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Impulse of deflected ball

  1. Feb 23, 2008 #1
    [SOLVED] Impulse of deflected ball

    1. Recent studies have raised concern about `heading' in youth soccer (i.e., hitting the ball with the head). A soccer player `heads' a 0.421 kg ball, deflecting it by 50.0 degrees, and keeps its speed of 10.40m/s constant. (The deflection angle is the angle between the ball's initial and final velocity vectors.) What is the magnitude of the impulse which the player must impart to the ball?

    m = 0.421 kg
    v[tex]_{}i[/tex] = -10.4 m/s
    v[tex]_{}f[/tex] = 10.4 m/s at 50 degrees

    2. The equations

    impulse = [tex]\Delta[/tex]p = m(v[tex]_{}i[/tex]) - m(v[tex]_{}f[/tex])

    so impulse = m(v[tex]_{}f[/tex] - v[tex]_{}i[/tex])

    So I multiplied the mass by the change in velocity. Namely:

    3. The solution

    0.421 *{sqrt[ (10.40*sin50)^2 + (10.40*(cos50 +1))^2) ]}

    So I got about 7.936. But this is wrong. What's up?
  2. jcsd
  3. Feb 23, 2008 #2
    I think you typed it into the calculator wrong. I just plugged it into google and got
    Code (Text):
    .421 * sqrt(((10.40 * sin(50))^2) + ((10.40 * (cos(50) + 1))^2)) = 8.67976478
    or, in [tex]\LaTeX[/tex]

    [tex].421 \sqrt{ \left(10.4\sin{50}\right)^2 + \left(10.4\left(\cos{50}+1\right)\right)^2 } = 8.67976478[/tex]
  4. Feb 23, 2008 #3
    I think that might be in radians...

    Thanks, though. My teacher went over the homework, and it seems that I was measuring the wrong angle. The angle is SUPPOSED to be between the tails of the two vectors, but I was measuring the angle between the tip of the initial and the tail of the final. So my calculations should have treated the angle as 130 degrees. Alternatively, I could have made both vectors' x-components have the same signs.

    Thanks anyway.
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