question_answer

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 12-Jan-2019 Morning]

A) \[\frac{m}{2}\left( \frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}} \right)\]        

B)      \[m\left( \frac{{{\theta }_{0}}+{{\theta }_{1}}}{{{\theta }_{0}}-{{\theta }_{1}}} \right)\]

C) \[\frac{m}{2}\left( \frac{{{\theta }_{0}}+{{\theta }_{1}}}{{{\theta }_{0}}-{{\theta }_{1}}} \right)\]        

D)      \[m\left( \frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}} \right)\]

Correct Answer: D

Solution :

[d]

Apply conservation of energy,

\[u=\sqrt{2gl(1-cos{{\theta }_{0}})}\]                              (i)

v = velocity of bob after collision

\[v=\left( \frac{m-M}{m+M} \right)u\]Bob rises up to angle \[{{\theta }_{1}}\]

\[v=\sqrt{2gl(1-\cos {{\theta }_{1}})}\]

\[v=\left( \frac{m-M}{m+M} \right)u=\sqrt{2gl(1-\cos {{\theta }_{1}})}\]                     (ii)

from eq. (i) and (ii)

\[\frac{m-M}{m+M}=\sqrt{\frac{1-\cos {{\theta }_{1}}}{1-\cos {{\theta }_{0}}}}\left\{ \cos \theta =1-2{{\sin }^{2}}\frac{\theta }{2} \right\}\]

\[\frac{m-M}{m+M}=\frac{\sin \left( \frac{{{\theta }_{1}}}{2} \right)}{\sin \left( \frac{{{\theta }_{0}}}{2} \right)}\] \[(\because \theta \simeq small)\]

\[\frac{M}{m}=\frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}},M=\left( \frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}} \right)m\]