter 32

2π rad = 1.45 ×107 rad/m . ω = (2π rad)(6.94 ×1014 Hz) = 4.36 ×1015 rad/s . 432 ×10−9 m E = Emax cos( kx − ω t ) = (375 V/m)cos([1.45 ×107 rad/m]x − [4.36 ×1015 rad/s] t )

(c) k =

32.6.

2π

=

λ

B = Bmax cos(kx − ωt ) = (1.25 × 10−6 T)cos([1.45 × 107 rad/m]x − [4.36 × 1015 rad/s]t ) EVALUATE: The cos(kx − ω t ) factor is common to both the electric and magnetic field expressions, since these two fields are in phase. IDENTIFY: c = f λ . Emax = cBmax . Apply Eqs.(32.17) and (32.19). The speed of the wave is c = 3.00 × 108 m/s. c 3.00 ×108 m/s = 6.90 ×1014 Hz EXECUTE: (a) f = = λ 435 ×10−9 m E 2.70 ×10−3 V/m = 9.00 ×10−12 T (b) Bmax = max = 3.00 ×108 m/s c ! 2π (c) k = = 1.44 ×107 rad/m . ω = 2π f = 4.34 ×1015 rad/s . If E ( z, t ) = iˆEmax cos(kz + ωt ) , then λ ! ! ! B ( z , t ) = − ˆjBmax cos(kz + ωt ) , so that E × B will be in the − kˆ direction. ! E ( z, t ) = iˆ(2.70 × 10−3 V/m)cos([1.44 × 107 rad/s) z + [4.34 × 1015 rad/s] t ) and ! B ( z , t ) = − ˆj (9.00 × 10−12 T)cos([1.44 × 107 rad/s) z + [4.34 × 1015 rad/s] t ) . ! ! EVALUATE: The directions of E and B and of the propagation of the wave are all mutually perpendicular. The argument of the cosine is kz + ω t since the wave is traveling in the − z -direction . Waves for visible light have very high frequencies. ! IDENTIFY and SET UP: The equations are of the form of Eqs.(32.17), with x replaced by z. B is along the y-axis; ! deduce the direction of E . EXECUTE: ω = 2π f = 2π (6.10 ×1014 Hz) = 3.83 ×1015 rad/s SET UP:

32.7.

2π f ω 3.83 × 1015 rad/s = = = 1.28 ×107 rad/m λ c c 3.00 × 108 m/s Bmax = 5.80 ×10−4 T

k=

2π

=

Emax = cBmax = (3.00 × 108 m/s)(5.80 ×10−4 T) = 1.74 ×105 V/m ! ! ! B is along the y-axis. E × B is in the direction of propagation (the +z-direction). From this we can deduce the ! direction of E , as shown in Figure 32.7.

! E is along the x-axis.

32.8.

Figure 32.7 ! E = Emax iˆ cos(kz − ω t ) = (1.74 ×105 V/m)iˆ cos[(1.28 × 107 rad/m)z − (3.83 × 1015 rad/s)t ] ! B = Bmax ˆj cos( kz − ω t ) = ( 5.80 × 10−4 T ) ˆj cos[(1.28 × 107 rad/m)z − (3.83 × 1015 rad/s)t ] ! ! EVALUATE: E and B are perpendicular and oscillate in phase. IDENTIFY: For an electromagnetic wave propagating in the negative x direction, E = Emax cos( kx + ω t ) . ω = 2π f

and k = SET UP:

2π

λ

.T=

1 . Emax = cBmax . f

The wave specified in the problem has a different phase, so E = − Emax sin(kx + ω t ) . Emax = 375 V/m ,

k = 1.99 × 107 rad/m and ω = 5.97 ×1015 rad/s . E EXECUTE: (a) Bmax = max = 1.25 μ T . c

Electromagnetic Waves

ω 2π 1 = 9.50 ×1014 Hz . λ = = 3.16 ×10−7 m = 316 nm . T = = 1.05 × 10 −15 s . This wavelength is too short 2π k f

(b) f =

32.9.

32-3

to be visible. (c) c = f λ = (9.50 × 1014 Hz)(3.16 × 10 −7 m) = 3.00 × 108 m/s . This is what the wave speed should be for an electromagnetic wave propagating in vacuum. ⎛ ω ⎞⎛ 2π ⎞ ω EVALUATE: c = f λ = ⎜ ⎟⎜ ⎟ = is an alternative expression for the wave speed. ⎝ 2π ⎠⎝ k ⎠ k ! IDENTIFY and SET UP: Compare the E ( y , t ) given in the problem to the general form given by Eq.(32.17). Use ! ! the direction of propagation and of E to find the direction of B. (a) EXECUTE: The equation for the electric field contains the factor sin(ky − ω t ) so the wave is traveling in the ! +y-direction. The equation for E ( y , t ) is in terms of sin(ky − ω t ) rather than cos(ky − ωt ); the wave is shifted in phase by 90° relative to one with a cos(ky − ω t ) factor. ! (b) E ( y, t ) = −(3.10 ×105 V/m)kˆ sin[ ky − (2.65 ×1012 rad/s)t ] Comparing to Eq.(32.17) gives ω = 2.65 ×1012 rad/s 2π c 2π c 2π (2.998 × 108 m/s) = = 7.11 × 10−4 m ω = 2π f = so λ = λ ω (2.65 × 1012 rad/s) (c) ! ! E × B must be in the +ydirection (the direction in which the wave is traveling). ! When E is in the –z-direction ! then B must be in the –xdirection, as shown in Figure 32.9. Figure 32.9

2π

ω

2.65 × 1012 rad/s = 8.84 ×103 rad/m λ c 2.998 ×108 m/s Emax = 3.10 × 105 V/m

k=

=

=

Emax 3.10 ×105 V/m = = 1.03 ×10−3 T c 2.998 ×108 m/s ! ! Using Eq.(32.17) and the fact that B is in the − iˆ direction when E is in the − kˆ direction, ! B = −(1.03 ×10−3 T)iˆ sin[(8.84 ×103 rad/m)y − (2.65 ×1012 rad/s)t ] ! ! EVALUATE: E and B are perpendicular and oscillate in phase. IDENTIFY: Apply Eqs.(32.17) and (32.19). f = c / λ and k = 2π / λ . Then Bmax =

32.10.

SET UP:

32.11.

The wave in this problem has a different phase, so By ( z , t ) = Bmax sin(kx + ωt ).

EXECUTE: (a) The phase of the wave is given by kx + ω t , so the wave is traveling in the − x direction. 2π 2π f kc (1.38 ×104 rad m)(3.0 ×108 m s) (b) k = = . f = = = 6.59 ×1011 Hz. c λ 2π 2π (c) Since the magnetic field is in the + y -direction, and the wave is propagating in the − x -direction, then the ! ! electric field is in the + z -direction so that E × B will be in the − x -direction. ! E ( x, t ) = + cB ( x, t )kˆ = cBmax sin(kx + ωt )kˆ. ! E ( x, t ) = (c(3.25 × 10−9 T))sin ( (1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ. ! E ( x, t ) = + (2.48 V m)sin ( (1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ. ! ! EVALUATE: E and B have the same phase and are in perpendicular directions. IDENTIFY and SET UP: c = f λ allows calculation of λ . k = 2π / λ and ω = 2π f . Eq.(32.18) relates the electric and magnetic field amplitudes. c 2.998 ×108 m/s EXECUTE: (a) c = f λ so λ = = = 361 m f 830 ×103 Hz

32-4

Chapter 32

2π rad = = 0.0174 rad/m λ 361 m (c) ω = 2π f = (2π )(830 ×103 Hz) = 5.22 ×106 rad/s (b) k =

32.12.

2π

(d) Eq.(32.18): Emax = cBmax = (2.998 ×108 m/s)(4.82 × 10−11 T) = 0.0144 V/m EVALUATE: This wave has a very long wavelength; its frequency is in the AM radio braodcast band. The electric and magnetic fields in the wave are very weak. IDENTIFY: Emax = cBmax .

The magnetic field of the earth is about 10−4 T. E 3.85 ×10−3 V/m = 1.28 ×10−11 T. EXECUTE: B = = c 3.00 ×108 m/s EVALUATE: The field is much smaller than the earth's field. IDENTIFY and SET UP: v = f λ relates frequency and wavelength to the speed of the wave. Use Eq.(32.22) to calculate n and K. v 2.17 ×108 m/s = 3.81×10−7 m EXECUTE: (a) λ = = f 5.70 ×1014 Hz SET UP:

32.13.

c 2.998 × 108 m/s = = 5.26 ×10−7 m f 5.70 ×1014 Hz

(b) λ = (c) n =

c 2.998 × 108 m/s = = 1.38 v 2.17 × 108 m/s

(d) n = KK m ≈ K so K = n 2 = (1.38) 2 = 1.90

32.14.

EVALUATE: In the material v < c and f is the same, so λ is less in the material than in air. v < c always, so n is always greater than unity. IDENTIFY: Apply Eq.(32.21). Emax = cBmax . v = f λ . Apply Eq.(32.29) with μ = K m μ 0 in place of μ 0 .

K = 3.64 . K m = 5.18

SET UP:

(a) v =

EXECUTE: (b) λ =

(3.00 ×108 m s) c = = 6.91×107 m s. (3.64)(5.18) KK m

v 6.91×107 m s = = 1.06 ×106 m. f 65.0 Hz

(c) Bmax =

Emax 7.20 ×10−3 V m = = 1.04 ×10−10 T. 6.91× 107 m s v

Emax Bmax (7.20 ×10−3 V m)(1.04 ×10−10 T) = = 5.75 ×10−8 W m 2 . 2Km μ0 2(5.18) μ 0 EVALUATE: The wave travels slower in this material than in air. 2 . Emax = cBmax . IDENTIFY: I = P / A . I = 12 P0cEmax (d) I =

32.15.

SET UP:

32.16.

The surface area of a sphere of radius r is A = 4π r 2 . P0 = 8.85 × 10−12 C 2 /N ⋅ m 2 . P (0.05)(75 W) = = 330 W/m 2 . A 4π (3.0 ×10−2 m) 2

EXECUTE:

(a) I =

(b) Emax =

2I 2(330 W/m 2 ) E = = 500 V/m . Bmax = max = 1.7 ×10−6 T = 1.7 μ T . −12 c P0c (8.85 × 10 C 2 /N ⋅ m 2 )(3.00 × 108 m/s)

EVALUATE: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb. Our calculation approximates the filament as a point source that radiates uniformly in all directions. ! ! IDENTIFY and SET UP: The direction of propagation is given by E × B . EXECUTE: (a) Sˆ = iˆ × (− ˆj ) = − kˆ. (b) Sˆ = ˆj × iˆ = −kˆ. (c) Sˆ = ( − kˆ ) × ( − iˆ) = ˆj .

(d) Sˆ = iˆ × ( − kˆ ) = ˆj. EVALUATE:

! ! In each case the directions of E , B and the direction of propagation are all mutually perpendicular.


32.17.

Emax

IDENTIFY: SET UP:

c = 3.00 ×108 m/s . Emax = 4.00 V/m.

! ! ! ! Bmax = Emax c = 1.33 × 10−8 T . For E in the +x-direction, E × B is in the +z-direction when B is in

EXECUTE:

32.18.

32.19.

the +y-direction. ! ! EVALUATE: E , B and the direction of propagation are all mutually perpendicular. 2 2 IDENTIFY: The intensity of the electromagnetic wave is given by Eq.(32.29): I = 12 P0cEmax = P0cErms . The total energy passing through a window of area A during a time t is IAt. SET UP: P0 = 8.85 × 10−12 F/m 2 EXECUTE: Energy = P0cErms At = (8.85 × 10 −12 F m)(3.00 × 108 m s)(0.0200 V m) 2 (0.500 m 2 )(30.0 s) = 15.9 μ J EVALUATE: The intensity is proportional to the square of the electric field amplitude. IDENTIFY and SET UP: Use Eq.(32.29) to calculate I, Eq.(32.18) to calculate Bmax , and use I = Pav / 4π r 2 to

calculate Pav . (a) EXECUTE:

2 I = 12 P0 Emax ; Emax = 0.090 V/m, so I = 1.1× 10−5 W/m 2

(b) Emax = cBmax so Bmax = Emax / c = 3.0 ×10−10 T

32.20.

(c) Pav = I (4π r 2 ) = (1.075 ×10−5 W/m 2 )(4π )(2.5 ×103 m) 2 = 840 W (d) EVALUATE: The calculation in part (c) assumes that the transmitter emits uniformly in all directions. 2 IDENTIFY and SET UP: I = Pav / A and I = P0cErms . (a) The average power from the beam is Pav = IA = (0.800 W m 2 )(3.0 × 10−4 m 2 ) = 2.4 × 10−4 W .

EXECUTE:

0.800 W m 2 I = =17.4 V m −12 (8.85 × 10 F m)(3.00 × 108 m s) P0c

(b) Erms = 32.21:

32.22.

EVALUATE: The laser emits radiation only in the direction of the beam. IDENTIFY: I = Pav / A SET UP: At a distance r from the star, the radiation from the star is spread over a spherical surface of area A = 4π r 2 . EXECUTE: Pav = I (4π r 2 ) = (5.0 ×103 W m 2 )(4π )(2.0 ×1010 m) 2 = 2.5 ×1025 J EVALUATE: The intensity decreases with distance from the star as 1/ r 2 . IDENTIFY and SET UP: c = f λ , Emax = cBmax and I = Emax Bmax / 2μ 0 (a) f =

EXECUTE: (b) Bmax =

EVALUATE:

3.00 ×108 m s = 8.47 ×108 Hz. 0.354 m

Emax Bmax (0.0540 V m)(1.80 × 10−10 T) = = 3.87 × 10−6 W m 2 . 2 μ0 2μ0 2 Alternatively, I = 12 P0cEmax .

The surface area of a sphere is A = 4π r 2 .

⎛ E2 ⎞ Pav = Sav A = ⎜ max ⎟ (4π r 2 ) . Emax = ⎝ 2cμ 0 ⎠ E 12.0 V m = max = = 4.00 × 10−8 T. c 3.00 × 108 m s

EXECUTE:

Bmax

λ

=

2 Pav = IA and I = 12 P0cEmax

IDENTIFY: SET UP:

c

0.0540 V m Emax = = 1.80 ×10−10 T. 3.00 ×108 m s c

(c) I = Sav =

32.23.

Pavcμ 0 (60.0 W)(3.00 ×108 m s) μ 0 = = 12.0 V m. 2 2π r 2π (5.00 m) 2

Emax and Bmax are both inversely proportional to the distance from the source. ! ! ! IDENTIFY: The Poynting vector is S = E × B. SET UP: The electric field is in the +y-direction, and the magnetic field is in the +z-direction. cos 2 φ = 12 (1 + cos 2φ ) EVALUATE:

32.24.

32-5

! ! = cBmax . E × B is in the direction of propagation.

EXECUTE: (a) Sˆ = Eˆ × Bˆ = (− ˆj ) × kˆ = − iˆ. The Poynting vector is in the –x-direction, which is the direction of propagation of the wave.

32-6

Chapter 32

(b) S ( x, t ) =

E ( x, t ) B ( x, t )

μ0

=

Emax Bmax

μ0

cos 2 (kx + ωt ) =

Emax Bmax (1 + cos(2(ωt + kx)) ). But over one period, the 2μ0

Emax Bmax . This is Eq.(32.29). 2μ0 EVALUATE: We can also show that these two results also apply to the wave represented by Eq.(32.17). IDENTIFY: Use the radiation pressure to find the intensity, and then Pav = I (4π r 2 ). cosine function averages to zero, so we have Sav =

32.25.

SET UP:

I c = 2.70 × 103 W/m 2 . Then

For a perfectly absorbing surface, prad = prad = I c so I = cprad

EXECUTE:

Pav = I (4π r ) = (2.70 ×103 W m 2 )(4π )(5.0 m) 2 = 8.5 ×105 W. EVALUATE: Even though the source is very intense the radiation pressure 5.0 m from the surface is very small. IDENTIFY: The intensity and the energy density of an electromagnetic wave depends on the amplitudes of the electric and magnetic fields. 2 SET UP: Intensity is I = Pav / A , and the average power is Pav = 2I / c, where I = 12 P0cEmax . The energy density is 2

32.26.

u = P0 E 2 . EXECUTE:

(a) I = Pav /A =

316,000 W 2(0.00201 W/m 2 ) = 1.34 × 10−11 Pa = 0.00201 W/m2. Pav = 2I / c = 2 3.00 ×108 m/s 2π (5000 m)

2 (b) I = 12 P0cEmax gives

Emax =

2(0.00201 W/m 2 ) = 1.23 N/C (8.85 ×10 −12 C 2 /N ⋅ m 2 )(3.00 ×108 m/s)

2I = P0c

Bmax = Emax /c = (1.23 N/C)/(3.00 × 108 m/s) = 4.10 × 10−9 T (c) u = P0 E 2 , so uav = P0 ( Eav ) 2 and Eav =

Emax , so 2

2 (8.85 ×10−12 C2 /N ⋅ m2 ) (1.23 N/C)2 = 6.69 × 10−12 J/m3 P0 Emax = 2 2 (d) As was shown in Section 32.4, the energy density is the same for the electric and magnetic fields, so each one has 50% of the energy density. EVALUATE: Compared to most laboratory fields, the electric and magnetic fields in ordinary radiowaves are extremely weak and carry very little energy. IDENTIFY and SET UP: Use Eqs.(32.30) and (32.31). dp Sav I = = EXECUTE: (a) By Eq.(32.30) the average momentum density is dV c 2 c 2 dp 0.78 × 103 W/m 2 = = 8.7 × 10−15 kg/m 2 ⋅ s dV (2.998 × 108 m/s) 2

uav =

32.27.

Sav I 0.78 ×103 W/m 2 = = = 2.6 × 10−6 Pa c c 2.998 × 108 m/s The radiation pressure that the sunlight would exert on an absorbing or reflecting surface is very

(b) By Eq.(32.31) the average momentum flow rate per unit area is

32.28.

EVALUATE: small. IDENTIFY: Apply Eqs.(32.32) and (32.33). The average momentum density is given by Eq.(32.30), with S replaced by Sav = I . SET UP: EXECUTE:

prad =

1 atm = 1.013 ×105 Pa (a) Absorbed light: prad =

8.33 × 10−6 Pa = 8.23 × 10−11 atm. 1.013 × 105 Pa atm

(b) Reflecting light: prad = prad =

I 2500 W m 2 = = 8.33 ×10−6 Pa. Then c 3.0 ×108 m s

2 I 2(2500 W m 2 ) = = 1.67 ×10−5 Pa. Then 3.0 ×108 m s c

1.67 × 10−5 Pa = 1.65 × 10−10 atm. 1.013 × 105 Pa atm


(c) The momentum density is

32.29.

2500 W m 2 dp Sav = 2 = = 2.78 ×10−14 kg m 2 ⋅ s. dV c (3.0 ×108 m s) 2

EVALUATE: The factor of 2 in prad for the reflecting surface arises because the momentum vector totally reverses direction upon reflection. Thus the change in momentum is twice the original momentum. IDENTIFY: Apply Eq.(32.4) and (32.9). SET UP: Eq.(32.26) is S = P0cE 2 . EXECUTE:

S=

P0 E2 = P0 μ0

P0

μ0

E2 =

P0

μ0

Ec

P E 1 = c 0 EB = μ0 c P0 μ0

32.32.

32.33.

μ0

EB =

EB

μ0

=

E2 = P0cE 2 μ 0c

λ

3

2

3.00 × 108 m s c = = 0.200 m = 20.0 cm. There must be nodes at the planes, which 2 2 f 2(7.50 ×108 Hz) are 80.0 cm apart, and there are two nodes between the planes, each 20.0 cm from a plane. It is at 20 cm, 40 cm, and 60 cm from one plane that a point charge will remain at rest, since the electric fields there are zero. EVALUATE: The magnetic field amplitude at these points isn’t zero, but the magnetic field doesn’t exert a force on a stationary charge. IDENTIFY and SET UP: Apply Eqs.(32.36) and (32.37). ! EXECUTE: (a) By Eq.(32.37) we see that the nodal planes of the B field are a distance λ / 2 apart, so λ / 2 = 3.55 mm and λ = 7.10 mm. ! (b) By Eq.(32.36) we see that the nodal planes of the E field are also a distance λ / 2 = 3.55 mm apart. (c) v = f λ = (2.20 ×1010 Hz)(7.10 ×10−3 m) = 1.56 ×108 m/s. ! ! EVALUATE: The spacing between the nodes of E is the same as the spacing between the nodes of B. Note that v < c, as it must. ! ! IDENTIFY: The nodal planes of E and B are located by Eqs.(32.26) and (32.27). c 3.00 ×108 m/s SET UP: λ = = = 4.00 m f 75.0 ×106 Hz EXECUTE:

32.31.

P0

EVALUATE: We can also write S = P0c(cB) = P0c B . S can be written solely in terms of E or solely in terms of B. IDENTIFY: The electric field at the nodes is zero, so there is no force on a point charge placed at a node. SET UP: The location of the nodes is given by Eq.(32.36), where x is the distance from one of the planes. λ = c/ f . 2

32.30.

32-7

Δxnodes =

EXECUTE:

(a) Δx =

EXECUTE:

λ=

=

λ

= 2.00 m. 2 (b) The distance between the electric and magnetic nodal planes is one-quarter of a wavelength, so is λ Δx 2.00 m = = =1.00 m. 4 2 2 ! ! EVALUATE: The nodal planes of B are separated by a distance λ / 2 and are midway between the nodal planes of E . ! (a) IDENTIFY and SET UP: The distance between adjacent nodal planes of B is λ / 2. There is an antinodal plane ! of B midway between any two adjacent nodal planes, so the distance between a nodal plane and an adjacent antinodal plane is λ / 4. Use v = f λ to calculate λ .

λ

v 2.10 × 108 m/s = = 0.0175 m f 1.20 ×1010 Hz

0.0175 m = 4.38 ×10−3 m = 4.38 mm 4 4 ! ! (b) IDENTIFY and SET UP: The nodal planes of E are at x = 0, λ / 2, λ , 3λ /2, . . . , so the antinodal planes of E ! are at x = λ / 4, 3λ /4, 5λ /4, . . . . The nodal planes of B are at x = λ / 4, 3λ / 4, 5λ /4, . . . , so the antinodal planes ! of B are at λ / 2, λ , 3λ /2, . . . . ! ! EXECUTE: The distance between adjacent antinodal planes of E and antinodal planes of B is therefore λ / 4 = 4.38 mm. ! ! (c) From Eqs.(32.36) and (32.37) the distance between adjacent nodal planes of E and B is λ / 4 = 4.38 mm. ! ! ! EVALUATE: The nodes of E coincide with the antinodes of B , and conversely. The nodes of B and the nodes ! of E are equally spaced. =

32-8

Chapter 32

32.34.

IDENTIFY:

Evaluate the derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) that are given in Eqs.(32.34) and

(32.35). ∂ ∂ ∂ ∂ sin kx = k cos kx , sin ω t = ω cos ω t . cos kx = −k sin kx , cos ω t = −ω sin ω t . ∂t ∂t ∂x ∂x ∂ 2 E y ( x, t ) ∂ 2 ∂ = 2 ( −2 Emax sin kx sin ωt ) = (−2kEmax cos kx sin ωt ) and EXECUTE: (a) ∂x 2 ∂x ∂x 2 ∂ 2 E y ( x, t ) ∂ 2 E y ( x, t ) ω 2 2 k E sin kx sin t 2 E sin kx sin t . P ω ω μ = = = max max 0 0 ∂x 2 ∂t 2 c2 ∂ 2 Bz ( x, t ) ∂ 2 ∂ = 2 (−2 Bmax cos kx cos ωt ) = (+2kBmax sin kx cos ωt ) and Similarly: ∂x 2 ∂x ∂x ∂ 2 Bz ( x, t ) ω2 ∂ 2 Bz ( x, t ) = 2k 2 Bmax cos kx cos ωt = 2 2 Bmax cos kx cos ωt = P0 μ0 . 2 ∂x ∂t 2 c ∂E y ( x, t ) ∂ = ( −2 Emax sin kx sin ωt ) = −2kEmax cos kx sin ωt . (b) ∂x ∂x ∂E y ( x, t ) ω E = − 2 Emax cos kx sin ωt = −ω 2 max cos kx sin ωt = −ω 2 Bmax cos kx sin ωt . ∂x c c ∂E y ( x, t ) ∂ ∂Bz ( x, t ) = + (2 Bmax cos kx cos ωt ) = − . ∂x ∂t ∂t ∂B ( x, t ) ∂ = (+2 Bmax cos kx cos ωt ) = −2kBmax sin kx cos ωt . Similarly: − z ∂x ∂x ∂Bz ( x, t ) ω ω − = − 2 Bmax sin kx cos ωt = − 2 2cBmax sin kx cos ωt . c c ∂x ∂E ( x, t ) ∂B ( x, t ) ∂ − z = −P0 μ0ω 2 Emax sin kx cos ωt = P0 μ0 (−2 Emax sin kx sin ωt ) = P0 μ0 y . ∂x ∂t ∂t EVALUATE: The standing waves are linear superpositions of two traveling waves of the same k and ω . IDENTIFY: The nodal and antinodal planes are each spaced one-half wavelength apart. SET UP: 2 12 wavelengths fit in the oven, so ( 2 12 ) λ = L, and the frequency of these waves obeys the equation fλ = c.

SET UP:

32.35.

EXECUTE:

(a) Since ( 2 12 ) λ = L, we have L = (5/2)(12.2 cm) = 30.5 cm.

(b) Solving for the frequency gives f = c/λ = (3.00 × 108 m/s)/(0.122 m) = 2.46 × 109 Hz. (c) L = 35.5 cm in this case. ( 2 12 ) λ = L, so λ = 2L/5 = 2(35.5 cm)/5 = 14.2 cm.

32.36.

f = c/λ = (3.00 × 108 m/s)/(0.142 m) = 2.11 × 109 Hz EVALUATE: Since microwaves have a reasonably large wavelength, microwave ovens can have a convenient size for household kitchens. Ovens using radiowaves would need to be far too large, while ovens using visible light would have to be microscopic. IDENTIFY: Evaluate the partial derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) . SET UP:

∂ ∂ ∂ sin( kx − ω t ) = k cos(kx − ω t ) , sin(kx − ω t ) = −ω cos(kx − ω t ) . cos(kx − ω t ) = −k sin( kx − ω t ) , ∂x ∂t ∂x

∂ cos( kx − ω t ) = ω sin( kx − ω t ) ∂t ! ! EXECUTE: Assume E = Emax ˆj sin( kx − ω t ) and B = Bmax kˆ sin( kx − ω t + φ ), with − π < φ < π . Eq. (32.12) is

∂E y

∂Bz . This gives kEmax cos(kx − ω t ) = +ω Bmax cos( kx − ω t + φ ) , so φ = 0 , and kEmax = ω Bmax , so ∂x ∂t ∂E y ∂B ω 2π f gives Emax = Bmax = Bmax = fω Bmax = cBmax . Similarly for Eq.(32.14), − z = P0 μ0 k 2π / ω ∂x ∂t − kBmax cos( kx − ωt + φ ) = −P0 μ0ω Emax cos( kx − ωt ) , so φ = 0 and kBmax = P0 μ0ω Emax , so =−

P0 μ0ω 2π f 1 fω Emax = 2 Emax = 2 Emax = Emax . k c 2π / ω c c ! ! EVALUATE: The E and B fields must oscillate in phase.

Bmax =


32.37.

32-9


Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem. ∂E y ∂B =− z EXECUTE: Eq.(32.12): ∂x ∂t ∂2Ey ∂E ∂2B ∂B ∂ ∂ = − 2 z . Eq.(32.14) says − z = P0 μ 0 y . Taking Taking of both sides of this equation gives of ∂x∂t ∂t ∂x ∂t ∂x ∂t ∂2Ey ∂2Ey ∂ 2 Ey ∂ 2 Ey ∂2B 1 ∂ 2 Bz =− = both sides of this equation gives − 2z = P0 μ 0 . But (The order in , so P0 μ 0 ∂x 2 ∂x ∂t ∂x ∂t ∂x ∂x∂t ∂t∂x which the partial derivatives are taken doesn't change the result.) So −

∂ 2 Bz 1 ∂ 2 Bz ∂ 2 Bz ∂2B and = P0 μ 0 2 z , =− 2 2 2 ∂t P0 μ 0 ∂x ∂x ∂t

as was to be shown. EVALUATE: Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that both fields propagate with the same speed v = 1/ P0 μ 0 . 32.38.

The average energy density in the electric field is u E ,av = 12 P0 ( E 2 )av and the average energy density in

IDENTIFY:

the magnetic field is u B ,av = SET UP:

( cos (kx − ωt ) )

EXECUTE:

2

av

1 ( B 2 )av . 2μ0

= 12 .

2 2 E y ( x, t ) = Emax cos(kx − ωt ) . u E = 12 P0 E y2 = 12 P0 Emax . cos 2 (kx − ωt ) and u E ,av = 14 P0 Emax

Bz ( x, t ) = Bmax cos( kx − ωt ) , so u B = 2 u E ,av = 14 P0c 2 Bmax . c=

1 1 2 , so u E ,av = Bmax , which equals u B ,av . 2μ0 P0 μ0

1 2 Bmax . 2μ 0 IDENTIFY: The intensity of an electromagnetic wave depends on the amplitude of the electric and magnetic fields. Such a wave exerts a force because it carries energy. 2 SET UP: The intensity of the wave is I = Pav / A = 12 P0cEmax , and the force is F = Pav A where Pav = I / c . 2 EXECUTE: (a) I = Pav /A = (25,000 W)/[4π(575 m) ] = 0.00602 W/m2 EVALUATE:

32.39.

1 2 1 2 1 2 Bz = Bmax cos 2 (kx − ω t ) and u B ,av = Bmax . Emax = cBmax , so 4μ0 2μ 0 2μ0

2 Our result allows us to write uav = 2u E ,av = 12 P0 Emax and uav = 2u B ,av =

2 (b) I = 12 P0cEmax , so Emax =

2I = P0c

2(0.00602 W/m 2 ) = 2.13 N/C. (8.85 ×10−12 C2 /N ⋅ m 2 )(3.00 ×108 m/s)

Bmax = Emax/c = (2.13 N/C)/(3.00 × 108 m/s) = 7.10 × 10−9 T

32.40.

(c) F =Pav A = ( I / c) A = (0.00602 W/m2)(0.150 m)(0.400 m)/(3.00 × 108 m/s) = 1.20 × 10−12 N EVALUATE: The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth worrying about! I 2 IDENTIFY: c = f λ . Emax = cBmax . I = 12 P0cEmax . For a totally absorbing surface the radiation pressure is . c SET UP: The wave speed in air is c = 3.00 ×108 m/s . c 3.00 ×108 m/s EXECUTE: (a) f = = = 7.81×109 Hz λ 3.84 ×10−2 m E 1.35 V/m (b) Bmax = max = = 4.50 ×10−9 T c 3.00 ×108 m/s 2 (c) I = 12 P0cEmax = 12 (8.854 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)(1.35 V/m) 2 = 2.42 × 10−3 W/m 2

IA (2.42 × 10−3 W/m 2 )(0.240 m 2 ) = = 1.94 ×10−12 N 3.00 ×108 m/s c EVALUATE: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the wavelength of the light. (d) F = (pressure)A =

32-10

Chapter 32

32.41.

(a) IDENTIFY and SET UP: EXECUTE: I=

Calculate I and then use Eq.(32.29) to calculate Emax and Eq.(32.18) to calculate Bmax .

The intensity is power per unit area: I =

3.20 ×10−3 W P = = 652 W/m 2 . A π (1.25 ×10−3 m) 2

2 Emax , so Emax = 2 μ 0cI 2μ 0 c

Emax = 2(4π ×10−7 T ⋅ m/A)(2.998 ×108 m/s)(652 W/m 2 ) = 701 V/m 701 V/m Emax = = 2.34 ×10−6 T 2.998 ×108 m/s c EVALUATE: The magnetic field amplitude is quite small. (b) IDENTIFY and SET UP: Eqs.(24.11) and (30.10) give the energy density in terms of the electric and magnetic field values at any time. For sinusoidal fields average over E 2 and B 2 to get the average energy densities. EXECUTE: The energy density in the electric field is uE = 12 P0 E 2 . E = Emax cos( kx − ω t ) and the average value of Bmax =

cos 2 (kx − ω t ) is 12 . The average energy density in the electric field then is 2 u E ,av = 14 P0 Emax = 14 (8.854 ×10−12 C2 / N ⋅ m 2 )(701 V/m) 2 = 1.09 ×10−6 J/m3 . The energy density in the magnetic field

B2 B2 (2.34 ×10−6 T) 2 . The average value is u B ,av = max = = 1.09 ×10−6 J/m 3 . 2μ 0 4μ 0 4(4π ×10−7 T ⋅ m/A) EVALUATE: Our result agrees with the statement in Section 32.4 that the average energy density for the electric field is the same as the average energy density for the magnetic field. (c) IDENTIFY and SET UP: The total energy in this length of beam is the total energy density uav = u E ,av + u B ,av = 2.18 × 10−6 J/m3 times the volume of this part of the beam.

is u B =

32.42.

EXECUTE: U = uav LA = (2.18 ×10−6 J/m3 )(1.00 m)π (1.25 ×10−3 m) 2 = 1.07 ×10−11 J. EVALUATE: This quantity can also be calculated as the power output times the time it takes the light to travel L = 1.00 m ⎛ L⎞ ⎛ ⎞ −11 1.00 m: U = P ⎜ ⎟ = (3.20 ×10−3 W) ⎜ ⎟ = 1.07 ×10 J, which checks. 8 × ⎝c⎠ ⎝ 2.998 10 m/s ⎠ IDENTIFY: Use the gaussian surface specified in the hint. ! ! SET UP: The wave is in free space, so in Gauss’s law for the electric field, Qencl = 0 and AE ⋅ dA = 0. Gauss’s law ! ! for the magnetic field says AB ⋅ dA = 0 EXECUTE: Use a gaussian surface such that the front surface is ahead of the wave front (no electric or magnetic ! ! Q fields) and the back face is behind the wave front, as shown in Figure 32.42. AE ⋅ dA = Ex A = encl = 0 , so Ex = 0.

ε0

! ! AB ⋅ dA = Bx A = 0 and Bx = 0. EVALUATE: The wave must be transverse, since there are no components of the electric or magnetic field in the direction of propagation.

Figure 32.42 32.43.

I c SET UP: Assume the electromagnetic waves are formed at the center of the sun, so at a distance r from the center of the sun I = Pav /(4π r 2 ).

IDENTIFY:

I = Pav / A . For an absorbing surface, the radiation pressure is prad =


EXECUTE:

(a) At the sun’s surface: I =

32-11

3.9 ×1026 W Pav = = 6.4 ×107 W m 2 and 2 4π R 4π (6.96 ×108 m) 2

I 6.4 ×107 W m 2 = = 0.21 Pa. c 3.00 ×108 m s (b) Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the radiation pressure: I = 2.6 × 108 W/m 2 and prad = 0.85 Pa. At the top of the earth’s atmosphere, the measured sunlight intensity is prad =

32.44.

1400 W m 2 = 5 × 10−6 Pa , which is about 100,000 times less than the values above. EVALUATE: (b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and halfway out of the sun the gas pressure is believed to be about 6 × 1013 times greater than the radiation pressure. Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure. P 2 IDENTIFY: I = . I = 12 P0cEmax . A SET UP: 3.00 ×108 m/s P 2.80 ×103 W EXECUTE: I = = = 77.8 W/m 2 . 36.0 m 2 A Emax =

32.45.

2I 2(77.8 W/m 2 ) = = 242 N/C . −12 (8.854 × 10 C 2 /N ⋅ m 2 )(3.00 × 108 m/s) P0c

EVALUATE: This value of Emax is similar to the electric field amplitude in ordinary light sources. IDENTIFY: The same intensity light falls on both reflectors, but the force on the reflecting surface will be twice as great as the force on the absorbing surface. Therefore there will be a net torque about the rotation axis. SET UP: For a totally absorbing surface, F = Pav A = ( I/c) A, while for a totally reflecting surface the force will be 2 twice as great. The intensity of the wave is I = 12 P0cEmax . Once we have the torque, we can use the rotational form

of Newton’s second law, τnet = Iα, to find the angular acceleration. 2 P cEmax A 1 2 = 2 P0 AEmax c 2 For a totally reflecting surface, the force will be twice as great, which is P0cEmax . The net torque is therefore

EXECUTE:

The force on the absorbing reflector is FAbs = pav A = ( I/c) A =

1 2 0

2 τ net = FRefl ( L/2) − FAbs ( L/2) = P0 AEmax L/4 2 Newton’s 2nd law for rotation gives τ net = Iα . P0 AEmax L/4 = 2m( L/2) 2 α 2 Solving for α gives α = P0 AEmax /(2mL) =

32.46.

(8.85 ×10

−12

C 2 /N ⋅ m 2 ) (0.0150 m) 2 (1.25 N/C) 2

= 3.89 × 10−13 rad/s 2 (2)(0.00400 kg)(1.00 m) EVALUATE: This is an extremely small angular acceleration. To achieve a larger value, we would have to greatly increase the intensity of the light wave or decrease the mass of the reflectors. IDENTIFY: For light of intensity I abs incident on a totally absorbing surface, the radiation pressure is I abs 2I . For light of intensity I refl incident on a totally reflecting surface, prad,refl = refl . c c SET UP: The total radiation pressure is prad = prad,abs + prad,refl . I abs = wI and I refl = (1 − w) I prad,abs =

I abs 2 I refl wI 2(1 − w) I (2 − w) I . + = + = c c c c c I 2I (b) (i) For totally absorbing w = 1 so prad = . (ii) For totally reflecting w = 0 so prad = . These are just equations c c 32.32 and 32.33. (2 − 0.9)(1.40 × 103 W m 2 ) = 5.13 × 10−6 Pa. For w = 0.1 and (c) For w = 0.9 and I = 1.40 × 102 W/m 2 , prad = 3.00 × 108 m/s (2 − 0.1)(1.40 × 102 W m 2 ) = 8.87 × 10−6 Pa. I = 1.40 × 103 W m 2 , prad = 3.00 × 108 m/s EVALUATE: The radiation pressure is greater when a larger fraction is reflected. IDENTIFY and SET UP: In the wire the electric field is related to the current density by Eq.(25.7). Use Ampere’s ! law to calculate B. The Poynting vector is given by Eq.(32.28) and the equation that follows it relates the energy ! flow through a surface to S .

EXECUTE:

32.47.

(a) prad = prad,abs + prad,refl =

32-12

Chapter 32

! EXECUTE: (a) The direction of E is parallel to the axis of the cylinder, in the direction of the current. From Eq.(25.7), E = ρ J = ρ I / π a 2 . (E is uniform across the cross section of the conductor.) (b) A cross-sectional view of the conductor is given in Figure 32.47a; take the current to be coming out of the page. Apply Ampere’s law to a circle of radius a. ! ! AB ⋅ dl = B(2π a ) I encl = I Figure 32.47a

! ! μ0I AB ⋅ dl = μ 0 I encl gives B (2π a) = μ 0 I and B = 2π a ! The direction of B is counterclockwise around the circle. ! ! (c) The directions of E and B are shown in Figure 32.47b. ! 1 ! ! The direction of S = E×B

μ0

is radially inward. 1 1 ⎛ ρ I ⎞⎛ μ I ⎞ S= EB = ⎜ 2 ⎟⎜ 0 ⎟ μ0 μ 0 ⎝ π a ⎠⎝ 2π a ⎠ ρI 2 S= 2 3 2π a Figure 32.47b

Since S is constant over the surface of the conductor, the rate of energy flow P is given by S ρI 2 ρ lI 2 ρl . But R = 2 , so the times the surface of a length l of the conductor: P = SA = S (2π al ) = 2 3 (2π al ) = 2π a π a2 πa result from the Poynting vector is P = RI 2 . This agrees with PR = I 2 R, the rate at which electrical energy is being ! dissipated by the resistance of the wire. Since S is radially inward at the surface of the wire and has magnitude equal to the rate at which electrical energy is being dissipated in the wire, this energy can be thought of as entering through the cylindrical sides of the conductor. IDENTIFY: The intensity of the wave, not the electric field strength, obeys an inverse-square distance law. SET UP: The intensity is inversely proportional to the distance from the source, and it depends on the amplitude of the electric field by I = Sav = 12 P0 cEmax2. (d) EVALUATE:

32.48.

EXECUTE:

32.49.

Since I = 12 P0 cEmax2, Emax ∝ I . A point at 20.0 cm (0.200 m) from the source is 50 times closer to

the source than a point that is 10.0 m from it. Since I ∝ 1/r2 and (0.200 m)/(10.0 m) = 1/50, we have I0.20 = 502 I10. Since Emax ∝ I , we have E0.20 = 50E10 = (50)(1.50 N/C) = 75.0 N/C. EVALUATE: While the intensity increases by a factor of 502 = 2500, the amplitude of the wave only increases by a factor of 50. Recall that the intensity of any wave is proportional to the square of its amplitude. dΦB IDENTIFY and SET UP: The magnitude of the induced emf is given by Faraday’s law: E = . To calculate dt d Φ B / dt we need dB / dt at the antenna. Use the total power output to calculate I and then combine Eq.(32.29) and (32.18) to calculate Bmax . The time dependence of B is given by Eq.(32.17). Φ B = Bπ R 2 , where R = 0.0900 m is the radius of the loop. (This assumes that the magnetic field is dB uniform across the loop, an excellent approximation.) E = π R 2 dt dB B = Bmax cos(kx − ω t ) so = Bmaxω sin( kx − ω t ) dt

EXECUTE:

The maximum value of

dB is Bmaxω , so E max = π R 2 Bmaxω . dt

R = 0.0900 m, ω = 2π f = 2π (95.0 ×106 Hz) = 5.97 ×108 rad/s


32-13

Calculate the intensity I at this distance from the source, and from that the magnetic field amplitude Bmax : I=

2 P 55.0 ×103 W Emax (cBmax ) 2 c 2 −4 2 = = = = = 7.00 × 10 W/m . I Bmax 4π r 2 4π (2.50 × 103 m) 2 2μ 0c 2μ 0c 2μ0

Thus Bmax =

2 μ0 I 2(4π × 10−7 T ⋅ m/A)(7.00 × 10−4 W/m 2 ) = = 2.42 × 10−9 T. Then c 2.998 × 108 m/s

E max = π R 2 Bmaxω = π (0.0900 m) 2 (2.42 ×10−9 T)(5.97 ×108 rad/s) = 0.0368 V.

32.50.

32.51.

EVALUATE: An induced emf of this magnitude is easily detected. IDENTIFY: The nodal planes are one-half wavelength apart. SET UP: The nodal planes of B are at x = λ/4, 3λ/4, 5λ/4, …, which are λ/2 apart. EXECUTE: (a) The wavelength is λ = c/ f = (3.000 × 108 m/s)/(110.0 × 106 Hz) = 2.727 m. So the nodal planes are at (2.727 m)/2 = 1.364 m apart. (b) For the nodal planes of E, we have λn = 2L/n, so L = nλ/2 = (8)(2.727 m)/2 = 10.91 m EVALUATE: Because radiowaves have long wavelengths, the distances involved are easily measurable using ordinary metersticks. IDENTIFY and SET UP: Find the force on you due to the momentum carried off by the light. Express this force in terms of the radiated power of the flashlight. Use this force to calculate your acceleration and use a constant acceleration equation to find the time. (a) EXECUTE: prad = I / c and F = prad A gives F = IA / c = Pav / c

ax = F / m = Pav /(mc) = (200 W)/[(150 kg)(3.00 × 108 m/s)] = 4.44 ×10−9 m/s 2 Then x − x0 = v0 xt + 12 axt 2 gives t = 2( x − x0 )/ax = 2(16.0 m)/(4.44 × 10−9 m/s 2 ) = 8.49 × 104 s = 23.6 h

32.52.

EVALUATE: The radiation force is very small. In the calculation we have ignored any other forces on you. (b) You could throw the flashlight in the direction away from the ship. By conservation of linear momentum you would move toward the ship with the same magnitude of momentum as you gave the flashlight. 2 IDENTIFY: Pav = IA and I = 12 P0cEmax . Emax = cBmax SET UP: The power carried by the current i is P = Vi . EXECUTE: Bmax =

Pav 1 = P0cE 2 and Emax = A 2

2 Pav = AP0c

2Vi 2(5.00 × 105 V)(1000 A) = = 6.14 × 104 V m. AP0c (100 m 2 )P0 (3.00 × 108 m s)

Emax 6.14 × 104 V m = = 2.05 × 10−4 T. c 3.00 × 108 m s

EVALUATE: 32.53.

I=

I = Vi / A =

(5.00 ×105 V)(1000 A) = 5.00 ×106 W/m 2 . This is a very intense beam spread over a 100 m 2

large area. IDENTIFY: The orbiting satellite obeys Newton’s second law of motion. The intensity of the electromagnetic waves it transmits obeys the inverse-square distance law, and the intensity of the waves depends on the amplitude of the electric and magnetic fields. SET UP: Newton’s second law applied to the satellite gives mv2/R = GmM/r2, where M is the mass of the Earth and m is the mass of the satellite. The intensity I of the wave is I = Sav = 12 P0 cEmax2, and by definition, I = Pav /A. EXECUTE: (a) The period of the orbit is 12 hr. Applying Newton’s 2nd law to the satellite gives mv2/R = GmM/r2, m ( 2π r / T ) GmM = . Solving for r, we get r r2 2

which gives

1/ 3 ⎡ ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )( 5.97 × 1024 kg ) (12 × 3600 s )2 ⎤ ⎛ GMT 2 ⎞ ⎢ ⎥ = 2.66 × 107 m r =⎜ = ⎟ 4π 2 ⎝ 4π 2 ⎠ ⎣ ⎦ The height above the surface is h = 2.66 × 107 m – 6.38 × 106 m = 2.02 × 107 m. The satellite only radiates its energy to the lower hemisphere, so the area is 1/2 that of a sphere. Thus, from the definition of intensity, the intensity at the ground is 1/ 3

I = Pav /A = Pav /(2πh2) = (25.0 W)/[2π(2.02 × 107 m)2] = 9.75 × 10−15 W/m2 (b) I = Sav = 12 P0 cEmax2, so Emax =

2I 2(9.75 × 10−15 W/m 2 ) = = 2.71 × 10−6 N/C (8.85 × 10−12 C 2 /N ⋅ m 2 )(3.00 × 108 m/s) P0c

Bmax = Emax /c = (2.71 × 10−6 N/C)/(3.00 × 108 m/s) = 9.03 × 10−15 T t = d/c = (2.02 × 107 m)/(3.00 × 108 m/s) = 0.0673 s

32-14

32.54.

Chapter 32

(c) Pav = I / c = (9.75 × 10–15 W/m2)/(3.00 × 108 m/s) = 3.25 × 10 – 2 3 Pa (d) λ = c / f = (3.00 × 108 m/s)/(1575.42 × 106 Hz) = 0.190 m EVALUATE: The fields and pressures due to these waves are very small compared to typical laboratory quantities. 2I IDENTIFY: For a totally reflective surface the radiation pressure is . Find the force due to this pressure and c mM sun express the force in terms of the power output P of the sun. The gravitational force of the sun is Fg = G . r2 SET UP: The mass of the sum is M sun = 1.99 × 1030 kg. G = 6.67 ×10−11 N ⋅ m 2 /kg 2 . EXECUTE: (a) The sail should be reflective, to produce the maximum radiation pressure. P ⎛ 2I ⎞ (b) Frad = ⎜ ⎟ A , where A is the area of the sail. I = , where r is the distance of the sail from the sun. 4π r 2 ⎝ c ⎠

PA PA mM sun ⎛ 2 A ⎞⎛ P ⎞ = Frad = ⎜ . Frad = Fg so . =G ⎟⎜ 2 ⎟ 2 2π r 2c r2 ⎝ c ⎠⎝ 4π r ⎠ 2π r c 2π cGmM sun 2π (3.00 ×108 m/s)(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10,000 kg)(1.99 × 1030 kg) = . P 3.9 ×1026 W A = 6.42 × 106 m 2 = 6.42 km 2 . (c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set Frad = Fg . A=

32.55.

EVALUATE: A very large sail is needed, just to overcome the gravitational pull of the sun. IDENTIFY and SET UP: The gravitational force is given by Eq.(12.2). Express the mass of the particle in terms of its density and volume. The radiation pressure is given by Eq.(32.32); relate the power output L of the sun to the intensity at a distance r. The radiation force is the pressure times the cross sectional area of the particle. mM EXECUTE: (a) The gravitational force is Fg = G 2 . The mass of the dust particle is m = ρV = ρ 43 π R 3 . Thus r 4 ρ Gπ MR 3 Fg = . 3r 2 I (b) For a totally absorbing surface prad = . If L is the power output of the sun, the intensity of the solar radiation c L L a distance r from the sun is I = . Thus prad = . The force Frad that corresponds to prad is in the 2 4π cr 2 4π r direction of propagation of the radiation, so Frad = prad A⊥ , where A⊥ = π R 2 is the component of area of the particle

LR 2 ⎛ L ⎞ perpendicular to the radiation direction. Thus Frad = ⎜ (π R 2 ) = . 2 ⎟ 4cr 2 ⎝ 4π cr ⎠ (c) Fg = Frad 4 ρ Gπ MR 3 LR 2 = 3r 2 4cr 2 L 3L ⎛ 4 ρ Gπ M ⎞ and R = ⎜ ⎟R = 3 4 c 16 c ρ Gπ M ⎝ ⎠ R=

3(3.9 ×1026 W) 16(2.998 ×10 m/s)(3000 kg/m3 )(6.673 ×10−11 N ⋅ m 2 / kg 2 )π (1.99 × 1030 kg) 8

R = 1.9 ×10−7 m = 0.19 μ m. EVALUATE: The gravitational force and the radiation force both have a r −2 dependence on the distance from the sun, so this distance divides out in the calculation of R. ⎛ LR 2 ⎞⎛ ⎞ F 3r 2 3L . Frad is proportional to R 2 and Fg is proportional to R 3 , so this (d) rad = ⎜ ⎟= 2 ⎟⎜ Fg ⎝ 4cr ⎠⎝ 4 ρ Gπ mR 3 ⎠ 16c ρ Gπ MR

ratio is proportional to 1/R. If R < 0.20 μ m then Frad > Fg and the radiation force will drive the particles out of the solar system. 32.56.

IDENTIFY: SET UP:

v2 . R C . An electron has q = e = 1.60 ×10−19 C .

The electron has acceleration a = 1 eV = 1.60 ×10−19


32.57.

32-15

EXECUTE: For the electron in the classical hydrogen atom, its acceleration is v 2 1 mv 2 2(13.6 eV)(1.60 ×10−19 J eV ) a = = 21 = = 9.03 ×1022 m/s 2 . Then using the formula for the rate of energy R 2 mR (9.11× 10−31 kg)(5.29 ×10−11 m) emission given in Problem 32.57: dE q 2a 2 (1.60 × 10−19 C) 2 (9.03 × 1022 m s 2 ) 2 dE = = = 4.64 × 10−8 J s = 2.89 × 1011 eV s. This large value of 3 dt dt 6π P0c 6π P0 (3.00 × 108 m s)3 would mean that the electron would almost immediately lose all its energy! EVALUATE: The classical physics result in Problem 32.57 must not apply to electrons in atoms. v2 IDENTIFY: The orbiting particle has acceleration a = . R SET UP: K = 12 mv 2 . An electron has mass me = 9.11×10−31 kg and a proton has mass mp = 1.67 × 10−27 kg . 2 ⎡ q 2a 2 ⎤ C2 ( m s )2 N⋅m J ⎡ dE ⎤ (a) ⎢ = = = = W = ⎢ ⎥. 3⎥ 2 2 3 ⋅ π 6 c (C N m )( m s) s s P ⎣ dt ⎦ 0 ⎣ ⎦ (b) For a proton moving in a circle, the acceleration is v 2 1 mv 2 2(6.00 ×106 eV) (1.6 × 10−19 J eV ) 2 a = = 21 = = 1.53 × 1015 m s . The rate at which it emits energy because of R 2 mR (1.67 × 10−27 kg) (0.75 m) dE q 2a 2 (1.6 × 10 −19 C) 2 (1.53 × 1015 m s 2 ) 2 its acceleration is = = = 1.33 × 10 −23 J s = 8.32 × 10−5 eV s. dt 6π P0c 3 6π P0 (3.0 × 108 m s)3 (dE dt )(1 s) 8.32 ×10−5 eV Therefore, the fraction of its energy that it radiates every second is = = 1.39 ×10−11. E 6.00 ×106 eV (c) Carry out the same calculations as in part (b), but now for an electron at the same speed and radius. That means the electron’s acceleration is the same as the proton, and thus so is the rate at which it emits energy, since they also have the same charge. However, the electron’s initial energy differs from the proton’s by the ratio of their masses: m (9.11×10−31 kg) Ee = Ep e = (6.00 ×106 eV) = 3273 eV. Therefore, the fraction of its energy that it radiates every mp (1.67 ×10−27 kg)

EXECUTE:

second is

( dE dt )(1 s) 8.32 ×10 −5 eV = = 2.54 × 10−8. E 3273 eV

EVALUATE:

32.58.

The proton has speed v =

2E 2(6.0 ×106 eV)(1.60 × 10−19 J/eV) = = 3.39 ×107 m/s . The electron mp 1.67 ×10−27 kg

has the same speed and kinetic energy 3.27 keV. The particles in the accelerator radiate at a much smaller rate than the electron in Problem 32.56 does, because in the accelerator the orbit radius is very much larger than in the atom, so the acceleration is much less. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) E y ( x, t ) = Emax e − kC x sin ( kC x − ω t ). ∂E y

= Emax (− kC )e − kC x sin( kC x − ω t ) + Emax ( + kC )e − kC x cos(kC x − ω t ) ∂x ∂ 2 Ey = Emax (+ kC2 )e − kC x sin(kC x − ωt ) + Emax (− kC2 )e − kC x cos(kC x − ωt ) ∂x 2 + Emax (− kC2 )e − kC x cos(kC x − ω t ) + Emax (− kC2 )e − kC x sin(kC x − ω t ) .

∂2Ey ∂x 2

= −2 Emax kC2e − kC x cos(kC x − ω t ) .

Setting 2kC2

ω

=

∂ 2 Ey ∂x

2

=

∂E y ∂t

= Emax e − kC x ω cos(kC x − ω t ).

μ∂E y gives 2 Emax kC2e − k x cos(kC x − ω t ) = Emax e − k x ω cos(kC x − ω t ) . This will only be true if ρ∂t C

C

μ ωμ , or kC = . 2ρ ρ

(b) The energy in the wave is dissipated by the i 2 R heating of the conductor. E 1 2ρ 2(1.72 ×10−8 Ω ⋅ m) (c) E y = y 0 ⇒ kC x = 1, x = = = = 6.60 ×10−5 m. e kC ωμ 2π (1.0 ×106 Hz)μ 0 EVALUATE: The lower the frequency of the waves, the greater is the distance they can penetrate into a conductor. A dielectric (insulator) has a much larger resistivity and these waves can penetrate a greater distance in these materials.

THE NATURE AND PROPAGATION OF LIGHT

33.1.

33

IDENTIFY: For reflection, θ r = θ a . SET UP: The desired path of the ray is sketched in Figure 33.1. 14.0 cm EXECUTE: tan φ = , so φ = 50.6° . θ r = 90° − φ = 39.4° and θ r = θ a = 39.4° . 11.5 cm EVALUATE: The angle of incidence is measured from the normal to the surface.

Figure 33.1 33.2

IDENTIFY: For reflection, θ r = θ a . SET UP: The angles of incidence and reflection at each reflection are shown in Figure 33.2. For the rays to be perpendicular when they cross, α = 90° . α EXECUTE: (a) θ + φ = 90° and β + φ = 90° , so β = θ . + β = 90° and α = 180° − 2θ . 2 (b) θ = 12 (180° − α ) = 12 (180° − 90°) = 45° . EVALUATE: As θ → 0° , α → 180° . This corresponds to the incident and reflected rays traveling in nearly the same direction. As θ → 90° , α → 0° . This corresponds to the incident and reflected rays traveling in nearly opposite directions.

Figure 33.2 33-1

33-2

33.3.

33.4.

33.5.

33.6.

Chapter 33

IDENTIFY and SET UP: Use Eqs.(33.1) and (33.5) to calculate v and λ. c c 2.998 × 108 m/s EXECUTE: (a) n = so v = = = 2.04 × 108 m/s v n 1.47 λ 650 nm (b) λ = 0 = = 442 nm 1.47 n EVALUATE: Light is slower in the liquid than in vacuum. By v = f λ , when v is smaller, λ is smaller. λ IDENTIFY: In air, c = f λ0 . In glass, λ = 0 . n SET UP: c = 3.00 × 108 m/s c 3.00 × 108 m/s = 517 nm EXECUTE: (a) λ0 = = f 5.80 × 1014 Hz λ 517 nm (b) λ = 0 = = 340 nm 1.52 n EVALUATE: In glass the light travels slower than in vacuum and the wavelength is smaller. c λ IDENTIFY: n = . λ = 0 , where λ0 is the wavelength in vacuum. v n SET UP: c = 3.00 × 108 m/s . n for air is only slightly larger than unity. c 3.00 × 108 m/s = 1.54. EXECUTE: (a) n = = v 1.94 × 108 m/s (b) λ0 = nλ = (1.54)(3.55 × 10−7 m) = 5.47 × 10−7 m. EVALUATE: In quartz the speed is lower and the wavelength is smaller than in air. IDENTIFY: SET UP:

λ=

λ0

. n From Table 33.1, nwater = 1.333 and nbenzene = 1.501 .

⎛ n ⎞ ⎛ 1.333 ⎞ (a) λwater nwater = λbenzene nbenzene = λ0 . λbenzene = λwater ⎜ water ⎟ = (438 nm) ⎜ ⎟ = 389 nm . n ⎝ 1.501 ⎠ ⎝ benzene ⎠ (b) λ0 = λwater nwater = (438 nm)(1.333) = 584 nm EVALUATE: λ is smallest in benzene, since n is largest for benzene. IDENTIFY: Apply Eqs.(33.2) and (33.4) to calculate θ r and θ b . The angles in these equations are measured with respect to the normal, not the surface. (a) SET UP: The incident, reflected and refracted rays are shown in Figure 33.7. EXECUTE:

33.7.

EXECUTE: θ r = θ a = 42.5° The reflected ray makes an angle of 90.0° − θ r = 47.5° with the surface of the glass. Figure 33.7

33.8.

(b) na sinθ a = nb sinθ b , where the angles are measured from the normal to the interface. n sin θ a (1.00)(sin 42.5°) sin θ b = a = = 0.4070 nb 1.66 θ b = 24.0° The refracted ray makes an angle of 90.0° − θ b = 66.0° with the surface of the glass. EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index. c IDENTIFY: Use the distance and time to find the speed of light in the plastic. n = . v SET UP: c = 3.00 × 108 m/s c 3.00 × 108 m/s d 2.50 m EXECUTE: v = = = 1.38 . = 2.17 × 108 m s . n = = −9 v 2.17 × 108 m/s t 11.5 × 10 s 2.50 m EVALUATE: In air light travels this same distance in = 8.3 ns . 3.00 × 108 m/s

The Nature and Propagation of Light

33.9.

33.10.

33-3

IDENTIFY and SET UP: Use Snell’s law to find the index of refraction of the plastic and then use Eq.(33.1) to calculate the speed v of light in the plastic. EXECUTE: na sinθ a = nb sinθ b

⎛ sin θ a ⎞ ⎛ sin 62.7° ⎞ nb = na ⎜ ⎟ = 1.00 ⎜ ⎟ = 1.194 sin θ sin 48.1° ⎠ ⎝ b ⎠ ⎝ c c n = so v = = (3.00 × 108 m/s) /1.194 = 2.51 × 108 m/s v n EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent toward the normal. IDENTIFY: Apply Snell’s law at both interfaces. SET UP: The path of the ray is sketched in Figure 33.10. Table 33.1 gives n = 1.329 for the methanol. EXECUTE: (a) At the air-glass interface (1.00)sin 41.3° = nglass sin α . At the glass-methanol interface

nglass sin α = (1.329)sinθ . Combining these two equations gives sin 41.3° = 1.329sin θ and θ = 29.8° . (b) The same figures applies as for part (a), except θ = 20.2° . (1.00)sin 41.3° = n sin 20.2° and n = 1.91. EVALUATE: The angle α is 25.2° . The index of refraction of methanol is less than that of the glass and the ray is bent away from the normal at the glass → methanol interface. The unknown liquid has an index of refraction greater than that of the glass, so the ray is bent toward the normal at the glass → liquid interface.

33.11.

33.12.

Figure 33.10 IDENTIFY: Apply Snell’s law to each refraction. SET UP: Let the light initially be in the material with refractive index na and let the final slab have refractive index nb . In part (a) let the middle slab have refractive index n1. EXECUTE: (a) 1st interface: na sin θ a = n1 sinθ1 . 2nd interface: n1 sin θ1 = nb sin θ b . Combining the two equations gives na sinθ a = nb sinθ b . This is the equation that would apply if the middle slab were absent. (b) For N slabs, na sin θ a = n1 sinθ1 , n1 sinθ1 = n2 sin θ 2 , …, nN − 2 sinθ N − 2 = nb sin θ b . Combining all these equations gives na sinθ a = nb sinθ b . EVALUATE: The final direction of travel depends on the angle of incidence in the first slab and the refractive indices of the first and last slabs. IDENTIFY: Apply Snell's law to the refraction at each interface. SET UP: nair = 1.00. nwater = 1.333.

⎛ n ⎞ ⎛ 1.00 ⎞ (a) θ water = arcsin ⎜ air sinθ air ⎟ = arcsin ⎜ sin 35.0° ⎟ = 25.5°. n 1.333 ⎝ ⎠ ⎝ water ⎠ EVALUATE: (b) This calculation has no dependence on the glass because we can omit that step in the chain: nair sin θ air = nglass sin θ glass = nwater sin θ water . EXECUTE:

33.13.

IDENTIFY: When a wave passes from one material into another, the number of waves per second that cross the boundary is the same on both sides of the boundary, so the frequency does not change. The wavelength and speed of the wave, however, do change. λ SET UP: In a material having index of refraction n, the wavelength is λ = 0 , where λ0 is the wavelength in n vacuum, and the speed is c . n λ EXECUTE: (a) The frequency is the same, so it is still f. The wavelength becomes λ = 0 , so λ0 = nλ. The speed n is v = c , so c = nv. n (b) The frequency is still f. The wavelength becomes λ ′ = λ0 = nλ = ⎜⎛ n ⎟⎞ λ and the speed becomes n′ n′ ⎝ n′ ⎠ c nv ⎛ n ⎞ v′ = = = ⎜ ⎟v n′ n′ ⎝ n′ ⎠ EVALUATE: These results give the speed and wavelength in a new medium in terms of the original medium without referring them to the values in vacuum (or air).

33-4

Chapter 33

33.14.

IDENTIFY: Apply the law of reflection. SET UP: The mirror in its original position and after being rotated by an angle θ are shown in Figure 33.14. α is the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled φ are equal and the two angles labeled φ ′are equal because of the law of reflection. The two angles labeled θ are equal because the lines forming one angle are perpendicular to the lines forming the other angle. EXECUTE: From the diagram, α = 2φ ′ − 2φ = 2(φ ′ − φ ) and θ = φ ′ − φ . α = 2θ , as was to be shown. EVALUATE: This result is independent of the initial angle of incidence.

33.15.

Figure 33.14 IDENTIFY: Apply na sin θ a = nb sin θ b . SET UP: The light refracts from the liquid into the glass, so na = 1.70 , θ a = 62.0° . nb = 1.58 . ⎛n ⎞ ⎛ 1.70 ⎞ sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 62.0° = 0.950 and θ b = 71.8° . ⎝ 1.58 ⎠ ⎝ nb ⎠ EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal. IDENTIFY: Apply Snell's law. SET UP: θ a and θ b are measured relative to the normal to the surface of the interface. θ a = 60.0° − 15.0° = 45.0° .

EXECUTE: 33.16.

33.17.

⎛n ⎞ ⎛ 1.33 ⎞ EXECUTE: θ b = arcsin ⎜ a sinθ a ⎟ = arcsin ⎜ sin 45.0° ⎟ = 38.2° . But this is the angle from the normal to the surface, ⎝ 1.52 ⎠ ⎝ nb ⎠ so the angle from the vertical is an additional 15° because of the tilt of the surface. Therefore, the angle is 53.2°. EVALUATE: Compared to Example 33.1, θ a is shifted by 15° but the shift in θ b is only 53.2° − 49.3° = 3.9° . IDENTIFY: The critical angle for total internal reflection is θ a that gives θ b = 90° in Snell's law. SET UP: In Figure 33.17 the angle of incidence θ a is related to angle θ by θ a + θ = 90° . EXECUTE: (a) Calculate θ a that gives θ b = 90°. na = 1.60 , nb = 1.00 so na sin θ a = nb sin θ b gives 1.00 and θ a = 38.7° . θ = 90° − θ a = 51.3° . (1.60)sin θ a = (1.00)sin 90° . sin θ a = 1.60 1.333 (b) na = 1.60 , nb = 1.333 . (1.60)sin θ a = (1.333)sin 90°. sin θ a = and θ a = 56.4° . θ = 90° − θ a = 33.6° . 1.60 n EVALUATE: The critical angle increases when the ratio a increases. nb

Figure 33.17


33.18.

33.19.

33-5

IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will occur at the cube surface if the angle of incidence is greater than or equal to the critical angle. SET UP: At the critical angle θc, Snell’s law gives nglass sin θc = nair sin 90° and likewise for water. EXECUTE: (a) At the critical angle θc , nglass sin θc = nair sin 90°. 1.53 sin θc = (1.00)(1) andθc = 40.8°. (b) Using the same procedure as in part (a), we have 1.53 sin θc = 1.333 sin 90° and θc = 60.6°. EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air. IDENTIFY: Use the critical angle to find the index of refraction of the liquid. SET UP: Total internal reflection requires that the light be incident on the material with the larger n, in this case the liquid. Apply na sinθ a = nb sinθ b with a = liquid and b = air, so na = nliq and nb = 1.0. EXECUTE:

nliq =

θ a = θ crit when θ b = 90°, so nliq sin θ crit = (1.0)sin 90°

1 1 = = 1.48. sin θ crit sin 42.5°

(a) na sinθ a = nb sinθ b (a = liquid, b = air)

sin θ b =

na sinθ a (1.48)sin 35.0° = = 0.8489 and θ b = 58.1° 1.0 nb

(b) Now na sin θ a = nb sin θ b with a = air, b = liquid

na sin θ a (1.0)sin 35.0° = = 0.3876 and θ b = 22.8° 1.48 nb EVALUATE: For light traveling liquid → air the light is bent away from the normal. For light traveling air → liquid the light is bent toward the normal. IDENTIFY: The largest angle of incidence for which any light refracts into the air is the critical angle for water → air . SET UP: Figure 33.20 shows a ray incident at the critical angle and therefore at the edge of the ring of light. The radius of this circle is r and d = 10.0 m is the distance from the ring to the surface of the water. EXECUTE: From the figure, r = d tan θ crit . θ crit is calculated from na sinθ a = nb sinθ b with na = 1.333 , θ a = θ crit , (1.00)sin 90° nb = 1.00 and θ b = 90°. sinθ crit = and θ crit = 48.6°. r = (10.0 m) tan 48.6° = 11.3 m . 1.333 A = π r 2 = π (11.3 m) 2 = 401 m 2 . EVALUATE: When the incident angle in the water is larger than the critical angle, no light refracts into the air. sin θ b =

33.20.

Figure 33.20 33.21.


For glass → water, θ crit = 48.7°. Apply Snell’s law with θ a = θ crit to calculate the index

of refraction na of the glass. nb 1.333 = = 1.77 sin θ crit sin 48.7° EVALUATE: For total internal reflection to occur the light must be incident in the material of larger refractive index. Our results give nglass > nwater , in agreement with this.

EXECUTE:

na sin θ crit = nb sin 90°, so na =

33-6

Chapter 33

33.22.

IDENTIFY: If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface is θ crit . SET UP: The ray has an angle of incidence of 0° at the first surface of the glass, so enters the glass without being bent, as shown in Figure 33.22. The figure shows that α + θ crit = 90°. EXECUTE: (a) For the glass-air interface θ a = θ crit , na = 1.52, nb = 1.00 and θb = 90°. na sinθ a = nb sinθ b gives (1.00)(sin 90°) sinθ crit = and θ crit = 41.1°. α = 90° − θ crit = 48.9°. 1.52 (1.333)(sin 90°) (b) Now the second interface is glass → water and nb = 1.333 . na sinθ a = nb sinθ b gives sinθ crit = 1.52 and θ crit = 61.3°. α = 90° − θ crit = 28.7° . EVALUATE: The critical angle increases when the air is replaced by water and rays are bent as they refract out of the glass.

Figure 33.22 33.23.

Apply na sinθ a = nb sinθ b .

IDENTIFY: SET UP:

The light is in diamond and encounters an interface with air, so na = 2.42 and nb = 1.00 . The largest

θ a is when θ b = 90° .

33.24.

1 and θ a = 24.4° . 2.42 EVALUATE: Diamond has an usually large refractive index, and this results in a small critical angle. c IDENTIFY: Snell's law is na sin θ a = nb sin θ b . v = . n SET UP: a = air , b = glass . EXECUTE:

(2.42)sin θ a = (1.00)sin 90° . sin θ a =

EXECUTE:

(a) red: nb =

na sinθ a (1.00)sin 57.0° (1.00)sin 57.0° = = 1.36 . violet: nb = = 1.40 . sin θ b sin 38.1° sin 36.7°

c 3.00 × 108 m/s c 3.00 × 108 m/s = = 2.21 × 108 m/s ; violet: v = = = 2.14 × 108 m/s . 1.36 1.40 n n EVALUATE: n is larger for the violet light and therefore this light is bent more toward the normal, and the violet light has a smaller speed in the glass than the red light. IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12 and the (b) red: v =

33.25.

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos 2 φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. SET UP: For the second polarizer φ = 60° . For the third polarizer, φ = 90° − 60° = 30° . EXECUTE:

(a) At point A the intensity is I 0 / 2 and the light is polarized along the vertical direction. At point B

the intensity is ( I 0 / 2)(cos60°) 2 = 0.125I 0 , and the light is polarized along the axis of the second polarizer. At point C the intensity is (0.125 I 0 )(cos30°) 2 = 0.0938 I 0 .

33.26.

(b) Now for the last filter φ = 90° and I = 0 . EVALUATE: Adding the middle filter increases the transmitted intensity. IDENTIFY: Apply Snell's law. SET UP: The incident, reflected and refracted rays are shown in Figure 33.26. sin θ a sin 53° EXECUTE: From the figure, θ b = 37.0° and nb = na = 1.33 = 1.77. sin θ b sin 37°


EVALUATE: refracts.

33-7

The refractive index of b is greater than that of a, and the ray is bent toward the normal when it

Figure 33.26 33.27.

IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals the polarizing angle, so θ p = 54.5°. Use Eq.(33.8) to calculate the refractive index of the glass. Then use Snell’s

law to calculate the angle of refraction. n EXECUTE: (a) tanθ p = b gives nglass = nair tanθ p = (1.00) tan 54.5° = 1.40. na (b) na sinθ a = nb sinθ b

na sinθ a (1.00)sin 54.5° = = 0.5815 and θ b = 35.5° 1.40 nb EVALUATE:

sin θ b =

Note: φ = 180.0° − θ r − θ b and θ r = θ a . Thus φ = 180.0° − 54.5° − 35.5° = 90.0°; the reflected ray and the refracted ray are perpendicular to each other. This agrees with Fig.33.28. Figure 33.27 33.28.

IDENTIFY: SET UP:

Set I = I 0 /10 , where I is the intensity of light passed by the second polarizer. When unpolarized light passes through a polarizer the intensity is reduced by a factor of

1 2

and the

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos 2 φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. I EXECUTE: (a) After the first filter I = 0 and the light is polarized along the vertical direction. After the second 2 I0 I0 ⎛ I0 ⎞ filter we want I = , so = ⎜ ⎟ (cos φ ) 2 . cos φ = 2 /10 and φ = 63.4° . 10 ⎝ 2 ⎠ 10 I (b) Now the first filter passes the full intensity I 0 of the incident light. For the second filter 0 = I 0 (cos φ ) 2 . 10

33.29.

cos φ = 1/10 and φ = 71.6° . EVALUATE: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the same overall reduction in intensity than when the incident light is unpolarized. IDENTIFY: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters. SET UP: If the angle between the two axes is θ, the intensity of the emerging light is I = Imax cos2θ. 1 EXECUTE: At angle θ, I = Imax cos2θ, and at the new angle α, I = Imax cos2α. Taking the ratio of the intensities 2 cosθ I max cos 2 α 12 I ⎛ cosθ ⎞ = , which gives us cosα = . Solving for α yields α = arccos ⎜ gives ⎟. I max cos 2 θ I 2 ⎝ 2 ⎠ EVALUATE: Careful! This result is not cos2θ.

33-8

Chapter 33

33.30.

IDENTIFY:

The reflected light is completely polarized when the angle of incidence equals the polarizing angle n θ p , where tanθ p = b . na

nb = 1.66 .

SET UP: EXECUTE:

(a) na = 1.00 . tan θ p =

1.66 and θ p = 58.9° . 1.00

1.66 and θ p = 51.2° . 1.333 EVALUATE: The polarizing angle depends on the refractive indicies of both materials at the interface. IDENTIFY: When unpolarized light of intensity I 0 is incident on a polarizing filter, the transmitted light has (b) na = 1.333 . tan θ p =

33.31.

intensity

1 2

I 0 and is polarized along the filter axis. When polarized light of intensity I 0 is incident on a polarizing

filter the transmitted light has intensity I 0 cos 2 φ . SET UP: EXECUTE:

For the second filter, φ = 62.0° − 25.0° = 37.0° . After the first filter the intensity is

1 2

I 0 = 10.0 W m 2 and the light is polarized along the axis of the

first filter. The intensity after the second filter is I = I 0cos 2φ , where I 0 = 10.0 W m 2 and φ = 37.0° . This

33.32.

gives I = 6.38 W m 2 . EVALUATE: The transmitted intensity depends on the angle between the axes of the two filters. IDENTIFY: After passing through the first filter the light is linearly polarized along the filter axis. After the second filter, I = I max (cos φ ) 2 , where φ is the angle between the axes of the two filters. SET UP: The maximum amount of light is transmitted when φ = 0 . EXECUTE:

(a) I = I 0 (cos 22.5°) 2 = 0.854 I 0

(b) I = I 0 (cos 45.0°) 2 = 0.500 I 0 (c) I = I 0 (cos67.5°) 2 = 0.146 I 0

33.33.

EVALUATE: As φ increases toward 90° the axes of the two filters are closer to being perpendicular to each other and the transmitted intensity decreases. IDENTIFY and SET UP: Apply Eq.(33.7) to polarizers #2 and #3. The light incident on the first polarizer is unpolarized, so the transmitted light has half the intensity of the incident light, and the transmitted light is polarized. (a) EXECUTE: The axes of the three filters are shown in Figure 33.33a.

I = I max cos 2 φ

Figure 33.33a

After the first filter the intensity is I1 = 12 I 0 and the light is linearly polarized along the axis of the first polarizer. After the second filter the intensity is I 2 = I1 cos 2 φ = ( 12 I 0 )(cos 45.0°) 2 = 0.250 I 0 and the light is linearly polarized along the axis of the second polarizer. After the third filter the intensity is I 3 = I 2 cos 2 φ = 0.250 I 0 (cos 45.0°) 2 = 0.125I 0 and the light is linearly polarized along the axis of the third polarizer. (b) The axes of the remaining two filters are shown in Figure 33.33b. After the first filter the intensity is I1 = 12 I 0 and the light is linearly polarized along the axis of the first polarizer.

Figure 33.33b

After the next filter the intensity is I 3 = I1 cos 2 φ = ( 12 I 0 )(cos90.0°) 2 = 0. No light is passed. EVALUATE: Light is transmitted through all three filters, but no light is transmitted if the middle polarizer is removed.


33.34.

33-9

IDENTIFY: Use the transmitted intensity when all three polairzers are present to solve for the incident intensity I 0 . Then repeat the calculation with only the first and third polarizers. SET UP:

For unpolarized light incident on a filter, I = 12 I 0 and the light is linearly polarized along the filter axis.

For polarized light incident on a filter, I = I max (cos φ ) 2 , where I max is the intensity of the incident light, and the emerging light is linearly polarized along the filter axis. EXECUTE: With all three polarizers, if the incident intensity is I 0 the transmitted intensity is I 75.0 W/cm 2 = = 293 W/cm 2 . With only the first 0.256 0.256 and third polarizers, I = ( 12 I 0 )(cos62.0°) 2 = 0.110 I 0 = (0.110)(293 W/cm 2 ) = 32.2 W/cm 2 . EVALUATE: The transmitted intensity is greater when all three filters are present. IDENTIFY: The shorter the wavelength of light, the more it is scattered. The intensity is inversely proportional to the fourth power of the wavelength. SET UP: The intensity of the scattered light is proportional to 1/λ4, we can write it as I = (constant)/ λ4. EXECUTE: (a) Since I is proportional to 1/λ4, we have I = (constant)/ λ4. Taking the ratio of the intensity of the 4 4 I R (constant) / λR4 ⎛ λG ⎞ ⎛ 520 nm ⎞ = 0.374, so IR = 0.374I. = = = red light to that of the green light gives ⎜ ⎟ ⎜ ⎟ I (constant) / λG4 ⎝ λR ⎠ ⎝ 665 nm ⎠

I = ( 12 I 0 )(cos 23.0°) 2 (cos[62.0° − 23.0°]) 2 = 0.256 I 0 . I 0 =

33.35.

4

33.36.

33.37.

33.38.

4

⎛ λ ⎞ ⎛ 520 nm ⎞ I (b) Following the same procedure as in part (a) gives V = ⎜ G ⎟ = ⎜ ⎟ = 2.35, so IV = 2.35I. I ⎝ λV ⎠ ⎝ 420 nm ⎠ EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about 7 times as great as that of the red light, so this scattered light will have a blue-violet color. IDENTIFY: As the wave front reaches the sharp object, every point on the front will act as a source of secondary wavelets. SET UP: Consider a wave front that is just about to go past the corner. Follow it along and draw the successive wave fronts. EXECUTE: The path of the wavefront is drawn in Figure 33.36. EVALUATE: The wave fronts clearly bend around the sharp point, just as water waves bend around a rock and light waves bend around the edge of a slit.

Figure 33.36 IDENTIFY: Reflection reverses the sign of the component of light velocity perpendicular to the reflecting surface but leaves the other components unchanged. SET UP: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the (x,z)-plane. EXECUTE: A light ray reflecting from M1 changes the sign of the z-component of the velocity, reflecting from M2 changes the x-component, and from M3 changes the y-component. Thus the velocity, and hence also the path, of the light beam flips by 180° EVALUATE: Example 33.3 discusses some uses of corner reflectors. IDENTIFY: The light travels slower in the jelly than in the air and hence will take longer to travel the length of the tube when it is filled with jelly than when it contains just air. SET UP: The definition of the index of refraction is n = c/v, where v is the speed of light in the jelly. EXECUTE: First get the length L of the tube using air. In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m. c L The speed in the jelly is v = = (2.616 m)/(8.72 ns + 2.04 ns) = 2.431 × 108 m/s. n = = v t (3.00 × 108 m/s)/(2.431 × 108 m/s) = 1.23 EVALUATE: A high-speed timer would be needed to measure times as short as a few nanoseconds.

33-10

Chapter 33

33.39.

IDENTIFY and SET UP: Apply Snell's law at each interface. EXECUTE: (a) n1 sinθ1 = n2 sinθ 2 and n2 sinθ 2 = n3 sinθ 3 , so n1 sin θ1 = n3 sinθ 3 and sin θ 3 = ( n1 sinθ1 ) / n3 . (b) n3 sinθ 3 = n2 sinθ 2 and n2 sinθ 2 = n1 sinθ1 , so n1 sin θ1 = n3 sinθ 3 and the light makes the same angle with respect to the normal in the material that has refractive index n1 as it did in part (a). (c) For reflection, θ r = θ a . These angles are still equal if θ r becomes the incident angle; reflected rays are also reversible. EVALUATE: Both the refracted and reflected rays are reversible, in the sense that if the direction of the light is reversed then each of these rays follow the path of the incident ray. c λ IDENTIFY: Use the change in transit time to find the speed v of light in the slab, and then apply n = and λ = 0 . n v SET UP: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam. 0.840 m 0.840 m 0.840 m EXECUTE: − = ( n − 1) = 4.2 ns. We can now solve for the index of refraction: c/n c c −9 8 490 nm (4.2 × 10 s) (3.00 × 10 m s) = 196 nm . n= + 1 = 2.50. The wavelength inside of the glass is λ = 2.50 0.840 m EVALUATE: Light travels slower in the slab than in air and the wavelength is shorter. IDENTIFY: The angle of incidence at A is to be the critical angle. Apply Snell’s law at the air to glass refraction at the top of the block. SET UP: The ray is sketched in Figure 33.41. EXECUTE: For glass → air at point A, Snell's law gives (1.38)sin θ crit = (1.00)sin 90°and θ crit = 46.4° . θ b = 90° − θ crit = 43.6° . Snell's law applied to the refraction from air to glass at the top of the block gives (1.00)sinθ a = (1.38)sin(43.6°) and θ a = 72.1° . EVALUATE: If θ a is larger than 72.1° then the angle of incidence at point A is less than the initial critical angle and total internal reflection doesn’t occur.

33.40.

33.41.

33.42.

Figure 33.41 IDENTIFY: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell’s law. SET UP: Snell’s law is na sin θa = nb sin θb and n = 1.000 for air. At point B the angle of the prism is 30.0° . EXECUTE: Apply Snell’s law at AB. The prism angle at A is 60.0°, so for the upper ray, the angle of incidence at AB is 60.0° + 12.0° = 72.0°. Using this value gives n1 sin 60.0° = sin 72.0° and n1 = 1.10. For the lower ray, the angle of incidence at AB is 60.0° + 12.0° + 8.50° = 80.5°, giving n2 sin 60.0° = sin 80.5° and n2 = 1.14. EVALUATE: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater index of refraction than the upper ray.


33.43.

33.44.

33.45.

33-11

IDENTIFY: Circularly polarized light consists of the superposition of light polarized in two perpendicular directions, with a quarter-cycle ( 90° ) phase difference between the two polarization components. SET UP: A quarter-wave plate shifts the relative phase of the two perpendicular polarization components by 90° . EXECUTE: In the circularly polarized light the two perpendicular polarization components are 90° out of phase. The quarter-wave plate shifts the relative phase by ±90° and then the two components are either in phase or 180° out of phase. Either corresponds to linearly polarized light. EVALUATE: Either left circularly polarized light or right circularly polarized light is converted to linearly polarized light by the quarter-wave plate. λ d IDENTIFY: Apply λ = 0 . The number of wavelengths in a distance d of a material is where λ is the n λ wavelength in the material. SET UP: The distance in glass is d glass = 0.00250 m . The distance in air is

d air = 0.0180 m − 0.00250 m = 0.0155 m EXECUTE: number of wavelengths = number in air + number in glass. d d 0.0155 m 0.00250 m + number of wavelengths = air + glass n = (1.40) = 3.52 × 104 . λ λ 5.40 × 10 −7 m 5.40 × 10−7 m EVALUATE: Without the glass plate the number of wavelengths between the source and screen is 0.0180 m = 3.33 × 104 . The wavelength is shorter in the glass so there are more wavelengths in a distance in 5.40 × 10−3 m glass than there are in the same distance in air. IDENTIFY: Find the critical angle for glass → air. Light incident at this critical angle is reflected back to the edge of the halo. SET UP: The ray incident at the critical angle is sketched in Figure 33.45.

Figure 33.45 2.67 mm = 0.8613; θ crit = 40.7°. 3.10 mm Apply Snell’s law to the total internal reflection to find the refractive index of the glass: na sinθ a = nb sinθ b

EXECUTE:

From the distances given in the sketch, tan θ crit =

nglass sin θ crit = 1.00sin 90° nglass =

1 1 = = 1.53 sinθ crit sin 40.7°

EVALUATE: 33.46.

Light incident on the back surface is also totally reflected if it is incident at angles greater than θ crit .

If it is incident at less than θ crit it refracts into the air and does not reflect back to the emulsion. IDENTIFY: Apply Snell's law to the refraction of the light as it passes from water into air. ⎛ 1.5 m ⎞ SET UP: θ a = arctan ⎜ ⎟ = 51° . na = 1.00 . nb = 1.333 . ⎝ 1.2 m ⎠ EXECUTE:

⎛ na ⎞ ⎛ 1.00 ⎞ sinθ a ⎟ = arcsin ⎜ sin 51° ⎟ = 36°. Therefore, the distance along the bottom of the n 1.333 ⎝ ⎠ b ⎝ ⎠

θ b = arcsin ⎜

pool from directly below where the light enters to where it hits the bottom is x = (4.0 m)tanθb = (4.0 m)tan36° = 2.9 m. xtotal = 1.5 m + x = 1.5 m + 2.9 m = 4.4 m. EVALUATE: The light ray from the flashlight is bent toward the normal when it refracts into the water.

33-12

Chapter 33

33.47.

IDENTIFY: Use Snell’s law to determine the effect of the liquid on the direction of travel of the light as it enters the liquid. SET UP: Use geometry to find the angles of incidence and refraction. Before the liquid is poured in the ray along your line of sight has the path shown in Figure 33.47a.

8.0 cm = 0.500 16.0 cm θ a = 26.57°

tan θ a =

Figure 33.47a After the liquid is poured in, θ a is the same and the refracted ray passes through the center of the bottom of the glass, as shown in Figure 33.47b.

4.0 cm = 0.250 16.0 cm θ b = 14.04°

tanθ b =

Figure 33.47b

33.48.

EXECUTE: Use Snell’s law to find nb , the refractive index of the liquid: na sinθ a = nb sinθ b n sin θ a (1.00)(sin 26.57°) nb = a = = 1.84 sin θ b sin14.04° EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal. IDENTIFY: Apply Snell’s law to each refraction and apply the law of reflection at the mirrored bottom. SET UP: The path of the ray is sketched in Figure 33.48. The problem asks us to calculate θ b′ . EXECUTE: Apply Snell's law to the air → liquid refraction. (1.00)sin(42.5°) = (1.63)sin θ b and θ b = 24.5° .

θ b = φ and φ = θ a′ , so θ a′ = θ b = 24.5°. Snell's law applied to the liquid → air refraction gives (1.63)sin(24.5°) = (1.00)sin θ b′and θ b′ = 42.5° .


EVALUATE:

33.49.

33-13

The light emerges from the liquid at the same angle from the normal as it entered the liquid.

Figure 33.48 IDENTIFY: Apply Snell’s law to the water → ice and ice → air interfaces. (a) SET UP: Consider the ray shown in Figure 33.49.

We want to find the incident angle θ a at the water-ice interface that causes the incident angle at the ice-air interface to be the critical angle. Figure 33.49 EXECUTE:

ice-air interface: nice sin θ crit = 1.0sin 90°

nice sin θ crit = 1.0 so sin θ crit =

1 nice

But from the diagram we see that θ b = θ crit , so sin θ b =

1 . nice

water-ice interface: nw sin θ a = nice sin θ b 1 1 1 so nw sin θ a = 1.0. sin θ a = = = 0.7502 and θ a = 48.6°. nice nw 1.333 (b) EVALUATE: The angle calculated in part (a) is the critical angle for a water-air interface; the answer would be the same if the ice layer wasn’t there! IDENTIFY: The incident angle at the prism → water interface is to be the critical angle. SET UP: The path of the ray is sketched in Figure 33.50. The ray enters the prism at normal incidence so is not bent. For water, nwater = 1.333 .

But sin θ b =

33.50.

EXECUTE: nglass =

From the figure, θ crit = 45° . na sinθ a = nb sinθ b gives nglass sin 45° = (1.333)sin 90° .

1.333 = 1.89 . sin 45°

33-14

Chapter 33

EVALUATE: For total internal reflection the ray must be incident in the material of greater refractive index. nglass > nwater , so that is the case here.

33.51.

Figure 33.50 IDENTIFY: Apply Snell’s law to the refraction of each ray as it emerges from the glass. The angle of incidence equals the angle A = 25.0°. SET UP: The paths of the two rays are sketched in Figure 33.51.

EXECUTE: na sinθ a = nb sinθ b nglas sin 25.0° = 1.00sin θ b

Figure 33.51

sin θ b = nglass sin 25.0°

sin θ b = 1.66sin 25.0° = 0.7015 θ b = 44.55° β = 90.0° − θ b = 45.45° Then δ = 90.0° − A − β = 90.0° − 25.0° − 45.45° = 19.55°. The angle between the two rays is 2δ = 39.1°. 33.52.

EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle. SET UP: θ b = 90° . The incident angle for the ray in the figure is 60° . ⎛ n sin θ a ⎞ ⎛ 1.62sin 60° ⎞ na sinθ a = nb sinθ b gives nb = ⎜ a ⎟=⎜ ⎟ = 1.40. ⎝ sin θ b ⎠ ⎝ sin 90° ⎠ EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater refractive index. IDENTIFY: No light enters the gas because total internal reflection must have occurred at the water-gas interface. SET UP: At the minimum value of S, the light strikes the water-gas interface at the critical angle. We apply Snell’s law, na sinθa = nb sinθb, at that surface. S EXECUTE: (a) In the water, θ = = (1.09 m)/(1.10 m) = 0.991 rad = 56.77°. This is the critical angle. So, using R the refractive index for water from Table 33.1, we get n = (1.333) sin 56.77° = 1.12 (b) (i) The laser beam stays in the water all the time, so

EXECUTE:

33.53.

Dnwater ⎛ c ⎞ t = 2R/v = 2R/ ⎜ = = (2.20 m)(1.333)/(3.00 × 108 m/s) = 9.78 ns ⎟ c n ⎝ water ⎠


33.54.

33.55.

33-15

(ii) The beam is in the water half the time and in the gas the other half of the time. Rngas tgas = = (1.10 m)(1.12)/(3.00 × 108 m/s) = 4.09 ns c The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12. IDENTIFY: No light enters the water because total internal reflection must have occurred at the glass-water surface. SET UP: A little geometry tells us that θ is the angle of incidence at the glass-water face in the water. Also, θ = 59.2° must be the critical angle at that surface, so the angle of refraction is 90.0°. Snell’s law, na sin θa = nb sin θb, c applies at that glass-water surface, and the index of refraction is defined as n = . v c EXECUTE: Snell’s law at the glass-water surface gives n sin 59.2° = (1.333)(1.00), which gives n = 1.55. v = = n (3.00 × 108 m/s)/1.55 = 1.93 × 108 m/s. EVALUATE: Notice that θ is not the angle of incidence at the reflector, but it is the angle of incidence at the glasswater surface. (a) IDENTIFY: Apply Snell’s law to the refraction of the light as it enters the atmosphere. SET UP: The path of a ray from the sun is sketched in Figure 33.55.

δ = θ a − θb From the diagram sin θ b =

R R+h

⎛ R ⎞

θ b = arcsin ⎜ ⎟ ⎝ R+h⎠ Figure 33.55

33.56.

EXECUTE: Apply Snell’s law to the refraction that occurs at the top of the atmosphere: na sinθ a = nb sinθ b (a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n) ⎛ R ⎞ ⎛ nR ⎞ sin θ a = n sin θ b = n ⎜ ⎟ so θ a = arcsin ⎜ ⎟ + R h ⎝ ⎠ ⎝R+h⎠ ⎛ nR ⎞ ⎛ R ⎞ δ = θ a − θ b = arcsin ⎜ ⎟ − arcsin ⎜ ⎟ + R h ⎝ ⎠ ⎝ R+h⎠ R 6.38 × 106 m (b) = = 0.99688 R + h 6.38 × 106 m + 20 × 103 m nR = 1.0003(0.99688) = 0.99718 R+h ⎛ R ⎞ θ b = arcsin ⎜ ⎟ = 85.47° ⎝ R+h⎠ ⎛ nR ⎞ θ b = arcsin ⎜ ⎟ 85.70° ⎝ R+h⎠ δ = θ a − θ b = 85.70° − 85.47° = 0.23° EVALUATE: The calculated δ is about the same as the angular radius of the sun. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) The distance traveled by the light ray is the sum of the two diagonal segments: d ( x 2 + y12 )1/ 2 + ((l − x ) 2 + y22 )1/ 2 d = ( x 2 + y12 )1/ 2 + ((l − x)2 + y22 )1/ 2 . Then the time taken to travel that distance is t = = . c c (b) Taking the derivative with respect to x of the time and setting it to zero yields 1 2 dt −1 2 dt 1 d ⎡ 2 1 ⎤ = 0 . This gives = ( x + y12 )1 2 + (l − x)2 + y22 ⎤⎥ = ⎡⎢ x( x 2 + y12 )−1 2 − (l − x) (l − x)2 + y22 ⎦ dx c ⎣ ⎦⎥ dx c dt ⎣⎢ x (l − x) = , sinθ1 = sinθ 2 and θ1 = θ 2 . x 2 + y12 (l − x)2 + y22

(

)

(

)

EVALUATE: For any other path between points 1 and 2, that includes a point on the reflective surface, the distance traveled and therefore the travel time is greater than for this path.

33-16

Chapter 33

33.57.

IDENTIFY and SET UP: Find the distance that the ray travels in each medium. The travel time in each medium is the distance divided by the speed in that medium. (a) EXECUTE:

The light travels a distance

h12 + x 2 in traveling from point A to the interface. Along this path

the speed of the light is v1 , so the time it takes to travel this distance is t1 =

h22 + (l − x) 2 in traveling from the interface to point B. Along this path the speed of the light is v2 ,

distance

so the time it takes to travel this distance is t2 =

h22 + (l − x ) 2 . The total time to go from A to B is v2

h12 + x 2 h 2 + (l + x ) 2 + 2 . v1 v2

t = t1 + t2 =

(b)

h12 + x 2 . The light travels a v1

1 ⎛1⎞ dt 1 ⎛ 1 ⎞ 2 = ⎜ ⎟ (h1 + x 2 )−1/ 2 (2 x) + ⎜ ⎟ (h22 + (l − x)2 )−1/ 2 2(l − x)(−1) = 0 dx v1 ⎝ 2 ⎠ v2 ⎝ 2 ⎠ x

v1 h + x 2 1

2

=

l−x v2 h + (l − x)2 2 2

Multiplying both sides by c gives

c x c = v1 h12 + x 2 v2

l−x h + (l − x)2 2 2

c c = n1 and = n2 (Eq.33.1) v1 v2 From Fig.33.55 in the textbook, sin θ1 =

33.58.

33.59.

x h12 + x 2

and sin θ 2 =

l−x h22 + (l − x) 2

.

So n1 sin θ1 = n2 sin θ 2 , which is Snell’s law. EVALUATE: Snell’s law is a result of a change in speed when light goes from one material to another. IDENTIFY: Apply Snell's law to each refraction. SET UP: Refer to the angles and distances defined in the figure that accompanies the problem. EXECUTE: (a) For light in air incident on a parallel-faced plate, Snell’s Law yields: n sin θ a = n′ sin θ b′ = n′ sin θ b = n sin θ a′ ⇒ sin θ a = sin θ a′ ⇒ θ a = θ a′ . (b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out. The requirement of parallel faces ensures that the angle θ n′ = θ n and the chain of equations can continue. (c) The lateral displacement of the beam can be calculated using geometry: t t sin(θ a − θ b′ ) ⇒d = . d = L sin(θ a − θ b′ ) and L = cosθ b′ cosθ b′ (2.40 cm)sin(66.0° − 30.5°) ⎛ n sinθ a ⎞ ⎛ sin 66.0° ⎞ (d) θ b′ = arcsin ⎜ = 1.62 cm. ⎟ = arcsin ⎜ ⎟ = 30.5° and d = ′ cos30.5° ⎝ n ⎠ ⎝ 1.80 ⎠ EVALUATE: The lateral displacement in part (d) is large, of the same order as the thickness of the plate. IDENTIFY: Apply Snell’s law to each refraction and apply the law of reflection to each reflection. SET UP: The paths of rays A and B are sketched in Figure 33.59. Let θ be the angle of incidence for the combined ray. EXECUTE: For ray A its final direction of travel is at an angle θ with respect to the normal, by the law of reflection. Let the final direction of travel for ray B be at angle φ with respect to the normal. At the upper surface,

Snell’s law gives n1 sin θ = n2 sin α . The lower surface reflects ray B at angle α . Ray B returns to the upper surface of the film at an angle of incidence α . Snell’s law applied to the refraction as ray B leaves the film gives n2 sin α = n1 sin φ . Combining the two equations gives n1 sin θ = n1 sin φ and θ = φ ; the two rays are parallel after they emerge from the film.


