Question

A bullet of mass 15 g is horizontally fired with a velocity 100 ms^-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Answer 1

The mass of the bullet, m₁ = 15 g = 0.015 kg 

Mass of the pistol, m₂ = 2 kg 

Initial velocity of the bullet, u₁ = 0 

Initial velocity of the pistol, u₂ = 0 

Final velocity of the bullet, v₁ = + 100 ms^-1 

(The direction of the bullet is taken from left to rightpositive, by convention) 

Recoil velocity of the pistol = v₂ 

Total momentum of the pistol and bullet before firing.

= m₁ u₁ + m₂ u₁ 

= (0.015 × 0) + (2 × 0)

= 0

Total momentum of the pistol and bullet after firing. 

= m₁v₁ + m₂v₂ 

= (0.015 × 100) + (2 × v₂) 

= 1.5 + 2v₂

According to the law of conservation of momentum, Total momentum after firing = Total momentum before firing.

1.5 + 2v₂ = 0

2v₂ = -1.5

v₂ = – 0.75 ms^-1 

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