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Conservation laws

Use a defined system and a named quantity to track transfers, distinguish changing parts from conserved totals, and check the conditions of energy and momentum balances.

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After reading, you can

  • Name the system, quantity, boundary and interval used in a conservation argument.
  • Calculate an energy change from transfers into and out of a system.
  • Distinguish total energy from its mechanical and internal contributions.
  • Use signed horizontal momentum in a stated Newtonian model.
  • Explain why thermodynamic closure alone does not guarantee constant energy or momentum.
  • Keep fixed-mass approximations and other model conditions attached to a calculation.
01

Name what is being conserved

A conservation law concerns a specified physical quantity. It can constrain a total even while the system's parts change. To use one, name the quantity, select the included objects or region, and declare the boundary and time interval. System choice determines which interactions are internal and which involve the surroundings.

This lesson uses energy and horizontal linear momentum in ordinary classical models. Energy balances track storage and transfers. Momentum balances track a quantity associated with motion and direction. They are different balances, with different units and conditions; neither says that every property of every object stays unchanged.

Energy is expressed here in joules, symbol J. Values being added must concern the same quantity and use compatible units. All numerical cases below are invented, with exact stipulated model inputs and consistent fixed energy references. Their arithmetic does not establish the precision of real measurements.

02

Balance energy storage and transfers

In a closed thermodynamic system, matter does not cross the boundary, but energy can. For the classical accounts here, include every relevant energy contribution in E. With nonnegative amounts received and delivered over the stated interval, write E_final = E_initial + energy received - energy delivered. The change is E_final - E_initial.

Heating and work can transfer energy across the boundary. A motor turning a shaft into a sealed vessel can transfer energy by work even when no matter crosses. Heat and work describe transfers, rather than extra stores to add again after the transferred energy has entered the system.

For a stationary closed-system process with no change of its overall kinetic or potential energy, the net transferred energy changes internal energy. Internal energy concerns the microscopic energy within the materials. If the whole system's motion or position changes, include the associated energy changes too; do not automatically assign every transfer to internal energy alone.

No net energy transfer gives equal initial and final energy in this balance. An isolated model excludes energy and matter exchange; a merely closed model need not. Equal incoming and outgoing amounts can also give zero net change over an interval even while transfers occur. The system's energy need not have stayed constant at every intermediate time.

03

A conserved total can have changing contributions

In these mechanical models, kinetic energy is associated with motion. Potential energy is associated with an interaction and its arrangement, such as the deformation of a spring. Mechanical energy is their sum. A conversion between those contributions need not change their sum under the appropriate mechanical conditions.

Mechanical energy is not every energy contribution. Friction or deformation can change mechanical energy into internal energy within an included system. A decline in mechanical energy therefore need not be a decline in total energy. Include the bodies receiving the energy, or account for transfer across the selected boundary.

For a cart, spring and track taken together, rubbing between cart and track is internal. For the cart alone, the track is outside. If spring potential energy is counted separately, the accompanying internal-energy entry must exclude it. We call that entry remaining internal energy. Use one accounting scheme consistently so its categories do not overlap.

Different totals require different checks in the models used here
Quantity being followedWhat can changeCondition to check
Energy of the selected systemEnergy can enter or leaveInclude every relevant contribution and boundary transfer
Mechanical energyIt can become internal energyDo not identify it automatically with total energy
Momentum of one cartAnother cart can change it during interactionThe other cart is external to a one-cart system
Combined horizontal momentum of the two cartsIndividual cart momenta can changeUse the stated Newtonian model and check the net external horizontal force
04

Add momentum with direction included

For our slow-moving carts with constant masses in a Newtonian model, a horizontal momentum component is p = m v: mass multiplied by the signed horizontal velocity. Its unit is kg m/s. Choose positive to the right in one fixed reference frame treated as non-accelerating. A leftward velocity then has a negative sign. Add signed components, not speeds, to obtain total horizontal momentum.

When the included carts interact, each can change the other's momentum. In the stated Newtonian model, the changes due to their mutual interaction are equal and opposite. If the net external horizontal force is zero throughout the interval, their combined horizontal momentum stays constant. External forces need not all vanish; their net component in the direction being checked is what matters.

If outside interactions change the total horizontal momentum, include that change in the balance instead of setting it to zero. In a real short interaction, neglecting a small external effect may support an approximation. Check its size for the chosen interval and required accuracy; a track described as frictionless does not by itself exclude every other external force.

05

State the conditions behind a constant total

Closed has different uses in different discussions. Here the thermodynamic meaning excludes matter exchange and permits energy transfer. Some mechanics texts use closed for a system meeting momentum-conservation conditions. State the actual exchanges and external-force conditions rather than relying on the label alone.

A sealed vessel can receive energy. A cart can be pushed from outside without exchanging matter. Neither case justifies a constant total merely because the system is closed in the thermodynamic sense. Likewise, zero energy change over an interval does not by itself establish zero momentum change.

