Download Problem Set 6 - Cabrillo College

Problem Set 6
Chapter 9: Momentum
Questions: 5, 8, 11, 13, 14
Exercises & Problems: 4, 11, 15, 23, 39, 44, 45, 47, 54, 55
Q9.5: A stationary firecracker explodes into three pieces. One piece travels off to the
east; a second travels to the north. Which of the vectors of the figure could be the
velocity of the third piece? Explain.
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.
Q9.8: Automobiles are designed with "crumpled zones" intended to collapse in a
collision. Why would a manufacturer design part of a car so that it collapses in a
collision?
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.
Q9.11: Two ice skaters, Megan and Jason, push off from each other on frictionless ice,
Jason's mass is twice that of Megan.
a. Which skater, if either, experiences the greater impulse during the push? Explain.
b. Which skater, if either, experiences the greater speed after the push? Explain
Q9.11. Reason: See Example 9.5. The two skaters interact with each other, but they form an isolated system
because, for each skater, the upward normal force of the ice balances their downward weight force to make
Fnet  0. Thus the total momentum of the system of the two skaters will be conserved. Assume that both skaters
are at rest before the push so that the total momentum before they push off is Pi  0. Consequently, the total
momentum will still be 0 after they push off.
(a) Because the total momentum of the two-skater system is 0 after the push off, Megan and Jason each have
momentum of the same magnitude but in the opposite direction as the other. Therefore the magnitude p is the
same for each: Favg t  p.
From the impulse-momentum theorem each experiences the same amount of impulse.
(b) They each experience the same amount of impulse because they experience the same magnitude force
over the same time interval. However, over that time interval they do not experience the same
acceleration. 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 t.
Assess: It is important to think about both results until you are comfortable with them.
Q9.13: While standing still on a basketball court, you throw the ball to a teammate. Why
do you not move backward as a result? Is the law of conservation of momentum violated?
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Q9.13. Reason: If you do not move backward when passing the basketball it is because the ball-you system
is not isolated: There is a net external force on the system to keep you from moving backward that changes the
momentum of the system. If the ball-you system is isolated (say you are on frictionless ice), then you do move
backward when you pass the ball.
Assess: If the friction force of the floor on you keeps you from moving backward (relative to the floor), then
the law of conservation of momentum doesn’t apply because the system isn’t isolated. But you could then
include the floor, building, and the earth in the system so it (the system) is isolated; then momentum of the
system is conserved—and that means the earth does recoil ever so slightly when you pass the basketball.
Q9.14: To win a prize at the county fair, you're trying to knock down a heavy bowling pin
by hitting it with a thrown object. Should you choose to throw a rubber ball or a beanbag
of equal size and weight? Explain.
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  pproj. The rubber ball and beanbag would start with the same
momentum, but the rubber ball would have a greater change in momentum because it would bounce off and so
its direction would change. The beanbag would continue to move in the same direction. Thus the rubber ball
would exert a greater impulse on the bowling pin.
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.
P9.4: In the figure, what value of Fmax gives an impulse of 6.0 N∙s?
P9.4. Prepare: Please refer to Figure P9.4. To get time in seconds, we note
that 1 ms = 10–3 s.
Solve:
The impulse as defined through Equation 9.1 is
J x  Area under the Fx(t) curve between ti and
tf  6.0 N s  1 ( Fmax )(8 ms)  Fmax  1500 N
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Assess: Note that the impulsive force shown is a linear function of time. Therefore, the impulsive force can
also be written as 0.5 Fmax(6 ms)  0.5 Fmax(2 ms) by looking at the first 6 ms and the next 2 ms separately.
Equating this to the change in momentum should also give the same answer, Fmax  1500 N.
P9.11: As part of a safety investigation, two 1400 kg cars traveling at 20 m/s are crashed
into different barriers. Find the average forces exerted on (a) the car that hits a line of
water barrels and takes 1.5 s to stop, and (b) the car that hits a concrete barrier and
takes 0.1 s to stop.
P9.11. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as
Favg 
p
t
In fact, this is much closer to what Newton actually wrote than F  ma.
Solve: This allows us to find the average force on the cars.
(a) Water barrels
Favg 
(b) Concrete barrier
p mvf  mvi m(vf  vi ) (1400 kg)(0 m/s  20 m/s)



 19,000 N
t
t
t
1.5 s
Favg 
p mvf  mvi m(vf  vi ) (1400 kg)(0 m/s  20 m/s)



 280,000 N
t
t
t
0.1s
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.
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P9.15: A 2.7 kg block of wood sits on a table. A 3.0 g bullet, fired horizontally at a speed of
500 m/s, goes completely through the block, emerging at a speed of 220 m/s. What is the
speed of the block immediately after the bullet exits?
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)(v bl ) f .
We can solve for the final velocity of the block: (vbl )f  0.31 m/s .
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.
P9.23: A kid at the junior high cafeteria wants to propel an empty milk carton along a
lunch table by hitting it with a 3.0 g spit ball. If he wants the speed of the 20 g carton just
after the spit ball hits it to be 0.30 m/s, at what speed should his spit ball hit the carton?
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.
(m  mCTN )(vBOTHx )f  mCTN (vCTNx )i (0.0030 kg  0.020 kg)(0.30 m/s)
(vSBx )i  SB

