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2-D Collion question involving a cue ball and a numbered ball.

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A 0.17kg cue ball is moving at 4.0m/s when it strikes a 0.16kg stationary numbered ball. After the collision the cue ball moves 60 degrees to the left of its original direction while the object ball moves 30 degrees to the right of the cue ball's original path. Determine the speed of the

    a) cue ball after the collision
    b) object ball after the collision


    2. Relevant equations

    Px - mv = mv +mv


    3. The attempt at a solution

    Ive marrowed it down to where you have

    (0.17kg)(4.0m/s) = (0.17kg)vCos60 + (0.16)vCos30

    I dont know what to do with the 2 unknown variables.

    Your help is greatly appreciated!

    Cheers.
     
  2. jcsd
  3. Oct 8, 2008 #2
    Well look at it like this: In the system you've defined, momentum is always conserved.
    You are completely right in that [tex] (0.17)(4.0) = (0.17)v_1 cos(60) + (0.16)v_2 cos(30) [/tex]

    You must conisder the plane perpendicular to this one though. If you define that as the y axis, you will see that initially there was no y momentum input, therefore there can be no y momentum output.

    This implies that:

    [tex] (0.17)v_3 sin(60) + (0.16)v_4 sin(30) = 0 [/tex]

    Where [tex] v_3 [/tex] is the y component of the cue ball and [tex] v_4 [/tex] is the y component of the ball that got struck.
     
  4. Oct 8, 2008 #3
    Your absolutly right. I hadnt gotten that far ahead beucase id run into the same problem i did when I did the x-axis equations. I still dont know how to solve for multiple unknown variables in the same equations.
     
  5. Oct 8, 2008 #4
    Okay I'll help you out a little more here then. Sorry I made an error too, v3 = v1 and v4 = v2
    I said they are different components, but they only become the individual componenets once multiplied by their respective sine or cosine.

    [tex]
    (0.17)v_1 cos(60) + (0.16)v_2 cos(30) = (0.17)(4.0) [/tex]

    [tex] (0.17)v_1 sin(60) + (0.16)v_2 sin(30) = 0 [/tex]

    By simple re-arranging of the first equation, we can see that:

    [tex] v_1 = \frac{(0.17)(4.0) - (0.16)v_2 cos(30)}{(0.17) cos(60)} [/tex]

    Now sub this value of [tex]v_1[/tex] into equation two so that the only variable in the equation is [tex]v_2[/tex].

    Then you can solve that equation so you have a numerical value for [tex]v_2[/tex]. Sub that numerical value in for equation 1, then you can solve for the numerical value of [tex]v_1[/tex]
     
  6. Oct 8, 2008 #5
    Okay I'll help you out a little more here then. Sorry I made an error too, v3 = v1 and v4 = v2
    I said they are different components, but they only become the individual componenets once multiplied by their respective sine or cosine.

    [tex]
    (0.17)v_1 cos(60) + (0.16)v_2 cos(30) = (0.17)(4.0) [/tex]

    [tex] (0.17)v_1 sin(60) + (0.16)v_2 sin(30) = 0 [/tex]

    By simple re-arranging of the first equation, we can see that:

    [tex] v_1 = \frac{(0.17)(4.0) - (0.16)v_2 cos(30)}{(0.17) cos(60)} [/tex]

    Now sub this value of [tex]v_1[/tex] into equation two so that the only variable in the equation is [tex]v_2[/tex].

    Then you can solve that equation so you have a numerical value for [tex]v_2[/tex]. Sub that numerical value in for equation 1, then you can solve for the numerical value of [tex]v_1[/tex]
     
  7. Oct 9, 2008 #6
    oh god thats brilliant. thank you!!
     
  8. Oct 9, 2008 #7
    Its been a while since ive done algerbra. With that being said ive gotten it down to the y axis equation with Vc being the cue ball and Vs being the stationary ball ive figured Vc = 6.3698Vs

    0=(0.17)(6.3698)VsSin60 + (0.16)VsSin30 as my equation to find Vs

    Could I get some assistance on how to fudge with this to single out Vs?

    Thanks!
     
  9. Oct 9, 2008 #8
    Sorry to break it to you but I think you've gotten the value of Vc wrong.

    Using this: [tex] v_c = \frac{(0.17)(4.0) - (0.16)v_s cos(30)}{(0.17) cos(60)} [/tex]

    You can see it is the same as [tex] v_c = \frac{(0.17)(4.0)}{(0.17) cos(60)} - \frac{(0.16)v_s cos(30)}{(0.17)cos(60)} [/tex]

    This can be solved to: [tex] v_c = 0.8 + 1.63v_s [/tex]

    Sub this value of Vc into original equation so you have: [tex] (0.17)(0.8 + 1.63v_s) sin(60) + (0.16)v_s sin(30) = 0 [/tex]

    This equals : [tex] (0.17)(0.8)sin(60) + v_s[(1.63)(0.17)sin(60) + (0.16)sin(30)] = 0 [/tex]

    Then just subtract the term without Vs on it from both sides of the equation and divide both sides by the terms in the square brackets.
     
  10. Oct 10, 2008 #9
    This can be solved to: [tex] v_c = 0.8 + 1.63v_s [/tex]

    The first half would = 8

    But wouldnt the final equation for Vc = 8 - 1.63Vs?

    Why the flip in signs? Unless of course your accounting for the fact the this part of the y axis is negative, so your making it positive ahead of time?
     
  11. Oct 10, 2008 #10
    Yeah shame on me, I clearly wasn't concentrating enough when I wrote this. It's both 8.0 (not 0.8) and it should be positive (not negative).
     
  12. Oct 10, 2008 #11
    Yeah thats no problem :) IM just curious as to why its positive instead of negative since in the original Vc equation is half an equation - the other half?
     
  13. Oct 10, 2008 #12
    Yeah that's a typo. I'm cursing myself at the moment. It's meant to be negative (not positive)

    fixed.
     
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