A ball weighing `10g` 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 speed The ball remains in contact with the surface for `0.01s` The average force exerted by the surface on the ball is .A. `100N`B. `10N`C. `1N`D. `0.1N`

A ball weighing `10g` hits a hard surface vertically with a speed of `

1.	A ball weighing `10g` 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 speed The ball remains in contact with the surface for `0.01s` The average force exerted by the surface on the ball is .A. `100N`B. `10N`C. `1N`D. `0.1N`
Answer» Correct Answer - b Here, `m = 10 g = (10)/(1000)kg = 0.01kg` `upsilon_(1) = -5m//s, upsilon_(2) = + 5m//s, t = 0.01sec` According to impulse momentum theorem Impulse = change in momentum `F _(av) xx t = m upsilon_(2) - m (-upsilon_(1)) =m upsilon_(2) + m upsilon_(1)` `F_(av) xx 0.01 = 0.01 xx 5 + 0.01 xx 5` `F_(av) = (0.1)/(0.01) = 10N` .

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