Conservation Of Momentum

Since no external horizontal force acts on the gun-bullet system during firing, total linear momentum is conserved and must remain zero, as it was before firing. Let the bullet move forward with momentum $m_b{v}_b$ and the gun recoil backward with momentum $m_g{v}_g$. Conservation requires $m_g{v}_g=m_b{v}_b$, so $v_g=\frac{m_b{v}_b}{m_g}=\frac{0.02\times 250}{5}=\frac{5}{5}=1.0\text{ m/s}$. The 0.5 m/s value would result from doubling the gun's mass mistakenly. The 2.0 m/s value comes from using half the gun mass. The 5.0 m/s value equals the bullet's momentum without dividing by the gun's mass, which is dimensionally inconsistent for a speed. It is instructive to note that although total momentum stays zero, kinetic energy is not conserved here, because the chemical energy of the propellant is converted into the kinetic energy of both bodies, with the light bullet carrying the overwhelming share. This is the NCERT recoil illustration of momentum conservation, also the basis of rocket propulsion, where continuously expelled exhaust drives the rocket the opposite way. As a check, the heavy gun recoils far slower than the light bullet flies, exactly as the inverse mass ratio of 5 kg to 0.02 kg predicts.