A simple pendulum, made of a string of length l and a bob of mass m , is released from a small angle θ…

1. $m\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)$
2. $m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$
3. $\frac{m}{2}\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$
4. $\frac{m}{2}\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)$

Solution

$u=\sqrt{2gl\left(1-\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_0\right)}$ .......(i)

$v=$ velocity of ball after collision

$v=\left(\frac{m-M}{m+M}\right)u$

Since ball rises up to angle ${\theta }_1$

$v=\sqrt{2gl\left(1-\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_1\right)}=\left(\frac{m-M}{m+M}\right)u$ ......(ii)

From (i) and (ii)

$\frac{m-M}{m+M}=\sqrt{\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_1}{1-\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_0}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{{\theta }_1}{2}\right)}{\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{{\theta }_0}{2}\right)}$

$\Rightarrow \frac{M}{m}=\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\Rightarrow M=\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)m$

Asked in: JEE Main 2019 (12 Jan Shift 1)

Practice more Center of Mass Momentum and Collision questions on Aicharya