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Q.

In a fixed hemispherical smooth bowl of Radius R, a particle is dropped at point B somewhere else other than point A. Collision with the bowl is perfectly elastic. For second collision to take place at point A, the maximum value of $∠ B O A$ is $\frac{π}{K}$ radius. Find the value of K.

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answer is 3.

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Detailed Solution

$T = \frac{2 v_{0} \cos (\frac{3 θ}{2})}{g \cos (\frac{θ}{2})}$
$= \frac{2 v_{0}}{g} [4 \cos^{2} (\frac{θ}{2}) - 3]$
Also $R \sin θ = V_{0} \sin 2 θ T$
$R \sin θ = \frac{2 v_{0}^{2}}{g} (2 \sin θ \cos θ) [2 (1 + \cos θ) - 3]$
As $v_{0}^{2} 2 g h$
$∴ h = \frac{R}{8 [\cos θ [2 \cos θ - 1]]} h \geq 0 2 \cos^{2} θ - \cos θ \geq 0 \cos θ (2 \cos θ - 1) \geq 0 \cos θ \geq \frac{1}{2} θ \leq \frac{π}{3} K = 3$
2nd Method:

Question Image

$2 θ + α + (90 - \frac{θ}{2}) = 180^{0}$
$α = 90 - \frac{3 θ}{2}$
Now $α > 0$
$θ < 60^{0}$

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$θ_{\max} = \frac{π}{3}$ radian

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