JEE Main 12 Jan 2019 Morning Question 17
Question: A simple pendulum, made of a string of length I and a bob of mass m, is released from a small angle $ {\theta_0}. $ It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $ {\theta_1}. $ Then M is given by [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{m}{2}( \frac{{\theta_0}-{\theta_1}}{{\theta_0}+{\theta_1}} ) $
B) $ m( \frac{{\theta_0}+{\theta_1}}{{\theta_0}-{\theta_1}} ) $
C) $ \frac{m}{2}( \frac{{\theta_0}+{\theta_1}}{{\theta_0}-{\theta_1}} ) $
D) $ m( \frac{{\theta_0}-{\theta_1}}{{\theta_0}+{\theta_1}} ) $
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Answer:
Correct Answer: D
Solution:
Apply conservation of energy, $ u=\sqrt{2gl(1-cos{\theta_0})} $ -(i)
v = velocity of bob after collision
$ v=( \frac{m-M}{m+M} )u $
Bob rises up to angle $ {\theta_1} $ $ v=\sqrt{2gl(1-\cos {\theta_1})} $
$ v=( \frac{m-M}{m+M} )u=\sqrt{2gl(1-\cos {\theta_1})} $ -(ii)
from eq. (i) and (ii)
$ \frac{m-M}{m+M}=\sqrt{\frac{1-\cos {\theta_1}}{1-\cos {\theta_0}}}{ \cos \theta =1-2{{\sin }^{2}}\frac{\theta }{2} } $
$ \frac{m-M}{m+M}=\frac{\sin ( \frac{{\theta_1}}{2} )}{\sin ( \frac{{\theta_0}}{2} )} $
$ (\because \theta \simeq small) $ $ \frac{M}{m}=\frac{{\theta_0}-{\theta_1}}{{\theta_0}+{\theta_1}},M=( \frac{{\theta_0}-{\theta_1}}{{\theta_0}+{\theta_1}} )m $
