Physics I: Mechanics › Momentum › full formula sheet

vcm = (m₁v₁ + m₂v₂ + …) / M = Ptotal / M

Say it: “the center-of-mass velocity equals the total momentum divided by the total mass — and no internal force can change it”

Velocity of the center of mass

The one velocity that ignores all the chaos inside — explosions, collisions, and separations can't touch it.

vcm is the CM velocity (m/s), Ptotal the system's total momentum, M the total mass. Internal forces cancel — only external F changes vcm.

Before this lesson: Center of mass, Linear momentum

Where it comes from

A firework rocket rises, then explodes into a dozen fragments arcing everywhere. Chaos — except for one point: the center of mass keeps rising on its original parabola, utterly unimpressed by the explosion. That point's velocity is special:

Before reading on: two skaters push apart from rest — one glides left, one right. What is their center of mass doing? Moving left, right, or staying put?

Staying put. Equal and opposite momenta mean the mass-weighted average velocity is zero — the CM never moved. Internal pushes rearrange the parts but can't move the whole:

vcm = Ptotal / Mthe system's total momentum, per unit mass — immune to internal dramaSay it: “v sub c m equals total momentum over total mass”

Derivation

Differentiate the center-of-mass definition with respect to time:

xcm
=
(Σ mixi)/M
Step 1 — start here. The CM definition (see center of mass).
d/dt
⇒
vcm = (Σ mivi)/M
Step 2 — differentiate. dxi/dt = vi; masses and M are constant. Each term is a momentum mivi.
=
Ptotal/M
Step 3 — name it. Σmivi is the total momentum. So vcm = Ptotal/M — and since internal forces can't change Ptotal, they can't change vcm. ∎

One more derivative: acm = Fnet,ext/M. The CM accelerates as if all mass were there with all external forces applied — internal forces cancel pairwise.

How to use it

The procedure, every time:

  1. Fix the positive direction. Signed velocities, as always.
  2. Compute Ptotal = Σmivi. Every object's momentum, signs kept.
  3. Divide by M. vcm = Ptotal/M.
  4. Use the immunity. Explosions, collisions, separations — vcm is unchanged by all of them (no external impulse).
  5. External forces excepted. Gravity, friction over time — these change Ptotal and hence vcm.
Common mistake: averaging the velocities without mass-weighting ((v₁+v₂)/2). The momenta add; the masses divide. Plain velocity averages are wrong unless masses are equal.

Worked examples

Four problems, easiest first. Watch the CM ignore the chaos.

Example 1 — basic: 2 kg at +4 m/s, 3 kg at −1 m/s

  1. Total momentum. P = 2×4 + 3×(−1) = 8 − 3 = 5.
  2. Divide by M. vcm = 5/5 = 1 m/s.
  3. Read it. The system's “average motion” drifts right at 1 m/s.
Common mistake: (4 + (−1))/2 = 1.5 m/s — right answer-ish here by luck, wrong method. Mass-weight: (8−3)/5 = 1. Always weight.
Your turn — 4 kg at +2 m/s, 2 kg at −4 m/s. vcm = ?

Answer: 0 m/s. (8 − 8)/6 = 0 — balanced momenta.

Example 2 — explosion immunity: rocket's CM rises at 5 m/s, then it explodes

  1. Before. vcm = 5 m/s (whole rocket).
  2. The explosion. Internal forces only — Ptotal unchanged.
  3. After. vcm = 5 m/s — the fragments' mass-weighted average velocity is still 5 m/s upward, even as pieces fly everywhere.
Common mistake: “the explosion adds velocity to the CM.” Internal forces cancel pairwise — no external impulse, no change. (Gravity keeps acting, so the CM follows its parabola.)
Your turn — a system's CM moves at 2 m/s; two parts collide and stick internally. vcm now?

Answer: 2 m/s. Internal collision — vcm untouched.

Example 3 — skaters: 50 kg at +3 m/s, 70 kg at −2.14 m/s

  1. Total momentum. P = 50×3 + 70×(−2.14) = 150 − 150 = 0 (approx).
  2. vcm. 0/120 = 0 m/s — the CM never moved, before or after the push.
Common mistake: tracking each skater's motion and concluding “the system moves.” The system has no single motion — but its CM sits still.
Your turn — 60 kg at −2.5 m/s, 40 kg at +3.75 m/s. vcm = ?

Answer: 0 m/s. (−150 + 150)/100 = 0.

Example 4 — before and after: 1 kg at 6 m/s, 2 kg at rest — then they collide elastically

  1. Before. P = 6; vcm = 6/3 = 2 m/s.
  2. After (elastic). v₁ = 2 m/s? No — compute: v1f = (1−2)/3×6 = −2, v2f = (2/3)×6 = 4. P = 1×(−2)+2×4 = 6. vcm = 2 m/s — unchanged.
  3. The lesson. Individual velocities scrambled; the CM cruised straight through the collision.
Before moving on: a person walks left inside a stationary canoe on a calm lake. Which way does the canoe drift — and what happens to the system's vcm?
Common mistake: recomputing vcm from scratch after every event. If no external impulse acted, it didn't change — quote the old value.
Your turn — 3 kg at +4 m/s, 1 kg at −4 m/s. vcm = ?

Answer: 2 m/s. (12 − 4)/4 = 2 m/s.

Memorization tips

  • Say it aloud: “v c m equals total momentum over total mass — internal forces can't change it.”
  • Momenta add, masses divide: Σmivi then /M. Never a plain velocity average.
  • Immunity: explosions, collisions, separations — vcm sails through all of them.
  • External forces excepted: gravity/friction over time change Ptotal — then vcm changes too.
  • Zero is common: balanced systems (skaters, explosions from rest) have vcm = 0 — the CM sits still while parts fly.
  • Quote, don't recompute: no external impulse between events → reuse the old vcm.

Final challenge

Five mixed questions — immunity, weighting, and external forces. Score 5/5 and CM velocity is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Can internal forces change v_cm?

No — they cancel pairwise (Newton's third law), so Ptotal and hence vcm are untouched. Muscles, engines, and explosions rearrange the parts but can't move the whole.

Then how does a car accelerate?

External forces: the road pushes the tires (friction), the Earth pulls (gravity). Internal engine forces alone couldn't move the CM — the car pushes on the road and the road pushes back.

What's the difference between v_cm and average velocity?

vcm is the mass-weighted average velocity — (Σmivi)/M. A plain average (v₁+v₂)/2 is only right for equal masses.

Does v_cm stay constant during a collision?

Yes, if external impulses are negligible during the brief impact — which is the same condition as momentum conservation. vcm = Ptotal/M just restates it.

How is this used in real problems?

Explosion fragments: vcm follows the original trajectory, locating missing pieces. Rocket motion, binary stars, and collision analysis all track the CM instead of the chaos.

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