• # question_answer A particle of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed v at right angles to each other. The kinetic energy released in the process is A)  $2\,m{{v}^{2}}$                B)  $\frac{3}{2}\,m{{v}^{2}}$ C)  $\frac{1}{2}\,m{{v}^{2}}$                       D)  $3\,m{{v}^{2}}$

The resultant momentum of two particles which fly-off at right angles is $\Rightarrow$. By law of conservation of momentum $\Rightarrow$ $|{{M}^{-1}}|\,|adj\,(adjM)|\,=\,|kI|$    $=100\,\left( \frac{\Delta \phi }{\Delta t} \right)$ Total energy released E is the sum of kinetic energies of all the three particles $=5\times {{10}^{-3}}\times 100=0.5\,\,V$ $R=v\sqrt{\frac{2h}{g}}$ $u=v,$            $u=v+at$