A ball is dropped from a height h on to a floor. If, in each collision, its speed becomes e times of its striking vaue (a) Find the total change in momentum of the ball (b) Find the average force exerted by the ball on the floor.

Explanation

(a) Change in momentum in I collision
`=mv_(1)- (-mv_(0)) = m(v_(1) + v_(0))`
Change in momentum in II collision `=m (v_(2) + v_(1))`
Change in momentum in `n^(th)` collision `= m(v_(n) + v_(n-1))`
Adding these all, total change in momentum of the ball is `Delta p= m [v_(0) + 2v_(1) + ...+ 2v_(n-1) + v_(n)]`
or `Delta p= mv_(0) [1 + 2e + 2e^(2) + ....] [ " as " v_(1)= ev_(0), v_(2) = e^(2) v_(0)...]`
`Delta p= mv_(0)[1 + 2e ((1)/(1-e))]= m sqrt(2gh) [1 + e)/(1-e)]` ..(1)
(b) Now as `vec(F )= ((vec(dp)),(dt)) "so", F_(av)= ((Delta p),(Delta T))`
we know `Delta T = sqrt((2h)/(g)) ((1 + e)/(1-e))`
From (1) `Delta p= m sqrt(2gh) [(1 + e)/(1-e)]`
`F_(av)= m sqrt(2gh) [(1 + e)/(1-e)] xx sqrt((g)/(2h)) [(1-e)/(1+e)] = mg`