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Homework Help: Conservation of Momentum problem

  1. Mar 24, 2008 #1
    [SOLVED] Conservation of Momentum problem

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

    A 12.5 g wad of sticky clay is hurled horizontally at a 95 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?



    2. Relevant equations
    Ki = Kf + loss
    Pi = Pf

    3. The attempt at a solution

    To solve this problem here is what I did:
    Ki = Kf + loss
    m1vi = (m1 + m2)vf

    Using equation 2, I solved for vf. I then plugged that into the vf in the Kf in equation 1. Using this method, however, I did not get a correct answer. I feel like I am missing something at the beginning. Can someone tell me if this is the proper set up or if I am missing something? Thanks.
     
  2. jcsd
  3. Mar 24, 2008 #2

    Doc Al

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    Staff: Mentor

    Separate the problem into two parts:
    (1) The collision itself. Use your momentum conservation equation here, just like you did.
    (2) Post collision. What's the KE of the block+clay immediately after the collision? What's the work done by the friction force?
     
  4. Mar 24, 2008 #3
    Would my 2 equations not be this:

    1) (m1)(Vbefore) = (m1 + m2)(Vafter)

    2) .5m1(Vbefore)^2 = .5(m1 + m2)(Vafter)^2 + loss
     
  5. Mar 24, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes.

    No.

    Instead use: KE(after) = Work done by friction.
     
  6. Mar 24, 2008 #5
    Ok, that worked.

    Im not sure I completely understand why KE(after) = work done by friction.
     
  7. Mar 24, 2008 #6

    Doc Al

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    Staff: Mentor

    Because the work done on the block+clay equals the change in its KE. After the collision, it starts out with KE(after) and ends up with 0 as it comes to rest.
     
  8. Mar 25, 2008 #7
    Simple enough. Thanks
     
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