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Elastic Collision of Two Balls

  1. Jun 29, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data

    billiardballsnew2.png

    A white billiard ball with mass mw = 1.53 kg is moving directly to the right with a speed of v = 3.25 m/s and collides elastically with a black billiard ball with the same mass mb = 1.53 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 51° and the black ball ends up moving at an angle below the horizontal of θb = 39°.

    1) What is the final speed of the white ball?

    2. Relevant equations
    p=mv
    k=0.5mv2
    Vcm = (M1V1 + M2V2)/ (M1+M2)

    3. The attempt at a solution

    I don't really have any idea how to solve this.

    I started by first finding the velocity of the center of mass:
    Vcm = (M1V1 + M2V2)/ (M1+M2)
    Vcm = (1.53*3.25 + 1.53*0) / (1.53 + 1.53)
    Vcm = 4.9725/3.06 = 1.625 m/s

    And the initial momentum: P1i = M1V1i = 1.53*3.25 = 4.97 kg*m/s

    Since the collision is elastic, I know the final kinetic energy is equal to the initial, which means that K1f + K2f = K1i
    K1f + K2f = 0.5M1V1i2 = 0.5(1.53)(3.252) = 8.08 J

    Since momentum is conserved, I also know that the vertical momentum M1V1fy + M2V2fy = 0
    And that the horizontal momentum of the two balls after the collision has to equal the initial momentum: M1V1fx + M2V2fx = 4.97 kg*m/s

    In the CoM frame, the 2 balls collide and then head off at the same speed, but for the life of me I can't figure out how to transform the speed in the CoM frame (1.625 m/s for each ball) back to the original frame without knowing what angle they are moving at after the collision in the CoM frame. And I don't know how to get the angles in the CoM frame from the angles in the original frame.
     
  2. jcsd
  3. Jun 29, 2015 #2

    Drakkith

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    Oh look... I see two equations with V1 and V2... I bet I can solve for V1 and V2 now... :rolleyes:
     
  4. Jun 29, 2015 #3

    Nathanael

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    The problem seems to give more information than needed, which means that once you solve it you should check that all the information is consistent.

    There are two unknowns: the final velocity of each ball. But there are three constraints: conservation of momentum in the x-direction, in the y-direction, and conservation of energy.

    This is the bulk of the problem done here. All you need to do is write the x and y components in terms of the respective angles and solve. Then I would check that the collision really is elastic.

    I do like the idea, but in this case it makes things considerably more difficult.
     
  5. Jun 29, 2015 #4

    Nathanael

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    It seems I'm late :oldshy:
     
  6. Jun 29, 2015 #5

    Drakkith

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    I've typed up about 2-3 times more posts asking for homework help than I've actually posted. I usually get everything typed up and organized and then see some connection I didn't see on my notebook that allows me to solve it. So infuriating!
     
  7. Jun 29, 2015 #6

    Nathanael

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    Oh, I've done exactly the same thing several times myself! (Almost posted a question today and another one last night.)
    It just goes to show that explaining the problem to someone else can often make you understand it better.

    It also doesn't help that I'm extremely disorganized on paper o0) (I like to use every inch of space in my notebook.)
     
  8. Jun 30, 2015 #7

    Drakkith

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    Anyways, I did the following to get the answer, but I in the 4th paragraph, I accidentally put 0.8V2 instead of -0.8V2, which somehow gave me the right answer. Now the math isn't working out if I keep it -0.8V2...

    M1V1fy + M2V2fy = 0
    M1V1fx + M2V2fx = 4.97 kg*m/s

    M1(sin51*v1) + M2(sin39*V2) = 0
    M1(cos51*v1) + M2(cos39*V2) = 4.97

    1.2V1 + 0.96V2 = 0
    0.96V1 + 1.2V2 = 4.97

    1.2V1 = -0.96V2, V1 = -0.8V2 (Originally I forgot the negative and put 0.8V2, which ended up working out fine)
    0.96(-0.8V2) + 1.2V2 = 4.97, 0.432V2= 4.97, V2 = 11.5 (Not correct according to my homework program)

    What have I done wrong here?
     
  9. Jun 30, 2015 #8

    Nathanael

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    One of the angles should technically be negative.

    The way I think about it is just that Vwy should equal Vby in magnitude (because we already know they are opposite in direction).
    So I would put: Vwy=Vby (Where both Vy's are positive.)

    But your equation is: Vwy=-Vby which means one of these Vy's should be negative (because they are in opposite directions).

    [The problem takes care of itself if you take one of the angles to be negative. The sin(-θ) becomes -sin(θ) and the cos(θ) in the other equation is unchanged because cos(-θ)=cos(θ)]
     
  10. Jun 30, 2015 #9

    Drakkith

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    Thanks!
     
  11. Jun 30, 2015 #10

    ehild

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    You can also notice that 39° + 51°= 90°. You need to work with a single angle. Do the derivation symbolically and plug in the data at the end.
     
  12. Jun 30, 2015 #11

    Drakkith

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    I noticed that, but I don't know what you mean when you say work with a single angle.
     
  13. Jun 30, 2015 #12

    ehild

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    sin(90-α)=cos(α) and cos(90-α)=sin(α). You can use the sine and cosine of a single angle, either 59°or 31°.
    It can be shown that when one ball collides with an other one in rest, and both balls have the same mass, and the collision is elastic, the velocities after the collision enclose 90°angle, or the balls exchange their velocities. The problem is solvable because the angles were correctly given, their sum was 90°.
     
    Last edited: Jun 30, 2015
  14. Jun 30, 2015 #13

    Drakkith

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    Oh I see. Thanks, Ehild.
     
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