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Physics: AP® Edition 2e

Student Solutions Manual

Chapter 1

CHAPTER 1: INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 1.2 PHYSICAL QUANTITIES AND UNITS 1.

Solution

The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?

(a)

100 km 1000 m 1h    27.77 m/s  27.8 m/s h 1 km 3600 s

(b)

100 km 1 mi   62 mi/h h 1.609 km

Show that 1.0 m/s  3.6 km/h . Hint: Show the explicit steps involved in converting 1.0 m/s  3.6 km/h .

Solution

1.0 m 1.0 m 3600 s 1 km     3.6 km/h s s hr 1000 m

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)

Physics: AP® Edition 2e

Solution

Student Solutions Manual

Chapter 1

1 ft  377.3 ft  377 ft long 0.3048 m 12 in. 377.3 ft   4528 in.  4.53  10 3 in. long 1.0 ft

115 m 

1 ft  278.9 ft  2.8  10 2 ft wide 0.3048 m 12 in. 278.9 ft   3346 in.  3.3  10 3 in. wide 1.0 ft

85 m 

Solution

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

29,028 ft 

1 km  8.847 km 3281 ft

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?

Solution

4.0 cm  1m 1y 1d 1h      1.3  10 -9 m (a)   y  1s  100 cm 365.25 d 24.0 h 3600 s   (b)

4.0 cm 1m 1 km 10 6 y     40 km/My y 100 cm 1000 m 1 My

1.3 ACCURACY, PRECISION, AND SIGNIFICANT FIGURES 11.

Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?

Physics: AP® Edition 2e

Solution

13.

Solution

äm

Student Solutions Manual

Chapter 1

3%  65 kg  2 kg 100%

(a) A car speedometer has a 5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h ? (b) Convert this range to miles per hour. (1 km  0.6214 mi)

5.0%  90.0 km/h  4.5 km/h 100% Thus, the range  90.0  5 km/h  85 to 95 km/h.

(a) äv 

(b) 85.5 km  0.6214 mi  53.1 mi/h ; 94.5 km  0.6214 mi  58.7 mi/h h 1 km 1h 1 km So the range is 53 to 59 mi/h.

15.

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats do they have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?

Solution

72.0 beats 60.0 min 24.0 h 365.25 d     2.0 y  7.5738  10 7 beats 1 min 1.00 h 1.00 d 1.00 y

(a) 7.6  10 7 beats (limited by 2.0 y) (b) 7.57  10 7 beats (limited by 2.00 y) (c) 7.57  10 7 beats (limited by 72.0 beats/min)

17.

State how many significant figures are proper in the results of the following calculations: (a) 106.7 98.2/ 46.2101.01 (b) 18.7  (c) 1.60  10 19 3712. 2

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 1

Solution (a) 3 (limited by 98.2 and 1.01) (b) 3 (limited by 18.7) (c) 3 (limited by 1.60)

19.

Solution

(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h , what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h , what is the range of speeds you could be going?

(a) % unc 

(b) ä v 

2.0 km/h  100%  2.2% 90 km/h

2.2 %  60 km/h  1 km/h 100 %

So the range is 60  1 km/h or 59 to 61 km/h.

21.

A person measures their heart rate by counting the number of beats in 30 s . If 40  1 beats are counted in 30.0  0.5 s , what is the heart rate and its uncertainty in beats per minute?

Solution

beats 40 beats 60.0 s    80 beats/min. minute 30.0 s 1.00 min 1 beat 0.5 s  100%   100%  2.5%  1.7%  4.2%  4% 40 beats 30.0 s % unc 4.2% äA  A  80 beats/min  3.3 beats/min  3 beats/min 100% 100%

% unc 

The heart rate is 80  3 beats/min.

23.

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 1

Solution

26.22 mi  2.8 h 9.5 mi/h

25.

The sides of a small rectangular box are measured to be 1.80  0.01 cm , 2.05  0.02 cm , and 3.1 0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.

Solution

V  1.80 cm  2.05 cm  3.1 cm  11.4 cm 3 . Use the methods of adding percents.

0.01 cm  100%  0.556% 1.80 cm 0.02 cm 2.05  0.02 cm   100%  0.976% 2.05 cm 0.1 cm 3.1  0.1 cm   100%  3.226% 3.1 cm 1.80  0.01 cm 

Adding these values and rounding to 1 sig. fig., the percent uncertainty of the volume is 5%. The uncertainty in the volume is therefore (11.4)(0.05) = 0.6. The volume is thus 11.4  0.6 cm3 .

27.

The length and width of a rectangular room are measured to be 3.955  0.005 m and 3.050  0.005 m . Calculate the area of the room and its uncertainty in square meters.

Solution The area is 3.995 m  3.050 m  12.06 m 2 . Now use method of adding percents to get uncertainty in the area.

% unc width 

0.005 m  100%  0.16% 3.050 m

0.005 m  100%  0.13% 3.955 m % unc area  0.13%  0.16%  0.29%  0.3%

% unc length 

ä area 

0.29%  12.06 m 2  0.035 m 2  0.04 m 2 100%

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 1

So the area is 12.06  0.04 m 2 .

1.4 APPROXIMATION 29.

How many heartbeats are there in a lifetime?

Solution

10 9 s 1 heartbeat 1 lifetime    2  10 9 heartbeats 0.5 lifetime 1s

31.

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10 22 s .)

Solution

Th 2  109 s   2  1031 times Tn 10-22 s

33.

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

Solution

dm dm 10 -8 m    50 atoms d a 2d H (2)(10 -10 m/atom)

35.

(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?

Solution

(a)

10 -2 kg/hummingbird  1012 cells/hummingbird 10  10 -15 kg/cell

Physics: AP® Edition 2e

Student Solutions Manual

10 2 kg/person (b)  1016 cells/person -15 10  10 kg/cell

Chapter 1

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 2

CHAPTER 2: KINEMATICS 2.1 DISPLACEMENT 1.

Solution

Find the following for path A in Figure 2.57: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

(a) 7 m (b) 7 m (c)

x  7 m 0 m   7 m

Find the following for path C in Figure 2.57: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

Solution (a) (b) (c)

8 m  2 m  2 m  12 m 8m

x  10 m 2 m  8 m

2.3 TIME, VELOCITY, AND SPEED 5.

(a) Calculate Earth’s average speed relative to the Sun. (b) What is its average velocity over a period of one year?

Physics: AP® Edition 2e

Solution

Student Solutions Manual

Chapter 2

Dist. Traveled 2r  Time t 11 2 (1.50  10 m) 1 day 1 hour    365.25 days 24 hours 3600 s

Avg. Speed of Earth =

(a)

 2.99  10 4 m/s  3.0  10 4 m/s

(b) After one year, Earth has returned to its original position with respect to the Sun. Thus, v  0 m/s. 7.

