The center of mass of a system comprising N masses is just the average location of the mass. Since the masses of the parts are not necessarily equal, it must be a mass weighted average. A mass weighted average is just like a weighted average of your score in your classes. Suppose you have an 80% quiz average, a 90% exam average and a 70% lab average, and quizzes are worth 100 points, exams are worth 300 points and labs are worth 500 points. How do you figure your grade in the course? I imagine you know that your score is [br][br][center][math]\text{your course \%}=\frac{100points(80\%)+300points(90\%)+500points(70\%)}{100points+300points+500points}.[/math][/center]

The center of mass is exactly the same thing except that the points (weightings) are the masses, and the scores or values are the position vectors of the masses. The equation for three masses 1, 2 and 3 is [br][br][center][math]\vec{r}_C=\frac{m_1\;\vec{r}_1+m_2\;\vec{r}_2+m_3\;\vec{r}_3}{m_1+m_2+m_3}. \\[br]\text{In general for N masses we have } \\[br]\vec{r}_C=\frac{\sum_{i=1}^N m_i\vec{r}_i}{\sum_{i=1}^N m_i} \\[br]\vec{r}_C=\frac{\sum_{i=1}^N m_i\vec{r}_i}{M} \\[br]\text{where $M=\sum_{i=1}^N m_i$.} [br][/math][/center][br][br]A three-body (N=3) system is illustrated in the interactive graphic below. The points can be moved and the masses changed, and the black point is the center of mass.

To get the velocity and the acceleration of the center of mass of a system we simply take a time derivative or two. In other words, [math]\vec{v}_C=\frac{d\vec{r}_C}{dt}[/math] and [math]\vec{a}_C=\frac{d^2\vec{r}_C}{dt^2}=\frac{d\vec{v}_C}{dt}.[/math] The velocity of the center of mass will be important for our analysis of collisions. The velocity of the center of mass is[br][br][center][math]\vec{v}_C=\frac{m_1\;\vec{v}_1+m_2\;\vec{v}_2+m_3\;\vec{v}_3}{m_1+m_2+m_3}.[/math][/center][br][br]The important thing about this velocity is that it is constant during collisions - in any reference frame. While in a two-body collision both individual masses have velocities that abruptly change, the center of mass velocity is nonetheless perfectly constant right through the collision. This can be easily shown using the definition of center of mass velocity above. The momentum may always be expressed as either [math]\vec{p}_{system}[/math] or as the sum of the momenta of the parts [math]\sum_i\vec{p}_i.[/math] The two expressions are equal. But if the momentum of the system is constant in the absence of external forces (which is true even during collisions), then [br][br][center][math]\vec{p}_{system}=m_{system}\vec{v}_C=\text{constant}. \\[br]\text{From the velocity of the center of mass above, that means: }\\[br]\vec{v}_C=\frac{m_1\;\vec{v}_1+m_2\;\vec{v}_2+m_3\;\vec{v}_3}{m_1+m_2+m_3} \\[br](m_1+m_2+m_3)\vec{v}_C=m_1\;\vec{v}_1+m_2\;\vec{v}_2+m_3\;\vec{v}_3. \\[br]m_{system}\vec{v}_C = m_1\;\vec{v}_1+m_2\;\vec{v}_2+m_3\;\vec{v}_3 \\[br]\text{The line above says the system's momentum is just the sum of its parts: } \\[br]\vec{p}_{system}=\sum_i\vec{p}_i \\[br]\therefore\vec{v}_{C}=constant.[br][/math][/center] [br]According to this equation, the center of mass velocity is constant if the sum of the momenta of the parts is constant... two ways of saying the same thing. [br][br]What's is even more useful in terms of analyzing complicated collisions is that if we view the system from the center of mass frame C, then the velocity of the center of mass itself must be zero since every object is at rest with respect to itself. This implies that the momentum is zero also in the center of mass frame! We use this fact to derive the consequences of a collision in the next section.