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Let $\mathbf{N}$ be the normal reaction of the stationary smooth surface of revolution. Its normal lies in the span of ${\mathbf{e}}_r,{\mathbf{e}}_z$, so it has no ${\mathbf{e}}_{\theta }$ component. The transverse component of Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{N}-mg{\mathbf{e}}_z$, is thereforewhich gives $\boxed{h=r^2\dot{\theta }=\text{constant}}$.

The velocity is tangent to the stationary surface, so $\mathbf{N}\cdot \dot{\mathbf{r}}=0$. Take the dot product of Newton's second law with $\dot{\mathbf{r}}$:Using $z=f(r)$ proves conservation of mechanical energy per unit mass:The ideal constraint does no work. These equations describe motion while the particle remains constrained to the surface.

Take $a>0$ and the intended horizontal launch around a parallel of the surface of revolution: $\dot{r}(0)=\dot{z}(0)=0$, $a\dot{\theta }(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot{z}=f^{\prime }(r)\dot{r}$, the conserved energy becomesHere $U$ is the radial effective potential on a surface of revolution. Rearrangement givesFor a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalentlyThe interval between the radial bounds must be dynamically accessible: $D\ge 0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\le E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the centrifugal barrier already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.

The initial radial acceleration, obtained by projecting Newton's second law along ${\mathbf{e}}_r+f^{\prime }(a){\mathbf{e}}_z$, isThus $v^2/g>a{f}^{\prime }(a)$ starts the particle outward, and $v^2/g<a{f}^{\prime }(a)$ starts it inward. Equality gives the exact circular orbit $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this separatrix is bounded, although the radial motion is not periodic. If $v=0$, the centrifugal barrier disappears and the particle may pass through the axis, still with bounded $r$.

The source leaves a direction ambiguity at points where $f^{\prime }(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f^{\prime }(a)\ne 0$ that direction is forced by horizontality and tangency.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

The kinetic energy is $K=\frac{1}{2}m(3{\dot{x}}^2+3{\dot{y}}^2+4{\dot{z}}^2)$. Differentiating the given potential and applying Newton's second law givesThese are the normal modes of a symmetric three-mass string, with mass matrix $M=m\mathrm{diag}(3,3,4)$ and stiffness matrix $m{\Omega }^2\left(\begin{array}{ccc}2& 0& -1\\ 0& 2& -1\\ -1& -1& 2\end{array}\right)$. The coordinate order here is $(x,y,z)$, although the spatial order along the string is $x,z,y$.

Reflection symmetry splits the generalized eigenvalue problem into an antisymmetric mode $x=-y$, $z=0$, with ${\omega }^2=2{\Omega }^2/3$, and symmetric modes $x=y=q$. Writing $\mu ={\omega }^2/{\Omega }^2$, these obey $z=(2-3\mu )q$ and $(2-4\mu )z=2q$. Thus $(2-4\mu )(2-3\mu )-2=2(6\mu -1)(\mu -1)=0$. The three results areThe displacement vectors are mutually orthogonal under the mass-weighted inner product $u^{T}Mv$, as required for these conservative normal modes.

With downward position as the vertical coordinate, the full potential energy isThe equilibrium tensions are $2k(a-\ell )=6mg$, $k(b-\ell )=3mg$ and $k(c-\ell )=mg$. On expanding, these relations cancel every term linear in the displacements, leaving $V=k{x}^2+k(x-y{)}^2/2+k(y-z{)}^2/2+V_0$. Newton's second law therefore givesFor a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, set $s=m{\omega }^2/k$. The generalized eigenvalue problem isIts determinant is $2(1-s)[3(1-s{)}^2-2]$, so the angular frequencies and mode directions areThe square roots of these positive values give the requested frequencies. Only the lowest mode has all components of the same sign, so all three particles oscillate in phase at the lowest frequency. The mode shapes are shown below.

For release from rest with $Y$ remaining at equilibrium, chooseThis excites only the middle-frequency mode, so $y(t)=0$ for all time. It is a stationary middle mass in a three-mass spring chain: the equal and opposite adjacent spring forces cancel on $Y$.

