Using that common collector configuration, the motor voltage will never be higher than the base voltage minus about 0.6V. So, about 4.4V, give or take.
That's a horrible circuit, junk it!
With only the 12V supply: Connect the emitter to PS (-) which we will call ground. Connect the motor between the PS (+) and the collector. Put the diode across the motor with the cathode (bar end) pointing back to the PS (+) (reverse biased).
Use the 4.7k (keep reading) resistor from the +12V (collector) and the base.
There will be limitiations. The maximum base current is (12-0.6) / 4700 or about 2 mA. If the transistor current gain is 100 then the maximum collector current is 200 mA. It's pretty obvious this isn't going to work. Try 1k... Then the maximum base current is (12-0.6)/1000 or 11 mA and if the gain is 100, the maximum collector current is 1.1 Amps.
All assuming of course that the transistor can handle 1.1 Amp of collector current while maintaining a current gain of 100.
For the 2N3055 (big power transistor), the gain at 1 Amps is about 90. So, maybe we have to kick the base current up a bit. Perhaps the resistor should be 470 Ohms.
Now, when it comes time to shut the transistor, we need to pull the base and the 470 Ohm resistor down to 0V. Well, that takes a current of 12V / 470 Ohm = 26 mA. We might use another transistor to do that. Perhaps a 2N3904. We would connect the emitter to ground, the collector to the base of the 2N3055 and then when we apply base voltage through a 1K resistor, a high voltage will turn the 2N3904 ON and this will turn the 2N3055 OFF. When we pull the base of the 2N3904 to 0V, the transistor is off and the 470 Ohm resistor turns the 2N3055 ON.
Something like that!
The base current issue (high base current for big power transistors) is one of the reasons we use MOSFETs.
https://www.onsemi.com/pub/Collateral/2N3055-D.PDFhttps://www.onsemi.com/pub/Collateral/2N3903-D.PDF