• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Scattering Angle

  • Thread starter e(ho0n3
  • Start date
1,356
0
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
 

Doc Al

Mentor
44,741
1,031
e(ho0n3 said:
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.
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...
 
1,356
0
Doc Al said:
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.
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
 

Doc Al

Mentor
44,741
1,031
e(ho0n3 said:
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.
Show me what you've got. You should have two momentum equations (horizontal and vertical components). Plus the KE equation.
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].
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.
 
1,356
0
Doc Al said:
[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.
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
 

Doc Al

Mentor
44,741
1,031
you're getting closer!

e(ho0n3 said:
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]​
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.
 
1,356
0
Doc Al said:
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.
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
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top