A bullet of mass 15 g is horizontally fired with a velocity `100 ms^(-1)` from a pistol of mass 2 kg. What is the recoil velocity of the pistol ?

Explanation

The mass of the bullet, `m_(1) = 15g = 0.015 kg`
Mass of the pistol , `m_(2) = 2kg`
Initial velocity of the bullet , `u_(1) =0`
Initial velocity of the pistol , `u_(2) = 0`
Final velocity of the bullet, `v_(1) = +100 ms^(-1)`
(The direction of the bullet is taken from left to right -positive, by convention)
Recoil velocity of the pistol = `v_(2)`
Total momentum of the pistol and the bullet before firing.
`m_(1)u_(1)+m_(2)u_(2)`
`= (0.015xx0)+(2xx0)`
= 0
Total momentum of the pistol and bullet after firing.
`=m_(1)v_(1)+m_(2)v_(2)`
`=(0.015xx100)+(2xxv_(2))`
`= 1.5+2v_(2)`
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing.
1.5+`2v_(2)` = 0
`2v_(2) = -1.5`
`v_(2) = 0.75 ms^(-1)`
Negative sign indicates that the direction in which the pistol would recoil is opposite to that of the bullet , that is , right to left.