- Head-on collision between moving ball and stationary ball
- Head-on collision between two moving balls
- Glancing collision with a stationary ball
- Glancing collision between two moving balls
- Some limit-approaching versions of the above equations
- Coefficient of resitution
- Updating our general equations to include a coefficient of restitution
- Some limit-approaching situations, with a coefficient of restitution

Consider the case of one ball of mass $m$ approaching and striking a stationary ball of mass $M$. As we'll see, solving this problem sets us up to solve arbitrary collisions.

By the conservation of momentum, we know:

\begin{align} mv_i = mv_f + MV_f \end{align}And by conservation of kinetic energy:

\begin{align} \frac{1}{2}mv_i^2 &= \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2\\ \end{align}We have two equations with two unknowns ($v_f$ and $V_f$). Let us solve for the unknowns. From the conservation of momentum, we can solve for $v_f$ in terms of $v_i$ and $V_f$:

\begin{align} v_i - v_f &= \frac{M}{m}V_f \end{align}We want to substitute an expression for $V_f$ that is a function of $v_i$ and $v_f$. Let's use the conservation of kinetic energy:

\begin{align} V_f^2 &= \frac{m}{M}\left(v_i^2 - v_f^2\right)\\ &= \frac{m}{M}\left(v_i - v_f\right)\left(v_i + v_f\right)\\ \end{align}Substitute from the previous expression, noting that we must add $2v_f$ in our substitution for the last term:

\begin{align} V_f^2 &= \frac{m}{M} \left(\frac{M}{m}V_f\right)\left(\frac{M}{m}V_f + 2v_f\right)\\ V_f^2 &= V_f\left(\frac{M}{m}V_f + 2v_f\right)\\ V_f &= \frac{M}{m}V_f + 2v_f \end{align}Rearrange to solve for $V_f$:

\begin{align} 2v_f &= V_f \left(1 - \frac{M}{m}\right) \\ V_f &= \frac{2}{1 - \frac{M}{m}}v_f\\ \end{align}Substitute this equation for $V_f$ into the earlier expression for $v_f$:

\begin{align} v_i - v_f &= \frac{M}{m}V_f\\ &= \frac{M}{m}\left(\frac{2}{1 - \frac{M}{m}}v_f\right)\\ &= \frac{2M}{m - M}v_f \end{align}Solve for $v_f$:

\begin{align} v_i &= v_f + \frac{2M}{m - M}v_f\\ v_i &= v_f \left(1 + \frac{2M}{m - M}\right)\\ v_i &= v_f \left(\frac{m-M}{m-M} + \frac{2M}{m - M}\right)\\ v_i &= v_f \frac{m+M}{m-M}\\ \end{align}\begin{align} \boxed{v_f = \frac{m-M}{m+M}v_i} \end{align}Sometimes it is more convenient to represent this as a *change* in velocity, $\delta V = v_f - v_i$. This is easily-enough computed:

And with expression for $v_f$ we can solve for $V_f$:

\begin{align} V_f &= \frac{2}{1 - \frac{M}{m}}v_f\\ &= \frac{2m}{m-M} \frac{m-M}{m+M}v_i\\ \end{align}So, finally:

\begin{align} \boxed{V_f = \frac{2m}{m+M}v_i} \end{align}Here too, this is sometimes more conveniently represented as a change in velocity $\Delta V = V_f - V_i$.

\begin{align} \Delta V &= V_f - V_i\\ &= \frac{2m}{m+M}v_i \end{align}We can perform a few quick sanity checks. If $M>m$, we see that $v_f$ is negative, which makes sense (the ball bounces backward). If $m=M$, we see that $v_f$ is zero, and $V_f = v_i$. This is the "Newton's cradle case" in which the first ball gives all its momentum to the second.

Now suppose that our two balls are both moving with some initial velocity, $v_i$ *and* $V_i$. What will their velocities be after a collision in this case? As before, let us write down our conservation equations.

