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]
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]
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]
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.