Question

A gun of mass 2kg fired a bullet of mass 1.6×10-2kg due east if the bullet leaves the gun with a velocity of 15m/s calculate the recoil of the gun and it's direction

Answer 1

Given,

M = 2 kg

m = 1.6 × 10^(-2) kg

v = 15 m s^(-1) due East

We have to find V.

Before firing, both were in rest so the total linear momentum is 0 because u = U = 0.

After firing, according to the law of conservation of linear momentum,

mv + MV = mu + MU = 0

V = - mv / M

V = - 1.6 × 10^(-2) × 15 / 2

V = - 0.12 m s^(-1)

The negative sign shows that the velocity of the gun is in the direction opposite to that of the bullet.

Hence, the velocity of the gun is 0.12 m s^(-1) due West.