Work, Energy & Momentum — Print · Physics & Environmental Science, Grade 12

Account for work, power, stored energy, collisions, and force-time evidence.

Printable chapter lessons and review.

Work & Power

Learning goals

Readings for this unit

Before this unit

Force components, displacement, and time.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why can equal lifting jobs require different motor power?

Illustrative evidence case. Two motors raise identical loads to the same height in different times.

Work connects one force with a displacementWork connects one force with a displacementdisplacement Δxforce FW = F Δx cosθ for a constant forceOnly the force component along displacement does work.
Original projection geometry; arrow lengths for force and displacement have separate units.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why can equal lifting jobs require different motor power?

Your first model

Record force, height, time and the selected useful output.

Build the language from the idea

work
The energy transferred to an object when a force acts on it through a displacement: W = Fd·cosθ, measured in joules (J).
joule (J)
The SI unit of work and energy; one joule is one newton-metre (1 J = 1 N·m).
work-energy theorem
The net work done on an object equals its change in kinetic energy: W_net = ΔKE.
power
The rate at which work is done or energy is transferred: P = W/t, measured in watts (W).
watt (W)
The SI unit of power; one watt is one joule per second (1 W = 1 J/s).

In everyday speech 'work' means almost any effort, but in physics it has a sharp, specific meaning: work is the energy a force transfers to an object as that object moves. When a force pushes something through a displacement in the same direction, the work done is simply W = Fd — force times distance. Because a joule is a newton-metre, pushing with 10 N through 3 m transfers 30 J of energy. Work is the bridge between force (Unit 2) and energy: a force by itself does nothing to an object's energy until the object actually moves.

Direction matters, which is why the full formula is W = Fd·cosθ, where θ is the angle between the force and the displacement. When you carry a heavy backpack horizontally across a room, gravity pulls straight down while you move sideways — the angle is 90°, cos 90° = 0, and gravity does zero work on the pack even though holding it feels exhausting. When friction drags backward against motion, the angle is 180°, cos 180° = −1, and friction does negative work, removing energy from the object. Only the part of a force that lies along the motion actually transfers energy.

Work connects one force with a displacementWork connects one force with a displacementdisplacement Δxforce FW = F Δx cosθ for a constant forceOnly the force component along displacement does work.
Original projection geometry; arrow lengths for force and displacement have separate units.

The payoff of tracking work is the work-energy theorem: the *net* work done on an object equals its change in kinetic energy, W_net = ΔKE. Push a cart and the positive net work speeds it up; let friction act and the negative net work slows it down. This theorem is a shortcut of real power — if you know the net work, you know exactly how much the object's motion energy changed, without having to follow the messy details of acceleration and time in between.

Doing work is one thing; how *fast* you do it is another, and that is power. Power is the rate of doing work, P = W/t, measured in watts (one watt is one joule per second). Two cranes might lift identical loads to the same rooftop — doing exactly the same amount of work — yet the crane that finishes in half the time delivers twice the power. Power can also be written P = Fv, the product of force and speed, which is why a car engine must supply more power to hold a high speed against air resistance than to cruise slowly.

Power measures the rate of energy transferPower measures the rate of energy transferTransferred work (J)Machine A: 3 s600Machine B: 6 s600PA = 600/3 = 200 WPB = 600/6 = 100 WEqual work can be transferred at different rates.
Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.

These ideas quietly rate almost every machine around you. The 'horsepower' on an engine, the wattage on a motor or a light bulb, the output of a power plant — all are statements about how quickly energy is being transferred. Engineers care about both totals and rates: a small motor and a huge one might eventually do the same job, but only the powerful one can do it fast enough to matter. Master work and power and you hold the vocabulary for describing every energy transfer in the rest of this unit.

Model check: A force perpendicular to displacement does no mechanical work in the point-particle model. W=30 J; P=W/t=15W, if that force is constant.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Define work by force and displacement

Work is energy transferred by a force through displacement. For a constant force, W=FΔx cosθ uses the angle between the vectors. Positive work adds energy through that interaction; negative work removes it. A perpendicular force does zero work in this model even though it can turn the motion. Specify which force’s work you calculate and distinguish it from the net work of all forces.

Work connects one force with a displacementWork connects one force with a displacementdisplacement Δxforce FW = F Δx cosθ for a constant forceOnly the force component along displacement does work.
Original projection geometry; arrow lengths for force and displacement have separate units.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

A 10 N force acts perpendicular to a 3 m displacement. What work does it do?

Visual reasoning — after your attempt

W=10×3×cos 90°=0 J.

For a constant force, work is Fd cosθ, with θ between force and displacement. Work describes energy transfer, not effort or elapsed time alone. A perpendicular force does no work in this ideal account even while it changes motion direction.

Holding a stationary load can require bodily energy while doing zero mechanical work on that load through displacement. The system and transfer pathway matter.

Pause and explain

Why can a force do zero work while present?

Check your reasoning after answering

The object may have zero displacement or the force may be perpendicular to displacement, giving a zero dot product.

Part 2 · Use a force-position area when force varies

If force changes with position, multiply-and-add small contributions or find the signed area under its force–position graph. For a linear spring stretched slowly from x=0 to 0.4 m with k=50 N/m, applied force rises from 0 to 20 N. The triangular area is 4 J. Multiplying the final 20 N by the whole extension gives twice the correct work because that final force did not act over the entire interval.

A changing force requires an area or incremental accountA changing force requires an area or incremental accountBalancing applied force (N)0510152000.10.20.30.4Extension (m)Area = ½ × 0.4 m × 20 N = 4 J
Calculated quasistatic linear-spring work; shaded triangle is an area, not another force.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What average applied force produces the same 4 J over 0.4 m?

Visual reasoning — after your attempt

Favg=W/Δx=10 N, the average of 0 and 20 N for this linear ramp.

For one-dimensional motion, signed area under a force-versus-position curve gives work. A force-versus-time area gives impulse instead. Match axes and units before using geometric areas. Negative work can reduce the selected object’s kinetic energy when no other work offsets it.

A linear force increasing from 0 to 20 N over 0.50 m has triangular work area 5.0 J.

Pause and explain

Why is force-time area not work?

Check your reasoning after answering

Its units are N·s and it accounts for momentum change; work requires displacement and has units N·m or joules.

Part 3 · Distinguish power from total transferred energy

Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.

Power measures the rate of energy transferPower measures the rate of energy transferTransferred work (J)Machine A: 3 s600Machine B: 6 s600PA = 600/3 = 200 WPB = 600/6 = 100 WEqual work can be transferred at different rates.
Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

How much work does a steady 100 W device transfer in 12 s?

Visual reasoning — after your attempt

W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.

Average power is energy transferred divided by elapsed time, commonly W/Δt for work. A larger power can transfer the same energy in less time. For a force parallel to velocity, instantaneous mechanical power can be Fv; direction matters in the general dot product.