For a breathing displacement take ${\mathbf{u}}_i=q(t){\mathbf{r}}_i$. Every spring has equilibrium length $\sqrt{2}$ in the given length units, so its linear extension is $\sqrt{2}q$. At the vertex ${\mathbf{r}}_i$, the four adjacent equilibrium vertices sum to zero. Therefore the spring force there isIt remains radial and proportional to the original displacement at every vertex, proving that this symmetric displacement is a normal mode. Newton's second law givesThe first is angular frequency and the second is cycles per unit time. For $k$ in $\mathrm{N}{\mathrm{m}}^{-1}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad}{\mathrm{s}}^{-1}$ and $\mathrm{Hz}$ respectively. An energy check gives $T=3m{\dot{q}}^2$ and $V=12k{q}^2$, leading to the same ratio ${\omega }_b^2=24k/(6m)$.

For constant force $\mathbf{F}$ and displacement $\mathbf{s}$, the work isFor a variable force, $W={\int }_C\mathbf{F}\cdot d\mathbf{r}$. The work-energy theorem explains this choice: Newton's second law for fixed mass gives $\mathbf{F}\cdot d\mathbf{r}=m\mathbf{a}\cdot \mathbf{v}dt=d(m{v}^2/2)$. The component along the motion changes kinetic energy; a perpendicular force can change direction without doing work. Exerting a force on a stationary object likewise does no work on it.

Walking on level ground still requires muscular work. The centre of mass rises and falls during each step; the limbs gain and lose kinetic energy; and muscles perform positive work that replaces energy dissipated in impacts and tissue motion. Air resistance can add another cost. Returning to the same height makes the net change of gravitational potential energy zero, not every intermediate work contribution.

At ideal no-slip contact with an immovable road, the contact point is instantaneously stationary, so the road's force does no work through that point. The muscles convert internal chemical energy into motion. Zero net gain of height is compatible with positive muscular work and energy expenditure.

For a material fluid volume, mass conservation yields ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$. Constant mass density reduces this to $\nabla \cdot \mathbf{u}=0$. For inviscid flow, the Cauchy stress tensor is $-pI$; its net surface force is the volume integral of $-\nabla p$ by the divergence theorem. Newton's second law for each material volume therefore gives $\rho D\mathbf{u}/Dt=-\nabla p$. Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$, these are exactly the two stated equations.

In two dimensions they readDifferentiate the second momentum equation in $x$ and subtract the $y$ derivative of the first. The pressure terms cancel. For $\omega =v_x-u_y$, collecting terms givesIncompressibility removes the last term, givingThus each fluid parcel retains its scalar vorticity as it moves. In particular an initially irrotational flow remains irrotational in this smooth two-dimensional inviscid evolution. There is no vortex stretching in this two-dimensional setting.

Explicit reference to laws is not necessary for every good scientific explanation. Laws remain indispensable in many explanations, but their occurrence alone is not sufficient either. The issue turns partly on whether a law means a contingent regularity of nature, any supporting generalization, or even a mathematical necessity.

The deductive-nomological model offers a strong positive case. Laws and circumstances entail the phenomenon, and at least one law is essential to the derivation. For example, Newton's second law and an inverse-square gravitational force, with appropriate initial conditions, yield predictions for an idealized two-body orbit. A universal relation unifies cases that otherwise look like unrelated observations. Carl Hempel and Paul Oppenheim explicitly develop adequacy conditions for such deductive explanations; their 1948 paper should not be read as a finished treatment of every statistical explanation.^[ref]

However, derivability does not always give the right explanatory direction. In a simplified geometrical optics model, a vertical pole of height $h$ on level ground casts a shadow of length $\ell =h/\tan \theta$ at solar elevation $\theta$. Given $h$ and $\theta$, one derives $\ell$; given $\ell$ and $\theta$, one derives $h$. The same geometrical relations enter both derivations. Yet the shadow does not normally explain why the pole was built to that height. The reverse calculation is a way to measure its height. The difference concerns which features produce or constrain which others, rather than deductive validity alone.

