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.

Initial situation: Just before collision

Final situation: Just after collision

Assume little change in the speed of the Ping-Pong ball, and assume that the collision is elastic.

System: Ping-Pong ball and bowling ball

Surroundings: Nothing that exerts significant forces

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:

$$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M} = \dfrac{2p_{1i}}{M} = \dfrac{2mv_{1i}}{M} = 2\dfrac({m}{M})v_{1i}$$