When dealing with all but the simplest collisions, analysis is much simpler in the center of mass reference frame. I will denote this frame 'C'. For the sake of doing physics, observing from C is not based on observation from the [i]position [/i]of the center of mass, but rather from a vantage point that is moving with the [i]velocity[/i] of the center of mass. Clearly if you are at the position of the center of mass, and that position is moving, you should be moving right along with it. But really what is important is the velocity from which you are observing. It is fine to imagine the position to also coincide with the center of mass position, but that is more a matter of preference than necessity. We will take that view.[br][br]The reason we go through all the trouble of using different reference frames is that if we view collisions from C, the calculations are greatly simplified as compared to viewing from any other frame such as from earth's frame of reference E. The reason for this simplification is that in C, the system's momentum is zero at all times before, during and after the collision.[br][br]So let's imagine observing a collision from the location of the center of mass, C. If C is moving, you are moving with it.  Recall that the velocity of the center of mass is arrived at by taking the derivative of the position of the center of mass, or [math]\vec{v}_{c,E}=\frac{d}{dt}\left(\frac{1}{M}\sum_im_i\vec{r_{i,E}}\right)=\frac{1}{M}\sum_im_i\vec{v_{i,E}}[/math]. Notice the dual subscripts. [b]The first subscript designates the object we are observing, and the second subscript designates the reference frame from which the observation is being made[/b]. Occasionally in this chapter there will be a subscript on the second subscript such as [math]\vec{v}_{2,E_i}[/math]. This vector designates the velocity of object 2 as measured from the earth (E) reference frame, initially - or before the collision. We will use an [i]f[/i] to designate a final velocity. [br][br]In the diagram below you can see all the vectors that may be needed to calculate the outcome of a collision. Notice that the initial velocities are linked. That is not to say that the velocity of a baseball has any influence on the velocity of a bat in the earth frame E, but once those velocities are expressed in the center of mass frame C, they are inextricably linked such that they always point in opposite directions.  If one of the velocities gets larger the other will as well. If you change one mass the other mass's velocity is affected. This is because both objects must carry equal momentum toward C so that total momentum is always zero in C.

When viewed in the center of mass reference frame, collisions are always symmetric.  When viewed in C, if you can find the component of each velocity carrying each object toward the center of mass C, then those component just get multiplied by -e upon colliding.[br][br]Notice that in collisions that are not head-on, like the grazing impact of a sharp cut shot in billiards, only the velocity components along the direction connecting the centers of mass get reversed and multiplied by e while the other orthogonal component will not be altered unless we were to discuss other factors like friction and angular momentum - which we will not do right now.  Since billiards balls are smooth and relatively slippery in contact with one another, ignoring such effects will be pretty accurate. Realize, however, that expert billiards players take advantage of the small friction forces and use them to affect the outcomes of their shots. It is too advanced a topic to consider such effects in this course.[br][br]The math that takes into account what we just said about the velocity vectors will give us velocities after collision viewed in C in terms of velocities before impact in C.  Notice that the expressions are identical for the two objects, except that one refers to object 1 and the other to object 2:[br][br][center][math][br]\vec{v}_{1,C_f}=\vec{v}_{1,C_i}-(1+e)(\vec{v}_{1,C_i}\cdot \hat{r})\hat{r} \\[br]\vec{v}_{2,C_f}=\vec{v}_{2,C_i}-(1+e)(\vec{v}_{2,C_i}\cdot \hat{r})\hat{r}[br][/math][/center][br][br]Also note that since [math]\vec{p}=m\vec{v}[/math], that we can write these expressions in terms of momenta as well:[br][br][center][math][br]\vec{p}_{1,C_f}=\vec{p}_{1,C_i}-(1+e)(\vec{p}_{1,C_i}\cdot \hat{r})\hat{r} \\[br]\vec{p}_{2,C_f}=\vec{p}_{2,C_i}-(1+e)(\vec{p}_{2,C_i}\cdot \hat{r})\hat{r}[br][/math][/center][br][br]Furthermore, since the total momentum in the center of mass frame is always zero, we know that we can also write the two equations below. Thus we have the ability to relate the momenta of the two objects before and after collision: [br][br][center][math]\vec{p}_{1,C_i}+\vec{p}_{2,C_i}=\vec{0} \\[br]\vec{p}_{1,C_f}+\vec{p}_{2,C_f}=\vec{0}[/math][/center] Since this is true, we can save calculation efforts by realizing that in the center of mass frame C, the two objects must have equal and opposite momenta to one another both in the moments before, during and after the collision.

