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The Center-of-Mass Frame

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The center of mass obeys Newton's second law

For any system of particles, however complicated its internal motion, the center of mass moves as if all the mass were concentrated there and all external forces acted on that single point: ΣF_ext = M a_cm. Internal forces come in third-law pairs and cancel exactly, so they cannot accelerate the center of mass. A wrench tumbling through the air rotates chaotically while its center of mass traces a clean parabola — that is this theorem made visible.

Center of mass position and velocity
x_cm = Σm_i x_i / M v_cm = Σm_i v_i / M = p_total/M
The second form is the useful one: the center-of-mass velocity is just the total momentum divided by the total mass, so a system with zero total momentum has a stationary center of mass.

Why collisions leave it untouched

During a collision the forces between the colliding bodies are internal to the system, so they cannot change the total momentum and therefore cannot change v_cm. This is a powerful check: whatever happens in a collision — sticking, bouncing, shattering into pieces — the center of mass continues at exactly the velocity it had before. An answer in which the center of mass changes velocity without an external force is wrong regardless of how the individual pieces came out.

The zero-momentum frame

Shift to a reference frame moving at v_cm and the total momentum becomes zero by construction. In that frame two colliding objects always approach with equal and opposite momenta and separate with equal and opposite momenta, which makes the symmetry of a collision obvious and the algebra far lighter. A perfectly inelastic collision is especially transparent there: the combined object is simply at rest, so the kinetic energy lost is exactly the total kinetic energy the system had in that frame. Solve in the center-of-mass frame, then add v_cm back to return to the ground frame.

Worked example

A 3.0 kg cart moves right at 4.0 m/s toward a 1.0 kg cart moving left at 2.0 m/s. Find v_cm, and state the velocity of the combined object if they stick together.

  1. 1.Total momentum, taking right as positive: p = (3.0)(+4.0) + (1.0)(−2.0) = 12.0 − 2.0 = 10.0 kg·m/s.
  2. 2.Total mass M = 4.0 kg, so v_cm = p/M = 10.0/4.0 = 2.5 m/s to the right.
  3. 3.If they stick, the combined 4.0 kg object carries the whole momentum: v = 10.0/4.0 = 2.5 m/s.
  4. 4.These are the same number, and necessarily so: a perfectly inelastic collision leaves the objects moving together, which is precisely at v_cm.
Answer: v_cm = 2.5 m/s to the right both before and after; the stuck-together carts move at exactly v_cm, which is why that case requires no further algebra
Checkpoint

Two skaters push off from each other on frictionless ice, starting at rest. After the push, their center of mass —

Tip

In a perfectly inelastic collision the final velocity is v_cm. Recognizing that turns a two-equation problem into a single division: total momentum over total mass. It works in one, two or three dimensions.

Checkpoint

An exploding firework shell bursts into many fragments in mid-flight. Neglecting air resistance, the center of mass of the fragments —

Answer the 2 checkpoints as you read.

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