Two balls impact e = 0.9, 30 and 50 kg

Assumptions (figure description is ambiguous): the line of impact is the horizontal x-axis. Ball A (30 kg) moves to the right at $30$ m/s. Ball B (50 kg) moves to the left and upward at $30$ m/s, $45^\circ$ to the line of impact. $e=0.9$, smooth balls.

Components before impact

$$v_{Ax}=30,\ v_{Ay}=0;\qquad v_{Bx}=-30\cos45^\circ=-21.21,\ v_{By}=30\sin45^\circ=21.21\ \text{m/s}$$

Tangential (y) components do not change: $v'_{Ay}=0$, $v'_{By}=21.21$ m/s.

Along the line of impact

Momentum:

$$30(30)+50(-21.21)=30v'_{Ax}+50v'_{Bx}\ \Rightarrow\ 30v'_{Ax}+50v'_{Bx}=-160.7$$

Restitution:

$$v'_{Bx}-v'_{Ax}=0.9\,(30+21.21)=46.09$$

Substituting $v'_{Bx}=v'_{Ax}+46.09$:

$$80v'_{Ax}+2304.6=-160.7\ \Rightarrow\ v'_{Ax}=-30.82\ \text{m/s},\qquad v'_{Bx}=15.28\ \text{m/s}$$

Final velocities

Ball A: $v'_A=30.82$ m/s, directed along the negative x-axis (it rebounds straight back, $180^\circ$).

Ball B:

$$v'_B=\sqrt{15.28^{2}+21.21^{2}}=26.14\ \text{m/s},\qquad \theta=\tan^{-1}\frac{21.21}{15.28}=54.2^\circ\ \text{above the positive x-axis}$$

Answer: $v'_A=30.8$ m/s at $180^\circ$; $v'_B=26.1$ m/s at $54.2^\circ$ above the +x direction. (If B instead moves down-left, the sign of B's y-component reverses and its direction becomes $54.2^\circ$ below +x.)