Question

A gun of mass 4.5 kg fires a bullet of mass 20 g with a velocity of 30 m/s. What will be it’s recoil velocity....

Answer 1

Answer:

$\bigstar{\bold{Recoil\:velocity=-0.13\:m/s}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Mass of gun (m₁) = 4.5 kg
• Mass of the bullet (m₂) = 20 g = 0.02 kg
• Velocity of bullet (v₂) = 30 m/s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Recoil velocity of the gun (v₁)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Here we have to find the recoil velocity of the gun

→ We know that momentum is conserved in an isolated system

→ Hence the momentum before firing = momentum after firing

→ Recoil velocity is given by the formula,

  v₁ = -(m₂ v₂)/ m₁

→ Substituting the given datas we get,

  v₁ = -(0.02 × 30)/4.5

  v₁ = -0.6/4.5

  v₁ = -0.13 m/s

→ Hence the recoil velocity of the gun is -0.13 m/s

→ Here recoil velocity is negative since it is in the direction opposite to the direction of the bullet.

$\boxed{\bold{Recoil\:velocity=-0.13\:m/s}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Momentum can neither be created nor be destroyed.

→ The total momentum in an isolated system is always 0. That is the momentum before collissio = momentum after collission.