EVALUATE: the film.

33-17

Ray B is bent toward the normal as it enters the film and away from the normal as it refracts out of

Figure 33.59 33.60.

IDENTIFY: Apply Snell's law and the results of Problem 33.58. SET UP: From Figure 33.58 in the textbook, nr = 1.61 for red light and nv = 1.66 for violet. In the notation of Problem 33.58, t is the thickness of the glass plate and the lateral displacement is d. We want the difference in d for (1.00)sin 70.0° the two colors of light to be 1.0 mm. θ a = 70.0° . For red light, na sinθ a = nb sinθ b′ gives sin θ b′ = 1.61 (1.00)sin 70.0° and θ b′ = 34.48° . and θ b′ = 35.71° . For violet light, sin θ b′ = 1.66 EXECUTE: (a) n decreases with increasing λ , so n is smaller for red than for blue. So beam a is the red one. sin(θ a − θ b′ ) sin(70° − 35.71°) (b) Problem 33.58 says d = t . For red light, d r = t = 0.6938t and for violet light, cosθ b′ cos35.71° sin(70° − 35.48°) 0.10 cm = 0.7048t . d v − d r = 1.0 mm gives t = = 9.1 cm . cos35.48° 0.7048 − 0.6958 EVALUATE: Our calculation shown that the violet light has greater lateral displacement and this is ray b. IDENTIFY: Apply Snell's law to the two refractions of the ray. SET UP: Refer to the figure that accompanies the problem. A + 2α A A A ⎛A ⎞ = n sin . EXECUTE: (a) na sinθ a = nb sinθ b gives sin θ a = nb sin . But θ a = + α , so sin ⎜ + α ⎟ = sin 2 2 2 2 ⎝2 ⎠ dv = t

33.61.

A+δ A = n sin . 2 2 A⎞ 60.0° ⎞ ⎛ ⎛ (b) From part (a), δ = 2arcsin ⎜ n sin ⎟ − A . δ = 2arcsin ⎜ (1.52)sin ⎟ − 60.0° = 38.9°. 2⎠ 2 ⎠ ⎝ ⎝ (c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ: At each face of the prism the deviation is α , so 2α = δ and sin

⎛ ⎝

δ red = 2arcsin ⎜ (1.61)sin ⎛ ⎝

60.0° ⎞ ⎟ − 60.0° = 47.2° 2 ⎠

δ violet = 2arcsin ⎜ (1.66)sin

33.62.

60.0° ⎞ ⎟ − 60.0° = 52.2° ⇒ Δδ = 52.2° − 47.2° = 5.0° 2 ⎠

EVALUATE: The violet light has a greater refractive index and therefore the angle of deviation is greater for the violet light. IDENTIFY: The reflected light is totally polarized when light strikes a surface at Brewster’s angle. SET UP: At the plastic wall, Brewster’s angle obeys the equation tan θp = nb /na, and Snell’s law, na sinθa = nb sinθb, applies at the air-water surface. EXECUTE: To be totally polarized, the reflected sunlight must have struck the wall at Brewster’s angle. tan θp = nb /na = (1.61)/(1.00) and θp = 58.15° This is the angle of incidence at the wall. A little geometry tells us that the angle of incidence at the water surface is 90.00° – 58.15° = 31.85°. Applying Snell’s law at the water surface gives (1.00) sin31.85° = 1.333 sin θ and θ = 23.3° EVALUATE: We have two different principles involved here: Reflection at Brewster’s angle at the wall and Snell’s law at the water surface.

33-18

Chapter 33

33.63.


The polarizer passes

1 2

of the intensity of the unpolarized component, independent of φ .

Out of the intensity I p of the polarized component the polarizer passes intensity I p cos 2 (φ − θ ), where φ − θ is the angle between the plane of polarization and the axis of the polarizer. (a) Use the angle where the transmitted intensity is maximum or minimum to find θ . See Figure 33.63.

Figure 33.63

The total transmitted intensity is I = 12 I 0 + I p cos 2 (φ − θ ). This is maximum when θ = φ and from the

EXECUTE:

table of data this occurs for φ between 30° and 40°, say at 35° and θ = 35°. Alternatively, the total transmitted intensity is minimum when φ − θ = 90° and from the data this occurs for φ = 125°. Thus, θ = φ − 90° = 125° − 90° = 35°, in agreement with the above. (b) IDENTIFY and SET UP:

I = 12 I 0 + I p cos 2 (φ − θ )

Use data at two values of φ to determine the two constants I 0 and I p . Use data where the I p term is large (φ = 30°) and where it is small (φ = 130°) to have the greatest sensitivity to both I 0 and I p :

φ = 30° gives 24.8 W/m 2 = 12 I 0 + I p cos 2 (30° − 35°)

EXECUTE:

24.8 W/m 2 = 0.500 I 0 + 0.9924 I p

φ = 130° gives 5.2 W/m 2 = 12 I 0 + I p cos 2 (130° − 35°) 5.2 W/m 2 = 0.500 I 0 + 0.0076 I p Subtracting the second equation from the first gives 19.6 W/m 2 = 0.9848I p and I p = 19.9 W/m 2 . And then I 0 = 2(5.2 W/m 2 − 0.0076(19.9 W/m 2 )) = 10.1 W/m 2 .

Now that we have I 0 , I p and θ we can verify that I = 12 I 0 + I p cos 2 (φ − θ ) describes that data in the

EVALUATE: 33.64.

table. IDENTIFY: The number of wavelengths in a distance D of material is D / λ , where λ is the wavelength of the light in the material. D D 1 = + , where we have assumed n1 > n2 so λ2 > λ1 . SET UP: The condition for a quarter-wave plate is λ1 λ2 4 EXECUTE:

λ0

n1D

λ0

=

n2 D

λ0

+

1 λ0 and D = . 4 4(n1 − n2 )

5.89 × 10−7 m = 6.14 × 10−7 m. 4(n1 − n2 ) 4(1.875 − 1.635) EVALUATE: The thickness of the quarter-wave plate in part (b) is 614 nm, which is of the same order as the wavelength in vacuum of the light. IDENTIFY: Follow the steps specified in the problem. SET UP: cos(α − β ) = sin α sin β + cosα cos β . sin(α − β ) = sin α cos β − cosα sin β . EXECUTE: (a) Multiplying Eq.(1) by sin β and Eq.(2) by sin α yields: (b) D =

33.65.

(a)

=

x y sin β = sin ω t cosα sin β − cos ω t sin α sin β and (2): sinα = sin ω t cos β sin α − cos ω t sin β sin α . a a x sin β − y sin α Subtracting yields: = sin ω t (cosα sin β − cos β sin α ). a

(1):


33-19

(b) Multiplying Eq. (1) by cos β and Eq. (2) by cos α yields:

x y (1) : cos β = sin ω t cosα cos β − cos ω t sin α cos β and (2) : cosα = sin ω t cos β cosα − cos ω t sin β cosα . a a x cos β − y cosα Subtracting yields: = − cos ω t (sin α cos β − sin β cosα ). a (c) Squaring and adding the results of parts (a) and (b) yields: ( x sin β − y sin α ) 2 + ( x cos β − y cosα ) 2 = a 2 (sin α cos β − sin β cosα ) 2 (d) Expanding the left-hand side, we have:

x 2 (sin 2 β + cos 2 β ) + y 2 (sin 2 α + cos 2 α ) − 2 xy (sin α sin β + cosα cos β )

= x 2 + y 2 − 2 xy (sin α sin β + cosα cos β ) = x 2 + y 2 − 2 xy cos(α − β ). The right-hand side can be rewritten: a 2 (sin α cos β − sin β cosα ) 2 = a 2 sin 2 (α − β ). Therefore, x 2 + y 2 − 2 xy cos(α − β ) = a 2 sin 2 (α − β ). Or, x 2 + y 2 − 2 xy cos δ = a 2 sin 2 δ , where δ = α − β .

EVALUATE:

δ= δ= 33.66.

π 4 π 2

(e) δ = 0 : x 2 + y 2 − 2 xy = ( x − y ) 2 = 0 ⇒ x = y , which is a straight diagonal line

: x 2 + y 2 − 2 xy =

a2 , which is an ellipse 2

: x 2 + y 2 = a 2 ,which is a circle. This pattern repeats for the remaining phase differences.

IDENTIFY: Apply Snell's law to each refraction. SET UP: Refer to the figure that accompanies the problem. EXECUTE: (a) By the symmetry of the triangles, θ bA = θ aB , and θ aC = θ rB = θ aB = θ bA . Therefore,

sinθ bC = n sinθ aC = n sinθ bA = sinθ aA = θ bC = θ aA . (b) The total angular deflection of the ray is Δ = θ aA − θ bA + π − 2θ aB + θ bC − θ aC = 2θ aA − 4θ bA + π .

⎛1 ⎞ (c) From Snell’s Law, sin θ aA = n sin θ bA ⇒ θ bA = arcsin ⎜ sinθ aA ⎟ . n ⎝ ⎠ ⎛1 ⎞ Δ = 2θ aA − 4θ bA + π = 2θ aA − 4arcsin ⎜ sin θ aA ⎟ + π . ⎝n ⎠

(d)

dΔ d ⎛ ⎛1 ⎞⎞ = 0 = 2 − 4 A ⎜ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ 0 = 2 − dθ aA dθ a ⎝ n ⎝ ⎠⎠

2 2 ⎛ cosθ1 ⎞ ⎛ sin θ1 ⎞ ⎛ 16cos θ1 ⎞ ⋅⎜ =⎜ . 4 ⎜1 − ⎟ ⎟. ⎟ 2 2 n ⎠ ⎝ n sin 2 θ1 ⎝ n ⎠ ⎝ ⎠ 1− n2

4

1 4cos 2 θ1 = n 2 − 1 + cos 2 θ1 . 3cos 2 θ1 = n 2 − 1 . cos 2 θ1 = (n 2 − 1). 3 ⎛ 1 2 ⎞ ⎛ 1 ⎞ (e) For violet: θ1 = arccos ⎜⎜ ( n − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 58.89° . ⎝ 3 ⎠ ⎝ 3 ⎠ Δ violet = 139.2° ⇒ θ violet = 40.8°.

33.67.

⎛ 1 ⎞ ⎛ 1 ⎞ For red: θ1 = arccos ⎜⎜ ( n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 59.58° . Δ red = 137.5° ⇒ θ red = 42.5°. ⎝ 3 ⎠ ⎝ 3 ⎠ EVALUATE: The angles we have calculated agree with the values given in Figure 37.20d in the textbook. θ1 is larger for red than for violet, so red in the rainbow is higher above the horizon. IDENTIFY: Follow similar steps to Challenge Problem 33.66. SET UP: Refer to Figure 33.20e in the textbook. EXECUTE: The total angular deflection of the ray is Δ = θ aA − θ bA + π − 2θ bA + π − 2θ bA + θ aA − θ bA = 2θ aA − 6θ bA + 2π , where we have used the fact from the previous problem that all the internal angles are equal and the two external equals are equal. Also using the Snell’s Law ⎛1 ⎞ ⎛1 ⎞ relationship, we have: θ bA = arcsin ⎜ sinθ aA ⎟ . Δ = 2θ aA − 6θ bA + 2π = 2θ aA − 6arcsin ⎜ sin θ aA ⎟ + 2π . ⎝n ⎠ ⎝n ⎠

33-20

Chapter 33

(b)

dΔ d ⎛ ⎛1 ⎞⎞ = 0 = 2 − 6 A ⎜ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ 0 = 2 − dθ aA dθ a ⎝ ⎝n ⎠⎠

6 sin θ 1− 2 2 n

⎛ cosθ 2 ⎞ .⎜ ⎟. ⎝ n ⎠

⎛ sin 2 θ 2 ⎞ 1 2 2 2 2 2 n 2 ⎜1 − ⎟ = (n − 1 + cos θ 2 ) = 9cos θ 2 . cos θ 2 = (n − 1) . 2 8 n ⎠ ⎝ ⎛ 1 ⎞ ⎛ 1 ⎞ (c) For violet, θ 2 = arccos ⎜⎜ ( n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 71.55° . Δ violet = 233.2° and θ violet = 53.2°. ⎝ 8 ⎠ ⎝ 8 ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ For red, θ 2 = arccos ⎜⎜ (n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 71.94°. Δ red = 230.1° and θ red = 50.1°. ⎝ 8 ⎠ ⎝ 8 ⎠ EVALUATE: The angles we calculated agree with those given in Figure 37.20e in the textbook. The color that appears higher above the horizon is violet. The colors appear in reverse order in a secondary rainbow compared to a primary rainbow.

34

GEOMETRIC OPTICS

34.1.

IDENTIFY and SET UP: Plane mirror: s = − s′ (Eq.34.1) and m = y′ / y = − s′ / s = +1 (Eq.34.2). We are given s and y and are asked to find s′ and y′. EXECUTE: The object and image are shown in Figure 34.1. s′ = − s = −39.2 cm y′ = m y = ( +1)(4.85 cm)

y′ = 4.85 cm Figure 34.1

34.2.

The image is 39.2 cm to the right of the mirror and is 4.85 cm tall. EVALUATE: For a plane mirror the image is always the same distance behind the mirror as the object is in front of the mirror. The image always has the same height as the object. h d IDENTIFY: Similar triangles say tree = tree . hmirror d mirror SET UP:

d tree 28.0 m + 0.350 m = 0.040 m = 3.24 m. d mirror 0.350 m EVALUATE: The image of the tree formed by the mirror is 28.0 m behind the mirror and is 3.24 m tall. IDENTIFY: Apply the law of reflection. SET UP: If up is the +y-direction and right is the +x-direction, then the object is at (− x0 , − y0 ) and P2′ is at EXECUTE:

34.3.

d mirror = 0.350 m, hmirror = 0.0400 m and d tree = 28.0 m + 0.350 m. htree = hmirror

( x0 , − y0 ) . EXECUTE: 34.4.

34.5.

Mirror 1 flips the y-values, so the image is at ( x0 , y0 ) which is P3′.

EVALUATE: Mirror 2 uses P1′ as an object and forms an image at P3′ . IDENTIFY: f = R / 2 SET UP: For a concave mirror R > 0. R 34.0 cm EXECUTE: (a) f = = = 17.0 cm 2 2 EVALUATE: (b) The image formation by the mirror is determined by the law of reflection and that is unaffected by the medium in which the light is traveling. The focal length remains 17.0 cm. IDENTIFY and SET UP: Use Eq.(34.6) to calculate s′ and use Eq.(34.7) to calculate y′. The image is real if s′ is positive and is erect if m > 0. Concave means R and f are positive, R = +22.0 cm; f = R / 2 = +11.0 cm. EXECUTE: (a)

Three principal rays, numbered as in Sect. 34.2, are shown in Figure 34.5. The principal ray diagram shows that the image is real, inverted, and enlarged. Figure 34.5

34-1

34-2

Chapter 34

1 1 1 + = s s′ f 1 1 1 s− f sf (16.5 cm)(11.0 cm) = − = so s′ = = = +33.0 cm s′ f s sf s − f 16.5 cm − 11.0 cm s′ > 0 so real image, 33.0 cm to left of mirror vertex s′ 33.0 cm = −2.00 (m < 0 means inverted image) y′ = m y = 2.00(0.600 cm) = 1.20 cm m=− =− s 16.5 cm EVALUATE: The image is 33.0 cm to the left of the mirror vertex. It is real, inverted, and is 1.20 cm tall (enlarged). The calculation agrees with the image characterization from the principal ray diagram. A concave mirror used alone always forms a real, inverted image if s > f and the image is enlarged if f < s < 2f. s′ 1 1 1 IDENTIFY: Apply + = and m = − . s s′ f s R SET UP: For a convex mirror, R < 0 . R = −22.0 cm and f = = −11.0 cm . 2 EXECUTE: (a) The principal-ray diagram is sketched in Figure 34.6. s′ −6.6 cm 1 1 1 sf (16.5 cm)(−11.0 cm) = +0.400 . = = −6.6 cm . m = − = − (b) + = . s′ = s − f 16.5 cm − (−11.0 cm) s s′ f s 16.5 cm y′ = m y = (0.400)(0.600 cm) = 0.240 cm . The image is 6.6 cm to the right of the mirror. It is 0.240 cm tall. (b)

34.6.

s′ < 0 , so the image is virtual. m > 0 , so the image is erect. EVALUATE: The calculated image properties agree with the image characterization from the principal-ray diagram.

Figure 34.6 34.7.

y′ 1 1 1 s′ . Find m and calculate y′ . + = . m=− . m = s s′ f s y SET UP: f = +1.75 m . EXECUTE: s ! f so s′ = f = 1.75 m .

IDENTIFY:

1.75 m s′ =− = −3.14 × 10−11. y′ = m y = (3.14 × 10−11 )(6.794 × 106 m) = 2.13 × 10−4 m = 0.213 mm . 5.58 × 1010 m s EVALUATE: The image is real and is 1.75 m in front of the mirror. 1 1 1 s′ IDENTIFY: Apply + = and m = − . s s′ f s SET UP: The mirror surface is convex so R = −3.00 cm . s = 24.0 cm − 3.00 cm = 21.0 cm . R 1 1 1 sf (21.0 cm)( −1.50 cm) EXECUTE: f = = −1.50 cm . + = . s′ = = = −1.40 cm . The image is 2 s − f 21.0 cm − (−1.50 cm) s s′ f 1.40 cm behind the surface so it is 3.00 cm − 1.40 cm = 1.60 cm from the center of the ornament, on the same side s′ −1.40 cm as the object. m = − = − = +0.0667 . y′ = m y = (0.0667)(3.80 mm) = 0.253 mm . s 21.0 cm EVALUATE: The image is virtual, upright and smaller than the object. IDENTIFY: The shell behaves as a spherical mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . s s′ f s m=−

34.8.

34.9.

Geometric Optics

EXECUTE:

1 1 1 1 2 1 + = ⇒ = − ⇒ s = 18.0 cm from the vertex. s s′ f s −18.0 cm −6.00 cm

s′ −6.00 cm 1 1 =− = ⇒ y′ = (1.5 cm) = 0.50 cm . The image is 0.50 cm tall, erect, and virtual. s 18.0 cm 3 3 EVALUATE: Since the magnification is less than one, the image is smaller than the object. IDENTIFY: The bottom surface of the bowl behaves as a spherical convex mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . ′ s s s f m=−

34.10.

EXECUTE:

1 1 1 1 −2 1 + = ⇒ = − ⇒ s′ = −15 cm behind bowl. ′ ′ s s f s 35 cm 90 cm

s′ 15 cm = = 0.167 ⇒ y′ = (0.167)(2.0 cm) = 0.33 cm . The image is 0.33 cm tall, erect, and virtual. s 90 cm EVALUATE: Since the magnification is less than one, the image is smaller than the object. IDENTIFY: We are dealing with a spherical mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . s s′ f s 1 1 1 1 1 1 s− f sf s′ f EXECUTE: (a) + = ⇒ = − = ⇒ s′ = . Also m = − = . s s′ f s′ f s fs s− f s f −s (b) The graph is given in Figure 34.11a. (c) s′ > 0 for s > f , s < 0. (d) s′ < 0 for 0 < s < f . (e) The image is at negative infinity, “behind” the mirror. (f ) At the focal point, s = f. (g) The image is at the mirror, s′ = 0 . (h) The graph is given in Figure 34.11b. (i) Erect and larger if 0 < s < f. (j) Inverted if s > f . (k) The image is smaller if s > 2 f or s < 0. (l) As the object is moved closer and closer to the focal point, the magnification increases to infinite values. EVALUATE: As the object crosses the focal point, both the image distance and the magnification undergo discontinuities. m=−

34.11.

Figure 34.11

34-3

34-4

34.12.

Chapter 34

sf f and m = . s− f f −s s f f SET UP: With f = − f , s′ = − and m = . s+ f s+ f EXECUTE: The graphs are given in Figure 34.12. (a) s′ > 0 for − f < s < 0. (b) s′ < 0 for s < − f and s < 0. IDENTIFY:

s′ =

(c) If the object is at infinity, the image is at the outward going focal point. (d) If the object is next to the mirror, then the image is also at the mirror (e) The image is erect (magnification greater than zero) for s > − f . (f) The image is inverted (magnification less than zero) for s < − f . (g) The image is larger than the object (magnification greater than one) for −2 f < s < 0. (h) The image is smaller than the object (magnification less than one) for s > 0 and s < − 2 f . EVALUATE: object.

For a real image ( s > 0) , the image formed by a convex mirror is always virtual and smaller than the

Figure 34.12 34.13.

1 1 1 y′ s′ + = and m = = − . ′ y s s s f SET UP: m = +2.00 and s = 1.25 cm . An erect image must be virtual. sf f EXECUTE: (a) s′ = and m = − . For a concave mirror, m can be larger than 1.00. For a convex mirror, s− f s− f f and m is always less than 1.00. The mirror must be concave ( f > 0) . f = − f so m = + s+ f 1 s′ + s ss′ s′ s (−2.00s ) = (b) . f = . m = − = +2.00 and s′ = −2.00s . f = = +2.00s = +2.50 cm . s s − 2.00s f ss′ s + s′ R = 2 f = +5.00 cm . (c) The principal ray diagram is drawn in Figure 34.13. IDENTIFY:

Geometric Optics

EVALUATE:

34-5

The principal-ray diagram agrees with the description from the equations.

Figure 34.13 34.14.

IDENTIFY: SET UP:

Apply

1 1 1 s′ + = and m = − . s s′ f s

For a concave mirror, R > 0 . R = 32.0 cm and f =

R = 16.0 cm . 2

1 1 1 sf (12.0 cm)(16.0 cm) s′ −48.0 cm = = −48.0 cm . m = − = − = +4.00 . + = . s′ = s 12.0 cm s − f 12.0 cm − 16.0 cm s s′ f (b) s′ = −48.0 cm , so the image is 48.0 cm to the right of the mirror. s′ < 0 so the image is virtual. (c) The principal-ray diagram is sketched in Figure 34.14. The rules for principal rays apply only to paraxial rays. Principal ray 2, that travels to the mirror along a line that passes through the focus, makes a large angle with the optic axis and is not described well by the paraxial approximation. Therefore, principal ray 2 is not included in the sketch. EVALUATE: A concave mirror forms a virtual image whenever s < f . EXECUTE:

(a)

Figure 34.14 34.15.

IDENTIFY:

Apply Eq.(34.11), with R → ∞. s′ is the apparent depth.

SET UP The image and object are shown in Figure 34.15.

na nb nb − na + = ; s s′ R R → ∞ (flat surface), so na nb + =0 s s′ Figure 34.15

nb s (1.00)(3.50 cm) =− = −2.67 cm na 1.309 The apparent depth is 2.67 cm. EVALUATE: When the light goes from ice to air (larger to smaller n), it is bent away from the normal and the virtual image is closer to the surface than the object is. EXECUTE:

s′ = −

34-6

34.16.

Chapter 34

na nb + =0. s s′ SET UP: The light travels from the fish to the eye, so na = 1.333 and nb = 1.00 . When the fish is viewed, s = 7.0 cm . The fish is 20.0 cm − 7.0 cm = 13.0 cm above the mirror, so the image of the fish is 13.0 cm below the mirror and 20.0 cm + 13.0 cm = 33.0 cm below the surface of the water. When the image is viewed, s = 33.0 cm . IDENTIFY:

The surface is flat so R → ∞ and

⎛n ⎞ ⎛ 1.00 ⎞ (a) s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (7.0 cm) = −5.25 cm . The apparent depth is 5.25 cm. ⎝ 1.333 ⎠ ⎝ na ⎠ ⎛n ⎞ ⎛ 1.00 ⎞ (b) s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (33.0 cm) = −24.8 cm . The apparent depth of the image of the fish in the mirror is 24.8 cm. n ⎝ 1.333 ⎠ ⎝ a⎠ EVALUATE: In each case the apparent depth is less than the actual depth of what is being viewed. n s′ na nb nb − na + = . m = − a . Light comes from the fish to the person’s eye. IDENTIFY: nb s s s′ R EXECUTE:

34.17.

SET UP: R = −14.0 cm . s = +14.0 cm . na = 1.333 (water). nb = 1.00 (air). Figure 34.17 shows the object and the refracting surface. 1.333 1.00 1.00 − 1.333 (1.333)( −14.0 cm) = +1.33 . + = EXECUTE: (a) . s′ = −14.0 cm . m = − 14.0 cm s′ −14.0 cm (1.00)(14.0 cm) The fish’s image is 14.0 cm to the left of the bowl surface so is at the center of the bowl and the magnification is 1.33. n n − na (b) The focal point is at the image location when s → ∞ . b = b . na = 1.00 . nb = 1.333 . R = +14.0 cm . s′ R 1.333 1.333 − 1.00 = . s′ = +56.0 cm . s′ is greater than the diameter of the bowl, so the surface facing the sunlight s′ 14.0 cm does not focus the sunlight to a point inside the bowl. The focal point is outside the bowl and there is no danger to the fish. EVALUATE: In part (b) the rays refract when they exit the bowl back into the air so the image we calculated is not the final image.

34.18.

Figure 34.17 na nb nb − na IDENTIFY: Apply + = . s s′ R SET UP: For a convex surface, R > 0 . R = +3.00 cm . na = 1.00 , nb = 1.60 . EXECUTE:

34.19.

(a) s → ∞ .

⎛ nb ⎞ nb nb − na ⎛ 1.60 ⎞ = . s′ = ⎜ ⎟R = ⎜ ⎟ ( +3.00 cm) = +8.00 cm . The image is s′ R n − n ⎝ 1.60 − 1.00 ⎠ a ⎠ ⎝ b

8.00 cm to the right of the vertex. 1.00 1.60 1.60 − 1.00 + = (b) s = 12.0 cm . . s′ = +13.7 cm . The image is 13.7 cm to the right of the vertex. 12.0 cm s′ 3.00 cm 1.00 1.60 1.60 − 1.00 (c) s = 2.00 cm . + = . s′ = −5.33 cm . The image is 5.33 cm to the left of the vertex. 2.00 cm s′ 3.00 cm EVALUATE: The image can be either real ( s′ > 0 ) or virtual ( s′ < 0 ), depending on the distance of the object from the refracting surface. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on the curvature of the surface and the indices of refraction of the glass and oil. SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the n n n − na equation a + b = b . s s′ R

Geometric Optics

34-7

na nb nb − na 1.45 1.60 0.15 + = ⇒ + = ⇒ s = 0.395 cm s s′ R s 1.20 m 0.0300 m EVALUATE: The presence of the oil changes the location of the image. na nb nb − na n s′ + = IDENTIFY: . m=− a . ′ s s R nb s EXECUTE:

34.20.

SET UP: EXECUTE:

R = +4.00 cm . na = 1.00 . nb = 1.60 . s = 24.0 cm . (1.00)(14.8 cm) 1 1.60 1.60 − 1.00 = −0.385 . + = . s′ = +14.8 cm . m = − (1.60)(24.0 cm) 24.0 cm s′ 4.00 cm

y′ = m y = (0.385)(1.50 mm) = 0.578 mm . The image is 14.8 cm to the right of the vertex and is 0.578 mm tall.

34.21.

m < 0 , so the image is inverted. EVALUATE: The image is real. IDENTIFY: Apply Eqs.(34.11) and (34.12). Calculate s′ and y′. The image is erect if m > 0. SET UP: The object and refracting surface are shown in Figure 34.21.

Figure 34.21

na nb nb − na + = s s′ R 1.00 1.60 1.60 − 1.00 + = 24.0 cm s′ −4.00 cm

EXECUTE:

Multiplying by 24.0 cm gives 1.00 +

38.4 = −3.60 s′

38.4 cm 38.4 cm = −4.60 and s′ = − = −8.35 cm s′ 4.60 n s′ (1.00)( −8.35 cm) = +0.217 Eq.(34.12): m = − a = − nb s (1.60)( +24.0 cm) y′ = m y = (0.217)(1.50 mm) = 0.326 mm

34.22.

34.23.

EVALUATE: The image is virtual ( s′ < 0) and is 8.35 cm to the left of the vertex. The image is erect (m > 0) and is 0.326 mm tall. R is negative since the center of curvature of the surface is on the incoming side. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on the curvature of the surface and the indices of refraction of the glass and liquid. SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the n n n − na equation a + b = b . s s′ R na nb nb − na na 1.60 1.60 − na + = ⇒ + = ⇒ na = 1.24 . EXECUTE: ′ s s R 14.0 cm 9.00 cm 4.00 cm EVALUATE: The result is a reasonable refractive index for liquids. ⎛1 1 1 1 y′ s′ 1 1 ⎞ and m = = − . = ( n − 1) ⎜ − ⎟ to calculate f. The apply + = IDENTIFY: Use y s f s s′ f ⎝ R1 R2 ⎠ SET UP:

R1 → ∞ . R2 = −13.0 cm . If the lens is reversed, R1 = +13.0 cm and R2 → ∞ .

1 1 1 s− f 1 1 0.70 ⎛1 ⎞ = − = . = (0.70) ⎜ − and f = 18.6 cm . ⎟= f s′ f s sf ⎝ ∞ −13.0 cm ⎠ 13.0 cm sf (22.5 cm)(18.6 cm) s′ 107 cm s′ = = = 107 cm . m = − = − = −4.76 . s− f 22.5 cm − 18.6 cm s 22.5 cm y′ = my = ( −4.76)(3.75 mm) = −17.8 mm . The image is 107 cm to the right of the lens and is 17.8 mm tall. The image is real and inverted. 1 1 1⎞ ⎛ = ( n − 1) ⎜ − ⎟ and f = 18.6 cm . The image is the same as in part (a). (b) f ⎝ 13.0 cm ∞ ⎠ EVALUATE: Reversing a lens does not change the focal length of the lens. EXECUTE:

(a)

34-8

34.24.

Chapter 34

1 1 1 + = . The sign of f determines whether the lens is converging or diverging. s s′ f SET UP: s = 16.0 cm . s′ = −12.0 cm . ss′ (16.0 cm)(−12.0 cm) = = −48.0 cm . f < 0 and the lens is diverging. EXECUTE: (a) f = s + s′ 16.0 cm + ( −12.0 cm) s′ −12.0 cm (b) m = − = − = +0.750 . y′ = m y = (0.750)(8.50 mm) = 6.38 mm . m > 0 and the image is erect. s 16.0 cm (c) The principal-ray diagram is sketched in Figure 34.24. EVALUATE: A diverging lens always forms an image that is virtual, erect and reduced in size. IDENTIFY:

Figure 34.24 34.25.


The liquid behaves like a lens, so the lensmaker’s equation applies. ⎛1 1 1 1 ⎞ s′ The lensmaker’s equation is + = (n − 1) ⎜ − ⎟ , and the magnification of the lens is m = − . s s′ R R s 2 ⎠ ⎝ 1 (a)

⎛1 ⎛ ⎞ 1 1 1 ⎞ 1 1 1 1 + = ( n − 1) ⎜ − ⎟ ⇒ + = (1.52 − 1) ⎜ − ⎟ ′ s s′ R R 24.0 cm s − 7.00 cm − 4.00 cm ⎝ ⎠ 2 ⎠ ⎝ 1 ⇒ s′ = 71.2 cm , to the right of the lens.

s′ 71.2 cm =− = −2.97 s 24.0 cm EVALUATE: Since the magnification is negative, the image is inverted. y′ s′ 1 1 1 IDENTIFY: Apply m = = − to relate s′ and s and then use + = . s s′ f y s SET UP: Since the image is inverted, y′ < 0 and m < 0 . (b) m = −

34.26.

EXECUTE:

34.27.

34.28.

m=

s′ 1 1 1 y′ −4.50 cm gives = = −1.406 . m = − gives s′ = +1.406 s . + = s s s′ f 3.20 cm y

1 1 1 + = and s = 154 cm . s′ = (1.406)(154 cm) = 217 cm . The object is 154 cm to the left of the s 1.406 s 90.0 cm lens. The image is 217 cm to the right of the lens and is real. EVALUATE: For a single lens an inverted image is always real. IDENTIFY: The thin-lens equation applies in this case. 1 1 1 s′ y ′ SET UP: The thin-lens equation is + = , and the magnification is m = − = . s s′ f s y y′ 34.0 mm s′ −12.0 cm EXECUTE: m = = = 4.25 = − = − ⇒ s = 2.82 cm . The thin-lens equation gives y 8.00 mm s s 1 1 1 + = ⇒ f = 3.69 cm . s s′ f EVALUATE: Since the focal length is positive, this is a converging lens. The image distance is negative because the object is inside the focal point of the lens. s′ 1 1 1 IDENTIFY: Apply m = − to relate s and s′ . Then use + = . s s s′ f SET UP: Since the image is to the right of the lens, s′ > 0 . s′ + s = 6.00 m . EXECUTE: (a) s′ = 80.0 s and s + s′ = 6.00 m gives 81.00s = 6.00 m and s = 0.0741 m . s′ = 5.93 m .

Geometric Optics

34-9

(b) The image is inverted since both the image and object are real ( s′ > 0, s > 0).

1 1 1 1 1 = + = + ⇒ f = 0.0732 m, and the lens is converging. f s s′ 0.0741 m 5.93 m EVALUATE: The object is close to the lens and the image is much farther from the lens. This is typical for slide projectors. ⎛1 1 1 ⎞ IDENTIFY: Apply = ( n − 1) ⎜ − ⎟ . f R R 2 ⎠ ⎝ 1 (c)

34.29.

SET UP:

For a distant object the image is at the focal point of the lens. Therefore, f = 1.87 cm . For the double-

convex lens, R1 = + R and R2 = − R , where R = 2.50 cm . EXECUTE: EVALUATE: 34.30.

1 1 ⎞ 2( n − 1) R 2.50 cm ⎛1 +1 = + 1 = 1.67 . = ( n − 1) ⎜ − . n= ⎟= f R 2f 2(1.87 cm) ⎝ R −R ⎠ f > 0 and the lens is converging. A double-convex lens is always converging.

⎛1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ f R R 2 ⎠ ⎝ 1 EXECUTE: We have a converging lens if the focal length is positive, which requires ⎛1 ⎛1 1 ⎞ 1 1 ⎞ = ( n − 1) ⎜ − ⎟ > 0 ⇒ ⎜ − ⎟ > 0. This can occur in one of three ways: f ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ IDENTIFY and SET UP:

Apply

(i) R1 and R2 both positive and R1 < R2 . (ii) R1 ≥ 0, R2 ≤ 0 (double convex and planoconvex). (iii) R1 and R2 both negative and R1 > R2 (meniscus). The three lenses in Figure 35.32a in the textbook fall into these categories. We have a diverging lens if the focal length is negative, which requires ⎛1 ⎛1 1 1 ⎞ 1 ⎞ = ( n − 1) ⎜ − ⎟ < 0 ⇒ ⎜ − ⎟ < 0. This can occur in one of three ways: f R R R R 2 ⎠ 2 ⎠ ⎝ 1 ⎝ 1 (i) R1 and R2 both positive and R1 > R2 (meniscus). (ii) R1 and R2 both negative and R2 > R1 . (iii) R1 ≤ 0, R2 ≥ 0

34.31.

34.32.

(planoconcave and double concave). The three lenses in Figure 34.32b in the textbook fall into these categories. EVALUATE: The converging lenses in Figure 34.32a are all thicker at the center than at the edges. The diverging lenses in Figure 34.32b are all thinner at the center than at the edges. 1 1 1 s′ IDENTIFY and SET UP: The equations + = and m = − apply to both thin lenses and spherical mirrors. s s′ f s EXECUTE: (a) The derivation of the equations in Exercise 34.11 is identical and one gets: 1 1 1 1 1 1 s− f sf s′ f + = ⇒ = − = ⇒ s′ = , and also m = − = . s s′ f s′ f s fs s− f s f −s (b) Again, one gets exactly the same equations for a converging lens rather than a concave mirror because the equations are identical. The difference lies in the interpretation of the results. For a lens, the outgoing side is not that on which the object lies, unlike for a mirror. So for an object on the left side of the lens, a positive image distance means that the image is on the right of the lens, and a negative image distance means that the image is on the left side of the lens. (c) Again, for Exercise 34.12, the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above. EVALUATE: Concave mirrors and converging lenses both have f > 0 . Convex mirrors and diverging lenses both have f < 0 . IDENTIFY: SET UP:

1 1 1 y′ s′ + = and m = = − . s s′ f y s f = +12.0 cm and s′ = −17.0 cm . Apply

1 1 1 1 1 1 + = ⇒ = − ⇒ s = 7.0 cm. s s′ f s 12.0 cm −17.0 cm (−17.0) s′ y′ 0.800 cm m=− =− = +2.4 ⇒ y = = = +0.34 cm, so the object is 0.34 cm tall, erect, same side as the 7.2 s m +2.4 image. The principal-ray diagram is sketched in Figure 34.32.

EXECUTE:

34-10

Chapter 34

EVALUATE:

When the object is inside the focal point, a converging lens forms a virtual, enlarged image.

Figure 34.32 34.33.

IDENTIFY: Use Eq.(34.16) to calculate the object distance s. m calculated from Eq.(34.17) determines the size and orientation of the image. SET UP: f = −48.0 cm. Virtual image 17.0 cm from lens so s′ = −17.0 cm. EXECUTE:

1 1 1 1 1 1 s− f + = , so = − = s s′ f s′ f s sf

(−17.0 cm)(−48.0 cm) s′f = = +26.3 cm s′ − f −17.0 cm − (−48.0 cm) s′ −17.0 cm m=− =− = +0.646 s +26.3 cm y′ 8.00 mm y′ so y = m= = = 12.4 mm 0.646 y m The principal-ray diagram is sketched in Figure 34.33. EVALUATE: Virtual image, real object (s > 0) so image and object are on same side of lens. m > 0 so image is erect with respect to the object. The height of the object is 12.4 mm. s=

34.34.

Figure 34.33 1 1 1 IDENTIFY: Apply + = . s s′ f SET UP: The sign of f determines whether the lens is converging or diverging. s = 16.0 cm . s′ = +36.0 cm . Use s′ m = − to find the size and orientation of the image. s (16.0 cm)(36.0 cm) ss′ EXECUTE: (a) f = = = 11.1 cm . f > 0 and the lens is converging. s + s′ 16.0 cm + 36.0 cm s′ 36.0 cm (b) m = − = − = −2.25 . y′ = m y = (2.25)(8.00 mm) = 18.0 mm . m < 0 so the image is inverted. s 16.0 cm (c) The principal-ray diagram is sketched in Figure 34.34. EVALUATE: The image is real so the lens must be converging.

Figure 34.34

Geometric Optics

34.35.

34.36.

34-11

1 1 1 + = . s s′ f SET UP: The image is to be formed on the film, so s′ = +20.4 cm . 1 1 1 1 1 1 EXECUTE: + = ⇒ + = ⇒ s = 1020 cm = 10.2 m. s s′ f s 20.4 cm 20.0 cm EVALUATE: The object distance is much greater than f, so the image is just outside the focal point of the lens. 1 1 1 y′ s′ IDENTIFY: Apply + = and m = = − . ′ s s f y s SET UP: s = 3.90 m . f = 0.085 m . Apply

IDENTIFY:

1 1 1 1 1 1 0.0869 s′ 1750 mm = −39.0 mm, so + = ⇒ + = ⇒ s′ = 0.0869 m. y′ = − y = − 3.90 s s′ f 3.90 m s′ 0.085 m s it will not fit on the 24-mm × 36-mm film. EVALUATE: The image is just outside the focal point and s′ ≈ f . To have y′ = 36 mm , so that the image will fit EXECUTE:

on the film, s = − 34.37.

IDENTIFY: SET UP:

s′ . s s ! f , so s′ ≈ f .

EXECUTE:

m =

(a) m =

s′ ≈ s s′ (c) m = ≈ s EVALUATE: (b) m =

34.38.

28 mm s′ f ≈ ⇒ m = = 1.4 × 10−4. 200,000 mm s s

105 mm f ⇒m = = 5.3 × 10−4. 200,000 mm s

300 mm f ⇒m = = 1.5 × 10−3. 200,000 mm s The magnitude of the magnification increases when f increases. y′ s′ IDENTIFY: m = = s y SET UP: s ! f , so s′ ≈ f . 5.00 m s′ f (70.7 m) = 0.0372 m = 37.2 mm. y≈ y= 9.50 × 103 m s s EVALUATE: A very long focal length lens is needed to photograph a distant object. IDENTIFY and SET UP: Find the lateral magnification that results in this desired image size. Use Eq.(34.17) to relate m and s′ and Eq.(34.16) to relate s and s′ to f. 24 × 10−3 m 36 × 10−3 m EXECUTE: (a) We need m = − = −1.5 × 10−4. Alternatively, m = − = −1.5 × 10−4. 160 m 240 m s ! f so s′ ≈ f EXECUTE:

34.39.

(0.085 m)(1.75 m) s′y ≈− = 4.1 m . The person would need to stand about 4.1 m from the lens. y′ −0.036 m

y′ =

s′ f = − = −1.5 × 10−4 and f = (1.5 × 10−4 )(600 m) = 0.090 m = 90 mm. s s A smaller f means a smaller s′ and a smaller m, so with f = 85 mm the object’s image nearly fills the picture area. 36 × 10−3 m f (b) We need m = − = 3.75 × 10−3 and = −3.75 × 10−3. Then, as in part (a), s 9.6 m f = (40.0 m)(3.75 × 10−3 ) = 0.15 m = 150 mm. Therefore use the 135 mm lens. EVALUATE: When s ! f and s′ ≈ f , y′ = − f ( y / s ). For the mobile home y/s is smaller so a larger f is needed. Note that m is very small; the image is much smaller than the object. 1 1 1 IDENTIFY: Apply + = to each lens. The image of the first lens serves as the object for the second lens. s s′ f SET UP: For a distant object, s → ∞ EXECUTE: (a) s1 = ∞ ⇒ s1′ = f1 = 12 cm. Then m = −

34.40.

(b) s2 = 4.0 cm − 12 cm = −8 cm.

34-12

Chapter 34

(c)

1 1 1 1 1 1 + = ⇒ + = ⇒ s2′ = 24 cm, to the right. s s′ f −8 cm s2′ −12 cm

(d) s1 = ∞ ⇒ s′1 = f1 = 12 cm. s2 = 8.0 cm − 12 cm = −4 cm. 34.41.

34.42.

1 1 1 1 1 1 + = ⇒ + = ⇒ s2′ = 6 cm. s s′ f −4 cm s2′ −12 cm

EVALUATE: In each case the image of the first lens serves as a virtual object for the second lens, and s2 < 0 . IDENTIFY: The f-number of a lens is the ratio of its focal length to its diameter. To maintain the same exposure, the amount of light passing through the lens during the exposure must remain the same. SET UP: The f-number is f/D. 180.0 mm f EXECUTE: (a) f -number = ⇒ f -number = ⇒ f -number = f /11 . (The f-number is an integer.) 16.36 mm D (b) f/11 to f/2.8 is four steps of 2 in intensity, so one needs 1/16th the exposure. The exposure should be 1/480 s = 2.1 × 10−3 s = 2.1 ms. EVALUATE: When opening the lens from f/11 to f/2.8, the area increases by a factor of 16, so 16 times as much light is allowed in. Therefore the exposure time must be decreased by a factor of 1/16 to maintain the same exposure on the film or light receptors of a digital camera. IDENTIFY and SET UP: The square of the aperture diameter is proportional to the length of the exposure time required. 2

⎛ 1 ⎞ ⎛ 8 mm ⎞ ⎛ 1 ⎞ s ⎟⎜ s⎟ ⎟ ≈⎜ ⎜ ⎝ 30 ⎠ ⎝ 23.1 mm ⎠ ⎝ 250 ⎠ EVALUATE: An increase in the aperture diameter decreases the exposure time. 1 1 1 IDENTIFY and SET UP: Apply + = to calculate s′ . s s′ f EXECUTE: (a) A real image is formed at the film, so the lens must be convex. 1 1 1 1 s− f sf (b) + = so = and s′ = , with f = +50.0.0 mm . For s = 45 cm = 450 mm, s′ = 56 mm. For s s′ f s′ sf s− f s = ∞, s′ = f = 50 mm. The range of distances between the lens and film is 50 mm to 56 mm. EVALUATE: The lens is closer to the film when photographing more distant objects. IDENTIFY: The projector lens can be modeled as a thin lens. 1 1 1 s′ SET UP: The thin-lens equation is + = , and the magnification of the lens is m = − . s s′ f s EXECUTE:

34.43.

34.44.

EXECUTE:

(a)

1 1 1 1 1 1 + = ⇒ = + ⇒ f = 147.5 mm , so use a f = 148 mm lens. s s′ f f 0.150 m 9.00 m

s′ ⇒ m = 60 ⇒ Area = 1.44 m × 2.16 m . s EVALUATE: The lens must produce a real image to be viewed on the screen. Since the magnification comes out negative, the slides to be viewed must be placed upside down in the tray. (a) IDENTIFY: The purpose of the corrective lens is to take an object 25 cm from the eye and form a virtual image at the eye’s near point. Use Eq.(34.16) to solve for the image distance when the object distance is 25 cm. 1 1 SET UP: m = +0.3636 m (converging lens) = +2.75 diopters means f = + 2.75 f f = 36.36 cm; s = 25 cm; s′ = ? (b) m = −

34.45.

1 1 1 + = so s s′ f (25 cm)(36.36 cm) sf s′ = = = −80.0 cm 25 cm − 36.36 cm s− f The eye’s near point is 80.0 cm from the eye. (b) IDENTIFY: The purpose of the corrective lens is to take an object at infinity and form a virtual image of it at the eye’s far point. Use Eq.(34.16) to solve for the image distance when the object is at infinity. 1 1 SET UP: m = −0.7692 m (diverging lens) = −1.30 diopters means f = − 1.30 f f = −76.92 cm; s = ∞; s′ = ? EXECUTE:

EXECUTE:

1 1 1 1 1 + = and s = ∞ says = and s′ = f = −76.9 cm. The eye’s far point is 76.9 cm from the eye. s s′ f s′ f

Geometric Optics

34.46.

34-13

EVALUATE: In each case a virtual image is formed by the lens. The eye views this virtual image instead of the object. The object is at a distance where the eye can’t focus on it, but the virtual image is at a distance where the eye can focus. na nb nb − na IDENTIFY: + = s s′ R SET UP: na = 1.00 , nb = 1.40 . s = 40.0 cm , s′ = 2.60 cm .

1 1.40 0.40 + = and R = 0.710 cm . 40.0 cm 2.60 cm R EVALUATE: The cornea presents a convex surface to the object, so R > 0 . IDENTIFY: In each case the lens forms a virtual image at a distance where the eye can focus. Power in diopters equals 1/ f , where f is in meters. SET UP: In part (a), s = 25 cm and in part (b), s → ∞ . 1 1 1 1 1 1 EXECUTE: (a) = + = + ⇒ power = = +2.33 diopters. ′ f s s 0.25 m −0.600 m f EXECUTE:

34.47.

1 1 1 1 1 1 = + = + ⇒ power = = −1.67 diopters. ′ f s s ∞ −0.600 m f EVALUATE: A converging lens corrects the near vision and a diverging lens corrects the far vision. 25.0 cm 1 1 1 to calculate s for IDENTIFY: When the object is at the focal point, M = . In part (b), apply + = s s′ f f s′ = −25.0 cm . SET UP: Our calculation assumes the near point is 25.0 cm from the eye. 25.0 cm 25.0 cm EXECUTE: (a) Angular magnification M = = = 4.17. f 6.00 cm (b)

34.48.

(b)

1 1 1 1 1 1 + = ⇒ + = ⇒ s = 4.84 cm. s s′ f s −25.0 cm 6.00 cm

y y 25.0 cm 25.0 cm ,θ= and M = = = 5.17 . M is greater when the image s 25.0 cm s 4.84 cm is at the near point than when the image is at infinity. IDENTIFY: Use Eqs.(34.16) and (34.17) to calculate s and y′. (a) SET UP: f = 8.00 cm; s′ = −25.0 cm; s = ?

EVALUATE: 34.49.

34.50.

34.51.

In part (b), θ ′ =

1 1 1 1 1 1 s′ − f + = , so = − = s s′ f s f s′ s′f ( −25.0 cm)(+8.00 cm) s′f EXECUTE: s = = = +6.06 cm s′ − f −25.0 cm − 8.00 cm s′ −25.0 cm (b) m = − = − = +4.125 s 6.06 cm y′ so y′ = m y = (4.125)(1.00 mm) = 4.12 mm m = y EVALUATE: The lens allows the object to be much closer to the eye than the near point. The lense allows the eye to view an image at the near point rather than the object. s′ y ′ y′ y IDENTIFY: For a thin lens, − = , so = , and the angular size of the image equals the angular size of the s y s′ s object. y SET UP: The object has angular size θ = , with θ in radians. f y y 2.00 mm EXECUTE: θ = ⇒ f = = = 80.0 mm = 8.00 cm. θ 0.025 rad f 2.00 mm EVALUATE: If the insect is at the near point of a normal eye, its angular size is = 0.0080 rad . 250 mm IDENTIFY: The thin-lens equation applies to the magnifying lens. 1 1 1 SET UP: The thin-lens equation is + = . s s′ f

34-14

Chapter 34

EXECUTE:

34.52.

34.53.

The image is behind the lens, so s′ < 0 . The thin-lens equation gives 1 1 1 1 1 1 + = ⇒ = − ⇒ s = 4.17 cm , on the same side of the lens as the ant. s s′ f s 5.00 cm −25.0 cm

EVALUATE: Since s′ < 0 , the image will be erect. IDENTIFY: Apply Eq.(34.24). SET UP: s1′ = 160 mm + 5.0 mm = 165 mm (250 mm)s1′ (250 mm)(165 mm) EXECUTE: (a) M = = = 317. (5.00 mm)(26.0 mm) f1 f 2 0.10 mm 0.10 mm (b) The minimum separation is = = 3.15 × 10−4 mm. 317 M EVALUATE: The angular size of the image viewed by the eye when looking through the microscope is 317 times larger than if the object is viewed at the near-point of the unaided eye. (a) IDENTIFY and SET UP:

Figure 34.53

Final image is at ∞ so the object for the eyepiece is at its focal point. But the object for the eyepiece is the image of the objective so the image formed by the objective is 19.7 cm – 1.80 cm = 17.9 cm to the right of the lens. Apply Eq.(34.16) to the image formation by the objective, solve for the object distance s. f = 0.800 cm; s′ = 17.9 cm; s = ? 1 1 1 1 1 1 s′ − f + = , so = − = s s′ f s f s′ s′f s′f (17.9 cm)(+0.800 cm) EXECUTE: s = = = +8.37 mm 17.9 cm − 0.800 cm s′ − f (b) SET UP: Use Eq.(34.17). s′ 17.9 cm EXECUTE: m1 = − = − = −21.4 0.837 cm s The linear magnification of the objective is 21.4. (c) SET UP: Use Eq.(34.23): M = m1M 2 25 cm 25 cm EXECUTE: M 2 = = = 13.9 1.80 cm f2 M = m1M 2 = (−21.4)(13.9) = −297 EVALUATE: M is not accurately given by (25 cm) s1′ / f1 f 2 = 311, because the object is not quite at the focal point of the objective ( s1 = 0.837 cm and f1 = 0.800 cm). 34.54.

Eq.(34.24) can be written M = m1 M 2 =

IDENTIFY: SET UP:

s1′ M2. f1

s1′ = f1 + 120 mm

EXECUTE:

f = 16 mm : s′ = 120 mm + 16 mm = 136 mm; s = 16 mm . m1 =

s′ 136 mm = = 8.5. s 16 mm

s′ 124 mm = = 31. 4 mm s s′ 122 mm = 64 . f = 1.9 mm : s′ = 120 mm + 1.9 mm = 122 mm; s = 1.9 mm ⇒ m1 = = s 1.9 mm The eyepiece magnifies by either 5 or 10, so: (a) The maximum magnification occurs for the 1.9-mm objective and 10x eyepiece: M = m1 M e = (64)(10) = 640. f = 4 mm : s′ = 120 mm + 4 mm = 124 mm; s = 4 mm ⇒ m1 =

(b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece: M = m1 M e = (8.5)(5) = 43. EVALUATE:

The smaller the focal length of the objective, the greater the overall magnification.

Geometric Optics

34.55.

34.56.

IDENTIFY: f -number = f / D SET UP: D = 1.02 m f EXECUTE: = 19.0 ⇒ f = (19.0) D = (19.0)(1.02 m) = 19.4 m. D EVALUATE: Camera lenses can also have an f-number of 19.0. For a camera lens, both the focal length and lens diameter are much smaller, but the f-number is a measure of their ratio. f IDENTIFY: For a telescope, M = − 1 . f2 SET UP:

34.57.

34-15

f 2 = 9.0 cm . The distance between the two lenses equals f1 + f 2 .

EXECUTE:

f1 + f 2 = 1.80 m ⇒ f1 = 1.80 m − 0.0900 m = 1.71 m . M = −

EVALUATE:

For a telescope, f1 ! f 2 .


f1 171 =− = −19.0. 9.00 f2

Use Eq.(34.24), with f1 = 95.0 cm (objective) and f 2 = 15.0 cm (eyepiece).

f1 95.0 cm =− = −6.33 15.0 cm f2 (b) IDENTIFY and SET UP: Use Eq.(34.17) to calculate y′. M =−

EXECUTE:

SET UP: s = 3.00 × 103 m s′ = f1 = 95.0 cm (since s is very large, s′ ≈ f )

s′ 0.950 m =− = −3.167 × 10−4 3.00 × 103 m s y′ = m y = (3.167 × 10−4 )(60.0 m) = 0.0190 m = 1.90 cm m=−

EXECUTE:

(c) IDENTIFY: Use Eq.(34.21) and the angular magnification M obtained in part (a) to calculate θ ′. The angular size θ of the image formed by the objective (object for the eyepiece) is its height divided by its distance from the objective. 0.0190 m EXECUTE: The angular size of the object for the eyepiece is θ = = 0.0200 rad. 0.950 m 60.0 m (Note that this is also the angular size of the object for the objective: θ = = 0.0200 rad. For a thin lens 3.00 × 103 m the object and image have the same angular size and the image of the objective is the object for the eyepiece.) θ′ M= (Eq.34.21) so the angular size of the image is θ ′ = M θ = −(6.33)(0.0200 rad) = −0.127 rad (The minus

θ

34.58.

34.59.

sign shows that the final image is inverted.) EVALUATE: The lateral magnification of the objective is small; the image it forms is much smaller than the object. But the total angular magnification is larger than 1.00; the angular size of the final image viewed by the eye is 6.33 times larger than the angular size of the original object, as viewed by the unaided eye. IDENTIFY: The angle subtended by Saturn with the naked eye is the same as the angle subtended by the image of Saturn formed by the objective lens (see Fig. 34.53 in the textbook). diameter of Saturn y′ SET UP: The angle subtended by Saturn is θ = = . distance to Saturn f1 y′ 1.7 mm 0.0017 m EXECUTE: Putting in the numbers gives θ = = = = 9.4 × 10−5 rad = 0.0054° 18 m 18 m f1 EVALUATE: The angle subtended by the final image, formed by the eyepiece, would be much larger than 0.0054°. f IDENTIFY: f = R / 2 and M = − 1 . f2 SET UP:

For object and image both at infinity, f1 + f 2 equals the distance d between the two mirrors.

f 2 = 1.10 cm . R1 = 1.30 m . R1 = 0.650 m ⇒ d = f1 + f 2 = 0.661 m. 2 f 0.650 m (b) M = 1 = = 59.1. f 2 0.011 m EXECUTE:

EVALUATE:

(a) f1 =

For a telescope, f1 ! f 2 .

34-16

Chapter 34

34.60.

IDENTIFY: The primary mirror forms an image which then acts as the object for the secondary mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 + = . s s′ f EXECUTE: For the first image (formed by the primary mirror):

1 1 1 1 1 1 + = ⇒ = − ⇒ s′ = 2.5 m . s s′ f s′ 2.5 m ∞ For the second image (formed by the secondary mirror), the distance between the two vertices is x. Assuming that the image formed by the primary mirror is to the right of the secondary mirror, the object distance is s = x – 2.5 m and the image distance is s′ = x + 0.15 m. Therefore we have 1 1 1 1 1 1 + = ⇒ + = s s′ f x − 2.5 m x + 0.15 m −1.5 m

34.61.

34.62.

The positive root of the quadratic equation gives x = 1.7 m, which is the distance between the vertices. EVALUATE: Some light is blocked by the secondary mirror, but usually not enough to make much difference. ds ds′ IDENTIFY and SET UP: For a plane mirror s′ = − s. v = and v′ = , so v′ = −v. dt dt EXECUTE: The velocities of the object and image relative to the mirror are equal in magnitude and opposite in direction. Thus both you and your image are receding from the mirror surface at 2.40 m/s, in opposite directions. Your image is therefore moving at 4.80 m/s relative to you. EVALUATE: The result derives from the fact that for a plane mirror the image is the same distance behind the mirror as the object is in front of the mirror. IDENTIFY: Apply the law of reflection. SET UP: The image of one mirror can serve as the object for the other mirror. EXECUTE: (a) There are three images formed, as shown in Figure 34.62a. (b) The paths of rays for each image are sketched in Figure 34.62b. EVALUATE: Our results agree with Figure 34.9 in the textbook.

Figure 34.62 34.63.

IDENTIFY: Apply the law of reflection for rays from the feet to the eyes and from the top of the head to the eyes. SET UP: In Figure 34.63, ray 1 travels from the feet of the woman to her eyes and ray 2 travels from the top of her head to her eyes. The total height of the woman is h. EXECUTE: The two angles labeled θ1 are equal because of the law of reflection, as are the two angles labeled

θ 2 . Since these angles are equal, the two distances labeled y1 are equal and the two distances labeled y2 are equal. The height of the woman is hw = 2 y1 + 2 y2 . As the drawing shows, the height of the mirror is hm = y1 + y2 . Comparing, we find that hm = hw / 2 . The minimum height required is half the height of the woman.

Geometric Optics

34-17

EVALUATE: The height of the image is the same as the height of the woman, so the height of the image is twice the height of the mirror.

34.64.

IDENTIFY:

1 1 2 s′ Apply + = and m = − . s s′ R s

Figure 34.63

s′ so m is negative and s m = −2.25 . The object, mirror and wall are sketched in Figure 34.64. The sketch shows that s′ − s = 400 cm . s′ EXECUTE: m = −2.25 = − and s′ = 2.25s . s′ − s = 2.25s − s = 400 cm and s = 320 cm . s 1 1 2 1 1 2 + = . s′ = 400 cm + 320 cm = 720 cm . The mirror should be 7.20 m from the wall. + = . s s′ R 320 cm 720 cm R R = 4.43 m. EVALUATE: The focal length of the mirror is f = R / 2 = 222 cm . s > f , as it must if the image is to be real. SET UP:

34.65.

Since the image is projected onto the wall it is real and s′ > 0 . m = −

Figure 34.64 IDENTIFY: We are given the image distance, the image height, and the object height. Use Eq.(34.7) to calculate the object distance s. Then use Eq.(34.4) to calculate R. (a) SET UP: Image is to be formed on screen so is real image; s′ > 0. Mirror to screen distance is 8.00 m, so s′ s′ = +800 cm. m = − < 0 since both s and s′ are positive. s

34-18

Chapter 34

y′ 36.0 m s′ s′ 800 cm = = 60.0 and m = −60.0. Then m = − gives s = − = − = +13.3 cm. s y 0.600 cm m −60.0 1 1 2 2 s + s′ (b) + = , so = s s′ R R ss′ ⎛ ss′ ⎞ ⎛ (13.3 cm)(800 cm) ⎞ R = 2⎜ ⎟ = 2⎜ ⎟ = 26.2 cm ⎝ s + s′ ⎠ ⎝ 800 cm + 13.3 cm ⎠ EVALUATE: R is calculated to be positive, which is correct for a concave mirror. Also, in part (a) s is calculated to be positive, as it should be for a real object. 1 1 1 s′ y ′ IDENTIFY: Apply + = to calculate s′ and then use m = − = to find the height of the image. s s′ f s y R SET UP: For a convex mirror, R < 0 , so R = −18.0 cm and f = = −9.00 cm . 2 1 1 1 sf (1300 cm)(−9.00 cm) s′ −8.94 cm EXECUTE: (a) + = . s′ = = = −8.94 cm . m = − = − = 6.88 × 10−3. 1300 cm s s′ f s s − f 1300 cm − (−9.00 cm) y′ = m y = (6.88 × 10−3 )(1.5 m) = 0.0103 m = 1.03 cm . EXECUTE:

34.66.

34.67.

34.68.

34.69.

m =

(b) The height of the image is much less than the height of the car, so the car appears to be farther away than its actual distance. EVALUATE: The image formed by a convex mirror is always virtual and smaller than the object. 1 1 2 s′ IDENTIFY: Apply + = and m = − . ′ s s R s SET UP: R = +19.4 cm . 1 1 2 1 1 2 EXECUTE: (a) + = ⇒ + = ⇒ s′ = −46 cm, so the image is virtual. 8.0 cm s′ 19.4 cm s s′ R s′ −46 (b) m = − = − = 5.8, so the image is erect, and its height is y′ = (5.8) y = (5.8)(5.0 mm) = 29 mm. 8.0 s EVALUATE: (c) When the filament is 8 cm from the mirror, the image is virtual and cannot be projected onto a wall. 1 1 2 s′ IDENTIFY: Combine + = and m = − . s s′ R s SET UP: m = +2.50 . R > 0 . s′ 1 1 2 0.600 2 EXECUTE: m = − = +2.50. s′ = −2.50s . + = . = and s = 0.300 R . s −2.50 s R s R s s′ = −2.50 s = (−2.50)(0.300 R ) = −0.750 R . The object is a distance of 0.300R in front of the mirror and the image is a distance of 0.750 R behind the mirror. EVALUATE: For a single mirror an erect image is always virtual. IDENTIFY and SET UP: Apply Eqs.(34.6) and (34.7). For a virtual object s < 0. The image is real if s′ > 0. EXECUTE: (a) Convex implies R < 0; R = −24.0 cm; f = R / 2 = −12.0 cm 1 1 1 1 1 1 s− f + = , so = − = s s′ f s′ f s sf sf (−12.0 cm)s s′ = = s− f s + 12.0 cm (12.0 cm) s s is negative, so write as s = − s ; s′ = + . Thus s′ > 0 (real image) for s < 12.0 cm. Since s is negative 12.0 cm − s

this means −12.0 cm < s < 0. A real image is formed if the virtual object is closer to the mirror than the focus. s′ (b) m = − ; real image implies s′ > 0; virtual object implies s < 0. Thus m > 0 and the image is erect. s

Geometric Optics

34-19

(c) The principal-ray diagram is given in Figure 34.69.

34.70.

Figure 34.69 EVALUATE: For a real object, only virtual images are formed by a convex mirror. The virtual object considered in this problem must have been produced by some other optical element, by another lens or mirror in addition to the convex one we considered. n n n − na IDENTIFY: Apply a + b = b , with R → ∞ since the surfaces are flat. s s′ R SET UP: The image formed by the first interface serves as the object for the second interface. EXECUTE: For the water-benzene interface to get the apparent water depth:

na nb 1.33 1.50 + =0⇒ + = 0 ⇒ s′ = −7.33 cm. 6.50 cm s s′ s′ For the benzene-air interface, to get the total apparent distance to the bottom:

na nb 1.50 1 + =0⇒ + = 0 ⇒ s′ = −6.62 cm. (7.33 cm + 2.60 cm) s′ s s′

34.71.

EVALUATE: At the water-benzene interface the light refracts into material of greater refractive index and the overall effect is that the apparent depth is greater than the actual depth. ⎛1 1 ⎞ 1 IDENTIFY: The focal length is given by = ( n − 1) ⎜ − ⎟ . f ⎝ R1 R2 ⎠ SET UP: R1 = ±4.0 cm or ±8.0 cm . R2 = ±8.0 cm or ±4.0 cm . The signs are determined by the location of the center of curvature for each surface. ⎛ ⎞ 1 1 1 EXECUTE: = (0.60) ⎜ − ⎟ , so f = ±4.44 cm, ± 13.3 cm. The possible lens shapes are f ⎝ ±4.00 cm ±8.00 cm ⎠

sketched in Figure 34.71. f1 = +13.3 cm; f 2 = +4.44 cm; f3 = 4.44 cm; f 4 = −13.3 cm; f5 = −13.3 cm; f 6 = +13.3 cm;

f 7 = −4.44 cm; f8 = −4.44 cm. EVALUATE: f is the same whether the light travels through the lens from right to left or left to right, so for the pairs (1,6), (4,5) and (7,8) the focal lengths are the same.

Figure 34.71 34.72.

1 1 1 + = and the concept of principal rays. s s′ f SET UP: s = 10.0 cm . If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens and the ray is parallel to the optic axis after it passes through the lens. EXECUTE: (a) The ray is bent toward the optic axis by the lens so the lens is converging. (b) The ray is parallel to the optic axis after it passes through the lens so it comes from the focal point; f = 18.0 cm . IDENTIFY:

Apply

34-20

Chapter 34

(c) The principal ray diagram is drawn in Figure 34.72. The diagram shows that the image is 22.5 cm to the left of the lens. 1 1 1 sf (10.0 cm)(18.0 cm) (d) + = gives s′ = = = −22.5 cm . The calculated image position agrees with the s s′ f s − f 10.0 cm − 18.0 cm principal ray diagram. EVALUATE: The image is virtual. A converging lens produces a virtual image when the object is inside the focal point.

Figure 34.72 34.73.

IDENTIFY: Since the truck is moving toward the mirror, its image will also be moving toward the mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 + = , where f = R/2. s s′ f EXECUTE: Since the mirror is convex, f = R/2 = (–1.50 m)/2 = –0.75 m. Applying the equation for a spherical 1 1 1 fs . mirror gives + = ⇒ s′ = s s′ f s− f

Using the chain rule from calculus and the fact that v = ds/dt, we have v′ = 2

34.74.

34.75.

ds′ ds′ ds f2 = =v ( s − f )2 dt ds dt

2

⎛s− f ⎞ ⎡ 2.0 m − ( −0.75 m) ⎤ Solving for v gives v = v′ ⎜ ⎟ = (1.5 m/s) ⎢ ⎥ = 20.2 m/s . −0.75 m ⎣ ⎦ ⎝ f ⎠ This is the velocity of the truck relative to the mirror, so the truck is approaching the mirror at 20.2 m/s. You are traveling at 25 m/s, so the truck must be traveling at 25 m/s + 20.2 m/s = 45 m/s relative to the highway. EVALUATE: Even though the truck and car are moving at constant speed, the image of the truck is not moving at constant speed because its location depends on the distance from the mirror to the truck. n n IDENTIFY: In this context, the microscope just looks at an image or object. Apply a + b = 0 to the image s s′ formed by refraction at the top surface of the second plate. In this calculation the object is the bottom surface of the second plate. SET UP: The thickness of the second plate is 2.50 mm + 0.78 mm , and this is s. The image is 2.50 mm below the top surface, so s′ = −2.50 mm . na nb n 1 s 2.50 mm + 0.780 mm EXECUTE: + =0⇒ + =0⇒n=− =− = 1.31. s s′ s s′ s′ −2.50 mm EVALUATE: The object and image distances are measured from the front surface of the second plate, and the image is virtual. 1 1 1 s′ IDENTIFY and SET UP: In part (a) use + = to evaluate ds′ / ds . Compare to m = − . In part (b) use s s′ f s 1 1 2 + = to find the location of the image of each face of the cube. s s′ R 1 1 1 d ⎛1 1 1 ⎞ 1 1 ds′ and taking its derivative with respect to s we have 0 = ⎜ + − ⎟ = − 2 − 2 EXECUTE: (a) + = ds ⎝ s s′ f ⎠ s s′ ds s s′ f ds′ s′ 2 ds′ and = − 2 = − m 2 . But = m′ , so m′ = − m 2 . Images are always inverted longitudinally. ds s ds 1 1 2 1 1 2 (b) (i) Front face: + = ⇒ + = ⇒ s′ = 120.00 cm. s s′ R 200.000 cm s′ 150.000 cm

Geometric Optics

34-21

1 1 2 1 1 2 + = ⇒ + = ⇒ s′ = 119.96 cm. ′ ′ s s R 200.100 cm s 150.000 cm s′ 120.000 2 (ii) m = − = − = −0.600 . m′ = − m 2 = − ( −0.600 ) = −0.360. s 200.000 (iii) The two faces perpendicular to the axis (the front and rear faces): squares with side length 0.600 mm. The four faces parallel to the axis (the side faces): rectangles with sides of length 0.360 mm parallel to the axis and 0.600 mm perpendicular to the axis. EVALUATE: Since the lateral and longitudinal magnifications have different values the image of the cube is not a cube. n s′ IDENTIFY: m′ = ds′ / ds and m = − a . nb s na nb nb − na SET UP: Use + = to evaluate ds′ / ds . s s′ R na nb nb − na EXECUTE: + = and taking its derivative with respect to s we have s s′ R ⎛ s′ 2 n 2 ⎞ n d ⎛ n n n − na ⎞ n n ds′ ds′ s′ 2 n n = − a2 − b2 and 0= ⎜ a + b − b = − 2 a = − ⎜ 2 a2 ⎟ b = − m 2 b . ⎟ ′ ′ ds s nb na ds ⎝ s s R ⎠ s s ds ⎝ s nb ⎠ na n ds′ But = m′, so m′ = − m 2 b . na ds EVALUATE: m′ is always negative. This means that images are always inverted longitudinally. s′ IDENTIFY and SET UP: Rays that pass through the hole are undeflected. All other rays are blocked. m = − . s EXECUTE: (a) The ray diagram is drawn in Figure 34.77. The ray shown is the only ray from the top of the object that reaches the film, so this ray passes through the top of the image. An inverted image is formed on the far side of the box, no matter how far this side is from the pinhole and no matter how far the object is from the pinhole. 20.0 cm s′ (b) s = 1.5 m . s′ = 20.0 cm . m = − = − = −0.133 . y′ = my = ( −0.133)(18 cm) = −2.4 cm . The image is 150 cm s 2.4 cm tall. EVALUATE: A defect of this camera is that not much light energy passes through the small hole each second, so long exposure times are required. Rear face:

34.76.

34.77.

34.78.

Figure 34.77 na s′ na nb nb − na to each refraction. The overall magnification is m = m1m2 . IDENTIFY: Apply + = and m = − s s′ R nb s SET UP: For the first refraction, R = +6.0 cm , na = 1.00 and nb = 1.60 . For the second refraction, R = −12.0 cm , na = 1.60 and nb = 1.00 . EXECUTE: (a) The image from the left end acts as the object for the right end of the rod. n n n − na 1 1.60 0.60 ⇒ + = ⇒ s′ = 28.3 cm. (b) a + b = b s s′ R 23.0 cm s′ 6.0 cm n s′ 28.3 So the second object distance is s2 = 40.0 cm − 28.3 cm = 11.7 cm. m1 = − a = − = −0.769. nb s 1.60 ( )( 23.0 ) (c) The object is real and inverted. n n n − na 1.60 1 −0.60 ⇒ + = ⇒ s′ = −11.5 cm. (d) a + b = b s2 s′2 R 11.7 cm s′2 −12.0 cm n s′ (1.60 )( −11.5) = 1.57 ⇒ m = m m = −0.769 1.57 = −1.21. m2 = − a = − ( )( ) 1 2 nb s 11.7 (e) The final image is virtual, and inverted.

34-22

Chapter 34

(f) y′ = (1.50 mm )( −1.21) = −1.82 mm.

34.79.

EVALUATE: The first image is to the left of the second surface, so it serves as a real object for the second surface, with positive object distance. IDENTIFY: Apply Eqs.(34.11) and (34.12) to the refraction as the light enters the rod and as it leaves the rod. The image formed by the first surface serves as the object for the second surface. The total magnification is mtot = m1m2 , where m1 and m2 are the magnifications for each surface. SET UP: The object and rod are shown in Figure 34.79.

Figure 34.79 (a) image formed by refraction at first surface (left end of rod): s = +23.0 cm; na = 1.00; nb = 1.60; R = +6.00 cm na nb nb − na + = s s′ R 1 1.60 1.60 − 1.00 + = 23.0 cm s′ 6.00 cm 1.60 1 1 23 − 10 13 = − = = s′ 10.0 cm 23.0 cm 230 cm 230 cm ⎛ 230 cm ⎞ s′ = 1.60 ⎜ ⎟ = +28.3 cm; image is 28.3 cm to right of first vertex. ⎝ 13 ⎠ This image serves as the object for the refraction at the second surface (right-hand end of rod). It is 28.3 cm − 25.0 cm = 3.3 cm to the right of the second vertex. For the second surface s = −3.3 cm (virtual object). (b) EVALUATE: Object is on side of outgoing light, so is a virtual object. (c) SET UP: Image formed by refraction at second surface (right end of rod): s = −3.3 cm; na = 1.60; nb = 1.00; R = −12.0 cm

EXECUTE:

na nb nb − na + = s s′ R 1.60 1.00 1.00 − 1.60 EXECUTE: + = −3.3 cm s′ −12.0 cm s′ = +1.9 cm; s′ > 0 so image is 1.9 cm to right of vertex at right-hand end of rod. (d) s′ > 0 so final image is real. Magnification for first surface: n s′ (1.60)(+28.3 cm) m=− a =− = −0.769 nb s (1.00)(+23.0 cm) Magnification for second surface: n s′ (1.60)(+1.9 cm) m=− a =− = +0.92 nb s (1.00)( −3.3 cm)

34.80.

The overall magnification is mtot = m1m2 = (−0.769)(+0.92) = −0.71 mtot < 0 so final image is inverted with respect to the original object. (e) y′ = mtot y = ( −0.71)(1.50 mm) = −1.06 mm The final image has a height of 1.06 mm. EVALUATE: The two refracting surfaces are not close together and Eq.(34.18) does not apply. 1 1 1 y′ s′ and m = = − . The type of lens determines the sign of f. The sign of IDENTIFY: Apply + = y s s s′ f s′ determines whether the image is real or virtual. SET UP: s = +8.00 cm . s′ = −3.00 cm . s′ is negative because the image is on the same side of the lens as the object. 1 s + s′ ss′ (8.00 cm)(−3.00 cm) = = −4.80 cm . f is negative so the lens is = and f = EXECUTE: (a) ′ ′ s+s 8.00 cm − 3.00 cm f ss diverging.

Geometric Optics

34-23

s′ −3.00 cm =− = +0.375 . y′ = my = (0.375)(6.50 mm) = 2.44 mm . s′ < 0 and the image is virtual. s 8.00 cm EVALUATE: A converging lens can also form a virtual image, if the object distance is less than the focal length. But in that case s′ > s and the image would be farther from the lens than the object is. (b) m = −

34.81.

34.82.

1 1 1 y′ s′ + = . The type of lens determines the sign of f. m = = − . The sign of s′ depends on y s s s′ f whether the image is real or virtual. s = 16.0 cm . SET UP: s′ = −22.0 cm ; s′ is negative because the image is on the same side of the lens as the object. 1 s + s′ ss′ (16.0 cm)(−22.0 cm) and f = EXECUTE: (a) = = = +58.7 cm . f is positive so the lens is converging. 16.0 cm − 22.0 cm s + s′ f ss′ s′ −22.0 cm = 1.38 . y′ = my = (1.38)(3.25 mm) = 4.48 mm . s′ < 0 and the image is virtual. (b) m = − = − s 16.0 cm EVALUATE: A converging lens forms a virtual image when the object is closer to the lens than the focal point. n n n − na . Use the image distance when viewed from the flat end to determine the IDENTIFY: Apply a + b = b s s′ R refractive index n of the rod. SET UP: When viewing from the flat end, na = n , nb = 1.00 and R → ∞ . When viewing from the curved end, IDENTIFY:

na = n , nb = 1.00 and R = −10.0 cm . EXECUTE:

na nb n 1 15.0 + =0⇒ + =0⇒n= = 1.58. When viewed from the curved end of the s s′ 15.0 cm −9.50 cm 9.50

na nb nb − na n 1 1− n 1.58 1 −0.58 + = ⇒ + = ⇒ + = , and s′ = −21.1 cm . The image is 21.1 cm s s′ R s s′ R 15.0 cm s′ −10.0 cm within the rod from the curved end. EVALUATE: In each case the image is virtual and on the same side of the surface as the object. (a) IDENTIFY: Apply Snell’s law to the refraction of a ray at each side of the beam to find where these rays strike the table. SET UP: The path of a ray is sketched in Figure 34.83. rod

34.83.

Figure 34.83

The width of the incident beam is exaggerated in the sketch, to make it easier to draw. Since the diameter of the beam is much less than the radius of the hemisphere, angles θ a and θb are small. The diameter of the circle of light formed on the table is 2 x. Note the two right triangles containing the angles θ a and θb . r = 0.190 cm is the radius of the incident beam. R = 12.0 cm is the radius of the glass hemisphere. r x′ x EXECUTE: θ a and θb small imply x ≈ x′; sin θ a = , sin θ b = ≈ R R R Snell’s law: na sin θ a = nb sin θb Using the above expressions for sin θ a and sin θb gives na

r x = nb R R

na r 1.00(0.190 cm) = = 0.1267 cm nb 1.50 The diameter of the circle on the table is 2 x = 2(0.1267 cm) = 0.253 cm. (b) EVALUATE: R divides out of the expression; the result for the diameter of the spot is independent of the radius R of the hemisphere. It depends only on the diameter of the incident beam and the index of refraction of the glass. na r = nb x so x =

34-24

Chapter 34

34.84.


Treating each of the goblet surfaces as spherical surfaces, we have to pass, from left to n n n − na to each surface. The image formed by one surface serves as right, through four interfaces. Apply a + b = b ′ s s R the object for the next surface. EXECUTE: (a) For the empty goblet: na nb nb − na 1 1.50 0.50 + = ⇒ + = ⇒ s1′ = 12 cm . ′ ′ s s R s1 4.00 cm ∞ s2 = 0.60 cm − 12 cm = −11.4 cm ⇒

1.50 1 −0.50 + = ⇒ s′2 = −64.6 cm. ′ −11.4 cm s2 3.40 cm

s3 = 64.6 cm + 6.80 cm = 71.4 cm ⇒

1 1.50 0.50 + = ⇒ s3′ = −9.31 cm. ′ 71.4 cm s3 −3.40 cm

1.50 1 −0.50 + = ⇒ s′4 = −37.9 cm. The final image is ′ 9.91 cm s4 −4.00 cm 37.9 cm − 2(4.0 cm) = 29.9 cm to the left of the goblet. (b) For the wine-filled goblet: na nb nb − na 1 1.50 0.50 + = ⇒ + = ⇒ s1′ = 12 cm . s s′ R s1′ 4.00 cm ∞ s4 = 9.31 cm + 0.60 cm = 9.91 cm ⇒

s2 = 0.60 cm − 12 cm = −11.4 cm ⇒

1.50 1.37 −0.13 + = ⇒ s′2 = 14.7 cm. s2′ 3.40 cm −11.4 cm

s3 = 6.80 cm − 14.7 cm = −7.9 cm ⇒

1.37 1.50 0.13 + = ⇒ s3′ = 11.1 cm. s3′ −7.9 cm −3.40 cm

s4 = 0.60 cm − 11.1 cm = −10.5 cm ⇒

34.85.

34.86.

1.50 1 −0.50 + = ⇒ s′4 = 3.73 cm . The final image is 3.73 cm to the −10.5 cm s′4 −4.00 cm

right of the goblet. EVALUATE: If the object for a surface is on the outgoing side of the light, then the object is virtual and the object distance is negative. n n n − na IDENTIFY: The image formed by refraction at the surface of the eye is located by a + b = b . s s′ R 1 SET UP: na = 1.00 , nb = 1.35 . R > 0 . For a distant object, s ≈ ∞ and ≈ 0 . s 1.35 1.35 − 1.00 = and R = 0.648 cm = 6.48 mm . EXECUTE: (a) s ≈ ∞ and s′ = 2.5 cm : 2.5 cm R 1.00 1.35 1.35 − 1.00 1.35 + = (b) R = 0.648 cm and s = 25 cm : . = 0.500 and s′ = 2.70 cm = 27.0 mm . The s′ s′ 25 cm 0.648 image is formed behind the retina. 1.35 1.35 − 1.00 = (c) Calculate s′ for s ≈ ∞ and R = 0.50 cm : . s′ = 1.93 cm = 19.3 mm . The image is formed in s′ 0.50 cm front of the retina. EVALUATE: The cornea alone cannot achieve focus of both close and distant objects. n s′ n n n − na IDENTIFY: Apply a + b = b and m = − a to each surface. The overall magnification is m = m1m2 . The nb s s s′ R image formed by the first surface is the object for the second surface. SET UP: For the first surface, na = 1.00, nb = 1.60 and R = +15.0 cm . For the second surface, na = 1.60, nb = 1.00 and R =→ ∞ . n n n − na 1 1.60 0.60 EXECUTE: (a) a + b = b ⇒ + = ⇒ s′ = −36.9 cm. The object distance for the far end ′ ′ s s R 12.0 cm s 15.0 cm of the rod is 50.0 cm − (−36.9 cm) = 86.9 cm. The final image is 4.3 cm to the left of the vertex of the n n n − na 1.60 1 hemispherical surface. a + b = b ⇒ + = 0 ⇒ s′ = −54.3 cm. s s′ R 86.9 cm s′ (b) The magnification is the product of the two magnifications: n s′ −36.9 m1 = − a = − = 1.92, m2 = 1.00 ⇒ m = m1m2 = 1.92. nb s (1.60)(12.0) EVALUATE: The final image is virtual, erect and larger than the object.

Geometric Optics

34.87.

34.88.

34-25

IDENTIFY: Apply Eq.(34.11) to the image formed by refraction at the front surface of the sphere. SET UP: Let ng be the index of refraction of the glass. The image formation is shown in Figure 34.87. s=∞ s′ = +2r , where r is the radius of the sphere na = 1.00, nb = ng , R = + r Figure 34.87 na nb nb − na + = s s′ R 1 ng ng − 1.00 + = EXECUTE: ∞ 2r r n g ng 1 ng 1 = − ; = and ng = 2.00 2r r r 2r r EVALUATE: The required refractive index of the glass does not depend on the radius of the sphere. n n n − na to each surface. The image of the first surface is the object for the second IDENTIFY: Apply a + b = b ′ s s R surface. The relation between s1′ and s2 involves the length d of the rod. SET UP:

For the first surface, na = 1.00 , nb = 1.55 and R = +6.00 cm . For the second surface, na = 1.55 ,

nb = 1.00 and R = −6.00 cm . EXECUTE: We have images formed from both ends. From the first surface: na nb nb − na 1 1.55 0.55 + = ⇒ + = ⇒ s′ = 30.0 cm. s s′ R 25.0 cm s′ 6.00 cm This image becomes the object for the second end:

34.89.

d − 30.0 cm = 20.3 cm ⇒ d = 50.3 cm. EVALUATE: The final image is real. The first image is 20.3 cm to the right of the second surface and serves as a real object. IDENTIFY: The first lens forms an image which then acts as the object for the second lens. 1 1 1 s′ SET UP: The thin-lens equation is + = and the magnification is m = − . s s s′ f

EXECUTE:

34.90.

na nb nb − na 1.55 1 −0.55 + = ⇒ + = . s s′ R d − 30.0 cm 65.0 cm −6.00 cm

(a) For the first lens:

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −3.75 cm , to the left of the lens s s′ f 5.00 cm s′ −15.0 cm

(virtual image). (b) For the second lens, s = 12.0 cm + 3.75 cm = 15.75 cm. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 315 cm , or 332 cm from the object. s s′ f 15.75 cm s′ 15.0 cm (c) The final image is real. s′ (d) m = − , m1 = 0.750, m2 = −20.0, mtotal = −15.0 ⇒ y′ = −6.00 cm, inverted. s EVALUATE: Note that the total magnification is the product of the individual magnifications. ⎛1 1 ⎞ 1 IDENTIFY and SET UP: Use = ( n − 1) ⎜ − ⎟ to calculate the focal length of the lenses. The image formed f ⎝ R1 R2 ⎠ 1 1 1 sf by the first lens serves at the object for the second lens. mtot = m1m2 . + = gives s′ = . s s′ f s− f 1 1 1 ⎞ ⎛ EXECUTE: (a) = (0.60) ⎜ − ⎟ and f = +35.0 cm . f 12.0 cm 28.0 cm ⎝ ⎠ s f (45.0 cm)(35.0 cm) Lens 1: f1 = +35.0 cm . s1 = +45.0 cm . s1′ = 1 1 = = +158 cm . s1 − f1 45.0 cm − 35.0 cm s1′ 158 cm =− = −3.51 . y1′ = m1 y1 = (3.51)(5.00 mm) = 17.6 mm . The image of the first lens is 158 cm s1 45.0 cm to the right of lens 1 and is 17.6 mm tall. m1 = −

34-26

Chapter 34

(b) The image of lens 1 is 315 cm − 158 cm = 157 cm to the left of lens 2. f 2 = +35.0 cm . s2 = +157 cm .

s2 f 2 (157 cm)(35.0 cm) s′ 45.0 cm = −0.287 . = = +45.0 cm . m2 = − 2 = − s2 − f 2 157 cm − 35.0 cm s2 157 cm mtot = m1m2 = ( −3.51)(−0.287) = +1.00 . The final image is 45.0 cm to the right of lens 2. The final image is 5.00 s2′ =

34.91.

mm tall. mtot > 0 . So the final image is erect. EVALUATE: The final image is real. It is erect because each lens produces an inversion of the image, and two inversions return the image to the orientation of the object. IDENTIFY and SET UP: Apply Eq.(34.16) for each lens position. The lens to screen distance in each case is the image distance. There are two unknowns, the original object distance x and the focal length f of the lens. But each lens position gives an equation, so there are two equations for these two unknowns. The object, lens and screen before and after the lens is moved are shown in Figure 34.91. s = x; s′ = 30.0 cm 1 1 1 + = s s′ f 1 1 1 + = x 30.0 cm f

34.92.

34.93.

Figure 34.91 s = x + 4.00 cm; s′ = 22.0 cm 1 1 1 1 1 1 + = gives + = s s′ f x + 4.00 cm 22.0 cm f EXECUTE: Equate these two expressions for 1/f: 1 1 1 1 + = + x 30.0 cm x + 4.00 cm 22.0 cm 1 1 1 1 − = − x x + 4.00 cm 22.0 cm 30.0 cm x + 4.00 cm − x 30.0 − 22.0 4.00 cm 8 = and = x( x + 4.00 cm) 660 cm x( x + 4.00 cm) 660 cm 1 x 2 + (4.00 cm)x − 330 cm 2 = 0 and x = (−4.00 ± 16.0 + 4(330)) cm 2 1 x must be positive so x = (−4.00 + 36.55) cm = 16.28 cm 2 1 1 1 1 1 1 Then + = + = and f 16.28 cm 30.0 cm x 30.0 cm f f = +10.55 cm, which rounds to 10.6 cm. f > 0; the lens is converging. EVALUATE: We can check that s = 16.28 cm and f = 10.55 cm gives s′ = 30.0 cm and that s = (16.28 + 4.0) cm = 20.28 cm and f = 10.55 cm gives s′ = 22.0 cm. n n n − na IDENTIFY and SET UP: Apply a + b = b . s s′ R n n n − na n n n −n n n n −n ⇒ a + b = b a and a + b = b a . EXECUTE: (a) a + b = b ′ ′ s s R f ∞ R ∞ f R na nb − na nb nb − na na nb f = = . Therefore, and = and na / nb = . f R f′ R f f′ f′ na nb nb − na nb f nb nb (1 − f f ′) f f ′ f ′(1 − f f ′) f ′ − f + = ⇒ + = . Therefore, + = (b) = = 1. s s′ R sf ′ s′ R s s′ R R EVALUATE: For a thin lens the first and second focal lengths are equal. (a) IDENTIFY: Use Eq.(34.6) to locate the image formed by each mirror. The image formed by the first mirror serves as the object for the 2nd mirror.

Geometric Optics

SET UP:

34-27

The positions of the object and the two mirrors are shown in Figure 34.93a.

R = 0.360 m f = R / 2 = 0.180 m

Figure 34.93a EXECUTE: Image formed by convex mirror (mirror #1): convex means f1 = −0.180 m; s1 = L − x s1 f1 ( L − x )( −0.180 m) ⎛ 0.600 m − x ⎞ = = −(0.180 m) ⎜ ⎟ L = 0.600 m) or x = 0.24 m. (b) SET UP: Which mirror is #1 and which is #2 is now reversed form part (a). This is shown in Figure 34.93b.

Figure 34.93b EXECUTE: Image formed by concave mirror (mirror #1): concave means f1 = +0.180 m; s1 = x s f (0.180 m)x s1′ = 1 1 = s1 − f1 x − 0.180 m (0.180 m)x (0.180 m) x (0.420 m)x − 0.180 m 2 The image is to the left of mirror #1, so s2 = 0.600 m − = x − 0.180 m x − 0.180 m x − 0.180 m Image formed by convex mirror (mirror #2): convex means f 2 = −0.180 m rays return to the source means s′2 = L − x = 0.600 m − x 1 1 1 + = gives s s′ f x − 0.180 m 1 1 + =− 2 (0.420 m)x − 0.180 m 0.600 m − x 0.180 m

⎛ ⎞ x − 0.180 m 0.780 m − x = −⎜ ⎟ 2 2 (0.420 m)x − 0.180 m ⎝ 0.180 m − (0.180 m)x ⎠ 0.600 x 2 − (0.576 m)x + 0.1036 m 2 = 0 This is the same quadratic equation as obtained in part (a), so again x = 0.24 m.

34-28

34.94.

Chapter 34

EVALUATE: For x = 0.24 m the image is at the location of the source, both for rays that initially travel from the source toward the left and for rays that travel from the source toward the right. 1 1 1 sf + = gives s′ = IDENTIFY: , for both the mirror and the lens. s s′ f s− f SET UP: For the second image, the image formed by the mirror serves as the object for the lens. For the mirror, f m = +10.0 cm . For the lens, f = 32.0 cm . The center of curvature of the mirror is R = 2 f m = 20.0 cm to the right of the mirror vertex. EXECUTE: (a) The principal-ray diagrams from the two images are sketched in Figures 34.94a-b. In Figure 34.94b, only the image formed by the mirror is shown. This image is at the location of the candle so the principal ray diagram that shows the image formation when the image of the mirror serves as the object for the lens is analogous to that in Figure 34.94a and is not drawn. (b) Image formed by the light that passes directly through the lens: The candle is 85.0 cm to the left of the lens. sf (85.0 cm)(32.0 cm) s′ 51.3 cm s′ = = = +51.3 cm . m = − = − = −0.604 . This image is 51.3 cm to the right of s− f 85.0 cm − 32.0 cm s 85.0 cm the lens. s′ > 0 so the image is real. m < 0 so the image is inverted. Image formed by the light that first reflects off the mirror: First consider the image formed by the mirror. The candle is 20.0 cm to the right of the mirror, so sf (20.0 cm)(10.0 cm) s′ 20.0 cm = −1.00 . The image formed by s = +20.0 cm . s′ = = = 20.0 cm . m1 = − 1 = − s− f 20.0 cm − 10.0 cm s1 20.0 cm

the mirror is at the location of the candle, so s2 = +85.0 cm and s2′ = 51.3 cm . m2 = −0.604 . mtot = m1m2 =

(−1.00)(−0.604) = 0.604 . The second image is 51.3 cm to the right of the lens. s2′ > 0, so the final image is real. mtot > 0 , so the final image is erect. EVALUATE: The two images are at the same place. They are the same size. One is erect and one is inverted.

34.95.

Figure 34.94 na nb nb − na IDENTIFY: Apply to each case. + = s s′ R SET UP: s = 20.0 cm . R > 0 . Use s′ = +9.12 cm to find R. For this calculation, na = 1.00 and nb = 1.55 . Then

repeat the calculation with na = 1.33 . EXECUTE:

na nb nb − na 1.00 1.55 1.55 − 1.00 + = gives + = . R = 2.50 cm . s s′ R 20.0 cm 9.12 cm R

Geometric Optics

34-29

1.33 1.55 1.55 − 1.33 + = gives s′ = −72.1 cm . The image is 72.1 cm to the left of the surface vertex. 20.0 cm s′ 2.50 cm EVALUATE: With the rod in air the image is real and with the rod in water the image is virtual. 1 1 1 to each lens. The image formed by the first lens serves as the object for the second IDENTIFY: Apply + = s s′ f Then

34.96.

lens. The focal length of the lens combination is defined by

⎛1 1 ⎞ 1 1 1 1 = ( n − 1) ⎜ − ⎟ to + = . In part (b) use ′ s1 s2 f f ⎝ R1 R2 ⎠

calculate f for the meniscus lens and for the CCl 4 , treated as a thin lens. SET UP: With two lenses of different focal length in contact, the image distance from the first lens becomes exactly minus the object distance for the second lens. 1 1 1 1 1 1 1 1 1 1 ⎛1 1⎞ 1 1 + = + = ⎜ − ⎟ + = . But overall for the lens EXECUTE: (a) + = ⇒ = − and ′ ′ ′ ′ s2 s2 − s1 s′2 ⎝ s1 f1 ⎠ s′2 f 2 s1 s1 f1 s1 f1 s1 1 1 1 1 1 1 + = ⇒ = + . s1 s2′ f f f 2 f1 (b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact. All we need in order to calculate the system’s focal length is calculate the individual focal lengths, and then use the formula from part (a). ⎛1 ⎛ ⎞ 1 1 ⎞ 1 1 −1 = (nb − na ) ⎜ − ⎟ = (0.55)⎜ − For the meniscus lens ⎟ = 0.061 cm and f m = 16.4 cm . fm ⎝ 4.50 cm 9.00 cm ⎠ ⎝ R1 R2 ⎠ system,

For the CCl 4 :

⎛1 ⎛ 1 1 ⎞ 1 1⎞ = (nb − na )⎜ − ⎟ = (0.46)⎜ − ⎟ = 0.051 cm −1 and f w = 19.6 cm . fw ⎝ 9.00 cm ∞ ⎠ ⎝ R1 R2 ⎠

1 1 1 = + = 0.112 cm −1 and f = 8.93 cm . f fw fm f1 f 2 , so f for the combination is less than either f1 or f 2 . f1 + f 2 IDENTIFY: Apply Eq.(34.11) with R → ∞ to the refraction at each surface. For refraction at the first surface the point P serves as a virtual object. The image formed by the first refraction serves as the object for the second refraction. SET UP: The glass plate and the two points are shown in Figure 37.97. plane faces means R → ∞ and na nb + =0 s s′ n s′ = − b s na EVALUATE:

34.97.

f =

Figure 34.97 EXECUTE: refraction at the first (left-hand) surface of the piece of glass: The rays converging toward point P constitute a virtual object for this surface, so s = −14.4 cm. na = 1.00, nb = 1.60.

1.60 (−14.4 cm) = +23.0 cm 1.00 This image is 23.0 cm to the right of the first surface so is a distance 23.0 cm − t to the right of the second surface. This image serves as a virtual object for the second surface. refraction at the second (right-hand) surface of the piece of glass: n The image is at P′ so s′ = 14.4 cm + 0.30 cm − t = 14.7 cm − t. s = −(23.0 cm − t ); na = 1.60; nb = 1.00 s′ = − b s na s′ = −

⎛ 1.00 ⎞ gives 14.7 cm − t = − ⎜ ⎟ (−[23.0 cm − t ]). 14.7 cm − t = +14.4 cm − 0.625t. ⎝ 1.60 ⎠ 0.375t = 0.30 cm and t = 0.80 cm EVALUATE: The overall effect of the piece of glass is to diverge the rays and move their convergence point to the right. For a real object, refraction at a plane surface always produces a virtual image, but with a virtual object the image can be real.

34-30

Chapter 34

34.98.

IDENTIFY: SET UP:

na nb nb − na n n n −n and b + c = c b . + = s1 s1′ R1 s2 s2′ R2 ′ , n = n , and s = − s . na = nliq = nc b 1 2

Apply the two equations

⎛1 1 ⎞ n n − nliq n nliq nliq − n 1 1 1 1 1 + = + = = (n nliq − 1)⎜ − ⎟ . = + = and . s1 s′2 s s′ f ′ R R s1′ R1 − s1′ s2′ R2 ⎝ 1 2 ⎠ (b) Comparing the equations for focal length in and out of air we have: EXECUTE:

(a)

nliq s1

+

⎛ n − nliq ⎞ ⎡ n (n − 1) ⎤ f ( n − 1) = f ′(n nliq − 1) = f ′ ⎜ ⇒ f ′ = ⎢ liq ⎥ f. ⎜ nliq ⎟⎟ ⎢⎣ n − nliq ⎥⎦ ⎝ ⎠ EVALUATE:

When nliq = 1 , f ′ = f , as it should.

