A bar of mass $M = 2 k g$ and length $L = 0.20 m$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10 k g$ is moving on the same horizontal surface with $5.00 m s^{- 1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $\frac{L}{2}$ from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity $ω$ . Which of the following statement is correct ?

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$ω = 6.98 {rad s}^{- 1} a n d v = 4.30 m s^{- 1}$

$ω = 3.61 {rad s}^{- 1} a n d v = 4.63 m s^{- 1}$

$ω = 3.75 {rad s}^{- 1} a n d v = 10.0 m s^{- 1}$

$ω = 6.80 {rad s}^{- 1} a n d v = 4.10 m s^{- 1}$

answer is B.

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Detailed Solution

Before collision
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After collision
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Conserving angular momentum about O.
$\begin{array}{l} {mv}_{0} \frac{L}{2} = - mv \frac{L}{2} + \frac{{ML}^{2}}{3} ω \\ 0.2 \times 5 \times \frac{0.2}{2} = - 0.2 v \times \frac{0.2}{2} + \frac{2 (0.2)^{2}}{3} ω 5 = - v + \frac{4 ω}{3} \end{array}$
$∵$ since collision is elastic,
Velocity of approach = velocity of separation