Two motors can lift the same load to the same height with the same ideal work while taking different times and delivering different average power.

Pause and explain

Does greater power necessarily mean greater total work?

Check your reasoning after answering

No. The duration also matters; power describes a transfer rate, while work describes the transferred amount.

Part 4 · Connect output and input with efficiency

Efficiency compares useful output with total input for a defined process and service. If 1000 J enters and 350 J is delivered as the selected useful output, efficiency is 35%; the remaining 650 J must be accounted for as other transfers or storage. Energy is not destroyed by inefficiency. What counts as useful can differ between a lamp and a heater, so comparisons require the same service and a consistent system boundary.

An efficiency account names useful output and the boundaryAn efficiency account names useful output and the boundaryInput energy1000 JUseful output350 JOther transfers650 JEfficiency = useful output / input = 35%Define useful service before comparing.
Original numerical energy account; branch arrows show separate destinations whose energies sum to the input.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Does 35% efficiency mean 650 J disappeared?

Visual reasoning — after your attempt

No. It left through other transfers or remained stored within the defined accounting interval.

Efficiency compares useful output to input for a stated job and interval. Mechanical output is not automatically equal to electrical input; other outputs and changes in stores must be accounted for. Define the useful service before reporting a percentage.

An 80%-efficient lift needing 100 J of useful output requires 125 J input in that stated steady account.

Pause and explain

Why must efficiency name the useful output?

Check your reasoning after answering

Different useful services and boundaries can produce different ratios. A percentage without the job or interval is ambiguous.

Apply the whole unit · Compare lifting work and motor input

Illustrative motor lifts a 20.0 kg load vertically by 3.00 m in 5.00 s at constant speed. Use g=10.0 m/s² and overall efficiency 0.750.

Your task

Find ideal useful work, output power, input energy, and average input power. Explain why constant speed does not mean the motor transfers no energy.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Useful work=mgh=20×10×3=600 J.

  2. Output power=600/5=120 W.

  3. Input energy=600/0.750=800 J; average input power=160 W.

The motor’s upward force does positive work while gravity does negative work, so load kinetic energy can remain constant. The useful increase is gravitational potential energy of the load-Earth system. The remaining 200 J enters other outputs or stores within the assumed efficiency account; it is not destroyed.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask students to name the system for potential energy and the interval for power. Check that input exceeds useful output for the supplied efficiency. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Work = force × displacementboxFd (displacement)W = F·din joules
Selected constant force aligned with displacement. In general work is Fdcosθ for a constant force or signed force–position area for changing force.

The key idea

Work is energy transferred by a force acting through a displacement (W = Fd·cosθ); the net work on an object changes its kinetic energy; power (P = W/t) is how fast that work is done.

Work

W = Fd·cosθ

F is the force in newtons (N); d is the displacement in metres; θ is the angle between the force and the displacement; W is in joules (J). When the force is along the motion, cosθ = 1 and W = Fd.

Work-Energy Theorem

W_net = ΔKE = ½mv_f² − ½mv_i²

the net work done on an object equals its change in kinetic energy; v_i and v_f are the initial and final speeds.

Power

P = W / t (also P = Fv)

W is work in joules; t is time in seconds; P is power in watts (1 W = 1 J/s); v is speed in m/s.

Worked examples

Example 1

A worker pulls a wagon with a 50 N force directed 60° above the horizontal, moving it 4.0 m along level ground. How much work does the pulling force do?
  1. Only the part of the force along the motion does work, so use W = Fd·cosθ.
  2. W = 50 × 4.0 × cos 60°.
  3. cos 60° = 0.5, so W = 200 × 0.5 = 100.
Answer: 100 J

Example 2 · Worked application

A 40 N force acts at 60° to a 5 m displacement. Find its work and average power over 4 s.
  1. Only the aligned force component contributes: W=Fd cosθ.
  2. W=40×5×0.5=100 J.
  3. P_avg=W/t=100/4=25 W.
Answer: 100 J; 25 W.
Check the reasoning

State which force's work is calculated.

Example 3

An escalator motor does 9,000 J of work lifting passengers in 6.0 s. Find its power output.
  1. Power is the rate of doing work: P = W/t.
  2. P = 9,000 ÷ 6.0.
  3. = 1,500.
Answer: 1,500 W

Example 4 · Find and fix the mistake

Context: A 40 N force acts at 60° to a 5 m displacement. Find its work and average power over 4 s. A student claims: “Power and energy have the same units because both describe a machine.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: State which force's work is calculated.
  2. Power is energy transfer per time, measured in watts; work/energy is in joules.
Answer: Power is energy transfer per time, measured in watts; work/energy is in joules.
Check the reasoning

State which force's work is calculated.

Example 5 · Reason from the visual model

Power measures the rate of energy transferPower measures the rate of energy transferTransferred work (J)Machine A: 3 s600Machine B: 6 s600PA = 600/3 = 200 WPB = 600/6 = 100 WEqual work can be transferred at different rates.
Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.
How much work does a steady 100 W device transfer in 12 s?
  1. Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.
  2. W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.
Answer: W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.
Check the reasoning

Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.

Avoid these mistakes

Check your understanding

A 10N parallel force moves a box 3 m in 2 s. Find work and average power.

Discuss the answer

W=30 J; P=W/t=15W, if that force is constant.

Return to the phenomenon

Return to the initial case. Why can equal lifting jobs require different motor power? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Work depends on the force–displacement account; power divides transfer by time, and input exceeds useful output for an inefficient system.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why can equal lifting jobs require different motor power? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Work depends on the force–displacement account; power divides transfer by time, and input exceeds useful output for an inefficient system.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A motor raises a 20 kg load by 3 m in 6 s at 70% efficiency. Find useful work, output power and input energy.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total 252 J under the stated boundary.

Words to know

work
The energy transferred to an object when a force acts on it through a displacement: W = Fd·cosθ, measured in joules (J).
joule (J)
The SI unit of work and energy; one joule is one newton-metre (1 J = 1 N·m).
work-energy theorem
The net work done on an object equals its change in kinetic energy: W_net = ΔKE.
power
The rate at which work is done or energy is transferred: P = W/t, measured in watts (W).
watt (W)
The SI unit of power; one watt is one joule per second (1 W = 1 J/s).

James Watt and the invention of 'horsepower'

When Scottish engineer James Watt was selling his improved steam engines in the 1770 s, customers wanted to know how many horses each machine could replace. So Watt measured how much work a strong draft horse could do — hauling loads out of a mine — and defined one horsepower as about 550 foot-pounds per second, which works out to roughly 746 watts. It was brilliant marketing dressed as physics, and it stuck: car engines are still rated in horsepower today, even though the SI unit of power, the watt, is named after the very same man.

