Two-Dimensional Elastic Collisions in the Laboratory Frame

Set the four unknowns $v_{1,f}$, $v_{2,f}$, $\theta_{1,f}$, $\theta_{2,f}$ against two momentum components and one energy equation, leaving one quantity, conventionally $\theta_{1,f}$, to be specified.

Square and add the two momentum equations to remove $\theta_{2,f}$, substitute into the energy condition, and solve $0 = (1+\alpha)v_{1,f}^2 - 2\alpha v_{1,i}v_{1,f}\cos\theta_{1,f} - (1-\alpha)v_{1,i}^2$.

Divide the sine momentum equation by the cosine one for $\tan\theta_{2,f} = \frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i} - v_{1,f}\cos\theta_{1,f}}$, then write $v_{2,f} = \frac{\alpha v_{1,f}\sin\theta_{1,f}}{\sin\theta_{2,f}}$.

Put $\alpha = 1$ and $v_{1,i} = 3.0$ m/s, reduce the quadratic to $v_{1,f} = v_{1,i}\cos\theta_{1,f} = 2.6$ m/s, then obtain $\theta_{2,f} = 60°$ and $v_{2,f} = 1.5$ m/s.