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A ball of mass 10 gm hits vertically a hard surface with a speed of $5 {ms}^{- 1}$ and rebounds with the same speed. The ball remains in contact with the surface for 1/100 sec . The average force exerted by the surface on the ball is.

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0 N

20 N

10 N

40 N

answer is C.

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Detailed Solution

$a = \frac{ΔV}{t} = \frac{5 - (- 5)}{1 / 100} = 10^{3} m / \sec^{2}$

$F = Ma$

$= 10 (10^{- 3}) (10^{3}) = 10 N$

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