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9
MOMENTUM
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.
p1x = (20 g)(1 m/s) = 20 g ⋅ m/s
p2 x = (20 g)(2 m/s) = 40 g ⋅ m/s
p3 x = (10 g)(2 m/s) = 20 g ⋅ m/s
p4 x = (10 g)(1 m/s) = 10 g ⋅ m/s
p5 x = (200 g)(0.1 m/s) = 20 g ⋅ m/s
So the answer is p2 x > p1x = p3 x = p5 x > p4 x .
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 Δ p = J . Since they start at rest with zero momentum, we can write mvf = pf = Δ p = J = F Δ 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
(v1 )f = (12 N)(2.0 s)/ m = (24 N ⋅ s)/ m
For the second object, the velocity is given by
(v2 )f = (15 N)(3.0 s)/(2 m) = (22.5 N ⋅ s)/ m
The speed 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 speed
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 impulsemomentum theorem, which tells us that to change the momentum of an object we must exert a net external force on it
G
G
over a time interval: Favg Δt = Δp.
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.
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9-1
9-2
Chapter 9
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
Δ x = 12 ax ( Δt ) 2
Solving for the time,
Δt =
2Δ x
ax
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. The larger puck has the greater
momentum after completing the distance.
Assess: The final speed of the larger puck is also twice the final speed 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 ( p1 ,0), where p1 is a positive number. Since the second piece is
traveling north, its momentum will have the form (0, p2 ), where p2 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 (− p1 , − p2 ). 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.
Assess: 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. It’s interesting to note that whether the student catches a ball moving to the right, or
throws a ball to the left, her speed to the right increases.
Q9.7. Reason: Since both carts are stationary after the collision the final total momentum of the two-cart system is
zero. By conservation of momentum, the total momentum of the system must have also been zero before the
collision. Therefore the momentum of the 3 kg cart must have been the same magnitude (and opposite direction) as
the momentum of the 2 kg cart. Since the momentum of the 2 kg cart was 6 kg m/s, then the speed of the 3 kg cart
must have been 2 m/s.
Assess: If the carts stick together the collision is inelastic.
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.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
G
G
G
Δ p is the same in either case. The impulse-momentum theorem also tells us Favg Δt = Δp.
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 Δ 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
G
G
because, for each skater, the upward normal force of the ice balances their downward weight force to make Fnet = 0.
Thus the total momentum of the system of the two skaters will be conserved. Assume that both skaters are at rest
G G
before the push so that the total momentum before they push off is Pi = 0. Consequently, the total momentum will
G
still be 0 after they push off.
G
(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 Δ p is the same
G
G
for each: Favg Δt = Δp.
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
G
G
same time interval. However, over that time interval they do not experience the same acceleration. Fnet = 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 Δ 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: You do not move backward when passing the basketball because the ball-you system is not
isolated: There is a net external force on the system—the friction force of the floor on your feet—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: J on pin = −Δpproj . 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
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9-4
Chapter 9
change. The beanbag would continue to move in the same direction. Thus the rubber ball would exert a greater
impulse on the bowling pin.
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. I for such a situation is mr 2, and ω = v/r. Putting this all together gives the
angular momentum for a particle in uniform circular motion: L = I ω = (mr 2 )(v/r ) = rmv .
So we compute L = rmv for each of the five situations.
L1 = (2 m)(2 kg)(2 m/s) = 8 kg ⋅ m 2/s
L2 = (2 m)(3 kg)(1 m/s) = 6 kg ⋅ m 2/s
L3 = (2 m)(1 kg)(3 m/s) = 6 kg ⋅ m 2/s
L4 = (4 m)(2 kg)(1 m/s) = 8 kg ⋅ m 2/s
L5 = (4 m)(2 kg)(2 m/s) = 16 kg ⋅ m 2/s
Finally, comparison gives L5 > L1 = L4 > L2 = L3 .
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 7.10).
Q9.16. Reason: Consider the Monica-platform system to be isolated (hence the frictionless axle) so that the
angular momentum is conserved. As Monica walks toward the center the moment of inertia of the system decreases
so the angular velocity increases because L = I ω stays constant. Once she passes the center and gets closer to the
opposite edge the moment of inertia increases until it is the same value as initially. So the angular velocity also then
decreases until reaching is initial value.
Assess: How much the moment of inertia decreases as she approaches the center depends on her mass relative to the
mass of the platform.
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 7.21. 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 9.20 gives the angular momentum L = I ω. For a solid disk I = 12 mr 2 around the axis of
symmetry. We’ll answer this question with ratios, using m2 = m1 , r2 = 2r1, and ω2 = 12 ω1.
L2 I 2ω2 12 m2 r 22ω2 12 m1 (2r1 ) 2 ( 12 ω1 )
⎛1⎞
=
= 1
=
= (22 ) ⎜ ⎟ = 2
2
2
1
ω
ω
L1 I1ω1
m
r
m
r
⎝2⎠
2 1 1 1
2 1 1 1
Therefore L2 = 2 L1, 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 ω2 = 12 ω1, 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.
Q9.19. Reason: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as
G
G
Δp
Favg =
Δt
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Momentum
9-5
This allows us to directly solve the question.
G
G
G
G G
G
Δp mvf − mvi m(vf − vi ) (20 kg)(0 m/s − 1 m/s)
Favg =
=
=
=
= − 10 000 N
Δt
Δt
Δt
0.002 s
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
m1vi = (m1 + m2 )vf
Solving for the mass of the second ball,
m v − m1vf (1.0 kg)(2.0 m/s) − (1.0 kg)(1.2 m/s)
m2 = 1 i
=
= 0.67 kg
vf
1.2 m/s
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.
G G
Pi = Pf
(m)(2vx )i + (3m)(vx )i = ( m + 3m)(vx )f
where the two blocks stick together to make one compound object after the collision. We want to know vf :
(v x ) f =
(m)(2(vx )i ) + (3m)(vx )i 5m(vx )i 5(vx )i
=
=
m + 3m
4m
4
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 vi 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.
Q9.23.
Reason: The initial momentum of the first block is ( p1x )i = (2.5 kg)(12.0 m/s) = 30 kg ⋅ m/s and the
initial momentum of the second block is ( p2 x )i = (14 kg)(–3.4 m/s) = –47.6 kg ⋅ m/s. 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:
(vx )f = ptotal x /(m1 + m2 ) = ( −17.6 kg ⋅ m/s)/(16.5 kg) = − 1.07 m/s.
Now the final momentum of block 2 is the product of its mass and final velocity:
( p2 x )f = (14 kg)( − 1.07 m/s) = − 15.0 kg ⋅ m/s. The impulse of block 1 on block 2 equals the change in momentum
of block 2:
( J1 on 2 ) x = Δp2x = −15.0 kg ⋅ m/s– (– 47.6 kg ⋅ m/s) = 32.6 kg ⋅ m/s.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9-6
Chapter 9
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 25 ms for an area of about:
A=
1
(25 × 10 −3 s)(3000 N) = 37.5 N ⋅ s = 37.5 kg ⋅ m/s.
2
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.
G
G
Q9.24. Reason: The impulse-momentum theorem F Δt = Δp tells us the change in momentum must be in the same
G G G
G
G
G
direction as the impulse. The change in momentum is pf − pi = pf + ( − pi ). − pi is to the left. Mentally put pf
G
and − pi 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
( px )i = mB (vBx )i + mG (vGx )i = m(3.0 m/s ) + m(−3.0 m/s ) = 0 kg ⋅ m/s
( p y )i = 0 kg ⋅ m/s
since the blue ball is initially at rest. The total initial momentum is zero.
