A) 25%
B) 125%
C) 150%
D) 50%
Correct Answer: B
Solution :
The relation between kinetic energy K, and momentum p, is \[p=\sqrt{2mK}\] where, m is mass. Given, \[{{p}_{1}}=p,{{p}_{2}}={{p}_{1}}+50%\]of \[{{p}_{1}}\] \[{{p}_{2}}={{p}_{1}}+\frac{{{p}_{1}}}{2}=\frac{3}{2}{{p}_{1}}=1.5{{p}_{1}}\] \[\therefore \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{p_{1}^{2}}{p_{2}^{2}}\] \[\Rightarrow \] \[{{K}_{2}}=\frac{p_{2}^{2}}{p_{1}^{2}}{{K}_{1}}\] \[\Rightarrow \] \[{{K}_{2}}=\frac{{{(1.5)}^{2}}}{1}\times K=2.25K\] \[\therefore \]Change in \[KE=2.25-1\] \[=1.25=125%\]
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