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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