A bullet of mass $m$ and velocity $v$ when fired at a sand bag of mass $M$, suspended by a string, gets…
A bullet of mass $m$ and velocity $v$ when fired at a sand bag of mass $M$, suspended by a string, gets embedded into the bag. The loss of kinetic energy in this process is
$\frac{m v^2}{2}$
$\frac{m v^2}{2(M+m)}$
$\frac{M v^2}{2}$
$\frac{m M N^2}{2(M+m)}$
Solution
Mass of the buIlet $=m$
Speed of bullet $=v$
According to question, bullets gets embedded into the bag, then they will move with common velocity $v_1$ (say).
This is the case of perfectly enelastic collision.
$\therefore$ Initial kinetic energy of bullet,
$
K_i=\frac{1}{2} m v^2
$
By the conservation of linear momentum,
$
\Rightarrow \quad \begin{aligned}
m v & =(M+m) v_1 \\
\Rightarrow \quad & v_1=\frac{m v}{M+m}
\end{aligned}
$
$\therefore$ Final kinetic energy, $K_f=\frac{1}{2}(M+m) v_1^2$
$
\begin{aligned}
& =\frac{1}{2}(M+m) \frac{m^2 v^2}{(M+m)^2} \quad \text { [from Eq. (ii)] } \\
& =\frac{1}{2} \frac{m^2 v^2}{M+m}
\end{aligned}
$
Loss in kine tic energy $=K_i-K_f$
$
\begin{aligned}
& =\frac{1}{2} m v^2-\frac{1}{2} \frac{m^2 v^2}{M+m} \\
& =\frac{1}{2} m v^2\left(1-\frac{m}{M+m}\right)=\frac{m M v^2}{2(M+m)}
\end{aligned}
$