Question

A 60 gram Bullet fired from a 5 kilogram gun leaves with a speed of 500 meter/second.
find the speed with which the gun recoils

Answer 1

Mass of gun (M) = 5 kg.
Mass of bullet (m) = 60 g = 0.06 kg.
Velocity of gun after firing (V) = V.
Velocity of bullet after firing (v) = 500 m/s.

V = (-m/M)v.
V = (-0.06/5)*500.
V = (-0.012) * 500.
V = -6 m/s.

Therefore, the gun recoils with a velocity of -6 m/s. Negative(-) sign here indicates that the gun is moving in the direction opposite to that of the bullet.

By using the law of conservation of momentum:

m₁u₁+m₂u₂=m₁v₁+m₂v₂.
5*0+0.06*0=5*V+0.06*500. [u₁ = 0, u₂ = 0 as initial velocity is 0 of both gun and bullet).
0+0=5V+30.
0-30=5V.
-30/5=V.
V=-6 m/s.

Thus, we can see that the recoil of the gun is -6 m/s.