A) \[\text{ME}_{k}^{2}\]
B) \[\text{2mE}_{k}^{{}}\]
C) \[\sqrt{\frac{\text{2E}_{k}^{{}}}{m}}\]
D) \[\sqrt{\text{2mE}_{k}^{{}}}\]
Correct Answer: D
Solution :
\[{{p}_{\max }}=m{{v}_{\max }}\] ?(i) \[{{E}_{k}}=\frac{1}{2}mv_{m}^{2}\] ?(ii) Squaring equation (i), we get, \[{{p}^{2}}={{m}^{2}}\upsilon _{\max }^{2}=m\cdot m\upsilon _{\max }^{2}=2{{E}_{k}}m\] hence \[p=\sqrt{2m{{E}_{k}}}\]
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