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Scattering and impulse model boundaries — Examples

1 · Prerequisites

2 · Summary

The mathematical incident-area and solid-angle measures used by the cross-section checks inherit countable choice from their suppliers. These examples exercise the scattering, impact and open-flow items of cm15-scattering-and-impulse-model-boundaries on declared model data and probe the hypotheses and boundaries of the page's statements.

Elastic reflection from a fixed sphere: the branch $b = a \cos (θ / 2)$ , the uniform density $a^{2} / 4$ and the total $π a^{2}$ of a contact model computes the elastic reflection of a reduced label from a fixed impenetrable sphere: the branch is $b = a \cos (θ / 2)$ , the locally defined incident-area pushforward has differential contact cross-section $a^{2} / 4$ and the total is $π a^{2}$ . The model is a contact (impulse) rule, outside the $C^{2}$ finite-range hypothesis of the central-scattering theorem; its example-specific fixed-wall rule is only analogous to the $e = 1$ normal reversal and is not the finite-mass restitution law. Its finite total neither contradicts nor tests the Coulomb forward divergence. A declared-data Rutherford check: $α = 1$ , $θ = π / 2$ , $(d σ / d Ω) = 4 \times 10^{- 24} m^{2}$ , the band values $π b (θ_{1})^{2}$ and the growth of the truncated total as $θ_{1} ↓ 0$ evaluates the Rutherford prediction on declared data in SI units --- reduced mass $2 \times 10^{- 26} k g$ , energy $10^{- 18} J$ , angular momentum $4 \times 10^{- 34} k g m^{2} s^{- 1}$ and coupling $κ = 4 \times 10^{- 30} J m$ --- obtaining $v_{\infty} = 10^{4} m s^{- 1}$ , $b = 2 \times 10^{- 12} m$ , $α = 1$ , $θ = π / 2$ and $(d σ / d Ω) (π / 2) = 4 \times 10^{- 24} m^{2}$ , and checks the band values $π (b (θ_{1})^{2} - b (θ_{2})^{2})$ and the small-angle growth of the truncated total.

The counterexample Smooth finite-range free scattering refutes global strict decrease; as a separate contact comparison, an attractive refracting shell with index $3 / 2$ has $θ (b) = 2 (\arcsin (b / a) - \arcsin (2 b / (3 a)))$ strictly increasing, with a jump down to $0$ at grazing incidence refutes the unqualified claim of global strict decrease with a smooth zero-potential witness whose deflection is identically zero. Separately, the adopted attractive refracting-shell rule with index $3 / 2$ has $θ (b) = 2 (\arcsin (b / a) - \arcsin (2 b / (3 a)))$ on $(0, a)$ , strictly increasing with $θ^{'} (b) > 0$ and jumping down to $0$ at grazing incidence. The endpoint extension at $b = 0$ and the grazing value $θ = 0$ are not values of this branch, whose image lies in $(0, π)$ ; branch data rather than a global monotone inverse are needed for the cross-section.

A declared-data impact check: $m_{1} = 2 k g$ , $m_{2} = 6 k g$ , $v_{1}^{-} = 3 m s^{- 1}$ , $v_{2}^{-} = - 1 m s^{- 1}$ , with elastic, semi-elastic and sticking outcomes checks three outcomes of one declared impact ( $m_{1} = 2$ , $m_{2} = 6 k g$ , velocities $3$ and $- 1 m s^{- 1}$ ): the elastic case reverses each velocity, giving outgoing velocities $- 3$ and $+ 1 m s^{- 1}$ along the contact direction, with $Δ K = 0$ ; the semi-elastic case $e = \frac{1}{2}$ gives impulse $- 9 N s$ and $Δ K = - 9 J$ , and the sticking case gives $Δ K = - 12 J$ , all with total momentum continuous across the impact. Its companion counterexamples isolate the model boundaries: Reciprocal impulses do not determine a collinear impact: the same incoming data admit the sticking outcome $v^{+} = 0$ and the elastic outcome $(v_{1}^{+}, v_{2}^{+}) = (- 3, 1) m s^{- 1}$ with different impulses and energies shows two admissible outcomes with different impulses for one data set, so a contact law is required, and A coefficient of restitution above $1$ manufactures kinetic energy: $e = 2$ on $m_{1} = m_{2} = 1 k g$ with velocities $\pm 1 m s^{- 1}$ raises the kinetic energy by $3 J$ while the configuration is unchanged shows that the explicitly hypothetical extension to $e = 2$ , derived locally rather than from the restricted impact theorems, raises the kinetic energy by $3 J$ while the configuration is unchanged, so coefficients above $1$ are incompatible with the conservative energy account.

Finally A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails verifies the open-body law on the declared linear mass loss $m (t) = 3 - t k g$ with constant exhaust speed $2 m s^{- 1}$ relative to the body, obtaining $v (t) = 2 \ln (3 / (3 - t)) m s^{- 1}$ and exhibiting the failure of the closed-system law $(m v)^{'} = F$ , and A declared smooth open flow: the uniform scaling $ρ = ρ_{0} e^{- 3 α t}$ , $u = α x$ satisfies the control-volume identities on fixed and material regions, and elbow port data illustrate the inlet/outlet decomposition checks the control-volume identities on the exact uniform-expansion field and on its explicitly mapped material ball, and illustrates the inlet/outlet decomposition with an ideal steady elbow whose balance requires a net external resultant on the fluid of about $3.578 N$ . No pressure or wall-force decomposition is supplied. All data are declared model data; no measurement, uncertainty budget or empirical comparison is asserted.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Elastic reflection from a fixed sphere: the branch $b = a \cos (θ / 2)$ , the uniform density $a^{2} / 4$ and the total $π a^{2}$ of a contact model

Physical scope

Assuming countable choice for the locally defined incident and solid-angle measures, a separately adopted fixed-wall contact model of a reduced label of mass $μ$ reflecting elastically from an impenetrable sphere of radius $a$ : the example-specific rule reverses the normal velocity component and preserves the tangential component, so speed and angular momentum are conserved across contact. The example computes the branch contribution $b = a \cos (θ / 2)$ , the constant density $a^{2} / 4$ and the total $π a^{2}$ . This fixed-wall rule is outside both the $C^{2}$ finite-range hypothesis of the central-scattering theorem and the finite-mass two-label restitution postulate; the $e = 1$ comparison concerns only the normal-velocity reversal. The finite total does not contradict the Coulomb forward divergence. The data are declared model data; the example reports no observation or measured value.

Example

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the mathematical incident and solid-angle measures below.

Declared model. Fix a length $a > 0$ (the sphere radius) and a reduced mass $μ > 0$ . In a fixed inertial chart (Inertial frames, stationary observers and the adopted Galilean transformations between frames), let the sphere $S_{a} : = {x \in R^{3} : ∥ x ∥_{2} = a}$ be immovable and centred at the origin. For $b \in (0, a)$ the label position is a continuous curve $γ_{b} : R \to R^{3}$ , affine on each side of one contact time $t_{*}$ , with one-sided velocity limits and $P : = γ_{b} (t_{*}) \in S_{a}$ . Its incoming affine line has speed $v_{\infty} > 0$ , direction $\hat{u}$ and perpendicular distance $b$ from the origin. Separately adopt, for this fixed-wall model, the elastic reflection rule at the contact point with outward unit normal $n : = P / a$ : $v^{+} = v^{-} - 2 ⟨ v^{-}, n ⟩ n,$ that is, the normal component reverses and the tangential component is unchanged. Its normal-velocity reversal is analogous to the $e = 1$ case of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (ii), but it is not a formal member of that law: the postulate assumes a finite-mass two-label collinear contact, whereas this example prescribes an immovable wall and a vector reflection rule. The inward unit normal $- n$ makes the angle $α : = \arcsin \frac{b}{a} \in (0, \frac{π}{2})$ with $\hat{u}$ ; equivalently $b = a \sin α$ is the perpendicular distance of the incoming line from the centre.

Incident preparation and contact cross-section. Choose orthonormal vectors $e_{1}, e_{2}$ perpendicular to the common incoming direction $\hat{u}$ . For an incident offset $b e_{φ}$ , where $e_{φ} = \cos φ e_{1} + \sin φ e_{2}$ , the first contact is $P = b e_{φ} - \sqrt{a^{2} - b^{2}} \hat{u}$ . The transverse component of the reflected direction is a positive multiple of $e_{φ}$ , so its outgoing azimuth is $φ$ . Let $D = (0, a) \times (0, 2 π)$ and $Y = (0, π) \times (0, 2 π)$ . Adopt uniform incident area measure $ν (B) = \int_{B} b d λ_{2} (b, φ)$ on Borel subsets of $D$ and solid-angle measure $Ω (B) = \int_{B} \sin θ d λ_{2} (θ, φ)$ on $Y$ , using the abstract mathematical measures of Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density. Define the contact cross-section measure by $σ (B) : = ν (T^{- 1} (B))$ , where $T (b, φ) = (θ (b), φ)$ is the outgoing direction map. For an angular band $A \subseteq (0, π)$ abbreviate $σ (A) : = σ (A \times (0, 2 π))$ . A differential contact cross-section is a density $f$ satisfying $σ (B) = \int_{B} f d Ω$ for every Borel $B \subseteq Y$ . This is a local definition for the fixed-wall contact model, independent of the smooth central-model class of Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation. The centre, grazing circle and azimuth seam have zero incident area and supply no atoms; free incoming lines with $b > a$ do not hit the sphere and are excluded from this contact cross-section.

Assertion to check. The motion is a single reflection with deflection $θ (b) = π - 2 α = π - 2 \arcsin \frac{b}{a} \in (0, π),$ so the branch is $b (θ) = a \cos (θ / 2)$ on $0 < θ < π$ , the differential cross-section is the constant $a^{2} / 4$ , and the truncated total cross-section over the whole branch is $π a^{2}$ .

Facts & Assumptions

Given: Countable choice for the mathematical measures; the declared sphere radius $a > 0$ , mass $μ > 0$ , speed $v_{\infty} > 0$ , impact parameter $b \in (0, a)$ , incoming direction $\hat{u}$ , contact point $P = γ_{b} (t_{*})$ with outward normal $n = P / a$ , the piecewise-affine trajectory of the Example and its reflection rule, and the locally defined incident measure $ν$ , solid-angle measure $Ω$ , direction map $T$ and contact cross-section $σ$ .

[L1]

The example's separately adopted reflection rule and Moore's normal/tangential split: the reflected velocity is $v^{+} = v^{-} - 2 ⟨ v^{-}, n ⟩ n$ , so $∣ v^{+} ∣ = ∣ v^{-} ∣$ and the tangential component $⟨ v^{-}, t ⟩$ is unchanged for $t ⊥ n$ ; the angular momentum about the centre is unchanged across contact because the impulse $- 2 μ ⟨ v^{-}, n ⟩ n$ passes through the origin: $μ P \times (v^{+} - v^{-}) = μ a n \times (- 2 ⟨ v^{-}, n ⟩ n) = 0$ (The cross product in $R^{3}$ ).

