Question

A particle of mass $4\,m$ which is at rest explodes into three fragments. Two of the fragments each of mass $m $ are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is.............

Answer 1

From conservation of linear momentum $p_{3}$ should be $\sqrt{2} mv$ in a direction opposite to $p _{12}$ (resultant of $p _{1}$ and $p _{2}$ ).
Let $v^{\prime}$ be the speed of third fragment, then

$(2 m) v^{\prime} =\sqrt{2} m v $
$\therefore v^{\prime} =\frac{v}{\sqrt{2}}$
$\therefore$ Total energy released is,
$ E =2\left(\frac{1}{2} m v^{2}\right)+\frac{1}{2}(2 m) v^{\prime 2} $
$=m v^{2}+m\left(\frac{v}{\sqrt{2}}\right)^{2}=\frac{3}{2} m v^{2} $