The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?

Solution

500 km 5.00  10 5 m 100 cm t     1.67  10 7 years  2  10 7 y rate 3 cm/y 1m

On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?

Solution

3600 s   60 s   t  13 h   4 min   58 s 1h   1 min   1h  4.7098  10 4 s  4.7098  10 4 s   13.083 hours 3600 s average speed in km/h  average speed in m/s 

distance traveled 1633.8 km   124.88 km/h time elapsed 13.0828 h 1.6338  10 6 m  34.689 m/s 4.7098  10 4 s

Physics: AP® Edition 2e

11.

Student Solutions Manual

Chapter 2

A student drove to the university from their home and noted that the odometer reading of their car increased by 12.0 km. The trip took 18.0 min. (a) What was their average speed? (b) If the straight-line distance from their home to the university is 10.3 km in a direction 25.0 south of east, what was their average velocity? (c) If they returned home by the same path 7 h 30 min after they left, what were their average speed and velocity for the entire trip?

Solution average speed 

(a) v

(b)

12.0 km 60 min   40.0 km/h 18.0 min 1h

x 10.3 km 60 min    34.3 km/h, 25 S of E. t 18.0 min 1h

average speed  2 

(c)

12.0 km  3.20 km/h 7.5 h

v  0 , since the total displacement was 0. 13.

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of 8 light (3.00  10 m/s) .

Solution

dist. traveled = avg. speed  elapsed time = 3.00  10 8 m/s  2.56 sec  767,000 km E M dist. 

dist. traveled 767,000 km   384,000 km 2 2

Physics: AP® Edition 2e

15.

Student Solutions Manual

Chapter 2

The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as 1.06  10 10 m in diameter. (a) If the having a single electron in a circular orbit  6 average speed of the electron in this orbit is known to be 2.20  10 m/s , calculate the number of revolutions per second it makes about the nucleus. (b) What is the  electron’s average velocity per revolution?

Solution

distance  2.20  10 6 m/s time distance/revolution  2r/revolution  d / revolution average speed 

  (1.06  10 10 m)/revolution  3.33  10 10 m/revolution revolutions average speed 2.20  10 6 m/s   1s distance/revolution 3.33  10 10 m/revolution  6.61  1015 rev/s

(a)

(b) v  0 m/s , since there is no net displacement per revolution.

2.4 ACCELERATION 16.

Solution

18.

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

a

vf v0 30.0 m/s 0 m/s   4.29 m/s 2 t 7s

2 A commuter backs her car out of her garage with an acceleration of 1.40 m/s . (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Solution t

v v0 2.00 m/s 0 m/s   1.43 s a 1.40 m/s 2

a

v v 0 0 m/s 2.00 m/s   2.50 m/s 2 t 0.800 s

(a)

(b)

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 2

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 2

2.5 MOTION EQUATIONS FOR CONSTANT ACCELERATION IN ONE DIMENSION 20.

Solution

An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

(a) v  v0  at  0 m/s  (4.50 m/s )(2.40 s)  10.8 m/s 2 2 (b) Assuming the acceleration is constant, we know x  (½)at  2.25t . We can create a graph by plugging in a few different t-values, say t = 1, 2, 3, 4, 5:

Time (s) 0 1 2 3 4 5 21.

Position (m) 0 2.25 9 20.25 36 56.25

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 3 2.10  10 4 m/s 2 , and 1.85 ms (1 ms  10 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Physics: AP® Edition 2e

Solution

Student Solutions Manual

v 0  v at  0 m/s (2.10  10 4 m/s 2 )(1.85  10 3 s)  38.9 m/s

Chapter 2

(about 87 miles

per hour) 23.

2 (a) A light-rail commuter train accelerates at a rate of 1.35 m/s . How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train 2 ordinarily decelerates at a rate of 1.65 m/s . How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in

m/s2 ?

Solution

t

v v0 (80.0 km/h 0 km/h) 1h 1000 m     16.5 s 2 a 1.35 m/s 3600 s 1 km

t

v v0 0 km/h 80.0 km/h 1h 1000 m     13.5 s 2 a 1.65 m/s 3600 s 1 km

a

v v0 0 km/h 80.0 km/h 1h 1000 m     2.68 m/s 2 t 8.30 s 3600 s 1 km

(a)

(b)

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2 . (a) How far does she travel in the next 5.00 s? (b) What is her final

velocity? (c) Evaluate the result. Does it make sense? Solution (a)

x  x0  v0t 

1 2 1 at  0 m  (9.00 m/s)(5.00 s)  (2.00 m/s 2 )(5.00 s) 2  20.0 m 2 2 2

(b) v  v0  at  9.00 m/s  (2.00 m/s )(5.00 s)  1.00 m/s (c) This result does not really make sense. If the runner starts at 9.00 m/s and 2 decelerates at 2.00 m/s , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards. 27.

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 2 40.0 m/s in the same direction. If this shot takes 3.33  10 s , calculate the distance over which the puck accelerates.

Physics: AP® Edition 2e

Solution

29.

Student Solutions Manual

x x0  vavg t 

Chapter 2

1 1 (v0  v)t  (8.00 m/s  40.0 m/s)(3.33  10 2 s)  0.799 m 2 2

Freight trains can produce only relatively small accelerations and decelerations. (a) 2 What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow

2 down at a rate of 0.550 m/s , how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?

Solution (a)

v  v0  at  4.00 m/s  (0.0500 m/s 2 )(8.00 min 

t

(b)

60 s )  28.0 m/s 1 min

v v0 0 m/s 28.0 m/s   50.9 s a 0.550 m/s 2

x x0  For part (b), 31.

v 2 v02 (0 m/s) 2 (28.0 m/s) 2   713 m 2a 2(0.550 m/s 2 )

A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it 2 accelerates from rest at an average rate of 0.350 m/s , how far will it travel before becoming airborne? (b) How long does this take?

Solution

x x0  (a) t

(b) 33.

v 2 v02 (6.00 m/s) 2 (0 m/s) 2   51.4 m 2a 2(0.350 m/s 2 )

v v0 6.00 m/s 0 m/s   17.1 s a 0.350 m/s 2

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?

Physics: AP® Edition 2e

Solution

a

v 2 v02 (0 m/s) 2 (7.50 m/s) 2   80.4 m/s 2 2( x x0 ) 2(0.350 m)

t

2( x x 0 ) 2(0.350 m)   9.33  10 2 s v  v0 0 m/s  7.50 m/s

(a) (b) 35.

Student Solutions Manual

Chapter 2

Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution 2 2 2 (a) v  v0  2ax  v  2ax  2(9.8 m/s )(3.0 m)  7.7 m/s

a (b)

v 2 v02 (0 m/s) 2 (7.7 m/s) 2   1.5  10 3 m/s 2 2x 2(0.02 m)

This is ~ 3 times the deceleration of the pilots, who were falling from thousands of meters high! 37.

