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The spring shows in the figure is in natural length initially and the blocks are at rest. Suddenly block $m_{2}$ receives horizetal velocity $V$ rightward then the maximum extension in the spring is (surface is smooth)

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$V \sqrt{\frac{m_{1} m_{2}}{m_{1} + m_{2}}}$

$V \sqrt{\frac{2 m_{1} m_{2}}{(m_{1} + m_{2}) K}}$

$V \sqrt{\frac{m_{1} m_{2}}{2 (m_{1} + m_{2}) K}}$

$V \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) K}}$

answer is D.

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Detailed Solution

$m_{2} V = m_{1} V^{'} + m_{2} V^{'} \Rightarrow V^{'} = \frac{m_{2} V}{m_{1} + m_{2}}$

$\frac{1}{2} m_{2} V^{2} = \frac{1}{2} m_{1} {(V^{'})}^{2} + \frac{1}{2} m_{2} {(V^{'})}^{2} + \frac{1}{2} K x^{2}$

$m_{2} V^{2} = (m_{1} + m_{2}) {(V^{'})}^{2} + K x^{2}$

$m_{2} V^{2} = (m_{1} + m_{2}) \times (\frac{m_{2} V}{m_{1} + m_{2}}) + K x^{2}$

$m_{2} V^{2} - \frac{m_{2} V}{(m_{1} + m_{2})} = K x^{2}$

$x = \sqrt[V]{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) K}}$

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