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Conservation of Energy Problem - Need Help

  1. Apr 8, 2008 #1
    [SOLVED] Conservation of Energy Problem - Need Help

    1. The problem statement, all variables and given/known data

    A 2.0 kg cart has a sping with k= 5000 N/m attached to its front, parallel to the ground. This cart rolls at 4.0 m/s toward a stationary 1.0 kg cart.

    a) What is the maximum compression of the spring during the collision?

    b) What is the speed of each cart after the collision?

    2. Relevant equations

    KE=(1/2)mv^2 Us=(1/2)kx^2

    3. Attempt at a solution

    I got the first part:

    Using first conservation of momentum and assuming the cars stick together momentarily
    then using conservation of energy

    v1= 2.67 m/s
    (.5)(2kg)(4m/s)^2=(.5)(5000N/m)(x)^2 + (.5)(3kg)(2.67m/s)^2
    x= .046 m

    But I'm not sure what to do with the second part. Any suggestions?
  2. jcsd
  3. Apr 8, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Apply the conservation laws. What's conserved? (The difference here is that there's no spring PE before and after the collision.)
  4. Apr 8, 2008 #3
    Ok, I see.

    So I would use conservation of momentum first:

    (3kg)(2.67 m/s) = (2kg)v1 + (1kg)v2 Solving to get an expression for v1
    v1= 4.005 - .5v2

    Then I would substitute that into conservation of energy:

    (.5)(3kg)(2.67 m/s)^2 + (.5)(5000 N/m)(.046 m)^2 = (.5)(2kg)(4.005-.5v2)^2 + (.5)(1kg)(v2)^2

    Then I would solve for v2 using the quadratic equation. I come up with values for v2:
    .016 & 5.3

    Plugging them into v1= 4.005- .5v2 I see that 5.3 m/s must be the velocity for the cart that was initially at rest because the velocity of the cart that was intially moving cannot be greater than the velocity of the cart initially at rest.

    So: v1= -1.3 m/s & v2= 5.3 m/s

    Does this look right to you?
  5. Apr 8, 2008 #4

    Doc Al

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    Staff: Mentor

    Looks OK, except for v1 being negative. (Double check that.)

    I would have simply compared the initial momentum and energy (prior to collision) to the final momentum and energy (after the collision). That way you don't need the results from the first part and you don't have to worry about spring compression. (You get the same answer either way, of course.)
  6. Apr 8, 2008 #5
    Ok, v1 shouldn't be negative. Not sure why I put that there.
    I also tried it the way you suggested, and it is much easier.

    Thanks for your help!
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