[L2]

Plane geometry (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): at the first contact of the incoming line with the sphere, $n \cdot \hat{u} < 0$ and the acute incidence angle is $α : = \arccos (- n \cdot \hat{u}) \in (0, π / 2)$ . Its impact parameter satisfies $b = ∣ P \times \hat{u} ∣ = a \sin α$ , so $α = \arcsin (b / a)$ . The reflection law gives $⟨ v^{-}, v^{+} ⟩ / ∣ v^{-} ∣^{2} = 1 - 2 (n \cdot \hat{u})^{2} = 1 - 2 \cos^{2} α = \cos (π - 2 α)$ , hence the deflection angle is $π - 2 α \in (0, π)$ .

[L3]

Trigonometry (Principal inverse sine and inverse cosine, Signs, monotonicity intervals, and ranges of sine and cosine, Parity and the Pythagorean identity for sine and cosine, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ): $\arcsin$ maps $[0, 1]$ increasingly onto $[0, π / 2]$ ; $\sin^{2} + \cos^{2} = 1$ ; $\cos (θ / 2) > 0$ and $\sin (θ / 2) > 0$ for $0 < θ < π$ ; $\sin θ = 2 \sin (θ / 2) \cos (θ / 2)$ ; derivatives of $\cos$ and of compositions are the usual ones.

[L4]

Abstract branch pushforward (Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density, One-dimensional substitution on a deflection branch: the outgoing angular measure has density $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$ ): for the mathematical measures $ν = b d λ_{2}$ and $Ω = \sin θ d λ_{2}$ , any $C^{1}$ bijective branch with nonzero derivative and inverse $b (θ)$ satisfies $T_{*} ν (B) = \int_{B} b (θ) ∣ b^{'} (θ) ∣ d λ_{2}$ on every Borel set. The supplier proves this by one-dimensional substitution on interior rectangles and Borel measure uniqueness; its branch and measures are abstract data independent of any smooth potential or physical trajectory. Thus it applies to the locally defined contact pushforward after the branch hypotheses have been verified. Truncated-band integration uses The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ and Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length.

Verification

technique · direct

1.1givenL1algebra

Reflection and conservation: by [L1], $∣ v^{+} ∣ = ∣ v^{-} ∣ = v_{\infty}$ and the angular momentum about the centre is conserved across the contact; hence the two free pieces of the motion are straight lines and the motion has a constant angular momentum $j$ with $∣ j ∣ = μ b v_{\infty}$ . This verifies directly for the incoming line the same geometric identity $b = J / (μ v_{\infty})$ used in Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation clause (ii); the contact trajectory itself is outside that definition's smooth central-model class.

1.2givenL1L2algebra

Deflection: let $α$ be the acute angle between the inward normal $- n$ at first contact and the incoming direction $\hat{u}$ , so that $\sin α = b / a$ by [L2]. The reflection turns the velocity by the angle computed in [L2], namely $π - 2 α$ ; since $b \in (0, a)$ we have $α \in (0, π / 2)$ and $θ = π - 2 α \in (0, π)$ .

2.1step 1.2L3algebra

Branch and inverse: the map $b \mapsto θ (b) = π - 2 \arcsin (b / a)$ is strictly decreasing on $(0, a)$ by [L3] and its image is $(0, π)$ ; solving $\sin ((π - θ) / 2) = b / a$ gives $b (θ) = a \cos (θ / 2)$ on $(0, π)$ , which is $C^{1}$ with $b^{'} (θ) = - \frac{a}{2} \sin (θ / 2) \neq 0$ . Thus the model realises a single deflection branch with image $(0, π)$ , and no branch sum is needed for it.

3.1step 2.1L3L4algebra

Contact pushforward: step 2.1 verifies the abstract branch hypotheses of [L4], and substituting $n = (b e_{φ} - \sqrt{a^{2} - b^{2}} \hat{u}) / a$ into the reflection rule gives $v^{+} / v_{\infty} = (2 b^{2} / a^{2} - 1) \hat{u} + [2 b \sqrt{a^{2} - b^{2}} / a^{2}] e_{φ}$ . The transverse coefficient is positive for $0 < b < a$ , verifying that the outgoing azimuth is $φ$ and the direction map is $T$ . Therefore, for every Borel $B \subseteq Y$ , $σ (B) = \int_{B} b (θ) ∣ b^{'} (θ) ∣ d λ_{2}$ . Since $\sin θ > 0$ , the locally defined differential contact cross-section is $\frac{d σ}{d Ω} (θ) = \frac{b (θ) ∣ b^{'} (θ) ∣}{\sin θ} = \frac{a \cos (θ / 2) \cdot \frac{a}{2} \sin (θ / 2)}{\sin θ}$ . Using $\sin θ = 2 \sin (θ / 2) \cos (θ / 2)$ this equals $\frac{a^{2}}{4}$ for every $θ \in (0, π)$ : the contact pushforward is $(a^{2} / 4) Ω$ , so the outgoing angular distribution of uniformly incident contacts is uniform, and the density is finite, positive and independent of $θ$ .

4.1step 3.1L3L4algebra

Truncated total: by [L4] and the fundamental theorem, for the full image interval $σ ((0, π)) = \int_{0}^{π} \frac{a^{2}}{4} 2 π \sin θ d θ = \frac{a^{2}}{4} \cdot 2 π \cdot [- \cos θ]_{0}^{π} = \frac{a^{2}}{4} \cdot 4 π = π a^{2},$ and for $0 < θ_{1} < θ_{2} < π$ the band value is $\frac{a^{2}}{4} \cdot 2 π (\cos θ_{1} - \cos θ_{2})$ , finite; in particular the branch has a finite ideal total $π a^{2}$ .

5.1givenL1L4∎

Model boundary: the motion just computed is piecewise linear with a contact rule, not a central-force trajectory of class $C^{2}$ with a finite-range potential; the quadrature theorem Finite-range central scattering in the model class: the scattering motion on all of time, the asymptotic velocities and impact parameter $b = J / \sqrt{2 μ E}$ , the advance $Δ$ with principal deflection $\arccos (\cos (π - 2 Δ))$ , and the cross-section of a deflection branch does not apply to it, and the finite total $π a^{2}$ is a property of this contact model, not a counterexample to the forward divergence of the ideal Coulomb branch. The declared data are model data: no measurement is reported, and the agreement of the algebraic form with the cited impenetrable-sphere computation is a consistency check of the model, not evidence about nature.

Remarks

Units. $[a] = [b] = ℓ$ , $[μ] = M$ , $[v_{\infty}] = ℓ τ^{- 1}$ , $[d σ / d Ω] = ℓ^{2}$ , $[σ] = ℓ^{2}$ , and $[j] = M ℓ^{2} τ^{- 1}$ with $j = μ b v_{\infty}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).
Why the rule is not a potential. The impenetrable sphere corresponds to no $C^{2}$ finite-range potential; its force is a contact impulse at $∣ x ∣ = a$ . This example-specific fixed-wall law differs from the finite-mass two-label extension in Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting, so the example illustrates the model boundary rather than an instance of its restitution law or the smooth-potential theorem.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A declared-data Rutherford check: $α = 1$ , $θ = π / 2$ , $(d σ / d Ω) = 4 \times 10^{- 24} m^{2}$ , the band values $π b (θ_{1})^{2}$ and the growth of the truncated total as $θ_{1} ↓ 0$

Physical scope

Assuming countable choice for the scattering measures and vocabulary, A numerical check of the repulsive-Coulomb model prediction on declared model data in SI units: reduced mass $2 \times 10^{- 26} k g$ , energy $10^{- 18} J$ , angular momentum $4 \times 10^{- 34} k g m^{2} s^{- 1}$ and coupling $κ = 4 \times 10^{- 30} J m$ give $v_{\infty} = 10^{4} m s^{- 1}$ , $b = 2 \times 10^{- 12} m$ , $α = 1$ , scattering angle $π / 2$ and density $4 \times 10^{- 24} m^{2}$ ; the angular band values $π (b (θ_{1})^{2} - b (θ_{2})^{2})$ and the small-angle growth $\approx (16 π \times 10^{- 24} m^{2}) / θ_{1}^{2}$ are checked. All numbers are declared model data fed into the model formulas; the example asserts no measurement, no uncertainty budget and no comparison with any experiment.

Example

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited scattering measures and vocabulary.

Declared model data. For a central scattering model of Repulsive Coulomb scattering in the model class: global non-colliding motion, the deflection map $θ = 2 \arctan (κ / (2 E b))$ , the Rutherford density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ , finite angular cutoffs and the divergent ideal total declare the repulsive Coulomb coupling and the reduced data $μ = 2 \times 10^{- 26} k g, E = 10^{- 18} J, J = 4 \times 10^{- 34} k g m^{2} s^{- 1}, κ = 4 \times 10^{- 30} J m,$ all read as model data, not as measured values of any physical pair. Abbreviate $κ_{4} : = κ / (4 E) = 10^{- 12} m$ and $κ_{2} : = κ / (2 E) = 2 \times 10^{- 12} m$ .

Assertions to check. The model quantities of Repulsive Coulomb scattering in the model class: global non-colliding motion, the deflection map $θ = 2 \arctan (κ / (2 E b))$ , the Rutherford density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ , finite angular cutoffs and the divergent ideal total take the values $v_{\infty} = 10^{4} m s^{- 1}, b = 2 \times 10^{- 12} m, α : = \frac{κ}{2 E b} = 1, θ = \frac{π}{2},$ the branch density at $θ = π / 2$ is $(d σ / d Ω) (π / 2) = κ_{4}^{2} \csc^{4} (π / 4) = 4 \times 10^{- 24} m^{2}$ , the angular band $(θ_{1}, θ_{2})$ carries the truncated total $σ = π (b (θ_{1})^{2} - b (θ_{2})^{2})$ with $b (θ) = κ_{2} \cot (θ / 2)$ , and as $θ_{1} ↓ 0$ the band value $π b (θ_{1})^{2}$ grows like $(16 π \times 10^{- 24} m^{2}) / θ_{1}^{2}$ , so no finite ideal total is assigned.

Facts & Assumptions

Given: Countable choice for the scattering measures. The declared model data $μ, E, J, κ$ and the vocabulary of Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation and The differential and truncated total cross-sections of a deflection branch: densities, units, branch sums and the empirical qualifications of the uniform-incident reading.

[L1]

Coulomb model predictions (Repulsive Coulomb scattering in the model class: global non-colliding motion, the deflection map $θ = 2 \arctan (κ / (2 E b))$ , the Rutherford density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ , finite angular cutoffs and the divergent ideal total clauses 2-4, The Rutherford deflection map, its inverse, the cross-section density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ and the divergence of the total cross-section clauses 1-3): for the repulsive $κ / r$ potential, $v_{\infty} = \sqrt{2 E / μ}$ , $b = J / \sqrt{2 μ E}$ , $Δ = π / 2 - \arctan (κ / (2 E b))$ , $θ = π - 2 Δ = 2 \arctan (κ / (2 E b))$ , $b (θ) = (κ / 2 E) \cot (θ / 2)$ , $\frac{d σ}{d Ω} (θ) = (κ / (4 E))^{2} \csc^{4} (θ / 2)$ , $σ ((θ_{1}, θ_{2})) = π (b (θ_{1})^{2} - b (θ_{2})^{2})$ , and the truncated integral over $(0, θ_{0})$ diverges.

