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Vectors and impulse force

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data
    A fast ball is pitched at a batter who hits a pop fly vertically into the air. The ball was moving horizontally at 38.45 m/s when it was hit by the bat. As it left the bat it was moving vertically and reached a maximum height of 44.196m. The mass of a baseball is 0.1417 kg. If the ball was in contact with the bat for 3/1000 of a second, determine the magnitude and direction of the impulse force with which it was struck.


    I need to find the change in momentum. So I need to find the initial and final velocity. The only thing left that I can't do is find the final velocity. How do I calculate it when all that I know is the vertical height?
     
  2. jcsd
  3. Sep 21, 2011 #2

    PeterO

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    For vertical motion, there is a constant downwards acceleration of g [we are told to take g= 9.8, 9.81 or 10 depending what level Physics you are taking]

    From that , and the height reached, you can calculate the initial velocity.
     
  4. Sep 22, 2011 #3
    but I don't have the time and the formula is v=vo + at
     
  5. Sep 22, 2011 #4
    or s=vot + a(1/2 a) t2
     
  6. Sep 22, 2011 #5

    PeterO

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    ever heard of V2 = Vo2 + 2as
     
  7. Sep 22, 2011 #6
    OK, but the acceleration would be negative, so how can I find the square root?
     
  8. Sep 22, 2011 #7

    PeterO

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    Yes, it is negative, but you have to move it to the other side of the equation [whence it will become positive] since you are trying to find Vo.

    V = 0 because it has finally stopped going up when it reaches maximum height.
     
  9. Sep 22, 2011 #8
    Ohh, I thought I was looking for V. Vo = 29.43 j m/s

    So I found Pi= 5.448i kgm/s
    and Pf= 4.17j kgm/s

    I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?
     
  10. Sep 22, 2011 #9

    PeterO

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    A pythagorus solution is what you are after.

    The change in momentum has to get rid of a lot of horizontal momentum, and produce a lot of vertical momentum at the same time, so it will have an up component, and a "back towards the pitcher" component
     
  11. Sep 22, 2011 #10
    So Vo is 29.43 m/s, and the back towards pitcher component is -29.43 m/s correct?

    And I would just need to find the horizontal velocity ?
     
  12. Sep 22, 2011 #11

    PeterO

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    No. The ball arrived at 38+ m/s, then went up at 29+ m/s, so the change is 38+ back, and 29+ up.
     
  13. Sep 22, 2011 #12
    OK, so what would the variation of momentum be ?

    P = mv

    I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.

    I need either two horizontal vectors or two vertical vectors

    Then I just need to divide the change in momentum by 3/1000
     
  14. Sep 22, 2011 #13
    OK so I used the pythagorea theorem but that gave me the resulting momentum when I only need the difference in momentum.

    It's probably really simple, I'm just not seeing it right now...
     
  15. Sep 22, 2011 #14
    I ended up just doing (4.17j-5.448i) / 3/1000
    Fnet=(1390j-1816i) N

    does that work?
     
  16. Sep 22, 2011 #15

    PeterO

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    NO.

    This question is basically a test of whether you can add and subtract vectors. Looks like you can't.
    It is done by drawing arrows which forms triangles which you use trigonometry and/or pythagorus to solve. Google search if you can't find it in your own text.
     
  17. Sep 22, 2011 #16

    PeterO

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    The bit I highlighted is a real problem. You have to learn how to do that.
     
  18. Sep 22, 2011 #17
    I actually do know how to subtract vectors, but my professor had said that we did not have time to cover that section, so we wouldn't see vector addition/subtraction.
    So I assumed we wouldn't have any assignments on it, and thought there was another way to do this problem.

    Anyways, I ended up doing it anyway and found 2286.57N (N52.6degresW)
     
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