Misconception: holding a heavy weight is 'hard work' but does no physics work

Stand holding a heavy box perfectly still and your arms will soon burn — yet in the language of physics you are doing zero work on the box. Work is W = Fd, and if the box does not move, d = 0, so no energy is transferred to it. Why does it feel so tiring? Your muscle fibres are rapidly twitching and relaxing to hold the tension, doing internal biological work and generating heat inside you, but none of that energy goes into the box. It is a sharp reminder that the physics definition of work is about energy transferred to the object, not about human effort.

Negative work and the angle in W = Fd·cosθ

The cosine in W = Fd·cosθ is not a technicality — it decides the sign of the work. When a force points partly against the motion, cos θ is negative and the force removes energy: friction sliding a box to a stop (θ = 180°, cos = −1) does negative work, draining kinetic energy into heat. When a force is exactly perpendicular to the motion (θ = 90°, cos = 0), it does no work at all — which is why the tension in a string does zero work on a ball whirling in a circle, and why gravity does no work on a satellite in a circular orbit even though it is always pulling. Only the component of force along the displacement transfers energy.

Real-world power: from a light bulb to a rocket

Power ratings span an astonishing range. A human can sustain about 100 W of useful mechanical output for hours (roughly the power of an old light bulb), while a trained sprinter can hit over 1,000 W for a few explosive seconds. A family car's engine peaks near 100,000 W (about 130 hp), and the first stage of a Saturn V rocket delivered on the order of 10¹¹ W at liftoff. Since all of these do work by transferring energy, they can be compared on a single scale of watts — the same unit whether you are rating a nightlight or a moon rocket.

In the real world

Engineers rate every motor, engine, and power plant by its power — how fast it can transfer energy — because a job done too slowly (charging a car, pumping floodwater, lifting an elevator) can be as useless as not doing it at all.

Exit check

A10N parallel force moves a box 3 m in 2 s. Find work and average power.

Check your explanation after answering

W=30 J; P=W/t=15W, if that force is constant.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Kinetic & Potential Energy

Learning goals

Readings for this unit

Before this unit

Squared speed, height reference, and spring energy.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Where does a falling object’s kinetic energy come from?

Illustrative evidence case. A 2 kg object falls from a higher gravitational position and speeds up.

Kinetic energy depends on mass, speed and reference frameKinetic energy depends on mass, speed and reference frameKinetic energy for m=2 kg (J)048121601234Speed (m/s)K=½mv²; doubling speed quadruples K.
Calculated curve for a 2 kg object; not a relativistic high-speed formula.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Where does a falling object’s kinetic energy come from?

Your first model

Define object–Earth boundary, reference frame and height zero.

Build the language from the idea

kinetic energy
The energy an object has because of its motion: KE = ½mv², measured in joules (J).
gravitational potential energy
Stored energy an object has because of its height in a gravitational field: PE = mgh.
elastic potential energy
Energy stored in a stretched or compressed elastic object such as a spring: PE = ½kx².
spring constant (k)
A measure of a spring's stiffness, in newtons per metre (N/m); it appears in Hooke's law F = kx.
reference level
The chosen height where gravitational potential energy is defined as zero; only changes in PE have physical meaning.

The energy of a moving object is its kinetic energy, KE = ½mv². The most important feature of this formula is that speed is *squared*, so kinetic energy is exquisitely sensitive to how fast something moves. Double an object's speed and its kinetic energy does not double — it quadruples; triple the speed and it grows nine-fold. This single fact explains why a highway crash releases so much more energy than a parking-lot fender-bender, and why even a light object, like a hailstone or a fast-pitched ball, can strike with surprising force when its speed is high.

Kinetic energy depends on mass, speed and reference frameKinetic energy depends on mass, speed and reference frameKinetic energy for m=2 kg (J)048121601234Speed (m/s)K=½mv²; doubling speed quadruples K.
Calculated curve for a 2 kg object; not a relativistic high-speed formula.

Gravitational potential energy is the energy an object stores simply by being up high, PE = mgh, where m is mass, g = 9.8 m/s² near Earth's surface, and h is height above some chosen reference level. Raise a mass higher, or choose a heavier mass, and you bank more energy that gravity can later cash in as motion. A subtle but important point: the value of h depends on where you decide to call 'zero.' Only *changes* in height — and therefore changes in potential energy — carry physical meaning, so you are free to place the reference wherever makes a problem simplest.

Potential energy requires an interaction and a chosen referencePotential energy requires an interaction and a chosen referenceh=2 mU=0 chosen at ground; m=3 kgUg=mgh=58.8 J relative to this zero.
Original 3 kg near-surface height model; ground zero is a choice.

Springs and other elastic objects store a third kind: elastic potential energy, PE = ½kx², where k is the spring constant (the spring's stiffness) and x is how far it is stretched or compressed from its natural length. This comes straight from Hooke's law, F = kx: because the restoring force grows steadily as you stretch a spring, the energy stored is the area under that force-versus-stretch line — a triangle, which is where the factor of ½ comes from. Compress a spring twice as far and you store four times the energy, ready to launch a dart, a pinball, or a valve.

Spring force and stored elastic energy are different quantitiesSpring force and stored elastic energy are different quantitiesElastic potential energy (J)00.511.5200.050.10.150.2Extension x (m)Us=½kx²; force magnitude is kx, not ½kx².
Calculated linear-spring energy curve, valid only over the stated elastic range.

These forms are not isolated curiosities; they are constantly converting into one another. A ball at the top of its arc holds mostly gravitational potential energy and almost no kinetic energy, then trades one for the other as it falls. A drawn bow stores elastic potential energy that becomes the arrow's kinetic energy at release. Tracking how much energy lives in each form at each moment — motion energy versus stored positional energy — is exactly the modelling skill this unit is building toward, and it sets up the conservation law of the next lesson.

Because all three are measured in the same unit, the joule, you can add them, compare them, and follow them as a single running total. That is what makes energy such a powerful accounting tool: an engineer can look at a roller-coaster car, a compressed spring in a mattress, or water held behind a dam and put a single number — in joules — on the energy each holds, then predict what that energy will do when it is released.

Model check: Gravitational potential energy needs a chosen reference; negative values can be valid. U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Identify the reference frame for kinetic energy

Kinetic energy is K=½mv² in ordinary nonrelativistic mechanics. Speed is measured in a specified frame, so an object can have zero kinetic energy relative to a moving vehicle but positive kinetic energy relative to the road. Squaring speed makes K nonnegative; direction still matters for momentum and force calculations. The graph’s quadratic shape means doubling speed quadruples kinetic energy for unchanged mass.

Kinetic energy depends on mass, speed and reference frameKinetic energy depends on mass, speed and reference frameKinetic energy for m=2 kg (J)048121601234Speed (m/s)K=½mv²; doubling speed quadruples K.
Calculated curve for a 2 kg object; not a relativistic high-speed formula.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

If mass and speed both double, how does K change?

Visual reasoning — after your attempt

It grows by 2×2²=8 times in the same frame.