By conservation of momentum:

\begin{align} mv_i + MV_i = mv_f + MV_f \end{align}And by conservation of kinetic energy:

\begin{align} \frac{1}{2}mv_i^2 + \frac{1}{2}MV_i^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2 \end{align}We could start turning the crank as we did previously, now with an extra term, but let's think about this for a moment. The two above conservation laws hold in any inertial (non-accelerating) reference frame. None are more correct than any other, and some make solving the above problem easier (though, ultimately, we will likely want to express our answer in a particular frame). Let us consider an inertial frame that is "riding along" with the mass $M$. That is to say, the frame itself is moving with velocity $V_i$. This is a perfectly valid frame, and in this frame the mass $M$ is stationary. We just need to represent all of our velocities in this new frame. I'll use the tilde to represent the new frame.

\begin{align} \tilde{v_i} &= v_i - V_i \\ \tilde{v_f} &= v_f - V_i \\ \tilde{V_i} &= V_i - V_i = 0\\ \tilde{V_f} &= V_f - V_i \end{align}We can re-write our conservation laws in this frame. By conservation of momentum:

\begin{align} m\tilde{v_i} = m\tilde{v_f} + M\tilde{V_f} \end{align}And by conservation of kinetic energy:

\begin{align} \frac{1}{2}m\tilde{v}_i^2 = \frac{1}{2}m\tilde{v}_f^2 + \frac{1}{2}M\tilde{V}_f^2 \end{align}We're left with *exactly* the problem that we just solved! From the previous section, we know that:

So we could either solve for all the $\tilde{v}$'s, and then transform back to our original frame by adding $V_i$. Or, if we wanted, we could perform some substitutions into the above equations to put them back in the original frame.

\begin{align} v_f &= \frac{m-M}{m+M}\left(v_i - V_i\right) + V_i\\ V_f &= \frac{2m}{m+M}\left(v_i - V_i\right) + V_i \end{align}By way of some algebraic manipulation, we can represent the above equations in a few different ways. Some of these will be more convenient than others for subsequent analysis:

\begin{align} v_f &= \frac{2M}{M+m}\left(V_i - v_i\right) + v_i\\ V_f &= \frac{2m}{m+M}\left(v_i - V_i\right) + V_i \end{align}Or alternatively:

\begin{align} v_f &= \frac{1}{m+M}\left(mv_i + MV_i + M(V_i - v_i)\right)\\ V_f &= \frac{1}{m+M}\left(MV_i + mv_i + m(v_i - V_i)\right) \end{align}Or, represented as changes in velocity:

\begin{align} \Delta v &= \frac{m-M}{m+M}\left(v_i - V_i\right) + V_i - v_i\\ \Delta V &= \frac{2m}{m+M}\left(v_i - V_i\right) \end{align}Algebraically equivalent:

\begin{align} \Delta v &= \frac{2M}{M+m}\left(V_i - v_i\right)\\ \Delta V &= \frac{2m}{m+M}\left(v_i - V_i\right) \end{align}Suppose that a ball strikes an initially-stationary second ball with a glancing impact. That is, the line which connects the two balls at the moment of impact is not parallel with the velocity of the moving ball. What happens in that case?

We're now in two dimensions, since the balls will move in the plane rather than along a line, but we've actually already solved most of this problem in the previous sections.

As before, momentum and kinetic energy are both conserved through the elastic collision. Because we're now in two dimensions, we'll represent conservation equations using vector arithmetic. By conservation of momentum and kinetic energy:

\begin{align} m\bf{v}_i &= m\bf{v}_f + M\bf{V}_f\\ \frac{1}{2}m\bf{v}_i^T\bf{v}_i &= \frac{1}{2}m\bf{v}_f^T \bf{v}_f + \frac{1}{2} m \bf{V}_f^T\bf{V}_f \end{align}Remember that our total momentum is a *vector*. The fact that this vector is conserved through the collision implies that the *components* of that vector, as represented in any orthogonal basis, are also conserved. In other words, if $\bf{p}$ = $p_x \hat{x} + p_y \hat{y}$, then $p_x$ and $p_y$ are also conserved. But we are free to choose any orthogonal basis that we want! In other words, we could break the above vector equation into the two equations below:

When the balls impact one another, it is a *normal force* that affects their motion. This normal force must be perpendicular to each of their surfaces (i.e., "normal"). In order for this to be possible, the force exerted must act along the line which connects the centers of each ball.

