Is the following statement true?
If $\frac{\partial \dot{q}}{\partial q}=0$, then the conjugate momentum $p_q$ is conserved.
We know that conjugate momentum of $q$ is conserved if $\frac{\partial L}{\partial q} = 0$. If $\frac{\partial \dot{q}}{\partial q}=0$, applying chain rule implies $$ \frac{\partial L}{\partial q} = \frac{\partial L}{\partial \dot{q}} \frac{\partial \dot{q}}{\partial q} = 0$$ Hence, $p_q$ is conserved.
Am I correct?
The partial derivative of $\dot{q}$ with respect to $q$ does not make any sense in this context because they are independent variables (thus it is $0$ independent of whether conservation of a momentum holds). We know this because the Lagrangian in this case $$L(q,\dot{q},t)$$ and so any partials between the variables are $0$ in this context. The total derivatives make sense (e.g. $\frac{dq}{dt}$) and we typically do take them to find the equation of motion.
As you noted, conjugate momentum of $q$ is $$p = \frac{\partial L}{\partial \dot{q}}$$. The sufficient and necessary condition for momentum conservation is $$\frac{\partial L}{\partial q} = 0$$ because by the Euler Lagrange equation of this system, this implies $$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$$and so$$\frac{d}{dt} p = 0$$ It is generally true that you don't take partials between these variables, they are regarded as independent.
