Question

A bullet of mass 20 g
horizontally fired with a velocit
150 m s' from a pistol of mass 2 kg .What is the recoil velocity of the pistol ?​

Answer 1

Answer:

According to the Law of Conservation of Linear Momentum,

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

 ∴ 0 + 0 = m₁v₁ + m₂v₂

⇒ v₂ = - (m₁v₁)/m₂

         = - (0.02 × 150)/2

         = - 1.5 m/s

∴ The recoil velocity of the pistol is 1.5 m/s.                      Ans.

Answer 2

Answer:

here we use law of conservation of momentum.

Explanation:

0+0=m1 v1+m2v2

0= m1v1/m2v2

0= 0.02×150/2

0= 1.5m/s

recoil velocity =1.5m/s