For ordinary low-speed motion, K=½mv² measures kinetic energy in a specified frame. It is nonnegative and depends on speed squared. Changing the velocity sign alone does not change K, but changing reference frame can change the speed and energy.

A 2 kg object at either +3 or −3 m/s has 9 J in that frame. Its momentum changes sign, so energy cannot replace a directional momentum account.

Pause and explain

Why can equal kinetic energies describe different directions?

Check your reasoning after answering

K uses speed squared and contains no direction sign; the two velocities can therefore have equal energy and opposite momentum.

Part 2 · Choose a potential-energy reference

Near Earth, gravitational potential energy of the object–Earth system can be written Ug=mgh relative to a chosen zero. Changing the zero shifts all potential values by a constant but leaves energy differences and predictions unchanged. The simple formula assumes g is approximately constant over the height interval. Potential energy belongs to an interaction/system account, rather than being a substance hidden inside the lifted object.

Potential energy requires an interaction and a chosen referencePotential energy requires an interaction and a chosen referenceh=2 mU=0 chosen at ground; m=3 kgUg=mgh=58.8 J relative to this zero.
Original 3 kg near-surface height model; ground zero is a choice.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

If the zero moves 1 m upward, what happens to this 3 kg object’s Ug?

Visual reasoning — after your attempt

Its new h is 1 m, so Ug=29.4 J; the physical state and energy differences between fixed events are unchanged.

Near Earth with nearly uniform g, gravitational potential-energy change is mgΔh for the object-Earth interaction. The zero level is a convenient reference; changes matter for mechanical predictions. Negative potential energy relative to that choice does not imply an impossible physical state.

Choosing the floor or tabletop as zero changes U values but leaves the change for the same lift unchanged.

Pause and explain

Does changing the zero of potential energy change required lifting work?

Check your reasoning after answering

No. It changes the numerical reference, not the potential-energy difference between the same initial and final states.

Part 3 · Distinguish force and elastic energy

For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.

Spring force and stored elastic energy are different quantitiesSpring force and stored elastic energy are different quantitiesElastic potential energy (J)00.511.5200.050.10.150.2Extension x (m)Us=½kx²; force magnitude is kx, not ½kx².
Calculated linear-spring energy curve, valid only over the stated elastic range.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

For k=100 N/m and x=0.10 m, find both energy and force magnitude.

Visual reasoning — after your attempt

Us=0.50 J; force magnitude=10 N. Their units and meanings differ.

An ideal spring stores elastic potential energy ½kx². The energy depends on displacement squared, while restoring force is proportional to signed displacement. The formulas answer different questions and have different units. Use them only within the assumed elastic range.

A spring compressed or stretched by the same magnitude can have equal ideal stored energy, with restoring force in opposite directions.

Pause and explain

Why is elastic energy not simply kx?

Check your reasoning after answering

kx has force units. Energy requires integrating the varying spring force over displacement, giving ½kx² in the linear model.

Part 4 · Track transfers between compatible stores

An energy account uses one boundary, frame and reference convention throughout. For a 2 kg object falling from rest through a 3 m height change with negligible resistance, gravitational potential decreases by 58.8 J while kinetic energy increases by the same amount. The selected object–Earth mechanical total stays 98 J with ground as zero. Changing the height reference midway or excluding an important interaction would make the bookkeeping appear inconsistent.

Track compatible energy stores in one consistent accountTrack compatible energy stores in one consistent accountStateKineticGravitationalTotalStart at rest, h=5 m0 J98 J98 JLater, h=2 m58.8 J39.2 J98 Jm=2 kg; negligible losses; object–Earth system.
Calculated loss-free falling example with constant g=9.8 m/s².

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What speed corresponds to K=58.8 J for 2 kg?

Visual reasoning — after your attempt

v=√(2K/m)=√58.8≈7.67 m/s.

An energy model must identify interactions and boundaries. A falling object can gain kinetic energy while object-Earth gravitational potential energy decreases. Drag or friction can transfer energy to thermal stores; an ideal prediction that omits them should be labeled accordingly.

A frictionless spring launcher gives a useful reference speed, but measured launch speed may be smaller if other stores gain energy.

Pause and explain

What does an energy shortfall suggest?

Check your reasoning after answering

It suggests omitted transfers, changed assumptions, or measurement error; total energy has not necessarily failed to be conserved.

Apply the whole unit · Predict a spring launch speed

Illustrative 0.500 kg cart starts at rest against a spring with k=200 N/m compressed 0.100 m. Assume a horizontal loss-free launch and negligible spring mass.

Your task

Calculate spring energy and cart speed after release, compare a measured 1.80 m/s, and identify the energy difference and plausible next measurements.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Initial spring energy=½×200×0.100²=1.00 J.

  2. ½×0.500×v²=1.00 gives v=2.00 m/s.

  3. At 1.80 m/s, K=0.810 J; difference from prediction=0.190 J.

The 0.190 J difference is an account relative to the ideal model, not proof of one particular loss mechanism. Track rolling resistance, spring motion, sound, and measurement uncertainty before assigning a cause. The result assumes all modeled elastic energy becomes cart translational kinetic energy.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require students to distinguish an energy account from a causal diagnosis. A plausible loss mechanism still needs evidence. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Stored energy becomes energy of motionElastic PE = ½kx²launchKE = ½mv²Grav PE= mgh
A compressed spring stores elastic potential energy (½kx²) that launches the ball into kinetic energy (½mv²); a raised mass stores gravitational potential energy (mgh).

The key idea

Kinetic energy depends on speed/frame; potential energy depends on interaction and chosen reference. Conserve a consistent energy account rather than mixing force with energy.

Kinetic Energy

KE = ½mv²

m is mass in kg; v is speed in m/s; KE is in joules (J). Speed is squared, so KE grows fast as v increases.

Gravitational Potential Energy

PE = mgh

m is mass in kg; g = 9.8 m/s² near Earth's surface; h is height above the reference level in metres; PE is in joules (J).

Elastic Potential Energy

PE = ½kx²

k is the spring constant in N/m; x is the stretch or compression from the natural length in metres; PE is in joules (J).

Worked examples

Example 1

Find the kinetic energy of a 1,200 kg car moving at 20 m/s.
  1. Use KE = ½mv².
  2. KE = ½ × 1,200 × 20².
  3. = ½ × 1,200 × 400 = 240,000.
Answer: 240,000 J (240 kJ)

Example 2 · Worked application

A 2 kg object moves at 5 m/s and sits 3 m above a chosen reference. Use g=10 m/s² to find KE and PE.
  1. KE=½mv²=½×2×25=25 J.
  2. PE=mgh=2×10×3=60 J.
  3. Potential energy depends on the chosen reference; changes are physically useful.
Answer: KE=25 J; PE=60 J relative to the stated level.
Check the reasoning

Do not add height to speed or confuse their squares.

