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  4. The bob of a simple pendulum of mass and total energy textE textk will have maximum linear momentum equal to:

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Q. The bob of a simple pendulum of mass and total energy $ {{\text{E}}_{\text{k}}} $ will have maximum linear momentum equal to:

JIPMERJIPMER 2000

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}}} $