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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 :

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41%

200%

68%

100%

answer is B.

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Detailed Solution

Momentum $P = m v = m \sqrt{2 g h}$

$(∵ v^{2} = u^{2} + 2 gh; Here u = 0)$

When stone hits the ground momentum

$P = m \sqrt{2 g h}$

when same stone dropped from 2h (100% of initial) then momentum

$P^{'} = m \sqrt{2 g (2 h)} = \sqrt{2} P$

Which is changed by 41% of initial.

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