A) 36%
B) 20%
C) 8%
D) 6%
Correct Answer: B
Solution :
The momentum \[p=\sqrt{2mE}\] \[\therefore \] \[p\propto \sqrt{E}\] In given problem \[KE\] becomes \[64%\] of the original value. \[\frac{{{p}_{2}}}{{{p}_{1}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}=\sqrt{\frac{64E}{100E}}=0.8\] \[\Rightarrow \] \[{{p}_{2}}=0.8\,\,{{p}_{1}}\] \[\therefore \] \[{{p}_{2}}=80%\]of the original value. \[ie,\]decrease in momentum is\[20%\].
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