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Momentum and net external force

  1. Feb 21, 2015 #1
    1. The problem statement, all variables and given/known data
    In the picture

    2. Relevant equations
    In the picture attached. 0219151551.jpg

    3. The attempt at a solution
    For the firecracker problem, I believe the only net external force is gravity. Therefore using, Fnet * delta t = delta p. Change in momentum is 12 kg * m/s. Thus, the final momentum of the system is zero. Is this reasoning correct? Please let me know of any errors in my reasoning. Thank you.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2015 #2

    Nathanael

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    Right.

    How did you conclude that?
     
  4. Feb 21, 2015 #3
    The initial momentum of the system is 3kg * -4m/s = -12 kg * m/s. The change in momentum from gravity is 12 as I explained.
    Pf = Pi + Delta P
    Pf = -12 + 12 = 0
     
  5. Feb 21, 2015 #4

    Nathanael

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    By saying "-4 m/s" you have defined a direction for positive and negative.
    The magnitude of the change in momentum is 12 kg.m/s, but Δp is a vector; what is it's direction?
     
  6. Feb 21, 2015 #5
    Gravity then should be negative since it too acts downward. So, delta p should be -30 * 0.4 or -12? So final momentum of system is actually -24 kg * m/s. The bottom piece of mass 2kg must therefore be moving at -14 m/s, since the top piece of mass 1kg is moving at +4m/s, and the sum of these two final momenta must add up to -24 kg * m/s.
     
  7. Feb 22, 2015 #6
    Is this reasoning correct?
     
  8. Feb 22, 2015 #7

    Nathanael

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    Yes, it is correct.
     
  9. Feb 22, 2015 #8
    0222151508.jpg
    I know in part b there is a net external force (gravity) that does positive work, fnet > 0, thus dp/dt is positive. Does the spring in part c increase or decrease the momentum of the system? I am not too sure of my answer.
     
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