A bullet with a mass of 20 g travelling horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string. The bullet passes through the block and emerges on the opposite side. Determine the speed of the bullet as it emerges from the block, given that the centre of mass of the block rises through a height of 20.0 cm .

Step-by-step solution

Given mass of the bullet m = 20 g = 0.02 kg and its initial velocity u = 300 m/s . Mass of the wooden block M = 500 g = 0.5 kg . Let v be the velocity of the bullet as it emerges and V be the velocity of the block just after the bullet passes through it. By conservation of mechanical energy for the block after the collision: 1 2 M V^2 = M g h V = 2 g h Substituting g = 10 m/s ^2 and h = 0.2 m : V = 2 10 0.2 = 2 m/s By conservation of linear momentum during the collision: m u = m v + M V 0.02 300 = 0.02 v + 0.5 2 6