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Law of Conservation of Mechanical Energy

A ball is dro...

Question

A ball is dropped along smooth inclined surfaces $A B, A C$ and $A D$ . The velocity acquired at $B, C, D$ be $v_{B}, v_{C}$ and $v_{D}$ respectively. Then

v_{B} > v_{C} > v_{D}

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v_{B} < v_{C} < v_{D}

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v_{B} > v_{C} < v_{D}

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v_{B} = v_{C} = v_{D}

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Solution

The correct option is D $v_{B} = v_{C} = v_{D}$
Given velocity at point $B, C, D$ is $v_{B}, v_{C}$ and $v_{D}$ respectively
Since, total mechanical energy is constant
$P E_{i} + K E_{i} = P E_{f} + K E_{f}$
Here, Initial point is $A$ and final points are $B, C, D$
Ball have only potential energy at point $A$ and kinetic energy at point $B, C, D$
$v = \sqrt{2 g h}$ , where $h$ is the vertical height from initial point to final point
So change in velocity depends only on height difference.
As height is same for all, $v_{B} = v_{C} = v_{D}$

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