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Physics: AP® Edition 2e

Solution a

(a) 2

Student Solutions Manual

Chapter 2

v v0 145 m/s 0 m/s   32.6 m/s 2 t 4.45 s 2

(b) v  v0  2a ( x x0 ) , and rearranging v  (0 m/s) 2  2(32.6 m/s 2 )(402 m)  162 m/s

(c) v  vmax because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be 2 greatest at the beginning, so it would not be accelerating at 32.6 m/s during the last few meters, but substantially less, and the final velocity would be less than 162 m/s . 39.

Solution

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

There are two parts to the race: an acceleration part and a constant speed part. First, we need to determine how long (both in distance and time) it takes the 60 mph a 4s . motorcycle to finish accelerating. During acceleration,  4s  v max  183 mph  at1  t1  183 mph 60 mph   12.2 s   183 mph  0 mph    1h  x1  vavg t1   (12.2 s)  (91.5 mph) (12.2 s)  0.31 mi   2   3600 s 

At constant velocity, x2  5 mi 0.31 mi  4.7 mi . Now, we complete the calculation by determining how much time is spent on the course at max speed. t2 

x2 4.7 mi 3600 s   0.026 h  92 s, so v max 183 mph 1h

t total  t1  t 2  12.2 s  92 s  104 s

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 2

40.

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution

(a) There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into x1 (while accelerating) and x2 (with constant speed), where x1  x2  100 m . When the speed is constant, t  6.69 s , so x2  vt  (6.69 s)v  100 m x1 . x1 

1 2 1 at  vt  (1.50 s)v 2 2 . Plugging x1 into

When accelerating, t  3.00 s , so the previous equation, we get his maximum speed: 100 m 1.50 s v  6.69 s v  v 

100 m  12.2 m/s 6.69 s  1.50 s

a

Therefore, his acceleration was

v 12.2 m/s   4.07 m/s 2 t 3.00 s

(b) Similar to part (a), we can plug in the different values for time and total distance: 200 m 1.50 s v  16.30 s v  v 

200 m  11.2 m/s 16.30 s  1.50 s

2.7 FALLING OBJECTS 41.

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0  0 .

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Chapter 2

Solution 2

Knowns: a  acceleration due to gravity  g  9.8 m/s ; y0  0 m; v0   15.0 m/s 1 y  y0  v0t  at 2 2 To find displacement we use , and to find velocity we use v  v0  at .

y1  y0  v0t1 

1 2 at1 2

1 (9.8 m/s 2 )(0.500 s) 2  6.28 m 2 (a) v1  v0  at1  (15.0 m/s)  (9.8 m/s 2 )(0.500 s)  10.1 m/s  0 m  (15.0 m/s)(0.500 s) 

y 2  y 0  v0 t 2 

1 2 at 2 2

1 (9.8 m/s 2 )(1.00 s) 2  10.1 m 2 (b) v2  v0  at 2  (15.0 m/s)  (9.8 m/s 2 )(1.00 s)  5.20 m/s  0 m  (15.0 m/s)(1.00 s) 

y 3  y 0  v0 t 3 

1 2 at3 2

1 (9.8 m/s 2 )(1.50 s) 2  11.5 m 2 (c) v3  v0  at3  (15.0 m/s)  (9.8 m/s 2 )(1.50 s)  0.300 m/s  0 m  (15.0 m/s)(1.50 s) 

The ball is almost at the top. y 4  y 0  v0 t 4 

1 2 at 4 2

1 (9.8 m/s 2 )(2.00 s) 2  10.4 m 2 (d) v4  v0  at 4  (15.0 m/s)  (9.8 m/s 2 )(2.00 s)  4.60 m/s  0 m  (15.0 m/s)(2.00 s) 

The ball has begun to drop.

43.

A basketball referee tosses the ball straight up for the starting tip-off. At what minimum velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Physics: AP® Edition 2e

Solution

Student Solutions Manual

Chapter 2

v 2  v02  2a ( y y0 )  v02  v 2 2a ( y y0 ) v0  v 2 2a ( y y0 )  (0 m/s) 2 2(9.80 m/s 2 )(1.25 m)  4.95 m/s

45.

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.

Solution

(a) Knowns: a  9.80 m/s ; v0  13.0 m/s; y0  0 m

(b) At the highest point in the jump, v  0 m/s . We can use the equation v 2 v02 y y 0  2a because the only unknown it includes is y , which is what we want to solve for. First we substitute the knowns into the equation and then we solve for y .

v 2 v02 v 2 v02 (0 m/s) 2 (13.0 m/s) 2 y y0  y y0  0 m  8.62m 2a 2a 2(9.80 m/s 2 ) . Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result. (c) If t is the time for the dolphin to reach its peak height, then 2t is the time the dolphin is out of the water. v  v0  at  t 

47.

v v0 0 m/s 13.0 m/s   1.3265 s, and 2t  2.65 s a 9.8 m/s 2

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

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Solution

(a) Knowns: t  2.35 s; y  0 m; v0   8.00 m/s; a  9.8 m/s

Since we know t , y , v0 , and a and want to find y0 , we can use the equation 1 y  y 0  v 0 t  at 2 2 . y  (0 m)  (8.00 m/s)(2.35 s)  

1 (9.80 m/s 2 )(2.35 s) 2  8.26 m 2 , so the cliff

is 8.26 m high. (b) Knowns: y  0 m; y0  8.26 m; v0  8.00 m/s; a  9.80 m/s

Now we know y , y0 , v0 , and a and want to find t , so we use the equation 1 y  y 0  v 0 t  at 2 2 again. Rearranging,



t

v0  v02 4(0.5a )( y0 y ) 2(0.5a )

(8.00 m/s)  (8.00 m/s) 2  2(9.80 m/s 2 )(8.26 m 0 m) t (9.80 m/s 2 ) 8.00 m/s  15.03 m/s  9.80 m/s 2 t  0.717 s or 2.35 s  t  0.717 s 49.

Solution

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Knowns: a  9.80 m/s ; v0  15.0 m/s; y  7.00 m

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Student Solutions Manual

y  v0 t  t 

Chapter 2

1 2 at  2

v0  v02  2ay a 15.0 m/s  (15.0 m/s) 2  2(9.80 m/s 2 )(7.00 m) 9.80 m/s



15.0 m/s  9.37 m/s 9.80 m/s 2

So t1  0.58 s and t 2  2.49 s , so the total time between passing the branch is

1.91 s .

51.