I want to be sure that you know how to use these equations. The most confusing term is likely the [math]\hat{r}.[/math] That term is the vector that points toward the collision plane which is shown as a dotted line in the diagram. Think about the fact that when two balls collide, it is as if they are each bouncing off a wall which would be oriented like the dotted line. In that sense the [math]\hat{r}[/math] vector is the surface normal vector of the collision plane. It doesn't actually matter whether you make it the outward or inward normal from the ball's perspective since in the equations, two minus signs would cancel anyway. [br][br]I want you to further recognize that [math]\hat{r}[/math] doesn't depend on the velocities of the objects, but rather only on their relative positions upon impact. Since it's a unit vector, you may use the unit circle from trigonometry and define [math]\hat{r}=\cos\phi\hat{i}+\sin\phi\hat{j}[/math], where [math]\phi[/math] is the angle with respect to the x-axis of the [math]\hat{r}[/math] vector.[br][br]In the event that an object like a ball bounces off a planar surface like a wall or a windshield or the face of a golf club, [math]\hat{r}[/math] is defined as the surface normal vector of the planar surface. Again, it will not matter if you reverse it since the negative signs will cancel in the expression. It will also have no dependence on the incoming trajectory of the ball.

The symmetry of the equations above is lost in other coordinate systems. It becomes so difficult to handle the equations in other frames that it is never worth doing. In other words, what we should do instead is take vectors known in other reference frames and convert them to vectors measured in C, treat the collision in C, and then reconvert vectors back to the other reference frame if need be.[br][br]The most common situation is that we measure velocities in the earth's reference frame, E.  That being true, our procedure looks like this:[br][list=1]1. Find the center of mass velocity with respect to earth.[br]2. Take the two earth-frame velocities and convert them to the center of mass frame using the relative velocity addition equation.[br]3. Apply the equations above to find the final velocities in C.[br]4. Convert those velocities back to frame E.[/list][br]

Let's do an example in which a baseball (b) is pitched toward a batter who swings a wooden bat (w) and strikes the ball. We would like to predict both how the bat and the ball move after impact given how they were moving before impact, how they were oriented during impact and assuming a typical coefficient of restitution, e, for a baseball and bat combination. We will assume that we only know the initial velocities in the earth (E) frame.[br][br]We will assume the following known values:[br][math]\vec{v}_{w,E_i}=25m/s\hat{i}+5m/s\hat{j} \\[br]\vec{v}_{b,E_i} = -40m/s\hat{i}-2m/s\hat{j} \\[br]e=0.5 \\[br]m_b=0.145kg \\[br]m_w=0.880kg \\[br]\text{Angle of impact }\theta=15^o. [/math][br][br]The first step is to find these velocities in the center of mass frame. To do this we need to find the center of mass velocity of the system (with respect to E). This is the weighted average of velocity vectors, as discussed in an earlier section:[br][math]\vec{v}_{C,E}=\frac{m_b\vec{v}_{b,E}+m_w\vec{v}_{w,E}}{m_b+m_w} \\[br]\vec{v}_{C,E} = 15.8\frac{m}{s}\hat{i}+4.0\frac{m}{s}\hat{j}.[/math][br][br]Next we use the relative velocity equation to find the ball's and wooden bat's velocities in C:[br][br][math]\vec{v}_{b,C_i}=\vec{v}_{b,E_i}+\vec{v}_{E,C} \\[br]\text{Here we need to flip the center of mass velocity due to inverted subscripts} \\[br]\vec{v}_{b,C_i}=\vec{v}_{b,E_i}-\vec{v}_{C,E} \\[br]\vec{v}_{b,C_i}=-55.8\frac{m}{s}\hat{i}-6\frac{m}{s}\hat{j}. \\[br]\text{and} \\[br]\vec{v}_{w,C_i}=\vec{v}_{w,E_i}+\vec{v}_{E,C} \\[br]\vec{v}_{w,C_i}=\vec{v}_{w,E_i}-\vec{v}_{C,E} \\[br]\vec{v}_{w,C_i}=9.2\frac{m}{s}\hat{i}+1.0\frac{m}{s}\hat{j}.[/math][br][br]Next we apply the equations to find the final velocities in C given the initial ones, and the value of the coefficient of restitution e. Notice that the bat's final velocity is opposite that of the ball in this frame. If we multiply by mass to get momenta, then they are exactly equal and opposite one another:[br][math]\vec{v}_{b,C_f}=\vec{v}_{b,C_i}-(1+e)(\vec{v}_{b,C_i}\cdot \hat{r})\hat{r} \\[br]\text{We need to use } \hat{r}=\cos(15^o)\hat{i}+\sin(15^o)\hat{j}. \\[br]\vec{v}_{b,C_f}=24.5\frac{m}{s}\hat{i}+15.5\frac{m}{s}\hat{j} \\[br]\vec{v}_{w,C_f}=-4.1\frac{m}{s}\hat{i}-2.5\frac{m}{s}\hat{j} \\.[/math][br]The last step is to convert these back to the frame E so that we know what they would look like to spectators at rest with respect to E (earth):[br][math][br]\vec{v}_{b,E_f}=40.3\frac{m}{s}\hat{i}+19.5\frac{m}{s}\hat{j} \\[br]\vec{v}_{w,E_f}=11.7\frac{m}{s}\hat{i}+1.5\frac{m}{s}\hat{j}. [/math]

When you watch a game of billiards or imagine air molecules colliding, it looks complicated. The reality is that in nearly every instance there are only pairs of objects colliding... it's just that there are many pairs. It's possible for more than two to be involved, but that occurs much more seldom. Below is an animation of inelastic collisions among many balls. Watch one of the balls for a long duration and observe its collisions with the other balls.