1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    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

    User Avatar

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?