91Ó°ÊÓ

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

We can start solution with momentum conservation:
$m_{\mathrm{Î\pm }}{v}_{\mathrm{Î\pm }i}=m_{\mathrm{Î\pm }}{v}_{\mathrm{Î\pm }f}\cos \mathrm{θ}+m_{nucleus}{v}_{nucleusf}\cos \mathrm{θ}$

$0=m_{\mathrm{Î\pm }}{v}_{\mathrm{Î\pm }f}\sin \mathrm{θ}−m_{nucleus}{v}_{nucleusf}\sin \mathrm{θ}$

$v_{nucleus}\cos \mathrm{Ï•}=\frac{m_{\mathrm{Î\pm }}}{m_{nucleus}}{v}_{\mathrm{Î\pm }i}−\frac{m_{\mathrm{Î\pm }}}{m_{nucleus}}{v}_{\mathrm{Î\pm }f}\cos \mathrm{Ï•}$

$v_{nucleusf}\cos \mathrm{ϕ}=\frac{4u}{197u}×1.5×{10}^7−\frac{4u}{197u}×1.49×{10}^7×\cos 49°$

localid="1651080986204" $v_{nucleusf}\cos \mathrm{ϕ}=1.06×{10}^5\frac{\text{m}}{\text{s}}$

localid="1651080990129" $v_{nucleusf}\sin \mathrm{Ï•}=\frac{m_{\mathrm{Î\pm }}}{m_{nucleus}}{v}_{\mathrm{Î\pm }f}\sin \mathrm{θ}$

$v_{nucleusf}\sin \mathrm{ϕ}=\frac{4u}{197u}×1.49×{10}^7×\sin {49}^{∘}$

localid="1651080994529" $v_{nucleusf}\sin \mathrm{Ï•}=2.28â‹\dots {10}^5\frac{\text{m}}{\text{s}}$

Now from simple geometry as:

$\begin{array}{rr}\tan \mathrm{Ï•}& =\frac{v_{nucleusf}\sin \mathrm{Ï•}}{v_{nucleusf}\cos \mathrm{Ï•}}\\ \mathrm{Ï•}& =\arctan \frac{v_{nucleusf}\sin \mathrm{Ï•}}{v_{nucleusf}\cos \mathrm{Ï•}}\\ \mathrm{Ï•}& =\arctan \frac{2.28â‹\dots {10}^5}{1.06â‹\dots {10}^5}\\ \mathrm{Ï•}& ={65.07}^{∘}\end{array}$

The final speed of the nucleus finding from the expression:

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