one of my lecturers sent me this.so what do you think?
"the 555's output is normally low, and goes high during the pulse. So you need to remove the PNP transistor, and feed the 555's output to the MOSFET's gate. Usually a small series resistor e.g. 22 ohms is used, to reduce the peak current (the MOSFET's gate has significant capacitance). You can keep the 10k resistor from the gate to the 0V rail.
You can also delete R3 and C1, and connect the pushbutton from pin 2 to 0V. This means that you have to release the pushbutton before the pulse time expires, but I expect you would be doing that anyway.
Also you need to use a MOSFET that will saturate (turn fully ON) with only 6V of gate-to-source voltage. In fact, it will probably only see 4.5~5.0V because of losses in the 555. Not all MOSFETs will do this. Ones that do are called "logic level gate" MOSFETs, although this term is not exactly defined and even those MOSFETs will "saturate more" if given more gate voltage.
Here are a few suitable MOSFETs that are available from Digikey:
NTD4906N:
https://www.digikey.com/product-detai...GOS-ND/2194521 USD 0.57; ON-resistance 0.008 ohms with 4.5V gate voltage; through-hole "IPAK" package.
PSMN022-30PL:
**broken link removed** USD 0.73; ON-resistance 0.034 ohms with 4.5V gate voltage; standard through-hole TO-220 package.
FDP8880:
https://www.digikey.com/product-detai...80FS-ND/976840 USD 0.92; ON-resistance 0.0145 ohms with 4.5V gate voltage; standard through-hole TO-220 package.
Lower ON-resistance values mean less voltage dropped in the device. For a small motor like this one, the difference is not important. But the TO-220 devices have the standard 0.1-inch pin spacing and are easier to use with breadboards and stripboard."