Can Momentum Transfer Exceed Double the Initial Value in Elastic Collisions?

[greswd]

Here's an interesting puzzle:

The whole scenario takes place in one dimension of space.

Ball B is at rest. Ball A moves with momentum $$p_a$$

Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum $$p_b$$


Prove that $$|\frac{p_b}{p_a}|<2$$

 

[A.T.]

Here's an interesting puzzle:

The whole scenario takes place in one dimension of space.

Ball B is at rest. Ball A moves with momentum $$p_a$$

Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum $$p_b$$Prove that $$|\frac{p_b}{p_a}|<2$$

Momentum conservation :
$${p_a} = {p_a}' + {p_b}$$
$$1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}$$Energy conservation (Ball B had intially no KE, so speed of ball A cannot increase):
$$|\frac{{p_a}'}{p_a}| < 1$$

 

[haruspex]

$$|\frac{{p_a}'}{p_a}| < 1$$

All true, but I don't see how that gets you to the answer required.
Using energy conservation more completely:
$$\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}$$
Combining with momentum eqn. we get
$$\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2$$

 

[A.T.]

All true, but I don't see how that gets you to the answer required.

You have to combine the two of course:
$$1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}$$
$$|\frac{{p_a}'}{p_a}| < 1$$
This leads to:
$$0 < \frac{p_b}{p_a} < 2$$

Using energy conservation more completely:
$$\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}$$
Combining with momentum eqn. we get
$$\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2$$

Yes, the clean & complete way is to derive the ratio as function of the masses. Mine was just showing that the ratio is < 2.