Centre Of Mass Ques 5
- 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
(2019 Main, 12 Jan I)
(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) $\frac{m}{2}\left(\frac{\theta_0+\theta_1}{\theta_0-\theta_1}\right)$
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Answer:
Correct Answer: 5.( a )
Solution:
8 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,
$ \begin{aligned} m g h & =\frac{1}{2} m v^2 \\ h & =l-l \cos \theta_0 \\ v & =\sqrt{2 g l\left(1-\cos \theta_0\right)} \end{aligned} $
Here,
So,
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 pendulum is deflected back upto angle $\theta_1$, then
$ v_1=\sqrt{2 g l\left(1-\cos \theta_1\right)} \ldots $
Using definition of coefficient of restitution to get
$ \begin{array}{r} e=\frac{\mid \text { velocity of separation } \mid}{\mid \text { velocity of approach } \mid} \\ 1=\frac{v_2-\left(-v_1\right)}{v-0} \Rightarrow v=v_2+v_1 \quad \text {…(iii) } \end{array} $
From Eqs. (i), (ii) and (iii), we get
$ \begin{aligned} & \Rightarrow \sqrt{2 g l\left(1-\cos \theta_0\right)}=v_2+\sqrt{2 g l\left(1-\cos \theta_1\right)} \\ & \Rightarrow \quad v_2=\sqrt{2 g l}\left(\sqrt{1-\cos \theta_0}-\sqrt{1-\cos \theta_1}\right) \end{aligned} $
According to the momentum conservation, initial momentum of the system $=$ final momentum of the system
$ \begin{aligned} \Rightarrow \quad m v & =M v_2-m v_1 \\ \Rightarrow M v_2 & =m\left(v+v_1\right) \\ M v_2 & =m \sqrt{2 g l}\left(\sqrt{1-\cos \theta_0}+\sqrt{1-\cos \theta_1}\right) \end{aligned} $
Dividing Eq. (v) and Eq. (iv), we get
$ \begin{aligned} \Rightarrow \frac{M}{m} & =\frac{\sqrt{1-\cos \theta_0}+\sqrt{1-\cos \theta_1}}{\sqrt{1-\cos \theta_0}-\sqrt{1-\cos \theta_1}} \\ & =\frac{\sqrt{\sin ^2\left(\frac{\theta_0}{2}\right)}+\sqrt{\sin ^2\left(\frac{\theta_1}{2}\right)}}{\sqrt{\sin ^2\left(\frac{\theta_0}{2}\right)-\sqrt{\sin ^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)} \end{aligned} $
For small $\theta_0$, we have
$ \begin{aligned} & \frac{M}{m}=\frac{\frac{\theta_0}{2}+\frac{\theta_1}{2}}{\frac{\theta_0}{2}-\frac{\theta_1}{2}} \\ & M=m\left(\frac{\theta_0+\theta_1}{\theta_0-\theta_1}\right) \end{aligned} $
BETA
