Question

A simple pendulum, made of a string of length $l$ 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 :

• A. $\frac{m}{2}\left(\frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}}\right)$
• B. $\frac{m}{2}\left(\frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}}\right)$
• C. $m\left(\frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}}\right)$
• D. $m\left(\frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}}\right)$

Answer 1

Answer: (D)

By momentum conservation
$m\sqrt{2g\ell \left(1-\cos\theta_{0}\right) } = MV_{m} -m \sqrt{2g1\left(1-\cos\theta\right)} $
$ \Rightarrow m\sqrt{2g\ell} \left\{\sqrt{1-\cos\theta_{0}} + \sqrt{1-\cos\theta_{1}}\right\} =MV_{m} $
and $ e=1 = \frac{V_{m} +\sqrt{2g\ell\left(1-\cos\theta_{1}\right)} }{\sqrt{2g\ell\left(1-\cos\theta_{0}\right)}} $
$\sqrt{2g\ell} \left(\sqrt{1-\cos\theta_{0} } - \sqrt{1-\cos\theta_{1}}\right) =V_{m} $ ...(I)
$m\sqrt{2g\ell} \left(\sqrt{1-\cos\theta_{0}} + \sqrt{1-\cos\theta_{0}} \right) =MV_{M} $ ....(II)
Dividing
$ \frac{\left(\sqrt{1-\cos\theta_{0}} + \sqrt{1-\cos\theta_{1}}\right)}{\left(\sqrt{1-\cos\theta_{0}} -\sqrt{1-\cos\theta_{1}}\right)} = \frac{M}{m}$
By componendo divided
$ \frac{m-M}{m+M} = \frac{\sqrt{1-\cos\theta_{1}}}{\sqrt{1-\cos\theta_{0}}} = \frac{\sin\left(\frac{\theta_{1}}{2}\right)}{\sin\left(\frac{\theta_{0}}{2}\right)} $
$ \Rightarrow \frac{M}{m} = \frac{\theta_{0} -\theta_{1}}{\theta_{0} +\theta_{1}} \Rightarrow M = \frac{\theta_{0} -\theta_{1}}{\theta_{0} +\theta_{1}} $