A bullet of mass 20 g is horizontally fired with a velocity 150 m s^-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Given, mass of bullet, m₁ = 20 g (= 0.02 kg)

The mass of the pistol, m₂ = 2 kg

Initial velocities of the bullet (u₁) and pistol (u₂) = 0

The final velocity of the bullet, v₁ = + 150 m s^-1

Let v be the recoil velocity of the pistol

Total momenta of the pistol and bullet before the fire, when the gun is at rest

= (2 + 0.02) kg × 0 m s^–1

= 0 kg m s^–1

Total momenta of the pistol and bullet after it is fired

= 0.02 kg × (+ 150 m s^–1) + 2 kg × v m s^–1

= (3 + 2v) kg m s^–1

According to the law of conservation of momentum,

Total momenta after the fire = Total momenta before the fire

Þ 3 + 2v = 0

⇒ v = − 1.5 m s^–1

The negative sign indicates that the direction in which the pistol would recoil is opposite to that of the bullet, that is, right to left.