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Momentum of a Tennis Ball at an Angle

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A tennis ball of mass = 0.060 kg and speed = 26 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° .

    2. Relevant equations

    Δp = pfinal - pinitial
    horizontal velocity = (26 m/s)cos 45°

    There's a picture if you all care.

    3. The attempt at a solution

    pintial = mv
    pintial = (0.060 kg)cos 45°
    pinitial = 1.1028 kgm/s

    pfinal = - pintial

    I made this relationship because it is now going in the opposite direction.

    pfinal = -1.1028 kgm/s

    Δp = -1.1028 kgm/s - 1.1028 kgm/s = -2.2056 kgm/s

    The answer is positive and I cannot understand why.

    What is the direction of the impulse given to the ball?

    A. Upward
    B. Downward
    C. Left
    D. Right

    I've eliminated B and C but I'm still unsure. The sine and cosine of forty-five degrees are the same.
     

    Attached Files:

  2. jcsd
  3. Mar 6, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Momentum is a vector. It's magnitude is m*v, where v is the speed.

    The change in momentum would be found by finding the momentum vectors for before and after and finding the difference.
     
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