In a smooth circular tube of radius `R` , a particle of mass `m` moving with speed `V_(0)` hits another particle of mass `3 m` at rest as shown. The time after which the next collision takes place (assume elastic collision)

Options

A. `(pi R)/(v_(0))`
B. `( 2pi R)/( v_(0))`
C. `(pi R)/(2 v_(0))`
D. `(pi R)/(4 v_(0))`

Answer

B

Explanation

`mv_(0) = mv_(1) + 3 mv_(2) rArr v_(1) + 3v_(2) = v_(0) ` (i)
`v_(1) + v_(0) = v_(2) + 0 rArr v_(1) - v_(2) = - v_(0)` (iii)
Solving ,`v_(2) = (v_(0))/(2) , v_(1) = -(v_(0))/(2)`
Velocity of `B` relative to `A = v_(2) + v_(1) = v_(0)`
Time taken by `B` to complete one revolution
`t = (2 pi R)/(v_(0))`