A stone is dropped from a height `h` . It hits the ground with a certain momentum `P` . If the same stone is dropped from a height `100%` more thanthe preyiious height, the momentum when it hits the ground will change by

Options

A. `68%`
B. `41%`
C. `200%`
D. `100%`

Answer

b

Explanation

Velocity of stone hitting the ground when dropped
from a height h is `upsilon = sqrt(2 gh)` Therefore
`p = m upsilon =m sqrt(2 gh)`
New height `= h+ h =2 h`
`:.` New velocity `upsilon = sqrt( 2 g (2h)) =sqrt2 upsilon`
New momentum `p = m upsilon' = m sqrt(2 upsilon)`
Percentage increase in momentum
`= ((p' -p)100)/(p) = ((sqrt2-1)m upsilon xx 100)/(m upsilon)`
`=41.4%` .