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From my understanding, the coefficient of restitution, $e$, between two colliding particles, $A$ and $B$ is given by: $$ -e = \frac{ \left( \vec{v_A} - \vec{v_B} \right) \cdot \hat{I} }{ \left( \vec{u_A} - \vec{u_B} \right) \cdot \hat{I} $$ where $\vec{u}$ and $\vec{v}$ represent the initial and final velcoities of a particle, respectively, and $\hat{I}$ is a unit vector parallel to the impulse that $A$ experiences from $B$ - or that $B$ experiences from $A$ - (it actually doesn't matter because they are parallel to one another).

I have heard it said that when the collision between $A$ and $B$ is perfectly elastic (defined as meaning that the total kinetic energy, from before the collision to after it, remains the same), $e=1$.

However, I have never seen a proof for this and I was wondering if anyone could generally demonstrate why this is true.

I would start by applying conservation of linear momentum: $$ m_A\vec{u_A} + m_B\vec{u_B} = m_A\vec{v_A} + m_B\vec{v_B}$$ and conservation fo kinetic energy: $$\frac12 m_Au_A^2 + \frac12 m_Bu_B^2 = \frac12 m_Av_A^2 + \frac12 m_Av_B^2$$ but wouldn't know where to go from there!

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starting with Newton equation immediately after the collision

\begin{align*} &m_1\,\frac{d\mathbf v}{dt}= -F_c\,\mathbf n\quad\Rightarrow\\ &m_1\,\int_{\mathbf u_1}^{\mathbf v_1}d\mathbf v=-\int F_c\,\mathbf n\,dt=-\lambda\mathbf{n}\\ &m_1\,(\mathbf v_1-\mathbf u_1)=-\lambda\,\mathbf n \quad \text{and}\\ &m_2\,(\mathbf v_2-\mathbf u_2)=+\lambda\,\mathbf n \quad\Rightarrow\\ &\text{conservation of linear momentum}\\ &m_1\,(\mathbf v_1-\mathbf u_1)+ m_2\,(\mathbf v_2-\mathbf u_2)=0 \end{align*} the conservation of the energy \begin{align*} &E=\frac{1}{2}\left(m_1\,(\mathbf{v}_1)^2+m_2\,(\mathbf{v}_2)^2- m_1\,(\mathbf{u}_1)^2-m_2\,(\mathbf{u}_2)^2\right)=0\\ &2\,E=\left(m_1\,\left [(\mathbf{v}_1)^2- (\mathbf{u}_1)^2\right] +m_2\,\left[(\mathbf{v}_2)^2- (\mathbf{u}_2)^2\right]\right)=0\\ &2\,E=\left(m_1\,\left [\mathbf{v}_1- \mathbf{u}_1\right]\cdot \left [\mathbf{v}_1+ \mathbf{u}_1\right] +m_2\,\left[\mathbf{v}_2-\mathbf{u}_2\right] \cdot \left[\mathbf{v}_2+\mathbf{u}_2\right]\right)=0\\ &\text{with}\quad \mathbf{v}_1- \mathbf{u}_1=-\frac{\lambda}{m_1}\,\mathbf n \quad, \mathbf{v}_2- \mathbf{u}_2=\frac{\lambda}{m_2}\,\mathbf n\quad\Rightarrow\\ &2\,E=\left[(\mathbf v_2-\mathbf{v}_1)+(\mathbf u_2-\mathbf u_1)\right]\cdot\mathbf n=0 \end{align*} and with the coefficient of restitution $~\epsilon~$ \begin{align*} &\left[(\mathbf v_2-\mathbf{v}_1)+\epsilon\,(\mathbf u_2-\mathbf u_1)\right]\cdot\mathbf n=0 \end{align*}

thus for $~\epsilon=1~$ you obtain the conservation of the energy

  • $~\mathbf v_i~$ final after the collision
  • $~\mathbf u_i~$ initial velocities
  • $~\mathbf n~$ unit normal vector
  • $~F_c~$ constraint force towards the normal vector
  • $~\epsilon~$ coefficient of restitution
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If, in a collision between two spheres, the coefficient of restitution is unity, then for the velocity components parallel to the line $O_AO_B$ joining the spheres' centres at the collision...

Relative velocity of separation = – Relative velocity of approach

that is:

$$\vec {v_{\parallel B}}-\vec {v_{\parallel A}}=\vec {u_{\parallel A}}-\vec {u_{\parallel B}}$$

Re-arranging

$$\vec {v_{\parallel B}}+\vec {u_{\parallel B}}=\vec {v_{\parallel A}}+\vec {u_{\parallel A}}$$

Re-arranging the momentum conservation equation for these components:

$$m_B\left(\vec {v_{\parallel B}}-\vec {u_{\parallel B}}\right)=m_A\left(\vec {v_{\parallel A}}-\vec {u_{\parallel A}}\right)$$

The scalar product of the left hand sides of the previous two equations must therefore be equal to that of their right hand sides, so ...

$$m_B\left({v_{\parallel B}}^2-{u_{\parallel B}}^2\right)=m_A\left({v_{\parallel A}}^2-{u_{\parallel A}}^2\right)$$

But if no tangential forces act, the spheres retain their own velocity components at right angles to $O_AO_B$, so

$$\vec{v_{\perp A}}=\vec{u_{\perp A}}\ \ \ ,\ \ \ \ \vec{v_{\perp B}}=\vec{u_{\perp B}}$$

Therefore

$$m_A\left({v_{\perp A}}^2-{u_{\perp A}}^2\right)=0\ \ \ \ ,\ \ \ \ m_B\left({v_{\perp B}}^2-{u_{\perp B}}^2\right)=0$$

Remembering that $\ {u_{\parallel A}}^2+{u_{\perp A}}^2={u_A}^2$ and so on, we can add the equations with squared quantities in them and re-arrange to give

$$m_A{v_A}^2+m_B{v_B}^2=m_A{u_A}^2+m_B{u_B}^2$$

We see that translational kinetic energy is conserved. As we are assuming that no tangential forces act, neither sphere will acquire rotational energy, so the collision will be elastic.

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  • $\begingroup$ This answer has been extensively altered, because the first version made the mistaken claim that if the coefficient of restitution is unity then relative velocity of separation is equal to –(relative velocity of approach). This is the case only for the velocity component parallel to the line joining the centres at impact. $\endgroup$ Commented Feb 14 at 19:17

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