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F=m*a - Mindthinking gone wrong.

  1. Oct 8, 2011 #1
    I'm startled due to a little thinking, the thing is. (I'm still a beginner when it comes to mechanics so please don't give me way to advanced answers :wink:)

    Think of a mass m, that accelerates with acceleration a. The accelerating force upon the mass m results into a force, F. I'm totally down with that but.

    * The object m is accelerating, which means the speed is constantly increasing.
    Now let's say an object accelerates with 1m/s^2. Give it enough time and it will reach the speed of light, right?
    Then take another object with the same mass m, accelerating at 2m/^2. It will also reach the speed of light if you give it enough time. But, the force F, is twice as much (right?). And when they both are at 99.99% speed of light, and imagine they collide (moving opposite directions). What will really happen? They both were at the same speed. But the second object m*2=F... They are moving towards each other and when they collide, they have exactly the same speed (=same amount of kinetic energy?), BUT, the second object had an acceleration that was twice as much as the first object. But will the (2F) do any difference during the impact?

    * Let's say now I push the first object and give the constant speed 99.99% speed of light. At the same time the opposite direction object two comes towards it. The collide with the same speed once again, but the first object was not accelerating, the second one was. They both had the same speed as I mentioned which means the same amount of kinetic energy? But! The second object had this equation along, m*a=F.

    Does this result into an impact when the second object as the dominating force? Or are they both equal?

    Gosh now it's getting tough in the head :shy:... I hope you understand what I'm trying to understand here.
  2. jcsd
  3. Oct 8, 2011 #2


    Staff: Mentor

    You are asking about relativistic mechanics. I don't think a non-advanced answer is possible. About the best I can do is to state that they will never both be at 99.99% of the speed of light, the one with the higher acceleration will always be travelling faster than the other.
    Last edited: Oct 8, 2011
  4. Oct 8, 2011 #3
    Like DaleSpam said...you need relativity to answer that.
  5. Oct 8, 2011 #4
    Serious? :eek:
  6. Oct 8, 2011 #5
    How is that possible? I mean compare one car with another one, both starting at zero velocity. They are accelerating with different accelerations, and at certain moments during the acceleration they will be at the same velocities? Then I don't mean they start at the same time. There for example will be one moment when the faster car "goes by" the first car with the lower acceleration force.

    If then try to answer with a more advanced reply, I'll do my best understanding :)
  7. Oct 8, 2011 #6
    nope....as posted, once you get a head start with constant acceleration, you keep it.

    It depends on the situation. If you assume an instantaneous collision, the 2F has no effect.
    If the collision extends over a period of time, then yes. Think of two cars colliding: If one has it's accelerator floored and continues to "accelerate" during the collison, extra energy
    must be dissipated...the energy supplied by the engine during the collision.
  8. Oct 8, 2011 #7


    Staff: Mentor

    Sorry, that wasn't clear to me in the original description.

    If they don't start at the same time then the details of when they each start matters. If the lower acceleration one has a head start then there will be one instant when they are both going the same speed. At that point they will have the same energy. The higher force will have acted over a smaller distance giving the same energy.
  9. Oct 8, 2011 #8
    If two cars are on a collision course and they collide at the moment they both have the same speed, then there different accelerations will not matter. The only thing that matters is their momentum (P=mv).
  10. Oct 8, 2011 #9
    Ahh I didn't know. I had honestly mixed up momentum with torque... Because in swedish torque is called "moment", tricky one...
    Until now, I see the formula you just stated here on Wiki:



    And sorry DaleSpam for not being clear regarding the time of the both objects starting.

    But I don't understand why the "2F" won't do any difference if they collide at the same speed.

    They got the same momentum during the collison, but the 2nd object is (2F) instead of just F (for object one). How will that effect the collision? For me, there must be some difference since the other object is accelerating faster. :/
    Last edited: Oct 8, 2011
  11. Oct 8, 2011 #10
    No. The key is that the normal formula you use to determine speed doesn't work for relativistic speeds. The normal formula being:
    [itex]v = at[/itex]
    Where v is velocity, a is acceleration, and t is time.

    This formula isn't totally accurate though. It turns out the error increases as you get closer to the speed of light, and so, it is fine for every day use. The more accurate formula is:
    [itex]v = \frac{at}{\sqrt{1 + (\frac{at}{c})^2}}[/itex]
    Where c is the speed of light.

    Note that this formula is the same as the other, but simply divided by a conversion factor:
    [itex]\sqrt{1 + (\frac{at}{c})^2}[/itex]
    Play around with this factor and it becomes clear that at low accelerations and times it is very close to 1, and thus close to no change to the basic formula.

    This site may be of interest:
    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html" [Broken]
    Last edited by a moderator: May 5, 2017
  12. Oct 8, 2011 #11
    Awh, can't believe I was thinking of mechanics and now I'm at relativity, not my day!
  13. Oct 8, 2011 #12


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    Just to add to DaleSwanson's post: the equation F=ma does not hold true at relativistic speeds.
  14. Oct 8, 2011 #13
    Haha, actually relativity isn't that hard if you know mechanics and calculus (although you could see a derivation of the Lorentz Transformations without it) and ~maybe~ some linear algebra (knowing how to multiply matrices is sufficient). Relativistic phenomena in Electromagnetics are also interesting, but calculus is required for E&M. If you are interested in relativistic E and M and know the background material (i.e. mechanics at a university level, and calculus--multi-variable stuff like divergence, curl, surface volume and line integration would be nice to know but you ~might~ be able to get by on just singe-variable), then you should consult Edward M. Purcell's book on E&M.

    Hope that helps.
  15. Oct 9, 2011 #14
    Interesting, I've tried to read some relativity on my own actually a while ago. It goes good as long as the math doesn't show up. And I have just started (2 months ago) studying physics at a university level. That's why I'm to mechanics at the moment. I must say it's very fun to learn. And a course in General Relativity will show for me in 3 years (I have just started my way to a PhD in physics).

    So I have a lot to learn :).

    Thanks though for the replies, I never thought a simple equation like F=ma could be so interesting.
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