Question

A simple pendulum is 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. $m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$
• 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. $\frac{m}{2}\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$

Answer 1

The correct option is A $m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$

Pendulum's velocity at lowest point just before striking mass m is found by equating it's initial potential energy (PE) with final kinetic energy (KE).
Initially, when pendulum is released from angle ${\theta }_0$ as shown in the figure below,

We have,
$mgh=\frac{1}{2}m{v}^2$
Here , $h=l-l\text{cos}{\theta }_0$
So, $v=\sqrt{2gl(1-\text{cos}{\theta }_0})....(i)$
With velocity $v$, bob of pendulum collides with block. After collision, let $v_1$ and $v_2$ are final velocities of masses m and M respectively as shown

Then if pednulum is deflected back upto angle ${\theta }_1$, then
$v_1=\sqrt{2gl(1-\text{cos}{\theta }_1})....(ii)$
Using definition of coefficient of restitution to get
$e=\frac{|\text{velocity of separation}|}{|\text{velocity of approach}|}$
$1=\frac{v_2-(-v_1)}{v-0}\Rightarrow v=v_2+v_1.....\left(iii\right)$
From Eqs. (i) , (ii) and (iii), we get
$\Rightarrow \sqrt{2gl(1-\text{cos}{\theta }_0)}=v_2+\sqrt{2gl(1-\text{cos}{\theta }_1)}$
$\Rightarrow v_2=\sqrt{2gl}(\sqrt{1-\text{cos}{\theta }_0}-\sqrt{1-\text{cos}{\theta }_1})....\left(iv\right)$
According to the momentum conservation, initial momentum of the system = final momentum of the system
$\Rightarrow mv=M{v}_2-m{v}_1$
$\Rightarrow M{v}_2=m(v+v_1)$
$M{v}_2=m\sqrt{2gl}(\sqrt{1-\text{cos}{\theta }_0}+\sqrt{1-\text{cos}{\theta }_1})$
Dividing Eq. (v) and (iv), we get
$\Rightarrow \frac{M}{m}=\frac{\sqrt{1-\text{cos}{\theta }_0}+\sqrt{1-\text{cos}{\theta }_1}}{\sqrt{1-\text{cos}{\theta }_0}-\sqrt{1-\text{cos}{\theta }_1}}$
$=\frac{\sqrt{si{n}^2\left(\frac{{\theta }_0}{2}\right)}+\sqrt{si{n}^2\left(\frac{{\theta }_1}{2}\right)}}{\sqrt{si{n}^2\left(\frac{{\theta }_0}{2}\right)}-\sqrt{si{n}^2\left(\frac{{\theta }_1}{2}\right)}}$
$\frac{M}{m}=\frac{sin\left(\frac{{\theta }_0}{2}\right)+sin\left(\frac{{\theta }_1}{2}\right)}{sin\left(\frac{{\theta }_0}{2}\right)-sin\left(\frac{{\theta }_1}{2}\right)}$
For small ${\theta }_0$, we have
$\frac{M}{m}=\frac{\frac{{\theta }_0}{2}+\frac{{\theta }_1}{2}}{\frac{{\theta }_0}{2}-\frac{{\theta }_1}{2}}orM=m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$