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Mastering Physics, Finding the Compresssion of a Spring using Momentum.

  1. Nov 10, 2009 #1
    This is another problem from the same homework set that I was working on earlier. Again I believe conservation of momentum is the key formula.

    1. The problem statement, all variables and given/known data
    An 8000kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant k=3.2*10^5. The car is hit by a second car of 7600kg mass moving at 6.8 m/s, and the two cars couple together.

    1.) What is the maximum compression of the spring?

    and

    2.) What is the speed of the two cars together when they rebound from the spring?

    2. Relevant equations
    M1V1+M2V2=(M1+M2)Vfinal
    F=-kx.

    3. The attempt at a solution
    I am having trouble bringing the compression of the spring into the equation. In order to find the compression I need the Force that the collision of the cars exert on it using F=m*a. I can kind the mass easily it the combination of both of the cars or 8000kg+7600kg=15600kg. How do I find the acceleration. I know that acceleration is change of velocity over time. I could find the velocity of the coupled cars using the conservation of momentum equation but I still don't have time. This is were I reach my dead end in my logic. A nudge in the right direction would help a ton. Thanks.
     
    Last edited: Nov 10, 2009
  2. jcsd
  3. Nov 10, 2009 #2
    Do you remember the formula for momentum?
     
  4. Nov 10, 2009 #3
    Being no expert, as I am only taking physics now myself, I would find how much kinetic energy the moving car has.
     
  5. Nov 10, 2009 #4
    Ok:

    Part 1: Think kinetic energy and potential elastic energy.

    Part 2: Think conservation of energy

    My answers are below in a spoiler tag for reference.

    1)

    KE = (1/2)mv2
    KE = (1/2)(7600)(6.8)2 = 175712 J

    PEElastic = (1/2)kx2
    And, since all of the kinetic energy from the moving car will be transferred to the spring:
    PE = 175712 J
    175712 = (1/2)(3.2 x 105)x2
    1.048 m = x

    2)

    I'm not sure about this method, but it is rather elegant so I'll post it.

    Since all of the initial kinetic energy of the moving car is being transferred to the spring and then into the two cars

    KEMoving car = KEBoth cars after the spring has decompressed

    (1/2)(7600)(6.82) = (1/2)(7600+8000)(vf)2

    4.764 m/s = vf
     
  6. Nov 10, 2009 #5

    ideasrule

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    Homework Helper

    This is actually an energy question, not a momentum question. It's easy to prove that momentum ISN'T conserved: when the spring reaches its maximum compression, the cars must have a velocity of 0. The whole system now has 0 momentum, which isn't equal to its initial momentum of (humongous mass)*(terrific speed). For momentum to be conserved, you'd have to consider the earth as part of the system.

    So, energy's roughly conserved and the trains have 0 speed when the spring reaches min. length. What's the min. length?
     
  7. Nov 10, 2009 #6
    F(x) your answers were incorrect according to Mastering Physics. I am having trouble understand how momentum does not come into play in this problem since it is in my momentum chapter. I blame my professor. Anyways, it is getting late and I don't have time to finish the problem. Thanks for your help though.
     
  8. Nov 10, 2009 #7

    ideasrule

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    Homework Helper

    Momentum does come into play, just not in the final step. After the two cars collide and stick together, you have to determine their final speed before you can use E=1/2mv^2 to calculate kinetic energy. That final speed can only be arrived at using the conservation of energy. (Sorry if I meant to imply that momentum isn't used at all.)
     
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