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System Boundaries & Interactions

Key idea: Choose a system boundary, classify interactions as internal or external, and connect external force, momentum change and energy transfer.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Track energy stores and transfers, then apply conservation of energy.
  • Define work and derive and apply the kinetic-energy relationship.
  • Derive Eₖ = ½mv² from the definition of work done by a force and the uniformly accelerated motion equations.
  • Represent fields and relate work done by a field to potential-energy change.
  • Draw field-line representations of uniform and radial gravitational and electric fields.
  • Use force–extension graphs to determine elastic potential energy.
  • Apply power, mechanical power and efficiency relationships.

1. The boundary decides the classification

An internal interaction occurs between parts that are both inside the chosen system. An external interaction occurs between the system and an agent outside it.

The same physical force may change classification when the boundary changes. Friction from a track is external for the system “trolley only”, but the trolley–track frictional interaction is internal for “trolley + track”.

State the system first

Never label a force internal or external in isolation. Write “For the system consisting of …” before classifying the interaction.

2. Momentum of a system

For a system of bodies:

∑ vector Fₑₓₜₑᵣₙₐₗ = (d vector p_system)/dt

Internal third-law forces cancel in the vector sum over the whole system. They may change the momenta of individual parts, but they do not change total system momentum.

Therefore total momentum is constant when the resultant external force is zero, or when the external impulse is negligible over the interval:

Δ vector p_system = vector Jₑₓₜₑᵣₙₐₗ

3. Energy of a system

Internal interactions can redistribute energy among the system’s stores. For example, an internal collision can transfer kinetic energy to internal energy through deformation while total momentum remains constant.

External forces can transfer energy across the system boundary by doing work. For the system “trolley only”, work done by an external pulling force can increase the trolley’s kinetic energy.

These statements are different:

  • Momentum condition: determined by resultant external impulse.
  • Energy condition: determined by transfers across the boundary and changes among all stores.
  • Kinetic-energy condition: kinetic energy may change even when total momentum and total energy are conserved.

4. Classification workflow

  1. Draw or state the system boundary.
  2. Name each interaction and both participating bodies.
  3. If both bodies are inside, classify it as internal.
  4. If one is outside, classify it as external.
  5. Apply the relevant momentum or energy statement; do not assume the two conditions are identical.

5. Common mistakes

  • Saying internal forces “do no work”. They can do work on individual parts and transfer energy among stores.
  • Assuming zero resultant external force means kinetic energy is constant.
  • Calling gravity external for an “object + Earth” system.
  • Treating the normal contact force and weight as a third-law pair; both act on the same body.
  • Changing the system halfway through a calculation.

6. Worked Examples

Modelled example 1

Connected blocks

Core

Problem

Two blocks are connected by a light string and pulled across a horizontal surface. Classify the string interaction when the system is both blocks together.
Study the worked solution

  1. State the boundary

    Method

    Both blocks lie inside the chosen system.

    Reason

    Classification depends on which interacting bodies are enclosed.

    Working

    system = {block 1, block 2}

  2. Classify the string interaction

    Method

    The tension interaction between the blocks is internal.

    Reason

    It transmits a force between two parts that are both inside the boundary.

    Working

    block 1 ↔ block 2: internal

  3. Distinguish external interactions

    Method

    The applied pull and surface friction are external horizontal interactions.

    Reason

    The pulling agent and surface are outside the stated two-block system.

    Working

    outside agent or surface → block system: external

Common misconception 2

Inelastic collision

Find and correct the mistake

Learner claim

Two carts collide on a smooth track and stick. For the two-cart system, a learner says negligible external horizontal impulse means both momentum and kinetic energy are conserved. Diagnose the claim.

Try this before viewing the solution

Correct comparison

View solution step by step

  1. Apply the momentum condition

    Method

    Total horizontal momentum is conserved.

    Reason

    The external horizontal impulse is negligible during the short collision.

    Working

    Δ p_system = Jₑₓₜₑᵣₙₐₗ ≈ 0

  2. Track kinetic energy separately

    Method

    Kinetic energy decreases.

    Reason

    Internal collision interactions transfer kinetic energy into deformation and internal-energy stores when the carts stick.

    Working

    Δ Eₖ < 0

  3. Retain total-energy conservation

    Method

    Total energy remains conserved.

    Reason

    The kinetic-energy decrease appears as increases in other stores within the system and surroundings included in the balance.

    Working

    Δ Eₜₒₜₐₗ = 0

Challenge 3

Ball rebounding from a wall

Minimal support

Independent transfer

A ball strikes a fixed wall and rebounds. Decide whether the ball’s momentum is conserved for the system “ball only”, then explain how enlarging the boundary changes the interaction classification.

Try this before viewing the solution

Ball-only momentum

Hints

Hint 1: name both interacting bodies
Ask whether the wall is inside or outside each proposed boundary.
View solution step by step

  1. Analyse the ball-only system

    Method

    The ball’s momentum is not conserved.

    Reason

    The wall is external and gives the ball the impulse that reverses its momentum.

    Working

    Δ p_ball = J_(wall on ball) ≠ 0

  2. Enlarge the boundary

    Method

    Choose ball + wall + Earth.

    Reason

    Earth supports the fixed wall and must be included to capture the recoil partner.

    Working

    system = {ball, wall, Earth}

  3. Reclassify and state the condition

    Method

    The ball–wall contact is internal, and total system momentum may be conserved if other external impulses are negligible.

    Reason

    Internal impulses redistribute momentum among system parts without changing the total.

    Working

    Jₑₓₜₑᵣₙₐₗ ≈ 0 ⇒ Δ p_system ≈ 0

7. Mind Stretchers

When an answer says “conserved”, name the quantity, the system and the condition. For example: “The horizontal momentum of the two-cart system is conserved because the external horizontal impulse is negligible.”

Mind stretcher 1: Trolley friction under two boundariesExtension

A trolley slows on a rough track. Classify the frictional interaction for (i) the trolley-only system and (ii) the trolley + track system. In each case, describe how the trolley’s lost kinetic energy is accounted for.

Show answer

For the trolley-only system, the track lies outside the boundary, so friction is external and transfers energy out of the trolley system.

For trolley + track, the frictional interaction is internal. The trolley’s kinetic-energy decrease is balanced mainly by an increase in the internal-energy stores of the trolley and track. Total energy of that enlarged system remains constant if other boundary transfers are negligible.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027