A bullet of mass `20g` is fired horizontly with a velocity of `150ms^(-1)` from a pistol of maass `2kg`. What is the recoil velocity of the pistol?

Explanation

We have the mass of bullet,
`m_(1)` = 20 g (= 0.02 kg) and the mass of the pistol, `m_(2)` = 2 kg, initial velocities of the bullet `(u_(1))` and pistol `(u_(2))` = 0, respectively. The final velocity of the bullet, `v_(1)` = + 150 m `s^(-1)`. The direction of bullet is taken from left to right (positive, by convention, Fig. 9.17). Let v be the recoil velocity of the pistol.
Total momenta of the pistol and bullet before the fire, when the gun is at rest
`= (2 + 0.02) kg xx 0 m s^(–1)`
`= 0 kg m s^(–1)`Total momenta of the pistol and bullet after it is fired
`= 0.02 kg xx (+ 150 m s^(–1))` `+ 2 kg xx v m s^(–1)`
`= (3 + 2v) kg m s^(–1)`
According to the law of conservation of momentum
Total momenta after the fire = Total momenta before the fire
3 + 2v = 0
`implies v = − 1.5 m s^(–1)`. Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left.