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.) 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 ms^(-1)`
` =0 kg ms^(-1)`
Total momenta of the pistol and bullet after it is fired
` = -0.02kg xx (+ 150 ms^(-1))`
+ 2kg xx vms^(-1) = (3 + 2v) kg ms^(-1)`
According to the law of conservation of momentum
Total momenta after the fire = Total momenta before the fire
` = 3 + 2v = 0`
`rArr v = - 1.5 ms^(-1)`
Negative sign indicates that the direction in which the pistol would is opposite to that of bullet, that is, right to left.