physics chapter 9 answers

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9 – 1

Chapter 9: Linear Momentum and Collisions

Answers to Even-Numbered Conceptual Questions

2. Doubling an object’s speed increases its kinetic energy by a factor of four, and its momentum by a factor oftwo.

4. No. Consider, for example, a system of two particles. The total momentum of this system will be zero if theparticles move in opposite directions with equal momentum. The kinetic energy of each particle is positive,however, and hence the total kinetic energy is also positive.

6. Yes. Just point the fan to the rear of the boat. The resulting thrust will move the boat forward.

8. (a) The force due to braking—which ultimately comes from friction with the road—reduces the momentumof the car. The momentum lost by the car does not simply disappear, however. Instead, it shows up as anincrease in the momentum of the Earth. (b) As with braking, the ultimate source of the force accelerating thecar is the force of static friction between the tires and the road.

10. Yes. For example, we know that in a one-dimensional elastic collision of two objects of equal mass theobjects “swap” speeds. Therefore, if one object is at rest before the collision, it is possible for one object to beat rest after the collision as well. See Figure 9-7(a).

12. No. Any collision between cars will be at least partially inelastic, due to denting, sound production, heating,and other effects.

14. The center of mass of the hourglass starts at rest in the upper half of the glass and ends up at rest in thelower half. Therefore, the center of mass accelerates downward when the sand begins to fall—to get itmoving downward—and then accelerates upward when most of the sand has fallen—to bring it to restagain. It follows from equation 9-18 that the weight read by the scale is less than Mg when the sand beginsfalling, but is greater than Mg when most of the sand has fallen.

16. (a) Assuming a very thin base, we conclude that the center of mass of the glass is at its geometric center ofthe glass. (b) In the early stages of filling, the center of mass is below the center of the glass. When the glassis practically full, the center of mass is again at the geometric center of the glass. Thus, as water is added, thecenter of mass first moves downward, then turns around and moves back upward to its initial position.

18. As this jumper clears the bar, a significant portion of his body extends below the bar due to the extremearching of his back. Just as the center of mass of a donut can lie outside the donut, the center of mass of the

jumper can be outside his body. In extreme cases, the center of mass can even be below the bar at all timesduring the jump.

Solutions to Problems and Conceptual Exercises

1. Picture the Problem: The car and the baseball each travel in a straight line at constant speed. The data given in the

Exercise 9-1 include car 15,800 kg m/s p = ⋅ and ball 0.142 kg.m =

Strategy: Use equation 9-1 to set the magnitude of the momentum of the baseball equal to the magnitude of the

momentum of the car, and solve for allm .

Solution: Set all car p p=

and solve for all :v

ball ball ball car

65car

ball

ball

15,800 kg m/s 1.11 10 m 3600 s 1 mi2.49 10 mi/h

0.142 kg s h 1609 m

p m v p

pv

m

= =

⋅ ×= = = × × = ×

Insight: The huge required speed for the baseball (3,240 times the speed of sound!) is due to the huge mass difference

between the two objects.

Chapter 9: Linear Momentum and Collisions James S. Walker, Physics, 4th Edition


9 – 2

2. Picture the Problem: The ducks approach each other and the goose

recedes as indicated in the figure at right.

Strategy: Sum the velocity vectors from the three birds using the

component method to find the total momentum.

Solution: 1. Find the momen-tum of the left duck:

( ) ( )

( )

d1 d dˆ ˆ4.00 kg 1.10 m/s

ˆ4.40 kg m/s

m v= =

= ⋅

p x x

x

2. Find the momentumof the top duck:

( ) ( )

( )

d2 d dˆ ˆ4.00 kg 1.10 m/s

ˆ4.40 kg m/s

m v= = −

= − ⋅

p y y

y

3. Find the momentum

of the goose:( ) ( )

( )

g g gˆ ˆ9.00 kg 1.30 m/s

ˆ11.7 kg m/s

m v= = −

= − ⋅

p y y

y

4. Add the momenta:

( ) ( )

total d1 d2 g

total

ˆ ˆ ˆ4.40 kg m/s 4.40 kg m/s 11.7 kg m/s

ˆ ˆ4.40 kg m/s 16.1 kg m/s

= + +

= ⋅ − ⋅ − ⋅

= ⋅ + − ⋅

p p p p

x y y

p x y

Insight: The total momentum vector now has a magnitude of 16.7 kg·m/s and points 74.7° below the positive x axis.

3. Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward and the dog runs

northward.

Strategy: Sum the momenta of the dog and cat using the component method. Use the known components of the total

momentum to find its magnitude and direction. Let north be in the y direction, east in the x direction. Use the

momentum together with the owner’s mass to find the velocity of the owner.

Solution: 1. Use the component

method of vector addition to find

the owner’s momentum:( ) ( ) ( ) ( )

( ) ( )

total cat dog cat cat dog dog

total

ˆ ˆ5.30 kg 3.04 m/s 26.2 kg 2.70 m/s

ˆ ˆ16.1 kg m/s 70.7 kg m/s

m m= + = +

= +

= ⋅ + ⋅

p p p v v

x y

p x y

2. Divide the owner’s momentum

by his mass to get the componentsof the owner’s velocity:

( ) ( )

( ) ( )

owner owner owner total

total

owner

0

ˆ ˆ16.1 kg m/s 70.7 kg m/s

74.0 kg

ˆ ˆ0.218 m/s 0.955 m/s

m

m

= =

⋅ + ⋅= =

= +

p v p

x ypv

x y

3. Use the known components to find

the direction and magnitude of the owner’svelocity:

( ) ( )2 2

owner

1

0.218 m/s 0.955 m/s 0.980 m/s

0.955tan 77.1 north of east

0.218

v

θ −

= + =

⎛ ⎞= = °⎜ ⎟

⎝ ⎠

Insight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The owner is movingmuch slower than either the cat or the dog because of his larger mass.



9 – 3

4. Picture the Problem: The two carts approach each other on a frictionless track at different speeds.

Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting expression for 2v . Then use

equation 7-6 to find the total kinetic energy of the two-cart system. Let cart 1 travel in the positive direction.

Solution: 1. (a) Set 0=∑p

and solve for 2 :v

( )( )

1 1 2 2

1 12

2

0

0.35 kg 1.2 m/s0.69 m/s

0.61 kg

m m

m vv

m

= + =

= − = =

∑p v v

2. (b) No, kinetic energy is always greater than or equal to zero.

3. (c) Use equation 7-6 to sum the kinetic

energies of the two carts:( ) ( ) ( ) ( )

2 21 11 1 2 22 2

2 21 1

2 20.35 kg 1.2 m/s 0.61 kg 0.69 m/s

0.40 J

K m v m v= +

= +

=

∑

Insight: If cart 1 is traveling in the positive x direction, then its momentum is ( ) ˆ0.42 kg m/s⋅ x and the momentum of

cart 2 is ( ) ˆ0.42 kg m/s− ⋅ x .

5. Picture the Problem: The baseball drops straight down, gaining momentum due to the acceleration of gravity.

Strategy: Determine the speed of the baseball before it hits the ground, then use equation 2-12 to find the height from

which it was dropped.

Solution: 1. Use equation 9-1 to find

the speed of the ball when it lands:0.780 kg m/s

5.20 m/s0.150 kg

pv

m

⋅= = =

2. (b) Use equation 2-12 to solve

for 0 y . Let 0 y = and 0 0 :v = ( )

( )

( )

2 2

0 0

22

0 2

2

5.20 m/s1.38 m

2 2 9.81 m/s

v v g y y

v y

g

= − −

= = =

Insight: Another way to find the initial height is to use conservation of energy, setting 210 2

mgy mv= and solving for 0. y

6. Picture the Problem: The ball falls vertically downward, landing with a speed of 2.5 m/s and rebounding upward witha speed of 2.0 m/s.

Strategy: Use equation 9-1 to find the change in momentum of the ball when it rebounds.

Solution: 1. (a) Use equation 9-1

to find Δp

:

( )

( ) ( ) ( ) ( )

f i f i

ˆ ˆ ˆ0.285 kg 2.0 m/s 2.5 m/s 1.3 kg m/s

1.3 kg m/s

mΔ = − = −

= − − = ⋅⎡ ⎤⎣ ⎦

Δ = ⋅

p p p v v

y y y

p

2. (b) Subtract the magnitudesof the momenta:

( )

( ) ( )

f i f i

f i

0.285 kg 2.0 m/s 2.5 m/s

0.14 kg m/s

p p m v v

p p

− = −

= −

− = − ⋅

3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its collision with the

floor, first of all because t = Δ Δ∑F p

(equation 9-3) and as we can see from above that f i p pΔ ≠ −p

. Secondly, we

expect the floor to exert an upward force on the ball but we calculated a downward (negative) value in part (b).

Insight: If the ball were to rebound at 2.5 m/s upward we would find 2 1.1 kg m/smvΔ = = ⋅p

and f i 0 p p− = . Such a

collision with the floor would be called elastic, as discussed in section 9-6.



9 – 4

7. Picture the Problem: The individual momenta and final momentum vectors are

depicted at right.

Strategy: The momenta of the two objects are perpendicular. Because of this we

can say that the momentum of object 1 is equal to the x-component of the totalmomentum and the momentum of object 2 is equal to the y-component of the total

momentum. Find the momenta of objects 1 and 2 in this manner and divide by their

speeds to determine the masses.

Solution: 1. Find total, x p and divide by 1v : ( ) ( )1 total, total

1

1

1

cos 17.6 kg m/s cos66.5 7.02 kg m/s

7.02 kg m/s2.51 kg

2.80 m/s

x p p p

pm

v

θ = = = ⋅ ° = ⋅

⋅= = =

2. Find total, y p and divide by 2v :( ) ( )

total2

2

2 2

17.6 kg m/s sin66.5sin5.21 kg

3.10 m/s

p pm

v v

θ ⋅ °= = = =

Insight: Note that object 2 has the larger momentum because the total momentum points mostly in the y direction. The

two objects have similar speeds, so object 2 must have the larger mass in order to have the larger momentum.

8. Picture the Problem: Your car rolls slowly in a parking lot and bangs into the metal base of a light pole.

Strategy: Use the concept of impulse during a collision to answer the conceptual question.

Solution: If the collision were inelastic the light pole gives your car only enough impulse to bring it to rest. If thecollision were elastic, the impulse given to your car is about twice as much; the car must be brought to rest and then

accelerated to its initial speed but in the opposite direction. We conclude that in terms of safety, it is better for your

collision with the light pole to be inelastic.

Insight: The additional impulse required for an elastic collision, which acts over a very short period of time, means thedriver will experience a larger force that could cause injury.

9. Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble.

Strategy: Use Newton’s Second Law in terms of momentum to answer the conceptual question.

Solution: 1. (a) One way to write Newton’s Second Law is to equate the net force with the change in momentum per

time (equation 9-3). Both the boulder and the pebble have the same rate of momentum change because the same force

acts on both objects. We conclude that the change of the boulder’s momentum in one second is equal to the change of

the pebble’s momentum in the same time period.

2. (b) The best explanation is III. Equal force means equal change in momentum for a given time. Statement I is false,

and statement II is partly true (the pebble’s speed is larger) but the pebble’s change in momentum is the same because it

has a smaller mass than the boulder.

Insight: Newton originally formulated his second law by asserting that the net force is equal to the rate of change ofmomentum. Later textbooks introduced the more familiar F = ma equation.

10. Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g

pebble.

Strategy: Use Newton’s Second Law in terms of momentum to answer the conceptual question.

Solution: 1. (a) Both the boulder and the pebble have the same rate of momentum change because the same force acts

on both objects (equation 9-3). As a result, the rate of change in velocity for the less massive object (the pebble) must be

greater than it is for the more massive object (the boulder). We conclude that the change of the boulder’s speed in one

second is less than the change of the pebble’s speed in the same time period.

2. (b) The best explanation is I. The large mass of the boulder results in a small acceleration. Statement II is false (the

same force results in the same change in momentum, not speed), and statement III is true but irrelevant.

Insight: We know that the acceleration (rate of change of velocity) of an object is proportional to the force acting on it

and inversely proportional to its mass. These objects experience the same force, and therefore the less massive object

has the greater acceleration.

y

x

totalp

1p

2p



9 – 5

11. Picture the Problem: A friend tosses a ball of mass m to you with a speed v. The force you exert on the ball when you

catch it produces a noticeable sting in your hand.

Strategy: Use the concept of impulse to answer the conceptual question.

Solution: 1. (a) The two balls have the same momentum. If they are stopped in the same amount of time by your hand,

your hand will exert an equal force on each ball. We conclude that the sting you feel is equal to the sting you felt when

you caught the first ball.

2. (b) The best explanation is II. The two balls have the same momentum, and hence they produce the same sting.

Statement I is true if the two balls are stopped in the same distance (see the Insight statement below) and statement III isfalse (the momentum, not just the mass, needs to be considered).

Insight: The first ball has a kinetic energy equal to 21

2 mv and the second ball has half that much kinetic energy:

( )( ) ( )2 21 1 1 1

2 2 2 22 .m v mv= If each ball is stopped in the same distance d , a smaller force F will be required to stop the

second ball because Fd = Δ K and Δ K is smaller for the second ball. A smaller force would produce a smaller “sting.”

12. Picture the Problem: Several forces of different magnitudes act for different time intervals.

Strategy: Use the concept of impulse to rank the forces in terms of the impulses they produce.

Solution: Multiply the force by the time interval in order to calculate the impulse. Using this approach we find that

impulse A is , F t Δ impulse B is 2

3, F t Δ impulse C is 1

2, F t Δ and impulse D is 1

10. F t Δ We arrive at the ranking

D < C < B < A.

Insight: The impulse of a force 3 F acting over a time interval 2Δt is exactly the same as a force F acting over a time

interval 6Δt .

13. Picture the Problem: The force of the kick acting over a period of time imparts an impulse to the soccer ball.

Strategy: Multiply the force by the time of impact to find the impulse according to equation 9-5.

Solution: Apply equation 9-5 directly: ( )( )3

av 1250 N 5.95 10 s 7.44 kg m/s I F t −= Δ = × = ⋅

Insight: The 1250-N force is equivalent to 281 lb! The same impulse could be delivered to the ball with 7.44 N

(1.67 lb) of force acting over a period of one second.

14. Picture the Problem: The golf club exerts an impulse on the ball, imparting momentum.

Strategy: Find the change of momentum of the golf ball and use it to find the force exerted on it according toequation 9-3.

Solution: Apply equation 9-3 directly:( ) ( ) ( )f i

av

0.045 kg 67 m/s 03.0 kN

0.0010 s

m v v p F

t t

− −Δ= = = =

Δ Δ

Insight: This is a force equivalent to 670 lb! A ball launched with a speed of 67 m/s launched at 45° will land 457 m

(500 yd) down range in the absence of air friction. A par 5 hole in one?

