Say I have a linear motor [aka rail-gun] and use a x amount of electrical power. I fire the gun and the object exits at velocity v. I then reuse the same object as my projectile and fire the rail-gun a second time this time with 2x the electrical power. My lessons on momentum suggest it will go 2v. The Work-Energy Theorem says 1.414v. Which is right and why?

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    $\begingroup$ First of all, not all linear motors are rail guns, so you should clarify this point. Also, you should explain how you got the answers you mention in your question. Try to identify the physical concepts which lead you to use the methods you've chosen, and tell us what it is about those concepts that confuse you. $\endgroup$ – Colin K Jan 11 '12 at 22:09

Your railgun has some fixed length. If you apply some voltage to generate some force, then the energy gained by the projectile will simply be force times distance (i.e. the length of the railgun).

If you now double the voltage, to get twice the force, the work done will be twice as great so the projectile will have twice the kinetic energy, and this is $\sqrt{2}$ times the velocity.

I think what's puzzling you is that you say "My lessons on momentum suggest it will go 2v". Remember that the change of momentum is force times time (called "impulse"). When you double the force the projectile goes faster so it spends less time in the railgun, so even though the force doubles the product $F\times t$ does not.

  • $\begingroup$ Talking about barrel length adds unnecessary complications. Time, not distance, is important here. N $\endgroup$ – Sammy Jan 14 '12 at 23:09
  • $\begingroup$ The distance and the applied force determine the amount of energy given to the projectile. Because the distance is constant, the energy is just proprtional to the force, which is (probably) proportional to the applied voltage. So knowing that the distance is constant simplifies things. By contrast the time is inversely proportional to the square root of the force/voltage, and this seems less simple to me. $\endgroup$ – John Rennie Jan 15 '12 at 17:27
  • $\begingroup$ From what I can see, time and force go hand in hand. Newton's 3rd Law states as much. One object pushes on another; neither accelerates unless the other one does. They both accelerate for the same time but the distance each travels during the event is apt to be different. Momentum relates to the time force acts, ke relates to the distance. From this it looks like there are instances where the ke formula cannot produce consistent values. Add Newton's 2nd, and things get worse. Accelerate 2 objects of different mass in the same way and one of them winds up with more ke. Think about it. $\endgroup$ – Sammy Jan 15 '12 at 19:54
  • $\begingroup$ The point is that your railgun has a barrel of a fixed length. That means every projectile accelerated along it travels the same distance, so when you're calculating what speed the projectiles leave the gun you can treat distance as a constant. You're quite correct that you can freely interchange distance and time using Newton's laws, but for any particular system it makes sense to choose whatever statement of the laws makes the calculation easiest. $\endgroup$ – John Rennie Jan 17 '12 at 18:44

Do you mean power or energy?

If you fire the rail gun with twice the power and (depending on the design of the rail gun) that causes twice the force and so the projectile accelerates faster and so spends less time in the barrel then it doesn't receive twice as much energy. The details will depend on the gun

If you fire with twice the total energy (you have a rail gun that gives a short impulse to the projectile) then the projectile has twice the total kinetic energy and so will be going \sqrt(2) faster

  • $\begingroup$ I'm sorry about how I worded the question. I thought I would hear that it would double its speed. That would happen if you fired an object and wi $\endgroup$ – Sammy Jan 14 '12 at 23:17

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