Question

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate:
(a) the velocity with which the gun recoils.
​(b) the force exerted on gunman due to recoil of the gun

Answer 1

(a) Applying conservation of momentum, we can write here,

Momentum before bullet leaves the gun = Momentum after bullet leaves the gun

Mass of the gun = 3 kg, Mass of the bullet = 30 g

The initial velocity of the gun = Initial velocity of the bullet = 0

The final velocity of the bullet = 100 m/s

The initial momentum of the system = Final momentum of the system

0 + 0 = 30 $\times$ 10^-3 kg $\times$ 100 m/s + 3kg $\times$ velocity of the gun

The velocity of the gun = $-\frac{3}{3}=-1\mathrm{m}/\mathrm{s}$

Hence, the recoil velocity of the gun is 1 m/s.

(b) Change in the momentum of the gun after the bullet leaves the barrel = 3 kg $\times$ 1 m/s = 3 kg m/s

Time take by the bullet to leave the barrel = 0.003 s

Force exerted by the recoil of the gun on the gunman = Rate of change of momentum of the gun = $\frac{3\mathrm{kg}\mathrm{m}/\mathrm{s}}{0.003\mathrm{s}}=1000\mathrm{N}$