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Homework Help: 2D elastic collision

  1. Jan 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball. If the first ball moves away with angle 30° to the original path, determine:

    a. the speed of the first ball after the collision.

    b. the speed and direction of the second ball after the collision.

    2. Relevant equations
    px = p'x (momentum conservation in the x direction)
    v1x = v1'cosθ + v2'cosθ

    piy = pfy (momentum conservation in the y direction)
    0 = v1'sinθ + v2'sinθ

    Ki = Kf (conservation of kinetic energy)
    v1^2 = v1'^2 + v2'^2

    3. The attempt at a solution
    I've attempted solving it in a number of values, but am finding the math challenging and unable to isolate variables. First I attempted to eliminate theta:
    v1x = v1'cosθ + v2'cosθ
    v1x - v1'cosθ = v2'cosθ ---square--
    v1x^2 - v1'^2cos^2θ = v2'^2cos^2θ

    0 = v1'sinθ + v2'sinθ
    -v1'^2sin^2θ = v2'^2sin^2

    add x and y
    v1x^2 - v1'^2cos^2θ - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
    v1x^2 - v1'^2(cos^2θ + sin^2θ) = v2'^2(cos^2θ + sin^2θ)
    v1^2 - v1'^2 = v2'^2
    (v1 - v1')(v1 + v1') = v2'^2

    Arranging Energy Conservation, v1^2 = v1'^2 + v2'^2 :
    v1^2- v1'^2 = v2'^2
    (v1 - v1')(v1 + v1') = v2'^2

    I feel like I've done something wrong up until this point because everything I get after just doesn't make sense to me. I'm not sure what to equate with what.

    Help is greatly appreciated!
  2. jcsd
  3. Jan 11, 2016 #2

    Ray Vickson

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    These types of problems are typically much easier to do by first transforming from the original frame of reference (the lab frame) into the so-called center-of-mass (CM) [or center-of-momentum (COM)] frame, in which the total initial momentum = 0. For identical masses, that means that the initial velocities of balls 1 and 2 are exactly opposite, and their kinetic energies are equal in the CM frame. For a perfectly elastic collision, their individual kinetic energies remain the same after the collision, and the new velocities are exact opposites again. So, if you know the deflection angle of ball 1 in the lab frame, you can set up an equation to determine the deflection angle of ball 1 in the CM frame, so find the final (vector) velocities of balls 1 and 2 in the CM frame. Then you can transform those back into the lab frame.

    OK, that method takes a bit more work, than the all-lab-frame method (the one you are using), but it is typically more straightforward.

    See. eg., http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node52.html or
  4. Jan 11, 2016 #3
    yes i highly support the CM method, deduction are much more straightforward, and btw the angle on an elastic colision between v2' and v1' when particles have the same mass is 90º, allways you seemed to have figured that out anyway...

    But the Cm method is defenitely the best, i dont think it takes more work. the lab frame method has a ton of formulas and deductions.
    but yeah your way seems fine
    this should be (v1x - v1'cosθ)^2 = v2'^2cos^2θ
    then you do everything else you have done, add the sine and cosine equations, solve for v2' and then substitute that into the kinectic energy and solve for v1', and everything else is just a matter of substitution now i think.

    but i highly recomend understanding and using the CM method, as @Ray Vickson said... it's so much simpler, in my opinion
  5. Jan 11, 2016 #4
    Thank you for your reply. For some reason the prof omitted CM/COM from the course. However, if it makes it easier than I am going to give it a try because what I am doing is not getting me too far.

    Thanks for the reference/examples as well!
  6. Jan 11, 2016 #5
    Wicked, thanks WrongMan. I'm going to correct my math and see if that answers that. But then definitely exploring the CM method!
  7. Jan 11, 2016 #6
    Right now I am not completely following the CM context - probably because my brain is too tired for new material. However, I tried the math again and it looks ugly, but I am hoping it's on the right track:

    (v1 - v1'cosθ)^2 - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
    v1^2 - 2v1v1'cosθ + v1'^2cos^2θ - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
    v1^2 - 2v1v1'cosθ - v1'^2(cos^2θ + sin^2θ) = v2'^2(cos^2θ + sin^2θ)
    v1^2 - 2v1v1'cosθ - v1'^2 = v2'^2
  8. Jan 11, 2016 #7
    If v2^2 = v1^2 - 2v1v1'cosθ - v1'^2

    and (v1 - v1')(v1 + v1') = v2'^2

    Can I equate them and say:

    v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1')


    I'm quite certain I will chuckle once I figure out the other method.
  9. Jan 11, 2016 #8
    I may have done something wrong again because I have:

    v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1')
    - 2v1v1'cosθ = 0
  10. Jan 11, 2016 #9
    yes, something is wrong, im trying to see what... might have to bust out the old pen and paper :D
  11. Jan 11, 2016 #10
    i think i've figured it out... source is way back in your formulas...

