9-1

MOMENTUM 9 Q9.1. Reason: The velocities and masses vary from object to object, so there is no choice but to compute px = mvx for each one and then compare.

1

2

3

4

5

(20 g)(1 m/s) 20 g m/s(20 g)(2 m/s) 40 g m/s(10 g)(2 m/s) 20 g m/s(10 g)(1 m/s) 10 g m/s(200g)(0.1m/s) = 20g m/s

x

x

x

x

x

ppppp

= = ⋅= = ⋅= = ⋅= = ⋅= ⋅

So the answer is 2 1 3 5 4 .x x x x xp p p p p> = = > Assess: The largest, most massive object did not have the greatest momentum because it was moving slower than the rest.

Q9.2. Reason: We can find the change in momentum of the objects by computing the impulse on them and using the equation .p J∆ = Since they start at rest with zero momentum, we can write

f f .mv p p J F t= = ∆ = = ∆ So the final velocity of either object equals the impulse on the object divided by its mass. For the first object, this will be

1 f( ) (12 N)(2.0 s) / = (24 N s)/v m m= ⋅

For the second object, the velocity is given by

2 f( ) (15 N)(3.0 s) /(2 ) = (22.5 N s)/v m m= ⋅

The velocity of the first object is seen to be higher, because no matter what the value of m is, the numerator is greater in the first expression than in the second expression. Assess: The second object was subject to a greater force exerted for a greater time. But it ended with less velocity because it was more massive. All three factors are important in determining the final speed.

Q9.3. Reason: When the question talks about forces, times, and momenta, we immediately think of the impulse-momentum theorem, which tells us that to change the momentum of an object we must exert a net external force on it over a time interval: avg .F t p∆ = ∆

Because equal forces are exerted over equal times, the impulses are equal and the changes in momentum are equal. Because both carts start from rest, their changes in momentum are the same as the final momentum for each, so their final momenta are equal. Assess: Notice that we did not need to know the mass of either cart, or even the specific time interval (as long as it was the same for both carts) to answer the question.

Q9.4. Reason: (a) The acceleration of the puck with the smaller mass will be four times greater than the acceleration of the puck with the larger mass. The puck with the larger mass takes longer to travel the distance. The time it takes an object to move a given distance from rest can be calculated from the kinematic equation

21 ( )2 xx a t∆ = ∆

9-2 Chapter 9

Solving for the time,

2

x

xta∆

∆ =

Since the acceleration of the smaller puck is four times that of the larger puck, the time it takes the smaller puck to travel the distance is half the time it takes the larger puck. (b) The force on each puck is the same. Since the smaller puck takes half the time to travel the distance, the impulse of the smaller puck is half the impulse of the larger puck. From the impulse-momentum theorem, the change in momentum of the smaller puck is half the change in momentum of the larger puck. Assess: The final velocity of the larger puck is also twice the final velocity of the smaller puck.

Q9.5. Reason: The sum of the momenta of the three pieces must be the zero vector. Since the first piece is traveling east, its momentum will have the form 1( ,0),p where 1p is a positive number. Since the second piece is traveling north, its momentum will have the form 2(0, ),p where 2p is a positive number. If a third momentum is to be added to these and the result is to be (0,0), then the third momentum must be 1 2( , ).p p− − Since its east-west and north-south components are both negative, the momentum of the third piece must point south west and so the velocity must be south west. The answer is D. Assess: It makes sense that the third piece would need to travel southwest. It needs a western component of momentum to cancel the eastern component of the first piece and it needs a southern component to cancel the northern component of the second piece.

Q9.6. Reason: Suppose the students always throw the ball with the same horizontal component of velocity. Each time a student throws the ball to his friend, he gains momentum in the opposite direction, that is, in the direction away from his friend. Also, each time a student catches the ball, he gains momentum in the direction of the ball’s motion, also away from his friend. So the students will start to drift apart. Each throw will increase the speed of the student throwing the ball and each catch will increase the speed of the student catching the ball. Theoretically, there will come a time when a student will attempt to throw the ball to his friend but will be unable because they will be moving apart too quickly. In other words, when the one student throws the ball, then from the point of view of the other student, the ball will be moving away from him, rather than toward him. The more massive the ball is, the fewer will be the number of throws which are possible because a more massive ball has more momentum and would change the students’ momenta more when caught and when thrown. However, it is likely that before this situation arises, they will be prevented from passing the ball by their large distance from one another. Assess: We see that objects can influence one another’s motion indirectly by exchanging a mass back and forth. As you will learn in chapter 30, modern physics teaches that all forces result from the back and forth exchange of something called a boson.

Q9.7. Reason: (a) Both particles cannot be at rest immediately after the collision. If they were both at rest, then the sum of the momenta after the collision would be zero, and since momentum is conserved in collisions, it would have had to be zero before as well (and it wasn’t). (b) If the masses are equal and the collision elastic, the moving particle will stop and give all of its momentum to the previously resting particle. A good example of this appears when a billiard ball rolls directly into another resting billiard ball. Assess: We say momentum is conserved in all collisions because we assume that both colliding objects are part of the system and we assume the “impulse approximation” that other forces can be neglected during the short time interval of the collision. In part (a) if the system contained a third particle that participated in the collision, then it is possible for the first two particles to end up at rest if the momentum is carried off by the third.

Q9.8. Reason: The impulse needed to stop a car without crumple zones will be the same as for a car with crumple zones since the change in momentum is same. The average force needed to stop the car is the impulse divided by the time the car takes to stop. A car designed with crumple zones will take longer to stop than a stiff car, so the force exerted on the car during a collision is smaller for a car with crumple zones. Assess: See Section 9.2 for a discussion of bridge abutments, which are used for impact-lessening and work on the same principle.

Momentum 9-3

Q9.9. Reason: The ball must change speed from its original speed to zero whether you wear the glove or not. So p∆

is the same in either case. The impulse-momentum theorem also tells us avg .F t p∆ = ∆

Given that the right side is the same in either case, the left side must also be the same in either case. But if we can increase t∆ then F will be decreased correspondingly. The padding of the glove increases the collision time, thereby decreasing the force. Assess: This is also how air bags work in car collisions.

Q9.10. Reason: Consider an astronaut floating in outer space. Imagine the astronaut throwing a heavy object away from his body. The astronaut will start moving in the direction opposite the throw. This is a simpler situation that is analogous to what happens in rocket propulsion. The astronaut has pushed against the object during the throw, and the reaction force on him causes his acceleration. Rockets “push against” the gases that they expel. Assess: Another analogous situation would be throwing a heavy object away from yourself while standing on slick ice. You will be propelled backward.

Q9.11. Reason: See Example 9.5. The two skaters interact with each other, but they form an isolated system because, for each skater, the upward normal force of the ice balances their downward weight force to make

net 0.F =

Thus the total momentum of the system of the two skaters will be conserved. Assume that both skaters

are at rest before the push so that the total momentum before they push off is i 0.P =

Consequently, the total

momentum will still be 0

after they push off. (a) Because the total momentum of the two-skater system is 0

after the push off, Megan and Jason each have momentum of the same magnitude but in the opposite direction as the other. Therefore the magnitude p∆ is the

same for each: avg .F t p∆ = ∆

From the impulse-momentum theorem each experiences the same amount of impulse. (b) They each experience the same amount of impulse because they experience the same magnitude force over the same time interval. However, over that time interval they do not experience the same acceleration. netF ma=

says that since Megan and Jason experience the same force but Megan’s mass is half of Jason’s, then Megan’s acceleration during push off will be twice Jason’s. So she will have the greater speed at the end of the push off .t∆ Assess: It is important to think about both results until you are comfortable with them.

Q9.12. Reason: The impulse on each object is equal to the object’s change in momentum. During the collision total momentum is conserved. The change in momentum of the rubber ball must be exactly equal to the negative of the change in momentum of the steel ball since the total change in momentum must be zero. Both balls receive the same impulse. Assess: This can also be seen from the fact that the duration of the collision is the same for both objects, and the forces on each are action/reaction pairs. So the impulses are equal and opposite.

Q9.13. Reason: If you do not move backward when passing the basketball it is because the ball-you system is not isolated: There is a net external force on the system to keep you from moving backward that changes the momentum of the system. If the ball-you system is isolated (say you are on frictionless ice), then you do move backward when you pass the ball. Assess: If the friction force of the floor on you keeps you from moving backward (relative to the floor), then the law of conservation of momentum doesn’t apply because the system isn’t isolated. But you could then include the floor, building, and the earth in the system so it (the system) is isolated; then momentum of the system is conserved—and that means the earth does recoil ever so slightly when you pass the basketball.

Q9.14. Reason: You should throw the rubber ball rather than the beanbag to increase your chances of knocking down the bowling pin. This is because the rubber ball will exert more impulse on the pin. The impulse on the bowling pin equals the negative of the impulse on the projectile, which in turn equals the negative of the change in momentum of the projectile: on pin proj.J p= −∆ The rubber ball and beanbag would start with the same momentum, but the rubber ball would have a greater change in momentum because it would bounce off and so its direction would change. The beanbag would continue to move in the same direction. Thus the rubber ball would exert a greater impulse on the bowling pin.

9-4 Chapter 9

Assess: This result make sense because we see that a collision with the rubber ball is more violent than a collision with the beanbag in that the former collision turns the projectile around.

Q9.15. Reason: Assume that each angular momentum is to be calculated about the axis of symmetry. It will be useful to derive a general formula for the angular momentum of a particle in uniform circular motion of radius ,r calculated around the axis of symmetry. Equation 7.14 reminds us that I for such a situation is 2,mr and Equation 6.7 tells us that / .v rω = Putting this all together gives the angular momentum for a particle in uniform circular motion: 2( )( / ) .L I mr v r rmvω= = = So we compute L rmv= for each of the five situations.

21

22

23

24

25

(2 m)(2 kg)(2 m/s) 8 kg m /s(2 m)(3 kg)(1 m/s) 6 kg m /s(2 m)(1 kg)(3 m/s) 6 kg m /s(4 m)(2 kg)(1 m/s) 8 kg m /s(4 m)(2 kg)(2 m/s) 16 kg m /s

LLLLL

= = ⋅

= = ⋅

= = ⋅

= = ⋅

= = ⋅

Finally, comparison gives 5 1 4 2 3.L L L L L> = > = Assess: Since p = mv the angular momentum for a particle in uniform circular motion can also be written ,L rp= or, more generally, L rp⊥= (compare with Equation 7.3).

Q9.16. Reason: Consider the entire device as the system. Since there is no external torque on the system, angular momentum is conserved. If the thread breaks, the masses will move outward. The moment of inertia of the system will increase since the masses will be farther from the axis of rotation. From Equation 9.23 the angular velocity must decrease to keep the angular momentum constant. Assess: The masses move outward because there is no longer any centripetal force keeping them moving in a circle with fixed radius.

Q9.17. Reason: Since there is no net torque on the earth, the angular momentum of the earth is conserved. As water from the polar ice caps moves farther from the earth’s rotation axis, the moment of inertia of the earth with increase from consideration of Equation 7.14. In order to keep the angular momentum of the earth constant, the angular velocity of the earth must decrease. If the angular velocity of the earth decreases, the period of rotation will increase, so the length of the day will increase. Assess: Since the mass of the polar ice caps is very small compared to the mass of the entire earth, the effect on the length of the day will probably be small.

Q9.18. Reason: Equation 9.21 gives the angular momentum .L Iω= For a solid disk 212I mr= around the axis

of symmetry. We’ll answer this question with ratios, using 2 1,m m= 2 12 ,r r= and 12 12 .ω ω=

2 21 1 122 2 2 1 1 12 2 2 2 2 2

2 21 11 1 1 1 1 1 1 1 12 2

(2 ) ( ) 1(2 ) 22

m r m rL IL I m r m r

ω ωωω ω ω

= = = = =

Therefore 2 12 ,L L= so the angular momentum of disk 2 is larger than the angular momentum of disk 1. Assess: Even with the same mass, and even though 1

2 12 ,ω ω= the angular momentum of disk 2 is larger because more of the mass is farther away from the axis. Therefore, it would take a larger net torque to change L for the second disk.

