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Momentum and impulse

  1. Nov 9, 2008 #1
    OK I have 1 more question for today...

    1. The problem statement, all variables and given/known data

    A 0.45kg volleyball travels witha horizontal velocity of 3.2m/s over the net. You jump up an hit the ball back witha horizontal velocity 7 m/s. If the contact time is 0.047s, what is the average force on the ball?

    2. Relevant equations

    Here are all the equations that were in our notes for this section.
    p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

    f*"change in"t="change in"p
    f*"change in"t=mvf-mvi
    "change in" P=mvf-mvi

    3. The attempt at a solution
    Again, I wasnt really sure what equation to use...


    I got -67.0, but it wasn't right...

    Please help...=]
  2. jcsd
  3. Nov 9, 2008 #2
    [tex]\textbf{f}=\Delta \textbf{p}/\Delta t[/tex].
  4. Nov 9, 2008 #3
    Ok the change in p would be...
    Im not really sure if i could just use the numbers from the problem or if i had to change the velocities or if they were in the correct order...
  5. Nov 9, 2008 #4
    [tex]\Delta \textbf{p} = m(\textbf{v}_f-\textbf{v}_i)/t[/tex], where the final velocity is 7 m/s in the negative direction and the initial velocity is 3.2 m/s in the positive direction.
  6. Nov 9, 2008 #5
    so it would be:
  7. Nov 9, 2008 #6

    The final velocity is -7, the initial velocity is 3.2. But, the change in momentum is given by the change in velocity. Id est, final minus initial.
  8. Nov 9, 2008 #7
    Oh, so:
  9. Nov 9, 2008 #8
    Final, -7, minus initial, +3.2, =>

    0.45kg (-7m/s - +3.2m/s)/0.047s =

    0.45kg (-10.2m/s) / 0.047s =...
  10. Nov 9, 2008 #9
    Wow. Thank you again. You are amazing...=]
  11. Nov 9, 2008 #10
    No big deal.
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