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Momentum and the Conservation Laws

  1. Oct 23, 2008 #1
    A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?


    F[tex]\Delta[/tex]T = [tex]\Delta[/tex]p
    p = mv

    [tex]\Delta[/tex]p = pf - pi = mvf - mvi = m(vf - vi)
    [tex]\Delta[/tex]p = m(vf - vi)
    [tex]\Delta[/tex]p = 4(-9 - 9)
    [tex]\Delta[/tex]p = -72
    Faverage[tex]\Delta[/tex]t = [tex]\Delta[/tex]p
    Faverage(.2) = -72
    Faverage = -360

    Final answer is incorrect, any ideas?
  2. jcsd
  3. Oct 23, 2008 #2


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    Homework Helper

    The only thing that reversed is the x-component of velocity.
    The y-component remained the same and since you are taking differences in Vectors - Velocity is a vector after all - then the force should bbe determined off of the change in x-component of velocity.
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