Momentum 9-5

Q9.19. Reason: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as

avgpFt

∆=

∆

In fact, this is much closer to what Newton actually wrote than .F ma=

This allows us to directly solve the question.

f i f iavg

( ) (20 kg)(0 m/s 1 m/s) 10, 000 N0.002 s

p mv mv m v vFt t t

∆ − − −= = = = = −

∆ ∆ ∆

Since we only need the magnitude, the correct answer is D. Assess: This is a large force but the speeds aren’t very large. However, the time interval is so short that a large force is needed to change the momentum that much in so short a time.

Q9.20. Reason: Momentum is conserved, so the momentum before the collision must equal the momentum after the collision. The momentum conservation equation gives

1 i 1 2 f( )m v m m v= +

Solving for the mass of the second ball,

1 i 1 f2

f

(1.0 kg)(2.0 m/s) (1.0 kg)(1.2 m/s) 0.67 kg1.2 m/s

m v m vmv− −

= = =

The correct choice is A. Assess: This result makes sense. The second ball must have small mass relative to the first ball, since the velocity of the pair is not that much less than the initial velocity of the first ball.

Q9.21. Reason: The system consisting of both blocks is isolated and so the momentum of the system is conserved.

i f

i i f( )(2 ) (3 )( ) ( 3 )( )P P

m v m v m m v=

+ = +

where the two blocks stick together to make one compound object after the collision. We want to know f:v

i i i if

( )(2 ) (3 )( ) 5 53 4 4

m v m v mv vvm m m

+= = =

+

The correct answer is D. Assess: One could mentally confirm that the answer is reasonable by thinking of the larger block on the right. It gets a little kick from the other block and so will go a bit faster than the iv it had before. Only answer D fits that.

Q9.22. Reason: Neglect the drag force from the water on the canoe. Consider the two people, ball, and canoe as the system. The initial momentum of the system is zero. The person in the front throws the ball to the person in the rear. The ball has a negative momentum, so the rest of the system must gain a positive momentum. The canoe and people move forward. When the other person catches the ball, the canoe, people, and ball move with a common velocity. Since the momentum of the system must remain zero, the final velocity of the system must be zero. The correct choice is A. Assess: An important thing to note here is that when the person in the rear catches the ball, the entire system will move with a common velocity.

9-6 Chapter 9

Q9.23. Reason: The initial momentum of the first block is 1 i( ) (2.5 kg)(12.0 m/s) 30 kg m/sp = = ⋅ and the initial momentum of the second block is 2 i( ) (14 kg)(–3.4 m/s) = –47.6 kg m/s.p = ⋅ The sum of these,

17.6 kg m/s,− ⋅ gives the total momentum, before and after the collision. After the collision, the final velocity of the two is obtained by dividing the total momentum by the total mass:

f total 1 2/ ( ) ( 17.6 kg m/s)/(16.5 kg) 1.07 m/s.v p m m= + = − ⋅ = −

Now the final momentum of block 2 is the product of its mass and final velocity: 2 f( ) (14 kg)( 1.07 m/s) 15.0 kg m/s.p = − = − ⋅ The impulse of block 1 on block 2 equals the change in momentum

of block 2:

1 on 2 2 15.0 kg m/s– (– 47.6 kg m/s) 32.6 kg m/s.J p= ∆ = − ⋅ ⋅ = ⋅

Now impulse equals the area under a force versus time graph. So we need to choose the graph whose area is about 32.6 kg m/s.⋅ The graphs are approximately triangular, so each has an area that is approximately half its

base times its height. Choice D looks the best because the height is 3000 N and the base is about net / ,F p t= ∆ ∆

for an area of about:

31 (25 10 s)(3000 N) 37.5 N s 37.5 kg m/s.2

A −= × = ⋅ = ⋅

The other graphs differ from this by a factor of 10 or more. Assess: Even though we can’t determine the value of the force on block 2 at each time, we can find the area under the force curve which is just the impulse.

Q9.24. Reason: The impulse-momentum theorem F t p∆ = ∆

tells us the change in momentum must be in the same direction as the impulse. The change in momentum is f i f i( ).p p p p− = + −

ip− is to the left. Mentally

put fp and ip− head to tail to see which direction their sum points. It is in the direction of arrow B.

The correct answer is B. Assess: Do a thought experiment if you can’t do it in real life: Use a hammer to whack a rolling bowling ball to make it change direction by 90°. What direction should your whack (impulse) be?

Q9.25. Reason: Momentum is conserved in this process. Consider the figure below.

The total momentum of the system before the collision is

i B B i G G i

i

( ) ( ) ( ) (3.0 m/s ) ( 3.0 m/s ) 0 kg m/s( ) 0 kg m/s

x x x

y

p m v m v m mp

= + = + − = ⋅= ⋅

since the blue ball is initially at rest. The total initial momentum is zero.

Momentum 9-7

The final momentum of the system must be zero also. This gives

f R R f B B f G G f B f G f

f R R f B B f G G f B f R f

( ) ( ) ( ) ( ) ( ) ( ) 0 kg m/s( ) ( ) ( ) ( ) ( ) ( ) 0 kg m/s

x x x x x x

y y y y y y

p m v m v m v m v m vp m v m v m v m v m v

= + + = + = ⋅= + + = + = ⋅

Solving for the momentum of the blue ball,

B f G f

B f R f

( ) ( )( ) ( )

x x

y y

v vv v

= −= −

Since the red ball is moving south and the green ball is moving east, the blue ball must be moving west and north. The correct choice is C. Assess: The final direction of the blue ball could be inferred from the fact that the total initial momentum is zero and the velocities of the red and green balls on the vector diagram above.

Q9.26. Reason: Since there are no net external torques on the system, the angular momentum must be conserved.

f

i i f f

iL LI Iω ωΣ = Σ

=

We are given i 30 rpmω = (we do not need to change to SI units since we are looking for the answer in the same units). Also call the mass of the spherical shell ,M and the mass of the bug m.

Known

i

24 g4.0 g3.0 cm30 rpm

MmR

ω

====

Find

fω As for the moment of inertia, the bug will not contribute to I when it is at the top (because its distance to the axis of rotation is zero). So the moment of inertia is just due to the spherical shell: 2 22 2

i 3 3 (24 g)(3.0 cm)I MR= = = 144 g ⋅ cm2. However, the bug will contribute 2( )mR to the moment of inertia when it is at the equator; the total will be the sum

of the contributions from the sphere and the bug. 2 22f sphere bug 3I I I MR mR= + = + = 144 g ⋅ cm2 + (4.0 g)(3.0 cm)2 =

180 g ⋅ cm2. Solve the second equation above for f :ω

2i

f i 2f

144 g cm30 rpm 24 rpm180 g cm

II

ω ω ⋅

= = = ⋅

The correct answer is B. Assess: We expected ω to decrease as I increased, and it did. The units of g ⋅ cm2 canceled, leaving rpm.

Q9.27. Reason: Not all of the information given in this question is needed to find the change in angular momentum. In fact, Equation 9.20, net t Lt ∆ = ∆ , tells us that to find the change in angular momentum of the disk, all we need is the net torque and the time the torque is applied. The change in L is given by

2(0.040 N m)(5.0 s) 0.20 kg m /s.L tt∆ = ⋅ ∆ = ⋅ = ⋅

The answer is B. Assess: We do not need to know anything about the disk itself to find its change in angular momentum; we only need the torque and the time interval. Similarly, from the equation ,p F t∆ = ∆ to find the change in momentum of an object, we do not need any information about the object, but only the force and the time interval.

9-8 Chapter 9

Problems

P9.1. Prepare: Model the bicycle and its rider as a particle. Also model the car as a particle. We will use Equations 9.7 for momentum. Solve: From the definition of momentum,

car bicyclep p= carcar car bicycle bicycle bicycle car

bicycle

mm v m v v vm

⇒ = ⇒ =1500 kg (5.0 m/s) 75 m/s100 kg

= =

Assess: This is a very high speed (≈168 mph). This problem shows the importance of mass in comparing two momenta.

P9.2. Prepare: We can use Equation 9.6 to calculate the change in momentum of the ball, and then use Equation 9.8 to find the impulse. Solve: Since the ball is initially at rest, the change in the momentum of the ball is

f i f f (0.057 kg)(45 m/s) 2.6 kg m/sp p p p mv∆ = − = = = = ⋅

The change in momentum of the ball is equal to the impulse on the ball, so 2.6 kg m/s.J = ⋅ Assess: Note that since the ball starts from rest, it is hit at the top of its vertical motion in the serve.

P9.3. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as

avgpFt

∆=

∆

In fact, this is much closer to what Newton actually wrote than .F ma=

Solve: This allows us to find the force on the snowball. By Newton’s third law we know that the snowball exerts a force of equal magnitude on the wall.

f i f iavg

( ) (0.12 kg)(0 m/s 7.5 m/s) 6.0 N0.15 s

p mv mv m v vFt t t

∆ − − −= = = = = −

∆ ∆ ∆

where the negative sign indicates that the force on the snowball is opposite its original momentum. So the force on the wall is also 6.0 N. Assess: This is not a large force, but the snowball has low mass, a moderate speed, and the collision time is fairly long.

P9.4. Prepare: Please refer to Figure P9.4. To get time in seconds, we note that 1 ms = 10–3 s. Solve: The impulse as defined through Equation 9.1 is

xJ = Area under the Fx(t) curve between ti and tf max max16.0 N s ( )(8 ms) 1500 N2

F F⇒ = ⇒ =

Assess: Note that the impulsive force shown is a linear function of time. Therefore, the impulsive force can also be written as 0.5 Fmax(6 ms) + 0.5 Fmax(2 ms) by looking at the first 6 ms and the next 2 ms separately. Equating this to the change in momentum should also give the same answer, Fmax = 1500 N.

P9.5. Prepare: We use the equation .J p= ∆ Solve: The initial momentum of sled and rider is i i (80 kg)(4.0 m/s) 320 kg m/sp mv= = = ⋅ and the final momentum of sled and rider is f f (80 kg)(3.0 m/s) 240 kg m/s.p mv= = = ⋅ So the impulse is given by

f i 320 kg m/s 240 kg m/s 80 kg m/s 80 N sJ p p= − = ⋅ − ⋅ = ⋅ = ⋅

Assess: This is a reasonable impulse. It could result, for example, from a force of around nine pounds for two seconds.

Momentum 9-9

P9.6. Prepare: Represent the stone as a particle and use Equation 9.1 and the impulse-momentum theorem, Equation 9.8. Note that the falling object is acted on by the gravitational force, whose magnitude is mg.

Solve: Using the impulse-momentum theorem,

f i( ) ( )y y yp p J− = f i( ) ( )y ym v m v mg t⇒ − = − ∆ i f( ) ( )y yv vt

g−

⇒ ∆ = 2

5.5 m/s ( 10.4 m/s) 0.50 s9.8 m/s

− − −= =

Assess: Acceleration due to gravity is 9.8 m/s2 or (9.8 m/s)/s. That is, speed changes by 9.8 m/s every second. A speed change of 4.9 m/s will thus take only 0.5 seconds, as obtained above.

P9.7. Prepare: Please refer to Figure P9.7. Model the object as a particle and its interaction with the force as a collision. We will use Equations 9.1 and 9.9. Because p = mv, so v = p/m. Solve: (a) Using the equations

( )f i

f

( ) ( )1area under the force curve ( ) 1.0 m/s (area under the force curve)

2.0 kg1(1.0 m/s) (1.0 N s) 1.5 m/s

2.0 kg

xx x

x x

p p J

J v

= +

= ⇒ = +

= + =

(b) Likewise,

f1( ) (1.0 m/s)

2.0 kgxv

= +

(area under the force curve) 1(1.0 m/s) ( 1.0 N s) 0.5 m/s2.0 kg

= + − =

Assess: For an object with positive velocity, a negative impulse slows down an object and a positive impulse increases speed. The opposite is true for an object with negative velocity.