15. Picture the Problem: The croquet mallet exerts an impulse on the ball, imparting momentum.

Strategy: Find the change in momentum of the croquet ball and then use it to find t Δ using equation 9-3.

Solution: Solve equation 9-3 for :t Δ ( ) ( ) ( )f i

av av

0.50 kg 3.2 m/s 00.0070 s 7.0 ms

230 N

m v v pt

F F

− −ΔΔ = = = = =

Insight: The large force (52 lb) is exerted over a very brief time to give the ball its small velocity (7.2 mi/h).



9 – 6

16. Picture the Problem: The hand of the volleyball player exerts an impulse on the ball, imparting momentum.

Strategy: Use equation 9-6 together with the known impulse and change of velocity of the ball in order to find its mass.

Solution: Solve equation 9-6 for mass:

9.3 kg m/s0.33 kg

24 m/s 4.2 m/s

I p m v

I m

v

= Δ = Δ

− ⋅= = =Δ − −

Insight: A mass this large weighs about ¾ lb at sea level. This ball is 22% heavier than the 0.27 kg regulation ball.

17. Picture the Problem: The marble drops straight down from rest and rebounds from the floor.

Strategy: Use equation 2-12 to find the speed of the marble just before it hits the floor. Use the same equation and theknown rebound height to find the rebound speed. Then use equation 9-6 to find the impulse delivered to the marble by

the floor. Let upward be the positive direction, so that the marble hits the floor with speed iv− and rebounds upward

with speed f v .

Solution: 1. (a) Find iv

using equation 2-12: ( )( )

2 2

i 0

2 2 2

i 0

2

2 0 2 9.81 m/s 1.44 m 5.32 m/s

v v g y

v v g y

= − Δ

= − − Δ = − − − = −

2. Find f v using equation 2-12 again:

( )( )

2 2

f

2 2 2

f

2

2 0 2 9.81 m/s 0.640 m 3.54 m/s

v v g y

v v g y

= − Δ

= + Δ = + =

3. Use equation 9-6 to find I : ( )

( ) ( )

f i

0.0150 kg 3.54 5.32 m/s 0.133 kg m/s

I m v m v v= Δ = −

= − − = ⋅⎡ ⎤⎣ ⎦

4. (b) If the marble had bounced to a greater height, its rebound speed would have been larger and the impulse would

have been greater than the impulse found in part (a).

Insight: By Newton’s Third Law the marble also delivers a downward impulse on the floor. The Earth theoretically

moves in response to this impulse, but its change of velocity is tiny (2.2×10-26 m/s) due to its enormous mass.

18. Picture the Problem: The ball rebounds from the floor in the manner

indicated by the figure at right.

Strategy: The impulse is equal to the change in the y-component of the

momentum of the ball. Use equation 9-6 in the vertical direction to find

the impulse.

Solution: Apply equation 9-6

in the y direction:[ ] ( ) ( ) ( )0 0cos65 ( cos65 ) 0.60 kg 5.4 m/s 2cos65 2.7 kg m/s

y y I p m v

m v v

= Δ = Δ

= ° − − ° = ° = ⋅

Insight: There is no impulse in the x direction because the ball does not change its horizontal speed or momentum.

19. Picture the Problem: The ball rebounds from the bat in the manner indicated

by the figure at right.

Strategy: The impulse is equal to the vector change in the momentum.

Analyze the x and y components of Δp

separately, then use the components

to find the direction and magnitude of .I

Solution: 1. (a) Find x pΔ : ( ) ( ) ( )f i 0.14 kg 0 36 m/s 5.0 kg m/s x x x p m v vΔ = − = − − = ⋅⎡ ⎤⎣ ⎦

65°

iv

65°

f v

iv

f v



9 – 7

2. Find y pΔ : ( ) ( ) ( )f i 0.14 kg 18 0 m/s 2.5 kg m/s y y y p m v vΔ = − = − = ⋅

3. Use equation 9-6 to find :I

( ) ( )ˆ ˆ5.0 kg m/s 2.5 kg m/s= Δ = ⋅ + ⋅I p x y

4. Find the direction of :I

1 1 2.5tan tan 27

5.0

y

x

I

I θ

− −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠above the horizontal

5. Find the magnitude of :I

( ) ( )2 22 2 5.0 kg m/s 2.5 kg m/s 5.6 kg m/s x y I I I = + = ⋅ + ⋅ = ⋅

6. (b) If the mass of the ball were doubled the impulse would double in magnitude. There would be no change in the

direction.

7. (c) If Δp

of the ball is unchanged, the impulse delivered to the ball would not change, regardless of the mass of the

bat.

Insight: The impulse brings the ball to rest horizontally but gives it an initial horizontal speed. Verify for yourself that

this ball will travel straight upward 16.5 m (54 feet) before falling back to Earth. An easy popup!

20. Picture the Problem: The ball rebounds from the player’s head in the manner indicated by the figure at right.

Strategy: The impulse is equal to the vector change in the momentum. Analyze the x

and y components of Δp

separately, then use the components to find the direction and

magnitude of .I

Solution: 1. (a) Find x pΔ : ( ) ( ) ( )f i 0.43 kg 5.2 8.8 m/s

1.5 kg m/s

x x x p m v vΔ = − = −

= − ⋅

2. Find y pΔ : ( ) ( ) ( )f i 0.43 kg 3.7 2.3 m/s 2.6 kg m/s y y y p m v vΔ = − = − − = ⋅⎡ ⎤⎣ ⎦

3. Use equation 9-6 to find :I

( ) ( )ˆ ˆ1.5 kg m/s 2.6 kg m/s= Δ = − ⋅ + ⋅I p x y

4. Find the direction of :I

1 1 2.58tan tan 59 180 121

1.55

y

x

I

I θ

− −⎛ ⎞ ⎛ ⎞= = = − ° + ° = °⎜ ⎟ ⎜ ⎟

−⎝ ⎠⎝ ⎠from the positive x axis

5. (b) Find the magnitude: ( ) ( )2 22 2 1.55 kg m/s 2.58 kg m/s 3.0 kg m/s x y I I I = + = − ⋅ + ⋅ = ⋅

Insight: The ball delivers an equal and opposite impulse to the player’s head, which would exert a force of 300 N(67 lb) if the time of collision were 10 milliseconds.

21. Picture the Problem: The two canoes are pushed apart by the force exerted by a passenger.

Strategy: By applying the conservation of momentum we conclude that the total momentum of the two canoes after the

push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for 2m . Let the

velocity 1v

point in the negative direction, 2v

in the positive direction.

Solution: Set total 0 p = and solve for 2 :m

( ) ( )1 2 1 1 2 2

1 1

2

2

0

320 kg 0.58 m/s440 kg

0.42 m/s

x x x x

x

x

p p m v m v

m vm

v

+ = = +

− −−= = =

Insight: An alternative way to find the mass is to use the equations of kinematics in a manner similar to that described

in Example 9-3.

iv

f v

y

x



9 – 8

22. Picture the Problem: The two skaters push apart and move in opposite directions without friction.

Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the

push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for 2m . Let the

velocity 1v

point in the negative direction, 2v


Solution: Set total 0 p = and solve for 2 :m

( ) ( )1 2 1 1 2 2

1 1

2

2

0

45 kg 0.62 m/s31 kg

0.89 m/s

x x x x

x

x

p p m v m v

m vm

v

+ = = +

− −−= = =

Insight: An alternative way to find the mass is to use the equations of kinematics in a manner similar to that described

in Example 9-3.

23. Picture the Problem: The bee and the stick move in opposite directions without friction.

Strategy: By applying the conservation of momentum we conclude that the total momentum of the bee and the stick iszero, just as it was before the bee started walking. Set the total momentum of the system to zero and solve for the speed

of the stick. Let the velocity sv

point in the negative direction, v


Solution: Set total 0 p = and solve for s :v

( ) ( )1 2 b b s s

b b

s

s

0

0.175 g 14.1 mm/s0.519 mm/s

4.75 g

x x p p m v m v

m vv

m

+ = = +

−−= = = −

Insight: The larger mass of the stick means that it must have a slower speed in order to have the same momentum as the

bee. At scales this small the surface tension of the water plays an important role and the stick’s speed would be even

smaller than 0.512 mm/s.

24. Picture the Problem: The two pieces fly in opposite directions at different speeds.

Strategy: As long as there is no friction the total momentum of the two pieces must remain zero, as it was before the

explosion. Combine the conservation of momentum with the given kinetic energy ratio to determine the ratio of the

masses. Let 1m represent the piece with the smaller kinetic energy.

Solution: 1. Set 0=∑p

and solve for 1 2m m : 1 2 1 1 2 2

2 2

1 2 1 2

2 1 2 1

0

p p m v m v

m v m v

m v m v

+ = = +

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2. Set 2 1 2 K K = :

2212 22 2 2 1

211 1 21 12

2 2m v K v m

K v mm v

⎛ ⎞= = ⇒ =⎜ ⎟

⎝ ⎠

3. Combine the expressions from steps 1 and 2:

2 2

2 1 1 1

1 2 2 2

2 2v m m m

v m m m

⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4. The piece with the smaller kinetic energy has the larger mass.

Insight: The smaller mass carries the larger kinetic energy because kinetic energy increases with the square of the

velocity but is linear with mass. Its higher speed more than compensates for its smaller mass.

25. Picture the Problem: The astronaut and the satellite move in opposite directions after the astronaut pushes off. The

astronaut travels at constant speed a distance d before coming in contact with the space shuttle.

Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must remain zero, as it

was before the astronaut pushed off. Use the conservation of momentum to determine the speed of the astronaut, and

then multiply the speed by the time to find the distance. Assume the satellite’s motion is in the negative x-direction.

Solution: 1. Find the speed of the astronaut using

conservation of momentum:a s a a s s

s s

a

a

0 p p m v m v

m vv

m

+ = = +

= −



9 – 9

2. Find the distance to the space shuttle:( ) ( )

( ) ( )s s

a

a

1200 kg 0.14 m/s7.5 s 14 m

92 kg

m vd v t t

m

−= = − = − =

Insight: One of the tricky things about spacewalking is that whenever you push on a satellite or anything else, becauseof Newton’s Third Law you yourself get pushed ! Conservation of momentum makes it easy to predict your speed.

26. Picture the Problem: The lumberjack moves to the right while the log moves to the left.

Strategy: As long as there is no friction the total momentum of the lumberjack and the log remains zero, as it was before the lumberjack started trotting. Combine vector addition for relative motion (equation 3-8) with the expression

from the conservation of momentum to find L, sv = speed of lumberjack relative to the shore. Let L, logv = speed of

lumberjack relative to the log, and log, sv = speed of the log relative to the shore.

Solution: 1. (a) Write out the equation forrelative motion. Let the log travel in the

negative direction:

L, s L, log log, s

L, s L, log log, s

log, s L, log L, s

v v v

v v v

= +

= −

= −

v v v

2. Write out the conservation of momentum

with respect to the shore:L L, s log log, s

0 m v m v= = −∑p

3. Substitute the expression from step 1

into step 2 and solve for L, s :v

( )

( )

( )( ) ( )

( )

L L, s log log, s log L, log L, s

L, s L log log L, log

log L, log

L, s

L log

380 kg 2.7 m/s2.2 m/s

85 380 kg

m v m v m v v

v m m m v

m vv

m m

= = −

+ =

= = =++

4. (b) If the mass of the log had been greater, the lumberjack’s speed relative to the shore would have been greater than

that found in part (a), because the log would have moved slower in the negative direction.

5. (c) Use the expression from step 3 to

find the new speed of the lumberjack: ( )( ) ( )

( )log L , log

L, s

L log

450 kg 2.7 m/s2.3 m/s

85 450 kg

m vv

m m= = =

++

Insight: Taking the argument in (b) to its extreme, if the mass of the log equaled the mass of the Earth the lumberjack’sspeed would be exactly 2.7 m/s relative to the Earth (and the log). If the mass of the log were the same as the mass of

the lumberjack, the speed of each relative to the Earth would be half the lumberjack’s walking speed.

27. Picture the Problem: The vector diagram at right indicates the momenta of the

three pieces.

Strategy: Because the plate falls straight down its momentum in the xy plane iszero. That means the momenta of all three pieces must sum to zero. Choose the

motion of the two pieces at right angles to one another to be in the x and

y directions. Set the total momentum equal to zero and solve for 3v

.

Solution: 1. Set 0=∑p

and solve for 3v

:

( ) ( )( ) ( )

3

3

ˆ ˆ 0

ˆ ˆˆ ˆ

mv mv m

mv mvv v

m

= + + =

− + −= = − + −

∑p x y v

x yv x y

2. Find the speed 3 :v ( ) ( )2 2

3 2v v v v= − + − =

3. Find the direction of 3 :v 3,1 1

3,

tan tan 45 180 225 y

x

v v

v vθ

− −⎛ ⎞ −⎛ ⎞

= = = ° + ° = °⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠

Insight: As long as the first two pieces have equal masses the direction of 3v

will always be the same. For instance, if

the third piece has four times the mass of either piece 1 or 2, its speed would be 8v but θ would remain 225°.

x

3p

1p

2p

y

225°



9 – 10

28. Picture the Problem: The two carts collide on a frictionless track and stick together.

Strategy: The collision is completely inelastic because the two carts stick together. Momentum is conserved during the

collision because the track has no friction. The two carts move as if they were a single object after the collision. Usethe conservation of momentum to find the final speed of the carts and final kinetic energy of the system.

Solution: 1. Conserve momentum to find the

final speed of the carts:i f

f f (0) 22 2

p p

mv vmv m mv v

m

=

+ = ⇒ = =

2. Use equation 7-6 to find the final kinetic energy: ( )22 21 1 1

f f 2 2 4(2 ) K m v m v mv= = =

Insight: Half of the initial kinetic energy is gone, having been converted to heat, sound, and permanent deformation of

material during the inelastic collision.

29. Picture the Problem: The car and the minivan collide and stick together, as

indicated in the figure at right. The data given in Example 9-6 include

1 950 kgm = and 2 1300 kg .m =

Strategy: If we ignore frictional forces the total momentum of the two vehicles

before the crash equals the momentum after the collision. The vehicles stick

together so they behave as a single object with the sum of their masses. Use

conservation of momentum to find the initial speed 2v of the minivan and the

final speed f v of the two vehicles after the crash.

Solution: 1. Conserve momentum in the x

direction:( )1 1 1 2 f 0 cos x p m v m m v θ = + = +∑

2. Conserve momentum in the y direction: ( )2 2 1 2 f 0 sin y p m v m m v θ = + = +∑

3. Divide the equation from step 2 by the

equation for step 1 and solve for 2 :v

( )

( )

( ) ( )

( )

1 2 f 2 2

1 1 1 2 f

1

2 1

2

sintan

cos

950 kgtan 20.0 m/s tan 40.0 12 m/s

1300 kg

m m vm v

m v m m v

mv v

m

θ θ

θ

θ

+= =

+

= = ° =

4. Solve the equation in step 1 for f :v ( )

( ) ( )

( )1 1

f

1 2

950 kg 20.0 m/s11 m/s

cos 950 kg 1300 kg cos 40.0

m vv

m m θ = = =

+ + °

Insight: In real life the assumption that there is no friction is quite incorrect unless the collision occurs on very slick,icy pavement. Substantial deviations from the conservation of momentum will be observed if the pavement is dry.



