Question

A body of mass $(4m)$ is lying in $x-y$ plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass $(m)$ move perpendicular to each other with equal speeds $(v)$. The total kinetic energy generated due to explosion is

• A. $mv^2$
• B. $\frac{3}{2}mv^2$
• C. $2mv^2$
• D. $4mv^2$

Answer 1

The Correct Option is B

Let $\overrightarrow{v}$ be velocity of third piece of mass 2m.
Initial momentum, $\overrightarrow{p_i}=0$ (As the body is at rest)
Final momentum $\overrightarrow{p_f}=0=mc\widehat{i}+mv\widehat{j}+2m\overrightarrow{v}$
According to law of conservation of momentum
$\overrightarrow{p_i}=\overrightarrow{p_j}$
0=mv$\widehat{i}+mv\widehat{j}+2m\overrightarrow{v}$
$\overrightarrow{v}=-\frac{v}{2} \widehat{i}-\frac{v}{2}\widehat{j}$
The magnitude of v' is
$v'=\sqrt{\bigg(-\frac{v}{2}\bigg)^2+\bigg(-\frac{v}{2}\bigg)^2}=\frac{v}{\sqrt 2}$
Total kinetic energy generated due to explosion
$=\frac{1}{2}mv^2+\frac{1}{2}mv^2+\frac{1}{2}(2m)v'^{2}$
$=\frac{1}{2}mv^2+\frac{1}{2}mv^2+\frac{1}{2}(2m)\bigg(\frac{v}{\sqrt 2}\bigg)^2$

=$mv^2+\frac{mv^2}{2}=\frac{3}{2}mv^2$