Question

A 25 g bullet is fired horizontally with a velocity of 150 ms^-1 from a pistol of 2 kg. What is the expected velocity of the pistol?
1. - 1.25 ms^-1
2. - 1.5 ms-1
3. - 2.0 ms-1
4. - 1.87 ms-1

Answer 1

Correct Answer - Option 4 : - 1.87 ms-1

The correct answer is - 1.87 ms-1.

• Given Mass of bullet, m1 = 25 g = 0.025 kg
  • Mass of pistol, m2 = 2 kg
  • The initial velocity of the bullet (u1) and pistol, u2 = 0
  • The final velocity of the bullet, v1 = 150 ms-1
  • Let the recoil velocity of the pistol be v
  • 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
  • = \((0.025 kg * 150 ms^{-1}) + (2 kg * v {ms^{-1}})\)
  • On calculation, we get,
    • = (3.75 + 2v) kg ms-1
  • Total momentum after the fire = Total momentum before the fire
    • 3.75 + 2v = 0
    • 2v = -3.75
    • v = -3.75/2
    • v = -1.875 m/s
• Hence, the recoil velocity of the pistol is -1.875 m/s.