A) 100%
B) 150%
C) \[\sqrt{300}%\]
D) 175%
Correct Answer: A
Solution :
Let initial kinetic energy, \[{{E}_{1}}=E\] Final kinetic energy, \[{{E}_{2}}=E+300%\] of E = 4E As \[P\propto \sqrt{E}\]Þ\[\frac{{{P}_{2}}}{{{P}_{1}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}=\sqrt{\frac{4E}{E}}=2\]Þ \[{{P}_{2}}=2{{P}_{1}}\] Þ\[{{P}_{2}}={{P}_{1}}+100%\] of \[{{P}_{1}}\] i.e. Momentum will increase by 100%.
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