Mechanistic explanation is another challenge to a compulsory law-based format. A clock stopped because debris blocked a particular wheel, thereby preventing the mechanism from advancing. Inspecting the obstruction, removing it and observing restored operation can answer the causal question with considerable precision. The explanation describes organization and a relevant difference-making event without stating a universal law. It is more informative about this failure than the observation that every component obeyed Newtonian mechanics.

A defender of the covering-law model can reply that the causal account relies implicitly on regularities: debris that offers no resistance would not stop the wheel. This is a fair warning against treating a bare chronology as an explanation. But it does not establish that explanatory understanding requires identifying exceptionless fundamental laws, or that replacing the account with a complete microscopic derivation would improve the answer. Stable, tested relations in the actual operating regime may be sufficient.

An explanation can also turn on a mathematical constraint. Suppose an engineer asks why a connected network with four odd-degree junctions cannot be inspected by a route using every segment exactly once. Represent junctions as vertices and segments as edges of an undirected graph. Every intermediate visit pairs an incoming and an outgoing edge, so only the two endpoints of an open trail can have odd vertex degree. Four such vertices therefore exclude an Eulerian trail. The impossibility persists if segment lengths, construction materials or walking speeds change while the connections remain fixed. This is the obstruction in the Seven Bridges of Königsberg.^[ref]

The physical application requires that the graph represent the allowed routes correctly. Once that is granted, the explanatory necessity comes from parity, rather than a contingent physical law. Calling every mathematical constraint a law would preserve the original claim only by greatly widening its meaning. A critic may restrict “scientific” to causal explanations, but that restriction needs justification when applied mathematics answers a concrete engineering question.

Thus laws are powerful explanatory resources, and good explanations normally support relevant what-if judgements. Those judgements can be grounded in laws, mechanisms or mathematical constraints. Requiring explicit reference to laws imposes an unnecessarily narrow format; requiring some disciplined account of dependence is much more defensible.

There is no unconditional logical dependence of Newton's first law on Newton's third law. There is, however, a dependence if one applies the first to an extended, isolated composite and interprets its motion as the motion of its centre of mass. The difference is between the behaviour of an elementary body with zero total force and a theorem about cancellation of internal forces. Isaac Newton's own first-law discussion includes projectiles and spinning bodies, so the composite-body issue is physically significant.^[ref]

For a constant positive mass in an inertial frame, Newton's second law gives $m\ddot{\mathbf{x}}=\mathbf{F}$. Setting the total force to zero yields uniform rectilinear motion; it uses no action–reaction assumption. The first law additionally identifies a common class of inertial frames in which such a statement applies. One cannot establish that class by arbitrarily straightening a single trajectory, and one cannot test the claim by defining every observed acceleration to be an unspecified force.

A concrete countermodel separates these issues. Suppose two bodies obey the second-law equations with internal forces ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$, with nonzero constant $\mathbf{f}$. Every elementary body with zero total force still has constant velocity, but the two-body momentum satisfiesThus the elementary-body first law is compatible with failure of the third. Calling the pair isolated does not make its internal total force zero: that is precisely what has not been assumed. This countermodel establishes logical independence, not a proposal for a successful fundamental interaction.

Now strengthen the first-law claim: every isolated composite has an inertially moving centre of mass whenever the total external force vanishes. With $M={\sum }_i{m}_i$, Newton's second law givesFor an isolated pair, the strengthened claim requires ${\mathbf{F}}_{12}+{\mathbf{F}}_{21}=0$. This is the weak, equal-and-opposite form of Newton's third law. It does not by itself require collinearity with the bodies' separation; the central-force condition is stronger and bears on conservation of angular momentum. For many bodies, cancellation of the sum does not alone imply pairwise cancellation without assumptions about the decomposition into interactions.

