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. $m{v}^2$
• B. $\frac{3}{2}m{v}^2$
• C. $2m{v}^2$
• D. $4m{v}^2$

Answer 1

Answer: (B)

Let $\vec{v}$ be velocity of third piece of mass $2m$.
Initial momentum, $\overrightarrow{p_i}=0$ (As the body is at rest)
Final momentum $\overrightarrow{p_f}=0=mc\hat{i}+mv\hat{j}+2m\vec{v}$
According to law of conservation of momentum
$\overrightarrow{p_i}=\overrightarrow{p_j}$
$0=mv\hat{i}+mv\hat{j}+2m\vec{v}$
$\vec{v}=-\frac{v}{2}\hat{i}-\frac{v}{2}\hat{j}$
The magnitude of $v^{\prime }$ is
$v^{\prime }=\sqrt{{\left(-\frac{v}{2}\right)}^2+{\left(-\frac{v}{2}\right)}^2}=\frac{v}{\sqrt{2}}$
Total kinetic energy generated due to explosion
$=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}(2m)v^{\prime 2}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}(2m){\left(\frac{v}{\sqrt{2}}\right)}^2$
$=m{v}^2+\frac{m{v}^2}{2}=\frac{3}{2}m{v}^2$