A) 100%
B) 300%
C) 200%
D) 150%
Correct Answer: B
Solution :
Initial kinetic energy is \[K=\frac{{{p}^{2}}}{2m}\] ??..(1) When momentum increases by 100%. It becomes \[p=p+p=2p\] Final kinetic energy, \[K=\frac{p{{}^{2}}}{2m}=\frac{{{(2p)}^{2}}}{2m}=\frac{{{p}^{2}}}{m}\] ??..(2) From equations (1) and (2), we; get \[\frac{K}{K}=\frac{2{{p}^{2}}/m}{{{p}^{2}}/2m}=4\] or \[K=4K\] Percentage increases \[=\frac{(4K-K)}{K}\times 100\] \[=300%\]You need to login to perform this action.
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