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What is the speed of 1.4-kg cart right after the separation?

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  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A system consists of a 5.0-kg cart and a 1.4-kg cart attached to each other by a compressed spring. Initially, the system is at rest on a low-friction track. When the spring is released, an explosive separation occurs at the expense of the internal energy of the compressed spring. The change in the spring's internal energy during the separation is 1.0 kJ.
    What is the speed of 1.4-kg cart right after the separation?

    2. Relevant equations
    k=1/2mav2+ 1/2 mbv2

    3. The attempt at a solution
    -5v1=1.4v2
    v1=-1.4v2/5

    1000J=(mav12)/2 +(mbv22)/2
    2000J=5(-1.4v2/5)+1.4v22
    v2=√2000/(24/5)
    v2=20.4 m/s
     
  2. jcsd
  3. Oct 4, 2016 #2

    Doc Al

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    Staff: Mentor

    Redo that second equation, making sure to square things properly.
     
  4. Oct 4, 2016 #3
    2000J=5(-1.4v22/5)+1.4v22
     
  5. Oct 4, 2016 #4
    i caught my error the answer is 33.4 m/s
     
  6. Oct 4, 2016 #5

    Doc Al

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    Staff: Mentor

    I would write it as: 2000J=5(1.4/5)2v22 + 1.4v22
     
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