I am having trouble answering the following question, please could you help! Thank you in advance for any assistance you can give.
Consider classical Rutherford scattering of a particle with mass $m$ and initial speed $v_0$ from a potential
\begin{equation}
V(r) = \frac{\alpha}{r}
\end{equation}
where $\alpha$ is some constant and $r$ is the location of the particle from the origin. Starting from Newton's second law, show that
\begin{equation}
|\Delta {\bf{p}}| = \frac{2 \alpha}{v_0 b} \cos \Big(\frac{\Theta}{2} \Big).
\end{equation}
Note that $b$ is the impact parameter and $\Theta$ is the scattering angle.
Please note that I have already shown that "from geometry" the change in momentum is $|\Delta {\bf{p}}| = 2 p \sin(\Theta / 2)$, and that $b v_0 = r^2 \frac{d \theta}{dt}$ where $t$ is time and $\theta$ is the angle $\angle({\bf{r}},{\bf{r^*}})$ where ${\bf{r^*}}$ is the point of closest approach. I am unsure however if the the above two equations will be of assistance.
Best Answer
Certainly late for your homework problem, but figure below shows the schematic of elastic scattering. The scattering angle is $\Theta$. The momentum transfer vector is $\Delta \bf{p}$.
Since your question starts with potential, we shall obtain the central force first:
\begin{align} F(r) &= \frac{dV}{dr} \\ &= -\frac{\alpha}{r^2} \end{align}
Since the force is central, at any time the force in the direction of $\Delta \bf{p}$ is $F \cos\phi$, so the total momentum transfer is \begin{align} \int_{-\infty}^{\infty} F \cos{\phi}\,dt \end{align}
Using one of your information $bv_0=r^2\frac{d\phi}{dt}$ (which is obtainable from conservation of angular momentum) we make the change of variable
\begin{align} dt = \frac{r^2}{bv_0}d\phi \end{align}
. In the limit $t\rightarrow \pm \infty$, the angle $\phi$ approaches $\pm(\pi-\Theta)/2$ as measured from the direction of momentum transfer axis. The integration becomes,
\begin{align} \int_{-\frac{\pi-\Theta}{2}}^{\frac{\pi-\Theta}{2}} \left(-\frac{\alpha}{r^2}\right)\cos{\phi}\,\frac{r^2}{bv_0}d\phi &= \frac{\alpha}{bv_0}\left[\sin\phi\right]^{\frac{\pi-\Theta}{2}}_{-\frac{\pi-\Theta}{2}} \\ &= \frac{2\alpha}{bv_0}\cos{\left(\frac{\Theta}{2}\right)} \end{align}