Question

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

Answer 1

mass of of pistol m1=2kg
mass of bullet m2=20g=20/1000=0.2kg
initial velocity of the pistol and bullet = 0m/s
final velocity of the bullet v2=150m/s
let the recoil velocity of the pistol =v1
we know that,
m1v1+m2v2=0
v1=-m2/m1 x v2
V1=-(0.2/2)*150
V1=-15m/s
The -ve sign indicate that the recoil velocity of the pistol is opposite to the direction of the speed of the bullet.

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