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## Transfer of momentum problem?

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 $$|\frac{p_b}{p_a}|<2$$
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 Quote by 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 $$|\frac{p_b}{p_a}|<2$$
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):
$$|\frac{{p_a}'}{p_a}| < 1$$

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 Quote by A.T. $$|\frac{{p_a}'}{p_a}| < 1$$
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
$$\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2$$

## Transfer of momentum problem?

 Quote by haruspex All true, but I don't see how that gets you to the answer required.
You have to combine the two of course:
$$1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}$$
$$|\frac{{p_a}'}{p_a}| < 1$$
$$0 < \frac{p_b}{p_a} < 2$$
 Quote by haruspex 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 $$\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2$$