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution (a)

y  v0 t 

1 2 1 at  (9.80 m/s 2 )(1.50 s) 2  11.0 m 2 2

So the rock falls 11.0 m in the 1.50 s before the hiker sees the rock. When he finally sees the rock, it is 94.0 m above his head.

y (b)

1 2 at  t  2

2y 2(105 m)   4.63 s a 9.80 m/s 2 , so the rock will take 4.63 s to fall

the full distance. Thus, the hiker will have 4.63 s 1.50 s  3.13 s to move out of the way before the rock strikes the hiker's location, ignoring the height of person.

Physics: AP® Edition 2e

53.

Solution

Student Solutions Manual

Chapter 2

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

(a) Knowns: v0  0 m/s, y0  0 m, y  250 m, a  9.80 m/s .

v   v02  2a( y y0 )   (0 m/s) 2  2(9.80 m/s 2 )(250 m)  70.0 m/s

downward. (b) Let t s = the time for the sound to travel to the tourist. ts 

height of cliff 250 m   0.7463 s vs 335 m/s

Let t = the total time before the tourist can react. t  t s  reaction time  0.7463 s  0.300 s  1.046 s

Let t b = the time it takes the rock to reach the bottom. tb 

v v0 7.00 m/s 0 m/s   7.143s. g 9.8 m/s 2

Now subtract t b t  7.143 s 1.046 s  6.10 s 55.

Solution

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

1 y y0  v0 t  at 2  0.5(9.80 m/s 2 )(2.0000 s) 2  19.6 m. 2 (a)

(b) Let h be the depth of the well, y 0  0 m, y  h at the bottom of the well. Let

v2 = the speed of sound and t 2 = the time for the sound to travel from the

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Chapter 2

bottom to the top of the well. Let t1 = the time for the rock to reach the bottom.

t1 + t 2 = t = 2.0000 s

(i)

1 2 at1 2

(ii)

h 

h = v 2 t 2 = v 2 (2.0000 s t1 )

(iii)

Adding equations (ii) and (iii): 1 0  at12  v2 (2.0000 s t1 ) 2  0.5(9.80 m/s 2 )t12  (2.0000 s)(332.00 m/s) (232.00 m/s)t1 0  (4.90 m/s 2 )t12 (332.00 m/s)t1  664m

Using the quadratic formula: 332 m/s  (332 m/s) 2 4(4.90 m/s 2 )(664 m) t1   1.944 s. 9.80 m/s 2 1 h   at12  0.5(9.80 m/s 2 )(1.944 s) 2  18.5 m 2

57.

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Physics: AP® Edition 2e

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Chapter 2

Solution 2

(a) Knowns: v0 = 10.0 m/s; a = 9.80 m/s ; v = 0 m/s (at top of ascent) v 2 v02 0 m/s (10.0 m/s) 2 y   5.10 m 2a 2(9.80 m/s 2 ) . Maximum height = 300 m + 5.1 m = 305 m . 1 1 y  v0t  at 2  (10.0 m/s)(4.00 s)  (9.80 m/s 2 )(4.00 s) 2  38.4 m 2 2 (b) , which is a height of 262 m. v  v0  at  10.0 m/s  ( 9.80 m/s 2 )(4.00 s)  29.2 m/s

1 1 y  y0 + v0 t  at 2  ( y0 y )  v0 t  at 2  2 2 t

v0  v02 2a ( y0 y )



a 10 m/s  77 m/s t   8.91s. (c) 9.8 m/s 2

10.0 m/s  (10.0 m/s) 2 2(9.80 m/s 2 )(300 m) 9.80 m/s 2

2.8 GRAPHICAL ANALYSIS OF ONE-DIMENSIONAL MOTION 59.

(a) By taking the slope of the curve in Figure 2.58, verify that the velocity of the jet car is 115 m/s at t  20 s . (b) By taking the slope of the curve at any point in 2

Figure 2.59, verify that the jet car’s acceleration is 5.0 m/s . Solution

position (meters)

Position vs. Time 4000 3000 2000 1000 0 0

time (seconds)

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t (s) 0 10 20 30

v

velocity (meters per second)

(a)

Chapter 2

x (m) 200 600 1500 2900

rise (2900 600) m   115 m/s run (30 10) s

Velocity vs. Time 200 150 100 50 0 0

10 15 20 25 30 35

time (seconds)

t (s) 0 5 10 15 20 25 30 a

(b)

61.

v (m/s) 15 40 65 90 115 140 165

rise (140-65) m   5.0 m/s 2 run (25-10) s

Using approximate values, calculate the slope of the curve in Figure 2.60 to verify that the velocity at t  30.0 s is approximately 0.24 m/s.

Solution

v

(11.7 7) m  .235  0.238 m/s (40 20) s

Physics: AP® Edition 2e

63.

Student Solutions Manual

Chapter 2

Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). You will need to use the information on acceleration and velocity given in the examples for this figure.

Solution

position (km)

Position vs. Time 4.85 4.8 4.75 4.7 4.65 0

time (seconds)

65.

A graph of vt  is shown for a world-class track sprinter in a 100-m race. (See Figure 2.65). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t  5 s ? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Solution

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(a) For the first 4 s,

v avg = 6 m/s

Chapter 2

(b) At 5 s, v = 12 m/s . a

(c)

12 m/s  3 m/s 2 4s .

(d) For the first 4 s, we know a and v, so x = 24 m . For the second interval, distance remaining = 100 m 24 m = 76 m and 76 m t2   6.3 s. v = 12 m/s , so 12 m/s So t total = t1 + t 2 = 4 s + 6.3 s = 10 s .

67.

Critical Thinking Two cars are side by side on a path where friction is negligible. Car A starts from rest and has a positive, constant acceleration while car B travels at a constant speed. The cars reach the same check point down the road at the same time. (a) Which car is traveling faster at the check point? Briefly explain your reasoning. (b) Based on experimental data, an equation that fits the data is suggested for the speed of car A at the check point, V, which may not be correct: V  KV0 , where K is a constant with appropriate units and V0 is the initial velocity of car B. Is this equation consistent with your answer from part a? Explain why or why not. Does this equation make sense? Explain why or why not. (c) Graph the velocity of car A and the velocity of car B vs. distance along the path, starting when the cars are side by side.

Solution

(a) Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance. (b) i. ii.

Yes, the equation is consistent with the answer because the speed of car A is only a constant away from the correct answer. Yes, the equation makes sense because V  2V0 .

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Chapter 2

(c)

TEST PREP FOR AP® COURSES Which of the following statements comparing position, distance, and displacement is correct? a. An object may record a distance of zero while recording a non-zero displacement. b. An object may record a non-zero distance while recording a displacement of zero. c. An object may record a non-zero distance while maintaining a position of zero. d. An object may record a non-zero displacement while maintaining a position of zero.

