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Carts and a spring (find velocity of the cart)

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Two carts have a spring of negligible mass compressed between them. The spring has a force constant of 8000 N/m and is compressed 0.05 m from equilibrium. Cart 1 has a mass of 1.3 kg and cart 2 has a mass of 0.5 kg.
    What is the speed of cart 1?
    What is the speed of cart 2?


    2. Relevant equations

    p=mv
    F=ma
    U=(.5)kx^2
    K=(.5)mv^2
    F=kx

    3. The attempt at a solution

    F=kx
    F=400

    400 = (.5)m1v1^2 + (.5)kx^2
    800 = m1v1^2 + kx^2
    780 = m1v1^2
    v1 = 24.494 m/s

    but it's wrong.. anyone can help me?
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    gneill

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    Staff: Mentor

    You seem to be equating a force to the sum of two energies. That's not right, the units don't match.

    Actually, the spring force is not required for this problem. You can work with the spring energy, since that's what is going to be transformed into kinetic energy of the carts.

    When the spring expands it will dump the potential energy stored due to its compression into kinetic energy in the carts. What's the value of that potential energy?

    What will dictate the ratio by which the KE is split between the carts? (HINT: it's a conservation law).
     
  4. Nov 6, 2011 #3
    so what i'm getting from what u are saying is that

    U(spring) = KE
    .5kx^2 = .5mv^2
    kx^2 = mv^2
    8000(.05)^2 = 1.3v^2
    v = 3.92m/s?

    its still wrong :confused:
     
  5. Nov 6, 2011 #4
    Use the conservation of mechanical energy (potential energy of spring when it is compressed equals the sum of the kinetic energies of the carts as they are pushed away) and the conservation of momentum (the sum of momentum of the two carts should be zero).
     
  6. Nov 6, 2011 #5

    gneill

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    Staff: Mentor

    There are two masses. Both will be moving, so each will have some share of the kinetic energy. These shares will not be equal. You have to figure out how that energy will be divided between the carts. In order to do that you should first look at how the momentum will be distributed. What's the total momentum before the spring is released and everything is at rest? What is it after?
     
  7. Nov 6, 2011 #6
    did it but got it wrong again holy **** god damn it!

    this is what i did

    m1v1 + m2v2 = 0
    1.3v1 + .5v2 = 0
    v2 = -2.6v1

    kx^2=m1v1^2 + m2v2^2
    10 = .65v1^2 + .25v2^2
    10 = .65v1^2 + .25(-2.6v1)^2
    10 = 1.3v1^2
    v1 = 7.69 m/s

    and its wrong again hahahaha :confused:
     
  8. Nov 6, 2011 #7

    gneill

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    Staff: Mentor

    Check your value of kx2.
     
  9. Nov 6, 2011 #8
    kx^2=m1v1^2 + m2v2^2
    20 = .65v1^2 + .25v2^2
    20 = .65v1^2 + .25(-2.6v1)^2
    20 = 1.3v1^2
    v1 = 3.92m/s

    i was so excited, but.. it turns out that its still wrong :confused:
     
  10. Nov 6, 2011 #9

    gneill

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    Staff: Mentor

    You've also got to fix your masses ... the values you've got are still divided by two.
     
  11. Nov 6, 2011 #10
    true that... now its correct :biggrin: thank you so much
     
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