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Conservation of energy using orientation

  1. Nov 10, 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 accidentally hit a linesman who was just standing on the ice. The mass of the linesman is 90 kg and his velocity after the collision was 3 m/s [ N 30 E]. Find the velocity of the hockey player after the collision and the kinetic energy lost during the collision in percentage?


    2. Relevant equations
    I would be using the kinetic energy equation which is
    1/2mv^2=1/2mv1`+1/2mv2`


    3. The attempt at a solution
    The problem i have is that how would i get started i mean they are standing on the incline and i do not understand why the linesman's collision orientaion N 30 E. isn't that almost verticle?
     
  2. jcsd
  3. Nov 10, 2012 #2

    Doc Al

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    Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
     
  4. Nov 10, 2012 #3
    well the conservation of momentum is conserved...
     
  5. Nov 10, 2012 #4

    Doc Al

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    Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.
     
  6. Nov 10, 2012 #5
    so...

    m1v1 + m2v2 = m1v1` + m2v2`
    (80 kg)(7.5 m/s [E20S] = (80kg)(V1`) + (90 kg)(3 m/s[N30E])
    600 kgm/s [E20S] = (80kg)(V1`) + 270 kgm/s [N30E]
    600 kgm/s [E20S] - 270 kgm/s [N30E]
    563.8155 kgm/s[E]+205.212086 kgm/s+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1`)
    428.8155 kgm/s[E] + 439.038086 kgm/s = (80kg)(V1`)
    613.44 kgm/s [S44.3E] = (80kg)(V1`)
    7.668 m/s [S44.3E]

    But why can't i just simple use this equation
    1/2m1v1^2 = 1/2m1v1`^2 + 1/2m2v2`^2
    This way i don't have to use the orientation since its scaler
     
  7. Nov 10, 2012 #6

    Doc Al

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    You can't use it because it's not true. In general, kinetic energy is not conserved in a collision. (When it is conserved, it's called a perfectly elastic collision.) But momentum is always conserved.
     
  8. Nov 10, 2012 #7
    Im confused..... so for the kinetic energy it is the work being done...
    after the collision wouldn't the work be transferred into some other energy???
     
  9. Nov 10, 2012 #8

    Doc Al

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    The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.
     
  10. Nov 10, 2012 #9
    yes.... so
    why can't i use the equation for kinetic energy
    is it because some of the energy has been used for other energies???
     
  11. Nov 10, 2012 #10

    Doc Al

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    Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)
    That's right.
     
  12. Nov 10, 2012 #11
    oh ok ... thanks :D
     
  13. Nov 10, 2012 #12
    Umm how would i find the kinetic lost during the collision in percentage?
     
  14. Nov 10, 2012 #13

    Doc Al

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    Calculate the total kinetic energy before and after the collision. Subtract to find the change.
     
  15. Nov 10, 2012 #14
    so...

    Ek1 = 1/2m1v1
    Ek2 = 1/2m1v1`^2 + 1/2m2v2`^2

    Ek1= 1/2(80kg)(7.5m/s)^2
    Ek1= 2250 J

    Ek2= 1/2m1v1`^2 + 1/2m2v2`^2
    Ek2= 1/2(80kg)(7.668 m/s)^2 + 1/2(90kg)(3 m/s)^2
    EK2= 2351.92896 J + 405 J
    Ek2= 2756.92896 J

    Ek2 - Ek1
    506.929 J

    But this does not seem right..
    and how would i get this into percentage?
     
  16. Nov 11, 2012 #15

    Doc Al

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    I suspect that your calculation of the velocity is wrong. Redo your conservation of momentum work. I suggest that you write separate equations for North and East components of momentum.
     
  17. Nov 11, 2012 #16
    I am still getting the same answer...... i mean what do you mean by separate components i do break them apart.. as i did in the calculation above .....
     
  18. Nov 11, 2012 #17

    Doc Al

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    I did not quite follow the calculation you did in post #5.

    Here's what I would do. Map things onto an x-y axis, where east is in the positive x direction and north is in the positive y direction. Then set up two momentum conservation equations, one for the x components and one for the y components. Then you can solve for the components of the velocity of the first hockey player.
     
  19. Nov 11, 2012 #18
    But why can't i do it like i did in post 5? I mean thats how we have been thought in class to break the components in the equation itself, i have never seen of breaking them apart and then using two different equations,,,,,
     
  20. Nov 11, 2012 #19

    Doc Al

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    I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.
     
  21. Nov 11, 2012 #20
    m1v1 + m2v2 = m1v1` + m2v2`
    (80 kg)(7.5 m/s [E20S] = (80kg)(V1`) + (90 kg)(3 m/s[N30E])
    600 kgm/s [E20S] = (80kg)(V1`) + 270 kgm/s [N30E]
    600 kgm/s [E20S] - 270 kgm/s [N30E]
    563.8155 kgm/s[E]+205.212086 kgm/s+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1`)
    428.8155 kgm/s[E] + 439.038086 kgm/s = (80kg)(V1`)
    613.44 kgm/s [S44.3E] = (80kg)(V1`)
    7.668 m/s [S44.3E]

    so what i did was i multiplied the mass of player which is 80 kg by the initial velocity before the collision. The linesman is not moving so the velocity is 0m/s. And after the collision for the player we are trying to find the final velocity and so for the linesman we know the final velocity and the mass so i multiplied them.
    I took the linesman's momentum after collision and took it to the other side where it becomes negative. I broke the north and east compoenets
    Then i broke the kgm/s before and after into their separate components
     
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