9 – 11

30. Picture the Problem: The initial and final momentum vectors for this

collision are depicted at right.

Strategy: Assuming there is no friction between the players’ skates andthe ice, we can use conservation of momentum together with the fact that

the players stick together after the collision to find the final velocity. Let

the motion of player 1 be in the positive x-direction and the motion of

player 2 be at an angle of 115° measured counterclockwise from the

positive x-axis.

Solution: 1. Write out the conservation of

momentum and solve for f v

: ( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( )

1i 2i f

f

1 1f 2 2

1 1f 2 2

f

ˆ ˆ ˆcos sin 2

ˆ ˆ1 cos sin

ˆ ˆ5.45 m/s 1 cos115 5.45 m/s sin115

ˆ ˆ1.57 m/s 2.47 m/s

mv mv mv m

v v

θ θ

θ θ

+ =

+ + =

+ + =

+ ° + ° =

+ =

p p p

x x y v

x y v

x y v

x y v

2. Determine the magnitude of f v

: ( ) ( )2 22 2

f f f 1.57 m/s 2.47 m/s 2.93 m/s x yv v v= + = + =

Insight: The two players slide away from the collision at 57.5° above the positive x axis. Player 2’s initial momentum

provides the y component of the final momentum, but his x momentum in the ˆ−x direction is smaller than player 1’s

momentum in the x direction, and so the two players have a positive x component of their total momentum. Note that because the mass cancels out, in this particular case only (the case of two identical masses colliding) the final velocity is

the average of the two initial velocities: ( )11i 2i f f 1i 2i2

2m m m+ = ⇒ = +v v v v v v

.

31. Picture the Problem: A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at

6.2 m/s. The vehicles stick together after the collision and move together with a speed of 5.0 m/s.

Strategy: In an inelastic collision we expect a loss of kinetic energy. Use equation 7-6 to find the kinetic energies

before and after the collision and verify the loss.

Solution: 1. (a) The final kinetic energy of the car and truck together is less than the sum of their initial kinetic

energies. Some of the initial kinetic energy is converted to heat, sound, and the permanent deformations in the materials

of the car and truck.

2. (b) Use equation 7-6 to find

the initial kinetic energy:( ) ( ) ( ) ( )

2 22 21 1 1 1i c c t t2 2 2 2

1200 kg 2.5 m/s 2600 kg 6.2 m/s 54 kJ K m v m v= + = + =

3. Use equation 7-6 to find the

final kinetic energy:( ) ( ) ( )

221 1f c t f 2 2

1200 kg 2600 kg 5.0 m/s 48 kJ K m m v= + = + =

Insight: The kinetic energy loss is only 11% because the two vehicles are traveling in the same direction initially. If

this were a head-on collision, the final speed would be 3.5 m/s in the direction the truck was traveling, and f 23 kJ K = ,

a loss of 58%. In fact, two identical objects traveling at the same speed and colliding together will lose 100% of their

kinetic energy if they stick together after the collision.

x

f p

1ip

2ip

y

115°



9 – 12

32. Picture the Problem: The bullet collides with the block, passing right through it and continuing in a straight line but at

a slower speed than it had initially. Afterwards the block moves with constant speed in the same direction as the bullet.

Strategy: Use conservation of momentum to find the speed of the bullet after the collision. Because this is an inelasticcollision we expect a loss of kinetic energy. Use equation 7-6 to find the initial and final kinetic energies and confirm

the energy loss. Let the subscripts b and B denote the bullet and the block, respectively.

Solution: 1. (a) Let i f =p p

and solve for f v :

( ) ( ) ( ) ( )

b bi b bf B Bf

b bi B Bf

bf

b

2

0

0.0040 kg 650 m/s 0.095 kg 23 m/s

0.0040 kg

1.0 10 m/s

m v m v m v

m v m vv

m

+ = +−

=

−=

= ×

2. (b) The final kinetic energy is less than the initial kinetic energy because energy is lost to the heating and

deformation of the bullet and block.

3. (c) Use equation 7-6 to find i K : ( )( )221 1

i b bi2 20.0040 kg 650 m/s 850 J K m v= = =

4. Use equation 7-6 to find f K :

( ) ( ) ( ) ( )

2 21 1f b bf B BF2 2

2 21 1

2 20.0040 kg 104 m/s 0.095 kg 23 m/s 47 J

K m v m v= +

= + =

Insight: In this collision 94% of the bullet’s initial kinetic energy is converted to heat, sound, and permanentdeformation of materials.

33. Picture the Problem: The putty is thrown straight upward, strikes the bottom of the block, and sticks to it. The putty

and the block move together in the upward direction immediately after the collision.

Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after thecollision, and then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximum height

above the collision point to which the conglomerate will rise.

Solution: 1. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of

the putty will be converted to heat, sound, and permanent deformation of material during the inelastic collision.

2. (b) Set i f =p p

and solve for f v : ( ) p

p p b p f f p

b p

m

m v m m v v vm m

⎛ ⎞= + ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠

3. Set ottom top E E = after the collision:

( ) ( )

bottom top

2

p 21 p p b p top2

b p

0 0 K U

mm m v m m g y

m m

+ = +

⎛ ⎞+ = +⎜ ⎟⎜ ⎟+⎝ ⎠

4. Solve the resulting expression for top y ( )

( )

2 2 22

p p

top 2 b p

5.74 m/s0.0750 kg

2 0.420 0.0750 kg 2 9.81 m/s

0.0386 m 3.86 cm

m v y

m m g

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + ⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎣ ⎦

= =

Insight: The putty-block conglomerate will rise even higher if either pv is larger or pm is larger.

34. Picture the Problem: The putty is thrown horizontally, strikes the side of the block, and sticks to it. The putty and the

block move together in the horizontal direction immediately after the collision, compressing the spring.

Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the

collision, then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximumcompression of the spring after the collision.

Solution: 1. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of

the putty will be converted to heat, sound, and permanent deformation of material during the inelastic collision.



9 – 13

2. Set i f =p p

and solve for f v : ( ) p

p p b p f f p

b p

m

m v m m v v vm m

⎛ ⎞= + ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠

3. Set after rest E E = after the collision:

( )

after rest

2

p 2 21 1 p p max2 2

b p

0 0 K U

mm m v kx

m m

+ = +

⎛ ⎞+ =⎜ ⎟⎜ ⎟+⎝ ⎠

4. Solve the resulting expression for max : x ( )

( ) ( )

( ) ( )

22 22 2

p p

max

p b

0.0500 kg 2.30 m/s

20.0 N/m 0.430 0.0500 kg

0.0371 m 3.71 cm

m v x

k m m

⎡ ⎤ ⎡ ⎤⎢ ⎥= = ⎢ ⎥

++⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

= =

Insight: The putty-block conglomerate will compress the spring even farther if pv is larger or if pm is larger.

35. Picture the Problem: Two objects with different masses travel in opposite directions with the same speed, collide, and

stick together.

Strategy: Use equation 7-6 to find the ratio of the kinetic energies before and after the collision. Use conservation of

momentum to find the ratio of the masses.

Solution: 1. (a) Use equation 7-6 to find i K and f K :( ) ( ) ( )

( )

2 21 11 2 1 22 32f

2 2 21 1 1i 1 2 1 22 2 2

4 1

16

m m v m m v K

K m v m v m m v

+ += =

+ +

2. (b) Use i f =p p

to find 1 2m m : ( ) ( ) ( )( )

1 2 1 2

11 2 1 24

1 2 1 2

1 2 1 2

4

4 4

3 5 5 3

m v m v m m v

m m m m

m m m m

m m m m

+ − = +

− = +

− = +

= ⇒ =

Insight: Most of the kinetic energy is lost in the inelastic collision. If the same two masses collide and stick together

except they are traveling in the same direction, with 1m traveling at speed v and 2m at 4v , their final speed will be

( )23 32 v and the energy loss will be only 17%.

36. Picture the Problem: The hammer strikes the nail with an elastic collision. The collision drives the nail in the forward

direction and slows down the speed of the hammer.

Strategy: This is a one-dimensional, elastic collision where one of the objects (the nail) is initially at rest. Therefore,equation 9-12 applies and can be used to find the final speed of the nail. Once the nail speed is found, its kinetic energy

is determined by equation 7-6. Let 1m be the mass of the hammer, 2m be the mass of the nail, and 0v be the initial

speed of the hammer.

Solution: 1. Use equation 9-12 to find 2,f v :( )

( )1

2,f 0 nail

1 2

2 550 g24.5 m/s 8.8 m/s

550 12 g

mv v v

m m

⎡ ⎤⎛ ⎞= = = =⎜ ⎟ ⎢ ⎥

+ +⎝ ⎠ ⎣ ⎦

2. Use equation 7-6 to find nail K : ( ) ( )221 1

nail nail2 20.012 kg 8.8 m/s 0.46 J K mv= = =

Insight: If the collision were instead considered to be inelastic (the hammer remains in contact with the nail and they

move together) the final velocity would be 4.4 m/s and the nail’s kinetic energy would be 0.12 J, a reduction by a factor

of four.



9 – 14

37. Picture the Problem: The truck strikes the car from behind. The collision sends the car lurching forward and slows

down the speed of the truck.

Strategy: This is a one-dimensional, elastic collision where one of the objects (the car) is initially at rest. Therefore,

equation 9-12 applies and can be used to find the final speeds of the vehicles. Let 1m be the mass of the truck, 2m be

the mass of the car, and 0v be the initial speed of the truck.

Solution: 1. Use equation 9-12 to find 1,f v : ( )1 2

1,f 0 truck

1 2

1720 732 kg15.5 m/s 6.25 m/s

1720 732 kg

m mv v v

m m

⎛ ⎞ ⎛ ⎞− −= = = =⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠⎝ ⎠

2. Use equation 9-12 to find 2,f v :

( )( )1

2,f 0 car

1 2

2 1720 kg215.5 m/s 21.7 m/s

1720 732 kg

mv v v

m m

⎡ ⎤⎛ ⎞= = = =⎜ ⎟ ⎢ ⎥

+ +⎝ ⎠ ⎣ ⎦

Insight: The elastic collision produces a bigger jolt for the car. If the collision were instead inelastic and the two

vehicles stuck together, the final speed of the car (and the truck) would be 10.9 m/s.

38. Picture the Problem: You throw a rubber ball at an elephant that is charging directly toward you.

Strategy: Use the principle that the speed of separation after a head-on elastic collision is always equal to the speed ofapproach before the collision to answer the conceptual question.

Solution: Suppose the ball is thrown at velocity −v b and the elephant is charging at velocity +ve. The speed of approach

of the two objects is ve −(− v b) = ve + v b. After the collision the elephant’s speed will be essentially unchanged becauseit has so much more inertia and momentum than the ball. In order to maintain the same speed of separation (v bf − ve) as

the speed of approach, the ball’s speed must now be v bf = v b + 2 ve. We conclude that the speed of the ball after

bouncing off the elephant will be greater than the speed it had before the collision.

Insight: The situation is similar to that discussed in Conceptual Checkpoint 9-4, except that the ball has a nonzero

speed before the collision. From an energy standpoint the elephant exerts a force on the ball, doing work on it and

increasing its kinetic energy. The elephant loses a little kinetic energy in the process, but because its mass is so large its

speed decreases by only a tiny bit. If the elephant were at rest, the ball would rebound with its initial speed.

39. Picture the Problem: The ball and the elephant move horizontally toward each other, and then an elastic collision

occurs, sending the ball back along the direction it came from.

Strategy: This problem can be very difficult because both the ball and the elephant are moving before and after the

collision (see problem 88). However, if we adopt a frame of reference in which the observer is moving with the ball asit approaches the elephant, then in that frame of reference the ball is initially at rest, and equation 9-12 applies. We’ll

take that approach, finding 0v

of the elephant in the ball’s frame of reference, determining the ball’s final velocity in

that frame, and then switching back to the Earth frame of reference to report the ball’s final speed. Let the elephant

initially travel in the positive direction so egˆ4.45 m/s=v x

and bgˆ7.91 m/s .= −v x

Solution: 1. (a) Find 0v

using

equation 3-8: ( ) ( ) ( )

eg,i eb,i bg,i

eb,i eg,i bg,iˆ ˆ ˆ4.55 m/s 7.81 m/s 12.36 m/s

= +

= − = − − =

v v v

v v v x x x

2. Apply equation 9-12 to find b,f v for

the ball in this frame of reference:

( )( )e

bb,f eb,i

e

2 5240 kg212.36 m/s 24.7 m/s

5240 0.150 kgb

mv v

m m

⎡ ⎤⎛ ⎞= = =⎜ ⎟ ⎢ ⎥

+ +⎝ ⎠ ⎣ ⎦

3. Apply equation 3-8 again to find g,f v : ( ) ( ) ( ) bg,f bb,f bg,i

bg,f

ˆ ˆ ˆ24.7 m/s 7.81 m/s 16.9 m/s

16.9 m/sv

= + = + − =

=

v v v x x x

4. (b) The ball’s kinetic energy has increased because kinetic energy has been transferred from the elephant to the ball

as a result of the collision.

Insight: The ball’s kinetic energy has increased from 4.57 J to 21.4 J, almost a factor of five! The elephant’s speed and

kinetic energy hardly changes at all. This is the basic idea behind the gravitational slingshot effect (passage problems97-100). Note that the approach speed of 12.36 m/s is essentially the same as the separation speed of

16.9 − 4.55 = 12.4 m/s , as discussed in section 9-6 of the text.



9 – 15

40. Picture the Problem: The neutron collides elastically and head-on with a stationary particle and is slowed down.

Strategy: If we assume the target particle is initially stationary, we can use equation 9-12 to determine the final speed

of the neutron after the collision, and then use that speed to determine the ratio of kinetic energies in each case. Let m be the mass of the neutron and M be the mass of the target particle.

Solution: 1. (a) Use equation 9-12 to find f v for the neutron: f i

m M v v

m M

−⎛ ⎞= ⎜ ⎟

+⎝ ⎠

2. Calculate the ratio f i K K :( )

2 221i2f

21i i2

m m M m M v K m M

K m M mv

− + −⎛ ⎞= = ⎜ ⎟

+⎝ ⎠

3. Apply the expression in step 2 for an electron, M = 5.49×10−4 u:

24

f

4

i

1.009u 5.49 10 u0.9978

1.009u 5.49 10 u

K

K

−

−

⎛ ⎞− ×= =⎜ ⎟

+ ×⎝ ⎠

4. (b) Apply the expression in step 2 for a proton, M = 1.007 u:

2

6f

i

1.009u 1.007u1 10

1.009u 1.007u

K

K

−⎛ ⎞−= = ×⎜ ⎟

+⎝ ⎠

5. (c) Apply the expression in step 2 for an electron, M = 207.2 u:

2

f

i

1.009u 207.2u0.9807

1.009u 207.2u

K

K

⎛ ⎞−= =⎜ ⎟

+⎝ ⎠

Insight: In case (a) the neutron plows ahead and is hardly slowed by the tiny electron. In case (c) the neutron bounceselastically from the much heavier nucleus and hardly slows down. In case (b) the neutron transfers nearly all its kinetic

energy to the similarly sized proton and almost comes to rest. You might notice this phenomenon when a pool ball

collides with a second identical ball at rest, or with the apparatus of five suspended metal balls depicted on page 275.