Solution

(b)

3. A group of students has two carts, A and B, with wheels that turn with negligible friction. The two carts travel along a straight horizontal track and eventually collide. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together. a. Describe an experimental procedure to determine the velocities of the carts before and after the collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure. b. There will be sources of error in the measurements taken in the experiment both before and after the collision. Which velocity will be more greatly affected by this error: the velocity prior to the collision or the velocity after the collision? Or will both sets of data be affected equally? Justify your answer.

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Solution

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Chapter 2

a. Use tape to mark off two distances on the track — one for cart A before the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times. b. If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision’ and error that ‘equally affects both sets of measurement.’) Push a book across a table and observe it slow to a stop. Draw graphs showing the book’s position vs. time and velocity vs. time if the direction of its motion is considered positive. Draw graphs showing the book’s position vs. time and velocity vs. time if the direction of its motion is considered negative.

Solution

The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive y-intercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis. The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative y-intercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.

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Chapter 2

7. Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 100 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid? a. b. c. d. e. Solution

m/s2 0.03 m/s2 0.20 m/s2 0.65 m/s2 33 m/s2

(c)

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Chapter 3

CHAPTER 3: TWO-DIMENSIONAL KINEMATICS 3.2 VECTOR ADDITION AND SUBTRACTION: GRAPHICAL METHODS 1.

Find the following for path A in Figure 3.49: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

Solution

(a) d  (3  120 m)  (1  120 m)  480 m 2

s  sx  s y 

1  120 m 2  3  120 m 2  379 m

s  1  120 m   tan 1  x  tan 1   18.4, E of N s  3  120 m   (b) y  

Find the north and east components of the displacement for the hikers shown in Figure 3.47.

Solution

N-component: 5.00 km sin 40.0  3.21 km o

E-component: 5.00 km cos 40.0  3.83 km o

Suppose you first walk 12.0 m in a direction 20.0 west of north and then 20.0 m in a o

direction 40.0 south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.51, then this problem finds their sum R = A + B .) Solution

R  AB

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Chapter 3

Rx  Ax  Bx  A sin 20  B cos 40  12.0 m sin 20 20.0 m cos 40  19.425 m R y  Ay  B y  A cos 20 B sin 40  12.0 m cos 20 20.0 m sin 40  1.579 m 2

19.425 m 2  1.579 m 2  19.5 m

R  Rx  R y  è  tan 1

1.579 m   tan 1   4.65 S of W Rx 19.425 m  

Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a o

direction 40.0 north of east (which is equivalent to subtracting B from A —that is, to finding R'  A B ). (b) Repeat the problem two problems prior, but now you first o o walk 20.0 m in a direction 40.0 south of west and then 12.0 m in a direction 20.0 east of south (which is equivalent to subtracting A from B —that is, to finding  R  B A  R  ). Show that this is the case.

Solution

R  A  B R x Ax B x A cos110  B  cos40  12.0 m cos110  20.0 m cos 40  11.217 m     R y  A y  B y  A sin 110  B sin 40

(a)

 12.0 m sin 110  20.0 m sin 40  24.132 m



  26.6 m

2 1/ 2

R  Rx R y

R y '  24.132   1  tan 1   65.1 N of E R '   tan  11.217  x   R  B A  B x Ax  20.0 m cos 220 12.0 m cos110  11.217 m R x  R 24.132 m y  B y A y  20.0 m sin 220 12.0 m sin 110  



  26.6 m

2 1/ 2

R  R x R  y

R y "   tan 1  R "   65.1 S of W (b) x  This is consistent with part (a) because A B  B A .

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Show that the sum of the vectors discussed in the example Subtracting Vectors Graphically: A Woman Sailing a Boat gives the result shown in Figure 3.23.

Solution

R x  Ax  B x  27.5 m cos 66  30.0 m cos112  0.0529 m R y  Ay  B y  27.5 m sin 66  30.0 m sin 112  52.938 m 2

R  Rx  R y 

0.0529 m 2  52.938 m 2  52.9 m

R y   1 52.938 m è  tan 1  R   tan 0.0529 m  89.9 N of W  90.1w.r.t.  x - axis x

(note Rx  0 ) 11.

Find the components of v tot along the x- and y-axes in Figure 3.57.

Solution

v tot x  6.72 m/s  cos(49 o )  4.41 m/s v tot y  6.72 m/s  sin( 49 o )  5.07 m/s v tot x  4.41 m/s; v tot y  5.07 m/s

3.3 VECTOR ADDITION AND SUBTRACTION: ANALYTICAL METHODS 13.

Find the following for path C in Figure 3.53: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

d  1  120 m  5  120 m  2  120 m  1  120 m  1  120 m  3  120 m 

Solution

(a)

 1.56  10 3 m s x  0  600  0 120  0 360 m  120 m s y  120  0 240  0  120  0 m  0 m s  s 2 x  s 2 y  (120 m) 2  (0 m) 2  120 m

(b)



s  è  tan 1 y  tan 1 0 m s 120 m  x



Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

0  east , so S = 120 m, East

15.

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.54.

Solution

N-component: s x  123 km sin 45  86.97 km  87.0 km E-component:

17.

s y  123 km cos 45  86.97 km  87.0 km

Repeat Problem 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B  A  A  B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Solution

R  A  B  B  A by the commutative property of vector addition.

R  B 2  A2 

25.0 m 2  18.0 m 2  30.8 m

A  18.0 m  ö  tan 1   tan 1   35.8 W of N B  25.0 m  19.

Do Problem 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A —that is, finding R' = A - B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A  B  C . Is that consistent with your result?)

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Solution

(a) R x  18.0 m, R y  25.0 m 2

R'  R x  R y  (18.0 m) 2  (25.0 m) 2  30.8 m è  tan 1

opp 25.0 m   tan 1   54.2 S of W adj 18.0 m 

(b) R x  18.0 m, R y  25.0 m 2

R"  R x  R y  (18.0 m) 2  (25.0 m) 2  30.8 m è  tan 1

opp 25.0 m   tan 1   54.2 N of E adj 18.0 m 

which is consistent with part (a). 21.

You fly 32.0 km in a straight line in still air in the direction 35.0 south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0 south of west and then in a direction 45.0 west of north. These are the components of the displacement along a different set of axes—one rotated 45 .

Physics: AP® Edition 2e

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Chapter 3

DS  Rsin 32.0 km sin 35  18.4 km

Solution

(a) DW  Rcos 32.0 km cos 35  26.2 km

(b) Consider rotated axes, x and y :

DSW  Rcos  32.0 km cos10  31.5 km DNW  Rsin  32.0 km sin 10  5.56 km 23.

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0 north of west; then 4.70 km 60.0 south of east; then 1.30 km 25.0 south of west; then 5.10 km straight east; then 1.70 km 5.00 east of north; then 7.20 km 55.0 south of west; and finally 2.80 km 10.0 north of east. What is his final position relative to the island?

