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

1.	A ball is dropped from a heighth 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))` Charge 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= m v _(0) [1 + 2 e + 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 ` vecF = (vec(dp))/( dt) so, F _(av) = (Deltap)/( 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 `