A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates…

Let $m_1 = 4$ kg (gun) recoil with speed $v_1$ and $m_2 = 0.2$ kg (shell) move with speed $v_2$.
Conservation of momentum (system initially at rest): $m_1 v_1 = m_2 v_2$
$4\,v_1 = \dfrac{200}{1000}\,v_2 \Rightarrow v_1 = \dfrac{v_2}{20}$ ...(1)
The energy of the explosion appears as kinetic energy: $\dfrac{1}{2}m_1 v_1^2 + \dfrac{1}{2}m_2 v_2^2 = 1.05 \times 10^{3}$ J
$\dfrac{1}{2} \times 4 \times v_1^2 + \dfrac{1}{2} \times 0.2 \times v_2^2 = 1050$
$2v_1^2 + 0.1\,v_2^2 = 1050$ ...(2)
Substituting (1) in (2): $2 \times \dfrac{v_2^2}{400} + 0.1\,v_2^2 = 1050$
$0.005\,v_2^2 + 0.1\,v_2^2 = 1050 \Rightarrow 0.105\,v_2^2 = 1050$
$v_2^2 = 10000 \Rightarrow v_2 = 100$ m/s
So the initial velocity of the shell is 100 m s$^{-1}$.