[L2]

Trigonometric algebra (The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ , Tangent, cotangent, secant, and cosecant on their exact natural domains, Pythagorean and parity identities for all six trigonometric functions on their natural domains, The addition formulas for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Signs, monotonicity intervals, and ranges of sine and cosine): $\sin^{2} x + \cos^{2} x = 1$ , the addition and parity formulas hold, $\cos (π / 2) = 0$ , and sine and cosine are positive on $(0, π / 2)$ . Cotangent is $\cos / \sin$ and cosecant is $1 / \sin$ on their nonzero-denominator domains; the principal arctangent inverts tangent on $(- π / 2, π / 2)$ . The special values used below are derived in step 1.2.

[L3]

Limit tools (The derivatives of sine and cosine are cosine and minus sine, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, A function differentiable at $c$ is continuous at $c$ , Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): sine and cosine are differentiable on $R$ , $\sin 0 = 0$ , $\cos 0 = 1$ , and $\sin^{'} (0) = 1$ ; differentiability implies continuity. Quotients of functions with finite limits converge to the quotient of those limits when the limiting denominator is nonzero. Here $f (x) \sim g (x)$ means $f (x) / g (x) \to 1$ in the stated limit, with both expressions defined near the endpoint.

Verification

technique · direct

1.1givenL1L2algebra

Speed and impact parameter: $2 E / μ = 2 \times 10^{- 18} / (2 \times 10^{- 26}) = 10^{8} m^{2} s^{- 2}$ , so $v_{\infty} = \sqrt{10^{8}} = 10^{4} m s^{- 1}$ ; then $b = J / (μ v_{\infty}) = 4 \times 10^{- 34} / (2 \times 10^{- 26} \cdot 10^{4}) = 2 \times 10^{- 12} m$ .

1.2L2L3algebra

Special values: at $x = π / 4$ , the double-angle identity gives $0 = \cos (π / 2) = \cos^{2} x - \sin^{2} x$ ; positivity and the Pythagorean identity yield $\sin x = \cos x = 1 / \sqrt{2}$ , hence $\cot x = 1$ , $\csc^{4} x = 4$ and $\arctan 1 = π / 4$ . At $x = π / 6$ , the addition formulas give $0 = \cos (3 x) = 4 \cos^{3} x - 3 \cos x$ . Since $\cos x > 0$ , $\cos x = \sqrt{3} / 2$ and $\sin x = 1 / 2$ , so $\cot x = \sqrt{3}$ . Applying the difference formulas to $π / 12 = π / 4 - π / 6$ now gives $\cot (π / 12) = (\sqrt{6} + \sqrt{2}) / (\sqrt{6} - \sqrt{2}) = 2 + \sqrt{3}$ . Finally continuity and positivity give $\sin (π / 2) = 1$ ; thus $\cot (θ / 2) \to 0$ as $θ ↑ π$ .

2.1step 1.2step 1.1L1L2algebra

Angle: $α = κ / (2 E b) = 4 \times 10^{- 30} / (2 \times 10^{- 18} \cdot 2 \times 10^{- 12}) = 1$ , so $\arctan α = π / 4$ and $Δ = π / 2 - π / 4 = π / 4$ , giving $θ = 2 \arctan 1 = π / 2 \in (0, π)$ , consistent with $b (θ) ∣_{θ = π / 2} = κ_{2} \cot (π / 4) = 2 \times 10^{- 12} m = b$ .

2.2step 1.2step 1.1L1L2algebra

Band values: using $b (θ) = κ_{2} \cot (θ / 2)$ and the endpoint limit $\lim_{θ ↑ π} b (θ) = 0$ , for the band $(π / 2, π)$ the value is $σ = π (b (π / 2)^{2} - 0^{2}) = π (2 \times 10^{- 12})^{2} = 4 π \times 10^{- 24} \approx 1.257 \times 10^{- 23} m^{2}$ ; for $(π / 3, π)$ , $b (π / 3) = 2 \times 10^{- 12} \sqrt{3}$ , so $σ = 12 π \times 10^{- 24} \approx 3.770 \times 10^{- 23} m^{2}$ ; for $(π / 6, π)$ , $b (π / 6) = 2 \times 10^{- 12} (2 + \sqrt{3})$ , so $σ = π \cdot 4 \times 10^{- 24} (2 + \sqrt{3})^{2} \approx 1.750 \times 10^{- 22} m^{2}$ . These are exact expressions in $π$ and $\sqrt{3}$ evaluated to four significant digits; all are finite.

3.1step 1.2step 2.1L1L2algebra

Density at $π / 2$ : $κ_{4} = κ / (4 E) = 10^{- 12} m$ and $\csc^{4} (π / 4) = (\sqrt{2})^{4} = 4$ , so $\frac{d σ}{d Ω} (π / 2) = κ_{4}^{2} \cdot 4 = 4 \times 10^{- 24} m^{2}$ , with dimension $ℓ^{2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

3.2step 2.2L1L2L3algebra

Small-angle growth: the derivative definition and [L3] give $\sin x / x = (\sin x - \sin 0) / (x - 0) \to \sin^{'} (0) = 1$ as $x ↓ 0$ , and continuity gives $\cos x \to 1$ . Since $\sin x > 0$ on $(0, π / 2)$ , quotient limits yield $x \cot x = \cos x / (\sin x / x) \to 1$ . With $x = θ_{1} / 2$ , this proves $\cot (θ_{1} / 2) \sim 2 / θ_{1}$ , so $π b (θ_{1})^{2} = π κ_{2}^{2} \cot^{2} (θ_{1} / 2) \sim 4 π κ_{2}^{2} / θ_{1}^{2} = 16 π \times 10^{- 24} / θ_{1}^{2}$ in $m^{2}$ . At $θ_{1} = 10^{- 2}$ rad, $b (θ_{1}) = 2 \times 10^{- 12} \cot (5 \times 10^{- 3}) \approx 4.000 \times 10^{- 10} m$ and $σ \approx 5.026 \times 10^{- 19} m^{2}$ . The positive asymptotic coefficient implies unbounded band values as $θ_{1} ↓ 0$ , agreeing with Repulsive Coulomb scattering in the model class: global non-colliding motion, the deflection map $θ = 2 \arctan (κ / (2 E b))$ , the Rutherford density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ , finite angular cutoffs and the divergent ideal total clause 4.

4.1givenL1L2∎

Reading: every number above was obtained by substituting the declared model data into the model formulas; no measurement, uncertainty, detector acceptance or sampling model is supplied, so the computation is a check of the model's arithmetic and not a comparison with nature. In particular the divergence of the truncated total at forward angles is a feature of the declared idealisation, and any empirical statement would require a separate experiment item with measured inputs and uncertainties.

Remarks

Why the data are round. The values were chosen so that $α = 1$ and $θ = π / 2$ exactly, making the check arithmetic rather than numerical; this is a declared convenience of the model computation and carries no empirical meaning.
Units. $[κ] = M ℓ^{3} τ^{- 2}$ (here $J m$ ), $[κ / (2 E)] = ℓ$ , $[b] = ℓ$ ; the displayed $σ$ values have dimension $ℓ^{2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A declared-data impact check: $m_{1} = 2 k g$ , $m_{2} = 6 k g$ , $v_{1}^{-} = 3 m s^{- 1}$ , $v_{2}^{-} = - 1 m s^{- 1}$ , with elastic, semi-elastic and sticking outcomes

Physical scope

A declared-data check of the adopted collinear impact model: masses $2$ and $6 k g$ with one-dimensional velocities $3$ and $- 1 m s^{- 1}$ along the contact direction, for coefficients of restitution $e = 1, \frac{1}{2}, 0$ ; the example verifies total momentum continuity, the impulse $J = - 9 N s$ at $e = \frac{1}{2}$ , the post-impact velocities, and the kinetic-energy changes $0, - 9, - 12 J$ , all in SI units. Every number is declared model data substituted into the model formulas; the example reports no measurement and no empirical comparison.

Example

Declared model data. Adopt a collinear impact of two labels of the contact model of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clauses (i)-(ii) along the contact direction $n$ , with masses and incoming velocities $m_{1} = 2 k g, m_{2} = 6 k g, v_{1}^{-} = 3 n m s^{- 1}, v_{2}^{-} = - 1 n m s^{- 1} .$ Thus $M = 8 k g$ , $μ = \frac{2 \cdot 6}{8} = \frac{3}{2} k g$ , $V = \frac{2 \cdot 3 + 6 \cdot (- 1)}{8} = 0$ , and the relative velocity is $d^{-} : = v_{2}^{-} - v_{1}^{-} = - 4 n m s^{- 1}$ .

Assertions to check. For $e = 1$ (elastic): $v_{1}^{+} = - 3 n$ , $v_{2}^{+} = + 1 n$ , $Δ K = 0$ . For $e = \frac{1}{2}$ : $v_{1}^{+} = - \frac{3}{2} n$ , $v_{2}^{+} = + \frac{1}{2} n$ , impulse $J = - 9 n N s$ , $Δ K = - 9 J$ . For $e = 0$ (sticking): $v_{1}^{+} = v_{2}^{+} = 0$ , $Δ K = - 12 J$ . In every case the total momentum $P = 0$ is continuous across the impact, as Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 1 requires, and the energy changes are those of its clause 4.

Facts & Assumptions

Given: The declared masses, incoming velocities and contact direction; the collinear impact model of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting with coefficients $e \in {0, \frac{1}{2}, 1}$ ; the vocabulary and units of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework and Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame.

[L1]

Impact consequences (Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clauses 1, 3, 4, Isolated reciprocal impulse with full-vector restitution: momentum jumps $\pm J$ , conservation of total momentum, the restitution solution and the kinetic-energy change $- \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ clauses a-c): in the collinear model, $m_{1} (v_{1}^{+} - v_{1}^{-}) = J = - m_{2} (v_{2}^{+} - v_{2}^{-})$ with total momentum continuous across the impact; $v_{1}^{+} = V + \frac{m_{2}}{M} e d^{-}$ , $v_{2}^{+} = V - \frac{m_{1}}{M} e d^{-}$ ; and $Δ K = - \frac{1 - e^{2}}{2} μ ∥ d^{-} ∥_{2}^{2}$ .

[L2]

Kinetic split (Centre of mass, the identity $\sum_{i} m_{i} ρ_{i} = 0$ , and the splitting of the kinetic sum $\sum_{i} m_{i} ∥ x_{i}^{'} ∥_{2}^{2}$ into centre-of-mass and relative parts (with the two-body reduced-mass case), The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $K^{\pm} = \frac{M}{2} ∥ V ∥_{2}^{2} + \frac{μ}{2} ∥ d^{\pm} ∥_{2}^{2}$ on the respective sides of impact, where $d^{\pm} = v_{2}^{\pm} - v_{1}^{\pm}$ , with $V$ unchanged across the impact.

Verification

technique · direct

1.1givenL1L2algebra

Common data: $M = 8$ , $μ = \frac{3}{2}$ , $V = 0$ and $d^{-} = - 4 n$ , so $∥ d^{-} ∥_{2}^{2} = 16$ and $\frac{μ}{2} ∥ d^{-} ∥_{2}^{2} = 12 J$ ; the initial kinetic energy is $\frac{1}{2} \cdot 2 \cdot 9 + \frac{1}{2} \cdot 6 \cdot 1 = 9 + 3 = 12 J$ ; the incoming relative speed is $4 m s^{- 1}$ .

