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.

A ball is dropped from a height h on to a floor. If, in each collision

1.	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.
Answer» Solution :(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`