A simple pendalum is suspended from a peg on a verticle wall . The pendulum is pulled away from the well is a horizental position (see fig) and released . The bell his the well the coefficient of resitution being `(2)/(sqrt(5)`

what is the miximum number of colision after which the amplitube of secillections between less that `60` digree ?

Answer

4

Explanation

After `1^(st)` collision
`-(2)/(sqrt5) sqrt(2gl) = sqrt(2gl (1 - cos theta))`
`"cos" theta (1)/(5)`
Fro `II^(nd)` collision
`(2)/(sqrt5) . (2)/(sqrt5) sqrt(2gl) = sqrt(2gl (1 - cos theta))`
For `n^(th)` collision
`((2)/(sqrt5))^(n) = sqrt(1 - cos theta) rArr ((4)/(5))^(n) = 1 - cos theta`
`"cos" theta = 1 - ((4)/(5))^(n)`
Put `n = 0, 1, 2, 3` and get answer