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 :
(i) the velocity with which the gun recoils.

Answer 1

Answer:

Recoil velocity of gun will be 1 m/s.

Explanation:

Here, the momentum remains conserved.

So,

$m_{bullet}~*~v_{bullet} = m_{gun}~*~v_{gun}$

3 × 10⁻² × 100 = 3 × $v_{gun}$

$v_{gun}$ = 1 m/s

Answer 2

Answer: 1m/s

Explanation

Mass of the gun m1 = 3 kg

Mass of bullet m2 = 30 g = 0.03 kg

Velocity of bullet v2 =100 m/s

(i) According to the law of conservation of momentum

m1 x v1 = m2 x v2

3 x v1 =0.03 x 100

recoil velocity v1=(100x0.03)/3

=1m/s

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