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Homework Help: Find the velocity of the slower object after the collision

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A(n) 12 g object moving to the right at
    31 cm/s overtakes and collides elastically with
    a 29 g object moving in the same direction at
    15 cm/s.

    Find the velocity of the slower object after
    the collision.

    Answer in units of cm/s.


    2. Relevant equations



    3. The attempt at a solution

    I know it probably looks very confusing. I am not the best at keeping track of the work I am doing. Some steps may have changed or are irrelevant to the answer:

    m1*v1+m2*v2=m1*v1f+m2*v2f

    v2f=[(m1*v1+m2*v2)-m1*v1f]/m2


    .5(m1)(v1)^2 + .5(m2)(v2)^2 = .5(m1)v1f^2 + .5(m2)v2f^2

    -----------------------------------------------------------------

    v2f=(807-12v1f)/29


    18057= 12v1f^2 + 29v2f^2

    18057= 12v1f^2 + 29[(807-12v1f)/29]^2

    [(807-12v1f)/29][(807-12v1f)/29]

    (807-12v1f)(807-12v1f)

    651249 - 2(9684v1f) + 144v1f^2

    c.................b.................a

    -19368 (+-) [19368^2 - 4(144)(651249)]^.5
    ______________________________________
    2(144)


    ------------------------------------------------------------------------------

    v1f = 67.25

    v2f=(807-12v1f)/29

    v2f=(807-12[67.25])/29

    18057= 54270.75 + 29v2f^2

    -36213.75 = 29v2f^2
     
  2. jcsd
  3. Mar 28, 2010 #2
    Re: Impulse/Momentum/Aggravating

    Ok so your last correct step was this.

    18057= 12v1f^2 + 29[(807-12v1f)/29]^2

    And then jumped into the quadratic equation.

    Yeah, you can't do that in this case. You should expand the 29[(807-12v1f)/29]^2 term which yields

    (1/29)[(144 x^2)-(19368 x)+651249]

    Substitute this back to your previous equation.

    18057= 12v1f^2 + (1/29)[(144 x^2)-(19368 x)+651249]

    THEN you can move everything over to right, simplify, and use the quadratic equation.

    Best of luck, quite annoying numbers there.
     
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