A bomb of mass $3 m$ is kept inside a closed box of mass $3 m$ and length $4 L$ at it's centre. It explodes in two parts of mass $m$ & $2 m$ . The two parts move in opposite direction and stick to the apposite side of the walls of box. Box is kept on a smooth horizontal surface.
What is the distance moved by the box during this time interval.

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$0$

$\frac{L}{6}$

$\frac{L}{12}$

$\frac{L}{3}$

answer is D.

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Detailed Solution

$0 = m ∆ x_{1 G} + 2 m ∆ x_{2 G} + 3 m ∆ x_{3 G} = - 2 L + 2 (2 L) + 6 ∆ x_{3 G} ∆ x_{3 G} = L / 3$

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