A bullet of mass 20 gm is fired in the horizontal direction with a velocity 150 m/s from a pistol of mass 1 kg. Recoil velocity of the pistol is

The problem is solved using the law of conservation of linear momentum, which states that the total momentum of an isolated system remains constant. Initially, both the pistol and the bullet are at rest, so the total initial momentum is zero. Upon firing, the momentum of the bullet must be balanced by the recoil momentum of the pistol in the opposite direction. The mass of the bullet is 20 g, which converts to 0.02 kg. Using the formula m1v1 + m2v2 = 0, where m1 is the mass of the bullet (0.02 kg), v1 is its velocity (150 m/s), m2 is the mass of the pistol (1 kg), and v2 is the recoil velocity, we get: (0.02 kg * 150 m/s) + (1 kg * v2) = 0. This simplifies to 3 + v2 = 0, resulting in v2 = -3 m/s. The magnitude of the recoil velocity is 3 m/s.