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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.
ā€‹(b) the force exerted on gunman due to recoil of the gun

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Solution

(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 × 10-3 kg × 100 m/s + 3kg × velocity of the gun

The velocity of the gun = -33=-1 m/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 × 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 = 3 kg m/s0.003 s=1000 N

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