The final momentum of the system must be zero also. This gives
( px )f = mR (vRx )f + mB (vBx )f + mG (vGx )f = m(vBx )f + m(vGx )f = 0 kg ⋅ m/s
( p y )f = mR (vRy )f + mB (vBy )f + mG (vGy )f = m(vBy )f + m(vRy )f = 0 kg ⋅ m/s
Solving for the momentum of the blue ball,
(vBx )f = −(vGx )f
(vBy )f = −(vRy )f
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.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Momentum
9-7
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 the Ryan-merry-go-round system is stationary after he jumps on the final total angular
momentum of the system is zero. By conservation of angular momentum (because it “rotates freely”), the
total angular momentum of the system must have also been zero before the jump. Therefore the angular
momentum of Ryan before the jump must have been the same magnitude (and opposite direction) as the
angular momentum of the merry-go-round. The angular momentum of the merry-go-round was
L = I ω = (400 kg ⋅ m 2 )(2.0 rad/s) = 800 kg ⋅ m 2 /s. Ryan’s angular momentum around the same axis is
L
800 kg ⋅ m 2 /s
=
= 5.0 m/s
mr (80 kg)(2.0 m)
So the answer is C.
Assess: Ryan has angular momentum about a certain axis (such as the axis of the merry-go-round) even while
running in a straight line.
L = mvr ⇒ v =
Q9.27. Reason: The two-disk system is isolated, so the angular momentum is conserved: Lf = Li . The friction force
is not an external force on the system, but within the system. The moment of inertia of the identical disks is the same:
call it I.
Iω1 + I ω2 = I (ω1 + ω2 ) = I f ωf
Where I f is the combined moment of inertia of the disks stuck together; it equals twice the moment of inertia of one
disk.
I (ω1 + ω2 ) = 2 I ωf ⇒ ωf = 12 (ω1 + ω2 ) = 12 (30 rpm + 20 rpm) = 25 rpm
So the answer is C.
Assess: Intuition would have guided us to this answer. The faster disk speeds up the slower one and the slower one
slows down the faster one.
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,
⎛ 1500 kg ⎞
mcar
v car = ⎜
pcar = pbicycle ⇒ mcar v car = mbicyclevbicycle ⇒ vbicycle =
⎟ (1.0 m/s) = 15 m/s
mbicycle
⎝ 100 kg ⎠
Assess: This is a reasonable speed. 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
Δ p = pf − pi = pf = mvf = (0.057 kg)(45 m/s) = 2.6 kg ⋅ m/s
The change in momentum of the ball is equal to the impulse on the ball, so J = 2.6 kg ⋅ m/s.
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
G
G
Δp
Favg =
Δt
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9-8
Chapter 9
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.
G
G
G
G G
G
Δ p mvf − mvi m(vf − vi ) (0.12 kg)(0 m/s − 7.5 m/s)
Favg =
=
=
=
= − 6.0 N
Δt
Δt
Δt
0.15 s
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: To get time in seconds, we note that 1 ms = 10−3 s.
Solve: The impulse as defined through Equation 9.1 is
1
( Fmax )(8 ms) ⇒ Fmax = 1500 N
2
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
J x = Area under the Fx (t ) curve between ti and tf ⇒ 6.0 N ⋅ s =
the change in momentum should also give the same answer, Fmax = 1500 N.
P9.5.
Solve:
Prepare: We use the equation J = Δ p.
The initial momentum of sled and rider is ( px )i = m(vx )i = (80 kg)(4.0 m/s) = 320 kg ⋅ m/s and the final
momentum of sled and rider is ( px )f = m(vx )f = (80 kg)(3.0 m/s) = 240 kg ⋅ m/s. So the impulse is given by
J x = ( px )f − ( px )i = 320 kg ⋅ m/s − 240 kg ⋅ m/s = 80 kg ⋅ m/s
Assess: This is a reasonable impulse. It could result, for example, from a force of around nine pounds for two
seconds.
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,
( p y )f − ( p y )i = J y ⇒ m(v y )f − m(v y )i = −mg Δ t ⇒ Δ t =
( v y ) i − (v y ) f
g
=
− 5.5 m/s − (− 10.4 m/s)
= 0.50 s
9.8 m/s 2
2
Assess: Acceleration due to gravity is 9.8 m/s 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: 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.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Momentum
9-9
Solve: (a) Using the equations
( px )f = ( px )i + Jx
J x = area under the force curve ⇒ (vx )f = (1.0 m/s) +
= (1.0 m/s) +
1
(area under the force curve)
2.0 kg
1
(1.0 N ⋅ s) = 1.5 m/s
2.0 kg
The force is to the right.
(b) Likewise,
⎛ 1 ⎞
⎛ 1 ⎞
(vx )f = (1.0 m/s) + ⎜
⎟ (area under the force curve) = (1.0 m/s) + ⎜
⎟ (− 1.0 N ⋅ s) = 0.5 m/s
⎝ 2.0 kg ⎠
⎝ 2.0 kg ⎠
This force is also to the right.
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.
The graph shows that Fx is positive, so the force acts to the right.
Solve: Using the impulse-momentum theorem ( pf ) x = ( pi ) x + J x,
(0.06 kg)(32 m/s) = (0.06 kg)( −32 m/s) + area under force graph
Now,
1
1
Fmax (0.002 s) + Fmax (0.002 s) + Fmax (0.002 s) = (0.004 s) Fmax
2
2
(0.06 kg)(32 m/s) + (0.06 kg)(32 m/s)
⇒ Fmax =
= 960 N
0.004 s
Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as
G
G
Δp
Favg =
Δt
area under force curve =
P9.9.
Solve:
This allows us to find the force on the child and sled.
( Fx )ave =
Δ px m(vx )f − m(vx )i m((vx )f − (vx )i ) (35 kg)(0 m/s − 1.5 m/s)
=
=
=
= − 105 N ≈ − 110 N
0.50 s
Δt
Δt
Δt
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 110 N.
Assess: This result is neither too large nor too small. In some collision problems Δ t is quite a bit shorter and so the
force is correspondingly larger.
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9-10
Chapter 9
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 ( px )i = + mvx . After the
collision, the puck is moving with the same velocity, but in the opposite direction, so its final momentum
is ( px )f = − mvx . Using Equation 9.8 we obtain
J x = ( px )f − ( px )i = −mvx − (+ mvx ) = − 2mvx
Solving this equation for m,
m=
Jx
−4.0 kg ⋅ m/s
=
= 0.17 kg
− 2vx − 2(12.0 m/s)
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.
P9.11. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as
G
G
Δp
Favg =
Δt
Solve:
This allows us to find the average force on the cars.
(a) Water barrels ( Fx )avg =
Δ px m(vx )f − m(vx )i (m((vx )f − (vx )i ) (1400 kg)(0 m/s − 20 m/s)
=
=
=
= − 19 000 N
Δt
Δt
Δt
1.5 s
(b)
Concrete barrier ( Fx )avg =
Δpx m(vx )f − m(vx )i m((vx )f − (vx )i ) (1400 kg)(0 m/s − 20 m/s)
=
=
=
= − 280 000 N
0.10 s
Δt
Δt
Δt
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 (vx )i = +15.0 m/s. After the
collision, the velocity of the ball is (vx )f = − 20.0 m/s. The impulse on the ball is given by Equation 9.8:
Jx = ( px )f − ( px )i = (0.145 kg)(− 20.0 m/s) − (0.145 kg)(15.0 m/s) = − 5.08 kg ⋅ m /s
The magnitude is 5.08 kg ⋅ m /s.
(b) Given the impulse and duration of the collision, Equation 9.2 gives the average force on the ball.
J
− 5.08 kg ⋅ m/s
( Fx )avg = x =
= −3400 N
Δt
0.0015 s
The magnitude is 3400 N.
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:
( Px )i = ( Px )f
( p1x )i + ( p2 x )i = ( p1x )f + ( p2 x )f
m1 (v1x )i + m2 (v2 x )i = m1 (v1x )f + m2 (v2 x )f
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Momentum
9-11
We want to know (v2 x )f so we solve for it. Also recall that (v2 x )i = 0 m/s, so the middle term in the following
numerator drops out. The small cart recoils, which means its velocity after the collision is negative.
m (v ) + m2 (v2 x )i − m1 (v1x )f (0.100 kg)(1.20 m/s) − (0.100 kg)( −0.850 m/s)
(v2 x )f = 1 1x i
=
= 0.205 m/s
m2
1.00 kg
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.
