A ball is moving with speed $20 m / s$ collides with a smooth surface. The magnitude of change in velocity of the ball will be :

10 \sqrt{3} m / s

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20 \sqrt{3} m / s

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\frac{40}{\sqrt{3}} m / s

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40 m / s

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Solution

The correct option is D $40 m / s$
D. $40 m / s$

When a ball normally collides with smooth surface then ball rebound with the same velocity.

$v_{i} = 20^i m / s$

$v_{f} = - 20^i m / s$

Change in the velocity, $Δ v = v_{f} - v_{i}$

$Δ v = - 40^i m / s$

The magnitude of change in velocity of the ball is $40 m / s$

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