Question

A bullet of mass 0.02 kg is fired from a gun weighing 7.5 kg .if the intial velocity of the bullet is 200m/s .calculate the speed with which the gun recoils

Answer 1

m1v1 = m2v2

Mass of bullet = 0.02 kg

Velocity of bullet = 200 m/s

Mass of gun = 7.5 kg

Velocity of recoil = x m/s

0.02 * 200 = 7.5 * x     (Since m1v1 = m2v2)

4 = 15/2 * x

x = 4 * 2/15

x = 8/15 m/s = 0.5333 m/s

By sign convention , since the recoil is in the opposite direction , recoil velocity = -0.5333 m/s

Answer 2

Given :-

mass of the bullet (m₁) = 0.02 kg

mass of the gun (m₂) = 7.5 kg

Velocity of the bullet (v₁) = 200 m/s

and the speed of the gun (v₂) = ?  (not given)

So,

To be found :-

the recoil velocity of the gun

Now,

According to the law of conservation of momentum,

The total momentum of system after firing will be equal to total momentum of system before firing

⇒ m₁v₁ + m₂v₂ = 0

[ ∴ as initial velocities of the gun and bullet is 0 before firing ]

⇒ 0.02 × 200 + 7.5 × v₂ = 0

⇒ 4 + 7.5 × v₂ = 0

⇒ 7.5 × v₂ = -4

⇒ v₂ = $\bf \frac{-4}{7.5}$

⇒ v₂ = 0.53 m/s

Hence,

The recoil velocity of gun is 0.53 m/s