By definition, force is change in momentum with respect to time ($F = ma = m\frac{dv}{dt} = \frac{dp}{dt}$). The fact that the force between the balls must act along the line which connects the centers of each ball implies that the change in momentum (for each ball) must also be along this line, and *only* along this line. This implies that we can make our lives easier by strategically choosing the basis which we use to represent the components of our conserved total momentum vector.

Let us choose a set of basis vectors such that one basis vector points along the line which connects the centers of the two balls, and the other basis vector points along the line which is tangent to both balls at the point of impact. The perpendicular and parallel symbols below indicate the component of the momentum that is perpendicular to the surfaces of both balls (along the line which connects their centers) and the component that is orthogonal to this vector, along the tangent of both balls at the point of impact.

\begin{align} mv_{i\bot} &= mv_{f\bot} + MV_{f\bot}\\ mv_{i\parallel} &= mv_{f\parallel} + MV_{f\parallel} \end{align}Now let us apply some reason. We just said that the only force (the only change in momentum for each ball) acts along the line which connects their centers, in the direction of the $\bot$ basis vector. Thus, there can be no change in momentum, for either ball, in the $\parallel$ direction! That is:

\begin{align} v_{i\parallel} &= v_{f\parallel}\\ V_{i\parallel} &= 0 = V_{f\parallel} \end{align}The component of the non-stationary ball which is perpendicular to the line connecting its center to the center of the ball that it strikes at the moment of impact remains *unaffected*. This leaves only the second conservation equation:

Again, we already solved this in a previous section! We found that the answer is the following:

\begin{align} v_{f\bot} &= \frac{m-M}{m+M}v_{i\bot}\\ V_{f\bot} &= \frac{2m}{m+M}v_{i\bot} \end{align}With a little bit of work, we can represent these changes in velocity in our original frame of reference. To find $v_{i\bot}$, we dot the $\bf{v}_i$ vector with the unit vector which points from mass $M$ to $m$. If the vector to the center of mass $m$ is $\bf{r}_m$, and to the center of mass $M$ is $\bf{r}_M$, then the vector which points from the center of $M$ to the center of $m$ can be found by simply subtracting the two:

\begin{align} \bf{r}_{\bot} &= \bf{r}_m - \bf{r}_M\\ \end{align}Normalize this vector and dot it with $\bf{v}_i$, and we have $v_{i\bot}$:

\begin{align} v_{i\bot} &= \frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\\ \end{align}With $v_{i\bot}$, we can solve for $v_{f\bot}$:

\begin{align} v_{f\bot} &= \left(\frac{m-M}{m+M}\right)\left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\\ \end{align}And indeed we can now solve for $\Delta \bf{v}$:

\begin{align} \Delta{\bf{v}} &= \left(v_{f\bot} - v_{i\bot}\right)_{\bot} + \left(v_{f \parallel} - v_{i \parallel}\right)_{\parallel}\\ &= \left[\left(\frac{m-M}{m+M}\right)\left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right) - \frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right]_{\bot} + \left[v_{i\parallel} - v_{i\parallel}\right]_{\parallel}\\ &= \left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\left[\left(\frac{m-M}{m+M}\right) - 1\right]_{\bot}\\ &= \left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\left[\left(\frac{m-M}{m+M}\right) - 1\right]\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||}\\ &= \left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_i\right)\left[\left(\frac{m-M}{m+M}\right) - 1\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}By nearly identical analysis, we can also solve for $V_{f\bot}$:

\begin{align} V_{f\bot} &= \frac{2m}{m+M}v_{i\bot}\\ &= \frac{2m}{m+M}\left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\\ \end{align}And for $\Delta \bf{V}$:

\begin{align} \Delta \bf{V} &= \left(V_{f\bot} - V_{i\bot}\right)_{\bot} + \left(V_{f\parallel} - V_{i\parallel}\right)_{\parallel}\\ &= \left(V_{f\bot} - 0\right)_{\bot} + \left(0 - 0\right)_{\parallel}\\ &= \left[\frac{2m}{m+M}\left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\right]_{\bot}\\ &= \left[\frac{2m}{m+M}\left(\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||} \cdot \bf{v}_i\right)\right]\frac{\bf{r}_{\bot}}{||\bf{r}_{\bot}||}\\ &= \left[\frac{2m}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_i\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}In the case of two moving balls have a glancing collision, we can execute exactly the same maneuver that we used for the head-on collision between two moving balls.