2.1step 1.1L1L2algebra

Elastic case $e = 1$ : [L1] gives $v_{1}^{+} = V + \frac{m_{2}}{M} d^{-} = \frac{6}{8} (- 4) n = - 3 n$ and $v_{2}^{+} = V - \frac{m_{1}}{M} d^{-} = - \frac{2}{8} (- 4) n = + 1 n$ , so each velocity is reversed and the total momentum is $2 (- 3) + 6 (1) = 0 = P^{-}$ ; $Δ K = - \frac{1 - 1}{2} μ ∥ d^{-} ∥_{2}^{2} = 0$ , and directly $K^{+} = \frac{1}{2} \cdot 2 \cdot 9 + \frac{1}{2} \cdot 6 \cdot 1 = 12 J = K^{-}$ .

2.2step 1.1L1L2algebra

Semi-elastic case $e = \frac{1}{2}$ : $v_{1}^{+} = \frac{6}{8} \cdot \frac{1}{2} \cdot (- 4) n = - \frac{3}{2} n$ , $v_{2}^{+} = - \frac{2}{8} \cdot \frac{1}{2} \cdot (- 4) n = + \frac{1}{2} n$ ; the impulse is $J = m_{1} (v_{1}^{+} - v_{1}^{-}) = 2 (- \frac{3}{2} - 3) n = - 9 n N s$ , and indeed $m_{2} (v_{2}^{+} - v_{2}^{-}) = 6 (\frac{1}{2} + 1) n = 9 n = - J$ ; the total momentum is $2 (- \frac{3}{2}) + 6 (\frac{1}{2}) = 0 = P^{-}$ ; $Δ K = - \frac{1 - \frac{1}{4}}{2} \cdot \frac{3}{2} \cdot 16 = - \frac{3}{4} \cdot 12 = - 9 J$ , and directly $K^{+} = \frac{1}{2} \cdot 2 \cdot \frac{9}{4} + \frac{1}{2} \cdot 6 \cdot \frac{1}{4} = \frac{9}{4} + \frac{3}{4} = 3 J = 12 - 9$ .

2.3step 1.1L1L2algebra

Sticking case $e = 0$ : $v_{1}^{+} = v_{2}^{+} = V = 0$ ; the total momentum is $0 = P^{-}$ ; $Δ K = - \frac{1}{2} \cdot \frac{3}{2} \cdot 16 = - 12 J$ , so $K^{+} = 0$ ; the impulse is $J = 2 (0 - 3) n = - 6 n N s$ , consistent with $m_{2} (0 + 1) n = 6 n = - J$ .

3.1step 2.1step 2.2step 2.3givenL1∎

Reading: all three outcomes satisfy the reciprocal impulse equations with different impulses and different energy changes; the momentum data alone ( $P = 0$ ) are compatible with all of them, illustrating clause 5 of Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws: the coefficient $e$ is model data and must be supplied by the contact law. Every number above is a declared model datum substituted into the model formulas; no measurement, uncertainty or empirical inference is involved.

Remarks

Units. $[P] = k g m s^{- 1} = μ ℓ τ^{- 1}$ , $[J] = N s = μ ℓ τ^{- 1}$ , $[K] = J = μ ℓ^{2} τ^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); the impulse values displayed are the model's impulse vector components along $n$ .
Frame reading. The one-sided centre-of-mass velocities at the impact are both zero. Thus the centre-of-mass kinetic energy is zero on both sides of that instant: the elastic case reverses the relative velocity, and the inelastic cases decrease its kinetic energy only. The supplied data do not imply rest at other times; external forces can accelerate the centre of mass away from the impact.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails

Physical scope

A declared-data check of the open-body material accounting: a body of initial mass $m_{0} = 3 k g$ losing mass at $r_{0} = 1 k g s^{- 1}$ has $m (t) = m_{0} - r_{0} t$ . With no external force and ejecta of relative speed $u_{e} = 2 m s^{- 1}$ opposite a fixed unit direction $n$ , its velocity from rest is $v (t) = u_{e} \ln (m_{0} / m (t)) n$ . It satisfies the material balance identities and violates the naive closed-system law $(m v)^{'} = F$ ; the example verifies the identities, the thrust $2 N$ and the failure of the naive law on declared model data. It reports no measurement, no engine data and no empirical comparison.

Example

Declared model data. Adopt the open-body material accounting of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii) with, on the open interval $I = (- \frac{1}{2} s, \frac{5}{2} s)$ containing the initial time $0$ , set $m_{0} : = 3 k g$ , $r_{0} : = 1 k g s^{- 1}$ , and $m (t) : = m_{0} - r_{0} t, F \equiv 0, r_{i n} \equiv 0, r_{o u t} \equiv r_{0}, w_{o u t} (t) : = v (t) - u_{e} n,$ with a fixed unit vector $n$ , $u_{e} : = 2 m s^{- 1}$ , and initial velocity $v (0) = 0$ . The ejecta therefore leave with the body's velocity reduced by the constant exhaust speed $u_{e}$ along $n$ . Set $q : = t / (1 s)$ , the dimensionless numerical time, so $m (t) = (3 - q) k g$ . Primes denote differentiation with respect to physical frame time $t$ , read in seconds.

Assertions to check. The balance identities hold with $v (t) = u_{e} \ln \frac{m_{0}}{m (t)} n = 2 \ln \frac{3}{3 - q} n m s^{- 1} (t \in I),$ the thrust is $r_{o u t} u_{e} = 2 N$ along $n$ , and the naive closed-system law $(m v)^{'} = F$ fails: it would force $v \equiv 0$ on $I$ .

Facts & Assumptions

Given: The declared maps $m, F, r_{i n}, r_{o u t}, w_{o u t}$ and the vocabulary of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework.

[L1]

Material accounting and derived law (Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii), Variable-mass momentum balance: from $m^{'} = r_{i n} - r_{o u t}$ and $(m v)^{'} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ to $m v^{'} = F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v)$ , Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 6): the model requires $m^{'} = r_{i n} - r_{o u t}$ and $(m v)^{'} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ , equivalently $m v^{'} = F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v)$ .

[L2]

Calculus (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): for real-valued differentiable functions, $(\ln)^{'} = 1 / x$ on $(0, \infty)$ , the chain rule and product rule hold, and a curve with zero derivative on an interval is constant.

Verification

technique · direct

1.1givenL1L2algebra

Mass identity: $m^{'} = - r_{0} = r_{i n} - r_{o u t}$ on $I$ , in $k g s^{- 1}$ . Since $- 1 / 2 < q < 5 / 2$ , $m (t) = (3 - q) k g > 0$ , so the first balance identity and the positive-mass condition hold.

2.1step 1.1givenL2algebra

Velocity candidate: define $v (t) : = u_{e} \ln (m_{0} / m (t)) n$ . By [L2] and $m^{'} = - r_{0}$ , $\frac{d}{d t} \ln (m_{0} / m (t)) = r_{0} / m (t)$ , so $v^{'} (t) = u_{e} r_{0} n / m (t)$ , with acceleration units $m s^{- 2}$ , and $v (0) = u_{e} \ln 1 \cdot n = 0$ . The logarithm acts on the positive dimensionless ratio $m_{0} / m (t)$ .

3.1step 1.1step 2.1L1algebra

Strong form: $m v^{'} = r_{0} u_{e} n$ , while $F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v) = - r_{0} ((v - u_{e} n) - v) = r_{0} u_{e} n$ . Thus the derived law holds with thrust $r_{0} u_{e} = 2 N$ along $n$ .

4.1step 1.1step 3.1L1L2algebra

Material identities: the product rule gives $(m v)^{'} = m^{'} v + m v^{'} = - r_{0} v + r_{0} u_{e} n = - r_{0} w_{o u t} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ . Along with step 1.1 this verifies both balance identities of [L1], with every term in this momentum-rate identity having force units.

5.1step 4.1L1L2algebra

Failure of the naive law: $(m v)^{'} = F \equiv 0$ would make $m v$ constant and, since $m (0) v (0) = 0$ and $m > 0$ , would force $v \equiv 0$ by [L2]. But the computed $v (t)$ is nonzero for $0 < t < 2 s$ ; at $t = 2 s$ , $(m v)^{'} = (2 - 2 \ln 3) n N \approx - 0.197 n N \neq F$ . Hence the naive law fails for this open body, as Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 6(b) records.

6.1step 5.1L1L2algebra∎

Tsiolkovsky value: at $t = 1 s$ , $q = 1$ , $m = 2 k g$ and $v = 2 \ln (3 / 2) n m s^{- 1} \approx 0.8109 n m s^{- 1}$ . The momentum is approximately $1.6219 n k g m s^{- 1}$ , whereas the naive law gives zero. These values come from the exact logarithm before rounding; all are declared model computations, with no measurement or empirical inference.

Remarks

Signs and conventions. The ejecta velocity $w_{o u t} = v - u_{e} n$ is expressed in the same inertial frame as $v$ , as Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii) requires; $u_{e} > 0$ is the exhaust speed relative to the body, and the thrust $r_{o u t} (v - w_{o u t}) = r_{o u t} u_{e} n$ points along $+ n$ .
Units. $[m] = k g$ , $[r_{o u t}] = k g s^{- 1}$ , $[u_{e}] = m s^{- 1}$ , $[F] = N = k g m s^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); $m v^{'}$ and $r_{o u t} u_{e}$ both have the dimension of force.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A declared smooth open flow: the uniform scaling $ρ = ρ_{0} e^{- 3 α t}$ , $u = α x$ satisfies the control-volume identities on fixed and material regions, and elbow port data illustrate the inlet/outlet decomposition

Physical scope

Two declared checks of the finite-regularity open-flow bookkeeping: (i) the exact uniform-scaling model $ρ (t) = ρ_{0} e^{- 3 α t}$ , $u (t, x) = α x$ , $p = p_{0} - ρ α^{2} ∣ x ∣^{2} / 2$ , with the control maps specified and the units of $α, ρ_{0}, p_{0}$ declared; (ii) a conditional steady-elbow calculation with declared inlet/outlet data, where the balance gives mass flow $0.4 k g s^{- 1}$ and requires a net external resultant on the fluid of magnitude about $3.578 N$ . No pressure-free momentum-flux force decomposition is inferred, since no boundary-stress data are given. Both are computations on declared model data; the example reports no measurement and does not solve a boundary-value problem.

Example

(i) Declared uniform scaling. Fix constants $ρ_{0} > 0$ , $α \neq 0$ and $p_{0}$ , with $[ρ_{0}] = μ ℓ^{- 3}$ , $[α] = τ^{- 1}$ and $[p_{0}] = μ ℓ^{- 1} τ^{- 2}$ . On an open interval $I ∋ 0$ use uniform scaling, meaning expansion for $α > 0$ and contraction for $α < 0$ , and set $ρ (t) = ρ_{0} e^{- 3 α t}, u (t, x) = α x, σ (t, x) = - p (t, x) I d, p (t, x) = p_{0} - \frac{ρ (t) α^{2}}{2} ∣ x ∣^{2}, b \equiv 0,$ and fix radii $R_{0} > 0$ and $R > 0$ with units $ℓ$ . Let the material region be a ball $B_{R_{B}}$ with radius $R_{B} > \max (R, R_{0})$ . Define the material motion $χ (t, X) = e^{α t} X$ on a neighbourhood of $\overline{B_{R_{B}}}$ . More precisely, choose an open ball $O = B_{R_{O}}$ with $R_{O} > R_{B}$ and define $χ : I \times O \to R^{3}$ by this formula. Its restriction to $B_{R_{0}}$ has current radius $R (t) = R_{0} e^{α t}$ . Define the fixed-ball control motion by $ψ_{R} (t, X) = X$ for $X \in O$ , with control region $B_{R}$ , and the material-ball control motion by $ψ_{m a t} (t, X) = e^{α t} X$ for $X \in O$ , with control region $B_{R_{0}}$ . Each map is a $C^{2}$ motion on the open reference domain $O$ and has positive spatial Jacobian. Take the fields on a common open neighbourhood of the closed sweep of $χ$ and the closed sweeps of these controls. The fields satisfy the adopted balances of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws (the density is spatially uniform, the velocity is $α x$ , and the pressure is quadratic).

