To learn the definition of the center of mass for systems of particles and be able to locate it. Imagine throwing a rock upward and away from you. With negligible air resistance, the rock will follow a parabolic path before hitting the ground. Now imagine throwing a stick (or any other extended object). The stick will tend to rotate as it travels through the air, and the motion of each point of the stick (taken individually) will be fairly complex. However, there will be one point that will follow a simple parabolic path: the point about which the stick rotates. No matter how the stick is thrown, this special point will always be located at the same position within the stick. The motion of the entire stick can then be described as a combination of the translation of that single point (as if the entire mass of the stick were concentrated there) and the rotation of the stick about that point. Such a point, it turns out, exists for every rigid object or system of massive particles. It is called the center of mass.
To calculate the center of mass for a system of massive point particles that have coordinates (xi,yi) and masses mi, the following equations are used:
xcm= m1 + x1+m2x2 + m3x3+ .../m1 + m2 + m3+...
ycm= m1y1 + m2y2 + m3y3+ .../m1 + m2 + m3+...
Part A) Two particles of masses m1 and m2 (m1 < m2) are located 10 meters apart. Where is the center of mass of the system located ?

For a group of point masses, located along a straight line, we know, first of all, that the center of mass must be located on this line, as the y-coordinate of the center of mass is 0, due to no mass has a y-coordinate ≠ 0.

If we know that m₁ < m₂, we can choose to put m₁ just in the origin, so his x-coordinate, is 0 also.

With these premises, we can find mathematically the center of mass as follows:

Xcm = x₂*m₂ / (m₁+m₂)

So, as m₂/m₁+m₂ < 1, the x-coordinate of the center of mass must be located to the left of m₂.

An obvious question arises : How much to the left?

If (in the limit) m₁ =m₂, the factor m₂/ m₁+m₂ would be exactly 1/2.

⇒ Xcm = 1/2 * X₂ = 0.5* 10 m = 5 m

If m₁ < m₂, the center of mass will be more to the right (closer to m₂) than in the limit case when m₁=m₂, so it will be located at a distance longer than 5 m, and less than 10 m (which would be the case for m₁=0).