First, I shall refer you to this problem of elastic Rutherford scattering (your problem is identical to it because of you use the same force equation, and the angles in both repulsive and attractive are identical), where the question states two results:

$$|\Delta {\bf{p}}| = 2 p \sin\left(\frac{\Theta}{ 2}\right)$$ and $$|\Delta {\bf{p}}| = \frac{2 \alpha}{v_0 b} \cos \Big(\frac{\Theta}{2} \Big)$$

The first can be derived from triangular cosine rule, while the second has been worked out in my response to that same problem in that link.

Using $\sin^2(\Theta/2 )+\cos^2(\Theta/2 )=1$ and $p=\mu v_0$, you should be able to obtain $$|\Delta {\bf{p}}| = \frac{2 \mu v_0}{\sqrt{1+z^2}} $$

where $z=b\mu v^2_0/\alpha$. In your case, replacing $q=|\Delta {\bf{p}}|$ and $v=v_0$ will give your desired answer.

First, I shall refer you to this problem of elastic Rutherford scattering (your problem is identical to it because of you use the same force equation, and the angles in both repulsive and attractive are identical), where the question states two results:

$$|\Delta {\bf{p}}| = 2 p \sin\left(\frac{\Theta}{ 2}\right)$$ and $$|\Delta {\bf{p}}| = \frac{2 \alpha}{v_0 b} \cos \Big(\frac{\Theta}{2} \Big)$$

The first can be derived from triangular cosine rule, while the second has been worked out in my response to that same problem in that link.

Using $\sin^2(\Theta/2 )+\cos^2(\Theta/2 )=1$ and $p=\mu v_0$, you should be able to obtain $$|\Delta {\bf{p}}| = \frac{2 \mu v_0}{\sqrt{1+z^2}} $$

where $z=b\mu v^2_0/\alpha$. In your case, replacing $q=|\Delta {\bf{p}}|$ and $v=v_0$ will give your desired answer.