Writing the momentum of the man and gun as mM (vMx ), and the momentum of the bullet as mB (vBx ), the
momentum conservation equation is
mM (vMx )f + mB (vBx )f = mM (vMx )i + mB (vBx )i = 0
Solving for (vMx )f ,
⎛m ⎞
⎛ 0.01 kg ⎞
−2
(vMx )f = − ⎜ B ⎟ (vBx )f = − ⎜
⎟ (500 m/s) = − 7.1 × 10 m/s
70
kg
m
⎝
⎠
⎝ M⎠
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
mb (vb )i + mbl (vbl )i = mb (vb )f + mbl (vbl )f ⇒
(3.0 × 10−3 kg)(500 m/s) + (2.7 kg)(0 m/s) = (3.0 × 10−3 kg)(220 m/s) + (2.7 kg)(vbl )f .
We can solve for the final speed of the block: (vbl )f = 0.31 m/s .
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9-12
Chapter 9
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.
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
m1 (v1x )f + m2 (v2 x )f = m1 (v1x )i + m2 (v2 x )i = 0
Solving for (v1x )f ,
⎛m ⎞
⎛ 2.3 kg ⎞
(v1x )f = − ⎜ 2 ⎟ (v2 x )f = − ⎜
⎟ (6.0 m/s) = − 2.6 m/s
m
⎝ 5.3 kg ⎠
⎝ 1⎠
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.
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Momentum
9-13
Solve: There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This
means ( px )f = ( px )i . Hence,
(10 000 kg + 4000 kg)(vx )f = (10 000 kg)(2.0 m/s) + (4000 kg)(0.0 m/s) ⇒ (vx )f = 1.4 m/s
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.
Solve: Conservation of momentum is ( px )f = ( px )i . Hence,
(5000 kg + mrain )(20.0 m/s) = (5000 kg )(22.0 m/s) ⇒
(5000 kg )(22.0 m/s)
− 5000 kg = 500 kg
20.0 m/s
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.
mrain =
P9.19. Prepare: The ice is frictionless. Considering the hunter and bullet as the system, momentum will be
conserved because there are no external forces acting in the horizontal direction. Assume the hunter shoots to our
right.
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.
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9-14
Chapter 9
Writing the momentum of the man and gun as mM (vMx ), and the momentum of the bullet as mB (vBx ), the
momentum conservation equation is
mM (vMx )f + mB (vBx )f = mM (vMx )i + mB (vBx )i = 0
Solving for (vMx )f ,
⎛m ⎞
⎛ 0.042 kg ⎞
(vMx )f = − ⎜ B ⎟ (vBx )f = − ⎜
⎟ (620 m/s) = − 0.47 m/s
⎝ 55 kg ⎠
⎝ mM ⎠
The man moves to the left with a speed of 0.47 m/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.
P9.20. 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.
Writing the momentum of dog as mD (vDx ), and the momentum of the canoe as mC (vCx ), the momentum
conservation equation is
mD(vDx )f + mC (vCx )f = mD(vDx )i + mC (vCx )i = 0
Solving for mC,
mD(vDx )f
(9.5 kg)(0.50 m/s)
=−
= 32 kg
(vCx )f
( − 0.15 m/s)
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.
mC = −
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.
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Momentum
9-15
Solve: The conservation of momentum equation pf x = pi x is
(m1 + m2 )(vx )f = m1 (v1x )i + m2 (v2 x )f
⇒ (300 g + 10 g)(vx )f = (300 g)(6.0 m/s) + (10 g)( −30 m/s) ⇒ (vx )f = 4.8 m/s
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 (v2 x )i 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.
After the player catches the ball, they move together with a common final velocity. Writing the mass of the player as
mP , and the momentum of the ball as mB (vBx ), the momentum conservation equation is
(mP + mB )(vx )f = mB (vBx )i
Solving for (vx )f ,
mB (vBx )i (0.140 kg)(28 m/s)
=
= 5.5 cm/s
71 kg + 0.140 kg
mP + mB
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.
(v x ) f =
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:
( Px )i = ( Px )f
( pSBx )i + ( pCTNx )i = ( pBOTHx )f
mSB (vSBx )i + mCTN (vCTNx )i = (mSB + mCTN )(vBOTHx )f
We want to know (vSBx )i so we solve for it. Also recall that (vCTNx )i = 0 m/s so the last term in the following
numerator drops out.
(mSB + mCTN )(vBOTHx )f − mCTN (vCTNx )i (0.0030 kg + 0.020 kg)(0.30 m/s)
=
= 2.3 m/s
mSB
0.0030 kg
Assess: The answer of 2.3 m/s is certainly within the capability of an expert spitballer.
(vSBx )i =
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9-16
Chapter 9
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
(mC + mVW )(vx )f = mC (vC x )i + mVW (vVW x )i ⇒ 0 kg ⋅ mph
= (2000 kg)(1.0 mph) + (1000 kg)(vVW x )i ⇒ (vVW x )i = − 2.0 mph
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
m1 (v1x )i + m2 (v2x )i = (m1 + m2 )(vx )f ⇒
(2.0 kg)(1.0 m/s) + m2 (4.0 m/s) = (2.0 kg + m2 )(2.0 m/s).
We can solve this last equation for the second mass and obtain m2 = 1.0 kg.
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: Ignore the force of gravity since we are examining such a short time interval. Then we can use the
law of conservation of momentum.
Solve: All motion takes place in the y direction. The sum of the momenta of the two people before the grab equals
the momentum of the couple after the grab so we write
mE (vE )i + mD (vD )i = (mE + mD )vf ⇒
(36 kg)(0.0 m/s) + (47 kg)(4.1 m/s) = (36 kg + 47 kg)vf ⇒
vf =
(47 kg)(4.1 m/s)
= 2.3 m/s
36 kg + 47 kg
Assess: They would be moving upward still, right after the grab.
P9.27. Prepare: Ignore the force of gravity since we are examining such a short time interval. Then we can use the
law of conservation of momentum in both the x and y directions. Just before it gets hit the coin has zero velocity. The
bullet will deflect somewhat downward. Use subscript b for the bullet and c for the coin.
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Momentum
9-17
Solve: Using both directions, we’ll get two equations in two unknowns. If we are clever we can divide the two
equations to get what we want. First start with the x-direction; only the bullet moves in the x-direction.
Σ ( p x )i = Σ ( p x )i
mb (vb )i cosθ1 = mb (vb )f cosθ 2 ⇒
mb (vb )i cosθ1
mb
Now apply conservation of momentum in the y-direction with the initial momentum of the coin being zero.
Σ ( p y ) i = Σ ( p y )i
(vb )f cosθ 2 =
mb (vb )i sin θ1 = mb (vb )f sin θ 2 + mcvc ⇒
mb (vb )i sin θ1 − mcvc
mb
Now the clever step: divide the equations so one of the unknowns (the one we don’t want) cancels out.
mb (vb )i sin θ1 − mcvc
(vb )f sin θ 2
mb
=
mb (vb )i cosθ1
(vb )f cosθ 2
mb
The final velocity of the bullet cancels on the left side, as does the mass of the bullet in each denominator on the right.
m (v ) sin θ1 − mcvc
mcvc
tan θ 2 = b b i
= tan θ1 −
mb (vb )i cosθ1
mb (vb )i cosθ1
Now plug in all the values.
(12 g)(120 m/s)
tan θ 2 = tan 45° −
⇒ θ 2 = 37°
(15 g)(550 m/s)cos 45 °
The bullet ricochets away from the collision at 37° below the horizontal.
Assess: We expected the angle to be between 0 and 45 degrees below the horizontal.
(vb )f sin θ 2 =
P9.28. Prepare: This problem deals with the conservation of momentum in two dimensions in an inelastic
collision. We will thus use Equation 9.14.
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9-18
Chapter 9
G
G
Solve: The conservation of momentum equation pbefore = pafter is
m1 (v1x )i + m2 (v2x )i = (m1 + m2 )(vx )f
m1 (v1 y )i + m2 (v2 y )i = (m1 + m2 )(v y )f
Substituting in the given values,
(.02 kg)(3.0 m/s) + 0 kg ⋅ m/s = (.02 kg + .03 kg)vf cos θ
0 kg m/s + (.03 kg)(2.0 m/s) = (.02 kg + .03 kg)vf sin θ ⇒ vf cos θ = 1.2 m/s
vf sin θ = 1.2 m/s ⇒ vf = (1.2 m/s) 2 + (1.2 m/s) 2 = 1.7 m/s
θ = tan −1
vy
vx
= tan −1 (1) = 45 °
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.
