Problems
Multiple Choice:
First two questions: A ball of mass M is caught by someone wearing a baseball glove. The ball is
in contact with the glove for a time t; the initial velocity of the ball (just before the catcher
touches it) is v0.
1. If the time of the ball's collision with the glove is doubled, what happens to the force
necessary to catch the ball?
A. It doesn't change.
B. It is cut in half.
C. It is cut to one fourth of the original force.
D. It quadruples.
E. It doubles.
2. If the time of collision remains t, but the initial velocity is doubled, what happens to the
force necessary to catch the ball?
A. It doesn't change.
B. It is cut in half.
C. It is cut to one fourth of the original force.
D. It quadruples.
E. It doubles.
3. Two balls, of mass m and 2m, collide and stick together. The combined balls are at rest
after the collision. If the ball of mass m was moving 5 m/s to the right before the
collision, what was the velocity of the ball of mass 2m before the collision?
A. 2.5 m/s to the right
B. 2.5 m/s to the left
C. 10 m/s to the right
D. 10 m/s to the left
E. 1.7 m/s to the left
4. Two identical balls have initial velocities v1 = 4 m/s to the right and v2 = 3 m/s to the left,
respectively. The balls collide head on and stick together. What is the velocity of the
combined balls after the collision?
A. 1/7 m/s to the right
B. 3/7 m/s to the right
C. 1/2 m/s to the right
D. 4/7 m/s to the rightv
E. 1 m/s to the right
Free Response:
5. A 75-kg skier skis down a hill. The skier collides with a 40-kg child who is at rest on the
flat surface near the base of the hill, 100 m from the skier's starting point, as shown
above. The skier and the child become entangled. Assume all surfaces are frictionless.
a. How fast will the skier be moving when he reaches the bottom of the hill?
Assume the skier is at rest when he begins his descent.
b. What will be the speed of the skier and child just after they collide?
c. If the collision occurs in half a second, how much force will be experienced by
each person.
Solutions
1. B—Impulse is force times the time interval of collision, and is also equal to an object's
change in momentum. Solving for force, F = Δp/Δt. Because the ball still has the same
mass, and still changes from speed vo to speed zero, the ball's momentum change is the
same, regardless of the collision time. The collision time, in the denominator, doubled; so
the entire expression for force was cut in half.
2. E—Still use F = Δp/Δt, but this time it is the numerator that changes. The ball still is
brought to rest by the glove, and the mass of the ball is still the same; but the doubled
velocity upon reaching the glove doubles the momentum change. Thus, the force doubles.
3. B—The total momentum after collision is zero. So the total momentum before collision
must be zero as well. The mass m moved 5 m/s to the right, giving it a momentum of 5m
units; the right-hand mass must have the same momentum to the left. It must be moving
half as fast, 2.5 m/s because its mass it twice as big; then its momentum is (2m)(2.5) = 5m
units to the left.
4. C—Because the balls are identical, just pretend they each have mass 1 kg. Then the
momentum conservation tells us that
(1 kg)(+4 m/s) + (1 kg)(–3 m/s) = (2 kg)(v').
The combined mass, on the right of the equation above, is 2 kg; v' represents the speed of
the combined mass. Note the negative sign indicating the direction of the second ball's
velocity. Solving, v' = +0.5 m/s, or 0.5 m/s to the right.
5.
a. This part is not a momentum problem, it's a Newton's second law and kinematics
proplem. (Or it's an energy problem, if you've studied energy.) Break up forces on
the skier into parallel and perpendicular axes—the net force down the plane is
mg(sin 45°). So by Newton's second law, the acceleration down the plane is g (sin
45°) = 7.1 m/s,2. Using kinematics with intitial velocity zero and distance 100 m,
the skier is going 38 m/s (!).
b. Now use momentum conservation. The total momentum before collision is (75
kg)(38 m/s) = 2850 kg.m/s. This must equal the total momentum after collision.
The people stick together, with combined mass 115 kg. So after collision, the
velocity is 2850 kg .m/s divided by 115 kg, or about 25 m/s.
c. Change in momentum is force multiplied by time interval … the child goes from
zero momentum to (40 kg)(25 m/s) = 1000 kg·m/s of momentum. Divide this
change in momentum by 0.5 seconds, and you get 2000 N, or a bit less than a
quarter ton of force. Ouch!
AP Free Response Momentum Problem
An open-top railroad car (initially empty and of mass M0) rolls with negligible friction along a straight
horizontal track and passes under the spout of a sand conveyor. When the car is under the conveyor,
sand is dispensed from the conveyor in a narrow stream at a steady rate M / t  C and falls
vertically from an average height h above the floor of the railroad car. The car has initial speed v0 and
sand is filling it from time t = 0 and t = T. Express your answers to the following in terms of the given
quantities and g.
6. Determine the mass M of the car plus the sand that it catches as a function of time t for 0<t<T.
m  m0  ct
7. Determine the speed v of the car as a function of time t for 0<t<T
Pix  Pfx
m0 v0  m0  ct v
v
m0 v 0
m0  ct
8. i. Determine the initial kinetic energy Ki of the empty car.
ki 
1
2
m0 v 0
2
ii. Determine the final kinetic energy Kf of the car and its load.
 m0 v 0 
1
1

K f  mv 2  m0  cT 
2
2
 m0  cT 
2
2
2
m0 v 0
Kf 
2m0  cT 
iii. Is kinetic energy conserved? Explain why or why not.
This is an inelastic collision, so that kinetic energy is not conserved.
9. Determine expressions for the normal force exerted on the car by the tracks at the following
times.
Fn  m0 g
It’s NOT just Fn  m0  ct g ! We’re forgetting the fact that the grain is falling, and making it
increase the normal force.
J  p
J  mv g
J  ct 0  v g 
U i  Ki  U f  K f
1
mgh  mv 2
2
v g  2 gh
J  ct 2 gh
Ft  ct 2 gh
F  c 2 gh
Fn  m0 g  c 2 gh  ctg
J  ct 2 gh
i. Before t = 0
ii. For 0 < t < T
Fn  m0 g  c 2 gh  ctg
iii. After t=T
Fn  m0  ct g
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