1 1 1 + = . s s′ f SET UP: The image formed by the converging lens is 30.0 cm from the converging lens, and becomes a virtual object for the diverging lens at a position 15.0 cm to the right of the diverging lens. The final image is projected 15 cm + 19.2 cm = 34.2 cm from the diverging lens. 1 1 1 1 1 1 EXECUTE: + = ⇒ + = ⇒ f = −26.7 cm. s s′ f −15.0 cm 34.2 cm f EVALUATE: Our calculation yields a negative value of f, which should be the case for a diverging lens. 34.100. IDENTIFY: The spherical mirror forms an image of the object. It forms another image when the image of the plane mirror serves as an object. SET UP: For the convex mirror f = −24.0 cm . The image formed by the plane mirror is 10.0 cm to the right of the plane mirror, so is 20.0 cm + 10.0 cm = 30.0 cm from the vertex of the spherical mirror. EXECUTE: The first image formed by the spherical mirror is the one where the light immediately strikes its surface, without bouncing from the plane mirror. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −7.06 cm, and the image height s s′ f 10.0 cm s′ −24.0 cm s′ −7.06 is y′ = − y = − (0.250 cm) = 0.177 cm. s 10.0 The second image is of the plane mirror image is located 30.0 cm from the vertex of the spherical mirror. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −13.3 cm and the image height is s s′ f 30.0 cm s′ −24.0 cm s′ −13.3 y′ = − y = − (0.250 cm) = 0.111 cm. s 30.0 EVALUATE: Other images are formed by additional reflections from the two mirrors. 34.101. IDENTIFY: In the sketch in Figure 34.101 the light travels upward from the object. Apply Eq.(34.11) with R → ∞ to the refraction at each surface. The image formed by the first surface serves as the object for the second surface. SET UP: The locations of the object and the glass plate are shown in Figure 34.101. 34.99.

IDENTIFY:

Apply

For a plane (flat) surface n n R → ∞ so a + b = 0 s s′ n s′ = − b s na Figure 34.101 EXECUTE:

First refraction (air → glass):

na = 1.00; nb = 1.55; s = 6.00 cm nb 1.55 (6.00 cm) = −9.30 cm s=− na 1.00 The image is 9.30 cm below the lower surface of the glass, so is 9.30 cm + 3.50 cm = 12.8 cm below the upper surface. s′ = −

Geometric Optics

34-31

Second refraction (glass → air): na = 1.55; nb = 1.00; s = +12.8 cm nb 1.00 (12.8 cm) = −8.26 cm s=− na 1.55 The image of the page is 8.26 cm below the top surface of the glass plate and therefore 9.50 cm − 8.26 cm = 1.24 cm above the page. EVALUATE: The image is virtual. If you view the object by looking down from above the plate, the image of the page that you see is closer to your eye than the page is. 34.102. IDENTIFY: Light refracts at the front surface of the lens, refracts at the glass-water interface, reflects from the plane mirror and passes through the two interfaces again, now traveling in the opposite direction. SET UP: Use the focal length in air to find the radius of curvature R of the lens surfaces. ⎛1 1 1 ⎞ 1 ⎛2⎞ = ( n − 1)⎜ − ⎟ ⇒ = 0.52 ⎜ ⎟ ⇒ R = 41.6 cm. EXECUTE: (a) f R R 40 cm ⎝R⎠ ⎝ 1 2 ⎠ s′ = −

At the air–lens interface:

na nb nb − na 1 1.52 0.52 + = ⇒ + = and s1′ = −851 cm and s2 = 851 cm. 70.0 cm 41.6 cm s s′ R s1′

−0.187 1.52 1.33 and s′2 = 491 cm . + = 851 cm s2′ −41.6 cm The mirror reflects the image back (since there is just 90 cm between the lens and mirror.) So, the position of the image is 401 cm to the left of the mirror, or 311 cm to the left of the lens. 1.33 1.52 0.187 and s3′ = +173 cm . At the water–lens interface: ⇒ + = 41.6 cm −311 cm s3′

At the lens–water interface: ⇒

At the lens–air interface: ⇒

1.52 1 −0.52 and s′4 = +47.0 cm , to the left of lens. + = −173 cm s4′ −41.6 cm

⎛ n s′ ⎞⎛ n s′ ⎞⎛ n s′ ⎞⎛ n s′ ⎞ ⎛ −851 ⎞⎛ 491 ⎞⎛ +173 ⎞⎛ +47.0 ⎞ m = m1m2 m3m4 = ⎜ a1 1 ⎟⎜ a 2 2 ⎟⎜ a 3 3 ⎟⎜ a 4 4 ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = −1.06. ⎝ nb1s1 ⎠⎝ nb 2 s2 ⎠⎝ nb 3 s3 ⎠⎝ nb 4 s4 ⎠ ⎝ 70 ⎠⎝ −851 ⎠⎝ −311 ⎠⎝ −173 ⎠ (Note all the indices of refraction cancel out.) (b) The image is real. (c) The image is inverted. (d) The final height is y′ = my = (1.06)(4.00 mm) = 4.24 mm. EVALUATE: The final image is real even though it is on the same side of the lens as the object! 34.103. IDENTIFY: The camera lens can be modeled as a thin lens that forms an image on the film. 1 1 1 s′ SET UP: The thin-lens equation is + = , and the magnification of the lens is m = − . s s′ f s s′ y′ 1 (0.0360 m) ⇒ s′ = (7.50 × 10−4 ) s , EXECUTE: (a) m = − = = s y 4 (12.0 m)

1 1 1 1 1⎛ 1 1 ⎞ 1 + = + = ⎜1 + ⇒ s = 46.7 m . ⎟= = s s′ s (7.50 × 10−4 ) s s ⎝ 7.50 × 10−4 ⎠ f 0.0350 m (b) To just fill the frame, the magnification must be 3.00 × 10−3 so:

1⎛ 1 1 ⎞ 1 ⇒ s = 11.7 m . ⎜1 + ⎟= = s ⎝ 3.00 × 10−3 ⎠ f 0.0350 m Since the boat is originally 46.7 m away, the distance you must move closer to the boat is 46.7 m – 11.7 m = 35.0 m. EVALUATE: This result seems to imply that if you are 4 times as far, the image is ¼ as large on the film. However this result is only an approximation, and would not be true for very close distances. It is a better approximation for large distances. 1 1 1 s′ 34.104. IDENTIFY: Apply + = and m = − . s s′ f s ′ SET UP: s + s = 18.0 cm 1 1 1 + = . ( s′) 2 − (18.0 cm) s′ + 54.0 cm 2 = 0 so s′ = 14.2 cm or 3.80 cm . 18.0 cm − s′ s′ 3.00 cm s = 3.80 cm or 14.2 cm , so the screen must either be 3.80 cm or 14.2 cm from the object.

EXECUTE:

(a)

34-32

Chapter 34

s′ 3.80 s′ 14.2 =− = −0.268. s = 14.2 cm : m = − = − = −3.74. s s 14.2 3.80 EVALUATE: Since the image is projected onto the screen, the image is real and s′ is positive. We assumed this when we wrote the condition s + s′ = 18.0 cm . 34.105. IDENTIFY: Apply Eq.(34.16) to calculate the image distance for each lens. The image formed by the 1st lens serves as the object for the 2nd lens, and the image formed by the 2nd lens serves as the object for the 3rd lens. SET UP: The positions of the object and lenses are shown in Figure 34.105. 1 1 1 + = s s′ f (b) s = 3.80 cm : m = −

1 1 1 s− f = − = s′ f s sf s′ =

sf s− f

Figure 34.105 EXECUTE: lens #1 s = +80.0 cm; f = +40.0 cm

sf (+80.0 cm)( +40.0 cm) = = +80.0 cm +80.0 cm − 40.0 cm s− f The image formed by the first lens is 80.0 cm to the right of the first lens, so it is 80.0 cm − 52.0 cm = 28.0 cm to the right of the second lens. lens #2 s = −28.0 cm; f = +40.0 cm s′ =

sf (−28.0 cm)( +40.0 cm) = = +16.47 cm s− f −28.0 cm − 40.0 cm The image formed by the second lens is 16.47 cm to the right of the second lens, so it is 52.0 cm − 16.47 cm = 35.53 cm to the left of the third lens. lens #3 s = +35.53 cm; f = +40.0 cm s′ =

sf (+35.53 cm)( +40.0 cm) = = −318 cm s− f +35.53 cm − 40.0 cm The final image is 318 cm to the left of the third lens, so it is 318 cm − 52 cm − 52 cm − 80 cm = 134 cm to the left of the object. EVALUATE: We used the separation between the lenses and the sign conventions for s and s′ to determine the object distances for the 2nd and 3rd lenses. The final image is virtual since the final s′ is negative. 1 1 1 34.106. IDENTIFY: Apply + = and calculate s′ for each s. s s′ f SET UP: f = 90 mm s′ =

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 96.7 mm. s s′ f 1300 mm s′ 90 mm 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 91.3 mm. s s′ f 6500 mm s′ 90 mm ⇒ Δ s′ = 96.7 mm − 91.3 mm = 5.4 mm toward the film sf EVALUATE: s′ = . For f > 0 and s > f , s′ decreases as s increases. s− f near point near point 34.107. IDENTIFY and SET UP: The generalization of Eq.(34.22) is M = . , so f = f M EXECUTE: (a) age 10, near point = 7 cm 7 cm f = = 3.5 cm 2.0 (b) age 30, near point = 14 cm 14 cm f = = 7.0 cm 2.0 EXECUTE:

Geometric Optics

34-33

(c) age 60, near point = 200 cm 200 cm f = = 100 cm 2.0 (d) f = 3.5 cm (from part (a)) and near point = 200 cm (for 60-year-old)

200 cm = 57 3.5 cm (e) EVALUATE: No. The reason f = 3.5 cm gives a larger M for a 60-year-old than for a 10-year-old is that the eye of the older person can’t focus on as close an object as the younger person can. The unaided eye of the 60year-old must view a much smaller angular size, and that is why the same f gives a much larger M. The angular size of the image depends only on f and is the same for the two ages. 1 1 1 θ′ 34.108. IDENTIFY: Use + = to calculate s that gives s′ = −25 cm . M = . θ s s′ f M=

SET UP: EXECUTE:

y y and θ = . s 25 cm 1 1 1 1 1 1 f (25 cm) (a) + = ⇒ + . = ⇒s= ′ s s f s −25 cm f f + 25 cm

Let the height of the object be y , so θ ′ =

⎛ y ( f + 25 cm) ⎞ y ( f + 25 cm) ⎛ y⎞ . (b) θ ′ = arctan ⎜ ⎟ = arctan ⎜ ⎟≈ f (25 cm) ⎝s⎠ ⎝ f (25 cm) ⎠ θ ′ y ( f + 25 cm) 1 f + 25 cm = . (c) M = = θ f (25 cm) y / 25 cm f (d) If f = 10 cm ⇒ M =

10 cm + 25 cm = 3.5. This is 1.4 times greater than the magnification obtained if the image 10 cm

25 cm = 2.5). f EVALUATE: (e) Having the first image form just within the focal length puts one in the situation described above, where it acts as a source that yields an enlarged virtual image. If the first image fell just outside the second focal point, then the image would be real and diminished. 1 1 1 34.109. IDENTIFY: Apply + = . The near point is at infinity, so that is where the image must be formed for any s s′ f objects that are close. 1 SET UP: The power in diopters equals , with f in meters. f if formed at infinity ( M ∞ =

1 1 1 1 1 1 = + = + = = 4.17 diopters. f s s′ 24 cm −∞ 0.24 m EVALUATE: To focus on closer objects, the power must be increased. n n n − na 34.110. IDENTIFY: Apply a + b = b . s s′ R SET UP: na = 1.00 , nb = 1.40 . EXECUTE:

1 1.40 0.40 + = ⇒ s′ = 2.77 cm. s′ 36.0 cm 0.75 cm EVALUATE: This distance is greater than the normal eye, which has a cornea vertex to retina distance of about 2.6 cm. 34.111. IDENTIFY: Use similar triangles in Figure 34.63 in the textbook and Eq.(34.16) to derive the expressions called for in the problem. (a) SET UP: The effect of the converging lens on the ray bundle is sketched in Figure 34.111. EXECUTE: From similar triangles in Figure 34.111a, r0 r′ = 0 f1 f1 − d EXECUTE:

Figure 34.111a

34-34

Chapter 34

⎛ f −d ⎞ Thus r0′ = ⎜ 1 ⎟ r0 , as was to be shown. ⎝ f1 ⎠ (b) SET UP: The image at the focal point of the first lens, a distance f1 to the right of the first lens, serves as the

object for the second lens. The image is a distance f1 − d to the right of the second lens, so s2 = −( f1 − d ) = d − f1. EXECUTE:

s′2 =

s2 f 2 (d − f1 ) f 2 = s2 − f 2 d − f1 − f 2

( f1 − d ) f 2 , as was to be shown. f 2 − f1 + d (c) SET UP: The effect of the diverging lens on the ray bundle is sketched in Figure 34.111b. f 2 < 0 so f 2 = − f 2 and s′2 =

From similar r r′ triangles in the sketch, 0 = 0 f s2′

EXECUTE:

Thus

r0 f = r0′ s2′

Figure 34.111b

From the results of part (a),

r0 f1 f1 f . Combining the two results gives = = f1 − d s2′ r0′ f1 − d

( f1 − d ) f 2 f1 f1 f 2 ⎛ f ⎞ = f = s2′ ⎜ 1 ⎟ = , as was to be shown. − − + − f d f f d f d f ( ) ) 1 1 2 − f1 + d ⎝ 1 ⎠ ( 2 (d) SET UP: Put the numerical values into the expression derived in part (c). f1 f 2 EXECUTE: f = f 2 − f1 + d

216 cm 2 6.0 cm + d d = 0 gives f = 36.0 cm; maximum f d = 4.0 cm gives f = 21.6 cm; minimum f f1 = 12.0 cm, f 2 = 18.0 cm, so f =

216 cm 2 6.0 cm + d 6.0 cm + d = 7.2 cm and d = 1.2 cm EVALUATE: Changing d produces a range of effective focal lengths. The effective focal length can be both smaller and larger than f1 + f 2 . f = 30.0 cm says 30.0 cm =

34.112. IDENTIFY: SET UP:

M =

y′ y′ y′ f θ′ . θ = 1 , and θ ′ = 2 . This gives M = 2 . 1 . θ f1 s′2 s2′ y1′

Since the image formed by the objective is used as the object for the eyepiece, y1′ = y2 .

y′2 f1 y′ f s′ f f f 48.0 cm = 2 . 1 = 2 . 1 = 1 . Therefore, s2 = 1 = . = 1.33 cm, and this is just M 36 s′2 y2 y2 s′2 s2 s′2 s2 outside the eyepiece focal point. Now the distance from the mirror vertex to the lens is f1 + s2 = 49.3 cm, and so

EXECUTE:

M =

−1

⎛ 1 1 1 1 1 ⎞ + = ⇒ s2′ = ⎜ − ⎟ = 12.3 cm. Thus we have a final image which is real and 12.3 cm from s2 s2′ f 2 1.20 cm 1.33 cm ⎠ ⎝ the eyepiece. (Take care to carry plenty of figures in the calculation because two close numbers are subtracted.) EVALUATE: Eq.(34.25) gives M = 40 , somewhat larger than M for this telescope. 34.113. IDENTIFY and SET UP: The image formed by the objective is the object for the eyepiece. The total lateral magnification is mtot = m1m2 . f1 = 8.00 mm (objective); f 2 = 7.50 cm (eyepiece)

Geometric Optics

34-35

(a) The locations of the object, lenses and screen are shown in Figure 34.113.


Find the object distance s1 for the objective:

s1′ = +18.0 cm, f1 = 0.800 cm, s1 = ?

1 1 1 1 1 1 s1′ − f1 + = , so = − = s1 s1′ f1 s1 f1 s1′ s1′ f1 ′ s f (18.0 cm)(0.800 cm) = 0.8372 cm s1 = 1 1 = s1′ − f1 18.0 cm − 0.800 cm Find the object distance s2 for the eyepiece: s2′ = +200 cm, f 2 = 7.50 cm, s2 = ?

1 1 1 + = s2 s′2 f 2 s′ f (200 cm)(7.50 cm) = 7.792 cm s2 = 2 2 = s′2 − f 2 200 cm − 7.50 cm Now we calculate the magnification for each lens: s′ 18.0 cm m1 = − 1 = − = −21.50 0.8372 cm s1 s′ 200 cm = −25.67 m2 = − 2 = − s2 7.792 cm mtot = m1m2 = (−21.50)( −25.67) = 552. (b) From the sketch we can see that the distance between the two lenses is s1′ + s2 = 18.0 cm + 7.792 cm = 25.8 cm. EVALUATE: The microscope is not being used in the conventional way; it merely serves as a two-lens system. In particular, the final image formed by the eyepiece in the problem is real, not virtual as is the case normally for a microscope. Eq.(34.23) does not apply here, and in any event gives the angular not the lateral magnification. u′ 34.114. IDENTIFY: For u and u′ as defined in Figure 34.64 in the textbook, M = . u SET UP: f 2 is negative. From Figure 34.64, the length of the telescope is f1 + f 2 . u′ f y y y EXECUTE: (a) From the figure, u = and u′ = = − . The angular magnification is M = = − 1 . f1 f2 f2 u f2 f1 f1 95.0 cm (b) M = − ⇒ f 2 = − =− = −15.0 cm. f2 M 6.33 (c) The length of the telescope is 95.0 cm − 15.0 cm = 80.0 cm, compared to the length of 110 cm for the telescope in Exercise 34.57. EVALUATE: An advantage of this construction is that the telescope is somewhat shorter. 1 1 1 34.115. IDENTIFY: Use + = to calculate s′ (the distance of each point from the lens), for points A, B and C. s s′ f SET UP: The object and lens are shown in Figure 34.115a. 1 1 1 1 1 1 + = ⇒ s′ = 36.0 cm. EXECUTE: (a) For point C : + = ⇒ ′ ′ s s f 45.0 cm s 20.0 cm s′ 36.0 y′ = − y = − (15.0 cm) = −12.0 cm , so the image of point C is 36.0 cm to the right of the lens, and s 45.0 12.0 cm below the axis. For point A: s = 45.0 cm + 8.00 cm(cos 45°) = 50.7 cm . 1 1 1 1 1 1 s′ 33.0 + = ⇒ + = ⇒ s′ = 33.0 cm. y′ = − y = − (15.0 cm − 8.00 cm(sin 45°)) = −6.10 cm, s s s′ f 45.0 50.7 cm s′ 20.0 cm so the image of point A is 33.0 cm to the right of the lens, and 6.10 cm below the axis.

34-36

Chapter 34

For point B: s = 45.0 cm − 8.00 cm(cos 45°) = 39.3 cm .

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 40.7 cm. ′ ′ s s f 39.3 cm s 20.0 cm

s′ 40.7 y=− (15.0 cm + 8.00 cm(sin 45°)) = −21.4 cm, so the image of point B is 40.7 cm to the right of the s 39.3 lens, and 21.4 cm below the axis. The image is shown in Figure 34.115b. (b) The length of the pencil is the distance from point A to B: L = ( x A − xB ) 2 + ( y A − yB ) 2 = (33.0 cm − 40.7 cm) 2 + (6.10 cm − 21.4 cm) 2 = 17.1 cm y′ = −

EVALUATE:

The image is below the optic axis and is larger than the object.

Figure 34.115 34.116. IDENTIFY and SET UP:

Consider the ray diagram drawn in Figure 34.116. h sin θ sin α EXECUTE: (a) Using the diagram and law of sines, = but sin θ = = sin α (law of (R − f ) g R R2 R reflection), g = ( R − f ). Bisecting the triangle: cosθ = ⇒ R cosθ − f cosθ = . (R − f ) 2 R⎡ 1 ⎤ 1 ⎤ R ⎡ f = ⎢2 − = f0 ⎢ 2 − ⎥ . f 0 = 2 is the value of f for θ near zero (incident ray near the axis). When θ 2⎣ cosθ ⎥⎦ cos θ ⎣ ⎦ increases, (2 − 1/ cosθ ) decreases and f decreases. f − f0 f 1 1 = −0.02 ⇒ = 0.98 so 2 − = 0.98 . cosθ = (b) = 0.98 and θ = 11.4°. f0 f0 cosθ 2 − 0.98 EVALUATE: For θ = 45° , f = 0.586 f 0 , and f approaches zero as θ approaches 60° .

Figure 34.116

Geometric Optics

34-37

34.117. IDENTIFY: The distance between image and object can be calculated by taking the derivative of the separation distance and minimizing it. SET UP: For a real image s′ > 0 and the distance between the object and the image is D = s + s′ . For a real image must have s > f . EXECUTE:

D = s + s′ but s′ =

sf sf s2 ⇒D=s+ = . s− f s− f s− f

dD d ⎛ s 2 ⎞ 2s s2 s 2 − 2sf = ⎜ − = = 0 . s 2 − 2sf = 0 . s ( s − 2 f ) = 0 . s = 2 f is the solution for which ⎟= ds ds ⎝ s − f ⎠ s − f ( s − f ) 2 ( s − f ) 2 s > f . For s = 2 f , s′ = 2 f . Therefore, the minimum separation is 2 f + 2 f = 4 f . (b) A graph of D / f versus s / f is sketched in Figure 34.117. Note that the minimum does occur for D = 4 f . EVALUATE: If, for example, s = 3 f / 2 , then s′ = 3 f and D = s + s′ = 4.5 f , greater than the minimum value.

Figure 34.117 34.118. IDENTIFY and SET UP: For a plane mirror, s′ = − s . EXECUTE: (a) By the symmetry of image production, any image must be the same distance D as the object from the mirror intersection point. But if the images and the object are equal distances from the mirror intersection, they lie on a circle with radius equal to D. (b) The center of the circle lies at the mirror intersection as discussed above. (c) The diagram is sketched in Figure 34.118. EVALUATE: To see the image, light from the object must be able to reflect from each mirror and reach the person's eyes.

Figure 34.118 na nb nb − na + = to refraction at the cornea to find where the object for the cornea must be in 34.119. IDENTIFY: Apply s s′ R 1 1 1 order for the image to be at the retina. Then use + = to calculate f so that the lens produces an image of a s s′ f distant object at this point. SET UP: For refraction at the cornea, na = 1.33 and nb = 1.40 . The distance from the cornea to the retina in this model of the eye is 2.60 cm. From Problem 34.46, R = 0.71 cm . EXECUTE: (a) People with normal vision cannot focus on distant objects under water because the image is unable to be focused in a short enough distance to form on the retina. Equivalently, the radius of curvature of the normal eye is about five or six times too great for focusing at the retina to occur. (b) When introducing glasses, let’s first consider what happens at the eye: na nb nb − na 1.33 1.40 0.07 + = ⇒ + = ⇒ s2 = −3.02 cm. That is, the object for the cornea must be 3.02 cm s2 s′2 R s2 2.6 cm 0.71 cm behind the cornea. Now, assume the glasses are 2.00 cm in front of the eye, so s1′ = 2.00 cm + s2 = 5.02 cm .

34-38

Chapter 34

1 1 1 1 1 1 + = gives + = and f1′ = 5.02 cm. This is the focal length in water, but to get it in air, we use s1 s1′ f1′ ∞ 5.02 cm f1′ ⎡ n − nliq ⎤ ⎡ 1.52 − 1.333 ⎤ the formula from Problem 34.98: f1 = f1′⎢ ⎥ = (5.02 cm) ⎢ ⎥ = 1.35 cm . ⎢⎣ nliq (n − 1) ⎥⎦ ⎣1.333(1.52 − 1) ⎦ EVALUATE: A converging lens is needed.

35

INTERFERENCE

35.1.

35.2.

IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interference the path difference is mλ , m = 0, ± 1, ± 2, . . . Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. (b) For destructive interference the path difference is (m + 12 )λ , m = 0, ± 1, ± 2, . . . A path difference of ± λ / 2 = 3.00 m is possible but a path difference as large as 3λ / 2 = 9.00 m is not possible. For a point a distance x from A and 5.00 − x from B the path difference is x − (5.00 m − x). x − (5.00 m − x) = +3.00 m gives x = 4.00 m. x − (5.00 m − x) = −3.00 m gives x = 1.00 m . EVALUATE: The point of constructive interference is midway between the points of destructive interference. IDENTIFY: For destructive interference the path difference is (m + 12 )λ , m = 0, ±1, ±2, … . The longest wavelength is for m = 0 . For constructive interference the path difference is mλ , m = 0, ±1, ±2, … The longest wavelength is for m = 1 . SET UP: The path difference is 120 m.

35.3.

λ

= 120 m ⇒ λ = 240 m. 2 (b) The longest wavelength for constructive interference is λ = 120 m. EVALUATE: The path difference doesn't depend on the distance of point Q from B. IDENTIFY: Use c = f λ to calculate the wavelength of the transmitted waves. Compare the difference in the distance from A to P and from B to P. For constructive interference this path difference is an integer multiple of the wavelength. SET UP: Consider Figure 35.3 The distance of point P from each coherent source is rA = x and EXECUTE:

(a) For destructive interference

rB = 9.00 m − x. Figure 35.3 EXECUTE:

The path difference is rB − rA = 9.00 m − 2 x.

rB − rA = mλ , m = 0, ± 1, ± 2, …

λ=

c 2.998 × 108 m/s = = 2.50 m f 120 × 106 Hz

9.00 m − m(2.50 m) = 4.50 m − (1.25 m)m. x must lie in the range 0 to 2 9.00 m since P is said to be between the two antennas. m = 0 gives x = 4.50 m m = +1 gives x = 4.50 m − 1.25 m = 3.25 m m = +2 gives x = 4.50 m − 2.50 m = 2.00 m m = +3 gives x = 4.50 m − 3.75 m = 0.75 m m = −1 gives x = 4.50 m + 1.25 m = 5.75 m m = −2 gives x = 4.50 m + 2.50 m = 7.00 m m = −3 gives x = 4.50 m + 3.75 m = 8.25 m

Thus 9.00 m − 2 x = m(2.50 m) and x =

35-1

35-2

35.4.

Chapter 35

All other values of m give values of x out of the allowed range. Constructive interference will occur for x = 0.75 m, 2.00 m, 3.25 m, 4.50 m, 5.75 m, 7.00 m, and 8.25 m. EVALUATE: Constructive interference occurs at the midpoint between the two sources since that point is the same distance from each source. The other points of constructive interference are symmetrically placed relative to this point. IDENTIFY: For constructive interference the path difference d is related to λ by d = mλ , m = 0,1,2, … For destructive interference d = ( m + 12 )λ , m = 0,1,2, … SET UP: d = 2040 nm EXECUTE: (a) The brightest wavelengths are when constructive interference occurs: d 2040 nm 2040 nm d = mλm ⇒ λm = ⇒ λ3 = = 680 nm, λ4 = = 510 nm and m 3 4 2040 nm λ5 = = 408 nm. 5 (b) The path-length difference is the same, so the wavelengths are the same as part (a). d 2040 nm (c) d = ( m + 12 )λm so λm = . The visible wavelengths are λ3 = 583 nm and λ4 = 453 nm . = m + 12 m + 12

35.5.

EVALUATE: The wavelengths for constructive interference are between those for destructive interference. IDENTIFY: If the path difference between the two waves is equal to a whole number of wavelengths, constructive interference occurs, but if it is an odd number of half-wavelengths, destructive interference occurs. SET UP: We calculate the distance traveled by both waves and subtract them to find the path difference. EXECUTE: Call P1 the distance from the right speaker to the observer and P2 the distance from the left speaker to the observer. (a) P1 = 8.0 m and P2 = (6.0 m) 2 + (8.0 m) 2 = 10. 0 m . The path distance is

ΔP = P2 − P1 = 10.0 m – 8.0 m = 2.0 m (b) The path distance is one wavelength, so constructive interference occurs.

35.6.

(c) P1 = 17.0 m and P2 = (6.0 m) 2 + (17.0 m) 2 = 18.0 m . The path difference is 18.0 m – 17.0 m = 1.0 m, which is one-half wavelength, so destructive interference occurs. EVALUATE: Constructive interference also occurs if the path difference 2 λ , 3 λ , 4 λ , etc., and destructive interference occurs if it is λ /2, 3 λ /2, 5 λ /2, etc. IDENTIFY: At an antinode the interference is constructive and the path difference is an integer number of wavelengths; path difference = mλ , m = 0, ±1, ±2, … at an antinode. SET UP: The maximum magnitude of the path difference is the separation d between the two sources. EXECUTE: (a) At S1 , r2 − r1 = 4λ , and this path difference stays the same all along the y -axis, so

m = +4. At S 2 , r2 − r1 = −4λ , and the path difference below this point, along the negative y-axis, stays the same, so m = −4. (b) The wave pattern is sketched in Figure 35.6. d (c) The maximum and minimum m-values are determined by the largest integer less than or equal to .

λ

1 (d) If d = 7 λ ⇒ −7 ≤ m ≤ +7, so there will be a total of 15 antinodes between the sources. 2 EVALUATE: We are considering points close to the two sources and the antinodal curves are not straight lines.

Figure 35.6

Interference

35.7.

35.8.

IDENTIFY: At an antinodal point the path difference is equal to an integer number of wavelengths. SET UP: For m = 3 , the path difference is 3λ . EXECUTE: Measuring with a ruler from both S1 and S 2 to the different points in the antinodal line labeled m = 3 , we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diagram. EVALUATE: There is a whole curve of points where the path difference is 3λ . mλ IDENTIFY: The value of y20 is much smaller than R and the approximate expression ym = R is accurate. d SET UP: y20 = 10.6 × 10−3 m .

d=

EXECUTE:

20 Rλ (20)(1.20 m)(502 × 10−9 m) = = 1.14 × 10−3 m = 1.14 mm y20 10.6 × 10−3 m

y20 so θ 20 = 0.51° and the approximation sin θ 20 ≈ tan θ 20 is very accurate. R IDENTIFY and SET UP: The dark lines correspond to destructive interference and hence are located by Eq.(35.5): 1⎞ ⎛ ⎜ m + ⎟λ 1⎞ 2⎠ ⎛ ⎝ , m = 0, ± 1, ± 2,… d sin θ = ⎜ m + ⎟ λ so sinθ = 2⎠ d ⎝ Solve for θ that locates the second and third dark lines. Use y = R tan θ to find the distance of each of the dark lines from the center of the screen. EXECUTE: 1st dark line is for m = 0 3λ 3(500 × 10−9 m) 2nd dark line is for m = 1 and sin θ1 = = = 1.667 × 10−3 and θ1 = 1.667 × 10−3 rad 2d 2(0.450 × 10−3 m) EVALUATE:

35.9.

35-3

tan θ 20 =

3rd dark line is for m = 2 and sin θ 2 =

5λ 5(500 × 10−9 m) = = 2.778 × 10−3 and θ 2 = 2.778 × 10−3 rad 2d 2(0.450 × 10−3 m)

(Note that θ1 and θ 2 are small so that the approximation θ ≈ sin θ ≈ tan θ is valid.) The distance of each dark line from the center of the central bright band is given by ym = R tan θ , where R = 0.850 m is the distance to the screen. tan θ ≈ θ so ym = Rθ m

y1 = Rθ1 = (0.750 m)(1.667 × 10−3 rad) = 1.25 × 10−3 m y2 = Rθ 2 = (0.750 m)(2.778 × 10−3 rad) = 2.08 × 10−3 m Δy = y2 − y1 = 2.08 × 10−3 m − 1.25 × 10−3 m = 0.83 mm EVALUATE:

35.10.

1⎞ ⎛ interference: ym = R ⎜ m + ⎟ λ / d . 2⎠ ⎝ IDENTIFY: Since the dark fringes are eqully spaced, R " ym , the angles are small and the dark bands are located by ym + 1 = R 2

SET UP:

(m + 12 )λ . d

The separation between adjacent dark bands is Δy =

Rλ . d

Rλ Rλ (1.80 m) (4.50 × 10−7 m) ⇒d = = = 1.93 × 10−4 m = 0.193 m. d Δy 4.20 × 10−3 m EVALUATE: When the separation between the slits decreases, the separation between dark fringes increases. IDENTIFY and SET UP: The positions of the bright fringes are given by Eq.(35.6): ym = R (mλ / d ). For each EXECUTE:

35.11.

Since θ1 and θ 2 are very small we could have used Eq.(35.6), generalized to destructive

Δy =

fringe the adjacent fringe is located at ym +1 = R(m + 1)λ / d . Solve for λ. EXECUTE:

The separation between adjacent fringes is Δy = ym +1 − ym = Rλ / d .

d Δy (0.460 × 10−3 m)(2.82 × 10−3 m) = = 5.90 × 10−7 m = 590 nm R 2.20 m EVALUATE: Eq.(35.6) requires that the angular position on the screen be small. The angular position of bright fringes is given by sin θ = mλ / d . The slit separation is much larger than the wavelength (λ / d = 1.3 × 10−3 ), so θ is small so long as m is not extremely large.

λ=

35-4

Chapter 35

35.12.

The width of a bright fringe can be defined to be the distance between its two adjacent destructive ( m + 12 )λ minima. Assuming the small angle formula for destructive interference ym = R . d SET UP: d = 0.200 × 10−3 m . R = 4.00 m . EXECUTE: The distance between any two successive minima λ (400 × 10−9 m) is ym +1 − ym = R = (4.00 m) = 8.00 mm. Thus, the answer to both part (a) and part (b) is that the (0.200 × 10−3 m) d width is 8.00 mm. EVALUATE: For small angles, when ym # R , the interference minima are equally spaced.

35.13.

IDENTIFY:

1⎞ ⎛ The dark lines are located by d sin θ = ⎜ m + ⎟ λ . The distance of each line from the 2⎠ ⎝ center of the screen is given by y = R tan θ . IDENTIFY and SET UP:

First dark line is for m = 0 and d sin θ1 = λ / 2.

EXECUTE:

sin θ1 =

λ 2d

=

550 × 10−9 m = 0.1528 and θ1 = 8.789°. Second dark line is for m = 1 and d sin θ 2 = 3λ / 2. 2(1.80 × 10−6 m)

⎛ 550 × 10−9 m ⎞ 3λ = 3⎜ ⎟ = 0.4583 and θ 2 = 27.28°. −6 2d ⎝ 2(1.80 × 10 m) ⎠ y1 = R tan θ1 = (0.350 m) tan8.789° = 0.0541 m

sin θ 2 =

y2 = R tan θ 2 = (0.350 m) tan 27.28° = 0.1805 m The distance between the lines is Δy = y2 − y1 = 0.1805 m − 0.0541 m = 0.126 m = 12.6 cm.

35.14.

EVALUATE: sin θ1 = 0.1528 and tan θ1 = 0.1546. sin θ 2 = 0.4583 and tan θ 2 = 0.5157. As the angle increases, sin θ ≈ tan θ becomes a poorer approximation. mλ IDENTIFY: Using Eq.(35.6) for small angles: ym = R . d SET UP: First-order means m = 1 . EXECUTE: The distance between corresponding bright fringes is

Δy =

35.15.

Rm (5.00 m)(1) (660 − 470) × (10−9 m) = 3.17 mm. Δλ = d (0.300 × 10−3 m)

EVALUATE: The separation between these fringes for different wavelengths increases when the slit separation decreases. IDENTIFY and SET UP: Use the information given about the bright fringe to find the distance d between the two slits. Then use Eq.(35.5) and y = R tan θ to calculate λ for which there is a first-order dark fringe at this same place on the screen. Rλ Rλ (3.00 m)(600 × 10−9 m) EXECUTE: y1 = 1 , so d = 1 = = 3.72 × 10−4 m. (R is much greater than d, so Eq.35.6 d y1 4.84 × 10−3 m

1⎞ ⎛ is valid.) The dark fringes are located by d sin θ = ⎜ m + ⎟ λ , m = 0, ± 1, ± 2, … The first order dark fringe is located 2⎠ ⎝ by sin θ = λ2 / 2d , where λ2 is the wavelength we are seeking. y = R tan θ ≈ R sin θ =

λ2 R 2d

Rλ1 Rλ2 = and λ2 = 2λ1 = 1200 nm. d 2d EVALUATE: For λ = 600 nm the path difference from the two slits to this point on the screen is 600 nm. For this same path difference (point on the screen) the path difference is λ / 2 when λ = 1200 nm. mλ IDENTIFY: Bright fringes are located at ym = R , when ym # R . Dark fringes are at d sin θ = ( m + 12 )λ and d y = R tan θ . We want λ2 such that y = y1. This gives

35.16.

c 3.00 × 108 m/s = = 4.75 × 10−7 m . For the third bright fringe (not counting the central bright f 6.32 × 1014 Hz spot), m = 3 . For the third dark fringe, m = 2 . SET UP:

λ=

Interference

EXECUTE:

(a) d =

mλ R 3(4.75 × 10−7 m)(0.850 m) = = 3.89 × 10−5 m = 0.0389 mm ym 0.0311 m

λ

⎛ 4.75 × 10−7 m ⎞ = (2.5) ⎜ ⎟ = 0.0305 and θ = 1.75° . y = R tan θ = (85.0 cm) tan1.75° = 2.60 cm . −5 d ⎝ 3.89 × 10 m ⎠ EVALUATE: The third dark fringe is closer to the center of the screen than the third bright fringe on one side of the central bright fringe. IDENTIFY: Bright fringes are located at angles θ given by d sin θ = mλ . SET UP: The largest value sin θ can have is 1.00. d 0.0116 × 10−3 m d sin θ EXECUTE: (a) m = . For sin θ = 1 , m = = = 19.8 . Therefore, the largest m for fringes λ λ 5.85 × 10−7 m on the screen is m = 19 . There are 2(19) + 1 = 39 bright fringes, the central one and 19 above and 19 below it. (b) sin θ = (2 + 12 )

35.17.

λ

⎛ 5.85 × 10−7 m ⎞ = ±19 ⎜ ⎟ = ±0.958 and θ = ±73.3° . −3 d ⎝ 0.0116 × 10 m ⎠ EVALUATE: For small θ the spacing Δy between adjacent fringes is constant but this is no longer the case for larger angles. IDENTIFY: At large distances from the antennas the equation d sin θ = mλ , m = 0, ±1, ±2, … gives the angles where (b) The most distant fringe has m = ±19 . sin θ = m

35.18.

35-5

maximum intensity is observed and d sin θ = (m + 12 )λ , m = 0, ±1, ±2, … gives the angles where minimum intensity is observed. c SET UP: d = 12.0 m . λ = . f EXECUTE:

35.19.

(a) λ =

c 3.00 × 108 m/s mλ ⎛ 2.78 m ⎞ = m⎜ = = 2.78 m . sin θ = ⎟ = m(0.232) . 6 f 107.9 × 10 Hz d ⎝ 12.0 m ⎠

θ = ±13.4°, ± 27.6°, ± 44.1°, ± 68.1° . λ (b) sin θ = (m + 12 ) = (m + 12 )(0.232) . θ = ±6.66°, ± 20.4°, ± 35.5°, ± 54.3° . d EVALUATE: The angles for zero intensity are approximately midway between those for maximum intensity. IDENTIFY: Eq.(35.10): I = I 0 cos 2 (φ 2) . Eq.(35.11): φ = (2π / λ )(r2 − r1 ) . SET UP: φ is the phase difference and (r2 − r1 ) is the path difference. EXECUTE:

(a) I = I 0 (cos 30.0°) 2 = 0.750 I 0

(b) 60.0° = (π / 3) rad . (r2 − r1 ) = (φ / 2π )λ = [ (π / 3) / 2π ] λ = λ / 6 = 80 nm . EVALUATE: 35.20.

IDENTIFY:

φ = 360° / 6 and (r2 − r1 ) = λ / 6 . Δφ path difference relates the path difference to the phase difference Δφ . = 2π λ

The sources and point P are shown in Figure 35.20. ⎛ 524 cm − 486 cm ⎞ EXECUTE: Δφ = 2π ⎜ ⎟ = 119 radians 2 cm ⎝ ⎠

SET UP:

EVALUATE:

The distances from B to P and A to P aren't important, only the difference in these distances.

Figure 35.20 35.21.


The phase difference φ is given by φ = (2π d / λ )sin θ (Eq.35.13.)

EXECUTE: φ = [2π (0.340 × 10−3 m)/(500 × 10 −9 m)]sin 23.0° = 1670 rad EVALUATE: The mth bright fringe occurs when φ = 2π m, so there are a large number of bright fringes within 23.0° from the centerline. Note that Eq.(35.13) gives φ in radians.

35-6

Chapter 35

35.22.

IDENTIFY: The maximum intensity occurs at all the points of constructive interference. At these points, the path difference between waves from the two transmitters is an integral number of wavelengths. SET UP: For constructive interference, sin θ = mλ/d. EXECUTE: (a) First find the wavelength of the UHF waves: λ = c/f = (3.00 × 108 m/s)/(1575.42 MHz) = 0.1904 m For maximum intensity (πd sin θ )/λ = mπ, so sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m The maximum possible m would be for θ = 90°, or sin θ = 1, so mmax = d/λ = (5.18 m)/(0.1904 m) = 27.2 which must be ±27 since m is an integer. The total number of maxima is 27 on either side of the central fringe, plus the central fringe, for a total of 27 + 27 + 1 = 55 bright fringes. (b) Using sin θ = mλ/d, where m = 0, ±1, ±2, and ±3, we have sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m

m = 0: sin θ = 0, which gives θ = 0° m = ±1: sin θ = ±(0.03676)(1), which gives θ = ±2.11° m = ±2: sin θ = ±(0.03676)(2), which gives θ = ±4.22° m = ±3: sin θ = ±(0.03676)(3), which gives θ = ±6.33°

35.23.

⎛ π d sin θ ⎞ 2 2 2 ⎡ π (5.18 m)sin(4.65°) ⎤ (c) I = I 0 cos 2 ⎜ ⎟ = ( 2.00 W/m ) cos ⎢ ⎥ = 1.28 W/m . 0.1904 m ⎣ ⎦ ⎝ λ ⎠ EVALUATE: Notice that sinθ increases in integer steps, but θ only increases in integer steps for small θ. 1⎞ ⎛ (a) IDENTIFY and SET UP: The minima are located at angles θ given by d sin θ = ⎜ m + ⎟ λ . The first minimum 2⎠ ⎝ corresponds to m = 0. Solve for θ . Then the distance on the screen is y = R tan θ . sin θ =

EXECUTE:

λ 2d

=

660 × 10−9 m = 1.27 × 10−3 and θ = 1.27 × 10−3 rad 2(0.260 × 10−3 m)

y = (0.700 m) tan(1.27 ×10−3 rad) = 0.889 mm. (b) IDENTIFY and SET UP: Eq.(35.15) given the intensity I as a function of the position y on the screen: ⎛ π dy ⎞ I = I 0 cos 2 ⎜ ⎟ . Set I = I 0 / 2 and solve for y. ⎝ λR ⎠ I=

EXECUTE:

1 ⎛ π dy ⎞ 1 I 0 says cos 2 ⎜ ⎟= 2 ⎝ λR ⎠ 2

π dy π ⎛ π dy ⎞ 1 so = rad cos ⎜ ⎟= λR 4 2 ⎝ λR ⎠ λ R (660 × 10−9 m)(0.700 m) = = 0.444 mm y= 4d 4(0.260 × 10−3 m) EVALUATE: 35.24.


35.25.

I = I 0 / 2 at a point on the screen midway between where I = I 0 and I = 0. ⎛πd ⎞ Eq. (35.14): I = I 0 cos 2 ⎜ sin θ ⎟ . ⎝ λ ⎠

The intensity goes to zero when the cosine’s argument becomes an odd integer multiple of

πd sin θ = ( m + 1/ 2)π gives d sin θ = λ (m + 1/ 2), which is Eq. (35.5). λ

π 2

EVALUATE: Section 35.3 shows that the maximum-intensity directions from Eq.(35.14) agree with Eq.(35.4). IDENTIFY: The intensity decreases as we move away from the central maximum. ⎛ π dy ⎞ SET UP: The intensity is given by I = I 0 cos 2 ⎜ ⎟. ⎝ λR ⎠ EXECUTE: First find the wavelength: λ = c/f = (3.00 × 108 m/s)/(12.5 MHz) = 24.00 m At the farthest the receiver can be placed, I = I0/4, which gives I0 1 1 ⎛ π dy ⎞ ⎛ π dy ⎞ 2 ⎛ π dy ⎞ = I 0 cos 2 ⎜ ⎟ ⇒ cos ⎜ ⎟ = ⇒ cos ⎜ ⎟=± 4 2 ⎝ λR ⎠ ⎝ λR ⎠ 4 ⎝ λR ⎠

Interference

35.26.

35-7

The solutions are πdy/λR = π/3 and 2π/3. Using π/3, we get y = λR/3d = (24.00 m)(500 m)/[3(56.0 m)] = 71.4 m It must remain within 71.4 m of point C. EVALUATE: Using πdy/λR = 2π/3 gives y = 142.8 m. But to reach this point, the receiver would have to go beyond 71.4 m from C, where the signal would be too weak, so this second point is not possible. 2π IDENTIFY: The phase difference φ and the path difference r1 − r2 are related by φ = (r1 − r2 ) . The intensity is

λ

⎛φ ⎞ given by I = I 0 cos 2 ⎜ ⎟ . ⎝2⎠

SET UP: λ = EXECUTE:

35.27.

c 3.00 × 108 m/s = = 2.50 m . When the receiver measures zero intensity I 0 , φ = 0 . f 1.20 × 108 Hz

(a) φ =

2π

λ

( r1 − r2 ) =

2π (1.8 m) = 4.52 rad. 2.50 m

⎛φ ⎞ ⎛ 4.52 rad ⎞ (b) I = I 0 cos 2 ⎜ ⎟ = I 0 cos 2 ⎜ ⎟ = 0.404 I 0 . 2 ⎝2⎠ ⎝ ⎠ EVALUATE: (r1 − r2 ) is greater than λ / 2 , so one minimum has been passed as the receiver is moved. IDENTIFY: Consider interference between rays reflected at the upper and lower surfaces of the film. Consider phase difference due to the path difference of 2t and any phase differences due to phase changes upon reflection. SET UP: Consider Figure 35.27. Both rays (1) and (2) undergo a 180° phase change on reflection, so these is no net phase difference introduced and the condition for destructive interference is 1⎞ ⎛ 2t = ⎜ m + ⎟ λ. 2⎠ ⎝ Figure 35.27

35.28.

35.29.

1⎞ ⎛ ⎜ m + ⎟λ λ 2⎠ ⎝ ; thinnest film says m = 0 so t = EXECUTE: t = 2 4 −9 λ0 650 × 10 m λ = = 1.14 × 10−7 m = 114 nm λ = 0 and t = 1.42 4(1.42) 4(1.42) EVALUATE: We compared the path difference to the wavelength in the film, since that is where the path difference occurs. IDENTIFY: Require destructive interference for light reflected at the front and rear surfaces of the film. SET UP: At the front surface of the film, light in air ( n = 1.00 ) reflects from the film ( n = 2.62 ) and there is a 180° phase shift due to the reflection. At the back surface of the film, light in the film ( n = 2.62 ) reflects from glass ( n = 1.62 ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 505 nm wavelength in the film is λ = . 2.62 EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected ⎛ 505 nm ⎞ light occurs when 2t = mλ . t = m ⎜ ⎟ = (96.4 nm)m . The minimum thickness is 96.4 nm. ⎝ 2[2.62] ⎠ (b) The next three thicknesses are for m = 2 , 3 and 4: 192 nm, 289 nm and 386 nm. EVALUATE: The minimum thickness is for t = λ / 2n . Compare this to Problem 35.27, where the minimum thickness for destructive interference is t = λ / 4n . IDENTIFY: The fringes are produced by interference between light reflected from the top and bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do

35-8

Chapter 35

have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) is therefore 2t = (m + 12 )λ . SET UP: The geometry of the air wedge is sketched in Figure 35.29. At a distance x from the point of contact of the two plates, the thickness of the air wedge is t. t λ λ λ EXECUTE: tan θ = so t = x tan θ . tm = (m + 12 ) . xm = (m + 12 ) and xm +1 = ( m + 32 ) . The 2 2 tan θ 2 tan θ x λ 1.00 distance along the plate between adjacent fringes is Δx = xm +1 − xm = . 15.0 fringes/cm = and Δx 2 tan θ 1.00 λ 546 × 10−9 m Δx = = 0.0667 cm . tan θ = = = 4.09 × 10−4 . The angle of the wedge is 2Δx 2(0.0667 × 10−2 m) 15.0 fringes/cm 4.09 × 10−4 rad = 0.0234° . EVALUATE: The fringes are equally spaced; Δx is independent of m.

Figure 35.29 35.30.

IDENTIFY: The fringes are produced by interference between light reflected from the top and from the bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) therefore is 2t = (m + 12 )λ . SET UP: The geometry of the air wedge is sketched in Figure 35.30. t 0.0800 mm λ EXECUTE: tan θ = = 8.89 × 10−4 . tan θ = so t = (8.89 × 10−4 ) x . tm = (m + 12 ) . 2 x 90.0 mm

xm = (m + 12 )

λ 2(8.89 × 10−4 )

is Δx = xm +1 − xm =

and xm +1 = (m + 32 )

λ 2(8.89 × 10 −4 )

=

λ 2(8.89 × 10−4 )

. The distance along the plate between adjacent fringes

656 × 10−9 m = 3.69 × 10−4 m = 0.369 mm . The number of fringes per cm is 2(8.89 × 10−4 )

1.00 1.00 = = 27.1 fringes/cm . Δx 0.0369 cm EVALUATE: As t → 0 the interference is destructive and there is a dark fringe at the line of contact between the two plates.

35.31.

Figure 35.30 IDENTIFY: The light reflected from the top of the TiO2 film interferes with the light reflected from the top of the glass surface. These waves are out of phase due to the path difference in the film and the phase differences caused by reflection. SET UP: There is a π phase change at the TiO2 surface but none at the glass surface, so for destructive interference the path difference must be mλ in the film. EXECUTE: (a) Calling T the thickness of the film gives 2T = mλ0/n, which yields T = mλ0/(2n). Substituting the numbers gives T = m (520.0 nm)/[2(2.62)] = 99.237m

Interference

35.32.

35.33.

35-9

T must be greater than 1036 nm, so m = 11, which gives T = 1091.6 nm, since we want to know the minimum thickness to add. ΔT = 1091.6 nm – 1036 nm = 55.6 nm (b) (i) Path difference = 2T = 2(1092 nm) = 2184 nm = 2180 nm. (ii) The wavelength in the film is λ = λ0/n = (520.0 nm)/2.62 = 198.5 nm. Path difference = (2180 nm)/[(198.5 nm)/wavelength] = 11.0 wavelengths EVALUATE: Because the path difference in the film is 11.0 wavelengths, the light reflected off the top of the film will be 180° out of phase with the light that traveled through the film and was reflected off the glass due to the phase change at reflection off the top of the film. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. For destructive interference the total phase difference is an integer number of half cycles. SET UP: The reflection at the top surface of the film produces a half-cycle phase shift. There is no phase shift at the reflection at the bottom surface. EXECUTE: (a) Since there is a half-cycle phase shift at just one of the interfaces, the minimum thickness for λ λ 550 nm = 74.3 nm. constructive interference is t = = 0 = 4 4n 4(1.85) (b) The next smallest thickness for constructive interference is with another half wavelength thickness added: 3λ 3λ0 3 ( 550 nm ) = = = 223 nm. t= 4 4n 4(1.85) EVALUATE: Note that we must compare the path difference to the wavelength in the film. IDENTIFY: Consider the interference between rays reflected from the two surfaces of the soap film. Strongly reflected means constructive interference. Consider phase difference due to the path difference of 2t and any phase difference due to phase changes upon reflection. (a) SET UP: Consider Figure 35.33. There is a 180° phase change when the light is reflected from the outside surface of the bubble and no phase change when the light is reflected from the inside surface. Figure 35.33 EXECUTE: The reflections produce a net 180° phase difference and for there to be constructive interference the path difference 2t must correspond to a half-integer number of wavelengths to compensate for the λ / 2 shift due to 1⎞ ⎛ the reflections. Hence the condition for constructive interference is 2t = ⎜ m + ⎟ (λ0 / n), m = 0,1,2,… Here λ0 is 2⎠ ⎝

the wavelength in air and (λ0 / n) is the wavelength in the bubble, where the path difference occurs.

2tn 2(290 nm)(1.33) 771.4 nm = = 1 1 1 m+ m+ m+ 2 2 2 for m = 0, λ = 1543 nm; for m = 1, λ = 514 nm; for m = 2, λ = 308 nm;… Only 514 nm is in the visible region; the color for this wavelength is green. 2tn 2(340 nm)(1.33) 904.4 nm (b) λ0 = = = 1 1 1 m+ m+ m+ 2 2 2 for m = 0, λ = 1809 nm; for m = 1, λ = 603 nm; for m = 2, λ = 362 nm;… Only 603 nm is in the visible region; the color for this wavelength is orange. EVALUATE: The dominant color of the reflected light depends on the thickness of the film. If the bubble has varying thickness at different points, these points will appear to be different colors when the light reflected from the bubble is viewed. IDENTIFY: The number of waves along the path is the path length divided by the wavelength. The path difference and the reflections determine the phase difference.

λ0 =

35.34.

SET UP:

The path length is 2t = 17.52 × 10−6 m . The wavelength in the film is λ =

λ0 n

.

35-10

Chapter 35

2t 17.52 × 10−6 m 648 nm = = 36.5 . = 480 nm . The number of waves is 480 × 10−9 m 1.35 λ (b) The path difference introduces a λ / 2 , or 180° , phase difference. The ray reflected at the top surface of the film undergoes a 180° phase shift upon reflection. The reflection at the lower surface introduces no phase shift. Both rays undergo a 180° phase shift, one due to reflection and one due to reflection. The two effects cancel and the two rays are in phase as they leave the film. EVALUATE: Note that we must use the wavelength in the film to determine the number of waves in the film. IDENTIFY: Require destructive interference between light reflected from the two points on the disc. SET UP: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so they both have the same phase shift upon reflection and the condition for destructive interference (cancellation) is EXECUTE:

35.35.

(a) λ =

2t = (m + 12 )λ , where t is the depth of the pit. λ =

35.37.

35.38.

35.39.

λ

λ0

n

. The minimum pit depth is for m = 0 .

790 nm = 110 nm = 0.11 μ m . 2 4 4n 4(1.8) EVALUATE: The path difference occurs in the plastic substrate and we must compare the wavelength in the substrate to the path difference. IDENTIFY: Consider light reflected at the front and rear surfaces of the film. SET UP: At the front surface of the film, light in air ( n = 1.00 ) reflects from the film ( n = 2.62 ) and there is a 180° phase shift due to the reflection. At the back surface of the film, light in the film ( n = 2.62 ) reflects from glass ( n = 1.62 ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 505 nm wavelength in the film is λ = . 2.62 EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected ⎛ 505 nm ⎞ light occurs when 2t = mλ . t = m ⎜ ⎟ = (96.4 nm)m . The minimum thickness is 96.4 nm. ⎝ 2[2.62] ⎠ EXECUTE:

35.36.

λ

λ0

2t =

. t=

=

=

(b) The next three thicknesses are for m = 2 , 3 and 4: 192 nm, 289 nm and 386 nm. EVALUATE: The minimum thickness is for t = λ / 2n . Compare this to Problem 34.27, where the minimum thickness for destructive interference is t = λ / 4n . IDENTIFY and SET UP: Apply Eq.(35.19) and calculate y for m = 1800. EXECUTE: Eq.(35.19): y = m(λ / 2) = 1800(633 × 10−9 m) / 2 = 5.70 × 10−4 m = 0.570 mm EVALUATE: A small displacement of the mirror corresponds to many wavelengths and a large number of fringes cross the line. IDENTIFY: Apply Eq.(35.19). SET UP: m = 818 . Since the fringes move in opposite directions, the two people move the mirror in opposite directions. mλ 818(6.06 × 10−7 m) EXECUTE: (a) For Jan, the total shift was y1 = 1 = = 2.48 × 10−4 m. For Linda, the total shift 2 2 mλ2 818(5.02 × 10−7 m) was y2 = = = 2.05 × 10−4 m. 2 2 (b) The net displacement of the mirror is the difference of the above values: Δy = y1 − y2 = 0.248 mm − 0.205 mm = 0.043 mm. EVALUATE: The person using the larger wavelength moves the mirror the greater distance. IDENTIFY: Consider the interference between light reflected from the top and bottom surfaces of the air film between the lens and the glass plate. 1⎞ ⎛ SET UP: For maximum intensity, with a net half-cycle phase shift due to reflections, 2t = ⎜ m + ⎟ λ . 2⎠ ⎝ t = R − R2 − r 2 . (2m + 1)λ (2m + 1)λ = R − R2 − r 2 ⇒ R2 − r 2 = R − EXECUTE: 4 4 2

(2m + 1)λ R ⎡ (2m + 1)λ ⎤ ⎡ (2m + 1)λ ⎤ (2m + 1)λ R ⇒ R2 − r 2 = R2 + ⎢ − ⇒r= −⎢ ⎥ ⎥⎦ 4 2 2 4 ⎣ ⎦ ⎣ ⇒r≈

(2m + 1)λ R , for R " λ . 2

2

Interference

35-11

The second bright ring is when m = 1: r≈

(2(1) + 1) (5.80 × 10−7 m) (0.952 m) = 9.10 × 10−4 m = 0.910 mm. 2

So the diameter of the second bright ring is 1.82 mm. EVALUATE: The diameter of the m th ring is proportional to increases. This agrees with Figure 35.17b in the textbook. 35.40.

As found in Problem 35.39, the radius of the mth bright ring is r ≈

IDENTIFY:

λair n

n

, where n

.

EXECUTE: EVALUATE:

35.43.

λ

is the refractive index of the liquid. (2m + 1)λ R r 0.850 mm = = = 0.737 mm. EXECUTE: r (n) ≈ 2n n 1.33 EVALUATE: The refractive index of the water is less than that of the glass plate, so the phase changes on reflection are the same as when air is in the space. IDENTIFY: The liquid alters the wavelength of the light and that affects the locations of the interference minima. SET UP: The interference minima are located by d sin θ = (m + 12 )λ . For a liquid with refractive index n,

λliq =

35.42.

(2m + 1)λ R , for R " λ . 2

Introducing a liquid between the lens and the plate just changes the wavelength from λ to

SET UP:

35.41.

2m + 1 , so the rings get closer together as m

sin θ

λ

=

(m + 12 ) sin θ air sin θ liq sin θ air sin θ liq sin θ air sin 35.20° = . = and n = = = 1.730 . = constant , so sin θ liq sin19.46° λair λliq λair λair / n d

In the liquid the wavelength is shorter and sin θ = (m + 12 )

λ d

gives a smaller θ than in air, for the

same m. IDENTIFY: As the brass is heated, thermal expansion will cause the two slits to move farther apart. SET UP: For destructive interference, d sin θ = λ/2. The change in separation due to thermal expansion is dw = αw0 dT, where w is the distance between the slits. EXECUTE: The first dark fringe is at d sin θ = λ/2 ⇒ sin θ = λ/2d. Call d ≡ w for these calculations to avoid confusion with the differential. sin θ = λ/2w Taking differentials gives d(sin θ) = d(λ/2w) and cosθ dθ = − λ/2 dw/w2. λ α w0 dT λα dT =− . Solving for dθ gives For thermal expansion, dw = αw0 dT, which gives cosθ dθ = − 2 2 w0 2w0 λα dT dθ = − . Get λ: w0 sin θ0 = λ/2 → λ = 2w0 sinθ0. Substituting this quantity into the equation for dθ gives 2 w0 cosθ 0 2 w sin θ 0α dT dθ = − 0 = − tan θ 0 α dT . 2w0 cosθ 0 dθ = − tan(32.5°)(2.0 × 10−5 K −1 )(115 K) = −0.001465 rad = −0.084° The minus sign tells us that the dark fringes move closer together. EVALUATE: We can also see that the dark fringes move closer together because sinθ is proportional to 1/d, so as d increases due to expansion, θ decreases. IDENTIFY: Both frequencies will interfere constructively when the path difference from both of them is an integral number of wavelengths. SET UP: Constructive interference occurs when sinθ = mλ/d. EXECUTE: First find the two wavelengths. λ1 = v/f1 = (344 m/s)/(900 Hz) = 0.3822 m

λ2 = v/f2 = (344 m/s)/(1200 Hz) = 0.2867 m To interfere constructively at the same angle, the angles must be the same, and hence the sines of the angles must be equal. Each sine is of the form sin θ = mλ/d, so we can equate the sines to get m1λ1/d = m2λ2/d

m1(0.3822 m) = m2(0.2867 m) m2 = 4/3 m1

35-12

Chapter 35

Since both m1 and m2 must be integers, the allowed pairs of values of m1 and m2 are m1 = m2 = 0

m1 = 3, m2 = 4 m1 = 6, m2 = 8 m1 = 9, m2 = 12 etc.

35.44.

For m1 = m2 = 0, we have θ = 0. For m1 = 3, m2 = 4, we have sin θ1 = (3)(0.3822 m)/(2.50 m), giving θ1 = 27.3° For m1 = 6, m2 = 8, we have sin θ1 = (6)(0.3822 m)/(2.50 m), giving θ1 = 66.5° For m1 = 9, m2 = 12, we have sin θ1 = (9)(0.3822 m)/(2.50 m) = 1.38 > 1, so no angle is possible. EVALUATE: At certain other angles, one frequency will interfere constructively, but the other will not. 1⎞ ⎛ IDENTIFY: For destructive interference, d = r2 − r1 = ⎜ m + ⎟ λ . 2⎠ ⎝ SET UP:

r2 − r1 = (200 m) 2 + x 2 − x 2

⎡⎛ 1⎞ ⎤ 1⎞ ⎛ (200 m) 2 + x 2 = x 2 + ⎢⎜ m + ⎟ λ ⎥ + 2 x ⎜ m + ⎟ λ . 2 2⎠ ⎠ ⎦ ⎝ ⎣⎝ 2 20,000 m 1 ⎛ 1⎞ c 3.00 × 108 m s − ⎜ m + ⎟ λ . The wavelength is calculated by λ = = x= = 51.7 m. 1⎞ f 2⎝ 2⎠ 5.80 × 106 Hz ⎛ ⎜ m + ⎟λ 2⎠ ⎝ m = 0 : x = 761 m; m = 1: x = 219 m; m = 2 : x = 90.1 m; m = 3; x = 20.0 m. EVALUATE: For m = 3 , d = 3.5λ = 181 m . The maximum possible path difference is the separation of 200 m between the sources. IDENTIFY: The two scratches are parallel slits, so the light that passes through them produces an interference pattern. However the light is traveling through a medium (plastic) that is different from air. SET UP: The central bright fringe is bordered by a dark fringe on each side of it. At these dark fringes, d sin θ = ½ λ/n, where n is the refractive index of the plastic. EXECUTE: First use geometry to find the angles at which the two dark fringes occur. At the first dark fringe tanθ = [(5.82 mm)/2]/(3250 mm), giving θ = ±0.0513° For destructive interference, we have d sin θ = ½ λ/n and n = λ/(2dsin θ) = (632.8 nm)/[2(0.000225 m)(sin 0.0513°)] = 1.57 EVALUATE: The wavelength of the light in the plastic is reduced compared to what it would be in air. IDENTIFY: Interference occurs due to the path difference of light in the thin film. SET UP: Originally the path difference was an odd number of half-wavelengths for cancellation to occur. If the path difference decreases by ½ wavelength, it will be a multiple of the wavelength, so constructive interference will occur. EXECUTE: Calling ΔT the thickness that must be removed, we have path difference = 2ΔT = ½ λ/n and ΔT = λ/4n = (525 nm)/[4(1.40)] = 93.75 nm, At 4.20 nm/yr, we have (4.20 nm/yr)t = 93.75 nm and t = 22.3 yr. EVALUATE: If you were giving a warranty on this film, you certainly could not give it a “lifetime guarantee”! IDENTIFY and SET UP: If the total phase difference is an integer number of cycles the interference is constructive and if it is a half-integer number of cycles it is destructive. EXECUTE: (a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to the path difference equations Eq.(35.1) and Eq.(35.2). This exactly changes one for the other, for m → m + 12 and m + 12 → m, since m in any integer. φ (b) If one source leads the other by a phase angle φ , the fraction of a cycle difference is . Thus the path length 2π difference for the two sources must be adjusted for both destructive and constructive interference, by this amount. So for constructive inference: r1 − r2 = ( m + φ 2π )λ , and for destructive interference, r1 − r2 = ( m + 1 2 + φ 2π )λ , where in each case m = 0, ±1, ±2, … EVALUATE: If φ = 0 these results reduce to Eqs.(35.1) and (35.2). IDENTIFY: Follow the steps specified in the problem. SET UP: Use cos(ωt + φ / 2) = cos(ωt ) cos(φ / 2) − sin(ωt )sin(φ / 2) . Then 1 + cos(φ ) 2cos(φ / 2)cos(ωt + φ / 2) = 2cos(ωt )cos 2 (φ / 2) − 2sin(ωt )sin(φ / 2)cos(φ / 2) . Then use cos 2 (φ / 2) = and 2 EXECUTE:

35.45.

35.46.

35.47.

35.48.

Interference

35-13

2sin(φ / 2)cos(φ / 2) = sin φ . This gives cos(ωt ) + (cos(ωt )cos(φ ) − sin(ωt )sin(φ )) = cos(ωt ) + cos(ωt + φ ) , using again the trig identity for the cosine of the sum of two angles. EXECUTE: (a) The electric field is the sum of the two fields and can be written as EP (t ) = E2 (t ) + E1 (t ) = E cos(ωt ) + E cos(ωt + φ ) . EP (t ) = 2 E cos(φ / 2)cos(ωt + φ / 2) . (b) E p (t ) = A cos(ωt + φ / 2), so comparing with part (a), we see that the amplitude of the wave (which is always

positive) must be A = 2 E | cos(φ / 2) | . (c) To have an interference maximum,

φ = 2π m . So, for example, using m = 1, the relative phases are 2

φ = 2π , and all waves are in phase. 2 φ 1⎞ ⎛ (d) To have an interference minimum, = π ⎜ m + ⎟ . So, for example using m = 0, relative phases are E2 : 0; E1: φ = 4π ; E p :

2 2⎠ ⎝ E2 : 0; E1: φ = π ; E p : φ /2 = π /2, and the resulting wave is out of phase by a quarter of a cycle from both of the

original waves. (e) The instantaneous magnitude of the Poynting vector is $% | S |= ε 0cE p2 (t ) = ε 0c (4 E 2 cos 2 (φ 2)cos 2 (ωt + φ 2)).

35.49.

35.50.

1 For a time average, cos 2 (ωt + φ 2) = , so Sav = 2ε 0cE 2 cos 2 (φ 2). 2 EVALUATE: The result of part (e) shows that the intensity at a point depends on the phase difference φ at that point for the waves from each source. IDENTIFY: Follow the steps specified in the problem. SET UP: The definition of hyperbola is the locus of points such that the difference between P to S2 and P to S1 is a constant. EXECUTE: (a) Δr = mλ . r1 = x 2 + ( y − d ) 2 and r2 = x 2 + ( y + d ) 2 . Δr = x 2 + ( y + d ) 2 − x 2 + ( y − d ) 2 = mλ . (b) For a given m and λ , Δr is a constant and we get a hyperbola. Or, in the case of all m for a given λ , a family of hyperbolas. (c) x 2 + ( y + d ) 2 − x 2 + ( y − d ) 2 = ( m + 12 )λ . EVALUATE: The hyperbolas approach straight lines at large distances from the source. IDENTIFY: Follow the derivation of Eq.(35.7), but with different amplitudes for the two waves. SET UP: cos(π − φ ) = − cos φ

EXECUTE:

(a) E p2 = E12 + E22 − 2 E1E2 cos(π − φ ) = E 2 + 4 E 2 + 4 E 2 cos φ = 5E 2 + 4 E 2 cos φ

1 ⎡⎛ 5 ⎞ ⎛ 4 ⎞ ⎤ 9 ⎡5 4 ⎤ I = ε 0cE p 2 = ε 0c ⎢⎜ E 2 ⎟ + ⎜ E 2 ⎟ cos φ ⎥ . φ = 0 ⇒ I 0 = ε 0cE 2 . Therefore, I = I 0 ⎢ + cos φ ⎥ . 9 9 2 2 2 2 ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 1 (b) The graph is shown in Figure 35.50. I min = I 0 which occurs when φ = nπ (n odd). 9 EVALUATE: The maxima and minima occur at the same points on the screen as when the two sources have the same amplitude, but when the amplitudes are different the intensity is no longer zero at the minima.

Figure 35.50

35-14

Chapter 35

35.51.

IDENTIFY and SET UP: Consider interference between rays reflected from the upper and lower surfaces of the film to relate the thickness of the film to the wavelengths for which there is destructive interference. The thermal expansion of the film changes the thickness of the film when the temperature changes. EXECUTE: For this film on this glass, there is a net λ / 2 phase change due to reflection and the condition for destructive interference is 2t = m(λ / n), where n = 1.750. Smallest nonzero thickness is given by t = λ / 2n. At 20.0°C, t0 = (582.4 nm) /[(2)(1.750)] = 166.4 nm. At 170°C, t0 = (588.5 nm) /[(2)(1.750)] = 168.1 nm. t = t0 (1 + αΔT ) so

35.52.

35.53.

34.54.

35.55.

α = (t − t0 ) /(t0 ΔT ) = (1.7 nm) /[(166.4 nm)(150°C)] = 6.8 × 10−5 (C°) −1 EVALUATE: When the film is heated its thickness increases, and it takes a larger wavelength in the film to equal 2t.The value we calculated for α is the same order of magnitude as those given in Table 17.1. IDENTIFY and SET UP: At the m = 3 bright fringe for the red light there must be destructive interference at this same θ for the other wavelength. EXECUTE: For constructive interference: d sin θ = mλ1 ⇒ d sin θ = 3(700 nm) = 2100 nm. For destructive 1⎞ d sin θ 2100 nm ⎛ interference: d sin θ = ⎜ m + ⎟ λ2 ⇒ λ2 = = . So the possible wavelengths are 2⎠ m + 12 m + 12 ⎝ λ2 = 600 nm, for m = 3, and λ2 = 467 nm, for m = 4. EVALUATE: Both d and θ drop out of the calculation since their combination is just the path difference, which is the same for both types of light. ⎛πd ⎞ IDENTIFY: Apply I = I 0 cos ⎜ sin θ ⎟ . ⎝ λ ⎠ πd π 3π sin θ is rad , rad ,…. SET UP: I = I 0 / 2 when 4 4 λ EXECUTE: First we need to find the angles at which the intensity drops by one-half from the value of the m th πd π dθ m π ⎛πd ⎞ I bright fringe. I = I 0 cos 2 ⎜ = (m + 1 2) . sin θ ⎟ = 0 ⇒ sin θ ≈ 2 λ λ ⎝ λ ⎠ 2 λ 3λ λ − + m = 0 :θ = θm = ⇒ Δθ m = . ; m = 1: θ = θ m = 4d 4d 2d EVALUATE: There is no dependence on the m-value of the fringe, so all fringes at small angles have the same half-width. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. SET UP: There is just one half-cycle phase change upon reflection, so for constructive interference 2t = (m1 + 12 )λ1 = (m2 + 12 )λ2 , where these wavelengths are in the glass. The two different wavelengths differ by just one m-value, m2 = m1 − 1. 1⎞ 1⎞ λ + λ2 λ + λ2 ⎛ ⎛ EXECUTE: ⎜ m1 + ⎟ λ1 = ⎜ m1 − ⎟ λ2 ⇒ m1 (λ2 − λ1 ) = 1 . ⇒ m1 = 1 λ2 − λ1 ) 2 2 2 2( ⎝ ⎠ ⎝ ⎠ 477.0 nm + 540.6 nm 1⎞λ 17(477.0 nm) ⎛ m1 = = 8. 2t = ⎜ 8 + ⎟ 01 ⇒ t = = 1334 nm. n 2(540.6 nm − 477.0 nm) 2 4(1.52) ⎝ ⎠ EVALUATE: Now that we have t we can calculate all the other wavelengths for which there is constructive interference. IDENTIFY: Consider the phase difference due to the path difference and due to the reflection of one ray from the glass surface. (a) SET UP: Consider Figure 35.55 path difference = 2 h2 + x2 / 4 − x =

4h 2 + x 2 − x

Figure 35.55

Interference

35-15

Since there is a 180° phase change for the reflected ray, the condition for constructive interference is path 1⎞ ⎛ difference = ⎜ m + ⎟ λ and the condition for destructive interference is path difference = mλ . 2⎠ ⎝ (b) EXECUTE:

1⎞ 4h 2 + x 2 − x ⎛ Constructive interference: ⎜ m + ⎟ λ = 4h 2 + x 2 − x and λ = . Longest λ is for 1 2⎠ ⎝ m+ 2

m = 0 and then λ = 2 35.56.

35.57.

35.58.

(

) (

4h 2 + x 2 − x = 2

EVALUATE: For λ = 0.72 m the path difference is λ / 2. IDENTIFY: Require constructive interference for the reflection from the top and bottom surfaces of each cytoplasm layer and each guanine layer. SET UP: At the water (or cytoplasm) to guanine interface, there is a half-cycle phase shift for the reflected light, but there is not one at the guanine to cytoplasm interface. Therefore there will always be one half-cycle phase difference between two neighboring reflected beams, just due to the reflections. EXECUTE: For the guanine layers: 2t n 1 λ 2(74 nm) (1.80) 266 nm = ⇒ λ = 533 nm (m = 0). 2tg = ( m + ) ⇒ λ = g g1 = 2 ng (m + 2 ) (m + 12 ) (m + 12 )

For the cytoplasm layers: 1⎞ λ 2tc nc 2(100 nm) (1.333) 267 nm ⎛ = = ⇒ λ = 533 nm (m = 0). 2tc = ⎜ m + ⎟ ⇒ λ = 1 + 2 ( ) (m + 12 ) (m + 12 ) n m ⎝ ⎠ c 2 (b) By having many layers the reflection is strengthened, because at each interface some more of the transmitted light gets reflected back, increasing the total percentage reflected. (c) At different angles, the path length in the layers changes (always to a larger value than the normal incidence case). If the path length changes, then so do the wavelengths that will interfere constructively upon reflection. EVALUATE: The thickness of the guanine and cytoplasm layers are inversely proportional to their refractive ⎛ 100 1.80 ⎞ indices ⎜ = ⎟ , so both kinds of layers produce constructive interference for the same wavelength in air. ⎝ 74 1.333 ⎠ IDENTIFY: The slits will produce an interference pattern, but in the liquid, the wavelength of the light will be less than it was in air. SET UP: The first bright fringe occurs when d sin θ = λ/n. EXECUTE: In air: dsin18.0° = λ. In the liquid: dsin12.6° = λ/n. Dividing the equations gives n = (sin 18.0°)/(sin 12.6°) = 1.42 EVALUATE: It was not necessary to know the spacing of the slits, since it was the same in both air and the liquid. IDENTIFY: Consider light reflected at the top and bottom surfaces of the film. Wavelengths that are predominant in the transmitted light are those for which there is destructive interference in the reflected light. SET UP: For the waves reflected at the top surface of the oil film there is a half-cycle reflection phase shift. For the waves reflected at the bottom surface of the oil film there is no reflection phase shift. The condition for constructive interference is 2t = (m + 12 )λ . The condition for destructive interference is 2t = mλ . The range of visible wavelengths is approximately 400 nm to 700 nm. In the oil film, λ = EXECUTE:

35.59.

)

4(0.24 m) 2 + (0.14 m) 2 − 0.14 m = 0.72 m

λ0

. n λ 2tn 2(380 nm)(1.45) 1102 nm . (a) 2t = (m + 12 )λ = (m + 12 ) 0 . λ0 = = = 1 m+ 2 m + 12 m + 12 n

m = 0 : λ0 = 2200 nm . m = 1 : λ0 = 735 nm . m = 2 : λ0 = 441 nm . m = 3 : λ0 = 315 nm . The visible wavelength for which there is constructive interference in the reflected light is 441 nm. λ 2tn 1102 nm (b) 2t = mλ = m 0 . λ0 = = . m = 1 : λ0 = 1102 nm . m = 2 : λ0 = 551 nm . m = 3 : λ0 = 367 nm . m m n The visible wavelength for which there is destructive interference in the reflected light is 551 nm. This is the visible wavelength predominant in the transmitted light. EVALUATE: At a particular wavelength the sum of the intensities of the reflected and transmitted light equals the intensity of the incident light. (a) IDENTIFY: The wavelength in the glass is decreased by a factor of 1/ n, so for light through the upper slit a shorter path is needed to produce the same phase at the screen. Therefore, the interference pattern is shifted downward on the screen. (b) SET UP: Consider the total phase difference produced by the path length difference and also by the different wavelength in the glass.

35-16

Chapter 35

At a point on the screen located by the angle θ the difference in path length is d sin θ . This

EXECUTE:

⎛ 2π ⎞ introduces a phase difference of φ = ⎜ ⎟ (d sin θ ), where λ0 is the wavelength of the light in air or vacuum. ⎝ λ0 ⎠ L nL In the thickness L of glass the number of wavelengths is = . A corresponding length L of the path of the ray

λ

λ0

through the lower slit, in air, contains L / λ0 wavelengths. The phase difference this introduces is ⎛ nL

φ = 2π ⎜

⎝ λ0

−

L⎞ ⎟ and φ = 2π (n − 1)( L / λ0 ). The total phase difference is the sum of these two, λ0 ⎠

⎛ 2π ⎞ ⎜ ⎟ (d sin θ ) + 2π (n − 1)( L / λ0 ) = (2π / λ0 )( d sin θ + L(n − 1)). Eq.(35.10) then gives ⎝ λ0 ⎠ ⎡⎛ π ⎞ ⎤ I = I 0 cos 2 ⎢⎜ ⎟ (d sin θ + L( n − 1)) ⎥ . ⎣⎢⎝ λ0 ⎠ ⎦⎥ (c) Maxima means cos φ / 2 = ±1 and φ / 2 = mπ , m = 0, ± 1, ± 2, … (π / λ0 )(d sin θ + L( n − 1)) = mπ d sin θ + L(n − 1) = mλ0 mλ0 − L(n − 1) d EVALUATE: When L → 0 or n → 1 the effect of the plate goes away and the maxima are located by Eq.(35.4). IDENTIFY: Dark fringes occur because the path difference is one-half of a wavelength. ⎛ π d sin θ ⎞ SET UP: At the first dark fringe, dsinθ = λ/2. The intensity at any angle θ is given by I = I 0 cos 2 ⎜ ⎟. ⎝ λ ⎠ (a) At the first dark fringe, we have d sin θ = λ/2 sin θ =

35.60.

d/λ = 2/(2 sin 15.0°) = 1.93 1 ⎛ π d sin θ ⎞ I 0 ⎛ π d sin θ ⎞ (b) I = I 0 cos 2 ⎜ ⇒ cos ⎜ ⎟= ⎟= λ λ 10 10 ⎝ ⎠ ⎝ ⎠

π d sin θ ⎛ 1 ⎞ = arccos ⎜ ⎟ = 71.57° = 1.249 rad λ ⎝ 10 ⎠

35.61.

Using the result from part (a), that d/λ = 1.93, we have π(1.93)sin θ = 1.249. sin θ = 0.2060 and θ = ±11.9° EVALUATE: Since the first dark fringes occur at ±15.0°, it is reasonable that at ≈12° the intensity is reduced to only 1/10 of its maximum central value. IDENTIFY: There are two effects to be considered: first, the expansion of the rod, and second, the change in the rod’s refractive index.

λ=

λ0

and Δn = n0 (2.50 × 10−5 (C°) −1 ) ΔT . ΔL = L0 (5.00 × 10−6 (C°) −1 ) ΔT . n EXECUTE: The extra length of rod replaces a little of the air so that the change in the number of wavelengths due 2n ΔL 2n ΔL 2(nglass − 1) L0αΔT = and to this is given by: ΔN1 = glass − air SET UP:

λ0

λ0

λ0

−6

2(1.48 − 1)(0.030 m)(5.00 × 10 C°)(5.00 C°) = 1.22. 5.89 × 10−7 m The change in the number of wavelengths due to the change in refractive index of the rod is: 2Δnglass L0 2(2.50 × 10 −5 C°)(5.00 C° min)(1.00 min)(0.0300 m) ΔN 2 = = = 12.73. λ0 5.89 × 10−7 m So, the total change in the number of wavelengths as the rod expands is ΔN = 12.73 + 1.22 = 14.0 fringes/minute. EVALUATE: Both effects increase the number of wavelengths along the length of the rod. Both ΔL and Δnglass are very small and the two effects can be considered separately. ΔN1 =

35.62.

IDENTIFY:

Apply Snell's law to the refraction at the two surfaces of the prism. S1 and S2 serve as coherent

sources so the fringe spacing is Δy =

Rλ , where d is the distance between S1 and S2 . d

Interference

35-17

SET UP: For small angles, sin θ ≈ θ , with θ expressed in radians. EXECUTE: (a) Since we can approximate the angles of incidence on the prism as being small, Snell’s Law tells us that an incident angle of θ on the flat side of the prism enters the prism at an angle of θ n , where n is the index of refraction of the prism. Similarly on leaving the prism, the in-going angle is θ / n − A from the normal, and the outgoing angle, relative to the prism, is n(λ n − A). So the beam leaving the prism is at an angle of

θ ′ = n(θ n − A) + A from the optical axis. So θ − θ ′ = (n − 1) A. At the plane of the source S0 , we can calculate the d = tan(θ − θ ′) a ≈ (θ − θ ′)a = (n − 1) Aa ⇒ d = 2aA(n − 1). 2 (b) To find the spacing of fringes on a screen, we use (2.00 m + 0.200 m) (5.00 × 10−7 m) Rλ Rλ Δy = = = = 1.57 × 10−3 m. 2aA( n − 1) 2(0.200 m) (3.50 × 10−3 rad) (1.50 − 1.00) d EVALUATE: The fringe spacing is proportional to the wavelength of the light. The biprism serves as an alternative to two closely spaced narrow slits. height of one image above the source:

36

DIFFRACTION

36.1.

IDENTIFY: Use y = x tan θ to calculate the angular position θ of the first minimum. The minima are located by mλ , m = ±1, ± 2,… First minimum means m = 1 and sin θ1 = λ / a and λ = a sin θ1. Use this Eq.(36.2): sin θ = a equation to calculate λ . SET UP: The central maximum is sketched in Figure 36.1. EXECUTE: y1 = x tan θ1 y tan θ1 = 1 = x 1.35 × 10−3 m = 0.675 × 10−3 2.00 m θ1 = 0.675 × 10−3 rad Figure 36.1

λ = a sin θ1 = (0.750 × 10−3 m)sin(0.675 × 10−3 rad) = 506 nm EVALUATE: θ1 is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been used. 36.2.

IDENTIFY: The angle is small, so ym = x SET UP:

y1 = 10.2 mm

xλ xλ (0.600 m)(5.46 × 10−7 m) ⇒a = = 3.21 × 10−5 m. a y1 10.2 × 10−3 m EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit. mλ IDENTIFY: The dark fringes are located at angles θ that satisfy sin θ = , m = ±1, ± 2, .... a SET UP: The largest value of sin θ is 1.00. EXECUTE:

36.3.

mλ . a

y1 =

EXECUTE: (a) Solve for m that corresponds to sin θ = 1 : m =

a

λ

=

0.0666 × 10−3 m = 113.8. The largest value m 585 × 10−9 m

can have is 113. m = ±1 , ±2 , …, ±113 gives 226 dark fringes. ⎛ 585 × 10−9 m ⎞ (b) For m = ±113, sin θ = ±113 ⎜ ⎟ = ±0.9926 and θ = ±83.0° . −3 ⎝ 0.0666 × 10 m ⎠

36.4.

EVALUATE: When the slit width a is decreased, there are fewer dark fringes. When a < λ there are no dark fringes and the central maximum completely fills the screen. mλ IDENTIFY and SET UP: λ / a is very small, so the approximate expression ym = R is accurate. The distance a between the two dark fringes on either side of the central maximum is 2 y1 .

36.5.

λR

(633 × 10−9 m)(3.50 m) = 2.95 × 10 −3 m = 2.95 mm . 2 y1 = 5.90 mm . a 0.750 × 10 −3 m EVALUATE: When a is decreased, the width 2 y1 of the central maximum increases. mλ IDENTIFY: The minima are located by sin θ = a SET UP: a = 12.0 cm . x = 40.0 cm . EXECUTE:

y1 =

=

36-1

36-2

36.6.

Chapter 36

⎛ 9.00 cm ⎞ ⎛λ⎞ EXECUTE: The angle to the first minimum is θ = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 48.6°. a ⎝ ⎠ ⎝ 12.00 cm ⎠ So the distance from the central maximum to the first minimum is just y1 = x tan θ = (40.0 cm) tan(48.6°) = ±45.4 cm. EVALUATE: 2λ / a is greater than 1, so only the m = 1 minimum is seen. IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is given by λ 1 v sin θ = . The time between crests is the period T. f = and λ = . a T f SET UP: The time between crests is the period, so T = 1.0 h . 1 1 v 800 km/h EXECUTE: (a) f = = = 1.0 h −1 . λ = = = 800 km . T 1.0 h f 1.0 h −1 800 km and θ = 10.2° . 4500 km 800 km Australia-Antarctica: sin θ = and θ = 12.5° . 3700 km EVALUATE: Diffraction effects are observed when the wavelength is about the same order of magnitude as the dimensions of the opening through which the wave passes. IDENTIFY: We can model the hole in the concrete barrier as a single slit that will produce a single-slit diffraction pattern of the water waves on the shore. SET UP: For single-slit diffraction, the angles at which destructive interference occurs are given by sinθm = mλ/a, where m = 1, 2, 3, …. EXECUTE: (a) The frequency of the water waves is f = 75.0 min −1 = 1.25 s −1 = 1.25 Hz, so their wavelength is λ = v/f = (15.0 cm/s)/(1.25 Hz) = 12.0 cm. At the first point for which destructive interference occurs, we have tan θ = (0.613 m)/(3.20 m) ⇒ θ = 10.84°. a sin θ = λ and a = λ/sin θ = (12.0 cm)/(sin 10.84°) = 63.8 cm. (b) First find the angles at which destructive interference occurs. sin θ2 = 2λ/a = 2(12.0 cm)/(63.8 cm) → θ2 = ±22.1°

(b) Africa-Antarctica: sin θ =

36.7.

sin θ3 = 3λ/a = 3(12.0 cm)/(63.8 cm) → θ3 = ±34.3° sin θ4 = 4λ/a = 4(12.0 cm)/(63.8 cm) → θ4 = ±48.8°

36.8.

sin θ5 = 5λ/a = 5(12.0 cm)/(63.8 cm) → θ5 = ±70.1° EVALUATE: These are large angles, so we cannot use the approximation that θm ≈ mλ/a. mλ IDENTIFY: The minima are located by sin θ = . For part (b) apply Eq.(36.7). a SET UP: For the first minimum, m = 1 . The intensity at θ = 0 is I 0 . mλ mλ λ = sin 90.0° = 1 = = . Thus a = λ = 580 nm = 5.80 × 10 −4 mm. a a a (b) According to Eq.(36.7),

EXECUTE: (a) sinθ =

I ⎧⎪ sin [π a (sin θ ) λ ] ⎪⎫ ⎪⎧ sin [π (sin π / 4) ] ⎪⎫ =⎨ ⎬ =⎨ ⎬ = 0.128. I 0 ⎪⎩ π a (sin θ ) λ ⎭⎪ ⎪⎩ π (sin π / 4) ⎭⎪ 2

2

I ⎧⎪ sin [ (π / 2)(sin π / 4) ] ⎫⎪ =⎨ ⎬ = 0.81. As a / λ decreases, I 0 ⎪⎩ (π / 2)(sin π / 4) ⎪⎭ 2

EVALUATE: If a = λ / 2 , for example, then at θ = 45° , 36.9.

the screen becomes more uniformly illuminated. IDENTIFY and SET UP: v = f λ gives λ . The person hears no sound at angles corresponding to diffraction minima. The diffraction minima are located by sin θ = mλ / a, m = ±1, ± 2,… Solve for θ . EXECUTE: λ = v / f = (344 m/s) /(1250 Hz) = 0.2752 m; a = 1.00 m. m = ±1, θ = ±16.0°; m = ±2, θ = ±33.4°; m = ±3, θ = ±55.6°; no solution for larger m EVALUATE: λ / a = 0.28 so for the large wavelength sound waves diffraction by the doorway is a large effect. Diffraction would not be observable for visible light because its wavelength is much smaller and λ / a V 1.

Diffraction

36.10.

IDENTIFY: Compare E y to the expression E y = Emax sin( kx − ωt ) and determine k, and from that calculate λ . mλ . a SET UP: c = 3.00 × 108 m/s . The first dark band corresponds to m = 1. 2π 2π 2π EXECUTE: (a) E = Emax sin( kx − ωt ) . k = ⇒λ = = = 5.24 × 10−7 m . λ k 1.20 × 107 m −1 c 3.0 × 108 m s fλ =c⇒ f = = = 5.73 × 1014 Hz . λ 5.24 × 10−7 m λ 5.24 × 10−7 m a= = = 1.09 × 10−6 m . (b) a sin θ = λ . sin θ sin 28.6° −7 (c) a sin θ = mλ (m = 1, 2, 3, . . .). sin θ 2 = ±2 λ = ±2 5.24 × 10−6 m and θ 2 = ±74D . a 1.09 × 10 m mλ EVALUATE: For m = 3 , is greater than 1 so only the first and second dark bands appear. a IDENTIFY and SET UP: sinθ = λ / a locates the first minimum. y = x tan θ . EXECUTE: tanθ = y x = (36.5 cm) (40.0 cm) and θ = 42.38°.

f = c / λ . The dark bands are located by sin θ =

36.11.

36.12.

a = λ sinθ = (620 × 10 −9 m) (sin 42.38°) = 0.920 μ m EVALUATE: θ = 0.74 rad and sin θ = 0.67 , so the approximation sin θ ≈ θ would not be accurate. mλ IDENTIFY: The angle is small, so ym = x applies. a SET UP: The width of the central maximum is 2y1 , so y1 = 3.00 mm .

xλ xλ (2.50 m)(5.00 × 10−7 m) ⇒a= = = 4.17 × 10−4 m. a y1 3.00 × 10−3m xλ (2.50 m)(5.00 × 10−5 m) = = 4.17 × 10−2 m = 4.2 cm. (b) a = y1 3.00 × 10 −3 m xλ (2.50 m)(5.00 × 10−10 m) = = 4.17 × 10−7 m. (c) a = y1 3.00 × 10−3 m EVALUATE: The ratio a / λ stays constant, so a is smaller when λ is smaller. IDENTIFY: Calculate the angular positions of the minima and use y = x tan θ to calculate the distance on the screen between them. (a) SET UP: The central bright fringe is shown in Figure 36.13a. EXECUTE: The first minimum is located by EXECUTE: (a) y1 =

36.13.

sin θ1 =

λ

= a 633 × 10−9 m = 1.809 × 10−3 0.350 × 10−3 m θ1 = 1.809 × 10 −3 rad Figure 36.13a y1 = x tan θ1 = (3.00 m) tan(1.809 × 10−3 rad) = 5.427 × 10 −3 m w = 2 y1 = 2(5.427 × 10−3 m) = 1.09 × 10 −2 m = 10.9 mm (b) SET UP: The first bright fringe on one side of the central maximum is shown in Figure 36.13b. EXECUTE: w = y2 − y1 y1 = 5.427 × 10−3 m (part (a)) 2λ sin θ 2 = = 3.618 × 10−3 a θ 2 = 3.618 × 10−3 rad y2 = x tan θ 2 = 1.085 × 10−2 m

Figure 36.13b w = y2 − y1 = 1.085 × 10 −2 m − 5.427 × 10 −3 m = 5.4 mm EVALUATE: The central bright fringe is twice as wide as the other bright fringes.

36-3

36-4

Chapter 36

36.14.

IDENTIFY:

2

SET UP: EXECUTE:

⎛ sin( β / 2) ⎞ 2π I = I0 ⎜ a sin θ . ⎟ . β= λ ⎝ β /2 ⎠ The angle θ is small, so sin θ ≈ tan θ ≈ y / x .

β=

2π a

λ

(a) y = 1.00 × 10−3 m :

sin θ ≈

β 2

=

2π a y 2π (4.50 × 10−4 m) = y = (1520 m −1 ) y. λ x (6.20 × 10−7 m)(3.00 m)

(1520 m −1 )(1.00 × 10−3 m) = 0.760. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(0.760) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.822 I 0 ⎝ 0.760 ⎠ ⎝ β 2 ⎠

(b) y = 3.00 × 10−3 m :

β 2

=

2

(1520 m −1 )(3.00 × 10−3 m) = 2.28. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(2.28) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.111I 0 . β 2 ⎝ 2.28 ⎠ ⎝ ⎠

(c) y = 5.00 × 10−3 m :

β 2

=

2

(1520 m −1 )(5.00 × 10−3 m) = 3.80. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(3.80) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.0259 I 0 . β 2 ⎝ 3.80 ⎠ ⎝ ⎠

(6.20 × 10−7 m)(3.00 m) = 4.1 mm . The distances in parts (a) a 4.50 × 10−4 m and (b) are within the central maximum. y = 5.00 mm is within the first secondary maximum. (a) IDENTIFY: Use Eq.(36.2) with m = 1 to locate the angular position of the first minimum and then use y = x tan θ to find its distance from the center of the screen. SET UP: The diffraction pattern is sketched in Figure 36.15. EVALUATE: The first minimum occurs at y1 =

36.15.

λx

2

=

sin θ1 =

λ

= a 540 × 10−9 m = 2.25 × 10−3 0.240 × 10−3 m θ1 = 2.25 × 10−3 rad

Figure 36.15 y1 = x tan θ1 = (3.00 m) tan(2.25 × 10 −3 rad) = 6.75 × 10−3 m = 6.75 mm (b) IDENTIFY and SET UP: Use Eqs.(36.5) and (36.6) to calculate the intensity at this point. EXECUTE: Midway between the center of the central maximum and the first minimum implies 1 y = (6.75 mm) = 3.375 × 10−3 m. 2 y 3.375 × 10−3 m tan θ = = = 1.125 × 10−3 ; θ = 1.125 × 10−3 rad x 3.00 m The phase angle β at this point on the screen is 2π ⎛ 2π ⎞ (0.240 × 10−3 m)sin(1.125 × 10 −3 rad) = π . ⎟ a sin θ = −9 540 10 m λ × ⎝ ⎠

β =⎜

2

⎛ sin β / 2 ⎞ 2 ⎛ sin π / 2 ⎞ −6 Then I = I 0 ⎜ ⎟ = (6.00 × 10 W/m ) ⎜ ⎟ ⎝ π /2 ⎠ ⎝ β /2 ⎠ ⎛ 4 ⎞ I = ⎜ 2 ⎟ (6.00 × 10−6 W/m 2 ) = 2.43 × 10 −6 W/m 2 . ⎝π ⎠ EVALUATE: The intensity at this point midway between the center of the central maximum and the first minimum is less than half the maximum intensity. Compare this result to the corresponding one for the two-slit pattern, Exercise 35.23. 2

Diffraction

36.16.

36-5

IDENTIFY: In the single-slit diffraction pattern, the intensity is a maximum at the center and zero at the dark spots. At other points, it depends on the angle at which one is observing the light. SET UP: Dark fringes occur when sin θm = mλ/a, where m = 1, 2, 3, …, and the intensity is given by 2

⎛ sin β / 2 ⎞ π a sin θ . I0 ⎜ ⎟ , where β / 2 = λ ⎝ β /2 ⎠ EXECUTE: (a) At the maximum possible angle, θ = 90°, so

mmax = (asin90°)/λ = (0.0250 mm)/(632.8 nm) = 39.5 Since m must be an integer and sin θ must be ≤ 1, mmax = 39. The total number of dark fringes is 39 on each side of the central maximum for a total of 78. (b) The farthest dark fringe is for m = 39, giving sinθ39 = (39)(632.8 nm)/(0.0250 mm) ⇒ θ39 = ±80.8° (c) The next closer dark fringe occurs at sinθ38 = (38)(632.8 nm)/(0.0250 mm) ⇒θ38 = 74.1°. The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is I = 2

⎛ sin β / 2 ⎞ π a sin θ π (0.0250 mm)sin(77.45°) = = 121.15 rad, which I0 ⎜ ⎟ , where β / 2 = λ 632.8 nm β / 2 ⎝ ⎠ 2

36.17.

