===== 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?

[c] kinetic energy?

Of each object after the collision.



=== Facts ===

Situation occurring in an orbiting spacecraft.

Ping-Pong ball of mass m with initial momentum $\vec{p}_{1i}$ traveling in the +x direction.

Bowling ball of mass M is hit by the Ping-Pong ball while stationary.

Initial situation: Just before collision

Final situation: Just after collision


=== Lacking ===

What are the 

[a] momentum?

[b] speed?

[c] kinetic energy?

Of each object after the collision.


=== 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 

{{183_notes:examples:mi3e_10-007.jpg?350}}

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

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


=== 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 $\vec{p}_{1f} \approx -\vec{p}_{1i}$.

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

Add like terms and rearrange:

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

[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:

From the equation for momentum: $p_{2f} = M(v_{2f})$

Therefore:

$$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M}$$ 

Substitute $2\vec{p}_{1i}$ in for ${p_{2f}}$ from previous result in this example above.

$$\vec{v}_{2f} \approx \dfrac{2p_{1i}}{M}$$

Momentum is mass times velocity so substitute this in:

$$\vec{v}_{2f} \approx \dfrac{2mv_{1i}}{M}$$ 

Rearrange:

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

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:

From our representations we know that $K = \frac{1}{2}m(\frac{p^{2}}{m})$

But we also know that $p_{2f} = 2\vec{p}_{1i}$ from earlier so substituting this into $K = \frac{1}{2}m(\frac{p^{2}}{m})$ we get this equation for the final kinetic energy of the bowling ball

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

and the equation for the final kinetic energy of the Ping-Pong ball is 

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

As we assumed that the speed of the Ping-Pong ball does not change significantly after the collision.

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.