I'm trying to interpret a way to solve the following exercise:
My instinct is to note that the particle is moving to the right of frame $S$, so we use the formula:
$$u_y = \frac{u'_y}{\gamma (1+\frac{u'_x v}{c^2})}$$
Where $u_y$ is the $y$ component of the particle's velocity in $S$, $u'_y$ is the $y$ component of the particle's velocity in $S'$, and $u'_x$ is the particle's $x$ velocity in $S'$.
My question is dealing with resolving whether $u'_x$ is $0$ or $0.8c$.
$u'_x$, in the frame of the nucleus, would say it's equal to $0$ since in its rest frame, it is only moving in the $y'$ direction of $S'$.
I feel like this is definitely the correct logic, although I'm a bit confused as to why $u'_x$ in this case is $0$ while in question $1$ it's $0.8c$. It is perhaps obvious, but, just to be clear, am I correct in saying it's $0.8c$ in (a) because the particle is moving at $0.8c$ relative to $S'$ and $u'_x = 0$ in (b) because the particle is now moving only along $y$ relative to $S'$?
And thus the answer should be $u'_y =0.8/\gamma$.