As before, momentum and kinetic energy are both conserved through the elastic collision. *Before* we switch inertial frames, our conservation equations look like this:

But of course momentum and energy are conserved in any inertial frame. Let's switch to one which moves with a constant velocity $\bf{V}_i$, alongside the mass M before the collision. As before, we'll use the tilde to represent our velocities in this new (but equally valid) inertial frame:

\begin{align} \tilde{\bf{v}}_i &= \bf{v}_i - \bf{V}_i\\ \tilde{\bf{v}}_f &= \bf{v}_f - \bf{V}_i\\ \tilde{\bf{V}}_i &= \bf{V}_i - \bf{V}_i = 0\\ \tilde{\bf{V}}_f &= \bf{V}_f - \bf{V}_i \end{align}In this frame, our conservation equations now look like the following:

\begin{align} m\tilde{\bf{v}}_i &= m\tilde{\bf{v}}_f + M\tilde{\bf{V}}_f\\ \frac{1}{2}m\tilde{\bf{v}}_i^T\tilde{\bf{v}}_i &= \frac{1}{2}m\tilde{\bf{v}}_f^T \tilde{\bf{v}}_f + \frac{1}{2} m \tilde{\bf{V}}_f^T\tilde{\bf{V}}_f \end{align}We've already solved this system! From the previous section, we know that:

\begin{align} \tilde{v}_{f\bot} &= \frac{m-M}{m+M}\tilde{v}_{i\bot}\\ \tilde{V}_{f\bot} &= \frac{2m}{m+M}\tilde{v}_{i\bot} \end{align}And thus, from previously (using one of the algebraically equivalent forms from the head-on collision between two moving balls section):

\begin{align} v_{f\bot} &= \frac{2M}{M+m}\left(V_{i\bot} - v_{i\bot}\right) + v_{i\bot}\\ V_{f\bot} &= \frac{2m}{m+M}\left(v_{i\bot} - V_{i\bot}\right) + V_{i\bot} \end{align}As before, we can solve for $\Delta \bf{v}$:

\begin{align} \Delta{\bf{v}} &= \left(v_{f\bot} - v_{i\bot}\right)_{\bot} + \left(v_{f \parallel} - v_{i \parallel}\right)_{\parallel}\\ &= \left[\left(\frac{2M}{M+m}\left(V_{i\bot} - v_{i\bot}\right) + v_{i\bot}\right) - v_{i\bot}\right]_{\bot} + \left[v_{i\parallel} - v_{i\parallel}\right]_{\parallel}\\ &= \left[\frac{2M}{M+m}\left(V_{i\bot} - v_{i\bot}\right) \right]_{\bot}\\ &= \left[\frac{2M}{M+m}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{V}_{i} - \frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_{i}\right) \right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ &= \left[\frac{2M}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{V}_i - \bf{v}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ &= -\left[\frac{2M}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}Via nearly identical analysis, we can solve for $\Delta \bf{V}$:

\begin{align} \Delta{\bf{V}} &= \left(V_{f\bot} - V_{i\bot}\right)_{\bot} + \left(V_{f \parallel} - V_{i \parallel}\right)_{\parallel}\\ &= \left[\left(\frac{2m}{M+m}\left(v_{i\bot} - V_{i\bot}\right) + V_{i\bot}\right) - V_{i\bot}\right]_{\bot} + \left[V_{i\parallel} - V_{i\parallel}\right]_{\parallel}\\ &= \left[\frac{2m}{M+m}\left(v_{i\bot} - V_{i\bot}\right) \right]_{\bot}\\ &= \left[\frac{2m}{M+m}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_{i} - \frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{V}_{i}\right) \right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ &= \left[\frac{2m}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}These are the most general forms of the equations, from which all previous equations can be extracted.