P9.29. Prepare: We assume that the momentum is conserved in the collision.
Solve: The conservation of momentum Equation 9.14 yields
( p1x )f + ( p2 x )f = ( p1x )i + ( p2 x )i ⇒ ( p1x )f + 0 kg ⋅ m/s = 2 kg ⋅ m/s − 4 kg ⋅ m/s ⇒ ( p1x )f = − 2 kg ⋅ m/s
( p1 y )f + ( p2 y )f = ( p1 y )i + ( p2 y )i ⇒ ( p1 y )f − 1 kg ⋅ m/s = 2 kg ⋅ m/s + 1 kg ⋅ m/s ⇒ ( p1 y )f = 4 kg ⋅ m/s
The final momentum vector of particle 1 that has the above components is shown below.
P9.30. 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.
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Momentum
9-19
Solve: The x-component of the final momentum is
( px )f = ( px )i = m1 (v1x )i + m2 (v2 x )i = (0.020 kg)(2.0 m/s) − (0.030 kg)(1.0 m/s) cos 30 ° = 0.0140 kg ⋅ m/s
The y-component of the final momentum is
( px )f = ( px )i = m1 (v1 y )i + m2 (v2 y )i = (0.02 kg)(0 m/s) − (0.03 kg)(1.0 m/s) sin 30° = − 0.0150 kg ⋅ m/s
⇒ pf = (0.014 kg ⋅ m/s) 2 + (− 0.015 kg ⋅ m/s) 2 = 0.0205 kg ⋅ m/s
Since pf = (m1 + m2 )vf = 0.0205 kg ⋅ m/s, the final speed is
vf =
0.0205 kg ⋅ m/s
= 0.41 m/s
(0.02 + 0.03) kg
and the direction is
|( p y )f |
0.015
= tan −1
= 47 ° south of east
( px )f
0.014
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.
θ = tan −1
P9.31. 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
+ m1 (v1x )f + m3 (v3f )cos θ = 0 kg m/s ⇒ (v3f )cos θ =
−m1 (vfx )1 −m(− 20 m/s)
=
= 10 m/s
m3
2m
From ( p y )f = ( p y )i , we get
+ m2 (v2y )f + m3 (v3f )sin θ = 0 kg m/s ⇒ (v3f )sin θ =
− m2 (vfy ) 2
⇒ (v3f ) = (10 m/s) 2 + (10 m/s) 2 = 14 m/s
m3
=
−m(− 20 m/s)
= 10 m/s
2m
θ = tan −1 (1) = 45°
Assess: The obtained speed of the third piece is of similar order of magnitude as the other two pieces, which is
physically reasonable.
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9-20
Chapter 9
P9.32. 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: v = 2π r /T .
Solve:
2π 2
2π
m2
L = mvr = m
r = 7.4 × 1022 kg
(3.8 × 108 m) 2 = 2.8 × 1034 kg
6
T
2.36 × 10 sec
s
Assess: Note that the rotation of moon or the earth about their respective axes does not come into these calculations.
P9.33. Prepare: We will model the mother as a particle with m = 47 kg, v = 4.2 m/s, and r = 2.6 m.
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 7.21 reminds us that I for such a situation is mr 2 , 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:
L = I ω = (mr 2 )(v/r ) = rmv
Solve:
L = rmv = (2.6 m)(47 kg)(4.2 m/s) = 510 kg ⋅ m 2/s
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 = mvr is straightforward and we checked our multiplication twice, so it is
probably correct.
P9.34. Prepare: The bar is a rotating rigid body. From Table 7.1, the moment of inertial of a rod about its center is
I = 121 ML2 . We will work with SI units and use Equation 9.20 to find the angular momentum.
Solve: The angular velocity ω = 120 rpm = (120)(2π )/60 rad/s = 4π rad/s. The angular momentum is
⎛ 1 ⎞
L = I ω = ⎜ ⎟ (0.50 kg)(2.0 m) 2 (4π rad/s) = 2.1 kg ⋅ m 2/s
⎝ 12 ⎠
Assess: For objects of size ~1 m and mass ~1 kg spinning at ~1 rad/s, an answer on order of ~1 kg· m 2 /s is reasonable.
Prepare:
The disk is a rotating rigid body. The angular velocity ω is 600 rpm =
600 × 2 π /60 rad/s = 20 π rad/s. rad/s. From Table 7.1, the moment of inertial of the disk about its center is (1/2)
MR 2 , which can be used with L = I ω to find the angular momentum.
Solve:
1
1
I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 × 10−4 kg ⋅ m 2
2
2
Thus, L = I ω = (4.0 × 10−4 ⋅ kg m 2 )(20π rad/s) = 0.025 kg m 2 /s.
P9.35.
Assess: For objects of size ~1 m and mass ~1 kg spinning at ~1 rad/s, an answer on order of ~1 kg· m 2 /s is
reasonable.
P9.36. 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 9.22 applies.
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Momentum
9-21
Solve: Using Equation 9.22,
I iωi = I f ωf
Solving for ωf / ωi ,
ωf I i 14 kg ⋅ m 2
= =
= 3.5
ωi I f 4.0 kg ⋅ m 2
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.
P9.37. Prepare: We neglect any small frictional torque the ice may exert on the skater and apply the law of
conservation of angular momentum.
Li = Lf
I iωi = I f ωf
Even though the data for ωi = 5.0 rev/s is not in SI units, it’s okay because we are asked for the answer in the same
units. We are also given I i = 0.80 kg ⋅ m 2 and I f = 3.2 kg ⋅ m 2 .
Solve:
I iωi (0.80 kg ⋅ m 2 )(5.0 rev/s)
=
= 1.25 rev/s ≈ 1.3 rev/s
If
3.2 kg ⋅ m 2
Assess: I increased by a factor of 4, so we expect ω to decrease by a factor of 4.
ωf =
P9.38. Prepare: 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
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.
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9-22
Chapter 9
P9.39. Prepare: 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 Δ t.
Solve: Using the impulse-momentum theorem ( px )f − ( px )i = J x ,
(0.60 kg)(3 m/s) − (0.60 kg)( −3 m/s) = area under force curve = 12 (36 N)(Δ t ) ⇒ Δ t = 0.2 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/s 2 . 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 0.2 s found in the previous
figure.
P9.40. Prepare: 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 impulsemomentum theorem.
Solve: (a) The impulse is
J x = 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 N ⋅ s. That is, the momentum or velocity increases
as long as J x increases. When J x increases no more, the speed will be a maximum. This happens at t = 30 s. At
this time,
m(vx )f − m(vx )i = + 15 000 N ⋅ s ⇒ (425 kg)(vx )f = (425 kg)(75 m/s) + 15 000 N ⋅ s ⇒ (vx )f = 110 m/s
P9.41. 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.
Solve: To find the ball’s velocity just before and after it hits the floor:
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2( −9.8 m/s 2 )(0 − 2.0 m) ⇒ v1 y = −6.261 m/s
v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2(− 9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 y = 5.422 m/s
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Momentum
9-23
The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
J y = area under the force curve = Δp y = mv2 y − mv1 y
or
1
Fmax Δt = mv2 y − mv1 y
2
so that
2m(v2 y − v1 y )
2(0.20 kg)(5.42 m/s − (− 6.26 m/s))
=
= 930 N
Δt
5.0 × 10−3 s
Assess: A force of 930 N exerted by the floor is typical of such collisions.
Fmax =
P9.42. 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
(1000 N)(4 × 10−3 s)
= 2 kg ⋅ m/s
Jy =
2
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
(vf ) 2 = (vi ) 2 + 2a y Δ 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,
vf = − 2 g Δ y = − 2(9.80 m/s 2 )(− 2.0 m) = 6.26 m/s
The magnitude of the momentum of the ball just before it hits the ground is mvf = (0.2 kg)(6.26 m/s)=
1.25 kg ⋅ m/s. This is the initial momentum of the ball just before the collision, ( p y )i = −1.25 kg ⋅ m/s. 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,
( p y )f = J y + ( p y )i = 2 kg ⋅ m/s+(−1.25 kg ⋅ m/s) = 0.75 kg ⋅ m/s
We can now find the velocity of the ball right after it bounces up from the floor,
(p )
(v y )f = y f = 3.75 m/s
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
−(v y )i 2
−(3.75 m/s) 2
Δy =
=
= 0.72 m
2a y
2(− 9.80 m/s 2 )
Assess: This height seems reasonable for a ball dropped from 2.0 m.
