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Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with ΔK=−1−e22μ∥v2−−v1−∥22, and the open-body and rocket laws

Physical scope

The conditional consequences of the page's adopted impact and open-mass extensions: total momentum is continuous across an isolated reciprocal impulse with the impulse removed from the second label; a smooth-force motion with continuous force maps cannot carry a nonzero impulse, so the contact model is an extension rather than a smooth-force theorem; a collinear contact with coefficient of restitution e∈[0,1] has the stated post-impact velocities and kinetic-energy change, is dissipative for e<1, and its contact law is an independent assumption not derivable from momentum balance; and the open-body material accounting yields the derived law mv′=F+rin(win−v)−rout(wout−v) with the rocket thrust as the special case. The result is a conditional deduction from adopted model data; it asserts no measured value of e, the impulse or the mass rates, and identifies no observed dissipation mechanism.

Statement

Assume the Axiom of Countable Choice ACω (The Axiom of Countable Choice (ACω)), inherited from the statement of Isolated reciprocal impulse with full-vector restitution: momentum jumps ±J, conservation of total momentum, the restitution solution and the kinetic-energy change −1−e22μ∥v2−−v1−∥22, whose distributional jump lemma uses it; the algebraic clauses used here do not otherwise use choice. Adopt the impact and open-mass extensions 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 m1,m2>0, an interval I with nonempty interior, t∗∈int⁡I, an impulse motion x=(x1,x2):I→Q run in the isolated reciprocal impulse model at t∗ with impulse J and one-sided velocities vi±, and write M:=m1+m2,μ:=m1m2M,P(t):=m1x1′(t)+m2x2′(t)(t∈I∖{t∗}). Only the one-sided limits of P at t∗ are used; no derivative of the position curves at the contact time is assumed. Then:

  1. (Momentum continuity.) The total momentum is continuous across the impact, P(t∗+)=P(t∗−),equivalently∑i=12mi(vi+−vi−)=0, and the impulse delivered to label 1 is removed from label 2: J=m1(v1+−v1−)=−m2(v2+−v2−).
  2. (Smooth-force exclusion.) Suppose this same candidate motion x from clause 1 also satisfies the smooth force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance and The associated first-order system, motions, maximal motions and the Newtonian flow of a force model on I, with the force maps continuous there. Then the momentum curves pi=mixi′ are of class C1 on I; in particular pi(t∗+)=pi(t∗−)(i=1,2). Clause 1 identifies the jump of p1 with the same impulse J, so J=0: no candidate motion carrying a nonzero isolated impulse can also be a motion of that smooth-force class. The finite-jump 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) is therefore a genuine extension of the smooth-force branch, not one of its cases.
  3. (Collinear restitution solution.) Suppose in addition that the contact is collinear with contact direction n and coefficient of restitution e∈[0,1] (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)). Then J=μ(1+e)(v2−−v1−), J is parallel to n, the centre-of-mass velocity V:=M−1(m1v1−+m2v2−) is unchanged across the impact, and the post-impact velocities are v1+=V+m2M e (v2−−v1−),v2+=V−m1M e (v2−−v1−).
  4. (Kinetic energy of the impact.) The kinetic-energy change across the impact is ΔK:=m1∥v1+∥22+m2∥v2+∥222−m1∥v1−∥22+m2∥v2−∥222=−1−e22 μ ∥v2−−v1−∥22≤0, with equality if and only if e=1 or v2−=v1−. For 0≤e<1 the contact model is dissipative: the total kinetic energy of the two labels is not conserved across the impact, and the deficit 1−e22μ∥v2−−v1−∥22 is a quantity of the adopted contact law. This item identifies it with no measured heat, temperature, sound or deformation observable.
  5. (Momentum balance alone does not determine the impact.) In the collinear case with d−:=v2−−v1− and dn:=⟨d−,n⟩<0, the impulse equations of clause 1 alone leave the scalar impulse magnitude free: for every real Jn with Jn≤μdn (so that the contact is separating or grazing) the velocities v1+=v1−+Jnm1n,v2+=v2−−Jnm2n satisfy the reciprocal impulse equations with impulse J=Jnn and preserve the total momentum, and their kinetic-energy change is ΔK(Jn)=−Jn dn+Jn22μ, whose minimum over the admissible range is ΔK(μdn)=−μ2dn2 at the perfectly inelastic value Jn=μdn (that is, e=0). The value of Jn is fixed only by 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), which selects Jn=μ(1+e)dn; in particular momentum balance alone does not determine the post-impact velocities.
  6. (Open-body and rocket laws.) Under 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 differentiable m:I→(0,∞) and v:I→R3, F:I→R3, rin,rout:I→[0,∞), and win,wout:I→R3 satisfying the material balance identities, one has on I m v′=F+rin(win−v)−rout(wout−v). In particular: (a) if rin=rout=0 then m v′=F and (mv)′=F; (b) if rin=0 and rout=−m′>0 (mass loss) then m v′=F+rout(v−wout)=F−m′(v−wout), so the thrust is T:=rout(v−wout)=−m′(v−wout), and the closed-system law (mv)′=F fails unless wout=0; (c) if rout=0 and win=0 then m v′=F−rinv: capture contributes the acceleration −(rin/m)v, opposing v, while (mv)′=F still holds.

