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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 than the previous height, the momentum when it hits the ground will change by
A
100%
B
68%
C
41%
D
200%
Detailed Solution
Speed on hitting the ground: $v = \sqrt{2gh}$, so momentum $P = m\sqrt{2gh} \propto \sqrt h$
The new height is 100% more, i.e. 2h.
$\frac{P'}{P} = \sqrt{\frac{2h}{h}} = \sqrt2 \Rightarrow P' = \sqrt2P = 1.41P$
Change in momentum $= \frac{P' - P}{P}\times100 = 41\%$
The new height is 100% more, i.e. 2h.
$\frac{P'}{P} = \sqrt{\frac{2h}{h}} = \sqrt2 \Rightarrow P' = \sqrt2P = 1.41P$
Change in momentum $= \frac{P' - P}{P}\times100 = 41\%$