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Conservation of momentum after a collision

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball of mass 3m collides head-on with a initially stationary ball of mass m. No kinetic energy is transformed into heat, or sound. In what direction is the mass-3m ball moving after the collision, and how fast is it going compared to its original velocity?


    2. Relevant equations

    p=mv

    3. The attempt at a solution

    I initially set it up as p1i +p2i = p1f +p2f

    p2i being zero it was discarded from the equation

    so p1i = p1f + p2f

    3mv = 3mv1f + mv2f since m is a unknown constant I can remove it from the equation

    3v = 3v1f + v2f

    unfortunately, this is where I am stuck :(

    or should I have calculated this via Kinetic energy?
     
  2. jcsd
  3. Oct 27, 2011 #2

    hage567

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    You can use conservation of energy to get another equation that will be useful.
     
  4. Oct 27, 2011 #3
    so 3m(v1i)^2 + m(0) = 3m(v1f)^2 + m(v2f)^2

    same thing I can remove the m variable

    to 3(v1i)^2 = 3(v1f)^2 + (v2f)^2
     
  5. Oct 27, 2011 #4
    You need two equations.
    Use conservation of momentum to set one eq. And conservation of energy to set the other.
    The speed of the ball with mass 'm' will be the same, they can be used to equate the two equations together for you to solve for speed and direction of the ball with mass '3m'.

    I believe you'll have 'm' canceling out somewhere as well.
     
  6. Oct 28, 2011 #5
    That is true the "m" constant is the same in a conservation of energy, and momentum equation. However, we don't know any of the velocity values. So I'm stuck
     
  7. Oct 28, 2011 #6

    hage567

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    You don't need to know the actual values. You have two equations that are related by the same variables, you can combine them and come up with an expression for the final velocity of the 3m mass.
     
  8. Oct 28, 2011 #7
    Yep, I don't think I explained it quite clearly above. The final speed of the ball with mass 'm' is the same in the momentum and energy conservation equations. That is the variable you set equal to each equation to then solve for the only other remaining velocity.
     
  9. Oct 28, 2011 #8
    I have a bunch of unknown velocities, and I don't see any way of changing that
     
  10. Oct 28, 2011 #9

    hage567

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    It seems that you are hung up on a "have no numbers" thing and it is preventing you from trying the question. No numbers are required.

    These are your equations from the posts you wrote:

    3v = 3v1f + v2f
    3(v1i)^2 = 3(v1f)^2 + (v2f)^2

    One thing I notice is that you weren't consistent in labeling your variables, the v and v1i are both representing the initial velocity of the 3m mass. Maybe that is confusing you?

    So your equations should look like

    3v1i = 3v1f + v2f
    3(v1i)^2 = 3(v1f)^2 + (v2f)^2

    In the end, you want to get v1f in terms of v1i. So you can rearrange your first equation for v2f and then substitute it into your second equation. This new equation can be solved for v1f. Give this a try and show how far you get.
     
  11. Oct 28, 2011 #10
    Originally I just used "v" and in the second case when I balanced my energy equation I needed two initial starting points.

    I'm stuck on the what to do next phase not the lack of numbers.

    making the momentum formula for the smaller ball's velocity gives: 3v1i -3v1f = v2f

    and when I plug that into the energy formula it gets messy.

    3(v1i)^2 = 3(v1f)^2 + (3v1i -3v1f)^2 from this point I could try trying to get rid of the squaring, or factor it all out. Both expressions lead to jibberish.

    The crux of the problem is that I don't know how much energy is transferred in the collision. I just don't see it.
     
    Last edited: Oct 28, 2011
  12. Oct 28, 2011 #11
    All that you need to know is its speed compared to the original.

    So you need vf om terms of vi.


    The substitution you used works just find, you just need to solve for vf now.
     
  13. Oct 28, 2011 #12
    yes, that is true but I have so far been unable to even remotely to figure what that ratio is.

    And trying unsuccessfully to plug the conservation of momentum into the conservation of energy formula in theory would give me the answer. However getting 0.573333, and 1 are not correct.

    Logically, 1 doesn't even make sense as zero energy would have been transferred.
     
  14. Oct 28, 2011 #13
    3(v1i)^2 = 3(v1f)^2 + (3v1i -3v1f)^2

    You found this equation, did you try to expand out that binomial and then isolate vf?
     
  15. Oct 28, 2011 #14
    Yes, and that lead to a complete mess, and terms that wouldn't factor out.
     
  16. Oct 28, 2011 #15

    hage567

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    You need to show your work so we can see where you could be making errors. Expanding and grouping terms in that equation should not lead to a complete mess. Don't worry about isolating v1f, you can't do it directly anyway*. Just show what you get when you expand and simplify that expression.


    *This is because if you've done it correctly, you should have a quadratic equation (it might be difficult to recognize at first glance, but that's what it is).
     
  17. Oct 28, 2011 #16
    3(v1i)^2 = 3(v1f)^2 + (3v1i -3v1f)(3v1i-3v1f)
    3(v1i)^2 = 3(v1f)^2 + 9v1i^2 -9v1iv1f-9v1iv1f +9v1f^2
    6v1i^2 -18v1iv1f +12v1f^2
    v1i^2 -3v1iv1f +2v1f^2
    (v1i-2v1f)(v1i-v1f)
     
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