P9.43. Prepare: We combine Equation 9.1 J = Favg Δt with the impulse-momentum theorem in the y-direction
J y = Δp y . See Example 9.1. This tells us that Δp y 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/s 2 ) = 61.2 kg.
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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 t = 0.475 s.
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
A = 12 × height × base
Abig = 12 (1800 N)(0.275 s) = 247.5 N ⋅ s
Asmall = 12 (600 N)(0.025 s) = 7.5 N ⋅ s
The total area (with the triangle above the axis positive and the triangle below the axis negative) is
247.5 N ⋅ s − 7.5 N ⋅ s = 240 N ⋅ s; this is the vertical impulse on the woman, and is also equal to her change in
vertical momentum. Since Δ p y = mΔ v y we simply divide by her mass to find her change in velocity:
Δv y = Δp y / m = (240 N ⋅ s)/(61.2 kg) = 3.9 m/s
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: (v y )f ≈ 4 m/s. The answer of ≈ 4 m/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.
P9.44. Prepare: We will use the particle model for the sled and the impulse-momentum theorem. Note that the
force of kinetic friction f k imparts a negative impulse to the sled. The force f k = μk n where μ k is the coefficient of
G G
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 Δ px = J x, we have
( px )f − ( px )i = − f k Δt ⇒ m(vx )f − m(vx )i = − μ k nΔ t = − μk mg Δ t
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Momentum
9-25
Thus,
1
1
((vx )i − (vx )f ) =
(8.0 m/s − 5.0 m/s) = 1.2 s
(0.25)(9.8 m/s 2 )
μk g
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.
Δt =
P9.45. 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 = Δ p. Since the direction of the ball changes, we need to use vector components. The
x-component of the ball’s final velocity is
(vx )f = (− 55 m/s)cos(25°) = −49.8 m/s and the y-component is (v y )f = ( −55 m/s)cos(25°) = −23.2 m/s
Solve: The initial velocity of the ball is (vx )i = 35 m/s and its initial momentum is obtained by multiplying by the
mass of the ball:
( px )i = (0.140 kg)(35 m/s) = 4.90 kg ⋅ m/s
(p y )i = 0 kg ⋅ m/s
The initial final momentum is the final velocity of the ball times its mass:
( px )f = (0.140 kg)(− 49.8 m/s) = − 6.97 kg ⋅ m/s
( p y )f = (0.140 kg)(23.2 m/s) = 3.25 kg ⋅ m/s
Finally, the impulse on the ball equals the change in the ball’s momentum:
J x = ( px )f − ( px )i = −11.9 kg ⋅ m/s
J y = ( p y )f − ( p y )i = 3.25 kg ⋅ m/s
The magnitude of the impulse can be obtained from the Pythagorean theorem: J = 12 kg ⋅ m/s and we can find the
angle, θ , above the horizontal using inverse tangent: θ = tan −1 (3.25/11.9) = 15 °. 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.
P9.46. 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.
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9-26
Chapter 9
Writing the momentum of the squid as mS (vSx ), and the momentum of the water as mW (vWx ), the momentum
conservation equation is
mS (vSx )f + mW (vWx )f = mS (vSx )i + mW (vWx )i = 0
Solving for (vWx )f ,
⎛m ⎞
⎛ 1.5 kg ⎞
(vWx )f = − ⎜ S ⎟ (vSx )f = − ⎜
⎟ (3.0 m/s) = 45 m/s
⎝ 0.10 kg ⎠
⎝ mW ⎠
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.
P9.47. Prepare: We are asked for the impulse given to the pollen grain; impulse is defined in Equation 9.1:
J = Favg Δ t. We are given that Δt = 3.0 × 10−4 s, 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 Favg? Given
m = 1.0 × 10−10 kg and a = 2.5 × 104 m/s 2 guides us to use Newton’s second law to find Favg (assuming a is constant
over Δ t ). The Favg in the impulse equation is the same as the Fnet in Newton’s law because we are ignoring any
other forces on the grain.
Favg = Fnet = ma = (1.0 × 10−10 kg)(2.5 × 104 m/s 2 ) = 2.5 × 10−6 N
Solve:
J = Favg Δt = (2.5 × 10−6 N)(3.0 × 10−4 s) = 7.5 × 10−10 N ⋅ s = 7.5 × 10−10 kg ⋅ m/s
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.
P9.48. Prepare: Given a = 2.5 × 104 m/s 2 and Δt = 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, (vx)f = (vx)i + ax Δ 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 monegrain = 1.0 × 10−10 kg so for 1000 ejected pollen grains mp = 1.0 × 10−17 kg. We are also given
mb = 0.0050 kg.
Solve: (a)
(vx )f + (vx )i + ax Δt = 0 + (2.5 × 10 4 m/s 2 )(3.0 × 10−4 s) = 7.5 m/s
(b) We employ the conservation of momentum. Since the collision is inelastic the final momentum will be
pf = (mP + mb )vf , where vf is the answer we seek. That the bee is hovering indicates (vbx )i = 0 m/s.
G G
Pf = Pi
(mp + mb )vf = mp (vpx )i + mb (vbx )i
vf =
mp (vpx )i + mb (vbx )
mp / mb
=
(1.0 × 10−7 kg)(7.5 m/s) + (0.0050 kg)(0 m/s)
= 1.5 × 10−4 m/s = 0.15 mm/s
1.0 × 10−7 kg + 0.0050 kg
The bee would not notice such a tiny speed.
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Momentum
9-27
P9.49. 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
mR (vRx )f + mB(vBx )f = mR (vRx )i + mB(vBx )i
(1.0 kg)(vRx )f + (0.06 kg)(40 m/s) = (1.0 kg)(10 m/s) + (0.06 kg)( − 20 m/s) ⇒ (vRx )f = 6.4 m/s
(b) The impulse on the ball is calculated from ( pBx )f = ( pBx )i + J x as follows:
(0.06 kg)(40 m/s) = (0.06 kg)( − 20 m/s) + J x ⇒ J x = 3.6 N ⋅ s = Favg Δ t
3.6 Ns
= 360 N
10 ms
Assess: Let us now compare this force with the ball’s weight wB = mB g = (0.06 kg)(9.8 m/s 2 ) = 0.588 N. Thus,
⇒ Favg =
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.
P9.50. 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
(1.0 kg + 0.02 kg)(vx )f = (0.02 kg)(30 m/s) + (1.0 kg)(0 m/s) ⇒ (vx )f = 0.588 m/s = 0.59 m/s
(b) The impulse of the ball of clay on the block is calculated as follows:
( pBx )f = ( pBx )i + ( J Bx ) ⇒ ( J Bx ) = mB (vBx )f − mB (vBx )i = (1.0 kg)(vx )f − 0 N ⋅ s = 0.588 N ⋅ s = 0.59 kg ⋅ m/s
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9-28
Chapter 9
(c) The impulse of the block on the ball of clay is calculated as follows:
( pCx )f = ( pCx )i + ( J Cx ) ⇒ ( J Cx ) = mC (vCx )f − mC (vCx )i = (0.02 kg)(0.588 m/s) − (0.02 kg)(30 m/s) = −0.59 N ⋅ s
(d) Yes, J Bx = − J Cx
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.
P9.51. Prepare: We will define our system to be Dan + skateboard. The system has nonzero initial momentum
pi x . 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
G
G
because Fext = 0 on the system.
Solve: We have mS (vSx )f + mD (vDx )f = (mS + mD )(vx )i . Hence,
(5.0 kg)(8.0 m/s) + (50 kg)(vDx )f = (5.0 kg + 50 kg)(4.0 m/s) ⇒ (vDx )f = 3.6 m/s
Assess: A speed of 3.6 m/s or 8 mph is reasonable.
