Question

A small ball is suspended from a fixed point $O$ by means of an ideal string of length $l$. The ball is first taken aside such that the string becomes horizontal and then released from rest. At the bottom, it collides with a fixed obstacle. The coefficient of restitution is $e$. The maximum angular deflection of the string after $n^{\text{th}}$ collision is

• A. ${\sin }^{-1}(1-e^{2n})$
• B. ${\cos }^{-1}(1-e^{2n})$
• C. ${\sin }^{-1}\left(e^{2n}\right)$
• D. ${\cos }^{-1}\left(e^{2n}\right)$

Answer 1

The correct option is B ${\cos }^{-1}(1-e^{2n})$

Let, $v_1,v_2,v_3,\cdots ,v_n$ be the velocities acquired by the ball immediately after $1^{\text{st}},2^{\text{nd}},3^{\text{rd}},\cdots n^{\text{th}}$ collision respectively.

So, the velocities after collisions be

$v_1=e{v}_o$ (after 1st collision)
where, $v_o=(2gl{)}^{\frac{1}{2}}$

$v_2=e^2{v}_0$ (after 2nd collision)
$\cdot$
$\cdot$
$v_n=e^n{v}_0$ (after n collision)

After $n^{\text{th}}$ collision, let $\theta$ be the maximum angular deflection.


From the law of conservation of energy,
$\frac{1}{2}m{v}_n^2=mgl(1-\cos \theta )$

$\Rightarrow \frac{1}{2}{e}^{2n}{v}_0^2=gl(1-\cos \theta )$

$\Rightarrow \frac{1}{2}{e}^{2n}\cdot 2gl=gl(1-\cos \theta )$

$\Rightarrow e^{2n}=1-\cos \theta$

$\Rightarrow \cos \theta =1-e^{2n}$

$\therefore \theta ={\cos }^{-1}(1-e^{2n})$

Hence, $\left(C\right)$ is the correct answer.