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Spring loaded collisionneed some guidance

  1. Nov 18, 2003 #1
    A cart of mass m = 6 kg carrying a spring of spring constant k = 48 N/m and moving at speed v = 1.5 m/s hits a stationary cart of mass M = 12 kg. Assume all motion is along a line.

    a) What is the speed of the center of mass of this system?
    b) When the spring is at its maximum compression, with what speed are the carts moving in the lab frame?
    c) How far will the spring be compressed?
    d) After the collision is over, what is the velocity of m in the center-of-mass frame? (right is positive)
    e) After the collision is over, what is the velocity of m in the lab frame?
    f) After the collision is over, what is the velocity of M in the lab frame?
    ======================================================
    Ok, I figured out a, and b, but now I'm stuck on c.

    This is what I've done so far.

    a) Initial momentum=final momentum
    (m1v1+m2v2)=(m1+m2)Vcom

    where m1=6kg
    v1=1.5m/s
    m2=12kg
    v2=0
    Vcom=velocity of center of mass

    I figured out the answer of Vcom=.5m/s

    b) I know that the answer is .5 m/s, but I'm sure why it is though. Can someone elaborate on this please?

    c) This is where I get stuck so far. I used the idea of conservation of momentum and energy. In the end, I get this equation
    KE=SPE or
    .5(m1+m2)(Vcom)^2=.5kx^2
    So I substituted my numbers, and the x I get isn't right. Is this the right way of apporaching it?
     
  2. jcsd
  3. Nov 18, 2003 #2
    Here is a pic of the thing.
     

    Attached Files:

  4. Nov 18, 2003 #3

    NateTG

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    Science Advisor
    Homework Helper

    Part B:

    When the spring is fully compressed, then the carts are (momentarily) at rest with respect to each other right? What does that mean about their motion with respect to the center of gravity?

    Part C:
    You can calculate the energy before the colision, and when the spring is fully compressed. That shoud give you the answer.

    Part D:
    One way to do this would be to solve the colision as perfectly elastic, and then do the rest of the work. Perhaps you can come up with a more elegant solution.

    Part E:
    See part D

    Part F:
    See part D
     
  5. Nov 18, 2003 #4

    Doc Al

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    Staff: Mentor

    As long as the spring is still compressing, there is relative speed between the two carts. When maximum compression is reached, the two carts---momentarily---move as one: relative speed is zero. At that point, Vcart 1 = Vcart 2 = Vcm.
    Consider this: initial KE = KE(at max compression) + PE(of spring at max compression)
     
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