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,

M_{1} =3 kg(Given)

V_{1} =?

For bullet,

M_{2}= 30 g = 0.03kg (As 1kg=1000g)

V_{2} = 100m/s

By law of conservation of momentum

M_{1}× V

_{1}= M

_{2}× V

_{2}

3 × V_{1} = 0.03 × 100

_{1}= 0.03 × 100/ 3

**V**

_{1}= 1m/sHence, the velocity of recoil of gun is 1m/s

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

100-0/0.003 = 33333.33m/s^{2}

We know that:-

Force, F = ma

= 0.03 × 33333.33

= **1000 N**

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