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Impulse momentum method for linear motion

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I was told by a friend that M1+I1-2=M2 is actually FT=mVf-mVi.

    So using FT=mVf-mVi to solve the following question....

    3. The attempt at a solution
    FT=mVf-mVi
    (F)(0.15)=800(0)-800(5000/3.6)
    F=-7407.4N

    Ans is 7407.4N positive. Can I ask did I get my sign wrong?
     

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  2. jcsd
  3. Jul 18, 2012 #2

    Doc Al

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    All they want is the magnitude of the force.

    If you choose to make the initial velocity positive, then the force will indeed be negative.
     
  4. Jul 18, 2012 #3
    So -ve or +ve doesn't matter?
     
  5. Jul 18, 2012 #4

    Doc Al

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    Not in presenting your final answer.
     
  6. Jul 18, 2012 #5
    So how do i determine where the values that i am going to put in is going to be +ve or -ve?

    The formula is FT=mVf-mVi, i am just subbing in the values in, but in the end i got negative.
     
  7. Jul 18, 2012 #6

    Doc Al

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    There's nothing wrong with your solution. I'm just pointing out that often when they ask for some quantity they just want the magnitude of that quantity.

    If you had assumed the car was initially moving to the left, then using your sign convention you would have found that the force was positive. The direction of motion (or sign convention) wasn't specified so presumably they just want the magnitude of that force.
     
  8. Jul 18, 2012 #7
    Ok, sorry I would like to ask another question attached below. Actually I tried this question:

    FT=mVf-mVi
    -441.45t=150(0)-150(1.2)
    -441.45t=-180
    t=0.407s

    I don't understand this question why do they consider them individually? For work energy method is about setting of datum, is there anything special for impulse momentum that I need to know?


    Thanks...
     

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  9. Jul 18, 2012 #8

    Doc Al

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    Your error is thinking that the package ends up with zero final velocity. It just stops moving with respect to the flatcar. After the package stops sliding, the package and flatcar end up moving with the same nonzero speed.
     
  10. Jul 18, 2012 #9
    why did the question consider the package and flatcar individually?
     
  11. Jul 18, 2012 #10

    Doc Al

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    In order to analyze the effect of the friction force between package and flatcar, you need to treat them separately so that the friction is an external force.
     
  12. Jul 18, 2012 #11
    So anything with friction between them I must treat them separately if not I can treat them as a whole?
     
  13. Jul 18, 2012 #12

    Doc Al

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    It all depends on the particular problem. The key thing is to realize that you can apply the impulse momentum to them separately or together, as needed.
     
  14. Jul 18, 2012 #13
    So for this question I have to apply impulse momentum separately because there is a friction force between them if not I won't be able to solve them. Correct?
     
  15. Jul 19, 2012 #14

    Doc Al

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    Right.
     
  16. Jul 19, 2012 #15
    Roger.
     
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