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