Electronic – Are additional smoothing capacitors required off of a voltage regulator

Appropriate bleeder resistors are the usual solution. They aren't usually switched, although they can be.

The value depends on the time you require to discharge the capacitors. The formula is

$$ V_{t} = V_{0} \, e^{ -t / RC } $$

where \$V_{t}\$ is the voltage at time t and \$V_{0}\$ is the initial voltage at time 0. It's an exponential function, so I'd just assume 1/10 of the initial voltage.

It isn't a power function, as someone edited it!

You should find that the power taken by the bleeder resistors is negligible compared to the 120W capability of the supply.

All digital ICs should have decoupling caps between their power and ground pins. These should be ceramic and physically as close as possible to the IC. You want to minimize the loop length from power pin to cap to ground pin thru the IC and back to the power pin.

Decoupling caps deal with short term current spikes the IC draws. They must therefore be high frequency. Large capacitance is not necessary, and since large capacitors usually have poor high frequency response, they are worse. A 100µF electrolytic cap is pretty useless for decoupling. 1µF to 100nF ceramic is good.

As for the cap on the motor, the idea is good but I think 100nF is too large. That could cause excessive or unnecessary current to flow in the H bridge every time it switches. If you're only reversing motor direction occasionally, then this isn't a big deal. If you're using the H bridge to modulate the apparent motor drive with PWM, then you should lower the cap. Something like 1nF should still cut down the noise the motor is making while not getting in the way of switching.