Question

A body of mass $4 m$ at rest explodes into three fragments. Two of the fragments each of mass $m$ move with speed $v$ in mutually perpendicular directions. Total energy released in the process is

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

Answer 1

Answer: (B)

Here initial momentum $\vec{p}=0$. Since no external force exists, hence momentum must remain conserved
i.e, $ \vec{p}_{1}+\vec{p}_{2}+\vec{p}_{3}=0$
As two fragments of mass $m$ each are moving with speed $v$ each at right angles, so
$\left|\vec{p}_{1}+\vec{p}_{2}\right|=m \sqrt{v^{2}+v^{2}}=\sqrt{2} m v $
$\therefore \left|\vec{p}_{3}\right|=\left|\vec{p}_{1}+\vec{p}_{2}\right|=\sqrt{2} m v$
The mass of third fragment is $2 m$.
$\therefore $ Kinetic energies of three fragments are
$K_{1}=\frac{p_{1}^{2}}{2 m}=\frac{1}{2} m v^{2}, K_{2}=\frac{p_{2}^{2}}{2 m}=\frac{1}{2} m v^{2} $
and $ K_{3}=\frac{p_{3}^{2}}{2(2 m)}=\frac{1}{2} m v^{2}$
Total energy released during explosion
$=K_{1}+K_{2}+K_{3}=\frac{3}{2} m v^{2}$