Studied by 3 peopleImpulse
________ can be expressed both as force times a time interval and as a change in momentum.
Momentum
________ is a vector quantity abbreviated with a p.
Units of momentum
kg.m/s, same as N·s
Elastic collision
A collision in which kinetic energy is conserved
Inelastic Collision
A collision in which kinetic energy is not conserved
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
C—Because the balls are identical, just pretend they each have mass 1 kg.
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.
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.
What will be the speed of the skier and child just after they collide?
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
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.
If the collision occurs in half a second, how much force will be experienced by each person?
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.
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 .
. 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.
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 v0 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.
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 .
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
E— 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.
Note
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