Example 3

A spring with spring constant k = 400 N/m is compressed 0.10 m. How much elastic potential energy is stored?
  1. Use PE = ½kx².
  2. PE = ½ × 400 × 0.10².
  3. = ½ × 400 × 0.01 = 2.0.
Answer: 2.0 J

Example 4 · Find and fix the mistake

Context: A 2 kg object moves at 5 m/s and sits 3 m above a chosen reference. Use g=10 m/s² to find KE and PE. A student claims: “Potential energy is absolute and independent of reference level.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Do not add height to speed or confuse their squares.
  2. Its zero is chosen; differences between states are the meaningful quantity in the model.
Answer: Its zero is chosen; differences between states are the meaningful quantity in the model.
Check the reasoning

Do not add height to speed or confuse their squares.

Example 5 · Reason from the visual model

Spring force and stored elastic energy are different quantitiesSpring force and stored elastic energy are different quantitiesElastic potential energy (J)00.511.5200.050.10.150.2Extension x (m)Us=½kx²; force magnitude is kx, not ½kx².
Calculated linear-spring energy curve, valid only over the stated elastic range.
For k=100 N/m and x=0.10 m, find both energy and force magnitude.
  1. For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.
  2. Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
Answer: Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
Check the reasoning

Calculated linear-spring energy curve, valid only over the stated elastic range.

Avoid these mistakes

Check your understanding

A 2 kg mass at 3 m has what U usingg=9.8 andU=0 atground?

Discuss the answer

U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.

Return to the phenomenon

Return to the initial case. Where does a falling object’s kinetic energy come from? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Gravitational potential decreases while kinetic energy rises; resistance would require additional energy destinations.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Where does a falling object’s kinetic energy come from? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Gravitational potential decreases while kinetic energy rises; resistance would require additional energy destinations.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A 1.5 kg cart has speed 4 m/s. It compresses a loss-free spring with k=600 N/m on a level track. Find initial K and maximum compression.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

K=½×1.5×16=12 J. Set 12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.

Words to know

kinetic energy
The energy an object has because of its motion: KE = ½mv², measured in joules (J).
gravitational potential energy
Stored energy an object has because of its height in a gravitational field: PE = mgh.
elastic potential energy
Energy stored in a stretched or compressed elastic object such as a spring: PE = ½kx².
spring constant (k)
A measure of a spring's stiffness, in newtons per metre (N/m); it appears in Hooke's law F = kx.
reference level
The chosen height where gravitational potential energy is defined as zero; only changes in PE have physical meaning.

The 'vis viva' controversy: was motion measured by mv or mv²?

In the early 1700 s, Europe's best thinkers fought over how to quantify a moving object's 'force.' Followers of Descartes backed quantity of motion, mv (what we now call momentum); Leibniz argued for vis viva, or 'living force,' proportional to mv². Experiments by Willem 's Gravesande — dropping brass balls into soft clay and measuring the depth of the dent — showed the effect grew with the square of the speed, and the mathematician Émilie du Châtelet helped make the case decisively. The resolution was that both quantities are real and useful: mv is conserved as momentum, and ½mv² is the kinetic energy. This whole unit is, in a sense, the happy ending to that 300-year-old argument.

Misconception: potential energy is not an absolute number

It is tempting to think an object 'has' a definite amount of gravitational potential energy, but PE = mgh depends entirely on where you set h = 0. A book on a second-floor desk has one PE value measured from the floor, a larger one measured from the ground outside, and a negative one measured from the ceiling. None is more 'correct' than the others. What is physically real — and always comes out the same — is the change in PE as the object rises or falls. Choosing a convenient reference level is a freedom that often makes a hard problem easy.

Where the ½ in ½kx² comes from: the area under Hooke's law

A spring obeys Hooke's law, F = kx: the more you stretch it, the harder it pulls back, in a straight-line relationship. To find the energy stored you cannot just multiply force by distance, because the force keeps changing as you stretch. Instead you take the area under the force-versus-stretch graph — and since that graph is a straight line from 0 up to kx, the area is a triangle: ½ × base × height = ½ × x × (kx) = ½kx². That same 'area under a changing force' reasoning is why kinetic energy also carries a ½, and it is your first taste of the calculus that underlies all of energy physics.

Storing energy at grid scale: pumped hydro and giant springs

The formulas in this lesson run some of the world's largest machines. Pumped-storage plants use surplus night-time electricity to pump enormous volumes of water uphill, banking it as gravitational PE = mgh; at peak demand they let it fall back through turbines, recovering about 80% of the energy. Elastic potential energy is stored at scale too — from the coiled springs in mechanical watches to experimental flywheel and spring systems that smooth out power grids. Whenever engineers need to save energy now and release it later, they reach for exactly the potential-energy equations you are practicing here.

In the real world

Pumped-storage power stations bank surplus electricity by pumping water uphill into a reservoir as gravitational potential energy, then release it through turbines at peak demand — a giant rechargeable battery built out of PE = mgh.

Exit check

A2 kg mass at 3 m has what U usingg=9.8 andU=0 atground?

Check your explanation after answering

U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Conservation of Energy

Learning goals

Readings for this unit

Before this unit

Work, energy types, and system boundaries.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Does friction destroy energy?

Illustrative evidence case. A sliding cart slows while the surface warms and some energy paths remain unmeasured.

A boundary determines which transfers cross the accountA boundary determines which transfers cross the accountChosen system: object + Earth + surfaceinputoutputΔEstored = Ein − EoutDraw the boundary before deciding what is external work.
Original boundary diagram; arrows denote energy transfer, not physical force directions.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Does friction destroy energy?

Your first model

Separate mechanical stores from the complete boundary account.

Build the language from the idea

mechanical energy
The sum of an object's kinetic and potential energy: E = KE + PE.
conservation of mechanical energy
In the absence of friction and other non-conservative forces, KE + PE stays constant: KE_i + PE_i = KE_f + PE_f.
law of conservation of energy
The total energy of an isolated system never changes; energy is only transformed or transferred, never created or destroyed.
dissipation
The spreading of useful mechanical energy into scattered thermal energy and sound, usually through friction or drag.
efficiency
The fraction of input energy that comes out as useful output: efficiency = (useful output ÷ total input) × 100%.

Mechanical energy includes kinetic and appropriate potential-energy stores for a selected system. It can remain constant when external transfers and nonconservative effects are negligible in the stated account. Absence of friction alone is not sufficient if an external motor does work or energy leaves by another pathway. For an ideal loss-free descent, gravitational potential energy can transfer into kinetic energy while their sum stays constant.

A boundary determines which transfers cross the accountA boundary determines which transfers cross the accountChosen system: object + Earth + surfaceinputoutputΔEstored = Ein − EoutDraw the boundary before deciding what is external work.
Original boundary diagram; arrows denote energy transfer, not physical force directions.

This is not just elegant; it is a ruthlessly efficient problem-solving tool. Suppose a ball is dropped from a height h and you want its speed at the bottom. Setting mgh = ½mv², the mass cancels and you get v = √(2gh) — the answer in a single line, with no need to track acceleration or time at all. Conservation lets you leap straight from a starting condition to a final one, ignoring the complicated journey in between, which is why physicists reach for it constantly.

