A) 22%
B) 44%
C) 10%
D) 300%
Correct Answer: C
Solution :
\[P=\sqrt{2mE}.\] If m is constant then \[\frac{{{P}_{2}}}{{{P}_{1}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}=\sqrt{\frac{1.22E}{E}}\]Þ \[\frac{{{P}_{2}}}{{{P}_{1}}}=\sqrt{1.22}=1.1\] Þ\[{{P}_{2}}=1.1{{P}_{1}}\]Þ \[{{P}_{2}}={{P}_{1}}+0.1{{P}_{1}}={{P}_{1}}+10%\ \text{of }\ {{P}_{1}}\] So the momentum will increase by 10%You need to login to perform this action.
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