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UY1: Motion Of A System Of Particles

Derive how the centre of mass of a particle system moves under external force, and connect it to total momentum.

Reading settings
  • University Physics Year 1
On this page

Enlarged diagram

Cluster role + pathway links

This page is the core concept page for centre-of-mass dynamics in the UY1 mechanics pathway.

1) At a glance

  • Prerequisite: centre of mass definition (What Is Centre of Mass?), Newton’s laws, vectors.
  • Outcomes: treat a multi-particle system’s translation as if all mass were concentrated at the CM; relate CM motion to total momentum and net external force.
  • Key result:
    ∑ vector Fₑₓₜ = M vector a_CM = (d vector P)/dt
  • Common trap: including internal forces in net force on the system.

2) Setup (model, assumptions, coordinates/signs)

Model

  • System has particles mᵢ at positions vector rᵢ in an inertial frame.
  • Total mass M = ∑ᵢ mᵢ is constant for the derivation below (see the rocket note further down).
  • CM position:
    vector r_CM = (∑ᵢ mᵢ vector rᵢ)/M

Modelling workflow (what to write before the algebra)

  • Choose the system boundary (which bodies are included).
  • Choose axes and a sign convention (then keep them for the whole question).
  • Draw a free-body diagram of the system as a whole and list only external forces that cross the system boundary.
What CM equations can and cannot do
  • The CM equation predicts the translation of the system: ∑ vector Fₑₓₜ = M vector a_CM.
  • It does not tell you the internal relative motion (tension forces, explosions, how the pieces separate). Internal forces rearrange the parts while leaving CM motion controlled only by external forces.
Pitfall: “internal forces cancel” needs the right system boundary

Internal forces cancel only if both members of each action–reaction pair are inside your chosen system. If mass leaves/enters your system (variable-mass problems), you must be more careful. See: Rocket Propulsion.

3) Core method / derivation

Differentiate CM position:

vector v_CM = (d vector r_CM)/dt = 1/M∑ᵢ mᵢ vector vᵢ

Multiply by M:

M vector v_CM = ∑ᵢ mᵢ vector vᵢ = ∑ᵢ vector pᵢ = vector P

So total momentum equals momentum of a single particle of mass M moving at vector v_CM.

Differentiate again:

M vector a_CM = (d vector P)/dt

Using Newton’s 2nd law per particle and summing:

(d vector P)/dt = ∑ᵢ vector Fᵢ = ∑ vector Fₑₓₜ + ∑ vector Fᵢₙₜ

Internal-force pairs cancel, so:

∑ vector Fₑₓₜ = M vector a_CM

Checks:

  • Units: both sides are N = kg m s⁻².
  • Sign: if ∑ F_(ext,x) < 0, then a_(CM,x) < 0.
  • Limit: if ∑ vector Fₑₓₜ = 0, CM moves at constant velocity.

4) Worked example

Worked example 1

Two skaters push off

Problem

Two skaters at rest on frictionless ice push off from each other. What happens to their centre of mass?

Show full solution

  1. Find the external force

    Method

    Identify the horizontal forces from outside the system.

    Reason

    The push between the skaters is internal: its two forces cancel in the total.

    Working

    Answer

    ∑ F_(ext,x) = 0.

  2. Apply the centre-of-mass law

    Method

    Use ∑ Fₑₓₜ = Ma_CM.

    Reason

    Zero external force means zero centre-of-mass acceleration.

    Working

    Answer

    a_(CM,x) = 0, so v_(CM,x) stays at its initial value, 0.

  3. State the answer

    Working

    Answer

    The centre of mass stays fixed while the skaters move apart.

Worked example 2

Two blocks pulled by an external force

Problem

Two blocks m₁ and m₂ are joined by a light string on a frictionless surface. You pull block 1 to the right with force F. Find the acceleration of the centre of mass.

Show full solution

  1. Choose the system

    Method

    Take both blocks together.

    Reason

    The string tension is then internal and cancels in the total.

    Working

    Answer

    The only external horizontal force is F.

  2. State the answer

    Working

    Answer

    a_CM = F/(m₁ + m₂).

5) Practice set

Check your understanding 1

A system of total mass 5 kg feels a net external force (10,0) N. Find vector a_CM.

Show hint

Divide the net external force by the total mass.

Show answer

vector a_CM = vector Fₑₓₜ/M = (10,0)/5 = (2,0) m s⁻².

Check your understanding 2

A system’s total momentum is constant. What can you conclude about the net external force?

Show hint

Use ∑ vector Fₑₓₜ = d vector P/dt.

Show answer

It is zero, because ∑ vector Fₑₓₜ = d vector P/dt = 0.

Check your understanding 3

True or false: an explosion in deep space, with no external force, can change the velocity of the centre of mass.

Show hint

Do internal forces change the total momentum?

Show answer

False. The explosion forces are internal and cancel in pairs, so total momentum, and hence vector v_CM, is unchanged. The fragments fly apart around a centre of mass that keeps moving uniformly.

6) Summary + next steps

  • CM motion isolates the effect of external forces on a many-body system.
  • Internal forces can change relative motion, but not total momentum.
  • Use CM equations to simplify collisions, explosions, and recoil problems.

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

Last reviewed:

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