= Solution
In the <centre-of-momentum frame> the two equal <masses> have opposite <velocities>. Conservation of <kinetic energy> in an elastic collision fixes their common <speed>, so scattering changes their directions only. For incoming laboratory <velocities> $\mathbf v$ and $\mathbf0$, the frame <velocity> is $\mathbf V=\mathbf v/2$ and the incoming relative <velocities> are $\pm\mathbf v/2$. If the outgoing relative <velocities> are $\pm\mathbf w$, then $|\mathbf w|=v/2$ and the laboratory <velocities> are $\mathbf V\pm\mathbf w$. Their dot product is zero: \b[nonzero outgoing laboratory <velocities> are perpendicular].
For this <smooth equal-disc oblique collision>, let $\mathbf n$ point from A to B at contact. The <impulse> acts only along $\mathbf n$; the tangential <velocity> of A is unchanged. With $\mathbf v\cdot\mathbf n=v\cos\theta$, equal <masses> exchange their normal <velocity> components:
$$
\mathbf v_{Af}=\mathbf v-v\cos\theta\,\mathbf n,\qquad
\mathbf v_{Bf}=v\cos\theta\,\mathbf n.
$$
Therefore
$$
\boxed{v_{Af}=v\sin\theta,\qquad\angle(\mathbf v,\mathbf v_{Af})=\frac\pi2-\theta.}
$$
No rotation about a disc centre is generated by these central <impulses>. At exactly head-on impact A stops, so its outgoing direction is undefined; the angle formula then denotes a limiting direction.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-ia2013-disc-velocities.png]
{title=Original velocity-vector construction for an equal-mass elastic collision, before and after impact in the laboratory and centre-of-momentum frames}
For this <adhesive collision of two equal discs>, use the <parallel axis theorem> about the midpoint of the touching centres:
$$
I_{\rm pair}=2(\tfrac12ma^2+ma^2)=3ma^2,\qquad
\mathbf L_{\rm CM}=(-a\mathbf n)\mathbin\times m\mathbf v.
$$
Its magnitude is $mav\sin\theta$. Linear and <angular momentum> conservation give
$$
\boxed{\mathbf V_{\rm CM}=\frac{\mathbf v}{2},\qquad
|\Omega|=\frac{v\sin\theta}{3a},\qquad
K_f=mv^2\left(\frac14+\frac{\sin^2\theta}{6}\right).}
$$
The rotation has the sign of $\mathbf L_{\rm CM}$, clockwise for the printed geometry. The <energy> loss is $mv^2(1/4-\sin^2\theta/6)>0$. After impact the midpoint translates uniformly and the rigid pair rotates uniformly about it: there are no external horizontal <forces> or axial <torques>.
Rough unglued edges can transmit a tangential <impulse> and generate <angular momentum> about the individual disc centres. The centre-of-mass <velocity> and total <angular momentum> still have the values above, but individual translational <speeds>, <angular velocities> and <energy> loss require a <friction>/contact law and normal restitution. \b[Roughness alone does not imply permanent sticking or the adhesive result.]
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