Explanation

(c) Mass of the shell $= m_{1} = 0.2\ kg$
Mass of the gun $= m_{2} = 4kg$
Let energy of shell $= E_{1},$ energy of gun $= E_{2}$
Total energy liberated
$= E_{1} + E_{2} = 1050\ Joule$ …(i)
As $E = \frac{P^{2}}{2m}$
$\therefore\frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}} = \frac{4}{0.2} = 20 \Rightarrow E_{2} = \frac{E_{1}}{20}$ …(ii)
From equation (i) and (ii) we get $E_{1} = 1000\ Joule$
$\therefore$ Kinetic energy of the shell $= \frac{1}{2}\ m_{1}v_{1}^{2} = 1000$
$\Rightarrow \frac{1}{2}\ (0.2)v_{1}^{2} = 1000 \Rightarrow v_{1} = \sqrt{10000} = 100\ m/s$