With the general form of the equations given above, we can derive some simpler-equations in some limit-approaching situations. These include bouncing off a wall, bouncing off a round peg (a "ball" that doesn't move), and two equal-mass balls bouncing off one another.

A flat wall is the *same thing* as a "ball" with infinite mass and infinite radius. Let us consider how the equations above simplify as as we approach these two limits. We'll start with $\Delta \bf{v}$, the change in velocity of the ball which strikes the wall. Here's the most general equation for this change in velocity:

Let us first consider the term $\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}$. Recall that this is the *unit vector* which points from the center of one "ball" ($M$) to the center of the other ball ("m"). As the radius of $M$ approaches infinity, this unit vector approaches the *normal vector* to the wall. That is, the vector which points directly *away* from the surface of the wall. For a ball of infinite radius, this is the line which passes through its "center" at infinity! Let's call this normal vector $\hat{\bf{n}}$.

Furthermore, the initial velocity of our wall is zero (the wall isn't moving). So $\bf{V}_i$ is also zero:

\begin{align} \Delta \bf{v} &= -\left[\frac{2M}{m+M}\left(\hat{\bf{n}} \cdot \bf{v}_i \right)\right]\hat{\bf{n}}\\ \end{align}And furthermore, the *mass* of our wall is infinite (it doesn't move when we hit it). So, $M$ is infinity, and the above equation simplifies even further:

The above equation should hopefully match your intuition. It says that, when a ball strikes a wall, the component of the ball's velocity which is normal to the ball reverses. Suppose, for instance, that the ball's velocity is entirely in the $\hat{x}$ direction, and that it strikes a vertical wall with a normal vector in the $-\hat{x}$ direction. In that case:

\begin{align} \Delta \bf{v} &= -2 \begin{bmatrix}v_x \\ 0 \end{bmatrix} \cdot \begin{bmatrix}-1\\0\end{bmatrix} &= \begin{bmatrix}-2v_x\\0\end{bmatrix} \end{align}The x-velocity simply reverses, which makes sense.

Suppose that the ball bounces off a round peg, which can be modeled as another ball with infinite mass. How do the general equations above simplify in this case? Here again is the most general expression:

\begin{align} \Delta \bf{v} &= -\left[\frac{2M}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}If we let $M\rightarrow \infty$, then the above expression becomes:

\begin{align} \Delta \bf{v} &= -2\left[\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}And the peg is stationary, so $V_i = 0$:

\begin{align} \Delta \bf{v} &= -2 \left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_i\right)\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}And of course, the final velocity of the peg reduces to zero as $M\rightarrow \infty$:

\begin{align} \Delta \bf{V} &= \left[\frac{2m}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ &= \bf{0} \end{align}What if $M=m$? Consider the most general expression below:

\begin{align} \Delta \bf{v} &= -\left[\frac{2M}{m+M}\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}If $M=m$, this becomes:

\begin{align} \Delta \bf{v} &= -\left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}And the expression for $\Delta \bf{V}$ becomes:

\begin{align} \Delta \bf{V} &= \left(\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_i - \bf{V}_i\right)\right)\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}Note that $\Delta \bf{v} = -\Delta \bf{V}$.

If you drop a ball in real life, and it bounces off the floor, it *never* returns to its original height. This is because the collision with the floor is not perfectly elastic, but instead some energy is lost. This energy goes into sound, heat, and vibration of the ball and floor. All these are quite tricky to model, but we can lump all of them together into a parameter called the *coefficient of restitution*.

The coefficient of restitution is the ratio between the relative velocities of the two balls before and after collision, in the direction parallel to the line which connects their centers at the moment of collision. It is a dimensionless paramter that describes *how elastic* (or how inelastic) a collision is. If the parameter is 1, the collision is perfectly elastic. If it's 0, it's perfectly inelastic. Anywhere inbetween models *some* energy loss. A coefficient of restitution that is greater than 1 models an explosive collision in which the balls gain energy.

