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Elastic collision between particles of equal mass

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central impact.


    2. Relevant equations
    Assume particle 1 travels in the x direction with speed v1(before) and strikes particle 2, which is at rest. After the collision, particle 1 makes an angle theta1 with the x axis, and particle 2 makes angle theta2. We need to show that theta1 + theta2 = pi/2.

    Conservation of linear momentum
    mv1(before) = SUM(mvi(after))
    x component: mv1(before) = mv1(after) cos(theta1) + mv2(after) cos(theta2)
    y component: 0 = mv2(after) sin(theta2) - mv1(after) sin(theta1)

    Conservation of kinetic energy
    .5mv1(before)^2 = .5mv1(after)^2 + .5mv2(after)^2


    3. The attempt at a solution
    I'm afraid I really don't know what to do with this. I notice that if theta1 + theta2 form a right angle, then by the Pythagorean theorem,

    v1(after)^2 + v2(after)^2 = [v1(after) sin(theta1) + v2(after) sin(theta2)]^2
    = v1(after)^2 sin^2(theta1) + (2)(v1(after))(v2(after)) sin(theta1) sin(theta2) + v2(after)^2 sin^2(theta2)

    But this doesn't really look very useful. Hint please?
     
  2. jcsd
  3. Aug 25, 2011 #2

    PeterO

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    In might be better if you use V for the initial velocity, then v1 ad v2 for the two parts of the velocity afterwards.
    Conservation of momentum
    mV = mv1 + mv2

    divide by mass and

    V = v1 + v2

    You could draw the triangle that shows that vector sum

    Now:
    Conservation of kinetic energy
    .5mV^2 = .5mv1^2 + .5mv2^2

    divide by .5m

    V^2 = v1^2 + v2^2

    so here is an interesting connection between the sides of your triangle. Note: the squaring removes the vector nature, so this expression relates the magnitudes only.
     
  4. Aug 26, 2011 #3
    Thanks for this reply -- it certainly helped to reduce the clutter in what I wrote!

    Obviously, from V^2 = v1^2 + v2^2, we can see that IF theta1 + theta2 = pi/2, then the magnitude V is the length of the hypotenuse of the right triangle formed from sides v1 and v2. And the converse is also true: if we could somehow know that the distance between the endpoints of the vectors v1 and v2 is V, that would show that theta1 + theta 2 = pi/2.

    So now we want to show that the distance between the endpoints of the vectors v1 and v2 is V.

    We can find this distance by adding the squares of the differences between the x and y components of v1 and v2:

    distance^2 = [v2sin(theta2) - (-v1sin(theta1)]^2 + [v2cos(theta2) - v1cos(theta1)]^2
    = v2^2sin^2(theta2) + 2v1sin(theta1)v2sin(theta2) + v1^2sin^2(theta1) + v2^2cos^2(theta2) - 2v1cos(theta1)v2cos(theta2) + v1^2cos^2(theta1)
    = v1^2(sin^2(theta1) + cos^2(theta1)) + v2^2(sin^2(theta2) + cos^2(theta2)) + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)
    = v1^2 + v2^2 + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)

    (The last transition is because sin^2(theta) + cos^2(theta) = 1.)

    Now, we want distance^2 = V^2, and we've seen that V^2 = v1^2 + v2^2. So we can get what we want if 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2) = 0.

    Is it true? Unfortunately, I don't know how to show this. Can somebody help? Am I even proceeding the right way?
     
  5. Aug 26, 2011 #4

    PeterO

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    isn't the vector sum of the momentums sufficient to show that the distance between the endpoints of the vectors v1 and v2 is V. or, in other words vectors v1 and v2 add up to the vector V? [I am pretty sure that should have been momenta]
     
  6. Aug 26, 2011 #5
    I don't know. It would certainly be nice.

    v1 = v1cos(theta1) - v1sin(theta1)
    v2 = v2cos(theta2) + v2sin(theta2)

    So

    v1 + v2 = v1cos(theta1) + v2cos(theta2) - v1sin(theta1) + v2sin(theta2)

    Because we have v2sin(theta2) - v1sin(theta1) = 0, the last two terms sum to 0.
    And because we have V = v1cos(theta1) + v2cos(theta2), we have V = v1 + v2, QED.

    And I guess this must be the right way to do it, but I don't really see the logic of it, because it is not in general true that the distance between the endpoints of two vectors can be found from their sum. Indeed, it is usually NOT true. It is true in the special case that the angle between the vectors is pi/2. So if we use this technique, aren't we assuming what we're trying to prove?

    A general procedure for finding the distance between the endpoints would be take the difference between the vectors. That's what I was trying to do, but it led to a mess I couldn't see my way out of.

    Put it this way. From V^2 = v1^2 + v2^2, we know that if v1 and v2 form a right angle, the distance between their endpoints is V. But there is no necessity in this. v1 and v2 could form any angle, and V^2 would just be the arbitrary sum of their squared magnitudes. Again, if v1 and v2 form a right angle, their vector sum will be the distance between their endpoints. But if not -- if, say, the angle is pi/4 -- then the vector sum will not be the distance between their endpoints. And if we then find that the vector sum is V, that will show nothing about the distance between their endpoints. Why can't it just be a coincidence that the vector sum V, if squared, is equal to v1^2 + v2^2?
     
