Three identical balls are connected by light inextensible strings with each other as shown and rest over a smooth horizontal table. Length of each string is `l`.
At moment `t = 0`, ball `B` is imparted a velocity `v_(0)` perpendicular to the strings and then the system is left on its own.

Calculate the velocity of `B` just before `A` collides with ball `C`.

Explanation

from collision
`mv_(0)=3mv_(y)`
`v_(y)=(v_(0))/(3)`
From COE
`(1)/(2)mv_(0)^(2)=3(1)/(2)mv_(y)^(2)+2(1)/(2)mv_(x)^(2)`
`(1)/(2)mv_(x)^(2)(1-2(v_(0)^(2)2)/(9))=mv_(x)`
`v_(x)=(v_(0))/(sqrt(3))`
From frame of ball `B`,
`3T=(mv_(x)^(2))/(l)`
`:. T=(mv_(x)^(2))/(3l)=((3)(v_(0)^(2)))/(3xx3xx1)=1`