Two identical blocks of mass $2 M$ are joined by means of a light spring of spring constant $k$ . A man of mass $M$ is standing on one of the block as shown in the diagram. If man jumps horizontally with a velocity $V$ relative to block and horizontal surface is smooth, then

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Velocity of centre of mass of two blocks after $2 M$ loses contact with wall is $\frac{V}{6}$

The maximum compression in the spring is $\sqrt{\frac{2 M}{k}} (\frac{V}{3})$

Man lands at horizontal distance $V \sqrt{\frac{2 h}{g}}$ from initial position of the block

Right block loses contact with wall when the elongation in spring is maximum

answer is A, D.

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Detailed Solution

$M (V - v) = 2 Mv \Rightarrow v = V / 3 ∴ \frac{1}{2} {kx}_{0}^{2} = \frac{1}{2} 2 {Mv}^{2} \Rightarrow k = \sqrt{\frac{2 M}{k}} (\frac{V}{3})$

At the instant when block looses contact with wall

$∴$ $V_{c m} = \frac{2 M V / 3}{4 M} = \frac{V}{6}$

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