Question

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position (see fig.) and released. The ball hits the wall, the coefficient of restitution being $\frac{2}{\sqrt{5}}$. What is the minimum number of collisions after which the amplitude of oscillations becomes less than $60$ degree?

Answer 1

When pendulum is released from vertical position it poses some potential energy and finally when it reaches at lowest point its potential energy is minimum and kinetic energy is maximum.

So to find velocity of pendulum just before collision with the wall Law of conservation of energy is applied.

assuming vertical position as datum.

$MgL+0=0+\frac{1}{2}M{v}^2$

$v=\sqrt{2gL}$

Now velocity of pendulum after collision is calculated by the using following equation $\frac{velocityofsepration}{velocityofapproach}=e$

$\frac{V_f-0}{v_i-0}=e=\frac{2}{\sqrt{5}}$

$V_f=\frac{2}{\sqrt{5}}.\sqrt{2gL}$

after n number of collision$V_f=[\frac{2}{\sqrt{5}}{]}^n.\sqrt{2gL}$

Now for $\theta <60,$ applying law of conservation of energy

$V_f<\sqrt{gL}$

so from here we can say that $[\frac{2}{\sqrt{5}}{]}^n.\sqrt{2gL}<\sqrt{gL}$

this gives $n>3.125$.

So after 4 collision amplitude if oscillation will be less than 60 degree.