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Energy conservation (2 carts and a spring)

  1. Aug 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 2 kg. The carts are pushed toward one another until the spring is compressed a distance 1.5 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?


    2. Relevant equations
    I am using K_f + U_f = K-i + U_i
    Why doesnt it work?


    3. The attempt at a solution
    OK heres what I know

    U_i = 1/2k delta s^2
    = 22.5J

    K_i = 0 (because object initially not moving)

    U_f = 0 (because carts now moving)

    K_f = 1/2 mv^2
    Energy is conserved, so 1/2mv^2 must = 22.5J.
    Also, 22.5J is the kinetic energy of BOTH carts

    So 22.5J = K (of cart 1) + K (of cart 2)

    so that leaves me with 2 unknowns in each equation for each cart???

    1/2mv^2 (cart 1) = (some amount of 22.5J)

    1/2mv^2 (cart 2) = (some amount of 22.5J)

    Please Help,

    (Thanks)
    (ALSO, I do understand that potential E is converting to kinetic E and the amount is conseved
    and also that momentum is conserved)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Aug 1, 2007
  2. jcsd
  3. Aug 1, 2007 #2
    Conservaiton of energy always works... so does Conservation of momentum...
    Since you already have one equation with two unknown, i.e.
    [tex] 22.5 = 1/2m_1v_1^2 + 1/2m_2v_2^2 [/tex]
    why don't you apply the conservation of momentum to add one more equation to the problem and make it become 2eqn 2 unkn...
     
  4. Aug 1, 2007 #3
    OK but how could it have initial momentum if its not moving initially ?/
    p=mv (v=0)
    cause that would mean momentum not conserved right?
     
    Last edited: Aug 1, 2007
  5. Aug 1, 2007 #4
    Okay... that's mean the initial momentum p = 0...
    And the final momentum depends on two object moving toward opposite direction, that's mean
    [tex] m_1v_1 = m_2v_2 [/tex]

    understand?>
     
  6. Aug 1, 2007 #5
    OH, so in opposite directions so they must add to zero?
     
  7. Aug 1, 2007 #6
    They are in opposite direction, so one must be positive while the other one must be negative, let's say v1 is moving to the positive direction, that's mean v2 must be negative... i.e.

    [tex] m_1v_1 + m_2(-v_2) = 0[/tex]
    [tex] m_1v_1 = m_2v_2 [/tex]
     
  8. Aug 1, 2007 #7
    SO if I have
    22.5 = 1/2m1v1^2 + 1/2m2v2^2
    and
    m1v1 = m2v2

    and set them equal to each other:

    m1v1^2 + m2v2^2 - 45 = m1v1 - m2v2

    I am still left with v1 + v2 = (some number) after all the algebra (unless my algebras bad)
    ??
     
  9. Aug 1, 2007 #8
    Bad algebra indeed.
    You have two equations and two unknowns, this can be solved.
    You can easily use the second equation (momentum) to eliminate one of the variables in the first equation (energy) and to solve this equation then.

    It is a common problem in physics: you have the total energy in a frame of reference where the total momentum is zero.
    Another example of this problem is the recoil of a gun. If you think of it, you see it is the same problem.
    In particle physics, when a particle (U238 e.g.) decays in two fragments, the total energy of the reaction (spontaneous fission) is split in the fragments, but the total momentum is not changed (zero if in the proper frame).

    Write down the general solution to this problem (not only your specific numbers) and analyse it, it is interresting.
     
    Last edited: Aug 1, 2007
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