 Solution

Choose the heavier particle's outgoing velocity above the initial direction. Let its speed be $V$, and let the lighter particle have components $(v\cos \theta ,-v\sin \theta )$. Thus $\theta$ and $\varphi$ are nonnegative deflections on opposite sides. Conservation of momentum givesThe elastic collision conserves kinetic energy:Taking the squared magnitude of the vector momentum equation yieldsEliminate $v^2$ with the kinetic energy equation. For a nontrivial collision, $V\ne 0$, soSubstituting back into the two momentum components givesThereforeThe ratio alone does not determine the quadrant: for $M>m$, the lighter particle can move backwards. The two component equations select the correct $0\le \theta \le \pi$. At $m/M=\cos 2\varphi$, the outgoing direction is perpendicular to the initial direction; the displayed tangent has a pole there.