41. Picture the Problem: The apple and orange collide in the mannerdepicted in the figure at right. The mass of the apple is 0.130 kg and the

mass of the orange is 0.160 kg.

Strategy: Use conservation of momentum in the y direction to find the

component of the apple’s speed, and conservation of momentum in the

x direction to find the x component of the apple’s speed. The knowncomponents can then be used to find the total speed and direction of the

apple after the collision.

Solution: 1. Conserve momentum

in the y direction to find 1f, yv : ( ) ( )

( )

1 1f,y 2 2f,

2 2f,

1f,y

1

0

0.160 kg 1.03 m/s1.268 m/s

0.130 kg

y y

y

p m v m v

m vv

m

= = −

= = =

∑

2. Conserve momentum in the x

direction to find 1f, xv :

( ) ( )

1 1i, 2 2i, 1 1f,

2

1i, 2i, 1f,

1

0

0.160 kg1.11 m/s 1.21 m/s

0.130 kg

0.38 m/s

x x x x

x x x

p m v m v m v

mv v v

m

= + = +

⎛ ⎞+ = = + −⎜ ⎟

⎝ ⎠

= −

∑

3. Find the apple’s speed: ( ) ( )2 22 2

1f 1f, 1f, 1.268 m/s 0.38 m/s 1.32 m/s x yv v v= + = + − =

4. Find the apple’s direction:1f,1 1

1f,

1.27 m/stan tan 73 180 107

0.38 m/s

y

x

v

vθ

− −⎛ ⎞ ⎛ ⎞

= = = − ° + ° = °⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠

This angle is measured counterclockwise from the positive x axis.

Insight: Because the collision is elastic you can also set 2 21 1f 1 1f 2 2f i2 2

K m v m v K = + = and use the i 0.197 J K = given in

the example to find 1f 1.31 m/sv = , which is correct to within rounding error. Note that we bent the rules of significant

figures in steps 1 and 3 to avoid such rounding error. This approach leaves an ambiguity in the x component of the

apple’s final velocity, however, and you still need conservation of momentum in the x direction to resolve it.



9 – 16

42. Picture the Problem: The cart 4m collides with the cart 2m, which is

given kinetic energy as a result and later collides with the cart m.

Strategy: In each case a moving cart collides with a cart that is at rest,so application of equation 9-12 will yield the final velocities of all the

carts. First apply equation 9-12 to the collision between carts 4m and

2m, then to the collision between 2m and m. Let the 4m cart be called

cart 4, the 2m cart be called cart 2, and the m cart be called cart 1:

Solution: 1. (a) Apply equation 9-12 to thefirst collision:

( )

4 2 14,f 4,i 0 03

4 2

4 42,f 4,i 0 03

4 2

4 2

4 2

2 42

4 2

m m m mv v v v

m m m m

mmv v v v

m m m m

⎛ ⎞− −⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠⎝ ⎠

⎡ ⎤⎛ ⎞= = =⎜ ⎟ ⎢ ⎥

+ +⎝ ⎠ ⎣ ⎦

2. Apply equation 9-12 to the second collision.

In this case cart 2 has an initial speed of 403

v :( )

( )( )

2 1 4 42,f 2,i 0 03 9

2 1

2 1641,f 2,i 0 03 9

2 1

2

2

2 22

2

m m m mv v v v

m m m m

mmv v v v

m m m m

⎛ ⎞− −⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

⎡ ⎤⎛ ⎞= = =⎜ ⎟ ⎢ ⎥

+ +⎝ ⎠ ⎣ ⎦

3. (b) Verify that i f K K = by writing using

equation 7-6 and dividing both sides by 2

0mv :

( ) ( ) ( ) ( ) ( ) ( )( )? 2 2 22 161 1 1 1 4 1

0 0 0 02 2 3 2 9 2 94 4 2

2 16 256 36 32 256 3242 2

9 81 162 162 162 162 162

m v m v m v m v= + +

= + + = + + = =

Insight: Note that due to the transfer of kinetic energy via collisions, the cart with the smallest mass ends up with the

largest speed.

43. Picture the Problem: The balls collide elastically along a single direction.

Strategy: Set the initial momentum equal to the final momentum and solve for the final speed of the two balls. Then set

the initial kinetic energy equal to the final kinetic energy and solve for the final speed of the two balls again.

Solution: 1. (a) Set i f =p p

and solve for f v : ( ) 0

0 f f 22

vmv m v v= ⇒ =

2. (b) Set i f K K = and solve for f v : ( )2 2 01 10 f f 2 2

22

vmv m v v= ⇒ =

Insight: If you conserve the momentum and kinetic energy during the collision, assuming the balls that are going into

the collision come to rest afterwards, you can show that in outm m= no matter how many balls are sent into the collision.

44. Picture the Problem: A stalactite in a cave has drops of water falling from it to the cave floor below. The drops are

equally spaced in time and come in rapid succession, so that at any given moment there are many drops in midair.

Strategy: Consider the center of mass as the “average” location of the system’s mass in order to answer the conceptualquestion.

Solution: 1. (a) The center of mass is a weighted average of the location of the mass of a system. As the drops fall, theirseparations increase (see Conceptual Checkpoint 2-5). With the drops more closely spaced on the upper half of their

fall, the center of mass is shifted above the halfway mark. We conclude that the center of mass of the midair drops is

higher than the halfway distance between the tip of the stalactite and the cave floor.

2. (b) The best explanation is III. Though equally spaced in time, the drops are closer together higher up. Statements I

and II are each false.

Insight: It does not matter whether you place the origin of the coordinate system at the tip of the stalactite or at the floorof the cave. In either case there are more droplets closer to the stalactite tip and the center of mass is higher than the

midpoint between the tip of the stalactite and the cave floor.



9 – 17

45. Picture the Problem: The bricks are configured as shown at right.

Strategy: Use equation 9-14 to calculate the x-coordinate of the center of

mass. Assume that the mass of each brick is m and that the mass of each

brick is distributed uniformly. Reference all positions from the left side of

the bottom brick, as indicated in the figure.

Solution: Apply equation 9-14 directly:( )5

2 41 1 2 2 3 3

cm

1 2 3

1 11 11

3 3 4 12

L Lmx m Lm x m x m x L X L

M m m m m

+ ++ + ⎛ ⎞= = = = =⎜ ⎟

+ + ⎝ ⎠

∑

Insight: The bricks will be balanced as long as the center of mass remains over the point of support. That means the

table could end at 11 12 the length of the bottom brick and the stack would still be stable.

46. Picture the Problem: The cereal cartons are at the left end of the basket and the milk

will be placed toward the right end.

Strategy: Place the origin at the left end of the 0.75 m L = basket. Then, the two

cartons of cereal are at c 0 m . x = Put the milk at position x and set the center of mass

equal to 2 L , then solve for x. That will be the correct position of the milk relative to

the left end of the basket to put the center of mass at the center of the basket.

Solution: Use equation 9-14 and set cm2 X L= :

( ) ( ) ( )

( )

c c m m m

cm

c m c m

c m

m

2 0

2 2 2

0.71 m 2 0.56 kg 1.8 kg20.58 m

2 2 1.8 kg

mx m x m x m x L X

M m m m m

L m m x

m

+ += = = =

+ +

+⎡ ⎤+ ⎣ ⎦= = =

∑

Insight: Another approach is to put the origin at the center of the basket and find the position of the milk that puts thecenter of mass at zero. You should get 0.22 m to the right of center, or about 0.58 m from the left end.

47. Picture the Problem: The Earth and Moon lie along a line connecting their centers of mass.

Strategy: Place the x-axis along the line connecting the centers of the Earth and Moon, and put the origin at the center

of the Earth. Use equation 9-14 to find the location of the center of mass of the Earth-Moon system.

Solution: 1. Find the distance

between the center of mass of theEarth and the center of mass of

the Earth-Moon system:

( )( )22 8

6E E m mcm 24 22

E m

0 7.35 10 kg 3.85 10 m4.67 10 m

5.98 10 kg 7.35 10 kg

mx m x m x X

M m m

+ × ×+= = = = ×

+ × + ×

∑

2. Subtract cm X from6

Earth 6.37 10 m: R = ×

6 6 66.37 10 m 4.67 10 m 1.70 10 m below the surface of the Earth .× − × = ×

Insight: In Chapter 12 we’ll explore how the Earth and Moon orbit about their common center of mass, producing a

“wobble” in the Earth’s path around the Sun.

48. Picture the Problem: A triangular piece of sheet metal of mass M is markedwith a vertical dashed line at the point where the mass to the left of the line

( M /2) is equal to the mass to the right of the line (also M /2).

Strategy: Use the definition of the center of mass to answer the question.

Solution: 1. (a) You can think of the center of mass as the “average” location of the system’s mass. The mass M /2 on

the right, on average, is located closer to the fulcrum than the mass M /2 on the left. The center of mass must therefore

be to the left of the dashed line, and we conclude that the metal sheet will tip to the left.

2. (b) The best explanation is II. The metal sheet extends for a greater distance to the left, which shifts the center of

mass to the left of the dashed line. Statements I and III are each false.

Insight: The center of mass of a system is a combination of how much mass it contains and where it is located. If the

sheet metal were cut into a rectangular shape, the center of mass would be located right at the dashed line.

C e r e a l

x

y

M i l k



9 – 18

49. Picture the Problem: A pencil standing upright on its eraser end falls over and lands on a table. As the pencil falls, its

eraser does not slip.

Strategy: Consider the location of the pencil’s center of mass when answering the conceptual question. Let the positive

x direction be in the direction the pencil falls, and the positive y direction be vertically upward.

Solution: 1. (a) As the pencil falls its center of mass moves along an arc in the positive x direction and in the negative y

direction (downward). By Newton’s Second Law the net force must point in the direction of the acceleration. We

conclude that the x component of the contact force is positive.

2. (b) Because the center of mass of the pencil has a nonzero, downward acceleration, it follows that the net verticalforce acting on it is downward. The net vertical force is the vector sum of the upward contact force and the downward

weight of the pencil. We conclude that the y component of the contact force is less than the weight of the pencil.

Insight: When the vertical contact force equals the weight the pencil is in equilibrium in the vertical direction.

50. Picture the Problem: The box with no top rests on a flat surface. Its center of mass is measured relative to the

geometric center of the box, a distance 2 L above the bottom surface.

Strategy: Place the origin at the center of the box with the plane of the missing top perpendicular to the positive z -axis.

Due to symmetry, X cm = Y cm = 0. Use a version of equation 9-14 written in the z direction to determine the location of

the center of mass. Assume each side of the box has mass m. Let z = 0 correspond to the center of the box.

Solution: Apply equation 9-14

directly:( )21 2 3 4 bottom

cm

0 0 0 0

5 5 10

Lmmz mz mz mz mz mz L Z

M m m

+ + + −+ + + +Σ= = = = −

The center of mass is L/10 units below the center of the box.

Insight: Placing some other masses at the bottom of the box will lower the center of mass even farther.

51. Picture the Problem: The partially eaten pizza is depicted at right.

Strategy: We expect the center of mass of the second quadrant to be at a

negative x value that is greater in magnitude than −1.4 in, and a positive y value. Adding the third and fourth quadrants back in would shift the center of

mass to the given (−1.4 in,−1.4 in) position. By symmetry the positions of the

centers of mass of each quadrant must be the same distances x and y from themiddle of the pizza. Use this fact to find those positions.

Solution: 1. Use equation 9-14

to set cm1.4 X = − in:

( ) ( )

( )

2 3 4

cm3

1.4 in3

3 1.4 in 4.2 in

mx mx mx X

m x x x

x

+ +=

− + − += = −

= − − =

2. Use equation 9-15 to set

cm1.4Y = − in:

( ) ( )

( )

2 3 4

cm 1.4 in3 3

3 1.4 in 4.2 in

y y ymy my myY

m y

+ − + −+ += = = −

= − − =

3. This tells us the distances from the centers of mass of each quadrant to the origin. For the second quadrant x must be

negative and y must be positive: ( ) ( ), 4.2 in, 4.2 in x y = −

Insight: We can therefore conclude the center of mass of the third quadrant is (− 4.2 in, − 4.2 in) and the center of mass

of the fourth quadrant is (4.2 in, − 4.2 in).



9 – 19

52. Picture the Problem: The geometry of the sulfur dioxide molecule is

shown at right.

Strategy: The center of mass of the molecule will lie somewhere along the

axis because it is symmetric in the x direction. Find cmY using equation

9-15. Both oxygen atoms will be the same vertical distance O y from the

origin. Let m represent the mass of an oxygen atom, sm the sulfur atom.

Solution: 1. Use equation 9-15 to find cmY

( )( )

( )

O O s s O

cm

s s

2 0

2 2

2 16 u 0.143 nm sin 300.036 nm

2 16 u 32 u

my my my m y myY

m m m m

+ + += = =

+ +

°= =

+

∑

2. Recalling that 1 nm = 1×10−9 m, we can write ( ) ( )11

cm cm, 0, 3.6 10 m X Y −= ×

Insight: If the angle were to decrease from 120° the center of mass would move upward. For instance, if the bond

angle were only 90°, the center of mass would be located at (0, 5.1×10−11 m).

53. Picture the Problem: The sticks are arranged as depicted at right.

Strategy: Use equations 9-14 and 9-15 to find the position of the center of

mass.

Solution: 1. (a) Apply equation 9-14 directly:

( )

1 2 3

cm

1

3

cm

30 0.50 m 1.5 m

0.67 m

Mx Mx Mx X

M

X

+ +=

= + +

=

2. Apply equation 9-15 directly: 1 2 3

cm

1(0.50 m 0 0) 0.17 m

3 3

My My MyY

M

+ += = + + =

3. The center of mass of the meter sticks is therefore located at ( ) ( )cm cm, 0.67 m, 0.17 m . X Y =

4. (b) The location of the center of mass would not be affected. The mass drops out of the equations.

