1. Set up the system

A uniform thin bar has:

• length $L$
• mass $8m$

Two particles collide and stick to the bar:

• particle $m$ with speed $2v$
• particle $2m$ with speed $v$

They strike at distances from the center of the bar:

• $m$ at $L/3$
• $2m$ at $L/6$

Since they come from opposite sides and the table is smooth, we use conservation of angular momentum about the final center of mass of the system.

2. Linear momentum before collision

Take rightward direction as positive.

Let:

• mass $m$ move rightward with speed $2v$
• mass $2m$ move leftward with speed $v$

Then total linear momentum is $P = m(2v) + 2m(-v) = 2mv - 2mv = 0$

So the final center of mass remains at rest. Hence the system rotates about its final COM, which is also fixed.

3. Find the position of final center of mass

Take the center of the bar as origin.

To make the bar rotate after collision, the two particles must hit on opposite sides of the center. Let:

• mass $m$ stick at $x=+L/3$
• mass $2m$ stick at $x=-L/6$

Then the COM of the combined system is $x_{cm} = \frac{8m(0)+m\left(\frac{L}{3}\right)+2m\left(-\frac{L}{6}\right)}{8m+m+2m}$

$x_{cm} = \frac{mL/3 - 2mL/6}{11m} = \frac{mL/3 - mL/3}{11m} = 0$

So the final center of mass is at the original center of the bar.

Thus we can calculate angular momentum about the bar center itself.

4. Initial angular momentum about the center

Angular momentum magnitude for a particle moving perpendicular to radius line is $L = mvr$

Here each particle moves horizontally, and the perpendicular distance from the center is just the given impact distance.

For particle $m$: $L_1 = m(2v)\left(\frac{L}{3}\right)=\frac{2mvL}{3}$

For particle $2m$: $L_2 = 2m(v)\left(\frac{L}{6}\right)=\frac{mvL}{3}$

Both contribute in the same rotational sense, so total initial angular momentum is $L_i = \frac{2mvL}{3}+\frac{mvL}{3}=mvL$

5. Moment of inertia after collision

The final body consists of:

• the bar of mass $8m$
• point mass $m$ at distance $L/3$
• point mass $2m$ at distance $L/6$

Moment of inertia of the bar about its center: $I_{\text{bar}} = \frac{1}{12}(8m)L^2 = \frac{2mL^2}{3}$

Moment of inertia of particle $m$: $I_1 = m\left(\frac{L}{3}\right)^2 = \frac{mL^2}{9}$

Moment of inertia of particle $2m$: $I_2 = 2m\left(\frac{L}{6}\right)^2 = 2m\cdot \frac{L^2}{36}=\frac{mL^2}{18}$

So total moment of inertia is $I = \frac{2mL^2}{3}+\frac{mL^2}{9}+\frac{mL^2}{18}$

Taking LCM $18$: $I = \frac{12mL^2+2mL^2+mL^2}{18} = \frac{15mL^2}{18} = \frac{5mL^2}{6}$

6. Apply conservation of angular momentum

$L_i = I\omega$

$mvL = \frac{5mL^2}{6}\,\omega$

Therefore, $\omega = \frac{mvL}{(5mL^2/6)} = \frac{6v}{5L}$

7. Match with options

$\boxed{\omega = \frac{6v}{5L}}$

So the correct option is: $\boxed{\text{B}}$