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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.

Two initially equal buckets of total mass $M_0$ share an ideal Atwood machine. If one loses sand at constant rate $\alpha$ and the sand leaves with zero initial velocity relative to that bucket, the leakage supplies no rocket thrust. The remaining masses satisfy $(M+M_0)\dot{v}=g(M_0-M)$ for upward velocity of the leaking side. Using $\dot{M}=-\alpha$ and integrating from rest gives $v=(g/\alpha )[M-M_0-2M_0\log ((M+M_0)/(2M_0))]$.

If the initial sand fraction is $ϵ$, emptying occurs at $M=(1-ϵ)M_0$. This gives the displayed speed, which tends to zero for $ϵ\to 0$ and to $(g{M}_0/\alpha )(2\log 2-1)$ for $ϵ\to 1^{-}$. The latter is a finite limiting result, not a physical claim about an exactly massless empty bucket. The derivation is an application of a variable-mass system, rather than of $d(Mv)/dt=F$ for a closed body.

Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

Let $m=4\pi \rho a^3/3$ be the drop's mass, with constant mass density $\rho$. Its volume-growth rule gives $4\pi a^2\dot{a}=c\pi a^{2}v$, hence $\dot{a}=cv/4$ and $\dot{m}=\rho c\pi a^{2}v$. The incoming mist is stationary, so the variable-mass system momentum balance isExpanding the derivative and substituting the accretion rate gives the accreting raindrop with quadratic drag equationsThe extra $c$ contribution is the momentum needed to accelerate collected matter from rest; omitting it would treat the growing drop as a constant-mass particle.

Set $q=v^2/a$ and $B=7c/8+3k/4$. The chain rule givesTo justify the limiting value as well as its stability sign, use $c>0$, as appropriate to volume growth. Divide by $\dot{a}=cv/4$ and integrate the resulting first-order equation:Thus $q$ remains between two positive constants, its initial value and $g/B$. Since $d\sqrt{a}/dt=c\sqrt{q}/8$, the radius is defined for all increasing time and tends to infinity. The correction vanishes, proving the speed-radius attractor of an accreting raindrop:The strict inequality uses $k>0$. The growing radius prevents a finite terminal velocity; instead the acceleration approaches this positive constant.

For constant-area capture of stationary dust, let $x=0$ at cloud entry, $M(0)=M_p$, $v(0)=v_{\infty }$, and define $k=\rho A$ with fixed collecting area $A$. Every swept volume $Adx$ adds stationary mass, soThere is no external longitudinal force, and the incoming dust has zero momentum in the cloud frame. The variable-mass system therefore satisfies $d(Mv)/dt=0$, rather than $M\dot{v}=0$. Using $Mv=M_p{v}_{\infty }$ givesHenceThe kinetic energy decreases during sticking: $dK/dt=-\frac{1}{2}\dot{M}{v}^2$. The lost energy becomes internal energy. Radiation, gravity, sputtering and changes in area are excluded by the ideal complete-collection model. The zero-density limit recovers $M=M_p$, $v=v_{\infty }$ and $x=v_{\infty }t$.

The snow remains aboard, so the moving mass grows as $M+\sigma t$. Incoming snow contributes no horizontal momentum, and there is no horizontal external force. ThereforeThe combined sled–snow system follows a variable-mass system balance; setting $(M+\sigma t)\dot{v}=0$ would omit the incoming mass term.

Let $\lambda =1{\mathrm{s}}^{-1}$ be the launch rate and $\mu =\lambda m$. Treat many small impacts as a continuous momentum flux, justified by $m\ll M$. Write $x(t)$ for truck displacement from its initial position and $U=\dot{x}$, with $t=0$ at the first impact. First neglect vertical motion, so every projectile has horizontal velocity $v$.

A fixed launch rate is different from a fixed impact rate on a receding truck. Projectiles reaching the truck at time $t$ were launched at retarded time $s=t-x(t)/v$, after an irrelevant constant initial-flight offset. The moving-target interception rate is thereforeFor an elastic collision with a much heavier truck, each ball transfers momentum approximately $2m(v-U)$. The elastic-ball truck keeps mass $M$, soSolving from rest and integrating once more givesThe rebound reverses the ball's motion relative to the truck; its laboratory velocity $2U-v$ can still be forwards once $U>v/2$. The truck approaches $v$ asymptotically, while its decreasing impact rate prevents exponential approach in laboratory time. The approximation omits the velocity jumps of individual finite-mass impacts.

For sticking snowballs, let $A(t)$ be the instantaneous truck-plus-snow mass, with $A(0)=M$. The variable-mass system obeysIntegrating the second equation gives $AU=v(A-M)$. Therefore $A\dot{A}=\mu M$, and with $z=\sqrt{1+2\mu t/M}$,Retaining the accumulated mass is necessary even though each individual snowball is small.

For the range calculation, take the intended idealization of horizontal launch from height $h$, no air resistance, and interception possible until the projectile reaches the ground. Its flight time is $\tau =\sqrt{2h/g}$ and horizontal range is $D=v\tau$. If the initial truck distance from the boy is $X_0<D$, substitute $x=D-X_0$ in the last expression. The speed at the range limit isFor an initially nearby truck, $X_0=0$ in this point-target model and $a=\sqrt{2\mu \tau /M}$. A different launch angle or finite interception height changes the available range and must be specified separately.