Insight: If three more meter sticks were added to make a rectangle, the center of mass would be at (1.00 m, 0.50 m).

x

y

0.5 m

0.50 m 1.5 m

Center of Mass



9 – 20

54. Picture the Problem: One end of the rope is lifted at constant speed.

Strategy: Use equation 9-15 to find the position of the center of mass andequation 9-16 to find the velocity of the center of mass of the rope. When

writing subscripts for the variables, let f = on the floor, nf = not on the floor,

and L = length of the rope. The graphs should all end at

( )2.00 m 0.710 m/s 2.82 s,t L v= = = at which time the rope will be

entirely off the floor and the velocity of its center of mass will be 0.710 m/s

upward. The top of the rope is at position vt during the lift, so the center of

mass of the above floor portion of the rope is halfway between zero and vt ,

and the fraction of the rope that is above the floor is ( )nf ,m vt L M = where

is the total mass of the rope.

Solution: 1. Use equation 9-15 to

find the position of the center of

mass of the rope:

( )

( ) ( )

( )

( )

( )

1nf 2nf nf f f

cm

nf f

1 2 22

2 2

2 2

cm

0

2

0.710 m/s

2 2.00 m

0.126 m/s , 0 2.82 s

m vt m y m yY

m m M

vt L M vt v t

M L

t

Y t t

++= =

+

⎡ ⎤⎣ ⎦= =

=

= < <

2. Use equation 9-16 to find the

velocity of the center of mass of

the rope:

( )

( )

( )

nf nf f f cm

nf f

22

2

cm

0

0.710 m/s

2.00 m

0.252 m/s

vt L M vm v m vV

m m M

v t t

L

V t

+⎡ ⎤+ ⎣ ⎦= =+

= =

=

3. The plots of cmY and cmV as a function of time are shown above.

Insight: The position of the center of mass varies as the square of the time at first but would become linear with a slope

of 0.710 m/s as soon as the last bit of rope left the floor (at t = 2.82 s).



9 – 21

55. Picture the Problem: One end of the rope is lowered at constant speed.

Strategy: Use equation 9-15 to find the position of the center of mass andequation 9-16 to find the velocity of the center of mass of the rope. When

writing subscripts for the variables, let f = on the floor, nf = not on the floor,

and L = length of the rope. The time t is zero the instant the first part of the

rope touches the floor. The graphs should all end at

( )2.00 m 0.710 m/s 2.82 s,t L v= = = at which time the rope will be entirely

on the floor and the velocity of its center of mass will be zero. The top of the

rope is at position L vt − during the drop, so the center of mass of the above

floor portion of the rope is halfway between zero and L vt − , and the fraction

of the rope that is above the floor is ( )nf ,m L vt L M = −⎡ ⎤⎣ ⎦ where M is the

total mass of the rope.

Solution: 1. Use

equation 9-15 tofind the position of

the center of mass

of the rope:

( )

( ){ } ( ) ( )

( )

( )

( )

1nf 2nf nf f f

cm

nf f

212

2

21

cm

0

2

2.00 m 0.710 m/s

2 2.00 m

1.00 0.355 s m, 0 2.82 s

m L vt m y m yY

m m M

L vt L M L vt L vt

M L

t

Y t t −

− +⎡ ⎤+ ⎣ ⎦= =+

− −⎡ ⎤ ⎡ ⎤ −⎣ ⎦ ⎣ ⎦= =

⎡ ⎤−⎢ ⎥=⎢ ⎥⎣ ⎦

⎡ ⎤= − < <⎣ ⎦

2. Use equation 9-16

to find the velocity of

the center of mass of

the rope:

( ){ }( )

( )

( )

nf nf f f cm

nf f

22

2

cm

0

0.710 m/s0.710 m/s

2.00 m

0.252 m/s 0.710 m/s

L vt L M vm v m vV

m m M

t v t vL

L

V t

− − +⎡ ⎤+ ⎣ ⎦= =

+

−= = −

= −

3. The plots of cmY and cmV as a function of time are shown above.

Insight: The position of the center of mass varies linearly as the rope is lowered until the end of the rope hits the floor

(at t = 0), at which time it begins to vary with the square of the time.

56. Picture the Problem: The physical arrangement for this problem is

depicted in the figure at right.

Strategy: Before the string breaks, the reading on the scale is the totalweight of the ball and the liquid. After the string breaks, the ball falls with

constant speed, so that the center of mass of the ball–liquid system does

not accelerate.

Solution: 1. (a) Find the total weight of the ball and the liquid: ( )( )21.20 0.150 kg 9.81 m/s 13.2 N . Mg = + =

2. (b) After the string breaks, the reading is 13.2 N. Because the center of mass of the ball-liquid system undergoes no

net acceleration, the scale must not exert any net force on it (Newton’s Second Law), and by Newton’s Third Law itexerts no net force on the scale. Therefore, the reading will not change.

Insight: If the ball were to accelerate, the center of mass of the ball-liquid system would accelerate downward, and the

scale force would decrease. Another way of looking at the problem is to realize that the water must exert exactly the

same upward force on the ball (1.47 N) when it is at rest as when it is falling at constant speed.



9 – 22

57. Picture the Problem: The two blocks are connected by a string and hang vertically from a spring.

Strategy: Use Newton’s Second Law to determine the force exerted by the spring before the string is cut. Then sum theforces on the two blocks immediately after the string is cut to find the net force. Finally, use Newton’s Second Law

again to find the acceleration of the two-block system just after the string is cut.

Solution: 1. (a) Before the string is cut, the force of gravity is countered by the force of the spring, so the spring force at

this stretch distance is 2 .mg Just after the string is cut, the upper block experiences a force of

s g 2 , F F mg mg mg + = − = and the lower block experiences a force of g . F mg = − The net force acting on the two-block

system is ( )net,ext 0 F mg mg = + − = .

2. (b) Because net,ext cm 0, F MA= = we conclude that cm 0 A = .

Insight: While the top block moves upward it experiences a smaller and smaller force from the spring. When it reaches

the unstretched position of the spring, the force on it is ,mg − so that the net force on the two-block system is 2mg −

and the acceleration of the two blocks is . g −

58. Picture the Problem: The helicopter hovers in midair.

Strategy: Use Newton’s Second Law to set the upward lift of the helicopter blades equal to the downward weight of theaircraft because the acceleration of the helicopter is zero. Solve for the rate of mass ejection by the blades.

Solution: Use Newton’s Second Law

and equation 9-19 to find m t Δ Δ :

( )( )2

2

thrust 0

thrust

5300 kg 9.81 m/s840 kg/s 8.4 10 kg/s

62 m/s

y F mg

mv mg

t

m mg

t v

= − =

Δ⎛ ⎞= =⎜ ⎟

Δ⎝ ⎠

Δ= = = = ×

Δ

∑

Insight: 840 kg/s is about 1100 cubic meters or 39,000 cubic feet of air every second! That is why helicopter rotors

need to be so large.

59. Picture the Problem: The child throws rocks in the backward direction, propelling the wagon in the forward direction.

Strategy: Use Newton’s Second Law to set the forward thrust equal to the backward friction force because the

acceleration of the wagon is zero. Solve for the rate of mass ejection by the rock-throwing girl.

Solution: Use Newton’s Second

Law and equation 9-19 to find

m t Δ Δ :

k

k

k

thrust 0

thrust

3.4 N kg 1 rock 60 s0.31 29 rocks/min

11 m/s s 0.65 kg min

x F f

mv f

t

f m

t v

= − =

Δ⎛ ⎞= =⎜ ⎟Δ⎝ ⎠

Δ= = = × × =

Δ

∑

Insight: If the girl had a 40-lb (18-kg) pile of rocks she could keep the wagon moving for about 1 minute.

60. Picture the Problem: The person throws the bricks in the backward direction, propelling the skateboard in the forward

direction.

Strategy: Use conservation of momentum to determine the recoil speed. The total momentum is initially zero and

remains zero after the bricks are thrown because there is no friction. The speed of the bricks is given relative to the

person ( pv ), not the ground ( gv ), so we must first use equation 3-8 to reference the speed relative to the ground. Let

the bricks be thrown in the negative direction and the person be propelled in the positive direction.

Solution: 1. Use equation 3-8 to relate thespeed of the bricks relative to the ground:

g bp pg= +v v v



9 – 23

2. Set i f =p p

, substitute the expression

from step 1, and solve for the recoil speed

pgv of the person relative to the ground:

( )

( ) ( ) ( )

( )( ) ( )

( )

b bg p s pg

b bp pg p s pg b bp p s b pg

b bp

pg

p s b

0 2

2 2

2 0.880 kg 17.0 m/s0.485 m/s

57.8 2.10 2 0.880 kg2

m v m m v

m v v m m v m v m m m v

m vv

m m m

= + +

= + + + = + + +

× −= − = − =

+ + ×+ +

Insight: The concept of rocket thrust is more useful when mass is being continually ejected, and when the problem asks

about forces and acceleration. In this case the mass is ejected all at once, and recoil speed is requested, so conservationof momentum is the way to go.

61. Picture the Problem: The person throws the bricks in the backward direction, propelling the skateboard in the forward

direction.

Strategy: Use conservation of momentum to determine the recoil speed. This must be done separately for each brick,and care must be taken to ensure the speeds are measured relative to the ground, not the person on the skateboard. The

momentum is initially zero and remains zero after the first brick is thrown because there is no friction. The speed of the

bricks is given relative to the person ( pv ), not the ground ( gv ), so we must use equation 3-8 to reference the speed

relative to the ground. Let the bricks be thrown in the negative direction and the person be propelled in the positive

direction.

Solution: 1. Use equation 3-8 to relate the

speed of the bricks relative to the ground:g bp pg

= +v v v

2. For the first brick, set i f =p p

,

substitute the expression from step 1,

and solve for the recoil speed pg1v of the

person relative to the ground afterthrowing one brick:

( )

( ) ( ) ( )

( )( ) ( )

( )

b bg p s b pg1

bp pg1 p s b pg1 b bp p s b pg1

b bp

pg1

p s b

0

2

0.880 kg 17.0 m/s0.243 m/s

57.8 2.10 2 0.880 kg2

m v m m m v

m v v m m m v m v m m m v

m vv

m m m

= + + +

= + + + + = + + +

−= − = − =

+ + ×+ +

3. If we concentrate only on the person

and the remaining brick, the initialmomentum is not zero. Repeat step 2,

setting i f =p p

and substituting the

expression from step 1to find pg2v :

( ) ( )

( ) ( )

( )

( )

( ) ( ) ( )

( )

p s b pg1 b bg p s pg2

b bp pg2 p s pg2

b bp p s b pg2

b bp

pg2 pg1

p s b

0.880 kg 17.0 m/s0.243 m/s 0.489 m/s

57.8 2.10 0.880 kg

m m m v m v m m v

m v v m m v

m v m m m v

m vv v

m m m

+ + = + +

= + + +

= + + +

= −+ +

−= − =

+ +

Insight: The person ends up going slightly faster if the bricks are thrown one at a time instead of all at once. This is

because when the second brick is thrown the person has a smaller initial mass (he is holding only one brick, not two) so

that the momentum of the second brick represents a larger fraction of the initial momentum and he gets a slightly larger

boost in his speed.



9 – 24

62. Picture the Problem: Sand is poured at a constant rate into a bucket that is sitting on a scale.

Strategy: The reading of the scale is given by the total weight of the sand and bucket plus the force of impact due to the

pouring sand. Use the concept of thrust (equation 9-19) to determine the total force S F on the scale.

Solution: 1. (a) Add the sand thrust

to the weight:( )

( )( ) ( ) ( )

S b s

20.540 0.750 kg 9.81 m/s 0.0560 kg/s 3.20 m/s 12.8 N

m F m m g v

t

Δ= + +

Δ= + + =

2. (b) Calculate W : ( ) ( )( )2

b s 0.540 0.750 kg 9.81 m/s 12.7 NW m m g = + = + =

Insight: The thrust of the sand is 0.179 N, but it appears to make only a 0.1-N difference in the scale reading because of

the rounding required by proper attention to significant figures.

63. Picture the Problem: The rope is lowered onto the scale, continuously adding to the mass that is resting on the scale.

Strategy: Calculate the thrust due to the mass per time of rope being added to the scale at the given speed. The mass

per time of the rope is related to the mass per length and the speed by: ( ) ( ) ( )m t m x x t m x vΔ Δ = Δ Δ Δ Δ = Δ Δ . Add the

thrust to the weight to find the total force on the scale.

Solution: 1. (a) Use equation 9-19 to find the thrust: ( ) ( )2

thrust 0.13 kg/m 1.4 m/s 0.25 Nm m

v v vt x

Δ Δ⎛ ⎞= = = =⎜ ⎟

Δ Δ⎝ ⎠

2. (b) The scale reads more than 2.5 N. It reads the weight of the rope on the scale plus the thrust due to the falling rope.

3. (c) Add the thrust to the weight: Scale reading thrust 0.25 N 2.5 N 2.8 Nmg = + = + =

Insight: The thrust in part (c) is only 10% of the weight. It represents 100% of the scale reading at the instant the rope

first touches the scale, but becomes less and less significant as more of the rope’s mass rests on the scale.

64. Picture the Problem: Three objects have different masses but the same momenta.

Strategy: Write the kinetic energy of an object in terms of its momentum: 2 2 2 21

22 2 . K mv m v m p m= = = Use this

equation to determine the ranking of the kinetic energies.

Solution: The kinetic energy is inversely proportional to the mass when the momentum is constant. The smallest mass

will therefore have the highest kinetic energy. We therefore arrive at the ranking B < A < C.

Insight: The smaller mass will have a higher speed in order to have the same momentum as a larger mass. This will

give the smaller mass a larger kinetic energy because K is proportional to the square of the speed.

65. Picture the Problem: Three objects have different masses but the same kinetic energy.

Strategy: Write the momentum of an object in terms of its kinetic energy: .222 mK p K m p =⇒= Use this

equation to determine the ranking of the momenta.

Solution: The momentum is proportional to the square root of the mass when the kinetic energy is constant. The

smallest mass will therefore have the smallest momentum. We therefore arrive at the ranking C < A < B.

Insight: Object A has four times more mass and twice the momentum of object C.



9 – 25

66. Picture the Problem: A block of wood is struck by a bullet. The bullet either embeds itself in the wood or it rebounds.

Strategy: Consider the change in the momentum of the bullet for the two scenarios. The larger the change inmomentum of the bullet, the larger the impulse the wood must deliver to the bullet, and by Newton's Third Law the

larger the impulse the bullet delivers to the wood.

Solution: 1. (a) If the bullet bounces off the wood then its change in momentum is greater than if it embeds itself into

the wood and comes to rest. We conclude that block is more likely to be knocked over if the bullet is rubber and

bounces off the wood.

2. (b) The best explanation is I. The change in momentum when a bullet rebounds is larger than when it is brought to

rest. Statement II is true, but irrelevant, and statement III is false.

Insight: If the bullet were to bounce elastically, with the same rebound speed as its initial speed, the impulse delivered

to the wood is exactly twice the impulse delivered if the bullet embeds itself in the wood.

67. Picture the Problem: A juggler performs a series of tricks with three bowling balls while standing on a bathroom scale.

Strategy: Consider the forces on the juggler and the scale while he performs the tricks.