 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.
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P9.39: A 200 g ball is dropped from a height of 2.0 m,
bounces on a hard floor, and rebounds to a height of
1.5 m. The figure shows the impulse received from the
floor. What maximum force does the floor exert on the
ball?
P9.39. Prepare: Model the ball as a particle that is subjected to an
impulse when it is in contact with the floor. We will also use constantacceleration kinematic equations.
Solve:
To find the ball’s velocity just before and after it hits the floor:
v12y  v02y  2a y ( y1  y0 )  0 m2 /s 2  2(9.8 m/s 2 )(0  2.0 m)  v1 y  6.261 m/s
v32y  v22y  2a y ( y3  y2 )  0 m2 /s 2  v22y  2(9.8 m/s 2 )(1.5 m  0 m)  v2 y  5.422 m/s
The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
mv2 y  mv1 y  area under the force curve  (0.2 kg)(5.422 m/s)
 (0.2 kg)(6.261 m/s)  12 Fmax (5 103 s)  Fmax  940 N
Assess: A force of 940 N exerted by the floor is typical of such collisions.
P9.44: Squids rely on jet propulsion, a versatile technique to move around in water. A
1.5 kg squid at rest suddenly expels 0.10 kg of water backward to quickly get itself
moving forward at 3.0 m/s. If other forces (such as the drag force on the squid ) are
ignored, what is the speed with which the squid expels the water?
P9.44. Prepare: Consider the system to be the squid and the water it takes in and expels. Ignoring all other
forces, there is no net external force on the system, so momentum of the system is conserved. Assume the system
is at rest before the squid expels the water.
Solve: The momentum of the system before the squid expels the water is zero because everything is at rest. The
momentum of the entire system must also be zero after the squid expels the water since momentum is conserved
in the system. See the following diagram.
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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.
P9.45: The flowers of the bunchberry plant open with astonishing force and speed,
causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an
acceleration of 2.5 × 104 m/s2. If the acceleration is constant, what impulse is delivered
to a pollen grain with a mass of 1.0 × 10-7 g?
P9.45. Prepare: We are asked for the impulse given to the pollen grain; impulse is defined in Equation 9.1:
J  Favg t. We are given that t  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 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 t  (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. Note that N  s  kg  m/s.
P9.47: A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g
tennis ball that was approaching her at a speed of 20 m/s. The ball rebounds at 40 m/s.
a. How fast is her racket moving immediately after the impact? You can ignore the
interaction of the racket with her hand for the brief duration of the collision.
b. If the tennis ball and racket are in contact for 10 ms, what is the average force that
the racket exerts on the ball?
P9.47. Prepare: Let the system be ball + racket. During the collision of the ball and the racket, momentum is
conserved because all external interactions are insignificantly small. We will also use the momentum-impulse
theorem.
Solve: (a) The conservation of momentum equation (px)f  (px)i is
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
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(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 t
 Favg 
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/s2 )  0.588 N. Thus, Favg  610 wB. This is a significant force and is
reasonable because the impulse due to this force changes the direction as well as the speed of the
ball from approximately 45 mph to 90 mph.
P9.54: A 110 kg linebacker running at 2.0 m/s and an 82 kg quarterback running at 3.0
m/s have a head-on collision in midair. The linebacker grabs and holds onto the
quarterback. Who ends up moving forward after they hit?
P9.54. Prepare: We arbitrarily pick the direction the linebacker was running as the positive x-direction since
he was mentioned first. If, after we solve the conservation of momentum equation for vf, the answer is positive, then
we know the linebacker ends up moving forward; if vf is negative, then the quarterback ends up moving forward.
We will use subscripts l for linebacker and q for quarterback.
Known
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:
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.‖
P9.55: Most geologists believe that the dinosaurs became extinct 65 million years ago
when a large comet or asteroid struck the earth, throwing up so much dust that the sun
was blocked out for a period of many months. Suppose an asteroid with a diameter of
2.0 km and a mass 1.0 × 1013 kg hits the earth with an impact speed of 4.0 × 104 m/s.
a. What is the earth's recoil speed after the collision? (Use a reference frame in which
the earth was initially at rest.)
b. What percentage is this of the earth's speed around the sun? (Use the astronomical
data inside the back cover.)
P9.55. Prepare: Model the earth (E) and the asteroid (A) as particles. Earth asteroid is our system. Since the
two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in
the collision since no significant external force acts on the system.
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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
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(b) The speed of the earth going around the sun is
vE 
2 r 2 (1.50 1011 m)

 3.0  104 m/s
T
3.15 107 s
Hence, (vx )f /vE  2 1012  2 1010%.
Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
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