# 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?

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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?

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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.