In all of our previous analyses, we solved for the final velocities of the balls as a function of their initial velocities by means of conservation of momentum and conservation of energy. From these two equations, we can solve for a *third* equation (the restitution equation). Any two of these three equations (conservation of momentum, conservation of kinetic energy, and the restitution equation) can be used to solve collision problems. For incorporating a coefficient of restoration into our general collision equations, it will be easiest to use the conservation of momentum and the restitution equation.

Consider again the conservation of momentum and kinetic energy equations for a collision between two moving balls:

\begin{align} m\bf{v}_i + M\bf{V}_i &= m\bf{v}_f + M\bf{V}_f\\ \frac{1}{2}m\bf{v}_i^T\bf{v}_i + \frac{1}{2}M\bf{V}_i^T\bf{V}_i &= \frac{1}{2}m\bf{v}_f^T \bf{v}_f + \frac{1}{2} m \bf{V}_f^T\bf{V}_f \end{align}We've shown that, because the force between the balls occurs only along the line which connects their centers, we can consider the one-dimensional conservation equations along this line:

\begin{align} mv_i + MV_i &= mv_f + MV_f\\ \frac{1}{2}mv_i^2 + \frac{1}{2}MV_i^2 &= \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2 \end{align}From the conservation of energy:

\begin{align} \frac{1}{2}mv_i^2 + \frac{1}{2}MV_i^2 &= \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2\\ mv_i^2 + MV_i^2 &= mv_f^2 + MV_f^2\\ m\left(v_i^2 - v_f^2\right) &= M\left(V_f^2 - V_i^2\right)\\ m(v_i - v_f)(v_i + v_f) &= M(V_f - V_i)(V_f+V_i) \end{align}And from the conservation of momentum:

\begin{align} mv_i + MV_i &= mv_f + MV_f\\ m(v_i - v_f) &= M(V_f - V_i) \end{align}Combining these two equations:

\begin{align} v_i + v_f &= V_i + V_f\\ (v_i - V_i) &= -(v_f - V_f) \end{align}From which we can obtain the restitution equation, shown below. The restitution equation tells us that, for a perfectly elastic collision, the relative velocities of the balls before and after the collision (along the direction of the force) is equal.

\begin{align} \frac{|v_i - V_i|}{|v_f - V_f|} = 1 \end{align}To add a coefficient of restitution, which will allow for us to adjust the elasticity/inelasticity of the collision, we set the restituion equation equal to a number $C_R$, which is not necessarily equal to 1:

\begin{align} \frac{|v_i - V_i|}{|v_f - V_f|} = C_R \end{align}Or, written alternatively:

\begin{align} C_R(v_i - V_i) &= -(v_f - V_f) \end{align}Solve the conservation of momentum equation for $v_f$:

\begin{align} v_f &= \frac{mv_i + MV_i - MV_f}{m} \end{align}Solve the restitution equation for $V_f$:

\begin{align} V_f &= C_R(v_i - V_i) + v_f \end{align}Substitute:

\begin{align} v_f &= \frac{mv_i + MV_i - MV_f}{m}\\ &= \frac{mv_i + MV_i - M(C_R(v_i - V_i) + v_f)}{m}\\ &= \frac{mv_i + MV_i - MC_Rv_i + MC_RV_i - Mv_f}{m} \end{align}Rearrange:

\begin{align} v_f\left(1 + \frac{M}{m}\right) &= \frac{mv_i + MV_i + MC_R\left(V_i - V_i\right)}{m} \end{align}And solve for $v_f$:

\begin{align} v_f &= \frac{1}{m+M}\left(mv_i + MV_i + MC_R(V_i - v_i)\right) \end{align}By nearly identical analysis for $V_f$:

\begin{align} V_f &= \frac{1}{m+M}\left(MV_i + mv_i + mC_R(v_i - V_i)\right) \end{align}Let us solve for the modified $\Delta \bf{v}$ with a coefficient of resitution $C_R$:

