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

Answer: (B)

Let $\vec{v}$ be velocity of third piece of mass $2 m$.
Initial momentum, $\vec{p_{i}}=0$ (As the body is at rest)
Final momentum $\vec{p_{f}}=0=m c \hat{i}+m v \hat{j}+2 m \vec{v}$
According to law of conservation of momentum
$\vec{p_{i}}=\vec{p_{j}} $
$ 0 = mv \hat{i}+m v \hat{j}+2 m \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}(2 m) v^{\prime 2} $
$=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(2 m)\left(\frac{v}{\sqrt{2}}\right)^{2} $
$=m v^{2}+\frac{m v^{2}}{2}=\frac{3}{2} m v^{2}$