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Homework Help: Final velocity after repulsion.

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider 2 magnetic masses, m1 and m2; these 2 repel each other for a distance s with force f (relative to each other)...which applies from m1 to m2.

    Predict final velocity of m1 and m2.

    Assume final distance to be v1 and v2

    Assume the force that f acts is constants, i.e not a function of distance.

    Assume the 2 masses to have an initial velocity as per 2 cases -
    1)The 2 masses have initial velocity in the same direction
    2)or in a different direction.

    Take any values pointing to the direction of motion of m1 as positive; that is the final velocity of m1 will be positive and m2 will be negative.

    So the f by the above definition will be negative (when real world values are taken).

    2. Relevant equations

    We'll need 4 equations

    3. The attempt at a solution

    Well...all have failed.

    As said before we need 4 equations, make it by the following facts -

    2as = v^2 - u^2 (2 equations by this)

    Distance traveled by each mass will be different, but the summation will be equal to s.

    Distance traveled by each mass is also inversely proportional to its mass. I replaced this equation with law of conservation of energy's; but no go...wrong results (after checking out dynamics in spreadsheets).
    Last edited: Mar 12, 2009
  2. jcsd
  3. Mar 12, 2009 #2


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    What's a "magnetic mass"?

    Is f constant? Are we ignoring Newton's third law here (i.e. m1>>m2)?

    How did you come to that conclusion?
  4. Mar 12, 2009 #3
    They are repelling each other, that's the point.

    Yes f is a constant, not a function of distance (see the edit).

    No the force that applies to both the bodies in the same but opposite in direction.

    F = ma
    [tex]F = m \frac{dv}{dt}[/tex]

    [tex]F = m \frac{ds}{dt^2}[/tex]

    [tex]ds = \frac{F*dt^2}{m}[/tex]
  5. Mar 17, 2009 #4


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    I appologize for not responding sooner.

    Firstly, you need to understand that, EITHER ...
    ... the F that you are using in the following equations is not the same as the f that is given in the problem (as I understand the problem), and likewise the m (v) is neither m1 (v1) nor m2 (v2), OR ...
    ... the s that you are using in the following equations is not the same as the s that you defined in your original post:
    Aside from that, there still may be a subtle problem:
    I don't know if you are having a caculus problem in the second line here, but at least you are using nonstandard notation (maybe just a typo - god knows I make them all the time). Using a standard notation:

    [itex]F = m \frac{d^2s}{dt^2}[/itex] ( i.e. [itex]F = m \frac{d}{dt}\frac{ds}{dt}[/itex] )

    I know the physicists do this one a lot, but the math guys have told me that expressions like the this third line don't make any sense. Anyway, in order to save confusion, you should always do it this way:

    [itex]\frac{d^2s}{dt^2} = \frac{F}{m}[/itex]

    which you can integrate to

    [itex]\frac{ds}{dt} = \right.\frac{ds}{dt}\left|_{t=t_0}+\frac{F}{m}t[/itex]

    As you can see, this gives a speed ("velocity"), not a distance (separation). If you want distance, you must integrate again. At any rate, the distance is NOT inversely proportional to mass unless you are chosing special values of the initial separation, initial velocity and final time, for which I see no motivation.
  6. Mar 23, 2009 #5
    All the fs are the same following the 3rd law.

    That derivation was just a proof of the relation between s and m, it doesn't have to do directly with the problem; moreover it was just a suggestions by me, you can discard it always while doing the original question.

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