A uniform bar of length 12 cm and mass 20m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with same speed of $$\nu$$ and in the same plane as the bar, as shown in the figure below. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $$\omega$$. The ratio of $$\nu$$ and $$\omega$$ is :

$$COM$$ of the system after the collision relative to the center of the bar ($$x=0$$):

$$x_{cm} = \frac{(20m \cdot 0) + (2m \cdot -2) + (m \cdot 4)}{20m + 2m + m} = \frac{-4m + 4m}{23m} = 0$$

Initial Angular Momentum ($$L_i$$):

For mass $$2m$$: $$L_1 = (2m) \cdot v \cdot (2 \text{ cm}) = 4mv$$ (Counter-clockwise)

For mass $$m$$: $$L_2 = (m) \cdot v \cdot (4 \text{ cm}) = 4mv$$ (Counter-clockwise)

$$L_i = 4mv + 4mv = 8mv$$

Final Angular Momentum ($$L_f$$):

Bar: $$I_{bar} = \frac{1}{12}ML^2 = \frac{1}{12}(20m)(12)^2 = 240m$$

Mass $$2m$$: $$I_{2m} = (2m)(2)^2 = 8m$$

Mass $$m$$: $$I_m = (m)(4)^2 = 16m$$

$$I_f = 240m + 8m + 16m = 264m$$

Applying conservation of angular momentum ($$L_i = I_f \omega$$):

$$8mv = 264m \cdot \omega$$

$$\frac{v}{\omega} = \frac{264}{8}$$

$$\frac{v}{\omega} = 33$$