A ball is attached to a string of length L. The ball is released when the string is horizontal. The ball collides with wall and e = 2//sqrt(5). After how many minimum number of collisions , the amplitude of oscillation becomes less than 60^(@).

A ball is attached to a string of length L. The ball is released when

1.	A ball is attached to a string of length L. The ball is released when the string is horizontal. The ball collides with wall and e = 2//sqrt(5). After how many minimum number of collisions , the amplitude of oscillation becomes less than 60^(@).
Answer» Solution :When the ball strikes the first time , its speed `v = sqrt( 2gL)` Speed of ball after the `N^(th)` collision `= e^(n) v` `0 = (e^(n)v)^(2) - 2gL( 1 - cos theta)` `= e^(2n) v^(2) - 2 gL(1 - cos theta)` `cos theta = 1 - e^(2n)` ` = 1 - ((2)/(sqrt(5)))^(2n) = 1 - ((4)/(5))^(n)` If `theta lt 60^(@) , cos theta gt (1)/(2)` `1 - ((4)/(5))^(n) gt (1)/(2) rArr ((4)/(5))^(n) lt (1)/(2)` If `n = 1 , ((4)/(5))^(1) = 0.8` `n = 2, ((4)/(5))^(2) = 0.64` `n = 3 , ((4)/(5))^(3) = 0.512` `n = 4 , ((4)/(5))^(4) = 0.41` Minimum number of collisions `= 4`