P9.8. Prepare: Model the tennis ball as a particle, and its interaction with the wall as a collision. The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during the last two ms. Please refer to Figure P 9.8. The graph shows that Fx is positive, so the force acts to the right.

9-10 Chapter 9

Solve: Using the impulse-momentum theorem f i ,x x xp p J= +

(0.06 kg)(32 m/s) (0.06 kg)( 32 m/s)= − + area under force graph

Now,

max max max max

max

1 1area under force curve (0.002 s) (0.002 s) (0.002 s) (0.004 s)2 2(0.06 kg)(32 m/s) (0.06 kg)(32 m/s) 960 N

0.004 s

F F F F

F

= + + =

+⇒ = =

P9.9. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as

avgpFt

∆=

∆

In fact, this is much closer to what Newton actually wrote than .F ma=

Solve: This allows us to find the force on the child and sled.

f i f iavg

( ) (35 kg)(0 m/s 1.5 m/s) 105 N 110 N0.50 s

p mv mv m v vFt t t

∆ − − −= = = = = − ≈ −

∆ ∆ ∆

where the negative sign indicates that the force is in the direction opposite the original motion, as stated in the problem. So the amount (magnitude) of the average force you need to exert is 105 N. Assess: This result is neither too large nor too small. In some collision problems t∆ is quite a bit shorter and so the force is correspondingly larger.

P9.10. Prepare: We can use Equation 9.8 along with Equation 9.6 to calculate the mass of the puck. Solve: Consider the puck initially moving along the positive x-axis. Its initial momentum is i .p mv= + After the collision, the puck is moving with the same velocity, but in the opposite direction, so its final momentum is f .p mv= − Using Equation 9.8 we obtain

f i ( ) 2J p p mv mv mv= − = − − + = −

Solving this equation for m,

4.0 kg m/s 0.17 kg2 2(12.0 m/s)Jm

v− ⋅

= = = −− −

Assess: Momentum and impulse are vectors. Note that the impulse is in the negative direction because the force creating the impulse is the negative direction.

Momentum 9-11

P9.11. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as

avgpFt

∆=

∆

In fact, this is much closer to what Newton actually wrote than .F ma=

Solve: This allows us to find the average force on the cars.

(a) f i f iavg

( ) (1400 kg)(0 m/s 20 m/s)Water barrels 19,000 N1.5 s

p mv mv m v vFt t t

∆ − − −= = = = = −

∆ ∆ ∆

(b) f i f iavg

( ) (1400 kg)(0 m/s 20 m/s)Concrete barrier 280,000 N0.1s

p mv mv m v vFt t t

∆ − − −= = = = = −

∆ ∆ ∆

where the negative sign indicates that the force is in the direction opposite the original motion. Assess: We clearly see that a shorter collision time dramatically affects the magnitude of the average force. From a practical standpoint, find something like water barrels or a haystack if you have to crash your car.

P9.12. Prepare: We can use Equations 9.6 and 9.8 to calculate the impulse and Equation 9.2 to calculate the average force. Solve: (a) Consider the ball to be moving along the positive x-axis before it is hit. Then i( ) 15.0 m/s.xv = + After the collision, the velocity of the ball is f( ) 20.0 m/s.xv = − The impulse on the ball is given by Equation 9.8:

f i( ) ( ) (0.145 kg)( 20.0 m/s) (0.145 kg)(15.0 m/s) 5.08 kg m/sx x xJ p p= − = − − = − ⋅

(b) Given the impulse and duration of the collision, Equation 9.2 gives the average force on the ball.

avg5.08 kg m/s( ) 3400 N

0.0015 sx

xJF

t− ⋅

= = = −∆

Assess: Note that impulse and momentum are vectors. The impulse and force are directed in the negative x direction.

P9.13. Prepare: We’ll call the system the two carts and consider it an isolated system so we can apply the law of conservation of momentum. The action all takes place in one dimension, so we don’t need y-components. Let the subscript 1 stand for the small cart and 2 for the large. Solve:

i f

1 i 2 i 1 f 2 f

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

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

( ) ( ) ( ) ( )

x x

x x x x

x x x x

P Pp p p p

m v m v m v m v

=+ = +

+ = +

We want to know 2 f( )xv so we solve for it. Also recall that 2 i( ) 0 m/s,xv = so the middle term in the following numerator drops out. The small cart recoils, which means its velocity after the collision is negative.

1 1 i 2 2 i 1 1 f2 f

2

( ) ( ) ( ) (0.100 kg)(1.20 m/s) (0.100 kg)( 0.850 m/s)( ) 0.205 m/s1.00 kg

x x xx

m v m v m vvm

+ − − −= = =

Assess: The large cart does not move quickly, but the answer is reasonable because of the greater mass of the large cart. We have followed the significant figure rules and kept three significant figures.

P9.14. Prepare: Since the ice is slick, we will assume the ice is frictionless. Considering the man, gun, and bullet as the system, momentum will be conserved because there are no external forces acting in the horizontal direction. Solve: The momentum of the system before the man fires the gun is zero because everything is at rest. The momentum of the entire system must also be zero after the man fires the gun since momentum is conserved in the system. See the diagram below.

9-12 Chapter 9

Writing the momentum of the man and gun as M M( ),xm v and the momentum of the bullet as B B( ),xm v the momentum conservation equation is

M M f B B f M M i B B i( ) ( ) ( ) ( ) 0x x x xm v m v m v m v+ = + =

Solving for M f( ) ,xv

2BM f B f

M

0.01 kg( ) ( ) (500 m/s) 7.1 10 m/s70 kgx x

mv vm

− = − = − = − ×

The man moves to the left with a speed of 7.1 cm/s. Assess: This result seems reasonable. Though the bullet has a high speed, the mass of the bullet is much smaller than the mass of the man.

P9.15. Prepare: This is a problem with no external forces so we can use the law of conservation of momentum. Solve: The total momentum before the bullet hits the block equals the total momentum after the bullet passes through the block so we can write

b b i bl bl i b b f bl bl f3 3

bl f

( ) ( ) ( ) ( )

(3.0 10 kg)(500 m/s) (2.7 kg)(0 m/s) (3.0 10 kg)(220 m/s) (2.7 kg)(v ) .

m v m v m v m v− −

+ = + ⇒

× + = × +

We can solve for the final velocity of the block: bl f( ) 0.31 m/sv = .

Assess: This is reasonable since the block is about one thousand times more massive than the bullet and its change in speed is about one thousand times less.

Momentum 9-13

P9.16. Prepare: We will assume there is no friction. Considering the two weights as the system, momentum will be conserved because there is no net external force acting in the horizontal direction. Solve: The momentum of the system before the man loses his grip is zero because both weights are at rest. The momentum of the entire system must also be zero after the man releases the weights since momentum is conserved. See the diagram below.

The momentum conservation equation is

1 1 f 2 2 f 1 1 i 2 2 i( ) ( ) ( ) ( ) 0x x x xm v m v m v m v+ = + =

Solving for 1 f( ) ,xv

21 f 2 f

1

2.3 kg( ) ( ) (6.0 m/s) 2.6 m/s5.3 kgx x

mv vm

= − = − = −

The more massive weight moves to the left with a speed of 2.6 m/s. Assess: The heavier weight is almost twice the mass of the lighter one, so it makes sense that it ends up with about half the velocity of the lighter one.

P9.17. Prepare: We will choose car + gravel to be our system. The initial x-velocity of the car is 2 m/s and that of the gravel is 0 m/s. To find the final x-velocity of the system, we will apply the momentum conservation Equation 9.15.

Solve: There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This means (px)f = (px)i. Hence,

f(10,000 kg 4,000 kg)( ) (10,000 kg)(2.0 m/s) (4,000 kg)(0.0 m/s)xv+ = + f( ) 1.4 m/sxv⇒ =

Assess: The motion of railroad has to be on a level track for conservation of linear momentum to hold. As we would have expected, the final speed is smaller than the initial speed.

P9.18. Prepare: Choose car + rainwater to be the system. There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved. We will use Equation 9.13.

9-14 Chapter 9

Solve: Conservation of momentum is (px)f = (px)i. Hence,

Assess: The motion of railroad has to be on a level track for linear momentum to be constant. Because the train car’s speed decreases slightly from 22.0 m/s to 20.0 m/s we also expected a relatively small amount of rain water in the car.

P9.19. Prepare: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is equal to the force of the arrow on the archer. These are internal forces within the system. The archer is standing on frictionless ice, and the normal force by ice on the system balances the weight force. Thus ext 0F =

on the system, and momentum is conserved. The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment pfx must also be zero.

Solve: We have MAvA + mava = 0 kg m/s. Therefore,

AA

(0.04 kg)(60 m/s) 0.48 m/s50 kg

a am vvm

− −= = = −

The archer’s recoil speed is 0.48 m/s. Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow has forward momentum, the archer will have backward momentum.

P9.20. Prepare: Assume there is no friction on the canoe and take the dog and canoe as the system. Momentum is conserved since there is no net external force on the system. Solve: The momentum of the system before the dog wakes up is zero because both the dog and canoe are at rest. The momentum of the entire system must also be zero after the dog starts walking since momentum is conserved. See the following diagram.

Momentum 9-15

Writing the momentum of dog as D D( ),xm v and the momentum of the canoe as C C( ),xm v the momentum conservation equation is

D D f C C f D D i C C i( ) ( ) ( ) ( ) 0x x x xm v m v m v m v+ = + =

Solving for C,m

D D fC

C f

( ) (9.5 kg)(0.50 m/s) 32 kg( ) ( 0.15 m/s)

x

x

m vmv

= − = − =−

Assess: This result makes sense. The mass of the canoe must be larger than the mass of the dog, since the canoe has a smaller velocity than the dog. Note that the signs of the momenta and velocities are important. Momentum is a vector.

P9.21. Prepare: We will define our system to be bird + bug. This is the case of an inelastic collision because the bird and bug move together after the collision. Horizontal momentum is conserved because there are no external forces acting on the system during the collision.

Solve: The conservation of momentum equation pfx = pix is

1 2 f 1 1 i 2 2 f( )( ) ( ) ( )x x xm m v m v m v+ = +

f f(300 g 10 g)( ) (300 g)(6.0 m/s) (10 g)( 30 m/s) ( ) 4.8 m/sx xv v⇒ + = + − ⇒ =

Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both sides of the equation. Note that 2 i( )xv is negative. As would have been expected, the final speed is a little lower than the initial speed because (1) the bug has finite mass and (2) the bug has relatively large speed compared to the bird.

P9.22. Prepare: There is no external force on the player in the horizontal direction. Momentum is conserved in the horizontal direction. At the highest point in his leap the player has no velocity. Consider the player and the ball as the system. Solve: The momentum of the system before the catch is entirely due to the motion of the ball. Momentum is conserved in the system. See the following diagram.

9-16 Chapter 9

After the player catches the ball, they move together with a common final velocity. Writing the mass of the player as P ,m and the momentum of the ball as B B( ),xm v the momentum conservation equation is

P B f B B i( )( ) ( )x xm m v m v+ =

Solving for f( ) ,xv

B B if

P B

( ) (0.140 kg)(28 m/s)( ) 5.5 cm/s71 kg 0.140 kg

xx

m vvm m

= = =+ +

The player moves with a speed of 5.5 cm/s in the same direction the ball was originally moving. Assess: This result seems reasonable, since the mass of the ball is so small relative to the mass of the player.

P9.23. Prepare: Even though this is an inelastic collision, momentum is still conserved during the short collision if we choose the system to be spitball plus carton. Let SB stand for the spitball, CTN the carton, and BOTH be the combined object after impact (we assume the spitball sticks to the carton). We are given mSB = 0.0030 kg, mCTN = 0.020 kg, and (vBOTHx)f = 0.30 m/s. Solve:

i f

SB i CTN i BOTH f

SB SB i CTN CTN i SB CTN BOTH f

( ) ( )( ) ( ) ( )

( ) ( ) ( )( )

x x

x x x

x x x

P Pp p p

m v m v m m v

=+ =

+ = +

We want to know SB i( )xv so we solve for it. Also recall that CTN i( ) 0 m/sxv = so the last term in the following numerator drops out.

