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?

Solution:
We have the mass of bullet,
m₁ = 20 g (= 0.02 kg) and the mass of
the pistol, m₂ = 2 kg; initial velocities of the bullet (u₁ ) and pistol (u₂ ) = 0,respectively. The final velocity of the bullet, v₁ = + 150 m s^-1. The direction
of bullet is taken from left to right (positive, by convention. 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
3 + 2v = 0
⇒ v = − 1.5 m s^–1
.
Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left.