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Homework Help: More than just a momentum problem

  1. Feb 21, 2007 #1
    When a bullet traveling at 721 m/s strikes a block of wood originally at rest on a frictionless surface, the bullet emerges from the other side of the block of wood traveling at 349 m/s. If the mass of the bullet is 5.38 g and the mass of the block is 744 g, what is the speed of the block after the collision?


    I don't know what to do, I tried doing m1v1 + m2v2 = m1v1' + m2v2', and I couldn't get the answer. Any help would be greatly appreciated.
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Feb 21, 2007 #2

    Doc Al

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    Staff: Mentor

    But this is "just" a conservation of momentum problem, so that should work. Show exactly what you did. What values did you use?
     
  4. Feb 21, 2007 #3
    I did the following:

    (0.0058kg)(721 m/s) - (0.0058kg)(349m/s) \ 0.744 kg

    and that didnt work

    and thnks for the fast reply
     
  5. Feb 21, 2007 #4

    Doc Al

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    That looks perfectly OK to me:
    Speed of block = [(0.0058kg)(721 m/s) - (0.0058kg)(349m/s)]/(0.744 kg)

    Check your arithmetic; perhaps you made an error there.
     
  6. Feb 26, 2007 #5
    I have tried that many many times, are you sure there arent any tricks to this question or something you have overlooked?
     
  7. Feb 26, 2007 #6

    Doc Al

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    This is as straightforward a momentum conservation problem as you are likely to find. What answer did you get and why do you think it's wrong?
     
  8. Feb 26, 2007 #7
    I got it, it was just a calculational error. Thanks for the clarification!
     
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