Variable-mass system ($m\dot{\mathbf{v}}=\mathbf{F}+(\mathbf{u}-\mathbf{v})\dot{m}$)

A body exchanging mass with material of inertial velocity $\mathbf{u}$ satisfies the displayed momentum conservation law for one stream, with signed $\dot{m}$ positive for entry and negative for exit. The force $\mathbf{F}$ excludes the incoming or outgoing momentum flux. A short time interval gives $d(m\mathbf{v})/dt=\mathbf{F}+\mathbf{u}\dot{m}$, proving the formula. For stationary incoming matter, $d(m\mathbf{v})/dt=\mathbf{F}$; for matter leaving at the body's own velocity, $m\dot{\mathbf{v}}=\mathbf{F}$. The rocket equation and accreting-drop motion are applications.

A body of initial mass $M_0$ and speed $v_0$ collects all stationary dust swept out by constant area $A$ in a uniform cloud of mass density $\rho$. Then $M=M_0+\rho Ax$ and $Mv=M_0{v}_0$ by conservation of momentum. Integration gives the displayed mass law, $v=M_0{v}_0/M$ and $x=(M-M_0)/(\rho A)$. Sticking dissipates kinetic energy at rate $\dot{M}{v}^2/2$. The model assumes no external longitudinal force and negligible relativistic corrections.

A body moving at constant velocity $V$ incorporates material at rate $\dot{M}$ with zero initial momentum along its motion. The required external force is $F=\dot{M}V$, so the supplied power is $P=\dot{M}{V}^2$. Only $\dot{M}{V}^2/2$ becomes the added material's translational kinetic energy; the other half is dissipated during inelastic incorporation. This conclusion concerns the component along the body's motion; an incoming transverse velocity can add further dissipation.

A sled intercepting snow with zero horizontal velocity in the ground frame is a simple momentum flux example. At arrival rate $\sigma$, immediate removal with outgoing ground-frame horizontal velocity $v_{\mathrm{out}}$ gives $M\dot{v}=-\sigma v_{\mathrm{out}}$ for constant moving mass $M$. Removal sideways in the sled frame has $v_{\mathrm{out}}=v$ and exponential deceleration; removal sideways in the ground frame has $v_{\mathrm{out}}=0$ and constant sled velocity. Retained snow instead gives $(M+\sigma t)v=M{v}_0$. The outgoing reference frame must be specified.

Let a massless bucket approach a wall under constant tension $T>0$, with remaining distance $x$ and remaining sand mass $m=Mx/L$. Assume zero-relative-velocity mass shedding horizontally. Then $m\ddot{x}=-T$ and, from rest at $x=L$, the speed satisfies $u^2=(2TL/M)\ln (L/x)$. Its kinetic energy is $K=Tx\ln (L/x)$ and its momentum magnitude is $p=x\sqrt{(2MT/L)\ln (L/x)}$. Differentiating gives $x_K=L/e$ and $x_p=L/\sqrt{e}$: the kinetic energy peaks later, closer to the wall, because its dependence on the increasing speed is stronger. Both tend to zero as the remaining mass vanishes, despite divergent ideal speed.

Let fuel of remaining mass $M_f$ be carried at radius $a$ on a rotor whose dry moment of inertia is $I_d$. Then $I=I_d+a^2{M}_f$. Fuel leaves tangentially backwards with speed $v_e$ relative to the rim, so its inertial tangential velocity is $a\omega -v_e$. Conserving angular momentum of rotor plus newly emitted fuel during $d{M}_f<0$ givesSince $dI=a^{2}d{M}_f$, integration yields ${\omega }_f-{\omega }_i=(v_e/a)\log (I_i/I_f)$. Including the fuel's co-rotation is essential: the rotor alone is an open system and its angular momentum is not conserved. For three radial rods of mass $m$ and length $a$, each carrying a point rocket of dry mass $m$ and fuel mass $m_f$, $I_i=(4m+3m_f)a^2$ and $I_f=4m{a}^2$.

For a sphere with constant mass density $\rho$ collecting mass at rate $\beta$ per unit surface area, $\dot{m}=4\pi r^2\beta$ and $m=4\pi \rho r^3/3$ give $\dot{r}=\beta /\rho$. If it falls through stationary material under constant gravitational acceleration $g$ without drag, its velocity satisfies $\dot{v}+3\dot{r}v/r=g$. With $v(0)=0$, an integrating factor $r^3$ gives $v=g[r-r_0^4/r^3]/(4\dot{r})$.

For surface-fed growth of an accreting sphere with $\mu =\dot{r}>0$, stationary incoming material, and quadratic drag $F_D=C_D{\rho }_a\pi r^2{v}^2/2$, the variable-mass system obeys $\dot{v}+3\mu v/r=g-3C_D{\rho }_a{v}^2/(8\rho r)$. For constant positive parameters, $r\sim \mu t$ and the late balance gives the displayed speed. The omitted acceleration and accretion terms are both $O(t^{-1/2})$, consistent with the balance. Thus this growing sphere has unbounded speed with vanishing acceleration, rather than a finite terminal velocity. This model has a different growth law from the existing accreting raindrop with quadratic drag, whose growth rate is proportional to speed.

For a spherical drop of constant mass density $\rho$, radius $a>0$ and downward velocity $v>0$, suppose its volume grows at rate $c\pi a^{2}v$ by collecting stationary mist and its upward quadratic drag is $k\rho \pi a^2{v}^2$, with $c,k>0$. The drop mass is $4\pi \rho a^3/3$. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional $c$ term alongside drag.

For the accreting raindrop with quadratic drag, set $q=v^2/a$ and $B=7c/8+3k/4$. Its exact equation is $\dot{q}=2(v/a)(g-Bq)$. Dividing by $\dot{a}=cv/4$ gives $q(a)=g/B+(q(a_0)-g/B)(a_0/a{)}^{7+6k/c}$. Thus $q$ stays between its positive initial value and positive equilibrium. Since $d\sqrt{a}/dt=c\sqrt{q}/8$, the radius grows without finite-time divergence and tends to infinity. The power correction then vanishes, proving both limits. Positive drag makes the limiting acceleration strictly less than $g/7$.