Solution

After collision they will move as a unit, rotating around the centre of mass, which is at a distance $l/4$ from the contact point of the balls

Conservation of angular momentum

$$mv\cdot\frac{L}{4}=m\frac{\omega L}{4}\cdot\frac{L}{4}+I\omega\tag{1}$$

Determening the moment of inertia, using parallel axis theorem

$$I=\frac{1}{2}mL^2+m\left(\frac{L}{4}\right)^2=\frac{7}{48}mL^2\tag{2}$$

From the $(1)$ and $(2)$

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

Also, in addition to rotational motion, translational motion will be observed. From the law of conservation of momentum

$$mv=2mu\Rightarrow u=v/2$$

$u$ — centre of mass velocity

THe initial energy was as a kinetic energy

$$E_0=\frac{mv^2}{2}$$

The total energy after the collision:

$$E_f=\frac{2mu^2}{2}+\frac{I\omega^2}{2}+\frac{m(\omega\cdot L/u)^2}{2}$$

Conservation of energy

$$E_0=E_f+Q$$

Where, $Q$ — the released energy

$$Q=E_0-E_f$$

$$Q=\frac{mv^2}{2}-\frac{mv^2}{2\cdot2}-\frac{7mL^2}{48}\cdot \left(\frac{6v}{5L}\right)^2\cdot \frac{1}{2}+\frac{m\cdot(6v/5L\cdot L/4)^2}{2}$$

After the mathematical transformations

$$\boxed{Q=\frac{mv^2}{10}}$$