# Homework Help: Scattering Angle

1. Jun 1, 2004

### e(ho0n3

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

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

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. Jun 1, 2004

### 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:
$$mv = mv_1cos\theta + Mv_2cos\alpha$$
where θ is the scattering angle.

etc...

3. Jun 2, 2004

### e(ho0n3

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 $$\alpha$$ in your expression varies from 0 to 90 degrees with respect to $$\vec{v}$$, but that still doesn't help me much because I don't know $$v_1$$ and $$v_2$$. I'll go over what I have again, but I'm just not getting it.

e(ho0n3

4. Jun 2, 2004

### 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:
$$mv - mv_1cos\theta = Mv_2cos\alpha$$
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.

5. Jun 2, 2004

### e(ho0n3

Good tip. That never came to mind.
From conservation of momentum:
$$mv - mv'cos(\theta) = Mucos(\alpha)$$​
$$m^2(v^2 - 2vv'cos(\theta) + v'^2(cos(\theta)^2) = (Mucos(\alpha))^2$$​
$$0 = mv'sin(\theta) + Musin(\alpha) \rightarrow -mv'sin(\theta) = Musin(\alpha)$$​
$$m^2(v^2 - 2vv'cos(\theta)) + (mv')^2(1 - sin(\theta)^2) = (Mucos(\alpha))^2$$​
$$m^2(v^2 - 2vv'cos(\theta)) + (mv')^2 = (Mu)^2$$​
From conservation of kinetic energy:
$$m(v^2- v'^2) = Mu^2$$​
Putting it all together:
$$m^2(v^2 - 2vv'cos(\theta)) + (mv')^2 = Mm(v^2 - v'^2)$$​
$$cos(\theta)= \frac{v'^2(M + m) - v^2(M - m)}{2mvv'}$$​

Assuming this is right, $$v'$$ 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, $$cos(\theta)$$ varies from -1 to 1, and thus $$\theta$$ 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

6. Jun 2, 2004

### 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.

7. Jun 3, 2004

### e(ho0n3

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
$$v^2(1-\frac{M}{m}) - 2vv'cos(\theta) + v'^2(1+\frac{M}{m}) = 0.$$​
has a solution if
$$cos^2(\theta) \geq (1-\frac{M}{m}) (1+\frac{M}{m}) = 1-\frac{M^2}{m^2}$$​
or
$$1 - sin^2(\theta) \geq 1 - \frac{M^2}{m^2} \rightarrow sin(\theta) < \frac{M}{m}$$​
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 $$\theta$$ must satisfy
$$cos^2(\theta) = 1-\frac{M^2}{m^2}$$​

Thanks you Doc Al,
e(ho0n3