Isolated reciprocal impulse with full-vector restitution: momentum jumps $\pm J$, conservation of total momentum, the restitution solution and the kinetic-energy change $-\frac{1-e^2}{2}\mu \parallel v_2^{-}-v_1^{-}{\parallel }_2^2$

For $k\in \{0,1,2\}$, let $e_k\in {\mathbb{R}}^3$ be the vector with $k$th coordinate $1$ and other coordinates $0$, and let ${\delta }_{kl}=1$ if $k=l$ and $0$ otherwise. Fix a real-valued $\phi \in C_c^{\mathrm{\infty }}(I)$; the vector test function $\mathrm{\Phi }:=\phi e_k$ has components $(\phi e_k{)}_l={\delta }_{kl}\phi$ and lies in $C_c^{\mathrm{\infty }}(I;{\mathbb{R}}^3)$. Reading the impulse pairing of the model for this $\mathrm{\Phi }$ gives $-{\int }_I{p}_{1,k}(t){\phi }^{\mathrm{\prime }}(t)d\lambda (t)={\int }_I{F}_{1,k}(t)\phi (t)d\lambda (t)+J_k\phi (t_{\ast }),-{\int }_I{p}_{2,k}(t){\phi }^{\mathrm{\prime }}(t)d\lambda (t)={\int }_I{F}_{2,k}(t)\phi (t)d\lambda (t)-J_k\phi (t_{\ast })\mathrm{.}$

Each component $p_{i,k}$ satisfies the hypotheses of the jump lemma [L1]: it is continuous and $C^1$ off $t_{\ast }$, its one-sided limits exist, and its one-sided derivative limits exist too, because off $t_{\ast }$ one has $p_{i,k}^{\mathrm{\prime }}=m_i{x}_{i,k}^{\mathrm{\prime }\mathrm{\prime }}=F_{i,k}$ by the regular-dynamics requirement of the model and $F_{i,k}$ is continuous on $I$. Moreover the off-$t_{\ast }$ derivative of $p_{i,k}$ is $F_{i,k}$, so $(p_{i,k}{)}_{\mathrm{r}\mathrm{e}\mathrm{g}}^{\mathrm{\prime }}=F_{i,k}$ except possibly at $t_{\ast }$, which does not change the regular distribution or its integral. Hence [L1] gives, for every real $\phi \in C_c^{\mathrm{\infty }}(I)$, $-{\int }_I{p}_{i,k}{\phi }^{\mathrm{\prime }}d\lambda ={\int }_I{F}_{i,k}\phi d\lambda +[p_{i,k}]\phi (t_{\ast })\mathrm{.}$

Comparing the identity of step 2.1 with the model identity of step 1.1 for the same $\phi$, the terms ${\int }_I{F}_{i,k}\phi d\lambda$ cancel and leave $([p_{1,k}]-J_k)\phi (t_{\ast })=0$ and $([p_{2,k}]+J_k)\phi (t_{\ast })=0$ for every $\phi \in C_c^{\mathrm{\infty }}(I)$; choosing by [L3] a $\phi$ with $\phi (t_{\ast })=1$ gives $[p_{1,k}]=J_k$ and $[p_{2,k}]=-J_k$. As $k$ is arbitrary, $m_1(v_1^{+}-v_1^{-})=J$ and $m_2(v_2^{+}-v_2^{-})=-J$, and adding these two vector equations gives $p_1(t_{\ast }^{+})+p_2(t_{\ast }^{+})=p_1(t_{\ast }^{-})+p_2(t_{\ast }^{-})$; this is claim a.

For claim b, let $V:=\frac{m_1{v}_1^{-}+m_2{v}_2^{-}}{M}$ and $d:=v_2^{-}-v_1^{-}$. By step 3.1, $V=\frac{m_1{v}_1^{+}+m_2{v}_2^{+}}{M}$ as well. The restitution law reads $v_2^{+}-v_1^{+}=-ed$. Writing $d^{+}:=v_2^{+}-v_1^{+}$ and solving the pair of linear equations ($v_2^{+}=v_1^{+}+d^{+}$ and $m_1{v}_1^{+}+m_2{v}_2^{+}=MV$) gives $v_1^{+}=V-\frac{m_2}{M}{d}^{+}=V+\frac{m_2}{M}ed$ and $v_2^{+}=V+\frac{m_1}{M}{d}^{+}=V-\frac{m_1}{M}ed$; conversely these displayed vectors satisfy both equations, so they are the unique solution and claim b holds.

For claim c, apply the kinetic identity [L2] to the pairs $(v_1^{-},v_2^{-})$ and $(v_1^{+},v_2^{+})$, whose centre-of-mass velocities are both $V$: $\frac{m_1\parallel v_1^{\pm }{\parallel }_2^2+m_2\parallel v_2^{\pm }{\parallel }_2^2}{2}=\frac{M\parallel V{\parallel }_2^2}{2}+\frac{\mu \parallel v_2^{\pm }-v_1^{\pm }{\parallel }_2^2}{2}\mathrm{.}$ Subtracting the minus identity from the plus identity cancels the centre-of-mass term, and the restitution law gives $\parallel v_2^{+}-v_1^{+}{\parallel }_2^2=e^2\parallel v_2^{-}-v_1^{-}{\parallel }_2^2$ by homogeneity of the norm [L2], so $\mathrm{\Delta }K=\frac{\mu }{2}(e^2-1)\parallel v_2^{-}-v_1^{-}{\parallel }_2^2=-\frac{1-e^2}{2}\mu \parallel v_2^{-}-v_1^{-}{\parallel }_2^2\le 0$ because $0\le e\le 1$; the cases $e=1$ and $e=0$ are the stated substitutions. This is claim c.