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Velocity of cart launched by spring

  1. Feb 18, 2009 #1

    TG3

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    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 = 3 kg. The carts are pushed toward one another until the spring is compressed a distance 1.3 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
    K = 1/2 MV^2
    F = -kX

    3. The attempt at a solution
    Once I find out how to solve for one I'll know how to solve for both, so let's just deal with the 5 kg cart.
    Force in the Spring = 1.3 x 20 = 26
    26 = 1/2 x (5) x V^2
    10.4 = V^2
    3.22 = V
    Wrong. I suspect I made this much simpler than it really is... where did I go wrong?
     
  2. jcsd
  3. Feb 18, 2009 #2
    Once I find out how to solve for one I'll know how to solve for both, so let's just deal with the 5 kg cart.
    That statement is incorrect, you have to deal with both of them at the same time.

    So think of them in terms of Energy,

    1/2kx^2 = 1/2m1v1^2 + 1/2m2v2^2
     
  4. Feb 19, 2009 #3

    TG3

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    What is the x in 1/2 kx^2? Is that the distance the spring is stretched? That can't be right, but I'm not sure what else to put in there...
     
  5. Feb 19, 2009 #4
    yes, x is the distance a spring is compressed (from equilibrium)
     
  6. Feb 20, 2009 #5

    TG3

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    Okay, so:
    1/2 x 20 x 1.3^2 = 1/2 (5) v1 ^2 + (3) v2 ^2
    16.9 = 1/2 5 v1^2 + 1/2 3 v2^2
    33.8 = 5 v1^2 + 3 v2^2

    Also, 5v1 = 3v2 because of conservation of momentum.
    So substituting 5/3 V1 in for V2 will hopefully result in being able to solve for V1.

    33.8 = 5V1^2 + 3 (5/3V1)^2 right?
    I don't see that I'm any closer to solving for V1...
     
  7. Feb 20, 2009 #6

    Nabeshin

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    Science Advisor

    Right. You can substitute one variable into the other equation which will allow you to solve for the other variable. (substituting in v1 solves for v2 and vice versa). You have one equation only with V1, so just solve it for V1. It's really just algebra at this point, you've got the physics down pat.
     
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