\begin{align} \Delta{\bf{v}} &= \left(C_Rv_{f\bot} - v_{i\bot}\right)_{\bot} + \left(v_{f \parallel} - v_{i \parallel}\right)_{\parallel}\\ &= \left[\frac{m}{m+M}v_{i\bot} + \frac{M}{m+M}V_{i\bot} + \frac{MC_R}{m+M}V_{i\bot} - \frac{MC_R}{m+M}v_{i\bot} - \frac{m+M}{m+M}v_{i\bot}\right]_{\bot} + \left[v_{i\parallel} - v_{i\parallel}\right]_{\parallel}\\ &= \left[-\frac{M}{m+M}v_{i\bot} + \frac{M}{m+M}V_{i\bot} + \frac{MC_R}{m+M}V_{i\bot} - \frac{MC_R}{m+M}v_{i\bot}\right]_{\bot}\\ &= \left[\frac{M}{m+M}\left(-v_{i\bot} + V_{i\bot} + C_RV_{i\bot} - C_Rv_{i\bot}\right)\right]_{\bot}\\ &= \left[\frac{M}{m+M}\left(\left(V_{i\bot} - v_{i\bot}\right) + C_R\left(V_{i\bot} - v_{i\bot}\right)\right)\right]_{\bot}\\ &= \left[\frac{M}{m+M}\left(V_{i\bot} - v_{i\bot}\right)\left(1 + C_R\right)\right]_{\bot}\\ &= \left[\frac{M(1+C_R)}{m+M}\left(V_{i\bot} - v_{i\bot}\right)\right]_{\bot}\\ &= -\left[\frac{M(1+C_R)}{m+M}\left(v_{i\bot} - V_{i\bot}\right)\right]_{\bot}\\ &= -\frac{M(1+C_R)}{m+M}\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||}\\ \end{align}And by nearly identical analysis, the modified equation for $\Delta \bf{V}$:

\begin{align} \Delta{\bf{V}} &= \frac{m(1+C_R)}{m+M}\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}Let us revisit the same limit-approaching situations as before, now with a coefficient of resitution.

Start with the most general form of the equation:

\begin{align} \Delta{\bf{v}} &= -\frac{M(1+C_R)}{m+M}\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}As before, the vector from $M$ to $m$ approaches the normal to the wall as the radius of $M$ approaches infinity . . .

\begin{align} \Delta{\bf{v}} &= -\frac{M(1+C_R)}{m+M}\left[\hat{\bf{n}} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\hat{\bf{n}} \end{align}The mass of the wall, $M$, is infinite . . .

\begin{align} \Delta{\bf{v}} &= -(1+C_R)\left[\hat{\bf{n}} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\hat{\bf{n}} \end{align}And the velocity of the wall is zero . . .

\begin{align} \Delta{\bf{v}} &= -(1+C_R)\left[\hat{\bf{n}} \cdot \bf{v}_{i} \right]\hat{\bf{n}} \end{align}If $C_R$ is 1, the collision is fully elastic and the ball bounces off the wall with a velocity in the direction parallel to the normal of the wall that is of equal and opposite direction. If $C_R$ is 0, the collision is fully inelastic and the ball stops dead.

Start with the most general form of the equation:

\begin{align} \Delta{\bf{v}} &= -\frac{M(1+C_R)}{m+M}\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}The mass of the peg, $M$, is infinite . . .

\begin{align} \Delta{\bf{v}} &= -(1+C_R)\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}And the velocity of the peg is 0 . . .

\begin{align} \Delta{\bf{v}} &= -(1+C_R)\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \bf{v}_{i} \right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}Start with the most general form of the equation . . .

\begin{align} \Delta{\bf{v}} &= -\frac{M(1+C_R)}{m+M}\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}Set $M=m$ . . .

\begin{align} \Delta{\bf{v}} &= -\frac{1}{2}(1+C_R)\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}And nearly identically for $\Delta \bf{V}$:

\begin{align} \Delta{\bf{V}} &= \frac{1}{2}(1+C_R)\left[\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \cdot \left(\bf{v}_{i} - \bf{V}_{i}\right)\right]\frac{\bf{r}_m - \bf{r}_M}{||\bf{r}_m - \bf{r}_M||} \end{align}