Solution: 1. When the juggler exerts an upward force on a bowling ball, by Newton's Third Law the ball exerts a

downward force on the juggler and the scale reading is higher than when the juggler simply holds the ball at rest. A

similar effect occurs when a ball lands in the juggler’s hand. However, when one or more balls are in the air, the scalesupports a weight that is less than the weight of the juggler plus the balls. One way to think about the problem is to

realize that the center of mass of the juggler plus balls remains more or less at rest. Therefore, we conclude that the

average reading of the scale is equal to the weight of the juggler plus the weight of the three balls.

Insight: If the juggler were tossing only two balls in the air with one hand, the scale force would be large when a ball is

either being launched or caught by the juggler’s hand (requiring an upward scale force to accelerate it) but equal to theuggler’s weight during those instants when both balls are simultaneously in the air. The average reading of the scale

would equal the weight of the juggler plus both balls because the center of mass of the juggler-ball system does not

accelerate on average.

68. Picture the Problem: The tourist climbs the staircase, placing her center of mass farther from the center of the Earth.

Strategy: Because the net external force on the person-Earth system is zero (gravity is an internal force to this system)

the center of mass must remain stationary. That means that as the tourist moves away a distance t x from the center of

mass, the Earth must move a distance E x to keep the center of mass stationary. Set the center of mass of the person-

Earth system equal to zero and determine the displacement of the Earth.

Solution: Set cm 0 X = and solve for E x :

( )

t t E E

cm

t E

21t

E t 24

E

0

72.5 kg555 ft 6.73 10 ft

5.97 10 kg

m x m x X

m m

m x x

m

−

+= =

+

⎛ ⎞= − = = ×⎜ ⎟

×⎝ ⎠

Insight: The distance the Earth moves is 2.05×10−21 m or about one-millionth of the width of the nucleus of a single

atom inside the Earth!



9 – 26

69. Picture the Problem: The figure shows two blocks connected by a string

that runs over a pulley. The table is frictionless, and the two blocksaccelerate in a clockwise direction when they are released. The initial and

final centers of mass are marked in the diagram.

Strategy: Consider Newton's Second Law in terms of the entire system of blocks when answering the conceptual question.

Solution: 1. (a) The net force acting on the system of the two connected blocks—including gravity, the normal force,

and the force exerted by the pulley—is constant in both magnitude and direction. Therefore, because the center of mass

starts at rest, it will accelerate in a straight line in the direction of the net force. Only the green path satisfies this

condition.

2. (b) The best explanation is II. The center of mass starts at rest, and moves in a straight line in the direction of the net

force. Statement I is true but does not distinguish between the paths. Statement III is false.

Insight: You must consider all the forces on a system in order to determine the net force and apply Newton’s Second

Law.

70. Picture the Problem: The moving car collides with a stationary car on a frictionless surface.

Strategy: Conserve momentum in order to find the speed of the second car. Then calculate the initial and final kinetic

energies of the entire system to determine if any energy is lost. If energy is lost the collision is inelastic.


and solve for 2f v : ( )

( )( )

11i 1f 2f 2

1i 1f

2f

0

42 3

2 3

mv mv m v

m v vv v v v

m

+ = +

−= = − =

2. (b) Use equation 7-6 to find i K : 21i 2

K mv=

3. Use equation 7-6 to find f : K ( )

( ) ( )( ) ( )

2 21 1 1f 1f 2f 2 2 2

2 2 2 2161 1 1 1 4 1 1 1

2 3 2 2 3 2 9 18 2

K mv m v

m v m v mv mv

= +

= + = + =

4. Because i f K K = we conclude the collision is elastic.

Insight: The cars bounce off each other and separate because the final speed of the second car ( )4 3 v is greater than the

final speed of the first car 3.v The first car transfers 8 9 of its initial kinetic energy to the second car.

71. Picture the Problem: The bullet strikes the block and together they arelaunched horizontally off the edge of the table.

Strategy: Use conservation of momentum to determine the horizontalspeed of the bullet and block, then apply equation 4-9 to find the landing

site of the bullet-block combination. Let m = the mass of the bullet,

= the mass of the block, 0v = the initial speed of the bullet, and h = the

height of the table.

Solution: 1. Set i f =p p

and solve for f :v ( ) ( )

( )

0 f

f 0

0

0.0105 kg715 m/s

0.0105 1.35 kg

5.52 m/s

mv M m M v

mv v

m M

+ = +

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠=



9 – 27

2. Now apply equation 4-9: ( ) ( )

f 2

2 0.782 m25.52 m/s 2.20 m

9.81 m/s

h x v

g = = =

Insight: If the block were twice as massive it would land about half as far, 1.11 m. In this way the landing spot is ameasure of the mass of the block, as long as the block is much more massive than the bullet.

72. Picture the Problem: The egg carton’s dimensions are indicated in the

figure at right. Either egg 1 or egg 2 is removed.

Strategy: The center of mass changes when an egg is removed becausethe center of mass is an average mass weighted by the distance each egg

is from the origin. Use equations 9-14 and 9-15 to calculate the centerof mass position. In each case we expect the center of mass to move

down and to the left.

Solution: 1. (a) Because the center of mass is an average quantity weighted by the distance each egg is from the origin,

its location will change more if egg 2 is removed.

2. (b) Apply equation 9-14

with egg 1 removed:

( ) ( )cm

2 15.0 9.0 3.0 9.0 15.0 cm 3.0 cm 3.0 cm0.27 cm

11 11

m m X

m

− − − + + + −= = = −

3. Apply equation 9-15

with egg 1 removed:

( ) ( )cm

6 3.5 cm 5 3.5 cm 3.5 cm0.32 cm

11 11

m mY

m

− + −= = = −

4. When egg 1 is removed the center of mass moves to ( ) ( )cm cm, 0.27 cm, 0.32 cm X Y = − −

5. (c) Apply equation 9-14

with egg 2 removed:

( ) ( )cm

2 15.0 9.0 3.0 3.0 9.0 cm 15.0 cm 15.0 cm1.36 cm

11 11

m m X

m

− − − + + + −= = = −

6. Apply equation 9-15

with egg 2 removed:

( ) ( )cm

6 3.5 cm 5 3.5 cm 3.5 cm0.32 cm

11 11

m mY

m

− + −= = = −

7. When egg 2 is removed the center of mass moves to ( ) ( )cm cm, 1.36 cm, 0.32 cm X Y = − −

Insight: In Chapter 11 we will learn how the weight of egg 2 produces a larger torque about the center of mass than

egg 1 because of its greater moment arm. It’s another way to think about how removing egg 2 produces a larger effect.

73. Picture the Problem: The falling raindrops are stopped by the upward force from the ground.

Strategy: Calculate the rate m t Δ Δ at which water is delivered to a square meter of the ground, and then use the thrust

equation (9-19) to estimate the force.

Solution: 1. Find the rate at which

the water fell over 1.0 m2: ( )2

3

31 in 0.0254 m 1 h 1000 kg1.0 m 0.024 kg/s

9.0 h in 3600 s m

m

t

Δ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= × =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

Δ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠

2. Find the thrust from equation 9-19: ( ) ( )thrust 0.024 kg/s 10 m/s 0.24 N

mv

t

Δ⎛ ⎞= = =⎜ ⎟Δ⎝ ⎠

Insight: Because a Newton is about ¼ lb, this force is about 1 16 lb or about one ounce on a square meter. However, if

you allow all 31 inches to pool on top of the square meter it will weigh 7.7 kN = 1700 lb = 0.87 ton!



9 – 28

74. Picture the Problem: The apple falls vertically downward due to gravity.

Strategy: The rate of change of the apple’s momentum is the net force acting on it, according to equation 9-3. The onlyforce acting on the apple is its weight mg . Calculate the change in momentum produced by gravity using equation 9-3.

Solution: 1. (a) Solve equation 9-3 for p t Δ Δ : 2.7 N p

F t

Δ= =

Δ

2. (b) Solve equation 9-3 for pΔ : ( ) ( )2.7 N 1.4 s 3.8 kg m/s p F t Δ = Δ = = ⋅

Insight: We could also write the units of the answer to part (a) as ( )2.7 kg m/s s.⋅

75. Picture the Problem: The placement of the lead weight moved the center of mass of the tire from a certain location tothe geometric center.

Strategy: Let the coordinate system be placed so that the origin is at the geometric center of the wheel. Treat the entire

mass of the wheel as a point mass at some small distance d from the origin. Together with the lead weight the center of

mass of the system is exactly at zero, so write out equation 9-14 and solve for d .

Solution: 1. Solve equation 9-14 for d :

( )

lw lw wh

cm

lw wh

lw

lw

wh

0

0.0502kg25.0 cm 10 mm/cm 0.354 mm

35.5 kg

m x m d X

m m

md x

m

+= =

+

⎛ ⎞= − = − × = −⎜ ⎟

⎝ ⎠

2. Before the lead weight was added, the center of mass was 0.354 mm from the center of the wheel.

Insight: Treating the entire wheel as if it were a point mass located at its center of mass is an important concept inmechanics. The attempt to prove this concept led Isaac Newton to develop the mathematical method we call calculus!

76. Picture the Problem: The locations of the hoop and the ball in

each configuration are represented in the figure at right.

Strategy: Because there are no external forces acting on the

system in the x-direction, cm X of the system will not change.

Use equation 9-14 to set cm X before the ball moves equal to

cm X after the ball moves and solve for x . Note that after the

ball moves it is directly under the center of the hoop, so that both

the hoop and the ball are at position x .

Solution: Set cm,before cm,after X X = and solve for x :( ) ( )

( )

( ) ( )

b b

b

2 2 1 1 b 3 3 4 2

0 2 2

2 2

2 3

M M R r Mx Mx

M M M M

R r x

x R r R R R

+ − +=

+ +

− =

= − = − =

Insight: If there were an unbalanced external force on the system, such as a child pushing on the hoop, the center of

mass of the system would accelerate in accordance with Newton’s Second Law.



9 – 29

77. Picture the Problem: The locations of the canoeist and the canoe relative

to shore in each configuration are depicted in the figure at right.

Strategy: Because there are no external forces, the center of mass of the

canoe-person system remains stationary relative to the shore. Because the

canoe is 3.0 m long, its center is 2.5 1.5 m 4.0 m+ = from the shore. Put

the shore at the origin of the coordinate system. Force the center of mass

to remain at 4.0 m and find the location of the person when the center of

the canoe is 1.5 m farther from shore than the person.

Solution: 1. (a) When the canoeist walks toward the shore, the canoe will move away from the shore according to

conservation of momentum. Therefore, the end of the canoe closest to the shore will be a distance greater than 2.5 m

from shore when the canoeist arrives at that end.

2. (b) By inspection of the diagram the position of the

canoe will be 1.5 m farther from shore than the person:c p 1.5 m x x= +

3. Use equation 9-14 to set cm 4.0 m. X =

Substitute the expression for c x from step 2,

and solve for p x : ( ) ( ) ( )

( )( ) ( )

( ) ( ) ( )

( )

c c p p

cm

c p

c p p p c p

c p c

p

c p

4.0 m

1.5 m 4.0 m

4.0 m 1.5 m

22 kg 1.5 m4.0 m 3.6 m

22 63 kg

m x m x X

m m

m x m x m m

m m m x

m m

+= =

+

+ + = +

+ −=

+

= − =+

Insight: Another way to solve this problem is to set the origin of the coordinate system at the center of mass of the

canoe (set cm 0 X = ), and solve for p 0.39 m x = when she has walked to the end of the canoe. She is then a distance

4.0 − 0.39 = 3.6 m from the shore.

78. Picture the Problem: The locations of the canoeist and the canoe relative

to shore in each configuration are depicted in the figure at right.

Strategy: Because there are no external forces the center of mass of the

canoe-person system remains stationary relative to the shore. Because the

canoe is 3.0 m long, its center is 2.5 1.5 m 4.0 m+ = from the shore. Put

the shore at the origin of the coordinate system. The center of mass of thecanoe-person system remains at 4.0 m, the person is at 3.4 m, and the

center of the canoe is 3.4 m + 1.5 m = 4.9 m from shore. Use these data to

solve equation 9-14 for the mass of the canoe cm .

Solution: Solve eq. 9-14 for cm :

( )( )

( )( )

( ) ( )

( )

c c p p

cm

c p

c c p p c p

p p

c

c

4.0 m

4.0 m

4.0 m 63 kg 4.0 3.4 m42 kg

4.0 m 4.9 4.0 m

m x m x X

m m

m x m x m m

m xm

x

+= =

+

+ = +

− −= = =

− −

Insight: The heavier canoe moves less than the lighter canoe of problem 77, so the person can get slightly closer toshore while keeping the center of mass of the person-canoe system at 4.0 m. There is only one significant figure in the

answer because of the rules of subtraction, in this case for both numerator and denominator. However, it is awkward to

report the answer as 4×101 kg, so we bent the rules a little bit and reported 42 kg.

3.0 m 2.5 m

3.4 m

4.0 m

3.0 m 2.5 m

3.6 m

4.0 m



9 – 30

79. Picture the Problem: The physical arrangement for this problem is

depicted in the figure at right.

Strategy: Before the string breaks, the reading on the scale is the total

weight of the ball and the liquid. After the string breaks, the ball falls

with constant acceleration, so that the center of mass of the ball–liquid

system also accelerates. Use equation 9-17 to find the acceleration of thecenter of mass and then Newton’s Second Law to find the scale force. Let

m be the mass of the ball, f m be the mass of the fluid and bucket, and M

be the combined mass of 1.20 0.150 kg 1.35 kg.+ =

Solution: 1. (a) Find the totalweight of the ball and the liquid:

( )( )21.35 kg 9.81 m/s 13.2 N . Mg = =

2. (b) Apply equation 9-17 to find cm A :( ) ( ) ( )2

f b b 2

cm

0.150 kg 0.250 9.81 m/s00.273 m/s

1.35 kg

m m a A

M

− ×+= = = −

3. Use Newton’s Second Law

to find the scale force s F :( ) ( ) ( )

s cm

2

s cm 1.35 kg 9.81 0.273 m/s 12.9 N

F Mg MA

F M g A

= − =

= + = − =

∑F

Insight: The scale force must decrease as the ball accelerates downward because if it did not, the forces on the center ofmass would be balanced and it could not accelerate.

80. Picture the Problem: The hockey player tosses the helmet in one direction and recoils in the other.

Strategy: The total horizontal momentum of the hockey player pm and helmet hm is conserved because there is no

friction. The total horizontal momentum remains zero both before and after the helmet is tossed. Use this fact to find

the mass of the player. Let the helmet be tossed in the positive direction.

Solution: Set i, f, x x p p= and solve for pm :

( ) ( ) p p, h h,

h h,

p

p,

0

1.3 kg 6.5 m/s cos1133 kg

0.25 m/s

x x x

x

x

p m v m v

m vm

v

= = +

°= − = − =

−

∑

Insight: This must be a junior player (a 33 kg player weighs 73 lb). If a 70-kg adult tossed the helmet in the samemanner, the recoil velocity would be − 0.118 m/s. Note that the momentum in the vertical direction is not conserved;

the normal force from the ice is an external force that allows the vertical momentum to change from zero to

h h sin11 1.6 kg m/sm v ° = ⋅ .

