Coefficient of restitution for a perfectly inelastic collision The coefficient of restitution is defined as the ratio of the differences in velocities of colliding objects after and before the collision: $$k_{COR}=\frac{v_{1,after}-v_{2,after}}{v_{1,before}-v_{2,before}}.$$ There also exists a second definition, where $$k_{COR}=\sqrt \frac{E_{k,after}}{E_{k,before}}.$$
As such, in a perfectly inelastic collision, where the colliders stick (have equal velocity), $$k_{COR}=\frac{v_{after}-v_{after}}{v_{1,before}-v_{2,before}}=0.$$ However, according to the second definition, this means that all kinetic energy is lost ($E_{k,after}=0$). This cannot be true, as were there no kinetic energy, there could be motion at all.
Both definitions are from Wikipedia. Is the page wrong, or there exists an explanation for this?
 A: The article specifies the equation dealing with kinetic energy is looking at the relative kinetic energy. For a perfectly inelastic collision, the bodies are not moving relative to each other, so the relative kinetic energy is $0$. Thus there is no contradiction.

To add more detail to this, the best thing to do is to work in the center of momentum frame, which is the frame where the total momentum of the system is $0$. This can be done by first noting that, by definition, the center of mass of two objects (which we treat as point particles) is
$$x_\text{COM}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$$
which means the velocity of the center of mass is
$$v_\text{COM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$
where $v_1$ and $v_2$ are the velocities observed in some inertial frame of reference.
Therefore, to move to the center of momentum frame, all we need to do is change our velocities to $v_1\to v_1-v_\text{COM}$ and $v_2\to v_2-v_\text{COM}$. You can easily show that in this frame, $p_\text{total}=0$, i.e.
$$m_1(v_1-v_\text{COM})+m_2(v_2-v_\text{COM})=0$$
The kinetic energy in this center of momentum frame is the "relative kinetic energy".
$$K_r=\frac12m_1(v_1-v_\text{COM})^2+\frac12m_2(v_2-v_\text{COM})^2=\frac12\cdot\frac{m_1m_2}{m_1+m_2}\cdot(v_1-v_2)^2$$
As you can see, this kinetic energy involves the relative velocity between the two objects, as well as the reduced mass $\mu=m_1m_2/(m_1+m_2)$. You can then easily show from here that for a collision between two objects
$$k_\text{COR}=\frac{v_{1,\text{after}}-v_{2,\text{after}}}{v_{1,\text{before}}-v_{2,\text{before}}}=\sqrt{\frac{(v_{1,\text{after}}-v_{2,\text{after}})^2}{(v_{1,\text{before}}-v_{2,\text{before}})^2}}=\sqrt{\frac{K_{r,\text{after}}}{K_{r,\text{before}}}}$$
