I know that when a pool ball hits another ball, if there is no spin involved, they should move away at an angle of 90 degrees.

Not true. Consider the case of two identical balls. If the line of approach is aligned with the centers, they will move off in the same direction. As the skew angle increases, so does the departure angle.

The cue ball is often smaller than the other balls though. How would this change the angle at which they move away from each other?

Same principle applies, because the balls are constrained to move in two dimensions only, but the arithmetic is changed slightly.

The underlying principle is conservation of momentum. if the balls have mass m and velocity

** v** then

m

_{1}**v**_{1} + m

_{2}**v**_{2} = m

_{1}'

**v**_{1}' + m

_{2}'

**v**_{2}'

where ' indicates "after collision". The important point is that

**v** is a vector so the components in any two orthogonal directions (say the speed along and across the table) must be conserved.

The fact that the cue ball is smaller than the target simply alters the true angle of approach, means that m

_{1} ≠ m

_{2}, and in practice adds a bit of spin.