Question

A particle of mass $4m$ which is at rest explodes into four equal fragments. All four fragments scatter in the same horizontal plane. Three fragments are found to move with velocity $v$ as shown in the figure. The total energy released in the process is

• A. $m{v}^2(3-\sqrt{2})$
• B. $\frac{1}{2}m{v}^2(3-\sqrt{2})$
• C. $2m{v}^2$
• D. $\frac{1}{2}m{v}^2(1+\sqrt{2})$

Answer 1

The correct option is A $m{v}^2(3-\sqrt{2})$

Net Momentum of three fragments
${\overrightarrow{P}}_i={\overrightarrow{P}}_f=0$
$\therefore$ Momentum of fourth part is $(\sqrt{2}-1)mv$ in opposite direction.
So, velocity of fourth part is $(\sqrt{2}-1)v$

$\therefore KE=3\left(\frac{1}{2}m{v}^2\right)+\frac{1}{2}m\left[\right(\sqrt{2}-1)v{]}^2$

$=(3-\sqrt{2})m{v}^2$