A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular directions each with speed v. The energy released during the process is [NEET (Odisha) 2019]

Solution:

The particle of mass 5m breaks into three fragments of masses m, m and 3m, respectively. Two fragments of 
mass m each move in perpendicular directions with velocity v and the left fragment will move in a direction with velocity v' such that the total momentum of the system must remain conserved.

By law of conservation of momentum, 

                   $5m\times 0=mv\hat{\mathrm{i}}+mv\hat{\mathrm{j}}+3m\mathrm{v}\text{'}$

$\Rightarrow$                           $\mathrm{v}\text{'}=-\frac{\mathrm{v}}{3}\hat{\mathrm{i}}-\frac{\mathrm{v}}{3}\hat{\mathrm{j}}$

$\therefore$                  $\left|{\mathbf{v}}^{\text{'}}\right|=\sqrt{{\left(-\frac{v}{3}\right)}^2+{\left(-\frac{v}{3}\right)}^2}=\frac{v\sqrt{2}}{3}$

$\therefore$  Energy released, 

                      $E=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\times 3m{\left(\frac{v\sqrt{2}}{3}\right)}^2$

                         $=m{v}^2+\frac{m{v}^2}{3}=\frac{4}{3}m{v}^2$