A bullet of mass 20 g moving 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 crosses the block and emerges on the other side. If the center of mass of the block rises through a height of 20.0 cm, find the speed of the bullet as it emerges from the block.

Mass of bullet is m = 20gm = 0.02kg,


Mass of wooden block is M = 500gm = 0.5kg,


Initial velocity of block is u’ = 0 m/sec,


Velocity of the bullet with which it strikes u = 300 m/sec.


Let the bullet emerges out with velocity is v and the velocity of block is v’.


According to law of conservation of momentum



______ (1)


Again applying work-energy principle for the block after the collision,








Substituting the value of v’ in the equation (1) we get


0.02 × 300 = 0.5 × 2 + 0.2 × v


v= 250m/sec


Thus, the speed of the bullet as it emerges from the block is v= 250 m/s.


1