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Momentum Question

  1. Oct 30, 2005 #1
    Here is a link to the image: http://img500.imageshack.us/img500/2681/171812nv.gif

    I found the initial momentum to be [itex]\left(50\cos{30^o},-50\sin{30^o}\right)[/itex] and the final momentum to be [itex]\left(50\cos{25^o},50\sin{25^o}\right)[/itex]. To find the change, I just subtracted these, but the answer is incorrect. Note: The method I used to find the initial and final vectors was to multiply the magnitude of the velocity by sine and cosine respectively, and then by 2 (the mass).

    Can someone give me a hint as to where I went wrong?

    Thank you.
     
  2. jcsd
  3. Oct 30, 2005 #2

    Chi Meson

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    What values did you get when you subtracted these momenta?

    (Did you subtract the negative)
     
  4. Oct 30, 2005 #3
    Yes, I did. This is what I came up with (rounded off):

    [tex]\Delta\vec{p}=\left<2.01,46.1\right>\text{N}-\text{s}[/tex]

    ...I hope it's just an arithmetic error. Please let me know what you think if you get a chance.

    Thanks again! :smile:
     
  5. Oct 30, 2005 #4
    ∆px = px`-px = 50cos25[E]-50cos30[E] ≈ 2.014[E]

    ∆py = py`-py = 50sin25[N]-50sin30 = 50sin25+50sin30[N] ≈ 46.131 [N]

    Now you know the horizontal and verticial components of the impulse, but you need to add the vectors together to find the total impulse. Drawing these 2 vectors will form a right triangle in which you can use pythagorean theorem to find the total impulse.
     
    Last edited: Oct 30, 2005
  6. Oct 30, 2005 #5
    I need to find the vector change though, not the magnitude.
     
  7. Oct 30, 2005 #6
    You mean the horizontal and vertical changes in momentum? Your answer looks correct then. What answer does your book give?
     
  8. Oct 30, 2005 #7
    I don't have an answer for it, but it does give me a hint. It says that the impulse delivered to the mass by the plate only has a y-component. Does this mean I can get rid of my x component and use the y? For this problem, does that hint mean I would use (0,46.1) instead of (2.01,46.1)?

    Thanks again.
     
  9. Oct 30, 2005 #8
    Hmm, that's strange. If ∆px = 0, then px` = px but that clearly isn't the case here since the angle of incidence and the angle of deflection are not equal. I don't see how they can make that claim. I'm just as stumped as you are now. :yuck:
     
  10. Oct 30, 2005 #9
    Hmm... I found the answer, which is the one I suggested above (0,46.1). I find that strange as well, but I guess I'll just have to work with some similar problems.

    Thanks a lot for your help.
     
  11. Oct 30, 2005 #10

    Chi Meson

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    Dang! I was just coming back to this to point out that the frictionless steel surface could not have changed the x-momentum of the ball.

    Therefore, the final speed could not be 50 m/s because 50 cos30 has to equal 2*v'*cos 25. From this you find v'.
     
  12. Oct 30, 2005 #11
    Oh okay, I see the problem we both made here. We assumed that the final velocity of the ball was equal to the inital velocity but the question didn't state that. Tricky...
     
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