SB CTN BOTH f CTN CTN iSB i

S

( )( ) ( ) (0.0030 kg 0.020 kg)(0.30 m/s)( ) 2.3 m/s0.0030 kg

x xx

B

m m v m vvm

+ − += = =

Assess: The answer of 2.3 m/s is certainly within the capability of an expert spitballer.

P9.24. Prepare: The two cars are not an isolated system because of external frictional forces. But during the collision friction is not going to be significant. Within the impulse approximation, the momentum of the Cadillac + Volkswagen system will be conserved in the collision. We will use Equation 9.14 for momentum conservation along the x direction.

Solve: The momentum conservation equation (px )f = (px )i is

Momentum 9-17

C VW f C C i VW VW i

VW i VW i

( )( ) ( ) ( ) 0 kg mph

(2000 kg)(1.0 mph) (1000 kg)( ) ( ) 2.0 mphx x x

x x

m m v m v m v

v v

+ = + ⇒ ⋅

= + ⇒ = −

You need a speed of 2.0 mph. Assess: Usually, a car rolls down on an inclined road. However, we had to assume that the sloping is very small (or the road is level) to apply conservation of linear momentum along the direction of the road. Without this assumption we wouldn’t have been able to solve this problem.

P9.25. Prepare: Since there are no external forces, we can use the law of conservation of momentum. Solve: The sum of the momenta of the two blocks before the collision equals the momentum of the coupled blocks after the collision so we write

1 1 i 2 2 i 1 2 f

2 2

( ) ( ) ( )(2.0 kg)(1.0 m/s) (4.0 m/s) (2.0 kg )(2.0 m/s).m v m v m m v

m m+ = + ⇒

+ = +

We can solve this last equation for the second mass and obtain 2 1.0 kg.m = Assess: It is reasonable that the second mass is less than the first because the final speed, 2.0 m/s, is closer to the initial speed of the first block, 1.0 m/s, than it is to the initial speed of the second block, 4.0 m/s.

P9.26. Prepare: This problem deals with the conservation of momentum in two dimensions in an inelastic collision. We will thus use Equation 9.14.

Solve: The conservation of momentum equation before afterp p=

is

1 1 i 2 2 i 1 2 f 1 1 i 2 2 i 1 2 f( ) ( ) ( )( ) ( ) ( ) ( )( )x x x y y ym v m v m m v m v m v m m v+ = + + = +

Substituting in the given values,

f

f f

2 2f f

1 1

(.02 kg)(3.0 m/s) 0 kg m/s (.02 kg .03 kg) cos0 kg m/s (.03 kg)(2.0 m/s) (.02 kg .03 kg) sin cos 1.2 m/s

sin 1.2 m/s (1.2 m/s) (1.2 m/s) 1.7 m/s

tan tan (1) 45y

x

vv v

v vvv

θθ θ

θ

θ − −

+ ⋅ = ++ = + ⇒ =

= ⇒ = + =

= = = °

The ball of clay moves 45° north of east at 1.7 m/s. Assess: Noting that the magnitude of momenta of both particles have the same value 0.6 kg ⋅ m/s. This information, along with the fact that they are coming at each other at 90 degrees, may be used to reason out that the outgoing path will be right in the middle, that is, at 45 degrees as shown above.

P9.27. Prepare: We assume that the momentum is conserved in the collision. Please refer to Figure P9.27. Solve: The conservation of momentum Equation 9.14 yields

1 f 2 f 1 i 2 i( ) ( ) ( ) ( )x x x xp p p p+ = + 1 f( ) 0 kg m/s 2 kg m/s 4 kg m/sxp⇒ + ⋅ = ⋅ − ⋅ 1 f( ) 2 kg m/sxp⇒ = − ⋅

1 f 2 f 1 i 2 i( ) ( ) ( ) ( )y y y yp p p p+ = + 1 f( ) 1 kg m/s 2 kg m/s 1 kg m/syp⇒ − ⋅ = ⋅ + ⋅ 1 f( ) 4 kg m/syp⇒ = ⋅

The final momentum vector of particle 1 that has the above components is shown below.

9-18 Chapter 9

P9.28. Prepare: Model the two balls of clay as particles. Our system is comprised of these two balls. As the two balls stick together, this constitutes a perfectly inelastic collision and momentum is conserved. It is clear from the pictorial representation below that we are dealing with a two-dimensional motion and that momentum will be conserved along both the x and y directions.

Solve: The x-component of the final momentum is

f i 1 1 i 2 2 i( ) ( ) ( ) ( ) (0.020 kg)(2.0 m/s) (0.030 kg)(1.0 m/s) cos 30 0.0140 kg m/sx x x xp p m v m v= = + = − ° = ⋅

The y-component of the final momentum is

f i 1 1 i 2 2 i

2 2f

( ) ( ) ( ) ( ) (0.02 kg)(0 m/s) (0.03 kg)(1.0 m/s) sin 30 0.0150 kg m/s

(0.014 kg m/s) ( 0.015 kg m/s) 0.0205 kg m/s

x x y yp p m v m v

p

= = + = − ° = − ⋅

⇒ = ⋅ + − ⋅ = ⋅

Since f 1 2 f( ) 0.0205 kg m/s,p m m v= + = ⋅ the final speed is

f0.0205 kg m/s 0.41 m/s(0.02 0.03) kg

v ⋅= =

+

and the direction is

f1 1

f

|( ) | 0.015tan tan 47( ) 0.014

y

x

pp

θ − −= = = ° south of east

Assess: Before the collision, the momentum component in the southward direction comes only from the 30 g ball of clay. Therefore, it must be exactly equal to the southward component of momentum of the 50 g blob of clay after collison.

Momentum 9-19

P9.29. Prepare: This problem deals with a case that is the opposite of a collision. Our system is comprised of three coconut pieces that are modeled as particles. During the blow up or “explosion,” the total momentum of the system is conserved in the x-direction and the y-direction. We can thus apply Equation 9.14.

Solve: The initial momentum is zero. From (px)f = (px)i we get

1 1 f 3 3f( ) ( )cos 0 kg m/sxm v m v θ+ + = 1 f 13f

3

( ) ( 20 m/s)( )cos 10 m/s2

xm v mvm m

θ − − −⇒ = = =

From (py)f = (py)i, we get

2 2 f 3 3f( ) ( )sin 0 kg m/sym v m v θ+ + = 2 f 23f

3

( ) ( 20 m/s)( )sin 10 m/s2

ym v mvm m

θ− − −

⇒ = = =

2 2 13f( ) (10 m/s) (10 m/s) 14 m/s tan (1) 45v θ −⇒ = + = = = °

Assess: The obtained speed of the third piece is of similar order of magnitude as the other two pieces, which is physically reasonable.

P9.30. Prepare: The moon is treated as a particle. The period of the moon is T = 27.3 days = 2.36 × 106 s, its orbit radius is r = 3.8 × 108 m, and its mass is m = 7.4 × 1022 kg. A particle moving in a circular orbit of radius r with velocity v has an angular momentum L = mvr. While m and r are given, we will determine v using: 2 / .v r Tπ= Solve:

22L mvr m rTπ

= = 22 8 26

27.4 10 kg (3.8 10 m)2.36 10 sec

π= × ×

×

234 m2.8 10 kg

s= ×

Assess: Note that the rotation of moon or the earth about their respective axes does not come into these calculations.

P9.31. Prepare: We will model the mother as a particle with 47 kg, 4.2 m/s,m v= = and 2.6 m.r = It will be useful to derive a general formula for the angular momentum of a particle in uniform circular motion of radius ,r calculated around the axis of symmetry. Equation 7.14 reminds us that I for such a situation is mr2, and Equation 6.7 tells us that ( ) ( )55 m/s cos 25 23.2 m/s.− = − Putting this all together gives the angular momentum for a particle in uniform circular motion:

2( )( / )L I mr v r rmvω= = =

9-20 Chapter 9

Solve:

2(2.6 m)(47 kg)(4.2 m/s) 510 kg m /sL rmv= = = ⋅

to two significant figures. Assess: Until one gets a feel for how much angular momentum objects have it is difficult to know if our answer is reasonable, but the derivation of L = rmv is straightforward and we checked our multiplication twice, so it is probably correct.

P9.32. Prepare: The bar is a rotating rigid body. Please refer to Figure P9.32. From Table 7.4, the moment of inertial of a rod about its center is 21

12 .I ML= We will work with SI units and use Equation 9.21 to find the angular momentum. Solve: The angular velocity ω = 120 rpm = (120)(2π)/60 rad/s = 4π rad/s. The angular momentum is

2 21 (0.50 kg)(2.0 m) (4 rad/s) 2.1 kg m /s12

L Iω π = = = ⋅

If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page. Consequently,

2(2.1 kg m /s, out of the page)L = ⋅

Assess: Don’t forget the direction of angular momentum.

P9.33. Prepare: The disk is a rotating rigid body. Please refer to Figure P9.33. The angular velocity ω is 600 rpm = 600 × 2π/60 rad/s = 20π rad/s. From Table 7.4, the moment of inertial of the disk about its center is (1/2) MR2, which can be used with L = Iω to find the angular momentum. Solve:

2 2 4 21 1 (2.0 kg)(0.020 m) 4.0 10 kg m2 2

I MR −= = = × ⋅

Thus, L = Iω = (4.0 × 10−4 kg m2)(20π rad/s) = 0.025 kg m2/s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the –x direction. Consequently,

2(0.025 kg m /s, into page)L = ⋅

Assess: Don’t forget the direction of the angular momentum because it is a vector quantity.

P9.34. Prepare: Once the diver has left the diving board her angular momentum is conserved. Her weight acts at her center of gravity, so there is no net external torque about her center of gravity. Equation 9.23 applies.

Momentum 9-21

Solve: Using Equation 9.23,

i i f fI Iω ω= Solving for f i/ ,ω ω

2f i

2i f

14 kg m 3.54.0 kg m

II

ωω

⋅= = =

⋅

Her final angular velocity is 3.5 times her initial angular velocity. Assess: This result makes sense. Her final angular velocity is greater than her initial angular velocity.

P9.35. Prepare: We neglect any small frictional torque the ice may exert on the skater and apply the law of conservation of angular momentum.

i f

i i f f

L LI Iω ω

==

Even though the data for i 5.0 rev/sω = is not in SI units, it’s okay because we are asked for the answer in the

same units. We are also given 2i 0.80 kg mI = ⋅ and 2

f 3.2 kg m .I = ⋅ Solve:

2i i

f 2f

(0.80 kg m )(5.0 rev/s) 1.25 rev/s 1.3 rev/s3.2 kg m

IIωω ⋅

= = = ≈⋅

Assess: I increased by a factor of 4, so we expect ω to decrease by a factor of 4.

P9.36. Prepare: Please refer to Figure P9.36. The particle is subjected to an impulsive force and the impulse due to this force is the area under the force graph. Solve: Using Equation 9.1, the impulse from 0 ms to 2 ms is

(0 N)(2 ms) 0.0 N s= ⋅

From 2 ms to 6 ms, the impulse is

(1500 N)(6 ms 2 ms) 6.0 N s− = + ⋅

From 6 ms to 12 ms, the impulse is

( 1000 N)(12 ms 6 ms) 6.0 N s− − = − ⋅

9-22 Chapter 9

Thus, from 0 ms to 12 ms, the impulse is (0.0 6.0 6.0) N s 0.0 N s+ − ⋅ = ⋅ Assess: The total impulse is zero because the area above the time axis equals the area below the time axis. From the impulse-momentum theorem, the 3.0 kg particle will have no overall change in momentum.

P9.37. Prepare: Please refer to Figure P9.37. Model the glider cart as a particle, and its interaction with the spring as a collision. The initial and final speeds of the glider are shown on the velocity graph and the mass of the glider is known. We can thus find the momentum change of the glider, which is equal to the impulse. Impulse is also given by the area under the force graph, which we will find from the force graph in terms of ∆t.

