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Exploding spring

  1. Aug 3, 2005 #1
    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 = 3.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.8 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

    i know both momentum and energy conservation applies here, but don't know where to start...
    both carts are initially at rest,

    so I'm guessing : PE(initial) = PE(final) + KE (final)
  2. jcsd
  3. Aug 3, 2005 #2


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    You are correct: Conservation of energy and momentum is the way to deal with this problem. So your starting point should be obvious: Write down the initial energy and momentum (these are known) and equate them to the final energy and momentum respectively. This will give you 2 equations in 2 unknowns (v1 and v2).
  4. Aug 3, 2005 #3
    also, wouldn't PE final be zero too? after the carts are released, the spring would be relaxed again, thus no compression, thus zero..

    that'll make it: initial PE = final KE

    but i don't know where to go from here
    Last edited: Aug 3, 2005
  5. Aug 4, 2005 #4


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    That's right, the final potential energy of the spring is zero.
    What are the expressions for the initial energy, initial momentum, final momentum and final energy?
  6. Aug 4, 2005 #5
    now we have initial PE = final KE,

    it goes....

    .5kx^2 = .5m(1)v(1)^2 + .5m(2)v(2)^2

    does it make sense? but the problem now is i have two variables to solve...one equation...
  7. Aug 4, 2005 #6

    Doc Al

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    Don't forget conservation of momentum. That will give you the second equation that you need.
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