⎡ sin(121.15 rad) ⎤ -4 2 gives I = ( 8.50 W/m 2 ) ⎢ ⎥ = 5.55 × 10 W/m ⎣ 121.15 rad ⎦ EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible. IDENTIFY and SET UP: Use Eq.(36.6) to calculate λ and use Eq.(36.5) to calculate I. θ = 3.25°, β = 56.0 rad, a = 0.105 × 10 −3 m. ⎛ 2π ⎞ ⎟ a sin θ so ⎝ λ ⎠

β =⎜

(a) EXECUTE:

λ=

2π a sin θ

β

=

2π (0.105 × 10−3 m)sin 3.25° = 668 nm 56.0 rad 2

⎛ sin β / 2 ⎞ ⎛ 4 ⎞ 4 2 (b) I = I 0 ⎜ [sin(28.0 rad)]2 = 9.36 × 10 −5 I 0 ⎟ = I 0 ⎜ 2 ⎟ (sin( β / 2)) = I 0 (56.0 rad) 2 ⎝ β /2 ⎠ ⎝β ⎠ EVALUATE: At the first minimum β = 2π rad and at the point considered in the problem β = 17.8π rad, so the point is well outside the central maximum. Since β is close to mπ with m = 18, this point is near one of the minima. The intensity here is much less than I 0 .

36.18.

IDENTIFY: Use β =

2π a

λ

sin θ to calculate β .

SET UP: The total intensity is given by drawing an arc of a circle that has length E0 and finding the length of the chord which connects the starting and ending points of the curve. E 2 2π a 2π a λ EXECUTE: (a) β = = π . From Figure 36.18a, π p = E0 ⇒ E p = E0 . sin θ = λ λ 2a 2 π 2

4I ⎛2⎞ The intensity is I = ⎜ ⎟ I 0 = 20 = 0.405 I 0 . This agrees with Eq.(36.5). π ⎝π ⎠ 2π a 2π a λ (b) β = = 2π . From Figure 36.18b, it is clear that the total amplitude is zero, as is the intensity. sin θ = λ λ a This also agrees with Eq.(36.5). E 2 2π a 2π a 3λ E0 . The intensity is (c) β = = 3π . From Figure 36.18c, 3π p = E0 ⇒ E p = sin θ = λ λ 2a 2 3π 2

4 ⎛ 2 ⎞ I = ⎜ ⎟ I 0 = 2 I 0 . This agrees with Eq.(36.5). 9π ⎝ 3π ⎠

36-6

Chapter 36

EVALUATE: In part (a) the point is midway between the center of the central maximum and the first minimum. In part (b) the point is at the first maximum and in (c) the point is approximately at the location of the first secondary maximum. The phasor diagrams help illustrate the rapid decrease in intensity at successive maxima.

Figure 36.18 36.19.

IDENTIFY: The space between the skyscrapers behaves like a single slit and diffracts the radio waves. SET UP: Cancellation of the waves occurs when a sin θ = mλ, m = 1, 2, 3, …, and the intensity of the waves is 2

⎛ sin β / 2 ⎞ π a sin θ . given by I 0 ⎜ ⎟ , where β / 2 = λ β / 2 ⎝ ⎠ EXECUTE: (a) First find the wavelength of the waves: λ = c/f = (3.00 × 108 m/s)/(88.9 MHz) = 3.375 m For no signal, a sin θ = mλ. m = 1: sin θ1 = (1)(3.375 m)/(15.0 m) ⇒ θ1 = ±13.0° m = 2: sin θ2 = (2)(3.375 m)/(15.0 m) ⇒ θ2 = ±26.7° m = 3: sin θ3 = (3)(3.375 m)/(15.0 m) ⇒ θ3 = ±42.4° m = 4: sin θ4 = (4)(3.375 m)/(15.0 m) ⇒ θ4 = ±64.1° 2

⎛ sin β / 2 ⎞ π a sin θ π (15.0 m)sin(5.00°) = = 1.217 rad (b) I 0 ⎜ ⎟ , where β / 2 = λ 3.375 m β / 2 ⎝ ⎠ 2

⎡ sin(1.217 rad) ⎤ 2 I = ( 3.50 W/m 2 ) ⎢ ⎥ = 2.08 W/m ⎣ 1.217 rad ⎦

36.20.

EVALUATE: The wavelength of the radio waves is very long compared to that of visible light, but it is still considerably shorter than the distance between the buildings. IDENTIFY: The net intensity is the product of the factor due to single-slit diffraction and the factor due to double slit interference. 2

φ⎞ ⎛ sin β / 2 ⎞ ⎛ SET UP: The double-slit factor is I DS = I 0 ⎜ cos 2 ⎟ and the single-slit factor is ISS = ⎜ ⎟ . 2⎠ ⎝ ⎝ β /2 ⎠ EXECUTE: (a) d sinθ = mλ ⇒ sinθ = mλ/d. sinθ1 = λ/d, sinθ2 = 2λ/d, sinθ3 = 3λ/d, sinθ4 = 4λ/d π a sin θ π (d / 3)sin θ (b) At the interference bright fringes, cos2φ/2 = 1 and β / 2 = = .

λ λ π ( d / 3)(λ / d ) = π / 3 . The intensity is therefore At θ1, sin θ1 = λ/d, so β / 2 = λ 2

φ ⎞ ⎛ sin β / 2 ⎞ ⎛ sin π / 3 ⎞ ⎛ I1 = I 0 ⎜ cos 2 ⎟ ⎜ ⎟ = I 0 (1) ⎜ ⎟ = 0.684 I0 β 2 / 2 ⎝ ⎠⎝ ⎝ π /3 ⎠ ⎠ At θ2, sin θ2 = 2λ/d, so β / 2 =

2

π ( d / 3)(2λ / d ) = 2π / 3 . Using the same procedure as for θ1, we have I2 = λ

2

⎛ sin 2π / 3 ⎞ I 0 (1) ⎜ ⎟ = 0.171 I0 ⎝ 2π / 3 ⎠ At θ3, we get β / 2 = π , which gives I3 = 0 since sin π = 0. 2

⎛ sin 4π / 3 ⎞ At θ4, sin θ4 = 4λ/d, so β / 2 = 4π / 3 , which gives I 4 = I 0 ⎜ ⎟ = 0.0427 I0 ⎝ 4π / 3 ⎠ (c) Since d = 3a, every third interference maximum is missing. (d) In Figure 36.12c in the textbook, every fourth interference maximum at the sides is missing because d = 4a.

Diffraction

36.21.

36-7

EVALUATE: The result in this problem is different from that in Figure 36.12c because in this case d = 3a, so every third interference maximum at the sides is missing. Also the “envelope” of the intensity function decreases more rapidly here than in Figure 36.12c because the first diffraction minimum is reached sooner, and the decrease in intensity from one interference maximum to the next is faster for a = d/3 than for a = d/4. (a) IDENTIFY and SET UP: The interference fringes (maxima) are located by d sin θ = mλ , with 2

⎛ sin β / 2 ⎞ ⎛ 2π ⎞ m = 0, ± 1, ± 2, …. The intensity I in the diffraction pattern is given by I = I 0 ⎜ ⎟ , with β = ⎜ ⎟ a sin θ . β / 2 ⎝ λ ⎠ ⎝ ⎠ We want m = ±3 in the first equation to give θ that makes I = 0 in the second equation. ⎛ 2π ⎞ ⎛ 3λ ⎞ EXECUTE: d sin θ = mλ gives β = ⎜ ⎟ a ⎜ ⎟ = 2π (3a / d ). ⎝ λ ⎠ ⎝ d ⎠ sin β / 2 = 0 so β = 2π and then 2π = 2π (3a / d ) and ( d / a ) = 3. β /2 (b) IDENTIFY and SET UP: Fringes m = 0, ± 1, ± 2 are within the central diffraction maximum and the m = ±3 fringes coincide with the first diffraction minimum. Find the value of m for the fringes that coincide with the second diffraction minimum. EXECUTE: Second minimum implies β = 4π .

I = 0 says

⎛ 2π ⎞ ⎛ 2π ⎞ ⎛ mλ ⎞ β = ⎜ ⎟ a sin θ = ⎜ ⎟ a ⎜ ⎟ = 2π m(a / d ) = 2π (m / 3) ⎝ λ ⎠ ⎝ λ ⎠ ⎝ d ⎠ Then β = 4π says 4π = 2π ( m / 3) and m = 6. Therefore the m = ±4 and m = +5 fringes are contained within the

36.22.

36.23.

first diffraction maximum on one side of the central maximum; two fringes. EVALUATE: The central maximum is twice as wide as the other maxima so it contains more fringes. IDENTIFY and SET UP: Use Figure 36.14b in the textbook. There is totally destructive interference between slits whose phasors are in opposite directions. EXECUTE: By examining the diagram, we see that every fourth slit cancels each other. EVALUATE: The total electric field is zero so the phasor diagram corresponds to a point of zero intensity. The first two maxima are at φ = 0 and φ = π , so this point is not midway between two maxima. (a) IDENTIFY and SET UP: If the slits are very narrow then the central maximum of the diffraction pattern for each slit completely fills the screen and the intensity distribution is given solely by the two-slit interference. The maxima are given by d sin θ = mλ so sin θ = mλ / d . Solve for θ . λ 580 × 10−9 m EXECUTE: 1st order maximum: m = 1, so sin θ = = = 1.094 × 10−3 ; θ = 0.0627° d 0.530 × 10−3 m 2λ 2nd order maximum: m = 2, so sin θ = = 2.188 × 10−3 ; θ = 0.125° d 2

⎛ sin β / 2 ⎞ (b) IDENTIFY and SET UP: The intensity is given by Eq.(36.12): I = I 0 cos (φ / 2) ⎜ ⎟ . Calculate φ and ⎝ β /2 ⎠ β at each θ from part (a). 2

⎛ 2π d ⎞ ⎛ 2π d ⎞⎛ mλ ⎞ 2 2 ⎟ sin θ = ⎜ ⎟⎜ ⎟ = 2π m, so cos (φ / 2) = cos ( mπ ) = 1 d λ λ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (Since the angular positions in part (a) correspond to interference maxima.) ⎛ 2π a ⎞ ⎛ 2π a ⎞⎛ mλ ⎞ ⎛ 0.320 mm ⎞ β =⎜ ⎟ sin θ = ⎜ ⎟⎜ ⎟ = 2π m(a / d ) = m 2π ⎜ ⎟ = m(3.794 rad) ⎝ λ ⎠ ⎝ λ ⎠⎝ d ⎠ ⎝ 0.530 mm ⎠ EXECUTE:

φ =⎜

2

⎛ sin(3.794 / 2) rad ⎞ 1st order maximum: m = 1, so I = I 0 (1) ⎜ ⎟ = 0.249 I 0 ⎝ (3.794 / 2) rad ⎠ 2

36.24.

⎛ sin 3.794 rad ⎞ 2nd order maximum: m = 2, so I = I 0 (1) ⎜ ⎟ = 0.0256 I 0 ⎝ 3.794 rad ⎠ EVALUATE: The first diffraction minimum is at an angle θ given by sin θ = λ / a so θ = 0.104°. The first order fringe is within the central maximum and the second order fringe is inside the first diffraction maximum on one side of the central maximum. The intensity here at this second fringe is much less than I 0 . IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes occur when d sin θ = m′ λ.

36-8

Chapter 36

EXECUTE: (a) The angles are the same for cancellation, so dividing the equations gives d/a = m′ /m ⇒ m′ /m = 7 ⇒ m′ = 7m

36.25.

When m = 1, m′ = 7; when m = 2, m′ = 14, and so forth, so every 7th bright fringe is missing from the double-slit interference pattern. EVALUATE: (b) The result is independent of the wavelength, so every 7th fringe will be cancelled for all wavelengths. But the bright interference fringes occur when d sin θ = mλ, so the location of the cancelled fringes does depend on the wavelength. IDENTIFY and SET UP: The phasor diagrams are similar to those in Fig.36.14. An interference minimum occurs when the phasors add to zero. EXECUTE: (a) The phasor diagram is given in Figure 36.25a

Figure 36.25a

There is destructive interference between the light through slits 1 and 3 and between 2 and 4. (b) The phasor diagram is given in Figure 36.25b.

Figure 36.25b

There is destructive interference between the light through slits 1 and 2 and between 3 and 4. (c) The phasor diagram is given in Figure 36.25c.

Figure 36.25c

36.26.

36.27.

There is destructive interference between light through slits 1 and 3 and between 2 and 4. EVALUATE: Maxima occur when φ = 0, 2π , 4π , etc. Our diagrams show that there are three minima between the maxima at φ = 0 and φ = 2π . This agrees with the general result that for N slits there are N − 1 minima between each pair of principal maxima. IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes occur when d sin θ = m′ λ. EXECUTE: (a) The angle at which the first bright fringe occurs is given by tan θ1 = (1.53 mm)/(2500 mm) ⇒ θ1 = 0.03507°. d sin θ1 = λ and d = λ/(sinθ1) = (632.8 nm)/sin(0.03507°) = 0.00103 m = 1.03 mm (b) The 7th double-slit interference bright fringe is just cancelled by the 1st diffraction dark fringe, so sinθdiff = λ/a and sinθinterf = 7λ/d The angles are equal, so λ/a = 7λ/d → a = d/7 = (1.03 mm)/7 = 0.148 mm. EVALUATE: We can generalize that if d = na, where n is a positive integer, then every nth double-slit bright fringe will be missing in the pattern. mλ IDENTIFY: The diffraction minima are located by sin θ = d and the two-slit interference maxima are located a miλ . The third bright band is missing because the first order single slit minimum occurs at the same by sinθ = d angle as the third order double slit maximum. 3 cm SET UP: The pattern is sketched in Figure 36.27. tan θ = , so θ = 1.91° . 90 cm

λ 500 nm = = 1.50 × 10 4 nm = 15.0 μ m (width) sinθ sin1.91° 3λ 3(500 nm) Double-slit bright fringe: d sin θ = 3λ and d = = = 4.50 × 104 nm = 45.0 μ m (separation) . sinθ sin1.91° EXECUTE: Single-slit dark spot: a sin θ = λ and a =

Diffraction

36-9

EVALUATE: Note that d / a = 3.0 .

Figure 36.27 36.28.

36.29.

IDENTIFY: The maxima are located by d sin θ = mλ . SET UP: The order corresponds to the values of m. EXECUTE: First-order: d sin θ1 = λ . Fourth-order: d sin θ 4 = 4λ . d sin θ 4 4λ , sinθ 4 = 4sin θ1 = 4sin8.94° and θ 4 = 38.4° . = d sin θ1 λ EVALUATE: We did not have to solve for d. IDENTIFY and SET UP: The bright bands are at angles θ given by d sin θ = mλ. Solve for d and then solve for θ for the specified order. EXECUTE: (a) θ = 78.4° for m = 3 and λ = 681 nm, so d = mλ / sin θ = 2.086 × 10−4 cm The number of slits per cm is 1/ d = 4790 slits/cm (b) 1st order: m = 1, so sin θ = λ / d = (681× 10−9 m) /(2.086 × 10−6 m) and θ = 19.1°

36.30.

2nd order: m = 2, so sin θ = 2λ / d and θ = 40.8° (c) For m = 4, sinθ = 4λ / d is greater than 1.00, so there is no 4th-order bright band. EVALUATE: The angular position of the bright bands for a particular wavelength increases as the order increases. IDENTIFY: The bright spots are located by d sin θ = mλ . SET UP: Third-order means m = 3 and second-order means m = 2 . mλ mλ mλ = d = constant , so r r = v v . EXECUTE: sin θ sin θ r sin θ v

36.31.

⎛ m ⎞⎛ λ ⎞ ⎛ 2 ⎞⎛ 400 nm ⎞ sin θ v = sin θ r ⎜ v ⎟⎜ v ⎟ = (sin 65.0°) ⎜ ⎟⎜ ⎟ = 0.345 and θ v = 20.2° . m λ 3 ⎠⎝ 700 nm ⎠ ⎝ ⎝ r ⎠⎝ r ⎠ EVALUATE: The third-order line for a particular λ occurs at a larger angle than the second-order line. In a given order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm). IDENTIFY and SET UP: Calculate d for the grating. Use Eq.(36.13) to calculate θ for the longest wavelength in the visible spectrum and verify that θ is small. Then use Eq.(36.3) to relate the linear separation of lines on the screen to the difference in wavelength. ⎛ 1 ⎞ −5 EXECUTE: (a) d = ⎜ ⎟ cm = 1.111 × 10 m ⎝ 900 ⎠ For λ = 700 nm, λ / d = 6.3 × 10−2. The first-order lines are located at sin θ = λ / d ; sin θ is small enough for sin θ ≈ θ to be an excellent approximation. (b) y = xλ / d , where x = 2.50 m. The distance on the screen between 1st order bright bands for two different wavelengths is Δy = x( Δλ ) / d , so

36.32.

Δλ = d ( Δy ) / x = (1.111 × 10 −5 m)(3.00 × 10−3 m) /(2.50 m) = 13.3 nm EVALUATE: The smaller d is (greater number of lines per cm) the smaller the Δλ that can be measured. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 350 slits mm ⇒ d = = 2.86 × 10 −6 m 3.50 × 105 m −1

36-10

Chapter 36

⎛ 4.00 × 10−7 m ⎞ ⎛λ ⎞ D m = 1: θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 8.05 . −6 × 2.86 10 m ⎝d⎠ ⎝ ⎠ ⎛ 7.00 × 10−7 m ⎞ ⎛λ ⎞ D = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 14.18 . Δθ1 = 14.18° − 8.05° = 6.13°. −6 ⎝d⎠ ⎝ 2.86 × 10 m ⎠

EXECUTE:

θ 700

⎛ 3(4.00 × 10 −7 m) ⎞ ⎛ 3λ ⎞ m = 3 : θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 24.8° . −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ ⎛ 3(7.00 × 10−7 m) ⎞ ⎛ 3λ ⎞ ⎟ = 47.3° . Δθ1 = 47.3° − 24.8° = 22.5°. ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ EVALUATE: Δθ is larger in third order. IDENTIFY: The maxima are located by d sin θ = mλ . SET UP: d = 1.60 × 10 −6 m

θ 700 = arcsin ⎜

36.33.

EXECUTE:

36.34.

⎛ m[6.328 × 10−7 m] ⎞ ⎛ mλ ⎞ ⎟ = arcsin([0.396]m) . For m = 1, θ1 = 23.3° . For ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 1.60 × 10 m ⎠

θ = arcsin ⎜

m = 2 , θ 2 = 52.3° . There are no other maxima. EVALUATE: The reflective surface produces the same interference pattern as a grating with slit separation d. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 5000 slits cm ⇒ d = = 2.00 × 10−6 m. 5.00 × 105 m −1 d sin θ (2.00 × 10−6 m)sin13.5D = = 4.67 × 10−7 m. m 1 ⎛ 2(4.67 × 10−7 m) ⎞ ⎛ mλ ⎞ D (b) m = 2 : θ = arcsin ⎜ ⎟ = 27.8 . ⎟ = arcsin ⎜ −6 × d 2.00 10 m ⎝ ⎠ ⎝ ⎠ EXECUTE: (a) λ =

36.35.

EVALUATE: Since the angles are fairly small, the second-order deviation is approximately twice the first-order deviation. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 350 slits mm ⇒ d = = 2.86 × 10 −6 m 3.50 × 105 m −1

⎛ m(5.20 × 10−7 m) ⎞ ⎛ mλ ⎞ ⎟ = arcsin((0.182) m) . ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ m = 1: θ = 10.5°; m = 2 : θ = 21.3°; m = 3 : θ = 33.1°. EVALUATE: The angles are not precisely proportional to m, and deviate more from being proportional as the angles increase. EXECUTE:

36.36.

θ = arcsin ⎜

IDENTIFY: The resolution is described by R = SET UP:

λ = Nm . Maxima are located by d sin θ = mλ . Δλ

For 500 slits/mm, d = (500 slits mm) −1 = (500,000 slits m) −1 .

EXECUTE: (a) N =

λ mΔλ

=

6.5645 × 10−7 m = 1820 slits. 2(6.5645 × 10 −7 m − 6.5627 × 10−7 m)

⎛ mλ ⎞ −1 −7 −1 (b) θ = sin −1 ⎜ ⎟ ⇒ θ1 = sin ((2)(6.5645 × 10 m)(500,000 m )) = 41.0297° and ⎝ d ⎠

θ 2 = sin −1 ((2)(6.5627 × 10−7 m)(500,000 m −1 )) = 41.0160° . Δθ = 0.0137° EVALUATE: d cosθ dθ = λ / N , so for 1820 slits the angular interval Δθ between each of these maxima and the λ 6.56 × 10 −7 m first adjacent minimum is Δθ = = = 0.0137°. This is the same as the angular Nd cosθ (1820)(2.0 × 10 −6 m)cos 41° 36.37.

separation of the maxima for the two wavelengths and 1820 slits is just sufficient to resolve these two wavelengths in second order. IDENTIFY: The resolving power depends on the line density and the width of the grating. SET UP: The resolving power is given by R = Nm = = λ/Δλ. EXECUTE: (a) R = Nm = (5000 lines/cm)(3.50 cm)(1) = 17,500 (b) The resolving power needed to resolve the sodium doublet is R = λ/Δλ = (589 nm)/(589.59 nm – 589.00 nm) = 998

Diffraction

36-11

so this grating can easily resolve the doublet. (c) (i) R = λ/Δλ. Since R = 17,500 when m = 1, R = 2 × 17,500 = 35,000 for m = 2. Therefore Δλ = λ/R = (587.8 nm)/35,000 = 0.0168 nm

λmin = λ +Δλ = 587.8002 nm + 0.0168 nm = 587.8170 nm (ii) λmax = λ − Δλ = 587.8002 nm − 0.0168 nm = 587.7834 nm EVALUATE: (iii) Therefore the range of resolvable wavelengths is 587.7834 nm < λ < 587.8170 nm. 36.38.


λ 587.8002 nm 587.8002 N 3302 slits . = = = 3302 slits . = = 2752 0.178 1.20 cm 1.20 cm cm mΔλ (587.9782 nm − 587.8002 nm) EVALUATE: A smaller number of slits would be needed to resolve these two lines in higher order. IDENTIFY and SET UP: The maxima occur at angles θ given by Eq.(36.16), 2d sin θ = mλ , where d is the spacing between adjacent atomic planes. Solve for d. EXECUTE: second order says m = 2. mλ 2(0.0850 × 10−9 m) d= = = 2.32 × 10−10 m = 0.232 nm 2sin θ 2sin 21.5° EVALUATE: Our result is similar to d calculated in Example 36.5. IDENTIFY: The maxima are given by 2d sin θ = mλ , m = 1 , 2, … SET UP: d = 3.50 × 10−10 m . 2d sin θ EXECUTE: (a) m = 1 and λ = = 2(3.50 × 10−10 m)sin15.0° = 1.81 × 10−10 m = 0.181 nm . This is an x ray. m ⎛ 1.81 × 10 −10 m ⎞ ⎛ λ ⎞ (b) sin θ = m ⎜ ⎟ = m(0.2586) . m = 2 : θ = 31.1° . m = 3 : θ = 50.9° . The equation ⎟ = m⎜ −10 ⎝ 2d ⎠ ⎝ 2[3.50 × 10 m] ⎠ EXECUTE:

36.39.

36.40.

λ = Nm Δλ

N=

doesn't have any solutions for m > 3 . EVALUATE: In this problem λ / d = 0.52. 36.41.

IDENTIFY: Rayleigh's criterion says sin θ = 1.22

λ

D SET UP: The best resolution is 0.3 arcseconds, which is about (8.33 × 10 −5 )° .

1.22λ 1.22(5.5 × 10−7 m) = = 0.46 m sin θ sin(8.33 × 10−5 °) EVALUATE: (b) The Keck telescopes are able to gather more light than the Hale telescope, and hence they can detect fainter objects. However, their larger size does not allow them to have greater resolution⎯atmospheric conditions limit the resolution. EXECUTE: (a) D =

36.42.

IDENTIFY: Apply sin θ = 1.22 SET UP:

36.44.

.

θ = (1/ 60)°

1.22λ 1.22(5.5 × 10−7 m) = = 2.31 × 10−3 m = 2.3 mm sin θ sin(1/ 60)D EVALUATE: The larger the diameter the smaller the angle that can be resolved. λ IDENTIFY: Apply sin θ = 1.22 . D W SET UP: θ = , where W = 28 km and h = 1200 km . θ is small, so sin θ ≈ θ . h 1.22λ h 1.2 × 106 m EXECUTE: D = = 1.22λ = 1.22(0.036 m) = 1.88 m W sin θ 2.8 × 10 4 m EVALUATE: D must be significantly larger than the wavelength, so a much larger diameter is needed for microwaves than for visible wavelengths. λ IDENTIFY: Apply sin θ = 1.22 . D SET UP: θ is small, so sin θ ≈ θ = 1.00 × 10 −8 rad . D sin θ Dθ (8.00 × 106 m)(1.00 × 10−8 ) EXECUTE: λ = ≈ = = 0.0656 m = 6.56 cm 1.22 1.22 1.22 EVALUATE: λ corresponds to microwaves. EXECUTE:

36.43.

λ D

D=

36-12

Chapter 36

36.45.

IDENTIFY and SET UP: The angular size of the first dark ring is given by sin θ1 = 1.22λ / D (Eq.36.17). Calculate

θ1 , and then the diameter of the ring on the screen is 2(4.5 m) tan θ1. ⎛ 620 × 10 −9 m ⎞ sin θ1 = 1.22 ⎜ ⎟ = 0.1022; θ1 = 0.1024 rad −6 ⎝ 7.4 × 10 m ⎠ The radius of the Airy disk (central bright spot) is r = (4.5 m) tan θ1 = 0.462 m. The diameter is 2r = 0.92 m = 92 cm. EVALUATE: λ / D = 0.084. For this small D the central diffraction maximum is broad. IDENTIFY: Rayleigh’s criterion limits the angular resolution. SET UP: Rayleigh’s criterion is sin θ ≈ θ = 1.22 λ/D. EXECUTE: (a) Using Rayleigh’s criterion sinθ ≈ θ = 1.22 λ/D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10–5 rad On the bear this angle subtends a distance x. θ = x/R and x = Rθ = (11.5 m)(1.99 × 10–5 rad) = 2.29 × 10–4 m = 0.23 mm (b) At f/22, D is 4/22 times as large as at f/4. Since θ is proportional to 1/D, and x is proportional to θ, x is 1/(4/22) = 22/4 times as large as it was at f/4. x = (0.229 mm)(22/4) = 1.3 mm EVALUATE: A wide-angle lens, such as one having a focal length of 28 mm, would have a much smaller opening at f/22 and hence would have an even less resolving ability. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means angular separation θ of the objects equals 1.22λ / D. The angular separation θ of the objects is their linear separation divided by their distance from the telescope. 250 × 103 m EXECUTE: θ = , where 5.93 × 1011 m is the distance from earth to Jupiter. Thus θ = 4.216 × 10−7. 5.93 × 1011 m λ 1.22λ 1.22(500 × 10−9 m) Then θ = 1.22 and D = = = 1.45 m θ D 4.216 × 10 −7 EVALUATE: This is a very large telescope mirror. The greater the angular resolution the greater the diameter the lens or mirror must be. EXECUTE:

36.46.

36.47.

36.48.

IDENTIFY: Rayleigh’s criterion says θ res = 1.22

.

y , where s is the distance of the object from the lens and y = 4.00 mm . s y λ yD (4.00 × 10 −3 m)(7.20 × 10−2 m) EXECUTE: = 1.22 . s = = = 429 m . s D 1.22λ 1.22(550 × 10−9 m) EVALUATE: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve resolution that is at the diffraction limit. IDENTIFY and SET UP: Let y be the separation between the two points being resolved and let s be their distance λ y from the telescope. Then the limit of resolution corresponds to 1.22 = . D s EXECUTE: (a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the crater. For the moon, s = 3.8 × 108 m. y = 1.22λ s D . SET UP:

36.49.

λ D

D = 7.20 cm . θ res =

Hubble: D = 2.4 m and λ = 400 nm gives the maximum resolution, so y = 77 m Arecibo: D = 305 m and λ = 0.75 m; y = 1.1 × 106 m yD . Let y ≈ 0.30 (the size of a license plate). s = (0.30 m)(2.4 m) [(1.22)(400 × 10−9 m)] = 1500 km . 1.22λ EVALUATE: D / λ is much larger for the optical telescope and it has a much larger resolution even though the diameter of the radio telescope is much larger.

(b) s =

36.50.

IDENTIFY: Apply sin θ = 1.22

λ

D

.

θ is small, so sin θ ≈ θ . Smallest resolving angle is for short-wavelength light (400 nm). λ 400 × 10−9 m 10,000 mi EXECUTE: θ ≈ 1.22 = (1.22) , where R is the distance to the star. = 9.61× 10−8 rad . θ = SET UP:

R=

R D 5.08 m 16,000 km = = 1.7 × 1011 km . 9.6 × 10−8 rad This is less than a light year, so there are no stars this close.

10,000 mi

θ EVALUATE:

Diffraction

36.51.

IDENTIFY: Let y be the separation between the two points being resolved and let s be their distance from the telescope. The limit of resolution corresponds to 1.22 λ D = y s . s = 4.28 ly = 4.05 × 1016 m . Assume visible light, with λ = 400 m .

SET UP:

y = 1.22 λ s D = 1.22(400 × 10−9 m) (4.05 × 1016 m (10.0 m) = 2.0 × 109 m

EXECUTE:

36.52.

36-13

EVALUATE: The diameter of Jupiter is 1.38 × 108 m, so the resolution is insufficient, by about one order of magnitude. IDENTIFY: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the wavelength

λ → λ′ =

λ

. n SET UP: For y 1.00. The smaller λ in the liquid reduces θ that located the first dark band. 1 IDENTIFY: d = , so the bright fringes are located by 1 sin θ = λ N N SET UP: Red: 1 sin λR = 700 nm . Violet: 1 sin λV = 400 nm . N N sin θ R 7 sin(θ V + 15°) 7 EXECUTE: = . θ R − θ V = 15° → θ R = θ V + 15°. = . Using a trig identify from Appendix B, sin θ V 4 sin θ V 4 sin θ V cos15° + cosθ V sin15° = 7 4 . cos15° + cot θ V sin15° = 7 4 . sin θ V tan θ V = 0.330 ⇒ θ V = 18.3°and θ R = θ V + 15° = 18.3° + 15° = 33.3°. Then 1 sin θ R = 700 nm gives N sin θ R sin 33.3D = = 7.84 × 105 lines m = 7840 lines cm . The spectrum begins at 18 .3D and ends at 33.3D . N= 700 nm 700 × 10−9 m EVALUATE: As N is increased, the angular range of the visible spectrum increases. (a) IDENTIFY and SET UP: The angular position of the first minimum is given by a sin θ = mλ (Eq.36.2), with m = 1. The distance of the minimum from the center of the pattern is given by y = x tan θ .

λ

540 × 10−9 m = 1.50 × 10 −3 ; θ = 1.50 × 10−3 rad a 0.360 × 10−3 m y1 = x tan θ = (1.20 m) tan(1.50 × 10−3 rad) = 1.80 × 10−3 m = 1.80 mm. (Note that θ is small enough for θ ≈ sin θ ≈ tan θ , and Eq.(36.3) applies.) (b) IDENTIFY and SET UP: Find the phase angle β where I = I 0 / 2. Then use Eq.(36.6) to solve for θ and y = x tan θ to find the distance. sin θ =

=

1 From part (a) of Problem 36.53, I = I 0 when β = 2.78 rad. 2 2 π βλ ⎛ ⎞ β = ⎜ ⎟ a sin θ (Eq.(36.6)), so sin θ = . 2π a ⎝ λ ⎠ βλ x (2.78 rad)(540 × 10−9 m)(1.20 m) y = x tan θ ≈ x sin θ ≈ = = 7.96 × 10−4 m = 0.796 mm 2π a 2π (0.360 × 10 −3 m) EVALUATE: The point where I = I 0 / 2 is not midway between the center of the central maximum and the first minimum; see Exercise 36.15. EXECUTE:

Diffraction

36-15

2

36.58.

⎛ sin γ ⎞ I = I0 ⎜ ⎟ . The maximum intensity occurs when the derivative of the intensity function with ⎝ γ ⎠ respect to γ is zero.

IDENTIFY:

SET UP:

d ⎛1⎞ 1 d sin γ = cos γ . ⎜ ⎟=− 2 . dγ ⎝ γ ⎠ γ dγ 2

⎛ sin γ ⎞⎛ cos γ sin γ ⎞ dI d ⎛ sin γ ⎞ cos γ sin γ = I0 − 2 ⎟ =0. − 2 ⇒ γ cos γ = sin γ ⇒ γ = tan γ . ⎜ ⎟ = 2⎜ ⎟⎜ γ γ γ dγ dγ ⎝ γ ⎠ γ γ ⎝ ⎠⎝ ⎠ (b) The graph in Figure 36.58 is a plot of f( γ ) = γ − tan γ . When f (γ ) equals zero, there is an intensity maximum. Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that the three smallest γ -values are γ = 4.49 rad 7.73 rad, and 10.9 rad. EVALUATE: γ = 0 is the central maximum. The three values of γ we found are the locations of the first three secondary maxima. The first four minima are at γ = 3.14 rad , 6.28 rad, 9.42 rad, and 12.6 rad. The maxima are between adjacent minima, but not precisely midway between them.

EXECUTE:

Figure 36.58 36.59.

IDENTIFY and SET UP: Relate the phase difference between adjacent slits to the sum of the phasors for all slits. The 2π d 2π dθ λφ when θ is small and sin θ ≈ θ . Thus θ = . sin θ ≈ phase difference between adjacent slits is φ = 2π d λ λ EXECUTE: A principal maximum occurs when φ = φ max = m 2π , where m is an integer, since then all the phasors

add. The first minima on either side of the mth principal maximum occur when φ = φ ±min = m 2π ± (2π / N ) and the phasor diagram for N slits forms a closed loop and the resultant phasor is zero. The angular position of a principal ⎛ λ ⎞ ⎛ λ ⎞ ± ± maximum is θ = ⎜ ⎟φ max . The angular position of the adjacent minimum is θ min = ⎜ ⎟φ min. ⎝ 2π d ⎠ ⎝ 2π d ⎠ 2π ⎞ λ ⎛ λ ⎞⎛ ⎛ λ ⎞⎛ 2π ⎞ θ +min = ⎜ ⎟⎜ φ max + ⎟ =θ +⎜ ⎟⎜ ⎟ =θ + 2 π d N 2 π d N Nd ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 2π ⎞ λ ⎛ λ ⎞⎛ ⎟⎜ φ max − ⎟ =θ − N ⎠ Nd ⎝ 2π d ⎠⎝

θ −min = ⎜

2λ , as was to be shown. Nd EVALUATE: The angular width of the principal maximum decreases like 1/ N as N increases. IDENTIFY: The change in wavelength of the Hα line is due to a Doppler shift in the wavelength due to the motion of the galaxy. c−v fS . SET UP: From Equation 16.30, the Doppler effect formula for light is f R = c+v EXECUTE: First find the wavelength of the light using the grating information. λ = d sin θ1 = [1/(575,800 lines/m)] sin 23.41° = 6.900 × 10–7 m = 690.0 nm The angular width of the principal maximum is θ

36.60.

+ min

−θ

− min

=

c−v fS . In this case, fR is the frequency of the 690.0-nm light that the c+v cosmologist measures, and fS is the frequency of the 656.3-nm light of the Hα line obtained in the laboratory. Using Equation 16.30, we have f R =

36-16

Chapter 36

Solving for v gives v =

1 − ( f R / fS )

1 + ( f R / fS )

2 2

c. Since fλ = c, f = c/λ, which gives fR/fS = λS/λR. Substituting this into the

equation for v, we get ⎛λ 1− ⎜ S λ v= ⎝ R ⎛λ 1+ ⎜ S ⎝ λR

36.61.

2

2 ⎞ ⎛ 656.3 nm ⎞ 1− ⎜ ⎟ ⎟ ⎝ 690.0 nm ⎠ 3.00 × 108 m/s = 1.501 × 107 m/s, ⎠ c= ) 2 ( 2 ⎞ ⎛ 656.3 nm ⎞ 1+ ⎜ ⎟ ⎟ ⎝ 690.0 nm ⎠ ⎠

which is 5.00% the speed of light. EVALUATE: Since v is positive, the galaxy is moving away from us. We can also see this because the wavelength has increased due to the motion. IDENTIFY and SET UP: Draw the specified phasor diagrams. There is totally destructive interference between two slits when their phasors are in opposite directions. EXECUTE: (a) For eight slits, the phasor diagrams must have eight vectors. The diagrams for each specified value of φ are sketched in Figure 36.61a. In each case the phasors all sum to zero. (b) The additional phasor diagrams for φ = 3π / 2 and 3π / 4 are sketched in Figure 36.61b. 3π 5π 7π 3π ,φ = , and φ = , totally destructive interference occurs between slits four apart. For φ = , For φ = 4 4 4 2 totally destructive interference occurs with every second slit. EVALUATE: At a minimum the phasors for all slits sum to zero.

Figure 36.61 36.62.

IDENTIFY: Maxima are given by 2d sin θ = mλ . SET UP: d is the separation between crystal planes. ⎛ 0.125 nm ⎞ ⎛ mλ ⎞ EXECUTE: (a) θ = arcsin ⎜ ⎟ = arcsin(0.2216m) . ⎟ = arcsin ⎜ m ⎝ 2d ⎠ ⎝ 2(0.282 nm) ⎠ For m = 1: θ = 12.8°, m = 2 : θ = 26.3°, m = 3: θ = 41.7°, and m = 4 : θ = 62.4°. No larger m values yield answers. ⎛ 2mλ ⎞ a , then θ = arcsin ⎜⎜ ⎟⎟ = arcsin(0.3134m). 2 ⎝ 2a ⎠ So for m = 1: θ = 18.3°, m = 2: θ = 38.8°, and m = 3: θ = 70.1°. No larger m values yield answers. EVALUATE: In part (b), where d is smaller, the maxima for each m are at larger θ IDENTIFY and SET UP: In each case consider the relevant phasor diagram. EXECUTE: (a) For the maxima to occur for N slits, the sum of all the phase differences between the slits must add to zero (the phasor diagram closes on itself). This requires that, adding up all the relative phase shifts, 2π m N φ = 2π m, for some integer m . Therefore φ = , for m not an integer multiple of N , which would give a N maximum. 2π m (b) The sum of N phase shifts φ = brings you full circle back to the maximum, so only the N − 1 previous N phases yield minima between each pair of principal maxima. EVALUATE: The N − 1 minima between each pair of principal maxima cause the maxima to become sharper as N increases. IDENTIFY: Set d = a in the expressions for φ and β and use the results in Eq.(36.12). SET UP: Figure 36.64 shows a pair of slits whose width and separation are equal

(b) If the separation d =

36.63.

36.64.

Diffraction

36-17

EXECUTE: Figure 36.64 shows that the two slits are equivalent to a single slit of width 2a . 2π d 2π a φ= sin θ , so β = sin θ = φ . So then the intensity is

λ

λ

⎛ sin 2 ( β /2) ⎞ (2sin( β /2)cos( β /2)) 2 sin 2 β sin 2 ( β ′/2) 2π (2a) = = = , where β ′ = I = I 0 cos 2 ( β /2) ⎜ I I I sin θ , ⎟ 0 0 0 2 2 2 2 λ β β ( β ′/2) ⎝ ( β /2) ⎠ which is Eq. (35.5) with double the slit width. EVALUATE: In Chapter 35 we considered the limit where a d is not possible.

Figure 36.64 36.65.

36.66.

IDENTIFY and SET UP: The condition for an intensity maximum is d sin θ = mλ , m = 0, ± 1, ± 2,… Third order means m = 3. The longest observable wavelength is the one that gives θ = 90° and hence θ = 1. 1 EXECUTE: 6500 lines/cm so 6.50 × 105 lines/m and d = m = 1.538 × 10−6 m 6.50 × 105 d sin θ (1.538 × 10−6 m)(1) λ= = = 5.13 × 10−7 m = 513 nm m 3 EVALUATE: The longest wavelength that can be obtained decreases as the order increases. IDENTIFY and SET UP: As the rays first reach the slits there is already a phase difference between adjacent slits of 2π d sin θ ′ . This, added to the usual phase difference introduced after passing through the slits, yields the

λ

condition for an intensity maximum. For a maximum the total phase difference must equal 2π m . 2π d sin θ 2π d sin θ ′ EXECUTE: + = 2π m ⇒ d (sin θ + sin θ ′) = mλ

λ

(b) 600 slits mm ⇒ d =

λ

1 = 1.67 × 10−6 m. 6.00 × 105 m −1

For θ ′ = 0D , m = 0 : θ = arcsin(0) = 0. ⎛ 6.50 × 10−7 m ⎞ ⎛λ⎞ D m = 1: θ = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 22.9 . −6 × 1.67 10 m d ⎝ ⎠ ⎝ ⎠ ⎛ 6.50 × 10−7 m ⎞ ⎛ λ⎞ D m = −1: θ = arcsin ⎜ − ⎟ = arcsin ⎜ − ⎟ = −22.9 . −6 × 1.67 10 m ⎝ d⎠ ⎝ ⎠

For θ ′ = 20.0D , m = 0 : θ = arcsin(− sin 20.0D ) = −20.0D.

36.67.

⎛ 6.50 × 10−7 m ⎞ − sin 20.0D ⎟ = 2.71D. m = 1: θ = arcsin ⎜ −6 × 1.67 10 m ⎝ ⎠ ⎛ 6.50 × 10−7 m ⎞ − sin 20.0D ⎟ = −47.0D. m = −1: θ = arcsin ⎜ − −6 ⎝ 1.67 × 10 m ⎠ EVALUATE: When θ ′ > 0 , the maxima are shifted downward on the screen, toward more negative angles. mλ IDENTIFY: The maxima are given by d sin θ = mλ . We need sin θ = ≤ 1in order for all the visible d wavelengths are to be seen. 1 SET UP: For 650 slits mm ⇒ d = = 1.53 × 10−6 m. 6.50 × 105 m −1

36-18

Chapter 36

λ1 2λ 3λ = 0.26; m = 2 : 1 = 0.52; m = 3 : 1 = 0.78. d d d λ 2 λ 3 λ λ2 = 7.00 × 10 −7 m : m = 1: 2 = 0.46; m = 2 : 2 = 0.92; m = 3 : 2 = 1.37. So, the third order does not contain the violet d d d end of the spectrum, and therefore only the first and second order diffraction patterns contain all colors of the spectrum. EVALUATE: θ for each maximum is larger for longer wavelengths. λ IDENTIFY: Apply sin θ = 1.22 . D Δx SET UP: θ is small, so sin θ ≈ , where Δx is the size of the detail and R = 7.2 ×108 ly. 1 ly = 9.41×1012 km . λ = c/f R λ Δx 1.22λ R (1.22)cR (1.22)(3.00 × 105 km s)(7.2 × 108 ly) EXECUTE: sin θ = 1.22 ≈ ⇒ Δx = = = = 2.06 ly . D R D Df (77.000 × 103 km)(1.665 × 109 Hz) (9.41 × 1012 km ly)( 2.06 ly) = 1.94 × 1013 km. Δx EVALUATE: λ = 18 cm . λ / D is very small, so is very small. Still, R is very large and Δx is many orders of R magnitude larger than the diameter of the sun. IDENTIFY and SET UP: Add the phases between adjacent sources. EXECUTE: (a) d sin θ = mλ . Place 1st maximum at ∞ or θ = 90D. d = λ . If d < λ , this puts the first maximum “beyond ∞. ” Thus, if d < λ there is only a single principal maximum. (b) At a principal maximum when δ = 0 , the phase difference due to the path difference between adjacent slits ⎛ d sin θ ⎞ is Φ path = 2π ⎜ ⎟ . This just scales 2π radians by the fraction the wavelength is of the path difference between ⎝ λ ⎠ adjacent sources. If we add a relative phase δ between sources, we still must maintain a total phase difference of zero to keep our principal maximum. 2π d sin θ ⎛ δλ ⎞ Φ path ± δ = 0 ⇒ = ±δ or θ = sin −1 ⎜ ⎟ λ ⎝ 2π d ⎠ 0.280 m (c) d = = 0.0200 m (count the number of spaces between 15 points). Let θ = 45D. Also recall f λ = c, so 14 2π (0.0200 m)(8.800 × 109 Hz)sin 45D δ max = ± = ±2.61 radians. (3.00 × 108 m s) EXECUTE:

36.68.

36.69.

36.70.

36.71.

λ1 = 4.00 × 10 −7 m : m = 1:

EVALUATE: δ must vary over a wider range in order to sweep the beam through a greater angle. IDENTIFY: The wavelength of the light is smaller under water than it is in air, which will affect the resolving power of the lens, by Rayleigh’s criterion. SET UP: The wavelength under water is λ = λ0/n, and for small angles Rayleigh’s criterion is θ = 1.22λ/D. EXECUTE: (a) In air the wavelength is λ0 = c/f = (3.00 × 108 m/s)/(6.00 × 1014 Hz) = 5.00 × 10–7 m. In water the wavelength is λ = λ0/n = (5.00 × 10–7 m)/1.33 = 3.76 × 10–7 m. With the lens open all the way, we have D = f/2.8 = (35.0 mm)/2.80 = (0.0350 m)/2.80. In the water, we have sin θ ≈ θ = 1.22 λ/D = (1.22)(3.76 × 10–7 m)/[(0.0350 m)/2.80] = 3.67 × 10–5 rad Calling w the width of the resolvable detail, we have θ = w/R → w = Rθ = (2750 mm)(3.67 × 10–5 rad) = 0.101 mm (b) θ = 1.22 λ/D = (1.22)(5.00 × 10–7 m)/[(0.0350 m)/2.80] = 4.88 × 10–5 rad w = Rθ = (2750 mm)(4.88 × 10–5 rad) = 0.134 mm EVALUATE: Due to the reduced wavelength underwater, the resolution of the lens is better under water than in air. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means the angular separation θ of the objects is given by θ = 1.22λ / D. θ = y / s, where y = 75.0 m is the distance between the two objects and s is their distance from the astronaut (her altitude). y λ = 1.22 EXECUTE: s D yD (75.0 m)(4.00 × 10−3 m) s= = = 4.92 × 105 m = 492 km 1.22λ 1.22(500 × 10−9 m) EVALUATE: In practice, this diffraction limit of resolution is not achieved. Defects of vision and distortion by the earth’s atmosphere limit the resolution more than diffraction does.

Diffraction

36.72.

IDENTIFY: Apply sin θ = 1.22

36-19

λ

. D Δx , where Δx is the size of the details and R is the distance to the earth. SET UP: θ is small, so sin θ ≈ R 1 ly = 9.41 × 1015 m . EXECUTE: (a) R =

DΔx (6.00 × 106 m)(2.50 × 105 m) = = 1.23 × 1017 m = 13.1 ly 1.22λ (1.22)(1.0 × 10−5 m)

1.22λ R (1.22)(1.0 × 10−5 m)(4.22 ly)(9.41 × 1015 m ly) = = 4.84 × 108 km . This is about 10,000 times the 1.0 m D diameter of the earth! Not enough resolution to see an earth-like planet! Δx is about 3 times the distance from the earth to the sun. (1.22)(1.0 × 10−5 m)(59 ly)(9.41× 1015 m ly) = 1.13 × 106 m = 1130 km. (c) Δx = 6.00 × 106 m (b) Δx =

Δx 1130 km = = 8.19 × 10−3 ; Δx is small compared to the size of the planet. Dplanet 1.38 × 105 km 36.73.

EVALUATE: The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) From the segment dy′, the fraction of the amplitude of E0 that gets through is

⎛ dy′ ⎞ ⎛ dy′ ⎞ E0 ⎜ ⎟ ⇒ dE = E0 ⎜ ⎟ sin( kx − ωt ). a ⎝ ⎠ ⎝ a ⎠ (b) The path difference between each little piece is E dy′ y′ sin θ ⇒ kx = k ( D − y′ sin θ ) ⇒ dE = 0 sin(k ( D − y′ sin θ ) − ωt ). This can be rewritten as a E0 dy′ dE = (sin( kD − ωt )cos(ky′ sin θ ) + sin( ky′ sin θ )cos( kD − ωt )). a (c) So the total amplitude is given by the integral over the slit of the above. a 2 E a2 ⇒ E = ∫ dE = 0 ∫ dy′ (sin(kD − ωt ) cos(ky′ sin θ ) + sin(ky′ sin θ ) cos( kD − ωt )). −a 2 a −a 2 But the second term integrates to zero, so we have: a 2

E=

a 2 ⎡⎛ sin(ky′ sin θ ) ⎞ ⎤ E0 sin( kD − ωt ) ∫ dy′(cos( ky′ sin θ )) = E0 sin ( kD − ωt ) ⎢⎜ ⎟⎥ a − 2 a ⎣⎝ ka sin θ 2 ⎠ ⎦ − a 2

⎛ sin( ka (sin θ ) 2) ⎞ ⎛ sin(π a (sin θ ) λ ) ⎞ ⇒ E = E0 sin( kD − ωt ) ⎜ ⎟ = E0 sin( kD − ωt ) ⎜ ⎟. ka (sin ) 2 θ ⎝ ⎠ ⎝ π a (sin θ ) λ ⎠ sin [. . .] At θ = 0, = 1 ⇒ E = E0 sin(kD − ωt ). [. . .] 2

36.74.

2

⎛ sin( ka (sin θ )/2) ⎞ ⎛ sin( β 2) ⎞ 2 2 (d) Since I ∝ E 2 ⇒ I = I 0 ⎜ ⎟ = I0 ⎜ ⎟ , where we have used I 0 = E0 sin ( kx − ωt ). ka (sin )/ 2 2 θ β ⎝ ⎠ ⎝ ⎠ EVALUATE: The same result for I (θ ) is obtained as was obtained using phasors. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at x = R with a maximum amplitude of E0 . However, each successive source will pick up an extra phase from its respective pathlength to point ⎛ d sin θ ⎞ P . φ = 2π ⎜ ⎟ which is just 2π , the maximum phase, scaled by whatever fraction the path difference, ⎝ λ ⎠ d sin θ , is of the wavelength, λ . By adding up the contributions from each source (including the accumulating phase difference) this gives the expression provided. (b) ei ( kR −ωt + nφ ) = cos( kR − ωt + nφ ) + i sin(kR − ωt + nφ ). The real part is just cos ( kR − ωt + nφ ). So, ⎡ N −1 ⎤ N −1 Re ⎢ ∑ E0ei ( kR −ω t + nφ ) ⎥ = ∑ E0 cos( kR − ωt + nφ ). (Note: Re means “the real part of . . . .”). But this is just ⎣ n=0 ⎦ n =0 E0 cos( kR − ωt ) + E0 cos( kR − ωt + φ ) + E0 cos( kR − ωt + 2φ ) + " + E0 cos(kR − ωt + ( N − 1)φ )

36-20

Chapter 36 N −1

(c)

∑E

0

ei ( kR −ωt + nφ ) = E0

n=0

N −1

∑e

− iω t

e + ikR einφ = E0ei ( kR −ω t )

n=0

N −1

∞

N −1

N −1

∑ e φ . ∑ e φ = ∑ (e φ ) . But recall ∑ x n=0

in

n=0

in

n =0

i

n

n=0

n

=

xN −1 . x −1

eiN φ − 1 eiNφ − 1 eiNφ / 2 (eiNφ / 2 − e −iNφ / 2 ) (eiN φ / 2 − e −iN φ / 2 ) Let x = e so ∑ (e ) = iφ (nice trick!). But iφ = iφ / 2 iφ / 2 − iφ / 2 = ei ( N −1)φ / 2 iφ / 2 −iφ / 2 . e −1 e −1 e (e − e ) (e − e ) n=0 Putting everything together: N −1 (eiN φ / 2 − e − iNφ / 2 ) E0ei ( kR −ωt + nφ ) = E0ei ( kR −ωt + ( N −1)φ / 2) iφ / 2 − iφ / 2 ∑ (e − e ) n=0 N −1

iφ

iφ n

⎡ cos Nφ /2 + i sin Nφ /2 − cos Nφ /2 + i sin Nφ /2 ⎤ = E0 [ cos( kR − ωt + ( N − 1)φ /2) + i sin( kR − ωt + ( N − 1)φ / 2) ] ⎢ ⎥ cosφ /2 + i sin φ /2 − cosφ /2 + i sin φ /2 ⎣ ⎦ sin( Nφ /2) Taking only the real part gives ⇒ E0 cos( kR − ωt + ( N − 1)φ /2) = E. sin φ /2 2

(d) I = E av = I 0

I0 ∝

sin 2 ( Nφ / 2) . (The cos 2 term goes to sin 2 (φ / 2)

in the time average and is included in the definition of I 0 .)

E02 . 2

EVALUATE: (e) N = 2. I = I 0

36.75.

1 2

sin 2 (2φ / 2) I 0 (2sin φ / 2cos φ / 2) 2 φ = = 4 I 0 cos 2 . Looking at Eq.(35.9), sin 2 φ / 2 sin 2 φ / 2 2

I 0′ ∝ 2 E02 but for us I 0 ∝

E02 I 0′ = . 2 4


From Problem 36.74, I = I 0

sin 2 ( Nφ / 2) . Use this result to obtain each result specified sin 2 φ / 2

in the problem. ⎛ N 2 ⎞ cos( Nφ / 2) 0 sin ( Nφ / 2) ˆ rule: lim . Use l'Hopital's = lim ⎜ = N . So lim I = N 2 I 0 . ⎟ φ →0 φ →0 φ →0 sin φ / 2 1 2 cos( φ / 2) 0 ⎝ ⎠ N 2π . The (b) The location of the first minimum is when the numerator first goes to zero at φmin = π or φmin = 2 N 1 width of the central maximum goes like 2φmin , so it is proportional to . N Nφ (c) Whenever = nπ where n is an integer, the numerator goes to zero, giving a minimum in intensity. That is, 2 2nπ . This is true assuming that the denominator doesn’t go to zero as well, which I is a minimum wherever φ = N EXECUTE: (a) lim I → φ →0

occurs when

φ 2

= mπ , where m is an integer. When both go to zero, using the result from part(a), there is a

n is an integer, there will be a maximum. N n n (d) From part (c), if is an integer we get a maximum. Thus, there will be N − 1 minima. (Places where is N N not an integer for fixed N and integer n .) For example, n = 0 will be a maximum, but n = 1, 2. . ., N − 1 will be minima with another maximum at n = N . φ ⎛ π 3π ⎞ , etc.) ⎟ and if N is odd then (e) Between maxima is a half-integer multiple of π ⎜ i.e. , 2 2 2 ⎝ ⎠ maximum. That is, if

sin 2 ( Nφ / 2) → 1, so I → I 0 . sin 2 φ / 2 EVALUATE: These results show that the principal maxima become sharper as the number of slits is increased.

37

RELATIVITY

37.1.

IDENTIFY and SET UP: Consider the distance A to O′ and B to O′ as observed by an observer on the ground (Figure 37.1).

Figure 37.1

37.2.

EXECUTE: Simultaneous to observer on train means light pulses from A′ and B′ arrive at O′ at the same time. To observer at O light from A′ has a longer distance to travel than light from B′ so O will conclude that the pulse from A( A′) started before the pulse at B ( B′). To observer at O bolt A appeared to strike first. EVALUATE: Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on the train measures that the bolt at B′ struck first. 1 (a) γ = = 2.29. t = γτ = (2.29) (2.20 × 10−6 s) = 5.05 × 10 −6 s. 1 − (0.9) 2 (b) d = vt = (0.900) (3.00 × 108 m s) (5.05 × 10−6 s) = 1.36 × 103 m = 1.36 km.

37.3.

1 IDENTIFY and SET UP: The problem asks for u such that Δt0 / Δt = . 2

37.4.

37.5.

2

Δt0

u 2 ⎛1⎞ gives u = c 1 − ( Δt0 / Δt ) = (3.00 × 108 m/s) 1 − ⎜ ⎟ = 2.60 × 108 m/s ; = 0.867 c ⎝ 2⎠ 1 − u 2 / c2 Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower speeds than we calculated for u. IDENTIFY: Time dilation occurs because the rocket is moving relative to Mars. SET UP: The time dilation equation is Δt = γΔt0 , where t0 is the proper time. EXECUTE: (a) The two time measurements are made at the same place on Mars by an observer at rest there, so the observer on Mars measures the proper time. 1 (75.0 μ s) = 435 μ s (b) Δt = γΔt0 = 1 − (0.985) 2 EXECUTE:

Δt =

EVALUATE: The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer. (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10 −8 s; Δt = 4.20 × 10−7 s. In the lab frame the pion is created and decays at different points, so this time is not the proper time. 2 Δt0 u 2 ⎛ Δt ⎞ EXECUTE: Δt = says 1 − 2 = ⎜ 0 ⎟ c ⎝ Δt ⎠ 1 − u 2 / c2 2

⎛ 2.60 × 10−8 s ⎞ u ⎛ Δt ⎞ = 1− ⎜ 0 ⎟ = 1− ⎜ ⎟ = 0.998; u = 0.998c −7 c ⎝ Δt ⎠ ⎝ 4.20 × 10 s ⎠ EVALUATE: u < c, as it must be, but u/c is close to unity and the time dilation effects are large. (b) IDENTIFY and SET UP: The speed in the laboratory frame is u = 0.998c; the time measured in this frame is Δt , so the distance as measured in this frame is d = u Δt 2

EXECUTE: d = (0.998)(2.998 × 108 m/s)(4.20 × 10−7 s) = 126 m EVALUATE: The distance measured in the pion’s frame will be different because the time measured in the pion’s frame is different (shorter). 37-1

37-2

37.6.

Chapter 37 γ = 1.667

1.20 × 108 m = 0.300 s. γ γ(0.800c) (b) (0.300 s) (0.800c) = 7.20 × 107 m. (c) Δt0 = 0.300 s γ = 0.180 s. (This is what the racer measures your clock to read at that instant.) At your origin

(a) Δt0 =

Δt

=

you read the original 37.7.

37.8.

37.9.

37.10.

order of events! IDENTIFY and SET UP: A clock moving with respect to an observer appears to run more slowly than a clock at rest in the observer’s frame. The clock in the spacecraft measurers the proper time Δt0 . Δt = 365 days = 8760 hours. EXECUTE: The clock on the moving spacecraft runs slow and shows the smaller elapsed time. Δt0 = Δt 1 − u 2 / c 2 = (8760 h) 1 − (4.80 × 106 / 3.00 × 108 ) 2 = 8758.88 h . The difference in elapsed times is 8760 h − 8758.88 h = 1.12 h . IDENTIFY and SET UP: The proper time is measured in the frame where the two events occur at the same point. EXECUTE: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time. Δt0 gives u = c 1 − ( Δt0 / Δt ) 2 = c 1 − (12.0 × 10−3 / 0.190) 2 = 0.998c . (b) Δt = 2 2 1− u / c EVALUATE: The observer at rest with respect to the searchlight measures a much shorter duration for the event. IDENTIFY and SET UP: l = l0 1 − u 2 / c 2 . The length measured when the spacecraft is moving is l = 74.0 m; l0 is the length measured in a frame at rest relative to the spacecraft. l 74.0 m = = 92.5 m. EXECUTE: l0 = 2 2 1− u / c 1 − (0.600c / c ) 2 EVALUATE: l0 > l. The moving spacecraft appears to an observer on the planet to be shortened along the direction of motion. IDENTIFY and SET UP: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m, and that is its proper length, l0 . l = 0.3048 m . EXECUTE:

37.11.

l = l0 1 − u 2 / c 2 gives u = c 1 − (l / l0 ) 2 = c 1 − (0.3048 /1.00) 2 = 0.9524c = 2.86 × 108 m/s .

IDENTIFY and SET UP: The 2.2 μs lifetime is Δt0 and the observer on earth measures Δt. The atmosphere is moving relative to the muon so in its frame the height of the atmosphere is l and l0 is 10 km. EXECUTE: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 2.2 × 10 −6 s is d = vt = (3.00 × 108 m/s)(2.2 × 10−6 s) = 660 m = 0.66 km . (b) Δt =

37.12.

1.20 × 108m = 0.5 s. Clearly the observers (you and the racer) will not agree on the (0.800) (3 × 108 m s)

Δt0 1 − u 2 / c2

=

2.2 × 10−6 s 1 − (0.999) 2

= 4.9 × 10−5 s

d = vt = (0.999)(3.00 × 108 m/s)(4.9 × 10−5 s) = 15 km In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime. (c) l = l0 1 − u 2 / c 2 = (10 km) 1 − (0.999) 2 = 0.45 km In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime. IDENTIFY and SET UP: The scientist at rest on the earth’s surface measures the proper length of the separation between the point where the particle is created and the surface of the earth, so l0 = 45.0 km . The transit time

measured in the particle’s frame is the proper time, Δt0 . EXECUTE: (a) t =

l0 45.0 × 103 m = = 1.51 × 10−4 s v (0.99540)(3.00 × 108 m/s)

(b) l = l0 1 − u 2 / c 2 = (45.0 km) 1 − (0.99540) 2 = 4.31 km (c) time dilation formula: Δt0 = Δt 1 − u 2 / c 2 = (1.51 × 10−4 s) 1 − (0.99540) 2 = 1.44 × 10−5 s

l 4.31 × 103 m = = 1.44 × 10−5 s v (0.99540)(3.00 × 108 m/s) The two results agree. (a) l0 = 3600 m . from Δl : t =

37.13.

l = l0 1 −

u2 (4.00 × 107 m s) 2 = l (3600 m) 1 − = (3600 m)(0.991) = 3568 m. 0 c2 (3.00 × 108 m s) 2

Relativity

37-3

l0 3600 m = = 9.00 × 10 −5 s. u 4.00 × 107 m s l 3568 m (c) Δt = = = 8.92 × 10−5 s. u 4.00 × 107 m s Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives (b) Δt0 =

37.14.

⎛ u2 ⎞ 1 x′ + ut ′ = γ x ⎜1 − 2 ⎟ = x, ⎝ c ⎠ γ and multiplying the first by

u and adding to the last to eliminate x gives c2 t′ +

37.15.

37.16.

⎛ u2 ⎞ 1 u x′ = γt ⎜1 − 2 ⎟ = t , 2 c ⎝ c ⎠ γ

so x = γ( x′ + ut ′) and t = γ(t ′ + ux′ c 2 ), which is indeed the same as Eq. (37.21) with the primed coordinates replacing the unprimed, and a change of sign of u. v′ + u 0.400c + 0.600c (a) v = = = 0.806c 1 + uv′ c 2 1 + (0.400) (0.600) v′ + u 0.900c + 0.600c (b) v = = = 0.974c 1 + uv′ c 2 1 + (0.900)(0.600) v′ + u 0.990c + 0.600c (c) v = = = 0.997c. 2 ′ 1 + uv c 1 + (0.990)(0.600) γ = 1.667( γ = 5 3 if u = (4 5)c ). (a) In Mavis’s frame the event “light on” has space-time coordinates x′ = 0 and t′ = 5.00 s, so from the result of ux′ ⎞ ⎛ Exercise 37.14 or Example 37.7, x = γ( x′ + ut′) and t = γ ⎜ t ′ + 2 ⎟ ⇒ x = γut ′ = 2.00 × 109 m, t = γt ′ = 8.33 s . c ⎠ ⎝

(b) The 5.00-s interval in Mavis’s frame is the proper time Δt0 in Eq.(37.6), so Δt = γΔt0 = 8.33 s, as in part (a). 37.17.

(c) (8.33 s) (0.800c ) = 2.00 × 109 m, which is the distance x found in part (a). IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. v −u . SET UP: The relativistic velocity addition formula is v′x = x uv 1 − 2x c EXECUTE: (a) For the pursuit ship to catch the cruiser, the distance between them must be decreasing, so the velocity of the cruiser relative to the pursuit ship must be directed toward the pursuit ship. (b) Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We want the velocity v′ of the cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine). v −u 0.600c − 0.800c = = −0.385c v′x = x uvx 1 − (0.600) (0.800) 1− 2 c

37.18.

The result implies that the cruiser is moving toward the pursuit ship at 0.385c. EVALUATE: The nonrelativistic formula would have given –0.200c, which is considerably different from the correct result. Let u y be the y-component of the velocity of S ′ relative to S. Following the steps used in the derivation of v′y + u y . Eq.(37.23) we get v y = 1 + u y v′y / c 2

37.19.

IDENTIFY and SET UP: Reference frames S and S ′ are shown in Figure 37.19.

Frame S is at rest in the laboratory. Frame S ′ is attached to particle 1. Figure 37.19 u is the speed of S ′ relative to S; this is the speed of particle 1 as measured in the laboratory. Thus u = +0.650c. The speed of particle 2 in S ′ is 0.950c. Also, since the two particles move in opposite directions, 2 moves in the − x′ direction and v′x = −0.950c. We want to calculate vx , the speed of particle 2 in frame S; use Eq.(37.23).

37-4

Chapter 37

v′x + u −0.950c + 0.650c −0.300c = = = −0.784c. The speed of the second particle, 2 2 ′ 1 + uvx / c 1 + (0.950c)(−0.650c) / c 1 − 0.6175 as measured in the laboratory, is 0.784c. EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle in the lab frame is 0.300c. The correct relativistic calculation gives a result more than twice this. IDENTIFY and SET UP: Let S be the laboratory frame and let S ′ be the frame of one of the particles, as shown in Figure 37.20. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1 is moving in the +x direction and particle 2 is moving in the − x direction. Then u = 0.9520c and v = −0.9520c . v′ is the velocity of particle 2 relative to particle 1. v−u −0.9520c − 0.9520c EXECUTE: v′ = = = −0.9988c . The speed of particle 2 relative to particle 1 1 − uv / c 2 1 − (0.9520c )( −0.9520c) / c 2 is 0.9988c . v′ < 0 shows particle 2 is moving toward particle 1. EXECUTE:

37.20.

vx =

Figure 37.20 37.21.

37.22.

37.23.

IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. v −u SET UP: The relativistic velocity addition formula is v′x = x . uv 1 − 2x c EXECUTE: In the relativistic velocity addition formula for this case, vx′ is the relative speed of particle 1 with respect to particle 2, v is the speed of particle 2 measured in the laboratory, and u is the speed of particle 1 measured in the laboratory, u = – v. v′ v − (−v) 2v v′x = = . x v 2 − 2v + vx′ = 0 and (0.890c)v 2 − 2c 2v + (0.890c 3 ) = 0 . 1 − ( − v )v c 2 1 + v 2 c 2 c 2 This is a quadratic equation with solution v = 0.611c (v must be less than c). EVALUATE: The nonrelativistic result would be 0.445c, which is considerably different from this result. IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S ′ . Let the positive x direction for both frames be from the enemy spaceship toward the starfighter. Then u = +0.400c . v′ = +0.700c . v is the velocity of the missile relative to you. v′ + u 0.700c + 0.400c = = 0.859c EXECUTE: (a) v = 1 + uv′ / c 2 1 + (0.400)(0.700) (b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it 8.00 × 109 m takes in your frame. t = = 31.0 s . (0.859)(3.00 × 108 m/s) IDENTIFY and SET UP: The reference frames are shown in Figure 37.23.

S = Arrakis frame S ′ = spaceship frame The object is the rocket. Figure 37.23

u is the velocity of the spaceship relative to Arrakis. vx = +0.360c; v′x = +0.920c (In each frame the rocket is moving in the positive coordinate direction.)

Relativity

Use the Lorentz velocity transformation equation, Eq.(37.22): v′x =

37-5

vx − u . 1 − uvx / c 2

vx − u ⎛ v v′ ⎞ ⎛ v v′ ⎞ so v′x − u ⎜ x 2 x ⎟ = vx − u and u ⎜ 1 − x 2 x ⎟ = vx − v′x c ⎠ 1 − uvx / c 2 ⎝ c ⎠ ⎝ vx − v′x 0.360c − 0.920c 0.560c u= = =− = −0.837c 1 − vx v′x / c 2 1 − (0.360c)(0.920c ) / c 2 0.6688

EXECUTE:

37.24.

v′x =

The speed of the spacecraft relative to Arrakis is 0.837c = 2.51 × 108 m/s. The minus sign in our result for u means that the spacecraft is moving in the − x-direction, so it is moving away from Arrakis. EVALUATE: The incorrect Galilean expression also says that the spacecraft is moving away from Arrakis, but with speed 0.920c – 0.360c = 0.560c. IDENTIFY: We need to use the relativistic Doppler shift formula. SET UP: The relativistic Doppler shift formula, Eq.(37.25), is f =

c+u f0 . c−u

c+u 2 f 0 . (c − u ) f 2 = (c + u ) f 02 . cf 2 − uf 2 = cf 02 + uf 02 . cf 2 − cf 02 = uf 2 + uf 02 and c−u c( f 2 − f 02 ) ( f / f 0 ) 2 − 1 = u= c. f 2 + f 02 ( f / f0 )2 + 1 (a) For f/f0 = 0.95, u = – 0.051c moving away from the source. (b) For f/f0 = 5.0, u = 0.923c moving towards the source. EVALUATE: Note that the speed required to achieve a 10 times greater Doppler shift is not 10 times the original speed. EXECUTE:

37.25.

f2=

IDENTIFY and SET UP: Source and observer are approaching, so use Eq.(37.25): f =

c+u f 0 . Solve for u, the c−u

speed of the light source relative to the observer. ⎛c+u⎞ 2 (a) EXECUTE: f 2 = ⎜ ⎟ f0 ⎝ c −u ⎠ (c − u ) f 2 = (c + u ) f 02 and u =

⎛ ( f / f0 )2 − 1 ⎞ c( f 2 − f 02 ) c = ⎜ ⎟ 2 f 2 + f 02 ⎝ ( f / f0 ) + 1 ⎠

λ0 = 675 nm, λ = 575 nm ⎛ (675 nm/575 nm) 2 − 1 ⎞ 8 7 u =⎜ ⎟ c = 0.159c = (0.159)(2.998 × 10 m/s) = 4.77 × 10 m/s; definitely speeding 2 (675 nm/575 nm) 1 + ⎝ ⎠ (b) 4.77 × 107 m/s = (4.77 × 107 m/s)(1 km/1000 m)(3600 s/1 h) = 1.72 × 108 km/h. Your fine would be $1.72 × 108 (172 million dollars). EVALUATE: The source and observer are approaching, so f > f 0 and λ < λ0 . Our result gives u < c, as it must. 37.26.

Using u = −0.600c = − ( 3 5 ) c in Eq.(37.25) gives 1 − ( 3 5) 25 f0 = f 0 = f 0 2. 1 + ( 3 5) 85 G G G G IDENTIFY and SET UP: If F is parallel to v then F changes the magnitude of v and not its direction. ⎞ dp d ⎛ mv = ⎜ F= ⎟ dt dt ⎝ 1 − v 2 / c 2 ⎠ d df dv Use the chain rule to evaluate the derivative: f (v (t )) = . dt dv dt m mv ⎛ dv ⎞ ⎛ 1 ⎞⎛ 2v ⎞⎛ dv ⎞ EXECUTE: (a) F = ⎜ ⎟+ ⎜ − ⎟⎜ − ⎟⎜ ⎟ (1 − v 2 / c 2 )1/ 2 ⎝ dt ⎠ (1 − v 2 / c 2 )3 / 2 ⎝ 2 ⎠⎝ c 2 ⎠⎝ dt ⎠ f =

37.27.

F=

⎛ v 2 v 2 ⎞ dv dv m m ⎜1 − + ⎟ = dt (1 − v 2 / c 2 )3 / 2 ⎝ c 2 c 2 ⎠ dt (1 − v 2 / c 2 )3 / 2

dv = a, so a = ( F / m)(1 − v 2 / c 2 )3 / 2 . dt EVALUATE: Our result agrees with Eq.(37.30).

But

37-6

37.28.

Chapter 37

G G G G (b) IDENTIFY and SET UP: If F is perpendicular to v then F changes the direction of v and not its magnitude. G G d⎛ ⎞ mv F= ⎜ ⎟. 2 2 dt ⎝ 1 − v / c ⎠ G G a = dv / dt but the magnitude of v in the denominator of Eq.(37.29) is constant. ma EXECUTE: F = and a = ( F / m)(1 − v 2 / c 2 )1/ 2 . 2 2 1− v / c EVALUATE: This result agrees with Eq.(37.33). 1 IDENTIFY and SET UP: γ = . If γ is 1.0% greater than 1 then γ = 1.010 , if γ is 10% greater than 1 1 − v2 / c2 then γ = 1.10 and if γ is 100% greater than 1 then γ = 2.00 .

EXECUTE:

v = c 1 − 1/ γ 2

(a) v = c 1 − 1/(1.010) 2 = 0.140c (b) v = c 1 − 1/(1.10) 2 = 0.417c (c) v = c 1 − 1/(2.00) 2 = 0.866c 37.29.

(a) p =

mv

= 2mv .

1 − v2 c2

1 v2 3 3 = 1 − 2 ⇒ v2 = c2 ⇒ v = c = 0.866c. 4 c 4 2 v 1 (b) F = γ 3ma = 2ma ⇒ γ3 = 2 ⇒ γ = (2)1/ 3 so = 22 / 3 ⇒ = 1 − 2−2 / 3 = 0.608 v2 c 1− 2 c The force is found from Eq.(37.32) or Eq.(37.33). (a) Indistinguishable from F = ma = 0.145 N. ⇒ 1 = 2 1 − v2 c2 ⇒

37.30.

(b) γ3ma = 1.75 N. (c) γ3ma = 51.7 N. (d) γma = 0.145 N, 0.333 N, 1.03 N. 37.31.

(a) K =

mc 2 1− v c 2

⇒

(b) K = 5mc 2 ⇒ 37.32. 37.33.

2

− mc 2 = mc 2

1 1− v c 2

2

1 1 − v2 c2

=2⇒

v2 1 3 =1− 2 ⇒ v = c = 0.866c. c 4 4

=6⇒

v2 1 35 =1− 2 ⇒ v = c = 0.986c. c 36 36

E = 2mc 2 = 2(1.67 ×10−27 kg)(3.00×108 m s) 2 = 3.01×10 −10 J = 1.88×109 eV. IDENTIFY and SET UP: Use Eqs.(37.38) and (37.39). EXECUTE: (a) E = mc 2 + K , so E = 4.00mc 2 means K = 3.00mc 2 = 4.50 × 10 −10 J

(b) E 2 = (mc 2 ) 2 + ( pc) 2 ; E = 4.00mc 2 , so 15.0( mc 2 ) 2 = ( pc ) 2

p = 15mc = 1.94 × 10−18 kg ⋅ m/s (c) E = mc 2 / 1 − v 2 / c 2

37.34.

37.35.

E = 4.00mc 2 gives 1 − v 2 / c 2 = 1/16 and v = 15/16c = 0.968c EVALUATE: The speed is close to c since the kinetic energy is greater than the rest energy. Nonrelativistic expressions relating E, K, p and v will be very inaccurate. (a) W = ΔK = ( γ f − 1) mc 2 = (4.07 × 10−3 ) mc 2 . (b) ( γ f − γ i ) mc 2 = 4.79mc 2 . (c) The result of part (b) is far larger than that of part (a). IDENTIFY: Use E = mc 2 to relate the mass increase to the energy increase. (a) SET UP: Your total energy E increases because your gravitational potential energy mgy increases.

Relativity

37-7

ΔE = mg Δy

EXECUTE:

ΔE = ( Δm)c so Δm = ΔE / c 2 = mg (Δy ) / c 2 2

Δm / m = ( g Δy ) / c 2 = (9.80 m/s 2 )(30 m)/(2.998 × 108 m/s) 2 = 3.3 × 10 −13% This increase is much, much too small to be noticed. (b) SET UP: The energy increases because potential energy is stored in the compressed spring. EXECUTE: ΔE = ΔU = 12 kx 2 = 12 (2.00 × 104 N/m)(0.060 m) 2 = 36.0 J

37.36.

Δm = ( ΔE ) / c 2 = 4.0 × 10 −16 kg Energy increases so mass increases. The mass increase is much, much too small to be noticed. EVALUATE: In both cases the energy increase corresponds to a mass increase. But since c 2 is a very large number the mass increase is very small. m0 = 2m0 . (a) E0 = m0c 2 . 2 E = mc 2 = 2m0c 2 . Therefore, m = 2m0 ⇒ 1 − v2 / c2

1 v2 v2 3 = 1 − 2 ⇒ 2 = ⇒ v = c 3 4 = 0.866c = 2.60 × 108 m s 4 c c 4 m0 (b) 10 m0c 2 = mc 2 = c2 . 1 − v2 c2 1− 37.37.

v2 1 v 2 99 99 = . v=c ⇒ 2= = 0.995c = 2.98 × 108 m s . 2 c 100 c 100 100

IDENTIFY and SET UP: The energy equivalent of mass is E = mc 2 . ρ = 7.86 g/cm3 = 7.86 × 103 kg/m3 . For a

cube, V = L3 . 1.0 × 1020 J E = = 1.11 × 103 kg 2 (3.00 × 108 m/s) 2 c m m 1.11 × 103 kg (b) ρ = so V = = = 0.141 m3 . L = V 1/ 3 = 0.521 m = 52.1 cm V ρ 7.86 × 103 kg/m3 EVALUATE: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities of antimatter. (5.52 × 10 −27 kg)(3.00 × 108 m s) 2 = 4.97 × 10−10 J = 3105 MeV. IDENTIFY and SET UP: The total energy is given in terms of the momentum by Eq.(37.39). In terms of the total energy E, the kinetic energy K is K = E − mc 2 (from Eq.37.38). The rest energy is mc 2 . EXECUTE: (a) m =

37.38. 37.39.

EXECUTE: (a) E = (mc 2 ) 2 + ( pc) 2 =

[(6.64 × 10−27 )(2.998 × 108 ) 2 ]2 + [(2.10 × 10−18 )(2.998 × 108 )]2 J E = 8.67 × 10−10 J (b) mc 2 = (6.64 × 10 −27 kg)(2.998 × 108 m/s) 2 = 5.97 × 10 −10 J

37.40.

37.41.

K = E − mc 2 = 8.67 × 10 −10 J − 5.97 × 10−10 J = 2.70 × 10−10 J K 2.70 × 10−10 J (c) = = 0.452 mc 2 5.97 × 10−10 J EVALUATE: The incorrect nonrelativistic expressions for K and p give K = p 2 / 2m = 3.3 × 10 −10 J; the correct relativistic value is less than this. 12 ⎛ ⎛ p ⎞2 ⎞ 2 4 2 2 12 2 E = ( m c + p c ) = mc ⎜1 + ⎜ ⎜ ⎝ mc ⎟⎠ ⎟⎟ ⎝ ⎠ 2 2 ⎛ 1 p ⎞ p 1 = mc 2 + = mc 2 + mv 2 , the sum of the rest mass energy and the classical kinetic energy. E ≈ mc 2 ⎜ 1 + 2 2 ⎟ 2 m c 2 m 2 ⎝ ⎠ 1 1 (a) v = 8 × 107 m s ⇒ γ = = 1.0376 . For m = mp , K nonrel = mv 2 = 5.34 × 10−12 J . 2 1 − v2 c2 K rel = ( γ − 1) mc 2 = 5.65 × 10 −12 J.

K rel = 1.06. K nonrel

(b) v = 2.85 × 108 m s; γ = 3.203. 1 K rel = mv 2 = 6.78 × 10−11 J; K rel = (γ − 1)mc 2 = 3.31 × 10−10 J; K rel K nonrel = 4.88. 2

37-8

Chapter 37

37.42.

IDENTIFY: Since the speeds involved are close to that of light, we must use the relativistic formula for kinetic energy. ⎛ ⎞ 1 SET UP: The relativistic kinetic energy is K = (γ − 1) mc 2 = ⎜ − 1 ⎟ mc 2 . 2 2 ⎝ 1− v / c ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (1.67 × 10−27 kg)(3.00 × 108 m s) 2 ⎜ − 1⎟ (a) K = (γ − 1)mc 2 = ⎜ 2 2 ⎜ 1 − 0.100c / c 2 ⎝ 1− v / c ⎠ ( ) ⎟⎠ ⎝

37.43.

37.44.

1 ⎛ ⎞ K = (1.50 × 10−10 J) ⎜ − 1⎟ = 7.56 × 10−13 J = 4.73 MeV ⎝ 1 − 0.0100 ⎠ ⎛ ⎞ 1 (b) K = (1.50 × 10−10 J) ⎜ − 1⎟ = 2.32 × 10−11 J = 145 MeV ⎜ 1 − (0.500) 2 ⎟ ⎝ ⎠ ⎛ ⎞ 1 (c) K = (1.50 × 10−10 J) ⎜ − 1⎟ = 1.94 × 10−10 J = 1210 MeV ⎜ 1 − (0.900) 2 ⎟ ⎝ ⎠ (d) ΔE = 2.32 × 10−11 J − 7.56 × 10−13 J = 2.24 × 10−11 J = 140 MeV (e) ΔE = 1.94 × 10−10 J − 2.32 × 10−11 J = 1.71× 10−10 J = 1070 MeV 1 (f) Without relativity, K = mv 2 . The work done in accelerating a proton from 0.100c to 0.500c in the 2 1 1 nonrelativistic limit is ΔE = m(0.500c) 2 − m(0.100c) 2 = 1.81 × 10−11 J = 113 MeV . 2 2 The work done in accelerating a proton from 0.500c to 0.900c in the nonrelativistic limit is 1 1 ΔE = m(0.900c) 2 − m(0.500c) 2 = 4.21× 10−11 J = 263 MeV . 2 2 EVALUATE: We see in the first case the nonrelativistic result is within 20% of the relativistic result. In the second case, the nonrelativistic result is very different from the relativistic result since the velocities are closer to c. IDENTIFY and SET UP: Use Eq.(23.12) and conservation of energy to relate the potential difference to the kinetic energy gained by the electron. Use Eq.(37.36) to calculate the kinetic energy from the speed. EXECUTE: (a) K = qΔV = eΔV ⎛ ⎞ 1 K = mc 2 ⎜ − 1⎟ = 4.025mc 2 = 3.295 × 10−13 J = 2.06 MeV 2 2 ⎝ 1− v / c ⎠ 6 ΔV = K / e = 2.06 × 10 V (b) From part (a), K = 3.30 × 10−13 J = 2.06 MeV EVALUATE: The speed is close to c and the kinetic energy is four times the rest mass. (a) According to Eq.(37.38) and conservation of mass-energy m 9.75 2 Mc 2 + mc 2 = γ 2 Mc 2 ⇒ γ = 1 + =1+ = 1.292. 2M 2(16.7) Note that since γ =

v 1 1 1 = 0.6331. , we have that = 1 − 2 = 1 − 2 2 γ c (1.292) 2 1− v c

(b) According to Eq.(37.36), the kinetic energy of each proton is ⎛ 1.00 MeV ⎞ K = (γ − 1) Mc 2 = (1.292 − 1)(1.67 × 10−27 kg)(3.00 × 108 m s ) 2 ⎜ ⎟ = 274 MeV. −13 ⎝ 1.60 × 10 J ⎠

⎛ 1.00 MeV ⎞ (c) The rest energy of η 0 is mc 2 = (9.75 × 10−28 kg)(3.00 × 108 m s) 2 ⎜ ⎟ = 548 MeV. −13 ⎝ 1.60 × 10 J ⎠ (d) The kinetic energy lost by the protons is the energy that produces the η 0 , 548 MeV = 2(274 MeV). 37.45.

IDENTIFY: The relativistic expression for the kinetic energy is K = (γ − 1) mc 2 , where γ =

1 The Newtonian expression for the kinetic energy is K N = mv 2 . 2 3 SET UP: Solve for v such that K = K N . 2

1 and x = v 2 / c 2 . 1− x

Relativity

37-9

2

3 1 3 1 ⎛ 3 ⎞ (γ − 1) mc 2 = mv 2 . −1 = x. = ⎜ 1 + x ⎟ . After a little algebra this becomes 4 4 1− x ⎝ 4 ⎠ 1− x 1 −15 ± (15) 2 + 4(9)(8) . The positive root is x = 0.425 . x = v 2 / c 2 , so 9 x 2 + 15 x − 8 = 0 . x = 18 v = x c = 0.652c . EVALUATE: The fractional increase of the relativistic expression above the nonrelativistic one increases as v increases. (4.0015 u) The fraction of the initial mass (a) that becomes energy is 1 − = 6.382 × 10−3 , and so the energy released 2(2.0136 u) per kilogram is (6.382 × 10 −3 )(1.00 kg)(3.00 × 108 m s) 2 = 5.74 × 1014 J. EXECUTE:

)

(

37.46.

1.0 × 1019 J = 1.7 × 104 kg. 5.74 × 1014 J kg (a) E = mc 2 , m = E c 2 = (3.8 × 1026 J) (2.998 × 108 m s ) 2 = 4.2 × 109 kg .

(b) 37.47.

1 kg is equivalent to 2.2 lbs, so m = 4.6 × 106 tons (b) The current mass of the sun is 1.99 × 1030 kg, so it would take it 37.48.

(1.99 × 1030 kg) (4.2 × 109 kg s) = 4.7 × 1020s = 1.5 × 1013 years to use up all its mass. IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the relativistic and nonrelativistic results. 1 1 SET UP: The nonrelativistic work-energy theorem is F Δx = mv 2 − mv02 , and the relativistic formula for a 2 2 constant force is F Δx = (γ − 1)mc 2 . (a) Using the classical work-energy theorem and solving for Δx , we obtain

Δx =

m(v 2 − v02 ) (0.100 × 10−9 kg)[(0.900)(3.00 × 108 m s)]2 = = 3.65 m. 2F 2(1.00 × 106 N)

(b) Using the relativistic work-energy theorem for a constant force, we obtain (γ − 1) mc 2 Δx = . F 1 For the given speed, γ = = 2.29, thus 1 − 0.9002

Δx =

37.49.

(2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m s) 2 = 11.6 m. (1.00 × 106 N)

EVALUATE: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is required to accelerate an object to speeds close to c, so that force must act over a greater distance. (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10−8 s is the proper time, measured in the pion’s frame. The time measured in the lab must satisfy d = cΔt , where u ≈ c. Calculate Δt and then use Eq.(37.6) to calculate u. d 1.20 × 103 m = 4.003 × 10 −6 s EXECUTE: Δt = = c 2.998 × 108 m/s 2 Δt0 Δt ⎛ Δt ⎞ Δt = so (1 − u 2 / c 2 )1/ 2 = 0 and (1 − u 2 / c 2 ) = ⎜ 0 ⎟ Δt ⎝ Δt ⎠ 1 − u 2 / c2 Write u = (1 − Δ )c so that (u / c) 2 = (1 − Δ ) 2 = 1 − 2Δ + Δ 2 ≈ 1 − 2Δ since Δ is small.

⎛ Δt ⎞ Using this in the above gives 1 − (1 − 2Δ) = ⎜ 0 ⎟ ⎝ Δt ⎠ 2

2

1 ⎛ Δt ⎞ 1 ⎛ 2.60 × 10−8 s ⎞ −5 Δ= ⎜ 0⎟ = ⎜ ⎟ = 2.11 × 10 2 ⎝ Δt ⎠ 2 ⎝ 4.003 × 10 −6 s ⎠ EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving pion so that the pion travels that length before decaying. The contracted length must be l = cΔt0 = (2.998 × 108 m/s)(2.60 × 10−8 s) = 7.79 m. 2

⎛l ⎞ l = l0 1 − u 2 / c 2 so 1 − u 2 / c 2 = ⎜ ⎟ ⎝ l0 ⎠

2

37-10

Chapter 37 2

1 ⎛ l ⎞ 1 ⎛ 7.79 m ⎞ −5 Then u = (1 − Δ)c gives Δ = ⎜ ⎟ = ⎜ ⎟ = 2.11 × 10 , which checks. 2 ⎝ l0 ⎠ 2 ⎝ 1.20 × 103 m ⎠ 2

(b) IDENTIFY and SET UP: E = γ mc 2 (Eq.(37.38). 1 1 1 EXECUTE: γ = = = = 154 2 2 2Δ 1− u /c 2(2.11 × 10−5 )

37.50.

37.51.

E = 154(139.6 MeV) = 2.15 × 104 MeV = 21.5 GeV EVALUATE: The total energy is 154 times the rest energy. IDENTIFY and SET UP: The proper length of a side is l0 = a . The side along the direction of motion is shortened

to l = l0 1 − v 2 / c 2 . The sides in the two directions perpendicular to the motion are unaffected by the motion and still have a length a. EXECUTE: V = a 2l = a 3 1 − v 2 / c 2 IDENTIFY and SET UP: There must be a length contraction such that the length a becomes the same as b; l0 = a, l = b. l0 is the distance measured by an observer at rest relative to the spacecraft. Use Eq.(37.16) and solve for u. b l EXECUTE: = 1 − u 2 / c 2 so = 1 − u 2 / c 2 ; a l0 a = 1.40b gives b /1.40b = 1 − u 2 / c 2 and thus 1 − u 2 / c 2 = 1/(1.40) 2

37.52.

u = 1 − 1/(1.40) 2 c = 0.700c = 2.10 × 108 m/s EVALUATE: A length on the spacecraft in the direction of the motion is shortened. A length perpendicular to the motion is unchanged. IDENTIFY and SET UP: The proper time Δt0 is the time that elapses in the frame of the space probe. Δt is the time that elapses in the frame of the earth. The distance traveled is 42.2 light years, as measured in the earth frame. c ⎛ ⎞ EXECUTE: (a) Light travels 42.2 light years in 42.2 yr, so Δt = ⎜ ⎟ (42.2 yr) = 42.6 yr . 0.9910 c ⎝ ⎠

Δt0 = Δt 1 − u 2 / c 2 = (42.6 yr) 1 − (0.9910) 2 = 5.7 yr . She measures her biological age to be 19 yr + 5.7 yr = 24.7 yr. (b) Her age measured by someone on earth is 19 yr + 42.6 yr = 61.6 yr .

37.53.

(a) E = γ mc 2 and γ = 10 =

1 1 − (v c )

2

⇒

2 γ −1 99 v v = ⇒ = = 0.995. 2 γ 100 c c

⎞ ⎛ ⎛ v ⎞2 (b) ( pc) 2 = m 2v 2 γ 2c 2 , E 2 = m 2c 4 ⎜ ⎜ ⎟ γ 2 + 1⎟ ⎜⎝ c ⎠ ⎟ ⎝ ⎠ ⇒

37.54.

E 2 − ( pc) 2 = E2

1 ⎛v⎞ 1+ γ 2 ⎜ ⎟ ⎝c⎠

2

=

1 = 0.01 = 1%. 1 + (10 (0.995)) 2

IDENTIFY and SET UP: The clock on the plane measures the proper time Δt0 . Δt = 4.00 h = 4.00 h (3600 s/1 h) = 1.44 × 10 4 s. Δt0 Δt = and Δt0 = Δt 1 − u 2 / c 2 2 2 1− u / c ⎛ 1 u2 ⎞ u 1 u2 ; EXECUTE: small so 1 − u 2 / c 2 = (1 − u 2 / c 2 )1/ 2 ≈ 1 − thus Δ t = Δ t ⎜1 − 0 2 ⎟ 2 c2 c ⎝ 2c ⎠ 2

The difference in the clock readings is Δt − Δt0 =

37.55.

1 u2 1⎛ 250 m/s ⎞ −9 4 Δt = ⎜ ⎟ (1.44 × 10 s) = 5.01 × 10 s. The 2 c2 2 ⎝ 2.998 × 108 m/s ⎠

clock on the plane has the shorter elapsed time. EVALUATE: Δt0 is always less than Δt ; our results agree with this. The speed of the plane is much less than the speed of light, so the difference in the reading of the two clocks is very small. IDENTIFY: Since the speed is very close to the speed of light, we must use the relativistic formula for kinetic energy.

Relativity

37-11

⎛ ⎞ 1 SET UP: The relativistic formula for kinetic energy is K = mc 2 ⎜ − 1⎟ and the relativistic mass is ⎜ 1 − v2 c2 ⎟ ⎝ ⎠ m mrel = . 1 − v2 c2 EXECUTE: (a) K = 7 × 1012 eV = 1.12 × 10−6 J . Using this value in the relativistic kinetic energy formula and ⎛ ⎞ 1 substituting the mass of the proton for m, we get K = mc 2 ⎜ − 1⎟ ⎜ 1 − v2 c2 ⎟ ⎝ ⎠ 2 v 1 1 which gives = 7.45 × 103 and 1 − 2 = . Solving for v gives 2 2 × (7.45 103 ) 2 c 1− v c

1−

v 2 (c + v )(c − v) 2(c − v) = = , since c + v ≈ 2c. Substituting v = (1 − Δ )c , we have. c2 c2 c 1

3 1 − v 2 / c 2 ( 7.45 × 10 ) v 2 2(c − v) 2 [ c − (1 − Δ )c ] 1− 2 = = = 2Δ . Solving for Δ gives Δ = = = 9 × 10−9 , to one 2 2 c c c significant digit. 1 (b) Using the relativistic mass formula and the result that = 7.45 × 103 , we have 2 2 1− v c 2

⎛ ⎞ 1 ⎟ = (7 × 103 )m , to one significant digit. = m⎜ 2 2 ⎜ 1− v c ⎟ 1− v c ⎝ ⎠ EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass. E ⎛ 1 ⎞ IDENTIFY and SET UP: The energy released is E = ( Δm)c 2 . Δm = ⎜ 4 ⎟ (8.00 kg) . Pav = . The change in t ⎝ 10 ⎠ gravitational potential energy is mg Δy . m

mrel =

37.56.

2

2

⎛ 1 ⎞ EXECUTE: (a) E = ( Δm)c 2 = ⎜ 4 ⎟ (8.00 kg)(3.00 × 108 m/s) 2 = 7.20 × 1013 J ⎝ 10 ⎠ E 7.20 × 1013 J = = 1.80 × 1019 W t 4.00 × 10−6 s E 7.20 × 1013 J = = 7.35 × 109 kg (c) E = ΔU = mg Δy . m = g Δy (9.80 m/s 2 )(1.00 × 103 m) (b) Pav =

37.57.

c IDENTIFY and SET UP: In crown glass the speed of light is v = . Calculate the kinetic energy of an electron that n has this speed. 2.998 × 108 m/s = 1.972 × 108 m/s. EXECUTE: v = 1.52 K = mc 2 (γ − 1) mc 2 = (9.109 × 10 −31 kg)(2.998 × 108 m/s) 2 = 8.187 × 10 −14 J(1 eV/1.602 × 10−19 J) = 0.5111 MeV

γ=

37.58.

37.59.

1 1− v /c 2

2

=

1 1 − ((1.972 × 10 m/s)/(2.998 × 108 m/s)) 2 8

= 1.328

K = mc 2 (γ − 1) = (0.5111 MeV)(1.328 − 1) = 0.168 MeV EVALUATE: No object can travel faster than the speed of light in vacuum but there is nothing that prohibits an object from traveling faster than the speed of light in some material. p ( E c) E , where the atom and the photon have the same magnitude of momentum, E c . (a) v = = = m m mc E (b) v = c, so E mc 2 . mc IDENTIFY and SET UP: Let S be the lab frame and S ′ be the frame of the proton that is moving in the +x direction, so u = + c / 2 . The reference frames and moving particles are shown in Figure 37.59. The other proton moves in

37-12

Chapter 37

the − x direction in the lab frame, so v = −c / 2 . A proton has rest mass mp = 1.67 × 10−27 kg and rest energy

mp c 2 = 938 MeV . EXECUTE: (a) v′ =

−c / 2 − c / 2 v −u 4c = =− 1 − uv / c 2 1 − (c / 2)( −c / 2) / c 2 5

4 c. 5 (b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c. mc 2 (c) K = − mc 2 1 − v2 / c2 (i) Relative to the lab frame each proton has speed v = c / 2 . The total kinetic energy of each proton is 938 MeV K= − (938 MeV) = 145 MeV . 2 ⎛1⎞ 1− ⎜ ⎟ ⎝ 2⎠ 4 (ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed c . In this frame 5 938 MeV the kinetic energy of the moving proton is K = − (938 MeV) = 625 MeV 2 ⎛ 4⎞ 1− ⎜ ⎟ ⎝5⎠ (d) (i) Each proton has speed v = c / 2 and kinetic energy mc 2 938 MeV 1 2 ⎛1 ⎞ K = mv 2 = ⎜ m ⎟ ( c / 2 ) = = = 117 MeV 2 8 8 ⎝2 ⎠ (ii) One proton has speed v = 0 and the other has speed c. The kinetic energy of the moving proton 1 938 MeV is K = mc 2 = = 469 MeV 2 2 EVALUATE: The relativistic expression for K gives a larger value than the nonrelativistic expression. The kinetic energy of the system is different in different frames. The speed of each proton relative to the other is

Figure 37.59 37.60.

