Q. 334.4( 18 Votes )

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:

(i) The velocity with which the gun recoils.

(ii) The force exerted on gunman due to recoil of the gun

Answer :

(i) For gun,

M1 =3 kg(Given)


V1 =?


For bullet,

M2 = 30 g = 0.03kg (As 1kg=1000g)

V2 = 100m/s


By law of conservation of momentum

M1 × V1 = M2 × V2

3 × V1 = 0.03 × 100

V1 = 0.03 × 100/ 3
V1 = 1m/s

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


(ii) Acceleration of bullet, a = v-u/t


100-0/0.003 = 33333.33m/s2


We know that:-

Force, F = ma


= 0.03 × 33333.33


= 1000 N

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