= 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 $\phi$ are nonnegative deflections on opposite sides. <Conservation of momentum> gives
$$
mu=mv\cos\theta+MV\cos\phi,
\qquad mv\sin\theta=MV\sin\phi.
$$
The <elastic collision> conserves <kinetic energy>:
$$
mu^2=mv^2+MV^2.
$$
Taking the squared magnitude of the vector <momentum> equation yields
$$
m^2v^2=m^2u^2+M^2V^2-2mMuV\cos\phi.
$$
Eliminate $v^2$ with the <kinetic energy> equation. For a nontrivial <collision>, $V\ne0$, so
$$
(M+m)V=2mu\cos\phi.
$$
Substituting back into the two <momentum> components gives
$$
v\cos\theta=\frac{u(m-M\cos2\phi)}{m+M},
\qquad
v\sin\theta=\frac{uM\sin2\phi}{m+M}.
$$
Therefore
$$
\boxed{\tan\theta=\frac{\sin2\phi}{m/M-\cos2\phi}.}
$$
The ratio alone does not determine the quadrant: for $M>m$, the lighter particle can move backwards. The two component equations select the correct $0\leq\theta\leq\pi$. At $m/M=\cos2\phi$, the outgoing direction is perpendicular to the initial direction; the displayed <tangent> has a pole there.
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