IDENTIFY and SET UP: Let S be the lab frame and let S ′ the frame of the proton that is moving in the +x direction in the lab frame, as shown in Figure 37.60. In S ′ the other proton moves in the − x′ direction with speed c / 2 , so v′ = −c / 2 . In the lab frame each proton has speed α c , where α is a constant that we need to solve for. v′ + u −0.50c + α c and EXECUTE: (a) v = with v = −α c , u = +α c and v′ = −0.50c gives −α c = 2 1 + uv′ / c 1 + (α c)(−0.50c) / c 2

−0.50 + α . α 2 − 4α + 1 = 0 and α = 0.268 or α = 3.73 . Can’t have v > c , so only α = 0.268 is physically 1 − 0.50α allowed. The speed measured by the observer in the lab is 0.268c. (b) (i) v = 0.269c . γ = 1.0380 . K = (γ − 1)mc 2 = 35.6 MeV . −α =

Relativity

37-13

(ii) v = 0.500c . γ = 1.1547 . K = (γ − 1)mc = 145 MeV . 2

37.61.

37.62.

37.63.

x′ = c t ′ ⇒ ( x − ut ) γ = c γ ( t − ux c 2

2 2

2

2

2

2

)

Figure 37.60

2 2

⎛ u⎞ 1 ⇒ x − ut = c(t − ux c 2 ) ⇒ x ⎜ 1 + ⎟ = x(u + c ) = t (u + c) ⇒ x = ct ⇒ x 2 = c 2t 2 . ⎝ c⎠ c IDENTIFY and SET UP: Let S be the lab frame and let S ′ be the frame of the nucleus. Let the +x direction be the direction the nucleus is moving. u = 0.7500c . v′ + u 0.9995c + 0.7500c EXECUTE: (a) v′ = +0.9995c . v = = = 0.999929c 1 + uv′ / c 2 1 + (0.7500)(0.9995) −0.9995c + 0.7500c (b) v′ = −0.9995c . v = = −0.9965c 1 + (0.7500)( −0.9995) (c) emitted in same direction: ⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 42.4 MeV (i) K = ⎜ 2 2 2 ⎜ 1 − (0.999929) ⎟ ⎝ 1− v / c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 (ii) K ′ = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV 2 2 2 ⎜ ⎟ ⎝ 1− v /c ⎠ ⎝ 1 − (0.9995) ⎠ (d) emitted in opposite direction: ⎛ ⎞ ⎛ ⎞ 1 1 (i) K = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 5.60 MeV 2 2 ⎜ 1 − (0.9965) 2 ⎟ ⎝ 1− v / c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 2 1 1 (ii) K ′ = ⎜ − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV 2 2 2 ⎜ ⎟ ⎝ 1− v /c ⎠ ⎝ 1 − (0.9995) ⎠ IDENTIFY and SET UP: Use Eq.(37.30), with a = dv / dt , to obtain an expression for dv / dt. Separate the variables v and t and integrate to obtain an expression for v (t ). In this expression, let t → ∞. dv F EXECUTE: a = = (1 − v 2 / c 2 )3 / 2 . (One-dimensional motion is assumed, and all the F, v, and a refer to xdt m components.) dv ⎛F⎞ = ⎜ ⎟ dt (1 − v 2 / c 2 )3 / 2 ⎝ m ⎠ Integrate from t = 0, when v = 0, to time t, when the velocity is v. v t⎛ F ⎞ dv ∫ 0 (1 − v 2 / c 2 )3 / 2 = ∫ 0 ⎜⎝ m ⎟⎠ dt t⎛ F ⎞ Ft Since F is constant, ∫ ⎜ ⎟ dt = . In the velocity integral make the change of variable y = v / c; then dy = dv / c. 0 m m ⎝ ⎠ v/c

v/c ⎡ ⎤ dv dy y v ∫ 0 (1 − v 2 / c 2 )3 / 2 = c ∫ 0 (1 − y 2 )3 / 2 = c ⎢⎣ (1 − y 2 )1/ 2 ⎥⎦ = 1 − v 2 / c 2 0 v Ft Thus = . 1 − v2 / c2 m v

37-14

Chapter 37

37.64.

Solve this equation for v: 2 2 v2 ⎛ Ft ⎞ ⎛ Ft ⎞ 2 2 2 v = = and ⎜ ⎟ (1 − v / c ) ⎜ ⎟ 1 − v2 / c2 ⎝ m ⎠ ⎝m⎠ ⎛ ⎛ Ft ⎞2 ⎞ ⎛ Ft ⎞ 2 ( Ft / m) Ft so v = v 2 ⎜1 + ⎜ = =c 2 2 2 ⎜ ⎝ mc ⎟⎠ ⎟⎟ ⎜⎝ m ⎟⎠ 1 + ( Ft / mc) m c + F 2t 2 ⎝ ⎠ Ft Ft → → 1, so v → c. As t → ∞, 2 2 2 2 m c +F t F 2t 2 Ft is always less than 1, so v < c always and v approaches c only when t → ∞. EVALUATE: Note that 2 2 m c + F 2t 2 Setting x = 0 in Eq.(37.21), the first equation becomes x′ = −γ ut and the last, upon multiplication by c, becomes

37.65.

ct ′ = γ ct.Squaring and subtracting gives c 2t′2 − x′2 = γ 2 (c 2t 2 − u 2t 2 ) = c 2t 2 , or x′ = c t ′2 − t 2 = 4.53 × 108 m. (a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq.37.21) for ( x1 , t1 ) and ( x2 , t2 ) : x1 − ut1 x − ut2 x1′ = , x2′ = 2 2 2 1− u /c 1 − u 2 / c2 2 t − ux1 / c t − ux2 / c 2 t1′ = 1 , t2′ = 2 1 − u 2 / c2 1 − u 2 / c2 Same point in S ′ implies x1′ = x2′ . What then is Δt ′ = t2′ − t1′ ? EXECUTE: x1′ = x′2 implies x1 − ut1 = x2 − ut2 x − x Δx u (t2 − t1 ) = x2 − x1 and u = 2 1 = t2 − t1 Δt From the time transformation equations, 1 ( Δt − uΔx / c 2 ) Δt ′ = t2′ − t1′ = 1 − u 2 / c2 Δx gives Using the result that u = Δt 1 Δt ′ = ( Δt − ( Δx) 2 /(( Δt )c 2 )) 2 1 − ( Δx ) /(( Δt ) 2 c 2 )

Δt ′ =

Δt ′ =

Δt ( Δt ) − ( Δx ) 2 / c 2 2

( Δt ) 2 − ( Δx ) 2 / c 2 ( Δt ) 2 − ( Δx ) 2 / c 2

( Δt − ( Δx) 2 /((Δt )c 2 ))

= ( Δt ) 2 − ( Δx / c ) 2 , as was to be shown.

This equation doesn’t have a physical solution (because of a negative square root) if ( Δx / c ) 2 > (Δt ) 2 or Δx ≥ cΔt. (b) IDENTIFY and SET UP: Now require that t2′ = t1′ (the two events are simultaneous in S ′ ) and use the Lorentz coordinate transformation equations. EXECUTE: t2′ = t1′ implies t1 − ux1 / c 2 = t2 − ux2 / c 2 c 2 Δt ⎛x −x ⎞ ⎛ Δx ⎞ t2 − t1 = ⎜ 2 2 1 ⎟ u so Δt = ⎜ 2 ⎟ u and u = Δx ⎝ c ⎠ ⎝c ⎠ From the Lorentz transformation equations, ⎛ ⎞ 1 Δx′ = x′2 − x1′ = ⎜ ⎟ (Δx − u Δt ). 2 2 ⎝ 1− u / c ⎠ Using the result that u = c 2 Δt / Δx gives 1 Δx′ = (Δx − c 2 ( Δt ) 2 / Δx) 2 1 − c (Δt ) 2 /( Δx) 2

Δx′ = Δx′ =

Δx ( Δx) 2 − c 2 ( Δt ) 2 (Δx) 2 − c 2 ( Δt ) 2 ( Δx ) 2 − c 2 ( Δt ) 2

(Δx − c 2 (Δt ) 2 / Δx) = (Δx) 2 − c 2 (Δt ) 2

(c) IDENTIFY and SET UP: The result from part (b) is Δx′ = ( Δx) 2 − c 2 (Δt ) 2

Relativity

37-15

Solve for Δt : ( Δx′) 2 = ( Δx) 2 − c 2 ( Δt ) 2 ( Δx) 2 − ( Δx′) 2 (5.00 m) 2 − (2.50 m) 2 = = 1.44 × 10−8 s c 2.998 × 108 m/s EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first. 1 (a) 80.0 m s is non-relativistic, and K = mv 2 = 186 J. 2 (b) (γ − 1) mc 2 = 1.31 × 1015 J. Δt =

EXECUTE:

37.66.

(c) In Eq. (37.23), c) v′ = 2.20 × 108 m s, u = −1.80 × 108 m s,and so v = 7.14 × 107 m s.

20.0 m

(d) (e)

γ

20.0 m = 9.09 × 10−8 s. 2.20 × 108 m s

(f) t ′ = 37.67.

= 13.6 m.

t 13.6 m = 6.18 × 10−8 s, or t ′ = = 6.18 × 10−8 s. γ 2.20 × 108 m s

IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency ( f = c / λ ), so the

atoms are moving away from the earth. Receding, so use Eq.(37.26): f =

c−u f0 c+u

⎛ 1 − ( f / f0 )2 ⎞ EXECUTE: Solve for u: ( f / f 0 ) 2 (c + u ) = c − u and u = c ⎜ 2 ⎟ ⎝ 1 + ( f / f0 ) ⎠ f = c / λ , f 0 = c / λ0 so f / f 0 = λ0 / λ

37.68.

⎛ 1 − (λ0 / λ ) 2 ⎞ ⎛ 1 − (656.3/ 953.4) 2 ⎞ u = c⎜ c = = 0.357c = 1.07 × 108 m/s ⎟ ⎜ 2 2 ⎟ ⎝ 1 + (656.3/ 953.4) ⎠ ⎝ 1 + (λ0 / λ ) ⎠ EVALUATE: The relative speed is large, 36% of c. The cosmological implication of such observations will be discussed in Section 44.6. The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq.(37.25)) becomes f ≅ f 0 (1 − (u c)). In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and the baseball reradiates at f. But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected frequency is f (1 − (u c)) = f 0 (1 − (u c)) 2 ≈ f 0 (1 − 2(u c)), so Δf = 2 f 0 (u c) and the fractional frequency shift is

Δf = 2(u c ). In this case, f0 u= 37.69.

Δf (2.86 × 10 −7 ) c= (3.00 × 108 m) = 42.9 m s = 154 km h = 92.5 mi h. 2 f0 2

IDENTIFY and SET UP: 500 light years = 4.73 × 1018 m . The proper distance l0 to the star is 500 light years. The energy needed is the kinetic energy of the rocket at its final speed. d 4.73 × 1018 m EXECUTE: (a) u = 0.50c . Δt = = = 3.2 × 1010 s = 1000 yr u (0.50)(3.00 × 108 m/s)

The proper time is measured by the astronauts. Δt0 = Δt 1 − u 2 / c 2 = 866 yr

⎛ ⎞ 1 − mc 2 = (1000 kg)(3.00 × 108 m/s) 2 ⎜ − 1⎟ = 1.4 × 1019 J ⎜ 1 − (0.500) 2 ⎟ 1 − v2 / c2 ⎝ ⎠ This is 140% of the U.S. yearly use of energy. d 4.73 × 1018 m (b) u = 0.99c . Δt = = = 1.6 × 1010 s = 505 yr , Δt0 = 71 yr u (0.99)(3.00 × 108 m/s) K=

mc 2

⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 5.5 × 1020 J ⎜ 1 − (0.99) 2 ⎟ ⎝ ⎠ This is 55 times the U.S. yearly use.

37-16

Chapter 37

(c) u = 0.9999c . Δt =

37.70.

d 4.73 × 1018 m = = 1.58 × 1010 s = 501 yr , Δt0 = 7.1 yr u (0.9999)(3.00 × 108 m/s)

⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 6.3 × 1021 J 2 ⎜ 1 − (0.9999) ⎟ ⎝ ⎠ This is 630 times the U.S. yearly use. The energy cost of accelerating a rocket to these speeds is immense. (a) As in the hint, both the sender and the receiver measure the same distance. However, in our frame, the ship has moved between emission of successive wavefronts, and we can use the time T = 1 f as the proper time, with the result that f = γ f 0 > f 0 . 1/ 2

(b) Toward: f = f 0

c+u ⎛ 1 + 0.758 ⎞ = 345 MHz ⎜ ⎟ c−u ⎝ 1 − 0.758 ⎠

= 930 MHz

f − f 0 = 930 MHz − 345 MHz = 585 MHz. 1/ 2

Away: f = f 0

37.71.

c−u ⎛ 1 − 0.758 ⎞ = 345 MHz ⎜ ⎟ c+u ⎝ 1 + 0.758 ⎠

= 128 MHz and f − f 0 = −217 MHz.

(c) γf 0 = 1.53 f 0 = 528 MHz, f − f 0 = 183 MHz. The shift is still bigger than f 0 , but not as large as the approaching frequency. The crux of this problem is the question of simultaneity. To be “in the barn at one time” for the runner is different than for a stationary observer in the barn. The diagram in Figure 37.71a shows the rod fitting into the barn at time t = 0 , according to the stationary observer. The diagram in Figure 37.71b is in the runner’s frame of reference. The front of the rod enters the barn at time t1 and leaves the back of the barn at time t2 . However, the back of the rod does not enter the front of the barn until the later time t3 .

Figure 37.71 37.72.

(c/n) + V (c/n) + V = . For V non-relativistic, this is cV 1 + (V/nc) 1+ 2 nc c ⎛ 1 ⎞ 1 ⎞ ⎛ v ≈ ((cn) + V )(1 − (V/nc )) = (nc/n) + V − (V/n 2 ) − (V 2 /nc) ≈ + ⎜1 − 2 ⎟V , so k = ⎜1 − 2 ⎟ . For water, n = 1.333 n ⎝ n ⎠ ⎝ n ⎠ and k = 0.437. In Eq.(37.23), u = V , v′ = (c n), and so v =

Relativity

37.73.

37-17

dv dv v−u u dv . dt ′ = γ ( dt − udx c 2 ) . dv′ = + 2 2 2 dt ′ (1 − uv c ) (1 − uv c ) c 2 1 dv′ v −u ⎛u⎞ = + ⎜ ⎟. dv 1 − uv c 2 (1 − uv c 2 ) 2 ⎝ c 2 ⎠

(a) a′ =

⎛ ⎛ 1 − u 2 c2 ⎞ 1 (v − u ) u c 2 ⎞ dv′ = dv ⎜ + = dv ⎜ 2 2 2 ⎟ 2 2 ⎟ (1 − uv c ) ⎠ ⎝ 1 − uv c ⎝ (1 − uv c ) ⎠ (1 − u 2 c 2 ) 1 (1 − uv c 2 ) 2 dv (1 − u 2 c 2 ) a′ = = 2 γdt − uγ dx c dt (1 − uv c 2 ) 2 γ(1 − uv c 2 )

dv

= a (1 − u 2 c 2 )3 2 (1 − uv c 2 ) −3 . −3

37.74.

⎛ uv′ ⎞ (b) Changing frames from S ′ → S just involves changing a → a′, v → − v′ ⇒ a = a′(1 − u 2 c 2 )3 2 ⎜1 + 2 ⎟ . c ⎠ ⎝ (a) The speed v′ is measured relative to the rocket, and so for the rocket and its occupant, v′ = 0. The acceleration as seen in the rocket is given to be a′ = g , and so the acceleration as measured on the earth is 32

⎛ u2 ⎞ du = g ⎜1 − 2 ⎟ . dt ⎝ c ⎠ (b) With v1 = 0 when t = 0 , t1 1 du 1 v1 du v1 dt = . ∫ dt = ∫ . t1 = . 2 2 32 0 g (1 − u c ) g 0 (1 − u 2 c 2 )3 2 g 1 − v12 c 2

a=

(c) dt ′ = γ dt = dt / 1 − u 2 c 2 , so the relation in part (b) between dt and du, expressed in terms of dt ′ and du, is 1 du 1 du dt ′ = γ dt = . = 2 2 g (1 − u 2 c 2 ) 3 2 g (1 − u 2 c 2 ) 2 1− u c

c ⎛v ⎞ arctanh ⎜ 1 ⎟ . For those who wish to g ⎝c⎠ avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions; c ⎛ c + v1 ⎞ du 1 ⎡ du du ⎤ gdt ′ = 1n ⎜ = ⎢ + ⎟. ⎥ , which integrates to t1′ = 2 g ⎝ c − v1 ⎠ (1 + u c)(1 − u c) 2 ⎣1 + u c 1 − uc ⎦ (d) Solving the expression from part (c) for v1 in terms of t1 , (v1 c ) = tanh( gt1′ c), so that Integrating as above (perhaps using the substitution z = u c ) gives t1′ =

1 − (v1 c) 2 = 1 cosh ( gt1′ c), using the appropriate indentities for hyperbolic functions. Using this in the expression gt c tanh( gt1′ c) c ⎛ gt ′ ⎞ = sinh( gt1′ c), which may be rearranged slightly as 1 = sinh ⎜ 1 ⎟ . If c g 1 cosh( gt1′ c ) g ⎝ c ⎠ gt1′/c − gt1′/c v e −e hyperbolic functions are not used, v1 in terms of t′1 is found to be 1 = gt1′/c which is the same as c e + e− gt1′/c c gt1′ c − gt1′ c (e −e ), tanh( gt1′ c ). Inserting this expression into the result of part (b) gives, after much algebra, t1 = 2g which is equivalent to the expression found using hyperbolic functions. (e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on earth is c t1′ = sinh( gt1′ c) = 2.65 × 109 s = 84.0 yr. g

found in part (b), t1 =

37.75.

The elapsed time will be the same for each of the four parts of the voyage, so when Stella has returned, Terra has aged 336 yr and the year is 2436. (Keeping more precision than is given in the problem gives February 7 of that year.) (a) f 0 = 4.568110 × 1014 Hz; f + = 4.568910 × 1014 Hz; f − = 4.567710 × 1014 Hz c + (u + v ) ⎫ f0 ⎪ c − (u + v) ⎪ f +2 (c − (u + v)) = f 02 (c + (u + v)) ⎬⇒ 2 f − (c − (u − v)) = f 02 (c + (u − v)) c + (u − v) ⎪ f− = f0 ⎪ c − (u − v ) ⎭ f+ =

37-18

Chapter 37

where u is the velocity of the center of mass and v is the orbital velocity.

⇒ (u + v ) =

( f + f0 ) 2 − 1 ( f −2 f 02 ) − 1 and c ( u − v ) = c ( f + f0 )2 + 1 ( f −2 f 02 ) + 1

⇒ u + v = 5.25 × 10 4 m s and u − v = −2.63 × 104 m s .

This gives u = +1.31 × 104 m s (moving toward at 13.1 km s) and v = 3.94 × 10 4 m/s . (b) v = 3.94 × 104 m s; T = 11.0 days. 2π R = vt ⇒

(3.94 × 104 m s)(11.0 days)(24 hrs day)(3600 sec hr) = 5.96 × 109 m . This is about 2π 0.040 times the earth-sun distance. Also the gravitational force between them (a distance of 2R) must equal the centripetal force from the center of mass: (Gm 2 ) mv 2 4 Rv 2 4(5.96 × 109 m)(3.94 × 104 m s) 2 ⇒m= = = 5.55 × 10 29 kg = 0.279 m sun . = 2 (2 R) R G 6.672 × 10 −11 N ⋅ m 2 kg 2 For any function f = f ( x, t ) and x = x( x′, t′), t = t ( x′, t′), let F ( x′, t ′) = f ( x( x′, t ′), t ( x′, t ′)) and use the standard (but mathematically improper) notation F ( x′, t ′) = f ( x′, t ′). The chain rule is then R=

37.76.

∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t′ , = + ∂x ∂x′ ∂x ∂t ′ ∂x ∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t ′ . = + ∂t ∂x′ ∂t ∂t ′ ∂t In this solution, the explicit dependence of the functions on the sets of dependent variables is suppressed, and the ∂f ∂f ∂x′ ∂f ∂t′ ∂f ∂f ∂x′ ∂f ∂t ′ = + , = + . above relations are then ∂x ∂x′ ∂x ∂t′ ∂x ∂t ∂x′ ∂t ∂t ′ ∂t ∂x′ ∂x′ ∂t ′ ∂t ′ ∂E ∂E ∂2E ∂2E (a) = 1, = −v, = 0 and = 1. Then, = , and 2 = 2 . For the time derivative, ∂x ∂t ∂x ∂t ∂x ∂x′ ∂x ∂x′ ∂E ∂E ∂E = −v + . To find the second time derivative, the chain rule must be applied to both terms; that is, ∂t ∂x′ ∂t′

∂ ∂t ∂ ∂t

∂E ∂2E ∂2E = −v 2 + , ∂x′ ∂x′ ∂t ′∂x′ ∂E ∂2E ∂2E = −v + . ∂t′ ∂x′∂t ′ ∂t ′2

∂2E , collecting terms and equating the mixed partial derivatives gives ∂t 2 2 ∂2E ∂2E ∂2E ∂2E 2 ∂ E = v − 2 v + , and using this and the above expression for gives the result. ∂t 2 ∂x′2 ∂x′∂t ′ ∂t ′2 ∂x′2 ∂x′ ∂x′ ∂t′ ∂t ′ = γ, = γ v, = γ v / c 2 and = γ. (b) For the Lorentz transformation, ∂x ∂t ∂x ∂t The first partials are then Using these in

∂E ∂E v ∂E ∂E ∂E ∂E =γ −γ 2 = − γv +γ , ∂x ∂x′ c ∂t′ ∂t ∂x′ ∂t′ and the second partials are (again equating the mixed partials) 2 2 2 ∂2E ∂2E 2 ∂ E 2 v ∂ E 2 v = + − γ γ 2 γ ∂x 2 ∂x′2 c 4 ∂t ′2 c 2 ∂x′∂t ′ 2 2 2 ∂ E ∂ E ∂ E ∂2E = γ 2 v 2 2 + γ 2 2 − 2 γ 2v . 2 ∂t ∂x′ ∂t ′ ∂x′∂t′

Substituting into the wave equation and combining terms (note that the mixed partials cancel), 2 v2 ⎞ ∂2 E ∂2E 1 ∂2E 1 ⎞ ∂2E ∂2E 1 ∂2E 2⎛ 2⎛v γ 1 γ − = − + − = 2− 2 = 0. ⎜ ⎟ ⎜ 2 2 4 2 ⎟ 2 ∂x 2 c 2 ∂t 2 ∂x′ c ∂t ′2 ⎝ c ⎠ ∂x′ ⎝ c c ⎠ ∂t ′

Relativity

37.77.

37-19

(a) In the center of momentum frame, the two protons approach each other with equal velocities (since the protons have the same mass). After the collision, the two protons are at rest─but now there are kaons as well. In this situation the kinetic energy of the protons must equal the total rest energy of the two kaons ⇒ 2( γ cm − 1)mp c 2 = 2mk c 2 ⇒ γ cm = 1 + γ cm − 1

mk = 1.526. The velocity of a proton in the center of momentum frame is then mp

2

vcm = c

2 γ cm

= 0.7554c.

To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations. This is the same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is. Taking the lab frame to be the unprimed frame moving to the left, u = vcm and v′ = vcm (the velocity of the projectile proton in the center of momentum frame). v′ + u 2vcm 1 vlab = = = 0.9619c ⇒ γ lab = = 3.658 ⇒ K lab = ( γ lab − 1)mpc 2 = 2494 MeV. 2 2 uv′ v v 1 + 2 1 + cm2 1 − lab2 c c c

K lab 2494 MeV = = 2.526. 2mk 2(493.7 MeV) (c) The center of momentum case considered in part (a) is the same as this situation. Thus, the kinetic energy required is just twice the rest mass energy of the kaons. K cm = 2(493.7 MeV) = 987.4 MeV. This offers a substantial advantage over the fixed target experiment in part (b). It takes less energy to create two kaons in the proton center of momentum frame. (b)

PHOTONS, ELECTRONS, AND ATOMS

38.1.

38

IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 =

h φ f − . The e e

slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to φ by φ = hf th . EXECUTE: (a) From the graph, f th = 1.25 × 1015 Hz . Therefore, with the value of h from part (b),

φ = hf th = 4.8 eV . (b) From the graph, the slope is 3.8 × 10−15 V ⋅ s . h = (e)(slope) = (1.60 × 10−16 C)(3.8 × 10−15 V ⋅ s) = 6.1 × 10 −34 J ⋅ s (c) No photoelectrons are produced for f < f th .

38.2.

(d) For a different metal fth and φ are different. The slope is h/e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis. EVALUATE: As the frequency f of the light is increased above fth the energy of the photons in the light increases and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. IDENTIFY and SET UP: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a

photon. c = 3.00 × 108 m/s . 1 eV = 1.60 × 10 −16 J . c 3.00 × 108 m/s = 5.94 × 1014 Hz EXECUTE: (a) f = = λ 505 × 10−9 m (b) E = hf = (6.626 × 10 −34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV (c) K = 12 mv 2 so v = 38.3.

38.4.

c 3.00 × 108 m s = = 5.77 × 1014 Hz λ 5.20 × 10−7 m h 6.63 × 10 −34 J ⋅ s = 1.28 × 10−27 kg ⋅ m s p= = λ 5.20 × 10 −7 m f =

E = pc = (1.28 × 10−27 kg ⋅ m s) (3.00 × 108 m s) = 3.84 × 10−19 J = 2.40 eV. energy hc . 1 eV = 1.60 × 10−19 J . For a photon, E = hf = . h = 6.63 × 10−34 J ⋅ s. IDENTIFY and SET UP: Pav = t λ EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10 −2 J = 7.5 × 1016 eV

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = 3.05 × 10−19 J = 1.91 eV 652 × 10−9 m λ (c) The number of photons is the total energy in a pulse divided by the energy of one photon: 1.20 × 10−2 J = 3.93 × 1016 photons . 3.05 × 10 −19 J/photon EVALUATE: The number of photons in each pulse is very large. IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. c = f λ relates speed, frequency and wavelength for an electromagnetic wave. E (2.45 × 106 eV)(1.602 × 10−19 J/1 eV) = 5.92 × 1020 Hz EXECUTE: (a) E = hf so f = = 6.626 × 10−34 J ⋅ s h c 2.998 × 108 m/s = 5.06 × 10−13 m (b) c = f λ so λ = = f 5.92 × 1020 Hz (c) EVALUATE: λ is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was converted to the SI unit of Joules. (b) E =

38.5.

2K 2(3.94 × 10−19 J) = = 9.1 mm/s m 9.5 × 10−15 kg

hc

=

38-1

38-2

38.6.

Chapter 38

λth = 272 nm . c = f λ .


h = 4.136 × 10−15 eV ⋅ s . EXECUTE: (a) f th =

38.7.

c

λth

1 2 mvmax = hf − φ . At the threshold frequency, f th , vmax → 0. 2

3.00 × 108 m/s = 1.10 × 1015 Hz . 272 × 10−9 m eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV .

=

(b) φ = hf th = (4.136 × 10−15 1 2 (c) mvmax = hf − φ = (4.136 × 10−15 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV 2 EVALUATE: The threshold wavelength depends on the work function for the surface. hc 1 2 IDENTIFY and SET UP: Eq.(38.3): mvmax = hf − φ = − φ . Take the work function φ from Table 38.1. Solve 2 λ for vmax . Note that we wrote f as c / λ . EXECUTE:

1 2 (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) − (5.1 eV)(1.602 × 10−19 J/1 eV) mvmax = 2 235 × 10−9 m

1 2 mvmax = 8.453 × 10 −19 J − 8.170 × 10−19 J = 2.83 × 10−20 J 2 2(2.83 × 10 −20 J) vmax = = 2.49 × 105 m/s 9.109 × 10−31 kg

38.8.

EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with λ = 235 nm has energy greater then the work function for the surface. hc IDENTIFY and SET UP: φ = hf th = . The minimum φ corresponds to the minimum λ .

λth

EXECUTE: 38.9.

φ=

hc

λth

=

(4.136 × 10

−15

eV ⋅ s)(3.00 × 108 m/s) = 1.77 eV 700 × 10−9 m c = f λ . The source emits (0.05)(75 J) = 3.75 J of energy as visible light each second.


E = hf , with h = 6.63 × 10−34 J ⋅ s. EXECUTE: (a) f =

38.10.

c

λ

=

3.00 × 108 m/s = 5.00 × 1014 Hz 600 × 10 −9 m J ⋅ s)(5.00 × 1014 Hz) = 3.32 × 10−19 J . The number of photons emitted per second is

(b) E = hf = (6.63 × 10−34 3.75 J = 1.13 × 1019 photons . 3.32 × 10−19 J/photon (c) No. The frequency of the light depends on the energy of each photon. The number of photons emitted per second is proportional to the power output of the source. IDENTIFY: In the photoelectric effect, the energy of the photon is used to eject an electron from the surface, and any excess energy goes into kinetic energy of the electron. SET UP: The energy of a photon is E = hf, and the work function is given by φ = hf0, where f0 is the threshold frequency. EXECUTE: (a) From the graph, we see that Kmax = 0 when λ = 250 nm, so the threshold wavelength is 250 nm. Calling f0 the threshold frequency, we have f0 = c/λ0 = (3.00 × 108 m/s)/(250 nm) = 1.2 × 1015 Hz. (b) φ = hf0 = (4.136 × 10–15 eV ⋅ s )(1.2 × 1015 Hz) = 4.96 eV = 5.0 eV (c) The graph (see Figure 38.10) is linear for λ < λ0 (1/λ > 1/λ0), and linear graphs are easier to interpret than curves. EVALUATE: If the wavelength of the light is longer than the threshold wavelength (that is, if 1/λ < 1/λ0), the kinetic energy of the electrons is really not defined since no photoelectrons are ejected from the metal.

Figure 38.10

Photons, Electrons, and Atoms

38.11.

38-3

IDENTIFY: Protons have mass and photons are massless. (a) SET UP: For a particle with mass, K = p 2 / 2m. EXECUTE: p2 = 2 p1 means K 2 = 4 K1. (b) SET UP: For a photon, E = pc.

38.12.

EXECUTE: p2 = 2 p1 means E2 = 2 E1. EVALUATE: The relation between E and p is different for particles with mass and particles without mass. 1 2 IDENTIFY and SET UP: eV0 = mvmax , where V0 is the stopping potential. The stopping potential in volts equals 2 1 2 eV0 in electron volts. mvmax = hf − φ . 2 1 2 EXECUTE: (a) eV0 = mvmax so 2 (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) − 2.3 eV = 4.96 eV − 2.3 eV = 2.7 eV . The stopping potential eV0 = hf − φ = 250 × 10 −9 m is 2.7 electron volts. 1 2 (b) mvmax = 2.7 eV 2 (c) vmax =

38.13.

2(2.7 eV)(1.60 × 10 −19 J/eV) = 9.7 × 105 m/s 9.11 × 10−31 kg

First use Eq.(38.4) to find the work function φ . hc eV0 = hf − φ so φ = hf − eV0 = − eV0

(a) IDENTIFY: SET UP:

λ

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) EXECUTE: φ = − (1.602 × 10−19 C)(0.181 V) 254 × 10−9 m φ = 7.821 × 10−19 J − 2.900 × 10 −20 J = 7.531 × 10−19 J(1 eV/1.602 × 10 −19 J) = 4.70 eV IDENTIFY and SET UP: The threshold frequency f th is the smallest frequency that still produces photoelectrons.

It corresponds to K max = 0 in Eq.(38.3), so hf th = φ . f =

EXECUTE:

λ

says

hc

λth

=φ

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 2.64 × 10 −7 m = 264 nm φ 7.531 × 10 −19 J (b) EVALUATE: As calculated in part (a), φ = 4.70 eV. This is the value given in Table 38.1 for copper. IDENTIFY and SET UP: A photon has zero rest mass, so its energy and momentum are related by Eq.(37.40). Eq.(38.5) then relates its momentum and wavelength. EXECUTE: (a) E = pc = (8.24 × 10−28 kg ⋅ m/s)(2.998 × 108 m/s) = 2.47 × 10−19 J =

λth =

38.14.

c

hc

=

(2.47 × 10 −19 J)(1 eV/1.602 × 10−19 J) = 1.54 eV

h 6.626 × 10−34 J ⋅ s = = 8.04 × 10 −7 m = 804 nm λ p 8.24 × 10 −28 kg ⋅ m/s EVALUATE: This wavelength is longer than visible wavelengths; it is in the infrared region of the electromagnetic spectrum. To check our result we could verify that the same E is given by Eq.(38.2), using the λ we have calculated. 1 ⎛ 1 1 ⎞ IDENTIFY and SET UP: Balmer’s formula is = R ⎜ 2 − 2 ⎟ . For the Hγ spectral line n = 5. Once we have λ , λ ⎝2 n ⎠ calculate f from f = c / λ and E from Eq.(38.2). (b) p =

38.15.

h

so λ =

⎛1 1⎞ ⎛ 25 − 4 ⎞ ⎛ 21 ⎞ = R⎜ 2 − 2 ⎟ = R⎜ ⎟ = R⎜ ⎟. 2 5 100 ⎝ ⎠ ⎝ ⎠ ⎝ 100 ⎠ 100 100 m = 4.341 × 10 −7 m = 434.1 nm. = Thus λ = 21R 21(1.097 × 107 )

EXECUTE: (a)

(b) f =

c

λ

=

1

λ

2.998 × 108 m/s = 6.906 × 1014 Hz 4.341 × 10−7 m

38-4

Chapter 38

(c) E = hf = (6.626 × 10−34 J ⋅ s)(6.906 × 1014 Hz) = 4.576 × 10−19 J = 2.856 eV EVALUATE: Section 38.3 shows that the longest wavelength in the Balmer series (Hα ) is 656 nm and the

shortest is 365 nm. Our result for Hγ falls within this range. The photon energies for hydrogen atom transitions are 38.16.

in the eV range, and our result is of this order. IDENTIFY and SET UP: For the Lyman series the final state is n = 1 and the wavelengths are given by 1 ⎛1 1 ⎞ = R ⎜ 2 − 2 ⎟ , n = 2,3,.... For the Paschen series the final state is n = 3 and the wavelengths are given by λ ⎝1 n ⎠

⎛1 1⎞ = R ⎜ 2 − 2 ⎟ , n = 4,5,.... R = 1.097 × 107 m −1 . The longest wavelength is for the smallest n and the shortest λ ⎝3 n ⎠ wavelength is for n → ∞ . 1 4 ⎛ 1 1 ⎞ 3R EXECUTE: Lyman Longest: = R⎜ 2 − 2 ⎟ = . λ= = 121.5 nm . λ 3(1.097 × 107 m −1 ) ⎝1 2 ⎠ 4 1

Shortest:

1

λ

1 ⎛1 1 ⎞ = R⎜ 2 − 2 ⎟ = R . λ = = 91.16 nm × ∞ 1 1.097 107 m −1 ⎝ ⎠

Paschen Longest:

1 ⎞ R ⎛1 = R⎜ 2 − 2 ⎟ = . λ ⎝3 ∞ ⎠ 9 hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = = 2.31 × 10 −19 J = 1.44 eV. (a) Eγ = λ 8.60 × 10−7 m So the internal energy of the atom increases by 1.44 eV to E = −6.52 eV + 1.44 eV = −5.08 eV. Shortest:

38.17.

144 ⎛ 1 1 ⎞ 7R = R⎜ 2 − 2 ⎟ = . λ= = 1875 nm . λ 7(1.097 × 107 m −1 ) ⎝ 3 4 ⎠ 144 1

1

hc (6.63 × 10 −34 J ⋅ s) (3.00 × 108 m s) = = 4.74 × 10 −19 J = 2.96 eV. λ 4.20 × 10−7 m So the final internal energy of the atom decreases to E = −2.68 eV − 2.96 eV = −5.64 eV. IDENTIFY and SET UP: The ionization threshold is at E = 0 . The energy of an absorbed photon equals the energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom. EXECUTE: (a) ΔE = 0 − (−20 eV) = 20 eV (b) When the atom in the n = 1 level absorbs a 18 eV photon, the final level of the atom is n = 4 . The possible transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3 eV ; n = 4 → n = 2, 8 eV ; n = 4 → n = 1, 18 eV . Once the atom has gone to the n = 3 level, the following transitions can occur: n = 3 → n = 2, 5 eV ; n = 3 → n = 1, 15 eV . Once the atom has gone to the n = 2 level, the following transition can occur: n = 2 → n = 1, 10 eV . The possible energies of emitted photons are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV, and 18 eV. (c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed. (d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon energy for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV. IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy Ei − Ef of the atom by (b) Eγ =

38.18.

38.19.

Ei − Ef =

hc

λ

where hc = 1.240 × 10−6 eV ⋅ m .

EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0 , so 1.240 × 10 −6 eV ⋅ m = 16.79 eV . E1 = −17.50 eV . For n = 5 → n = 1 , λ = 73.86 nm and E5 − E1 = 73.86 × 10−9 m E5 = −17.50 eV + 16.79 eV = −0.71 eV . For n = 4 → n = 1 , λ = 75.63 nm and E4 = −1.10 eV . For

n = 3 → n = 1 , λ = 79.76 nm and E3 = −1.95 eV . For n = 2 → n = 1 , λ = 94.54 nm and E2 = −4.38 eV . hc 1.240 × 10−6 eV ⋅ m = = 378 nm Ei − Ef 3.28 eV EVALUATE: The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so the wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1 state. (b)

Ei − Ef = E4 − E2 = −1.10 eV − ( −4.38 eV) = 3.28 eV and λ =


38.20.

38.21.

38-5

(a) Equating initial kinetic energy and final potential energy and solving for the separation radius r, 1 (92e) (2e) 1 (184) (1.60 × 10 −19 C) = = 5.54 × 10−14 m. r= 4π P0 4π P0 (4.78 × 106 J C) K (b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force and the magnitude of the potential energy in a Coulombic field is K (4.78 × 106 eV) (1.6 × 10−19 J ev) = 13.8 N. F= = (5.54 × 10−14 m) r 1 q1q2 . (a) IDENTIFY: If the particles are treated as point charges, U = 4π P0 r SET UP: q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U. EXECUTE:

(2)(82)(1.602 × 10−19 C) 2 = 5.82 × 10 −13 J 6.50 × 10 −14 m J) = 3.63 × 106 eV = 3.63 MeV

U = (8.987 × 109 N ⋅ m 2 /C 2 )

U = 5.82 × 10 −13 J(1 eV/1.602 × 10−19

(b) IDENTIFY: Apply conservation of energy: K1 + U1 = K 2 + U 2 . SET UP: Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = 0. Alpha

particle stops implies K 2 = 0. EXECUTE: Conservation of energy thus says K1 = U 2 = 5.82 × 10−13 J = 3.63 MeV.

1 2K 2(5.82 × 10 −13 J) = = 1.32 × 107 m/s (c) K = mv 2 so v = 2 m 6.64 × 10−27 kg

38.22. 38.23.

38.24.

EVALUATE: v / c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v. When the alpha particle stops, all its initial kinetic energy has been converted to electrostatic potential energy. h (a), (b) For either atom, the magnitude of the angular momentum is = 1.05 × 10−34 kg ⋅ m 2 s. 2π IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to calculate L. EXECUTE: En = −(13.6 eV)/n 2 , so this state has n = 13.6 /1.51 = 3. In the Bohr model. L = nU so for this state L = 3U = 3.16 × 10−34 kg ⋅ m 2 /s. EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different result. 13.6 eV hc IDENTIFY and SET UP: For a hydrogen atom En = − . ΔE = , where ΔE is the magnitude of the λ n2 energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted. ⎛ 1 1⎞ EXECUTE: ΔE = E4 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = +12.75 eV . ⎝4 1 ⎠

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) c = = 97.3 nm . f = = 3.08 × 1015 Hz . ΔE λ 12.75 eV 1 Ze 2 , where Z = 4 is the nuclear IDENTIFY: The force between the electron and the nucleus in Be3+ is F = 4π P0 r 2

λ=

38.25.

charge. All the equations for the hydrogen atom apply to Be3+ if we replace e 2 by Ze 2 . (a) SET UP: Modify Eq.(38.18). 1 me 4 EXECUTE: En = − (hydrogen) becomes P0 8n 2h 2 ⎛ 1 me 4 ⎞ 1 m( Ze2 ) 2 ⎛ 13.60 eV ⎞ 3+ = Z2⎜− = Z2⎜− ⎟ (for Be ) 2 2 2 2 ⎟ 2 8 n h n P0 8n h P ⎝ ⎠ ⎝ 0 ⎠ ⎛ 13.60 eV ⎞ The ground-level energy of Be3+ is E1 = 16 ⎜ − ⎟ = −218 eV. 12 ⎝ ⎠ En = −

EVALUATE: The ground-level energy of Be3+ is Z 2 = 16 times the ground-level energy of H. (b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the n = 1 level energy.

38-6

Chapter 38

EXECUTE: The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV. EVALUATE: This is 16 times the ionization energy of hydrogen. ⎛ 1 1 ⎞ 1 (c) SET UP: = R ⎜ 2 − 2 ⎟ just as for hydrogen but now R has a different value. λ ⎝ n1 n2 ⎠ EXECUTE:

RH =

me4 = 1.097 × 107 m −1 for hydrogen becomes 8P0 h3c

me4 = 16(1.097 × 107 m −1 ) = 1.755 × 108 m −1 for Be3+ . 8P0 h3c 1 ⎛1 1⎞ For n = 2 to n = 1, = RBe ⎜ 2 − 2 ⎟ = 3R/4. λ ⎝1 2 ⎠ RBe = Z 2

λ = 4 /(3R ) = 4 /(3(1.755 × 108 m −1 )) = 7.60 × 10−9 m = 7.60 nm. EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding transition in the hydrogen atom. n 2h 2 (d) SET UP: Modify Eq.(38.12): rn = P0 (hydrogen). π me2 2 2 nh EXECUTE: rn = P0 (Be3+ ). π m( Ze 2 )

38.26.

38.27.

EVALUATE: For a given n the orbit radius for Be3+ is smaller by a factor of Z = 4 compared to the corresponding radius for hydrogen. (a) We can find the photon’s energy from Eq. 38.8 ⎛ 1 1 ⎞ ⎛ 1 1⎞ E = hcR ⎜ 2 − 2 ⎟ = (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s ) (1.097 × 107 m −1 ) ⎜ 2 − 2 ⎟ = 4.58 × 10 −19 J. The ⎝2 n ⎠ ⎝2 5 ⎠ E corresponding wavelength is λ = = 434 nm. hc (b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by h Eq. 38.10: L = n . Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following 2π loss in angular momentum (which we would assume is transferred to the photon): h 3(6.63 × 10 −34 J ⋅ s) ΔL = (2 − 5) =− = −3.17 × 10−34 J ⋅ s. 2π 2π However, this prediction of the Bohr model is wrong (as shown in Chapter 41). 1 e2 (1.60 × 10−19 C) 2 : n = 1 ⇒ v1 = = 2.18 × 106 m/s (a) vn = P0 2nh P0 2 (6.63 × 10−34 J ⋅ s)

h = 2 ⇒ v2 = (b) Orbital period =

v1 v = 1.09 × 106 m s. h = 3 ⇒ v3 = 1 = 7.27 × 105 m s. 2 3

2π rn 2P0 n 2h 2 me 2 4P02n3h3 = = vn 1 P0 ⋅ e2 2nh me4 n = 1 ⇒ T1 =

4P02 (6.63 × 10−34 J ⋅ s)3 = 1.53 × 10−16 s (9.11 × 10 −31 kg) (1.60 × 10−19 C) 4

n = 2 : T2 = T1 (2)3 = 1.22 × 10 −15 s. n = 3 : T3 = T1 (3)3 = 4.13 × 10 −15 s. (c) number of orbits = 38.28.

1.0 × 10−8 s = 8.2 × 106. 1.22 × 10−15 s

IDENTIFY and SET UP: EXECUTE: (a) En = −

En = −

13.6 eV n2

13.6 eV 13.6 eV and En +1 = − n2 ( n + 1) 2

⎡ 1 1⎤ n 2 − (n + 1) 2 ΔE = En +1 − En = (−13.6 eV) ⎢ − 2 ⎥ = −(13.6 eV) 2 2 n ⎦ (n )(n + 1) 2 ⎣ (n + 1) 2n 2 2n + 1 ΔE = (13.6 eV) 2 As n becomes large, ΔE → (13.6 eV) 4 = (13.6 eV) 3 n n (n )(n + 1) 2


38.29.

38-7

Thus ΔE becomes small as n becomes large. (b) rn = n 2r1 so the orbits get farther apart in space as n increases. IDENTIFY and SET UP: The number of photons emitted each second is the total energy emitted divided by the energy of one photon. The energy of one photon is given by Eq.(38.2). E = Pt gives the energy emitted by the laser in time t. EXECUTE: In 1.00 s the energy emitted by the laser is (7.50 × 10−3 W)(1.00 s) = 7.50 × 10 −3 J. The energy of each photon is E =

hc

λ

=

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 1.874 × 10−20 J. 10.6 × 10 −6 m

−3

7.50 × 10 J/s = 4.00 × 1017 photons/s 1.874 × 10−20 J/photon EVALUATE: The number of photons emitted per second is extremely large. IDENTIFY and SET UP: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each hc 1.99 × 10 −25 J ⋅ m photon is E = hf = . The power is the total energy per second and the total energy Etot is the =

Therefore 38.30.

λ

λ

number of photons N times the energy E of each photon. EXECUTE: (a) 193 nm is shorter than visible light so is in the ultraviolet. hc (b) E = = 1.03 × 10 −18 J = 6.44 eV

λ

38.31.

E NE Pt (1.50 × 10−3 W)(12.0 × 10 −9 s) (c) P = tot = so N = = = 1.75 × 107 photons t t E 1.03 × 10 −18 J EVALUATE: A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a large number of photons. n − ( E − E ) / kT IDENTIFY: Apply Eq.(38.21): 5 s = e 5 s 3 p n3 p SET UP: EXECUTE:

38.32.

From Fig.38.24a in the textbook, E5 s = 20.66 eV and E3 p = 18.70 eV

E5 s − E3 p = 20.66 eV − 18.70 eV = 1.96 eV(1.602 × 10−19 J/1 eV) = 3.140 × 10 −19 J

(a)

−19 −23 n5 s = e − (3.140×10 J)/[(1.38×10 J/K)(300 K)] = e −75.79 = 1.2 × 10−33 n3 p

(b)

−19 −23 n5 s = e− (3.140×10 J)/[(1.38×10 J/K)(600 K)] = e −37.90 = 3.5 × 10−17 n3 p

(c)

−19 −23 n5 s = e− (3.140×10 J)/[(1.38×10 J/K)(1200 K)] = e−18.95 = 5.9 × 10−9 n3 p

(d) EVALUATE: At each of these temperatures the number of atoms in the 5s excited state, the initial state for the transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases. n2 P3 2 ) KT −( E −E = e 2 P3 2 2 P1 2 . n2 P1/ 2

hc (6.626 × 10−34 J)(3.000 × 108 m s) = = 3.375 × 10−19 J. λ1 5.890 × 10−7 m hc (6.626 × 10−34 J)(3.000 × 108 m s) = = = 3.371 × 10−19 J. so ΔE3 / 2 −1/ 2 = 3.375 × 10−19 J − 3.371 × 10 −19 J = λ2 5.896 × 10−7 m

From the diagram ΔE3 / 2 − g = ΔE1 2 − g

38.33.

4.00 × 10−22 J.

n2 P3 / 2

eVAC = hf max =

hc λ min

n2 P1/ 2

= e − (4.00 × 10

−22

J) (1.38 × 10−23 J / K ⋅500 K).

⇒ λ min =

= 0.944. So more atoms are in the 2 p1 2 state.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s) = = 3.11× 10−10 m eVAC (1.60 × 10−19 C)(4000 V)

This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more easily produced and accelerated than proton beams.

38-8

Chapter 38

38.34.

IDENTIFY and SET UP: EXECUTE: (a) V =

hc

λ

= eV , where λ is the wavelength of the x ray and V is the accelerating voltage.

hc (6.63 × 10 −34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 kV eλ (1.60 × 10−19 C)(0.150 × 10 −9 m)

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 4.14 × 10−11 m = 0.0414 nm eV (1.60 × 10−19 C)(30.0 × 103 V) (c) No. A proton has the same magnitude of charge as an electron and therefore gains the same amount of kinetic energy when accelerated by the same magnitude of potential difference. IDENTIFY: The initial electrical potential energy of the accelerated electrons is converted to kinetic energy which is then given to a photon. SET UP: The electrical potential energy of an electron is eVAC, where VAC is the accelerating potential, and the energy of a photon is hf. Since the energy of the electron is all given to a photon, we have eVAC = hf. For any wave, fλ = v. EXECUTE: (a) eVAC = hfmin gives (b) λ =

38.35.

fmin = eVAC/h = (1.60 × 10–19 C)(25,000 V)/(6.626 × 10–34 J ⋅ s ) = 6.037 × 1018 Hz

38.36.

= 6.04 × 1018 Hz, rounded to three digits (b) λmin = c/fmax = (3.00 × 108 m/s)/(6.037 × 1018 Hz) = 4.97 × 10–11 m = 0.0497 nm (c) We assume that all the energy of the electron produces only one photon on impact with the screen. EVALUATE: These photons are in the x-ray and γ-ray part of the electromagnetic spectrum (see Figure 32.4 in the textbook) and would be harmful to the eyes without protective glass on the screen to absorb them. hc IDENTIFY and SET UP: The wavelength of the x rays produced by the tube is give by = eV .

λ

h h hc λ′ = λ + (1 − cosφ ) . = 2.426 × 10−12 m . The energy of the scattered x ray is . λ′ mc mc hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 6.91 × 10−11 m = 0.0691 nm EXECUTE: (a) λ = (1.60 × 10−19 C)(18.0 × 103 V) eV h (1 − cos φ ) = 6.91 × 10−11 m + (2.426 × 10−12 m)(1 − cos 45.0°) . mc λ ′ = 6.98 × 10 −11 m = 0.0698 nm . hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 17.8 keV (c) E = 6.98 × 10 −11 m λ′ EVALUATE: The incident x ray has energy 18.0 keV. In the scattering event, the photon loses energy and its wavelength increases. h IDENTIFY: Apply Eq.(38.23): λ ′ − λ = (1 − cosφ ) = λC (1 − cos φ ) mc SET UP: Solve for λ ′ : λ ′ = λ + λC (1 − cos φ ) The largest λ ′ corresponds to φ = 180°, so cosφ = −1. (b) λ ′ = λ +

38.37.

38.38.

EXECUTE: λ ′ = λ + 2λC = 0.0665 × 10 −9 m + 2(2.426 × 10 −12 m) = 7.135 × 10 −11 m = 0.0714 nm. This wavelength occurs at a scattering angle of φ = 180°. EVALUATE: The incident photon transfers some of its energy and momentum to the electron from which it scatters. Since the photon loses energy its wavelength increases, λ ′ > λ . Δλ (a) From Eq. (38.23), cosφ = 1 − , and so Δλ = 0.0542 nm − 0.0500 nm, ( h mc)

cos φ = 1 −

0.0042 nm = −0.731, and φ = 137°. 0.002426 nm

0.0021 nm = 0.134. φ = 82.3°. 0.002426 nm (c) Δλ = 0, the photon is undeflected, cosφ = 1 and φ = 0.

(b) Δλ = 0.0521 nm − 0.0500 nm. cosφ = 1 −

38.39.

IDENTIFY and SET UP: The shift in wavelength of the photon is λ ′ − λ =

wavelength after the scattering and

h (1 − cos φ ) where λ ′ is the mc

h = λc = 2.426 × 10−12 m . The energy of a photon of wavelength λ is mc


E=

38.40.

hc

λ

=

1.24 × 10

−6

eV ⋅ m

λ

38-9

. Conservation of energy applies to the collision, so the energy lost by the photon

equals the energy gained by the electron. EXECUTE: (a) λ ′ − λ = λc (1 − cos φ ) = (2.426 × 10−12 m)(1 − cos35.0°) = 4.39 × 10 −13 m = 4.39 × 10 −4 nm (b) λ ′ = λ + 4.39 × 10−4 nm = 0.04250 nm + 4.39 × 10−4 nm = 0.04294 nm hc hc (c) Eλ = = 2.918 × 104 eV and Eλ ′ = = 2.888 × 104 eV so the photon loses 300 eV of energy. λ λ′ (d) Energy conservation says the electron gains 300 eV of energy. The change in wavelength of the scattered photon is given by Eq. 38.23 Δλ = h (1 − cos φ ) ⇒ λ = h (1 − cosφ ). mcλ λ ⎛ Δλ ⎞ mc ⎜ ⎟ ⎝ λ ⎠ (6.63 × 10−34 J ⋅ s) (1 + 1) = 2.65 × 10 −14 m. (1.67 × 10−27 kg)(3.00 × 108 m/s)(0.100) The derivation of Eq.(38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of h p′ = h λ′ and p = h λ yielding λ′ − λ = (1 − cos φ ). mc 2.898 × 10−3 m ⋅ K c From Eq. (38.30), (a) λ m = = 0.966 mm, and f = = 3.10 × 1011 Hz. Note that a more precise 3.00 K λm value of the Wien displacement law constant has been used. (b) A factor of 100 increase in the temperature lowers λm by a factor of 100 to 9.66 μ m and raises the frequency

Thus, λ =

38.41.

38.42.

by the same factor, to 3.10 × 1013 Hz. (c) Similarly, λ m = 966 nm and f = 3.10 × 1014 Hz. 38.43.

(a) H = AeσT 4 ; A = π r 2l 14

⎛ ⎞ 100 W ⎛ H ⎞ T =⎜ ⎟ =⎜ 2 4 ⎟ −3 −8 ⎝ Aeσ ⎠ ⎝ 2π (0.20 × 10 m)(0.30 m)(0.26)(5.671 × 10 W m ⋅ K ) ⎠ 3 T = 2.06 × 10 K 14

38.44. 38.45.

(b) λ mT = 2.90 × 10−3 m ⋅ K; λ m = 1410 nm Much of the emitted radiation is in the infrared. 2.90 × 10−3 m ⋅ K 2.90 × 10−3 m ⋅ K T= = = 7.25 × 103 K. λm 400 × 10−9 m IDENTIFY and SET UP: The wavelength λm where the Planck distribution peaks is given by Eq.(38.30). 2.90 × 10−3 m ⋅ K = 1.06 × 10 −3 m = 1.06 mm. 2.728 K EVALUATE: This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation is often referred to as the “microwave background” (Section 44.7). Note that in Eq.(38.30), T must be in kelvins. IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law and Wien’s displacement law. SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power is P = σAT 4. Wien’s displacement law tells us that the peak-intensity wavelength is λm = (constant)/T. EXECUTE: (a) The hot and cool stars radiate the same total power, so the Stefan-Boltzmann law gives σAhTh4 =

EXECUTE:

38.46.

λm =

σAcTc4 ⇒ 4πRh2Th4 = 4πRc2Tc4 = 4π(3Rh)2Tc4 ⇒ Th4 = 9T 4 ⇒ Th = T 3 = 1.7T, rounded to two significant digits. (b) Using Wien’s law, we take the ratio of the wavelengths, giving λm (hot) Tc T 1 = = = = 0.58, rounded to two significant digits. λm (cool) Th T 3 3

38.47.

EVALUATE: Although the hot star has only 1/9 the surface area of the cool star, its absolute temperature has to be only 1.7 times as great to radiate the same amount of energy. (a) Let α = hc / kT . To find the maximum in the Planck distribution:

dI d ⎛ 2π hc 2 ⎞ (2π hc 2 ) 2π hc 2 ( −α λ 2 ) = ⎜ 5 αλ − 5 αλ ⎟ = 0 = −5 5 α λ dλ dλ ⎝ λ (e − 1) ⎠ λ (e − 1) λ (e − 1) 2 α hc ⇒ − 5(eα λ − 1) λ = α ⇒ − 5eα λ + 5 = α λ ⇒ Solve 5 − x = 5e x where x = = . λ λkT

38-10

Chapter 38

Its root is 4.965, so

α hc . = 4.965 ⇒ λ = λ (4.965) kT

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s) = = 2.90 × 10−3 m ⋅ K. (4.965)k (4.965)(1.38 × 10−23 J K) IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law. SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power is P = σAT 4. EXECUTE: (a) I = σT 4 = (5.67 × 10–8 W/m 2 ⋅ K 4 )(24,000 K)4 = 1.9 × 1010 W/m2 (b) Wien’s law gives λm = (0.00290 m ⋅ K )/(24,000 K) = 1.2 × 10–7 m = 120 nm This is not visible since the wavelength is less than 400 nm. (c) P = AI ⇒ 4πR2 = P/I = (1.00 × 1025 W)/(1.9 × 1010 W/m2) which gives RSirius = 6.51 × 106 m = 6510 km. RSirius/Rsun = (6.51 × 106 m)/(6.96 × 109 m) = 0.0093, which gives RSirius = 0.0093 Rsun ≈ 1% Rsun (d) Using the Stefan-Boltzmann law, we have (b) λ mT =

38.48.

2

38.49.

x V1 ⇒ I ( λ) ≈ 38.50.

4

2

4 2 4 ⎛ ⎞ ⎛ 5800 K ⎞ ⎛ R ⎞ ⎛ T ⎞ Psun σ AsunTsun 4π Rsun Tsun P Rsun = = = ⎜ sun ⎟ ⎜ sun ⎟ ⋅ sun = ⎜ ⎟ ⎜ ⎟ = 39 4 2 4 PSirius σ ASiriusTSirius 4π RSiriusTSirius R T P 0.00935 R Sirius sun ⎠ ⎝ 24,000 K ⎠ ⎝ Sirius ⎠ ⎝ Sirius ⎠ ⎝ EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates only 1/39 times as much energy per second as our sun because it is so small. 2π hc 2 x2 but e x = 1 + x + + " ≈ 1 + x for Eq. (38.32): I ( λ) = 5 hc λkT − 1) λ (e 2 4

2π hc 2 2π ckT = = Eq. (38.31), which is Rayleigh’s distribution. λ (hc λkT ) λ4 5

2.90 × 10−3 K ⋅ m = 9.7 × 10−8 m = 97 nm (a) Wien’s law: λ m = k . λ m = T 30,000 K This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible light radiated is in the blue violet part of the visible spectrum (b) P = σAT 4 (Stefan-Boltzmann law)

W ⎞ ⎛ (100, 000)(3.86 × 1026 W) = ⎜ 5.67 × 10−8 2 4 ⎟ (4π R 2 )(30,000 K) 4 mK ⎠ ⎝ 9 R = 8.2 × 10 m

Rstar Rsun =

38.51.

8.2 × 109 m = 12 6.96 × 108 m

(c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity. IDENTIFY and SET UP: Use c = f λ to relate frequency and wavelength and use E = hf to relate photon energy and frequency. EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a single molecule. The problem states that it requires 1.00 × 105 J to dissociate one mole of AgBr, and one mole contains Avogadro’s number (6.02 × 10 23 ) of molecules, so the energy required to dissociate one AgBr is

1.00 × 105 J/mol = 1.66 × 10−19 J/molecule. 6.02 × 1023 molecules/mol The photon is to have this energy, so E = 1.66 × 10−19 J(1eV/1.602 × 10−19 J) = 1.04 eV. hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.20 × 10 −6 m = 1200 nm E λ 1.66 × 10 −19 J c 2.998 × 108 m/s = 2.50 × 1014 Hz (c) c = f λ so f = = λ 1.20 × 10−6 m (d) E = hf = (6.626 × 10−34 J ⋅ s)(100 × 106 Hz) = 6.63 × 10−26 J (b) E =

hc

so λ =

E = 6.63 × 10−26 J(1 eV/1.602 × 10−19 J) = 4.14 × 10 −7 eV


38.52.

38-11

(e) EVALUATE: A photon with frequency f = 100 MHz has too little energy, by a large factor, to dissociate a AgBr molecule. The photons in the visible light from a firefly do individually have enough energy to dissociate AgBr. The huge number of 100 MHz photons can’t compensate for the fact that individually they have too little energy. h h (a) Assume a non-relativistic velocity and conserve momentum ⇒ mv = ⇒ v = . mλ λ 2 1 1 ⎛ h ⎞ h2 . = (b) K = mv 2 = m ⎜ ⎟ 2 2 ⎝ mλ ⎠ 2mλ 2 K h2 λ h = ⋅ = . Recoil becomes an important concern for small m and small λ since this ratio (c) 2 E 2mλ hc 2mcλ becomes large in those limits. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s ) = = 1.22 × 10−7 m = 122 nm. (d) E = 10.2 eV ⇒ λ = E (10.2 eV)(1.60 × 10 −19 J eV)

K=

(6.63 × 10 −34 J ⋅ s) 2 = 8.84 × 10−27 J = 5.53 × 10−8 eV. 2(1.67 × 10−27 kg)(1.22 × 10−7 m) 2

K 5.53 × 10−8 eV = = 5.42 × 10 −9. This is quite small so recoil can be neglected. E 10.2 eV 38.53.


f =

c

λ

. The ( f ,V0 ) values are: (8.20 × 1014 Hz, 1.48 V) , (7.41 × 1014 Hz, 1.15 V) ,

(6.88 × 1014 Hz, 0.93 V) , (6.10 × 1014 Hz, 0.62 V) , (5.49 × 1014 Hz, 0.36 V) , (5.18 × 1014 Hz, 0.24 V) . The graph

of V0 versus f is given in Figure 38.53. EXECUTE: (a) The threshold frequency, f th , is f where V0 = 0 . From the graph this is f th = 4.56 × 1014 Hz . c 3.00 × 108 m/s = = 658 nm f th 4.56 × 1014 Hz (c) φ = hf th = (4.136 × 10 −15 eV ⋅ s)(4.56 × 1014 Hz) = 1.89 eV (b) λth =

h ⎛h⎞ (d) eV0 = hf − φ so V0 = ⎜ ⎟ f − φ . The slope of the graph is . e ⎝e⎠ h ⎛ 1.48 V − 0.24 V ⎞ −15 =⎜ ⎟ = 4.11 × 10 V/Hz and e ⎝ 8.20 × 1014 Hz − 5.18 × 1014 Hz ⎠ h = (4.11 × 10 −15 V/Hz)(1.60 × 10 −19 C) = 6.58 × 10 −34 J ⋅ s .

Figure 38.53 38.54.

dN ( dE dt ) P (200 W)(0.10) = = = = 6.03 × 1019 photons sec. dt ( dE dN ) hf h(5.00 × 1014 Hz) (dN dt ) (b) Demand = 1.00 × 1011 photons sec ⋅ cm 2 . 4π r 2 (a)

1/ 2

38.55.

⎛ ⎞ 6.03 × 1019 photons sec Therefore, r = ⎜ ⎟ = 6930 cm = 69.3 m. 11 2 ⎝ 4π (1.00 × 10 photons sec ⋅ cm ) ⎠ (a) IDENTIFY: Apply the photoelectric effect equation, Eq.(38.4). SET UP: eV0 = hf − φ = ( hc / λ ) − φ . Call the stopping potential V01 for λ1 and V02 for λ2 . Thus eV01 = (hc / λ1 ) − φ and eV02 = (hc / λ2 ) − φ . Note that the work function φ is a property of the material and is independent of the wavelength of the light. ⎛λ −λ ⎞ EXECUTE: Subtracting one equation from the other gives e(V02 − V01 ) = hc ⎜ 1 2 ⎟ . ⎝ λ 1λ 2 ⎠

38-12

Chapter 38

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) ⎛ 295 × 10 −9 m − 265 × 10−9 m ⎞ ⎜ ⎟ = 0.476 V. −9 −9 1.602 × 10−19 C ⎝ (295 × 10 m)(265 × 10 m) ⎠ EVALUATE: eΔV0 , which is 0.476 eV, is the increase in photon energy from 295 nm to 265 nm. The stopping potential increases when λ deceases because the photon energy increases when the wavelength decreases. IDENTIFY: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any excess energy going into kinetic energy of the electron. SET UP: Conservation of energy gives hf = hc/λ = Kmax + φ. EXECUTE: (a) Using hc/λ = Kmax + φ, we solve for the work function: (b) ΔV0 =

38.56.

φ = hc/λ – Kmax = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(124 nm) – 4.16 eV = 5.85 eV

38.57.

(b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal per second. N × (energy of a photon) = 2.50 W. N(hc/λ) = 2.50 W. N = (2.50 W)(124 nm)/[(6.626 × 10–34 J ⋅ s )(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s (c) N is proportional to the power, so if the power is cut in half, so is N, which gives N = (1.56 × 1018 el/s)/2 = 7.80 × 1017 el/s (d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To maintain the same power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s. We could also see this from part (b), where N is proportional to λ. So if the wavelength is cut in half, so is N. EVALUATE: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only reduces the number of ejected electrons. In part (d), reducing the wavelength does change the maximum kinetic energy of the photoelectrons because we have increased the energy of each photon. IDENTIFY and SET UP: The energy added to mass m of the blood to heat it to Tf = 100 °C and to vaporize it is

Q = mc (Tf − Ti ) + mLv , with c = 4190 J/kg ⋅ K and Lv = 2.256 × 106 J/kg . The energy of one photon is

E=

38.58.

38.59.

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

.

EXECUTE: (a) Q = (2.0 × 10−9 kg)(4190 J/kg ⋅ K)(100°C − 33°C) + (2.0 × 10 −9 kg)(2.256 × 106 J/kg) = 5.07 × 10−3 J The pulse must deliver 5.07 mJ of energy. energy 5.07 × 10−3 J (b) P = = = 11.3 W t 450 × 10−6 s hc 1.99 × 10−25 J ⋅ m (c) One photon has energy E = = = 3.40 × 10−19 J . The number N of photons per pulse is the 585 × 10−9 m λ 5.07 × 10−3 J energy per pulse divided by the energy of one photon: N = = 1.49 × 1016 photons 3.40 × 10−19 J/photon hc (a) λ 0 = , and the wavelengths are: cesium: 590 nm, copper: 264 nm, potassium: 539 nm, zinc: 288 nm. E b) The wavelengths of copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome the threshold energy of these metals. 207 me mp mm (a) IDENTIFY and SET UP: Apply Eq.(38.20): mr = 1 2 = m1 + m2 207 me + mp EXECUTE:

mr =

207(9.109 × 10−31 kg)(1.673 × 10−27 kg) = 1.69 × 10 −28 kg 207(9.109 × 10−31 kg) + 1.673 × 10−27 kg

We have used me to denote the electron mass. (b) IDENTIFY: In Eq.(38.18) replace m = me by mr : En = −

1 mr e 4 . P02 8n 2 h 2

⎛ m ⎞ ⎛ 1 m e4 ⎞ 1 m e4 SET UP: Write as En = ⎜ r ⎟ ⎜ − 2 H2 2 ⎟ , since we know that 2 H 2 = 13.60 eV. Here mH denotes the P0 8h ⎝ mH ⎠⎝ P0 8n h ⎠ −31 reduced mass for the hydrogen atom; mH = 0.99946(9.109 × 10 kg) = 9.104 × 10−31 kg. ⎛ m ⎞ ⎛ 13.60 eV ⎞ En = ⎜ r ⎟ ⎜ − ⎟ n2 ⎠ ⎝ mH ⎠ ⎝ 1.69 × 10−28 kg E1 = ( −13.60 eV) = 186( −13.60 eV) = −2.53 keV 9.104 × 10−31 kg

EXECUTE:


38-13

⎛ m ⎞ ⎛ R ch ⎞ From part (b), En = ⎜ r ⎟ ⎜ − H 2 ⎟ , where RH = 1.097 × 107 m −1 is the Rydberg constant for the ⎝ mH ⎠ ⎝ n ⎠ hc hydrogen atom. Use this result in = Ei − E f to find an expression for 1/ λ . The initial level for the transition is (c) SET UP:

λ

the ni = 2 level and the final level is the n f = 1 level. EXECUTE:

hc

λ

=

mr ⎛ RHch ⎛ RHch ⎞ ⎞ − ⎜ − 2 ⎟⎟ ⎜− ⎟ mH ⎜⎝ ni2 ⎝ nf ⎠ ⎠

⎛ 1 1⎞ mr RH ⎜ 2 − 2 ⎟ λ mH ⎝ nf ni ⎠ 1 1.69 × 10−28 kg ⎛1 1⎞ (1.097 × 107 m −1 ) ⎜ 2 − 2 ⎟ = 1.527 × 109 m −1 = λ 9.104 × 10−31 kg ⎝1 2 ⎠ λ = 0.655 nm EVALUATE: From Example 38.6 the wavelength of the radiation emitted in this transition in hydrogen is 122 nm. m The wavelength for muonium is H = 5.39 × 10 −3 times this. The reduced mass for hydrogen is very close to the mr 1

=

electron mass because the electron mass is much less then the proton mass: mp / me = 1836. The muon mass is

38.60.

207 me = 1.886 × 10−28 kg. The proton is only about 10 times more massive than the muon, so the reduced mass is somewhat smaller than the muon mass. The muon-proton atom has much more strongly bound energy levels and much shorter wavelengths in its spectrum than for hydrogen. (a) The change in wavelength of the scattered photon is given by Eq. 38.23

h h (1 − cos φ ) ⇒ λ = λ′ − (1 − cosφ ) = mc mc (6.63 × 10−34 J ⋅ s) (0.0830 × 10−9 m) − (1 + 1) = 0.0781 nm. (9.11 × 10−31 kg)(3.00 × 108 m s) λ′ − λ =

(b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the magnitude of the change in the photon’s momentum. Thus,

1 ⎞ 9 −1 ⎛ 1 −1 ⎞ ⎛ 1 pe = h ⎜ − ⎟ = (6.63 × 10−34 J ⋅ s) ⎜ + ⎟ (10 m ) ′ λ λ 0.0781 0.0830 ⎝ ⎠ ⎝ ⎠ = 1.65 × 10−23 kg ⋅ m s ≈ 2 × 10−23 kg ⋅ m s. (c) Since the electron is non relativistic ( β = 0.06), K e = 38.61.


λ′ = λ +

pe2 = 1.49 × 10 −16 J ≈ 10−16 J. 2m

h (1 − cosφ ) mc

2h = 0.09485 m. Use Eq.(38.5) to calculate the momentum of the scattered photon. Apply mc conservation of energy to the collision to calculate the kinetic energy of the electron after the scattering. The energy of the photon is given by Eq.(38.2), EXECUTE: (a) p′ = h / λ ′ = 6.99 × 10 −24 kg ⋅ m/s.

φ = 180° so λ ′ = λ +

(b) E = E′ + Ee ; hc / λ = hc / λ ′ + Ee

38.62.

λ′ − λ ⎛1 1 ⎞ Ee = hc ⎜ − ⎟ = (hc) = 1.129 × 10−16 J = 705 eV λλ′ ⎝ λ λ′ ⎠ EVALUATE: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the electron. This corresponds to a fractional shift in the photon’s wavelength that is also 5%. 2h (a) φ = 180° so (1 − cosφ ) = 2 ⇒ Δλ = = 0.0049 nm, so λ′ = 0.1849 nm. mc ⎛1 1 ⎞ (b) ΔE = hc ⎜ − ⎟ = 2.93 × 10−17 J = 183 eV. This will be the kinetic energy of the electron. ⎝ λ λ′ ⎠ (c) The kinetic energy is far less than the rest mass energy, so a non-relativistic calculation is adequate; v = 2 K m = 8.02 × 106 m s.

38-14

Chapter 38

38.63.

IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the n = 3 to n = 2 transition.

En = −

13.6 eV hc . = ΔE . λ n2

⎛1 1⎞ EXECUTE: (a) The atom must be given an amount of energy E3 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = 12.1 eV . ⎝3 1 ⎠ hc (b) There are three possible transitions. n = 3 → n = 1 : ΔE = 12.1 eV and λ = = 103 nm ; ΔE ⎛1 1⎞ n = 3 → n = 2 : ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 1.89 eV and λ = 657 nm ; n = 2 → n = 1 : ⎝3 2 ⎠

38.64.

⎛ 1 1⎞ ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 10.2 eV and λ = 122 nm . ⎝2 1 ⎠ −( Eex − Eg ) n2 − ( E − E ) kT ⇒T = . = e ex g n1 kln ( n2 / n1 )

Eex = E2 =

−13.6 eV = −3.4 eV. Eg = −13.6 eV. Eex − Eg = 10.2 eV = 1.63 × 10 −18 J. 4

(a)

−(1.63 × 10−18 J) n2 = 10 −12. T = = 4275 K. (1.38 × 10 −23 J K ) ln(10 −12 ) n1

(b)

−(1.63 × 10−18 J) n2 = 10 −8. T = = 6412 K. (1.38 × 10 −23 J K ) ln(10−8 ) n1

−(1.63 × 10−18 J) n2 = 10 −4. T = = 12824 K. (1.38 × 10 −23 J K ) ln(10 −4 ) n1 (d) For absorption to take place in the Balmer series, hydrogen must start in the n = 2 state. From part (a), colder stars have fewer atoms in this state leading to weaker absorption lines. (a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom. Some of this energy overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron. Apply hf = Ef − Ei , the energy given to the electron in the atom when a photon is absorbed. (c)

38.65.

EXECUTE: The energy of one photon is

hc

λ

= 2.323 × 10

−18

J(1 eV/1.602 × 10

−19

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 85.5 × 10−9 m

J) = 14.50 eV.

The final energy of the electron is Ef = Ei + hf . In the ground state of the hydrogen atom the energy of the electron

38.66.

is Ei = −13.60 eV. Thus Ef = −13.60 eV + 14.50 eV = 0.90 eV. (b) EVALUATE: At thermal equilibrium a few atoms will be in the n = 2 excited levels, which have an energy of −13.6 eV/4 = −3.40 eV, 10.2 eV greater than the energy of the ground state. If an electron with E = −3.40 eV gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV − 3.4 eV = 11.1 eV of kinetic energy. IDENTIFY: The diffraction grating allows us to determine the peak-intensity wavelength of the light. Then Wien’s displacement law allows us to calculate the temperature of the blackbody, and the Stefan-Boltzmann law allows us to calculate the rate at which it radiates energy. SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ. Wien’s displacement law is 2.90 × 10 −3 m ⋅ K λpeak = , and the Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the T total radiated power is P = σAT 4. EXECUTE: (a) First find the wavelength of the light: λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10–7 m Now use Wien’s law to find the temperature: T = (2.90 × 10–3 m ⋅ K )/(5.22 × 10–7 m) = 5550 K. (b) The energy radiated by the blackbody is equal to the power times the time, giving U = Pt = IAt = σAT 4t, which gives t = U/(σAT 4) = (12.0 × 106 J)/[(5.67 × 10–8 W/m 2 ⋅ K 4 )(4π)(0.0750 m)2(5550 K)4] = 3.16 s. EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of energy.


38.67.

38-15

IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the StefanBoltzmann law apply to its radiation. 2.90 × 10 −3 m ⋅ K , and the Stefan-Boltzmann law says that the SET UP: Wien’s displacement law is λpeak = T 4 intensity of the radiation is I = σT , so the total radiated power is P = σAT 4. EXECUTE: (a) First use Wien’s law to find the peak wavelength: λm = (2.90 × 10–3 m ⋅ K )/(3000 K) = 9.667 × 10–7 m Call N the number osf photons/second radiated. N × (energy per photon) = IA = σAT 4. λ σ AT 4 N (hc/λm) = σAT 4. N = m . hc (9.667 × 10−7 m)(5.67 × 10−8 W/m 2 ⋅ K 4 )(4π )(600 × 6.96 × 108 m)2 (3000 K) 4 . N= (6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s) N = 5 × 1049 photons/s. 2

I B AB σ ABTB4 4π RB2TB4 ⎛ 600 RS ⎞ ⎛ 3000 K ⎞ 4 = = =⎜ ⎟ ⎜ ⎟ = 3 × 10 I S AS σ ASTS4 4π RS2TS4 ⎝ RS ⎠ ⎝ 5800 K ⎠ EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun! IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water. dQ 4 4 SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is = σ A (Tsphere − Twater ) . Since dt dQ dm = Lv . Wien’s displacement law is this heat evaporates water, the rate at which water evaporates is dt dt 2.90 × 10−3 m ⋅ K λm = , and the wavelength in the water is λw = λ0/n. T dQ dQ dm 4 4 EXECUTE: (a) The net radiated heat is = σ A (Tsphere − Twater and the evaporation rate is = Lv , where ) dt dt dt dm 4 4 = σ A (Tsphere − Twater dm is the mass of water that evaporates in time dt. Equating these two rates gives Lv ). dt 4

(b)

38.68.

dm σ ( 4π R = dt

2

)(T

4 sphere

Lv

4 − Twater )

.

−8 2 4 2 4 4 dm ( 5.67 × 10 W/m ⋅ K ) ( 4π ) (0.120 m) ⎡⎣ (498 K) − (373 K) ⎤⎦ = = 1.92 × 10 −4 kg/s = 0.193 g/s dt 2256 × 103 J/Kg (b) (i) Wien’s law gives λm = (0.00290 m ⋅ K )/(498 K) = 5.82 × 10–6 m But this would be the wavelength in vacuum. In the water the thermophile organism would measure λw = λ0/n = (5.82 × 10–6 m)/1.333 = 4.37 × 10–6 m = 4.37 µm (ii) The frequency is the same as if the wave were in air, so f = c/λ0 = (3.00 ×108 m/s)/(5.82 × 10–6 m) = 5.15 × 1013 Hz

EVALUATE: An alternative way is to use the quantities in the water: f =

38.69.

c/n

λ0 / n

= c/λ0, which gives the same

answer for the frequency. An organism in the water would measure the light coming to it through the water, so the wavelength it would measure would be reduced by a factor of 1/n. IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the n = 1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength. 13.6 eV SET UP: In the Bohr model, the energy of an electron in shell n is En = − , and Wien’s displacement law n2 2.90 × 10−3 m ⋅ K is λm = . The energy of a photon is E = hf = hc/λ. T EXECUTE: First find the energy (ΔE) that a photon would need to excite the atom. The ground state of the atom is n = 1 and the third excited state is n = 4. This energy is the difference between the two energy levels. Therefore

38-16

Chapter 38

⎛ 1 1⎞ ΔE = ( −13.6 eV ) ⎜ 2 − 2 ⎟ = 12.8 eV. Now find the wavelength of the photon having this amount of energy. ⎝4 1 ⎠ hc/λ = 12.8 eV and

λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(12.8 eV) = 9.73 ×10–8 m

38.70.

38.71.

Now use Wien’s law to find the temperature. T = (0.00290 m ⋅ K )/(9.73 × 10–8 m) = 2.98 × 104 K. EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in excited states during everyday conditions. IDENTIFY and SET UP: Electrical power is VI. Q = mcΔT . EXECUTE: (a) (0.010)VI = (0.010)(18.0 × 103 V)(60.0 × 10 −3 A) = 10.8 W = 10.8 J/s (b) The energy in the electron beam that isn’t converted to x rays stays in the target and appears as thermal energy. Q 1.07 × 103 J For t = 1.00 s , Q = (0.990)VI (1.00 s) = 1.07 × 103 J and ΔT = = = 32.9 K . The mc (0.250 kg)(130 J/kg ⋅ K) temperature rises at a rate of 32.9 K/s. EVALUATE: The target must be made of a material that has a high melting point. IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom plus photon. (a) SET UP: Let Etr be the transition energy, Eph be the energy of the photon with wavelength λ ′, and Er be

the kinetic energy of the recoiling atom. Conservation of energy gives Eph + Er = Etr . Eph =

hc hc hc so = Etr − Er and λ ′ = . λ′ λ′ Etr − Er

EXECUTE: If the recoil energy is neglected then the photon wavelength is λ = hc / Etr .

⎛ 1 ⎞ 1 ⎞ ⎛ hc ⎞⎛ 1 Δλ = λ ′ − λ = hc ⎜ − − 1⎟ ⎟=⎜ ⎟⎜ − − 1 / E E E E E E r tr ⎠ r tr ⎝ tr ⎝ tr ⎠⎝ ⎠ −1

⎛ 1 E ⎞ E E = ⎜ 1 − r ⎟ ≈ 1 + r since r V 1 1 − Er / Etr ⎝ E tr ⎠ Etr Etr (We have used the binomial theorem, Appendix B.) hc ⎛ Er ⎞ ⎛ Er ⎞ 2 Thus Δλ = ⎜ ⎟ , or since Etr = hc / λ , Δλ = ⎜ ⎟ λ . Etr ⎝ Etr ⎠ ⎝ hc ⎠ SET UP: Use conservation of linear momentum to find Er : Assuming that the atom is initially at rest, the

momentum pr of the recoiling atom must be equal in magnitude and opposite in direction to the momentum pph = h / λ of the emitted photon: h / λ = pr .

pr2 h2 , where m is the mass of the atom, so Er = . 2m 2mλ 2 ⎛ h 2 ⎞⎛ λ 2 ⎞ h ⎛E ⎞ Use this result in the above equation: Δλ = ⎜ r ⎟ λ 2 = ⎜ ; ⎟= 2 ⎟⎜ 2 m hc 2 mc λ ⎝ hc ⎠ ⎝ ⎠⎝ ⎠ EXECUTE:

Er =

note that this result for Δλ is independent of the atomic transition energy. 6.626 × 10−34 J ⋅ s h (b) For a hydrogen atom m = mp and Δλ = = = 6.61 × 10−16 m 2mpc 2(1.673 × 10−27 kg)(2.998 × 108 m/s)

38.72.

EVALUATE: The correction is independent of n. The wavelengths of photons emitted in hydrogen atom transitions are on the order of 100 nm = 10−7 m, so the recoil correction is exceedingly small. (a) Δλ1 = (h mc )(1 − cos θ1 ), Δλ 2 = (h mc)(1 − cos θ2 ), and so the overall wavelength shift is

Δλ = (h mc )(2 − cos θ1 − cos θ2 ). (b) For a single scattering through angle θ , Δλ s = ( h mc)(1 − cos θ ). For two successive scatterings through an angle of θ 2 for each scattering,

Δλ t = 2(h mc )(1 − cosθ 2). 1 − cos θ = 2(1 − cos 2 (θ 2)) and Δλ s = ( h mc)2(1 − cos 2 (θ 2)) cos(θ 2) ≤ 1 so 1 − cos 2 (θ 2) ≥ (1 − cos(θ 2)) and Δλ s ≥ Δλ t


38-17

Equality holds only when θ = 180°. (c) ( h mc)2(1 − cos30.0°) = 0.268( h mc). 38.73.

(d) ( h mc)(1 − cos 60°) = 0.500( h mc), which is indeed greater than the shift found in part (c). IDENTIFY and SET UP: Find the average change in wavelength for one scattering and use that in Δλ in Eq.(38.23) to calculate the average scattering angle φ . EXECUTE: (a) The wavelength of a 1 MeV photon is hc (4.136 × 10 −15 eV ⋅ s)(2.998 × 108 m/s) = 1 × 10−12 m λ= = E 1 × 106 eV The total change in wavelength therefore is 500 × 10 −9 m − 1 × 10 −12 m = 500 × 10−9 m. If this shift is produced in 10 26 Compton scattering events, the wavelength shift in each scattering event is 500 × 10−9 m Δλ = = 5 × 10−33 m. 1 × 1026 h (b) Use this Δλ in Δλ = (1 − cos φ ) and solve for φ . We anticipate that φ will be very small, since Δλ is mc much less than h / mc, so we can use cosφ ≈ 1 − φ 2 / 2. h h 2 Δλ = (1 − (1 − φ 2 / 2)) = φ mc 2mc

φ=

2Δλ 2(5 × 10−33 m) = = 6.4 × 10 −11 rad = (4 × 10−9 )° ( h / mc ) 2.426 × 10−12 m

φ in radians is much less than 1 so the approximation we used is valid. (c) IDENTIFY and SET UP: We know the total transit time and the total number of scatterings, so we can calculate the average time between scatterings. EXECUTE: The total time to travel from the core to the surface is (106 y)(3.156 × 107 s/y) = 3.2 × 1013 s. There are 3.2 × 1013 s = 3.2 × 10−13 s. 1026 The distance light travels in this time is d = ct = (3.0 × 108 m/s)(3.2 × 10 −13 s) = 0.1 mm EVALUATE: The photons are on the average scattered through a very small angle in each scattering event. The average distance a photon travels between scatterings is very small. hc (a) The final energy of the photon is E ′ = ,and E = E ′ + K , where K is the kinetic energy of the electron after λ′ the collision. Then, hc hc hc λ′ = = = . λ= ⎤ E ′ + K ( hc λ′) + K ( hc λ′) + (γ − 1) mc 2 λ′mc ⎡ 1 −1 1+ h ⎢⎣ (1 − v 2 c 2 )1 2 ⎥⎦ 10 26 scatterings during this time, so the average time between scatterings is t =

38.74.

( K = mc 2 (γ − 1) since the relativistic expression must be used for three-figure accuracy).

(b) φ = arccos(1 − Δ λ ( h mc)). (c) γ − 1 =

(1 − (

1

)

12 1.80 2 3.00

)

− 1 = 1.25 − 1 = 0.250,

⇒λ= 1+

(5.10 × 10

−12

5.10 × 10 −3 mm = 3.34 × 10−3 nm . m)(9.11 × 10 −31 kg)(3.00 × 108 m s)(0.250) (6.63 × 10−34 J ⋅ s)

⎛

φ = arccos ⎜1 − ⎝

38.75.

h = 2.43 × 10 −12 m mc

(5.10 × 10−12 m − 3.34 × 10−12 m) ⎞ ⎟ = 74.0°. 2.43 × 10−12 m ⎠

(a) IDENTIFY and SET UP: Conservation of energy applied to the collision gives Eλ = Eλ ′ + Ee , where Ee is the

kinetic energy of the electron after the collision and Eλ and Eλ ′ are the energies of the photon before and after the collision. The energy of a photon is related to its wavelength according to Eq.(38.2).

38-18

Chapter 38

⎛1 1 ⎞ ⎛ λ′ − λ ⎞ Ee = hc ⎜ − ⎟ = hc ⎜ ⎟ ⎝ λ λ′ ⎠ ⎝ λλ ′ ⎠

EXECUTE:

⎛ ⎞ 0.0032 × 10−9 m Ee = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) ⎜ ⎟ −9 −9 ⎝ (0.1100 × 10 m)(0.1132 × 10 m) ⎠ Ee = 5.105 × 10−17 J = 319 eV

1 2 Ee 2(5.105 × 10 −17 J) Ee = mv 2 so v = = = 1.06 × 107 m/s 2 m 9.109 × 10 −31 kg (b) The wavelength λ of a photon with energy Ee is given by Ee = hc / λ so

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 3.89 nm Ee 5.105 × 10−17 J EVALUATE: Only a small portion of the incident photon’s energy is transferred to the struck electron; this is why the wavelength calculated in part (b) is much larger than the wavelength of the incident photon in the Compton scattering. h IDENTIFY: Apply the Compton scattering formula λ ′ − λ = Δλ = (1 − cos φ ) = λc (1 − cos φ ) mc (a) SET UP: Largest Δλ is for φ = 180°. EXECUTE: For φ = 180°, Δλ = 2λc = 2(2.426 pm) = 4.85 pm.

λ=

38.76.

λ ′ − λ = λc (1 − cos φ ) Wavelength doubles implies λ ′ = 2λ so λ ′ − λ = λ . Thus λ = λC (1 − cosφ ). λ is related to E by Eq.(38.2). EXECUTE: E = hc / λ , so smallest energy photon means largest wavelength photon, so φ = 180° and (b) SET UP:

λ = 2λc = 4.85 pm. Then E =

38.77.

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 4.096 × 10−14 J(1 eV/1.602 × 10−19 J) = 4.85 × 10−12 m

0.256 MeV. EVALUATE: Any photon Compton scattered at φ = 180° has a wavelength increase of 2λc = 4.85 pm. 4.85 pm is near the short-wavelength end of the range of x-ray wavelengths. 2π hc 2 c (a) I (λ ) = 5 hc λkT but λ = f − 1) λ (e ⇒ I( f ) = (b)

=∫

∫ ∞ 0

∞ 0

2π hc 2 2π hf 5 = 3 hf kT 5 hf kT − 1) c (e − 1) (c f ) (e

0 ⎛ −c ⎞ I (λ ) d λ = ∫ I ( f ) df ⎜ 2 ⎟ ∞ ⎝ f ⎠

2π hf 3 df 2π (kT ) 4 ∞ x3 2π ( kT ) 4 1 (2π )5 (kT ) 4 2π 5k 4T 4 dx = (2π ) 4 = = = 2 3 2 3 hf kT x ∫ 0 c (e ch e −1 ch 240 240h3c 2 15c 2h3 − 1) 2

2π 5k 4T 4 = σ as shown in Eq. (38.36). Plugging in the values for the constants we get 15h3c 2 σ = 5.67 × 10−8 W m 2 ⋅ K 4 .

(c) The expression

38.78.

I = σT 4 , P = IA, and ΔE = Pt ; combining,

t= 38.79.

ΔE (100 J) = = 8.81 × 103 s = 2.45 hrs. Aσ T 4 (4.00 × 10−6 m 2 )(5.67 × 10−8 W m 2 ⋅ K 4 )(473 K) 4

(a) The period was found in Exercise 38.27b: T =

4P02n 3h3 1 me 4 and frequency is just f = = 2 3 3 . 4 me T 4P0 n h

1 me4 ⎛ 1 1 ⎞ (b) Eq. (38.6) tells us that f = ( E2 − E1 ). So f = 2 3 ⎜ 2 − 2 ⎟ (from Eq. (38.18)). h 8P0 h ⎝ n2 n1 ⎠ If n2 = n and n1 = n + 1, then =

1 1 1 1 − = − n22 n12 n 2 ( n + 1) 2

⎞ 1⎛ ⎛ 2 1⎛ 1 me 4 ⎞⎞ 2 1− ≈ 2 ⎜ 1 − ⎜1 − + " ⎟ ⎟ = 3 for large n ⇒ f ≈ 2 3 3 . 2 ⎜ 2 ⎟ n ⎝ (1 + 1 n) ⎠ n ⎝ ⎝ n 4P0 n h ⎠⎠ n


38.80.

38.81.

38-19

h Each photon has momentum p = , and if the rate at which the photons strike the surface is ( dN dt ) , the force λ on the surface is ( h λ )( dN dt ), and the pressure is ( h λ )( dN dt ) A. The intensity is I = (dN dt )( E ) A = ( dN dt )(hc λ ) A , and comparison of the two expressions gives the pressure as ( I c). G G G G Momentum: p + P = p′ + P ′ ⇒ p − P = − p′ − P′ ⇒ p′ = P − ( p + P′) energy: pc + E = p′c + E′ = p′c + ( P′c ) 2 + (mc 2 ) 2 ⇒ ( pc − p′c + E ) 2 = ( P′c) 2 + ( mc 2 ) 2 = (Pc) 2 + ((p + p′)c) 2 − 2 P (p + p′)c 2 + (mc 2 ) 2 ( pc − p′c) 2 + E 2 = E 2 + ( pc + p′c) 2 − 2( Pc 2 )( p + p′) + 2 Ec( p − p′) − 4 pp′c 2 + 2 Ec( p − p′) +2( Pc 2 )( p + p′) = 0

⇒ p′( Pc 2 − 2 pc 2 − Ec) = p ( − Ec − Pc 2 ) ⇒ p′ = p

Ec + Pc 2 E + Pc =p 2 pc 2 + Ec − Pc 2 2 pc + ( E − Pc )

2hc ⎛ 2 hc λ + ( E − Pc) ⎞ ⎛ E − Pc ⎞ ⇒ λ′ = λ ⎜ ⎟ = λ⎜ ⎟+ E + Pc ⎝ ⎠ ⎝ E + Pc ⎠ E + Pc (λ ( E − Pc) + 2hc) ⇒ λ′ = E + Pc ⎛ mc 2 ⎞ If E W mc 2 , Pc = E 2 − ( mc 2 ) 2 = E 1 − ⎜ ⎟ ⎝ E ⎠ ⇒ E − Pc ≈

2

⎛ 1 ⎛ mc 2 ⎞ 2 ⎞ ≈ E ⎜1 − ⎜ + "⎟ ⎟ ⎜ 2⎝ E ⎠ ⎟ ⎝ ⎠

1 ( mc 2 ) 2 λ( mc 2 ) 2 hc hc ⎛ m 2c 4 λ ⎞ ⇒ λ1 ≈ + = ⎜1 + ⎟ 2 E 2 E (2 E ) E E⎝ 4hcE ⎠

(b) If λ = 10.6 × 10 −6 m, E = 1.00 × 1010 eV = 1.60 × 10−9 J

⇒ λ′ ≈

hc 1.60 × 10−9

⎛ (9.11 × 10−31 kg) 2c 4 (10.6 × 10−6 m) ⎞ ⎜1 + ⎟ J⎝ 4hc (1.6 × 10−9 J) ⎠

= (1.24 × 10−16 m)(1 + 56.0) = 7.08 × 10−15 m.

(c) These photons are gamma rays. We have taken infrared radiation and converted it into gamma rays! Perhaps useful in nuclear medicine, nuclear spectroscopy, or high energy physics: wherever controlled gamma ray sources might be useful.

39

THE WAVE NATURE OF PARTICLES

39.1.

IDENTIFY and SET UP: EXECUTE: (a) λ =

h h = . For an electron, m = 9.11 × 10 −31 kg . For a proton, m = 1.67 × 10 −27 kg . p mv

6.63 × 10−34 J ⋅ s = 1.55 × 10−10 m = 0.155 nm (9.11 × 10−31 kg)(4.70 × 106 m/s)

(b) λ is proportional to 39.2.

λ=


⎛m ⎞ ⎛ 9.11 × 10 −31 kg ⎞ 1 −14 , so λp = λe ⎜ e ⎟ = (1.55 × 10−10 m) ⎜ ⎟ = 8.46 × 10 m . −27 ⎜ mp ⎟ m 1.67 10 kg × ⎝ ⎠ ⎝ ⎠ For a photon, E =

hc

λ

. For an electron or proton, p =

h

λ

and E =

p2 h2 , so E = . 2m 2mλ 2

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) EXECUTE: (a) E = = = 6.2 keV 0.20 × 10 −9 m λ hc

2

(b) E =

39.3.

⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ (c) Ep = Ee ⎜ e ⎟ = (38 eV) ⎜ ⎟ = 0.021 eV −27 ⎜ ⎟ ⎝ 1.67 × 10 kg ⎠ ⎝ mp ⎠ EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has more energy than a proton. h h (6.63 × 10−34 J ⋅ s) = 2.37 × 10−24 kg ⋅ m s. (a) λ = ⇒ p = = p λ (2.80 × 10 −10 m) (b) K =

39.4.

λ=

=

39.5.

⎛ 6.63 × 10−34 J ⋅ s ⎞ h2 1 = = 6.03 × 10−18 J = 38 eV ⎜ ⎟ 2mλ 2 ⎝ 0.20 × 10 −9 m ⎠ 2(9.11 × 10−31 kg)

p 2 (2.37 × 10 −24 kg ⋅ m s) 2 = = 3.08 × 10−18 J = 19.3 eV. 2m 2(9.11 × 10 −31 kg)

h h = p 2mE (6.63 × 10−34 J ⋅ s) 2(6.64 × 10 −27 kg) (4.20 × 106 eV) (1.60 × 10 −19 J e V)

= 7.02 × 10 −15 m.

h h = . In the Bohr model, mvrn = n( h / 2π ), p mv so mv = nh /(2π rn ). Combine these two expressions and obtain an equation for λ in terms of n. Then

IDENTIFY and SET UP: The de Broglie wavelength is λ =

⎛ 2π rn ⎞ 2π rn . ⎟= n ⎝ nh ⎠ EXECUTE: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10−10 m, so λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m

λ = h⎜

λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit. (b) For n = 4, λ = 2π r4 / 4. rn = n 2a0 so r4 = 16a0 .

λ = 2π (16a0 ) / 4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10 −9 m 1 1 = times the circumference of the orbit. n 4 EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For any n, the circumference of the orbits equals an integer number of de Broglie wavelengths.

λ = 2π r4 / 4; the de Broglie wavelength is

39-2

39.6.

39.7.

39.8.

Chapter 39

(a) For a nonrelativistic particle, K =

(b) (6.63 × 10−34 J ⋅ s) 2(800 eV)(1.60 × 10-19 J/eV)(9.11 × 10-31 kg) = 4.34 × 10 −11 m. IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave properties. SET UP: The de Broglie wavelength of the person is λ = h/mv. EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s. λ = h/mv = (6.626 × 10–34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10–36 m EVALUATE: (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence ordinary objects do not show wave behavior in everyday life.

Combining Equations 37.38 and 37.39 gives p = mc γ 2 − 1. (a) λ =

39.9.

p2 h h , so λ = = . 2m p 2 Km

h = (h mc) p

γ 2 − 1 = 4.43 × 10−12 m. (The incorrect nonrelativistic calculation gives 5.05 × 10−12 m.)

