A small box starts sliding from rest at the top of the frictionless track of radius $R$ . The initial height of the box equal to $2 R$ above a horizontal surface. At the moment, the box leaves the bottom of the track, a ball of the same mass as the box is dropped from the same height at the bottom of the track.
Calculate the speed of the box when it reaches the end of the track?

v = \sqrt{g R}

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v = \sqrt{2 g R}

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v = \sqrt{π g R}

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v = \sqrt{2 π g R}

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v = π \sqrt{2 g R}

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Solution

The correct option is D $v = \sqrt{2 g R}$
Consider conservation of energy of the box from the point when it starts sliding to the point when it reaches the end of the track.
Loss in gravitational potential energy=Gain in kinetic energy of the box
$⟹ m g (2 R) - m g (R) = \frac{1}{2} m v^{2}$
$⟹ v = \sqrt{2 g R}$

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The correct option is D v=√2gRConsider conservation of energy of the box from the point when it starts sliding to the point when it reaches the end of the track.Loss in gravitational potential energy=Gain in kinetic energy of the box⟹mg(2R)−mg(R)=12mv2⟹v=√2gR