= Solution
For a neutrally buoyant horizontal <turbulent round jet>, a segment's horizontal <momentum flux> changes only if an external horizontal <force> or ambient <momentum> input exists. Neither is present, so
$$
\boxed{\frac{dM_x}{ds}=0,\qquad M_x=M_0\quad(F_0=0,\ \theta_0=0)}.
$$
Positive <buoyancy> adds a vertical <force>, not a horizontal one. Therefore $dM_x/ds=0$ remains true for an <inclined forced plume> even when $F_0\ne0$; in general $M_x=M_0\cos\theta_0$.
The net upward <force> on a segment is $g(\rho_0-\rho)\pi b^2ds$. Since $F=g(\rho_0-\rho)\pi b^2V$ and $V=|\mathbf M|/Q$, vertical <momentum> balance gives
$$
\boxed{\frac{dM_z}{ds}=\frac FV=\frac{FQ}{|\mathbf M|}}.
$$
In <kinematic plume fluxes>, the equations are $m_x'=0$, $m_z'=fq/m$, where $m=(m_x^2+m_z^2)^{1/2}$. The centreline geometry satisfies
$$
\frac{dx}{ds}=\cos\theta=\frac{m_x}{m},\qquad
\frac{dz}{ds}=\sin\theta=\frac{m_z}{m}.
$$
For $F_0>0$, $Q>0$ beyond the source, and an initially nonvertical <turbulent round jet> pointing right, $M_z$ increases and the centreline turns toward the vertical. The magnitude of <momentum flux> is not conserved once the tangent has a vertical component.
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