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Elastic Head-On Collision, Spring Compression.

  1. Jul 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A 0.60kg cart moving at 5.0m/s[W] collides with a 0.80kg cart moving at 2.0m/s[E]. The collision is cushioned by a spring(k=1200N/m).
    2. Relevant equations
    [W] is Positive
    PTo=PTf
    ETo=ETf
    3. The attempt at a solution
    PTo=PTf
    m1v1o+m2v2o=(m1+m2)vf
    (0.6kg)(5m/s)+(0.8kg)(-2m/s)=(0.6kg+0.8kg)Vf
    3kgm/s-1.6kgm/s=(1.4kg)Vf
    (1.4kgm/s)/(1.4kg)=Vf
    Vf=1m/s[W]

    ETo=ETf
    1/2m1v1o2+1/2m1v2o2=1/2(m1+m2)vf2+1/2kx2
    1/2(0.6kg)(5m/s)+1/2(0.8kg)(-2m/s)=1/2(0.6kg+0.8kg)(1m/s2)+1/2(1200N/m)x2
    7.5kgm/s+1.6kgm/s=0.24kgm/s+600N/mx2
    9.1kgm/s-0.24kgm/s=600Nmx2
    0.0148=x2
    x=0.12m

    Im pretty sure this is correct. I think I actually figured out what my real question about this was while typing this out. Initially when I saw this problem I wanted to just take the combined initial energy of both carts 9.1J and use that equal to elastic potential energy, so 1/2kx2=9.1J. I assumed that since the max compression should happen in the moment before both carts change direction or the spring rebounds that the velocity of both carts should be 0. Now I think the kinetic energy in the ETf is actually the kinetic energy of the entire "system" of both carts and the spring moving since the cart moving West would push everything slightly west. And so the carts do have a velocity of 0 relative to the spring at max compression but the entire collection of stuff has a velocity relative to the earth or ground. Is that somewhat right? The only other thing that bothers me about this is why would the kinetic energy of all of them only be 1/2(m2+m2)Vf2 shouldn't this include the mass of the spring if the spring was also moving?
     
  2. jcsd
  3. Jul 26, 2015 #2

    haruspex

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    No, you can't assume that.
    That's right. The max compression is when the rate of change of spring length is zero, and that occurs when the carts are moving at the same velocity.
    It would, but you are not given a mass for the spring, nor how the spring was moving beforehand, so you have no choice but to assume that either it is massless or it is included in the mass of one of the carts.
     
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