(b) ( h mc) γ 2 − 1 = 7.07 × 10−13 m. IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An electron has mass. Its energy is related to its momentum by E = p 2 / 2m and its wavelength is related to its momentum by Eq.(39.1). EXECUTE: (a) photon hc hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) E= so λ = = = 62.0 nm E (20.0 eV)(1.602 × 10 −19 J/eV) λ electron

E = p 2 /(2m) so p = 2mE = λ = h / p = 0.274 nm

2(9.109 × 10−31 kg)(20.0 eV)(1.602 × 10 −19 J/eV) = 2.416 × 10 −24 kg ⋅ m/s

(b) photon E = hc / λ = 7.946 × 10−19 J = 4.96 eV electron λ = h / p so p = h / λ = 2.650 × 10 −27 kg ⋅ m/s E = p 2 /(2m) = 3.856 × 10 −24 J = 2.41 × 10 −5 eV (c) EVALUATE: You should use a probe of wavelength approximately 250 nm. An electron with λ = 250 nm has much less energy than a photon with λ = 250 nm, so is less likely to damage the molecule. Note that λ = h / p applies to all particles, those with mass and those with zero mass. E = hf = hc / λ applies only to photons and

39.10.

E = p 2 / 2m applies only to particles with mass. IDENTIFY: Any moving particle has a de Broglie wavelength. The speed of a molecule, and hence its de Broglie wavelength, depends on the temperature of the gas. SET UP: The average kinetic energy of the molecule is Kav = 3/2 kT, and the de Broglie wavelength is λ = h/mv = h/p.

EXECUTE: (a) Combining Kav = 3/2 kT and K = p2/2m gives 3/2 kT = pav2/2m and pav = 3mkT . The de Broglie 6.626 × 10−34 J ⋅ s h h wavelength is λ = = = = 1.08 × 10−10 m . p 3mkT 3( 2 × 1.67 × 10−27 kg )(1.38 × 10 −23 J/K ) ( 273 K ) (b) For an electron, λ = h/p = h/mv gives v=

h 6.626 × 10−34 J ⋅ s = = 6.75 × 106 m/s mλ ( 9.11 × 10 −31 kg )(1.08 × 10 −10 m )

This is about 2% the speed of light, so we do not need to use relativity. (c) For photon: E = hc/λ = (6.626 × 10–34 J ⋅ s)(3.00 × 108 m/s)/(1.08 × 10–10 m) = 1.84 × 10–15 J For the H2 molecule: Kav = (3/2)kT = 3/2 (1.38 × 10–23 J/K)(273 K) = 5.65 × 10–21 J For the electron: K = ½ mv2 = ½ (9.11 × 10–31 kg)(6.73 × 106 m/s)2 = 2.06 × 10–17 J EVALUATE: The photon has about 100 times more energy than the electron and 300,000 times more energy than the H2 molecule. This shows that photons of a given wavelength will have much more energy than particles of the same wavelength.

The Wave Nature of Particles

39.11.

39.12.

39-3

IDENTIFY and SET UP: Use Eq.(39.1). h h 6.626 × 10−34 J ⋅ s EXECUTE: λ = = = = 3.90 × 10−34 m p mv (5.00 × 10−3 kg)(340 m/s) EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties. (a) λ = h mv → v = h m λ

1 Energy conservation: eΔV = mv 2 2 2

⎛ h ⎞ m⎜ ⎟ mv h2 (6.626 × 10−34 J ⋅ s) 2 mλ ⎠ ΔV = = ⎝ = = = 66.9 V 2 −19 2e 2e 2emλ 2(1.60 × 10 C) (9.11× 10−31 kg) (0.15 × 10−9 m) 2 2

(b) Ephoton = hf =

hc

λ

=

(6.626 × 10−34 J ⋅ s) (3.0 × 108 m s) = 1.33 × 10 −15 J 0.15 × 10−9 m

eΔV = K = Ephoton and ΔV =

39.13.

39.14.

Ephoton e

=

1.33 × 10 −15 J = 8310 V 1.6 × 10−19 C

(a) λ = 0.10 nm . p = mv = h λ so v = h ( mλ ) = 7.3 × 106 m s .

1 (b) E = mv 2 = 150 eV 2 (c) E = hc / λ = 12 KeV (d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure being probed. IDENTIFY: The electrons behave like waves and are diffracted by the slit. SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find their de Broglie wavelength, which is λ = h/mv. Finally we know that the first dark fringe for single-slit diffraction occurs when a sin θ = λ. EXECUTE: (a) Use energy conservation to find the speed of the electron: ½ mv2 = eV. v=

2 (1.60 × 10−19 C ) (100 V) 2eV = = 5.93 × 106 m/s 9.11 × 10 −31 kg m

which is about 2% the speed of light, so we can ignore relativity. (b) First find the de Broglie wavelength:

λ=

h 6.626 × 10−34 J ⋅ s = = 1.23 × 10–10 m = 0.123 nm mv ( 9.11 × 10−31 kg )( 5.93 × 106 m/s )

For the first single slit dark fringe, we have a sin θ = λ, which gives

a=

39.15.

39.16.

λ sin θ

=

1.23 × 10−10 m = 6.16 × 10–10 m = 0.616 nm sin(11.5°)

EVALUATE: The slit width is around 5 times the de Broglie wavelength of the electron, and both are much smaller than the wavelength of visible light. h For m =1, λ = d sin θ = . 2mE h2 (6.63 × 10 −34 J ⋅ s) 2 = = 6.91 × 10−20 J = 0.432 eV. E= 2 2 −27 2md sin θ 2(1.675 × 10 kg) (9.10 × 10−11 m) 2 sin 2 (28.6°) h h mh Intensity maxima occur when d sin θ = mλ. λ = = so d sin θ = . (Careful! Here, m is the order p 2ME 2ME of the maxima, whereas M is the mass of the incoming particle.) mh (2)(6.63 × 10 −34 J ⋅ s) (a) d = = 2 ME sin θ 2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV) sin(60.6°) = 2.06 × 10−10 m = 0.206 nm.

(b) m = 1 also gives a maximum.

⎛ ⎞ (1) (6.63 × 10−34 J ⋅ s) ⎟ = 25.8° . θ = arcsin ⎜ ⎜ 2(9.11 × 10−31 kg) (188 eV) (1.60 × 10−19 J e V) (2.06 × 10−10 m) ⎟ ⎝ ⎠

39-4

Chapter 39

This is the only other one. If we let m ≥ 3, then there are no more maxima. (c) E =

m2h2 (1) 2 (6.63 × 10 −34 J ⋅ s) 2 = 2 Md 2 sin 2 θ 2(9.11 × 10−31 kg) (2.60 × 10−10 m) 2 sin 2 (60.6°)

= 7.49 × 10−18 J = 46.8 eV. Using this energy, if we let m = 2, then sin θ > 1. Thus, there is no m = 2 maximum in this case.

39.17.

h h ⎛ mh ⎞ = , so θ = arcsin ⎜ ⎟ . (Careful! Here, m is the order of p Mv ⎝ dMv ⎠ the maximum, whereas M is the incoming particle mass.) ⎛ h ⎞ (a) m = 1 ⇒ θ1 = arcsin ⎜ ⎟ ⎝ dMv ⎠ The condition for a maximum is d sinθ = mλ . λ =

⎛ ⎞ 6.63 × 10−34 J ⋅ s = arcsin ⎜ ⎟ = 2.07°. −6 −31 4 × × × (1.60 10 m) (9.11 10 kg) (1.26 10 m s) ⎝ ⎠ ⎛ ⎞ (2) (6.63 × 10−34 J ⋅ s) m = 2 ⇒ θ2 = arcsin ⎜ ⎟ = 4.14°. −6 −31 4 × × × (1.60 10 m) (9.11 10 kg) (1.26 10 m s) ⎝ ⎠ ⎛ π radians ⎞ (b) For small angles (in radians!) y ≅ Dθ , so y1 ≈ (50.0 cm) (2.07°) ⎜ ⎟ = 1.81 cm , ⎝ 180° ⎠

39.18.

⎛ π radians ⎞ y2 ≈ (50.0 cm) (4.14°) ⎜ ⎟ = 3.61 cm and y2 − y1 = 3.61 cm − 1.81 cm = 1.81 cm. ⎝ 180° ⎠ IDENTIFY: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum. SET UP: The uncertainty principle is Δ xΔpx ≥ = . EXECUTE: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its horizontal position is Δ x = 5.0 m. (b) The uncertainty principle gives Δ xΔpx ≥ = , and Δpx = mΔvx since we know the mosquito’s mass. This gives

Δ x mΔvx ≥ = , which we can solve for Δvx to get the minimum uncertainty in vx. Δvx =

39.19.

which is hardly a serious impediment! EVALUATE: For something as “large” as a mosquito, the uncertainty principle places a negligible limitation on our ability to measure its speed. (a) IDENTIFY and SET UP: Use Δ xΔpx ≥ h / 2π to calculate Δ x and obtain Δvx from this. EXECUTE: Δvx =

39.20.

= 1.055 × 10−34 J ⋅ s = 1.4 × 10–29 m/s = mΔx (1.5 × 10-6 kg)(5.0 m)

Δpx ≥

h 6.626 × 10−34 J ⋅ s = = 1.055 × 10−28 kg ⋅ m/s 2πΔ x 2π (1.00 × 10 −6 m)

Δp x 1.055 × 10−28 kg ⋅ m/s = = 8.79 × 10−32 m/s m 1200 kg

(b) EVALUATE: Even for this very small Δ x the minimum Δvx required by the Heisenberg uncertainty principle is very small. The uncertainty principle does not impose any practical limit on the simultaneous measurements of the positions and velocities of ordinary objects. IDENTIFY: Since we know that the marble is somewhere on the table, there is an uncertainty in its position. The Heisenberg uncertainty principle tells us that there is therefore an uncertainty in its momentum. SET UP: The uncertainty principle is Δ xΔpx ≥ = . EXECUTE: (a) Since the marble is somewhere on the table, the maximum uncertainty in its horizontal position is Δ x = 1.75 m. (b) Following the same procedure as in part (b) of problem 39.18, the minimum uncertainty in the horizontal velocity of the marble is Δvx =

= 1.055 × 10 −34 J ⋅ s = 6.03 × 10–33 m/s = mΔx ( 0.0100 kg ) (1.75 m)

(c) The uncertainty principle tells us that we cannot know that the marble’s horizontal velocity is exactly zero, so the smallest we could measure it to be is 6.03 × 10–33 m/s, from part (b). The longest time it could remain on the


table is the time to travel the full width of the table (1.75 m), so t = x/vx = (1.75 m)/(6.03 × 10 s = 9.20 × 1024 years Since the universe is about 14 × 109 years old, this time is about

–33

39-5

m/s) = 2.90 × 1032

9.0 × 1024 yr ≈ 6 × 1014 times the age of the universe! Don’t hold your breath! 14 × 109 yr

39.21.

39.22.

EVALUATE: For household objects, the uncertainty principle places a negligible limitation on our ability to measure their speed. h Heisenberg’s Uncertainty Principles tells us that Δ xΔpx ≥ . We can treat the standard deviation as a direct 2π h −10 = 1.05 × 10−34 J ⋅ s measure of uncertainty. Here Δ xΔpx = (1.2 × 10 m) (3.0 × 10−25 kg ⋅ m s) = 3.6 × 10−35 J ⋅ s but 2π h Therefore Δ xΔpx < so the claim is not valid . 2π (a) ( Δ x) (mΔvx ) ≥ h 2π , and setting Δvx = (0.010)vx and the product of the uncertainties equal to h / 2π (for the minimum uncertainty) gives vx = h (2πm(0.010)Δ x ) = 57.9 m s. (b) Repeating with the proton mass gives 31.6 mm s.

h (6.63 × 10−34 J ⋅ s) = = 2.03 × 10−32 J = 1.27 × 10−13 eV. 2π Δt 2π (5.2 × 10−3 s)

39.23.

ΔE >

39.24.

IDENTIFY and SET UP: The Heisenberg Uncertainty Principle says Δ xΔpx ≥

Δ xΔpx is h / 2π . Δpx = mΔvx .

h . The minimum allowed 2π

h h 6.63 × 10−34 J ⋅ s . Δvx = = = 3.2 × 104 m/s 2π mΔ x 2π (1.67 × 10−27 kg)(2.0 × 10−12 m) 2π h 6.63 × 10−34 J ⋅ s (b) Δ x = = = 4.6 × 10−4 m 2π mΔvx 2π (9.11× 10−31 kg)(0.250 m/s) EXECUTE: (a) mΔ xΔvx =

39.25.

ΔE Δt =

h (6.63 × 10−34 J ⋅ s) h = = 1.39 × 10−14 J = 8.69 × 104 eV = 0.0869 MeV. . ΔE = 2π Δt 2π (7.6 × 10−21 s) 2π ΔE 0.0869 MeV c 2 = = 2.81× 10−5. E 3097 MeV c 2

h . ΔE = Δmc 2 . Δm = 2.06 × 109 eV c 2 = 3.30 × 10−10 J c 2 . 2π h 6.63 × 10−34 J ⋅ s Δt = = = 3.20 × 10−25 s. 2π Δmc 2 2π (3.30 × 10−10 J)

39.26.

ΔE Δt =

39.27.


For a photon Eph =

39.29.

λ

=

1.99 × 10−25 J ⋅ m

λ

2

. For an electron Ee =

p2 1 ⎛h⎞ h2 = . ⎜ ⎟ = 2m 2m ⎝ λ ⎠ 2mλ 2

−25

1.99 × 10 J ⋅ m = 1.99 × 10−17 J 10.0 × 10−9 m (6.63 × 10−34 J ⋅ s) 2 electron Ee = = 2.41× 10−21 J 2(9.11 × 10−31 kg)(10.0 × 10−9 m) 2 Eph 1.99 × 10−17 J = = 8.26 × 103 Ee 2.41 × 10−21 J (b) The electron has much less energy so would be less damaging. EVALUATE: For a particle with mass, such as an electron, E ~ λ −2 . For a massless photon E ~ λ −1 . p 2 (h λ )2 (h λ )2 , so V = = = 419 V. (a) eV = K = 2m 2m 2me 9.11× 10−31 kg (b) The voltage is reduced by the ratio of the particle masses, (419 V) = 0.229 V. 1.67 × 10−27 kg EXECUTE: (a) photon Eph =

39.28.

hc

ψ ( x ) = A sin kx. The position probability density is given by ψ ( x ) = A2 sin 2 kx. EXECUTE: (a) The probability is highest where sin kx = 1 so kx = 2π x / λ = nπ / 2, n = 1, 3, 5,… x = nλ / 4, n = 1, 3, 5,… so x = λ / 4, 3λ / 4, 5λ /4,… IDENTIFY and SET UP:

2

39-6

Chapter 39

39.30.

(b) The probability of finding the particle is zero where ψ = 0, which occurs where sin kx = 0 and kx = 2π x / λ = nπ , n = 0, 1, 2,… x = nλ / 2, n = 0, 1, 2,… so x = 0, λ / 2, λ , 3λ / 2,… EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the amplitude of the standing wave. 2 2 2 Ψ ∗ = ψ ∗ sin ωt , so Ψ = Ψ *Ψ = ψ *ψ sin 2 ωt = ψ sin 2 ωt . Ψ is not time-independent, so Ψ is not the 2

39.31.

wavefunction for a stationary state. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form (b) Using the result from part (a), we have

∫ ( e ) dx = ∫ ∞

ax 2

−∞

∞ −∞

e2 ax dx =

e 2 ax 2a

∫

∞

−∞

| ψ ( x) |2 dx = 1.

∞

= ∞ . Hence this wave function cannot −∞

be normalized and therefore cannot be a valid wave function. (c) We only need to integrate this wave function of 0 to ∞ because it is zero for x < 0. For normalization we have ∞

1 = ∫ |ψ |2 dx = −∞

39.32.

∫ ( Ae ) dx = ∫ ∞

∞

− bx 2

0

0

A2e−2bx dx =

A2e −2bx −2b

∞

= 0

A2 A2 , which gives = 1, so A = 2b . 2b 2b

EVALUATE: If b were positive, the given wave function could not be normalized, so it would not be allowable. (a) The uncertainty in the particle position is proportional to the width of ψ ( x ) , and is inversely proportional to

α . This can be seen by either plotting the function for different values of α , finding the expectation value x 2 = ∫ ψ 2 x 2 dx for the normalized wave function or by finding the full width at half-maximum. The particle’s uncertainty in position decreases with increasing α . The dependence of the expectation value 〈 x 2 〉 on α may be found by considering ∞

〈 x2 〉 =

∫xe

2 −2α x 2

−∞ ∞

∫e

dx

=−

−2α x 2

dx

∞ 1 ∂ ⎡ −2α x2 ⎤ 1 ∂ ⎡ 1 dx ⎥ = − ln ⎢ ∫ e ln ⎢ 2 ∂α ⎣ −∞ 2 ∂α ⎣ 2α ⎦

∞

∫e

−∞

−u 2

⎤ 1 du ⎥ = , ⎦ 4α

−∞

39.33. 39.34.

where the substitution u = α x has been made. (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase. ⎛ x − iy ⎞ ⎛ x + iy ⎞ ⎛ x − iy ⎞ ⎛ x + iy ⎞ 2 * * f ( x, y ) = ⎜ ⎟ ⇒ f = f f =⎜ ⎟ and f ( x, y ) = ⎜ ⎟⋅⎜ ⎟ = 1. + − x iy x iy ⎠ ⎝ ⎠ ⎝ ⎝ x + iy ⎠ ⎝ x − iy ⎠ 2

The same. ψ ( x, y , z ) = ψ * ( x, y , z )ψ ( x, y, z ) 2

ψ ( x, y, z )eiφ = (ψ * ( x, y, z )e− iφ )(ψ ( x, y, z )e + iφ ) = ψ * ( x, y, z )ψ ( x, y, z ). 39.35.

The complex conjugate means convert all i’s to–i’s and vice-versa. eiφ ⋅ e −iφ = 1. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. EXECUTE: (a) For normalization of the one-dimensional wave function, we have ∞

1 = ∫ |ψ |2 dx = −∞

∫ ( Ae ) dx + ∫ ( Ae ) dx = ∫ 0

∞

bx 2

−∞

− bx 2

0

0 −∞

∞

A2e 2bx dx + ∫ A2e −2bx dx . 0

∞ ⎧⎪ e2bx 0 e −2bx ⎫⎪ A2 –1/2 −1 1 = A2 ⎨ + ⎬ = , which gives A = b = 2.00 m = 1.41 m b b b 2 2 − ⎪ −∞ 0 ⎭ ⎩⎪ (b) The graph of the wavefunction versus x is given in Figure 39.35.

(c) (i) P = ∫

+ 5.00 m − 0.500 m

|ψ |2 dx = 2 ∫

+ 5.00 m 0

A2e −2bx dx , where we have used the fact that the wave function is an even

function of x. Evaluating the integral gives − A2 −2b (0.500 m) −(2.00 m −1 ) −2.00 e − 1) = ( ( e − 1) = 0.865 b 2.00 m −1 There is a little more than an 86% probability that the particle will be found within 50 cm of the origin. P=


(ii) P =

∫ ( Ae ) dx = ∫

(iii) P =

∫

2

39-7

−1

A 2.00 m 1 = = = 0.500 −1 2b 2(2.00 m ) 2 There is a 50-50 chance that the particle will be found to the left of the origin, which agrees with the fact that the wave function is symmetric about the y-axis. 0

bx 2

−∞

1.00 m 0.500 m

0

−∞

A2e 2bx dx =

A2e −2bx dx

(

)

A2 −2(2.00 m-1 )(1.00 m) −2(2.00 m-1 )(0.500 m) 1 e −e = − ( e −4 − e−2 ) = 0.0585 2 −2b EVALUATE: There is little chance of finding the particle in regions where the wave function is small. =

39.36.

Figure 39.35 −= 2 d 2ψ + Uψ = Eψ . Let ψ = Aψ1 + Bψ 2 Eq. (39.18): 2m dx 2 −= 2 d 2 ⇒ ( Aψ1 + Bψ 2 ) + U ( Aψ1 + Bψ 2 ) = E ( Aψ1 + Bψ 2 ) 2m dx 2 ⎛ = 2 d 2ψ1 ⎞ ⎛ = 2 d 2ψ 2 ⎞ ⇒ A⎜ − + Uψ − Eψ + B + Uψ 2 − Eψ 2 ⎟ = 0. ⎟ ⎜− 1 1 2 2 ⎝ 2m dx ⎠ ⎝ 2m dx ⎠

But each of ψ1 and ψ2 satisfy Schrödinger’s equation separately so the equation still holds true, for any A or B. 39.37.

39.38.

= 2 d 2ψ + Uψ = BE1ψ1 + CE2ψ 2 . If ψ were a solution with energy E, then BE1ψ1 + CE2ψ 2 = BEψ1 + CEψ2 or 2m dx 2 B( E1 − E )ψ1 = C ( E − E2 )ψ 2 . This would mean that ψ1 is a constant multiple of ψ 2 , and ψ1 and ψ 2 would be wave functions with the same energy. However, E1 ≠ E2 , so this is not possible, and ψ cannot be a solution to Eq. (39.18). −

(a) λ =

h (6.63 × 10 −34 J ⋅ s) = 1.94 × 10−10 m. = −31 −19 2mK 2(9.11 × 10 kg)(40 eV)(1.60 × 10 J eV)

(2.5 m)(9.11 × 10 −31 kg)1 2 R R = = = 6.67 × 10−7 s. v 2E m 2(40 eV)(1.6 × 10−19 J eV) λ (c) The width w is w = 2R ' and w = Δv yt = Δp yt m, where t is the time found in part (b) and a is the slit width. a 2mλ R Combining the expressions for w, Δp y = = 2.65 × 10 −28 kg ⋅ m s. at h = 0.40 μm, which is the same order of magnitude. (d) Δy = 2π Δp y

(b)

39.39.

(a) E = hc λ = 12 eV (b) Find E for an electron with λ = 0.10 × 10 −6 m. λ = h p so p = h λ = 6.626 × 10 −27 kg ⋅ m s . E = p 2 (2m) = 1.5 × 10−4 eV . E = qΔV so ΔV = 1.5 × 10−4 V v = p m = (6.626 × 10 −27 kg ⋅ m s) (9.109 × 10 −31 kg) = 7.3 × 103 m s

(c) Same λ so same p. E = p 2 /(2m) but now m = 1.673 × 10−27 kg so E = 8.2 × 10−8 eV and ΔV = 8.2 × 10−8 V. v = p m = (6.626 × 10 −27 kg ⋅ m s) (1.673 × 10 −27 kg) = 4.0 m s

39.40.

(a) Single slit diffraction: a sin θ = mλ . λ = a sin θ = (150 × 10−9 m)sin20° = 5.13 × 10−8 m

λ = h mv → v = h mλ . v = (b) a sin θ2 = 2λ . sin θ2 = ±2

6.626 × 10−34 J ⋅ s = 1.42 × 104 m s (9.11× 10−31 kg)(5.13 × 10−8 m)

⎛ 5.13 × 10−8 m ⎞ = ±2 ⎜ ⎟ = ±0.684 . θ2 = ±43.2° −9 a ⎝ 150 × 10 m ⎠

λ

39-8

Chapter 39

39.41.

IDENTIFY: The electrons behave like waves and produce a double-slit interference pattern after passing through the slits. SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ/2. The de Broglie wavelength of each of the electrons is λ = h/mv. EXECUTE: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ/2, which gives (1.25 nm)(sin 18.0°) = λ/2 , and λ = 0.7725 nm. Now solve the de Broglie wavelength equation for the speed of the electron:

v=

39.42.

h 6.626 × 10−34 J ⋅ s = = 9.42 × 105 m/s mλ (9.11× 10−31 kg)(0.7725 × 10−9 m)

which is about 0.3% the speed of light, so they are nonrelativistic. (b) Energy conservation gives eV = ½ mv2 and V = mv2/2e = (9.11 × 10–31 kg)(9.42 × 105 m)2/[2(1.60 × 10–19 C)] = 2.52 V EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction. IDENTIFY: The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after passing through the hole. SET UP: For a round hole, the first dark ring occurs at the angle θ for which sinθ = 1.22λ /D, where D is the diameter of the hole. The de Broglie wavelength for a particle is λ = h/p = h/mv. EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives sin θα 1.22λα λα = = sin θ p 1.22λp λp The de Broglie wavelength gives λp = h/pp and λα = h/pα, so

sin θα h / pα pp . Using K = p2/2m, we have = = sin θ p h / pp pα

p = 2mK . Since the alpha particle has twice the charge of the proton and both are accelerated through the same potential difference, Kα = 2Kp. Therefore pp = 2mp K p and pα = 2mα Kα = 2mα (2 K p ) = 4mα K p . Substituting these quantities into the ratio of the sines gives 2mp K p mp sin θα pp = = = sin θ p pα 2 mα 4mα K p Solving for sin θα gives sin θα =

39.43.

1.67 × 10−27 kg sin15.0° and θα = 5.3°. 2(6.64 × 10 −27 kg)

EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha particles form their first dark ring at a smaller angle than the ring for the lighter protons. IDENTIFY: Both the electrons and photons behave like waves and exhibit single-slit diffraction after passing through their respective slits. SET UP: The energy of the photon is E = hc/λ and the de Broglie wavelength of the electron is λ = h/mv = h/p. Destructive interference for a single slit first occurs when a sin θ = λ. EXECUTE: (a) For the photon: λ = hc/E and a sinθ = λ. Since the a and θ are the same for the photons and electrons, they must both have the same wavelength. Equating these two expressions for λ gives a sin θ = hc/E. h h and a sin θ = λ. Equating these two expressions for λ gives a sin θ = . For the electron, λ = h/p = 2mK 2mK h , which gives E = c 2mK = (4.05 × 10−7 J1/2 ) K Equating the two expressions for asinθ gives hc/E = 2mK

E c 2mK 2mc 2 = = . Since v K, so the square root is >1. Therefore E/K > 1, meaning that the K K K photon has more energy than the electron. EVALUATE: As we have seen in Problem 39.10, when a photon and a particle have the same wavelength, the photon has more energy than the particle. d sin θ (40.0 × 10−6 m)sin(0.0300 rad) According to Eq.(35.4) λ = = = 600 nm. The velocity of an electron with m 2 this wavelength is given by Eq.(39.1) (b)

39.44.

v=

p h (6.63 × 10−34 J ⋅ s) = = = 1.21 × 103 m s. m mλ (9.11 × 10−31 kg)(600 × 10 −9 m)


39-9

Since this velocity is much smaller than c we can calculate the energy of the electron classically

39.45.

1 1 K = mv 2 = (9.11 × 10−31 kg)(1.21 × 103 m s) 2 = 6.70 × 10 −25 J = 4.19 μeV. 2 2 The de Broglie wavelength of the blood cell is λ=

h (6.63 × 10 −34 J ⋅ s) = = 1.66 × 10 −17 m. mv (1.00 × 10−14 kg)(4.00 × 10−3 m s)

We need not be concerned about wave behavior. 12

39.46.

⎛ v2 ⎞ h ⎜1 − 2 ⎟ c ⎠ h (a) λ = = ⎝ p mv

⎛ v2 ⎞ h 2v 2 v2 ⇒ λ 2 m 2 v 2 = h 2 ⎜ 1 − 2 ⎟ = h 2 − 2 ⇒ λ 2 m 2v 2 + h 2 2 = h 2 c c ⎝ c ⎠ ⇒ v2 =

(b) v =

c 12

⎛ ⎛ λ ⎞2 ⎞ ⎜1 + ⎜ ⎟ ⎜ ⎝ ( h mc) ⎟⎠ ⎟ ⎝ ⎠

h2 ⎛ 2 2 h2 ⎞ ⎜λ m + 2 ⎟ c ⎠ ⎝

=

c c2 ⇒v= . 12 2 2 2 ⎛λ m c ⎞ ⎛ ⎛ mcλ ⎞ 2 ⎞ 1 + ⎜ ⎟ 2 ⎜⎜1 + ⎜ ⎟ ⎟⎟ ⎝ h ⎠ ⎝ ⎝ h ⎠ ⎠

⎛ 1 ⎛ mcλ ⎞2 ⎞ m 2 c 2λ 2 ≈ c ⎜1 − ⎜ = − Δ (1 ) c . Δ = . ⎟ ⎜ 2 ⎝ h ⎟⎠ ⎟ 2h 2 ⎝ ⎠

h . mc (9.11 × 10−31 kg) 2 (3.00 × 108 m s) 2 (1.00 × 10−15m) 2 Δ= = 8.50 × 10−8 2(6.63 × 10−34 J ⋅ s) 2 ⇒ v = (1 − Δ )c = (1 − 8.50 × 10 −8 )c.

(c) λ = 1.00 × 10 −15 m
0, x = x so U = Ax and F = − = − A. For x < 0, x = − x so U = − Ax and dx d (− Ax) F =− = + A. We can write this result as F = − A x / x, valid for all x except for x = 0. dx

39-12

Chapter 39

(b) IDENTIFY and SET UP: Use the uncertainty principle, expressed as ΔpΔ x ≈ h, and as in Problem 39.50 estimate Δp by p and Δ x by x. Use this to write the energy E of the particle as a function of x. Find the value of x that gives the minimum E and then find the minimum E. p2 EXECUTE: E = K + U = +Ax 2m px ≈ h, so p ≈ h / x

h2 +Ax. 2mx 2 h2 + Ax. For x > 0, E = 2mx 2 Then E ≈

To find the value of x that gives minimum E set 0=

dE = 0. dx

−2h 2 +A 2mx 3 1/ 3

⎛ h2 ⎞ h2 x = and x = ⎜ ⎟ mA ⎝ mA ⎠ With this x the minimum E is 3

h2 ⎛ mA ⎞ E= ⎜ ⎟ 2m ⎝ h 2 ⎠

2/3

1/ 3

⎛ h2 ⎞ + A⎜ ⎟ ⎝ mA ⎠

1 = h 2 / 3m −1/ 3 A2 / 3 + h 2 / 3m −1/ 3 A2 / 3 2

1/ 3

39.62.

⎛ h 2 A2 ⎞ E = 32 ⎜ ⎟ ⎝ m ⎠ EVALUATE: The potential well is shaped like a V. The larger A is the steeper the slope of U and the smaller the region to which the particle is confined and the greater is its energy. Note that for the x that minimizes E, 2K = U. For this wave function, Ψ ∗ = ψ1∗eiω1t + ψ 2∗eiω2 t , so Ψ 2 = Ψ ∗Ψ = (ψ1∗eiω1t + ψ2∗eiω2 t )(ψ1e −iω1t + ψ 2e −iω2 t ) = ψ1∗ψ1 + ψ 2∗ψ2 + ψ1*ψ 2ei (ω1 − ω2 )t + ψ 2∗ψ1ei (ω2 − ω1 )t .

The frequencies ω1 and ω2 are given as not being the same, so Ψ 39.63.

2

is not time-independent, and Ψ is not the

wave function for a stationary state. The time-dependent equation, with the separated form for Ψ ( x, t ) as given becomes ⎛ = 2 d 2ψ ⎞ i=ψ (−iω) = ⎜ − + U ( x)ψ ⎟ . 2 2 m dx ⎝ ⎠

Since ψ is a solution of the time-independent solution with energy E , the term in parenthesis is Eψ , and so ω= = E , and ω = ( E =). 39.64.

p 2 (=k ) 2 =k 2 = ⇒ω= . 2m 2m 2m = 2 ∂ 2ψ ( x, t ) (b) From Problem 39.63 the time-dependent Schrödinger’s equation is − + 2m ∂x 2 ∂ψ ( x, t ) ∂ 2ψ ( x, t ) 2mi ∂ψ ( x, t ) U ( x )ψ ( x, t ) = i= . U ( x) = 0 for a free particle, so . =− ∂t ∂x 2 ∂t = Try ψ ( x, t ) = cos( kx − ωt ) : (a) ω = 2π f =

2π E E 2 π 2π p p= . = . k= = h h = λ =

=ω = E = K =

∂ψ ( x, t ) = Aω sin(kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ = − Ak sin(kx − ωt ) and 2 = Ak 2 cos( kx − ωt ). ∂x ∂x ⎛ 2mi ⎞ Putting this into the Schrödinger’s equation, Ak 2 cos(kx − ωt ) = − ⎜ ⎟ Aω sin(kx − ωt ). ⎝ = ⎠ This is not generally true for all x and t so is not a solution.


39-13

(c) Try ψ ( x, t ) = A sin( kx − ωt ) :

∂ψ ( x, t ) = − Aω cos( kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ ( x, t ) = Ak cos(kx − ωt ) and = − Ak 2 sin( kx − ωt ). ∂x ∂x 2 ⎛ 2mi ⎞ Again, − Ak 2 sin( kx − ωt ) = − ⎜ ⎟ Aω cos(kx − ωt ) is not generally true for all x and t so is not a solution. ⎝ = ⎠ (d) Try ψ ( x, t ) = A cos( kx − ωt ) + B sin(kx − ωt ) :

39.65.

∂ψ ( x, t ) = + Aω sin( kx − ωt ) − Bω cos(kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ ( x, t ) = − Ak sin(kx − ωt ) + Bk cos(kx − ωt ) and = − Ak 2 cos((kx − ωt ) − Bk 2 sin( kx − ωt ). ∂x ∂x 2 Putting this into the Schrödinger’s equation, 2mi − Ak 2 cos(kx − ωt ) − Bk 2 sin( kx − ωt ) = − ( + Aω sin(kx − ωt ) − Bω cos(kx − ωt )). = =k 2 . Collect sin and cos terms. Recall that ω = 2m ( A + iB ) k 2 cos( kx − ωt ) + (iA − B) k 2 sin (kx − ωt ) = 0. This is only true if B = iA. (a) IDENTIFY and SET UP: Let the y-direction be from the thrower to the catcher, and let the x-direction be horizontal and perpendicular to the y-direction. A cube with volume V = 125 cm3 = 0.125 × 10−3 m3 has side length l = V 1/ 3 = (0.125 × 10−3 m3 )1/ 3 = 0.050 m. Thus estimate Δ x as Δ x ≈ 0.050 m. Use the uncertainty principle to estimate Δpx .

h 0.0663 J ⋅ s = = 0.21 kg ⋅ m/s 2πΔ x 2π (0.050 m) (The value of h in this other universe has been used.) (b) IDENTIFY and SET UP: Δ x = (Δvx )t is the uncertainty in the x-coordinate of the ball when it reaches the EXECUTE:

Δ xΔpx ≥ h / 2π then gives Δpx ≈

catcher, where t is the time it takes the ball to reach the second student. Obtain Δvx from Δ px . EXECUTE: The uncertainty in the ball’s horizontal velocity is Δvx =

Δpx 0.21 kg ⋅ m/s = = 0.84 m/s 0.25 kg m

12 m = 2.0 s. The uncertainty in the x-coordinate 6.0 m/s of the ball when it reaches the second student that is introduced by Δvx is Δ x = (Δvx )t = (0.84 m/s)(2.0 s) = 1.7 m. The ball could miss the second student by about 1.7 m. EVALUATE: A game of catch would be very different in this universe. We don’t notice the effects of the uncertainty principle in everyday life because h is so small. The time it takes the ball to travel to the second student is t =

39.66.

(a) ψ 2 = A2 x 2e−2( αx

2

+ βy 2 + γz 2 )

. To save some algebra, let u = x 2 , so that ψ = ue−2α u f ( y, z ) . 2

∂ 2 1 1 2 , x0 = ± . ψ = (1 − 2α u ) ψ ; the maximum occurs at u0 = ∂u 2α 2α (b) ψ vanishes at x = 0, so the probability of finding the particle in the x = 0 plane is zero. The wave function vanishes for x = ±∞. 39.67.

(a) IDENTIFY and SET UP: The probability is P = ψ dV with dV = 4π r 2 dr 2

EXECUTE:

ψ = A2e −2α r so P = 4π A2 r 2e −2α r dr 2

2

2

(b) IDENTIFY and SET UP: P is maximum where EXECUTE:

dP =0 dr

d 2 −2α r 2 (r e )=0 dr

2re −2α r − 4α r 3e −2α r = 0 and this reduces to 2r − 4α r 3 = 0 r = 0 is a solution of the equation but corresponds to a minimum not a maximum. Seek r not equal to 0 so divide by r and get 2 − 4α r 2 = 0 2

2

39-14

Chapter 39

1 (We took the positive square root since r must be positive.) 2α

This gives r =

EVALUATE: This is different from the value of r, r = 0, where ψ

2

is a maximum. At r = 0, ψ

2

has a

maximum but the volume element dV = 4π r dr is zero here so P does not have a maximum at r = 0. 2

39.68.

(a) B (k ) = e −α

2 2

k

B (0) = Bmax = 1 B (kh ) =

2 2 1 1 = e −α kh ⇒ ln(1 2) = −α 2 kh2 ⇒ kh = ln(2) = ωk . 2 α

∞

(b) Using integral tables: ψ ( x) = ∫ e −α 0

2 2

k

cos kxdk =

π − x2 / 4α 2 (e ). ψ ( x) is a maximum when x = 0. 2α

1 − x2 ⇒ h2 = ln(1/2) ⇒ xh = 2α ln2 = ωx 4α 2 4α h ω h 1 h h ln 2 ⎛ ⎞ ⎛ ⎞ (d) ω p ωx = ⎜ k ⎟ ωx = ln2 ⎟ 2α ln2 = (2ln2) = . ⎜ π 2π ⎝ α 2π ⎝ 2π ⎠ ⎠ (c) ψ ( xh ) =

π

when e − xh / 4α = 2

2

(

39.69.

)

k0 ⎛ 1 ⎞ ∞ sin kx (a) ψ ( x) = ∫ B ( k )cos kxdk = ∫ ⎜ ⎟ cos kxdk = 0 0 k0 x ⎝ k0 ⎠

k0

= 0

sin k0 x k0 x

(b) ψ ( x) has a maximum value at the origin x = 0. ψ ( x0 ) = 0 when k0 x0 = π so x0 =

π . Thus the width of this k0

2π 2π . If k0 = , wx = L. B( k ) versus k is graphed in Figure 39.69a. The graph of ψ ( x) versus L k0 x is in Figure 39.69b.

function wx = 2 x0 =

(c) If k0 =

π L

wx = 2 L.

hk h ⎛ hw ⎞ ⎛ 2π ⎞ hw (d) wp wx = ⎜ k ⎟ ⎜ ⎟ = k = 0 = h. The uncertainty principle states that wp wx ≥ . For us, no matter what 2 2 π k k k π ⎝ ⎠⎝ 0 ⎠ 0 0 h k0 is, wp wx = h, which is greater than . 2π

Figure 39.69 39.70.

p 2 (h λ )2 n 2h 2 = = . 2m 2m 8mL2 m, E1 = 2.15 × 10−17 J = 134 eV.

(a) For a standing wave, nλ = 2 L, and En = (b) With L = a0 = 0.5292 × 10−10


39.71.

Time of flight of the marble, from a free-fall kinematic equation is just t =

2y 2(25.0 m) = = 2.26 s . 9.81 m s 2 g

ht ⎛ Δp ⎞ Δ x f = Δ xi + ( Δvx )t = Δ xi + ⎜ x ⎟ t = + Δ xi 2πΔ xi m ⎝ m ⎠ d (Δ x f ) − ht =0= +1 To minimize Δ x f with respect to Δ xi , 2πm( Δ xi ) 2 d ( Δ xi ) ⎛ ht ⎞ ⇒ Δxi (min) = ⎜ ⎟ ⎝ 2πm ⎠ ⇒ Δx f (min) =

ht ht 2ht 2(6.63 × 10−34 J ⋅ s)(2.26 s) + = = = 2.18 × 10−16 m = 2.18 × 10−7 nm. 2πm 2πm πm π (0.0200 kg)

39-15

40

QUANTUM MECHANICS

40.1.

IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =

n2h 2 . 8mL2

(1)(6.626 × 10−34 J ⋅ s) 2 = 1.2 × 10−67 J. 8(0.20 kg)(1.5 m) 2

1 2E 2(1.2 × 10−67 J) (b) E = mv 2 so v = = = 1.1 × 10−33 m/s. If the ball has this speed the time it would take it 2 m 0.20 kg to travel from one side of the table to the other is t =

1.5 m = 1.4 × 1033 s. 1.1× 10−33 m/s

h2 , E2 = 4 E1 , so ΔE = E2 − E1 = 3E1 = 3(1.2 × 10−67 J) = 3.6 × 10 −67 J 8mL2 (d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards. The discrete, quantized nature of the energy levels is completely unobservable. h L= 8mE1 (c) E1 =

40.2.

L=

40.3.

8(1.673 × 10

−27

kg)(5.0 × 10 eV)(1.602 × 10 6

−19

J eV )

= 6.4 × 10−15 m.

IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the ground state of hydrogen. nh 2 SET UP: The energy of the nth level of an electron in a box is En = . 8mL2 EXECUTE: An electron in the ground state of hydrogen has an energy of −13.6 eV, so find the width corresponding to an energy of E1 = 13.6 eV. Solving for L gives L=

40.4.

(6.626 × 10−34 J ⋅ s)

h (6.626 × 10−34 J ⋅ s) = = 1.66 × 10 −10 m. 8mE1 8(9.11 × 10−31 kg)(13.6 eV)(1.602 × 10−19 J eV )

EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the K shell. (a) The energy of the given photon is E = hf = h

c

λ

= (6.63 × 10−34 J ⋅ s)

(3.00 × 103 m/s) = 1.63 × 10−18 J. (122 × 10−9 m)

The energy levels of a particle in a box are given by Eq.40.9 h2 h 2 ( n12 − n2 ) (6.63 × 10−34 J ⋅ s) 2 (2 2 − 12 ) ( n 2 − n2 ). L = = = 3.33 × 10 −10 m. 2 8mL 8mΔ E 8(9.11 × 10−31 kg)(1.63 × 10−20 J) (b) The ground state energy for an electron in a box of the calculated dimensions is h2 (6.63 × 10−34 J ⋅ s) 2 E= = = 5.43 × 10−19 J = 3.40 eV (one-third of the original photon energy), 2 8mL 8(9.11 × 10−31 kg)(3.33 × 10 −10 m) 2 which does not correspond to the −13.6 eV ground state energy of the hydrogen atom. Note that the energy levels for a particle in a box are proportional to n 2 , whereas the energy levels for the hydrogen atom are proportional to − 12 . n

ΔE =

40-1

40-2

40.5.

Chapter 40

IDENTIFY and SET UP: Eq.(40.9) gives the energy levels. Use this to obtain an expression for E2 − E1 and use the value given for this energy difference to solve for L. 4h 2 h2 . The energy separation EXECUTE: Ground state energy is E1 = ; first excited state energy is E2 = 2 8mL 8mL2 3h2 3 = . This gives L = h between these two levels is Δ E = E2 − E1 = 8mΔ E 8mL2 3 = 6.1 × 10−10 m = 0.61 nm. 8(9.109 × 10 −31 kg)(3.0 eV)(1.602 × 10−19 J/1 eV) EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom. (a) The wave function for n = 1 vanishes only at x = 0 and x = L in the range 0 ≤ x ≤ L. (b) In the range for x, the sine term is a maximum only at the middle of the box, x = L / 2. (c) The answers to parts (a) and (b) are consistent with the figure. IDENTIFY and SET UP: For the n = 2 first excited state the normalized wave function is given by Eq.(40.13). 2 2 ⎛ 2π x ⎞ 2 2 2 ⎛ 2π x ⎞ sin ⎜ ψ 2 ( x) = ⎟ . ψ 2 ( x) dx = sin ⎜ ⎟ dx. Examine ψ 2 ( x) dx and find where it is zero and where it is L L L L ⎝ ⎠ ⎝ ⎠ maximum. 2 ⎛ 2π x ⎞ EXECUTE: (a) ψ 2 dx = 0 implies sin ⎜ ⎟=0 ⎝ L ⎠ 2π x = mπ , m = 0, 1, 2, . . . ; x = m( L/2) L For m = 0, x = 0; for m = 1, x = L/2; for m = 2, x = L The probability of finding the particle is zero at x = 0, L/2, and L. 2 ⎛ 2π x ⎞ (b) ψ 2 dx is maximum when sin ⎜ ⎟ = ±1 ⎝ L ⎠ 2π x = m(π /2), m = 1, 3, 5, . . . ; x = m( L/4) L For m = 1, x = L/4; for m = 3, x = 3L/4 The probability of finding the particle is largest at x = L/4 and 3L/4. L = 6.626 × 10 −34 J ⋅ s

40.6.

40.7.

(c) EVALUATE:

The answers to part (a) correspond to the zeros of ψ

answers to part (b) correspond to the two values of x where ψ

2

2

shown in Fig.40.5 in the textbook and the

in the figure is maximum.

dψ 8π 2 m 2m 2 2 = − k ψ , and for ψ to be a solution of Eq.(40.3), k = E =E 2 . 2 2 dx h = (b) The wave function must vanish at the rigid walls; the given function will vanish at x = 0 for any k , but to vanish at x = L, kL = nπ for integer n. 2

40.8.

40.9.

(a) IDENTIFY and SET UP: satisfied. EXECUTE: Eq.(40.3): −

ψ = A cos kx. Calculate dψ 2 /dx 2 and substitute into Eq.(40.3) to see if this equation is

h2 d 2ψ = Eψ 8π 2 m dx 2

dψ = A( −k sin kx) = − Ak sin kx dx 2 dψ = − Ak ( k cos kx) = − Ak 2 cos kx dx 2 h2 Thus Eq.(40.3) requires − 2 (− Ak 2 cos kx ) = E ( A cos kx). 8π m 2 2 hk 2mE 2mE = This says − 2 = E ; k = 8π m ( h/2π ) = 2mE . = (b) EVALUATE: The wave function for a particle in a box with rigid walls at x = 0 and x = L must satisfy the boundary conditions ψ = 0 at x = 0 and ψ = 0 at x = L. ψ (0) = A cos0 = A, since cos 0 = 1. Thus ψ is not 0 at x = 0 and this wave function isn't acceptable because it doesn't satisfy the required boundary condition, even though it is a solution to the Schrödinger equation.

ψ = A cos kx is a solution to Eq.(40.3) if k =

Quantum Mechanics

40.10.

(a) The third excited state is n = 4, so

Δ E = (42 − 1) (b) λ = 40.11.

40-3

h2 15(6.626 × 10 −34 J ⋅ s) 2 = = 5.78 × 10 −17 J = 361 eV. 2 8mL 8(9.11 × 10 −31 kg)(0.125 × 10 −9 m) 2

hc (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) = = 3.44 nm ΔE 5.78 × 10−17 J

Recall λ = (a) E1 =

h h = . p 2mE

h2 h ⇒ λ1 = = 2 L = 2(3.0 × 10−10 m) = 6.0 × 10 −10 m. The wavelength is twice the width of 2 8mL 2mh 2 /8mL2

the box. p1 =

h

λ1

=

(6.63 × 10−34 J ⋅ s) = 1.1 × 10−24 kg ⋅ m/s 6.0 × 10−10 m

2

(b) E2 =

p2 =

h

λ2

4h ⇒ λ2 = L = 3.0 × 10 −10 m. The wavelength is the same as the width of the box. 8mL2 = 2 p1 = 2.2 × 10−24 kg ⋅ m/s.

9h 2 2 ⇒ λ3 = L = 2.0 × 10−10 m. The wavelength is two-thirds the width of the box. 8mL2 3 −24 p3 = 3 p1 = 3.3 × 10 kg ⋅ m/s. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation.

(c) E3 =

40.12.

SET UP: We must substitute the equation Ψ ( x, t ) =

equation −

2 ⎛ nπ x ⎞ −iEn t/= sin ⎜ into the one-dimensional Schrödinger ⎟e L ⎝ L ⎠

= 2 d 2ψ ( x) + U ( x)ψ ( x) = Eψ ( x). 2m dx 2 2

EXECUTE: Taking the second derivative of Ψ ( x, t ) with respect to x gives

d 2 Ψ (x, t ) ⎛ nπ ⎞ = −⎜ ⎟ Ψ (x, t ) dx 2 ⎝ L ⎠

= 2 d 2ψ (x) = 2 ⎛ nπ ⎞ + = we get U ( x ) ψ ( x ) E ψ ( x ), ⎜ ⎟ Ψ (x, t ) = E Ψ (x, t ) which 2m dx 2 2m ⎝ L ⎠ 2

Substituting this result into − 2

= 2 ⎛ nπ ⎞ ⎜ ⎟ , the energies of a particle in a box. 2m ⎝ L ⎠ EVALUATE: Since this process gives us the energies of a particle in a box, the given wave function is a solution to the Schrödinger equation. −= 2 d 2ψ (a) Eq.(40.1): + Uψ = Eψ. 2m dx 2 ⎛ = 2k 2 ⎞ −= 2 d 2 = 2k 2 ( sin ) sin sin sin Left-hand side: A kx + U A kx = A kx + U A kx = + U 0 ⎟ ψ. ⎜ 0 0 2 2m dx 2m ⎝ 2m ⎠ gives En =

40.13.

= 2k 2 = 2k 2 + U 0 > U 0 > E for constant k . But + U 0 should equal E ⇒ no solution. 2m 2m = 2k 2 (b) If E > U 0 , then + U 0 = E is consistent and so ψ = A sin kx is a solution of Eq.(40.1) for this case. 2m According to Eq.(40.17), the wavelength of the electron inside of the square well is given by h 2mE k= . By an analysis similar to that used to derive Eq.40.17, we can show that outside ⇒ λin = = 2m(3U 0 )

But

40.14.

the box

λout =

Thus, the ratio of the wavelengths is

h h = . 2 m( E − U 0 ) 2m(2U 0 )

2m(3U 0 ) λout 3 = = . 2 λin 2m(2U 0 )

40-4

Chapter 40

40.15.

E1 = 0.625 E∞ = 0.625

π 2= 2 ; E1 = 2.00 eV = 3.20 × 10−19 J 2mL2 1/ 2

40.16.

40.17.

⎛ ⎞ 0.625 −10 L = π= ⎜ ⎟ = 3.43 × 10 m −31 −19 2(9.109 10 kg)(3.20 10 J) × × ⎝ ⎠ Since U 0 = 6 E∞ we can use the result E1 = 0.625 E∞ from Section 40.3, so U 0 − E1 = 5.375 E∞ and the maximum wavelength of the photon would be

Eq.(40.16): ψ = Asin

λ=

hc hc 8mL2c = = 2 2 U 0 − E1 (5.375)(h /8mL ) (5.375) h

λ=

8(9.11 × 10−31 kg)(1.50 × 10−9 m) 2 (3.00 × 108 m/s) = 1.38 × 10−6 m. (5.375)(6.63 × 10−34 J ⋅ s)

2mE 2mE x + B cos x = =

d 2ψ 2mE 2mE −2mE ⎛ 2mE ⎞ ⎛ 2mE ⎞ = − A ⎜ 2 ⎟ sin x − B ⎜ 2 ⎟ cos x= (ψ ) = Eq.(40.15). 2 dx = = =2 ⎝ = ⎠ ⎝ = ⎠ 40.18.

40.19.

dψ d 2ψ = κ (Ceκ x − De −κ x ), = κ 2 (Ceκ x + De −κ x ) = κ 2ψ for all constants C and D. Hence ψ is a solution to dx dx 2 =2 2 Eq.(40.1) for − κ + U 0 = E , or κ = [2m(U 0 − E )]1/ 2 =, and κ is real for E < U 0 . 2m IDENTIFY: Find the transition energy Δ E and set it equal to the energy of the absorbed photon. Use E = hc/λ to find the wavelength of the photon. π 2= 2 SET UP: U 0 = 6 E∞ , as in Fig.40.8 in the textbook, so E1 = 0.625 E∞ and E3 = 5.09 E∞ with E∞ = . In this 2mL2 problem the particle bound in the well is a proton, so m = 1.673 × 10 −27 kg. EXECUTE:

E∞ =

π 2= 2 2

2mL

=

π 2 (1.055 × 10−34 J ⋅ s) 2 2(1.673 × 10−27 kg)(4.0 × 10−15 m) 2

= 2.052 × 10−12 J. The transition energy is

Δ E = E3 − E1 = (5.09 − 0.625) E∞ = 4.465 E∞ . Δ E = 4.465(2.052 × 10 −12 J) = 9.162 × 10−12 J The wavelength of the photon that is absorbed is related to the transition energy by Δ E = hc/λ , so hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 2.2 × 10−14 m = 22 fm. ΔE 9.162 × 10−12 J EVALUATE: The wavelength of the photon is comparable to the size of the box. IDENTIFY: The longest wavelength corresponds to the smallest energy change. h2 SET UP: The ground level energy level of the infinite well is E∞ = , and the energy of the photon must be 8mL2 equal to the energy difference between the two shells. EXECUTE: The 400.0 nm photon must correspond to the n = 1 to n = 2 transition. Since U 0 = 6 E∞ , we have

λ=

40.20.

E2 = 2.43E∞ and E1 = 0.625 E∞ . The energy of the photon is equal to the energy difference between the two levels, and E∞ =

h2 hc 1.805 h 2 , which gives = − ⇒ = − = E E E (2.43 0.625) E 2 1 γ ∞ 8mL2 8mL2 λ

(1.805) hλ (1.805)(6.626 × 10 −34 J ⋅ s)(4.00 × 10−7 m) = = 4.68 × 10 −10 m = 0.468 nm. 8mc 8(9.11 × 10 −31 kg)(3.00 × 108 m s) EVALUATE: This width is approximately half that of a Bohr hydrogen atom. E⎛ E ⎞ −2 L 2 m (U 0 − E )/= E 6.0 eV = and E − U 0 = 5 eV = 8.0 × 10−19 J. T = 16 ⎜1 − . ⎟e U0 ⎝ U0 ⎠ U 0 11.0 eV

Solving for L gives L =

40.21.

⎛ 6.0 eV ⎞ ⎛ 6.0 ev ⎞ −2(0.80×10−9 m) (a) L = 0.80 × 10−9 m: T = 16 ⎜ ⎟ ⎜1 − ⎟e 11.0 eV 11.0 eV ⎠ ⎝ ⎠⎝ (b) L = 0.40 × 10−9 m: T = 4.2 × 10−4.

2(9.11×10−31 kg)(8.0×10−19 J) /1.055×10−34 J ⋅s

= 4.4 × 10 −8

Quantum Mechanics

40.22.

The transmission coefficient is T = 16

E⎛ E ⎞ −2 ⎜1 − ⎟e U0 ⎝ U0 ⎠

2 m (U 0 − E ) L/=

40-5

, with E = 5.0 eV, L = 0.60 × 10−9 m, and

m = 9.11 × 10 −31 kg

(a) U 0 = 7.0 eV ⇒ T = 5.5 × 10 −4 . (b) U 0 = 9.0 eV ⇒ T = 1.8 × 10−5 (c) U 0 = 13.0 eV ⇒ T = 1.1 × 10−7. 40.23.

IDENTIFY and SET UP: Use Eq.(39.1), where K = p 2 /2m and E = K + U .

λ = h/p = h/ 2mK , so λ K is constant

EXECUTE:

λ1 K1 = λ2 K 2 ; λ1 and K1 are for x > L where K1 = 2U 0 and λ2 and K 2 are for 0 < x < L where K2 = E − U0 = U 0

λ1 K2 U0 1 = = = K1 2U 0 λ2 2 40.24.

EVALUATE: When the particle is passing over the barrier its kinetic energy is less and its wavelength is larger. IDENTIFY: The probability of tunneling depends on the energy of the particle and the width of the barrier. E⎛ E ⎞ SET UP: The probability of tunneling is approximately T = Ge−2κ L , where G = 16 ⎜ 1 − ⎟ and U0 ⎝ U0 ⎠

κ=

2 m (U 0 − E ) =

EXECUTE:

κ=

.

G = 16

E⎛ E ⎞ 50.0 eV ⎛ 50.0 eV ⎞ ⎜1 − ⎟ = 16 ⎜1 − ⎟ = 3.27. U0 ⎝ U0 ⎠ 70.0 eV ⎝ 70.0 eV ⎠

2m(U 0 − E ) 2(1.67 × 10 −27 kg)(70.0 eV − 50.0 eV)(1.60 × 10 −19 J/eV) = = 9.8 × 1011 m −1 (6.63 × 10−34 J ⋅ s) 2π =

1 1 ⎛ 3.27 ⎞ −12 ln(G / T ) = ln ⎜ ⎟ = 3.6 × 10 m = 3.6 pm 2κ 2(9.8 × 1011 m −1 ) ⎝ 0.0030 ⎠ If the proton were replaced with an electron, the electron’s mass is much smaller so L would be larger. EVALUATE: An electron can tunnel through a much wider barrier than a proton of the same energy. Solving T = Ge−2κ L for L gives L =

40.25.

IDENTIFY and SET UP: The probability is T = Ae−2κ L , with A = 16

2m(U 0 − E ) E⎛ E ⎞ . ⎜1 − ⎟ and κ = U0 ⎝ U0 ⎠ =

E = 32 eV, U 0 = 41 eV, L = 0.25 × 10 −9 m. Calculate T.

EXECUTE: (a) A = 16

E⎛ E ⎞ 32 ⎛ 32 ⎞ ⎜1 − ⎟ = 16 ⎜ 1 − ⎟ = 2.741. U0 ⎝ U0 ⎠ 41 ⎝ 41 ⎠

2m(U 0 − E ) = 2(9.109 × 10−31 kg)(41 eV − 32 eV)(1.602 × 10 −19 J/eV) = 1.536 × 1010 m −1 κ= 1.055 × 10−34 J ⋅ s

κ=

−1

−9

T = Ae−2κ L = (2.741)e −2(1.536×10 m )(0.25×10 m) = 2.741e −7.68 = 0.0013 (b) The only change in the mass m, which appears in κ . 2m(U 0 − E ) κ= = 2(1.673 × 10 −27 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV) = 6.584 × 1011 m −1 κ= 1.055 × 10−34 J ⋅ s 11 -1 −9 Then T = Ae−2κ L = (2.741)e −2(6.584×10 m )(0.25×10 m) = 2.741e −392.2 = 10−143 EVALUATE: The more massive proton has a much smaller probability of tunneling than the electron does. 10

40-6

Chapter 40

40.26.

T = Ge −2κ L with G = 16

−2 2m(U 0 − E ) E⎛ E ⎞ E⎛ E ⎞ , so T = 16 ⎜ 1 − ⎜1 − ⎟ and κ = ⎟e U0 ⎝ U0 ⎠ = U0 ⎝ U0 ⎠

2 m (U 0 − E ) =

L

.

(a) If U 0 = 30.0 × 106 eV, L = 2.0 × 10 −15 m, m = 6.64 × 10 −27 kg and U 0 − E = 1.0 × 106 eV (E = 29.0 × 106 eV), T = 0.090.

(b) If U 0 − E = 10.0 × 106 eV (E = 20.0 × 106 eV), T = 0.014. 40.27.

IDENTIFY and SET UP: The energy levels are given by Eq.(40.26), where ω =

k′ 110 N/m = = 21.0 rad/s m 0.250 kg The ground state energy is given by Eq.(40.26): 1 1 E0 = =ω = (1.055 × 10−34 J ⋅ s)(21.0 rad/s) = 1.11 × 10 −33 J(1 eV/1.602 × 10 −19 J) = 6.93 × 10−15 eV 2 2 1⎞ 1⎞ ⎛ ⎛ En = ⎜ n + ⎟ =ω ; E( n +1) = ⎜ n + 1 + ⎟ =ω 2 2⎠ ⎝ ⎠ ⎝ The energy separation between these adjacent levels is ΔE = En +1 − En = =ω = 2 E0 = 2(1.11 × 10−33 J) = 2.22 × 10−33 J = 1.39 × 10−14 eV EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator. dψ d 2ψ = (4 x 2δ 2 − 2δ)ψ , and ψ is a solution of Eq.(40.21) if Let mk ′ 2= = δ , and so = −2 xδ ψ and dx dx 2 =2 1 1 E = δ = = k ′/m = =ω. 2 2 m IDENTIFY: We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the energy difference between the two levels of the oscillator. SET UP: The energy of a photon is Eγ = hf = hc/λ , and the energy levels of a harmonic oscillator are given by EXECUTE:

40.28.

40.29.

k′ . m

ω=

1 ⎞ k′ ⎛ 1⎞ ⎛ En = ⎜ n + ⎟ = = ⎜ n + ⎟ =ω . 2⎠ m ⎝ 2⎠ ⎝ (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = 0.21 eV λ 5.8 × 10−6 m k′ 2π =c k′ (b) The transition energy is Δ E = En +1 − En = =ω = = , which gives . Solving for k ′, we get == λ m m 4π 2c 2 m 4π 2 (3.00 × 108 m s ) 2 (5.6 × 10−26 kg) = = 5,900 N/m. k′ = λ2 (5.8 × 10−6 m) 2 EVALUATE: This would be a rather strong spring in the physics lab. According to Eq.(40.26), the energy released during the transition between two adjacent levels is twice the ground state energy E3 − E2 = =ω = 2 E0 = 11.2 eV. For a photon of energy E c hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s ) E = hf ⇒ λ = = = = 111 nm. f E (11.2 eV)(1.60 × 10−19 J/eV) EXECUTE: (a) The photon’s energy is Eγ =

40.30.

40.31.

hc

=

IDENTIFY and SET UP: Use the energies given in Eq.(40.26) to solve for the amplitude A and maximum speed vmax of the oscillator. Use these to estimate Δ x and Δpx and compute the uncertainty product Δ xΔpx . EXECUTE: The total energy of a Newtonian oscillator is given by E = 12 k ′A2 where k ′ is the force constant and A

is the amplitude of the oscillator. Set this equal to the energy E = (n + 12 ) =ω of an excited level that has quantum number n, where ω =

k′ , and solve for A: m

1 2

k ′A2 = ( n + 12 ) =ω

(2n + 1)=ω k′ 2 . Set this equal to E = (n + 12 ) =ω and The total energy of the Newtonian oscillator can also be written as E = 12 mvmax A=

solve for vmax : vmax =

1 2

2 mvmax = ( n + 12 ) =ω

(2n + 1)=ω m

Quantum Mechanics

40-7

Thus the maximum linear momentum of the oscillator is pmax = mvmax = (2n + 1)=mω . Assume that A represents the uncertainty Δ x in position and that pmax is the corresponding uncertainty Δ px in momentum. Then the (2n + 1)=ω m ⎛1⎞ (2n + 1)=mω = (2n + 1)=ω = (2n + 1)=ω ⎜ ⎟ = (2n + 1)=. k′ k′ ⎝ω ⎠ EVALUATE: For n = 1 this gives Δ xΔ px = 3=, in agreement with the result derived in Section 40.4. The uncertainty product Δ xΔ px increases with n. uncertainty product is Δ xΔ px =

40.32.

⎛ ω⎞ mk ′ 2 ⎞ ⎛ A ⎟⎟ = exp ⎜ − mk ′ ⎟ = e −1 = 0.368. = exp ⎜⎜ − = k′ ⎠ ψ (0) ⎝ ⎝ ⎠ This is consistent with what is shown in Figure 40.20 in the textbook. (a)

2

2

⎛ ⎞ mk ′ ω⎞ ⎛ = exp ⎜⎜ − (2 A) 2 ⎟⎟ = exp ⎜ − mk ′ 4 ⎟ = e −4 = 1.83 × 10−2. = k′ ⎠ ⎝ ψ (0) ⎝ ⎠ This figure cannot be read this precisely, but the qualitative decrease in amplitude with distance is clear. IDENTIFY: We model the atomic vibration in the crystal as a harmonic oscillator. 1 ⎞ k′ ⎛ 1⎞ ⎛ SET UP: The energy levels of a harmonic oscillator are given by En = ⎜ n + ⎟ = = ⎜ n + ⎟ =ω . 2 m 2⎠ ⎝ ⎠ ⎝ EXECUTE: (a) The ground state energy of a simple harmonic oscillator is (b)

40.33.

ψ ( A)

ψ (2 A) 2 2

1 1 k ′ (1.055 × 10 −34 J ⋅ s) 12.2 N/m E0 = =ω = = = = 9.43 × 10−22 J = 5.89 × 10−3 eV 2 2 m 2 3.82 × 10−26 kg (b) E4 − E3 = =ω = 2 E0 = 0.0118 eV, so λ =

40.34.

hc (6.63 × 10 −34 J ⋅ s)(3.00 × 108 m/s) = = 106 μ m E 1.88 × 10 −21 J

(c) En +1 − En = =ω = 2 E0 = 0.0118 eV EVALUATE: These energy differences are much smaller than those due to electron transitions in the hydrogen atom. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. SET UP: The given function is ψ ( x) = Aeikx , and the one-dimensional Schrödinger equation is

−

= d 2ψ ( x) + U ( x)ψ ( x) = Eψ ( x). 2m dx 2

EXECUTE: Start with the given function and take the indicated derivatives: ψ ( x) = Aeikx .

dψ ( x ) = Aikeikx . dx

= d 2ψ ( x) = 2 2 d 2ψ ( x) d 2ψ ( x) = Ai 2k 2eikx = − Ak 2eikx . = − k 2ψ ( x). − = k ψ ( x). Substituting these results into the 2 2 dx dx 2m dx 2 2m = 2k 2 ψ ( x) + U 0ψ ( x) = E ψ ( x). one-dimensional Schrödinger equation gives 2m = 2k 2 EVALUATE: ψ ( x ) = A eikx is a solution to the one-dimensional Schrödinger equation if E − U 0 = or 2m 2 m( E − U 0 ) . (Since U 0 < E was given, k is the square root of a positive quantity.) In terms of the particle’s =2 momentum p: k = p/=, and in terms of the particle’s de Broglie wavelength λ : k = 2π /λ . k=

40.35.

IDENTIFY: Let I refer to the region x < 0 and let II refer to the region x > 0, so ψ I ( x) = Aeik1 x + Be− ik1 x and

ψ II ( x) = Ceik x . Set ψ I (0) = ψ II (0) and 2

SET UP:

dψ I dψ II = at x = 0. dx dx

d ikx (e ) = ikeikx . dx

dψ I dψ II = at x = 0 gives ik1 A − ik1B = ik2C. Solving this pair of dx dx ⎛k −k ⎞ ⎛ 2k 2 ⎞ equations for B and C gives B = ⎜ 1 2 ⎟ A and C = ⎜ ⎟ A. ⎝ k1 + k2 ⎠ ⎝ k1 + k2 ⎠

EXECUTE:

ψ I (0) = ψ II (0) gives A + B = C.

40-8

Chapter 40

EVALUATE: The probability of reflection is R =

B 2 (k1 − k2 ) 2 = . The probability of transmission is A2 ( k1 + k2 ) 2

C2 4k12 = . Note that R + T = 1. A2 ( k1 + k2 ) 2 (n + 1) 2 − n 2 2n + 1 2 1 (a) Rn = = = + 2 . This is never larger than it is for n = 1, and R1 = 3. n2 n2 n n (b) R approaches zero; in the classical limit, there is no quantization, and the spacing of successive levels is vanishingly small compared to the energy levels. n2h 2 IDENTIFY and SET UP: The energy levels are given by Eq.(40.9): En = . Calculate Δ E for the transition 8mL2 and set Δ E = hc/λ , the energy of the photon. T=

40.36.

40.37.

EXECUTE: (a) Ground level, n = 1, E1 =

First excited level, n = 2, E2 =

4h 2 8mL2

The transition energy is ΔE = E2 − E1 =

3h 2 . Set the transition energy equal to the energy hc/λ of the emitted 8mL2

3h 2 . λ 8mL2 2 −31 8mcL 8(9.109 × 10 kg)(2.998 × 108 m/s)(4.18 × 10−9 m) 2 λ= = 3h 3(6.626 × 10 −34 J ⋅ s) −5 λ = 1.92 × 10 m = 19.2 μ m. 9h 2 9h 2 4h 2 5h 2 (b) Second excited level has n = 3 and E3 = . The transition energy is Δ E = E3 − E2 = − = . 2 2 2 8mL 8mL 8mL 8mL2 hc 5h 2 8mcL2 3 = so λ = = (19.2 μ m) = 11.5 μ m. 2 5h 5 λ 8mL EVALUATE: The energy spacing between adjacent levels increases with n, and this corresponds to a shorter wavelength and more energetic photon in part (b) than in part (a). L 4 2 L/4 πx 2 L/ 4 1 ⎛ 2πx ⎞ 1⎛ L 2πx ⎞ 1 1 (a) ∫ = − , which is about 0.0908. sin 2 dx = ∫ 1 − cos dx = ⎜ x − sin ⎜ ⎟ ⎟ 0 0 L L L 2⎝ L ⎠ L⎝ 2π L ⎠0 4 2π photon. This gives

40.38.

h2 8mL2

hc

=

L2

(b) Repeating with limits of L 4 and L 2 gives

1⎛ L 2πx ⎞ 1 1 , ⎜ x − sin ⎟ = + L ⎠ L 4 4 2π L⎝ 2π

about 0.409. (c) The particle is much likely to be nearer the middle of the box than the edge. (d) The results sum to exactly 1/2, which means that the particle is as likely to be between x = 0 and L 2 as it is to be between x = L 2 and x = L. (e) These results are represented in Figure 40.5b in the textbook. 40.39.

IDENTIFY: The probability of the particle being between x1 and x2 is

∫

x2 x1

2

ψ dx, where ψ is the normalized

wave function for the particle. 2 ⎛πx⎞ sin ⎜ ⎟. L ⎝ L ⎠ EXECUTE: The probability P of the particle being between x = L / 4 and x = 3L / 4 is 3L / 4 2 3L / 4 2 ⎛ π x ⎞ 2 P=∫ ψ 1 dx = ∫ sin ⎜ ⎟ dx. Let y = π x / L; dx = ( L / π ) dy and the integration limits become π / 4 and L/4 L L/4 ⎝ L ⎠ 3π / 4. 3π / 4 2 ⎛ L ⎞ 3π / 4 2 2 ⎡1 1 ⎤ P = ⎜ ⎟∫ sin y dy = ⎢ y − sin 2 y ⎥ L ⎝ π ⎠ π /4 4 π ⎣2 ⎦ π /4 (a) SET UP: The normalized wave function for the ground state is ψ 1 =

2 ⎡ 3π π 1 ⎛ 3π ⎞ 1 ⎛ π ⎞ ⎤ − − sin + sin π ⎢⎣ 8 8 4 ⎜⎝ 2 ⎟⎠ 4 ⎜⎝ 2 ⎟⎠ ⎥⎦ 2 ⎛π 1 1 ⎞ 1 1 1 1 P = ⎜ − ( −1) + (1) ⎟ = + = 0.818. (Note: The integral formula ∫ sin 2 y dy = y − sin 2 y was used.) 4 ⎠ 2 π 2 4 π⎝4 4 P=

Quantum Mechanics

(b) SET UP: The normalized wave function for the first excited state is ψ 2 = EXECUTE:

P=∫

3L / 4 L/4

2

ψ 2 dx =

40-9

2 ⎛ 2π x ⎞ sin ⎜ ⎟ L ⎝ L ⎠

2 3 L / 4 2 ⎛ 2π x ⎞ sin ⎜ ⎟ dx. Let y = 2π x / L; dx = ( L / 2π ) dy and the integration limits L ∫L/ 4 ⎝ L ⎠

become π / 2 and 3π / 2. 3π / 2 2 ⎛ L ⎞ 3π / 2 2 1 ⎡1 1 1 ⎛ 3π π ⎞ ⎤ P= ⎜ y dy y y sin sin 2 = − = ⎜ − ⎟ = 0.500 ⎟ ⎥ L ⎝ 2π ⎠ ∫ π / 2 4 π ⎣⎢ 2 ⎦ π /2 π ⎝ 4 4 ⎠ (c) EVALUATE:

These results are consistent with Fig.40.4b in the textbook. That figure shows that ψ

2

is more

concentrated near the center of the box for the ground state than for the first excited state; this is consistent with the answer to part (a) being larger than the answer to part (b). Also, this figure shows that for the first excited state half the area under ψ 40.40.

2

curve lies between L/4 and 3L/4, consistent with our answer to part (b).

Using the normalized wave function ψ1 = 2 L sin(πx L) , the probabilities | ψ |2 dx are (a) (2 L) sin 2 ( π 4)dx = dx / L (b) (2 L) sin 2 (π / 2) dx = 2dx / L (c) (2 L )sin 2 (3π 4) = dx L .

40.41.

IDENTIFY and SET UP: The normalized wave function for the n = 2 first excited level is ψ 2 =

2 ⎛ 2π x ⎞ sin ⎜ ⎟. L ⎝ L ⎠

P = ψ ( x) dx is the probability that the particle will be found in the interval x to x + dx. 2

EXECUTE: (a) x = L/4

2 ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠ P = (2/L) dx (b) x = L/2

2 ⎛π ⎞ sin ⎜ ⎟ = L ⎝2⎠

2 ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 2 ⎠ ⎠

2 sin(π ) = 0 L

ψ ( x) =

ψ ( x) =

2 . L

P=0 (c) x = 3L/4 2 ⎛ ⎛ 2π ⎞⎛ 3L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠ P = (2/L) dx

ψ ( x) =

2 2 ⎛ 3π ⎞ sin ⎜ ⎟ = − . L L ⎝ 2 ⎠

EVALUATE: Our results are consistent with the n = 2 part of Fig.40.5 in the textbook. ψ 40.42.

40.43.

2

is zero at the center

of the box and is symmetric about this point. G G hn G G G =nπ hn Δ p = pfinal − pinitial . p = =k = = . At x = 0 the initial momentum at the wall is pinitial = − iˆ and the final L 2L 2L G G hn ˆ hn ˆ ⎛ hn ˆ ⎞ hn ˆ momentum, after turning around, is pfinal = + i . So, Δp = + i − ⎜ − i ⎟ = + i . At x = L the initial 2L 2L ⎝ 2L ⎠ L G G hn hn momentum is pinitial = + iˆ and the final momentum, after turning around, is pfinal = − iˆ. So, 2L 2L G hn hn ˆ hn Δ p = − iˆ − i = − iˆ 2L 2L L d 2ψ ( x ) 2m (a) For a free particle, U ( x) = 0 so Schrodinger's = − 2 Eψ ( x ). The graph is given in equation becomes 2 dx h Figure 40.43. = 2κ 2 dψ ( x) d 2ψ ( x ) 2m . (b) For x < 0: ψ ( x) = e +κ x . = κ e +κ x . = κ 2e +κ x . So κ 2 = − 2 E ⇒ E = − dx dx 2m = dψ ( x) d 2ψ ( x) (c) For x > 0: ψ ( x ) = e −κ x . = − ke −κ x . = κ 2e−κ x dx dx

40-10

Chapter 40

So again κ 2 = −

2m −= 2κ 2 E E . Parts (c) and (d) show ψ ( x) satisfies the Schrodinger's ⇒ = equation, provided 2m =2

−= 2κ 2 . 2m dψ ( x) (d) Note is discontinuous at x = 0. (That is, negative for x > 0 and positive for x < 0.) dx

E=

40.44.

Figure 40.43 IDENTIFY: We start with the penetration distance formula given in the problem. = SET UP: The given formula is η = . 2m(U 0 − E ) EXECUTE: (a) Substitute the given numbers into the formula: = 1.055 × 10−34 J ⋅ s η= = = 7.4 × 10 −11 m 2m (U 0 − E ) 2(9.11 × 10 −31 kg)(20 eV − 13 eV)(1.602 × 10−19 J/eV)

(b) η = 40.45.

1.055 × 10−34 J ⋅ s 2(1.67 × 10

−27

kg)(30 MeV − 20 MeV)(1.602 × 10

−13

J/MeV)

= 1.44 × 10−15 m

EVALUATE: The penetration depth varies widely depending on the mass and energy of the particle. (a) We set the solutions for inside and outside the well equal to each other at the well boundaries, x = 0 and L. x = 0 : A sin(0) + B = C ⇒ B = C , since we must have D = 0 for x < 0.

2mEL 2mEL + B cos = + De−κ L since C = 0 for x > L. = = 2mE . This gives A sin kL + B cos kL = De −κ L , where k = = (b) Requiring continuous derivatives at the boundaries yields dψ x = 0: = kA cos(k ⋅ 0) − kB sin(k ⋅ 0) = kA = κ Ce k ⋅0 ⇒ kA = κ C dx x = L: kA cos kL − kB sin kL = −κ De −κ L . 2m(U 0 − E ) 1 ⎛T ⎞ E⎛ E ⎞ T = Ge−2κ L with G = 16 ⎜ 1 − ⇒ L = − ln ⎜ ⎟ . ⎟ and κ = = 2κ ⎝ G ⎠ U0 ⎝ U0 ⎠ x = L: A sin

40.46.

If E = 5.5 eV, U 0 = 10.0 eV, m = 9.11 × 10−31 kg, and T = 0.0010. Then

κ=

2(9.11× 10−31 kg)(4.5 eV)(1.60 × 10−19 J eV) 5.5 eV ⎛ 5.5 eV ⎞ = 1.09 × 1010 m −1 and G = 16 ⎜1 − ⎟ = 3.96 −34 10.0 eV ⎝ 10.0 eV ⎠ (1.054 × 10 J ⋅ s)

so L = − 40.47.

1 ⎛ 0.0010 ⎞ −10 ln ⎜ ⎟ = 3.8 × 10 m = 0.38 nm. 2(1.09 × 1010 m −1 ) ⎝ 3.96 ⎠

IDENTIFY and SET UP: When κ L is large, then eκ L is large and e −κ L is small. When κ L is small, sinh κ L → κ L. Consider both κ L large and κ L small limits.

⎡ (U sinh κ L) 2 ⎤ EXECUTE: (a) T = ⎢1 + 0 ⎥ 4E (U 0 − E ) ⎦ ⎣

−1

Quantum Mechanics

sinh κ L =

e

κL

−e 2

40-11

−κ L

−1

⎡ eκ L U 02e 2κ L ⎤ 16 E (U 0 − E ) and T → ⎢1 + ⎥ = − 2 16 E ( U E ) 16 E ( U 0 − E ) + U 02e 2κ L ⎣ ⎦ 0 For κ L W1, 16 E (U 0 − E ) + U 02e2κ L → U 02e 2κ L

For κ L W1, sinhκ L →

T→

⎛ E ⎞⎛ 16 E (U 0 − E ) E ⎞ − 2κ L = 16 ⎜ ⎟⎜ 1 − ⎟ e , which is Eq.(40.21). U 02e 2κ L U U 0 ⎠ ⎝ 0 ⎠⎝

L 2m(U 0 − E ) . So κ L W1 when L is large (barrier is wide) or U 0 − E is large. (E is small compared to U 0 . ) = 2m(U 0 − E ) ; κ becomes small as E approaches U 0 . For κ small, sinh κ L → κ L and (c) κ = = −1 −1 ⎡ ⎡ U 02 2m(U 0 − E ) L2 ⎤ U 02κ 2 L2 ⎤ = + T → ⎢1 + 1 ⎥ ⎢ ⎥ (using the definition of κ ) = 2 4 E (U 0 − E ) ⎦ ⎣ 4 E (U 0 − E ) ⎦ ⎣ (b) κ L =

⎡ 2U 2 L2 m ⎤ Thus T → ⎢1 + 0 2 ⎥ 4 E= ⎦ ⎣ U 0 → E so

−1

⎡ 2 EL2 m ⎤ U 02 → E and T → ⎢1 + ⎥ E 4= 2 ⎦ ⎣

−1

−1

⎡ ⎛ kL ⎞ 2 ⎤ 2mE But k = 2 , so T → ⎢1 + ⎜ ⎟ ⎥ , as was to be shown. = ⎢⎣ ⎝ 2 ⎠ ⎥⎦ EVALUATE: When κ L is large Eq.(40.20) applies and T is small. When E → U 0 , T does not approach unity. 1 (a) E = mv 2 = (n + (1 2))=ω = ( n + (1 2)) hf , and solving for n, 2 1 2 mv 1 (1/2)(0.020 kg)(0.360 m/s) 2 1 n= 2 − = − = 1.3 × 1030. hf 2 (6.63 × 10−34 J ⋅ s)(1.50 Hz) 2 2

40.48.

40.49.

40.50.

(b) The difference between energies is =ω = hf = (6.63 × 10−34 J ⋅ s)(1.50 Hz) = 9.95 × 10 −34 J. This energy is too small to be detected with current technology IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq.(40.26) for the energy levels. 2π 2π EXECUTE: ω = = = 4π s −1 T 0.500 s 1 1 The ground-state energy is E0 = =ω = (1.055 × 10−34 J ⋅ s)(4π s −1 ) = 6.63 × 10−34 J. 2 2 E0 = 6.63 × 10−34 J(1 eV/1.602 × 10−19 J) = 4.14 × 10 −15 eV

1⎞ ⎛ En = ⎜ n + ⎟ =ω 2⎠ ⎝ 1⎞ ⎛ En +1 = ⎜ n + 1 + ⎟ =ω 2⎠ ⎝ The energy difference between the adjacent energy levels is Δ E = En +1 − En = =ω = 2 E0 = 1.33 × 10−33 J = 8.30 × 10−15 eV EVALUATE: These energies are much too small to detect. Quantum effects are not important for ordinary size objects. IDENTIFY: We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the energy difference between the two energy states in the box. n2h 2 SET UP: The energy of an electron in the nth level is En = . We do not know the initial or final levels, but 8mL2 we do know they differ by 1. The energy of the photon, hc/λ , is equal to the energy difference between the two states. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = EXECUTE: The energy difference between the levels is Δ E = λ 1.649 × 10−7 m −18 1.206 × 10 J. Using the formula for the energy levels in a box, this energy difference is equal to h2 h2 Δ E = ⎡⎣ n2 − (n − 1) 2 ⎤⎦ = (2n − 1) . 2 8mL 8mL2

40-12

40.51.

Chapter 40

⎛ Δ E 8mL2 ⎞ 1 ⎛ (1.206 × 10−18 J)8(9.11 × 10−31 kg)(0.500 × 10−9 m) 2 ⎞ + 1⎟ = ⎜ + 1⎟ = 3. Solving for n gives n = ⎜ 2 (6.626 × 10−34 J ⋅ s) 2 ⎝ h ⎠ 2⎝ ⎠ The transition is from n = 3 to n = 2. EVALUATE: We know the transition is not from the n = 4 to the n = 3 state because we let n be the higher state and n − 1 the lower state. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. 2 2 SET UP: The given wave function is ψ 0 ( x ) = A0 e −α x / 2 and the Schrödinger equation is = d 2ψ ( x) k ′x 2 − + ψ ( x) = E ψ ( x). 2m dx 2 2 2 2 2 2 dψ 0 ( x ) EXECUTE: (a) Start by taking the derivatives: ψ 0 ( x ) = A0e −α x / 2 . = −α 2 xA0 e −α x / 2 . dx 2 d 2ψ 0 ( x) −α 2 x 2 / 2 d ψ 0 ( x ) 2 −α 2 x 2 / 2 2 2 2 2 2 2 2 . = − A0α e + (α ) x A0 e = [ −α + (α ) x ] ψ 0 ( x). dx 2 dx 2 = d 2ψ 0 ( x) =2 = d 2ψ ( x) k ′x 2 [ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x). Equation (40.22) is − + − =− ψ ( x) = E ψ ( x). Substituting 2 2m dx 2m 2m dx 2 2 2 2 = k ′x mω [ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x) + ψ 0 ( x ) = E ψ 0 ( x ). Since α 2 = the above result into that equation gives − and = 2m 2 2 =2 2 2 k′ = 2 ⎛ mω ⎞ mω 2 k′ = 0. (α ) + = − ω= , the coefficient of x2 is − ⎜ ⎟ + 2m 2 2m ⎝ = ⎠ 2 m

⎛ mω ⎞ (b) A0 = ⎜ ⎟ ⎝ =π ⎠

1/ 4

= 2 . The probability density function ψ is ωm 1/ 2 1/ 2 2 ⎛ mω ⎞ ⎛ mω ⎞ − mω x 2 /= =⎜ = . x = 0, At e ψ 0 ⎟ ⎜ ⎟ . ⎝ =π ⎠ ⎝ =π ⎠

(c) The classical turning points are at A = ±

ψ 0 ( x ) = A02e −α 2

x

d ψ 0 ( x) d ψ 0 ( x) mω ⎛ mω ⎞ ⎛ mω ⎞ −α 2 x 2 −α 2 x 2 2 =⎜ = −2 = 0. . At x = 0, ⎟ (−α 2 x)e ⎜ ⎟ xe dx = ⎝ =π ⎠ dx ⎝ =π ⎠ 2 2 1/ 2 d 2 ψ 0 ( x) d 2 ψ 0 ( x) mω ⎛ mω ⎞ −α 2 x 2 2 2 < 0. Therefore, at x = 0, the first derivative is = −2 . At x = 0, ⎜ ⎟ [1 − 2α x ]e 2 2 dx dx = ⎝ =π ⎠ zero and the second derivative is negative. Therefore, the probability density function has a maximum at x = 0. 2 2 =2 EVALUATE: ψ 0 ( x) = A0e−α x / 2 is a solution to equation (40.22) if − (−α 2 )ψ 0 ( x ) = E ψ 0 ( x) or 2m = 2α 2 =ω =ω E= = . E0 = corresponds to n = 0 in Equation (40.26). 2 2m 2 IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. 2 2 SET UP: The given wave function is ψ 1 ( x ) = A1 2 xe −α x / 2 and the Schrödinger equation is = d 2ψ ( x ) k ′x 2 − + ψ ( x ) = E ψ ( x ). 2m dx 2 2 2 2 EXECUTE: (a) Start by taking the indicated derivatives: ψ 1 ( x ) = A1 2 xe−α x / 2 . 2

40.52.

2 2

1/ 2

1/ 2

2

2 2 2 2 2 2 2 2 2 2 dψ 1 ( x ) d 2ψ 1 ( x) = −2α 2 x 2 A1e −α x / 2 + 2 A1e −α x / 2 . = −2 A1α 2 2 xe−α x / 2 − 2 A1α 2 x 2 (−α 2 x )e −α x / 2 + 2 A1 ( −α 2 x)e−α x / 2 . 2 dx dx d 2ψ 1 ( x) 2 2 2 2 2 = [−2α + (α ) x − α ] ψ 1 ( x ) = [ −3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x) dx 2 = d 2ψ 1 ( x) =2 [ −3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x ) − = − 2m dx 2 2m

Quantum Mechanics

40-13

= d ψ ( x) k ′x ψ ( x) = E ψ ( x). Substituting the above result into that equation gives + 2m dx 2 2 mω k ′x 2 =2 k′ [−3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x) + ψ 1 ( x) = E ψ 1 ( x). Since α 2 = and ω = , the coefficient of x2 is − 2m 2 = m 2 =2 2 2 k′ = 2 ⎛ mω ⎞ mω 2 − =0 (α ) + = − ⎜ ⎟ + 2m 2 2m ⎝ = ⎠ 2 2

2

Equation (40.22) is −

1 ⎛ mω ⎞ ⎜ ⎟ 2 ⎝ =π ⎠

1/ 4

(b) A1 =

1 ⎛ mω ⎞ = ⎜ ⎟ 2 ⎝ =π ⎠

1/ 2

(c) The probability density function ψ is ψ 1 ( x) = A12 4 x 2 e−α 2

2

d ψ 1 ( x)

At x = 0, ψ 1 = 0. 2

d ψ 1 ( x)

At x = 0, d 2 ψ 1 ( x) 2

x

4 x 2e

−

mω x 2 =

2

= A12 8 xe−α

dx

2

= 0. At x = ±

dx

2 2

1

α

,

2 2

x

+ A12 4 x 2 (−α 2 2 x )e −α

d ψ 1 ( x)

2 2

x

= A12 8 xe −α

2 2

x

− A12 8 x3α 2e −α

2 2

x

2

dx

= 0.

2

dx 2 d 2 ψ 1 ( x)

= A12 8e −α

2 2

+ A12 8 x( −α 2 2 x)e−α

= A12 8e −α

2 2

− A1216 x 2α 2e −α

x

2 2

2 2

x

− A12 8(3 x 2 )α 2e −α

− A12 24 x 2α 2e−α

2 2

2 2

x

− A12 8 x3α 2 ( −α 2 2 x)e−α

2 2

x

.

+ A1216 x 4 (α 2 ) 2 e −α x . At x = 0, 2 2

d 2 ψ 1 ( x)

2

> 0. So at dx 2 dx 2 x = 0, the first derivative is zero and the second derivative is positive. Therefore, the probability density function x

has a minimum at x = 0. At x = ±

1

α

,

x

d 2 ψ 1 ( x) dx 2

x

2

< 0. So at x = ±

1

α

, the first derivative is zero and the second

derivative is negative. Therefore, the probability density function has maxima at x = ± classical turning points for n = 0 as found in the previous question. EVALUATE:

ψ 1 ( x ) = A1 2 xe −α

2 2

x /2

is a solution to equation (40.22) if −

, corresponding to the

=2 (−3α 2 )ψ 1 ( x) = E ψ 1 ( x) or 2m

3= 2α 2 3=ω 3=ω . E1 = = corresponds to n = 1 in Equation (40.26). 2 2m 2 IDENTIFY and SET UP: Evaluate ∂ 2ψ / ∂x 2 , ∂ 2ψ / ∂y 2 , and ∂ 2ψ / ∂z 2 for the proposed ψ and put Eq.(40.29). Use E=

40.53.

1

α

that ψ nx , ψ ny , and ψ nz are each solutions to Eq.(40.22). ⎞ ⎟ + Uψ = Eψ ⎠ 2 = 2 d ψ nx 1 2 ψ nx , ψ ny , ψ nz are each solutions of Eq.(40.22), so − + k ′x ψ nx = Enxψ nx 2m dx 2 2 2 = 2 d ψ ny 1 2 − + k ′y ψ ny = Enyψ ny 2m dy 2 2 2 = 2 d ψ nz 1 2 − + k ′z ψ nz = Enzψ nz 2m dz 2 2 1 1 1 ψ = ψ nx ( x)ψ ny ( y )ψ nz ( z ), U = k ′x 2 + k ′y 2 + k ′z 2 2 2 2 2 2 2 2 2 ⎛ ⎞ d ψ ⎛ ⎞ d ψ ∂ψ ∂ψ ∂ 2ψ ⎛ d ψ nz ⎞ ny nx ⎜ ⎟ ψ ψ , ψ ψ , = = = ⎜ ⎟ ⎜ ⎟ψ n ψ n . n n n n ∂x 2 ⎜⎝ dx 2 ⎟⎠ y z ∂y 2 ⎜⎝ dy 2 ⎟⎠ x z ∂z 2 ⎜⎝ dz 2 ⎟⎠ x y ⎛ = 2 d 2ψ nx 1 2 ⎞ = 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞ + + + = − + k ′x ψ nx ⎟ψ nyψ nz U So − ψ ⎜ ⎜ ⎟ 2 ⎜ ⎟ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ 2 ⎝ 2m dx ⎠

EXECUTE: (a) −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

⎛ = 2 d 2ψ ny 1 2 +⎜ − + k ′y ψ ny ⎜ 2m dy 2 2 ⎝ −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

⎞ ⎛ = 2 d 2ψ nz 1 2 ⎞ ⎟ψ nxψ nz + ⎜ − + k ′z ψ nz ⎟ψ nxψ ny ⎜ 2m dz 2 ⎟ ⎟ 2 ⎝ ⎠ ⎠

⎞ ⎟ + Uψ = ( Enx + Eny + Enz )ψ ⎠

40-14

Chapter 40

Therefore, we have shown that this ψ is a solution to Eq.(40.29), with energy 3⎞ ⎛ Enx ny nz = Enx + Eny + Enz = ⎜ nx + n y + nz + ⎟ =ω 2⎠ ⎝ 3 (b) and (c) The ground state has nx = n y = nz = 0, so the energy is E000 = =ω. There is only one set of nx , n y and 2 nz that give this energy. 5 First-excited state: nx = 1, ny = nz = 0 or ny = 1, nx = nz = 0 or nz = 1, nx = ny = 0 and E100 = E010 = E001 = =ω 2 There are three different sets of nx , n y , nz quantum numbers that give this energy, so there are three different

40.54.

quantum states that have this same energy. EVALUATE: For the three-dimensional isotropic harmonic oscillator, the wave function is a product of onedimensional harmonic oscillator wavefunctions for each dimension. The energy is a sum of energies for three onedimensional oscillators. All the excited states are degenerate, with more than one state having the same energy. 1⎞ ⎛ ω1 = k1′ m , ω2 = k2′ m . Let ψnx ( x) be a solution of Eq.(40.22) with Enx = ⎜ nx + ⎟ =ω1 , ψ nx ( y ) be a similar 2⎠ ⎝ solution, ψ nz ( z ) be a solution of Eq.(40.22) but with z as the independent variable instead of x, and 1⎞ ⎛ energy Enz = ⎜ nz + ⎟ ω2. 2⎠ ⎝ (a) As in Problem 40.53, look for a solution of the form ψ ( x, y, z ) = ψ nx ( x )ψ ny ( y )ψ nz ( z ). Then, = 2 ∂ 2ψ ⎛ 1 ∂ 2ψ ∂ 2ψ ⎞ = ⎜ Enx − k1′x 2 ⎟ ψ with similar relations for and 2 . Adding, 2 2 2m ∂x 2 ∂y ∂z ⎝ ⎠ 2 2 2 2 ⎛ ⎞ = ∂ψ ∂ψ ∂ψ 1 2 1 2 1 2⎞ ⎛ − + + ⎜ ⎟ = ⎜ En + Eny + Enz − k1′x − k1′ y − k2′ z ⎟ ψ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ ⎝ x 2 2 2 ⎠

−

= ( Enx + Eny + Enz − U )ψ = ( E − U )ψ

40.55.

⎡ 1⎞ ⎤ ⎛ where the energy E is E = Enx + Eny + Enz = = ⎢( nx + n y + 1)ω12 + ⎜ nz + ⎟ ω22 ⎥ , with nx , n y and nz all nonnegative 2⎠ ⎦ ⎝ ⎣ integers. 1 ⎞ ⎛ (b) The ground level corresponds to nx = n y = nz = 0, and E = = ⎜ ω21 + ω 22 ⎟ . The first excited level corresponds to 2 ⎠ ⎝ 3 ⎛ ⎞ nx = ny = 0and nz = 1, since ω12 > ω 22 , and E = =ω ⎜ ω 21 + ω 22 ⎟ . There is only one set of quantum numbers for both 2 ⎠ ⎝ the ground state and the first excited state. (a) ψ ( x ) = A sin kx and ψ (− L 2) = 0 = ψ (+ L 2) 2nπ 2π ⎛ + kL ⎞ + kL ⇒ 0 = A sin ⎜ = nπ ⇒ k = = ⎟⇒ 2 L λ ⎝ 2 ⎠

L h nh p 2 n 2 h2 (2n) 2 h2 ⇒ pn = = ⇒ En = n = = , where n = 1, 2... λn L n 2m 2mL2 8mL2 (b) ψ ( x ) = A cos kx and ψ ( − L / 2) = 0 = ψ ( + L / 2) ⇒λ =

π (2n + 1)π 2π ⎛ kL ⎞ kL ⇒ 0 = A cos ⎜ ⎟ ⇒ = (2n + 1) ⇒ k = = 2 2 2 L λ ⎝ ⎠ 2L (2n + 1) h ⇒λ = ⇒ pn = (2n + 1) 2L (2n + 1) 2 h 2 n = 0, 1, 2... 8mL2 (c) The combination of all the energies in parts (a) and (b) is the same energy levels as given in Eq.(40.9), where n2 h 2 En = . 8mL2 (d) Part (a)’s wave functions are odd, and part (b)’s are even. (a) As with the particle in a box, ψ ( x) = A sin kx, where A is a constant and k 2 = 2mE = 2 . Unlike the particle in a box, however, k and hence E do not have simple forms. ⇒ En =

40.56.

Quantum Mechanics

40-15

(b) For x > L, the wave function must have the form of Eq.(40.18). For the wave function to remain finite as x → ∞, C = 0. The constant κ 2 = 2m(U 0 − E ) =, as in Eq.(14.17) and Eq.(40.18).

