Question

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 ?

• 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 i.e remains at a distance of $2L$ from the y-axis.

Using $X=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{M}$

$\therefore$ $2L=\frac{2m\left(x\right)+3m(2L+x)+m(4L+x)}{m+2m+3m}$

$⟹$ $x=\frac{L}{3}$