Please Answer This.

Dear Student,
let the velocity of mass m be v =16m/s (given)
and velocity of mass 2m be v'.

as no external force is applied, momentum of the system will be remain conserved.

$$\begin{gathered} \mathrm{From} \mathrm{law} \mathrm{of} \mathrm{conservation} \mathrm{of} \mathrm{momentum}, \\ \mathrm{Momentum} \mathrm{of} \mathrm{the} \mathrm{mass} \mathrm{m} = \mathrm{Momentum} \mathrm{of} \mathrm{the} \mathrm{mass} 2\mathrm{m} \\ \mathrm{mv}=2\mathrm{mv}\text{'} \\ \mathrm{v}\text{'}=\frac{\mathrm{v}}{2} \\ =\frac{16}{2} \\ =8 \mathrm{m}/\mathrm{s} \\ \mathbf{Total}\mathbf{ }\mathbf{kinetic}\mathbf{ }\mathbf{energy}\mathbf{ }\mathbf{released}\mathbf{:} \\ {\mathrm{E}}_{\mathrm{total}}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left[2\mathrm{m}\right]{\left[\mathrm{v}\text{'}\right]}^2 \\ =\frac{1}{2}\mathrm{m}\left[{16}^2+\left(2\times 8^2\right)\right] \\ \mathbf{=}\mathbf{192}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{joules} \end{gathered}$$

So, the correct option is option 1. 192 mJ

Regards,