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. `msqrt2g(h_(1)+h_(2))`
B. `msqrt(2g(m_(1)+m_(2))))`
C. `mg(h_(1)-h_(2))`
D. `m(sqrt(2gh_(1))-sqrt(2gh_(2)))`

Answer

D

Explanation

Let `v_(1)`= velocity when height of free fall is `h_(1)`
`v_(2)`= velocity when height of free rise is `h_(2)`
`:. V_(1)^(2)=u^(2)+2gh_(1)` for free fall
or
for free rise after impact on ground `0=v_(2)^(2)-2gh_(2) or v_(2)^(2)=2hg_(2)`
Initial momentum `=mv_(1)`
Final momentum `=mv_(2)`
`:.` Change in momentum `=m(v_(1)-v_(2))= m( sqrt(2gh_(1))- sqry(2hg_(2))`