Science a GoGo's Home Page Forums General Discussion Sci Fi Forum Orion, Mission to Alpha Centauri

I didn't have a train set - I did have an electric car set though. And I apparently paid more attention to my childhood toy than you did to yours.

When a car rocketed around a corner the whole track slid (if you got the cars up fast enough). The reason for this movement is simple - the track produced an inwards force which caused the car to turn, but in turn experienced an opposing force in the opposite direction. If you got things up fast enough that outward force would overcome friction and the track would then slide in the opposite direction of the force on the car.

But, when the car got to the other end of the track and hit the next corner, the track slid back - because once again the track applied a force to the car and thus experienced an opposing force.

Its an imperfect example (due to friction), but does reflect your system somewhat - both the car and the track undergo movement relative to each other, due to the internal forces. But the net product in the end was zero movement - every time the car turned at one end of the oval it created returned the track to its origin.

The major difference in your system is that every acceleration, change of direction, etc, will result in movement of the whole system - due to the absence of friction in space. The net is a sum of zero change in momentum - just as with my childhood toy.



Nope, that is not what we are saying. What we are saying is when you move either one of those you get equal forces in both directions. The net effect is zero change in momentum of the total system, since the movement of the mass in one direction is exactly countered by the opposing force which is exerted on the system in the opposing direction.

The problem you are having is quite simple - you're ignoring the "reaction" half of the equation. You forget that when you move one of those masses, you're both exerting the force which moves the mass (the action) as well as exerting a force on the internal environment of your closed system in the direction opposite (the reaction).



Nope, you've still got it wrong:

(reaction)<--mass-->(action)

<--450kg--> (equal and opposite force/momentum)
<--50kg--> (another equal and opposite force/momentum)

Net force/momentum of <--450kg--> is zero
Net force/momentum of <--50kg--> is zero
Net of the two combined is zero

Like I said earlier, you're missing half the equasion.

Bryan