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 Online Paper Held On 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 :

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