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

Explanation

The mass of bullet, m -20 g (= 0.02 kg) and the mass of the pistol, `m_(2)`=2 kg: initial velocities of the bullet and pistol are `u_(1)` =0 and `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 u be the recoll velocity of the pistol. Total momenta of the pistol and bullet before the fire, when the gun is at rest =0 kg m `s^(-1)`. Total momenta of the pistol and bullet after it is fired -0.02 kg x (+150 m`s^(-1)`) + `2 kg xxvms^(-1)` According to the law of conservation of momentum Total momenta after the fire - Total momenta before the fire 3 + 2 = 0`implies -1.5 ms^(-1)`
Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is right to left.