The fixed masses in our cart model are an ordinary classical approximation, not a universal separate conservation law for the sum of individual rest masses. Special relativity relates rest mass to rest energy; in particle or nuclear processes, individual rest masses and their sum can change while the full energy account includes rest energy. No relativistic calculation is needed for the supplied slow-cart model.

06

Use the balance to constrain and check a model

Write the initial and final totals, list boundary contributions during the same interval, and check units and signs. An unexplained difference may reflect an omitted transfer, a changing contribution, an unsuitable approximation or a measurement problem. A failed partial account does not by itself establish a failure of the conservation law.

A conserved sum also need not determine every part separately. Knowing a combined momentum gives one constraint on the individual momenta; further information about the interaction or one final motion may still be needed. Conservation narrows the possible accounts and helps expose inconsistencies, while a complete prediction can require additional physics.

A closed-system energy balance with boundary transfers
  1. 1. Initial internal energy: 20 J

    Start with the supplied internal-energy value for the defined system. Thermodynamic closure excludes matter exchange.

  2. 2. Energy received: 13 J

    Heating supplies 9 J and shaft work supplies 4 J, so the total received is 9 J + 4 J = 13 J.

  3. 3. Energy delivered: 6 J

    The system delivers 6 J through heating. Net energy received is 13 J - 6 J = +7 J.

  4. 4. Final internal energy: 27 J

    With the stated endpoint conditions, the internal-energy increase is 7 J: 20 J + 7 J = 27 J. Boundary transfers explain the change; no heat or work store is counted again.

Original invented classical process: the system is a sealed rigid vessel, its contents and stirrer. No matter crosses its boundary. Overall kinetic and potential energies have no net change between endpoints. Every boundary energy transfer is listed below, and internal-energy values use the same fixed reference. The stipulated values do not establish real measurement precision.

Worked examples

A closed vessel can gain internal energy

In an invented classical process, take a sealed rigid vessel, its contents and stirrer as the system. No matter crosses its boundary. Its overall kinetic and potential energies have no net change between the endpoints. Its initial internal energy is 20 J in a fixed reference. It receives 9 J through heating and 4 J through shaft work, and delivers 6 J through heating. These are all boundary energy transfers during the interval. What are the internal-energy change and final value?

  1. The system is closed in the thermodynamic sense, but energy crosses its boundary.
  2. Combine the received amounts: 9 J + 4 J = 13 J.
  3. Subtract delivered energy: 13 J - 6 J = +7 J. With no net overall kinetic or potential energy change, this is the internal-energy increase.
  4. Use the same internal-energy reference at both endpoints: 20 J + 7 J = 27 J.

Internal energy increases by 7 J to 27 J in the stipulated account. The gain is explained by net boundary transfer. Closure excludes matter exchange; it does not exclude heating or work. No heat or work store is added a second time.

Keep the total while its energy contributions change

An invented classical model includes a cart, deformable spring and track. During the chosen interval no energy or matter crosses the outer boundary. Non-overlapping mechanical and remaining internal-energy entries account for every changing energy contribution; other contributions are unchanged. The mechanical entry includes spring potential energy, which is excluded from the remaining internal entry. Initially the two entries are 18 J and 52 J, respectively, in consistent fixed references. Finally mechanical energy is 12 J. Find the final remaining internal-energy entry and the changes in both entries.

  1. The initial accounted total is 18 J + 52 J = 70 J.
  2. No boundary energy transfer occurs, so the accounted total remains 70 J in this model.
  3. Subtract the final mechanical entry: 70 J - 12 J = 58 J of remaining internal energy.
  4. Mechanical energy changes by 12 J - 18 J = -6 J; remaining internal energy changes by 58 J - 52 J = +6 J. Their changes sum to zero.

The final remaining internal-energy entry is 58 J. Mechanical energy falls by 6 J while that entry rises by 6 J. The stipulated total remains 70 J. This is consistent with an internal conversion, such as rubbing or deformation; the energy balance alone does not identify the detailed mechanism.

Individual momenta change while their sum stays fixed

In an invented slow-cart Newtonian model, use one non-accelerating track reference with positive to the right. Cart A has constant mass 2 kg and initial velocity +1 m/s; cart B has constant mass 1 kg and initial velocity -1 m/s. Take both carts as the system, with no matter exchange and zero net external horizontal force throughout the interaction interval. At the final endpoint A is at rest. What final velocity must B have, and how does each cart's momentum change?

  1. Initial horizontal momenta are 2 kg × (+1 m/s) = +2 kg m/s for A and 1 kg × (-1 m/s) = -1 kg m/s for B.
  2. The initial sum is +2 - 1 = +1 kg m/s. The stated conditions keep this combined horizontal momentum constant.
  3. A's final momentum is zero. B must therefore have final momentum +1 kg m/s, giving v_B = (+1 kg m/s)/(1 kg) = +1 m/s.
  4. A changes by 0 - (+2) = -2 kg m/s; B changes by +1 - (-1) = +2 kg m/s. These changes cancel in the two-cart sum.

