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General Question- Compuing velocity of a center of mass

  1. Nov 30, 2006 #1
    General Question-- Compuing velocity of a center of mass

    1. The problem statement, all variables and given/known data

    A ball of mass 0.198 kg has a velocity of 1.47 i m/s; a ball of mass 0.309 kg has a velocity of -0.401 i m/s.They meet in a head-on elastic collision.

    (a) Find their velocities after the collision.

    (b) Find the velocity of their center of mass before and after the collision.


    2. Relevant equations

    v1f = ((m1-m2)/(m1+m2))(v1i)+(2(m2)/(m1+m2))(v2i)

    (m1)(v1i)+(m2)(v2i)=(m1)(v1f)+(m2)(v2f)

    3. The attempt at a solution

    a.) Calculated velocities correctly: v1f = -.8106m/s, v2f = 1.0604m/s

    b.) This is my problem

    As i understand center of mass, it applies to an object as a whole, not a system. I have only been taught how to compute center of mass for an object with specific x and y dimensions, i have no idea how to find the center of mass of a system let alone the VELOCITY of the center of mass as is asked in this question.

    my only guess is that since p = m(delta v) and p = m(v center of mass), that maybe the velocity of the center of mass is equal to the change in velocity?
     
  2. jcsd
  3. Nov 30, 2006 #2
    Think about it this way..
    Center of mass means you're treating the TOTAL momentum of the system as if it's the momentum of ONE object which composes of both the objects' masses! (Since the total mass of the objects would be concentrated at the center of mass)
    I think I said too much..
     
  4. Nov 30, 2006 #3
    im still not quite sure how you would obtain the velocity of the center of mass from this. apparently its not as simple as i'd thought...
     
  5. Nov 30, 2006 #4

    OlderDan

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    What is the definition of center of mass? If you know how to find it in some cases, you must know its definition.

     
  6. Nov 30, 2006 #5
    What's the momentum of a single particle whose mass = (m1 + m2), and which moves with speed Vcm? Does this have any relationship to your system, in other words?

    Dorothy
     
  7. Dec 1, 2006 #6

    OlderDan

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    While this is certainly a valid observation, it is sort of using the answer to find the answer. At least once in a lifetime every physics student should have to actually compute the velocity of the CM from the individual velocities of the particles in the system to verify the conclusion to be reached in this exercise.
     
  8. Dec 1, 2006 #7
    Oh. Sorry :-(

    I'm not sure I have ever done that... When I solved this problem, I just took the derivative of the position vector of the CM, and that's what I ended up with, which I thought was interesting... Maybe I should have just suggested that..

    Sorry again.
    Dorothy
     
  9. Dec 1, 2006 #8

    OlderDan

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    No need to be sorry. Your observations are valid.

    The derivative is the approach to doing the direct calculation, and it is what I think he problem intended. I thought you were suggesting that since the total mometum has been calculated one only need to divide by the total mass, which is the conclusion the problem is trying to reach. It is a subtle difference.
     
  10. Dec 2, 2006 #9
    Well, sometimes when I work through these calculations, I end up with a result that I say, "Oh, of course. I should have seen that..." a sort of an "Ah hah" moment, and I think I should have been able to see it without doing the calculation, just from the concepts.

    I was trying to create an "Ah Hah!" moment for someone else, having missed it myself :-)

    Thanks for all the help you give all of us, btw.

    Dorothy
     
  11. Dec 2, 2006 #10

    OlderDan

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    And that is exactly the point of doing the exercise!! :smile: Those moments are memorable and help to drive home the concepts.
     
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