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Scattering Angle

  1. Jun 1, 2004 #1
    Here I go again,

    Problem: Suppose we have a particle of mass m moving with speed v that collides with another particle of mass M which is initially at rest. Show that the scattering angle of m after collision can be anywhere between 0 and 180 degress if m < M. Determine the maximum scattering angle when m > M.

    Well, first of all I have no idea what they mean by the 'scattering angle'. I assume it's the angle between the old and new velocity vectors of m.

    Let [tex]\vec{v} = (v, 0)[/tex] be the velocity vector of m before colliding, [tex]\vec{v'} = (v'_x, v'_y)[/tex] be the velocity vector of m after colliding, and [tex]\vec{u} = (u_x, u_y)[/tex] be the velocity vector of M after the collision.

    Using conservation of momentum:
    [tex](mv, 0) = m(v'_x, v'_y) + M(u_x, u_y)[/tex]​
    [tex]0 = mv'_y + Mu_y[/tex]​
    [tex]mv = mv'_x + Mu_x[/tex]​
    Using conservation of kinetic energy (since this is an elastic collion):
    [tex]mv^2/2 = mv'^2/2 + Mu^2/2 \rightarrow Mu^2/m = v^2 - v'^2[/tex]​

    Now, there are three equations and four unknowns (not counting m < M or M > m). I can derive an equation for the scattering angle in terms of of the velocities, but it doesn't tell me much. What can I do?

    e(ho0n3
     
  2. jcsd
  3. Jun 1, 2004 #2

    Doc Al

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    Staff: Mentor

    Yes. A scattering angle of 180° means it bounces straight backwards.

    If you want to deduce anything about the scattering angle, then you'd better write your conservation of momentum equations in terms of it. For example:
    [tex]mv = mv_1cos\theta + Mv_2cos\alpha[/tex]
    where θ is the scattering angle.

    etc...
     
  4. Jun 2, 2004 #3
    Unfortunately, changing my expression doesn't do much for me as I still end up with a strage equation that I can't seem to analyze very well. I suppose that [tex]\alpha[/tex] in your expression varies from 0 to 90 degrees with respect to [tex]\vec{v}[/tex], but that still doesn't help me much because I don't know [tex]v_1[/tex] and [tex]v_2[/tex]. I'll go over what I have again, but I'm just not getting it.

    e(ho0n3
     
  5. Jun 2, 2004 #4

    Doc Al

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    Staff: Mentor

    Show me what you've got. You should have two momentum equations (horizontal and vertical components). Plus the KE equation.
    Why are you supposing? The range of θ (the scattering angle) is what you're supposed to be figuring out. Eliminate α.

    Here's a hint for you. Write your two equations for momentum. (I already gave you one.) Move all the terms containing θ and α to separate sides. For example, one would be:
    [tex]mv - mv_1cos\theta = Mv_2cos\alpha[/tex]
    What happens if you square both sides of this equation and the other momentum equation, then add them? Things will simplify. Don't give up.
     
  6. Jun 2, 2004 #5
    Good tip. That never came to mind.
    From conservation of momentum:
    [tex]mv - mv'cos(\theta) = Mucos(\alpha)[/tex]​
    [tex]m^2(v^2 - 2vv'cos(\theta) + v'^2(cos(\theta)^2) = (Mucos(\alpha))^2[/tex]​
    [tex]0 = mv'sin(\theta) + Musin(\alpha) \rightarrow -mv'sin(\theta) = Musin(\alpha)[/tex]​
    [tex]m^2(v^2 - 2vv'cos(\theta)) + (mv')^2(1 - sin(\theta)^2) = (Mucos(\alpha))^2[/tex]​
    [tex]m^2(v^2 - 2vv'cos(\theta)) + (mv')^2 = (Mu)^2[/tex]​
    From conservation of kinetic energy:
    [tex]m(v^2- v'^2) = Mu^2[/tex]​
    Putting it all together:
    [tex]m^2(v^2 - 2vv'cos(\theta)) + (mv')^2 = Mm(v^2 - v'^2)[/tex]​
    [tex]cos(\theta)= \frac{v'^2(M + m) - v^2(M - m)}{2mvv'}[/tex]​

    Assuming this is right, [tex]v'[/tex] can be vary from -v (the particle bounces back) to v (the particle continues as if it had never collided (although this sounds kind of fishy so maybe I should say up to and excluding v)). Thus, [tex]cos(\theta)[/tex] varies from -1 to 1, and thus [tex]\theta[/tex] can vary from 0 to 180 degrees. Notice that I haven't made any assumptions about M versus m here. What is the significance if m < M or m > M?

    e(ho0n3
     
  7. Jun 2, 2004 #6

    Doc Al

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    Staff: Mentor

    you're getting closer!

    You're almost there. Here's the final hint. Instead of solving for cosθ (and then having to give a handwaving argument), realize that that first equation can be rewritten as a quadratic in v'. Now what had better be true in order for that quadratic to have a solution?

    If you do it right, you will see the significance of m versus M.
     
  8. Jun 3, 2004 #7
    I'm starting to like this problem. I'm not really sure what you mean by "a quadratic in v'", but anyways...rewriting the first equation into
    [tex]v^2(1-\frac{M}{m}) - 2vv'cos(\theta) + v'^2(1+\frac{M}{m}) = 0.[/tex]​
    has a solution if
    [tex]cos^2(\theta) \geq (1-\frac{M}{m}) (1+\frac{M}{m}) = 1-\frac{M^2}{m^2}[/tex]​
    or
    [tex]1 - sin^2(\theta) \geq 1 - \frac{M^2}{m^2} \rightarrow sin(\theta) < \frac{M}{m}[/tex]​
    So, since M/m > 1, then our scattering angle can take any value from 0 to 180 (I hope that did it). Suppose m > M. The maximum angle [tex]\theta[/tex] must satisfy
    [tex]cos^2(\theta) = 1-\frac{M^2}{m^2}[/tex]​

    Thanks you Doc Al,
    e(ho0n3
     
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