Question 20: Laws of Motion

Question

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 different height h ' such that percentage change in momentum is $41.4 \%$, then the height from which the stone is dropped is $\mathrm{h}^{\prime}=x \mathrm{~h}$, where $x$ is:

Accepted Answer

Let the stoneÃ¢ÂÂs mass be $m$.  When dropped from height $h$, its impact momentum is
$$ p = m\,v = m\sqrt{2gh}\,. $$
From a new height $h'$, the momentum is
$$ p' = m\sqrt{2gh'} = m\sqrt{2gh}\,\sqrt{\frac{h'}{h}} = p\;\sqrt{\frac{h'}{h}}\,. $$
The percentage increase in momentum is
$$ \frac{p' - p}{p}	imes100\% \;=\;\Bigl(\sqrt{	frac{h'}{h}} - 1\Bigr)	imes100\% \;=\;41.4\%\,. $$
Hence
$$ \sqrt{\frac{h'}{h}} - 1 = 0.414 \quad\Longrightarrow\quad \sqrt{\frac{h'}{h}} = 1.414 = \sqrt{2} \quad\Longrightarrow\quad \frac{h'}{h} = 2\,. $$
So $x=2$.

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4

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2

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6

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3

Laws of Motion

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 different height h ' such that percentage change in momentum is 41.4 \%, then the height from which the stone is dropped is \mathrm{h}^{\prime}=x \mathrm{~h}, where x is:

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