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Homework Help: Impulse Problem

  1. Jan 8, 2005 #1
    This problem is driving me crazy! I'm currently taking AP physics and my teacher gets confused with her own problems, anyways I want to see if anyone could give me a hand on this impulse problem.

    Here is a link to the problem:

    Thank you guys
  2. jcsd
  3. Jan 8, 2005 #2
    Impulse is the change in momentum.


    I=p1-p2 (impulse is the change in momentum)

    p1=(.06kg)(25 m/s)sin45=1.06Ns
    p2=(.06kg)(-25 m/s)sin45=-1.06Ms

  4. Jan 8, 2005 #3


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    what was your approach?

    this is what i see so far

    [tex] \vec{I} = \Delta \vec{p} [/tex]
  5. Jan 8, 2005 #4


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    Urban aren't your forgetting something?

    [tex] \vec{I} = I_{x} \vec{i} + I_{y} \vec{j} [/tex]
  6. Jan 8, 2005 #5
    The impulse of the wall has nothing to do with the "y direction"

    The wall only pushs in the x direction so [tex]I_{y} \vec{j}[/tex] is zero
  7. Jan 8, 2005 #6


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    I am saying this:

    [tex] \vec{p}_{o} = p_{xo} \vec{i} + p_{yo} \vec{j} [/tex]

    [tex] \vec{p} = p_{x} \vec{i} + p_{y} \vec{j} [/tex]

    You neglected the x component.
  8. Jan 8, 2005 #7
    The x componet is the "(25 m/s)sin45"
    The y componet is 0
  9. Jan 8, 2005 #8


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    Where do you get 0??

    [tex] \vec{p}_{o} = m|\vec{v_{o}}| \cos \theta \vec{i} + m|\vec{v_{o}}| \sin \theta \vec{j} [/tex]

    [tex] \vec{p} = m|\vec{v}| \cos \theta \vec{i} + m|\vec{v}| \sin \theta \vec{j} [/tex]

    What do you mean wall neglected? i am talking about the momentum of the ball, it's directed at a 45 degree angle, so it has a nonzero x and y component which happen to be the same because [itex] \sin \theta = \cos \theta [/itex] when theta is 45 degrees.
  10. Jan 8, 2005 #9


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    Depends on how you whose the axes.There's another thread with "Incline" in which u see that the axis cosing may alter your result,meaning that the modulus will be the same,but comonpents on two axis of coordinates will be generally different.
    In two favorable cases,one of the components of the impulse is zero.It's less relevant whether it's denoted by "x",or by "y".

  11. Jan 8, 2005 #10
    But the wall does not act in the y direction does it? When I throw a ball at the wall, the wall only supplies the push backwards, it doesnt push it up nor down. The impulse goven by the wall should only concern the x direction. That's why I only did sin45 of 25 m/s
  12. Jan 8, 2005 #11
    Dextercioby said you're right... as long as you defined your axis. Some people use the "up" direction for the x axis instead... But you should give a direction for your answer too since impulse is a vector...
  13. Jan 9, 2005 #12
    you guys are definitelly on top of it! This is my 1st year taking physics, my high school teacher is not a physics major, but a chemistry major, what is she doing in physics? Anyways, she usually get confused while explaining the problems, how could we possibly understand. :( I think we need one of you guys over in the school teaching some physics :)

    Thank you guys I now have a better picture, how couldn;t I with this great discussion. Thank you guys for all...
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