P9.52. 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
pballs = mJ vJ + msvs = (1.0 kg)(4.5 m/s) + (0.50 kg)(− 1.0 m/s) = 4.0 kg ⋅ m/s
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 (− 4.0 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: Since James throws a heavier, faster ball than Sarah, it makes sense that the cart would move opposite to
James’s ball.
P9.53. 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: (vEg )f = 8.0 m/s + (vcg )f .
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Momentum
9-29
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:
0 kg ⋅ m/s = mE (vEg )f + mc (vcg )f ⇒
0 kg ⋅ m/s = (80 kg)(8.0 m/s + (vcg )f ) + (500 kg)(v cg )f .
This equation can be solved for the velocity of the cart: (vcg )f = − 1.1 m/s. 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.
P9.54. Prepare: Since the cart rolls “freely” we will ignore friction and we’ll use conservation of momentum. No
numbers are given, so we must solve this algebraically. Let the number of balls that need to fall be called n.
Solve: Use conservation of momentum in the x-direction. The clay balls have no x momentum while falling.
Σ ( p x )i = Σ ( px ) f
mvi = ( m + n4 m)vf
vf =
mvi
1
=
vi
m + n4 m 1 + n4
We want n such that vf < 13 vi .
1
vi < 13 vi ⇒ 1 + n4 > 3 ⇒ n4 > 2 ⇒ n > 8
1 + n4
If n = 8 then the new speed would be equal to 1/3 the initial speed, but we must drop 9 balls before the new speed is
less than 1/3 the initial speed.
Assess: This makes sense; 8 balls of clay have a mass twice that of the cart, so the total mass with 8 balls of clay is
3 times the original mass, so the speed would be one third of the original speed.
P9.55. 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:
mv1x + (3m)v2 x = 4mv3 x ⇒ v1x + 3v2 x = 4v3 x ⇒ (4.0 m/s) + 3(2.0 m/s) = 4v3 x ⇒ v3 x = 2.5 m/s
In the collision between the four-car train and the stationary car:
(4 m)v3 x + mv4 x = (5 m)v5 x ⇒ 4v3 x + 0 m/s = 5 v5 x ⇒ v5 x =
4v3 x
= (0.8)(2.5 m/s) = 2.0 m/s
5
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9-30
Chapter 9
Assess: The motion of railroad has to be on a level track for linear momentum to be constant. The speed of the fivecar train, as expected, is of the same order of magnitude as the other speeds.
P9.56. 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
m1 = 110 Kg
mq = 82 Kg
(vlx ) = 2.0 m/s
(vqx )i = − 3.0 m/s
Find
vf
We employ the conservation of momentum. Since the collision is inelastic (the linebacker grabs and holds onto the
quarterback) the final momentum will be Pf = ( ml + mq )vf , where vf is the answer we seek.
Solve:
G G
Pf = Pi
(ml + mq )vf = ml (vlx )i mq (vqx )i
vf =
ml (vlx )i + mq (vqx )i
ml + mq
=
(110 kg)(2.0 m/s) + (82 kg)( −3.0 m/s)
= −0.14 m/s
110 kg + 82 kg
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.”
P9.57. 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.
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Momentum
9-31
Solve: (a) The conservation of momentum equation ( px )f = ( px )i is
mA(vAx )i + mE(vEx )i = (mA + mE )(vx )f
⇒ (1.0 × 10 kg)(4 × 10 m/s) + 0 kg m/s = (1.0 × 1013 kg + 5.98 × 1024 kg)(vx )f ⇒ (vx )f = 6.7 × 10−8 m/s
13
4
(b) The speed of the earth going around the sun is
2π r 2π (1.50 × 1011 m)
vE =
=
= 3.0 × 104 m/s
T
3.15 × 107 s
Hence, (vx )f /vE = 2.2 × 10−12 = 2.2 × 10−10%.
Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
P9.58. Prepare: Assume the ice is frictionless and use conservation of momentum in the x direction. Call the 75-kg
skater skater 1, and the lighter skater call skater 2.
Known
m1 = 75 Kg
m2 = 55 Kg
Δ x1 + Δ x2 = 15 m
Find
Δ x1
Solve:
Σ( px ) i = Σ( px ) f
0 = −m1 (v1 )f + m2 (v2 )f
(v2 )f m1
=
(v1 )f m2
Now use v = Δx / Δt and also that the time intervals for both skaters is the same so Δ t cancels.
Δx2 (v2 )f m1
m
=
=
⇒ Δ x2 = 1 Δx1
Δ x1 (v1 )f m2
m2
We still have two unknowns in the equation, so we employ another equation:
Δ x1 + Δ x2 = 15 m
Substitute in our expression for Δ x2 .
Δ x1 +
m1
15 m
15 m
=
= 6.3 m
Δx1 = 15 m ⇒ Δ x1 =
m1
75 kg
m2
1 + m2 1 + 55
kg
Assess: A similar computation for Δ x2 gives 8.7 m so the total is 15 m, as expected. We expected the heavier skater
to move slower and cover less distance than the lighter skater in the same amount of time.
P9.59. 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.
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9-32
Chapter 9
Solve: The initial momentum is zero. Thus the conservation of momentum equation ( px )f = ( px )i is
mH (vHx )f + mL (vLx )f = 0 kg ⋅ m/s ⇒ (75 kg)(vHx )f + (50 kg)(vLx )f = 0 kg ⋅ m/s
Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find (vHx )f = 30 m/20 s =
1.5 m/s. Thus,
(75 kg)(1.5 m/s) + (50 kg)(vLx )f = 0 kg ⋅ m/s ⇒ (vLx )f = − 2.25 m/s
Thus, the time for the lighter skater to reach the edge is
30 m
30 m
=
= 13 s
(vLx )f 2.25 m/s
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.
P9.60. Prepare: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m 2 (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:
m1 (v1x )f + m2 (v2x )f = m1 (v1x )i + m2 (v2x )i ⇒ m1 (v1 )f cos θ + 0 kg ⋅ m/s = m1 (v1x )i + m2 (v2x )i
⇒ (v1 )f cos θ = (v1x )i + (v2x )i = 2.0 m/s − 1.0 m/s = 1.0 m/s
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Momentum
9-33
The equation ( p y )f = ( p y )i yields:
m1 (v1 )f sin θ + m2 (v2y )f = 0 kg ⋅ m/s ⇒ (v1 )f sin θ = −(v2y )f = −1.41 m/s
⇒ (v1 )f = (1.0 m/s) 2 + ( −1.41 m/s) 2 = 1.7 m/s
⎛ 1.41 m/s ⎞
⎟ = 55 °
⎝ 1.0 m/s ⎠
θ = tan −1 ⎜
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.
P9.61. 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
(mB + mW )(vCx )f = mB (vBx )i + mW (vWx )i
1
(vBx )i
1001
Using − f k = − μ k n = − μ k (mB + mW ) g = ( mB + mW )ax ⇒ ax = − μ k g . Note that the negative sign appears in front of f k
⇒ (0.01 kg + 10 kg)(vCx )f = (0.01 kg)(vBx )i + (10.0 kg)(0 m/s) ⇒ (vCx )f =
because the force of friction points in the –x direction.
Using the kinematics equation (vx )f2 = (vx )i2 + 2ax ( xf − xi ),
2
2
⎛ 1 ⎞
⎛ 1 ⎞
2
2 2
2
0 = (vCx )f2 − 2 μ k g ( xf − xi ) ⇒ 0 m 2/s 2 = ⎜
⎟ (vBx )i − 2 μ k g xf ⇒ 0 m /s = ⎜ 1001 ⎟ (vBx )i − 2μ k g xf
1001
⎝
⎠
⎝
⎠
⇒ (vBx )i = 1001 2 μ k g xf = 1001 2(0.2)(9.8 m/s 2)(0.05 m) = 440 m/s
Assess: The bullet’s speed is reasonable (≈1000 mph).
P9.62. Prepare: Use conservation of momentum in the y direction. Because it is hovering, the initial y velocity of
the mosquito is 0 m/s.
Solve: (a)
Σ( p y ) i = Σ( p y ) f
md (vd )i = (md + mm )vf
md (vd )i (40mm )(8.2 m/s) 40
(8.2 m/s) = 8.0 m/s
=
=
md + mm
40mm + mm
41
(b) Use the definition of acceleration.
