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एक पत्थर को $h$ ऊँचाई से गिराया जाता है। यह एक निश्चित संवेग $P$ से भू-तल से टकराता है, यदि इसी पत्थर को, इस ऊँचाई से $100 \%$ अधिक ऊँचाई से गिराया जाये तो भू-तल से टकराते समय इसके संवेग में परिवर्तन होगा
$68$
$41$
$200$
$100$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,When\,a\,stone\,is\,dropped\,from\,a\\
height\,h,\,it\,hits\,the\,ground\,with\,a\,\\
momentum\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = m\sqrt {2gh} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
Where\,m\,is\,the\,mass\,of\,the\,stone.\\
When\,the\,same\,stone\,is\,dropped\,from\\
a\,height\,2h\,\left( {i.e,.\,100\% \,of\,initial} \right),\,then\\
its\,momentum\,with\,which\,it\,hits\,the\,\\
ground\,becomes
\end{array}$
$\begin{array}{l}
\,\,\,\,\,\,P' = m\sqrt {2g\left( {2h} \right)} = \sqrt {2P} \,\,\,\,\,({\rm{Using}}\left( i \right))…\left( {ii} \right)\\
\% \,chenge\,in\,momentum\, = \frac{{P' – P}}{P} \times 100\% \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 P – P}}{P} \times 100\% = 41\% \,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{array}$