(c) At x = L, A sin kL = De − κL and kA cos kL = − κDe − κL . Dividing the second of these by the first gives k cot kL = − κ , a transcendental equation that must be solved numerically for different values of the length L and

the ratio E U 0 . 40.57.

p2 h h + U ( x) ⇒ p = 2m( E − U ( x)). λ = ⇒ λ ( x) = . 2m p 2m( E − U ( x)) (b) As U ( x ) gets larger (i.e., U ( x) approaches E from below—recall k ≥ 0), E − U ( x) gets smaller, so λ ( x) gets larger. (c) When E = U ( x), E − U ( x) = 0, so λ ( x ) → ∞. b dx b b dx 1 b n hn 2m( E − U ( x )) dx = ⇒ ∫ 2m( E − U ( x)) dx = . (d) ∫ =∫ = a λ ( x) a a 2 2 h 2m( E − U ( x)) h ∫ a (e) U ( x ) = 0 for 0 < x < L with classical turning points at x = 0 and x = L. So, (a) E = K + U ( x) =

∫

b

2m( E − U ( x )) dx = ∫

a

L 0

L

2mEdx = 2mE ∫ dx = 2mEL. So, from part (d), 0

2

hn 1 ⎛ hn ⎞ hn ⇒E= ⎜ ⎟ = 2 2m ⎝ 2 L ⎠ 8mL2 . (f ) Since U ( x) = 0 in the region between the turning points at x = 0 and x = L, the results is the same as part (e). 2 2

2mE L =

The height U 0 never enters the calculation. WKB is best used with smoothly varying potentials U ( x). 40.58.

1 2 2E ⇒ xTP = ± . (a) At the turning points E = k ′xTP 2 k′ (b)

+

2 E/k ′

∫ − 2 E/k ′

1 nh ⎛ ⎞ 2m ⎜ E − k ′x 2 ⎟ dx = . To evaluate the integral, we want to get it into a form that matches the 2 2 ⎝ ⎠

standard integral given. Letting A2 = ⇒ mk ′ ∫

40.59.

b a

1 2mE 2E ⎛ ⎞ 2m ⎜ E − k ′x 2 ⎟ = 2mE − mk ′x 2 = mk ′ − x 2 = mk ′ − x2. 2 mk ′ k′ ⎝ ⎠

2E 2E 2E ,a=− ,and b = + k′ k′ k′ ⎛ x ⎞⎤ mk ′ ⎡ 2 2 2 A − x dx = 2 ⎢ x A − x + A arcsin ⎜⎜ ⎟⎟ ⎥ 2 ⎣⎢ ⎝ A ⎠ ⎥⎦ 2

b

2

0

⎡ 2E 2E 2E 2E ⎛ 2E k ′ ⎞⎤ 2E m⎛π⎞ arcsin ⎜ arcsin (1) = 2 E = mk ′ ⎢ − + ⎟⎟ ⎥ = mk ′ ⎜ ⎟. ⎜ ′ ′ ′ ′ ′ k k k k ′ k k′ ⎝ 2 ⎠ ⎝ 2 E k ⎠ ⎥⎦ ⎣⎢ hn m hn k′ h π = . Recall ω = , so E = ωn = =ωn. Using WKB, this is equal to , so E 2 k′ 2 m 2π =ω ⎛ 1 ⎞⎞ ⎛ (c) We are missing the zero-point-energy offset of ⎜ recall E = =ω ⎜ n + ⎟ ⎟ . However, our approximation isn’t 2 ⎝ 2 ⎠⎠ ⎝ bad at all! E (a) At the turning points E = A xTP ⇒ xTP = ± . A (b)

∫

+E / A −E / A

2m( E − A x ) dx = 2∫

dy = −2mA dx when x = 2∫

E A 0

2m( E − Ax)dx = −

E/A 0

2m( E − Ax) dx. Let y = 2m( E − Ax) ⇒

E , y = 0, and when x = 0, y = 2mE. So A 1 0 12 2 32 y dy = − y ∫ 2 mE mA 3mA 23

2 hn 1 ⎛ 3mAh ⎞ 23 ⇒E= (2mE )3 2 = ⎜ ⎟ n . 3mA 2 2m ⎝ 4 ⎠

0

= 2 mE

hn 2 . So, (2mE )3 2 . Using WKB, this is equal to 2 3mA

40-16

Chapter 40

(c) The difference in energy decreases between successive levels. For example: 12 3 − 02 3 = 1, 2 2 3 − 12 3 = 0.59, 33 2 − 23 2 = 0.49,...

•

A sharp ∞ step gave ever-increasing level differences (~ n 2 ).

•

A parabola (~ x 2 ) gave evenly spaced levels (~n).

• Now, a linear potential (~ x ) gives ever-decreasing level differences (~ n 2 3 ). Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the level differences will increase. If the curvature is less than a parabola, the differences will decrease.

41

ATOMIC STRUCTURE

41.1.


L = l (l + 1)=. Lz = ml = . l = 0, 1, 2,..., n − 1. ml = 0, ± 1, ± 2,..., ± l . cosθ = Lz / L .

EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2= , Lz = =,0, −= . l = 2 : L = 6= , Lz = 2=, =,0, −=, −2= .

41.2.

(b) In each case cosθ = Lz / L . L = 0 : θ not defined. L = 2= : 45.0°, 90.0°, 135.0° . L = 6= : 35.3°, 65.9°, 90.0°, 114.1°, 144.7°. G EVALUATE: There is no state where L is totally aligned along the z axis. IDENTIFY and SET UP: L = l (l + 1)= . Lz = ml = . l = 0,1, 2,..., n − 1. ml = 0, ±1, ±2,..., ±l . cosθ = Lz / L . EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2= , Lz = =,0, −= . l = 2 : L = 6= , Lz = 2=, =,0, −=, −2= . l = 3 : L = 2 3= , Lz = 3=,2=, =,0, −=, −2=, −3= . l = 4 : L = 2 5= , Lz = 4=,3=,2=, =,0, −=, −2=, −3=, −4= . (b) L = 0 : θ not defined. L = 2= : 45.0°,90.0°,135.0° . L = 6= : 35.3°,65.9°,90.0°,114.1°,144.7°. L = 2 3= :

41.3.

54.7°,73.2°,90.0°,106.8°,125.3°,150.0°. L = 2 5= : 26.6°,47.9°,63.4°,77.1°,90.0°,102.9°,116.6°,132.1°,153.4° . (c) The minimum angle is 26.6° and occurs for l = 4 , ml = +4 . The maximum angle is 153.4° and occurs for l = 4 , ml = −4 . IDENTIFY and SET UP: The magnitude of the orbital angular momentum L is related to the quantum number l by Eq.(41.4): L = l (l + 1)=, 1 = 0, 1, 2,… −34 2 ⎛ L ⎞ ⎛ 4.716 × 10 kg ⋅ m /s ⎞ l (l + 1) = ⎜ ⎟ = ⎜ ⎟ = 20 −34 ⎝ = ⎠ ⎝ 1.055 × 10 J ⋅ s ⎠ And then l (l + 1) = 20 gives that l = 4. EVALUATE: l must be integer. (a) ( ml ) max = 2, so (Lz ) max = 2=. 2

EXECUTE:

41.4.

(b)

l (l + 1)= = 6= = 2.45=.

⎛m ⎞ ⎛L ⎞ (c) The angle is arccos ⎜ z ⎟ = arccos ⎜ l ⎟ , and the angles are, for ml = −2 to ml = 2, 144.7°, L ⎝ ⎠ ⎝ 6⎠ 114.1°, 90.0°, 65.9°, 35.3°. The angle corresponding to ml = l will always be larger for larger l . 41.5.

IDENTIFY and SET UP: The angular momentum L is related to the quantum number l by Eq.(41.4), L = l (l + 1)=.

The maximum l, lmax , for a given n is lmax = n − 1. EXECUTE:

For n = 2, lmax = 1 and L = 2= = 1.414=.

For n = 20, lmax = 19 and L = (19)(20)= = 19.49=.

41.6.

41.7.

For n = 200, lmax = 199 and L = (199)(200)= = 199.5=. EVALUATE: As n increases, the maximum L gets closer to the value n= postulated in the Bohr model. The (l , ml ) combinations are (0, 0), (1, 0), (1, ± 1) , (2, 0), (2, ± 1), (2, ± 2), (3, 0), (3, ± 1), (3, ± 2), (3, ± 3), (4, 0), (4, ± 1), (4, ± 2), (4, ± 3), and (4, ± 4), a total of 25. 13.60 eV = −0.544 eV. (b) Each state has the same energy (n is the same), − 25 −19 2 1 q1q2 −1 (1.60 × 10 C) = = −2.3 × 10−18 J U= 4πP0 r 4πP0 1.0 × 10−10 m U=

−2.3 × 10−18 J = −14.4 eV. 1.60 × 10−19 J eV 41-1

41-2

41.8.

Chapter 41

(a) As in Example 41.3, the probability is a 2

P=∫

a/2 0

| ψ1s |2 4πr 2 dr =

4 ⎡⎛ ar 2 a 2 r a 3 ⎞ −2 r a ⎤ 5e −1 − − − ⎟e = 0.0803 . ⎥ =1− 3 ⎢⎜ a ⎣⎝ 2 2 4⎠ 2 ⎦0

(b) The difference in the probabilities is (1 − 5e −2 ) − (1 − (5 2)e −1 ) = (5 2)(e−1 − 2e −2 ) = 0.243. 41.9.

(a) | ψ |2 = ψ*ψ =| R (r ) |2 | Θ(θ ) |2 ( Ae− imlφ )( Ae + imlφ ) = A2 | R( r ) |2 | Θ(θ ) |2 , which is independent of φ . (b)

41.10.

∫

2π 0

En = −

| Φ (φ ) |2 dφ = A2

∫

2π 0

dφ = 2πA2 = 1 ⇒ A =

1 . 2π

1 mr e4 E Δ E12 = E2 − E1 = 21 − E1 = −(0.75) E1. 2 2 2 (4πP0 ) 2n = 2

(a) If mr = m = 9.11 × 10 −31 kg 2 mr e 4 (9.109 × 10−31 kg)(1.602 × 10−19 C) 4 = 8.988 × 109 N ⋅ m 2 C ) = 2.177 × 10 −18 J = 13.59 eV ( 2 2 2 −34 (4πP0 ) = 2(1.055 × 10 J ⋅ s) For 2 → 1 transition, the coefficient is (0.75)(13.59 eV) = 10.19 eV. m (b) If mr = , using the result from part (a), 2

mr e 4 ⎛ m 2 ⎞ ⎛ 13.59 eV ⎞ = (13.59 eV) ⎜ ⎟=⎜ ⎟ = 6.795 eV. (4πP0 ) 2 = 2 2 ⎝ m ⎠ ⎝ ⎠ ⎛ 10.19 eV ⎞ Similarly, the 2 → 1 transition, ⇒ ⎜ ⎟ = 5.095 eV. 2 ⎝ ⎠ (c) If mr = 185.8m, using the result from part (a),

mr e4 ⎛ 185.8m ⎞ = (13.59 eV) ⎜ ⎟ = 2525 eV, 2 2 (4πP0 ) = ⎝ m ⎠

41.11.

and the 2 → 1 transition gives ⇒ (10.19 eV)(185.8) = 1893 eV. 4π P0= 2 P0 h2 . IDENTIFY and SET UP: Eq.(41.8) gives a = = mr e 2 π mr e2 EXECUTE: (a) mr = m

a=

P0 h 2 (8.854 × 10−12 C2 /N ⋅ m 2 )(6.626 × 10−34 J ⋅ s) 2 = = 0.5293 × 10 −10 m π mr e 2 π (9.109 × 10−31 kg)(1.602 × 10−19 C) 2

(b) mr = m / 2

⎛ P h2 ⎞ a = 2 ⎜ 0 2 ⎟ = 1.059 × 10−10 m ⎝ π mr e ⎠ (c) mr = 185.8m a=

41.12.

41.13.

1 ⎛ P0 h 2 ⎞ −13 ⎜ ⎟ = 2.849 × 10 m 185.8 ⎝ π mr e 2 ⎠

EVALUATE: a is the radius for the n = 1 level in the Bohr model. When the reduced mass mr increases, a decreases. For positronium and muonium the reduced mass effect is large. eiml φ = cos( mlφ ) + i sin( mlφ ), and to be periodic with period 2π , ml 2π must be an integer multiple of 2π , so ml must be an integer. a a 1 P (a ) = ∫ ψ1s 2V = ∫ e −2 r a (4πr 2dr ) . 0 πa 3 0 a

4 a 4 ⎡⎛ − ar 2 a 2 r a 2 ⎞ −2 r a ⎤ 4 ⎡⎛ − a 3 a 3 a 3 ⎞ −2 a 3 0 ⎤ P (a ) = 3 ∫ r 2e −2 r a dr = 3 ⎢⎜ − − ⎟e = 3 ⎢⎜ − − ⎟e + e ⎥ ⎥ 2 4⎠ 2 4⎠ 4 ⎦ a o a ⎣⎝ 2 ⎦ 0 a ⎣⎝ 2 ⇒ P ( a ) = 1 − 5e −2 . 41.14.

(a) ΔE = μB B = (5.79 × 10 −5 e V T)(0.400 T) = 2.32 × 10−5 eV (b) ml = −2 the lowest possible value of ml .

Atomic Structure

41-3

(c) The energy level diagram is sketched in Figure 41.14.

41.15.

Figure 41.14 IDENTIFY and SET UP: The interaction energy between an external magnetic field and the orbital angular momentum of the atom is given by Eq.(41.18). The energy depends on ml with the most negative ml value having the lowest energy. EXECUTE: (a) For the 5g level, l = 4 and there are 2l + 1 = 9 different ml states. The 5g level is split into 9 levels by the magnetic field. (b) Each ml level is shifted in energy an amount given by U = ml μ B B. Adjacent levels differ in ml by one, so ΔU = μ B B. e= (1.602 × 10−19 C)(1.055 × 10−34 J ⋅ s) = = 9.277 × 10 −24 A ⋅ m 2 2m 2(9.109 × 10 −31 kg) ΔU = μ B B = (9.277 × 10−24 A/m 2 )(0.600 T) = 5.566 × 10−24 J(1 eV/1.602 × 10−19 J) = 3.47 × 10−5 eV

μB =

(c) The level of highest energy is for the largest ml , which is ml = l = 4; U 4 = 4 μ B B. The level of lowest energy is

for the smallest ml , which is ml = −l = −4; U −4 = −4 μ B B. The separation between these two levels is

41.16.

U 4 − U −4 = 8μ B B = 8(3.47 × 10−5 eV) = 2.78 × 10−4 eV. EVALUATE: The energy separations are proportional to the magnetic field. The energy of the n = 5 level in the absence of the external magnetic field is ( −13.6 eV)/52 = −0.544 eV, so the interaction energy with the magnetic field is much less than the binding energy of the state. (a) According to Figure 41.11 in the textbook there are three different transitions that are consistent with the selection rules. The initial ml values are 0, ±1; and the final ml value is 0. (b) The transition from ml = 0 to ml = 0 produces the same wavelength (122 nm) that was seen without the magnetic field. (c) The larger wavelength (smaller energy) is produced from the ml = −1 to ml = 0 transition. (d) The shorter wavelength (greater energy) is produced from the ml = +1 to ml = 0 transition.

41.17.

3 p ⇒ n = 3, l = 1, ΔU = μB B ⇒ B =

U (2.71 × 10 −5 eV) = = 0.468 T μB (5.79 × 10 −5 e V T)

(b) Three: ml = 0, ± 1. 41.18.

41.19.

(2.00232) ⎛ e ⎞ ⎛ −= ⎞ (a) U = + (2.00232) ⎜ μB B ⎟⎜ ⎟B = − 2 ⎝ 2m ⎠ ⎝ 2 ⎠ (2.00232) U =− (5.788 × 10 −5 e V T)(0.480 T) = −2.78 × 10−5 eV. 2 (b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if n ≠ 0 there could be since l < n allows for l ≠ 0. G G IDENTIFY and SET UP: The interaction energy is U = − μ ⋅ B , with μ z given by Eq.(41.22). G G EXECUTE: U = − μ ⋅ B = + μ z B, since the magnetic field is in the negative z-direction. ⎛ e ⎞ ⎛ e ⎞ ⎟ S z , so U = −(2.00232) ⎜ ⎟ Sz B 2 m ⎝ ⎠ ⎝ 2m ⎠ ⎛ e= ⎞ S z = ms =, so U = −2.00232 ⎜ ⎟ ms B ⎝ 2m ⎠ e= = μB = 5.788 × 10−5 eV/T 2m U = −2.00232 μ B ms B 1 The ms = + level has lower energy. 2 1⎞ 1⎞ ⎛ 1 ⎛ 1 ⎞⎞ ⎛ ⎛ ΔU = U ⎜ ms = − ⎟ − U ⎜ ms = + ⎟ = −2.00232 μB B ⎜ − − ⎜ + ⎟ ⎟ = +2.00232 μB B 2⎠ 2⎠ ⎝ ⎝ ⎝ 2 ⎝ 2 ⎠⎠

μ z = −(2.00232) ⎜

ΔU = +2.00232(5.788 × 10−5 eV/T)(1.45 T) = 1.68 × 10−4 eV

41-4

41.20. 41.21.

41.22.

41.23.

Chapter 41

EVALUATE: The interaction energy with the electron spin is the same order of magnitude as the interaction energy with the orbital angular momentum for states with ml ≠ 0. But a 1s state has l = 0 and ml = 0, so there is no orbital magnetic interaction. ⎛ 1⎞ ⎛ 1⎞ ⎛ 3⎞ ⎛ 3⎞ ⎛ 5⎞ The allowed (l , j ) combinations are ⎜ 0, ⎟ , ⎜1, ⎟ , ⎜1, ⎟ , ⎜ 2, ⎟ and ⎜ 2, ⎟ . 2 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2⎠ IDENTIFY and SET UP: j can have the values l + 1/ 2 and l − 1/ 2. EXECUTE: If j takes the values 7/2 and 9/2 it must be that l − 1/ 2 = 7 / 2 and l = 8/ 2 = 4. The letter that labels this l is g. EVALUATE: l must be an integer. hc (4.136 × 10 −15 eV ⋅ s)(300 × 108 m s ) c (3.00 × 108 m s ) (a) λ = f = = 21 cm, = = = 1.4 × 109 Hz, a short radio λ ΔE (5.9 × 10−6 eV) 0.21 m wave. (b) As in Example 41.6, the effective field is B ≅ Δ E 2 μB = 5.1 × 10−2 T, for smaller than that found in the example. IDENTIFY and SET UP: For a classical particle L = I ω. For a uniform sphere with mass m and radius R, 2 ⎛2 ⎞ I = mR 2 , so L = ⎜ mR 2 ⎟ ω. Solve for ω and then use v = rω to solve for v. 5 5 ⎝ ⎠ EXECUTE: (a) L =

ω=

41.24. 41.25.

41.26. 41.27.

3 2 3 = so mR 2ω = = 4 5 4

5 3/ 4= 5 3/ 4(1.055 × 10−34 J ⋅ s) = = 2.5 × 1030 rad/s 2 2mR 2(9.109 × 10 −31 kg)(1.0 × 10 −17 m) 2

(b) v = rω = (1.0 × 10−17 m)(2.5 × 1030 rad/s) = 2.5 × 1013 m/s. EVALUATE: This is much greater than the speed of light c, so the model cannot be valid. However the number of electrons is obtained, the results must be consistent with Table (41.3); adding two more electrons to the zinc configuration gives 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 2 . The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells. 1 n = 1, l = 0, ml = 0, ms = ± : 2 states. 2 1 n = 2, l = 0, ml = 0, ms = ± : 2 states. 2 1 n = 2, l = 1, ml = 0, ± 1, ms = ± : 6 states. 2 For the outer electrons, there are more inner electrons to screen the nucleus. IDENTIFY and SET UP: The energy of an atomic level is given in terms of n and Z eff by Eq.(41.27),

⎛ Z2 ⎞ En = − ⎜ eff2 ⎟ (13.6 eV). The ionization energy for a level with energy − En is + En . ⎝ n ⎠ (2.771) 2 EXECUTE: n = 5 and Zeff = 2.771 gives E5 = − (13.6 eV) = −4.18 eV 52 The ionization energy is 4.18 eV. 2 EVALUATE: The energy of an atomic state is proportional to Z eff . 41.28.

41.29.

For the 4s state, E = −4.339 eV and Z eff = 4 ( −4.339) ( −13.6) = 2.26. Similarly, Z eff = 1.79 for the 4p state and 1.05 for the 4d state. The electrons in the states with higher l tend to be further away from the filled subshells and the screening is more complete. IDENTIFY and SET UP: Use the exclusion principle to determine the ground-state electron configuration, as in Table 41.3. Estimate the energy by estimating Z eff , taking into account the electron screening of the nucleus. EXECUTE: (a) Z = 7 for nitrogen so a nitrogen atom has 7 electrons. N 2+ has 5 electrons: 1s 2 2 s 2 2 p. (b) Z eff = 7 − 4 = 3 for the 2p level.

⎛ Z2 ⎞ 32 En = − ⎜ eff2 ⎟ (13.6 eV) = − 2 (13.6 eV) = −30.6 eV 2 ⎝ n ⎠ (c) Z = 15 for phosphorus so a phosphorus atom has 15 electrons. P 2+ has 13 electrons: 1s 2 2 s 2 2 p 6 3s 2 3 p

Atomic Structure

41-5

(d) Z eff = 15 − 12 = 3 for the 3p level.

41.30.

41.31.

⎛ Z2 ⎞ 32 En = − ⎜ eff2 ⎟ (13.6 eV) = − 2 (13.6 eV) = −13.6 eV 3 ⎝ n ⎠ EVALUATE: In these ions there is one electron outside filled subshells, so it is a reasonable approximation to assume full screening by these inner-subshell electrons. 13.6 eV 2 (a) E2 = − Z eff , so Z eff = 1.26. 4 (b) Similarly, Z eff = 2.26. (c) Z eff becomes larger going down the columns in the periodic table. IDENTIFY and SET UP: Estimate Z eff by considering electron screening and use Eq.(41.27) to calculate the energy. Z eff is calculated as in Example 41.8. EXECUTE: (a) The element Be has nuclear charge Z = 4. The ion Be + has 3 electrons. The outermost electron sees the nuclear charge screened by the other two electrons so Z eff = 4 − 2 = 2.

⎛ Z2 ⎞ 22 En = − ⎜ eff2 ⎟ (13.6 eV) so E2 = − 2 (13.6 eV) = −13.6 eV 2 ⎝ n ⎠ 22 (13.6 eV) = −3.4 eV 42 EVALUATE: For the electron in the highest l-state it is reasonable to assume full screening by the other electrons, as in Example 41.8. The highest l-states of Be + , Mg + , Ca + , etc. all have a Z eff = 2. But the energies are different because for each ion the outermost sublevel has a different n quantum number. 7.46 × 103 eV = 28.0, which corresponds to the element Nickel (Ni). Ekx ≅ ( Z − 1) 2 (10.2 eV) . Z ≈ 1 + 10.2 eV

(b) The outermost electron in Ca + sees a Z eff = 2. E4 = −

41.32. 41.33.

(a) Z = 20 : f = (2.48 × 1015 Hz)(20 − 1) 2 = 8.95 × 1017 Hz . c 3.00 × 108 m s = = 3.35 × 10 −10 m. f 8.95 × 1017 Hz (b) Z = 27: f = 1.68 × 1018 Hz. E = 6.96 keV. λ = 1.79 × 10 −10 m. E = hf = (4.14 × 10−15 eV ⋅ s) (8.95 × 1017 Hz) = 3.71 keV. λ =

41.34.

(c) Z = 48 : f = 5.48 × 1018 Hz, E = 22.7 keV, λ = 5.47 × 10 −11 m. IDENTIFY: The orbital angular momentum is limited by the shell the electron is in. SET UP: For an electron in the n shell, its orbital angular momentum quantum number l is limited by 0 ≤ l < n, and its orbital angular momentum is given by L = l (l + 1) = . The z-component of its angular momentum is

Lz = ml =, where ml = 0, ±1, … , ±l, and its spin angular momentum is S = 3/ 4 = for all electrons. Its energy in the nth shell is En = −(13.6 eV)/n 2 . EXECUTE: (a) L = l (l + 1) = = 12= ⇒ l = 3. Therefore the smallest that n can be is 4, so En = – (13.6 eV)/n2 = – (13.6 eV)/42 = –0.8500 eV. (b) For l = 3, ml = ±3, ±2, ±1, 0. Since Lz = ml =, the largest Lz can be is 3 = and the smallest it can be is –3 = .

41.35.

(c) S = 3/ 4 = for all electrons. (d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is Lmax = 2(2 + 1) = = 6 = . The minimum L can be is zero when l = 0. EVALUATE: At the quantum level, electrons in atoms can have only certain allowed values of their angular momentum. IDENTIFY: The total energy determines what shell the electron is in, which limits its angular momentum. SET UP: The electron’s orbital angular momentum is given by L = l (l + 1) = , and its total energy in the nth shell

is En = −(13.6 eV)/n 2 . EXECUTE: (a) First find n: En = −(13.6 eV)/n 2 = − 0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0. Therefore the

possible values of L are given by L = l (l + 1) = , giving L = 0,

2 =,

6=,

12=,

20=.

(b) E6 = – (13.6 eV)/6 = –0.3778 eV. ΔE = E6 – E5 = –0.3778 eV – (–0.5440 eV) = +0.1662 eV This must be the energy of the photon, so ΔE = hc/λ, which gives λ = hc/ΔE = (4.136 × 10–15 eV ⋅ s )(3.00 ×108 m/s)/(0.1662 eV) = 7.47 × 10–6 m = 7470 nm, which is in the infrared and hence not visible. EVALUATE: The electron can have any of the five possible values for its angular momentum, but it cannot have any others. 2

41-6

Chapter 41

41.36.

IDENTIFY: For the N shell, n = 4, which limits the values of the other quantum numbers. SET UP: In the nth shell, 0 ≤ l < n, ml = 0, ±1, … , ±l, and ms = ±1/2. The orbital angular momentum of the electron is L = l (l + 1) = and its spin angular momentum is S = 3/ 4 = . EXECUTE: (a) For l = 3 we can have ml = ±3, ±2±, ±1, 0 and ms = ±½; for l = 2 we can have ml = ±2, ±1, 0 and ms = ±½; for l = 1, we can have ml = ±1, 0 and ms = ±1/2 ; for l = 0, we can have ml = 0 and ms = ±1/2. (b) For the N shell, n = 4, and for an f-electron, l = 3, giving L = l (l + 1) = = 3(3 + 1) = = 12= . Lz =

ml = = ±3=, ±2=, ± =, 0, so the maximum value is 3= . S = 3/ 4 = for all electrons.

41.37.

(c) For a d-state electron, l = 2, giving L = 2(2 + 1) = = 6= . Lz = ml =, and the maximum value of ml is 2, so the maximum value of Lz is 2 = . The smallest angle occurs when Lz is most closely aligned along the angular L 2= 2 = and θmin = 35.3°. The largest momentum vector, which is when Lz is greatest. Therefore cosθ min = z = L 6= 6 angle occurs when Lz is as far as possible from the L-vector, which is when Lz is most negative. Therefore −2= 2 and θ max = 144.7° . cosθ max = =− 6= 6 (d) This is not possible since l = 3 for an f-electron, but in the M shell the maximum value of l is 2. EVALUATE: The fact that the angle in part (c) cannot be zero tells us that the orbital angular momentum of the electron cannot be totally aligned along any specified direction. IDENTIFY: The inner electrons shield part of the nuclear charge from the outer electron. Z2 SET UP: The electron’s energy in the nth shell, due to shielding, is En = − eff2 (13.6 eV) , where Zeff e is the n effective charge that the electron “sees” for the nucleus. 2 Z eff (4 2 )( −1.947 eV) and n = 4 for the 4s state. Solving for Z (13.6 eV) gives Z = − eff eff n2 13.6 eV = 1.51. The nucleus contains a charge of +11e, so the average number of electrons that screen this nucleus must be 11 – 1.51 = 9.49 electrons (b) (i) The charge of the nucleus is +19e, but 17.2e is screened by the electrons, so the outer electron “sees” 19e – 17.2e = 1.8e and Zeff = 1.8. Z2 (1.8) 2 (ii) En = − eff2 (13.6 eV) = − 2 (13.6 eV) = −2.75 eV n 4 EVALUATE: Sodium has 11 protons, so the inner 10 electrons shield a large portion of this charge from the outer electron. But they don’t shield 10 of the protons, since the inner electrons are not totally equivalent to a uniform spherical shell. (They are lumpy.)

EXECUTE: (a) En = −

2

41.38. 41.39.

2

See Example 41.3; r 2 ψ = Cr 2e −2 r/a ,

d (r 2 ψ )

= Ce −2 r/a (2r − (2r 2 /a)), and for a maximum, r = a, the distance of dr the electron from the nucleus in the Bohr model. (a) IDENTIFY and SET UP: The energy is given by Eq.(38.18), which is identical to Eq.(41.3). The potential energy is given by Eq.(23.9), with q = + Ze and q0 = −e. EXECUTE:

E1s = −

E1s = U ( r ) gives −

1 me 4 1 e2 ; U (r ) = − 2 2 (4π P0 ) 2= 4π P0 r

1 me 4 1 e2 =− 2 2 (4π P0 ) 2= 4π P0 r

(4π P0 )2= 2 = 2a me 2 EVALUATE: The turning point is twice the Bohr radius. (b) IDENTIFY and SET UP: For the 1s state the probability that the electron is in the classically forbidden region r=

∞

∞

is P (r > 2a ) = ∫ ψ 1s dV = 4π ∫ ψ 1s r 2 dr. The normalized wave function of the 1s state of hydrogen is given in 2

2a

Example 41.3: ψ 1s (r ) = EXECUTE:

2

2a

1

π a3

e − r / a . Evaluate the integral; the integrand is the same as in Example 41.3.

⎛ 1 ⎞ ∞ P (r > 2a ) = 4π ⎜ 3 ⎟ ∫ r 2e−2 r / a dr ⎝ π a ⎠ 2a

Atomic Structure

Use the integral formula

∫r e

2 −α r

41-7

⎛ r 2 2r 2 ⎞ dr = −e −α r ⎜ + 2 + 3 ⎟ , with α = 2 / a. α ⎠ ⎝α α ∞

⎛ ar 2 a 2 r a3 ⎞ ⎤ 4⎡ 4 P (r > 2a ) = − 3 ⎢ e −2 r / a ⎜ + + ⎟ ⎥ = + 3 e −4 (2a 3 + a 3 + a3 / 4) a ⎣ 2 4 ⎠⎦ 2a a ⎝ 2

41.40.

P (r > 2a ) = 4e −4 (13/ 4) = 13e−4 = 0.238. EVALUATE: These is a 23.8% probability of the electron being found in the classically forbidden region, where classically its kinetic energy would be negative. (a) For large values of n, the inner electrons will completely shield the nucleus, so Z eff = 1 and the ionization

energy would be

13.60 eV . n2

13.60 eV = 1.11× 10−4 eV, r350 = (350) 2 a0 = (350) 2 (0.529 × 10 −10 m) = 6.48 × 10−6 m . 3502 13.60 eV (c) Similarly for n = 650, = 3.22 × 10−5 eV, r650 = (650) 2 (0.529 × 10−10 m) = 2.24 × 10 −5 m. (650) 2 (b)

41.41.

ψ 2 s (r ) =

r ⎞ − r/ 2 a ⎛ ⎜ 2 − ⎟e a⎠ 32π a ⎝ 1

3

∞

∞

(a) IDENTIFY and SET UP: Let I = ∫ ψ 2 s dV = 4π ∫ ψ 2 s r 2 dr. If ψ 2 s is normalized then we will find that 2

0

2

0

I = 1. r⎞ 1 ∞⎛ 4r 3 r 4 ⎞ − r / a ⎛ 1 ⎞ ∞⎛ I = 4π ⎜ + 2 ⎟ e dr 2 − ⎟ e − r/a r 2 dr = 3 ∫ ⎜ 4r 2 − 3 ⎟ ∫0 ⎜ a⎠ a a ⎠ 8a 0 ⎝ ⎝ 32π a ⎠ ⎝ ∞ n! Use the integral formula ∫ x ne −α x dx = n +1 , with α = 1/ a 2

EXECUTE:

α

0

1 ⎛ 4 1 ⎞ 1 I = 3 ⎜ 4(2!)(a 3 ) − (3!)( a) 4 + 2 (4!)( a)5 ⎟ = (8 − 24 + 24) = 1; this ψ 2 s is normalized. a a 8a ⎝ ⎠ 8 (b) SET UP: For a spherically symmetric state such as the 2s, the probability that the electron will be found at 4a

4a

r < 4a is P (r < 4a) = ∫ ψ 2 s dV = 4π ∫ ψ 2 s r 2 dr. 2

0

EXECUTE:

P (r < 4a ) =

Let P (r < 4a) =

2

0

1 4 a ⎛ 2 4r 3 r 4 ⎞ − r / a + 2 ⎟ e dr ⎜ 4r − 8a 3 ∫ 0 ⎝ a a ⎠

1 ( I1 + I 2 + I 3 ). 8a 3

4a

I1 = 4 ∫ r 2e − r / a dr 0

⎛ r 2 2r 2 ⎞ Use the integral formula ∫ r 2e −α r dr = −e −α r ⎜ + 2 + 3 ⎟ with α = 1/ a. α ⎠ ⎝α α −r / a 2 2 3 4a −4 3 I1 = −4[e ( r a + 2ra + 2a )]0 = ( −104e + 8)a .

I2 = −

4 4a 3 −r / a r e dr a ∫0

⎛ r 3 3r 2 6r 6 ⎞ Use the integral formula ∫ r 3e −α r dr = −e −α r ⎜ + 2 + 3 + 4 ⎟ with α = 1/ a. a α ⎠ ⎝α α 4 −r / a 3 I 2 = [e (r a + 3r 2 a 2 + 6ra3 + 6a 4 )] 04 a = (568e −4 − 24) a3 . a 1 4a I 3 = 2 ∫ r 4e − r / a dr a 0 ⎛ r 4 4r 3 12r 2 24r 24 ⎞ Use the integral formula ∫ r 4e −α r dr = −e −α r ⎜ + 2 + 3 + 4 + 5 ⎟ with α = 1/ a. α a a ⎠ ⎝α α

I3 = −

1 −r / a 4 [e (r a + 4r 3a 2 + 12r 2 a 3 + 24ra 4 + 24a 5 )] 04 a = ( −824e −4 + 24) a3 . a2

41-8

Chapter 41

Thus P (r < 4a) =

1 1 ( I1 + I 2 + I 3 ) = 3 a3 ([8 − 24 + 24] + e −4 [−104 + 568 − 824]) 3 8a 8a

1 P (r < 4a) = (8 − 360e −4 ) = 1 − 45e −4 = 0.176. 8 EVALUATE: There is an 82.4% probability that the electron will be found at r > 4a. In the Bohr model the electron is for certain at r = 4a; this is a poor description of the radial probability distribution for this state. 41.42.

(a) Since the given ψ ( r ) is real, r 2 | ψ |2 = r 2ψ 2 . The probability density will be an extreme when

d 2 2 dψ ⎞ dψ ⎞ ⎛ ⎛ ( r ψ ) = 2 ⎜ rψ 2 + r 2ψ ⎟ = 2rψ ⎜ ψ + r ⎟ = 0. This occurs at r = 0, a minimum, and when ψ = 0, also a dr dr dr ⎠ ⎝ ⎠ ⎝ dψ = 0. Within a multiplicative constant, ψ ( r ) = (2 − r a )e− r 2 a , minimum. A maximum must correspond to ψ + r dr dψ 1 = − (2 − r 2a)e − r 2 a , and the condition for a maximum is (2 − r a ) = (r a ) (2 − r 2a ), or r 2 − 6ra + 4a 2 = 0. dr a The solutions to the quadratic are r = a(3 ± 5). The ratio of the probability densities at these radii is 3.68, with

41.43.

the larger density at r = a(3 + 5) . (b) ψ = 0 at r = 2a Parts (a) and (b) are consistent with Figure 41.5 in the textbook; note the two relative maxima, one on each side of the minimum of zero at r = 2a. L ⎛L ⎞ IDENTIFY: Use Figure 41.2 in the textbook to relate θ L to Lz and L: cosθ L = z so θ L = arccos ⎜ z ⎟ L ⎝ L⎠ (a) SET UP: The smallest angle (θ L ) min is for the state with the largest L and the largest Lz . This is the state with

l = n − 1 and ml = l = n − 1. EXECUTE:

Lz = ml = = ( n − 1)=

L = l (l + 1)= = (n − 1) n= ⎛ (n − 1)= ⎞ ⎛ ( n − 1) ⎞ ⎛ n −1 ⎞ (θ L ) min = arccos ⎜ = arccos ⎜ = arccos ⎜⎜ ⎟ ⎟ = arccos( 1 − 1/ n ). ⎜ ( n − 1)n= ⎟ ⎜ (n − 1) n ⎟⎟ n ⎠⎟ ⎝ ⎝ ⎠ ⎝ ⎠ EVALUATE: Note that (θ L ) min approaches 0° as n → ∞. (b) SET UP: The largest angle (θ L ) max is for l = n − 1 and ml = −l = −(n − 1). EXECUTE: A similar calculation to part (a) yields (θ L ) max = arccos( − 1 − 1/ n ) EVALUATE: Note that (θ L ) max approaches 180° as n → ∞. 41.44.

(a) L2x + L2y = L2 − L2z = l (l + 1)= 2 − ml2= 2 so L2x + L2y = l (l + 1) − ml2 =. (b) This is the magnitude of the component of angular momentum perpendicular to the z-axis. (c) The maximum value is

l (l + 1)= = L, when ml = 0. That is, if the electron is known to have no z-component

of angular momentum, the angular momentum must be perpendicular to the z-axis. The minimum is ml = ±l. 41.45.

l= when

4 r4 ⎛ 1 ⎞ 4 − r 2 a dP ⎛ 1 ⎞ ⎛ 3 r ⎞ − r 2 a dP 4r − ⎟ e P( r ) = ⎜ r e . =⎜ . = 0 when 4r 3 − = 0; r = 4a. In the Bohr 5 ⎟⎜ 5 ⎟ dr ⎝ 24a ⎠ ⎝ a⎠ dr a ⎝ 24a ⎠

model, rn = n 2 a so r2 = 4a, which agrees. 41.46.

The time required to transit the horizontal 50 cm region is t =

Δ x 0.500 m = = 0.952 ms. The force required to vx 525 m s

⎛ ⎞ 2(0.50 × 10 −3 m) 2Δz 0.1079 kg mol =±⎜ = ⎟ −3 2 23 2 t ⎝ 6.022 × 10 atoms mol ⎠ (0.952 × 10 s) N. According to Eq.(41.22), the value of μz is | μ z | = 9.28 × 10 −24 A ⋅ m 2 . Thus, the required

deflect each spin component by 0.50 mm is Fz = maz = ± m ±1.98 × 10−22

magnetic-field gradient is

dBz F 1.98 × 10−22 N = z = = 21.3 T m. μz 9.28 × 10−24 J T dz

Atomic Structure

41.47.

41-9

Decay from a 3d to 2 p state in hydrogen means that n = 3 → n = 2 and ml = ±2, ± 1, 0 → ml = ±1, 0. However selection rules limit the possibilities for decay. The emitted photon carries off one unit of angular momentum so l must change by 1 and hence ml must change by 0 or ±1. The shift in the transition energy from the zero field e=B ( ml3 − ml2 ), where ml3 is the 3d ml value and ml2 is the 2 p ml value. Thus 2m there are only three different energy shifts. They and the transitions that have them, labeled by the ml names, are:

value is just U = ( ml3 − ml2 ) μB B =

e=B : 2 → 1, 2m 0 :1 → 1,

41.48.

1 → 0,

0 → −1

0 → 0,

− 1 → −1

e=B − : 0 → 1, − 1 → 0, − 2 → −1 2m IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the value of ml. SET UP: The selection rules tell us that for allowed transitions, Δl = 1 and Δml = 0 or ±1. EXECUTE: (a) E = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(475.082 nm) = 2.612 eV. (b) For allowed transitions, Δl = 1 and Δml = 0 or ±1. For the 3d state, n = 3, l = 2, and ml can have the values 2, 1, 0, –1, –2. In the 2p state, n = 2, l = 1, and ml can be 1, 0, –1. Therefore the 9 allowed transitions from the 3d state in the presence of a magnetic field are: l = 1, ml = 1; l = 2 , ml = 2 l = 1, ml = 0 l = 2 , ml = 1 l = 1, ml = 1 l = 2 , ml = 1 l = 1, ml = 0 l = 2 , ml = 0 l = 1, ml = 1 l = 2 , ml = 0 l = 1, ml = –1 l = 2 , ml = 0 l = 1, ml = 0 l = 2 , ml = –1 l = 1, ml = –1 l = 2 , ml = –1 l = 1, ml = –1 l = 2 , ml = –2 (c) ΔE = µ BB = (5.788 × 10–5 eV/T)(3.500 T) = 0.000203 eV So the energies of the new states are –8.50000 eV + 0 and –8.50000 eV ± 0.000203 eV, giving energies of: –8.50020 eV, –8.50000 eV, and –8.49980 eV (d) The energy differences of the allowed transitions are equal to the energy differences if no magnetic field were present (2.61176 eV, from part (a)), and that value ±ΔE (0.000203 eV, from part (c)). Therefore we get the following. For E = 2.61176 eV: λ = 475.082 nm (which was given) For E = 2.61176 eV + 0.000203 eV = 2.611963 eV:

→ → → → → → → → →

λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.611963 eV) = 475.045 nm For E = 2.61176 eV – 0.000203 eV = 2.61156 eV:

λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.61156 eV) = 475.119 nm

41.49.

41.50.

EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the wavelengths of the emitted light. IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the value of ml. SET UP: The energy difference due to the magnetic field is ΔE = µ BB and the energy of a photon is E = hc/λ. EXECUTE: For the p state, ml = 0 or ±1, and for the s state ml = 0. Between any two adjacent lines, ΔE = µ BB. Since the change in the wavelength (Δλ) is very small, the energy change (ΔE ) is also very small, so we can use hcΔλ hcΔλ hc hcΔλ . Since ΔE = µ BB, we get µ B B = and B = . differentials. E = hc/λ . | dE | = 2 d λ and ΔE = 2 2 λ λ λ µ Bλ 2 B = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)(0.0462 nm)/(5.788 × 10–5 eV/T)(575.050 nm)2 = 3.00 T EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the wavelengths of the emitted light. (a) The energy shift from zero field is ΔU 0 = ml μ B B. For ml = 2, ΔU 0 = (2) (5.79 × 10−5 e V T) (1.40 T) = 1.62 × 10−4 eV. For ml = 1, ΔU 0 = (1)(5.79 × 10 −5 e V T) (1.40 T) = 8.11 × 10−5 eV.

(b) | Δλ |= λ0

|ΔE | ⎛ 36 ⎞ 1 , where E0 = (13.6 eV)((1/ 4) − (1/ 9)), λ0 = ⎜ ⎟ = 6.563 × 10−7 m E0 ⎝ 5 ⎠R

41-10

41.51.

Chapter 41

and Δ E = 1.62 × 10−4 eV − 8.11 × 10−5 eV = 8.09 × 10 −5 eV from part (a). Then, | Δλ | = 2.81 × 10 −11 m = 0.0281 nm . The wavelength corresponds to a larger energy change, and so the wavelength is smaller. n IDENTIFY: The ratio according to the Boltzmann distribution is given by Eq.(38.21): 1 = e − ( E1 − E0 ) / kT , where 1 is n0 the higher energy state and 0 is the lower energy state. ⎛ e= ⎞ SET UP: The interaction energy with the magnetic field is U = − μ z B = 2.00232 ⎜ ⎟ ms B (Example 41.5.). The ⎝ 2m ⎠ 1 1 energy of the ms = + level is increased and the energy of the ms = − level is decreased. 2 2 n1/ 2 − (U1/ 2 −U −1/ 2 ) / kT =e n−1/ 2

EXECUTE:

⎛ e= ⎞ ⎛ 1 ⎛ 1 ⎞ ⎞ ⎛ e= ⎞ U1/ 2 − U −1/ 2 = 2.00232 ⎜ ⎟ B ⎜ − ⎜ − ⎟ ⎟ = 2.00232 ⎜ ⎟ B = 2.00232μB B m 2 2 2 ⎝ ⎠ ⎝ ⎝ ⎠⎠ ⎝ 2m ⎠

n1/ 2 = e − ( 2.00232) μB B / kT n−1/ 2 (a) B = 5.00 × 10−5 T −24 2 −5 −23 n1/ 2 = e −2.00232(9.274×10 A/m )(5.00×10 T)/([1.381×10 J/K][300 K ]) n−1/ 2 −7 n1/ 2 = e −2.24×10 = 0.99999978 = 1 − 2.2 × 10 −7 n−1/ 2

(b) B = 5.00 × 10 −5 T,

−3 n1/ 2 = e −2.24×10 = 0.9978 n−1/ 2

−2 n1/ 2 = e −2.24×10 = 0.978 n−1/ 2 EVALUATE: For small fields the energy separation between the two spin states is much less than kT for T = 300 K and the states are equally populated. For B = 5.00 T the energy spacing is large enough for there to be a small excess of atoms in the lower state.

(c) B = 5.00 × 10 −5 T,

41.52.

Using Eq.(41.4), L = mvr = l (l + 1)=, and the Bohr radius from Eq.(38.15), we obtain the following value for v :

l (l + 1)= 2(6.63 × 10−34 J ⋅ s) = = 7.74 × 105 m s. The magnetic field generated by the 2 m( n a0 ) 2π (9.11 × 10−31 kg) (4) (5.29 × 10−11 m) “moving” proton at the electrons position can be calculated from Eq.(28.1): μ | q | v sin φ (1.60 × 10−19 C) (7.74 × 105 m s) sin(90°) B= 0 = (10−7 T ⋅ m A) = 0.277 T. 2 4π (4) 2 (5.29 × 10−11 m) 2 r v=

41.53.

3 1 1 3 ms can take on 4 different values: ms = − , − , + , + . Each nlml state can have 4 2 2 2 2 electrons, each with one of the four different ms values. Apply the exclusion principle to determine the electron configurations. EXECUTE: (a) For a filled n = 1 shell, the electron configuration would be 1s 4 ; four electrons and Z = 4. For a IDENTIFY and SET UP:

filled n = 2 shell, the electron configuration would be 1s 4 2 s 4 2 p12 ; twenty electrons and Z = 20.

41.54.

(b) Sodium has Z = 11; 11 electrons. The ground-state electron configuration would be 1s 4 2 s 4 2 p 3. EVALUATE: The chemical properties of each element would be very different. (a) Z 2 ( −13.6 eV) = (7) 2 (−13.6 eV) = −666 eV. (b) The negative of the result of part (a), 666 eV. (c) The radius of the ground state orbit is inversely proportional to the nuclear charge, and a = (0.529 × 10−10 m) 7 = 7.56 × 10−12 m. Z hc hc (d) λ = = , where E0 is the energy found in part (b), and λ = 2.49 nm. ΔE E 1 − 1 0 2 2 1 2

(

)

Atomic Structure

41.55.

41-11

(a) IDENTIFY and SET UP: The energy of the photon equals the transition energy of the atom: Δ E = hc / λ . The energies of the states are given by Eq.(41.3). 13.60 eV 13.60 eV 13.60 eV EXECUTE: En = − so E2 = − and E1 = − n2 4 1 1 3 ⎛ ⎞ Δ E = E2 − E1 = 13.60 eV ⎜ − + 1⎟ = (13.60 eV) = 10.20 eV = (10.20 eV)(1.602 × 10−19 J/eV) = 1.634 × 10−18 J ⎝ 4 ⎠ 4

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.22 × 10−7 m = 122 nm ΔE 1.634 × 10 −18 J (b) IDENTIFY and SET UP: Calculate the change in ΔE due to the orbital magnetic interaction energy, Eq.(41.17), and relate this to the shift Δλ in the photon wavelength. EXECUTE: The shift of a level due to the energy of interaction with the magnetic field in the z-direction is U = ml μ B B. The ground state has ml = 0 so is unaffected by the magnetic field. The n = 2 initial state has

λ=

ml = −1 so its energy is shifted downward an amount U = ml μ B B = (−1)(9.274 × 10−24 A/m 2 )(2.20 T) = ( −2.040 × 10−23 J)(1 eV /1.602 × 10−19 J) = 1.273 × 10−4 eV Note that the shift in energy due to the magnetic field is a very small fraction of the 10.2 eV transition energy. Problem 39.56c shows that in this situation Δλ / λ = Δ E / E . This gives

⎛ 1.273 × 10 −4 eV ⎞ −3 Δλ = λ Δ E / E = 122 nm ⎜ ⎟ = 1.52 × 10 nm = 1.52 pm. 10.2 eV ⎝ ⎠ EVALUATE: The upper level in the transition is lowered in energy so the transition energy is decreased. A smaller ΔE means a larger λ ; the magnetic field increases the wavelength. The fractional shift in wavelength, Δλ / λ is 41.56.

41.57.

small, only 1.2 × 10 −5. The effective field is that which gives rise to the observed difference in the energy level transition, Δ E hc ⎛ λ1 − λ2 ⎞ 2πmc ⎛ λ1 − λ2 ⎞ −3 = B= ⎜ ⎟= ⎜ ⎟ . Substitution of numerical values gives B = 3.64 × 10 T, much smaller μB μB ⎝ λ1λ2 ⎠ e ⎝ λ1λ2 ⎠ than that for sodium. IDENTIFY: Estimate the atomic transition energy and use Eq.(38.6) to relate this to the photon wavelength. (a) SET UP: vanadium, Z = 23 minimum wavelength; corresponds to largest transition energy EXECUTE: The highest occupied shell is the N shell ( n = 4). The highest energy transition is N → K , with transition energy ΔE = EN − EK . Since the shell energies scale like 1/ n 2 neglect EN relative to EK , so Δ E = EK = ( Z − 1) 2 (13.6 eV) = (23 − 1) 2 (13.6 eV) = 6.582 × 103 eV = 1.055 × 10 −15 J. The energy of the emitted photon equals this transition energy, so the photon’s wavelength is given by Δ E = hc / λ so λ = hc / Δ E.

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 1.88 × 10 −10 m = 0.188 nm. 1.055 × 10 −15 J SET UP: maximum wavelength; corresponds to smallest transition energy, so for the Kα transition EXECUTE: The frequency of the photon emitted in this transition is given by Moseley’s law (Eq.41.29): f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(23 − 1) 2 = 1.200 × 1018 Hz

λ=

c 2.998 × 108 m/s = = 2.50 × 10−10 m = 0.250 nm f 1.200 × 1018 Hz (b) rhenium, Z = 45 Apply the analysis of part (a), just with this different value of Z. minimum wavelength Δ E = EK = ( Z − 1) 2 (13.6 eV) = (45 − 1) 2 (13.6 eV) = 2.633 × 104 eV = 4.218 × 10−15 J.

λ=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 4.71 × 10−11 m = 0.0471 nm. 4.218 × 10 −15 J maximum wavelength f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(45 − 1) 2 = 4.801× 1018 Hz

λ = hc / Δ E =

c 2.998 × 108 m/s = = 6.24 × 10−11 m = 0.0624 nm f 4.801 × 1018 Hz EVALUATE: Our calculated wavelengths have values corresponding to x rays. The transition energies increase when Z increases and the photon wavelengths decrease.

λ=

41-12

Chapter 41

41.58.

(a) Δ E = (2.00232)

41.59.

(a) To calculate the total number of states for the n th principal quantum number shell we must multiply all the possibilities. The spin states multiply everything by 2. The maximum l value is (n –1), and each l value has (2l + 1)ml values. So the total number of states is

e e= hc 2πmc ⇒B= BΔS z ≈ B = . λ λe 2m m 2π (9.11 × 10 −31 kg) (3.00 × 108 m s ) (b) B = = 0.307 T. (0.0350 m)(1.60 × 10−19 C)

n −1

n −1

n −1

l =0

l =0

l =0

N = 2∑ (2l + 1) = 2∑1 + 4∑ l = 2n + 41.60.

4( n − 1)(n) = 2n + 2n 2 − 2n = 2n 2 . 2

(b) The n = 5 shell (O-shell) has 50 states. IDENTIFY: We treat the Earth as an electron. SET UP: The intrinsic spin angular momentum of an electron is S =

3 = , and the angular momentum of the 4

spinning Earth is S = I ω , where I = 2/5 mR2. EXECUTE: (a) Using S = I ω =

41.61.

3 = and solving for ω gives 4

3 3 = (1.055 ×10−34 J ⋅ s ) 4 4 ω= = = 9.40 × 10−73 rad/s 2 2 2 2 24 6 mR ( 5.97 ×10 kg )( 6.38 × 10 m ) 5 5 (b) We could not use this approach on the electron because in quantum physics we do not view it in the classical sense as a spinning ball. EVALUATE: The angular velocity we have just calculated for the Earth would certainly be masked by its present angular spin of one revolution per day. 1 The potential U ( x ) = k ′x 2 is that of a simple harmonic oscillator. Treated quantum mechanically (see Section 40.4) 2 each energy state has energy En = =ω (n + 12 ). Since electrons obey the exclusion principle, this allows us to put two electrons (one for each ms = ± 12 ) for every value of n⎯each quantum state is then defined by the ordered pair of quantum numbers ( n, ms ). By placing two electrons in each energy level the lowest energy is then N −1 ⎛ N −1 ⎛ 1 ⎞⎞ 1⎤ ⎛ N −1 ⎞ ⎡ N −1 ⎡ ( N − 1)( N ) N ⎤ + ⎥= 2 ⎜ ∑ En ⎟ = 2 ⎜ ∑ =ω ⎜ n + ⎟ ⎟ = 2=ω ⎢ ∑ n + ∑ ⎥ = 2=ω ⎢ 2 ⎠⎠ 2 2⎦ ⎝ ⎣ n =0 2 ⎦ ⎝ n =0 ⎠ ⎣ n=0 ⎝ n=0

=ω [ N 2 − N + N ] = =ω N 2 = =N 2

41.62.

k′ . m

Here we used the hint from Problem 41.59 to do the first sum, realizing that the first value of n is zero and the last value of n is N – 1, giving us a total of N energy levels filled. (a) Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal force. There is an attractive (+2e)( −e) ( −e)(−e) force with charge +2e a distance r away and a repulsive force a distance 2r away. So, + = 4π P0 r 2 4π P0 (2r ) 2 − mv 2 = . But, from the quantization of angular momentum in the first Bohr orbit, L = mvr = = ⇒ v = . r mr 2

⎛ = ⎞ −m ⎜ 2 2 2 ⎟ e −2e −mv ⎝ mr ⎠ = − = ⇒ −7 e = − 4πP0= . So + = = 4 r2 mr 3 4πP0 r 2 4πP0 (4r ) 2 r r mr 3 2

2

2

4 ⎛ 4πP0= 2 ⎞ 4 4 −10 −11 r= ⎜ ⎟ = a0 = (0.529 × 10 m) = 3.02 × 10 m. 7 ⎝ me 2 ⎠ 7 7 −= 7 = 7 (1.054 × 10 −34 J ⋅ s) = = = 3.83 × 106 m s. And v = mr 4 ma0 4 (9.11 × 10−31 kg)(0.529 × 10−10 m) ⎛1 ⎞ (b) K = 2 ⎜ mv 2 ⎟ = 9.11 × 10−31 kg (3.83 × 106 m s) 2 = 1.34 × 10 −17 J = 83.5 eV. ⎝2 ⎠

Atomic Structure

41-13

⎛ −2e2 ⎞ −4e2 −7 ⎛ e 2 ⎞ e2 e2 −17 = + = (c) U = 2 ⎜ ⎟+ ⎜ ⎟ = −2.67 × 10 J = −166.9 eV 4 π r 4 π (2 r ) 4 π r 4 πE (2 r ) 2 4 π r P P P P ⎝ ⎝ 0 ⎠ 0 0 0 0 ⎠ (d) E∞ = −[ −166.9 eV + 83.5 eV] = 83.4 eV, which is only off by about 5% from the real value of 79.0 eV. 41.63.

(a) The radius is inversely proportional to Z, so the classical turning radius is 2a Z . (b) The normalized wave function is ψ1s ( r ) =

outside the classical turning point is P = ∫

∞

2a Z

1 πa Z 3

3

e − Z r a and the probability of the electron being found

2

ψ1s 4πr 2 dr = ∞

4 a Z3 3

∫

∞

2a Z

e −2 Zr a r 2dr. Making the change of variable

u = Zr a , dr = (a Z ) du changes the integral to P = 4∫ e −2uu 2du, which is independent of Z. The probability is 2

that found in Problem 41.39, 0.238, independent of Z.

42

MOLECULES AND CONDENSED MATTER

42.1.

3 2 K 2(7.9 × 10−4 eV)(1.60 × 10−19 J eV) (a) K = kT ⇒ T = = = 6.1 K 2 3k 3(1.38 × 10−23 J K) 2(4.48 eV) (1.60 × 10 −19 J eV) (b) T = = 34,600 K. 3(1.38 × 10−23 J K)

(c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of He 2 (calculated in part (a)), so the typical collision at room temperature will be more than enough to break up He2 .

42.2.

42.3.

However, the thermal energy at 300 K is much less than the bond energy of H 2 , so we would expect it to remain intact at room temperature. 1 e2 (a) U = − = −5.0 eV. 4πε0 r (b) −5.0 eV + (4.3 eV − 3.5 eV) = −4.2 eV. IDENTIFY: The energy given to the photon comes from a transition between rotational states. =2 SET UP: The rotational energy of a molecule is E = l (l + 1) and the energy of the photon is E = hc/λ. 2I EXECUTE: Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of the =2 5= 2 molecule is ΔE = [3(3 + 1) − 1(1 + 1) ] = . Since ΔE = hc/λ, we get hc/λ = 5 = 2 /I. Solving for I gives 2I I

I=

−34 5=λ 5 (1.055 × 10 J ⋅ s ) (1.780 nm) = = 4.981× 10−52 kg ⋅ m 2 . 2π c 2π ( 3.00 × 108 m/s )

Using I = mr r02, we can solve for r0: r0 =

I ( mN + mH ) = mN mH

( 4.981×10

−52

kg ⋅ m 2 )( 2.33 × 10−26 kg + 1.67 × 10−27 kg )

( 2.33 ×10

−26

kg )(1.67 × 10−27 kg )

42.4.

r0 = 5.65 × 10–13 m EVALUATE: This separation is much smaller than the diameter of a typical atom and is not very realistic. But we are treating a hypothetical NH molecule. The energy of the emitted photon is 1.01 × 10−5 eV, and so its frequency and wavelength are

42.5.

E (1.01 × 10−5 eV)(1.60 × 10−19 J eV) c (3.00 × 108 m s) and λ = = 2.44 GHz = = = 0.123 m. This frequency h (6.63 × 10−34 J ⋅ s) f (2.44 × 109 Hz) corresponds to that given for a microwave oven. Let 1 refer to C and 2 to O. m1 = 1.993 × 10−26 kg, m2 = 2.656 × 10−26 kg, r0 = 0.1128 nm . f =

⎛ m2 ⎞ ⎛ m1 ⎞ r1 = ⎜ ⎟ r0 = 0.0644 nm (carbon) ; r2 = ⎜ ⎟ r0 = 0.0484 nm (oxygen) ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠ (b) I = m1r12 + m2r22 = 1.45 × 10−46 kg ⋅ m 2 ; yes, this agrees with Example 42.2. 42.6.

Each atom has a mass m and is at a distance L 2 from the center, so the moment of inertia is 2( m)( L 2) 2 = mL2 2 = 2.21 × 10−44 kg ⋅ m 2 .

42.7.

IDENTIFY and SET UP: Set K = E1 from Example 42.2. Use K = 12 I ω 2 to solve for ω and v = rω to solve for v. EXECUTE: (a) From Example 42.2, E1 = 0.479 meV = 7.674 × 10−23 J and I = 1.449 × 10 −46 kg ⋅ m 2 K = 12 I ω 2 and K = E gives ω = 2 E1 / I = 1.03 × 1012 rad/s 42-1

42-2

Chapter 42

(b) v1 = r1ω1 = (0.0644 × 10 −9 m)(1.03 × 1012 rad/s) = 66.3 m/s (carbon) v2 = r2ω2 = (0.0484 × 10−9 m)(1.03 × 1012 rad/s) = 49.8 m/s (oxygen)

(c) T = 2π / ω = 6.10 × 10−12 s EVALUATE: From the information in Example 42.3 we can calculate the vibrational period to be T = 2π / ω = 2π mr / k ′ = 1.5 × 10−14 s. The rotational motion is over an order of magnitude slower than the

42.8. 42.9.

vibrational motion. 2 hc ⎛ 2πc ⎞ ΔE = = = k ′ mr , and solving for k ′, k ′ = ⎜ ⎟ mr = 205 N m. λ ⎝ λ ⎠ IDENTIFY and SET UP: The energy of a rotational level with quantum number l is El = l (l + 1)= 2 / 2 I (Eq.(42.3)). I = mr r 2 , with the reduced mass mr given by Eq.(42.4). Calculate I and Δ E and then use Δ E = hc / λ to find λ .

EXECUTE: (a) mr =

m1m2 mLi mH (1.17 × 10−26 kg)(1.67 × 10−27 kg) = = = 1.461 × 10−27 kg m1 + m2 mLi + mH 1.17 × 10 −26 kg + 1.67 × 10−27 kg

I = mr r 2 = (1.461 × 10−27 kg)(0.159 × 10 −9 m) 2 = 3.694 × 10 −47 kg ⋅ m 2

⎛ =2 ⎞ ⎛ =2 ⎞ l = 3 : E = 3(4) ⎜ ⎟ = 6 ⎜ ⎟ ⎝ 2I ⎠ ⎝ I ⎠ ⎛ =2 ⎞ ⎛ =2 ⎞ l = 4 : E = 4(5) ⎜ ⎟ = 10 ⎜ ⎟ ⎝ 2I ⎠ ⎝ I ⎠

42.10.

⎛ =2 ⎞ ⎛ (1.055 × 10−34 J ⋅ s) 2 ⎞ Δ E = E4 − E3 = 4 ⎜ ⎟ = 4 ⎜ = 1.20 × 10−21 J = 7.49 × 10−3 eV 2 ⎟ −47 I 3.694 10 kg m × ⋅ ⎝ ⎠ ⎝ ⎠ hc (4.136 × 10−15 eV)(2.998 × 108 m/s) = = 166 μ m (b) Δ E = hc / λ so λ = ΔE 7.49 × 10−3 eV EVALUATE: LiH has a smaller reduced mass than CO and λ is somewhat smaller here than the λ calculated for CO in Example 42.2 IDENTIFY: The vibrational energy of the molecule is related to its force constant and reduced mass, while the rotational energy depends on its moment of inertia, which in turn depends on the reduced mass. 1⎞ 1⎞ k′ =2 ⎛ ⎛ SET UP: The vibrational energy is En = ⎜ n + ⎟ =ω = ⎜ n + ⎟ = and the rotational energy is El = l (l + 1) . 2I 2⎠ 2 ⎠ mr ⎝ ⎝ EXECUTE:

For a vibrational transition, we have ΔEv = =

rotational transition is ΔER = mr r02 =

k′ , so we first need to find mr. The energy for a mr

=2 2= 2 [ 2(2 + 1) − 1(1 + 1)] = . Solving for I and using the fact that I = mrr02, we have 2I I

2= 2 , which gives ΔER

mr =

2 (1.055 × 10 −34 J ⋅ s )( 6.583 × 10 −16 eV ⋅ s ) 2= 2 = 2.0014 × 10–28 kg = 2 −9 −4 r02 ΔER 0.8860 10 m 8.841 10 eV × × ( )( )

Now look at the vibrational transition to find the force constant. ΔEv = =

42.11.

k′ mr

⇒ k′ =

2 ( 2.0014 ×10−28 kg ) (0.2560 eV) 2 = 30.27 N/m mr ( ΔEv ) = 2 =2 ( 6.583 ×10−16 eV ⋅ s )

EVALUATE: This would be a rather weak spring in the laboratory. l (l + 1)= 2 l (l − 1)= 2 =2 l= 2 , El −1 = (a) El = ⇒ ΔE = (l 2 + l − l 2 + l ) = 2I 2I 2I I Δ E ΔE l= = = . (b) f = h 2π= 2πI

Molecules and Condensed Matter

42.12.

IDENTIFY:

42-3

Find Δ E for the transition and compute λ from Δ E = hc / λ .

=2 =2 , with = 0.2395 × 10−3 eV. From Example 42.3, Δ E = 0.2690 eV 2I 2I is the spacing between vibrational levels. Thus En = ( n + 12 )=ω , with =ω = 0.2690 eV. By Eq.(42.9),

SET UP:

From Example 42.2, El = l (l + 1)

=2 . 2I (a) n = 0 → n = 1 and l = 1 → l = 2

E = En + El = ( n + 12 )=ω + l (l + 1)

EXECUTE:

⎛ =2 ⎞ For n = 0, l = 1, Ei = 12 =ω + 2 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 2, E f = 32 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω + 4 ⎜ ⎟ = 0.2690 eV + 4(0.2395 × 10−3 eV) = 0.2700 eV ⎝ 2I ⎠ hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s) hc = = 4.592 × 10−6 m = 4.592 μ m = Δ E so λ = ΔE 0.2700 eV λ (b) n = 0 → n = 1 and l = 2 → l = 1 ⎛ =2 ⎞ For n = 0, l = 2, Ei = 12 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 1, E f = 32 =ω + 2 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω − 4 ⎜ ⎟ = 0.2690 eV − 4(0.2395 × 10−3 eV) = 0.2680 eV ⎝ 2I ⎠ −15 hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s) = = 4.627 × 10−6 m = 4.627 μ m λ= ΔE 0.2680 eV (c) n = 0 → n = 1 and l = 3 → l = 2 ⎛ =2 ⎞ For n = 0, l = 3, Ei = 12 =ω + 12 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 2, E f = 32 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω − 6 ⎜ ⎟ = 0.2690 eV − 6(0.2395 × 10−3 eV) = 0.2676 eV ⎝ 2I ⎠ −15 hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s) = = 4.634 × 10−6 m = 4.634 μ m λ= ΔE 0.2676 eV EVALUATE: All three transitions are for n = 0 → n = 1. The spacing between vibrational levels is larger than the spacing between rotational levels, so the difference in λ for the various rotational transitions is small. When the transition is to a larger l, Δ E > =ω and when the transition is to a smaller l, Δ E < =ω. 42.13.

(a) IDENTIFY and SET UP: Use ω = k ′ / mr and ω = 2π f to calculate k ′. The atomic masses are used in

Eq.(42.4) to calculate mr . EXECUTE: mr =

f =

ω 2π

=

1 2π

k′ , so k ′ = mr (2π f ) 2 mr

m1m2 mH mF (1.67 × 10−27 kg)(3.15 × 10−26 kg) = = = 1.586 × 10−27 kg m1 + m2 mH + mF 1.67 × 10 −27 kg + 3.15 × 10−26 kg

k ′ = mr (2π f ) 2 = (1.586 × 10−27 kg)(2π [1.24 × 1014 Hz]) 2 = 963 N/m

(b) IDENTIFY and SET UP: The energy levels are given by Eq.(42.7). En = (n + 12 )=ω = ( n + 12 )hf , since =ω = (h / 2π )ω and (ω / 2π ) = f . The energy spacing between adjacent levels is

Δ E = En +1 − En = (n + 1 + 12 − n − 12 ) hf = hf , independent of n.

42-4

Chapter 42

EXECUTE: Δ E = hf = (6.626 × 10 −34 J ⋅ s)(1.24 × 1014 Hz) = 8.22 × 10 −20 J = 0.513 eV (c) IDENTIFY and SET UP: The photon energy equals the transition energy so Δ E = hc / λ . c 2.998 × 108 m/s = = 2.42 × 10−6 m = 2.42 μ m 1.24 × 1014 Hz f EVALUATE: This photon is infrared, which is typical for vibrational transitions. For an average spacing a, the density is ρ = m a 3 , where m is the average of the ionic masses, and so EXECUTE:

42.14.

hf = hc / λ so λ =

a3 =

42.15.

−26 −25 m ( 6.49 × 10 kg + 1.33 × 10 kg ) 2 = = 3.60 × 10 −29 m 3 , (2.75 × 103 kg m 3 ) ρ

and a = 3.30 × 10−10 m = 0.330 nm . (b) The larger (higher atomic number) atoms have the larger spacing. IDENTIFY and SET UP: Find the volume occupied by each atom. The density is the average mass of Na and Cl divided by this volume. EXECUTE: Each atom occupies a cube with side length 0.282 nm. Therefore, the volume occupied by each atom is V = (0.282 × 10−9 m)3 = 2.24 × 10−29 m 3. In NaCl there are equal numbers of Na and Cl atoms, so the average mass of the atoms in the crystal is m = 12 ( mNa + mCl ) = 12 (3.82 × 10−26 kg + 5.89 × 10−26 kg) = 4.855 × 10 −26 kg m 4.855 × 10−26 kg = = 2.17 × 103 kg/m3 . 2.24 × 10−29 m3 V EVALUATE: The density of water is 1.00 × 103 kg/m 3 , so our result is reasonable.

The density then is ρ =

42.16.

−34 8 hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) = = 0.200 nm. E ( 6.20 × 103 eV ) (1.60 × 10−19 J eV )

(a) As a photon, λ =

(b) As a matter wave,

λ=

( 6.63 ×10−34 J ⋅ s ) h h = = = 0.200 nm p 2mE 2 ( 9.11 × 10−31 kg ) ( 37.6 eV ) (1.60 × 10−19 J eV )

(c) As a matter wave,

λ= 42.17.

( 6.63 ×10−34 J ⋅ s ) h = = 0.200 nm . 2mE 2 (1.67 × 10−27 kg ) ( 0.0205 eV ) (1.60 × 10−19 J eV )

IDENTIFY: The energy gap is the energy of the maximum-wavelength photon. SET UP: The energy difference is equal to the energy of the photon, so ΔE = hc/λ. EXECUTE: (a) Using the photon wavelength to find the energy difference gives ΔE = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(1.11 × 10–6 m) = 1.12 eV

42.18.

42.19.

(b) A wavelength of 1.11 µm = 1110 nm is in the infrared, shorter than that of visible light. EVALUATE: Since visible photons have more than enough energy to excite electrons from the valence to the conduction band, visible light will be absorbed, which makes silicon opaque. hc (a) = 2.27 × 10−7 m = 227 nm , in the ultraviolet. ΔE (b) Visible light lacks enough energy to excite the electrons into the conduction band, so visible light passes through the diamond unabsorbed. (c) Impurities can lower the gap energy making it easier for the material to absorb shorter wavelength visible light. This allows longer wavelength visible light to pass through, giving the diamond color. hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) ΔE = = = 2.14 × 10−13 J = 1.34 × 106 eV . So the number of electrons that can be 9.31 × 10−13 m λ

excited to the conduction band is n = 42.20.

1.34 × 106 eV = 1.20 × 106 electrons 1.12 eV

1 = ∫ ψ dV 2

3 L ⎛L ⎛ n πy ⎞ ⎞ ⎛ L ⎛ n πx ⎞ ⎞ ⎛ ⎛ n πz ⎞ ⎞ ⎛ L⎞ = A2 ⎜ ∫ sin 2 ⎜ x ⎟ dx ⎟ ⎜⎜ ∫ sin 2 ⎜ y ⎟ dy ⎟⎟ ⎜ ∫ sin 2 ⎜ z ⎟ dz ⎟ = A2 ⎜ ⎟ ⎝ L ⎠ ⎠⎝0 ⎝ L ⎠ ⎠ ⎝2⎠ ⎝ L ⎠ ⎠⎝0 ⎝0

so A = ( 2 L )

32

(assuming A to be real positive).


42.21.

42-5

Density of states: g(E) =

( 2m )

32

2π =

V

2 3

E1 2 =

(2(9.11 × 10 −31 kg))3 2 (1.0 × 10 −6 m 3 )(5.0 eV)1 2 (1.60 × 10−19 J eV)1 2 2π 2 (1.054 × 10−34 J ⋅ s)3

g ( E ) = ( 9.5 × 1040 states J ) (1.60 × 10 −19 J eV ) = 1.5 × 1022 states eV. 42.22.

42.23.

vrms = 3kT m = 1.17 × 105 m s , as found in Example 42.9. The equipartition theorem does not hold for the electrons at the Fermi energy. Although these electrons are very energetic, they cannot lose energy, unlike electrons in a free electron gas. = 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞ + + (a) IDENTIFY and SET UP: The three-dimensional Schrödinger equation is − ⎜ ⎟ + Uψ = Eψ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ (Eq.40.29). For free electrons, U = 0. Evaluate ∂ 2ψ /∂x 2 , ∂ 2ψ /∂y 2 , and ∂ 2ψ / ∂z 2 for ψ as given by Eq.(42.10). Put the results into Eq.(40.20) and see if the equation is satisfied. ∂ψ nxπ ⎛ n πx⎞ ⎛ n π y ⎞ ⎛ n πz⎞ A cos ⎜ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟ EXECUTE: = L ∂x ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠ ∂ 2ψ ⎛nπ ⎞ ⎛n πx⎞ ⎛n π y⎞ ⎛ n πz ⎞ ⎛nπ ⎞ = − ⎜ x ⎟ A sin ⎜ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟ = − ⎜ x ⎟ ψ L L L L ∂x 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ L ⎠ ⎠ ⎝ 2

2

2

Similarly

⎛nπ ⎞ ∂ 2ψ ∂ 2ψ ⎛nπ ⎞ = − ⎜ y ⎟ ψ and = −⎜ z ⎟ ψ . 2 2 ∂y L z ∂ ⎝ L ⎠ ⎝ ⎠ 2

(nx2 + ny2 + nz2 )π 2= 2 ⎞ =2 ⎛ π 2 ⎞ 2 2 2 ( n n n ) = + + = ψ ψ ⎟ ⎜ 2⎟ x y z 2mL2 ⎠ 2m ⎝ L ⎠ ( n 2 + n y2 + nz2 )π 2= 2 This equals Eψ , with E = x , which is Eq.(42.11). 2mL2 EVALUATE: ψ given by Eq.(42.10) is a solution to Eq.(40.29), with E as given by Eq.(42.11). (b) IDENTIFY and SET UP: Find the set of quantum numbers nx , n y , and nz that give the lowest three values of Therefore, −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

E. The degeneracy is the number of sets nx , n y , nz and ms that give the same E. 3π 2= 2 . No other combination of nx , n y , and nz 2mL2 gives this same E, so the only degeneracy is the degeneracy of two due to spin. π 2= 2 6π 2= 2 = First excited level: next lower E so one n equals 2 and the others equal 1. E = (22 + 12 + 12 ) 2mL2 2mL2 There are three different sets of nx , n y , nz values that give this E:

EXECUTE: Ground level: lowest E so nx = n y = nz = 1 and E =

nx = 2, ny = 1, nz = 1; nx = 1, ny = 2, nz = 1; nx = 1, n y = 1, nz = 2 This gives a degeneracy of 3 so the total degeneracy, with the factor of 2 from spin, is 6. Second excited level: next lower E so two of nx , n y , nz equal 2 and the other equals 1. 9π 2= 2 2mL2 2mL2 There are different sets of nx , n y , nz values that give this E: E = (22 + 2 2 + 12 )

π 2= 2

=

nx = 2, ny = 2, nz = 1; nx = 2, ny = 1, nz = 2; nx = 1, ny = 2, nz = 2.

42.24. 42.25.

Thus, as for the first excited level, the total degeneracy, including spin, is 6. EVALUATE: The wavefunction for the 3-dimensional box is a product of the wavefunctions for a 1-dimensional box in the x, y, and z coordinates and the energy is the sum of energies for three 1-dimensional boxes. All levels except for the ground level have a degeneracy greater than two. Compare to the 3-dimensional isotropic harmonic oscillator treated in Problem 40.53. 12 Eq.(42.13) may be solved for nrs = ( 2mE ) ( L =π ), and substituting this into Eq. (42.12), using L3 = V , gives

Eq.(42.14). (a) IDENTIFY and SET UP: The electron contribution to the molar heat capacity at constant volume of a metal is ⎛ π 2 KT ⎞ CV = ⎜ ⎟ R. ⎝ 2 EF ⎠ π 2 (1.381 × 10 −23 J/K)(300 K) EXECUTE: CV = R = 0.0233R. 2(5.48 eV)(1.602 × 10−19 J/eV)

42-6

Chapter 42

(b) EVALUATE: The electron contribution found in part (a) is 0.0233R = 0.194 J/mol ⋅ K. This is 0.194 / 25.3 = 7.67 × 10−3 = 0.767% of the total CV . (c) Only a small fraction of CV is due to the electrons. Most of CV is due to the vibrational motion of the ions. 42.26.

3 (a) From Eq. (42.22), Eav = EF = 1.94 eV. 5 (b)

(c) 42.27.

42.28.

2E m =

2 (1.94 eV ) (1.60 × 10−19 J eV ) 9.11× 10−31 kg

= 8.25 × 105 m s.

−19 EF ( 3.23 eV ) (1.60 × 10 J eV ) = = 3.74 × 104 K. k (1.38 ×10−23 J K )

IDENTIFY: The probability is given by the Fermi-Dirac distribution. 1 SET UP: The Fermi-Dirac distribution is f ( E ) = ( E − EF ) / kT . e +1 EXECUTE: We calculate the value of f (E), where E = 8.520 eV, EF = 8.500 eV, k = 1.38 × 10–23 J/K = 8.625 × 10–5 eV/K, and T = 20°C = 293 K. The result is f (E) = 0.312 = 31.2%. EVALUATE: Since the energy is close to the Fermi energy, the probability is quite high that the state is occupied by an electron. (a) See Example 42.10: The probabilities are 1.78 × 10−7 , 2.37 × 10−6 , and 1.51 × 10 −5 . (b) The Fermi distribution, Eq.(42.17), has the property that f ( EF − E ) = 1 − f ( E ) (see Problem (42.48)), and so

42.29.

42.30.

the probability that a state at the top of the valence band is occupied is the same as the probability that a state of the bottom of the conduction band is filled (this result depends on having the Fermi energy in the middle of the gap). 1 IDENTIFY: Use Eq.(42.17), f ( E ) = ( E − EF ) / kT . Solve for E − EF . e +1 1 SET UP: e( E − EF ) / kT = −1 f (E) The problem states that f ( E ) = 4.4 × 10−4 for E at the bottom of the conduction band. 1 − 1 = 2.272 × 103. EXECUTE: e( E − EF ) / kT = 4.4 × 10−4 E − EF = kT ln(2.272 × 103 ) = (1.3807 × 10−23 J/T)(300 K)ln(2.272 × 103 ) = 3.201× 10−20 J = 0.20 eV EF = E − 0.20 eV; the Fermi level is 0.20 eV below the bottom of the conduction band. EVALUATE: The energy gap between the Fermi level and bottom of the conduction band is large compared to kT at T = 300 K and as a result f ( E ) is small. IDENTIFY: The current depends on the voltage across the diode and its temperature, so the resistance also depends on these quantities. SET UP: The current is I = IS (eeV/kT – 1) and the resistance is R = V/I. V V eV e(0.0850 V) = = EXECUTE: (a) The resistance is R = = . The exponent is I IS ( eeV / kT − 1) kT (8.625 × 10−5 eV/K ) (293 K) 3.3635, giving R =

85.0 mV = 4.06 Ω. (0.750 mA) ( e3.3635 − 1)

(b) In this case, the exponent is

which gives R =

42.31.

eV e(−0.050 V) = = −1.979 kT ( 8.625 × 10−5 eV/K ) (293 K)

−50.0 mV = 77.4 Ω (0.750 mA) ( e−1.979 − 1)

EVALUATE: Reversing the voltage can make a considerable change in the resistance of a diode. IDENTIFY and SET UP: The voltage-current relation is given by Eq.(42.23): I = Is (eeV / kT − 1). Use the current for

V = +15.0 mV to solve for the constant I s . EXECUTE: (a) Find Is : V = +15.0 × 10−3 V gives I = 9.25 × 10 −3 A eV (1.602 × 10−19 C)(15.0 × 10−3 V) = = 0.5800 kT (1.381 × 10 −23 J/K)(300 K) I 9.25 × 10 −3 A I s = eV / kT = = 1.177 × 10−2 = 11.77 mA e −1 e0.5800 − 1


Then can calculate I for V = 10.0 mV:

−19

42-7

−3

eV (1.602 × 10 C)(10.0 × 10 V) = = 0.3867 kT (1.381 × 10 −23 J/K)(300 K)

I = Is (eeV / kT − 1) = (11.77 mA)(e0.3867 − 1) = 5.56 mA eV eV has the same magnitude as in part (a) but not V is negative so is negative. kT kT eV V = −15.0 mV : = −0.5800 and I = I s (eeV / kT − 1) = (11.77 mA)(e −0.5800 − 1) = −5.18 mA kT eV V = −10.0 mV : = −0.3867 and I = I s (eeV / kT − 1) = (11.77 mA)(e−0.3867 − 1) = −3.77 mA kT EVALUATE: There is a directional asymmetry in the current, with a forward-bias voltage producing more current than a reverse-bias voltage of the same magnitude, but the voltage is small enough for the asymmetry not be pronounced. Compare to Example 42.11, where more extreme voltages are considered. (a) Solving Eq.(42.23) for the voltage as a function of current, (b)

42.32.

V=

42.33.

⎞ kT ⎛ 40.0 mA ⎞ kT ⎛ I ln ⎜ + 1⎟ = ln ⎜ + 1⎟ = 0.0645 V. e ⎝ 3.60 mA ⎠ ⎝ IS ⎠ e

(b) From part (a), the quantity eeV kT = 12.11 , so far a reverse-bias voltage of the same magnitude, ⎛ 1 ⎞ I = I S ( e − eV kT − 1) = I S ⎜ − 1 ⎟ = −3.30 mA . ⎝ 12.11 ⎠ IDENTIFY: During the transition, the molecule emits a photon of light having energy equal to the energy difference between the two vibrational states of the molecule. 1⎞ 1⎞ k′ ⎛ ⎛ SET UP: The vibrational energy is En = ⎜ n + ⎟ =ω = ⎜ n + ⎟ = . 2⎠ 2 ⎠ mr ⎝ ⎝ EXECUTE: (a) The energy difference between two adjacent energy states is ΔE = =

k′ , and this is the energy of mr 2

⎛ ΔE ⎞ the photon, so ΔE = hc/λ. Equating these two expressions for ΔE and solving for k ′ , we have k ′ = mr ⎜ ⎟ = ⎝ = ⎠ 2 ΔE hc / λ 2π c mH mO ⎛ ΔE ⎞ = = with the appropriate numbers gives us ⎜ ⎟ , and using = = λ mH + mO ⎝ = ⎠

k′ =

ω 1 = (b) f = 2π 2π

(1.67 ×10

−27

1.67 × 10−27

k′ 1 = mr 2π

kg )( 2.656 × 10−26 kg ) ⎡ 2π ( 3.00 × 108 m/s ) ⎤ ⎢ ⎥ = 977 N/m kg + 2.656 × 10−26 kg ⎢ 2.39 × 10−6 m ⎥ ⎣ ⎦ 2

mH mO mH + mO . Substituting the appropriate numbers gives us k′

(1.67 ×10 f =

42.34. 42.35.

1 2π

−27

kg )( 2.656 × 10−26 kg )

1.67 × 10−27 kg + 2.656 × 10 −26 kg = 1.25 × 1014 Hz 977 N/m

EVALUATE: The frequency is close to, but not quite in, the visible range. 2= 2 hλ I= = = 7.14 × 10−48 kg ⋅ m 2 . ΔE 2π 2c IDENTIFY and SET UP: Eq.(21.14) gives the electric dipole moment as p = qd , where the dipole consists of charges ± q separated by distance d. EXECUTE: (a) Point charges +e and −e separated by distance d, so p = ed = (1.602 × 10−19 C)(0.24 × 10−9 m) = 3.8 × 10−29 C ⋅ m p 3.0 × 10−29 C ⋅ m = = 1.3 × 10−19 C d 0.24 × 10−9 m q 1.3 × 10−19 C = 0.81 (c) = e 1.602 × 10−19 C

(b) p = qd so q =

42-8

Chapter 42

p 1.5 × 10 −30 C ⋅ m = = 9.37 × 10 −21 C d 0.16 × 10 −9 m q 9.37 × 10−21 C = = 0.058 e 1.602 × 10−19 C EVALUATE: The fractional ionic character for the bond in HI is much less than the fractional ionic character for the bond in NaCl. The bond in HI is mostly covalent and not very ionic. 1 e2 The electrical potential energy is U = −5.13 eV, and r = − = 2.8 × 10−10 m. 4πP0 U

(d) q =

42.36. 42.37.

(a) IDENTIFY:

E (Na) + E (Cl) = E (Na + ) + E (Cl− ) + U ( r ). Solving for U ( r ) gives

U ( r ) = −[ E (Na + ) − E (Na)] + [ E (Cl) − E (Cl− )].

SET UP:

[ E (Na + ) − E (Na)] is the ionization energy of Na, the energy required to remove one electron, and is

equal to 5.1 eV. [ E (Cl) − E (Cl− )] is the electron affinity of Cl, the magnitude of the decrease in energy when an electron is attached to a neutral Cl atom, and is equal to 3.6 eV. 1 e2 EXECUTE: U = −5.1 eV + 3.6 eV = −1.5 eV = −2.4 × 10−19 J, and − = −2.4 × 10−19 J 4π P0 r −19 ⎛ 1 ⎞ C) 2 e2 9 2 2 (1.602 × 10 = × ⋅ (8.988 10 N m /C ) r =⎜ ⎟ −19 2.4 × 10 −19 J ⎝ 4π P0 ⎠ 2.4 × 10 J r = 9.6 × 10−10 m = 0.96 nm (b) ionization energy of K = 4.3 eV; electron affinity of Br = 3.5 eV

Thus U = −4.3 eV + 3.5 eV = −0.8 eV = −1.28 × 10 −19 J, and −

42.38.

1 e2 = −1.28 × 10 −19 J 4π P0 r

⎛ 1 ⎞ (1.602 × 10−19 C) 2 e2 = (8.988 × 109 N ⋅ m 2 / C2 ) r =⎜ ⎟ −19 1.28 × 10−19 J ⎝ 4π P0 ⎠ 1.28 × 10 J r = 1.8 × 10−9 m = 1.8 nm EVALUATE: K has a smaller ionization energy than Na and the electron affinities of Cl and Br are very similar, so it takes less energy to make K + + Br − from K + Br than to make Na + + Cl− from Na + Cl. Thus, the stabilization distance is larger for KBr than for NaCl. The energies corresponding to the observed wavelengths are 3.29 × 10 −21 J, 2.87 × 10−21 J, 2.47 × 10 −21 J, 2.06 × 10−21 J and 1.65 × 10−21 J. The average spacing of these energies is 0.410 × 10−21 J and these are seen to

correspond to transition from levels 8, 7, 6, 5 and 4 to the respective next lower levels. Then,

42.39.

=2 = 0.410 × 10−21 J , I

from which I = 2.71 × 10 −47 kg ⋅ m 2 . (a) IDENTIFY: The rotational energies of a molecule depend on its moment of inertia, which in turn depends on the separation between the atoms in the molecule. SET UP: Problem 42.38 gives I = 2.71 × 10 −47 kg ⋅ m 2 . I = mr r 2 . Calculate mr and solve for r. EXECUTE:

mr =

mH mCl (1.67 × 10−27 kg)(5.81 × 10−26 kg) = = 1.623 × 10−27 kg mH + mCl 1.67 × 10−27 kg + 5.81 × 10−26 kg

I 2.71 × 10 −47 kg ⋅ m 2 = = 1.29 × 10 −10 m = 0.129 nm mr 1.623 × 10−27 kg EVALUATE: This is a typical atomic separation for a diatomic molecule; see Example 42.2 for the corresponding distance for CO. (b) IDENTIFY: Each transition is from the level l to the level l − 1. The rotational energies are given by Eq.(42.3). The transition energy is related to the photon wavelength by Δ E = hc / λ . r=

⎛ =2 ⎞ ⎛ =2 ⎞ El = l (l + 1)= 2 / 2 I , so Δ E = El − El −1 = [l (l + 1) − l (l − 1)] ⎜ ⎟ = l ⎜ ⎟ . ⎝ 2I ⎠ ⎝ I ⎠ 2 ⎛ = ⎞ hc EXECUTE: l ⎜ ⎟ = ⎝ I ⎠ λ SET UP:

l=

2π cI 2π (2.998 × 108 m/s)(2.71 × 10−47 kg ⋅ m 2 ) 4.843 × 10 −4 m = = (1.055 × 10 −34 J ⋅ s)λ λ =λ


42-9

4.843 × 10−4 m = 8. 60.4 × 10−6 m 4.843 × 10−4 m = 7. For λ = 69.0 μ m, l = 69.0 × 10−6 m 4.843 × 10−4 m = 6. For λ = 80.4 μ m, l = 80.4 × 10−6 m 4.843 × 10−4 m = 5. For λ = 96.4 μ m, l = 96.4 × 10−6 m 4.843 × 10−4 m = 4. For λ = 120.4 μ m, l = 120.4 × 10−6 m EVALUATE: In each case l is an integer, as it must be. (c) IDENTIFY and SET UP: Longest λ implies smallest Δ E , and this is for the transition from l = 1 to l = 0. For λ = 60.4 μ m, l =

⎛ =2 ⎞ (1.055 × 10−34 J ⋅ s) 2 Δ E = l ⎜ ⎟ = (1) = 4.099 × 10−22 J −47 2 2.71 10 kg m × ⋅ I ⎝ ⎠ hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) λ= = = 4.85 × 10−4 m = 485 μ m. ΔE 4.099 × 10−22 J EVALUATE: This is longer than any wavelengths in part (b). (d) IDENTIFY: What changes is mr , the reduced mass of the molecule. EXECUTE:

⎛ =2 ⎞ hc 2π cI SET UP: The transition energy is Δ E = l ⎜ ⎟ and Δ E = , so λ = (part (b)). I = mr r 2 , so λ is directly l= λ ⎝ I ⎠ λ (HCl) λ (DCl) m (DCl) proportional to mr . = so λ (DCl) = λ (HCl) r mr (HCl) mr (DCl) mr (HCl) EXECUTE: The mass of a deuterium atom is approximately twice the mass of a hydrogen atom, so mD = 3.34 × 10−27 kg. mr (DCl) =

m D m Cl (3.34 × 10−27 kg)(5.81 × 10 −27 kg) = = 3.158 × 10 −27 kg m D + m Cl 3.34 × 10−27 kg + 5.81 × 10 −26 kg

⎛ 3.158 × 10 −27 kg ⎞ ⎟ = (1.946)λ (HCl) −27 ⎝ 1.623 × 10 kg ⎠ l = 8 → l = 7; λ = (60.4 μ m)(1.946) = 118 μ m l = 7 → l = 6; λ = (69.0 μ m)(1.946) = 134 μ m l = 6 → l = 5; λ = (80.4 μ m)(1.946) = 156 μ m l = 5 → l = 4; λ = (96.4 μ m)(1.946) = 188 μ m l = 4 → l = 3; λ = (120.4 μ m)(1.946) = 234 μ m EVALUATE: The moment of inertia increases when H is replaced by D, so the transition energies decrease and the wavelengths increase. The larger the rotational inertia the smaller the rotational energy for a given l (Eq.42.3). = 2l hl λ From the result of Problem 42.11, the moment inertia of the molecule is I = = = 6.43 × 10−46 kg ⋅ m 2 and ΔE 4π 2c

λ (DCl) = λ (HCl) ⎜

42.40.

from Eq.(42.6) the separation is r0 = 42.41.

I = 0.193 nm. mr

L2 = 2l (l + 1) = . Eg = 0 (l = 0), and there is an additional multiplicative factor of 2l + 1 because for each l 2I 2I 2 n state there are really (2l + 1) ml -states with the same energy. So l = (2l + 1)e − = l ( l +1) /(2 IkT ) . n0 (a) Eex =

(b) T = 300 K, I = 1.449 × 10 −46 kg ⋅ m 2 .

= 2 (1) (1 + 1) E 7.67 × 10−23 J = 7.67 × 10−23 J. l =1 = = 0.0185. 2 −46 2(1.449 × 10 kg ⋅ m ) kT (1.38 × 10−23 J K) (300 K) n (2l + 1) = 3 , so l =1 = (3)e −0.0185 = 2.95. n0

(i) El =1 =

42-10

Chapter 42

El = 2 = 2 (2) (2 + 1) = = 0.0556. 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10 −23 J K) (300 K) kT n (2l + 1) = 5 , so l =1 = (5)(e −0.0556 ) = 4.73. n0

(ii)

(iii)

El =10 = 2 (10) (10 + 1) = = 1.02. kT 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

(2l + 1) = 21, so (iv)

nl =10 = (21) (e −1.02 ) = 7.57. n0

El = 20 = 2 (20) (20 + 1) = = 3.89. kT 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

(2l + 1) = 41 , so

(v)

nl = 20 = (41)e −3.89 = 0.838. n0

El = 50 = 2 (50) (50 + 1) = = 23.6. −46 kT 2(1.449 × 10 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

nl =50 = (101)e −23.6 = 5.69 × 10−9. n0 (c) There is a competing effect between the (2l + 1) term and the decaying exponential. The 2l + 1 term dominates for small l, while the exponential term dominates for large l. (a) I CO = 1.449 × 10−46 kg ⋅ m 2 . (2l + 1) = 101 , so

42.42.

El =1 =

= 2l (l + 1) (1.054 × 10−34 J ⋅ s) 2 (1) (1 + 1) = = 7.67 × 10−23 J . El = 0 = 0. 2I 2(1.449 × 10−46 kg ⋅ m 2 )

ΔE = 7.67 × 10−23 J = 4.79 × 10 −4 eV.

hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = = 2.59 × 10−3 m = 2.59 mm. ΔE (7.67 × 10−23 J) (b) Let’s compare the value of kT when T = 20 K to that of ΔE for the l = 1 → l = 0 rotational transition:

λ=

kT = (1.38 × 10−23 J K) (20 K) = 2.76 × 10−22 J. kT = 3.60. ΔE Therefore, although T is quite small, there is still plenty of energy to excite CO molecules into the first rotational level. This allows astronomers to detect the 2.59 mm wavelength radiation from such molecular clouds. IDENTIFY and SET UP: El = l (l + 1)= 2 / 2 I , so El and the transition energy Δ E depend on I. Different isotopic molecules have different I. mNa mCl (3.8176 × 10 −26 kg)(5.8068 × 10−26 kg) EXECUTE: (a) Calculate I for Na 35Cl: mr = = = 2.303 × 10 −26 kg mNa + mCl 3.8176 × 10−26 kg + 5.8068 × 10 −26 kg ΔE = 7.67 × 10−23 J (from part (a)). So

42.43.

I = mr r 2 = (2.303 × 10−26 kg)(0.2361× 10−9 m)2 = 1.284 × 10−45 kg ⋅ m 2 l = 2 → l = 1 transition ⎛ = 2 ⎞ 2= 2 2(1.055 × 10−34 J ⋅ s) 2 Δ E = E2 − E1 = (6 − 2) ⎜ ⎟ = = = 1.734 × 10−23 J −45 2 2 I I 1.284 10 kg m × ⋅ ⎝ ⎠ hc hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) ΔE = so λ = = = 1.146 × 10 −2 m = 1.146 cm λ ΔE 1.734 × 10 −23 J l = 1 → l = 0 transition ⎛ =2 ⎞ =2 1 Δ E = E1 − E0 = (2 − 0) ⎜ ⎟ = = (1.734 × 10−23 J) = 8.67 × 10−24 J 2 ⎝ 2I ⎠ I hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = = 2.291 cm λ= ΔE 8.67 × 10−24 J mNa mCl (3.8176 × 10 −26 kg)(6.1384 × 10−26 kg) (b) Calculate I for Na 37Cl: mr = = = 2.354 × 10−26 kg mNa + mCl 3.8176 × 10 −26 kg + 6.1384 × 10−26 kg

I = mr r 2 = (2.354 × 10−26 kg)(0.2361 × 10−9 m) 2 = 1.312 × 10−45 kg ⋅ m 2


42-11

l = 2 → l = 1 transition 2= 2 2(1.055 × 10−34 J ⋅ s) 2 = = 1.697 × 10 −23 J 1.312 × 10 −45 kg ⋅ m 2 I hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.171 × 10−2 m = 1.171 cm λ= ΔE 1.697 × 10−23 J ΔE =

l = 1 → l = 0 transition

=2 1 = (1.697 × 10−23 J) = 8.485 × 10−24 J I 2 hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 2.341 cm λ= ΔE 8.485 × 10−24 J The differences in the wavelengths for the two isotopes are: l = 2 → l = 1 transition: 1.171 cm − 1.146 cm = 0.025 cm l = 1 → l = 0 transition: 2.341 cm − 2.291 cm = 0.050 cm EVALUATE: Replacing 35 Cl by 37 Cl increases I, decreases Δ E and increases λ . The effect on λ is small but measurable. ΔE The vibration frequency is, from Eq.(42.8), f = = 1.12 × 1014 Hz. The force constant is h k ′ = (2πf ) 2 mr = 777 N m. ΔE =

42.44.

42.45.

