A particle of mass m is kept on a smooth cube of mass M and side L as shown in figure. Cube starts moving with a constant velocity v . Displacement of the center of mass along the horizontal direction when particle hits the ground is

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$\frac{m v}{m + M} [\frac{v}{L} + \sqrt{\frac{2 L}{g}}]$

$\frac{M v}{m + M} [\frac{v}{L} + \sqrt{\frac{2 L}{g}}]$

$\frac{m v}{m + M} [\frac{L}{v} + \sqrt{\frac{2 L}{g}}]$

$\frac{M v}{m + M} [\frac{L}{v} + \sqrt{\frac{2 L}{g}}]$

answer is C.

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Detailed Solution

Time taken by particle to hit the ground, $t = t_{1} + t_{2}$ here $t_{1}$ is the time to leave the block by the particle, and $t_{2}$ is the time to hit the ground after leaving the block.
Using $2^{nd}$ equation of motion
$\begin{array}{l} t_{1} = \frac{L}{v}, \\ A l s o, h = \frac{1}{2} g t_{2}^{2} \\ \Rightarrow t_{2} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 L}{g}} \\ ∴ t = \frac{L}{v} + \sqrt{\frac{2 L}{g}} \end{array}$

Now, in horizontal direction, $V_{c m} = \frac{M v}{m + M}$
The distance moved by the center of mass in this time,
$x_{c m} = v_{c m} t$
$∴ x_{c m} = \frac{M v}{m + M} [\frac{L}{v} + \sqrt{\frac{2 L}{g}}]$
Therefore, the correct answer is (C)