Question

A bullet of 40 gm is fired horizontally with a velocity of 160 ms^–1 from a pistol weighing 2 kg. What is the rebound velocity of the pistol?
1. –1.5 ms^–1
2. –3.2 ms^–1
3. –1.25 ms^–1
4. –2.0 ms^–1

Answer 1

Correct Answer - Option 2 : –3.2 ms^–1

The correct answer is –3.2 ms–1.

• Given, 
  • Mass of bullet, m₁ = 40g (= 0.04 kg)
  • Mass of pistol, m₂ = 2 kg
  • The initial velocity of the bullet (u₁) and pistol (u₂) = 0
  • Final velocity of the bullet, v₁ = +160m s^-1
  • Let, v₂ be the recoil velocity of the pistol.
  • The total momentum of the pistol and bullet is zero before the fire because both are at rest.
  • The total momentum of the pistol and bullet after it is fired is

                     = (0.0 kg 4x 160 m s^-1) + (2 kg x v₂ m s^-1)

                     = (6.4 + 2v₂) kg m s^-1

• Total momentum after the fire = Total momentum before the fire
•  6.4 + 2v₂ = 0
• →v₂ = 3.2 m/s  
• Thus, the recoil velocity of the pistol is 3.2 m/s.