A bullet of mass 20g is horizontally fired with a velocity 150 ms-1 from a pistol of mass 2kg. What is the recoil velocity of the pistol ?
We have the mass of bullet, m1 = 20 g (= 0.02 kg) and the mass of the pistol, m2 = 2 kg; initial velocities of the bullet (u1) and pistol (u2) = 0, respectively. The final velocity of the bullet, v1 = + 150 m s-1. The direction of bullet is taken from left to right (positive, by convention, Fig. 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 × 0 m s-1
= 0 kg m s-1
Total momenta of the pistol and bullet after it is fired
= 0.02 kg × (+ 150 m s-1)
+ 2 kg × 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
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.
