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Did I do this right?

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

    I was wondering if anybody might help me on this UNLESS i did this correct, but I'm not totally sure.

    Well here's the question.

    Jorge and Elizabeth are both going to the Antique Road show on PBS. Jorge pushes an antique desk weighing 300kg on a car to the right, while Elizabeth pushes a similar cart to the left with a dresser weighing 225kg. The mass of each cart is 500kg and Jorge pushes his with a speed of 3.0m/s and Elizabeth at a speed of 4.0m/s. If the cars collide and stick together then what is their final velocity and which direction are they headed?

    2. Relevant equations

    KE=1/2*m*v^2
    P=mv

    3. The attempt at a solution

    Car 1
    500kg+300kg=800kg
    3.0 m/s to the left

    KE=(.5)mv^2
    KE=(.5)(800)(3.0)^2
    KE=3600

    Car 2
    225kg+500kg=725kg
    4 m/s to the right

    KE=(.5)mv^2
    KE=(.5)(725)(4.0m/s)^2
    KE=5800

    5800-3600=2200

    KE=(.5)mv^2
    2200=(.5)(mass of Car1 and 2, 1525kg)( v )^2

    Solved using calculator

    Answer
    V=1.6986m/s to the right

    Am I right?
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    Delphi51

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    Homework Helper

    Looks like you are using conservation of kinetic energy when you should be using conservation of momentum. Stick-together collisions are not elastic - kinetic energy is lost.
     
  4. Feb 15, 2009 #3
    So should I be using P=mv?
     
  5. Feb 15, 2009 #4

    jgens

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    Gold Member

    You definately should be using conservation of momentum. In collisions that aren't completely elastic (this one's not as the carts stick together) kinetic energy is not conserved as some energy is lost to heat.
     
  6. Feb 15, 2009 #5
    What is conservation of momentum?

    formula please :P
     
  7. Feb 15, 2009 #6

    jgens

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    The law of conservation of momentum states that if no net external force acts on a system that the initial and final momentum of the system must be the same; hence, Pi = Pf
     
  8. Feb 15, 2009 #7
    So i should use these

    e26532f7b06dba85c58474ef61fb42ab.png
    807ef01bd2f00bae37644bc468c4b543.png

    If so, which one?

    i got 3.95082 with the first and
    2.95082 with the second
     
  9. Feb 15, 2009 #8

    Delphi51

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    I would start with "momentum before = momentum after".
    Then put an mv on each side for each object that is moving.
    mv + mv = mv (where all the m's and v's are different and perhaps should be numbered)
    Put in the numbers and solve for the one unknown speed.
     
  10. Feb 15, 2009 #9
    so is it
    ((M1*V1)+(M2*V2))/(M1+M2)
    that way velocity is isolated?

    Now I got -.000215

    I'm moving on to the next question now
     
  11. Feb 15, 2009 #10

    jgens

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    Gold Member

    Cooper43, the equations you posted only apply in the case of a completely elastic collision. The scenario you've described appears to be completely inelastic.

    Edit: I can't agree with your most recent answer.
     
    Last edited: Feb 15, 2009
  12. Feb 15, 2009 #11
    I (think I) FOUNT IT!!!

    cea19d6735094aa9af285bee2a873b47.png
     
    Last edited by a moderator: Apr 24, 2017
  13. Feb 15, 2009 #12

    jgens

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    Yes, that equation will suffice for the scenario described in the problem.
     
  14. Feb 15, 2009 #13
    (800) (3) + (725) (-4) = (1525)V

    V = -0.328 m/s?

    they move to the left?
     
  15. Feb 15, 2009 #14

    jgens

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    That's what I got. They move left with a velocity of v = 0.33 m/s. Good job!
     
  16. Feb 15, 2009 #15
    Yes, finally. Thanks jgens :)
     
  17. Feb 15, 2009 #16

    jgens

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    You're welcome!
     
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