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Homework Help: Combining Conservation Laws

  1. Nov 7, 2006 #1
    Having a difficult time determining which laws and equations to use?

    A 15.0 kg block is attached to a very light horizontal spring of force constant 350 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

    Find the maximum distance in meters that the block will compress the spring after the collision.
     
  2. jcsd
  3. Nov 7, 2006 #2

    Doc Al

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    Staff: Mentor

    Think of it as two problems. Take them one at a time:
    (1) The collision of stone with block--what's conserved there?
    (2) The compression of the spring--what's conserved here?
     
  4. Nov 7, 2006 #3
    Combining Conservation Laws; where'd I go wrong?

    A 15.0 kg block is attached to a very light horizontal spring of force constant 350 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

    Don't know where I went wrong?
    Here's my setup:

    Collision:

    Change in KE(stone) = (1/2)(m)(vfsquared-visquared)
    = (1/2)(3.00kg)(-2m/s squared - 8 m/s squared)
    = -90

    After the collision:

    Change in KE(stone) = (1/2)(Force constant)xsquared
    -90 = (1/2)(350N/m)xsquared
    x = 0.72m
     
  5. Nov 7, 2006 #4

    Doc Al

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    Please don't start a second thread on the same problem.

    Start by answering the questions I posed in my earlier response.
     
  6. Nov 7, 2006 #5
    1.) Kinetic energy
    2.) Kinetic energy of block is transferred to the spring, where it's stored as potential energy
     
  7. Nov 7, 2006 #6

    Doc Al

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    In general, KE is not conserved in a collision. But what is conserved in every collision?

    Right!
     
  8. Nov 7, 2006 #7
    1.) Momentum
     
  9. Nov 7, 2006 #8

    Doc Al

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    Staff: Mentor

    Right. Now use that to find the speed of the block just after the collision.
     
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