Work Energy And Power Question 141
Question: A heavy particle hanging from a string of length $ l $ is projected horizontally with speed $ \sqrt{2gl} $ . The speed of the particle at the point where the tension in the string equals weight of the particle
Options:
A) $ \sqrt{2gl} $
B) $ \sqrt{3gl} $
C) $ \sqrt{gl/2} $
D) $ \sqrt{gl/3} $
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Answer:
Correct Answer: D
Solution:
[d] $ T-mg\cos \theta =\frac{mv^{2}}{R} $ Given $ T=mg $
$ mg-mg\cos \theta =\frac{mv^{2}}{R} $
$ g(1-cos\theta )=\frac{v^{2}}{R} $ C.O.M.E. at A and B; $ \Delta K+\Delta U=0 $
$ ( \frac{1}{2}mv^{2}-\frac{1}{2}mu^{2} )+mg(R-Rcos\theta )=0 $
$ \Rightarrow v^{2}-u^{2}=-2gR(1-cos\theta ) $
$ \Rightarrow v^{2}-{{(\sqrt{gl})}^{2}}=-2v^{2} $
$ 3v^{2}=gl\Rightarrow v=\sqrt{\frac{gl}{3}} $
