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A ball of mass m is just disturbed from the top of a fixed smooth circular tube of radius a in a vertical plane and falls impinging on a ball of mass $2 m$ at the bottom. The coefficient of restitution is $1/2$ . Find the heights to which the balls rise after a second impact.

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\frac{a}{2}, \frac{a}{8}

\frac{a}{2}, \frac{a}{2}

\frac{a}{4}, \frac{a}{2}

\frac{a}{8}, \frac{a}{8}

answer is A.

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Detailed Solution

m_{1} = m, m_{2} = 2 m

$h = 2a, u_{2} = 0$

$e = \frac{1}{2}$

$u_{1} = \sqrt{2gh} = \sqrt{2 g (2a)} = 2 \sqrt{ga}$

$v_{1} = \frac{m_{1} - {em}_{2}}{m_{1} + m_{2}} . x_{1} + 0$

$= \frac{m - m}{3 m} (2 \sqrt{ga}) = 0$

$v_{2} = \frac{m [1 + \frac{1}{2}]}{3 m} \times 2 \sqrt{ga}$

$= \sqrt{ga}$

$v_{2} = v_{3}, v_{4} = ?$

$v_{4} = \frac{2 m - \frac{1}{2} m}{3 m} (\sqrt{ga})$

$= \frac{\sqrt{ga}}{2}$

$h_{1} = \frac{v_{3}^{2}}{2g} = \frac{ag}{2g} = \frac{a}{2}$

$h_{2} = \frac{v_{4}^{2}}{2g} = \frac{ga}{4} \times \frac{1}{2g} = \frac{a}{8}$

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