An alpha particle approaches a ${}^{197}Au$nucleus with a speed of $1.50\times {10}^{7}m/s$. As FIGURE CP $37.47$shows, the alpha particle is scattered at a $49°$angle at the slower speed of $1.49\times {10}^{7}m/s$. In what direction does the ${}^{197}Au$nucleus recoil, and with what speed?

We have to given ${}^{197}Au$nucleus with a speed of $1.50\times {10}^{7}m/s$and the alpha particle is scattered at a $49°$angle at the slower speed of $1.49\times {10}^{7}m/s$.

We can start solution with momentum conservation:
$m_{\alpha }{v}_{\alpha i}=m_{\alpha }{v}_{\alpha f}\cos \theta +m_{nucleus}{v}_{nucleusf}\cos \theta$

$0=m_{\alpha }{v}_{\alpha f}\sin \theta -m_{nucleus}{v}_{nucleusf}\sin \theta$

$v_{nucleus}\cos \varphi =\frac{m_{\alpha }}{m_{nucleus}}{v}_{\alpha i}-\frac{m_{\alpha }}{m_{nucleus}}{v}_{\alpha f}\cos \varphi$

$v_{nucleusf}\cos \varphi =\frac{4u}{197u}\times 1.5\times {10}^7-\frac{4u}{197u}\times 1.49\times {10}^7\times \cos 49°$

localid="1651080986204" $v_{nucleusf}\cos \varphi =1.06\times {10}^5\frac{\text{m}}{\text{s}}$

localid="1651080990129" $v_{nucleusf}\sin \varphi =\frac{m_{\alpha }}{m_{nucleus}}{v}_{\alpha f}\sin \theta$

$v_{nucleusf}\sin \varphi =\frac{4u}{197u}\times 1.49\times {10}^7\times \sin {49}^{\circ }$

localid="1651080994529" $v_{nucleusf}\sin \varphi =2.28\cdot {10}^5\frac{\text{m}}{\text{s}}$

Now from simple geometry as:

$\begin{array}{rr}\tan \varphi & =\frac{v_{nucleusf}\sin \varphi }{v_{nucleusf}\cos \varphi }\\ \varphi & =\arctan \frac{v_{nucleusf}\sin \varphi }{v_{nucleusf}\cos \varphi }\\ \varphi & =\arctan \frac{2.28\cdot {10}^5}{1.06\cdot {10}^5}\\ \varphi & ={65.07}^{\circ }\end{array}$

The final speed of the nucleus finding from the expression:

$$\begin{gathered} v_{nucleusf}=\sqrt{{\left(v_{nucleusf}\cos \varnothing \right)}^2+{\left(v_{nucleusf}\sin \varnothing \right)}^2} \\ {v}_{nucleusf}=\sqrt{{\left(1.06\times {10}^5\right)}^2+{\left(2.28\times {10}^5\right)}^2} \\ {v}_{nucleusf}=2.51\times {10}^5\frac{m}{s} \end{gathered}$$