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 degrees ?

Answer 1

By the definition of coefficient of restitution.after each impact velocity will be $\frac{2}{\sqrt{5}}$ time. hence K.E will change $\frac{4}{5}$ time before to make amplitude$\leqq {60}^0$ the maximum potential energy should be halved $\frac{4}{5}N<\frac{1}{2}$ So, $N=4$