A plate remains in equilibrium in air when n bullets are fired per second on it. The mass of each bullet is m and it strikes the plate with speed v. The coefficient of restitution is e=0.5. The mass of the plate is

The change in momentum of one bullet=final momentum-initial momentum

$$\begin{gathered} =\mathrm{m}\left(\frac{\mathrm{v}}{2}\right)-\left(-\mathrm{mv}\right)=\frac{3}{2}\mathrm{mv} \\ \mathrm{Also}, \mathrm{given} \mathrm{that} \mathrm{n} \mathrm{bullet} \mathrm{strikes} \mathrm{per} \mathrm{second}. \\ \therefore \mathrm{Force} \mathrm{exerted} \mathrm{by} \mathrm{the} \mathrm{bullets}=\frac{3}{2}\mathrm{mnv} \\ \mathrm{Since}, \mathrm{the} \mathrm{plate} \mathrm{remains} \mathrm{in} \mathrm{equilibrium} \\ \therefore \mathrm{Force} \mathrm{exerted} \mathrm{by} \mathrm{the} \mathrm{bullets}=\mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{plate} \\ \frac{3}{2}\mathrm{mnv}=\mathrm{Mg} \mathrm{or} \mathrm{M}=\frac{3\mathrm{mnv}}{2\mathrm{g}} (\mathrm{where} \mathrm{M}=\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{plate}) \end{gathered}$$