B must move at +1 m/s, to the right, under the stipulated model. A loses 2 kg m/s of horizontal momentum and B gains 2 kg m/s. Their combined momentum remains +1 kg m/s. This balance does not establish that each cart's momentum or the total kinetic energy is conserved.

Try it yourself

  1. Two included bodies exchange energy internally. For the combined system, no energy crosses the outer boundary and all relevant contributions are included. Must each body's energy stay constant? What does the combined balance say?

    Show the answer and reasoning

    Each body's energy can change. The combined total stays constant under the stated conditions.

    An internal transfer leaves one part and enters another, so it cancels in the combined account. For either body alone, the other is outside and that transfer crosses the smaller system's boundary.

  2. A stipulated classical energy account starts at 40 J, receives 5 J and delivers 11 J during an interval. Every relevant contribution and transfer is included, with the same reference at both endpoints. Find the final energy and its change. Has 6 J been destroyed?

    Show the answer and reasoning

    The final energy is 34 J and the change is -6 J. The account explains the change through net delivery to the surroundings.

    Calculate 40 J + 5 J - 11 J = 34 J, then 34 J - 40 J = -6 J. More energy leaves than enters. The selected system loses energy through its boundary; the balance supplies no destruction term.

  3. In an invented isolated classical account, the complete changing energy contributions are a mechanical entry and a non-overlapping remaining internal entry. Mechanical energy changes from 25 J to 19 J. What change of the remaining internal entry is required? Does this alone identify a particular friction process?

    Show the answer and reasoning

    The remaining internal entry must increase by 6 J. The balance alone does not identify the mechanism.

    Mechanical energy changes by 19 J - 25 J = -6 J. With no boundary transfer and no other changing contribution, the remaining internal change must be +6 J. Additional physical information is needed to identify how the conversion occurred.

  4. A stipulated Newtonian system has initial total horizontal momentum -3 kg m/s, with positive to the right. During the chosen interval, external interactions produce a signed total momentum change of +5 kg m/s. There is no matter exchange. What is the final total, and was it constant over the interval?

    Show the answer and reasoning

    The final total is +2 kg m/s. It was not constant over the interval.

    Add the specified external contribution: -3 kg m/s + 5 kg m/s = +2 kg m/s. No matter exchange alone does not remove the external force effect or justify setting the momentum change to zero.

  5. A closed thermodynamic system first receives 8 J and later delivers 8 J. These are its only energy transfers and all relevant storage is included. Are the initial and final energies equal? Was it isolated, and must its energy have been constant at every intermediate time?

    Show the answer and reasoning

    Its endpoint energies are equal. It was not isolated, and its energy was not constant throughout this sequential transfer process.

    The net transfer over the full interval is 8 J - 8 J = 0. After receiving the first 8 J and before delivering it, the system's energy is above its initial value. Equal endpoint totals do not erase the boundary crossings.

  6. A successful slow-cart calculation treats each cart's mass as constant and uses p = m v. Does that establish an exact law conserving the sum of individual rest masses in every particle or nuclear process?

    Show the answer and reasoning

    No. The calculation uses a stated classical approximation; it does not establish that universal mass claim.

    Relativistic energy accounts include rest energy, and the sum of individual rest masses can change in such processes. The conditions of the cart model must remain attached to its formula and result.

What to carry forward

  • Name the quantity, system, boundary, interval and reference before using a balance.
  • A selected system's energy change is accounted for by net boundary transfer when all relevant contributions are included.
  • Constant total energy does not require constant mechanical energy or constant energy of every part.
  • In the stated Newtonian model, add signed momenta and check the net external force in the chosen direction.
  • Thermodynamic closure excludes matter exchange; it does not by itself guarantee constant energy or momentum.
  • Keep domain, approximation and completeness conditions attached to a conservation calculation.

References

Sources for the concepts and teaching approaches in this article.

  1. The International System of Units (SI), 9th edition, version 4.01Bureau International des Poids et Mesures (BIPM)Reference guidebipm.orgRead source
  2. Classical Thermodynamics, Module 1, Lecture 1: Review of Thermodynamics — System approach and classification of systemNPTELCourse notesarchive.nptel.ac.inRead source
  3. University Physics Volume 2, 3.3 First Law of ThermodynamicsSamuel J. Ling, William Moebs and Jeff Sanny — OpenStax / Rice UniversityTextbook chapteropenstax.orgRead source
  4. Physics Part I, Class XI, Chapter 5: Work, Energy and PowerNational Council of Educational Research and Training (NCERT), IndiaTextbook chapter · PDFncert.nic.inRead source
  5. University Physics Volume 1, 9.3 Conservation of Linear MomentumWilliam Moebs, Samuel J. Ling and Jeff Sanny — OpenStax / Rice UniversityTextbook chapteropenstax.orgRead source
  6. University Physics Volume 3, 5.9 Relativistic EnergySamuel J. Ling, Jeff Sanny and William Moebs — OpenStax / Rice UniversityTextbook chapteropenstax.orgRead source