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Question

A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball is


A

1N

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B

10N

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C

100N

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D

0.1N

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Solution

The correct option is B

10N


Since impulse I = F x Δt

And also Impulse I = Δp (i.e. change in linear momentum)

Change in momentum

= 0.01 x 5 - (-5)

= 0.10 N s

Impulse = 0.10 N s

Therefore 0.10 = F x 0.01

Or F = 10 N


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