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Elastic collision in 2 dimensions

  1. Feb 23, 2012 #1
    [SOLVED] Elastic collision in 2 dimensions

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

    A ball with mass m and initial velocity v1i collides elastically with a second ball of the same mass that is initially at rest. After the collision, the first ball moves away at an angle of θ1 with respect to the initial velocity. What is the kinetic energy of the first ball after the collision, with respect to the variables m, v1i, and θ1?


    2. Relevant equations

    v1i - v1f*cos(θ1) = v2f*cos(θ2)
    v1f*sin(θ1) = v2f*sin(θ2)
    v1i2 = v1f2 + v2f2
    KE1f = 0.5*m*v1f2

    3. The attempt at a solution

    I know I have enough information to solve for the final kinetic energy of the first ball, but I'm stumped on how to isolate the unknown variables from the system of equations.
    So far I have:
    v1i2 = v2f2*((sin22) / sin21)) + 1)
    tan(θ2) = v1f*sin(θ1) / (v1i - v1f*cos(θ1))
    But I don't know where to go from here.

    EDIT: I squared the first and second equations and added them together, so now I have
    v1i2 - 2*v1i*v1f*cos(θ1) + v1f2 = v2f2

    I substituted using the kinetic energy balance equation, did the algebra, and came up with
    v1i2 - 2*v1i*v1f*cos(θ1) + v1f2 = v1i2 - v1f2

    Simplifying this gave:
    v1f = v1i*cos(θ1)

    Therefore the kinetic energy of ball 1 after the collision is
    KE1f = 0.5*m*(v1i*cos(θ1))2
     
    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 23, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Necropolitan! Welcome to PF! :wink:
    Yup, that should do it! :smile:
     
  4. Feb 23, 2012 #3
    Thanks tiny-tim!
     
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