(ii) Declared elbow control-volume data. Take a material region $B$ with motion $χ$ and associated spatial velocity field $u$ as in Continuum motions, spatial velocity, material acceleration and swept regions, and a material subregion $B^{'} \subseteq B$ with $C^{1}$ boundary. On an open reference neighbourhood $O \supset \overline{B^{'}}$ , define the fixed control motion $ψ : I \times O \to R^{3}$ by $ψ (t, X) = X$ ; its control region is $C_{t} = C : = B^{'}$ and its boundary velocity is $w = 0$ . Adopt the material balances of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws for this model and the local mass and momentum equations on an open field neighbourhood of the closed control region, as allowed by that theorem's local-equation alternative.

Assume $ρ = 1000 k g m^{- 3}$ is constant and the fields are time-independent on that neighbourhood. The bounded $C^{1}$ boundary of $C$ is partitioned into finitely many $C^{1}$ boundary pieces, up to surface-measure-zero edges; designate an inlet patch $Σ_{i n}$ , an outlet patch $Σ_{o u t}$ , and let the remaining pieces form an impermeable remainder $Σ_{0}$ . The outward unit normals on the ports are $- \hat{x}$ and $+ \hat{y}$ . On the ports take $u = q_{i n} \hat{x}$ and $u = q_{o u t} \hat{y}$ , respectively, where each $q$ is a nonnegative, nonzero $C^{1}$ profile supported compactly inside its port, with units $ℓ τ^{- 1}$ ; take $u \cdot ν = 0$ on the remainder. These traces vanish near the port edges and are compatible with a $C^{1}$ field and continuous boundary normal. For either port define $Q : = \int_{Σ} q d S > 0, v_{e f f} : = \frac{\int_{Σ} q^{2} d S}{Q}, A_{e f f} : = \frac{Q^{2}}{\int_{Σ} q^{2} d S},$ so $Q = A_{e f f} v_{e f f}$ , the mass rate is $ρ Q$ , and the flux-averaged velocity is $v_{e f f}$ in the port direction. Declare $A_{e f f, i n} = 10^{- 4} m^{2}$ , $A_{e f f, o u t} = 5 \times 10^{- 5} m^{2}$ and $v_{e f f, i n} = 4 m s^{- 1}$ ; the outlet effective speed is unknown. The effective areas are defined by these integrals, not the geometric patch areas. Assume the time-independent velocity field is $C^{1}$ on the common neighbourhood. Set the body-force field to $b = 0$ and assume the stress field is $C^{1}$ on the same neighbourhood. These are conditional model premises for a balance calculation; no existence claim for a boundary-value solution or decomposition of the boundary stress is made. The calculation determines only the net external resultant required by the steady balance.

Facts & Assumptions

Given: The declared fields and control motions of the Example, the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), assumed as in the statements of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws and Control-volume balance: mass and momentum rates for a moving non-material region, whose transport and divergence deductions use it, and the control-volume theorem The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws with its clauses 1-3.

[A1]

The Axiom of Countable Choice is assumed (The Axiom of Countable Choice ( ${A C}_{ω}$ )), inherited from The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws; the elementary verifications of steps 1.1-4.1 use no further choice.

[L1]

Balance and control-volume identities (The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws clauses 1-3, Control-volume balance: mass and momentum rates for a moving non-material region, Continuum motions, spatial velocity, material acceleration and swept regions, Euclidean spatial calculus: gradient, divergence and material derivative, Variable-mass momentum balance: from $m^{'} = r_{i n} - r_{o u t}$ and $(m v)^{'} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ to $m v^{'} = F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v)$ ): for a model satisfying the adopted local mass and momentum balances, (CV1) and (CV2) hold for every control motion with bounded $C^{1}$ control regions whose closed sweep lies inside the common field domain where those local equations hold; if the boundary splits into inlet and outlet with the stated signs, then $m^{'} = r_{i n} - r_{o u t}$ and $P^{'} = F_{e x t} + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ .

[L2]

Elementary calculus and algebra (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, $C^{k}$ Euclidean maps and diffeomorphisms, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): $\frac{d}{d t} e^{- 3 α t} = - 3 α e^{- 3 α t}$ ; for the field $u = α x$ , $div u = 3 α$ , $u \cdot \nabla u = α^{2} x$ and $\nabla ∣ x ∣^{2} = 2 x$ ; the surface integral of the unit normal over a sphere vanishes, and odd integrands over a centred ball integrate to zero.

Verification

technique · direct

1.1givenA1L1L2algebra

Local balances for (i): $\partial_{t} ρ + div (ρ u) = - 3 α ρ + ρ div u = - 3 α ρ + 3 α ρ = 0$ , and $ρ D_{t} u = ρ (0 + α^{2} x) = ρ α^{2} x$ while $ρ b + div σ = - \nabla p = ρ α^{2} x$ ; hence the declared fields satisfy the local balances (K) and (KM) of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws. For each admissible material subregion, applying Control-volume balance: mass and momentum rates for a moving non-material region with control motion $ψ = χ$ makes $w = u$ and gives the adopted material balances; thus the premises of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws hold. The fixed and material balls stay in the stipulated neighbourhoods of their closed sweeps, so its control-volume identities apply to them. For (ii), the fixed elbow control region stays in its declared common field domain, its boundary has nonpositive flux on the inlet patch, nonnegative flux on the outlet patch and zero flux on the remainder. The local equations are adopted there, so (CV1) and (CV2) of [L1] apply. Splitting their integrals over these patches remains valid when a profile vanishes: those subsets contribute zero.

1.2givenL1L2algebra

Inlet/outlet data for (ii): the fixed-region mass identity (CV1), steady density and velocity, and $u \cdot ν = 0$ on $Σ_{0}$ give equal inlet and outlet rates, $ρ A_{e f f, i n} v_{e f f, i n} = ρ A_{e f f, o u t} v_{e f f, o u t}$ . Hence $\dot{m} = 1000 \cdot 10^{- 4} \cdot 4 = \frac{2}{5} k g s^{- 1} = ρ A_{e f f, o u t} v_{e f f, o u t}$ , so $v_{e f f, o u t} = \frac{2}{5} / (1000 \cdot 5 \times 10^{- 5}) = 8 m s^{- 1}$ ; thus $r_{i n} = r_{o u t} = \frac{2}{5} k g s^{- 1}$ and the flux-averaged velocities are $w_{i n} = 4 \hat{x}$ , $w_{o u t} = 8 \hat{y}$ in $m s^{- 1}$ .

2.1step 1.1L1L2algebra

Fixed-ball check: for $C = B_{R}$ one has $m_{C} = \frac{4 π}{3} ρ (t) R^{3}$ and hence $m_{C}^{'} = \frac{4 π}{3} (- 3 α ρ) R^{3} = - 4 π α ρ R^{3}$ ; on the other hand $(u - 0) \cdot ν = α R$ on $\partial B_{R}$ , so $- \int_{\partial C} ρ (u - 0) \cdot ν d S = - ρ α R \cdot 4 π R^{2} = - 4 π α ρ R^{3}$ , which is (CV1). For (CV2): $P_{C} = \int_{B_{R}} ρ α x d x = 0$ (odd integrand) and $\int_{\partial C} ρ u (u \cdot ν) d S = ρ α^{2} R \int_{\partial C} x d S = 0$ and $\int_{\partial C} σ ν d S = - p (R) \int_{\partial C} ν d S = 0$ , with $\int_{B_{R}} ρ b = 0$ ; both sides of (CV2) vanish. At $ρ_{0} = 1000 k g m^{- 3}$ , $R = \frac{1}{10} m$ , $α = \frac{1}{10} s^{- 1}$ , $t = 0$ : $m_{C} = \frac{4 π}{3} k g$ and $m_{C}^{'} = - \frac{2 π}{5} k g s^{- 1}$ , agreeing with the flux value.

2.2step 1.2L1L2algebra

Net external resultant on the fluid: for steady fields on a fixed control region with zero body force, the fixed-control momentum identity (CV2) of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws, split over the three patches, gives $0 = F_{e x t} + \dot{m} w_{i n} - \dot{m} w_{o u t}$ . Hence the net external resultant required by the adopted balance is $F_{e x t} = \dot{m} (w_{o u t} - w_{i n}) = \frac{2}{5} (8 \hat{y} - 4 \hat{x}) N = (- \frac{8}{5} \hat{x} + \frac{16}{5} \hat{y}) N$ , with magnitude $\frac{8 \sqrt{5}}{5} N \approx 3.578 N$ . This is the resultant of all external tractions on the fluid under the stated model equations; no pressure or wall-traction decomposition, and no device-force value, is determined by the supplied data.

3.1step 1.1L1L2algebra

Material-ball check: use the material control motion $ψ_{m a t} (t, X) = e^{α t} X$ on $B_{R_{0}}$ , so $C_{t} = B_{R (t)}$ with $R (t) = R_{0} e^{α t}$ . For $x = ψ_{m a t} (t, X)$ on the boundary, $w = \partial_{t} ψ_{m a t} (t, X) = α x = u (t, x)$ , so $(u - w) \cdot ν = 0$ ; then (CV1) gives $m_{C}^{'} = 0$ , and indeed $m_{C} = \frac{4 π}{3} ρ (t) R (t)^{3} = \frac{4 π}{3} ρ_{0} R_{0}^{3}$ is constant. In (CV2), $P_{C}^{'} = \int_{C_{t}} ρ b + \int_{\partial C_{t}} σ ν$ with both sides zero as in step 2.1 (now at the moving radius), consistent with $P_{C} = 0$ . This is the material special case of The finite-regularity open-flow model: moving-control-region mass and momentum balances, their material and fixed special cases, and the inlet/outlet decomposition reproducing the open-body and rocket laws clause 2.

4.1givenL1L2∎

Reading: the uniform-scaling fields are an exact solution of the adopted balances, so steps 2.1 and 3.1 check the control-volume identities for the explicitly specified fixed and material control motions. In the elbow calculation, the vector and its magnitude are the conditional net external resultant required by the steady momentum balance; the example supplies no boundary stresses with which to decompose that resultant into pressure and wall contributions. It is not a solution of the full boundary-value problem. All data are model data; no measurement, apparatus or uncertainty is involved.

