Transfer of momentum problem?

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 [tex]p_a[/tex]

Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum [tex]p_b[/tex]


Prove that [tex]|\frac{p_b}{p_a}|<2[/tex]

A.T.

Momentum conservation :
[tex]{p_a} = {p_a}' + {p_b}[/tex]
[tex]1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex]


Energy conservation (Ball B had intially no KE, so speed of ball A cannot increase):
[tex]|\frac{{p_a}'}{p_a}| < 1[/tex]

haruspex

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

A.T.

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

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