According to question, the given situation is shown in the following figure.
Mass of block, M=2kg Mass of bullet, m1=0.01kg Speed of bullet, v=500ms−1 h=0.1 {m} If v1 and v2 be the velocities of the bullet and blocks after collision, then by conservation of energy,
1
2
Mv22=Mgh v2=√2gh=√2×9.8×0.1=1.4ms−1 According to the law of conservation of momentum, m1v=m1v1+Mv2 ⇒v1=