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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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68%

41%

200%

100%

answer is B.

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Detailed Solution

Velocity v = $\sqrt{2 gh} - - - (i)$

and momentum p = mv $- - - (ii)$

Form Eqs. (i) and (ii) we have p $\propto \sqrt{h}$

Here $\frac{p_{2}}{p_{1}}$ = $\sqrt{\frac{h_{2}}{h_{1}}}$

$\frac{p_{2}}{p_{1}}$ = $\sqrt{\frac{2 h}{h}}$ = $\sqrt{2}$

$p_{2}$ = 1.414 $p_{1}$

% change = $\frac{p_{2}}{p_{1}} - p_{1}$ $\times$ 100 = 41%

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