• # question_answer 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 is mutually perpendicular direction. The energy released in the process of explosion will be: A)  $1/4m{{V}^{2}}$             B)  $3/2m{{V}^{2}}$C)  $1/2m{{V}^{2}}$             D)  $3/4m{{V}^{2}}$

From the law of conservation of momentum From question, initial momentum = 0, so that final momentum should be zero. Let V be the velocity of mass 2m then $\therefore$    $\sqrt{{{(mV)}^{2}}+{{(mV)}^{2}}}=2m{{V}_{1}}$ or         ${{V}_{1}}=\frac{V}{\sqrt{2}}$                           .....(i) $\therefore$ K.E. Total = K.E (First piece) + K.E. (Second piece) + K.E (Third piece)             $K\,E=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}(2m){{V}_{1}}^{2}$             $=2\left( \frac{1}{2}m{{V}^{2}} \right)+m\left( \frac{{{V}^{2}}}{2} \right)$ $\Rightarrow$            $K.{{E}_{after\,\,\exp losion}}=\frac{3}{2}m{{V}^{2}}$ Also, $K.{{E}_{Before\,\,\exp losion}}=0$ Hence, Piece in K.E $=\frac{3}{2}m{{V}^{2}}$