Δ v 8.0 m/s
⎛ 1g
⎞
a=
=
= 1000 m/s 2 = (1000 m/s 2 ) ⎜
= 100 g
2 ⎟
Δ t 8.0 ms
9.8
m/s
⎝
⎠
Assess: This is an impressive acceleration to survive.
vf =
P9.63. Prepare: This is a straightforward problem if done one step at a time. First, use conservation of momentum
in the x direction to find the horizontal speed with which the block leaves the table. Then use kinematic equations to
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9-34
Chapter 9
find how far it will go as a projectile while falling 75 cm. Ignore air resistance. Use lower case b for the bullet and
upper case B for the block.
Solve: For the inelastic collision in the x direction:
mb (vb )i = ( mb + mB )vf
mb (vb )i (0.015 kg)(610 m/s)
=
= 2.28 m/s
mb + mB
0.015 kg + 4.0 kg
Use kinematic equations on the projectile motion. First determine how long it takes to fall 75 cm from rest.
2Δ y
2(0.75 m)
Δ y = 12 g (Δt ) 2 ⇒ Δt =
=
= 0.391 s
g
9.8 m/s 2
vf =
Now compute how far the projectile will travel horizontally in that time.
Δ x = vΔ t = (2.28 m/s)(0.391 s) = 0.89 m
Assess: This seems like a reasonable distance to go horizontally while falling 0.75 m.
P9.64. 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
(v y)1 = (v y )0 + a y (t1 − t0 ) = 0 m/s + (10 m/s 2)(2 s − 0 s) = 20 m/s
y1 = y0 + (v y)0 (t1 − t0 ) +
1
1
a y (t1 − t0 ) 2 = 0 m + 0 m + (10 m/s 2)(2 s) 2 = 20 m
2
2
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Momentum
9-35
At the explosion the equation ( p y )f = ( p y )i is
mL (vLy) 2 + mH (vHy) 2 = ( mL + mH )(v y)1 ⇒ (500 kg)(vLy ) 2 + (1000 kg)(vHy)2 = (1500 kg)(20 m/s)
To find (vHy ) 2 we must first find (vLy ) 2 , the velocity after the explosion of the upper section. Using kinematics,
(vLy)32 = (vLy) 22 + 2( −9.8 m/s 2 )(( yL)3 − ( yL) 2 ) ⇒ (vLy) 2 = 2(9.8 m/s 2)(530 m − 20 m) = 99.98 m/s
Now, going back to the momentum conservation equation we get
(500 kg)(99.98 m/s) + (1000 kg)(vHy) 2 = (1500 kg)(20 m/s) ⇒ (vHy) 2 = −20 m/s
The negative sign indicates downward motion.
P9.65. 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
mL (vLx )1 + mA (vAx )1 = mL (vLx )0 + mA (vAx )0
⇒ (0.01 kg)(vLx )1 + (0.500 kg)(6 m/s) = (0.01 kg)(400 m/s) + 0 kg m/s ⇒ (vLx)1 = 100 m/s
The bullet emerges from the first block at 100 m/s. For the second collision the equation ( px )f = ( px )i is
(mL + mB)(vx ) 2 = mL (vLx )1 ⇒ (0.01 kg + 0.5 kg)(vx ) 2 = (0.01 kg)(100 m/s) ⇒ (vx ) 2 = 2.0 m/s
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.
P9.66. 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
211 m/s relative to the cannon can be written (vBx )f = (vCx )f + 211 m/s. That is, the ball’s speed is 211 m/s more than
the cannon’s speed.
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9-36
Chapter 9
Solve: The initial momentum is zero, so the conservation of momentum equation ( px )f = ( px )i is
( px )f = mC (vCx )f + mB (vBx)f = mC (vCx )f + mB ((vCx)f + 211 m/s) = 0 kg m/s
mB
10.0 kg
(2 11 m/s) = −
(2 11 m/s) = − 4.18 m/s
mC + mB
505 kg
That is, the cannon recoils to the left (negative sign) at 4.18 m/s. Thus the cannonball’s speed relative to the ground is
(vBx )f = − 4.18 m/s + 211 m/s = 207 m/s.
⇒ (vCx )f = −
Assess: An expected relatively small recoil speed is essentially due to the large mass of the cannon.
P9.67. 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, (vLc )f , and not her final
speed relative to the water, (vLw )f , we will need to use the relative velocity formula. We are given the masses of
Laura and the canoe: mL = 35 kg and mc = 55 kg as well as the final speed of Laura relative to the canoe:
(vLc )f = 1.5 m/s.
Solve: Before she jumps, Laura and the canoe are stationary, so their total momentum is zero. Afterward, the canoe
has velocity (vcw )f and Laura has velocity:
(vLw )f = (vLc )f + (vcw )f = 1.5 m/s + (vcw )f
Now we can use the law of conservation of momentum:
0 kg ⋅ m/s = (mL )(vLw )f + ( mc )(vcw )f ⇒
0 kg ⋅ m/s = (35 kg)[1.5 m/s + (vcw )f ] + (55 kg)(vcw )f
This equation can be solved to obtain (vcw )f = − 0.58 m/s.
Assess: For perspective, Laura’s velocity relative to the water is given by (vLw )f = 1.5 m/s − 0.58 m/s = 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.
P9.68. Prepare: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector
G
quantity, so the direction of the initial velocity vector v1 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 m3 = mtotal − m1 − m2 = 7.0 × 105 kg.
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Momentum
Solve:
9-37
To conserve momentum along the x-axis, we require
[( px )i = mtotal (vx )i ] = [( px )f = ( p1x )f + ( p2 x )f + ( p3 x )f = m1 (v1x )f + m2 (v2 x )f + ( p3 x )f ] ⇒
( p3 x )f = mtotal (vx )i − m1 (v1x )f − m2 (v2 x )f = + 1.02 × 1013 kg ⋅m/s
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:
(v3 x )f =
( p3 x )f 1.02 × 1013 kg ⋅ m/s
=
= 1.5 × 107 m/s
m3
7.0 × 105 kg
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.
P9.69. Prepare: Use conservation of momentum in the x direction. Use lower case b for the ball and upper case B
for the bottle. mb = 0.15 kg, and mB = 2.0 kg
Solve:
Σ pi = Σ pf
mb (vb )i = mb ( −0.20(vb )i ) + mB (vB )f
(vB )f =
mb (vb )i + 0.20mb (vb )i
mB
(vB )f (1 + 0.20)mb 1.2(0.15 kg)
=
=
= 0.090 = 9.0%
(vb )i
mB
2.0 kg
Assess: Because of the relative masses, this seems reasonable.
P9.70. 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
G
G
( p1 )i = m1 (v1)i = (0.02 kg)(2 m/s) along + x = (0.04 kg ⋅ m/s, along + x)
G
G
( p2 )i = m2 (v2 )i = (0.03 kg)(3 m/s) along − x = (0.09 kg ⋅ m/s, along − x)
G
G
( p3 )i = m3 (v3)i = [(0.04 kg)(4 m/s)cos 45 °, along + x ] + [(0.04 kg)(4 m/s)sin 45 °, along − x ]
= (0.113 kg ⋅ m/s, along + x) + (0.113 kg ⋅ m/s, along − x )
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9-38
Chapter 9
G
G
G
G
G
Since pf = pi = ( p1)i + ( p2)i + ( p3 )i , we have
G
(m1 + m2 + m3)vf = (0.023 kg ⋅m/s, along + x) + (0.073 kg ⋅ m/s, along − x)
G
⇒ vf = (0.256 m/s, along + x) + (− 0.811 m/s, along + x )
⇒ vf = (0.256 m/s) 2 + (− 0.811 m/s) 2 = 0.85 m/s
| vfy|
0.811
= 72 ° below + x
vfx
0.256
Assess: The final speed is of the same order of magnitude as the initial speeds, as one would expect.
θ = tan −1
= tan −1
P9.71. Prepare: The 14 C 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 14 C 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 14 C 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.
