Motion of the Center of Mass

The center of mass position is a weighted average: x_cm = (m1 x1 + m2 x2 + ...) / (m1 + m2 + ...). Differentiate both sides with respect to time and you get v_cm = (m1 v1 + m2 v2 + ...) / (m1 + m2 + ...), the mass-weighted average of the velocities. Differentiate again and you get a_cm, tied to the net external force through F_net,ext = M a_cm. Internal forces, pushes and pulls between pieces of the system, always cancel in pairs by Newton's third law, so they never show up in this equation. Only external forces can change how the center of mass moves.

This is why an exploding firework's center of mass keeps tracing the same parabola it was on before the blast. The explosion is all internal force. Each fragment flies off in its own direction, but gravity is still the only external force acting, so the center of mass of the whole debris cloud falls exactly like a single projectile would, right up until pieces hit the ground. The same idea tells you something non-obvious about a high jumper: the center of mass is just a calculated point, not a piece of the body, so it can sit outside the body entirely. By arching backward over the bar, a jumper lets their actual mass clear the bar while their center of mass passes underneath it, which takes less energy to achieve.

This topic sets up everything else in the unit. Newton's second law for systems, F_net = M a_cm, is what lets you treat a complicated extended object or a cluster of interacting particles as one point mass, which is exactly the simplification momentum and collision problems rely on next.