Friction and other interactions can reduce translational mechanical energy while increasing thermal, sound, deformation, rotational, or other stores. A complete energy account includes transfers across the selected boundary. For an isolated system, total energy is conserved; for an open energy account, initial plus incoming equals final plus outgoing. An unexplained residual identifies missing information or a model limitation rather than proving energy has been destroyed.

Mechanical energy can decrease while total energy is conservedMechanical energy can decrease while total energy is conservedEnergy (J)Initial mechanical100Final mechanical70Thermal increase30100 = 70+30 J for the stated complete account.
Illustrative fully specified energy partition, not a measured universal friction fraction.

Efficiency is useful output divided by input for a specified service and boundary. A motor receiving 800 J and delivering 600 J useful work is 75% efficient; account for the remaining 200 J as other transfers or storage rather than assuming its exact destination without evidence. Thermal engines have a second-law limit under their operating conditions, but it is too broad to say every kind of defined efficiency must always be below 100%. An ideal electrical heater can convert electrical input to thermal output with a near-unity device account; a heat pump’s coefficient of performance is a different ratio that includes transferred environmental heat.

A Newton's cradle captures the whole story in one desktop toy. Lift a steel ball and release it, and the energy races through the row and out the far side, swing after swing, almost perfectly conserved. Almost — each click you hear is a sliver of energy leaving as sound, and each collision warms the steel a touch, so the swings slowly shrink and finally stop. Nothing was destroyed; the energy merely dissipated. The same accounting lets engineers design hydroelectric dams, regenerative brakes, and power plants, all of which live or die by how much of their energy they can steer into a useful form instead of losing it to heat.

Learning to write down every joule at the start, track where each one goes, and insist that the books balance at the end is the heart of energy analysis. Whether the 'missing' energy went into motion, into height, into a stretched spring, or into scattered heat, conservation guarantees it went *somewhere* — and that certainty is what turns a tangle of forces into a solvable problem.

A missing balance guides the next measurementA missing balance guides the next measurementMeasured/model quantityValueEnergy estimateInitial K80 JSupplied startFinal K45 JSupplied finishSurface thermal increase28 JMeasured/model estimateUnaccounted balance7 JInvestigate other pathwaysA residual identifies a question, not a new energy source.
Illustrative energy records with an intentionally incomplete measured account.

Model check: Mechanical energy can decrease while total energy remains conserved. Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · State an energy boundary before balancing

Energy conservation is a complete accounting principle: the change of energy stored in a system equals transfers into it minus transfers out. A boundary including Earth represents gravity with potential energy; a boundary around the moving object alone may instead count gravity’s external work. These are compatible choices when used consistently. Do not count the same gravitational transfer both as external work and as a potential-energy decrease in the same balance.

A boundary determines which transfers cross the accountA boundary determines which transfers cross the accountChosen system: object + Earth + surfaceinputoutputΔEstored = Ein − EoutDraw the boundary before deciding what is external work.
Original boundary diagram; arrows denote energy transfer, not physical force directions.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why must the system be stated before listing work and energy stores?

Visual reasoning — after your attempt

The boundary decides which interactions are internal stores and which are external transfers, preventing omissions or double counting.

Energy conservation applies to a complete account including stores and transfers across the chosen boundary. A system closed to matter can still exchange energy. Write initial energy plus incoming transfers equal to final energy plus outgoing transfers; identify what has deliberately been omitted.

A cooling sealed bottle loses energy to its surroundings although it retains its matter. Calling it closed does not make its energy constant.

Pause and explain

When can the selected total energy be treated as constant?

Check your reasoning after answering

When the account is complete and no net energy crosses the relevant boundary over the interval, or incoming and outgoing transfers balance.

Part 2 · Distinguish total and mechanical conservation

Mechanical energy combines selected kinetic and potential stores. Friction can reduce this sum while increasing internal thermal energy of bodies and surroundings. The 100 J account here ends with 70 J mechanical and 30 J thermal increase. Saying energy was “lost” is only shorthand for leaving the selected useful or mechanical category; it is not destruction. A complete account also checks sound, deformation and transfer outside the boundary where relevant.

Mechanical energy can decrease while total energy is conservedMechanical energy can decrease while total energy is conservedEnergy (J)Initial mechanical100Final mechanical70Thermal increase30100 = 70+30 J for the stated complete account.
Illustrative fully specified energy partition, not a measured universal friction fraction.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Can a 30 J mechanical decrease by itself establish where all 30 J went?

Visual reasoning — after your attempt

No. Additional evidence is needed to apportion thermal, sound, deformation and external transfers.

Mechanical energy K+U can remain constant when nonconservative transfers are negligible in the selected model. Friction can reduce mechanical energy while increasing thermal energy elsewhere. Total energy conservation and mechanical-energy conservation are different claims.

A sliding block stopping does not destroy its kinetic energy; contact interactions transfer energy into other stores.

Pause and explain

Does friction violate energy conservation?

Check your reasoning after answering

No. It can change the distribution among stores and transfers while reducing the part labeled mechanical energy.

Part 3 · Use the work-energy connection correctly

The work–energy theorem states Wnet=ΔK in this ordinary mechanics model. It concerns the sum of all forces’ work on the selected object, not necessarily total energy or one applied force’s contribution. A driving force doing 50 J while resistance does−20 J yields a 30 J kinetic increase. If gravity is among the forces, its work enters this net-work account; a potential-energy formulation reorganizes that term instead of adding it twice.

The work–energy theorem concerns net work and kinetic changeThe work–energy theorem concerns net work and kinetic changeTermWork on objectKinetic changeDriving interaction+50 JResistance−20 JSum / result+30 JΔK=+30 JNet work, not one force’s work, equals ΔK.
Original signed-work account on one object; internal-energy accounting may use a larger boundary.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

If initial K is 10 J, what is final K here?

Visual reasoning — after your attempt

40 J, because net work is+30 J.

Net work on a particle-like object equals its kinetic-energy change in the ordinary mechanics model. Individual forces may do nonzero work while net work is zero. Potential-energy formulations move suitable conservative interactions into an energy-store account, so avoid counting the same transfer twice.

An ideal constant-speed lift has positive work by support and negative work by gravity, leaving zero net work on the load.

Pause and explain

What error occurs if gravity’s work and the same mgΔh are both counted as separate losses?

Check your reasoning after answering

The same conservative interaction is double-counted. Use either a force-work account or a compatible system potential-energy account consistently.

Part 4 · Use evidence to account for nonmechanical outputs

A residual is the difference between a predicted account and the terms currently measured or estimated. If mechanical energy falls 35 J while only 28 J thermal increase is recorded,7 J remains unaccounted for in those terms. Sound, deformation, heat leaving the boundary or measurement bias are possible explanations. Do not force closure by inventing an observation; report uncertainty and identify a measurement that can distinguish plausible missing pathways.

