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Momentum question

  1. Jun 19, 2011 #1
    Hi I am not sure about the idea of momentum p=mv:

    3: A 0.5kg rubber ball bounces off a wall (below left). The speed of the ball is 8m/s before
    and after the collision. Use theta = 50.
    a: Determine the impulse given to the ball. In other words, find (delta)p.

    The diagram looks like
    bounce up to here
    Code (Text):

    |    /
    |t /
    |/
    |\
    |t \
    |    \
     
    start here
    the two t's represents theta.
    Thanks,
    Ted
     
  2. jcsd
  3. Jun 19, 2011 #2
    The initial momentum of the ball is [tex] p = mv = (.5)(8)=4[/tex] and this is at a fifty degree angle with the vertical. Therefore, the y-component is [tex]psin(\theta)=4sin(50)[/tex] and the x-component is [tex] pcos(\theta)=4cos(50)[/tex] When the ball bounces off the wall, the y-component of momentum is staying the same (not accounting for gravity in this case). Meaning, that only the x-component of momentum has changed. Since the final angle is the same as the initial angle, the change in momentum is enough for the x-component of momentum to have the same magnitude in the opposite direction, meaning it was two times the original x-component, so your [itex]\Delta P[/itex] should actually be [tex]2pcos(\theta)=8cos(50)[/tex]

    I'm pretty sure you're not supposed to account for gravity in this problem and assume that the y momentum is zero, but I could be thinking through this wrongly.
     
    Last edited: Jun 19, 2011
  4. Jun 19, 2011 #3
    thank you very much! this answer is perfectly matching the answer key 6.13kgm/s(x).
     
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