This problem isn't particularly elegant (although the CM frame is slick), but I want to write it out to cement my understanding of the CM frame.
We are tasked with proving that in a collision of "billiard balls" (two balls of equal mass colliding elastically, with one ball being initially stationary), the resulting trajectories
will be orthogonal to each other.
We can begin by setting up our problem: we have one ball moving towards the other at velocity $v_0$. We can calculate the velocity of the CM frame, relative to the lab frame:
$$u_x \ = \ \frac{P_x}{\sum M} \ = \ \frac{mv}{2m} \ = \ \frac{v}{2}$$
$$u_y \ = \ \frac{P_y}{\sum M} \ = \ 0$$
We can then calculate the velocities of the masses relative to the CM frame
$$v_{0x}' \ = \ v_0 \ - \ u_x \ = \ \frac{v}{2}$$
$$v_{1x}' \ = \ 0 \ - \ \frac{v}{2} \ = \ -\frac{v}{2}$$
Where it is clear that the $y$-direction velocities are still $0$. Now, consider this elastic collision in the CM frame: we have two equal masses
travelling towards each other at equal velocities. This is an example of the CM frame exposing nice symmetry in a physical system. Let $f_0'$ and
$f_1'$ be the speeds of the respective balls in the CM frame. Since momentum must sum to $0$ in the CM frame and the balls have equal mass, we have:
$$f_0' \sin \theta_0 \ = \ - f_1' \sin \theta_1$$
$$f_0' \cos \theta_0 \ = \ - f_1' \cos \theta_1$$
Where $\theta_0$ and $\theta_1$ are the angles that the ball trajectories make with the horizontal, in the CM frame. It follows that:
$$(f_0')^2 \big[ \sin^2 \theta_0 \ + \ \cos^2 \theta_0 \big] \ = \ (f_1')^2 \big[ \sin^2 \theta_1 \ + \ \cos^2 \theta_1 \big] \ \Rightarrow \ |f_0'| \ = \ |f_1'|$$
So the speeds are equal, which is obvious due to underlying symmetry in the system. Thus, in order for energy to be conserved, it must be true that each ball has a speed of $\frac{v}{2}$.
If we go back to the original equations, this implies that $\theta_0 \ = \ \theta_1$.
Now, let's switch back to the rest frame with our newly-found CM velocities:
$$f_{0y} \ = \ \frac{v}{2} \sin \theta$$
$$f_{0x} \ = \ \frac{v}{2} (1 \ - \ \cos \theta)$$
$$f_{1y} \ = \ -\frac{v}{2} \sin \theta$$
$$f_{1x} \ = \ \frac{v}{2} (1 \ + \ \cos \theta)$$
Now, if we treat each of these values as the components of velocity vectors, $\vec{f}_0$ and $\vec{f}_1$, we can take the dot product of the vectors
to determine the angle between them:
$$\vec{f}_0 \ \cdot \ \vec{f}_1 \ = \ \frac{v^2}{4} \Big[ -\sin^2 \theta \ - \ \cos^2 \theta \ + \ 1 \Big] \ = \ 0$$
Thus, the vectors are orthogonal and we have completedthe proof.