Here, mass of the shell, $m_{1}=0.02\ \mathrm{kg};$ mass of the gun, $m_{2}=100\ \mathrm{kg}$ Initial velocities of both the shell and the gun are zero i.e. $u_{1}=u_{2}=0$ After firing, speed of the shell, $v_{1}=80\ \mathrm{m}\ \mathrm{s}^{-1}$Let $v_{2}$ be the recoil speed of the gun. According to the principle of conservation of momentum,$m_{1} v_{1}+m_{2} v_{2}=m_{1} u_{2}+m_{2} u_{2}$$\therefore 0.02 \times 80+100 v_{2}$$=0.02\times 0+100\times 0$ or $v_{2}=-\frac{0.02 \times 80}{100}=-0.016\ \mathrm{ms}^{-1}$