Figure shows a circular frications trck of radius `R`, centred at point `O`. A particle of mass `M` is released from point `A (OA = R//2)`. After collision with the track, the particle moves along on the track. Then the coefficient of restitution `e` is.

Options

A. `0.2`
B. `0.5`
C. `1`
D. zero

Answer

D

Explanation

`cos theta = (1)/(2) rArr theta = (pi)/(3)`
(a) After collision the particle moves along the track. This means there is no normal component of velocity, Hence `e = 0`
(b) Just before collision, the compoenents of velocity along the normal and along the tangent are
`u_(n) = u sin theta`
`u_(t) = u cos theta`

During collision `u_(t)` does not change. The normal component of velocity becomes `eu_(n)` along `overset(vec)(B)O`.

Question says that velocity of the particle is horizontal after collision, which mean s`u_(t) cos theta = eu_(n) sin theta`
`u cos2 theta = eu sin2theta rArr e = cot^(2)'(pi)/(3) = (1)/(3)`