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Homework Help: Conservation of momentum and hockey puck

  1. Sep 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A hockey puck B rests on a smooth ice surface and is struck by a second puck A, which was originally travelling at 40.0m/s east and is deflected 30 degrees from its original direction. Puck B acquires a velocity of -45 degrees with the original velocity of A.
    THe pucks have the same mass.
    a.) Compute the speed of each puck after the collision.
    b.) What fraction of the original kinetic energy of puck A is dissipated during the collision?

    2. Relevant equations
    [tex]m_{A}v_{A1x} + m_{B}v_{B1x} = m_{A}v_{A2x} + m_{B}v_{B2x}[/tex]
    [tex]m_{A}v_{A1y} + m_{B}v_{B1y} = m_{A}v_{A2y} + m_{B}v_{B2y}[/tex]

    3. The attempt at a solution
    m_{A} = m_{B}

    For the x- component
    m_{A}(40.0m/s) = m_{A}(V_{A} cos(30) + m_{A} V_{B} cos(315)
    40.0m/s = V_{A} cos(30) + V_{B} cos(315) (Eq 1);

    For the y- component
    0 = m_{A}(V_{A}sin(30) + V_{B}sin(315))
    (V_{A}sin(30)) = (V_{B}sin(315))
    V_{A} = (V_{B}sin(315)) / [sin(30)] (Eq 2);

    substituting the x- component:
    40.0m/s = (V_{B}sin(315)) / [sin(30)] + V_{B} cos(315)
    V_{B} = 18.86 m/s

    V_{A} = 26.667 m/s
    Yep i got this problem right.... same as back of book

    b.) HEres my problem:
    What fraction of the original kinetic energy of puck A is dissipated during the collision?

    what does it mean when the kinetic energy is dissipated?
    K1 = 1/2(mA)(V_{A})^2 = 1/2 ( 40m/s)^2

    K2 = 1/2(mA)(V_{A})^2 + 1/2(mA)(V_{B})^2 = 1/2(mA)(18.86)^2 + 1/2(mB)(26.667)^2

    the answer is 0.196

    thank you very much
  2. jcsd
  3. Sep 18, 2007 #2


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    Science Advisor

    The total kinetic energy, before the two pucks collide is just the kinetic energy of the first puck. That is your K1. The total kinetic energy after the collision is the kinetic energy of two pucks added together. That is your K2. K1- K2 is the kinetic energy that is lost or "disappated". If that is not 0, the collision is not "perfectly elastic" and some energy goes into heat.
  4. Sep 18, 2007 #3
    K1 = 1/2mA(0.40m/s)^2 = 800mA
    K2 - K1 = 1/2(mA){(18.86)^2 + (26.667)^2} = 533.414 mA

    i divide K2- K1 / K1 and cancel mA
    i get 0.6667675 = 66.7 %

    the answer at back of book : 0.196 = 19.6 %
    i got it now . my answer to the velocities were all wrong:
    so that is 20.706 and (29.27)
    so that gives me K2 - K1 / K1 = -0.196
    Last edited: Sep 18, 2007
  5. Sep 18, 2007 #4


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    Homework Helper

    This isn't K2 - K1. This is K2.
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