A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{16}\]
Correct Answer: A
Solution :
As \[KE=\frac{{{p}^{2}}}{2m}\] \[\therefore \] \[\frac{p_{1}^{2}}{p_{2}^{2}}=\frac{{{m}_{1}}}{{{m}_{2}}}.\frac{KE}{K{{E}_{2}}}=\frac{1}{4}.\frac{2}{1}=\frac{1}{2}\] \[\Rightarrow \] \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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