Remarks

What the exact solution does and does not solve. The uniform scaling satisfies the local balances and therefore the control-volume identities; it does not satisfy any particular boundary condition on a wall, because the velocity field is defined on all of space. The material ball is a material region of the declared motion, consistent with Continuum motions, spatial velocity, material acceleration and swept regions.
Boundary forces in the elbow. The number $3.578 N$ is the magnitude of the net external resultant on the fluid required by the adopted steady balance. Equal inlet and outlet pressures and inviscid-wall language would not by themselves determine the wall pressure traction; no pressure-free flux force or device force is isolated here. The example illustrates the inlet/outlet decomposition and is not a design formula.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Smooth finite-range free scattering refutes global strict decrease; as a separate contact comparison, an attractive refracting shell with index $3 / 2$ has $θ (b) = 2 (\arcsin (b / a) - \arcsin (2 b / (3 a)))$ strictly increasing, with a jump down to $0$ at grazing incidence

Physical scope

Assuming countable choice for the scattering measures and vocabulary, the smooth finite-range model has zero deflection for every impact parameter beyond its range, so any globally defined deflection map is constant on this tail and cannot be strictly decreasing. The zero potential is a genuine smooth witness with unique deflection for every positive impact parameter. As a separate example-specific contact model, the adopted attractive refracting shell of index 3/2 has a strictly increasing branch on $0 < b < a$ and a jump to zero at grazing incidence $b = a$ ; the endpoint extension at $b = 0$ is not a branch value, and the contact model is outside the smooth central-scattering class and the restricted restitution postulate. All computations concern declared model data and report no observations.

Statement refuted

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited scattering measures and vocabulary.

False statement. Let a conservative central scattering model with a finite-range interaction, positive energy and nonzero angular momentum have a unique scattering angle for every impact parameter. Then the scattering angle is a strictly decreasing function of the impact parameter.

The smooth finite-range witness below refutes this claim within the exact central-scattering class of Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation: the zero potential has deflection $0$ at every positive impact parameter. More generally, beyond the range of any finite-range potential the scattering trajectories are free, so a globally defined deflection map has a constant tail. The separate adopted refracting-shell comparison retains its increasing branch; that contact rule is outside the smooth $C^{2}$ class and is not used as the witness for the stated claim. Finite-range central scattering in the model class: the scattering motion on all of time, the asymptotic velocities and impact parameter $b = J / \sqrt{2 μ E}$ , the advance $Δ$ with principal deflection $\arccos (\cos (π - 2 Δ))$ , and the cross-section of a deflection branch clause 5 makes no global monotonicity assertion, and The differential and truncated total cross-sections of a deflection branch: densities, units, branch sums and the empirical qualifications of the uniform-incident reading uses specified deflection branches instead of inverting a global decreasing map.

Facts & Assumptions

Given: Countable choice for the scattering measures. A length $a > 0$ , a reduced mass $μ > 0$ , a positive energy $E > 0$ , with $[a] = ℓ$ , $[μ] = μ$ and $[E] = [U_{0}] = μ ℓ^{2} τ^{- 2}$ ; the speeds $v_{\infty}, v_{i n}$ have units $ℓ τ^{- 1}$ and the angle parameters are dimensionless. Also adopt the following separate refracting-shell model: the label moves freely for $r \neq a$ ; at $r = a$ the example-specific radial contact rule preserves the tangential velocity component and conserves energy with a potential drop $U_{0} > 0$ inside, so the speed inside is $v_{i n} = \sqrt{v_{\infty}^{2} + 2 U_{0} / μ}$ with $v_{\infty} = \sqrt{2 E / μ}$ . Choose the index $n = \frac{3}{2}$ so that $v_{i n} = \frac{3}{2} v_{\infty}$ and $U_{0} = \frac{5}{4} E$ . This interface rule is adopted here and is not supplied by Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (ii), which covers only finite-mass collinear two-label restitution. The incoming motion is a straight line with impact parameter $b \geq 0$ and speed $v_{\infty}$ . The interface rule is applied at transverse crossings; at grazing $b = a$ adopt the free continuation. Separately, for the genuine smooth witness use the central model on time window $R$ with $u = 0$ and two equal label masses $2 μ$ .

Trajectory convention for the contact comparison. In the plane spanned by the orthonormal vectors $e_{x}, e_{y}$ , time is $t \in R$ in the fixed inertial chart and the label's position relative to the shell centre is a continuous curve $γ_{b} : R \to R^{3}$ . For $0 < b < a$ it is piecewise affine, with one incoming free segment, one interior chord and one outgoing free segment; its one-sided velocity limits at the two contact times obey the separately adopted interface rule. For $b > a$ it is a free line, and for $b = a$ the adopted free grazing continuation is used. The impact parameter is the perpendicular distance of the incoming affine line from the origin. At $b = 0$ the same rule gives the radial diameter path, while the extension $θ (0) = 0$ is only an endpoint convention for the branch. This contact path is not asserted to be a motion of the $C^{2}$ central-potential model.

[L1]

For the separately adopted interface rule of the Given clause, tangential-velocity preservation and the energy change imply Snell's relation $\sin α = n \sin β$ at entry; by chord symmetry the exit is the mirror configuration. The inside radial kinetic energy at $r = a$ is $\frac{μ}{2} (v_{i n}^{2} - v_{\infty}^{2} b^{2} / a^{2}) = \frac{μ}{2} v_{\infty}^{2} (n^{2} - b^{2} / a^{2}) > 0$ for $b < a$ . The outgoing radial energy is $\frac{μ}{2} v_{\infty}^{2} (1 - b^{2} / a^{2}) > 0$ , so the label reaches the second intersection and transmits without internal reflection. This separate rule is not the finite-mass collinear contact law of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (ii). (The cross product in $R^{3}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ )

[L2]

Chord and refraction geometry (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The addition formulas for sine and cosine): for an entry point $P = a e_{x}$ , incident direction $\hat{u} = - \cos α e_{x} + \sin α e_{y}$ with $\sin α = b / a$ , inside direction ${\hat{u}}_{i n} = - \cos β e_{x} + \sin β e_{y}$ and chord length $2 a \cos β$ , the exit point is $Q = a (- \cos 2 β e_{x} + \sin 2 β e_{y})$ with outward normal ${\hat{n}}_{Q} = Q / a$ , and the outgoing direction is ${\hat{u}}_{o u t} = \cos α {\hat{n}}_{Q} + \sin α {\hat{t}}_{Q}$ with ${\hat{t}}_{Q} : = - \sin 2 β e_{x} - \cos 2 β e_{y}$ ; expanding gives ${\hat{u}}_{o u t} = - \cos (2 β - α) e_{x} + \sin (2 β - α) e_{y}$ , so the angle between $\hat{u}$ and ${\hat{u}}_{o u t}$ is $2 (α - β)$ .

[L3]

Inverse trigonometry and calculus: $\arcsin : [- 1, 1] \to [- π / 2, π / 2]$ is continuous, strictly increasing and satisfies $\sin (\arcsin x) = x$ (Principal inverse sine and inverse cosine, Signs, monotonicity intervals, and ranges of sine and cosine). Sine and cosine are differentiable with $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine); $\sin^{2} s + \cos^{2} s = 1$ (Pythagorean and parity identities for all six trigonometric functions on their natural domains). The mean value theorem applies to sine on compact intervals; continuity and the intermediate value theorem control its cosine's sign (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Derivatives are difference-quotient limits (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), and the chain and algebra rules and positive-derivative monotonicity theorem apply (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed). The inverse derivative needed here is derived in step 3.1 rather than assumed from the inverse definition.

[L4]

Smooth central scattering (Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation clauses (i)-(iii), Finite-range central scattering in the model class: the scattering motion on all of time, the asymptotic velocities and impact parameter $b = J / \sqrt{2 μ E}$ , the advance $Δ$ with principal deflection $\arccos (\cos (π - 2 Δ))$ , and the cross-section of a deflection branch clauses 1-3 and 5): the adopted smooth model has masses $m_{1}, m_{2} > 0$ , reduced mass $μ$ and static $C^{2}$ potential $u$ . For $u = 0$ on $[R, \infty)$ , the effective potential on that region is $V (r) = J^{2} / (2 μ r^{2})$ . The theorem applies to a model trajectory attaining a simple outer turning radius with $V = E$ and $V < E$ beyond it; the scattering angle is the angle between the nonzero asymptotic relative velocities. No global monotonicity is asserted.

Counterexample

technique · direct

1.1givenL1L2

Motion: for $0 < b < a$ the model of the Facts admits the exact broken-straight motion: an incoming straight line, a chord inside $r < a$ (the label moves freely there with speed $v_{i n}$ ), and an outgoing straight line; the entry and exit data are related by the two conservation rules as in [L1], and [L2] computes the outgoing direction. For $b > a$ the line misses the shell; at $b = a$ it has the adopted free grazing continuation. In both cases the deflection is $0$ .

1.2givenL4algebra

Genuine smooth finite-range witness and the general obstruction: fix an adopted smooth central model on time window $R$ with $u (r) = 0$ for $r \geq R$ , and let $b > R$ , $v_{\infty} = \sqrt{2 E / μ}$ and $r_{b} (t) = (v_{\infty} t, b, 0)$ . Then $∣ r_{b} (t) ∣ \geq b > R$ , so $u = u^{'} = 0$ along it and $r_{b}^{''} = 0$ solves the reduced equation. With $M = m_{1} + m_{2}$ , the curves $x_{1} = (m_{2} / M) r_{b}$ , $x_{2} = - (m_{1} / M) r_{b}$ have separation $r_{b}$ , zero accelerations and zero force, hence are motions of the adopted model with zero centre of mass. Their angular momentum has magnitude $J = μ b v_{\infty} > 0$ and energy $E$ ; at the turning radius $b$ , $V (b) = E$ , $V^{'} (b) = - J^{2} / (μ b^{3}) < 0$ and $V (r) < E$ for every $r > b$ , matching [L4]. Both asymptotic relative velocities equal $(v_{\infty}, 0, 0)$ , so the deflection is uniquely $0$ . Any incoming line with impact parameter $b > R$ stays outside the interaction region and continues freely; thus every finite-range model has this constant deflection tail. In particular $u = 0$ is $C^{2}$ and finite-range, and the same construction works for every $b > 0$ , giving a unique angle $0$ for every positive impact parameter; two distinct positive parameters therefore violate the asserted strict decrease.

2.1step 1.1L2algebra

Deflection formula: by [L2], for $b \in (0, a)$ the angle between the incoming and outgoing directions is $θ (b) = 2 (α - β)$ with $\sin α = b / a$ and $\sin β = \sin α / n = b / (n a)$ , that is $θ (b) = 2 (\arcsin \frac{b}{a} - \arcsin \frac{2 b}{3 a}) (0 < b < a) .$

2.2step 1.1L1L2algebra

Consistency of the contact rule: at the entry the angular momentum about the centre is $μ b v_{\infty}$ , and inside it equals $μ a v_{i n} \sin β = μ a n v_{\infty} \sin β = μ b v_{\infty}$ , so angular momentum is conserved across the interface as the radial contact force requires; the energy is conserved by construction of $v_{i n}$ with $U_{0} = \frac{5}{4} E$ . This separate contact model is conservative and finite-range as an adopted radial rule; it is not a member of the smooth central-scattering class.

