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 = [tex] \frac{mv}{M+m} [/tex]
[tex]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)} [/tex]
