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A smooth hoop lie as on a smooth horizon...

A smooth hoop lie as on a smooth horizontal table and is fixed. A particle is projected on the table form a point A on the inner circumference of the hoop at angle `theta` with radius vector . If e be the coefficient of restitution and the particle returns to the point of projection after two successive impacts. The final angle `theta'` made by velocity vector with radius of hoop is

A

`tan theta' = ( e^(2))/( ( 1+ e^(2))) tan theta `

B

`tan theta' = e^(2) tan theta `

C

`tan theta ' = ( tan theta )/( e^(2))`

D

`tan theta' = ( tan theta)/( e )`

Text Solution

Verified by Experts

The correct Answer is:
C
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