Two masses m and `2m` are placed in fixed horizontal circular smooth hollow tube of radius r as shown. The mass m is moving with speed u and the mass `2m` is stationary. After their first collision, the time elapsed for next collision. (coefficient of restituation `e = 1//2` )

Options

A. `(2pir)/(u)`
B. `(4pir)/(u)`
C. `(3pir)/(u)`
D. `(12pir)/(u)`

Answer

(B)

Explanation

Let the speeds of balls of mass m and `2m` after collision be `v_(1)` and `v_(2)` as shown in figure. Applying conservation of momentum `mv_(1) + 2mv_(2) = m u` & `-v_(1) + v_(2) = (u)/(2)`. Solving we get `v_(1) = 0` and `v_(2) = (u)/(2)`
Hence the ball of mass m comes to rest and ball of mass `2m` moves with speed `(u)/(2). t = (2pir)/(u//2) = (4pir)/(u)`