Worked impulse and restitution computations and a variable-mass rocket check

Throughout, $e\in {\mathbb{R}}^3$ is a unit vector, $\parallel e{\parallel }_2=1$, for the Euclidean norm of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation. All numbers are pure real numbers. For the two collision computations, $v_i^{-}$ and $v_i^{+}$ denote pre- and post-collision vectors. Given positive masses and $e_r\in [0,1]$, require only the finite algebraic relations

$m_1{v}_1^{+}+m_2{v}_2^{+}=m_1{v}_1^{-}+m_2{v}_2^{-},v_2^{+}-v_1^{+}=-e_r(v_2^{-}-v_1^{-})\mathrm{.}$

Define $J:=m_1(v_1^{+}-v_1^{-})$ and

$\mathrm{\Delta }K:=\frac{m_1\parallel v_1^{+}{\parallel }_2^2+m_2\parallel v_2^{+}{\parallel }_2^2-m_1\parallel v_1^{-}{\parallel }_2^2-m_2\parallel v_2^{-}{\parallel }_2^2}{2}\mathrm{.}$

The two cases below solve these relations uniquely and compute $J$ and $\mathrm{\Delta }K$ directly. No distributional jump theorem or choice assumption is used in these finite calculations.

(i) Equal masses, half-elastic impulse. Take $m_1=m_2=1$, approach velocities $v_1^{-}=1e$, $v_2^{-}=-1e$ and restitution coefficient $e_r=\frac{1}{2}$. Then the unique post-collision vectors, impulse and kinetic-energy change are

$v_1^{+}=-\frac{1}{2}e,v_2^{+}=\frac{1}{2}e,J=-\frac{3}{2}e,\mathrm{\Delta }K=-\frac{3}{4}\mathrm{.}$

(ii) Elastic impulse with unequal masses. Take $m_1=1$, $m_2=3$, $v_1^{-}=2e$, $v_2^{-}=0$ and $e_r=1$. Then

$v_1^{+}=-1e,v_2^{+}=1e,J=-3e,\mathrm{\Delta }K=0.$

(iii) Rocket check. On $I=[0,1]$ take $m(t):=2-t$, $v(t):=\ln \text{\hspace{0.17em}⁣}\left(\frac{2}{2-t}\right)e$, $F\equiv 0$, $r_{\mathrm{i}\mathrm{n}}\equiv 0$, $r_{\mathrm{o}\mathrm{u}\mathrm{t}}\equiv 1$, $w_{\mathrm{i}\mathrm{n}}\equiv 0$, $w_{\mathrm{o}\mathrm{u}\mathrm{t}}(t):=v(t)-u_{e}e$ with $u_e:=1$. Then the two material balance identities of Variable-mass momentum balance: from $m^{\mathrm{\prime }}=r_{\mathrm{i}\mathrm{n}}-r_{\mathrm{o}\mathrm{u}\mathrm{t}}$ and $(mv{)}^{\mathrm{\prime }}=F+r_{\mathrm{i}\mathrm{n}}{w}_{\mathrm{i}\mathrm{n}}-r_{\mathrm{o}\mathrm{u}\mathrm{t}}{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $m{v}^{\mathrm{\prime }}=F+r_{\mathrm{i}\mathrm{n}}(w_{\mathrm{i}\mathrm{n}}-v)-r_{\mathrm{o}\mathrm{u}\mathrm{t}}(w_{\mathrm{o}\mathrm{u}\mathrm{t}}-v)$ hold, the derived law reads $m(t)v^{\mathrm{\prime }}(t)=1\cdot e$, and $v(1)=\ln (2)e$. The naive closed-system reading $(mv{)}^{\mathrm{\prime }}=F\equiv 0$ would instead force $v\equiv v(0)=0$ on $I$, which does not satisfy the material balance; the two readings differ as soon as $t>0$.

The finite collision relations used here are the momentum-conservation and restitution equations appearing in clauses a-b of Isolated reciprocal impulse with full-vector restitution: momentum jumps $\pm J$, conservation of total momentum, the restitution solution and the kinetic-energy change $-\frac{1-e^2}{2}\mu \parallel v_2^{-}-v_1^{-}{\parallel }_2^2$. That theorem derives momentum jumps from a full distributional impulse model and assumes the Axiom of Countable Choice through its jump-derivative supplier. It is contextual here: the present explicit vectors and kinetic changes are proved directly, without claiming to instantiate that full model or importing its choice assumption.