1⎞ k′ 1 2k ′ ⎛ En = ⎜ n + ⎟ = ⇒ E0 = = 2⎠ mr 2 mH ⎝ 1 2(576 N m ) ⇒ E0 = (1.054 × 10−34 J ⋅ s) = 4.38 × 10−20 J = 0.274 eV. 2 1.67 × 10−27 kg

42.46.

This is much less than the H 2 bond energy. (a) The frequency is proportional to the reciprocal of the square root of the reduced mass, and in terms of the atomic masses, the frequency of the isotope with the deuterium atom is 12

⎛ m m ( mH + mF ) ⎞ f = f0 ⎜ F H ⎟ ⎝ mF mD ( mD + mF ) ⎠ 42.47.

12

⎛ 1 + ( mF mD ) ⎞ = f0 ⎜ ⎟ . ⎝ 1 + (mF mH ) ⎠

Using f 0 from Exercise 42.13 and the given masses, f = 8.99 × 1013 Hz. IDENTIFY and SET UP: Use Eq.(42.6) to calculate I. The energy levels are given by Eq.(42.9). The transition energy Δ E is related to the photon wavelength by Δ E = hc / λ . EXECUTE: (a) mr =

mH mI (1.67 × 10−27 kg)(2.11 × 10−25 kg) = = 1.657 × 10−27 kg mH + mI 1.67 × 10 −27 kg + 2.11 × 10−25 kg

I = mr r 2 = (1.657 × 10 −27 kg)(0.160 × 10−9 m) 2 = 4.24 × 10−47 kg ⋅ m 2

⎛ =2 ⎞ k′ (b) The energy levels are Enl = l (l + 1) ⎜ ⎟ + ( n + 12 )= (Eq.(42.9)) I m 2 ⎝ ⎠ r ⎛ =2 ⎞ k′ = ω = 2π f so Enl = l (l + 1) ⎜ ⎟ + ( n + 12 )hf m ⎝ 2I ⎠

(i) transition n = 1 → n = 0, l = 1 → l = 0 ⎛ =2 ⎞ =2 Δ E = (2 − 0) ⎜ ⎟ + (1 + 12 − 12 )hf = + hf I ⎝ 2I ⎠ hc hc hc c ΔE = so λ = = = λ Δ E (= 2 / I ) + hf (= / 2π I ) + f =

2π I

λ=

=

1.055 × 10−34 J ⋅ s = 3.960 × 1011 Hz 2π (4.24 × 10−47 kg ⋅ m 2 )

c 2.998 × 108 m/s = = 4.30 μ m (= / 2π I ) + f 3.960 × 1011 Hz + 6.93 × 1013 Hz

42-12

Chapter 42

(ii) transition n = 1 → n = 0, l = 2 → l = 1 ⎛ =2 ⎞ 2= 2 Δ E = (6 − 2) ⎜ ⎟ + hf = + hf I ⎝ 2I ⎠ c 2.998 × 108 m/s λ= = = 4.28 μ m 2(= / 2π I ) + f 2(3.960 × 1011 Hz) + 6.93 × 1013 Hz (iii) transition n = 2 → n = 1, l = 2 → l = 3

42.48. 42.49.

⎛ =2 ⎞ 3= 2 Δ E = (6 − 12) ⎜ ⎟ + hf = − + hf I ⎝ 2I ⎠ c 2.998 × 108 m/s = = 4.40 μ m λ= −3(= / 2π I ) + f −3(3.960 × 1011 Hz) + 6.93 × 1013 Hz EVALUATE: The vibrational energy change for the n = 1 → n = 0 transition is the same as for the n = 2 → n = 1 transition. The rotational energies are much smaller than the vibrational energies, so the wavelengths for all three transitions don’t differ much. 1 1 1 e −ΔE / kT The sum of the probabilities is f ( EF + ΔE ) + f ( EF − ΔE ) = −ΔE kT + ΔE kT = −ΔE kT + = 1. e +1 e +1 e + 1 1 + e −ΔE kT 32 3 π 4 3= 2n 2 3 Since potassium is a metal we approximate EF = EF0 . ⇒ EF = . 2m ρ 851 kg m 3 = 1.31 × 1028 electron m3 But the electron concentration n = ⇒ n = m 6.49 × 10−26 kg

32 3 π 4 3 (1.054 × 10 −34 J ⋅ s) 2 (1.31 × 1028 /m3 ) 2 3 = 3.24 × 10−19 J = 2.03 eV. 2(9.11 × 10−31 kg) IDENTIFY: The only difference between the two isotopes is their mass, which will affect their reduced mass and hence their moment of inertia. =2 SET UP: The rotational energy states are given by E = l (l + 1) and the reduced mass is given by mr = 2I m1m2/(m1 + m2). EXECUTE: (a) If we call m the mass of the H-atom, the mass of the deuterium atom is 2m and the reduced masses of the molecules are H2 (hydrogen): mr(H) = mm/(m + m) = m/2 D2 (deuterium): mr(D) = (2m)(2m)/(2m + 2m) = m Using I = mr r02, the moments of inertia are IH = mr02/2 and ID = mr02. The ratio of the rotational energies is then ⇒ EF =

42.50.

2 EH l (l + 1) ( = / 2 I H ) I D mr02 = = = = 2. ED l (l + 1) ( = 2 / 2 I D ) I H m r 2 0 2

1⎞ k′ ⎛ ⎜ n + ⎟= m 2 E mr (D) m ⎝ ⎠ r (H) = = = 2. (b) The ratio of the vibrational energies is H = ED ⎛ mr (H) m/2 k′ 1⎞ ⎜ n + ⎟= 2 ⎠ mr (D) ⎝

41.51.

EVALUATE: The electrical force is the same for both molecules since both H and D have the same charge, so it is reasonable that the force constant would be the same for both of them. IDENTIFY and SET UP: Use the description of the bcc lattice in Fig.42.12c in the textbook to calculate the number of atoms per unit cell and then the number of atoms per unit volume. EXECUTE: (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells. So there are 1 + 8/ 8 = 2 atoms per unit cell. 2 n = = 4.66 × 10 28 atoms/m3 V (0.35 × 10−9 m)3 (b) EF0 =

32 / 3 π 4 / 3= 2 ⎛ N ⎞ ⎜ ⎟ 2m ⎝V ⎠

2/3

In this equation N/V is the number of free electrons per m3 . But the problem says to assume one free electron per atom, so this is the same as n/V calculated in part (a). m = 9.109 × 10 −31 kg (the electron mass), so EF0 = 7.563 × 10 −19 J = 4.7 eV EVALUATE: Our result for metallic lithium is similar to that calculated for copper in Example 42.8.


42.52.

(a)

d αe 2 1 1 8 A4π P0 U tot = − 8 A 9 . Setting this equal to zero when r = r0 gives r07 = dr 4πP0 r 2 r αe 2

and so U tot =

42.53.

42-13

αe 2 ⎛ 1 r07 ⎞ 7 αe 2 = −1.26 × 10−18 J = −7.85 eV. ⎜ − + 8 ⎟ . At r = r0 , U tot = − 4πP0 ⎝ r 8r ⎠ 32πP0r0

(b) To remove a Na + Cl− ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are formed from the Na + and Cl− atoms there is a net release of energy −5.14 eV + 3.61 eV = −1.53 eV, so the net energy required to remove a neutral Na, Cl pair from the crystal is 7.85 eV − 1.53 eV = 6.32 eV. dE (a) IDENTIFY and SET UP: p = − tot . Relate E tot to EF0 and evaluate the derivative. dV 3N 3 ⎛ 32 / 3 π 4 / 3= 2 ⎞ 5 / 3 −2 / 3 EF0 = ⎜ EXECUTE: Etot = NEav = ⎟N V 5 5⎝ 2m ⎠

⎛ 32 / 3 π 4 / 3= 2 ⎞ ⎛ N ⎞ dEtot 3 ⎛ 32 / 3 π 4 / 3= 2 ⎞ 5 / 3 ⎛ 2 −5 / 3 ⎞ = ⎜ ⎟N ⎜− V ⎟⎜ ⎟ ⎟ so p = ⎜ 5⎝ 2m dV 5m ⎝ 3 ⎠ ⎠ ⎝ ⎠⎝ V ⎠

5/3

, as was to be shown.

(b) N / V = 8.45 × 10 28 m −3 ⎛ 32 / 3 π 4 / 3 (1.055 × 10 −34 J ⋅ s) 2 ⎞ 28 10 5 −3 5 / 3 p=⎜ ⎟ (8.45 × 10 m ) = 3.81 × 10 Pa = 3.76 × 10 atm. −31 5(9.109 10 kg) × ⎝ ⎠

42.54.

(c) EVALUATE: Normal atmospheric pressure is about 105 Pa, so these pressures are extremely large. The electrons are held in the metal by the attractive force exerted on them by the copper ions. 53 ⎡ 5 32 3 π 4 3= 2 ⎛ N ⎞ 2 3 ⎛ − N ⎞ ⎤ 5 dp 32 3 π 4 3= 2 ⎛ N ⎞ = −V ⎢ ⋅ ⋅ ⎜ ⎟ ⎜ 2 ⎟ ⎥ = p. (a) From Problem 42.53, p = ⎜ ⎟ . B = −V dV 5m 5m ⎝ V ⎠ ⎝ V ⎠ ⎝ V ⎠ ⎦⎥ 3 ⎣⎢ 3

N 5 3 2 3 π 4 3= 2 = 8.45 × 1028 m −3 . B = ⋅ (8.45 × 10 28 m −3 )5 3 = 6.33 × 1010 Pa. V 3 5m 6.33 × 1010 Pa (c) = 0.45. The copper ions themselves make up the remaining fraction. 1.4 × 1011 Pa (b)

42.55.

(a) EF0 =

32 3 π 4 3= 2 2m

23

1 ⎛N⎞ mc 2 . ⎜ ⎟ . Let EF0 = 100 V ⎝ ⎠ 32

⎤ 2m 2c 2 23 2 m3c3 23 2 m3c 3 ⎛N⎞ ⎡ = = = = 1.67 × 1033 m −3 . ⎜ ⎟ ⎢ 23 43 2⎥ 1003 23π 2= 3 3000π 2= 3 ⎝ V ⎠ ⎣ (100)3 π = ⎦ 8.45 × 10 28 m −3 (b) = 5.06 × 10 −5. Since the real concentration of electrons in copper is less than one part in 10 −4 of the 1.67 × 1033 m −3 concentration where relativistic effects are important, it is safe to ignore relativistic effects for most applications. 6(2 × 1030 kg) (c) The number of electrons is N e = = 6.03 × 1056. The concentration is 1.99 × 10−26 kg Ne 6.03 × 1056 =4 = 6.66 × 1035 m −3 . V π (6.00 × 106 m)3 3 6.66 × 1035 m −3 ≅ 400 so relativistic effects will be very important. 1.67 × 1033 m −3 IDENTIFY: The current through the diode is related to the voltage across it. SET UP: The current through the diode is given by I = IS (eeV/kT – 1). EXECUTE: (a) The current through the resistor is (35.0 V)/(125 Ω) = 0.280 A = 280 mA, which is also the current through the diode. This current is given by I = IS (eeV/kT – 1), giving 280 mA = 0.625 mA(eeV/kT – 1) and 1 + −23 kT ln 449 (1.38 × 10 J/K ) (293 K)ln 449 (280/0.625) = 449 = eeV/kT. Solving for V at T = 293 K gives V = = = e 1.60 × 10−19 C 0.154 V (b) R = V/I = (0.154 V)/(0.280 A) = 0.551 Ω EVALUATE: At a different voltage, the diode would have different resistance.

(d) Comparing this to the result from part (a)

42.56.

42-14

Chapter 42

42.57.

(a) U =

qi q j q 2 ⎛ −1 1 1 1 1 1 1 ⎞ q2 ⎛ 2 2 1 1 ⎞ = + − − + − ⎟= − ∑ ⎟. ⎜ ⎜ − − 4πP0 i < j rij 4πP0 ⎝ d r r + d r − d r d ⎠ 4πP0 ⎝ r d r + d r − d ⎠

⎛ ⎞ 1 1 1⎜ 1 1 ⎟ 1⎛ d d2 d d 2 ⎞ 2 2d 2 But + = ⎜ + ≈ ⎜ 1 − + 2 + ⋅ ⋅⋅ + 1 + + 2 ⎟ ≈ + 3 ⎟ r + d r − d r ⎜1+ d 1− d ⎟ r ⎝ r r r r ⎠ r r ⎝ r r⎠ −2q 2 ⎛ 1 d 2 ⎞ −2 p 2 2 p2 ⇒U = − . ⎜ + 3 ⎟= 3 4πP0 ⎝ d r ⎠ 4πP0 r 4πP0 d 3

(b) U =

qi q j q 2 ⎛ −1 1 1 1 1 1 1 ⎞ q 2 ⎛ −2 2 2 2d 2 ⎞ = − + + − − ⎟= − + + 3 ⎟= ⎜ ∑ ⎜ r ⎠ 4πP0 i < j rij 4πP0 ⎝ d r r + d r − d r d ⎠ 4πP0 ⎝ d r r

−2 p 2 −2q 2 ⎛ 1 d 2 ⎞ 2 p2 + . ⎜ − 3 ⎟ ⇒U = 4πP0 d 3 4πP0 r 3 4πP0 ⎝ d r ⎠ If we ignore the potential energy involved in forming each individual molecule, which just involves a different choice for the zero of potential energy, then the answers are: −2 p 2 . The interaction is attractive. (a) U = 4πP0 r 3 (b) U = 42.58.

+2 p 2 . The interaction is repulsive. 4πP0 r 3

⎛ 1 e2 ⎞ 1 e2 1 e2 ′ and (a) Following the hint, k ′dr = − d ⎜ = = = dr = ω = 2 k m = = ⎟ 2 3 πP0 mr03 ⎝ 4πP0 r ⎠ r = r0 2πP0 r0 1.23 × 10−19 J = 0.77 eV, where ( m 2) has been used for the reduced mass.

(b) The reduced mass is doubled, and the energy is reduced by a factor of

2 to 0.54 eV.

NUCLEAR PHYSICS

43.1.

43.2.

(a)

28 14

Si has 14 protons and 14 neutrons.

(b)

85 37

Rb has 37 protons and 48 neutrons.

(c)

205 81

43

Tl has 81 protons and 124 neutrons.

(a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. (b) Using 4π R 2 for each of the radii in part (a), the areas are 163 fm 2 , 353 fm 2 and 633 fm 2 .

4 3 π R gives 195 fm3 , 624 fm 3 and 1499 fm3 . 3 (d) The density is the same, since the volume and the mass are both proportional to A: 2.3 × 1017 kg m3 (see Example 43.1). (c)

3

43.3.

(e) Dividing the result of part (d) by the mass of a nucleon, the number density is 0.14 fm3 = 1.40 × 1044 m . IDENTIFY: Calculate the spin magnetic energy shift for each spin state of the 1s level. Calculate the energy splitting between these states and relate this to the frequency of the photons. SET UP: When the spin component is parallel to the field the interaction energy is U = − μ z B. When the spin

component is antiparallel to the field the interaction energy is U = + μ z B. The transition energy for a transition between these two states is Δ E = 2μ z B, where μ z = 2.7928μ n . The transition energy is related to the photon frequency by Δ E = hf , so 2 μ z B = hf .

hf (6.626 × 10−34 J ⋅ s)(22.7 × 106 Hz) = = 0.533 T 2μ z 2(2.7928)(5.051× 10−27 J/T) EVALUATE: This magnetic field is easily achievable. Photons of this frequency have wavelength λ = c/f = 13.2 m. These are radio waves. G G (a) As in Example 43.2, ΔE = 2(1.9130)(3.15245 × 10 −8 eV T)(2.30 T) = 2.77 × 10−7 eV. Since μ and S are in opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons. ΔE c (b) f = = 66.9 MHz, λ = = 4.48 m. h f IDENTIFY: Calculate the spin magnetic energy shift for each spin component. Calculate the energy splitting between these states and relate this to the frequency of the photons. G G G (a) SET UP: From Example 43.2, when the z-component of S (and μ ) is parallel to B, U = − | μ z | B = G G G −2.7928μ n B. When the z-component of S (and μ ) is antiparallel to B , U = − μ z | B = +2.7928μ n B. The state B=

EXECUTE:

43.4.

43.5.

with the proton spin component parallel to the field lies lower in energy. The energy difference between these two states is Δ E = 2(2.7928μ n B ). Δ E 2(2.7928μ n ) B 2(2.7928)(5.051 × 10−27 J/T)(1.65 T) = = h h 6.626 × 10−34 J ⋅ s f = 7.03 × 107 Hz = 7.03 MHz

EXECUTE:

Δ E = hf so f =

And then λ = EVALUATE:

c 2.998 × 108 m/s = = 4.26 m f 7.03 × 107 Hz From Figure 32.4 in the textbook, these are radio waves.

43-1

43-2

Chapter 43

(b) SET UP: From Eqs. (27.27) and (41.22) and Fig.41.14 in the textbook, the state with the z-component of G G μ parallel to B has lower energy. But, since the charge of the electron is negative, this is the state with the G electron spin component antiparallel to B. That is, for the ms = − 12 state lies lower in energy. ⎛ e ⎞ ⎛ =⎞ ⎛ e= ⎞ 1 1 For the ms = + 12 state, U = +(2.00232) ⎜ ⎟ ⎜ + ⎟ B = + 2 (2.00232) ⎜ ⎟ B = + 2 (2.00232) μ B B. 2 m 2 ⎝ ⎠ ⎝ ⎠ ⎝ 2m ⎠ For the ms = − 12 state, U = − 12 (2.00232) μB B. The energy difference between these two states is Δ E = (2.00232) μ B B.

EXECUTE:

Δ E 2.00232 μ B B (2.00232)(9.274 × 10 −24 J/T)(1.65 T) = = = 4.62 × 1010 Hz = 46.2 GHz. And h h 6.626 × 10−34 J ⋅ s c 2.998 × 108 m/s λ= = = 6.49 × 10−3 m = 6.49 mm. f 4.62 × 1010 Hz EVALUATE: From Figure 32.4 in the textbook, these are microwaves. The interaction energy with the magnetic field is inversely proportional to the mass of the particle, so it is less for the proton than for the electron. The smaller transition energy for the proton produces a larger wavelength. (a) 146mn + 92mH − mU = 1.93 u Δ E = hf so f =

43.6.

43.7.

43.8.

43.9.

(b) 1.80 × 103 MeV (c) 7.56 MeV per nucleon (using 931.5 MeV/u and 238 nucleons). IDENTIFY and SET UP: The text calculates that the binding energy of the deuteron is 2.224 MeV. A photon that breaks the deuteron up into a proton and a neutron must have at least this much energy. hc hc E= so λ = λ E (4.136 × 10 −15 eV ⋅ s)(2.998 × 108 m/s) EXECUTE: λ = = 5.575 × 10−13 m = 0.5575 pm. 2.224 × 106 eV EVALUATE: This photon has gamma-ray wavelength. IDENTIFY: The binding energy of the nucleus is the energy of its constituent particles minus the energy of the carbon-12 nucleus. SET UP: In terms of the masses of the particles involved, the binding energy is EB = (6mH + 6mn – mC-12)c2. EXECUTE: (a) Using the values from Table 43.2, we get EB = [6(1.007825 u) + 6(1.008665 u) – 12.000000 u)](931.5 MeV/u) = 92.16 MeV (b) The binding energy per nucleon is (92.16 MeV)/(12 nucleons) = 7.680 MeV/nucleon (c) The energy of the C-12 nucleus is (12.0000 u)(931.5 MeV/u) = 11178 MeV. Therefore the percent of the mass 92.16 MeV = 0.8245% . that is binding energy is 11178 MeV EVALUATE: The binding energy of 92.16 MeV binds 12 nucleons. The binding energy per nucleon, rather than just the total binding energy, is a better indicator of the strength with which a nucleus is bound. IDENTIFY: Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the energy after the split (kinetic energy plus energy of the proton and neutron). Therefore the kinetic energy released is equal to the energy of the photon minus the binding energy of the deuteron. SET UP: The binding energy of a deuteron is 2.224 MeV and the energy of the photon is E = hc/λ. Kinetic energy is K = ½mv2. EXECUTE: (a) The energy of the photon is

Eph =

hc

λ

=

( 6.626 ×10

−34

J ⋅ s )( 3.00 × 108 m/s )

3.50 × 10−13 m

= 5.68 × 10−13 J .

The binding of the deuteron is EB = 2.224 MeV = 3.56 × 10 −13 J . Therefore the kinetic energy is K = (5.68 − 3.56) × 10−13 J = 2.12 × 10−13 J = 1.32 MeV . (b) The particles share the energy equally, so each gets half. Solving the kinetic energy for v gives v=

43.10.

2K 2(1.06 × 10−13 J) = = 1.13 × 107 m/s m 1.6605 × 10 −27 kg

EVALUATE: Considerable energy has been released, because the particle speeds are in the vicinity of the speed of light. (a) 7( mn + mH ) − mN = 0.112 u, which is 105 MeV, or 7.48 MeV per nucleon. (b) Similarly, 2( mH + mn ) − mHe = 0.03038 u = 28.3 MeV, or 7.07 MeV per nucleon, slightly lower (compare to Figure 43.2 in the textbook).

Nuclear Physics

43.11.

43-3

(a) IDENTIFY: Find the energy equivalent of the mass defect. SET UP: A 115 B atom has 5 protons, 11 − 5 = 6 neutrons, and 5 electrons. The mass defect therefore is

Δ M = 5mp + 6mn + 5me − M ( 115 B). EXECUTE:

Δ M = 5(1.0072765 u) + 6(1.0086649 u) + 5(0.0005485799 u) − 11.009305 u = 0.08181 u. The energy

equivalent is EB = (0.08181 u)(931.5 MeV/u) = 76.21 MeV. (b) IDENTIFY and SET UP: Eq.(43.11): EB = C1 A − C2 A2/3 − C3Z ( Z − 1) / A1/3 − C4 ( A − 2 Z ) 2 / A The fifth term is zero since Z is odd but N is even. A = 11 and Z = 5. EXECUTE: EB = (15.75 MeV)(11) − (17.80 MeV)(11) 2/3 − (0.7100 MeV)5(4)/111/3 − (23.69 MeV)(11 − 10) 2 /11.

EB = +173.25 MeV − 88.04 MeV − 6.38 MeV − 2.15 MeV = 76.68 MeV 76.68 MeV − 76.21 MeV = 0.6% 76.21 MeV EVALUATE: Eq.(43.11) has a greater percentage accuracy for 62 Ni. The semi-empirical mass formula is more accurate for heavier nuclei. (a) 34mn + 29mH − mCu = 34(1.008665) u + 29(1.007825) u − 62.929601 u = 0.592 u, which is 551 MeV,

The percentage difference between the calculated and measured EB is

43.12.

or 8.75 MeV per nucleon (using 931.5 MeV/u and 63 nucleons). (b) In Eq.(43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding energy is (29)(28) (5) 2 2 EB = (15.75 MeV)(63) − (17.80 MeV)(63) 3 − (0.7100 MeV) − (23.69 MeV) . 1 3 (63) (63)

43.13.

EB = 556 MeV . The fifth term is zero since the number of neutrons is even while the number of protons is odd, making the pairing term zero. This result differs from the binding energy found from the mass deficit by 0.86%, a very good agreement comparable to that found in Example 43.4. IDENTIFY In each case determine how the decay changes A and Z of the nucleus. The β + and β − particles have charge but their nucleon number is A = 0. (a) SET UP: α -decay: Z increases by 2, A = N + Z decreases by 4 (an α particle is a 42 He nucleus) EXECUTE:

239 94

β − decay: Z increases by 1, A = N + Z remains the same (a β − particle is an electron,

(b) SET UP: EXECUTE:

Pu → 42 He + 235 92 U

24 11

e)

Na → e + Mg 0 −1

24 12

(c) SET UP β + decay: Z decreases by 1, A = N + Z remains the same (a β + particle is a positron,

0 +1

e)

EXECUTE: O→ e+ N EVALUATE: In each case the total charge and total number of nucleons for the decay products equals the charge and number of nucleons for the parent nucleus; these two quantities are conserved in the decay. (a) The energy released is the energy equivalent of mn − mp − me = 8.40 × 10−4 u, or 783 keV. 15 8

43.14.

0 −1

0 +1

15 7

(b) mn > mp , and the decay is not possible. 43.15.

IDENTIFY: The energy of the photon must be equal to the difference in energy of the two nuclear energy levels. SET UP: The energy difference is ΔE = hc/λ.

43.16.

hc

( 6.626 ×10

−34

J ⋅ s )( 3.00 × 108 m/s )

= 8.015 × 10−15 J = 0.0501 MeV 0.0248 × 10−9 J EVALUATE: Since the wavelength of this photon is much shorter than the wavelengths of visible light, its energy is much greater than visible-light photons which are frequently emitted during electron transitions in atoms. This tells us that the energy difference between the nuclear shells is much greater than the energy difference between electron shells in atoms, meaning that nuclear energies are much greater than the energies of orbiting electrons. IDENTIFY: The energy released is equal to the mass defect of the initial and final nuclei. SET UP: The mass defect is equal to the difference between the initial and final masses of the constituent particles. EXECUTE: (a) The mass defect is 238.050788 u – 234.043601 u – 4.002603 u = 0.004584 u. The energy released is (0.004584 u)(931.5 MeV/u) = 4.270 MeV. (b) Take the ratio of the two kinetic energies, using the fact that K = p2/2m: EXECUTE:

ΔE =

λ

=

2 pTh K Th 2mTh mα 4 = = = . 2 p Kα mTh 234 α 2mα

43-4

Chapter 43

The kinetic energy of the Th is 4 4 (4.270 MeV) = 0.07176 MeV = 1.148 × 10−14 J K Total = 234 + 4 238 Solving for v in the kinetic energy gives

K Th =

v=

43.17.

2K 2(1.148 × 10−14 J) = = 2.431× 105 m/s −27 m (234.043601) (1.6605 × 10 kg )

EVALUATE: As we can see by the ratio of kinetic energies in part (b) , the alpha particle will have a much higher kinetic energy than the thorium. If β − decay of 14 C is possible, then we are considering the decay 146 C → 147 N + β − .

Δm = M ( 146 C) − M ( 147 N) − me Δm = (14.003242 u − 6(0.000549 u)) − (14.003074 u − 7(0.000549 u)) − 0.0005491 u Δm = +1.68 × 10−4 u. So E = (1.68 × 10−4 u)(931.5 MeV u ) = 0.156 MeV = 156 keV 43.18.

(a) A proton changes to a neutron, so the emitted particle is a positron ( β + ). (b) The number of nucleons in the nucleus decreases by 4 and the number of protons by 2, so the emitted particle is an alpha-particle. (c) A neutron changes to a proton, so the emitted particle is an electron ( β − ).

43.19.

(a) As in the example, (0.000898 u)(931.5 MeV u) = 0.836 MeV. (b) 0.836 MeV − 0.122 MeV − 0.014 MeV = 0.700 MeV.

43.20.

(a)

Sr → β − + 90 39 X . X has 39 protons and 90 protons plus neutrons, so it must be

90 39

(b) Use base 2 because we know the half life. A = A0 2

t=− 43.21.

T1 2 log 0.01 log 2

=−

−t T 1

2

and 0.01A0 = A0 2

−t T1 2

90

Y.

.

(28 yr)log 0.01 = 190 yr . log 2


T1/ 2 =

ln 2

λ

The mass of a single nucleus is 124mp = 2.07 × 10−25 kg .

ΔN / Δt = 0.350 Ci = 1.30 × 1010 Bq ; ΔN / Δt = λ N EXECUTE:

T1/ 2 = 43.22.

ln 2

λ

N=

6.13 × 10 −3 kg ΔN / Δt 1.30 × 1010 Bq = 2.96 × 1022 ; λ = = = 4.39 × 10−13 s −1 −25 2.07 × 10 kg N 2.96 × 10 22

= 1.58 × 1012 s = 5.01 × 104 yr

Note that Eq.(43.17) can be written as follows: N = N 0 2

− t / T1 2

. The amount of elapsed time since the source was

created is roughly 2.5 years. Thus, we expect the current activity to be N = (5000 Ci)2− (2.5 yr)/(5.271 yr) = 3600 Ci. The source is barely usable. Alternatively, we could calculate λ =

43.23.

ln(2) = 0.132(years) −1 and use the Eq. 43.17 directly T1 2

to obtain the same answer. IDENTIFY and SET UP: As discussed in Section 43.4, the activity A = dN / dt obeys the same decay equation as Eq. (43.17): A = A0e − λ t . For 14C, T1/2 = 5730 y and λ = ln2 / T1/ 2 so A = A0e− (ln 2)t / T1/ 2 ; Calculate A at each t; A0 = 180.0 decays/min. EXECUTE: (a) t = 1000 y, A = 159 decays/min (b) t = 50,000 y, A = 0.43 decays/min

43.24.

EVALUATE: The time in part (b) is 8.73 half-lives, so the decay rate has decreased by a factor or ( 12 )8.73. IDENTIFY and SET UP: The decay rate decreases by a factor of 2 in a time of one half-life. EXECUTE: (a) 24 d is 3T1/2 so the activity is (375 Bq) /(23 ) = 46.9 Bq (b) The activity is proportional to the number of radioactive nuclei, so the percent is

17.0 Bq = 36.2% 46.9 Bq

0 131 131 (c) 131 53 I → −1 e + 54 Xe The nucleus 54 Xe is produced. EVALUATE: Both the activity and the number of radioactive nuclei present decrease by a factor of 2 in one halflife.

Nuclear Physics

43.25.

(a) 31 H → −01 e + 23 He (b) N = N 0e − λt , N = 0.100 N 0 and λ = (ln 2) T1 0.100 = e

43.26.

43-5

− t (ln 2) T1 2

2

; −t (ln 2) T1

2

= ln(0.100); t =

− ln(0.100)T1 2 ln 2

= 40.9 y

dN = 500 μ Ci = (500 × 10−6 )(3.70 × 1010 s −1 ) = 1.85 × 107 decays s dt ln 2 ln 2 ln 2 →λ= = = 6.69 × 10−7 s . T1 2 = λ T1 2 12 d(86,400s d)

(a)

dN dN dt 1.85 × 107 decays / s = λN ⇒ N = = = 2.77 × 1013 nuclei . The mass of this many 131 Ba nuclei is dt λ 6.69 × 10 −7 s −1 m = 2.77 × 1013 nuclei × (131 × 1.66 × 10−27 kg nucleus ) = 6.0 × 10−12 kg = 6.0 × 10−9 g = 6.0 ng (b) A = A0e − λ t . 1 μ Ci = (500 μ Ci) e − λ t . ln(1/500) = −λt.

⎛ 1d ⎞ ln(1 500) = 9.29 × 106 s ⎜ ⎟ = 108 days −7 −1 λ 6.69 × 10 s ⎝ 86, 400 s ⎠ (ln 2)t − t (ln 2) / T1 / 2 = ln( A A0 ) . A = A0e − λ t = A0e . − T1 2

t=− 43.27.

ln(1 500)

T1 2 = − 43.28.

=−

(ln 2)t (ln 2)(4.00 days) =− = 2.80 days ln( A A0 ) ln(3091 8318)

dN ln 2 ln 2 = λN . λ = = = 1.36 × 10−11 s −1 . dt T1 2 1620 yr ( 3.15 × 107 s/yr ) ⎛ 6.022 × 1023 atoms ⎞ 25 N =1g⎜ ⎟ = 2.665 × 10 atoms . 226 g ⎝ ⎠

dN = λ N = (2.665 × 1025 )(1.36 × 10 −11 s −1 ) = 3.62 × 1010 decays/s = 3.62 × 1010 Bq dt ⎛ ⎞ 1 Ci Convert to Ci: 3.62 × 1010 Bq ⎜ ⎟ = 0.98 Ci 10 3.70 10 Bq × ⎝ ⎠ 43.29.

IDENTIFY and SET UP: Calculate the number N of 14 C atoms in the sample and then use Eq. (43.17) to find the decay constant λ. Eq. (43.18) then gives T1/ 2 . EXECUTE: Find the total number of carbon atoms in the sample. n = m/M; N tot = nN A = mN A / M = (12.0 × 10 −3 kg)(6.022 × 1023 atoms/mol)/(12.011 × 10−3 kg/mol) N tot = 6.016 × 1023 atoms, so (1.3 × 10−12 )(6.016 × 1023 ) = 7.82 × 1011 carbon-14 atoms ΔN / Δ t = −180 decays/min = −3.00 decays/s

−ΔN / Δt = 3.836 × 10−12 s −1 N T1/ 2 = (ln 2) / λ = 1.807 × 1011 s = 5730 y EVALUATE: The value we calculated agrees with the value given in Section 43.4. 360 × 106 decays = 4.17 × 103 Bq = 1.13 × 10−7 Ci = 0.113 μCi. 86,400 s Δ N / Δ t = −λ N ;

43.30. 43.31.

(a)

λ=

dN 0.693 0.693 = = 3.75 × 10−4 s −1. = 7.56 × 1011 Bq = 7.56 × 1011 decays s . λ = dt T1 2 (30.8 min)(60 s min)

N0 =

1 dN 7.56 × 1011 decays s = = 2.02 × 1015 nuclei. 3.75 × 10−4 s −1 λ dt

43-6

Chapter 43

(b) The number of nuclei left after one half-life is

N0 = 1.01 × 1015 nuclei, and the activity is half: 2

dN = 3.78 × 1011 decays s. dt (c) After three half lives (92.4 minutes) there is an eighth of the original amount, so N = 2.53 × 1014 nuclei, and an ⎛ dN ⎞ 10 eighth of the activity: ⎜ ⎟ = 9.45 × 10 decays s. dt ⎝ ⎠ 43.32.

The activity of the sample is

3070 decays min = 102 Bq kg, while the activity of atmospheric carbon is (60 sec min) (0.500 kg)

ln (102 255) = 7573 y. 1.21 × 10−4 y IDENTIFY and SET UP: Find λ from the half-life and the number N of nuclei from the mass of one nucleus and the mass of the sample. Then use Eq.(43.16) to calculate | dN / dt |, the number of decays per second. EXECUTE: (a) | dN / dt |= λ N 0.693 0.693 λ= = = 1.715 × 10−17 s −1 9 T1/ 2 (1.28 × 10 y)(3.156 × 107 s/1 y)

255 Bq kg (see Example 43.9). The age of the sample is then t = − 43.33.

The mass of 40 K atom is approximately 40 u, so the number of 1.63 × 10−9 kg 1.63 × 10 −9 kg N= = = 2.454 × 1016. 40 u 40(1.66054 × 10−27 kg)

40

ln (102 255)

λ

=−

K nuclei in the sample is

Then | dN / dt |= λ N = (1.715 × 10−17 s −1 )(2.454 × 1016 ) = 0.421 decays/s

43.34.

(b) | dN / dt |= (0.421 decays/s)(1 Ci/(3.70 × 1010 decays/s)) = 1.14 × 10−11 Ci EVALUATE: The very small sample still contains a very large number of nuclei. But the half life is very large, so the decay rate is small. (a) rem = rad × RBE. 200 = x(10) and x = 20 rad. (b) 1 rad deposits 0.010 J kg , so 20 rad deposit 0.20 J kg . This radiation affects 25 g (0.025 kg) of tissue, so the

total energy is (0.025 kg)(0.20 J kg ) = 5.0 × 10−3 J = 5.0 mJ (c) Since RBE = 1 for β -rays, so rem = rad. Therefore 20 rad = 20 rem. 43.35. 43.36.

1 rad = 10 −2 Gy, so 1 Gy = 100 rad and the dose was 500 rad. rem = (rad)(RBE) = (500 rad)(4.0) = 2000 rem. 1 Gy = 1 J kg, so 5.0 J kg . IDENTIFY and SET UP: For x rays RBE = 1 so the equivalent dose in Sv is the same as the absorbed dose in J/kg. EXECUTE: One whole-body scan delivers (75 kg)(12 × 10−3 J/kg) = 0.90 J . One chest x ray delivers

0.90 J = 900 chest x rays to deliver the same total energy. 1.0 × 10−3 J IDENTIFY and SET UP: For x rays RBE = 1 and the equivalent dose equals the absorbed dose. EXECUTE: (a) 175 krad = 175 krem = 1.75 kGy = 1.75 kSv (5.0 kg)(0.20 × 10−3 J/kg) = 1.0 × 10−3 J . It takes

43.37.

(1.75 × 103 J/kg)(0.150 kg) = 2.62 × 102 J (b) 175 krad = 1.75 kGy ; (1.50)(175 krad) = 262 krem = 2.62 kSv

43.38.

43.39.

The energy deposited would be 2.62 × 102 J , the same as in (a). EVALUATE: The energy required to raise the temperature of 0.150 kg of water 1 C° is 628 J, and 2.62 × 102 J is less than this. The energy deposited corresponds to a very small amount of heating. (a) 5.4 Sv (100 rem Sv) = 540 rem. (b) The RBE of 1 gives an absorbed dose of 540 rad. (c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy) (65 kg) = 351 J. The energy required to raise the temperature of 65 kg by 0.010° C is (65 kg) (4190 J kg ⋅ K) (0.01C°) = 3 kJ. (a) We need to know how many decays per second occur.

λ=

0.693 0.693 = = 1.79 × 10−9 s −1. (12.3 y) (3.156 × 107 s y) T1 2

The number of tritium atoms is N 0 =

1 dN (0.35 Ci) (3.70 × 1010 Bq Ci) = = 7.2540 × 1018 nuclei . 1.79 × 10−9 s −1 λ dt

Nuclear Physics

43-7

The number of remaining nuclei after one week is −9 −1 N = N 0e− λt = (7.25 × 1018 )e − (1.79 ×10 s ) (7) (24) (3600s) = 7.2462 × 1018 nuclei. Δ N = N 0 − N = 7.8 × 1015 decays. So the energy absorbed is Etotal = Δ N Eγ = (7.8 × 1015 ) (5000 eV) (1.60 × 10−19 J eV) = 6.24 J.

43.40.

43.41.

The absorbed dose is

(6.24 J) = 0.125 J kg = 12.5 rad. Since RBE = 1, then the equivalent dose is 12.5 rem. (50 kg) (b) In the decay, antinetrinos are also emitted. These are not absorbed by the body, and so some of the energy of the decay is lost (about 12 keV ). (0.72 × 10 −6 Ci) (3.7 × 1010 Bq Ci ) (3.156 × 107 s) = 8.41 × 1011 α particles. The absorbed dose is (8.41 × 1011 ) (4.0 × 106 eV) (1.602 × 10 −19 J eV ) = 1.08 Gy = 108 rad. The equivalent dose is (20) (108 rad) = 2160 rem. (0.50 kg) (a) IDENTIFY and SET UP: Determine X by balancing the charge and nucleon number on the two sides of the reaction equation. EXECUTE: X must have A = 2 + 14 − 10 = 6 and Z = 1 + 7 − 5 = 3. Thus X is 63 Li and the reaction is H + 147 N → 63 Li + 105 B (b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent. EXECUTE: The neutral atoms on each side of the reaction equation have a total of 8 electrons, so the electron masses cancel when neutral atom masses are used. The neutral atom masses are found in Table 43.2. mass of 21 H + 147 N is 2.014102 u + 14.003074 u = 16.017176 u 2 1

mass of 63 Li + 105 B is 6.015121 u + 10.012937 u = 16.028058 u The mass increases, so energy is absorbed by the reaction. The Q value is (16.017176 u − 16.028058 u)(931.5 MeV/u) = −10.14 MeV (c) IDENTIFY and SET UP: The available energy in the collision, the kinetic energy K cm in the center of mass reference frame, is related to the kinetic energy K of the bombarding particle by Eq. (43.24). EXECUTE: The kinetic energy that must be available to cause the reaction is 10.14 MeV. Thus K cm = 10.14 MeV. The mass M of the stationary target ( 147 N) is M = 14 u. The mass m of the colliding particle ( 21 H) is 2 u. Then by Eq. (43.24) the minimum kinetic energy K that the 21 H must have is ⎛M +m⎞ ⎛ 14 u + 2 u ⎞ K =⎜ ⎟ K cm = ⎜ ⎟ (10.14 MeV) = 11.59 MeV ⎝ M ⎠ ⎝ 14 u ⎠

EVALUATE: The projectile

43.42.

2 1

14 7

N ) so K is not much larger than K cm . The K

we have calculated is what is required to allow the mass increase. We would also need to check to see if at this energy the projectile can overcome the Coulomb repulsion to get sufficiently close to the target nucleus for the reaction to occur. m3 He + m 2 H − m 4 He − m1 H = 1.97 × 10−2 u, so the energy released is 18.4 MeV. 2

43.43.

( H ) is much lighter than the target (

1

2

1

IDENTIFY and SET UP: Determine X by balancing the charge and the nucleon number on the two sides of the reaction equation. EXECUTE: X must have A = +2 + 9 − 4 = 7 and Z = +1 + 4 − 2 = 3. Thus X is 73 Li and the reaction is

H + 94 Be = 37 Li + 24He (b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent. EXECUTE: If we use the neutral atom masses then there are the same number of electrons (five) in the reactants as in the products. Their masses cancel, so we get the same mass defect whether we use nuclear masses or neutral atom masses. The neutral atoms masses are given in Table 43.2. 2 9 1 H + 4 Be has mass 2.014102 u + 9.012182 u = 11.26284 u 2 1

Li + 42He has mass 7.016003 u + 4.002603 u = 11.018606 u The mass decrease is 11.026284 u − 11.018606 u = 0.007678 u. This corresponds to an energy release of 0.007678 u(931.5 MeV/1 u) = 7.152 MeV. (c) IDENTIFY and SET UP: Estimate the threshold energy by calculating the Coulomb potential energy when the 2 9 1 H and 4 Be nuclei just touch. Obtain the nuclear radii from Eq. (43.1). 7 3

EXECUTE: The radius RBe of the 94 Be nucleus is RBe = (1.2 × 10−15 m)(9)1/3 = 2.5 × 10−15 m.

The radius RH of the 12 H nucleus is RH = (1.2 × 10−15 m)(2)1/3 = 1.5 × 10−15 m. The nuclei touch when their center-to-center separation is R = RBe + RH = 4.0 × 10−15 m.

43-8

Chapter 43

The Coulomb potential energy of the two reactant nuclei at this separation is 1 q1q2 1 e(4e) U= = 4π P0 r 4π P0 r 4(1.602 × 10−19 C) 2 = 1.4 MeV (4.0 × 10−15 m)(1.602 × 10−19 J/eV) This is an estimate of the threshold energy for this reaction. EVALUATE: The reaction releases energy but the total initial kinetic energy of the reactants must be 1.4 MeV in order for the reacting nuclei to get close enough to each other for the reaction to occur. The nuclear force is strong but is very short-range. IDENTIFY and SET UP: 0.7% of naturally occurring uranium is the isotope 235 U . The mass of one 235 U nucleus is about 235mp. 1.0 × 1019 J = 3.13 × 1029 . The mass of EXECUTE: (a) The number of fissions needed is 6 (200 × 10 eV)(1.60 × 10 −19 J/eV) U = (8.988 × 109 N ⋅ m 2 / C2 )

43.44.

235

U required is (3.13 × 1029 )(235mp ) = 1.23 × 105 kg .

1.23 × 105 kg = 1.76 × 107 kg 0.7 × 10−2 EVALUATE: The calculation assumes 100% conversion of fission energy to electrical energy. IDENTIFY and SET UP: The energy released is the energy equivalent of the mass decrease. 1 u is equivalent to 931.5 MeV. The mass of one 235 U nucleus is 235mp. 1 144 89 1 EXECUTE: (a) 235 92 U + 0 n → 56 Ba + 36 Kr + 3 0 n We can use atomic masses since the same number of electrons are included on each side of the reaction equation and the electron masses cancel. The mass decrease is 1 89 1 ⎡ 144 ⎤ ΔM = m ( 235 92 U ) + m ( 0 n ) − ⎣ m ( 56 Ba ) + m ( 36 Kr ) + 3m ( 0 n ) ⎦ ΔM = 235.043930 u + 1.0086649 u − 143.922953 u − 88.917630 u − 3(1.0086649 u) ΔM = 0.1860 u . The energy released is (0.1860 u)(931.5 MeV/u) = 173.3 MeV . (b)

43.45.

(b) The number of

235

U nuclei in 1.00 g is

1.00 × 10−3 kg = 2.55 × 1021 . The energy released per gram is 235mp

(173.3 MeV/nucleus)(2.55 × 1021 nuclei/g) = 4.42 × 10 23 MeV/g .

43.46.

(a)

28 14

24 Si + γ ⇒ 12 Mg + ZA X. A + 24 = 28 so A = 4. Z + 12 = 14 so Z = 2. X is an α particle.

(b) Eγ = −Δmc 2 = (23.985042 u + 4.002603 u − 27.976927 u) (931.5 MeV u ) = 9.984 MeV 43.47. 43.48.

The energy liberated will be M (32 He) + M ( 42 He) − M (74 Be) = (3.016029 u + 4.002603 u − 7.016929 u)(931.5 MeV u) = 1.586 MeV. (a) Z = 3 + 2 − 0 = 5 and A = 4 + 7 − 1 = 10. (b) The nuclide is a boron nucleus, and mHe + mLi − mn − mB = −3.00 × 10−3 u, and so 2.79 MeV of energy is absorbed.

43.49.

Nuclei: ZA X Z + →

A−4 Z −2

Y ( Z − 2) + + 42 He2 + . Add the mass of Z electrons to each side and we find:

43.50.

Δm = M ( ZA X) − M ( ZA−− 42 Y) − M ( 42He), where now we have the mass of the neutral atoms. So as long as the mass of the original neutral atom is greater than the sum of the neutral products masses, the decay can happen. Denote the reaction as ZA X → Z +1A Y + e − . The mass defect is related to the change in the neutral atomic masses by

[ mX − Zme ] − [mY − ( Z + 1) me ] − me = ( mX − mY ), where mX and mY are the masses as tabulated in, for instance, Table (43.2). 43.51.

A Z

XZ + →

A Z −1

Y ( Z −1) + + β + . Adding (Z –1) electrons to both sides yields ZA X + →

(

)

A Z −1

Y + β + . So in terms of masses:

Δm = M ( ZA X + ) − M ( Z −1A Y ) − me = M ( ZA X ) − me − M ( Z −1A Y ) − me = M ( ZA X ) − M ( Z −1A Y ) − 2me . So the decay will 43.52.

occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses. IDENTIFY and SET UP: m = ρV . 1 gal = 3.788 L = 3.788 × 10−3 m 3 . The mass of a 235 U nucleus is 235mp. 1 MeV = 1.60 × 10 −13 J EXECUTE: (a) For 1 gallon, m = ρV = (737 kg/m3 )(3.788 × 10−3 m3 ) = 2.79 kg = 2.79 × 103 g

1.3 × 108 J/gal = 4.7 × 10 4 J/g 2.79 × 103 g/gal

Nuclear Physics

(b) 1 g contains

43-9

1.00 × 10 −3 kg = 2.55 × 1021 nuclei 235mp

(200 MeV/nucleus)(1.60 × 10−13 J/MeV)(2.55 × 1021 nuclei) = 8.2 × 1010 J/g (c) A mass of 6mp produces 26.7 MeV. (26.7 MeV)(1.60 × 10−13 J/MeV) = 4.26 × 1014 J/kg = 4.26 × 1011 J/g 6mp

(d) The total energy available would be (1.99 × 1030 kg)(4.7 × 107 J/kg) = 9.4 × 1037 J

energy energy 9.4 × 1037 J so t = = = 2.4 × 1011 s = 7600 yr t power 3.86 × 1026 W EVALUATE: If the mass of the sun were all proton fuel, it would contain enough fuel to last ⎛ 4.3 × 1011 J/g ⎞ 10 (7600 yr) ⎜ ⎟ = 7.0 × 10 yr . 4 ⎝ 4.7 × 10 J/g ⎠ power =

43.53.

24 Using Eq: (43.12): ZA M = ZM H + Nmn − EB c 2 ⇒ M (11 Na) = 11M H +13mn − EB / c 2 .

But EB = (15.75 MeV)(24) − (17.80 MeV)(24) 2 3 − (0.7100 MeV) (23.69 MeV)

(11)(10) − (24)1 3

(24 − 2(11)) 2 − (39 MeV)(24) −4 3 = 198.31 MeV. 24 24 Na) = 11(1.007825 u) + 13(1 .008665 u) − ⇒ M (11

% error =

(198.31 MeV) = 23.9858 u 931.5 MeV u

23.990963 − 23.9858 × 100 = 0.022%. 23.990963

24 Na) = 24.1987 u and the percentage error would be If the binding energy term is neglected, M (11

43.54.

43.55.

24.1987 − 23.990963 × 100 = 0.87%. 23.990963 226 226 The α -particle will have of the released energy (see Example 43.5). ( mTh − mRa − mα ) = 230 230 5.032 × 10−3 u or 4.69 MeV. (a) IDENTIFY and SET UP: The heavier nucleus will decay into the lighter one. 25 25 EXECUTE: 13 Al will decay into 12 Mg. (b) IDENTIFY and SET UP: Determine the emitted particle by balancing A and Z in the decay reaction. 25 25 EXECUTE: This gives 13 Al →12 Mg + +10 e. The emitted particle must have charge +e and its nucleon number must be zero. Therefore, it is a β + particle, a positron. (c) IDENTIFY and SET UP: Calculate the energy defect ΔM for the reaction and find the energy equivalent of 25 25 ΔM . Use the nuclear masses for 13 Al and 12 Mg, to avoid confusion in including the correct number of electrons if neutral atom masses are used. 25 25 EXECUTE: The nuclear mass for 13 Al is M nuc (13 Al) = 24.990429 u − 13(0.000548580 u) = 24.983297 u. 25 Mg is M nuc (1225 Mg) = 24.985837 u − 12(0.000548580 u) = 24.979254 u. The nuclear mass for 12 The mass defect for the reaction is 25 25 Δ M = M nuc (13 Al) − M nuc (12 Mg) − M (0+1 e) = 24.983297 u − 24.979254 u − 0.00054858 u = 0.003494 u

Q = (Δ M )c 2 = 0.003494 u(931.5 MeV/1 u) = 3.255 MeV EVALUATE: The mass decreases in the decay and energy is released. Note: 25 12 25 13

25 13

Al can also decay into

Mg by the electron capture. 25 Al + −10 e →12 Mg

25 Al atom. The mass defect can be calculated The −10 electron in the reaction is an orbital electron in the neutral 13 using the nuclear masses: 25 25 Δ M = M nuc ( 13 Al ) + M (0−1 e) − M nuc ( 12 Mg ) = 24.983287 u + 0.00054858 u − 24.979254 u = 0.004592 u.

Q = ( Δ M ) c 2 = (0.004592 u)(931.5 MeV/1 u) = 4.277 MeV 2

The mass decreases in the decay and energy is released.

43-10

Chapter 43

43.56.

(a) m210 Po − m 206 Pb − m4 He = 5.81 × 10−3 u, or Q = 5.41 MeV. The energy of the alpha particle is (206 210) times this, 84

82

2

or 5.30 MeV (see Example 43.5). (b) m210 Po − m209 Bi − m1 H = −5.35 × 10−3 u < 0, so the decay is not possible. 84

83

1

(c) m210 Po − m 209 Po − mn = −8.22 × 10−3 u < 0, so the decay is not possible. 84

84

(d) m 210 At > m 210 Po , so the decay is not possible (see Problem (43.50)). 85

84

(e) m 210 Bi + 2me > m 210 Po , so the decay is not possible (see Problem (43.51)). 83

43.57.

84

IDENTIFY and SET UP: The amount of kinetic energy released is the energy equivalent of the mass change in the decay. me = 0.0005486 u and the atomic mass of 147 N is 14.003074 u. The energy equivalent of 1 u is

931.5 MeV.

14

C has a half-life of T1/ 2 = 5730 yr = 1.81 × 1011 s . The RBE for an electron is 1.0.

EXECUTE: (a)

14 6

C → e − + 147 N + υe

(b) The mass decrease is ΔM = m ( 146 C ) − ⎡⎣ me + m ( 147 N ) ⎤⎦ . Use nuclear masses, to avoid difficulty in accounting for

atomic electrons. The nuclear mass of The nuclear mass of

14 7

14 6

C is 14.003242 u − 6me = 13.999950 u .

N is 14.003074 u − 7me = 13.999234 u .

Δ M = 13.999950 u − 13.999234 u − 0.000549 u = 1.67 × 10−4 u . The energy equivalent of Δ M is 0.156 MeV. (c) The mass of carbon is (0.18)(75 kg) = 13.5 kg . From Example 43.9, the activity due to 1 g of carbon in a

living organism is 0.255 Bq. The number of decay/s due to 13.5 kg of carbon is (13.5 × 103 )(0.255 Bq/g) = 3.4 × 103 decays/s .

(d) Each decay releases 0.156 MeV so 3.4 × 103 decays/s releases 530 MeV/s = 8.5 × 10−11 J/s . (e) The total energy absorbed in 1 yr is (8.5 × 10−11 J/s)(3.156 × 107 s) = 2.7 × 10 −3 J . The absorbed dose is 2.7 × 10−3 J = 3.6 × 10 −5 J/kg = 36 μ Gy = 3.6 mrad . With RBE = 1.0 , the equivalent dose is 36 μSv = 3.6 mrem . 75 kg

43.58.


mπ = 264me = 2.40 × 10−28 kg . The total energy of the two photons equals the rest mass

energy mπ c2 of the pion. EXECUTE: (a) Eph = 12 mπc 2 = 12 (2.40 × 10−28 kg)(3.00 × 108 m/s) 2 = 1.08 × 10−11 J = 67.5 MeV

Eph =

hc

λ

so λ =

hc 1.24 × 10−6 eV ⋅ m = = 1.84 × 10−14 m = 18.4 fm Eph 67.5 × 106 eV

These are gamma ray photons, so they have RBE = 1.0 . (b) Each pion delivers 2(1.08 × 10−11 J) = 2.16 × 10−11 J . The absorbed dose is 200 rad = 2.00 Gy = 2.00 J/kg .

43.59.

The energy deposited is (25 × 10−3 kg)(2.00 J/kg) = 0.050 J . 0.050 J = 2.3 × 109 mesons . The number of π 0 mesons needed is 2.16 × 10−11 J/meson EVALUATE: Note that charge is conserved in the decay since the pion is neutral. If the pion is initially at rest the photons must have equal momenta in opposite directions so the two photons have the same λ and are emitted in opposite directions. The photons also have equal energies since they have the same momentum and E = pc . IDENTIFY and SET UP: Find the energy equivalent of the mass decrease. Part of the released energy appears as the emitted photon and the rest as kinetic energy of the electron. 198 0 EXECUTE: 198 79 Au → 80 Hg + −1 e The mass change is 197.968225 u − 197.966752 u = 1.473 × 10 −3 u (The neutral atom masses include 79 electrons before the decay and 80 electrons after the decay. This one additional electron in the products accounts correctly for the electron emitted by the nucleus.) The total energy released in the decay is (1.473 × 10 −3 u)(931.5 MeV/u) = 1.372 MeV. This energy is divided between the energy of the emitted photon and the kinetic energy of the β − particle. Thus the β − particle has kinetic energy equal to 1.372 MeV − 0.412 MeV = 0.960 MeV. EVALUATE: The emitted electron is much lighter than the 198 80 Hg nucleus, so the electron has almost all the final kinetic energy. The final kinetic energy of the

198

Hg nucleus is very small.

Nuclear Physics

43-11

43.60.

(See Problem (43.51)) m11 C − m11 B − 2me = 1.03 × 10−3 u. Decay is energetically possible.

43.61.

IDENTIFY and SET UP: The decay is energetically possible if the total mass decreases. Determine the nucleus produced by the decay by balancing A and Z on both sides of the equation. 137 N → +10e + 136C. To avoid confusion in including the correct number of electrons with neutral atom masses, use nuclear masses, obtained by subtracting the mass of the atomic electrons from the neutral atom masses.

6

5

13 7

EXECUTE: The nuclear mass for

The nuclear mass for

13 6

N is M nuc

(

13 7

N ) = 13.005739 u − 7(0.00054858 u) = 13.001899 u.

( C ) = 13.003355 u − 6(0.00054858 u) = 13.000064 u. 13 6

C is M nuc

The mass defect for the reaction is Δ M = M nuc ( 137 N ) − M nuc ( 136 C ) − M ( +10 e ). Δ M = 13.001899 u − 13.000064 u − 0.00054858 u = 0.001286 u. 43.62.

EVALUATE: The mass decreases in the decay, so energy is released. This decay is energetically possible. ln 2 (a) A least-squares fit to log of the activity vs. time gives a slope of λ = 0.5995 h −1 , for a half-life of = 1.16 h.

λ

(2.00 × 104 Bq) (b) The initial activity is N 0λ , and this gives N 0 = = 1.20 × 108. (0.5995 hr −1 )(1 hr 3600 s) (c) N 0e− λt = 1.81 × 106. 43.63.

43.64.

dN (t ) dN (t ) but = − λN (t ) so − λN 0 = A0 . Taking the derivative of dt dt dN (t ) N (t ) = N 0e − λt ⇒ = −λN 0e − λt = A0e − λt , or A(t ) = A0e− λt . dt − (ln 2) ( t / T1 2 ) From Eq.43.17 N (t ) = N 0e − λt but N 0e − λt = N 0e

The activity A(t ) ≡

= N 0 ⎡⎣ e− (ln 2) ⎤⎦

43.65.

(t / T1 2 )

= N 0 ⎡⎣ eln ( 1/ 2) ⎤⎦

( t / T1 2 )

n

t ⎛1⎞ . So N (t ) = N 0 ⎜ ⎟ where n = . T1 2 ⎝ 2⎠

(We have used that a ln x = ln( x a ), eax = (e x ) a , and eln x = x. ) IDENTIFY and SET UP: One-half of the sample decays in a time of T1/2. 10 × 109 yr = 5.0 × 104 EXECUTE: (a) 200,000 yr 4

(b) ( 12 )5.0×10 . This exponent is too large for most hand-held calculators. But ( 12 ) = 10−0.301 so 4

4

( 12 )5.0×10 = (10−0.301 )5.0×10 = 10−15,000 43.66.


N=

T1/ 2 =

ln 2

λ

. The mass of a single nucleus is 149mp = 2.49 × 10−25 kg . ΔN / Δt = −λ N .

12.0 × 10−3 kg = 4.82 × 1022 . Δ N / Δ t = −2.65 decays/s 2.49 × 10−25 kg

Δ N / Δ t 2.65 decays/s ln 2 = = 5.50 × 10 −23 s −1 ; T1/ 2 = = 1.26 × 1022 s = 3.99 × 1014 yr 4.82 × 10 22 λ N IDENTIFY: Use Eq. (43.17) to relate the initial number of radioactive nuclei, N 0 , to the number , N, left after time t.

λ =−

43.67.

SET UP: We have to be careful; after 87 Rb has undergone radioactive decay it is no longer a rubidium atom. Let N85 be the number of 85 Rb atoms; this number doesn’t change. Let N 0 be the number of 87 Rb atoms on earth

when the solar system was formed. Let N be the present number of 87 Rb atoms. EXECUTE: The present measurements say that 0.2783 = N /( N + N85 ). ( N + N85 )(0.2783) = N , so N = 0.3856 N85 . The percentage we are asked to calculate is N 0 /( N 0 + N85 ). N and N 0 are related by N = N 0e − λ t so N 0 = e + λ t N .

Thus

N0 Neλ t (0.3855eλ t ) N85 0.3856eλ t . = λt = = λt N 0 + N85 Ne + N85 (0.3856e ) N85 + N85 0.3856eλ t + 1

t = 4.6 × 109 y; λ = eλ t = e(1.459×10

−11

0.693 0.693 = = 1.459 × 10−11 y −1 T1/2 4.75 × 1010 y

y−1 )(4.6×109 y)

= e0.16711 = 1.0694

43-12

Chapter 43

Thus

43.68.

N0 (0.3856)(1.0694) = = 29.2%. N 0 + N85 (0.3856)(1.0694) + 1

EVALUATE: The half-life for 87 Rb is a factor of 10 larger than the age of the solar system, so only a small fraction of the 87 Rb nuclei initially present have decayed; the percentage of rubidium atoms that are radioactive is only a bit less now than it was when the solar system was formed. (a) (6.25 × 1012 )(4.77 × 106 MeV)(1.602 × 10−19 J eV) (70.0 kg) = 0.0682 Gy = 0.682 rad (b) (20)(6.82 rad ) = 136 rem m ln(2) (c) N λ = = 1.17 × 109 Bq = 31.6 mCi . Amp T1 2

6.25 × 1012 = 5.34 × 103 s, about an hour and a half. Note that this time is so small in comparison with the 1.17 × 109 Bq half-life that the decrease in activity of the source may be neglected. IDENTIFY and SET UP: Find the energy emitted and the energy absorbed each second. Convert the absorbed energy to absorbed dose and to equivalent dose. EXECUTE: (a) First find the number of decays each second: ⎛ 3.70 × 1010 decays/s ⎞ 6 2.6 × 10 −4 Ci ⎜ ⎟ = 9.6 × 10 decays/s 1 Ci ⎝ ⎠ The average energy per decay is 1.25 MeV, and one-half of this energy is deposited in the tumor. The energy delivered to the tumor per second then is 1 (9.6 × 106 decays/s)(1.25 × 106 eV/decay)(1.602 × 10−19 J/eV) = 9.6 × 10−7 J/s. 2 (b) The absorbed dose is the energy absorbed divided by the mass of the tissue: 9.6 × 10−7 J/s = (1.9 × 10−6 J/kg ⋅ s)(1 rad/(0.01 J/kg)) = 1.9 × 10 −4 rad/s 0.500 kg (c) equivalent dose (REM) = RBE × absorbed dose (rad) In one second the equivalent dose is 0.70(1.9 × 10 −4 rad) = 1.3 × 10 −4 rem. (d)

43.69.

(d) (200 rem/1.3 × 10−4 rem/s) = 1/ 5 × 106 s(1 h/3600 s) = 420 h = 17 days. EVALUATE: The activity of the source is small so that absorbed energy per second is small and it takes several days for an equivalent dose of 200 rem to be absorbed by the tumor. A 200 rem dose equals 2.00 Sv and this is large enough to damage the tissue of the tumor. ⎛ 1

1

⎞

− (240) ⎜ ⎟ 2−240 122.2 = 2 ⎝ 26.9 122.2 ⎠ = 124. −240 26.9 2

43.70.

(a) After 4.0 min = 240 s, the ratio of the number of nuclei is

43.71.

(b) After 15.0 min = 900 s, the ratio is 7.15 × 107. IDENTIFY and SET UP: The number of radioactive nuclei left after time t is given by N = N 0e − λt . The problem

says N / N 0 = 0.21; solve for t. EXECUTE:

0.21 = e − λ t so ln(0.21) = −λt and t = −ln(0.21)/λ

Example 43.9 gives λ = 1.209 × 10−4 y −1 for 14C. Thus t = EVALUATE: The half-life of

remaining is less than 43.72.

( )

1 2 2

14

− ln(0.21) = 1.3 × 104 y. 1.209 × 10−4 y

C is 5730 y, so our calculated t is more than two half-lives, so the fraction

= 14 .

IDENTIFY: The tritium (H-3) decays to He-3. The ratio of the number of He-3 atoms to H-3 atoms allows us to calculate the time since the decay began, which is when the H-3 was formed by the nuclear explosion. The H-3 decay is exponential. SET UP: The number of tritium (H-3) nuclei decreases exponentially as N H = N 0,H e− λt , with a half-life of

12.3 years. The amount of He-3 present after a time t is equal to the original amount of tritium minus the number of tritium nuclei that are still undecayed after time t. EXECUTE: The number of He-3 nuclei after time t is N He = N 0,H − N H = N 0,H − N 0,H e − λt = N 0,H (1 − e − λt ) .

Taking the ratio of the number of He-3 atoms to the number of tritium (H-3) atoms gives − λt N He N 0,H (1 − e ) 1 − e − λ t = = − λt = eλt − 1. NH N 0,H e − λ t e

Nuclear Physics

43-13

ln (1 + N He / N H ) ln 2 , we have . Using the given numbers and T1/ 2 = λ λ ln (1 + 4.3) ln 2 ln 2 λ= = = 0.0563/ y and t = = 30 years. 0.0563/ y T1/ 2 12.3 y EVALUATE: One limitation on this method would be that after many years the ratio of H to He would be too small to measure accurately. (a) IDENTIFY and SET UP: Use Eq.(43.1) to calculate the radius R of a 21 H nucleus. Calculate the Coulomb potential energy (Eq.23.9) of the two nuclei when they just touch. EXECUTE: The radius of 21 H is R = (1.2 × 10−15 m)(2)1/3 = 1.51 × 10−15 m. The barrier energy is the Coulomb Solving for t gives t =

43.73.

potential energy of two 21 H nuclei with their centers separated by twice this distance: 1 e2 (1.602 × 10−19 C) 2 = (8.988 × 109 N ⋅ m 2 / C2 ) = 7.64 × 10−14 J = 0.48 MeV 4π P0 r 2(1.51 × 10−15 m) (b) IDENTIFY and SET UP: Find the energy equivalent of the mass decrease. EXECUTE: 12 H + 12 H → 32 He + 10 n If we use neutral atom masses there are two electrons on each side of the reaction equation, so their masses cancel. The neutral atom masses are given in Table 43.2. 2 2 1 H + 1 H has mass 2(2.014102 u) = 4.028204 u U=

3 2

He + 10 n has mass 3.016029 u + 1.008665 u = 4.024694 u

The mass decrease is 4.028204 u − 4.024694 u = 3.510 × 10−3 u. This corresponds to a liberated energy of (3.510 × 10 −3 u)(931.5 MeV/u) = 3.270 MeV, or (3.270 × 106 eV)(1.602 × 10 −19 J/eV) = 5.239 × 10−13 J. (c) IDENTIFY and SET UP: We know the energy released when two 12 H nuclei fuse. Find the number of reactions

obtained with one mole of 12 H. EXECUTE: Each reaction takes two 12 H nuclei. Each mole of D 2 has 6.022 × 1023 molecules, so 6.022 × 10 23 pairs

of atoms. The energy liberated when one mole of deuterium undergoes fusion is (6.022 × 10 23 )(5.239 × 10 −13 J) =

43.74.

3.155 × 1011 J/mol. EVALUATE: The energy liberated per mole is more than a million times larger than from chemical combustion of one mole of hydrogen gas. In terms of the number N of cesium atoms that decay in one week and the mass m = 1.0 kg, the equivalent dose is

N N N ((RBE) γ E γ + (RBE)e E e ) = ((1)(0.66 MeV) + (1.5)(0.51 MeV)) = (2.283 × 10−13 J), so m m m (1.0 kg)(3.5 Sv) N= = 1.535 × 1013 . The number N 0 of atoms present is related to (2.283 × 10−13 J)

3.5 Sv =

N by N 0 = Ne λt . λ =

ln 2 0.693 = 7.30 × 10−10 sec −1 . Ty2 (30.07 yr)(3.156 × 107 sec yr)

Then N 0 = Ne λt = (1.535 × 1013 )e(7.30×10 43.75.

(a) vcm = v

−10

s −1 ) (7 days) (8.64×104 s day)

= 1.536 × 1013.

m m vm ⎛ M ⎞ =⎜ . v′m = v − v . ⎟ v . v′M = m+M ⎝m+M ⎠ m+M m+M

1 1 1 mM 2 1 Mm 2 1 M ⎛ mM m2 2 2 K ′ = mv′m2 + Mv′M2 = v + v = + ⎜ 2 2 2 (m + M ) 2 2 (m + M ) 2 2 (m + M ) ⎝ m + M m + M K′ =

⎞ 2 ⎟v . ⎠

M ⎛1 2⎞ M K ≡ K cm . ⎜ mv ⎟ ⇒ K ′ = m+M ⎝2 m+M ⎠

(b) For an endoergic reaction K cm = −Q ( Q < 0 ) at threshold. Putting this into part (a) gives

−Q =

−( M + m) M K th ⇒ K th = Q M +m M

43-14

43.76.

43.77.

Chapter 43

Mα K ∞ , where K ∞ is the energy that the α -particle would have if the nucleus were infinitely massive. Mα + m 186 Then, M = M Os − M α − K ∞ = M Os − M α − ( 2.76 MeV c2 ) = 181.94821 u . 182 235 94 Δ m = M ( 92 U ) − M ( 140 54 Xe ) − M ( 38 Sr ) − mn Δ m = 235.043923 u − 139.921636 u − 93.915360 u − 1.008665 u = 0.1983 u K=

⇒ E = ( Δ m ) c 2 = ( 0.1983 u ) ( 931.5 MeV u ) = 185 MeV. 43.78.

(a) A least-squares fit of the log of the activity vs. time for the times later than 4.0 h gives a fit with correlation − (1 − 2 × 10−6 ) and decay constant of 0.361 h −1, corresponding to a half-life of 1.92 h. Extrapolating this back to time 0

gives a contribution to the rate of about 2500/s for this longer-lived species. A least-squares fit of the log of the activity vs. time for times earlier than 2.0 h gives a fit with correlation = 0.994, indicating the presence of only two species. − t 1.733 h ) − t 0.361 h ) (b) By trial and error, the data is fit by a decay rate modeled by R = ( 5000 Bq ) e ( + ( 2500 Bq ) e ( . This

43.79.

would correspond to half-lives of 0.400 h and 1.92 h. (c) In this model, there are 1.04 × 10 7 of the shorter-lived species and 2.49 × 10 7 of the longer-lived species. (d) After 5.0 h, there would be 1.80 × 103 of the shorter-lived species and 4.10 × 10 6 of the longer-lived species. (a) There are two processes occurring: the creation of 128 I by the neutron irradiation, and the decay of the newly dN = K − λ N where K is the rate of production by the neutron irradiation. Then produced 128 I . So dt K (1 − e− λt ) N t N dN ′ ′ . ln K − λ N = ln K − λ t dt = N t = . The graph is given in ln K λ N λ t . . ⎡ − ⎤ = − ( ) ( ) ( ) ∫0 K − λ N ′ ∫0 ⎣ ⎦0 λ Figure 43.79. ⎛ 0.693 ⎞ ⎛ −⎜ ⎟t ⎞ (b) The activity of the sample is λ N ( t ) = K (1 − e − λt ) = (1.5 × 106 decays s ) × ⎜ 1 − e ⎝ 25 min ⎠ ⎟ . So the activity is ⎟ ⎜ ⎝ ⎠ ′⎞ (1.5 × 106 decays s ) (1 − e−0.02772 t ) , with t in minutes. So the activity ⎛⎜ −dN ⎟ at various times is: dt ⎝ ⎠

− dN ′ − dN ′ (t = 1 min) = 4.1 × 104 Bq; (t = 10 min) = 3.6 × 105 Bq; dt dt − dN ′ − dN ′ (t = 25 min) = 7.5 × 105 Bq; (t = 50 min) = 1.1× 106 Bq; dt dt − dN ′ − dN ′ (t = 75 min) = 1.3 × 106 Bq; (t = 180 min) = 1.5 × 106 Bq; dt dt (1.5 × 106 ) (60) = 3.2 × 109 atoms . λ ( 0.02772 ) (d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals 1.5 × 106 decays s. (c) N max =

K

=

Figure 43.79 43.80.

The activity of the original iron, after 1000 hours of operation, would be (9.4 × 10−6 Ci) (3.7 × 1010 Bq Ci)2−(1000 h) (45 d×24 h d) = 1.8306 × 105 Bq . The activity of the oil is 84 Bq, or 4.5886 × 10−4 of the total iron activity, and this must be the fraction of the mass worn, or mass of 4.59 × 10−2 g .

The rate at which the piston rings lost their mass is then 4.59 × 10−5 g h .

44

PARTICLE PHYSICS AND COSMOLOGY

44.1.

44.2.

(a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K. ⎛ ⎞ 1 EXECUTE: K = mc 2 ⎜ − 1⎟ = 0.1547 mc 2 2 2 ⎝ 1− v / c ⎠ m = 9.109 × 10 −31 kg, so K = 1.27 × 10−14 J (b) IDENTIFY and SET UP: The total energy of the particles equals the sum of the energies of the two photons. Linear momentum must also be conserved. EXECUTE: The total energy of each electron or positron is E = K + mc 2 = 1.1547 mc 2 = 9.46 × 10 −13 J. The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite directions. Equal λ means equal energy, so each photon has energy 9.46 × 10 −14 J. (c) IDENTIFY and SET UP: Use Eq. (38.2) to relate the photon energy to the photon wavelength. EXECUTE: E = hc / λ so λ = hc / E = hc /(9.46 × 10−14 J) = 2.10 pm EVALUATE: The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less. The total energy of the positron is E = K + mc 2 = 5.00 MeV + 0.511 MeV = 5.51 MeV.

We can calculate the speed of the positron from Eq.(37.38): E=

mc 2 1−

44.3.

v2 c2

2

⇒

⎛ mc 2 ⎞ v ⎛ 0.511 MeV ⎞ = 1− ⎜ ⎟ = 1− ⎜ ⎟ = 0.996. c E ⎝ 5.51 MeV ⎠ ⎝ ⎠ 2

IDENTIFY and SET UP: By momentum conservation the two photons must have equal and opposite momenta. Then E = pc says the photons must have equal energies. Their total energy must equal the rest mass energy E = mc 2 of the pion. Once we have found the photon energy we can use E = hf to calculate the photon frequency and use λ = c / f to calculate the wavelength.

EXECUTE: The mass of the pion is 270me , so the rest energy of the pion is 270(0.511 MeV) = 138 MeV. Each

photon has half this energy, or 69 MeV. E = hf so f =

E (69 × 106 eV)(1.602 × 10−19 J/eV) = = 1.7 × 1022 Hz h 6.626 × 10−34 J ⋅ s

c 2.998 × 108 m/s = = 1.8 × 10−14 m = 18 fm. f 1.7 × 1022 Hz EVALUATE: These photons are in the gamma ray part of the electromagnetic spectrum. (a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27 × 1023 Hz and a

λ=

44.4.

wavelength of 1.32 × 10 −15 m. (b) The energy of each photon will be 938.3 MeV + 830 MeV = 1768 MeV, with frequency 42.8 × 1022 Hz and

wavelength 7.02 × 10 −16 m. 44.5.

(a) Δ m = mπ + − mμ + = 270 me − 207 me = 63 me ⇒ E = 63(0.511 MeV) = 32 MeV. (b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.

44-1

44-2

Chapter 44

hc hc h (6.626 × 10−34 J ⋅ s) = = = = 1.17 × 10−14 m = 0.0117 pm E mμ c 2 mμ c (207)(9.11 × 10−31 kg)(3.00 × 108 m s)

44.6.

(a) λ =

44.7.

In this case, the muons are created at rest (no kinetic energy). (b) Shorter wavelengths would mean higher photon energy, and the muons would be created with non-zero kinetic energy. IDENTIFY: The energy released comes from the mass difference. SET UP: The mass difference is the initial mass minus the final mass. Δ m = mμ − − me− − me+ EXECUTE: Using the masses from Table 44.2, we have Δ m = mμ − − me− − me+ = (105.7 MeV/c 2 ) − (0.511 MeV/c 2 ) − (0.511 MeV/c 2 ) = 105 MeV/c 2

44.8.

Multiplying these masses by c2 gives E = 105 MeV. EVALUATE: This energy is observed as kinetic energy of the electron and positron. IDENTIFY and SET UP: Calculate the mass change in each reaction, using the atomic masses in Table 44.2. A mass change of 1 u is equivalent to an energy of 931.5 MeV. EXECUTE: (a) and (b) Eq.(44.1): 42 He + 94 Be → 126 C + 01 n Δ M = m ( 4 He ) + m ( 9 Be ) − ⎡⎣ m ( 12 C ) + m ( 1 n ) ⎤⎦

Δ M = 4.00260 u + 9.01218 u − 12.00000 u − 1.00866 u = 0.00612 u The mass decreases and the energy liberated is 5.70 MeV. The reaction is exoergic. Eq.(44.2): 01 n + 105 B → 73 Li + 42 He

Δ M = m ( 1 n ) + m ( 10 B ) − ⎡⎣ m ( 7 Li ) + m ( 4 He ) ⎤⎦

44.9.

Δ M = 1.00866 u + 10.01294 u − 7.01600 u − 4.00260 u = 0.00300 u The mass decreases and the energy liberated is 2.79 MeV. The reaction is exoergic. (c) The reactants in the reactions of Eq.(44.1) have positive nuclear charges and a threshold kinetic energy is required for the reactants to overcome their Coulomb repulsion and get close enough for the reaction to occur. The neutron in Eq.(44.2) is neutral so there is no Coulomb repulsion and no threshold energy for this reaction. IDENTIFY: The antimatter annihilates with an equal amount of matter. SET UP: The energy of the matter is E = (Δ m)c 2 . EXECUTE: Putting in the numbers gives

E = (Δ m)c 2 = (400 kg + 400 kg)(3.00 × 108 m s)2 = 7.2 × 1019 J.

44.10.

This is about 70% of the annual energy use in the U.S. EVALUATE: If this huge amount of energy were released suddenly, it would blow up the Enterprise! Getting useable energy from matter-antimatter annihiliation is not so easy to do! IDENTIFY: With a stationary target, only part of the initial kinetic energy of the moving electron is available. Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there must also be left over kinetic energy. Therefore not all of the initial energy is available. SET UP: The available energy is given by Ea2 = 2mc 2 ( Em + mc 2 ) for two particles of equal mass when one is initially stationary. In this case, the initial kinetic energy (20.0 GeV = 20,000 MeV) is much more than the rest energy of the electron (0.511 MeV), so the formula for available energy reduces to Ea = 2mc 2 Em . EXECUTE: (a) Using the formula for available energy gives Ea = 2mc 2 Em = 2(0.511 MeV)(20.0 GeV) = 143 MeV

44.11.

(b) For colliding beams of equal mass, each particle has half the available energy, so each has 71.5 MeV. The total energy is twice this, or 143 MeV. EVALUATE: Colliding beams provide considerably more available energy to do experiments than do beams hitting a stationary target. With a stationary electron target in part (a), we had to give the moving electron 20,000 MeV of energy to get the same available energy that we got with only 143 MeV of energy with the colliding beams. (a) IDENTIFY and SET UP: Eq. (44.7) says ω = q B / m so B = mω / q . And since ω = 2π f , this becomes

B = 2π mf / q . EXECUTE: A deuteron is a deuterium nucleus

( H ). Its charge is q = +e. Its mass is the mass of the neutral 2 1

atom (Table 43.2) minus the mass of the one atomic electron: m = 2.014102 u − 0.0005486 u = 2.013553 u (1.66054 × 10−27 kg /1 u) = 3.344 × 10−27 kg B=

2π mf 2π (3.344 × 10−27 kg)(9.00 × 106 Hz) = = 1.18 T q 1.602 × 10−19 C

2 1

H

Particle Physics and Cosmology

(b) Eq.(44.8): K =

44-3

q 2 B 2 R 2 [(1.602 × 10−19 C)(1.18 T)(0.320 m)]2 = . 2m 2(3.344 × 10−27 kg)

K = 5.471 × 10 −13 J = (5.471 × 10 −13 J)(1 eV/1.602 × 10−19 J) = 3.42 MeV 2K = m

K = 12 mv 2 so v =

2(5.471 × 10−13 J) = 1.81 × 107 m/s 3.344 × 10−27 kg

v / c = 0.06, so it is ok to use the nonrelativistic expression for kinetic energy. ω eB (a) 2 f = = = 3.97 × 107 s. π mπ eBR (b) ω R = = 3.12 × 107 m/s m (c) For three-figure precision, the relativistic form of the kinetic energy must be used, ( γ − 1 )mc 2 = 5.11× 106 V. eV = ( γ − 1 )mc 2 , so eV = ( γ − 1 )mc 2 , so V = e (a) IDENTIFY and SET UP: The masses of the target and projectile particles are equal, so Eq. (44.10) can be used. Ea2 = 2mc 2 ( Em + mc 2 ). Ea is specified; solve for the energy Em of the beam particles.

EVALUATE: 44.12.

44.13.

Ea2 − mc 2 2mc 2 The mass for the alpha particle can be calculated by subtracting two electron masses from the 42 He atomic mass:

EXECUTE:

Em =

m = mα = 4.002603 u − 2(0.0005486 u) = 4.001506 u Then mc 2 = (4.001506 u)(931.5 MeV/u) = 3.727 GeV. Em =

44.14.

Ea2 (16.0 GeV) 2 − mc 2 = − 3.727 GeV = 30.6 GeV. 2 2mc 2(3.727 GeV)

(b) Each beam must have 12 Ea = 8.0 GeV. EVALUATE: For a stationary target the beam energy is nearly twice the available energy. In a colliding beam experiment all the energy is available and each beam needs to have just half the required available energy. 1000 × 103 MeV = 1065.8, so v = 0.999999559c. (a) γ = 938.3 MeV (b) Nonrelativistic: ω =

Relativistic: ω = 44.15.

eB = 3.83 × 108 rad s. m

eB 1 = 3.59 × 105 rad s. m γ

(a) IDENTIFY and SET UP: For a proton beam on a stationary proton target and since Ea is much larger than the

proton rest energy we can use Eq.(44.11): Ea2 = 2mc 2 Em . EXECUTE:

44.16.

Em =

Ea2 (77.4 GeV) 2 = = 3200 GeV 2 2mc 2(0.938 GeV)

(b) IDENTIFY and SET UP: For colliding beams the total momentum is zero and the available energy Ea is the total energy for the two colliding particles. EXECUTE: For proton-proton collisions the colliding beams each have the same energy, so the total energy of each beam is 12 Ea = 38.7 GeV. EVALUATE: For a stationary target less than 3% of the beam energy is available for conversion into mass. The beam energy for a colliding beam experiment is a factor of (1/83) times smaller than the required energy for a stationary target experiment. IDENTIFY: Only part of the initial kinetic energy of the moving electron is available. Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there must also be left over kinetic energy. SET UP: To create the η0, the minimum available energy must be equal to the rest mass energy of the products, which in this case is the η0 plus two protons. In a collider, all of the initial energy is available, so the beam energy is the available energy. EXECUTE: The minimum amount of available energy must be rest mass energy Ea = 2mp + mη = 2(938.3 MeV) + 547.3 MeV = 2420 MeV

44-4

Chapter 44

44.17.

Each incident proton has half of the rest mass energy, or 1210 MeV = 1.21 GeV. EVALUATE: As we saw in problem 44.10, we would need much more initial energy if one of the initial protons were stationary. The result here (1.21 GeV) is the minimum amount of energy needed; the original protons could have more energy and still trigger this reaction. Section 44.3 says m(Z0 ) = 91.2 GeV c 2 . E = 91.2 × 109 eV = 1.461 × 10 −8 J; m = E c 2 = 1.63 × 10−25 kg; m(Z0 ) m(p) = 97.2

44.18.

(a) We shall assume that the kinetic energy of the Λ 0 is negligible. In that case we can set the value of the photon’s energy equal to Q: Q = (1193 − 1116) MeV = 77 MeV = Ephoton .

(b) The momentum of this photon is

p=

Ephoton c

=

(77 × 106 eV)(1.60 × 10−18 J eV ) = 4.1 × 10−20 kg ⋅ m s (3.00 × 108 m s )

To justify our original assumption, we can calculate the kinetic energy of a Λ 0 that has this value of momentum K Λ0 =

44.19.

44.20.

p2 E2 (77 MeV) 2 = = = 2.7 MeV Fe > Fweak > Fg (d) Fe ≈ 1 × 1036 Fg . Fstr ≈ 100 Fe ≈ 1× 1038 Fg . Fweak ≈ 10−9 Fstr ≈ 1× 1029 Fg

44.46.

EVALUATE: The gravity force is much weaker than any of the other three forces. Gravity is important only when one very massive object is involved. In Eq.(44.9), Ea = ( mΣ0 + mK0 )c 2 , and with M = mp , m = mπ − and Em = ( mπ − )c 2 + K ,

K=

Ea2 − ( mπ − c 2 ) 2 − ( mp c 2 ) 2 2mpc 2

− ( mπ − )c 2

(1193 MeV + 497.7 MeV) 2 − (139.6 MeV) 2 − (938.3 MeV) 2 − 139.6 MeV = 904 MeV. 2(938.3 MeV) IDENTIFY: With a stationary target, only part of the initial kinetic energy of the moving proton is available. Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there must also be left over kinetic energy. Therefore not all of the initial energy is available. K=

44.47.

SET UP: The available energy is given by Ea2 = 2mc 2 ( Em + mc 2 ) for two particles of equal mass when one is

initially stationary. The minimum available energy must be equal to the rest mass energies of the products, which in this case is two protons, a K+ and a K − . The available energy must be at least the sum of the final rest masses. EXECUTE: The minimum amount of available energy must be Ea = 2mp + mK + + mK − = 2(938.3 MeV) + 493.7 MeV + 493.7 MeV = 2864 MeV = 2.864 GeV Solving the available energy formula for Em gives Ea2 = 2mc 2 ( Em + mc 2 ) and Em =

44.48.

Ea2 (2864 MeV) 2 − mc 2 = − 938.3 MeV = 3432.6 MeV 2 2mc 2(938.3 MeV)

Recalling that Em is the total energy of the proton, including its rest mass energy (RME), we have K = Em – RME = 3432.6 MeV – 938.3 MeV = 2494 MeV = 2.494 GeV Therefore the threshold kinetic energy is K = 2494 MeV = 2.494 GeV. EVALUATE: Considerably less energy would be needed if the experiment were done using colliding beams of protons. (a) The decay products must be neutral, so the only possible combinations are π 0π 0π 0 or π 0π +π − (b) mη0 − 3mπ 0 = 142.3 Me V c 2 , so the kinetic energy of the π 0 mesons is 142.3 MeV. For the other reaction,

K = (mη0 − mπ 0 − mπ + − mπ − )c 2 = 133.1 MeV. 44.49.

IDENTIFY and SET UP: Apply conservation of linear momentum to the collision. A photon has momentum hc . All the mass of the electron and p = h / λ , in the direction it is traveling. The energy of a photon is E = pc =

λ

positron is converted to the total energy of the two photons, according to E = mc 2 . The mass of an electron and of a positron is me = 9.11 × 10 −31 kg EXECUTE: (a) In the lab frame the initial momentum of the system is zero, since the electron and positron have equal speeds in opposite directions. According to momentum conservation, the final momentum of the system must also be zero. A photon has momentum, so the momentum of a single photon is not zero. (b) For the two photons to have zero total momentum they must have the same magnitude of momentum and move in opposite directions. Since E = pc , equal p means equal E.


44-9

(c) 2 Eph = 2mec 2 so Eph = me c 2

Eph =

44.50.

44.51.

44.52.

hc

λ

so

hc

λ

= mec 2 and λ =

h 6.63 × 10 −34 J ⋅ s = = 2.43 pm me c (9.11 × 10−31 kg)(3.00 × 108 m/s)

These are gamma ray photons. EVALUATE: The total charge of the electron/positron system is zero and the photons have no charge, so charge is conserved in the particle-antiparticle annihilation. (a) If the π − decays, it must end in an electron and neutrinos. The rest energy of π − (139.6 MeV) is shared between the electron rest energy (0.511 MeV) and kinetic energy (assuming the neutrino masses are negligible). So the energy released is 139.6 MeV – 0.511 MeV = 139.1 MeV. (b) Conservation of momentum leads to the neutrinos carrying away most of the energy. (a) The baryon number is 0, the charge is +e , the strangeness is 1, all lepton numbers are zero, and the particle is K + . (b) The baryon number is 0, the charge is −e , the strangeness is 0, all lepton numbers are zero, and the particle is π − . (c) The baryon numbers is –1, the charge is 0, the strangeness is zero, all lepton numbers are 0, and the particle is an antineutron. (d) The baryon number is 0, the charge is +e , the strangeness is 0, the muonic lepton number is –1, all other lepton numbers are 0, and the particle is μ + . Δ t = 7.6 × 10 −21 s ⇒ Δ E =

= 1.054 × 10−34 J ⋅ s = = 1.39 × 10 −14 J = 87 keV . 7.6 × 10 −21 s Δt

Δ E 0.087 MeV = = 2.8 × 10−5. mψ c 2 3097 MeV 44.53. 44.54.

= (1.054 × 10 −34 J ⋅ s) = = 1.5 × 10 −22 s. Δ E (4.4 × 106 eV)(1.6 × 10−19 J/eV)

φ → K + + K − . The total energy released is the energy equivalent of the mass decrease. (a) EXECUTE: The mass decrease is m(φ ) − m(K + ) − m(K − ). The energy equivalent of the mass decrease is IDENTIFY and SET UP:

mc 2 (φ ) − mc 2 (K + ) − mc 2 (K − ). The rest mass energy mc 2 for the φ meson is given in Problem 44.53, and the

values for K +and K − are given in Table 44.3. The energy released then is 1019.4 MeV − 2(493.7 MeV) = 32.0 MeV. The K + gets half this, 16.0 MeV. EVALUATE: (b) Does the decay φ → K + + K − + π 0 occur? The energy equivalent of the K + + K − + π 0 mass is 493.7 MeV + 493.7 MeV + 135.0 MeV = 1122 MeV. This is greater than the energy equivalent of the φ mass. The mass of the decay products would be greater than the mass of the parent particle; the decay is energetically forbidden. (c) Does the decay φ → K + + π − occur? The reaction φ → K + + K − is observed. K + has strangeness +1 and K − has strangeness −1, so the total strangeness of the decay products is zero. If strangeness must be conserved we deduce that the φ particle has strangeness zero. π − has strangeness 0, so the product K + + π − has strangeness

44.55.

−1. The decay φ → K + + π − violates conservation of strangeness. Does the decay φ → K + + μ − occur? μ − has strangeness 0, so this decay would also violate conservation of strangeness. (a) The number of protons in a kilogram is

⎛ 6.023 × 10 23 molecules mol ⎞ 25 (1.00 kg) ⎜ ⎟ (2 protons molecule) = 6.7 × 10 . −3 18.0 10 kg mol × ⎝ ⎠ Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy per decay is mp c 2 = 938.3 MeV = 1.503 × 10−10 J, and so the energy deposited in a year, per kilogram, is

44.56.

⎛ ln(2) ⎞ −10 −3 (6.7 × 1025 ) ⎜ ⎟ (1 y) (1.50 × 10 J) = 7.0 × 10 Gy = 0.70 rad 18 ⎝ 1.0 × 10 y ⎠ (b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem. IDENTIFY and SET UP: The total released energy is the equivalent of the mass decrease. Use conservation of linear momentum to relate the kinetic energies of the decay particles. EXECUTE: (a) The energy equivalent of the mass decrease is mc 2 (Ξ − ) − mc 2 ( Λ 0 ) − mc 2 (π − ) = 1321 MeV − 1116 MeV − 139.6 MeV = 65 MeV

44-10

Chapter 44

(b) The Ξ − is at rest means that the linear momentum is zero. Conservation of linear momentum then says that the Λ 0 and π − must have equal and opposite momenta: mΛ 0 vΛ 0 = mπ − vπ −

⎛m 0 ⎞ vπ − = ⎜ Λ ⎟ vΛ 0 ⎜m − ⎟ ⎝ π ⎠ Also, the sum of the kinetic energies of the Λ 0 and π − must equal the total kinetic energy K tot = 65 MeV calculated in part (a): K tot = K Λ0 + Kπ − K Λ0 + 12 mπ − vπ2 − = K tot Use the momentum conservation result: 2

⎛m 0 ⎞ K Λ0 + mπ − ⎜ Λ ⎟ vΛ2 0 = K tot ⎜m − ⎟ ⎝ π ⎠ ⎛ mΛ 0 ⎞ 1 K Λ0 + ⎜ ⎟ m v 2 = K tot ⎜ m − ⎟ ( 2 Λ0 Λ0 ) ⎝ π ⎠ ⎛ mΛ0 ⎞ K Λ0 ⎜ 1 + ⎟ = K tot ⎜ m−⎟ π ⎠ ⎝ K tot 65 MeV = = 7.2 MeV K Λ0 = 1+mΛ0 / mπ − 1 + (1116 MeV)/(139.6 MeV) 1 2

K Λ0 + Kπ − = K tot so Kπ − = K tot − K Λ 0 = 65 MeV − 7.2 MeV = 57.8 MeV

7.2 MeV = 11%. 65 MeV 57.8 MeV The fraction for the π − is = 89%. 65 MeV EVALUATE: The lighter particle carries off more of the kinetic energy that is released in the decay than the heavier particle does. dR dR dt HR (a) For this model, = HR, so = = H , presumed to be the same for all points on the surface. dt R R dr dR (b) For constant θ , = θ = HRθ = Hr. dt dt dR dt (c) See part (a), H 0 = . R dR (d) The equation = H 0 R is a differential equation, the solution to which, for constant H 0 , is R(t ) = dt R0e H 0 t , where R0 is the value of R at t = 0 . This equation may be solved by separation of variables, as

The fraction for the Λ 0 is

44.57.

dR dt d = ln ( R) = H 0 and integrating both sides with respect to time. R dt (e) A constant H 0 would mean a constant critical density, which is inconsistent with uniform expansion. 44.58.

r dR 1 dr r dθ 1 dr dθ since = − = = 0. From Problem 44.57, r = Rθ ⇒ R = . So θ dt θ dt θ 2 dt θ dt dt 1 dR 1 dr 1 dr dr ⎛ 1 dR ⎞ dv d ⎛ r dR ⎞ d ⎛ dR ⎞ So = = ⇒v= =⎜ =0= ⎟ r = H 0 r. Now ⎜ ⎟= ⎜θ ⎟ dt ⎝ R dt ⎠ dθ dθ ⎝ R dt ⎠ dθ ⎝ dt ⎠ R dt Rθ dt r dt dR dR K dθ 1 dR θ K 1 ⎛K⎞ ⇒θ = K where K is a constant. ⇒ = ⇒ R = ⎜ ⎟ t since = 0 ⇒ H0 = = = . So the dt θ dt R dt Kt θ t dt ⎝θ ⎠ 1 where T is the present age of the universe. T v0 − vcm (a) For mass m, in Eq. (37.23) u = −vcm , v′ = v0 , and so vm = . For mass 1 − v0 vcm c 2 M , u = −vcm , v′ = 0, so vM = −vcm .

current value of the Hubble constant is 44.59.


44-11

(b) The condition for no net momentum in the center of mass frame is mγ m vm + M γ M vM = 0, where

γ m and γ M correspond to the velocities found in part (a). The algebra reduces to β mγ m = ( β 0 − β ′)γ 0γ M , where v0 v , β ′ = cm , and the condition for no net momentum becomes m( β 0 − β ′)γ 0γ M = M β ′γ M , or c c β0 mv0 m β′ = . = β0 . vcm = 2 M m + M 1 − (v0 / c) 2 m + M 1 − β0 1+ mγ 0 (c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively M m . After some more algebra, ,vM = −v0γ 0 simple forms vm = v0γ 0 m + Mγ 0 mγ 0 + M

β0 =

γm =

m + Mγ 0 m + M + 2mM γ 0 2

2

,γM =

M + mγ 0 m + M 2 + 2mM γ 0 2

, from which mγ m + M γ M = m 2 + M 2 + 2mM γ 0 . This last

expression, multiplied by c 2 , is the available energy Ea in the center of mass frame, so that Ea2 = (m 2 + M 2 + 2mMγ0 )c 4 = (mc 2 ) 2 + (Mc 2 ) 2 + (2Mc 2 )(mγ0c 2 ) = ( mc 2 ) 2 + ( Mc 2 ) 2 + 2 Mc 2 Em , which is Eq.(44.9).

44.60.

Λ0 → n + π 0 (a) E = ( Δ m)c 2 = (mΛ0 )c 2 − (mn )c 2 ( mπ 0 )c 2 = 1116 MeV − 939.6 MeV − 135.0 MeV = 41.4 MeV

(b) Using conservation of momentum and kinetic energy; we know that the momentum of the neutron and pion must have the same magnitude, pn = pπ .

K n = En − mn c 2 = ( mn c 2 ) 2 + ( pn c) 2 − mn c 2 = ( mn c 2 ) 2 + ( pn c) 2 − mn c 2 K n = (mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ − mn c 2 = Kπ + K n = Kπ + ( mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ − mn c 2 = E. ( mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ = E 2 + ( mn c 2 ) 2 + Kπ2 + 2 Emπ c 2 − 2 EKπ − 2mn c 2 Kπ . Collecting terms we find: Kπ (2mπ c 2 + 2 E + 2mn c 2 ) = E 2 + 2 Emn c 2

⇒ Kπ =

(41.4 MeV) 2 + 2(41.4 MeV)(939.6 MeV) = 35.62 MeV. 2(135.0 MeV) + 2(41.4 MeV) + 2(939.6 MeV)

So the fractional energy carried by the pion is

35.62 = 0.86, and that of the neutron is 0.14. 41.4

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