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Reciprocal impulses do not determine a collinear impact: the same incoming data admit the sticking outcome v+=0 and the elastic outcome (v1+,v2+)=(−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 kg 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 m1(v1+−v1−)=J and m2(v2+−v2−)=−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 ΔK=−1−e22μ∥v2−−v1−∥22, 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∈R3. Take the two-label force model with time window I0=R, configuration domain Q=(R3)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 m1=2 kg, m2=6 kg and the incoming velocities are v1−=3n m s−1, v2−=−1n 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 ΔK=−1−e22μ∥v2−−v1−∥22, and the open-body and rocket laws clause 1): once a continuous, piecewise-C2 position curve in Q with the regular off-impact dynamics and the required one-sided velocities is supplied, its jump condition is m1(v1+−v1−)=J=−m2(v2+−v2−). 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=Jnn this endpoint equality is one scalar equation for two outgoing components.

[L2]

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), 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⟩=∑k<nxkyk on Rn): with V=m1v1+m2v2M and μ=m1m2M, K=M2∥V∥22+μ2∥v2−v1∥22; with momentum continuous, V is the same on both sides and ΔK=μ2(∥v2+−v1+∥22−∥v2−−v1−∥22).

Counterexample

technique · direct
1.1givenL2algebra

Common data: M=8, μ=32, V=2⋅3+6⋅(−1)8=0, d−=−4n, ∥d−∥22=16, K−=12 J.

1.2givenL1construct

For each candidate outgoing pair below, define xi(t)=tvi− for t≤0 and xi(t)=tvi+ for t>0. Both position curves are continuous on I, affine and hence C2 on either side of 0, and take values in the declared full domain Q. They have precisely the specified one-sided velocities, and mixi′′=0=Fi(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 ⟨v2−−v1−,n⟩=−4 m s−1<0, so the common incoming data approach along n.

2.1step 1.1L1algebra

Outcome (a), sticking: v1+=v2+=0. Then m1(v1+−v1−)=2(0−3)n=−6n and m2(v2+−v2−)=6(0+1)n=+6n; the impulse is J=−6n N s and the total momentum is 0=P−.

2.2step 1.1L1algebra

Outcome (b), elastic: v1+=−3n, v2+=+1n. Then m1(v1+−v1−)=2(−6)n=−12n and m2(v2+−v2−)=6(+2)n=+12n; the impulse is J=−12n 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 (−6n versus −12n) and outgoing velocities (0 versus −3n and +1n). 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 ΔK(a)=32⋅12(0−16)=−12 J and ΔK(b)=32⋅12(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 m1Δv1+m2Δv2=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 ΔK=−1−e22μ∥v2−−v1−∥22, 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≤e≤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.

Depends on

Dependency tree · 2 levels

Sources