81. Picture the Problem: The carts move without friction in the mannerindicated by the figure at right.

Strategy: Use equation 9-16 and the known velocity of the center of mass

to find the mass of cart 2.

Solution: Solve equation 9-16 for 2m :( )

( )

( ) ( )

1 0 2 0

cm 0

1 2

2 11 2 1 23 2

2 1 22 13 2 3

2 1

22

3

1

2 2

m v m vV v

m m

m m m m

m m

m m m

+= =

+

+ = +

− = −

= =

Insight: If cart 2 were even more massive, say, 4m, the center of mass of the system would move even slower, at

( ) 03 5 v . As the mass of cart 2 increases even more, the center of mass speed approaches 0 2,v the speed of cart 2.



9 – 31

82. Picture the Problem: One end of the rope is lifted at constant speed.

Strategy: The force you need to exert on the end of the rope equals the weight of the rope that is off the floor plus the

thrust required to add mass to the suspended portion of the rope at the given rate. No additional force is required

because the rope is not accelerating. Use equation 9-19 to find the thrust and the given information to find the mass andweight of the rope that is above the floor.

Solution: 1. Use equation 9-19 to

find the required thrust:( ) ( )

22thrust 0.135 kg/m 1.13 m/s 0.172 Nm m x m

v v vt x t x

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

Δ Δ Δ Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2. Find the total force required: ( )

( )( )2

thrust thrust

0.525 m 0.135 kg/m 9.81 m/s 0.172 N 0.868 N

F mg x m x g = + = ×Δ Δ +

= × + =

Insight: If you were to stop lifting when the end of the rope was 0.525 m above the floor you would still need to exert a

force of 0.695 N to support the weight of the 70.9 grams of rope suspended above the floor.

83. Picture the Problem: The geometry of the water molecule is shown at right.

Strategy: The center of mass of the molecule will lie somewhere along the x axis

because it is symmetric in the y direction. Find cm X using equation 9-14. Both

hydrogen atoms will be the same horizontal distance H x from the origin. Let m

represent the mass of a hydrogen atom, Om the oxygen atom.

Solution: 1. Use equation 9-14 to find cm X

( ) ( ) ( )

( )

H H O O O

cm

O s

12

2 0

2 2

2 1.0 u 0.096 nm cos 104.50.0065 nm

2 1.0 u 16 u

mx mx mx m x my X

M m m m m

+ + += = =

+ +

°= =

+

∑

2. Recalling that 1 nm = 1×10−9 m, we can write ( ) ( )12

cm cm, 6.5 10 m, 0 . X Y −= ×

Insight: If the angle were to increase from 104.5° the center of mass would move to the left. For instance, if the bond

angle were 135° the center of mass would be located at (0, 4.1×10−12 m).

84. Picture the Problem: The two moving carts collide with the4m cart that is at rest.

Strategy: Use conservation of momentum to find the speed of

the carts before and after the two collisions. Because themomentum always remains the same, it is not necessary to

calculate the intermediate speed of carts 2 and 3 before cart 1

collides with them. Finally, use equation 7-6 to find the ratio of

the final kinetic energy to initial kinetic energy.


and solve for f v : ( )0 0 f

30 f f 07

2 0 2 4

3 7

mv mv m m m v

v v v v

+ + = + +

= ⇒ =

2. (b) Calculate f i K K :( ) ( )

( )

231

02 7f

21i 02

7 3

73

m v K

K m v= =

Insight: If you calculate the intermediate speed of the 2m and 4m carts after they collide, you should get 0 3v . In part

(b) we learn that 4 7 or 57% of the initial energy is dissipated as heat, sound, and permanent deformation of material.



9 – 32

85. Picture the Problem: The coasting rocket explodes into two pieces of equal mass that are

ejected at 45.0° to the vertical.

Strategy: Assume gravity is the only force acting on the rocket after it is launched. Find itsspeed after rising for 2.50 s, then use conservation of momentum in the vertical direction and

the principles of center-of-mass motion to answer the questions.

Solution: 1. (a) Use equation 4-6

to find the speed of the rocket

before the explosion: ( ) ( )( )

i, 0

2

i,

44.2 m/s 9.81 m/s 2.50 s

19.68 m/s

y y

y

v v gt

v

= −

= −

=

2. Set i, f, y y p p= and solve for f v

of each piece:

i, f

i,

f

2 2 sin 45.0

19.68 m/s27.8 m/s

sin 45.0 sin 45.0

y

y

mv mv

vv

= °

= = =° °

3. (c) Before the explosion ( )cmˆ19.7 m/sV = y Because the momentum of the system is the same after the explosion,

and the total mass has not changed, ( )cmˆ19.7 m/sV = y after the explosion too.

4. (d) The only force acting on the system before and after the explosion is gravity. Therefore, ( )2

cmˆ9.81 m/s A = − y

Insight: Momentum is conserved in the x direction, as well, as guaranteed by the fact that each piece has the same

speed and the same angle from vertical. The momentum in the x direction is zero both before and after the explosion.

86. Picture the Problem: The momenta of the two cars as they approach the

intersection, as well as the total momentum, are depicted at right.

Strategy: Use the component method of vector addition to determine the

components of 2p

.

Solution: 1. (a) Use the

component method of

vector addition to find 2p

:

( ) ( ) ( ) ( )

( ) ( )

1 2 total

2

2

ˆ ˆ ˆ ˆ11,000 kg m/s 370 kg m/s 15,000 kg m/s 2100 kg m/s

ˆ ˆ4000 kg m/s 2470 kg m/s

+ =

⋅ + − ⋅ + = ⋅ + ⋅

= ⋅ + ⋅

p p p

x y p x y

p x y

2. (b) No, it does not matter which car is closer to the intersection because momentum depends only on mass and

velocity. It is independent of position.

Insight: The position of the cars is important if you are attempting to predict whether the momenta of the two cars will produce a collision at the intersection!

87. Picture the Problem: The books are arranged in a stack as depicted atright, with book 1 on the bottom and book 4 at the top of the stack.

Strategy: It is helpful to approach this problem from the top down. The

center of mass of each set of books must be above or to the left of the point

of support. Find the positions of the centers of mass for successive stacksof books to determine d . Measure the positions of the books from the right

edge of book 1 (right-hand dashed line in the figure).

Solution: 1. The center of mass of book 4

needs to be above the right end of book 3: 32

Ld =

2. The result of step 1 means that the center of mass of book 3 is located at 2 2 L L L+ = from the right edge of book 1.

3. The center of mass of books 4 and 3

needs to be above the right end of book 2:( ) ( )

2 cm,43

2 3

2 4

m L m Ld X L

m

+= = =

4. The result of step 3 means that the center of mass of book 2 is located at 3 4 2 5 4 L L L+ = .

45° 45°

f vf v

i, yv

1p

2p

totalp



9 – 33

5. The center of mass of books 4, 3, and 2

needs to be above the right end of book 1:( ) ( ) ( )

1 cm,432

2 5 4 11

3 12

m L m L m Ld X L

m

+ += = =

6. The result of step 3 means that the center of mass of book 1 is located at 11 12 2 17 12. L L L+ =

7. The center of mass of all four books needs

to be above the right edge of the table:( ) ( ) ( ) ( )

cm,4321

2 5 4 17 12 25

4 24

m L m L m L m Ld X L

m

+ + += = =

Insight: We will explore more about static equilibrium problems such as these in Chapter 11. If you examine the

overhang of each book you find an interesting series:25

2 4 6 8 24

L L L Ld L= + + + = . The series gives you a hint about how

to predict the overhang of even larger stacks of books!

88. Picture the Problem: The mass 1m has an initial velocity 1v and collides with 2m that has initial velocity 2v .

Strategy: Combine the equations of momentum conservation and energy conservation to find the final speeds of the

two masses.

Solution: 1. Set i f =p p

: 1 1 2 2 1 1f 2 2f m v m v m v m v+ = +

2. Set i f K K = : 2 2 2 21 1 1 11 1 2 2 1 1f 2 2f 2 2 2 2

m v m v m v m v+ = +

3. Rearrange the equation from step 1:( )

( )1 1 1f

2 2f 2

1m v v

m v v

−=

−

4. Rearrange the equation from step 2:( )( )

( ) ( )

( ) ( )

2 2

1 1 1f 1 1 1f 1 1f

2 22 2f 2 2f 22 2f 2

1m v v m v v v v

m v v v vm v v

− − += =

− +−

5. Set the expressions in steps 3 and 4 equal to

each other:

( )

( )

( ) ( )

( ) ( )1 1 1f 1 1 1f 1 1f

2 2f 2 2 2f 2 2f 2

2f 2 1 1f

2f 1 1f 2

m v v m v v v v

m v v m v v v v

v v v v

v v v v

− − +=

− − +

+ = +

= + −

6. Substitute the expression for 2f v in step 5

into the equation from step 1:

( )

( ) ( )

1 1 2 2 1 1f 2 1 1f 2

1 2 1 2 2 1 2 1f

1 2 21f 1 2

1 2 1 2

2

2

m v m v m v m v v v

m m v m v m m v

m m mv v v

m m m m

+ = + + −

− + = +

⎛ ⎞ ⎛ ⎞−= +⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

7. Now substitute the expression for 1f v into

the last equation of step 5 to find 2f v :

1 2 2

2f 1 1 2 2

1 2 1 2

1 2 21 2

1 2 1 2

1 2 12f 1 2

1 2 1 2

2

21 1

2

m m mv v v v v

m m m m

m m mv v

m m m m

m m mv v v

m m m m

⎡ ⎤⎛ ⎞ ⎛ ⎞−= + + −⎢ ⎥⎜ ⎟ ⎜ ⎟

+ +⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞−= + + −⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−= +⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

Insight: These complex formulae predict the outcome of any one-dimensional elastic collision. The equations get even

more complex if you include two dimensions!



9 – 34

89. Picture the Problem: Cart 2 moves with speed 2iv and collides with cart 1, which is at rest. The collision imparts

kinetic energy to cart 1, but the collision is elastic so the total energy is conserved.

Strategy: In the case where carts 1 and 2 have the same speed after the collision, conservation of momentum alone can be used to determine the speeds of the carts at the instant their speeds are equal. Because cart 1 is initially at rest we can

use equations 9-12 and 9-13, except all the index labels must be switched because in this problem it is cart 1, not cart 2,

that is initially at rest.


and

solve for f v ( ) ( ) ( )1 1 2 2 1 f 2 f

f

f

0 0.42 kg 0.68 m/s 0.84 0.42 kg

0.23 m/s

m v m v m v m v

v

v

+ = +

+ = +

=

2. (b) The energy stored in the

spring bumper is equal to the loss

of kinetic energy at that time.

( ) ( ) ( )( )2 21 1

f i 2 20.84 0.42 kg 0.227 m/s 0.42 kg 0.68 m/s

0.065 J

K K K Δ = − = + −

= −

The “missing” energy must have gone into the energy stored in the bumper, or

0.065 J.

3. (c) Use equation 9-12 with theindices switched:

( )( )2

1,f 0

1 2

2 0.42 kg20.68 m/s 0.45 m/s

0.84 0.42 kg

mv v

m m

⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟+ +⎝ ⎠ ⎣ ⎦

4. Use equation 9-12 with the

indices switched: ( )2 12,f 0

1 2

0.42 0.84 kg0.68 m/s 0.23 m/s

0.84 0.42 kg

m mv v

m m

⎡ ⎤− −⎛ ⎞= = = −⎢ ⎥⎜ ⎟+ +⎣ ⎦⎝ ⎠

Insight: The equations from problem 88 can also be used, but in this case 1i 0v = so that the first term of each equation

drops out and you end up with equation 9-12 again.

90. Picture the Problem: The geometry of the earring is indicated in the figure at right.

Strategy: In order for the earring to be balanced at point P, the center of mass must

reside there. The center of mass must lie along a vertical line through P because itis symmetric to the left and right of that line. Find the vertical coordinate of the

center of mass by subtracting the mass of a circle of diameter d from the mass of a

circle of diameter D. By inspection of the diagram we can see that the location of

the center of the circular hollow must be a distance 2d from the edge of the

earring, but we can’t assume anything about the relative sizes of d and D. Place the

origin of the coordinate system at the top of the figure and let y be positive in the

downward direction in the figure. Use equation 9-15 to find the center of mass andset it equal to d . From that equation we can find an expression for d and D and use

it to solve for φ .

Solution: 1. Write out equation 9-15 and set it equal to d :( ) ( ) set

gold circle

cm

gold circle

2 2m D m d Y d

m m

−= =

−

2. The masses of the gold and the circle cutout are propor-

tional to their areas. In these expressions the density of

gold ρ and the thickness t are constants:

( )

( )

2

gold 4

2

circle 4

m At D t

m At d t

π

π

ρ ρ

ρ ρ

= =

= =

3. Substituting these expressions into equation 9-15, the

constants cancel:

( )( ) ( )( )2 2

2 2

3 3 2 31 1

2 2

3 3 2

2 2

2

D D d d d

D d

D d d D d

D d d D

−=

−

− = −

+ =

4. Divide both sides by 2d D and introduce D d φ = :2 2

2

3 2

2

1 2

1 2

D d d D

φ φ

φ φ

+ =

+ =

+ =



9 – 35

5. From this point there are several directions we could go. We could simply test ( )1

21 5φ = + and verify that it is a

solution, or we could look up the roots of a cubic equation using a math reference book, or we could factor out a term

and solve a quadratic equation. I will do the latter, with help from a math computer program that suggested the factor.

6. Factor out ( )1φ − : ( ) ( )3 2 22 1 0 1 1φ φ φ φ φ − + = = − − −

7. The solutions are 1φ = and the roots of the quadratic term:( ) ( )

( )

22 1 1 4 1 14 1 5

2 2 1 2

b b ac

aφ

± − −− ± − ±= = =

8. The solution 1φ = is unphysical because it makes the hole the same size as the earring.

The solution ( )1

21 5φ = − doesn’t work because it is negative, so the only valid solution is ( )1

21 5φ = + .

Insight: We will discuss the density ρ in Chapter 15. Although this is a long and complex solution it shows that the basic concept of center-of-mass can help you solve difficult questions like this one without using calculus.

91. Picture the Problem: The mass 1m has an initial velocity 1v and collides with 2m that has initial velocity 2v .

Strategy: Subtract the expressions for 2f v and 1f v found in problem 88 to prove the indicated relation.

