Question

A bullet of mass m moving with a velocity u strikes a block of mass M at rest and gets embedded in the block. The loss of kinetic energy in the impact is

Answer 1

Dear student

A bullet of mass m moving with a velocity u strikes a block of mass M at rest

$TotalK.EofBlock+Masssystem=\frac{1}{2}m{u}^2+\frac{1}{2}M(0{)}^2=\frac{1}{2}m{u}^2$
Apply conservation of momentum on Block + Mass system before and after collision to find the final speed of Block+Mass

Total initial momentum = mu+M(0)= mu

let final speed of Block + Mass system be V........( since one gets embedded in other, both will have same speed).

Final momentum = (m+M)V

So, mu=(m+M)V

or, $V=\frac{mu}{m+M}$

$$\begin{gathered} So,finalK.E=\frac{1}{2}(m+M)V^2 \\ =\frac{1}{2}(m+M)(\frac{mu}{m+M}{)}^2 \\ =\frac{m^2{u}^2}{2(m+M)} \\ \Rightarrow lossinK.E=initialK.E-finalK.E \\ =\frac{1}{2}m{u}^2-\frac{m^2{u}^2}{2(m+M)}=\frac{1}{2}mu(u-\frac{u}{m+M})J \end{gathered}$$

Hence the answer.

Regards