A missing balance guides the next measurementA missing balance guides the next measurementMeasured/model quantityValueEnergy estimateInitial K80 JSupplied startFinal K45 JSupplied finishSurface thermal increase28 JMeasured/model estimateUnaccounted balance7 JInvestigate other pathwaysA residual identifies a question, not a new energy source.
Illustrative energy records with an intentionally incomplete measured account.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Does a 7 J residual prove energy conservation failed?

Visual reasoning — after your attempt

No. It may reflect omitted transfers, storage or uncertainty; test the account before making that inference.

A real-device energy budget may need thermal, sound, deformation, and external work terms. A residual can be recorded as unaccounted energy rather than assigned to an unsupported destination. Compare measured quantities with uncertainties and a clear interval.

If 100 J enters a device and 80 J is measured in named outputs, the remaining 20 J needs a store or transfer account, not a claim of destruction.

Pause and explain

How should an unexplained residual be reported?

Check your reasoning after answering

Keep it explicit, identify measurement limits and possible omitted pathways, and propose observations that can distinguish them.

Apply the whole unit · Account for a descending cart

Illustrative 2.00 kg cart starts at rest 1.00 m above a reference and reaches speed 3.00 m/s at that reference. Use g=10.0 m/s². Ignore cart rotation in the first account.

Your task

Find initial gravitational potential energy, final translational kinetic energy, and residual. Explain why the residual is not automatically all frictional heating and improve the measurement plan.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Initial U=2×10×1=20.0 J; initial K=0.

  2. Final translational K=½×2×3²=9.00 J.

  3. Residual=20.0−9.00=11.0 J in omitted stores, transfers, or measurement discrepancies.

Wheel rotation can be a significant store for a cart, so assigning all 11 J to friction without investigating it would be unjustified. Sound, track heating, deformation, and uncertainties may also contribute. A fuller model should include rotation or justify neglecting it, rather than treating a translational-only account as total mechanical energy.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask which energy stores belong to a cart rather than an ideal sliding point mass. Reward an explicit unresolved residual with a useful next measurement. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

A Newton's cradle transferring energy from one swinging steel ball through a row to the ball on the far end.

Energy passes almost perfectly from ball to ball — but each click of sound and faint warming of the steel dissipates a little, so the swings slowly fade. Total energy is always conserved.

The key idea

Total energy follows a complete boundary account. Mechanical energy remains constant only under appropriate transfer conditions; useful efficiency needs a defined service.

Mechanical Energy

E = KE + PE

the sum of an object's kinetic and potential energy.

Conservation of Mechanical Energy

KE_i + PE_i = KE_f + PE_f

valid when friction and other non-conservative forces are absent; the total of kinetic plus potential energy stays constant.

Efficiency

efficiency = (useful output ÷ total input) × 100%

the percentage of input energy that emerges as useful output; always less than 100% for a real device.

Worked examples

Example 1

A 1.5 kg ball is dropped from rest at a height of 10 m. Ignoring air resistance, how fast is it moving just before it lands? (g = 9.8 m/s²)
  1. All the gravitational potential energy converts to kinetic energy: mgh = ½mv².
  2. PE = mgh = 1.5 × 9.8 × 10 = 147 J, so 147 = ½ × 1.5 × v² = 0.75v².
  3. v² = 147 ÷ 0.75 = 196, so v = √196 = 14.
Answer: 14 m/s

Example 2 · Worked application

A 1 kg object drops through 5 m from rest. Use g=10 m/s², neglect drag, and find speed after the drop.
  1. Lost PE=mgΔh=50 J.
  2. Conservation gives ½mv²=50.
  3. v=√100=10 m/s.
Answer: 10 m/s.
Check the reasoning

Mass cancels in this ideal gravitational-energy calculation.

Example 3

An electric motor is supplied 800 J of electrical energy and delivers 600 J of useful mechanical work. What is its efficiency?
  1. Use efficiency = (useful output ÷ total input) × 100%.
  2. efficiency = (600 ÷ 800) × 100%.
  3. = 0.75 × 100% = 75%.
Answer: 75%

Example 4 · Find and fix the mistake

Context: A 1 kg object drops through 5 m from rest. Use g=10 m/s², neglect drag, and find speed after the drop. A student claims: “Mechanical energy must stay constant even when substantial friction acts.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Mass cancels in this ideal gravitational-energy calculation.
  2. Total energy is conserved, but mechanical energy can transfer to thermal energy or other stores.
Answer: Total energy is conserved, but mechanical energy can transfer to thermal energy or other stores.
Check the reasoning

Mass cancels in this ideal gravitational-energy calculation.

Example 5 · Reason from the visual model

The work–energy theorem concerns net work and kinetic changeThe work–energy theorem concerns net work and kinetic changeTermWork on objectKinetic changeDriving interaction+50 JResistance−20 JSum / result+30 JΔK=+30 JNet work, not one force’s work, equals ΔK.
Original signed-work account on one object; internal-energy accounting may use a larger boundary.
If initial K is 10 J, what is final K here?
  1. The work–energy theorem states Wnet=ΔK in this ordinary mechanics model. It concerns the sum of all forces’ work on the selected object, not necessarily total energy or one applied force’s contribution. A driving force doing 50 J while resistance does−20 J yields a 30 J kinetic increase. If gravity is among the forces, its work enters this net-work account; a potential-energy formulation reorganizes that term instead of adding it twice.
  2. 40 J, because net work is+30 J.
Answer: 40 J, because net work is+30 J.
Check the reasoning

Original signed-work account on one object; internal-energy accounting may use a larger boundary.

Avoid these mistakes

Check your understanding

A cart loses 20 J mechanical energy due to friction. Where is it?

Discuss the answer

Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries.

Return to the phenomenon

Return to the initial case. Does friction destroy energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Mechanical energy can decrease while total energy is conserved through thermal changes and transfers; residuals need evidence.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Does friction destroy energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Mechanical energy can decrease while total energy is conserved through thermal changes and transfers; residuals need evidence.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A system begins with 200 J mechanical energy. It ends with 120 J mechanical,50 J thermal increase and 20 J transferred outward as sound/heat. Find the unaccounted energy and a useful next check.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

The accounted final/output total is 190 J, leaving 10 J. Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.

Words to know

mechanical energy
The sum of an object's kinetic and potential energy: E = KE + PE.
conservation of mechanical energy
In the absence of friction and other non-conservative forces, KE + PE stays constant: KE_i + PE_i = KE_f + PE_f.
law of conservation of energy
The total energy of an isolated system never changes; energy is only transformed or transferred, never created or destroyed.
dissipation
The spreading of useful mechanical energy into scattered thermal energy and sound, usually through friction or drag.
efficiency
The fraction of input energy that comes out as useful output: efficiency = (useful output ÷ total input) × 100%.

