Glancing collisions: conserve momentum along x and along y

The main symbols used in this lesson:

In snooker you rarely hit a ball dead centre. The cue ball strikes the object ball off-centre, and the two shoot off at angles, like a pair of scissors opening.

Vector triangles get awkward when nothing is at right angles. The cleaner method: resolve everything into two perpendicular directions, $x$ and $y$.

Momentum is conserved in each direction separately, so one 2D problem becomes two 1D problems.

1. Choose $x$ along the original direction of motion and $y$ at right angles to it.
2. Resolve each momentum: a momentum $p$ at angle $\theta$ to the $x$-axis has components $p\cos\theta$ along $x$ and $p\sin\theta$ along $y$.
3. Write two equations, with signs: $\textstyle\sum p_x \text{ before} = \sum p_x \text{ after} \qquad \sum p_y \text{ before} = \sum p_y \text{ after}$
4. Solve, then recombine the components of the unknown velocity with Pythagoras and $\tan^{-1}$.

Bonus: if everything moved along $x$ before, the total $y$-momentum before is zero, so the $y$-momenta after must be equal and opposite. One ball heads off on each side of the $x$-axis.