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Magnitude of the Force of a kick to a basketball

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data
    A 0.5 kg basketball is rolling by you at 3.9 m/s. As it goes by, you give it a quick kick perpendicular to its path. Your foot is in contact with the ball for 0.002 s. The ball eventually rolls at an angle of theta = 24 degrees from its original direction.


    2. Relevant equations
    DeltaP=Fnet*DeltaT
    Pinitial=.5*0.5*3.9^2



    3. The attempt at a solution
    I know what Pinitial is, but what is Pfinal? Where does the theta come in, force or final velocity?
     
  2. jcsd
  3. Dec 11, 2006 #2

    radou

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    The impulse equals change in momentum, not kinetic energy, you may want to consider this, first.
     
  4. Dec 11, 2006 #3
    ok, impulse =Fnet*DeltaT=DeltaP(change in momentum), right?

    am i still assuming that p=mv, so m*Vfinal-m*Vinitial=DeltaP
    where would i find the Vfinal?
     
  5. Dec 11, 2006 #4

    radou

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    It is practical to work with vectors. Assume that the initial direction of the velocity of the ball is [tex]\vec{i}[/tex], and the direction of the force (i.e. impulse) [tex]\vec{j}[/tex]. Further on, use [tex]\vec{v}_{2} = v_{2}\cos(30)\vec{i}+v_{2}\sin(30)\vec{j}[/tex]. Form a vector equation based on [tex]\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}[/tex], where [tex]\vec{v}_{2}[/tex] is the final, and [tex]\vec{v}_{1}[/tex] the initial velocity. After 'solving for' [tex]\vec{i}[/tex] and for [tex]\vec{j}[/tex], you can calculate both the final velocity and the force.
     
  6. Dec 11, 2006 #5
    I don't think I'm following you on this. how and why are you using cos(30) and sin(30)?
     
  7. Dec 11, 2006 #6

    radou

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    I'm sorry, I meant 24 degrees, my mistake. Do you follow now?
     
  8. Dec 11, 2006 #7
    I'm still not understanding.
    To find Vf, I use Vi*sin(24)+Vi*sin(24), b/c previously, you said that i need to multiply the trigs by the Vf, but i don't know what that is.

    how do i know the direction of the force and velocity?
     
  9. Dec 11, 2006 #8

    radou

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    Ok, I'll assume you're familiar with basic vector algebra. First of all, do you agree with the equation I wrote: [tex]\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}[/tex]? (*)

    Further on, you know that the impulse equals [tex]\vec{I}=\vec{F}\cdot t[/tex]. Let the magnitude of the force (which is unknown, but that doesn't matter) be F. So now you have [tex]\vec{I}=F \vec{j} \cdot t[/tex]. The direction perpendicular to [tex]\vec{j}[/tex] is [tex]\vec{i}[/tex], so the initial momentum equals [tex]m \vec{v}_{1} = m 3.9 \vec{i}[/tex]. Now we have the condition satisfied that the force acted (i.e. the impulse) in a direction perpendicular to the initial direction of the ball's motion, right? Now, all we have to do is plug these values back into the equation (*), along with [tex]\vec{v}_{2} = v_{2}\cos(24)\vec{i}+v_{2}\sin(24)\vec{j}[/tex].

    You may want to draw a simple sketch, It should help, too.
     
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