  7. Aug 26, 2011 #6

    Dick

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    Your logic there is wrong. Like in v1 = v1cos(theta1) - v1sin(theta1). You aren't dealing with the quantities as vectors. V^2=v1^2+v2^2 comes from conservation of kinetic energy. Do you know what a dot product is?
     
  8. Aug 26, 2011 #7

    ehild

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    It is a bit shorter if you denote the original speed of the first ball by U, and the speeds after the collisions by v1 and v2. All your equations can be divided by m, and the energy equation multiplied by 2.Your equations become then

    U=v1cosθ1+v2cosθ2
    0=v1sinθ1-v2sinθ2
    U2=v12+v22
    Take the square of the first two equations and add them. Compare the result with the third one. Remember, you have to prove that θ1+θ2=pi/2, that is, cos(θ1+θ2)=0

    ehild
     
  9. Aug 28, 2011 #8
    Excellent. Thanks for this. (And I like the coding of your notation too!)

    So keeping your symbols, it goes like this:

    U2 = v12cos2θ1 + 2v1v2cosθ1cosθ2 + v22cos2θ2
    0 = v12sin2θ1 - 2v1v2sinθ1sinθ2 + v22sin2θ2

    So

    U2 = v12cos2θ1 + 2v1v2cosθ1cosθ2 + v22cos2θ2 + v12sin2θ1 - 2v1v2sinθ1sinθ2 + v22sin2θ2

    Since sin2θ + cos2θ = 1 and U2 = v12 + v22, we have

    U2 = U2 + 2v1v2cosθ1cosθ2 - 2v1v2sinθ1sinθ2

    Simplifying,

    sinθ1sinθ2 = cosθ1cosθ2

    Now in general, given f(Φ)f(Ψ) = g(Φ)g(Ψ), where Φ and Ψ range over multiple values and f(Φ) <> g(Φ), we can infer that f(Φ) = g(Ψ) and f(Ψ) = g(Φ). So sinθ1 = cosθ2 and sinθ2 = cosθ1. And this will be true just where θ2 = pi/2 - θ1, as is easy to see geometrically. Therefore, θ1 + θ2 = pi/2, QED.

    You remark that this could be done by showing that cos(θ1+θ2)=0. I haven't done that, so there must be some other way to do it as well. But this gets the job done, which is good enough for me.

    Concerning the idea that we could show that the distance between the endpoints of the vectors v1 and v2 = V (or in our new terminology, U) by showing that that is their vector sum, which was suggested by PeterO and supported by Dick, I think the procedure we just used can show that that won't work, as follows. Taking D to be the distance between the endpoints of v1 and v2, we have

    D2 = (v2sinθ2 - (-v1sinθ1))2 + (v2cosθ2 - v1cosθ1)2
    D2 = v12sin2θ1 + 2v1v2sinθ1sinθ2 + v22sin2θ2 + v12cos2θ1 - 2v1v2cosθ1cosθ2 + v22cos2θ2
    D2 = U2 + 2v1v2(sinθ1sinθ2 - cosθ1cosθ2)

    If D2 = U2, we could proceed as above to show that θ1 + θ2 = pi/2. But of course, whether D = U is just what we are trying to discover. Again, if we knew that θ1 + θ2 = pi/2, we could show that the second term of the last equation = 0 and therefore that D = U, but of course that θ1 + θ2 = pi/2 is what we are trying to prove to begin with. So this procedure begs the question, in line with my argument in an earlier post.
     
  10. Aug 29, 2011 #9

    ehild

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    It is the same as cosθ1cosθ2-sinθ1sinθ2=0

    You know the formula for the cosine of the sum of angles,

    cos(α+β) =cos(α)cos(β)-sin(α)sin(β)?

    You need the cosine of (θ1+θ2). And you also know that cos(90°)=0.

    ehild
     
  11. Aug 29, 2011 #10
    No, I didn't know that formula. Though I might have looked it up and spared myself a flight of logic! So from the formula and the fact that cosθ1cosθ2-sinθ1sinθ2=0, we have cos(θ12)=0, so θ12=pi/2, which is our goal.

    Nice!
     
  12. Aug 29, 2011 #11

    ehild

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    Just keep in mind that such formulae exist also for sin(α+β)=sin(α)cos(β)+cos(α)sin(β);
    and tan(α+β)=(tan(α)+tan(β))/(1-tan(α)tan(β)).:biggrin:
    The addition laws are not hard to prove geometrically. For the cosine, find the length AB in the picture.


    ehild
     

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    Last edited: Aug 29, 2011
  13. Aug 29, 2011 #12
    Yes, I did look these up in one of my math books and read the proof of the formula for cos(α+β). Thanks for the tip!
     
  14. Oct 8, 2011 #13
    that is not true, if you just consider that in that case direction of v1 is y-axis and v = 0
    "reductio ad absurdum" [[itex]\rightarrow\leftarrow[/itex]] and these precious hints give you an instant proof by contradiction that requires no math: angle of deflection cannot be [>/< than] different from 90°, as only π/ 2 meets those conditions
     
  15. Oct 12, 2011 #14

    PeterO

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    I am certainly happy that the two statements prove a right angle. No need to get too involved in trigonometry when a simple Pythagorus is offered.
     
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