A) 6%
B) 8%
C) 20%
D) 36%
Correct Answer: C
Solution :
The momentum and kinetic energy is related by \[p=\sqrt{2mE}\] \[\therefore \] \[p\propto \sqrt{E}\] According to question, kinetic energy becomes 64% of the original \[\therefore \] \[\frac{{{p}_{2}}}{{{p}_{1}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[=\sqrt{\frac{64E}{100E}}\] \[=\frac{8}{10}=0.8\] \[\Rightarrow \] \[{{p}_{2}}=0.8\,p\] \[\therefore {{p}_{2}}=80\,%\] of original value i.e., decrease in momentum is 20%.
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