A bar of mass M = 1.00kg and length L = 0.20m 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 = 0.10kg is moving on the same horizontal surface with 5.00ms-1 speed on a path perpendicular to the bar. It hits the bar at a distance L2 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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A bar of mass M = 1.00kg and length L = 0.20m 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 = 0.10kg is moving on the same horizontal surface with 5.00ms^-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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$ω = 6.98 {rads}^{- 1} and v = 4.30 {ms}^{- 1}$

$ω = 6.80 {rads}^{- 1} and v = 4.10 {ms}^{- 1}$

$ω = 3.75 {rads}^{- 1} and v = 4.30 {ms}^{- 1}$

$ω = 3.75 {rads}^{- 1} and v = 10.0 {ms}^{- 1}$

answer is A.

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Detailed Solution

Before collision
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After collision
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Conserving angular momentum about O.
$\begin{array}{l} {mv}_{0} \frac{L}{2} = - mv \frac{L}{2} + \frac{{ML}^{2}}{3} ω \\ 0.1 \times 5 \times \frac{0.2}{2} = - 0.1 v \times \frac{0.2}{2} + \frac{1 (0.2)^{2}}{3} ω 5 = - v + \frac{4 ω}{3} \end{array}$
$∵$ since collision is elastic,
Velocity of approach = velocity of separation
$\begin{array}{l} 5 + 0 = v + \frac{0.2 ω}{2} \\ 5 = v + \frac{ω}{10} ω = \frac{300}{43} = 6.98 v = 4.30 \end{array}$