3.1step 2.1L3algebra

Inverse derivative and monotonicity: fix $x \in (- 1, 1)$ and put $s = \arcsin x \in (- π / 2, π / 2)$ . Cosine is continuous and nonzero on that interval, since $\cos^{2} s = 1 - \sin^{2} s > 0$ there; the intermediate value theorem and $\cos 0 = 1$ therefore give $\cos s = \sqrt{1 - x^{2}} > 0$ . For $y \in (- 1, 1)$ with $y \neq x$ , put $q = \arcsin y$ . By inverse continuity $q \to s$ as $y \to x$ , and strict increase gives $q \neq s$ . The mean value theorem on the segment joining $q$ and $s$ gives $y - x = \cos ξ (q - s)$ for some $ξ$ between them. Consequently $ξ \to s$ and $\frac{\arcsin y - \arcsin x}{y - x} = \frac{1}{\cos ξ} ⟶ \frac{1}{\cos s} = \frac{1}{\sqrt{1 - x^{2}}} .$ Thus $\arcsin^{'} (x) = (1 - x^{2})^{- 1 / 2}$ on $(- 1, 1)$ . Applying the chain and algebra rules to step 2.1 gives $θ^{'} (b) = 2 (\frac{1}{\sqrt{a^{2} - b^{2}}} - \frac{1}{\sqrt{n^{2} a^{2} - b^{2}}}) > 0 (0 < b < a),$ because $n > 1$ gives $\sqrt{n^{2} a^{2} - b^{2}} > \sqrt{a^{2} - b^{2}} > 0$ . The positive-derivative theorem therefore makes $θ$ strictly increasing on $(0, a)$ , the opposite of the claimed decrease.

4.1step 3.1L3algebra

Endpoint limits and the grazing jump: the continuous endpoint extension at $b = 0$ is $θ (0) = 0$ , which is not a branch value because the branch domain is $(0, a)$ . As $b ↑ a$ , $α \to π / 2$ and $β \to \arcsin (1 / n) = \arcsin \frac{2}{3}$ , so $θ (b) \to π - 2 \arcsin \frac{2}{3} \approx 1.682$ rad $\approx {96.4}^{\circ}$ ; at grazing $b = a$ and for every $b > a$ the adopted free continuation has deflection $0$ . This value lies outside the branch's angular domain $(0, π)$ . The full contact-model deflection therefore increases on $(0, a)$ and then drops to $0$ , and it is not injective.

5.1step 1.2step 1.1step 4.1step 2.2L4∎

Conclusion: step 1.2 refutes the stated global strict-decrease claim with a genuine smooth central-scattering model satisfying its positive-energy, nonzero-angular-momentum and unique-angle hypotheses. More generally, any finite-range model whose angle is defined for every positive impact parameter has the constant free tail established there, so global strict decrease fails even when the interaction inside the range is nontrivial. Steps 1.1-4.1 separately retain the increasing refracting-shell branch and its adopted grazing jump; that contact comparison lies outside the smooth class and supplies no claim that $C^{2}$ regularity implies or is necessary for monotonicity. The central-scattering theorem itself records no global monotonicity claim and assumes a chosen nondegenerate branch for its cross-section formula.

Remarks

What is and is not claimed about the model. The refracting-shell rule is adopted, exactly as the reflection rule of Elastic reflection from a fixed sphere: the branch $b = a \cos (θ / 2)$ , the uniform density $a^{2} / 4$ and the total $π a^{2}$ of a contact model: the tangential-velocity and energy conditions are the model's contact law, consistent with a radial impulsive force at the shell. Unlike the impenetrable-sphere contact, this rule is not specular reflection; the inside speed exceeds the outside speed and the contact bends the path toward the normal.
The grazing discontinuity. As $b ↑ a$ , the interior chord length $2 a \cos β$ tends to $2 a \sqrt{5} / 3 > 0$ , since $\sin β \to 2 / 3$ . At $b = a$ the separately adopted free grazing path has no interior chord, and the deflection jumps from the branch's endpoint limit $π - 2 \arcsin \frac{2}{3}$ to $0$ . Neither $b = 0$ nor the zero-angle grazing outcome is in the defined branch, whose domain is $(0, a)$ and whose image lies in $(0, π)$ . This is a feature of the declared contact rule; no forward-angle cross-section contribution at $θ = 0$ is defined here.
No empirical claim. The index $3 / 2$ , the radius $a$ and the energy $E$ are declared model data; the computation checks an assertion about model functions, not a measurement of any material.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A coefficient of restitution above $1$ manufactures kinetic energy: $e = 2$ on $m_{1} = m_{2} = 1 k g$ with velocities $\pm 1 m s^{- 1}$ raises the kinetic energy by $3 J$ while the configuration is unchanged

Physical scope

A hypothetical extension of the adopted impact model, replacing only its restitution range by arbitrary real e; a refutation of the claim that every real coefficient of restitution is admissible in the adopted isolated-impact model: with the collinear restitution law and $e = 2$ on declared data, the kinetic energy rises by $3 J$ across an impact while the configuration is instantaneously unchanged, so the total mechanical energy of a conservative two-label model is not conserved; values $e > 1$ therefore signal an omitted internal energy reservoir and are excluded from the conservative contact model of this page, which restricts $e$ to $[0, 1]$ . The witness uses declared model data and asserts no measurement.

Statement refuted

False statement. If the collinear reciprocal-impulse equations of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (i) are retained but its clause (ii) is hypothetically extended to allow every real restitution coefficient $e$ , then momentum conservation and $v_{2}^{+} - v_{1}^{+} = - e (v_{2}^{-} - v_{1}^{-})$ guarantee conservation of total mechanical energy across the impact in a static central-potential model.

The adopted postulate restricts $e$ to $[0, 1]$ , as do the restitution and energy clauses of Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws and Isolated reciprocal impulse with full-vector restitution: momentum jumps $\pm J$ , conservation of total momentum, the restitution solution and the kinetic-energy change $- \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ . Those clauses are not applied at $e = 2$ here. The extension is a separate hypothetical law, and the post-impact velocities and energy difference below are derived locally from its equations. Only the elastic case $e = 1$ guarantees kinetic-energy conservation for nonzero relative velocity within the adopted range.

Facts & Assumptions

Given: A unit vector $n \in R^{3}$ ; two labels with declared masses $m_{1} = m_{2} = 1 k g$ ; incoming velocities $v_{1}^{-} = n m s^{- 1}$ , $v_{2}^{-} = - n m s^{- 1}$ ; the hypothetical momentum and restitution equations $m_{1} v_{1}^{+} + m_{2} v_{2}^{+} = m_{1} v_{1}^{-} + m_{2} v_{2}^{-}$ and $v_{2}^{+} - v_{1}^{+} = - e (v_{2}^{-} - v_{1}^{-})$ with $e = 2$ ; and continuous positions across the contact. Take a static central pair potential $u \equiv 0$ , so $U (x) = u (∣ x_{1} - x_{2} ∣) = 0$ and free motions away from the impact have constant mechanical energy. This is declared model data, not a measured collision.

[L1]

For positive masses and $M = m_{1} + m_{2}$ , $μ = m_{1} m_{2} / M$ , the two-body kinetic identity is $m_{1} ∣ v_{1} ∣^{2} + m_{2} ∣ v_{2} ∣^{2} = M ∣ V ∣^{2} + μ ∣ v_{2} - v_{1} ∣^{2}$ with $V = (m_{1} v_{1} + m_{2} v_{2}) / M$ . This is an algebraic identity for all real vectors, independent of any restitution range. (Centre of mass, the identity $\sum_{i} m_{i} ρ_{i} = 0$ , and the splitting of the kinetic sum $\sum_{i} m_{i} ∥ x_{i}^{'} ∥_{2}^{2}$ into centre-of-mass and relative parts (with the two-body reduced-mass case), The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame)

[L2]

A static central pair model permits the declared smooth potential $u \equiv 0$ ; the units of velocity and energy are metres per second and joules, respectively. (Adopted central-force model class: static central pair potentials, the reduced one-label problem, the fixed-centre idealization, reading conventions and exclusions, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length)

Counterexample

technique · direct

1.1givenL1algebra

In numerical SI coordinates, $M = 2$ , $μ = 1 / 2$ , $V = 0$ , $d^{-} : = v_{2}^{-} - v_{1}^{-} = - 2 n$ , and $∣ d^{-} ∣^{2} = 4$ , since $∣ n ∣ = 1$ . The incoming kinetic energy is $K^{-} = (∣ n ∣^{2} + ∣ - n ∣^{2}) / 2 = 1 J$ .

2.1givenstep 1.1algebra

For any real $e$ , put $d^{+} = - e d^{-}$ . The momentum equation gives $m_{1} v_{1}^{+} + m_{2} v_{2}^{+} = M V$ and the restitution equation gives $v_{2}^{+} = v_{1}^{+} + d^{+}$ . Substitution yields $M v_{1}^{+} + m_{2} d^{+} = M V$ , hence $v_{1}^{+} = V + (m_{2} / M) e d^{-}$ and $v_{2}^{+} = V - (m_{1} / M) e d^{-}$ . These vectors satisfy both equations, so they are the unique outcome of the hypothetical law. At $e = 2$ they are $v_{1}^{+} = - 2 n$ and $v_{2}^{+} = 2 n$ in metres per second; their total momentum is zero, and $J = m_{1} (v_{1}^{+} - v_{1}^{-}) = - 3 n = - m_{2} (v_{2}^{+} - v_{2}^{-})$ in kilogram metres per second.

3.1step 1.1step 2.1L1algebra

Directly $K^{+} = (4 ∣ n ∣^{2} + 4 ∣ n ∣^{2}) / 2 = 4 J$ , so $Δ K = 3 J$ . More generally, [L1] for the incoming and outgoing pairs and the equations of step 2.1 give $Δ K = μ (∣ d^{+} ∣^{2} - ∣ d^{-} ∣^{2}) / 2 = μ (e^{2} - 1) ∣ d^{-} ∣^{2} / 2$ for every real $e$ : the common center-of-mass term cancels and $∣ e d^{-} ∣^{2} = e^{2} ∣ d^{-} ∣^{2}$ by the Euclidean coordinate formula. This is a local derivation, not an application of a restricted impact theorem outside its hypotheses.

4.1givenstep 2.1step 3.1L2algebra

The declared potential is identically zero, hence unchanged across the continuous configuration. Consequently $Δ (K + U) = 3 J > 0$ , which refutes the universal across-impact conservation assertion. To realize the endpoint data by free paths, in numerical metres and seconds take $t_{*} = 0$ , $I = (- 1 / 4, 1 / 4)$ , $x_{1} (t) = t n$ for $t < 0$ and $x_{1} (t) = - 2 t n$ for $t \geq 0$ , and $x_{2} (t) = (1 - t) n$ for $t < 0$ and $x_{2} (t) = (1 + 2 t) n$ for $t \geq 0$ . Both positions are continuous, their derivatives are the stated velocities, their accelerations vanish off zero, and the separation is $1 - 2 t > 0$ before zero and $1 + 4 t > 0$ after zero. Thus the zero-potential free dynamics away from the impulse are consistent with the hypothetical endpoint law on a collision-excluded domain.

