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Change of momentum of a crate within a system

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok i am working through some course work questions and have been asked to calculate the change of momentum of the 200kg crate. I have the following information after working through the previous questions

    Mass of crate = 200Kg
    Distance moved (horizontally) = 20m
    time = 4.07s
    Initial Velocity = 0 m/s
    Final velocity = 9.82 m/s
    Aceleration = 2.41 m/s/s

    I realise i don't need all this information but wanted to include all the information i have gained about the system. I have also attached my initial free body diagram.

    2. Relevant equations

    Impulse = ft = mΔv

    3. The attempt at a solution

    So this is where i'm having difficulties i think i have got confused with all the information i have already gained about the system but my attempt is as below would appreciate any pointers in the right direction.

    Ft = mΔv

    F = mΔv/t

    F = (200Kg x 9.82m/s)/4.07s

    F = 482.56N
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 26, 2012 #2

    gneill

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    Staff: Mentor

    What's the initial momentum? What's the final momentum? What's the difference (change)?
     
  4. Mar 28, 2012 #3
    Momentum = Mass x Velocity

    Initial Momentum = 200 x 0 = 0
    Final Momentum = 200 x 9.82 = 1964

    So to summarise the initial momentum would be 0 kgm/s and the final momentum is 1964 kgm/s as such the change of moment would be 1964 kgm/s? so the change in momentum would be + 1964 kgm/s?

    I thought i needed to use the impulse = force applied x time equation? is this not the change of momentum. Thanks for you help.
     
  5. Mar 28, 2012 #4

    gneill

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    Staff: Mentor

    You could use the f*t version if you wanted to. You've provided the acceleration and time, and the force can be obtained from f=ma. But it amounts to the same thing, and really, by definition the impulse is the change in momentum so why not go directly there?
     
    Last edited: Mar 28, 2012
  6. Mar 28, 2012 #5
    Thanks for the reply. I see your point (why make life difficult! :D) i have just tried the force * time to work out the impulse and that confirms my answer for me. Thanks for the help
     
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