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Kinetic energy along with conservation of momentum

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    a hockey player of mass 80 kg was skating at a velocity of 7.5 m/s [E 20 S] when he accidently hit a linesman who was just standing on the ice. The mass of the linesmass is 90 kg and his velocity after the collision was 3 m/s [E 30 N] find.
    a) the velocity of the hockey player after the collision and
    b) the kinetic energy lost during the collision as a percentage.

    3. The attempt at a solution
    given:
    m1= 80 kg
    m2 = 90 kg
    v1 = 7.5 m/s [ E 20 S]
    v1` = ?
    v2= 0m/s
    v2` = 3m/s [ E 30 N]

    X component
    m1v1 = m1v1` + m2m2`
    (80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E])
    563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`)
    329.9887135 [E] / 80kg = v1`
    4.124858919 m/s [E] = v1`

    y components
    m1v1 = m1v1` + m2v2`
    (80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(1.5m/s[N])
    205.212086kgm/s-135m/s[N] = (80kg)(v1`)
    70.212086kgm/s/80kg = v1`
    4.252651075m/s = v1`

    4.124858919 m/s [E] + 4.252651075m/s = v1`
    5.924 m/s [S 44° E] = v1`

    b) Ektot = 1/2m1v1^2
    Ektot= 1/2(80kg)(7.5m/s)^2
    Ektot=2250 J

    Ektot` = 1/2m1v1`^2 + 1/2m2v2`
    Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
    Ektot` =1403.75104 J + 405 kg J
    Ektot` = 1808.75104 J

    Ektot - Wdef = Ektot`
    2250 J - Wdef = 1808.75104 J
    Wdef = 441.24896 J

    The Wdef is the energy lost during the collision

    (441.248896 J / 2250 J) x 100%
    19.6 %
     
  2. jcsd
  3. Nov 14, 2012 #2

    SammyS

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    This looks like a repeat of your earlier thread.

    https://www.physicsforums.com/showthread.php?t=651165&highlight=hockey+player

    Perhaps just a new attempt at a solution on your part.
     
  4. Nov 14, 2012 #3
  5. Nov 14, 2012 #4

    micromass

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    Please don't create multiple threads on the same topic.
     
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