Question

a bullet of mass 20 gm is fired from a pistol of 2 kg if velocity of bullet is.150m the find the recoil velocity

Answer 1

According to Law of conseravtion of momentum ,

Let recoil velocity be x


Momentum of Pistol = Momentum of Bullet
2x = 1/50 × 150
2x = 3
x = 3/2

Answer 2

Answer:

1.5m/s

Explanation:

Mass of bullet=M₁=20g=0.02kg

Mass of gun=M₂=2kg

velocity of bullet initially at rest=u₁=0m/s

velocity of gun initially=u₂=0m/s

velocity of bullet after firing=v₁=150m/s

velocity of gun after firing=v₂

according to law of conservation of momentum,

M₁u₁+M₂u₂=M₁v₁+M₂v₂

=0.02×0+2×0=0.02×150+2×v₂

=0+0=3+2v₂

= -3=2v₂

v₂=

v₂=-1.5m/s

the recoil velocity of the bullet is 1.5m/s

Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet