= Solution
For <masses> $m$ and $2m$, the constant <centre-of-mass velocity> is $(V/3)\mathbf e_x$. Initially the laboratory <velocities> are $V\mathbf e_x$ and zero, while the <centre-of-mass frame> <velocities> are $(2V/3)\mathbf e_x$ and $-(V/3)\mathbf e_x$. After the <elastic collision> they are
$$
\mathbf v_1^*=\frac{2V}{3}(\cos\chi,\sin\chi),\qquad
\mathbf v_2^*=-\frac V3(\cos\chi,\sin\chi).
$$
Adding $(V/3,0)$ produces the lower-left panel's laboratory <velocity vectors>. The given $30^\circ$ laboratory angle implies
$$
\tan30^\circ=\frac{2\sin\chi}{1+2\cos\chi},\qquad
\boxed{\chi=30^\circ+\arcsin(1/4)\simeq44.48^\circ.}
$$
The <centre-of-mass frame> scattering angle is therefore different from the laboratory angle. Reflection across the $x$-axis gives the equally possible negative-angle configuration.
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