 Avalanche front with linear mass entrainment

An idealized front at distance $x>0$ down a uniform slope entrains stationary material so that its moving mass is proportional to $x$. Its linear momentum balance is $d(x\dot{x})/dt=Ax$, where $A=g\sin \alpha >0$. Multiplying by $x\dot{x}$ gives $x^2{\dot{x}}^2=2A{x}^3/3+c$. Every forward-moving branch tends to $\dot{x}=\sqrt{2Ax/3}$ as $x\to \infty$, and $\ddot{x}=A/3-c/x^3\to A/3$. Acceleration is smaller than the constant-mass downslope value because some momentum accelerates newly entrained material.