A ball of mass `m` falls vertically to the ground from a height `h_1` and rebound to 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(2 gh_1) + sqrt(2 gh_2))`
C. `m sqrt(2g(h_1 + h_2))`
D. `m sqrt(2 g)(h_1 + h_2)`

Answer

B

Explanation

(b) When all falls vertically downward from height `h_1`
its velocity `vec v_1 = sqrt(2 gh_1)`
and its velocity after collision `vec v_2 = sqrt(2 gh_2)`
Change in momentum
`Delta vec P = m(vec v_2 - vec v_1) = m (sqrt(2 gh_1) + sqrt(2 gh_2))`
(because `vec v_2` and `vec v_2` are opposite in direction).