    its 0 = - v1'sinθ + v2'sinθ
    vertical velocity must cancel out
    Last edited: Jan 11, 2016
  12. Jan 11, 2016 #11
    Ok now:
    arrange equations so that all v1 coomponents are on the same side, and v2 are on the opposite.
    so you get:
    (x)=> (V1-V1'cos(θ))^2=V2'^2cos^2(θ) <=> V1^2 -2V1V1'cos(θ) +V1'^cos^2(θ)=V2'^2cos^2(θ)
    (y)=> V1'^2sen^2(θ)=V2'^2sen^2(θ)
    add them
    V1^2 - 2V1V1'cos(θ) + V1'^2cos^2(θ) V1'^2sen^2(θ)= V2'^2cos^2(θ)+V2'^2sen(θ)
    solve for v2'
    v1^2-2v1v1'cos(θ) +v1'^2 = v2'^2

    now its okay, not sure were you messed up

    ok your mistake was the one i stated above
    that made you commit this one:
    the sign on -v1'^2sin^2(θ) is not a "-"
    and bellow you had +v1'^2 and a -v1^2 multiplying with cos and sin respectivly, and you should have felt something was wrong when applying the distributivity property, this mistake could have passed unoticed if you had chosen to use +v1 instead of -v1 bellow so good thing it happened
  13. Jan 11, 2016 #12


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    Why did you want to eliminate theta, when it was given as 30°? substitute the values of sinθ and cosθ into your equations. They can be solved easily.
    The problem did not state that it was an elastic collision.
  14. Jan 11, 2016 #13


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    But it did state that it's elastic. (Easy to miss that.)
  15. Jan 11, 2016 #14


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    Oh, I see. But the angles are different. The first equations are not correct. The angle of the first ball is given, but the angle of the second ball is unknown.
    Last edited: Jan 11, 2016
  16. Jan 11, 2016 #15
    in elastic colisions if particles with the same mass the angle bettween final velocity vectors is 90º degrees is it not?
  17. Jan 11, 2016 #16


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    Yes, but the OP used the same angle for both balls.
    Last edited: Jan 11, 2016
  18. Jan 11, 2016 #17
    oh right, i don't know how i missed that... he can find it easly after finding v2' ...
  19. Jan 11, 2016 #18


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    @Jam51: The velocities of the balls enclose different angles with the original direction. It is easy to prove what WrongMan said, that the angles add up to 90° when particles of equal mass collide elastically.

    You have the following equations:
    v1=u1 cosθ +u2 cosΦ
    0=u1 sinθ-u2 sin Φ
    Square the first two equations, and add them.
    v12=(u1 cosθ +u2 cosΦ)2+(u1 sinθ-u2 sin Φ)2 ---> v12=u12+u22+2 u1 u2 (cosθ cosΦ-sinθ sinΦ)
    Because of conservation of energy, v12=u12+u22, so 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0. But the term in the parentheses is equal to cos(θ+Φ) =0 so
  20. Jan 12, 2016 #19
    @ehild I'm following everything until "But the term in the parentheses is equal to cos(θ+Φ) =0 so θ+Φ=90°." I understand you're proving what was stated in previous posts, which would make Φ = 60°. However, I have no idea how you went from 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0 to cos(θ+Φ) =0.
  21. Jan 12, 2016 #20
    that is a trig identitiy:
    cos(x +/- y) = cos(x)cos(y) +/- sen(x)sen(y)

    there is probably a way to deduce it but i dont know it...
  22. Jan 12, 2016 #21


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    We had the equations
    v12=u12+u22+2 u1 u2 (cosθ cosΦ-sinθ sinΦ) and
    They can be both true if cosθ cosΦ-sinθ sinΦ=0.
    The cosine of the sum of two angles can be derived according to the figure:
    There are three right triangles, yellow, blue and green. The hypotenuse of the yellow triangle is 1. Those of the blue one and green ones are cosΦ and sinΦ. Cos(θ+Φ) is the length of the red line, which is the difference between the blue line and green line.

  23. Jan 13, 2016 #22
    Oh, thank you. It's been a long time since I took any math, which makes application of physics theory at times challenging.
    I went back and solved for v2' and v1' using θ = 30 and Φ=60°.
    I hope this is ok:
    0 = v1' sinθ - v2' sin Φ
    0 = v1' sin30 - v2' sin 60
    0 = 0.5v1' - 0.866v2'
    v2' = 0.5v1' / 0.866

    v1 = v1' cosθ + v2' cosΦ
    3 = v1' cos30 + (0.5v1' / 0.866)cos60
    3 = 0.866v1' + (0.5v1' / 0.866)0.5
    3 = 0.999v1'/0.866
    v1' = 2.6 m/s

    v1'^2 + v2'^2 = v1^2
    v2'^2 = 3^2 - 2.6^2 = 1.5 m/s

    Does that look right?
  24. Jan 13, 2016 #23

    Yeah it looks correct...
    just two side notes, you should only substitute for values in the very end, its a good habit to keep, example: for v1' it would look like v1'=3/(cos(θ)+(sen(θ)/tg(Φ)))
    and, if the masses were diferent, and you knew them both, you would have 3 unknowns (phi v1' and v2') and 3 equations, and and thus you would be able to solve the problem, using the method we previously discussed in the thread.
  25. Jan 13, 2016 #24
    Yeah, that makes sense so that I'm not already rounding too before I even get to the answer...

    Thanks a lot for all your help!
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