Solution

Gilligan’s travels are composed of seven vectors. Their lengths in km and angles (ccw from due east) are: A = 2.50 km @ 135 , B = 4.70 km @ 300 , C = 1.30 km @ 205 ,

D = 5.10 km @ 0 , E = 1.70 km @ 85 , F = 7.20 km @ 235 , G = 2.80 km @ 10 The resultant vector is: R  A  B  C  D  E  F  G

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

2.50 cos135  4.70 cos 300  1.30 cos 205  5.10 cos 0   Rx    1.70 cos 85  7.20 cos 235  2.80 cos10 km  3.280 km   2.50 sin 135  4.70 sin 300  1.30 sin 205  5.10 sin 0   Ry    1.70 sin 85  7.20 sin 235  2.80 sin 10 km  - 6.570 km   R  (3.280 km) 2  (6.570 km) 2  7.34 km 6.570 km  á  tan 1   63.5  63.5 S of E 3.280 km 

3.4 PROJECTILE MOTION 25.

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0° above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Solution

R

Range of projectile on level ground:

v02 (50.0 m/s) 2 sin 2 sin 60.0  221 m g 9.80 m/s 2

The time in air is given as 3.00 s, so projectile landed above level ground. Find the position relative to the launching point:

x  v0 x t  50.0 m/s cos 30.03.00 s   1.30  10 2 m 1 1 2 y  v0 y t  at 2  50.0 m/s sin 30.03.00 s  9.80 m/s 2 3.00 s   30.9 m 2 2





Therefore, the projectile landed 1.30  10 m horizontally and 30.9 m vertically from the launching point. 27.

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Physics: AP® Edition 2e

Solution

Student Solutions Manual

Chapter 3

1 y  v0 t  at 2 2 1 60 m  0  9.8 m/s 2 t 2 2 (a) t  3.50 s





x  v0 x t 100 m  v0 x 3.50 s 

(b)

(c)

v0 x  28.6 m/s





v y  v0 y  at  0  9.8 m/s 2 3.50 s   34.3 m/s 2

v 2  v x2  v y2  28.6 m/s   34.3 m/s  v  44.7 m/s

v y  34.3    tan 1   tan 1   50.2 below x - axis   (d) 28.6  v x 

29.

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Solution (a) R  75.0 m, v0  35.0 m/s,  ? 2

v sin 2è 0 R 0 g Use the equation for a projectile on level ground: è0 

1 1 gR  1 1  (9.80 m/s 2 )(75.0 m)   sin   sin  (35.0 m/s) 2  18.4 v 2  2 2   0 

(b) The arrow will be at the tree when the vertical velocity is zero:

v y  v0 sin gt  t  v0 sin / g  1.127 s 1 y  v0 sin èt  gt 2 2 1  (35.0 m/s)sin18.4(1.13 s)  (9.80 m/s 2 )(1.13 s) 2  6.23 m 2 The arrow goes over the branch! 31.

Verify the ranges for the projectiles in Figure 3.38(a) for  45.0 and the given initial velocities.

Solution

v sin 2è 0 v0 R 0  for  45 : g g R  91.8 m for v0  30 m/s; R  163 m for v0  40 m/s; R  255 m for v0  50 m/s

33.

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat,

3 because the Earth is curved. Assume that the radius of the Earth is 6.37  10 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

Solution (a) The range is a maximum when  45 :

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Rmax 

v0 solving for the initial velocity g

v0  gRmax  (9.80 m/s 2 )(32.0  10 3 m)  560 m/s

h

(b) The height can be found from the equation

v0 y



(v0 sin 0 ) 2 2g

(560 m/s)(sin 45)2  8.00  103 m 2(9.80 m/s 2 )

á

32.0 km 32.0 km  360   360  0.2878 2ð R 2  (6.37  10 3 km)

R 6.37  10 3 km   6.37008  10 3 km cos á cos 0.2878 3 d = x - R = 0.00008  10 km  80 m. R  x cos á  x 

This error is not significant because it is only 1% of the answer in part (b). 35.

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Solution

vf2  v02  2ax  0  2 1.25  9.8 m/s 2 0.600 m   3.83 m/s





Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Assume the person leaves at 45 and on level ground, thus



vf y  v0 y  at



0  3.83 m/s sin 45 9.80 m/s 2 t t  0.276 s to the top of the jump, tTOTAL  2  0.276  0.553 s x  v0 x t  3.83 m/s cos 450.553 s   1.5 m

37.

Serving at a speed of 170 km/h a tennis player hits the ball at a height of 2.5 m and an angle below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line 6.40 m from the net?

Solution

v0  170 km/h  47.2 m/s 1 y  2.50 m 0.91 m  1.59 m and y  v0 y t  at 2  47.2 m/s sin t  4.9 m/s 2 t 2 2



In x direction x  11.9 m  47.2 m/s cos t , so the first equation,



t



0.252 cos è . Inserting this value into



1.59 m  11.9 m tan  0.311 m 1  tan 2   tan è  0.107 or è  6.1

Using this value, the time for the ball to fall the full 2.5 m can be calculated, and is t  0.366 s . The range of the ball in this time is R  v x t  17.2 m , so yes, the ball lands at 5.3 m from the net. 39.

Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Solution

(a) Use the 100 m data to calculate the release angle for the bullet. 2

R

v0 sin 2è 0 g so that

gR  1 1  1 (9.80 m/s 2 )(100 m)   è 0  sin 1   sin   0.3712 2 v 2  2 2 (275 m/s)   0   Next, calculate the time to travel.

x x0  150 m  v0 x t  v0 cos è 0 t t

x x0 150 m   0.5455 s v0 cos è 0 (275 m/s) cos 0.3712

Lastly, calculate the change in vertical position during the 150 m flight: 1 y y0  v0 y t  gt 2 2 1  (275 m/s) sin 0.3712(0.5455 s)  (9.80 m/s 2 )(0.5455 s) 2  0.486 m 2

(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation. 41.

An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0 below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

Solution The mouse will land in the nest if its horizontal, x , is > 3.85 m and < 4.15 m ( 4.0 m  the radius of the nest). y y0  12.0 m , and

v 0 ,y  (3.50 m/s)sin30  1.75 m/s. To calculate the horizontal position, we need to first calculate the time it takes to fall the distance of 12.0 m. 1 y y 0  v0, y t  gt 2 , 2 so solve for t :

9.80 m/s 2 2 12 m  (1.75 m/s)t  t , or t 2  (0.3571 s)t 2.449 s 2  0 so that 2 t

(0.3571 s)  (0.3571 s) 2 4(2.449 s 2 )  1.40 s 2

Next, we need to calculate the horizontal displacement. x  x0  v0 x t ,

so that x x0  (3.50 m/s) cos 30(1.40 s)  4.23 m

No, the owl is not lucky. The mouse just misses the nest. 43.

