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

Before:

After:

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

\left(m{v}_{0}\right)\dfrac{L}{2} = \left(\dfrac{M{L}^{2}}{3}\right)\omega - \left(mv\right)\dfrac{L}{2} . . . \left(i\right)$</p><p>As the collision is elastic,$e = 1$</p><p>$⇒ \dfrac{\text{velocity of separation(after collision)}}{\text{velocity of approach(before collision)}} = 1$</p><p>$⇒ \dfrac{\dfrac{L\omega}{2} - \left(- v\right)}{{v}_{0}} = 1

$⇒ v = {v}_{0} - \dfrac{L\omega}{2} . . . \left(ii\right)$

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

\left(m{v}_{0}\right)\dfrac{L}{2} = \left(\dfrac{M{L}^{2}}{3}\right)\omega - \left(m\left({v}_{0} - \dfrac{L\omega}{2}\right)\right)\dfrac{L}{2}$</p><p>Given:$m = 0 . 1kg, M = 1kg, L = 0 . 2m$and</p><p>${v}_{0} = 5m{s}^{- 1}$</p><p>Therefore,</p><p>\left(0 . 1 \times 5\right)\dfrac{0 . 2}{2} = \left(\dfrac{1 \times {\left(0 . 2\right)}^{2}}{3}\right)\omega - \left(0 . 1\left(5 - \dfrac{0 . 2\omega}{2}\right)\right)\dfrac{0 . 2}{2} ⇒ 0 . 05 = \dfrac{0 . 04}{3}\omega - \left(0 . 5 - 0 . 01\omega\right) \times 0 . 1 ⇒ 0 . 05 + 0 . 05 = \omega\left(\dfrac{0 . 04}{3} + 0 . 001\right) ⇒ \omega = \dfrac{0 . 1 \times 3}{0 . 04 + 0 . 003} \approx 6 . 98rad{s}^{- 1}$

Now,

$⇒ v = 5 - \dfrac{0 . 2 \times 6 . 98}{2} \approx 4 . 3m{s}^{- 1}$