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 |
\[\frac{\Delta p}{p}\times 100=\frac{p'-p}{p}\times 100%\] |
\[=\left( \frac{p'}{p}-1 \right)\times 100%\] |
\[=\left( \frac{2p}{p}-1 \right)\,\times 100%\] |
\[=100%\] |
You need to login to perform this action.
You will be redirected in
3 sec