Final velocity after repulsion.

In summary, the problem involves two magnetic masses, m1 and m2, repelling each other with a constant force f over a distance s. The final velocities of m1 and m2 can be predicted by taking into account their initial velocities and the direction of motion. However, there is a potential flaw in the equations used to solve the problem, as the distance traveled by each mass is not inversely proportional to its mass.
  • #1
dE_logics
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Homework Statement



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).

Homework Equations



We'll need 4 equations

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).
 
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  • #2
dE_logics said:
Consider 2 magnetic masses, ...
What's a "magnetic mass"?

dE_logics said:
these 2 repel each other ... with force f (relative to each other)...which applies from m1 to m2.
Is f constant? Are we ignoring Newton's third law here (i.e. m1>>m2)?

dE_logics said:
Distance traveled by each mass is also inversely proportional to its mass.
How did you come to that conclusion?
 
  • #3
turin said:
What's a "magnetic mass"?

They are repelling each other, that's the point.

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

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.

How did you come to that conclusion?

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]
 
  • #4
I appologize for not responding sooner.

dE_logics said:
turin said:
Are we ignoring Newton's third law here (i.e. m1>>m2)?
No the force that applies to both the bodies in the same but opposite in direction.
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:
dE_logics said:
Distance traveled by each mass will be different, but the summation will be equal to s.

Aside from that, there still may be a subtle problem:
dE_logics said:
[tex]F = m \frac{dv}{dt}[/tex]
[tex]F = m \frac{ds}{dt^2}[/tex]
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] )

dE_logics said:
[tex]ds = \frac{F*dt^2}{m}[/tex]
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.
 
  • #5
the F that you are using in the following equations is not the same as the f that is given in the problem

All the fs are the same following the 3rd law.

likewise the m (v) is neither m1 (v1) nor m2 (v2),

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.

the s that you are using in the following equations is not the same as the s that you defined in your original post:
You're right I realized that later, the s that I used in this equation is solely cause of F (and m), in the real question it should also be a function of u1 and u2.



And the ds and dt is an interval and not an infinitely small amount, doesn't have to do with calculus actually.


Anyway, problem's solved; I eliminated that equation from the picture.
 

1. What is "final velocity after repulsion?"

Final velocity after repulsion refers to the velocity of an object after it has been acted upon by a repulsive force. This force can cause the object to change direction, speed up, or slow down.

2. How is the final velocity after repulsion calculated?

The final velocity after repulsion can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to the repulsive force, and t is the time the force acts on the object.

3. Can the final velocity after repulsion ever be negative?

Yes, the final velocity after repulsion can be negative if the repulsive force slows down the object to a negative velocity. This means that the object is moving in the opposite direction of its initial velocity.

4. How does the mass of the object affect the final velocity after repulsion?

The mass of the object does not directly affect the final velocity after repulsion. However, it can affect the acceleration of the object due to the repulsive force. Objects with a larger mass will experience a smaller acceleration and therefore have a lower final velocity compared to objects with a smaller mass.

5. Can the final velocity after repulsion be greater than the initial velocity?

Yes, the final velocity after repulsion can be greater than the initial velocity if the repulsive force causes the object to speed up. This means that the object is moving faster in the direction of the repulsive force.

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