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Impulse graphing question

  1. May 22, 2012 #1
    1. The problem statement, all variables and given/known data
    [The problem is in the attached image]


    2. Relevant equations



    3. The attempt at a solution
    My best guess is to break up the graph into two triangles one positive, one negative and then calculate and add the impulse of both of these triangles... Is that what you're supposed to do?
     

    Attached Files:

  2. jcsd
  3. May 22, 2012 #2
    Impulse is defined as the change in momentum, or Δp=Δ(mv)=mΔv. This quantity is related to the force F by the following logic.

    F = dp/dt (you should memorize this equation; it's very useful.)

    = d(mv)/dt
    = m(dv)/dt
    = m(dv/dt)
    = ma, which is the definition of force.

    Given this simple relationship,

    ∫dp = Δp = ∫F dt

    What is an integral? An integral is the signed area between the x-axis (F=0) and the curve. So yes, calculate them separately, and then glue them together. Don't forget that everything below F=0 counts as negative impulse.
     
  4. May 23, 2012 #3
    Thank you :)
     
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