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Momentum fo two blocks and spring

  1. Nov 11, 2005 #1

    I have been set this question and I'm half-way through answering it but am stuck on where to go next...here's the question:

    The m1 = 1.1 kg block at rest on a horizontal frictionless surface is connected to an unstretched spring (k = 180 N/m) whose other end is fixed. (See Fig. 10-36.) The m2 = 2.2 kg block whose speed is 4.0 m/s collides with the 1.1 kg block. Assume that the two blocks stick together after the one-dimensional collision. What maximum compression of the spring occurs when the blocks momentarily stop?


    Using m1v1 = m2v2 I got:

    2.2 x 4.0 = 8.8kg/m/s
    The joint mass of the two blocks is 3.3kg so the velocity must be 2.66m/s

    But I don't understand how to calculate the spring compression knowing only the velocity and the "k" value of the spring?

    Any help appreciated,
  2. jcsd
  3. Nov 11, 2005 #2

    Good. You have the velocity of the composite mass which now presses against the spring.

    Hint: What happens to the (composite mass + spring system) as the composite mass moves forward pressing against the spring? What happens to the velocity of the mass at maximum compression? What happens to the initial kinetic energy? Can you relate these two parameters now?

    Hope that helps.

  4. Nov 11, 2005 #3


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    all of your composite mass K.E. is equal to all of the compressed springs ____ energy.

    (hint: You have equations that represent these energies...and they are just floating around out there waiting for you to use them....)
  5. Nov 14, 2005 #4
    Oh right...so at the springs maximum compression, the velocity of the two joint blocks will be 0m/s...so does the composite mass' KE equal the compressed springs PE?
  6. Nov 15, 2005 #5
    OK so I calculated the KE of the composite mass:

    KE = ½mv² = ½ x 3.3 x 2.6² = 11.674J

    So is this value supposed to equal the PE of the compressed spring? And if it does, I don't really see how that helps me answer the question, as mgh won't tell me the length of compression...is there a different formula I need to use?
  7. Nov 15, 2005 #6


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    Staff Emeritus
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    The key with the spring is

    an unstretched spring (k = 180 N/m)

    Spring force, F = kx, and PE in spring is just the integral of F dx = 1/2 kx2, where x is the displacement from fully relaxed (i.e. no force applied).

    The spring reaches maximum deflection when the velocity of blocks = 0, i.e. the KE of the blocks is completely transformed into the stored mechanical energy or PE of the spring.
    Last edited: Nov 15, 2005
  8. Nov 15, 2005 #7
    Did you mean to write something between "where" and "is"?
  9. Nov 15, 2005 #8


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    x, sorry about that. Post has been corrected.
  10. Nov 15, 2005 #9
    Ah. So the 11.674J will equal ½kx² so:

    ½kx² = 11.674 = ½ x 180 x x² = 90 x x²

    11.674/90 = x²

    x = 0.36m

    Is this right do you think?
  11. Nov 16, 2005 #10
    Help, the deadline for this question is tomorrow, and I only have one submission left so I don't want to waste it...does 0.36m seem correct?
  12. Nov 16, 2005 #11


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    Staff Emeritus
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    It appears to be correct.
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