A smooth ball falls vertically downward symmetrically on two other identical balls with speed $v$ as shown in figure. If it rebounds with velocity $\frac{v}{3}$ , find the velocity of other two balls moving horizontally.

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Solution

Loss of Kinetic Energy of 1 is, $\frac{1}{2} m V^{2} - \frac{1}{2} m ({\frac{V}{3}}^{2})$
= $\frac{8}{9} (\frac{1}{2} m V^{2})$
In horizontal direction,momentum is conservative ,so 2 and 3 have same velocity $u$ opposite to other.
$\frac{1}{2} m u^{2} + \frac{1}{2} m u^{2} = \frac{8}{9} (\frac{1}{2} m V^{2})$
$2 u^{2} = \frac{8}{9} V^{2}$
$u = \frac{2}{3} V$

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