Question

A bomb of mass $3\text{m}$ is kept inside a closed box of mass $3\text{m}$ and length $4\text{L}$ at its centre. It explodes in two parts of mass $m$ and $2m$. The two parts move in opposite direction and stick to the opposite 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?

• A. $0$
• B. $\frac{L}{6}$
• C. $\frac{L}{12}$
• D. $\frac{L}{3}$

Answer 1

The correct option is D $\frac{L}{3}$

Initially the centre of the system is at point $O$. Now the masses $m$ and $2m$ get stuck on the opposite walls of the box and thus, as a result the box moves by a distance $x$. The centre of the box is at point $O^{\prime }$ finally. As there is no external force, thus the box moves in such a way that the centre of mass of the whole system does not move ie. remains at a distance of $2L$ from the $y$ axis.


Using $x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{M}$
$2L=\frac{2m\left(x\right)+3m(2L+x)+m(4L+x)}{m+2m+3m}$
$x=\frac{L}{3}$