Physics, asked by lakshit07mehta, 2 days ago

A bullet of mass 25 gram fired at 100 m/s from a gun of mass 5 kg.Calculate recoil velocity of gun. In this process which quantity is constant.​

Answers

Answered by Johnsonmijo
0

Answer:

The recoil velocity of the gun is -0.5 m/s. In this process, the momentum of the system remains constant.

Explanation:

Mass of the bullet, m = 25 g = 0.025 Kg

Velocity of the bullet, v = 100 m/s

Mass of the gun, M = 5 Kg

Let the recoil velocity of the gun = V

The momentum of the bullet, P₁ = mv = 0.025 × 100 = 2.5 Kgm/s

The momentum of the gun, P₂ = MV = 5V Kgm/s

∴ Final momentum of the gun, P = mv + MV

                                                      = (2.5 + 5V) Kgm/s

Since the gun is at rest before firing, the initial momentum of the gun will be zero.

By the law of conservation of momentum,

Final momentum = Initial momentum

⇒ 2.5 + 5V = 0

             5V = -2.5

               V = -2.5 / 5

∴ Recoil velocity of the gun, V = -0.5 m/s

In this process, the momentum before and after firing remains constant.

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