Solve: Using the impulse-momentum theorem f i( ) ( ) ,x x xp p J− =

12(0.6 kg)(3 m/s) (0.6 kg)( 3 m/s) area under force curve (36 N)( )t− − = = ∆ ⇒∆t = 0.20 s

Assess: You can solve this problem using kinematics to check your answer. From the graph you have the average force during compression to be 18 N, and therefore the average acceleration to be 30 m/s2. Now calculate the time taken for the velocity to go from 3 m/s to 0 m/s, and twice this time should match the 0.20 s found in the previous figure.

P9.38. Prepare: Please refer to Figure P9.38. Model the rocket as a particle, and use the impulse-momentum theorem. The only external force acting on the rocket is due to its own thrust. Because the thrust force as a function of time is known, impulse due to the thrust is simply the area of the force graph. This impulse is then related to the speeds through the impulse-momentum theorem. Solve: (a) The impulse is

xJ = area of the Fx(t) graph between t = 0 s and t = 30 s 12 (1000 N)(30 s)= =15,000 N s

(b) From the impulse-momentum theorem, (px)f = (px)i + 15,000 Ns. That is, the momentum or velocity increases as long as Jx increases. When Jx increases no more, the speed will be a maximum. This happens at t = 30 s. At this time,

f i( ) ( )x xm v m v− = + 15,000 N s f(425 kg)( ) (425 kg)(75 m/s) 15,000 N sxv⇒ = + f( ) 110 m/sxv⇒ =

P9.39. Prepare: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations.

Momentum 9-23

Solve: To find the ball’s velocity just before and after it hits the floor: 2 2 2 2 21 0 1 0 1

2 2 2 2 2 23 2 3 2 2 2

2 ( ) 0 m /s 2( 9.8 m/s )(0 2.0 m) 6.261 m/s

2 ( ) 0 m /s 2( 9.8 m/s )(1.5 m 0 m) 5.422 m/sy y y y

y y y y y

v v a y y v

v v a y y v v

= + − = + − − ⇒ = −

= + − ⇒ = + − − ⇒ =

The force exerted by the floor on the ball can be found from the impulse-momentum theorem:

2 1

31max max2

area under the force curve (0.2 kg)(5.422 m/s)

(0.2 kg)(6.261 m/s) (5 10 s) 940 Ny ymv mv

F F−

= + ⇒

= − + × ⇒ =

Assess: A force of 940 N exerted by the floor is typical of such collisions.

P9.40. Prepare: There are three parts to the motion in this problem. The ball falls to the ground, the ball has a collision with the ground, and finally the ball travels upward under the influence of gravity. The floor exerts a force on the ball that creates an impulse on the ball. The change in momentum of the ball can be calculated with Equation 9.8 once the impulse is known. The velocity of the ball as it leaves the floor can be calculated from the momentum of the ball after the collision with the floor. Solve: Additional significant figures will be kept in intermediate results. The impulse due to a force is the area under the force-versus-time curve, from Equation 9.1. The curve on the graph is a triangle. Any triangle has an area that is equal to one half the length of the base multiplied by the height of the triangle. The “height” of the triangle is 1000 N and the base has a “length” equal to 4 ms. The impulse is then

3(1000 N)(4 10 s) 2 kg m/s2yJ

−×= = ⋅

The change in momentum of the ball is equal to the impulse, from Equation 9.8. We can find the velocity of the ball as it leaves the floor knowing its initial momentum. The velocity of the ball just before it hits the floor can be calculated with the kinematic equation

2 2f i( ) ( ) 2 yv v a y= + ∆

The initial velocity of the ball is zero, and the acceleration of the ball is the acceleration due to gravity. Solving for the magnitude of the velocity of the ball just before it hit the ground,

2f 2 2(9.80 m/s )( 2.0 m) 6.26 m/sv g y= − ∆ = − − =

The magnitude of the momentum of the ball just before it hits the ground is f (0.2 kg)mv = (6.26 m/s) = 1.25 kg ⋅ m/s. This is the initial momentum of the ball just before the collision, i( ) 1.25 kg m/s.yp = − ⋅ A negative sign has been inserted since the momentum is in the downward direction. Using Equation 9.8 to find the momentum of the ball right after the collision,

f i( ) ( ) 2 kg m/s+( 1.25 kg m/s) 0.75 kg m/sy y yp J p= + = ⋅ − ⋅ = ⋅

We can now find the velocity of the ball right after it bounces up from the floor,

ff

( )( ) 3.75 m/sy

y

pv

m= =

Using this velocity as the initial velocity for the upward part of the motion and using the same kinematics equation as above, we find the ball rebounds to a height of

2 2i

2

( ) (3.75 m/s) 0.717 m2 2( 9.80 m/s )

y

y

vy

a− −

∆ = = =−

Assess: This is a very special floor.

P9.41. Prepare: We combine Equation 9.1 J = Favg ∆t with the impulse-momentum theorem in the y-direction Jy = ∆py. See Example 9.1. This tells us that ∆py is the area under the curve of the net force in the vertical direction vs. time. And if we know the change in the woman’s vertical momentum we can figure out the speed with which she leaves the ground; to do this last step, however, we’ll need her mass. Look at the first part of the graph while the force exerted by the floor is constant. During that time she isn’t accelerating, so the force the floor exerts must be equal in magnitude to her weight; so she weighs 600 N and her mass is 600 N/(9.8 m/s2) = 61 kg.

9-24 Chapter 9

It should also be clear from the graph that she leaves the floor at t = 0.5 s when the force of the floor on her is zero. The graph we are given is not the graph of the net force. The hint warns us that the upward force of the floor is not the only force on the woman. We have just concluded that the earth is exerting a downward gravitational force of 600 N (her weight) on her. Therefore a graph of the net vertical force on her vs. time would simply be the same graph only 600 N lower on the force axis.

Note carefully that the graph now crosses the t-axis at = 0.475s.t Solve: What is the area under the new graph? We’ll take the area of the triangle above the t-axis to be positive and then subtract the area of the smaller triangle below the t-axis. The general formula for the area of a triangle is

121

big 2

1small 2

height base(1800 N)(0.275 s) 248 N s

(600 N)(0.025 s) 15 N s

AA

A

= × ×

= = ⋅

= = ⋅

The total area (with the triangle above the axis positive and the triangle below the axis negative) is 248 N⋅s 15 N⋅s = 233 N⋅s; this is the vertical impulse on the woman, and is also equal to her change in vertical momentum. Since y yp m v∆ = ∆ we simply divide by her mass to find her change in velocity:

/ (233 N s)/(61 kg) 3.8 m/sy yv p m∆ = ∆ = ⋅ =

Because she started from rest this value is also her final speed, just as she leaves the ground. Assess: We assumed the ability to read the data from the graph to two significant figures. If we were not confident in this we would report the result to just one significant figure: f( ) 4 m/s.yv ≈ The answer of 4m/s≈ does seem to be in the reasonable range. It is worth following the units in the last equation to see the answer end up in m/s.

P9.42. Prepare: We will use the particle model for the sled and the impulse-momentum theorem. Note that the force of kinetic friction fk imparts a negative impulse to the sled. The force fk = mkn where mk is the coefficient of kinetic friction and n is the normal (contact) force by the surface. As you see in the free-body diagram below .n w=

Solve: Using Equation 9.10 ,x xp J∆ = we have

f i k( ) ( )x xp p f t− = − ∆ f i k k( ) ( )x xm v m v n t mg tm m⇒ − = − ∆ = − ∆

Momentum 9-25

Thus,

i fk

1 (( ) ( ) )x xt v vgm

∆ = −2

1 (8.0 m/s 5.0 m/s)(0.25)(9.8 m/s )

= − = 1. 2 s

Assess: The coefficient of kinetic friction appearing in the denominator shows that smaller the friction force, the longer the travel time is, and the larger the friction the shorter the travel time. This is consistent with our everyday experience.

P9.43. Prepare: To find the impulse delivered by the bat to the ball, we need to know the change in the ball’s momentum and use J = ∆p. Since the direction of the ball changes, we need to use vector notation. The x-component of the ball’s final velocity is

( ) ( )55 m/s cos 25 49.8 m/s− ° = − and the y-component is ( ) ( )55 m/s cos 25 23.2 m/s− ° = −

Solve: The initial velocity of the ball is ( )i 35 m/s, 0v = and its initial momentum is obtained by multiplying by

the mass of the ball: ( )( ) ( )i 0.140 kg 35 m/s, 0 m/s 4.90 kg m/s, 0 kg m/sp = = ⋅ ⋅

The initial final momentum is the final velocity of the ball times its mass:

( )( ) ( )f 0.140 kg 49.8 m/s, 23.2 m/s 6.97 kg m/s, 3.25 kg m/sp = − = − ⋅ ⋅

Finally, the impulse on the ball equals the change in the ball’s momentum: ( )f i 11.9 kg m/s, 3.25 kg m/sJ p p= − = − ⋅ ⋅

The magnitude of the impulse can be obtained from the Pythagorean theorem: J =12 N⋅s and we can find the angle, ,θ above the horizontal using inverse tangent: ( )1tan 3.36 /12.30 15 .θ −= = ° The direction is to the left and 15° above the horizontal.

Assess: The angle 15° makes sense because the ball comes in at 0° with the horizontal and leaves the bat at 25° above the horizontal. We expect the force, and therefore the impulse, exerted by the bat to have an angle intermediate to these two.

P9.44. Prepare: Consider the system to be the squid and the water it takes in and expels. Ignoring all other forces, there is no net external force on the system, so momentum of the system is conserved. Assume the system is at rest before the squid expels the water. Solve: The momentum of the system before the squid expels the water is zero because everything is at rest. The momentum of the entire system must also be zero after the squid expels the water since momentum is conserved in the system. See the following diagram.

9-26 Chapter 9

Writing the momentum of the squid as S S( ),xm v and the momentum of the water as W W( ),xm v the momentum conservation equation is

S S f W W f S S i W W i( ) ( ) ( ) ( ) 0x x x xm v m v m v m v+ = + =

Solving for W f( ) ,xv

SW f S f

W

1.5 kg( ) ( ) (3.0 m/s) 45 m/s0.10 kgx x

mv vm

= − = − =

The water moves to the left with a speed of 45 m/s. Assess: This result makes sense. Since the squid is more massive than the water, the water must be expelled with a velocity greater than the final velocity of the squid.

P9.45. Prepare: We are asked for the impulse given to the pollen grain; impulse is defined in Equation 9.1:

avg .J F t= ∆ We are given that 43.0 10 s,t −∆ = × but we note that since we are not given any velocities, we will not

use momentum or the impulse-momentum theorem. With that approach eliminated, how will we find avg?F Given 101.0 10 kgm −= × and 4 22.5 10 m/sa = × guides us to use Newton’s second law to find avgF (assuming a is

constant over ).t∆ The avgF in the impulse equation is the same as the netF in Newton’s law because we are ignoring any other forces on the grain.

10 4 2 6avg net (1.0 10 kg)(2.5 10 m/s ) 2.5 10 NF F ma − −= = = × × = ×

Solve: 6 4 10 10

avg (2.5 10 N)(3.0 10 s) 7.5 10 N s 7.5 10 kg m/sJ F t − − − −= ∆ = × × = × ⋅ = × ⋅

Assess: This is certainly a small impulse, but the pollen grains have such small mass that the impulse is sufficient to give them the stated acceleration. Note that N s kg m/s.⋅ = ⋅

P9.46. Prepare: Given a = 2.5 × 104 m/s2 and ∆t = 0.30 ms, we are asked to find the final speed. We are told the acceleration is constant so we can use the kinematic equations. We assume the pollen grains start from rest, so we can use Equation 2.11, f i( ) ( ) .xvx vx a t= + ∆ For part (b) we will use subscripts p for pollen and b for bee. We are given the mass of one pollen grain in the previous problem as 10

onegrain 1.0 10 kgm −= × so for 1000 ejected pollen grains 17p 1.0 10 kg.m −= × We are also

given b 0.0050 kg.m = Solve: (a)

44 2f i x( ) ( ) 0 (2.5 10 m/s )(3.0 10 s) 7.5 m/sx xv v a t −+ + ∆ = + × × =

(b) We employ the conservation of momentum. Since the collision is inelastic the final momentum will be f P b f( ) ,p m m v= + where fv is the answer we seek. That the bee is hovering indicates b i( ) 0 m/s.xv =

Momentum 9-27

f i

p p bp b f i b i

7p p bi b 4

f 7p b

( ) ( ) ( )

( ) ( ) (1.0 10 kg)(7.5 m/s) (0.0050 kg) (0 m/s) 1.5 10 m/s 0.15 mm/s/ 1.0 10 kg 0.0050 kg

x x

x x

P Pm m v m v m v

m v m vv

m m

−−

−

=+ = +

+ × += = = × =

× +

The bee would not notice such a tiny speed.

