Two identical smooth discs, each of mass 0.20 kg and radius r, lie on a smooth horizontal table…

Step 1. The line of centres. At impact the centres are $2r$ apart, and A's line of motion passes $r$ from B's centre. So the line of centres makes an angle $\varphi$ with A's motion, where $\sin\varphi = \dfrac{r}{2r} = \dfrac12$, that is $\varphi = 30^\circ$.

Step 2. Split A's velocity. Along the line of centres: $u\cos 30^\circ = 3.464\ \text{m/s}$. Across it: $u\sin 30^\circ = 2.0\ \text{m/s}$. The smooth discs exert no force across the line, so that part of A's velocity stays, and B gets none of it.

Step 3. Along the line of centres. For equal masses with restitution $e$, momentum and $v_B - v_A = e\,u_n$ give

Step 4. Final velocities. B moves along the line of centres, at $2.60\ \text{m/s}$, $30^\circ$ from A's original direction. A has $0.866\ \text{m/s}$ along that line and $2.0\ \text{m/s}$ across it:

Its angle from the line of centres is $\tan^{-1}\dfrac{2.0}{0.866} = 66.6^\circ$. So A turns $66.6^\circ - 30^\circ = 36.6^\circ$ away from its original direction, on the other side from B.

Step 5. Angle between them. B moves along the line of centres, so the angle between the two velocities is A's angle from that line, $66.6^\circ$.

Step 6. Energy. Before: $\tfrac12 \times 0.20\ \text{kg} \times (4.0\ \text{m/s})^2 = 1.60\ \text{J}$. After: $\tfrac12 \times 0.20\ \text{kg} \times (2.179^2 + 2.598^2)\ \text{m}^2/\text{s}^2 = 1.15\ \text{J}$. Lost: $0.45\ \text{J}$, about $28\%$.