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

Solution For the maximum range, we use R

v02 sin( 20 ) (30 m/s) 2 sin(90)   91.8 m g 9.8 m/s 2

No, the maximum range is about 92 m (neglecting air resistance). The ball will not travel 95 m before hitting the ground.

Physics: AP® Edition 2e

45.

Student Solutions Manual

Chapter 3

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0 above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45 when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38 will give a longer range than 45 in the shot put.)

Solution Given:  38

x  x x0  24.77 m, x  (v0 cos )t , so that

y  y y 0  2.10 m

t

x v0 cos ,

1 y  (v0 sin )t  gt 2 2 , so that , substituting for t gives: 2

 x 1  x  y  (v0 sin ) v cos  2 g  v cos   0  0 

1   gx 2     y x tan   2  2 cos 2   v  , so 0  or

that  gx 2  1 v0   2 cos 2  ( y x tan )   

47.

(9.80 m/s 2 )(24.77 m) 2  15.0 m/s 2cos 2 38[2.10 m (24.77 m)tan38]

A football player punts the ball at a 45.0 angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

Solution (a)

gR è 0  45,R  60.0 m, v0   sin 2è 0

9.80 m/s 60.0 m   24.2 m/s 2

sin 2  45

(b) v0 ,x  v0 cos 45  (24.2 m/s)cos45  17.15 m/s

Physics: AP® Edition 2e

Student Solutions Manual

Chapter 3

With the wind gust, the horizontal velocity is decreased by 1.50 m/s: v x  v0, x 1.50 m/s  17.15 m/s 1.50 m/s  15.65 m/s

The time can be calculated for each half of the flight:

(The time is not changed by wind gust, since it only acts in x-direction.) Finally, the distance traveled after the wind gust is: x x0  v x t  15.65 m/s 1.750 s   27.38 m

So the ball travels a total of 57.4 m with the brief gust of wind. 49.

R

v 02 sin 20 g for the range of a projectile on level ground by finding the

Derive time t at which y becomes zero and substituting this value of t into the expression for x x0 , noting that R  x x0 . Solution

2(v0 sin ) 1 1 t y y 0  0  v0 y t  gt 2  (v0 sin )t  gt 2 g 2 2 so that x x0  v0 x t  (v0 cos )t  R, and substituting for t gives: 2

2v0 sin  2v0 sin cos   R  v0 cos   g  g   2

Since 2 sin cos  sin 2, the range is:

3.5 ADDITION OF VELOCITIES

v sin 2 R 0 g

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Chapter 3

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45 south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?

Solution

60 s   4 d  v0 t  3.53 m/s 169 min   3.579  10 m  35.8 km, 45 S of E 1 min   (a)

v PG  v PA  v AG  v PA  v PG v AG , so that

(b) v PA  3.53 m/s 2.0 m/s   5.53 m/s, 45 S of E 60 s   x A  v PA t  5.53 m/s 169 min   56.1 km, 45 S of E 1 min   (c)

54.

Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line?

Solution (a) Label F for the front runner and S for the second runner: v F  3.50 m/s, vS  4.20 m/s

vSF  vS v F  4.20 m/s 3.50 m/s  0.70 m/s (faster than the first runner)

(b)

tF 

xF 250 m   71.43 s v F 3.50 m/s

tS 

xS 250  45 m   70.24 s vS 4.20 m/s

The second runner will win. (c)

xSF  x0,SF  vSF tS  45.0 m  0.70 m/s 70.24 s   4.17 m

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Chapter 3

A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0 relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback?

Solution

vA  vB  v0  vA v Ax  v Bx  v0 x  v0 cos  v Ay  v By  v0 y  v0 sin 

2 m/s  vBx  (15.2 m/s)cos25.0  vBx  15.8 m/s 0  vBy  (15.2 m/s) sin 25.0  vBy  6.42 m/s 2

vB  vBx  vBy  17.0 m/s 6.42   tan 1   22.1 15.8  58.

(a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0 south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15 south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane’s path.

Solution (a) P = plane; A = air; G = ground

v PG , x  v PA , x  v AG , x v PG, x  (260 m/s)cos185  (35.0 m/s)cos(15)  225.20 m/s v PG , y  v PA , y  v AG , y v PG, y  (260 m/s)sin185  (35.0 m/s)sin(15)  31.72 m/s

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Chapter 3

v PG  v PG , x  v PG , y  (225.2 m/s) 2  (31.72 m/s) 2  227 m/s  230 m/s v BG,y  31.72 m/s   tan 1   tan 1    8.0 , S of W v   225.20 m/s   BG ,x  

(b) The wind should make the plane travel slower and more to the south, which is what was calculated. 60.

(a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20.0 south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is 45.0 south of west, while its direction of travel relative to the air is 5.00 south of west. What is the airplane’s speed relative to the air mass? (b) What is the airplane’s speed relative to the Earth?

Solution

(a) a = airplane, w = wind, g = ground va, x  va cos 5  (45.0 m/s) cos 20  vg cos 45 va,y  va sin 5  (45.0 m/s) sin 20  vg sin 45 va cos 5 sin 5  (45.0 m/s)(cos 20  sin 20) va  (45.0 m/s)

(since cos 45  sin 45)

cos 20  sin 20  63.5 m/s cos 5 sin 5

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Student Solutions Manual 2

Chapter 3

vg  va  v w 2va v w cos  1/ 2

  (63.45 m/s) 2  (45.0 m/s) 2   29.6 m/s vg     (b) 2(63.45 m/s)(45.0 m/s)cos25 

62.

The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0° east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0° south of west relative to the Earth. What is the velocity of the wind relative to the water?

Solution

v WO  v WE  v EO  v WE v OE

v WOx  v WEx v0Ex  4.50 m/s cos 50 2.20 m/s cos 60  3.993 m/s v WOy  v WEy vOEy  4.50 m/s sin 50 2.20 m/s sin 60  5.352 m/s 2

v  v x  v y  (3.993 m/s) 2  (5.352 m/s ) 2  6.68 m/s á  tan 1

64.

vy vx

 tan 1

-5.352 m/s  53.3 S of W -3.993 m/s

(a) Use the distance and velocity data in Figure 3.59 to find the rate of expansion as a function of distance. (b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.

Physics: AP® Edition 2e

Solution

Student Solutions Manual

Chapter 3

v , d 4500 km/s 1 ly H1    1.59  10 18 s 1 6 12 300  10 ly 9.46  10 km 2200 km/s 1 ly H2    1.55  10 18 s 1 6 12 150  10 ly 9.46  10 km 2830 km/s 1 ly H4    1.57  10 18 s 1 6 12 190  10 ly 9.46  10 km 6700 km/s 1 ly H5    1.57  10 18 s 1 6 450  10 ly 9.46  1012 km km/s H average  1.57  10 18 s 1  14.9 Mly (a) H

d 1 1   v H 1.57  10 18 s 1 1y 17 10 (b)  6.37  10 s  3.16  10 7 s  2.02  10 y  20.2 billion years t

66.