Facts & Assumptions

Given: ACω; the adopted extensions 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; the two-label motion run in the isolated reciprocal impulse model at t∗ with impulse J; for clause 3 the collinear contact direction n and coefficient e; for clause 6 the maps m,v,F,rin,rout,win,wout.

[L1]

Adopted jump and restitution identities (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)): m1(v1+−v1−)=J=−m2(v2+−v2−) and, when the contact law is invoked, v2+−v1+=−e(v2−−v1−) with e∈[0,1]. The velocity and energy formulas are derived below from these identities and [L4]; no continuity of the evaluated time-force across the impulse is assumed.

[L2]

Open-body algebra (Variable-mass momentum balance: from m′=rin−rout and (mv)′=F+rinwin−routwout to mv′=F+rin(win−v)−rout(wout−v)): from m′=rin−rout and (mv)′=F+rinwin−routwout follows mv′=F+rin(win−v)−rout(wout−v), together with the special cases by substitution.

[L4]

Two-body kinetic identity (Centre of mass, the identity ∑imiρi=0, and the splitting of the kinetic sum ∑imi∥xi′∥22 into centre-of-mass and relative parts (with the two-body reduced-mass case), The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn): with V=M−1(m1v1+m2v2) and μ=m1m2/M, m1∥v1∥22+m2∥v2∥22=M∥V∥22+μ∥v2−v1∥22; hence v1=V−m2M(v2−v1) and v2=V+m1M(v2−v1).

Proof

technique · direct
1.1givenL1algebra

Clause 1: 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 (i) and [L1], m1(v1+−v1−)=J and m2(v2+−v2−)=−J; adding gives m1(v1+−v1−)+m2(v2+−v2−)=0, which is exactly P(t∗+)=P(t∗−). This is clause 1.

1.2givenL3L5algebra

Clause 2: suppose the candidate motion x of clause 1 also satisfies the smooth-force equation on all of I with continuous force maps along it. By [L3] the curves xi′′=Fi/mi are continuous, hence xi is C2 and pi=mixi′ is C1 on I; a C1 function on an interval has equal one-sided limits at every interior point, so pi(t∗+)=pi(t∗−) for each label. Clause 1 gives p1(t∗+)−p1(t∗−)=J, hence J=0. Thus no candidate with a nonzero impulse is a smooth-force motion, as asserted.

1.3givenL1L4algebra

Clause 5: fix any real Jn≤μdn and put v1+:=v1−+Jnm1n, v2+:=v2−−Jnm2n; then m1(v1+−v1−)=Jnn and m2(v2+−v2−)=−Jnn, so the reciprocal impulse equations of clause 1 hold with impulse J=Jnn, and adding the two equations shows that the total momentum is preserved. The relative velocity becomes d+=d−−Jnμn; hence d+⋅n=dn−Jnμ≥0 precisely for Jn≤μdn, which is the stated admissibility (the contact separates or grazes). By the kinetic identity [L4], ΔK(Jn)=μ2(∥d+∥22−∥d−∥22)=μ2(−2Jnμdn+Jn2μ2)=−Jndn+Jn22μ, a quadratic in Jn with minimum −μ2dn2 at Jn=μdn; the restitution law selects Jn=μ(1+e)dn, which lies in the admissible range because 1+e≥1 and dn<0, and reproduces clause 3.

2.1step 1.1givenL1L4algebra

Clause 3: put d−=v2−−v1− and d+=v2+−v1+. The jump equations [L1] give d+=d−−J/μ, while restitution gives d+=−ed−. Hence J=μ(1+e)d−, parallel to n since d−∥n. By step 1.1, the two mass-weighted means are the same V. Solving m1v1++m2v2+=MV and v2+−v1+=d+ gives uniquely v1+=V−(m2/M)d+=V+(m2/M)ed− and v2+=V+(m1/M)d+=V−(m1/M)ed−.

3.1step 1.1step 2.1givenL4algebra

Clause 4: the kinetic split [L4] applied on both sides, with the same V by step 1.1, gives ΔK=(μ/2)(∥d+∥22−∥d−∥22). Substituting d+=−ed− from step 2.1 and expanding the inner product gives ΔK=−(1−e2)μ∥d−∥22/2≤0. Since μ>0 and 0≤e≤1, equality holds exactly when e=1 or d−=0; the approach condition excludes d−=0 for a contact. The computed deficit is not identified with any measured heat, sound or deformation observable.

4.1givenL2algebra∎

Clause 6: differentiability of m and v on I is part 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), so [L2] applies and gives the derived law from the material balance identities. The three special cases are its substitutions: (a) rin=rout=0 leaves mv′=F, while (mv)′=m′v+mv′=F because m′=0; (b) rin=0, rout=−m′ gives mv′=F−rout(wout−v)=F+rout(v−wout); and (mv)′=m′v+mv′=F+rout(v−wout)+m′v=F−routwout+routv−routv=F−routwout, which equals F exactly when wout=0 or rout=0; (c) rout=0, win=0 gives mv′=F−rinv, while (mv)′=m′v+mv′=rinv+F−rinv=F. This is clause 6.

Remarks

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