A bullet of mass 0.01 kg and travelling at a speed of 500 ms-1 strikes a block of mass 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to raise a vertical distance of 0.2 m. What is the speed of the bullet after it emerges from the block?

Conservation of momentum just before and after the collision yields          mu=mv1+Mv2            …(i)Conservation of energy of the block between the point 1 and 2 after the bullet pieces It yields ΔKE+ΔPE=0−12Mv22+Mgh=0⇒v2=2gh             …(ii)By using (i) and (ii), we obtain         u=v1+Mm2gh⇒v1=u−Mm2gh⇒ v1=500−20.012×10×0.2=100ms−1