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
$m\sqrt{2g\ell \left(1-\cos {\theta }_0\right)}=M{V}_m-m\sqrt{2g1\left(1-\cos \theta \right)}$
$\Rightarrow m\sqrt{2g\ell }\left\{\sqrt{1-\cos {\theta }_0}+\sqrt{1-\cos {\theta }_1}\right\}=M{V}_m$
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)
$m\sqrt{2g\ell }\left(\sqrt{1-\cos {\theta }_0}+\sqrt{1-\cos {\theta }_0}\right)=M{V}_M$ ....(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
$\frac{m-M}{m+M}=\frac{\sqrt{1-\cos {\theta }_1}}{\sqrt{1-\cos {\theta }_0}}=\frac{\sin \left(\frac{{\theta }_1}{2}\right)}{\sin \left(\frac{{\theta }_0}{2}\right)}$
$\Rightarrow \frac{M}{m}=\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\Rightarrow M=\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}$