Question

A particle of mass $4$ m which is at rest explodes into fragments.The fragment each of the mass M are found to move with speed V each in mutually perpendicular directions.The minimum energy released in the process of explosion is

Answer 1

$2m{v}^{\prime }=\sqrt{2}mv$

$V^{\prime }=\frac{v}{\sqrt{2}}$

$KE=2\frac{1}{2}m{v}^2$ + $\frac{1}{2}2m\frac{v^2}{2}$

$=m{v}^2+\frac{m{v}^2}{2}$

$=\frac{3}{2}m{v}^2$