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 ($\nu$). The total kinetic energy generated due to explosion is

Let ${\overrightarrow{v}}^{\mathrm{\prime }}$be velocity of third piece of mass 2 m. Initial momentum, ${\overrightarrow{p}}_i=0$(As the body is at rest)
Final momentum, ${\overrightarrow{p}}_f=mv\widehat{ı}+mv\widehat{ȷ}+2m{\overrightarrow{v}}^{\mathrm{\prime }}$

$\text{ According to law of conservation of momentum }$

$\begin{array}{l}{\overrightarrow{p}}_i={\overrightarrow{p}}_f\\ 0=mv\widehat{i}+mv\widehat{j}+2m{\overrightarrow{v}}^{\mathrm{\prime }}\\ {\overrightarrow{v}}^{\mathrm{\prime }}=-\frac{v}{2}\widehat{i}-\frac{v}{2}\widehat{j}\end{array}$

$\text{ The magnitude of }{\overrightarrow{v}}^{\mathrm{\prime }}\text{ is }$

$v\text{'}=\sqrt{{\left(-\frac{v}{2}\right)}^2+{\left(-\frac{v}{2}\right)}^2}=\frac{v}{\sqrt{2}}$

$\text{ Total kinetic energy generated due to explosion }$

$\begin{array}{l}=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right)v^{\mathrm{\prime }2}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right){\left(\frac{v}{\sqrt{2}}\right)}^2\\ =m{v}^2+\frac{m{v}^2}{2}=\frac{3}{2}m{v}^2\end{array}$