A bar of mass M = 1 . 00   kg and length L = 0 . 20   m is lying on a horizontal frictionless…

Before:

After:

Applying conservation of angular momentum about point $O$, we get

$\left(m{v}_0\right)\frac{L}{2}=\left(\frac{M{L}^2}{3}\right)\omega -\left(mv\right)\frac{L}{2} ...\left(\mathrm{i}\right)$

As the collision is elastic, $e=1$

$\Rightarrow \frac{\text{velocity of separation(after collision)}}{\text{velocity of approach(before collision)}}=1$

$\Rightarrow \frac{\frac{L\omega }{2}-\left(-v\right)}{v_0}=1$

$\Rightarrow v=v_0-\frac{L\omega }{2} ...\left(\mathrm{ii}\right)$

Using equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we can write

$\left(m{v}_0\right)\frac{L}{2}=\left(\frac{M{L}^2}{3}\right)\omega -\left(m\left(v_0-\frac{L\omega }{2}\right)\right)\frac{L}{2}$

Given: $m=0.1 \mathrm{kg}, M=1 \mathrm{kg}, L=0.2 \mathrm{m}$ and

$v_0=5 \mathrm{m} {\mathrm{s}}^{-1}$

Therefore,

$$\begin{gathered} \left(0.1\times 5\right)\frac{0.2}{2}=\left(\frac{1\times {\left(0.2\right)}^2}{3}\right)\omega -\left(0.1\left(5-\frac{0.2\omega }{2}\right)\right)\frac{0.2}{2} \\ \Rightarrow 0.05=\frac{0.04}{3}\omega -\left(0.5-0.01\omega \right)\times 0.1 \\ \Rightarrow 0.05+0.05=\omega \left(\frac{0.04}{3}+0.001\right) \\ \Rightarrow \omega =\frac{0.1\times 3}{0.04+0.003} \\ \approx 6.98 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow v=5-\frac{0.2\times 6.98}{2} \\ \approx 4.3 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$