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 .

Options

A. `40ms^(-1)`
B. `30ms^(-1)`
C. `20ms^(-1)`
D. `10ms^(-1)`

Answer

C

Explanation

Area under `F-t` graph = Change in momentum
`implies 1/2 F_(0)(4 xx 10^(-3))=(200)/(1000) [v+20]`
`implies v=20ms^(-1)`.