Two discs A and B are moving with their flat circular surface on a smooth horizontal surface. Mass, radius and velocity of the two discs are –  ${\mathrm{m}}_{\mathrm{A}}=2\mathrm{M},{\mathrm{m}}_{\mathrm{B}}=\mathrm{M},{\mathrm{r}}_{\mathrm{A}}=\mathrm{R},{\mathrm{r}}_{\mathrm{B}}=2\mathrm{R},{\mathrm{v}}_{\mathrm{A}}=\mathrm{v},\text{ and }{\mathrm{v}}_{\mathrm{B}}=$2v. The velocities of the two discs are oppositely directed so that they just cannot avoid collision and stick to each other (see figure).
 The angular speed of the composite system after collision is $\omega$ and 
the loss in kinetic energy due to collision is $△K.$Then

(a) Let’s first locate the COM of the system. Let it be at a distance x from the centre of disc A
$\mathrm{x}=\frac{\mathrm{M}\left(3\mathrm{R}\right)}{2\mathrm{M}+\mathrm{M}}=\mathrm{R}$
 

It means the COM is at point C where the two discs make contact.
Linear momentum of the system is zero. Hence the COM of the composite system will be at rest after collision. 
The system will rotate about C with angular speed $\omega$ after collision. We will apply conservation of angular 
momentum about C 
$$\begin{gathered} {\mathrm{L}}_{\mathrm{aftter}\text{ collision }}={\mathrm{L}}_{\text{before collision }} \\ {\mathrm{I}}_{{\mathrm{C}}^{\mathrm{\prime }}}\mathrm{\omega }=2\mathrm{MVR}+\mathrm{M}\cdot 2\mathrm{V}\cdot 2\mathrm{R} \\ \left[\frac{3}{2}\left(2\mathrm{M}\right){\mathrm{R}}^2+\frac{3}{2}\left(\mathrm{M}\right)(2\mathrm{R}{)}^2\right]\mathrm{\omega }=6\mathrm{MVR} \\ \Rightarrow 9{\mathrm{MR}}^2\mathrm{\omega }=6\mathrm{MVR}\Rightarrow \mathrm{\omega }=\frac{2}{3}\frac{\mathrm{V}}{\mathrm{R}} \\ \text{ (b) △K}=\left[\frac{1}{2}\left(2\mathrm{M}\right){\mathrm{V}}^2+\frac{1}{2}\left(\mathrm{M}\right)(2\mathrm{V}{)}^2\right]-\frac{1}{2}\left({\mathrm{I}}_{\mathrm{C}}\right){\mathrm{\omega }}^2 \\ =3{\mathrm{MV}}^2-\frac{1}{2}\left(9{\mathrm{MR}}^2\right){\left(\frac{2\mathrm{V}}{3\mathrm{R}}\right)}^2={\mathrm{MV}}^2 \end{gathered}$$