The right conclusion is therefore conditional. A first law for individual force-free bodies does not presuppose the third law. A first law for all isolated material composites, combined with the usual centre of mass and Newton's second law, imports an internal-momentum-balance condition. In theories with dynamical fields, material bodies can exchange momentum with those fields; conservation of momentum then concerns the combined system and need not imply instantaneous pairwise action–reaction between particles.

The familiar result follows if the force laws have Galilean invariance; the three abstract Newton's laws of motion do not guarantee that invariance. Isaac Newton's Corollary V correctly identifies a central feature of ordinary Newtonian mechanics, but its general physical scope requires assumptions about interactions as well as about the relation between force and acceleration.^[ref]

Take a constant velocity $\mathbf{U}$ and a Galilean transformationThen ${\ddot{\mathbf{x}}}_i^{\prime }={\ddot{\mathbf{x}}}_i$, ${\mathbf{x}}_i^{\prime }-{\mathbf{x}}_j^{\prime }={\mathbf{x}}_i-{\mathbf{x}}_j$, and ${\dot{\mathbf{x}}}_i^{\prime }-{\dot{\mathbf{x}}}_j^{\prime }={\dot{\mathbf{x}}}_i-{\dot{\mathbf{x}}}_j$. Any interaction depending only on these relative quantities, with unchanged parameters, supplies the same forces for the transformed initial data. Newton's second law consequently gives the same relative accelerations; uniqueness of the resulting initial value problem supplies the same relative histories. An inverse-square central interaction is a standard example. Newton's third law helps establish the common centre of mass result, but the boost argument principally needs invariant interactions and the second law.

Here is a counterexample to deduction from the three laws alone. Give two equal masses a preferred inertial frame and write $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $\mathbf{V}=({\dot{\mathbf{x}}}_1+{\dot{\mathbf{x}}}_2)/2$. For fixed $k>0$ and speed scale $u_{\ast }>0$, specifyThe forces are equal, opposite and central, the centre of mass is inertial, and each body obeys $m{\ddot{\mathbf{x}}}_i={\mathbf{F}}_i$. An elementary body with no force is inertial. Nevertheless,Boosting an isolated pair from $\mathbf{V}=0$ to $\mathbf{V}=\mathbf{U}$ changes its internal oscillation frequency. All three motion laws hold, but the proposed universal boost equivalence fails. The extra preferred structure in the force law is what makes the failure possible.

A passive coordinate transformation of this same model remains possible: transform the preferred frame's description too. That mathematical freedom does not establish an active dynamical symmetry which changes the physical pair's velocity relative to the preferred structure while holding the laws' background data fixed. Confusing those operations would hide the counterexample.

Finally, a change to rotating coordinates introduces centrifugal forces and Coriolis forces; their dependence on position or velocity can change internal experiments. Constant translational velocity has no analogous effect when the interactions have Galilean invariance. Corollary V expresses that physical symmetry, with the interaction assumption made explicit, rather than a consequence of action–reaction alone.

There are two different issues: whether a law follows logically from the other laws, and whether an experiment can test it without assuming anything else. Newton's laws of motion have empirical content when combined with rules for identifying inertial frames, measuring mass and specifying forces. An experiment tests predictions of that combination. Calling every observed acceleration an otherwise unspecified force would remove the empirical constraint rather than verify Newton's second law. The distinction is central to the empirical testing of Newton's laws of motion.

Newton's first law contributes the claim that a common family of inertial frames exists: bodies with zero net external force have uniform straight motion in those frames. It is not enough to choose coordinates which straighten the trajectory of just one body; an admissible frame must accommodate many independent bodies and interactions. Once that frame and the ordinary constant-positive-mass form of Newton's second law are assumed,This conditional deduction does not establish the existence of the required frame, or independently identify the physical circumstances in which the net force vanishes. Defining an inertial frame and then counting a straight trajectory as proof of the definition would not be a test.

