A horizontal frictionless rod is threaded through a bead of mass `m`. The length of the cart is `L` and the radius of the bead, `r`, is very small in comparison with `L(r lt lt L)`. Initially at (`t = 0`) the bead is at the right edge of the cart. The can is struck and as a result, it moves with velocity `v_(0)` towards right. When the bead collides with the cart's walls, the collisions are always completely elastic.

The first collision takes place at time ti and the second collision takes place at time `t_(2)`. Find `t_(2)-t_(1)`

Options

A. `(2L)/v_(0)`
B. `L/v_(0)`
C. `L/(2v_(0))`
D. `L/(3v_(0))`

Answer

B

Explanation

`Mv_(0)=Mv_(1)+mv_(2)`…………i
`(v_(2)-v_(1))/(v_(0))=`………..ii
Solve to get `v_(1)=(M-m)/(M+m)v_(0)`
`v_(2)=(2M)/(m+M)v_(0)`