Coulomb collision

Question

I was reading an article by N. Bohr and came upon the following problem (the following wording is actually taken from a book by Thompson - Conduction of Electricity Through Gases):

Let $M_1, M_2$ be the masses of the corpuscles $A$ and $B$ respectively; we shall suppose the velocities of the colliding corpuscles so great that in comparison the corpuscles in the atom may be regarded as at rest. Let $V$ be the velocity of $B$ before the collision, $b$ the perpendicular let fall from $A$ on $V$. Then if $2\theta$ be the angle through which the direction of relative motion is deflected by the collision, we can easily show that, taking the force between the corpuscles to be $e^2/r^2$, $$ \sin^2\theta=\frac{1}{1+\frac{b^2V^4}{e^4}\left(\frac{M_1M_2}{M_1+M_2}\right)^2}. $$ _{Original text.}

So I wanted to show the formula but as both Bohr and Thompson say that it is easy to show, I assume it is rather complicated. Anyway, can someone help me with graphical representation of the situation? Namely, what $b$ is? I had expected that the situation is like:

But after reading the text "perpendicular let fall" I doubt it.

Answer 1

I believe b is the impact parameter. So b should be the perpendicular distance between the asymtote you drew and the fixed ion. The Wikipedia link has a picture that is pretty clear, although the picture in the link is illustrating Coulomb REPULSION, while your problem involves Coulomb attraction..