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Elastic collision, conceptual

  1. Mar 29, 2007 #1
    Let two particles collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary. Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of particles 1 and 2?

    Express answer in terms of the initial velocity, v.

    My conservation of energy equation for the two particles is (I already cancelled out the masses):

    v1^1 + 0.5v2^2 = v^2

    My conservation of momentum equation is (masses already cancelled):

    2v1 + v2 = 2v


    v2 = 2v-2v1

    v1 = (2v-v2)/2

    I tried plugging these values for v1 and v2 into the conservation of energy formula to try and isolate v for each variable. For v1, I ran into difficulty because of the middle term of the binomial equation, -8vv1.

    v1^1 + 0.5v2^2 = v^2
    v1^2 + 0.5[(2v-2v1)(2v-2v1)] = v^2
    v1^2 + 0.5[4v^2 - 8vv1 + 4v1^2] = v^2
    -v^2 = v1^2 - 4vv1 + 2v1^2
    3v1^2 - 4vv1 = -v^2

    From here I spent like 10 lines trying to rearrange terms to separate v from that middle term (now -4vv1), but am unable to do so. Any ideas? Thanks!
  2. jcsd
  3. Mar 29, 2007 #2

    Doc Al

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    Staff: Mentor

    Recognize that you have a quadratic equation for v1. Solve it!
  4. Mar 29, 2007 #3
    Well, there are two different terms in the quadratic. How can you solve for two terms?
  5. Mar 29, 2007 #4

    Doc Al

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    Staff: Mentor

    Actually, there are three terms if you put it into standard form:
    [tex]ax^2 + bx + c = 0[/tex]

    Can you solve that for x? If so, you can solve for v1 in terms of v.
  6. Mar 29, 2007 #5
    sure go to a web site where the arithmetic is done for you.:cool:

    seriously, yopu should be able to derive these eqns, but it outlines good method and has good learning sim.
  7. Mar 29, 2007 #6
    You lost me. I assume you mean:

    3v1^2 - 4vv1 + v^2 = 0

    [-b +/- sqrt(b^2 - 4ac)]/2a = v1

    [4v +/- sqrt([-4v]^2 - 4(3v)(v^2))]/2(3v)

    [4v +/- sqrt(16v^2 - 12v^3)]/6v

    But now I'm left with

    (2/3)v +/- [sqrt(16v^2 - 12v^3)/6v]

    I can't combine terms in the square root because they have different exponents, so this doesn't seem like it makes much sense.
  8. Mar 29, 2007 #7


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    Homework Helper

    Those "v"s in red shouldn't be there. There is no v in the coefficient of your "a" term. Fix that up and it will be much easier to work with.
    By the way, you don't necessarily have to use the quadratic equation. You could just factor it.
  9. Mar 29, 2007 #8


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    Are you certain that all the information is given? Because the way I see it, you have three variables (v, v1, v2) and only two equations (conservation of momentum and conservation of KE).
  10. Mar 29, 2007 #9
    two unknowns right? since the question asks for Vf's in terms of Vi.
  11. Mar 29, 2007 #10
    all the info is given. i derived v1 and v2 from conservation of momentum. i have to do each individually, so really there are only two unknowns with two equations.

    i see what you mean, hage, about the extra v's i put in. ill try that.
  12. Mar 29, 2007 #11
    that works. i got (1/3)v = v1. Thanks everyone for all of your help
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