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Homework Help: Ball Collision Problem

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    a ball hits a wall at 60 degrees from the wall and leaves the wall at 60 degrees.

    v1 = 10 m/s
    v2 = 10 m/s
    m = 3.00 kg
    t = 0.200s

    2. Relevant equations

    p = mvcos(angle)
    J = p2 - p1


    3. The attempt at a solution

    p = (3.00)(10)(cos(60))
    p1 = 15
    p2 = 15

    and that is as far as i can get. I am not sure how time plays into this. Any help is appreciated. I have the text in a few hours. Thank you
     
  2. jcsd
  3. Nov 29, 2011 #2
    what is being asked for?
     
  4. Nov 29, 2011 #3

    As grzz said , I cannot understand what you are trying to find out. Anyways what's the question asking for ?
     
  5. Nov 29, 2011 #4
    My apologies, I always do that. Force on by the wall on ball
     
  6. Nov 29, 2011 #5
    *by wall on the ball
     
  7. Nov 29, 2011 #6
    Got that ! Ok so what is p1 ? Or I mean momentum of the ball before striking the wall ?

    Then you find out p2 or the momentum of ball after it stroke the wall.

    What do you get ?
     
  8. Nov 29, 2011 #7
    the same thing for p1 as p2?
     
  9. Nov 29, 2011 #8
  10. Nov 29, 2011 #9
    No p1 and p2 aren't same. Be careful regarding geometry.

    Hint : see the image : http://postimage.org/image/qvnhzhnsl/ [Broken]

    Above answer is wrong.
    Look at image
     
    Last edited by a moderator: May 5, 2017
  11. Nov 29, 2011 #10
    ok so then

    p1 = (3.00)(10)cos(30)
    p2 = (3.00)(10)cos(60)

    so p2 - p1 = -14.133 so then it would just be 14.133N?

    or is it sin(30) making it 14.5N
     
    Last edited by a moderator: May 5, 2017
  12. Nov 29, 2011 #11
    Correct.

    Right !

    Nope. What about time t ?
    From Newton's second law : F= Δp/Δt = p1-p2/t

    Plug in it. What do you get ?
    Do calculations properly. You got it wrong for p1-p2 also.
     
  13. Nov 29, 2011 #12
    OH! ok so then after plugging everything in i got 71.6N!
     
  14. Nov 30, 2011 #13
    Your method of evaluation is correct but calculations are wrong. Do you know the value of cos 60o and cos 30o ? What are they ?
     
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