P9.47. Prepare: Let the system be ball + racket. During the collision of the ball and the racket, momentum is conserved because all external interactions are insignificantly small. We will also use the momentum-impulse theorem.

Solve: (a) The conservation of momentum equation (px)f = (px)i is

R R f B B f R R i B B i

R f R f

( ) ( ) ( ) ( )(1.0 kg)( ) (0.06 kg)(40 m/s) (1.0 kg)(10 m/s) (0.06 kg)( 20 m/s) ( ) 6.4 m/s

x x x x

x x

m v m v m v m vv v

+ = ++ = + − ⇒ =

(b) The impulse on the ball is calculated from B f B i( ) ( )x x xp p J= + as follows:

avg

avg

(0.06 kg)(40 m/s) (0.06 kg)( 20 m/s) 3.6 N s

3.6 Ns 360 N10 ms

x xJ J F t

F

= − + ⇒ = = ∆

⇒ = =

Assess: Let us now compare this force with the ball’s weight 2B B (0.06 kg)(9.8 m/s ) 0.588 N.w m g= = = Thus,

Favg = 610 wB. This is a significant force and is reasonable because the impulse due to this force changes the direction as well as the speed of the ball from approximately 45 mph to 90 mph.

P9.48. Prepare: Use the particle model for the ball of clay (C) and the 1.0 kg block (B). The two objects are a system and it is a case of a perfectly inelastic collision. Since no significant external forces act on the system in the x-direction during the collision, momentum is conserved along the x-direction.

Solve: (a) The conservation of momentum equation (px)f = (px)i is

f(1.0 kg 0.02 kg)( ) (0.02 kg)(30 m/s) (1.0 kg)(0 m/s)xv+ = + f( ) 0.588 m/s 0.59 m/sxv⇒ = =

9-28 Chapter 9

(b) The impulse of the ball of clay on the block is calculated as follows:

B f B i B( ) ( ) ( )x x xp p J= + B B B f B B i( ) ( ) ( )x x xJ m v m v⇒ = − f(1.0 kg)( ) 0 N sxv= − = 0.588 N s = 0.59 kg m/s⋅

(c) The impulse of the block on the ball of clay is calculated as follows:

C f C i C( ) ( ) ( )x x xp p J= + C C C f C C i( ) ( ) ( )x x xJ m v m v⇒ = − (0.02 kg)(0.588 m/s) (0.02 kg)(30 m/s)= − = −0.59 N ⋅ s

(d) Yes, B Cx xJ J= − Assess: During the collision, the ball of clay and the block exert equal and opposite forces on each other for the same time. Impulse is therefore also equal in magnitude but opposite in direction.

P9.49. Prepare: We will define our system to be Dan + skateboard. The system has nonzero initial momentum pix. As Dan (D) jumps backward off the gliding skateboard (S), the skateboard will move forward in such a way that the final total momentum of the system is equal to the initial momentumx. This conservation of momentum occurs because ext 0F =

on the system.

Solve: We have S S f D D f S D i( ) ( ) ( )( ) .x x xm v m v m m v+ = + Hence,

D f(5.0 kg)(8.0 m/s) (50 kg)( ) (5.0 kg 50 kg)(4.0 m/s)xv+ = + D f( ) 3.6 m/sxv⇒ =

Assess: A speed of 3.6 m/s or 8 mph is reasonable.

P9.50. Prepare: The system comprised of a cart, James, Sarah, and balls is isolated in the sense that the net external force on the system is zero. Therefore we assume that the momentum of the system is conserved. In the initial conditions, the system is stationary and the total momentum is zero. Since total momentum is conserved in this problem, it must stay zero in both parts. Call the direction to the right positive and that to the left negative. Solve: (a) In this part it would be profitable to separate the system into two subsystems, one comprised of the cart and riders and the other of the two balls. The momenta of the two subsystems must be equal in magnitude and opposite in direction. The momentum of the balls is

balls J J s s (1.0 kg)(4.5 m/s) (0.50 kg)( 1.0 m/s) 4.0 kg m/sp m v m v= + = + − = ⋅

The momentum of the cart and riders must be – 4 kg ⋅ m/s. Since the mass of the cart-rider subsystem is 130 kg, then the speed and direction of the cart and riders is (– 4 kg ⋅ m/s)/(130 kg) = –0.031 m/s, where the negative sign indicates to the left. (b) After the balls are caught the speed of the cart and riders must be zero because the momentum of the whole system must be zero and no subparts are moving. Assess: This is certainly a small impulse, but the pollen grains have such small mass that the impulse is sufficient to give them the stated acceleration. Note that N ⋅ s = kg ⋅ m/s.

P9.51. Prepare: We can find the speed of the cart using the law of conservation of momentum. What makes this tricky is that Ethan’s speed is given relative to the cart but we need to find the speed of the cart relative to the ground. When using the conservation of momentum, all the velocities must be relative to the same observer. Ethan’s velocity relative to the ground is the sum of his velocity relative to the cart and the velocity of the cart relative to the ground. When he has reached his top speed, we have: Eg f cg f( ) 8.0 m/s ( ) .v v= + Solve: Initially, Ethan and the cart are at rest, so their total momentum is zero. We can now write down the equation for the conservation of momentum relative to the ground:

Momentum 9-29

E Eg f c cg f

cg f cg f

0 kg m/s ( ) ( )

0 kg m/s (80 kg)(8.0 m/s ( ) ) (500 kg)(v ) .

m v m vv

⋅ = + ⇒

⋅ = + +

This equation can be solved for the velocity of the cart: cg f( ) 1.1 m/s.v = − The speed of the cart when Ethan has reached his top speed is 1.1 m/s.

Assess: Relative to the ground, Ethan’s speed is 6.9 m/s. It is reasonable that Ethan is moving about six times as fast as the cart because the cart is about six times as massive as Ethan.

P9.52. Prepare: Impulse is defined in Equation 9.1: avg .J F t= ∆ We also want to use the impulse-momentum theorem: .J p= ∆ We are given 0.50 m/sv = and / 10,000 kg/s.m t∆ ∆ = Solve: Combine the two equations above.

avg ( )F t p mv∆ = ∆ = ∆

Now solve for ave.F Since constantv = it can be moved outside the parentheses.

2avg (0.50 m/s)(10,000 kg/s) 5000 kg m/s 5000 Nv m mF v

t t∆ ∆ = = = = ⋅ = ∆ ∆

Assess: This is not an extraordinarily large force for a locomotive. Note that the statement of Newton’s second law in the text, netF ma=

(Equation 4.6), doesn’t apply ( 0 buta =

net 0),F ≠

not because Newton was wrong, but because the correct statement of the law is net / ,F p t= ∆ ∆

as Newton

himself knew and said. netF ma=

is only equivalent to Newton’s formulation of his second law if the mass of the object isn’t changing, but it is in this case.

P9.53. Prepare: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectly inelastic collisions. Momentum is conserved in the collisions of this problem.

Solve: In the collision between the three-car train and the single car:

1 2 3(3 ) 4x x xmv m v mv+ = 1 2 33 4x x xv v v⇒ + = 3(4.0 m/s) 3(2.0 m/s) 4 xv⇒ + = 3 2.5 m/sxv⇒ = In the collision between the four-car train and the stationary car:

3 4 5(4 ) (5 )x x xm v mv m v+ = 3 54 0 m/s 5x xv v⇒ + = 35

4 (0.8)(2.5 m/s) 2.0 m/s5

xx

vv⇒ = = =

Assess: The motion of railroad has to be on a level track for linear momentum to be constant. The speed of the five-car train, as expected, is of the same order of magnitude as the other speeds.

9-30 Chapter 9

P9.54. Prepare: We arbitrarily pick the direction the linebacker was running as the positive x-direction since he was mentioned first. If, after we solve the conservation of momentum equation for vf, the answer is positive, then we know the linebacker ends up moving forward; if vf is negative, then the quarterback ends up moving forward. We will use subscripts l for linebacker and q for quarterback.

Known

1

q

l

q i

110 Kg82 Kg

( ) 2.0 m/s( ) 3.0 m/s

x

x

mm

vv

==

== −

Find

fv

We employ the conservation of momentum. Since the collision is inelastic (the linebacker grabs and holds onto the quarterback) the final momentum will be ff l q( ) ,P m m v= + where fv is the answer we seek. Solve:

f i

l q f l l i q q i

l l i q q if

l q

( ) ( ) ( )

( ) ( ) (110 kg)(2.0 m/s) (82 kg)( 3.0 m/s) 0.14 m/sm 110 kg 82 kg

x x

x x

P Pm m v m v m v

m v m vv

m

=+ =

+ + −= = = −

+ +

The answer is negative; this indicates that the quarterback ends up moving forward after the hit. (The linebacker gets knocked backward.) Assess: If all we want to know is the sign of the answer then we do not really need to compute or divide by the denominator—the total mass will certainly be positive and will not affect the sign of the answer. So we could have done a simple mental calculation of the numerator (82 × 3 > 110 × 2) to figure out which football players “wins.”

P9.55. Prepare: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in the collision since no significant external force acts on the system.

Solve: (a) The conservation of momentum equation (px)f = (px)i is

A A i E E i A E f( ) ( ) ( )( )x x xm v m v m m v+ = + 13 4 13 24

f(1.0 10 kg)(4 10 m/s) 0 kg m/s (1.0 10 kg 5.98 10 kg)( )xv⇒ × × + = × + × 8f( ) 6.7 10 m/sxv −⇒ = ×

Momentum 9-31

(b) The speed of the earth going around the sun is

114

E 7

2 2 (1.50 10 m) 3.0 10 m/s3.15 10 s

rvTπ π ×

= = = ××

Hence, 12 10f E( ) / 2 10 2 10 %.xv v − −= × = ×

Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.

P9.56. Prepare: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling west (W), are a system. Since the two skaters hold together after the “collision,” this is a case of a perfectly inelastic collision in two dimensions. Momentum is conserved since no significant external force in the x–y plane acts on the system in the “collision.”

Solve: (a) The x-component of the conservation of momentum is

N W f N N i W W i f

f

( )( ) ( ) ( ) (75 kg 60 kg)( )0 kg m/s (60 kg)( 3.5 m/s) ( ) 1.556 m/s

x x x x

x

m m v m v m v vv

+ = + ⇒ += + − ⇒ = −

The y-component of the conservation of momentum is

N W f N N i W W i f f( )( ) ( ) ( ) (75 kg 60 kg)( ) (75 kg)(2.5 m/s) 0 kg m/s ( ) 1.389 m/sy y y y ym m v m v m v v v+ = + ⇒ + = + ⇒ =

Thus 2 2f f f( ) ( ) 2.085 m/sx yv v v= + =

The time to glide to the edge of the rink is

f

radius of the rink 25 m2.085 m/sv

= = 12 s

(b) The location is 1f ftan ( ) /( ) 42y xv vθ −= = ° north of west.

Assess: A time of 12 s in covering a distance of 25 m at a speed of ≈ 2 m/s is reasonable.

9-32 Chapter 9

P9.57. Prepare: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier and lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other. During the “explosion,” the total momentum of the system is conserved.

