Question

a body of mass M is hung by a long thread and a bullet of mass m hits it horizontally with a velocity v and gets embedded in the body. Then for the body and the bullet system the momentum and kinetic energy are

Answer 1

body of mass M has velocity = 0
bullet of mass m has velocity = v

After bullet is embedded in the body,
let the velocity of the system = V

The momentum of the system is conserved
So momentum of system = m×v + M×0 = mv

Also;
mv + M×0 = (M+m)V
⇒mv= (M+m)V
⇒ V = $\frac{mv}{M+m}$

$KE= \frac{1}{2} \times (M+m) \times V^2 \\ \\KE=\frac{1}{2} \times (M+m) \times ( \frac{mv}{M+m} )^2\\ \\ KE= \frac{(mv)^2}{2(M+m)}$