5.1step 1.1step 3.1step 4.1algebra∎

For the adopted range $0 \leq e \leq 1$ , the locally derived formula in step 3.1 is nonpositive and vanishes precisely when $e = 1$ or $d^{-} = 0$ . At $e = 2$ , it is strictly positive for the nonzero incoming relative velocity of step 1.1. Therefore allowing every real $e$ does not give a conservative contact law for the two-label mechanical-energy account; it requires an additional energy input if conservation of an enlarged account is imposed. The example asserts no empirical energy reservoir or measured restitution coefficient.

Remarks

The hypothetical extension preserves the reciprocal momentum account but changes the adopted restitution range. Its energy gain comes from an explicit local computation. It does not claim that the adopted postulate admits $e > 1$ , or apply its restricted theorems at such values.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Reciprocal impulses do not determine a collinear impact: the same incoming data admit the sticking outcome $v^{+} = 0$ and the elastic outcome $(v_{1}^{+}, v_{2}^{+}) = (- 3, 1) m s^{- 1}$ with different impulses and energies

Physical scope

A refutation of the claim that the reciprocal impulse equations alone determine a collinear impact: with masses $2$ and $6 k g$ and incoming velocities $3$ and $- 1 m s^{- 1}$ , both the sticking outcome and the elastic outcome satisfy the equal-and-opposite momentum jumps with impulses $- 6 N s$ and $- 12 N s$ respectively and total momentum $0$ , while velocities and kinetic-energy changes differ. Hence the contact law fixing the impulse (or the coefficient of restitution) is an independent model datum. The witness uses declared model data and asserts no measurement, no determination of any real coefficient and no observation.

Statement refuted

False statement. In the collinear isolated-impact model of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (i), the reciprocal impulse equations $m_{1} (v_{1}^{+} - v_{1}^{-}) = J$ and $m_{2} (v_{2}^{+} - v_{2}^{-}) = - J$ , with the impulse $J$ determined by momentum balance of the two labels, fix the post-impact velocities for given incoming data; no additional contact law is needed.

Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (ii) adopts the restitution law as an independent contact assumption, and Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 5 records that the impulse equations leave the scalar impulse magnitude free. The witness exhibits two admissible outcomes for one data set.

Facts & Assumptions

Given: Fix a unit vector $n \in R^{3}$ . Take the two-label force model with time window $I_{0} = R$ , configuration domain $Q = (R^{3})^{2}$ , and all external and pair force maps identically zero. Coincident positions are permitted because these forces have no singularity. Let $I = (- 1 s, 1 s)$ and $t_{*} = 0$ . The masses are $m_{1} = 2 k g$ , $m_{2} = 6 k g$ and the incoming velocities are $v_{1}^{-} = 3 n m s^{- 1}$ , $v_{2}^{-} = - 1 n m s^{- 1}$ ; the two candidate outcomes are those below.

[L1]

Impulse equations (Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (i), Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 1): once a continuous, piecewise- $C^{2}$ position curve in $Q$ with the regular off-impact dynamics and the required one-sided velocities is supplied, its jump condition is $m_{1} (v_{1}^{+} - v_{1}^{-}) = J = - m_{2} (v_{2}^{+} - v_{2}^{-})$ . Existence of such a $J$ is equivalent to total-momentum equality for the endpoint velocities; it does not by itself establish the curve conditions. In the collinear case with $J = J_{n} n$ this endpoint equality is one scalar equation for two outgoing components.

[L2]

Kinetic identity (Centre of mass, the identity $\sum_{i} m_{i} ρ_{i} = 0$ , and the splitting of the kinetic sum $\sum_{i} m_{i} ∥ x_{i}^{'} ∥_{2}^{2}$ into centre-of-mass and relative parts (with the two-body reduced-mass case), Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): with $V = \frac{m_{1} v_{1} + m_{2} v_{2}}{M}$ and $μ = \frac{m_{1} m_{2}}{M}$ , $K = \frac{M}{2} ∥ V ∥_{2}^{2} + \frac{μ}{2} ∥ v_{2} - v_{1} ∥_{2}^{2}$ ; with momentum continuous, $V$ is the same on both sides and $Δ K = \frac{μ}{2} (∥ v_{2}^{+} - v_{1}^{+} ∥_{2}^{2} - ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2})$ .

Counterexample

technique · direct

1.1givenL2algebra

Common data: $M = 8$ , $μ = \frac{3}{2}$ , $V = \frac{2 \cdot 3 + 6 \cdot (- 1)}{8} = 0$ , $d^{-} = - 4 n$ , $∥ d^{-} ∥_{2}^{2} = 16$ , $K^{-} = 12 J$ .

1.2givenL1construct

For each candidate outgoing pair below, define $x_{i} (t) = t v_{i}^{-}$ for $t \leq 0$ and $x_{i} (t) = t v_{i}^{+}$ for $t > 0$ . Both position curves are continuous on $I$ , affine and hence $C^{2}$ on either side of $0$ , and take values in the declared full domain $Q$ . They have precisely the specified one-sided velocities, and $m_{i} x_{i}^{''} = 0 = F_{i} (t, x, x^{'})$ away from $0$ . Thus all trajectory conditions of the isolated reciprocal impulse model hold for both candidates; only the reciprocal jumps remain to check. Moreover $⟨ v_{2}^{-} - v_{1}^{-}, n ⟩ = - 4 m s^{- 1} < 0$ , so the common incoming data approach along $n$ .

2.1step 1.1L1algebra

Outcome (a), sticking: $v_{1}^{+} = v_{2}^{+} = 0$ . Then $m_{1} (v_{1}^{+} - v_{1}^{-}) = 2 (0 - 3) n = - 6 n$ and $m_{2} (v_{2}^{+} - v_{2}^{-}) = 6 (0 + 1) n = + 6 n$ ; the impulse is $J = - 6 n N s$ and the total momentum is $0 = P^{-}$ .

2.2step 1.1L1algebra

Outcome (b), elastic: $v_{1}^{+} = - 3 n$ , $v_{2}^{+} = + 1 n$ . Then $m_{1} (v_{1}^{+} - v_{1}^{-}) = 2 (- 6) n = - 12 n$ and $m_{2} (v_{2}^{+} - v_{2}^{-}) = 6 (+ 2) n = + 12 n$ ; the impulse is $J = - 12 n N s$ and the total momentum is $0 = P^{-}$ .

3.1step 1.2step 2.1step 2.2L1algebra

Together with the curves of step 1.2, steps 2.1 and 2.2 establish two complete runs of clause (i), with the same incoming data and different impulses ( $- 6 n$ versus $- 12 n$ ) and outgoing velocities ( $0$ versus $- 3 n$ and $+ 1 n$ ). The first also satisfies restitution with $e = 0$ , and the second with $e = 1$ . Hence reciprocal impulse accounting alone does not fix the outgoing velocities, and the false statement fails.

4.1step 3.1L2algebra

Energy: by [L2] the changes are $Δ K_{(a)} = \frac{3}{2} \cdot \frac{1}{2} (0 - 16) = - 12 J$ and $Δ K_{(b)} = \frac{3}{2} \cdot \frac{1}{2} (16 - 16) = 0$ ; the outcomes are therefore also distinguished by their energy accounting, so no choice of impulse can be recovered from the momentum data and the two outcomes are genuinely different model histories.

5.1givenL1∎

Consequence: in the collinear isolated-impact model the contact law of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (ii) (equivalently, the coefficient $e$ , or a dynamical contact model producing the impulse) is an independent model datum; the reciprocal impulse equations constrain only the sum $m_{1} Δ v_{1} + m_{2} Δ v_{2} = 0$ . The same conclusion is recorded as clause 5 of Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws.

Remarks

Consistency with the page's model. The witness does not show that both outcomes occur in one physical contact; it shows that the momentum balance of the model is compatible with both, so an additional assumption is required to select one. The page selects the restitution family $0 \leq e \leq 1$ , with $e$ left as model data.
Declared data. All masses, velocities and impulses are declared model data; nothing here is measured or inferred from an experiment.

Scattering and impulse model boundaries — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Elastic reflection from a fixed sphere: the branch b=acos⁡(θ/2)b=acos(θ/2), the uniform density a2/4a2/4 and the total πa2πa2 of a contact model

Example

Facts & Assumptions

Verification

Remarks

A declared-data Rutherford check: α=1α=1, θ=π/2θ=π/2, (dσ/dΩ)=4×10−24 m2(dσ/dΩ)=4×10−24m2, the band values πb(θ1)2πb(θ1​)2 and the growth of the truncated total as θ1↓0θ1​↓0

Example

Facts & Assumptions

Verification

Remarks

A declared-data impact check: m1=2 kgm1​=2kg, m2=6 kgm2​=6kg, v1−=3 m s−1v1−​=3ms−1, v2−=−1 m s−1v2−​=−1ms−1, with elastic, semi-elastic and sticking outcomes

Example

Facts & Assumptions

Verification

Remarks

A declared rocket check: initial mass 3 kg3kg, loss rate 1 kg s−11kgs−1 and relative exhaust speed 2 m s−12ms−1 give v=ueln⁡(m0/m)nv=ue​ln(m0​/m)n and show that (mv)′=F(mv)′=F fails

Example

Facts & Assumptions

Verification

Remarks

A declared smooth open flow: the uniform scaling ρ=ρ0e−3αtρ=ρ0​e−3αt, u=αxu=αx satisfies the control-volume identities on fixed and material regions, and elbow port data illustrate the inlet/outlet decomposition

Example

Facts & Assumptions

Verification

Remarks

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A coefficient of restitution above 11 manufactures kinetic energy: e=2e=2 on m1=m2=1 kgm1​=m2​=1kg with velocities ±1 m s−1±1ms−1 raises the kinetic energy by 3 J3J while the configuration is unchanged

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Reciprocal impulses do not determine a collinear impact: the same incoming data admit the sticking outcome v+=0v+=0 and the elastic outcome (v1+,v2+)=(−3,1) m s−1(v1+​,v2+​)=(−3,1)ms−1 with different impulses and energies

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

Elastic reflection from a fixed sphere: the branch $b = a \cos (θ / 2)$ , the uniform density $a^{2} / 4$ and the total $π a^{2}$ of a contact model

A declared-data Rutherford check: $α = 1$ , $θ = π / 2$ , $(d σ / d Ω) = 4 \times 10^{- 24} m^{2}$ , the band values $π b (θ_{1})^{2}$ and the growth of the truncated total as $θ_{1} ↓ 0$

A declared-data impact check: $m_{1} = 2 k g$ , $m_{2} = 6 k g$ , $v_{1}^{-} = 3 m s^{- 1}$ , $v_{2}^{-} = - 1 m s^{- 1}$ , with elastic, semi-elastic and sticking outcomes

A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails

A declared smooth open flow: the uniform scaling $ρ = ρ_{0} e^{- 3 α t}$ , $u = α x$ satisfies the control-volume identities on fixed and material regions, and elbow port data illustrate the inlet/outlet decomposition

A coefficient of restitution above $1$ manufactures kinetic energy: $e = 2$ on $m_{1} = m_{2} = 1 k g$ with velocities $\pm 1 m s^{- 1}$ raises the kinetic energy by $3 J$ while the configuration is unchanged

Reciprocal impulses do not determine a collinear impact: the same incoming data admit the sticking outcome $v^{+} = 0$ and the elastic outcome $(v_{1}^{+}, v_{2}^{+}) = (- 3, 1) m s^{- 1}$ with different impulses and energies