G
G
Solve: The conservation of momentum equation pf = pi = 0 kg ⋅ m/s is
G G
G
G
G G
G
G
pe + pn + pN = 0 N ⇒ pN = −( pe + pn ) = − meve − mn vn
= −[(9.11 × 10−31 kg)(5 × 107 m/s), along + x] − [(8.0 × 10 −24 kg ⋅ m/s), along + y ]
= −(45.55 × 10−24 kg ⋅ m/s, along + x) − (8.0 × 10−24 kg ⋅ m/s, along + y )
⇒ pN = mN vN = (45.55 × 10−24 ) 2 + (8.0 × 10−24 ) 2 kg ⋅ m/s = 4.62 × 10−23 kg ⋅ m/s
⇒ (2.34 × 10−26 kg)vN = 4.62 × 10−23 kg ⋅ m/s ⇒ vN = 1.97 × 103 m/s
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.
P9.72. 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.
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Momentum
9-39
Solve: The conservation of angular momentum equation Lf = Li is
2
2
⎛r ⎞
⎛ 30 cm ⎞
mvf rf = mvi ri ⇒ (rf ωf ) rf = (riωi)ri ⇒ ωf = ⎜ i ⎟ ωi = ⎜
⎟ (50 rpm) = 4.5 rpm
⎝ 100 cm ⎠
⎝ rf ⎠
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.
P9.73. Prepare: Model the puck as a particle.
Solve: The angular momentum is
L = mvr = 3.0 kg ⋅ m 2/s = (0.2 kg)v (0.5 m) ⇒ v = 30 m/s
G
The force that keeps the puck in circular motion is the tension T in the string. Thus,
mv 2 (0.2 kg)(30 m/s) 2
T=
=
= 360 N
r
0.5 m
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.
P9.74. Prepare: Because there’s no friction or other tangential forces, the angular momentum of puck is
conserved. The moment of inertia for the puck is mr 2 .
Solve: The conservation of angular momentum equation Li = Lf is
I iωi = I f ωf
Ii
mr
(20 cm) 2
100 rpm = 400 rpm
ωi = i 2 ωi =
If
mrf
(10 cm) 2
Assess: The increase in the angular speed by a factor of 4 comes because the radius is squared in the calculation of
the moment of inertia.
ωf =
2
P9.75. 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.
( I1 )f (ω1 )f + ( I 2 )f (ω2 )f + ... = ( I1 )i (ω1 )i + ( I 2 )i (ω2 )i + ...
The moment of inertia of the disk is I d = ( I d )i = ( I d )f = 12 mR 2 = 12 (2.0 kg)(0.10 m)2 = 0.010 kg ⋅ m 2.
The moment of inertia of the blocks is I b = ( I b )i = ( I b )f = 2mR 2 = 2(0.50 kg)(0.10 m) 2 = 0.010 kg ⋅ m 2.
Solve: The objects stick together after the blocks drop, so there is one combined ωtot afterwards. We solve the
conservation of angular momentum equation for the final angular speed.
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9-40
Chapter 9
( I d + I b )(ωtot )f = ( I d )(ωd )i + ( I b )(ωb )i ⇒
(ωtot )f =
( I d )(ωd )i + ( I b )(ωb )i
Id + Ib
(0.010 kg ⋅ m 2 )(100 rpm) + (0.010 kg ⋅ m 2 )(0 rpm)
= 50 rpm
(0.010 kg ⋅ m 2 ) + (0.010 kg ⋅ m 2 )
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.
=
P9.76. Prepare: Since there are no external torques, we can use the conservation of angular momentum to solve
this problem.
( I1 )f (ω1 )f + ( I 2 )f (ω2 )f + ... = ( I1 )i (ω1 )i + ( I 2 )i (ω2 )i + ...
Treat Joey as a particle and the merry-go-round as a disk. We need to use the moment of inertia of a rotating disk
1
I disk = MR 2 in order to write the final angular momentum of the merry-go-round. The moment of inertia of Joey is
2
I J = ( I J )f = ( I J )i = mJ 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
( I m )f (ωm )f + ( I J )f (ωJ )f = ( I m )i (ωm )i + ( I J )i (ωJ )i ⇒
(ωm )f =
( I m )i (ωm )i + ( I J )i (ωJ )i − ( I J )f (ωJ )f
( I m )f
=
( I m )i (0 rad/s) + ( I J )i (0 rad/s) − ( I J )f (ωJ )f
( I m )f
( )
vJ
2
−( I J )f (ωJ )f −(mJ R ) R
=
= 1
=
mm R 2
( I m )f
2
⎛ 5.0 m/s ⎞
−(36 kg)(2.0 m) 2 ⎜
⎟
⎝ 2.0 m ⎠ = − 0.90 rad/s
2
1
(200 kg)(2.0 m)
2
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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.
P9.77. 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.
( I1 )f (ω1 )f + ( I 2 )f (ω2 )f + ... = ( I1 )i (ω1 )i + ( I 2 )i (ω2 )i + ...
Assume John runs tangent to the rim of the merry-go-round.
The moment of inertia of the merry-go-round is I m = ( I m )i = ( I m )f = 12 MR 2 = 12 (250 kg)(1.5 m) 2 = 281 kg ⋅ m 2.
The moment of inertia of John is I J = ( I J )i = ( I J )f = mR 2 = 2(30 kg)(1.5 m) 2 = 67.5 kg ⋅ m 2.
Solve:
( ( I m )f + ( I J )f ) (ωtot )f = ( I m )i (ωm )i + ( I J )i (ωJ )i ⇒
(ωtot )f =
( I m )i (ωm )i + ( I J )i (ωJ )i
( I m )f + ( I J )f
=
( I m )i (20 rpm) + ( I J )i ( Rv )
( I m ) f + ( I J )f
⎛ 5.0 m/s ⎞⎛ 60 rpm ⎞
(281 kg ⋅ m 2 )(20 rpm) + (67.5 kg ⋅ m 2 ) ⎜
⎟
⎟⎜
⎝ 1.5 m ⎠⎝ 2 π rad/s ⎠ = 22 rpm
=
(281 kg ⋅ m 2 ) + (67.5 kg ⋅ m 2 )
P9.78.
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 ωf (which we seek).
We must remember to call counterclockwise angular speeds positive and clockwise angular speeds negative.
Known
MA = 2.0 kg
RA = 0.40 m
(ωA )i = − 30 rev/s
M B = 2.0 kg
RB = 0.20 m
(ωB)i = 30 rev/s
Find
ωf
Preliminarily compute the moments of inertia of the disks:
1
1
I A = M A RA2 = (2.0 kg)(0.40 m) 2 = 0.16 kg ⋅ m 2
2
2
1
1
I B = MB RB2 = (2.0 kg)(0.20 m) 2 = 0.040 kg ⋅ m 2
2
2
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9-42
Chapter 9
Solve: Since they stick together afterwards they will have a common angular speed.
( ( I A )f + ( I B )f ) (ωtot )f = ( I A )i (ωA )i + ( I B )i (ωB )i ⇒
(ωtot )f =
( I A )i (ωA )i + ( I B )i (ωB )i
( I A )f + ( I B )f
=
(0.16 kg ⋅ m 2 )( −30 rev/s) + (0.040 kg ⋅ m 2 )(30 rev/s)
= − 18 rev/s
0.16 kg ⋅ m 2 + 0.040 kg ⋅ m 2
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.
P9.79. 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
( px)i = mclub (vc lub x)i = (0.200 kg)(40.0 m/s) = 8.00 kg ⋅ m/s
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.
P9.80. 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.
P9.81. 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.
P9.82. Prepare: Because the system is isolated we can use the law of conservation of momentum. We are asked to
find (vCx)f − (vCx)i .
Solve:
( Px)i = ( Px)f
( pCx)i + ( pBx)i = ( pCx)f + ( pBx)f
mC (vCx)i + mB (vBx)i = mC (vCx)f + mB (vBx)f
Rearranging terms allows us to solve for (vCx)f − (vCx)i:
mB [(vBx )i − (vBx )f ] 0.0460 kg(0.0 m/s − 60.0 m/s)
=
= − 14 m/s
mC
0.200 kg
The correct choice is C. The negative sign indicates that the club slowed down.
Assess: Examining our equation for (vCx)f − (vCx)i confirms that we had a different (and maybe easier) approach
(vCx) f − (vCx)i =
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.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.