Nevertheless, the law of inertia has testable implications in a network of assumptions. Consider low-friction motion along an air track: determine positions at measured times, reduce the known residual influences, and ask whether the remaining acceleration approaches zero in the same laboratory frame for different bodies. Rotating the frame changes the description through Coriolis forces and centrifugal forces; this dependence must be accounted for. No finite experiment establishes that every unmeasured influence is absent, but varying the apparatus can discriminate an interaction from a frame effect. A failed prediction is evidence against some part of the package, not automatically the isolated verbal statement of Newton's first law.

Newton's second law is more than a definition if the force law and mass assignments constrain predictions before the new acceleration is measured. For instance, independently characterize a spring's reproducible extension response, specify the same interaction in several configurations, and compare the resulting accelerations for different bodies. If its strength and every mass are freely refitted to each outcome, agreement is empty. If they are fixed by an independently checked model and calibration, the law predicts held-out trajectories. Such a test remains conditional on the spring model, initial conditions and frame identification; this is an instance of the Duhem problem, not a reason to deny all empirical content.

Newton's third law supplies a different constraint. For two bodies exchanging forces, Newton's second law giveswhen there is no net external force. Equal-and-opposite interaction forces imply conservation of momentum for the pair. The two second-law equations alone do not require that cancellation: assigning ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$ is algebraically consistent with both second-law equations while violating the third law. This is a logical countermodel, not a proposed isolated physical interaction. Measuring paired impulses or checking the pair's total momentum therefore tests an additional constraint, conditional on the assumed absence of momentum exchange with anything else.

The system boundary matters. In electromagnetism, matter can exchange momentum with the electromagnetic field. Instantaneous mechanical forces between two particles need not satisfy the simple pairwise statement, although total matter-plus-field momentum is conserved under the appropriate closed-system conditions. Likewise, Noether theorem can derive conservation of momentum from a translationally invariant action; that uses an extra symmetry assumption and does not derive the third law from the second alone.

Thus the first law is not an additional differential consequence once the full second-law framework is fixed, and the third law is not a consequence of that framework. All three can contribute to empirical tests, but none is tested in total isolation from frame, interaction and measurement assumptions. This preserves the difference between logical dependence and experimental testability.^[ref]

Take the upward velocity as positive during ascent. Newton's second law with linear drag givesMultiplication by the integrating factor $e^{kt}$ and integration giveThe maximum height occurs at the first zero of this velocity, henceHere $k$ has units of inverse time. The limit $k\to 0$ gives $v_0/g$, and positive drag shortens the ascent relative to motion in a vacuum. The expression applies only up to the turning point; the drag force opposes the subsequent downward motion.

The lines of centres from either lower cylinder to the upper cylinder make $60^{\circ }$ with the horizontal. Let $N_L,N_R$ be the upward normal forces on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal force. While all three contacts remain closed, the triangular configuration translates horizontally with common acceleration $a$; frictionless normals exert no torques about the cylinder centres.

For the top cylinder, Newton's second law givesThus $N_L=m(g/\sqrt{3}+a)$ and $N_R=m(g/\sqrt{3}-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, soA contact may push but cannot pull: $N_L,N_R,Q\ge 0$. The limiting inequalities are thereforeAt the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal forces are strictly positive only inside the open interval. The ground normal forces are positive throughout this range. The applied horizontal force on the left cylinder is $F=3ma$, as follows from the whole system's horizontal momentum balance.

Take the initial internal pressure as negligible and the external atmospheric pressure as $p_0=1.01325\times 10^5\mathrm{Pa}$. The window has frontal area $A=\pi (0.400\mathrm{m}{)}^2=0.50265{\mathrm{m}}^2$. Before appreciable air has entered, its inward pressure force is approximately $p_{0}A=5.09\times 10^4\mathrm{N}$. Newton's second law therefore gives the initial collective inward accelerationHere $M=110\mathrm{kg}$ and $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$. Neglect gravity in this much larger pressure-driven acceleration, together with supports and elastic stresses once the glass has broken. The given thickness $d=0.080\mathrm{m}$ implies a plausible average glass mass density $M/(Ad)\simeq 2.74\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Equivalently, a fragment retaining the original thickness has acceleration $p_0/({\rho }_{g}d)$, independent of its frontal area while it experiences the full pressure difference.