Solve: The initial momentum is zero. Thus the conservation of momentum equation (px)f = (px)i is

H H f L L f( ) ( ) 0 kg m/sx xm v m v+ = ⋅ H f L f(75 kg)( ) (50 kg)( ) 0 kg m/sx xv v⇒ + = ⋅

Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find H f( ) 30 m/20 sxv = = 1.5 m/s. Thus,

L f(75 kg)(1.5 m/s) (50 kg)( ) 0 kg m/sxv+ = ⋅ L f( ) 2.25 m/sxv⇒ = −

Thus, the time for the lighter skater to reach the edge is

L f

30 m 30 m 13 s( ) 2.25 m/sxv

= =

Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to reach the edge of the ice rink. A time of 13 s at the speed of 2.3 m/s is reasonable.

P9.58. Prepare: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2 (moving west), together are our system. This is an isolated system because any frictional force during the brief collision period is going to be insignificant. Within the impulse approximation, the momentum of our system will be conserved in the collision. Note that m1 = m2 = m.

Solve: The equation (px)f = (px)i yields:

1 1 f 2 2 f 1 1 i 2 2 i( ) ( ) ( ) ( )x x x xm v m v m v m v+ = + 1 1 f 1 1 i 2 2 i( ) cos 0 kg m/s ( ) ( )x xm v m v m vθ⇒ + ⋅ = +

1 f 1 i 2 i( ) cos ( ) ( )x xv v vθ⇒ = + 2.0 m/s 1.0 m/s 1.0 m/s= − =

Momentum 9-33

The equation (py)f = (py)i yields:

1 1 f 2 2 f( ) sin ( ) 0 kg m/sym v m vθ + = ⋅ 1 f 2 f( ) sin ( ) 1.41 m/syv vθ⇒ = − = −

2 21 f( ) (1.0 m/s) ( 1.41 m/s) 1.7 m/sv⇒ = + − =

1 1.41 m/stan 551.0 m/s

θ − = = °

The angle is below +x axis. Assess: The speed obtained is of the same order of magnitude as the other speeds as would be expected.

P9.59. Prepare: This is a two part problem. First, we have an inelastic collision between the wood block and the bullet. The bullet and the wood block are an isolated system. Since any external force acting during the collision is not going to be significant, the momentum of the system will be conserved. The second part involves the dynamics of the block + bullet sliding on the wood table. We treat the block and the bullet as particles.

Solve: The equation (px)f = (px)i gives

B W C f B B i W W i( )( ) ( ) ( )x x xm m v m v m v+ = +

C f B i(0.01 kg 10 kg)( ) (0.01 kg)( ) (10.0 kg)(0 m/s)x xv v⇒ + = + C f B i1( ) ( )

1001x xv v⇒ =

Using k k k B W B W( ) ( ) xf n m m g m m am m− = − = − + = + k .xa gm⇒ = − Note that the negative sign appears in front of fk because the force of friction points in the –x direction. Using the kinematics equation 2 2

f i f i( ) ( ) 2 ( ),x x xv v a x x= + − 2 2

2 2 2 2 2 2 2C f k f i B i k f B i k f

2B i k f

1 10 ( ) 2 ( ) 0 m /s ( ) 2 0 m /s ( ) 21001 1001

( ) 1001 2 1001 2(0.2)(9.8 m/s )(0.05 m) 440 m/s

x x x

x

v g x x v g x v g x

v g x

m m m

m

= − − ⇒ = − ⇒ = −

⇒ = = =

Assess: The bullet’s speed is reasonable (≈1000 mph).

P9.60. Prepare: Model the package and the rocket as particles. This is a two-part problem. First we have an inelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved along the horizontal direction since no significant external force acts on the rocket and the package. However, as soon as the package + rocket (C) system leaves the cliff they become a projectile motion problem.

9-34 Chapter 9

Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is (vx)i, which can be found as follows:

2f i i f i f i

1( ) ( ) ( )2y yy y v t t a t t= + − + − 2 2

f1200 m 0 m 0 m ( 9.8 m/s )2

t⇒ − = + + − f 6.389 st⇒ =

With this time, we can now find (vx)i using 21f i i f i f i2( ) ( ) ( ) .x xx x v t t a t t= + − + − We obtain

i30 m 0 m ( ) (6.389 s) 0 mxv= + + i C f( ) 4.696 m/s ( )x xv v⇒ = =

We now use the momentum conservation equation (px)f = (px)i which can be written

R P C f R R i P P i

R i R i

( )( ) ( ) ( ) (1.0 kg 5.0 kg)(4.696 m/s)(1.0 kg)( ) (5.0 kg)(0 m/s) ( ) 28 m/s

x x x

x x

m m v m v m vv v

+ = + ⇒ += + ⇒ =

Assess: This is an example of a situation where the vertical momentum is not conserved, because the external force acting on the package plus rocket is not zero. Therefore, the law of conservation of momentum holds only in the horizontal direction, which is what you need to solve this problem.

P9.61. Prepare: We will model the two fragments of the rocket after the explosion as particles. We assume the explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use kinematics equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will apply conservation of momentum along the y direction to the system (that is, the two fragments) in the explosion. The momentum conservation “applies” because the forces involved during the explosion are much larger than the external force due to gravity during the small period that the explosion lasts. In the third part, we will again use kinematics equations to find the velocity of the heavier fragment just after the explosion.

Solve: The rocket accelerates for 2.0 s from rest, so

21 0 1 0( ) ( ) ( ) 0 m/s (10 m/s )(2 s 0 s) 20 m/sy y yv v a t t= + − = + − =

2 2 21 0 0 1 0 1 0

1 1( ) ( ) ( ) 0 m 0 m (10 m/s )(2 s) 20 m2 2y yy y v t t a t t= + − + − = + + =

At the explosion the equation (py)f = (py)i is

L L 2 H H 2 L H 1( ) ( ) ( )( )y y ym v m v m m v+ = + L 2 H 2(500 kg)( ) (1000 kg)( ) (1500 kg)(20 m/s)y yv v⇒ + =

To find (vHy)2 we must first find (vLy)2, the velocity after the explosion of the upper section. Using kinematics,

2 2 2L 3 L 2 L 3 L 2( ) ( ) 2( 9.8 m/s )(( ) ( ) )y yv v y y= + − − 2

L 2( ) 2(9.8 m/s )(530 m 20 m) 99.98 m/syv⇒ = − =

Now, going back to the momentum conservation equation we get

H 2(500 kg)(99.98 m/s) (1000 kg)( ) (1500 kg)(20 m/s)yv+ = H 2( ) 20 m/syv⇒ = −

The negative sign indicates downward motion.

Momentum 9-35

P9.62. Prepare: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B).

Solve: For the first collision the equation (px)f = (px)i is

L L 1 A A 1 L L 0 A A 0( ) ( ) ( ) ( )x x x xm v m v m v m v+ = +

L 1 L 1(0.01 kg)( ) (0.500 kg)(6 m/s) (0.01 kg)(400 m/s) 0 kg m/s ( ) 100 m/sx xv v⇒ + = + ⇒ =

The bullet emerges from the first block at 100 m/s. For the second collision the equation (px)f = (px)i is

L B 2 L L 1( )( ) ( )x xm m v m v+ = 2(0.01 kg 0.5 kg)( ) (0.01 kg)(100 m/s)xv⇒ + = 2( ) 2.0 m/sxv⇒ =

Assess: This problem invloves repeated application of the law of conservation of momentum. Also note that the actual value of 2 m for the separation between the blocks is not necessary for our calculations.

P9.63. Prepare: Choose the system to be cannon + ball. There are no significant external horizontal forces during the brief interval in which the cannon fires, so within the impulse approximation the horizontal momentum is conserved. We’ll ignore the very small mass loss of the exploding gunpowder. The statement that the ball travels at 200 m/s relative to the cannon can be written (vBx)f = (vCx)f + 200 m/s. That is, the ball’s speed is 200 m/s more than the cannon’s speed.

Solve: The initial momentum is zero, so the conservation of momentum equation (px)f = (px)i is

f C C f B B f C C f B C f

BC f

C B

( ) ( ) ( ) ( ) (( ) 200 m/s) 0 kg m/s10 kg( ) (200 m/s) (200 m/s) 3.9 m/s510 kg

x x x x x

x

p m v m v m v m vmv

m m

= + = + + =

⇒ = − = − = −+

That is, the cannon recoils to the left (negative sign) at 3.9 m/s. Thus the cannonball’s speed relative to the ground is (vBx)f = –3.9 m/s + 200 m/s = 196 m/s. Assess: An expected relatively small recoil speed is essentially due to the large mass of the cannon.

9-36 Chapter 9

P9.64. Prepare: In order to use the law of conservation of momentum, the velocities must all be measured relative to the same observer. Since the problem gives Laura’s final speed relative to the canoe, Lc f( ) ,v and not her final speed relative to the water, Lw f( ) ,v we will need to use the relative velocity formula. We are given the masses of Laura and the canoe: L 35 kgm = and c 55 kgm = as well as the final speed of Laura relative to the canoe: Lc f( ) 1.5 m/s.v = Solve: Before she jumps, Laura and the canoe are stationary, so their total momentum is zero. Afterward, the canoe has velocity cw f( )v and Laura has velocity:

Lw f Lc f cw f cw f( ) ( ) ( ) 1.5 m/s ( )v v v v= + = + No we can use the law of conservation of momentum:

L Lw f c cw f

cw f cw f

0 kg m/s ( )( ) ( )( )0 kg m/s (35 kg)[1.5 m/s ( ) ] (55 kg)(v )

m v m vv

⋅ = + ⇒⋅ = + +

This equation can be solved to obtain cw f( ) 0.58 m/s.v = − Assess: For perspective, Laura’s velocity relative to the water is given by Lw f( ) 1.5 m/s 0.58 m/sv = −

0.9 m/s.= This makes sense because Laura is about two thirds as massive as the canoe and the final speed of the canoe, 0.58 m/s, is about two thirds the final speed of Laura, 0.9 m/s.

P9.65. Prepare: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector quantity, so the direction of the initial velocity vector 1v establishes the direction of the momentum vector. The final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue to move along this line and have no y-components of momentum. The missing third piece cannot have a y-component of momentum if momentum is to be conserved, so it must move along the x-axis–either straight forward or straight backward. From the conservation of mass, the mass of piece 3 is 5

3 total 1 2 7.0 10 kg.m m m m= − − = ×

Solve: To conserve momentum along the x-axis, we require 13

i total i f 1f 2f 3f 1 1f 2 2f 3f 3f total i 1 1f 2 2f 1.02 10 kg m /sp m v p p p p m v m v p p m v m v m v= = = + + = + + ⇒ = − − = + × ⋅

Because p3f > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass m3, we can find the velocity of the third piece as follows:

1373f

3f 53

1.02 10 kg m/s 1.5 10 m/s7.0 10 kg

pvm

× ⋅= = = ×

×

Assess: Since this event is taking place in outer space, we don’t have to worry about any external forces, and naturally the total momentum has to be constant regardless of the direction.

P9.66. Prepare: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and momentum is conserved in the collision between the proton and the gold atom.

Solve: The conservation of momentum equation (px)f = (px)i is

Momentum 9-37

G G f P P f P P i G G i( ) ( ) ( ) ( )x x x xm v m v m v m v+ = + 7 7

P f G P P(197 )( ) (1 )( 0.90 5.0 10 m/s) (1 )(5.0 10 m/s) 0 m/sxm v m m⇒ + − × × = × + 5

G f( ) 4.8 10 m/sxv⇒ = ×

Assess: It is easy to estimate the answer. Roughly speaking, initial momentum is mv and final momentum is (MV-mv), where v is the apprioximate speed of the proton before and after collision, and V is the speed of gold nucleus after collision. m and M are the masses of proton and gold nucleus, where M is about 200 m. Conservation of momentum dictates that mv = MV – mv, or V = (2m/M)v = (2m/200 m)v or v/100, which is equal to about 5 × 105 m/s.

P9.67. Prepare: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the system is conserved. All the three masses and the three velocities before the collision are known; it is thus easy to find the speed and the direction of the resulting blob using momentum conservation equations in two dimensions.

