A bomb of mass 3m is kept inside a closed box of mass 3m and length 4L at its centre. It explodes in two parts of mass m and 2m. The two parts move in opposite directions and stick to the opposite sides of the walls of box. The box is kept on a smooth horizontal surface.
What is the distance moved by the box during this time interval.

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$\frac{L}{12}$

$\frac{L}{6}$

$\frac{L}{3}$

answer is D.

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Detailed Solution

Let box moves towards right by a distance x, then displacement of 2m is $x + 2 L$ and displacement of m is $x - 2 L$ towards right.

$\begin{array}{l} {Δx}_{cm} = \frac{m_{1} {Δx}_{1} + m_{2} {Δx}_{2} + m_{3} {Δx}_{3}}{m_{1} + m_{2} + m_{3}} \\ \Rightarrow 0 = 3 mx + 2 m (x + 2 L) + m (x - 2 L) \\ \Rightarrow x = - \frac{L}{3} \end{array}$

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