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Example: A Ping-Pong Ball Hits a Stationary Bowling Ball Head-on
In an orbiting spacecraft a Ping-Pong ball of mass m (object 1) traveling in the +x direction with initial momentum \vec{p}_1i hits a stationary bowling ball of mass M (object 2) head on, as shown in the figure in representations.
What are the
(a) momentum?
(b) speed?
© kinetic energy?
Of each object after the collision.
Facts
Initial situation: Just before collision
Final situation: Just after collision
Lacking
Approximations & Assumptions
Assume little change in the speed of the Ping-Pong ball, and assume that the collision is elastic.
Representations
System: Ping-Pong ball and bowling ball
Surroundings: Nothing that exerts significant forces
Solution
From the momentum principle
$$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$$
Assume that the speed of the Ping-Pong ball does not change significantly in the collision, so $\vec{p}_{1f} \approx -\vec{p}_{1i}$.
$$-\vec{p}_{1i} + \vec{p}_{2f} = \vec{p}_{1i}$$
$$\vec{p}_{2f} = 2\vec{p}_{1i}$$
(a) The final momentum of the bowling ball is twice the initial momentum of the Ping-Pong ball.
It may be surprising that the bowling ball ends up with about twice the momentum of the Ping-Pong ball. One way to understand this is that the final momentum of the Ping-Pong ball is approximately $-\vec{p}_{1i}$, so the change in the Ping-Pong ball's momentum is approximately
$-\vec{p}_{1i} -\vec{p}_{1i} = -2\vec{p}_{1i}$
The Ping-Pong ball's speed hardly changed, but its momentum changed a great deal. Because momentum is a vector, a change of direction is just as much a change of magnitude. This big change is of course due to the interatomic electric contact forces exerted on the Ping-Pong ball by the bowling ball. By reciprocity, the same magnitude of interatomic contact forces are exerted by the Ping-Pong ball on the bowling ball, which undergoes a momentum change of $+2\vec{p}_{1i}$
(b) Final speed of bowling ball:
$$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M} = \dfrac{2p_{1i}}{M} = \dfrac{2mv_{1i}}{M}$$ = 2\dfrac({m}{M})v_{1i}$$