If instead one idealizes one impact each second, regardless of the truck's changing position, the answers are different: elastic balls give $U=v(1-e^{-2\mu t/M})$ and $x=v[t-M(1-e^{-2\mu t/M})/(2\mu )]$; sticking balls give $U=v\mu t/(M+\mu t)$ and $x=vt-(Mv/\mu )\ln (1+\mu t/M)$. Those expressions assume controlled arrival times, whereas the boxed results respect the stated fixed launch rate.

The retained mass at time $t$ is $M(t)=M_0+kt$. Incoming rain has zero horizontal momentum in the ground reference frame. During $dt$, horizontal momentum conservation for the carriage and the rain about to enter givessoThis is a variable-mass system: the entering mass must be accelerated horizontally, even though no external horizontal force acts on the carriage-plus-rain system. Integrating the velocity,For $u_0\ge 0$ this is also the distance travelled. Its $k\to 0$ limit is $u_{0}t$. Vertical impact can dissipate kinetic energy; horizontal momentum conservation does not require kinetic energy conservation.

The diagram makes $x$ the remaining distance to the wall: $x$ decreases from $L$ to $0$. With the endpoint condition $m(0)=0$, integrating the supplied mass gradient givesAssume escaping sand has the same horizontal velocity as the bucket at release. There is then no horizontal recoil: the momentum flux of the outgoing sand cancels the $\dot{m}\dot{x}$ contribution to the bucket's momentum derivative. The variable-mass system equation is thereforeSet $u=-\dot{x}\ge 0$, the speed towards the wall. Since $\ddot{x}=udu/dx$, the initial condition $u(L)=0$ givesConsequently the remaining sand has kinetic energyIn terms of $z=x/L$, $K/(TL)=-z\ln z$. Its derivative is $-\ln z-1$ and its second derivative is $-1/z<0$, givingThe magnitude of the remaining sand's momentum isIt is easier to maximize its square, $p^2/(2MTL)=-z^2\ln z$. Its derivative is $-z(2\ln z+1)$, giving the unique interior maximumBoth kinetic energy and momentum tend to zero at $x\to 0^{+}$, even though the idealized speed diverges. There is no finite-mass bucket left at the endpoint, so that divergence signals the limit of the idealization.

The leaking bucket under constant tension has $x_K=L/e<L/\sqrt{e}=x_p$: the kinetic energy maximum is closer to the wall. As the bucket loses mass, its increasing speed continues to boost $K=pu/2$ after $p$ has begun to fall. Equivalently, $p\propto z\sqrt{-\ln z}$ and $K\propto z(-\ln z)$, so kinetic energy has the stronger dependence on the growing speed. The work $T(L-x)$ is shared between the remaining sand and the sand already lost; it is not all the bucket's instantaneous kinetic energy.

Choose downward velocity $v$ positive, and assume the collected material is initially at rest in the cloud's inertial frame. This assumption is required for the equation in the question. During $dt$, momentum conservation for the droplet plus the material it collects givesup to second-order increments. Thus the variable-mass system satisfiesSince $m\propto r^3$, $\dot{m}/m=3\dot{r}/r=3\mu /(r_0+\mu t)$, givingThe notation $dv/dt$ is appropriate because $v$ here depends only on time. The extra term accounts for the momentum needed to accelerate newly collected mass; using $m\dot{v}=mg$ would omit it. For incoming material with downward velocity $u$, the more general equation is $m\dot{v}=mg+(u-v)\dot{m}$.

An upward drag force $F_D\ge 0$ changes the variable-mass system equation toFor an ordinary drag force increasing with positive speed, the new curve lies below the no-drag curve. It still starts with slope $g$ if $F_D=0$ at rest. The dashed curve illustrates quadratic drag on a sphere with constant radial growth, using $F_D=\frac{1}{2}{C}_D{\rho }_{\mathrm{air}}\pi r^2{v}^2$ with constant drag coefficient $C_D$ and air mass density ${\rho }_{\mathrm{air}}$. The equation for these dimensionless quantities isThe figure uses $\eta =2$ solely to show the shape; it is not an additional datum of the paper.

There is a limitation to the usual terminal velocity sketch here. The examiner's commentary describes a curve approaching terminal velocity, but the paper maintains unbounded radius growth and specifies no drag law. A horizontal limit is justified if one additionally treats the radius as approximately fixed, or otherwise assumes a drag-to-mass ratio with a fixed limiting balance. With the growing radius and constant-$C_D$ quadratic drag just written, the late balance instead givesThis curve flattens in the sense that its slope tends to zero, but its speed remains unbounded. Thus a lower, more strongly flattening curve is warranted; a finite terminal speed requires an additional assumption. The no-growth approximation gives the familiar finite value $v_T=\sqrt{8\rho gr/(3C_D{\rho }_{\mathrm{air}})}$ for a specified fixed radius.^[ref]

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.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.