A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is $e = 1 / 2$ . The ball's velocity become horizontal after the collision.

The angle

θ = t a n^{- 1} (\frac{1}{\sqrt{2}})

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the speed of the ball after the collision

= 4 \sqrt{2} m s^{- 1}

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the total loss in kinetic energy during the collision is 8J.

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the ball hits the inclined plane again while travelling vertically downward.

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Solution

The correct options are
A The angle $θ = t a n^{- 1} (\frac{1}{\sqrt{2}})$ .
B the speed of the ball after the collision $= 4 \sqrt{2} m s^{- 1}$ .

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A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is e=1/2. The ball's velocity become horizontal after the collision.

The correct options are A The angle θ=tan−1(1√2). B the speed of the ball after the collision =4√2ms−1.

A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is $e = 1 / 2$ . The ball's velocity become horizontal after the collision.

The correct options are
A The angle $θ = t a n^{- 1} (\frac{1}{\sqrt{2}})$ .
B the speed of the ball after the collision $= 4 \sqrt{2} m s^{- 1}$ .