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

D

Explanation

The pendulum bob is beft from position `A` when it is at posion `C` the angular amplitube is `60^(@) ` in

The velocity of bob at `B` v_*(B) befor first collision is
`mgt = (1)/(2) mv_(p)^(2) rArr v_(B) = sqrt(2gt) `
Let after a collision , the angular amplilitube is `60^(@)` when the hot again move toward the wall from `(C) ,The velocity v^(1) g befor collision is
`mg (t)/(2) = (1)/(2) mv_(B)^(2) rArr v_(B)^(1) = sqrt(gt)`
The means that the velocity of the bob should redace from
`sqrt(2gt)` to `sqrt(gt) due to velocity after a collision with wall .
The final velocity afterc `x` collision is sqrt(gt)`
` :. x^(2) (sqrt(2gt)) = sqrt(gt)`
where `e` is coeffiicient of restinution .
((2)/(sqrt(5))^(0) xx sqrt(2gl) = sqrt(gt) rArr (2)/(sqrt5)) ^(0) = (1)/(sqrt(2)`
Taking log on both sides we get = ` n log ((2)/(sqrt(5)) = log (1)/(sqrt(2) rArr n = 3.1`
Therefor number of collision will be `4`