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Homework Help: Tennis Ball and Impulse!

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A tennis ball of mass m = 0.087 kg and speed v = 43 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° (Fig. 7-29). What is the impulse given the wall?

    2. Relevant equations

    Impulse = M x V
    Sin (theta) = opp/hyp

    3. The attempt at a solution

    Impulse = .087(43) = 3.741 (at 45 degrees)

    Sin(45) = x / 3.741
    x = 3.1832 Is the impulse given the wall. Is this corect?
  2. jcsd
  3. Mar 1, 2010 #2
    Impulse is change in momentum. Momentum is mass times velocity. What you've found is momentum in the y direction.
    Last edited: Mar 1, 2010
  4. Mar 1, 2010 #3
    Is there change in momentum though? Cause its going the same velocity.
  5. Mar 1, 2010 #4
    Same speed or same velocity?
  6. Mar 1, 2010 #5
    same speed, different velocity. But because the angles, speeds, and masses are the same don't they cancel?
  7. Mar 1, 2010 #6
    Which velocities will cancel and which ones won't? You've got two components.
  8. Mar 1, 2010 #7
    So the x and y components are both 30.4. That would be the x component cancels (30.4-30.4) but the y component would double because its going upwards..?
  9. Mar 2, 2010 #8
    Actually it's the other way around. Let's call up and right positive, and say the ball hits the wall traveling to the the right. It hits the wall and continues up
    [tex]\Delta p_y = mv_fy - mv_iy = m(30.4-30.4) = 0[/tex]
    while in the x direction it rebounds and goes the other way
    [tex]\Delta p_x = mv_fx - mv_ix = m(30.4-(-30.4))= m*60.8[/tex]
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