A uniform rod of length `l` and mass 2 m rests on a smooth horizontal table. A point mass `m` moving horizontally at right angles to the rod with velocity `v` collides with one end of the rod and sticks it. Then

Options

A. `v/l`
B. `(2v)/l`
C. `(5v)/(3l)`
D. `(3v)/(2l)`

Answer

A

Explanation

After collision, the cenre of mass of the composite system will move in a straight line with a constant speed and the system will rotate with a constant angular velocity about the vertical axis passing through (and translating with) the centre of mass of the system. During the collision, the external force and torque about any point (or axis) in an inertial reference frame is zero. The torque acting on the system in the centre of mass reference frame is also zero.

There are many (mathematically Infinite) points (axes) about which the angular momentum of the system remains constant during the collision. The centre of mass of the composite system is the most convenient point.
The distance of the centre of mass of the system from the end where m sticks is given by
`x_(CM) =(mxx0+2mxxl/2)/(m+2m)=l/3`
Let the velocity of the centre of mass of the system after colision be `v_(CM)` and the angular velocity of the system about the vertical axis through the centre of mass be `omega` as shown in the figure.
From the conservation of the angular momentum about the centre of mass axis
`(mvl)/3={m(1/3)^(2)+((2m)l^(2))/12+(2m)(l/6)^(2)}omega`........i
Moment of inertia of the composite system about the centre of mass axis is
`I_(CM)=m(l/3)^(2)+((2m)l^(2))/12+(2m)(l/6)^(2)=(ml^(2))/3a`.......ii
From eqn i and ii
`implies mvl/3=(ml^2)/3omegaimplies omega=v/l`