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Impulse from bat to baseball

  1. Nov 18, 2013 #1
    A 0.14 kg baseball moves horizontally with a speed of 29 m/s toward a bat. After striking the bat the ball moves vertically upward with 46% of its initial speed. Find the direction and magnitude of the impulse delivered to the ball by the bat.

    Find the ° (measured from the initial direction of the ball)
    Find the kg·m/s




    J = ΔP, Pf=Pi, P=mv



    I have the initial momentum as 4.06 kgm/s, and that is just the momentum of the baseball. Or would I put initial momentum as 0? I think it would be the 4.06 because at the beginning of the problem the baseball is moving. The momentum of the baseball after the hit is 1.86 kgm/s, which I got by multiplying the 46% by the mass of the baseball, giving me 13.34 m/s and then multiplying that by .14kg giving me 1.86 kgm/s. I'm unsure how to proceed from here.
     
  2. jcsd
  3. Nov 18, 2013 #2

    PhanthomJay

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    Impulse J is the change in momentum (mv_final - mv_initial). You have to vectorially subtract them. Familiar with vector addition and subtraction?
     
  4. Nov 19, 2013 #3
    Yupp. So I put the x vector as 29 and the y vector as 13.34 which gave me a hypotenuse of 31.92, then I figured out theta to be 24.7 by doing the arc tangent of 13.34/29 which I put as the direction of the impulse but it is telling me that it is wrong. What did I screw up?
     
  5. Nov 19, 2013 #4

    gneill

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    Make a sketch and determine which quadrant the angle should be in. arctan is insensitive to the placement of signs in its argument (it can't tell if a negative argument came from the numerator or denominator of the values comprising the argument). A better function is the atan2 function which takes two arguments, the y and the x (or the "rise" and the "run"), and always returns an unambiguous result.
     
  6. Nov 19, 2013 #5
    So how would I put it in the calculator then?
     
  7. Nov 19, 2013 #6

    gneill

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    You can use arctan, but you may have to adjust the result (generally involving a factor of 180 degrees). Make the sketch. Determine which angle you obtained. Adjust if required.

    Why don't you play with a few examples and see how your arctan function behaves? Choose some x and y values from each quadrant and see what arctan returns.

    Alternatively, if your calculator has built-in polar to rectangular and rectangular to polar conversions, use that.
     
  8. Nov 19, 2013 #7

    PhanthomJay

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    You are determining the magnitude of the velocity change when the problem asks for the magnitude of the impulse. Angle direction looks ok with respect to the initial direction of the thrown ball but you should round it off to 2 significant figures.
     
  9. Nov 19, 2013 #8
    So round it off to 25? Still wrong, and good point on the impulse. I'll try to work that out.
     
  10. Nov 19, 2013 #9

    gneill

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    Supposing the initial momentum is directed along the +x axis, and the final momentum directed along the +y axis, then:

    attachment.php?attachmentid=64097&stc=1&d=1384900419.gif
     

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  11. Nov 19, 2013 #10
    I got 335 for that angle, but it's still wrong. This is the only problem on the assignment that I had any trouble with. Very frustrating
     
  12. Nov 19, 2013 #11

    gneill

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    What quadrant is the angle in? What is the range of angles for that quadrant?
     
  13. Nov 19, 2013 #12
    1st quadrant and 0-90
     
  14. Nov 19, 2013 #13

    gneill

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    Which quadrant?

    attachment.php?attachmentid=64098&stc=1&d=1384902197.gif
     

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  15. Nov 19, 2013 #14
    Oh for some reason I was thinking it was starting from the other side. I got it now, thank you so much!
     
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