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?

[c] kinetic energy?

Of each object after the collision.

Initial situation: Just before collision

Final situation: Just after collision

What are the

[a] momentum?

[b] speed?

[c] kinetic energy?

Of each object after the 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

$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$

$K = \frac{1}{2}m(\frac{p^{2}}{m})$

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:

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

This is a very small speed since m«M/ For example, if the mass of the bowling ball is about 5 kg, and the gram Ping-Pong ball is initially traveling at 10 m/s, the final speed of the bowling ball will be 0.008 m/s.

[c] Kinetic energies:

$$K_{2f} = \dfrac{(2p_{1i})^2}{2M}$$

and

$$K_{1f} = \dfrac{p^2_{1i}}{2m}$$

Because the mass of the bowling ball is much larger than the mass of the Ping-Pong ball, the kinetic energy of the bowling ball is much smaller than the kinetic energy of the Ping-Pong ball. The kinetic energy of the 2g Ping-Pong ball, traveling at 10m/s, is about 0.1J, while the 5kg bowling ball has acquired a kinetic energy of $1.6 x 10^{-4} J$ - nearly 1000 times less.