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. `(5 pi)/(2)s`
D. `4 pi s`

Answer

C

Explanation

For first collision
`v = 10 m//s t_(1) = (pi(5))/(10)=pi//2 sec`
velocity of sep = e velocity of opp.
`v_(2)-v_(1)=(1)/(2)(10)rArr v_(2)-v_(2)=5m//s`
for second collision
`:. t_(2) = (2 pi (5))/(5) = 2 pi`
`:.` total time `t = t + t_(2) = pi//2 + 2pi`
`t = 2.5 pi`