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Finding Velocity of Particles in 'Free Space'?

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Two identical particles, each of mass 1300 kg, are coasting in free space along the same path. At one instant their separation is 15.0 m and each has precisely the same velocity of 900 m/s. What are their velocities when they are 2.00 m apart?

    m_1 = m_2 = 1300kg
    at 15m: v_1 = v_2 = 900m/s
    at 2m: ???


    2. Relevant equations

    Not sure, possibly conservation of energy?
    1/2mv^2 + mgh=0

    3. The attempt at a solution
    I thought I could use conservation of energy, but there is no gravity so there is no potential energy. I can plug in for m and v, but that obviously yields nothing?
    Solution is in ___ +/- ____ form, so I am guessing there will be some squaring and square rooting? I would have thought that in "free space" there would be no friction and the particles would keep going the same speed, but the distance between them changes, thus one must be going faster than the other. They both could be speeding up or slowing down (at different rates). I don't see how I could use kinematics? Maybe some relationship exists that should help me solve for their accelerations, then I could plug back in and find velocity or something? That's all I could think of...still not sure what to do? haha
     
  2. jcsd
  3. Nov 30, 2011 #2

    gneill

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    The particles have mass, so there is gravity; They will attract each other.

    The formula mgh for gravitational potential energy holds for masses moving in Earth's gravitational field, and only near the surface of the Earth. There's another formula based on Newton's law of gravity that should be applied for the general case.
     
  4. Nov 30, 2011 #3
    Ok, so I can use:

    E=1/2mv^2 - GMm/r
    where G=universal gravity constant
    M = m_1+m_2
    r=distance between them

    I solve for E, then plug back in and solve for v?
     
  5. Nov 30, 2011 #4

    gneill

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    Sort of. You want to find the change in PE that occurs for the starting and ending conditions (15m separation ---> 2.00m separation). This ΔPE ends up distributed to the two masses as changes in KE. You might consider doing the calculation in the center-of-mass frame of reference (where the objects are initially at rest), then converting the result to the original "stationary" frame.
     
  6. Nov 30, 2011 #5
    Yes, my described method will only result in one velocity and I need two...

    Center of mass = MƩmr, correct? A bit rusty...how do I "connect" what I have with C-of-M?
     
  7. Nov 30, 2011 #6

    gneill

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    Don't sweat the center of mass calc! They are two identical particles separated by 15m and moving at the same speed along the same path. All you need to do in order to be in the center of mass frame is subtract their common velocity -- that is, they are stationary at 15m separation to begin with. Halfway between them is the center (although you don't even need to use this fact).
     
  8. Nov 30, 2011 #7
    Alright, so that will yield a positive answer because m_1 is behind CM and m_2 is ahead? (Or whichever is going faster, haven't calculated it out yet).

    So, I am a bit pedantic so I will just restate what I think I am suppose to do...

    m_1(900m/s) - m_2(900m/s) = 0 because the masses are equal.
    we should then have a new equation when the velocities change?
    m(v_1 - v_2) = CM
    "This ΔPE ends up distributed to the two masses as changes in KE." Not entirely sure what is meant by this statement?
    ΔPE = (2Gm^2/2)_f - (2Gm^2/15)_i
    Sorry, am I still in the right direction or am I still missing the whole point?
     
  9. Nov 30, 2011 #8

    gneill

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    Okay, I'm not sure what you are trying to do there, or what it will be useful for. Perhaps you want to keep track of the momentum of the system to make sure it remains the same before and after the "operation"?
    In the center of mass frame the two identical particles start out stationary and separated by 15.0m. They will accelerate towards each other due to gravitational attraction. Since the particles are identical they will have identical accelerations towards the center of mass. So their velocities will also be identical with respect to the center of mass. That means that their changes in KE will be identical too.

    The energy for this change in KE comes from the gravitational PE. So by symmetry we must conclude that this change in PE will be divided equally between the two particles as changes in their KEs with respect to the center of mass.
    You're getting there! Question: why the 2's in the ΔPE formula? When you 'assemble' a gravitating system consisting of two bodies, the first one is "free", and the second one has magnitude [itex] \frac{G m1 m2}{r} [/itex].
     
  10. Nov 30, 2011 #9
    PE = GMm/r and M = 2m, so by substitution I came to
    2Gm^2/r

    Alright, I think we are getting somewhere, hopefully? :)
    Ok, so since they both have the same mass they will gravitate towards each other at the same rate, so we can conclude that they will both have the same velocity and will be "symmetrical" about the center of mass? So can't we just equate the two energy equations?
    1/2(1300kg)(900m/s)^2 - (6.67e-11N*m^2/kg^2)(13000kg)^2/15m = 1/2(13000kg)v^2 - (6.67e-11N*m^2/kg^2)(13000kg)^2/2m
    Then we solve for v? (This was using your Gm1m2/r equation...not sure I am familiar with that?)
     
  11. Nov 30, 2011 #10

    gneill

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    [tex] PE = \frac{G M m}{r} = \frac{G m^2}{r} ~~~~~\text{when M = m} [/tex]
    This would mix together the initial kinetic energy of the system (which is frame dependent) with the internal changes that are taking place within the system. What velocity would v represent? The velocity of particle 1? Particle 2? Your equation looks as though it's trying to account for some external energy source affecting the velocity of the center of mass. Remember that the velocity of the center of mass is not going to change.

    I think working strictly in the center of mass frame to find the velocity changes for each particle within that frame should be clearer and more straightforward.

    Can you calculate the system's change in PE that occurs between the starting separation and ending separation?
     
  12. Nov 30, 2011 #11
    Using the formula you gave, the change in PE would be:

    (Gm^2)/2 - (Gm^2)/15 Right?
     
  13. Nov 30, 2011 #12

    gneill

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    Right. So now take half of this value and "give" it to a mass m in the form of KE. What velocity will the particle have?
     
  14. Dec 1, 2011 #13
    Ok, calculating ΔPE:
    [(6.67e-11)(13000)^2]/2 - [(6.67e-11)(13000)^2]/15
    = 0.004884, half of that:
    = 0.002442

    Not sure what you mean by "give", however the change in PE should be proportional to to the KE, so we equate the two?

    0.002442 = (1/2)(13000)v^2
    solving we get:
    +/-11.26

    Though I feel like I just jumped something? *_*
     
  15. Dec 1, 2011 #14

    gneill

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    Good.
    Gravitational potential energy is being exchanged for kinetic energy. So the change in PE is "given" (mathematically) to the masses.
    That value looks a bit large. Calculator mishap? By the way, are the masses 13000 kg or 1300 kg? I've seen both used, and the original statement had 1300.
    Once you settle the issue with the mass value and the correctness of that last calculation, all that remains is to jump back to the original frame of reference. That is, add back the original 900m/s to both particles.
     
  16. Dec 1, 2011 #15
    Haha, oooops! I thought it was 13000, but it was suppose to be 1300. Thanks so much for your help and patience, I should be able to take it from here? :)
     
  17. Dec 1, 2011 #16
    Recalculated using the Correct mass value and got
    0.001938
    So, then the velocities should be:
    900 +/- 0.001938, correct?
     
  18. Dec 1, 2011 #17
    Ah, forgot to go back and change in PE, should be
    0.000048846/2 = 0.000024423
    Substituting accordingly, I got:
    900 +/- 1.938e-4
    I think that should be correct now.
     
  19. Dec 1, 2011 #18
    Oh, and thanks again! :)
     
  20. Dec 1, 2011 #19

    gneill

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    Yes, that looks good!
     
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