Particle 'A' moves with speed `10m//s` in a frictionless circular fixed horizontal pipe of radius 5m and strikes with 'B' of double mass that of A. Coefficient of restitution is `1/2` and particle 'A' -starts its journey at t=0. The time at which second collision occurs is:

Options

A. `(pi)/2 s`
B. `(2pi)/3 s`
C. `(5pi)/2 s`
D. `4pi s`

Answer

C

Explanation

For first collision v =10 m/s `t_(1)=(pi(5))/10 =pi//2 sec.`
velocity of separation =e. velocity of approach.
`v_(2)-v_(1)=1/2 (10)rArr v_(2)-v_(1)=5 m//s`
For second collision
`:. t_(2)=(2pi(5))/5=2pi`
`:.` Total time `t=t+t_(2)=pi//2 +2pirArr t=2.5 pi`