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}}
\\$\\$