A ship sailing in the Gulf Stream is heading 25.0 west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00 west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)

Solution

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Chapter 3

v WE  v WS  v SE  v SW  v SE v WEx  vSWx  vSEx  4.00 m/s cos115  4.80 m/s cos 95  1.272 m/s v WEy  vSWy  vSEy  4.00 m/s sin 115  4.80 m/s sin 95  1.157 m/s v WE  v 2 WE,x v 2 WE,y  (1.272 m/s) 2  (1.157 m/s) 2  1.72 m/s v WE, y   tan 1 1.157 m/s  42.3 N of E á  tan 1  v  1.272 m/s  WE,x  71.

Critical Thinking Two identical spring-loaded launchers are fixed to the end of a table. Launcher A is inclined 40 above the horizontal and launcher B is inclined 50 above the horizontal. (a) Which of these launchers will fire the ball the greatest vertical height above the launcher before it crashes into the ground? (b) Based on experimental data, an equation that fits the data is suggested for the vertical height above the launcher, y, which may not be correct: y  KV0 2 sin , where K is a constant with appropriate units, V0 is the initial velocity of the launcher and is the angle relative to the x-axis of the launcher. Is this equation consistent with your answer from part a? Explain why or why not. Does this equation make sense? Explain why or why not. (c) Graph the magnitude of the velocity vs. vertical height for both launcher A and B.

Solution (a) Since sin 2 50  sin 2 40 , then B reaches the greatest height. (b) i. Yes, it is consistent because sin 50  sin 40 . ii. No, it does not make sense because y is proportional to sin 2 .

(c)

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Chapter 3

A ball is thrown at an angle of 45 degrees above the horizontal. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower’s hand until the time it hits the ground? a) Always in the same direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground b) Initially positive in the upward direction, then zero at maximum height, then negative from there until it hits the ground c) Always in the opposite direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground d) Always in the downward direction with the same constant value

Solution

(d)

A student wishes to design an experiment to show that the acceleration of an object is independent of the object’s velocity. To do this, ball A is launched horizontally with some initial speed at an elevation 1.5 meters above the ground, ball B is dropped from rest, 1.5 meters above the ground, and ball C is launched vertically with some initial speed at an elevation 1.5 meters above the ground. What information would the student need to collect about each ball in order to test the hypothesis?

Solution

We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals.

[Table 03_02_02] Height (m) 0.490 0.882 1.176 1.372 1.470 1.470 1.372

Time (sec) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

A ball is launched at an angle of 60 degrees above the horizontal, and the vertical position of the ball is recorded at various points in time in the table shown, assuming the ball was at a height of 0 at time t = 0. a) Draw a graph of the ball’s vertical velocity vs. time. b) Describe how a graph of the ball’s horizontal velocity would look. Draw a graph of the ball’s vertical acceleration vs time.

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Solution

Student Solutions Manual

Chapter 3

The graph of the ball’s vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s2, crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec. The graph of the ball’s horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec.

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CHAPTER 4: DYNAMICS: FORCE AND NEWTON’S LAWS OF MOTION 4.3 NEWTON’S SECOND LAW OF MOTION: CONCEPT OF A SYSTEM 1.

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s . What is the net external force on him?

Solution

net F = ma = (63.0 kg)(4.20 m/s²) = 265 N

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

Solution

a = net F/m = 60.0 N/4.50 kg = 13.3 m/s 2 of the cart.

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.6 x 10 4 N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

Solution

(a) Use the thrust given for the rocket sled in Figure 4.7, T  2.59  10 N . With only one rocket burning, net F  T f or

a

net F T f 2.59  10 4 N 650 N    12.0 m/s 2 m m 2100 kg

(b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.

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Chapter 4

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calculate the acceleration. (d) What would the acceleration be if friction were 15.0 N?

Solution

(a) The system is the child in the wagon plus the wagon.

(b) (c) net F  Fr F1 f  ma so that

Fr F1 f 90.0 N 75.0 N 12.0 N   0.130 m/s 2 m 23.0 kg in the direction of the second child' s push

a

(d) net F  90.0 N 75.0 N 15.0 N  0  a  0 m/s 11.

Solution

The rocket sled shown in Figure 4.32 accelerates at a rate of 49.0 m/s . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body. Fh  ma  (75.0 kg)(49.0 m/s 2 )  3.68  10 3 N and w  mg  (75.0 kg)(9.80 m/s 2 )  7.35  10 2 N Fh ma a 49.0 m/s 2     5.00 times greater than weight (a) w mg g 9.80 m/s 2

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Chapter 4

(b) net F  [(Fh )2  (N)2 ]1/2 net F  [(3675 N) 2  (735 N) 2 ]1/2  3750 N N  è  tan 1  F   11.3 from horizontal h  13.

The weight of an astronaut plus their space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth?

Solution

WMoon  mg Moon m

WMoon 250N   150kg g Moon 1.625m/s 2

WEarth  mg Earth  (150kg)(9.8m/s2 )  1470N  1.5  103 N

Mass does not change, so the astronaut's mass on both Earth and the Moon will be 150 kg.

4.4 NEWTON’S THIRD LAW OF MOTION: SYMMETRY IN FORCES 15.

What net external force is exerted on a 1100-kg artillery shell fired from a 4

battleship if the shell is accelerated at 2.40  10 m/s ? What force is exerted on the ship by the artillery shell? Solution

net F  ma  (1100 kg )(2.40  10 4 m/s 2 )  2.64  10 7 N Force exerted on ship = 2.64  10 7 N opposite to shell’s direction of motion, by

Newton’s 3rd law.

4.5 NORMAL, TENSION, AND OTHER EXAMPLES OF FORCES

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Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?

Solution

(a) net F  Ma ; f1  1350 N ; f 2  1365 N 9( f 2 f1 )  9(m1  m2 )a ; m1  68 kg ; m2  73 kg a

f 2 f1 1365 N 1350 N   0.1064 m/s 2  0.11 m/s 2 ; m1  m2 68 kg  73 kg

Thus, the heavy team wins. Note that the difference 1365 N 1350 N  15 N limits the answer to two significant figures.

(b ) T 9 f1  9m1 a  T  9m1 a  9 f1 T  9(68 kg )(0.1064m/s 2 )  9(1350 N)  1.2  10 4 N 19.

(a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00  10 5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.16. The strand sags at an angle of 12 below the horizontal. Compare this with the tension in the vertical strand (find their ratio).