A ball of mass `m` falls vertically to the ground from a height `h_1` and rebound to a height `h_2`. The change in momentum of the ball on striking the ground is.

Options

A. `mg(h_(1)-h_(2))`
B. `m (sqrt(2gh_(1))+sqrt(gh_(2)))`
C. `m sqrt(2g(h_(1)+h_(2)))`
D. `m sqrt(2g)(h_(1)+h_(2))`

Answer

B

Explanation

When ball falls vertically downward from height `h_(1)` its velocity `vec(v)_(1)= sqrt(2gh_(1))`
and its velocity after collision `vec(v)_(2)= sqrt(2gh_(2))`
Change in momentum
`Delta vec(P)= m(vec(v)_(2)-vec(v)_(1))=m (sqrt(2gh_(1)) + sqrt(2gh_(2)))` (because `vec(v)_(1) and vec(v)_(2)` are opposite in direction)