Question

A box of mass '3m' explodes into two fregments of mass 'm' and '2m' on horizontal surface with coefficient of friction $\mu$ as shown in figure. Due to explosion mass '2m' gets speed 'v' then

• A. Speed of center of mass of box is zero just after explosion
• B. Speed of center of mass first increases then decreases
• C. Displacement of center of mass is $\frac{V^2}{3\mu g}$
• D. Center of mass displaces towards right

Answer 1

Answer: (A), (B), (C)

The correct options are
A Speed of center of mass of box is zero just after explosion
B Speed of center of mass first increases then decreases
C Displacement of center of mass is $\frac{V^2}{3\mu g}$

By Conservation of MOMENTUM $\to$

Just after the collision centre of mass will be at rest because of conservation of momentum and then friction will start acting on bothy the block and its position and velocity will start changing.

Due to friction retardation of both is $a=\mu g$

$V_2=V-\mu gt$ and $V_1=2V-\mu gt$

$V_{COM}=\frac{2mV-2m\mu gt-2mV+m\mu gt}{3m}=\frac{-\mu mgt}{3m}$

$V_{COM}=\frac{-\mu gt}{3}$

Let m and 2m stop after $d_1$ and $d_2$

$d_1=\frac{2V^2}{\mu g}$ and $d_2=\frac{V^2}{2\mu g}$

$X_{COM}=\frac{2m{d}_2-m{d}_1}{3m}=\frac{\frac{m{V}^2}{\mu g}-\frac{2m{V}^2}{\mu g}}{3m}=-\frac{V^2}{3\mu g}$

Negative sign in displacement indicates COM displaces toward left.

Initially, just after the collision, the COM is at rest and the net force on the two block system is $f_2-f_1$. Therefore, the speed of the COM will increase. After a while, both the blocks come to rest and does the COM, which means COM's speed got decreased and became zero.