Question

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

Answer 1

Answer:

MARK AS THE BRAINLIEST

Explanation:

Expert Answer:

Mass of bullet, m1 = 20g (= 0.02 kg)

Mass of pistol, m2 = 2 kg

Initial velocity of the bullet (u1) and pistol (u2) = 0

Final velocity of the bullet, v1 = +150m s-1

Let v be the recoil velocity of the pistol.

The total momentum of the pistol and bullet is zero before the fire. (Since both are at rest)

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)

= (3 + 2v) kg m s-1

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s

Answer 2

Answer:

Let,

    mass of bullet = m₁

   ................ pistol = m₂

  velocity of bullet = v₁

  ...................pistol = v₂

  ATQ,

           m₁ = 20 g

           m₂ = 2000 g

           v₁ = 150 ms⁻¹

 Now,

          m₁v₁ = m₂v₂

      ⇒ v₂ = 20×150÷2000 ms⁻¹

              = 1.5 ms⁻¹        

∵ pistol will move in the opp. direction w.r.t. bullet

∴ v₂ = - 1.5 ms⁻¹