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Deriving formula for final velocity in 2 repelling bodies

  1. Feb 28, 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 2 masses at rest initially.

    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

    2as = v^2 - u^2; this might be used...I used it



    3. The attempt at a solution

    Using standard formula 2as = v^2 – u^2

    Its to be noted that the formula is applicable only when the formula from where the normal reaction is derived is stationary with respect to the frame of the observer, so the distance df that the body and mass covers is partially due to the distance traveled by m1 and partially due to the the distance that the body travels (this will be observed by the observer).

    So the actual distance that the mass travels (while accelerating) is d1, and that for the body is d2 (new variables assumed, we do not know this value), however the sum of d1 and d2 is equal to the total distance traveled will be equal to the distance for which the force applies (s).

    Also the distance traveled by each the mass and the body is an inverse and direct function of mass.

    Here a1 (acceleration on m1) = -f/m1

    a2 (acceleration on m2) = f/m2

    By 2as = v^2 – u^2 -

    2*a2*d2 = v2^2............1

    2*a1*d1 = v1^2................2

    -d2 + d1 = s........3 (this is done cause the real value of d2 will be negative, following the coordinate system)

    d1/d2 = m2/m1............4 (as stated before, distance traveled is a direct and inverse function of mass).

    There are 4 equation and 4 unknown.

    Since the formula needs to be derived, they cannot be solved simultaneously.

    Making d2 the subject from equation 4 -

    (d1*m1)/m2 = d2

    Substituting in equation 3 -

    -((d1*m1)/m2) + d1 = s

    -d1((m1)/m2) - 1) = s

    -d1((m1-m2)/m2) = s

    d1 = -s*m2/(m1-m2)

    Substituting above value of d1 in equation 2 -

    2*a1*(-s*m2/(m1-m2)) = v1^2

    From equation 4 making d1 the subject -

    d1 = (d2*m2)/m1

    Substituting value of d1 in equation 3 -

    -d2 + ((d2*m2)/m1) = s

    -d2(1 - (m2)/m1) = s

    -d2((m1-m2)/m1) = s

    d2 = (-s*m1)/(m1-m2)

    Substituting value of d2 in equation 1

    2*a2*((-s*m1)/(m1-m2)) = v2^2


    Final equations -

    2*a2*((-s*m1)/(m1-m2)) = v2^2
    and
    2*a1*(-s*m2/(m1-m2)) = v1^2

    Putting real world values – f = -3000
    m1 = 610
    m2 = 10
    s = 2
    a2 = -300
    a1 = 300/61

    v2^2 = 1220.............ok
    v1^2 = -20/61..................the negativity needs to be removed

    Squaring v1^2 = -20/61 (is then step OK?)

    v1^4 = 400/3721..............ok
     
  2. jcsd
  3. Mar 1, 2009 #2
    I think this wont be called as a HW problem.
     
  4. Mar 3, 2009 #3
    Predict final velocity under repulsion.

    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 2 masses at rest initially.

    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).


    This is not a HW problem...officially too, cause I think the solution is pretty complex in relation to a HW (no one answered it anyway in that section, so I put it here).
     
  5. Mar 3, 2009 #4
    This formula wont work for both the mass being equal, m1-m2 will churn 0 in that case.
     
  6. Mar 3, 2009 #5
    Re: Predict final velocity under repulsion.

    Assume they are initially at rest.

    They repel each other for a distance s with force f = f(d), where d = s + a, a being the initial separation (perhaps zero).

    W = S f ds, and W = KE_f - KE_i = KE_f - 0 = (1/2) m v^2, where v is the final velocity.

    Note that m1 will have v > 0, and m2 will have v < 0, but |v| is the same for both, by symmetry.

    Does that make sense? Am I just totally wrong?
     
  7. Mar 3, 2009 #6
    Re: Predict final velocity under repulsion.

    Ok...assume force not to be a function of distance...just assume that to make things simple.

    Why has s been added to a?...I mean the distance traveled by both the bodies should be less than s.

    Can you pls define the variables? :smile:
     
  8. Mar 23, 2009 #7
    Problem solved.............
     
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