For a vacuum-window implosion energy estimate, the ideal pressure-work scale for filling vessel volume $V=1.5{\mathrm{m}}^3$ is $W\simeq p_{0}V$. Assigning all that pressure-volume work to the glass's kinetic energy givesThe same result follows from retaining acceleration $a_0$ over an effective swept distance $V/A\simeq 2.98\mathrm{m}$, so that $v_{\mathrm{est}}^2=2a_{0}V/A$. This is an ideal upper energy-budget estimate for the common speed, or mass-weighted root mean square speed, of all the glass. It assumes an atmospheric reservoir maintains the driving pressure and that the whole vessel's pressure-work budget can accelerate the glass. In practice, air motion, fragmentation, heating and collisions share that energy, the pressure difference falls, and fragments may hit the vessel after a much shorter path. The unspecified vessel shape and fragment trajectories prevent a unique actual maximum speed; concentrating energy into one small fragment also prevents interpreting this collective estimate as a strict bound on every individual fragment.

Descent equation. Take downward velocity $v$ and underwater depth $z$ as positive, with $z=0$ at the lake surface. For a sphere of radius $r$, the pebble mass is $m=4\pi r^3{\rho }_p/3$ and the frontal area is $A=\pi r^2$. Newton's second law includes downward gravitational weight, upward buoyancy and upward quadratic drag:ThusHere $b$ has units of inverse length. For arbitrary signed velocity, the last term is $-bv|v|$, so drag always opposes motion. The displayed downward form suffices for this trajectory.

Limiting speed. At terminal velocity, the net force and $\dot{v}$ vanish. For ${\rho }_p>{\rho }_w$,Use ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_w=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $r=0.020\mathrm{m}$, $C=0.7$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. ThenThe ${\rho }_p-{\rho }_w$ factor is essential: omitting buoyancy would overestimate the limiting speed.

For the numerical descent, assume the pebble is released from rest, neglect air drag, take $g$ and the water mass density as constant, and apply the given constant drag coefficient once it is submerged. Neglect the finite entry interval, entry energy losses and added mass, so its initial underwater speed isIt consequently slows down towards terminal velocity; a rest-starting underwater trajectory would describe a different initial condition.

Take upward velocity as positive. The linear drag force is $-mkv$, so it opposes the motion both during ascent and descent. Newton's second law gives $\dot{v}+kv=-g$. With $v(0)=u$, the integrating factor $e^{kt}$ yieldsThe terminal velocity is $-g/k$. The limit $k\to 0$ at fixed $t$ recovers $u-gt$; expanding the exponential is necessary to see the cancellation of the apparently singular terms.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.

Write the electron electric charge as $-e$, with $e>0$ the elementary charge, and choose $\mathbf{B}=B\hat{\mathbf{z}}$ and initial velocity $u\hat{\mathbf{x}}$. In the nonrelativistic approximation, Newton's second law with the magnetic Lorentz force givesThe magnetic force is perpendicular to the velocity, so it does no work: $d(m_e|\mathbf{v}{|}^2/2)/dt=0$. Setting ${\omega }_c=eB/m_e$, the solution with $x(0)=y(0)=0$ isConsequently $x^2+(y-r{)}^2=r^2$, where $r=u/{\omega }_c$: this is a cyclotron orbit. Its frequency and cyclotron radius areThe accelerating voltage changes $u$ and $r$ but cancels out of $f$. Using $m_e=9.1094\times 10^{-31}\mathrm{kg}$ and $e=1.6022\times 10^{-19}\mathrm{C}$ givesOnly the magnitude of the magnetic field is determined; its direction is perpendicular to the orbital plane, with the sense of rotation fixed by the negative electric charge. The claimed voltage independence is a nonrelativistic result: exactly, $\gamma =1+eV/(m_e{c}^2)$ and $f=eB/(2\pi \gamma m_e)$. At $1\mathrm{kV}$ the correction is about $0.2\%$. The stated model also neglects the energy lost to emitted electromagnetic radiation.