Velocities pictured in inertial frame G unless labeled otherwise ( there are a lot of symbols to juggle...if I messed up somewhere fundamentally I'll fix it later).
So continuing on ( continuity is a scalar equation at this point)
$$ V_1 A_1 = V_2 A_2 \sin \theta $$
Skipping to Energy ( neglecting system dynamics of flow between stations 1 and 2:
$$ \cancel{\dot Q}^0 - \cancel{\dot W_s}^0 = \cancel{\frac{d}{dt} \int_{cv} \rho \left( \frac{V^2}{2} + gz + u \right) d V\llap{-}}^{\text{neglecting}} + \int_{cs} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} $$
$$ \implies 0 = \int_{1} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} + \int_{2} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} $$
$$ \int_{1} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} $$
$$ = \int \left( \frac{V_1^2}{2} + gz_1 + u_1 + \frac{p_1}{\rho} \right) \rho \boldsymbol {V_1}\cdot d \boldsymbol{A_1} + \int \left( \frac{V_2^2}{2} + gz_2 + u_2 + \frac{p_2}{\rho} \right) \rho \boldsymbol {V_2}\cdot d \boldsymbol{A_2} $$
$$\boldsymbol {V_1}\cdot d \boldsymbol{A_1} = V_1 \langle -1 i + 0 j \rangle \cdot dA_1 \langle 1 i + 0 j \rangle = -V_1 dA_1 $$
$$\boldsymbol {V_2}\cdot d \boldsymbol{A_2} = V_1 \langle -\cos \theta i + \sin \theta j \rangle \cdot dA_2 \langle 0 i + 1 j \rangle = V_2 \sin \theta dA_2 $$
$$ \implies 0 = \int \left( \frac{V_1^2}{2} + gz_1 + u_1 + \frac{p_1}{\rho} \right) \rho (-V_1 dA_1)+ \int \left( \frac{V_2^2}{2} + gz_2 + u_2 + \frac{p_2}{\rho} \right) \rho (V_2 \sin \theta dA_2) $$
All properties are uniformly distributed over the stations 1 and 2, so they can come outside the integral:
$$\left( \frac{V_1^2}{2} + gz_1 + u_1 + \frac{p_1}{\rho} \right) \rho V_1 \int dA_1 = \left( \frac{V_2^2}{2} + gz_2 + u_2 + \frac{p_2}{\rho} \right) \rho V_2 \sin \theta \int dA_2$$
Assume ##u,p## are constant, and ##z## change is negligible they will cancel at each station:
$$ \left( \frac{V_1^2}{2} \right) \rho V_1 A_1 = \left( \frac{V_2^2}{2} \right) \rho V_2 \sin \theta A_2 $$
Substitute from continuity and we have:
$$ \left( \frac{V_1^2}{2} \right) \cancel{\rho V_1 A_1} = \left( \frac{V_2^2}{2} \right) \cancel{\rho V_2 \sin \theta A_2} $$
$$ \implies V_1 = V_2 $$
Finally Apply Momentum Equation to control volume:
$$ \sum \boldsymbol{F} = \cancel{\frac{ d}{dt} \int_{cv} \rho \boldsymbol{v} d V\llap{-}}^{\text{neg. dyn. for fluid}} + \int_{cs} \boldsymbol{v}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) $$
Also, I'm just focusing on horizontal components ( There would also be a normal force required ##\boldsymbol N## )
$$ F \langle 1i + 0j \rangle = M \frac{dv}{dt} \langle 1i \rangle + \int_{1} \boldsymbol{v}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) + \int_{2} \boldsymbol{v}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) $$
$$ F \langle 1i + 0j \rangle = M \frac{dv}{dt} \langle 1i \rangle + \cancel{\int \boldsymbol{v_1}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} )}^{\text{still air in inertial frame}} + \int \boldsymbol{v_2}\rho ( \boldsymbol {V_2}\cdot d \boldsymbol{A_2} ) $$
$$ \boldsymbol{v_2} = \langle (V_1 - V_2 \cos \theta)i+ V_2 \sin \theta j \rangle $$
$$ F \langle 1i + 0j \rangle = M \frac{dv}{dt} \langle i \rangle+ \int \langle ( V_1- V_2 \cos \theta ) i + V_2 \sin \theta j \rangle \rho \langle V_2 ( -\cos \theta i + \sin \theta j\rangle \cdot dA_2\langle 0 i + 1 j \rangle $$
$$ F \langle i \rangle = M \frac{dv}{dt} \langle i \rangle + \langle ( V_1- V_2 \cos \theta ) i \rangle V_2 A_2 \sin \theta $$
With ##V_1 = V_2## ( these are just scalars), Continuity, and the velocity of the cart is ##v = V_1## we get the scalar equation:
$$ F = M \frac{dv}{dt} + \rho A_1 v^2 ( 1 - \cos \theta ) $$
Making a really long story short, we get the same force on the cart if its accelerating or moving with constant velocity
so long as we ignore the dynamics of the fluid flow inside the control volume between stations 1 and 2.
