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Another Momentum Question

  1. Jan 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A 5.00 X 10^-3 kg steel ball moving 1.20m/s collides elastically (no loss of kinetic energy) with an identical, stationary steel ball. The incident ball is deflected 30 degrees from its original path.
    (a) Draw a vector diagram showing the paths of both balls after the perfectly elastic collision.
    (b) What is the velocity of the incident ball after the collision?


    2. Relevant equations

    All I used is the initial momentum = final momentum theorem and p=mv.

    3. The attempt at a solution

    a
    I drew the vector diagram: ----->
    \30d |
    \ |
    p=mv \ |
    =.006kgm/s \ |
    \ |
    \ |
    a=.005196kgm/s
    p=mv
    v= 1.04m/s

    Is this the correct way to do it? I think i mixed some things up as the velocity I found was for the horizonal vector, which should be the struck ball.. It feels like I made a real obvious mistake.. but its just out of reach.. :cry:
     
  2. jcsd
  3. Jan 13, 2008 #2
    Oh, no.. the vector diagram didn't turn out in the post..
    a
    ---->
    \ |
    \ |
    \ |
    \ |
    \ |

    the hypotnuse is where i put the p=mv equation and solved. I ended up with the velocity of the horizontal vector, which is not correct.
    I hope this diagram shows up!
     
  4. Jan 13, 2008 #3
    Apparently not.. next time I'll try a picture..
     
  5. Jan 13, 2008 #4

    olgranpappy

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    that "diagram" is impossible for me to read... sorry.

    anyways, you did well to recongize that you should use conservation of momentum, and energy. conservation of momentum is applied for each independant direction (each vector component), so you need to use the angles that the two balls make with the incident direction in your calculations. do you know what the angle that the second ball's velocity makes with the original direction?
     
  6. Jan 25, 2008 #5
    Here is the diagram I drew. The 30 degree angle is on the left side. On the hypotnuse side I got .006kgm/s. and solving it for the stationary ball, i got the velocity as 1.04 m/s. (Used p=mv) Shouldnt 1.04m/s be for the incident ball though?
     

    Attached Files:

  7. Jan 25, 2008 #6

    olgranpappy

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    nope. not the right diagram. angle between ball's velocities is wrong.
     
  8. Jan 26, 2008 #7
    Would that be the right diagram or am I doing this completely wrong.. :bugeye:
     

    Attached Files:

  9. Jan 26, 2008 #8
    The last diagram would be the same thing wouldn't it.. is this the right way?
     

    Attached Files:

  10. Jan 26, 2008 #9

    olgranpappy

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    it's wrong because the ball which was initially stationary does *not* move straight forward along the horizontal axis. It's velocity also makes an angle with the horizontal axis. It is well-known (and should be in your book somewhere) that after the collision the angle between the velocities must be 90 degrees. You are given the angle which one ball's velocity makes with the horizontal, thus what must the other angle be?
     
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