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Figure shows a circular frications trck ...

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.

A

`0.2`

B

`0.5`

C

`1`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D
`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)`
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