A ball is dropped from a height h on to a floor. If, in each collision, its speed becomes e times of its striking value (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 value (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 = 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`