Solution: Use the results of problem 88

to write an expression for 2f 1f v v− :

1 1 2 2 1 22f 1f 1i 2i

1 2 1 2 1 2 1 2

1 2 1 2

1i 2i

1 2 1 2

2f 1f 1i 2i

2 2m m m m m mv v v v

m m m m m m m m

m m m mv v

m m m m

v v v v

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −− = − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

+ + + +⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

⎛ ⎞ ⎛ ⎞+ − −= +⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

− = −

Insight: This shows that for a head-on, elastic collision the difference in speeds remains constant. If 2m were at rest

initially, then after the collision its speed relative to 1m will be 1iv . If the two masses are equal, and 2m is initially at

rest, then 1m stops completely and 2m leaves the collision with speed 1iv .

92. Picture the Problem: The large ball of mass M is moving upward with speed v and collides elastically at floor level

with the smaller ball of mass m that is moving with velocity −v. The small ball rebounds and rises to height mh .

Strategy: Use the results of problem 88 to find the speed of the smaller ball after the collision. In that equation let

1,m m= 2

,m M =1

,v v= − and 2v v= Let upward be the positive direction. Then use the speed of the smaller ball

together with the conservation of mechanical energy to find the height to which it will rise.

Solution: 1. Find the speed of the small ball

after the collision using the equation from problem 88:

( )

1 2 2

1f 1 2

1 2 1 2

2

2 3

m m mv v v

m m m m

m M M M mv v v

m M m M m M

⎛ ⎞ ⎛ ⎞−= +⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2. Now use the conservation of energy for

the drop and the rebound to relate the heights

and speeds:( ) ( )

i i f f i i f f

2 21 11f 2 2

2 2

1f

0 0 0 0

2 2

m

m

K U K U K U K U

m M gh m M v mv mgh

v g h v g h

+ = + + = +

+ + = + + + = +

= =

3. Substitute the expression from step 1 into step 2:

2 22 2

1f 3 3

2 2m

v M m v M mh h

g M m g M m

− −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

Insight: Note the impressive result of the transfer of kinetic energy. For instance, for a basketball ( M = 0.600 kg) and a

tennis ball (m = 0.0577 kg) the tennis ball will rise to a height 7.02mh h= . Try it!



9 – 36

93. Picture the Problem: The child and the sled are acted upon by a 40.0 N force but each object undergoes a different

acceleration.

Strategy: Use Newton’s Second Law to find the acceleration of the center of mass of the sled-child system. Then use

equation 9-17 to determine the acceleration of the child. Let cm be the mass of the child, sm be the mass of the sled, F

be the pulling force, and ca and sa be the accelerations of the child and sled, respectively.

Solution: 1. Use Newton’s Second

Law to find cm A relative to the ice:( )c s cm cm

c s

F

F m m A Am m

= = + ⇒ =+

∑F

2. Use equation 9-17 and the result

from step 1 to solve for c :a

( )( )

c c s s

cm

c s c s

2

2s s

c

c

40.0 N 9.75 kg 2.32 m/s0.828 m/s

21.0 kg

m a m a F A

m m m m

F m aa

m

+= =

+ +

−−= = =

Insight: If the child were also to accelerate at 2.32 m/s2 with the sled, the pulling force would have to be 71.3 N.

94. Picture the Problem: The object of mass m has a speed 0v and collides

elastically with another object of mass m that is at rest. The diagrams at rightindicate the directions the masses go after the collision and the relationship

between the initial and final momenta.

Strategy: Set the kinetic energies before and after the collision equal to each

other because the collision is elastic. Then use conservation of momentum in

two dimensions and the law of cosines to find an expression for the angle between the two final velocity vectors.

Solution: 1. Set i f K K = and find a

relation between the velocities:

2 2 21 1 10 1f 2f 2 2 2

2 2 2

0 1f 2f

0mv mv mv

v v v

+ = +

= +

2. Set i f =p p

in two dimensions: i 1f 2f = +p p p

3. Apply the law of cosines to the

triangle indicated in the lower diagram.

( ) ( )

2 2 2

i 1f 2f 1f 2f

2 2 2 2 2 2 2 2 22 2 2

0 1f 2f 0 1f 2f i 1f 2f

1f 2f 1f 2f 1f 2f

2 cos

cos2 2 2

p p p p p

m v m v m v v v v p p p

p p mv mv v v

α

α

= + −

− − − −− −= = =

4. From step 1, 2 2 2

0 1f 2f 0v v v− − = . Use this

together with step 3 to find α and φ :

cos 0 90

180 180 90 90

α α

φ α

= ⇒ = °

= °− = °− ° = °

Insight: The law of cosines is mentioned in most geometry books but is not included in Appendix A. There is an

algebraic way to solve this problem without using the law of cosines, but it is very lengthy.

95. Picture the Problem: The mass 2m slides down the incline as the reading on the

scale is observed.

Strategy: The reading on the scale is less than it would be if the two masses were

at rest because the center of mass of the two masses is accelerating downward.Find the acceleration of the center of mass and use it together with Newton’s

Second Law to find the reading on the scale. The acceleration of the block is

sin g θ in the direction indicated by the vector. Let upward be the positive

direction in this problem.

Solution: 1. (a) Find the vertical

component of a

: ( ) 2

sin

sin sin sin

ya a

g g

θ

θ θ θ

= −

= − = −

φ

1f p

2f p

ip

α

1 2

1

2

1f v

2f v

φ

0v

before after



9 – 37

2. Use equation 9-17 to find cm A :( ) ( )2 2

1 2 2cm

1 2 1 2

0 sin sinm m g m g A

m m m m

θ θ + −= = −

+ +

3. Now use Newton’s Second Law to

find the scale force F :( ) ( )

( ) ( )

( ) ( )

1 2 1 2 cm

2

2

1 2 1 2

1 2

2 2 2

1 2 1 2 1 2

sin

1 sin cos cos

y F F m m g m m A

m g F m m g m m

m m

m g m g m g m g m m g

θ

θ θ θ

= − + = +

⎡ ⎤= + + + −⎢ ⎥

+⎣ ⎦

= + − = + = +

∑

4. (b) The scale force must decrease as the block accelerates downward because if it did not, the forces on the center of

mass would be balanced and it could not accelerate. The net force on the two masses must point downward, so that the

scale force upward must be less than the weight of the masses, ( )1 2 .m m g +

5. (c) For the case where θ =0°, 2cos 1θ = and ( )1 2 . F m m g = + That is, the block 2m does not accelerate because it is

on a level surface. The center of mass does not accelerate, either, and the reading on the scale is just the weight of both

blocks. For the case where θ = 90°, 2cos 0θ = and 1 . F m g = In this case the block 2m is in freefall and does not exert

any force on the scale, so the scale reads the weight of the wedge only.

Insight: This problem is analogous to problem 79, where the ball accelerates downward through the fluid.

96. Picture the Problem: The rope is lifted by one end at constant speed until it is completely off the table, held steady for

a moment, then lowered at constant speed back onto the table.

Strategy: Use equation 9-16 to find the velocity of the center of mass of the rope. When writing subscripts for the

variables, let f = on the floor, nf = not on the floor, and L = length of the rope. The expression for cmV can then be used

to determine the acceleration of the center of mass. The top of the rope is at position vt during the lift, so the center of

mass of the above floor portion of the rope is halfway between zero and vt , and the fraction of the rope that is above the

floor is ( )nf ,m vt L M = where M is the total mass of the rope. When the rope is being lowered its top is at position

L vt − , so the center of mass of the above floor portion of the rope is halfway between zero and L vt − , and the fraction

of the rope that is above the floor is ( )nf .m L vt L M = −⎡ ⎤⎣ ⎦

Solution: 1. (a) Use equation 9-16 to find cmV :( )

( )2nf nf f f

cm

nf f

0vt L M vm v m vV v L t

m m M

+⎡ ⎤+ ⎣ ⎦= = =+

2. cm A is the slope of the cm

V vs. t graph: 2

cm upward A v L=

3. (b) The rope being lowered has downward momentum. Its downward momentum is decreasing as more and more of

its mass comes to rest. That means the rate of change of the momentum is upward. Therefore, there must be a net

upward force acting on the rope, resulting in an upward acceleration of the rope’s center of mass.

4. (c) Use equation 9-16 again for the case

when the rope is lowered to the table:

( ){ }( )

( )

nf nf f f cm

nf f

22

0 L vt L M vm v m vV

m m M

v t vLv L t v

L

− − +⎡ ⎤+ ⎣ ⎦= =

+

−= = −

5. cm A is the slope of the cm

V vs. t graph: 2

cm upward A v L= the same result as in part (a).

Insight: See problems 45 and 46 for plots of the position and velocity of the rope’s center of mass.



9 – 38

97. Picture the Problem: The spacecraft approaches the planet with speed

iv , interacts (“collides”) with it, and recedes in the opposite direction with

speed f v , as depicted in the figure at right.

Strategy: Subtract the initial velocities in order to find the speed of

approach.

Solution: The speed of approach can be found by subtracting the two

initial velocities: ( )app i app i i v u v u v= − ⇒ = − − = +v u v

Insight: The spacecraft appears to be approaching at a high rate of speed due to the motion of the planet itself. If thespacecraft were at rest relative to the Sun, it would still approach the observer at speed u.


iv , interacts (“collides”) with it, and recedes in the opposite direction with


Strategy: Subtract the final velocities in order to find the speed of

departure.

Solution: The speed of departure can be found by subtracting the two

final velocities: dep f dep f f v v u v u= − ⇒ = − = −v v u

Insight: The spacecraft appears to be receding at a relatively low rate of speed due to the motion of the planet itself. If

the spacecraft were at rest relative to the Sun, it would actually approach the observer at speed u.

99. Picture the Problem: The spacecraft approaches the planet with speed iv ,

interacts (“collides”) with it, and recedes in the opposite direction with


Strategy: Set the speed of approach from problem 97 to the speed of

departure from problem 98 and solve for f .v

Solution: Set app depv v=

and solve for f :v

app i f dep

f i 2

v u v v u v

v v u

= + = − =

= +

Insight: The spacecraft has gained velocity 2u at the expense of a tiny bit of the planet’s kinetic energy.


iv , interacts (“collides”) with it, and recedes in the opposite direction

with speed f v , as depicted in the figure at right.

Strategy: Use the results of problem 99, setting iv u= and calculating

the initial and final kinetic energies.

Solution: 1. Find the

initial and final speeds:f i 2 2 3v v u u u u= + = + =

2. Calculate the ratioof the kinetic energies:

( )221

f f 2

2 21i i2

39

umv K

K mv u= = =

Insight: The kinetic energy of the spacecraft has increased nine-fold at the expense of a tiny fraction of the planet’s

kinetic energy.



9 – 39

101. Picture the Problem: The bullet enters the bob and together they

rise to a height h.

Strategy: Use the expression from Example 9-5, which was

derived from conservation of momentum during the collision and

conservation of mechanical energy after the collision, to find thespeed of the bullet. Then use conservation of momentum to find

the speed of the bullet-bob combination after impact.

Solution: 1. (a) Solve the expression

from Example 9-5 for 0v :

( )( )

2 2

0

22

2

0

2

0.675 0.00675 kg2 2 9.81 m/s 0.128 m

0.00675 kg

160 m/s 0.160 km/s

vmh

M m g

M mv gh

m

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

+⎝ ⎠ ⎝ ⎠

⎛ ⎞+ +⎛ ⎞= = ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

= =

2. (b) Set i f p p= and solve for f

v : ( )

( )

0 f

f 0

0.00675 kg160 m/s 1.58 m/s

0.00675 0.675 kg

mv m M v

mv v

m M

= +

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

Insight: Another way to get the answer to part (b) is to use conservation of energy again for the bullet-bob combination.

You’ll find that f 2 1.58 m/sv gh= = for h = 0.128 m.

102. Picture the Problem: The bullet enters the bob and together they

rise to a height h.

Strategy: Use the expression from Example 9-5, which wasderived from conservation of momentum during the collision and

conservation of mechanical energy after the collision, to find the

appropriate mass of the bob.

Solution: Solve the expression

from Example 9-5 for M :

( ) ( )

( )( )

2 2

0

2

0

0

2

2

2

0.0081 kg 320 m/s0.0081 kg 1.6 kg

2 2 9.81 m/s 0.125 m

vmh

M m g

m gh v

M m

mv M m

gh

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

+⎝ ⎠ ⎝ ⎠

=+

= − = − =

Insight: If the bob mass is made smaller, the speed of the bullet-bob combination after the collision will be greater and

they will rise to a larger height h.



9 – 40

103. Picture the Problem: Cart 1 has initial speed 0v and collides with cart 2, which is at rest. The two carts stick together

and move with speed f v .

Strategy: Use conservation of momentum to determine the speed of the two carts after the collision. Then use equation

7-6 to find the kinetic energy before and after the collision to calculate . K Δ


and solve for f v : ( )

( )

1 0 1 2 f

1f 0

1 2

0

3.0 kg0.25 m/s 0.19 m/s

3.0+1.0 kg

m v m m v

mv v

m m

+ = +

= = =+

2. (b) Use equation 7-6 to find : K Δ ( )

( ) ( ) ( ) ( )

2 21 11 2 f 1 02 2

2 21 1

2 24.0 kg 0.19 m/s 3.0 kg 0.25 m/s

0.022 J 22 mJ

K m m v m v

K

Δ = + −

= −

Δ = − = −

Insight: The kinetic energy that is lost represents 25% of the initial kinetic energy of 94 mJ. There are some rounding

issues here; if we work with fractions instead we find f 3 16 m/s,v = i 3 32 J, K = and f 9 128 J. K = Thus the loss of

energy 3 128 J K Δ = − or −23 mJ.

104. Picture the Problem: The carts move without friction in the manner

indicated by the figure at right.

Strategy: Use equation 9-16 and the given information to find thevelocity of the center of mass. After the collision the two carts will

move with this same velocity because they stick together. Use that

information together with equation 7-6 to find the percentage of kinetic

energy lost during the collision. Finally, use the expression from problem 88 to find the speeds of the two carts after an elastic collision.

Solution: 1. (a) Apply equation 9-16 directly:( )0 0 3

cm 04

2mv m vV v

m m

+= =

+

2. (b) Use equation 7-6 to find i K : ( )22 251 1

i 0 0 02 2 82 K mv m v mv= + =

3. Use the result of step 1 to find f K : ( ) ( ) ( )22 23 91 1

f f 0 02 2 4 162 2 K m v m v mv= = =

4. Find the percentage loss of K :

2 29 50 016 8f i

25i 08

8100 100 100 10%

80

mv mv K K

K mv

−−× = × = − × = −

or 10% is lost

5. (c) Use the expression from

problem 88 to find1,f v :

0 0

1,f 0

2

2 2

v vm m mv v

m m m m

−⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

6. Use the expression from

problem 88 to find 2,f v :

02,f 0 0

2

2

vm m mv v v

m m m m

−⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟

+ +⎝ ⎠ ⎝ ⎠

Insight: Note that the two carts trade speeds after the collision. This is consistent with the result of problem 91 where

we asserted that the relative velocities of the two masses are unchanged for a head-on, elastic collision.

physics chapter 9 answers

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