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Collision with ice hockey puck

  1. Aug 24, 2008 #1
    1. The problem statement, all variables and given/known data
    ice hockey puck moving at speed V1 collides head on with a second identical puck moving towards it at speed V2. After the collision, the first puck slows down to speed v1 without changing direction.
    a)After the collision what is the speed v2 of the second puck?
    b) calculate the speed v2 of the second puck when the first puck had an initial speed of 18ms-1 that was changed to 2.0ms-1 by the collision, and the initial speed of the second puck was 12,0ms-1. both pucks have a mass of 0.16kg

    3. The attempt at a solution

    well the questions shows that the initial puck continues on its path at a lower speed to i guess this was an elastic collision

    not sure if the math is right tho

    m1V1+m2V2 = m1v1+m2v2
    v2 = m1V1+m2V2/m1v1 + m2

    when i factor in the given speeds and masses in the next part of the question i am not sure if it is right.

    v2 = (0.16kg)(18.oms-1)+(0.16kg)(12.0ms-1)/(0.16kg)(2.0ms-1)+0.16kg
    v2 = 15.16ms-1

    is this right???
     
  2. jcsd
  3. Aug 24, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi lilmissbossy! I forgot to say earlier … Welcome to PF! :smile:
    Nooo … v2 = (m1V1+m2V2 - m1v1)/m2.

    Try again! :smile:
     
  4. Aug 24, 2008 #3

    LowlyPion

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    The hockey pucks are identical. You can drop the mass out the equations. Might make it simpler.
     
  5. Aug 24, 2008 #4
    Re: Welcome to PF!


    Oh ok i see now i have made a mistake when rearranging the equation...
    thats great i will factor in the values and it should make more sense:smile:
     
  6. Apr 9, 2009 #5
    so the answer is 28 m/s that doesnt make much sence?
     
  7. Apr 25, 2009 #6
    Hey guys, had the same problem.
    Remember V is a vector quantity so if the pucks are heading toward each other one will be a negative quantity.

    eg: (.16x18)+(.16x-12)=(.16x2)+(.16xV2)

    so the answer is 4 not 28

    Momentum Conservation
    .96=.96
     
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