Question

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a gun of mass 40kgs fires a bullet of mass 0.01kg with a velocity of 400m/seconds, then the recoil velocity is?
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Answer 1

In the given question,

◉ Mass of gun, M = 40 Kg

◉ Mass of Bullet, m = 0.01 Kg

◉ Velocity at which bullet is fired from the gun, u = 400 m/s

We need to find the recoil velocity of the gun. In other words, Velocity at which the gun would go in opposite direction after firing the bullet.

So, by the Law of conservation of momentum,

⇒ Momentum of Gun & Bullet system before Firing = Momentum of Gun & Bullet system after firing

⇒ M × u₁ + m × u₂ = M × v₁ + m × v₂

Since, Before firing the bullet the gun and bullet were at rest simultaneously, Hence, Initial velocity of the two given bodies would be 0 m/s.

Also, We took the recoil velocity as v₁

⇒ 40 × 0 + 0.01 × 0 = 40v₁ + 0.01 × 400

⇒ 0 = 40v₁ + 4 kg. m/s

⇒ 40v₁ = -4

⇒ v₁ = -0.1 m/s

The negative sign indicates that the gun would move in the opposite direction due to which shooters usually put the gun in the support of thier shoulders control the gun.

∴ Recoil Velocity of Gun = 0.1 m/s