Question

a bullet of mass 30g is fired horizontally with a velocity of 120 m/s from a pistol of mass 2 kg , the rocoil velocity of the pistol is​

Answer 1

Explanation:

0.03 kg

v=120m/s

p=m*v

0.03*120

=3.6 kgm/s

3 kg *v=3.6kgm/s

v=3.6/3

1.2

Answer 2

Answer:

1.8 m/sec

Explanation:

Given :

Mass of bullet [m] = 30 gram

Velocity = 120 m/sec

Mass of pistol [M] = 2 kg = 2*10⁻³ grams

To Find :

recoil velocity of the pistol = ?

Solution :

Due to the law of conservation of momentum, the momentum of the bullet is equal to the recoil of the gun. So the sum of the momentum of both the objects is equal to zero.

Therefore,

[mass of bullet] [velocity of bullet] = - [mass of pistol] [velocity of pistol]

30*120 = - 2*10⁻³*[velocity of pistol]

[velocity of pistol] = - [30*120 ]/2*10⁻³

[velocity of pistol] = - 1.8 m/sec

−ve sign is ignored in the final answer as it denotes the direction of the velocity of the pistol.

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