Electronic – Voltage regulator circuit, efficient, low necessary drop

Steps to diagnose / eliminate the problem:

• Add a 10 μF, 6.3 Volts or higher capacitor in parallel with C2 from the 7805 output leg, as close as possible, to ground.
• Add a 100 uF, 10 Volts or higher capacitor to the 9 volt input, to act as a buffer / reservoir capacitor.
• Add a 100 Ω resistor on the line from the regulator to the MOSFET gate, and replace the pulldown resistor on the Gate with least 1 kOhms. This will reduce the current spike due to capacitance of the MOSFET gate - I believe this should not be the problem, but it is a simple and useful preventive step nevertheless.

If the above does not resolve the issue:

• Use a separate 9 volt supply for the motor, your existing supply is unable to cope with the starting load (stall current >> operating current, usually) of the motor.

Many supplies nowadays use both a switching regulator and a linear regulator together. The two are set up to track each other, so the voltage dropped by the linear regulator is never excessive.
So for example, say you set the supply to 5V, you might have a 24V input which the switching regulator drops down to 6-7V, then the linear regulator drops that down to the desired 5V. This way you get the benefit of the switchers efficiency with the quiet output of the linear regulator.

However, as mentioned by Passerby, efficiency and weight is not always so much of a concern compared to reliability and quietness, so many simply use a purely linear supply and deal with the losses involved.
To keep from being ridiculously inefficient however, generally many linear supplies will have more than one tap on the transformer which are switched between at appropriate points in the range (e.g. instead of dropping from 24V to 1V, it would switch to, say, a 5V tap so a lot less power is dissipated) You can usually hear the clicks of the relays at certain points in the voltage range.