Question

a bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg what is the recoil velocity of the pistol

Answer 1

law of conservation of momentum
20*150=2000*x is equals 1.50 m/s
if any doubt than ask me in comment section

Answer 2

Answer:

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