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I'm stcuk, sort of, not really, I need some resurance is all

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Ive got another question that I think I might have failed at.

    A dog weighing 250 kg jumps into an empty boat by the dock at a speed of 5m/s causing the boat to move away form the shore at 1.75m/s. Find the amount of kinetic energy dissipated.

    2. Relevant equations

    KE=1/2*m*v^2

    3. The attempt at a solution

    Before

    KE=1/2*m*v^2
    KE=(.5)(250kg)(5.0m/s)^2
    KE(1)=3125

    and after

    KE=1/2*m*v^2
    KE=(.5)(250)(1.75m/s)^2
    KE(2)=382.813

    KE1-KE2=2742.19

    KE dissipated = 2742

    So am I right?
     
  2. jcsd
  3. Feb 15, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi cooper43! Welcome to PF! :smile:

    You need to include the KE of the boat, so put the mass of the boat = M, and use conservation of momentum. :wink:
     
  4. Feb 15, 2009 #3
    Is conservation of momentum P=mv?

    And hat do you mean Mass of the boat. Should I leave it as M or do I replace it with the mass of the dog
     
  5. Feb 16, 2009 #4

    tiny-tim

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    Nooo … conservation of momentum is mvi + MVi = mvf + MVf

    ie total momentum before = total momentum after …

    it is always true for a collision. :smile:
    Now you're just being ridiculous.

    Stop asking questions and start answering the problem.
     
  6. Feb 16, 2009 #5
    LMAO! Well said!

    As Tiny-Tim said, Final Momentum = Initial Momentum

    Therefore,

    mv = mv and you can work out your mass from there if necessary...
     
  7. Feb 16, 2009 #6
    Okay, so i did the m1v1i + M2V2i = mvf + MVf
    and M2=464.286

    The KE of the dog is 3125

    but im not sure how to gt the other KE

    should it be the dog+boat and the final velocity?
     
  8. Feb 16, 2009 #7
    yes, the final value of the kinetic energy is found by considering the mass of the boat and the dog and the final velocity. They act as a single object (theoretically)
     
  9. Feb 16, 2009 #8
    So its KE1=3125 and KE2=1093.75

    what do i subtract from what?

    wouldnt it normaly be KE1-KE2,,,but it would end up negative
     
  10. Feb 16, 2009 #9
    never mind, that made me look like a dumb ***

    I figured it out

    The amount of kinetic energy dissipated is KE=2031.25
     
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