A coefficient of restitution above $1$ manufactures kinetic energy: $e=2$ on $m_1=m_2=1\mathrm{k}\mathrm{g}$ with velocities $\pm 1\mathrm{m}{\mathrm{s}}^{-1}$ raises the kinetic energy by $3\mathrm{J}$ while the configuration is unchanged

In numerical SI coordinates, $M=2$, $\mu =1\mathrm{/}2$, $V=0$, $d^{-}:=v_2^{-}-v_1^{-}=-2n$, and $\mathrm{\mid }{d}^{-}{\mathrm{\mid }}^2=4$, since $\mathrm{\mid }n\mathrm{\mid }=1$. The incoming kinetic energy is $K^{-}=(\mathrm{\mid }n{\mathrm{\mid }}^2+\mathrm{\mid }-n{\mathrm{\mid }}^2)\mathrm{/}2=1\mathrm{J}$.

For any real $e$, put $d^{+}=-e{d}^{-}$. The momentum equation gives $m_1{v}_1^{+}+m_2{v}_2^{+}=MV$ and the restitution equation gives $v_2^{+}=v_1^{+}+d^{+}$. Substitution yields $M{v}_1^{+}+m_2{d}^{+}=MV$, hence $v_1^{+}=V+(m_2\mathrm{/}M)e{d}^{-}$ and $v_2^{+}=V-(m_1\mathrm{/}M)e{d}^{-}$. These vectors satisfy both equations, so they are the unique outcome of the hypothetical law. At $e=2$ they are $v_1^{+}=-2n$ and $v_2^{+}=2n$ in metres per second; their total momentum is zero, and $J=m_1(v_1^{+}-v_1^{-})=-3n=-m_2(v_2^{+}-v_2^{-})$ in kilogram metres per second.

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

The declared potential is identically zero, hence unchanged across the continuous configuration. Consequently $\mathrm{\Delta }(K+U)=3\mathrm{J}>0$, which refutes the universal across-impact conservation assertion. To realize the endpoint data by free paths, in numerical metres and seconds take $t_{\ast }=0$, $I=(-1\mathrm{/}4,1\mathrm{/}4)$, $x_1(t)=tn$ for $t<0$ and $x_1(t)=-2tn$ for $t\ge 0$, and $x_2(t)=(1-t)n$ for $t<0$ and $x_2(t)=(1+2t)n$ for $t\ge 0$. Both positions are continuous, their derivatives are the stated velocities, their accelerations vanish off zero, and the separation is $1-2t>0$ before zero and $1+4t>0$ after zero. Thus the zero-potential free dynamics away from the impulse are consistent with the hypothetical endpoint law on a collision-excluded domain.

For the adopted range $0\le e\le 1$, 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.