Two masses $m_{1}$ and $m_{2} (m_{1} < m_{2})$ are positioned as shown in figure, $m_{1}$ being on the ground and $m_{2}$ at a height above the ground. When $m_{2}$ is released, the speed at which it hits the ground will be

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$\sqrt{\frac{2 g h (m_{1} + m_{2})}{(m_{1} - m_{2})}}$

$\sqrt{\frac{2 g h (m_{1} - m_{2})}{(m_{1} + m_{2})}}$

$\sqrt{\frac{2 g h m_{1}}{m_{2}}}$

$\sqrt{\frac{2 g h (m_{2} - m_{1})}{(m_{1} + m_{2})}}$

answer is D.

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Detailed Solution

$m_{2} g h = m_{1} g h + \frac{1}{2} (m_{1} + m_{2}) v^{2}$

$∴ v = \sqrt{\frac{2 g h (m_{2} - m_{1})}{m_{1} + m_{2}}}$

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