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Elastic collision of hockey puck

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially traveling at 15.2 m/s and is deflected 30.0^\circ from its initial direction. The pucks are made of superball-like material, so you may assume that the collision is perfectly elastic.

    Find the final speed of the puck A after the collision


    2. Relevant equations

    conservation of momentum for one object at rest (1)
    mvai = mvaf + mvbf

    conservation of kinetic energy (2)
    1/2mvai = 1/2mvaf + 1/2mvbf

    3. The attempt at a solution

    I have tried this about 10 times and cannot get it right :(

    i solved equation 1 for the x and y components of the velocity final for puck b

    vbxf = vaxi - vaxf

    vbyf = -vayf

    so

    vbf = sqrt [(vaxi - vaxf)2 + (-vayf)2]

    vbf = sqrt [vaxi2 - 2vaxf + vaxf2 + vayf2]

    then i plugged this into 1/2mvai = 1/2mvaf + 1/2mvbf for v_bf

    i would show all my steps of reducing but it would take forever.

    i end up getting vfa = cos(30) = .866 but this is wrong

    am i right up to this point of what i have done?

    any help would be great. Thank you
     
  2. jcsd
  3. Mar 31, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    The directional components of momentum before and after are the same. So:

    [tex]mv_{ai} = mv_{af}cos(\alpha) + mv_{bf}cos(\theta)[/tex] and

    [tex]mv_{af}sin(\alpha) + mv_{bf}sin(\theta) = 0[/tex]

    Now the trick is to realize that the conservation of energy equation gives you the angle between the two pucks after collision:

    [tex]v_{ai}^2 = v_{af}^2 + v_{bf}^2[/tex]

    so with that angle you can find [itex]\alpha[/itex]

    AM
     
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