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# Momentum Practice Problems for AP Physics

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By — McGraw-Hill Professional
Updated on Feb 10, 2011

Review the following concepts if necessary:

### 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?
1. It doesn't change.
2. It is cut in half.
3. It is cut to one fourth of the original force.
5. 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?
1. It doesn't change.
2. It is cut in half.
3. It is cut to one fourth of the original force.
5. 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?
1. 2.5 m/s to the right
2. 2.5 m/s to the left
3. 10 m/s to the right
4. 10 m/s to the left
5. 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?
1. 1/7 m/s to the right
2. 3/7 m/s to the right
3. 1/2 m/s to the right
4. 4/7 m/s to the rightv
5. 1 m/s to the right
5. ### Free Response:

6. 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.
1. 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.
2. What will be the speed of the skier and child just after they collide?
3. 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
5. (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.

6.
1. 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 (!).
2. 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.
3. 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!

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