If I understood correctly, the force $F$ related to a given mass flow rate $\dot{m}$ through orifice area $A$ (or, say, on a disk of area $A$) is given by Newton's 2nd law of motion - assuming a constant velocity - with
\begin{align} \vec{F} &= \tfrac{\mathrm{d}}{\mathrm{d} t} \left( m \vec{v} \right) \\ &= \tfrac{\mathrm{d} m}{\mathrm{d} t} \vec{v} \\ &= \dot{m} \vec{v} = \rho A \vec{v}^2 \end{align}
I am wondering what would happen if the area $A$ itself would be moving with a constant velocity $\vec{v}_0$ along the direction of $\dot{m}$.
If we assume
\begin{align} \vec{v}_{rel} &= \vec{v} - \vec{v}_0 \end{align}
would it be correct to say
\begin{align} \vec{F} &= \dot{m} \vec{v}_{rel} \\ &= \rho A \vec{v} \, \vec{v}_{rel} \\ &= \rho A \left( \vec{v}^2 - \vec{v} \,\vec{v}_0 \right) \end{align}
that is, the mass flow rate is still $\rho A v$ but the relative velocity of the disk is $v - v_0$?
Or should it be \begin{align} \vec{F} &= \rho A \vec{v}^2_{rel} \\ &= \rho A \left( \vec{v}^2 - 2 \vec{v} \,\vec{v}_0 + \vec{v}^2_0 \right) \end{align}
that is, the mass flow rate being $\rho A v_{rel}$?
Also, is this mathematically correct? It feels kind of weird to apply the Hadamard/Schur product to the velocity vectors just to obtain another velocity in the same direction.
