Question

A particle of mass $4m$ 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

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