The velocity of point R of the rod just after impact with the bumper is (in m/s)

1.2 \sqrt{5}

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1.2

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1.2 \sqrt{5 (2 - \sqrt{3})}

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1.2 \sqrt{5 (2 + \sqrt{3})}

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Solution

The correct option is D $1.2 \sqrt{5 (2 - \sqrt{3})}$
After impact, rod rebounds 30 $^{o}$ from vertical.
By conservation of energy,
$m g \frac{l}{2} (1 - c o s 30^{0}) = \frac{1}{2} \frac{m l^{2}}{3} w_{2}^{2}$
$\Rightarrow m g \frac{l}{2} (1 - \frac{\sqrt{3}}{2}) = \frac{1}{2} \frac{m l^{2}}{3} ω_{2}^{2}$
$\Rightarrow ω_{2} = \sqrt{\frac{3 (1 - \frac{\sqrt{3}}{2}) g}{l}}$
Velocity of point R after impact
$v_{2} = ω_{2} x$ = $⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{3 (1 - \frac{\sqrt{3}}{2}) X 10}{l} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \times 0.6$ (since $x = 0.6 m$ )
$\Rightarrow v_{2} = 1.2 \sqrt{5 (2 - \sqrt{3})} m / s$

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