Show that the $1$D momentum equation:

$\frac{\text{d}\text{e}\text{l}\text{v}}{\text{d}\text{e}\text{l}\text{t}}+\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}(\frac{v^2}{2})+g(\frac{\text{d}\text{e}\text{l}\text{h}}{\text{d}\text{e}\text{l}\text{x}}+S_f-S_0)=0$

Is equivalent to

$\frac{\text{d}\text{e}\text{l}\text{Q}}{\text{d}\text{e}\text{l}\text{t}}+\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}(Qv)+gA(\frac{\text{d}\text{e}\text{l}\text{h}}{\text{d}\text{e}\text{l}\text{x}}+S_f-S_0)=0$

Where $v$ is the flow velocity, $h$ is the flow depth, $Q$ is the volumetric flow rate, $S_f$ is the friction slope, $S_0$ is the bed slope, and $A$ is the cross$-$sectional flow area.

Hint $1$: You may assume that the continuity equation holds:

$\frac{\text{d}\text{e}\text{l}\text{A}}{\text{d}\text{e}\text{l}\text{t}}+\frac{\text{d}\text{e}\text{l}\text{Q}}{\text{d}\text{e}\text{l}\text{x}}=0$

Hint $2$: it is not true that $A\frac{\text{d}\text{e}\text{l}\text{v}}{\text{d}\text{e}\text{l}\text{t}}=\frac{\text{d}\text{e}\text{l}\text{Q}}{\text{d}\text{e}\text{l}\text{t}}$ in general so that will not work.