Question

On a smooth table two particles of mass of each, travelling with a velocity `v_(0)` in opposite directions, strike the ends of a rigid massless rod length `l`, kept perpendicular to their velocity. The particles stick to the rod after the collision. Find the tension in rod during subsequent motion.
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Answer 1

Correct Answer - `(2mv_(0)^(2))/(l)`
As no external torque is acting
:. Its angular momentum will remain conserve
`mv_(0) xx (l)/(2) xx 2 = I omega`
`mv_(0) l = [m ((l)/(2))^(2) xx 2] omega rArr omega = (2v_(0))/(l)`

Tensiom in the rod will be due to provide centripetal force
`T = m omega^(2) (l)/(2) rArr T = m xx((2v_(0))/(l))^(2) xx (l)/(2)`
`T = (2mv^(2))/(l)`.