Solve: The three initial momenta are

1 i 1 1 i( ) ( ) (0.02 kg)(2 m/s) along (0.04 kg m/s, along )p m v x x= = + = ⋅ +

2 i 2 2 i( ) ( ) (0.03 kg)(3 m/s) along (0.09 kg m/s, along )p m v x x= = − = ⋅ −

3 i 3 3 i( ) ( ) [(0.04 kg)(4 m/s)cos 45 , along ] [(0.04 kg)(4 m/s)sin 45 , along ](0.113kg m/s, along ) (0.113 kg m/s, along )

p m v x xx x

= = ° + + ° −= ⋅ + + ⋅ −

Since f i 1 i 2 i 3 i( ) ( ) ( ) ,p p p p p= = + + we have

1 2 3 f

f

2 2f

f1 1

f

( ) (0.023 kg m /s, along ) (0.073 kg m /s, along )(0.256 m/s, along ) + ( 0.811 m/s, along )

(0.256 m/s) ( 0.811 m/s) 0.85 m/s| | 0.811tan tan 73 below

0.256y

x

m m m v x xv x x

vv

xv

θ − −

+ + = ⋅ + + ⋅ −⇒ = + − +

⇒ = + − =

= = = ° +

Assess: The final speed is of the same order of magnitude as the initial speeds, as one would expect.

P9.68. Prepare: The 14C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino. Due to the lack of external forces acting on the carbon atom, momentum is conserved in the process of “explosion” or decay. The initial momentum of the 14C atom is zero. We will assume that explosion causes the electron to move along the x axis and the neutrino along the y axis. Also we note that the mass of the nucleus is essentially the same as the 14C atom because the masses of the electron and the neutrino are very small. From the given data, we will first calculate the momentum of the nucleus and then divide this quantity by neutron’s mass to obtain its speed.

9-38 Chapter 9

Solve: The conservation of momentum equation f i 0p p= = kg ⋅ m/s is

e n N N e n e e n n31 7 24

24 24

24 2 24 2N N N

0 N ( )

[(9.11 10 kg)(5 10 m/s), along ] [(8.0 10 kg m/s), along ](45.55 10 kg m/s, along ) (8.0 10 kg m/s, along )

(45.55 10 ) (8.0 10 ) kg

p p p p p p m v m vx y

x y

p m v

− −

− −

− −

+ + = ⇒ = − + = − −

= − × × + − × ⋅ +

= − × ⋅ + − × ⋅ +

⇒ = = × + × ⋅

23

26 23 3N N

m/s 4.62 10 kg m/s

(2.34 10 kg) 4.62 10 kg m/s 1.97 10 m/s

v v

−

− −

= × ⋅

⇒ × = × ⋅ ⇒ = ×

Assess: The nucleus speed is 4 orders of magnitude smaller than the electron’s speed. This difference is primarily due to 4 orders of magnitude larger mass of the nucleus compared to the electron.

P9.69. Prepare: Because there’s no friction or other tangential forces, the angular momentum of the block + rod system is conserved. The rod is massless, so the angular momentum is entirely that of a mass in circular motion.

Solve: The conservation of angular momentum equation f iL L= is

f f i imv r mv r= f f f i i i( ) ( )r r r rω ω⇒ = ⇒2 2

if i

f

30 cm (50 rpm)100 cm

rr

ω ω

= =

= 4.5 rpm

Assess: An angular speed of 4.5 rpm is reasonable, since the angular speed varies inversely as the square of the object’s distance from the rotation axis.

P9.70. Prepare: Model the puck as a particle.

Momentum 9-39

Solve: The angular momentum is

23.0 kg m /sL mvr= = ⋅ (0.2 kg) (0.5 m)v= ⇒ 30 m/sv =

The force that keeps the puck in circular motion is the tension T

in the string. Thus, 22 (0.2 kg)(30 m/s)

0.5 mmvT

r= = = 360 N

Assess: A tension of 360 N in the string whose farthest end has a puck moving at approximately 60 mph is large, but in view of the puck’s high speed the tension would be reasonable.

P9.71. Prepare: Model the puck and the 200 g weights as particles. The puck is in circular motion and the forces acting on the puck are its weight downward, the radial tension in the string, and a normal force upward. There is no tangential force acting on the puck, and thus the angular momentum of the puck is conserved. For the puck to move in a circle, the force causing the centripetal acceleration is provided by the two 200 g weights.

Solve: (a) Equating the two weights with the centripetal force

22i i2

pucki

(0.40 kg)(9.8 m/s ) (0.010 kg)(0.020 m)

v vmr

= = i 2.8 m/sv⇒ =

(b) The conservation of angular momentum equation f iL L= implies

puck f f puck i im v r m v r= ⇒ 2f f i i (2.8 m/s)(0.20 m) 0.56 m /sv r v r= = =

Now for the new weight to cause circular motion 2

puck f 2

f

(0.20 kg)(9.8 m/s )m v

r=

2 2puck f f

f 2 21.96 kg m/s 19.6 m/sm v vr⇒ = =

⋅

Substituting this expression into the conservation of angular momentum equation, we have

2f 2 3 3 3

f f f20.56 m /s 10.976 m /s 2.22 m/s 2.2 m/s

19.6 m/svv v v

= ⇒ = ⇒ = =

9-40 Chapter 9

Now we can obtain rf from the previous equation as 22

ff 2 2

(2.22 m/s) 0.252 m 25 cm19.6 m/s 19.6 m/s

vr = = = =

Assess: As expected, reduced weight decreases tension in the string, which implies a little lower speed and a little larger radius of trajectory.

P9.72. Prepare: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. The initial moment of inertia is I1 and the final moment of inertia is I2. The initial moment of inertia is 2 2 21 1

1 disk 2 2 (2.0 kg)(0.10 m) 0.010 kg m .I I mR= = = = ⋅ Solve: The final moment of inertia of the disk is

I2 = I1 + 2mR2 = 0.010 kg ⋅ m2 + 2(0.500 kg) × (0.10 m)2 = 0.010 kg ⋅ m2 + 0.010 kg ⋅ m2 = 0.020 kg ⋅ m2

Let ω1 and ω2 be the initial and final angular velocities. Then 2

1 1f i 2 2 1 1 2 2

2

(0.010 kg m )(100 rpm) 50 rpm0.020 kg m

IL L I IIωω ω ω ⋅

= ⇒ = ⇒ = = =⋅

Assess: Conservation of angular momentum says Iω = constant in the absence of external torques. In this problem the moment of inertia goes up and therefore the angular frequency has to decrease, which is consistent with what we found.

P9.73. Prepare: Since there are no external torques, we can use the conservation of angular momentum to

solve this problem. We need to use the moment of inertia of a rotating disk 2disk

12

I MR= in order to write the

final angular momentum of the merry-go-round: f disk f( ) .mL I ω= The final angular momentum of Joey is given by

J f J f( ) ( ) .L m v R= Solve: Before Joey begins running, both he and the merry-go-round are at rest, so the total angular momentum is 0. Let us say that he runs counterclockwise so that his final angular momentum is positive. Then the merry-go- round must rotate clockwise so that its final angular momentum is negative. In this way the angular momenta of Joey and the merry-go-round will still add to 0. The equation for conservation of angular momentum is

2 2J f f

2 2f

10 kg m / s ( )2

10 kg m / s (36 kg)(5.0 m/s)(2.0 m) (200 kg)(2.0 m)2

m v R MR ω

ω

⋅ = + ⇒

⋅ = +

This equation can be solved for the angular velocity: f 0.90 rad/s.ω = −

Momentum 9-41

Assess: The negative sign is as expected considering that the total angular momentum is zero. For this to be true, the two bodies, the merry-go-round and Joey, must move in opposite directions around the axis.

P9.74. Prepare: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to Ifinalωfinal. Solve: John’s initial angular momentum is that of a particle: J J J J Jsin .L m v R m v Rβ= = The angle β = 0 since John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is

2 2final final disk J i J J

22

finalfinal

2 2 2 2final disk J J

1 1 2 rad(250 kg)(1.5 m) (20 rpm) (30 kg)(5.0 m/s)(1.5 m)2 2 60 rpm

814 kg m /s814 kg m /s

1 1 (250 kg)(1.5 m) (30 kg)(1.5 m)2 2

I L L MR m v R

I

I I I MR m R

πω ω

ω

= + = + = +

⋅= ⋅ ⇒ =

= + = + = + 2

2

final 2

349 kg m

814 kg m /s 2.33 rad/s 22 rpm 349 kg m

ω

= ⋅

⋅= = =

⋅

P9.75. Prepare: Define the system to be Disk A plus Disk B so that during the time of the collision there are no net torques on this isolated system. This allows us to use the law of conservation of angular momentum. After the collision the combined object will have a moment of inertia equal to the sum of the moments of inertia of Disk A and Disk B and it will have one common angular speed fω (which we seek).

9-42 Chapter 9

We must remember to call counterclockwise angular speeds positive and clockwise angular speeds negative. Known

A

A

A i

B

B

B i

2.0 kg0.40 m

( ) 30 rev/s2.0 kg0.20 m

( ) 30 rev/s

MR

MR

ω

ω

=== −===

Find

fω Preliminarily compute the moments of inertia of the disks:

2 2 2A A A

2 2 2B B B

1 1 (2.0 kg)(0.40 m) 0.16 kg m2 21 1 (2.0 kg)(0.20 m) 0.040 kg m2 2

I M R

I M R

= = = ⋅

= = = ⋅

Solve: i f

A i B i A f

A A i B B i A B f

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

B

L LL L L

I I I Iω ω ω+

Σ = Σ+ =

+ = +

Solving for fω gives 2 2

A A i B B if 2 2

A B

( ) ( ) (0.16 kg m )( 30 rev/s) (0.040 kg m )(30 rev/s) 18 rev/s0.16 kg m 0.040 kg m

I II I

ω ωω + ⋅ − + ⋅= = = −

+ ⋅ + ⋅

That is, the angular speed is18 rev/s and the direction is clockwise. Assess: Since the two disks were rotating in opposite directions at the same speed we expect the angular speed afterwards to be less than the original speeds, and indeed it is.

P9.76. Prepare: We can use the definition of momentum, Equation 9.6 to calculate the initial momentum of the club-ball system. Solve: The ball is at rest, so the initial momentum of the system is entirely due to the head of the club. The initial momentum of the system is

i club c lub i( ) ( ) (0.200 kg)(40.0 m/s) 8.00 kg m/sx xp m v= = = ⋅

We assume the mass of the club is given to three significant figures since the mass of the ball is given to that many figures. The correct choice is B. Assess: In reality, the golfer exerts a force on the club, so there is a net external force on the system.

P9.77. Prepare: In this problem we make all the usual assumptions: that the collision happens quickly enough that we can ignore any net external forces and consider the system (club plus ball) to be isolated. Solve: Since the system is isolated the momentum of the system is conserved. The correct choice is B. Assess: Momentum is conserved in all collisions if we make the usual assumptions.

P9.78. Prepare: The average force during a collision can be calculated using Equation 9.2. Solve: If the ball compresses more during the collision, the duration of the collision will be longer. Since the impulse is the same, the average force decreases. The correct choice is A. Assess: This is similar to the idea behind crumple zones in cars. Compare to the solution of Question 9.8.

Momentum 9-43

P9.79. Prepare: Because the system is isolated we can use the law of conservation of momentum. We are asked to find C f C i( ) ( ) .x xv v− Solve:

i f

C i B i C f B f

C C i B B i C C f B B f

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

( ) ( ) ( ) ( )

x x

x x x x

x x x x

P Pp p p p

m v m v m v m v

=+ = +

+ = +

Rearranging terms allows us to solve for C f C i( ) ( ) :x xv v−

B B i B fC f C i

C

[( ) ( ) ] 0.0460 kg(0.0 m/s 60.0 m/s)( ) ( ) = 14 m/s0.200 kg

x xx x

m v vv vm

− −− = = −

The correct choice is C. The negative sign indicates that the club slowed down. Assess: Examining our equation for C f C i( ) ( )x xv v− confirms that we had a different (and maybe easier) approach from the very beginning. If the momentum is conserved for a two-body system, then the change in momentum of the club must be equal in magnitude (and opposite in direction) to the change in momentum of the ball. The numerator above is the change in momentum of the ball, so simply dividing by the mass of the club gives the difference in the club’s velocity.