= Solution
Set $q=1$ in the <laboratory scattering circle> formula. For $M_1>M_2$, a real outgoing <speed> requires
$$
\cos^2\theta_1-1+\left(\frac{M_2}{M_1}\right)^2\ge0,
\qquad \sin^2\theta_1\le\left(\frac{M_2}{M_1}\right)^2.
$$
Also $v_1\ge0$ excludes the backward-angle interval: the square root is smaller than $|\cos\theta_1|$ when $M_2/M_1<1$, so both roots would be negative for $\cos\theta_1<0$. Therefore
$$
\boxed{\theta_1^{\max}=\sin^{-1}\left(\frac{M_2}{M_1}\right)<\frac\pi2.}
$$
At this boundary the two possible <speeds> coincide, with $v_1/u_1=\sqrt{(M_1-M_2)/(M_1+M_2)}$. Geometrically the outgoing ray is tangent to the <laboratory scattering circle>. As $M_2/M_1\to0$ the maximum tends to zero, while it approaches $\pi/2$ as the masses approach equality from $M_1>M_2$.
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