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Homework Help: An Explosive separation of two carts already in motion

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    An assembled system consists of cart A of inertia mA, cart B of inertia mB, and a spring of negligible inertia, clamped together so that the fully compressed spring is aligned between the front end of cart B and the back end of cart A. The internal energy of the system, stored in the compressed spring, is Es. The system is put on a low-friction track and given a small shove so that it moves to the right at speed vi, with cart A in front. Once the system is moving, the clamp is released (by remote control, so that the motion of the system as a whole is not affected in any way). As the spring expands, the carts move apart.

    Find the final Velocity of a A and B respectively in terms of Espring, mA, Mb, and vi.

    2. Relevant equations
    dK + dEspring = Sum of Kinetic Energy + dEspring = 0

    3. The attempt at a solution
    Here is the solution for A: vA = vi + sqrt[ (2mB * Es) /( mA * (mA +mB))]

    Which was given to me after I gave up on the problem, (this is from online homework) from it I just changed a sign and a couple terms under the radical to get vB. So I've already turned in this home work, and did well on this section covering conservation of momentum and energy. The fact that this problem has been so hard for me to solve makes me feel like I'm missing something fundamental to my understanding of how this stuff works.
    vA = vi + sqrt[ (2mB * Es) /( mA * (mA +mB))]

    Any insight into how to solve, or think about this problem would be much appreciated! I've tried a bunch of different approaches that result in an ugly quadratic with way too many vA's, and Vb's to cancel out. I think it's going to come down to just making the right substitutions but I just keep coming up short. The other very simple explosive separation problems I've solved always start with an object that has no initial velocity. So how can I apply conservation of momentum mathematically when the final velocities are being increased by the internal energy of the spring? We already took the test on this section and I did well (there weren't any like this), but I still want to understand this problem! Unfortunately, I can't devote too much more time to it right now, since we're moving along quickly into new material! I'm sure I'm going to feel silly once it clicks, but I'll accept that humiliation if helps me grow my understanding! Thank you all for your help!
  2. jcsd
  3. Nov 9, 2014 #2


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    And how are we supposed to find the error if you don't show those approaches?

    The way you describe them sounds good. Energy and momentum conservation, and then just algebra.
    You can ignore the initial velocity by looking at the center of mass system. The initial velocity won't change anything, it is just something that gets added to both final velocities.
  4. Nov 9, 2014 #3
    Hi Eric. Welcome to Physics Forums.

    Suppose the two masses were not moving initially. Would you be able to solve the problem then? What result would you get?

  5. Nov 9, 2014 #4
    Hi mfb, Sorry I didn't sure more of the work, I was just intimidated by the prospect of typing out all the algebra. We just covered inertial reference frames so I'll give that approach a try. Thank you! Oh, and I just found the toolbar with a long list of mathematical expressions. That will be helpful!
  6. Nov 9, 2014 #5
    Thank you Chet, I wasn't expecting a reply so fast, but I went right to work when I saw your comment. Here's what I've got:
    ∑Kf + ΔEs = 0
    KA + KB = -ΔEs
    I hope I'm not wrong, but this seems right as well : -ΔEs = Es I'm just thinking I'm not going to like that negative sign, and since all of the spring's internal energy will be used the ΔEs = Es,f - Es,i = 0 Am I reasoning correctly?
    I would want to use their respective kinetic energies to find their velocities after the separation. I want to make an appropriate substitution to wind up with an equation that has only one unknown. So in my attempt to find VA I used a finding from an earlier explosive separation homework problem that I hope I'm not abusing: KA/KB = mB/mA
    Solving that expression for KB gives me a substitution that eliminates VB so I can solve for VA.
    KB = (mA/mB)KA

    Now I can write:
    ½mAVA2 + (mA/mB)*½mAVA2 = Es

    In second term on the left side of the equation the mA's cancel out. Which leaves me with:

    ½mAVA2 + 1/(2mB)VA2 = Es
    factoring out the VA2 I get:
    VA2*(mA/2 + 1/(2mB)) = Es
    Dividing both sides by VA2 and simplifying gives:
    VA2 = Es/(mA/2 + 1/(2mB))
    Then take the square root of both sides:
    VA = √[Es/(mA/2 + 1/(2mB))]

    I made up some values for the masses and the energy in the spring and came up with a reasonable answer with appropriate units. So I feel somewhat confident this is correct. I should probably simplify that last expression a bit more, but I can't seem to get the algebra to look right. What do you think? And if it is correct, how would I add in the initial velocity from when the carts where moving as one with vi?
  7. Nov 9, 2014 #6
    It looks like you have kinda the right idea, but I have trouble reading your equations and you have to do a better job of using the equation editor. Plus, your result does not match the answer provided in the first post, so maybe there are some algebra errors. Here are the equations that apply to the case where the initial velocity is zero:


    Is this what you wrote down? If so, then you should have gotten the same answer as in the first post (aside from the constant velocity term).

  8. Nov 9, 2014 #7
    I'll have to practice with the equation editor, I apologize, but yes, I used your first equation with the kinetic energy. I wasn't sure how to use the second equation equation you wrote down, with the sum of the momentums equal to zero. By that I mean I wasn't sure what to solve for to create a useful substitution. Thank you again for your help!
  9. Nov 10, 2014 #8
    You have two equations in two unknowns, one of which is linear in the unknowns. Are you saying you don't remember how to solve such a pair of equations?

    After you get your solution, I am recommending that you go back and set up the energy balance and the momentum balance in a way that includes the initial velocity in the formulation. Then I recommend that you manipulate these equations to a form that demonstrates that the initial velocity is just an add on.

  10. Nov 10, 2014 #9


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    That cannot be true, the left side has inconsistent units, you add an energy with something without a standard name (speed squared divided by mass). And I don't see how you got that.
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