A) 100%
B) 150%
C) 265%
D) 73.2%
Correct Answer: A
Solution :
Key idea: The relation between momentum p and kinetic energy K is \[K=\frac{1}{2\,m}({{p}^{2}})\] Kinetic energy \[K=\frac{1}{2m}({{p}^{2}})\] or \[p=\sqrt{2m\,K}\] If kinetic energy of a body is increased by 300%, let its momentum becomes p?. New kinetic energy \[K'=K+\frac{300}{100}K=4K\] Therefore, momentum is given by \[p'=\sqrt{2m\times 4K}=2\sqrt{2mK}=2p\] Hence, % change (increase) in momentum \[\frac{\Delta p}{p}\times 100=\frac{p'-p}{p}\times 100%\] \[=\left( \frac{p'}{p}-1 \right)\times 100%\] \[=\left( \frac{2p}{p}-1 \right)\,\times 100%\] = 100%
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