Joule's paddle wheel: proving heat is energy

In the 1840 s, English brewer and physicist James Prescott Joule ran a famous experiment: a falling weight turned a paddle wheel inside an insulated barrel of water, and Joule measured the tiny temperature rise as the paddles churned. He found a fixed exchange rate between mechanical energy lost and heat gained — the 'mechanical equivalent of heat.' This showed that the energy 'lost' to friction was not gone but had become thermal energy, and it was decisive evidence for the law of conservation of energy. Fittingly, the SI unit of energy, the joule, is named after him.

Why we can't recycle waste heat: the second law

Conservation of energy guarantees the total never changes, which raises a puzzle: if the heat wasted by an engine is still energy, why can't we just collect it and reuse it to get 100% efficiency? The answer is the second law of thermodynamics: energy spontaneously spreads from concentrated, organized forms into disordered, scattered heat, and reversing that spread perfectly is impossible. This is why every perpetual motion machine ever built has failed, and why patent offices now refuse to even consider them. Energy is always conserved (first law), but its usefulness steadily degrades (second law).

Conservative vs non-conservative forces

Mechanical energy is only conserved when the forces at work are conservative — forces like gravity and ideal springs, for which the energy depends only on position, not on the path taken. Lift a book straight up or along a winding ramp to the same shelf, and gravity stores the same PE either way. Friction and air resistance are non-conservative: they always oppose motion, drain energy along the whole path, and turn it into heat that cannot be recovered as motion. Recognizing which forces are present tells you instantly whether you can use the clean 'KE + PE = constant' shortcut or must also subtract the energy dissipated.

Regenerative braking: conservation of energy on the road

An ordinary car brake throws away kinetic energy as useless heat in the brake pads — you can smell it after a long downhill. Electric and hybrid vehicles do something cleverer with regenerative braking: the electric motor runs backward as a generator, using the car's kinetic energy to recharge the battery instead of just heating the brakes. It is conservation of energy turned into engineering profit, recovering a large fraction of the energy that would otherwise be lost and noticeably extending an EV's range, especially in stop-and-go city driving.

In the real world

Electric and hybrid cars use regenerative braking to capture the kinetic energy that would normally be wasted as brake heat, converting it back into stored electrical energy in the battery — a direct application of conservation of energy that noticeably extends driving range.

Exit check

A cart loses 20 J mechanical energy due to friction. Where is it?

Check your explanation after answering

Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Momentum & Collisions

Learning goals

Readings for this unit

Before this unit

Momentum vectors and external impulse.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why can colliding bodies conserve momentum while losing kinetic energy?

Illustrative evidence case. Two carts stick together after a short collision.

Momentum retains direction that kinetic energy does notMomentum retains direction that kinetic energy does notA:2 kg at+3 m/s → p=+6 kg·m/sB:1 kg at−4 m/s → p=−4 kg·m/sTotal p=+2 kg·m/s, not 10 kg·m/s.
Original two-body initial-state model; arrow lengths represent momentum magnitude.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why can colliding bodies conserve momentum while losing kinetic energy?

Your first model

Choose both carts as the system and name external impulses.

Build the language from the idea

momentum
The 'quantity of motion' of an object, p = mv; a vector measured in kilogram-metres per second (kg·m/s).
impulse
The product of force and the time it acts, J = FΔt; it equals the change in momentum, Δp.
conservation of momentum
When no net external force acts, the total momentum of a system is the same before and after an interaction.
elastic collision
A collision in which both momentum and kinetic energy are conserved.
inelastic collision
A collision in which momentum is conserved but some kinetic energy is lost; in a perfectly inelastic collision the objects stick together.

Momentum is an object's 'quantity of motion,' defined simply as p = mv — mass times velocity. Because it multiplies the two, a slow but massive object and a fast but light one can carry the same momentum: a loaded truck rolling at walking pace and a bullet in flight are both hard to stop. Momentum is a *vector*, so direction counts; something moving left has momentum opposite in sign to something moving right, and that bookkeeping of signs is what makes momentum such a precise tool. Its unit is the kilogram-metre per second (kg·m/s).

Momentum retains direction that kinetic energy does notMomentum retains direction that kinetic energy does notA:2 kg at+3 m/s → p=+6 kg·m/sB:1 kg at−4 m/s → p=−4 kg·m/sTotal p=+2 kg·m/s, not 10 kg·m/s.
Original two-body initial-state model; arrow lengths represent momentum magnitude.

To change an object's momentum you must apply a force for some time, and that combination is called impulse: J = FΔt. The impulse delivered exactly equals the change in momentum, J = Δp — the impulse-momentum theorem. This relationship has a life-saving twist. The momentum change in a car crash is fixed by how fast you were going, but F = Δp/Δt shows that stretching the collision over a longer time Δt shrinks the force. That is precisely how airbags, crumple zones, padded dashboards, and even bending your knees when you land all protect you: they lengthen the stopping time and soften the force.

The crown jewel of this lesson is conservation of momentum: if no net external force acts on a system, its total momentum is identical before and after any interaction. This follows straight from Newton's third law — in a collision the two objects push on each other with equal and opposite forces for the same time, so the impulses cancel and whatever momentum one gains, the other loses. It holds whether objects bounce apart, stick together, or explode outward, which is why it works for colliding billiard balls, coupling train cars, and a rocket hurling exhaust one way to drive itself the other.

Conservation requires negligible external impulse over the intervalConservation requires negligible external impulse over the intervalSystem A+B during a short collisionInternal impulses cancel; external impulse must be checked.
Original collision boundary and internal interaction arrows; no particular collision outcome is implied.

Collisions come in two flavours, and telling them apart hinges on energy. In an elastic collision, both momentum *and* kinetic energy are conserved; the objects bounce off with no energy lost to heat, the way hardened steel balls or gas molecules nearly do. In an inelastic collision, momentum is still conserved but some kinetic energy is converted into heat, sound, and permanent deformation. The extreme case is a perfectly inelastic collision, in which the objects stick and move off together — like a dart burying itself in a board — losing the most kinetic energy possible while still, always, conserving momentum.

Collision names classify kinetic energy and final motionCollision names classify kinetic energy and final motionCollision modelMomentumKinetic energyElastic, isolatedConservedConservedInelastic, isolatedConservedMay decreasePerfectly inelasticConservedBodies share finalvelocityKinetic decrease becomes other forms, not destroyed energy.
Original classification table assumes an isolated collision system for its momentum column.

A neat special case makes the ideas concrete: in a head-on elastic collision between two *equal* masses, they simply exchange velocities. A moving ball striking an identical stationary one stops dead while the target shoots off at the incoming speed — exactly what you see click across a Newton's cradle. Analyzing any collision then becomes a two-part checklist: momentum is *always* conserved, so write p_before = p_after; kinetic energy is conserved *only if* the collision is elastic. Keeping those two rules straight lets you predict the outcome of impacts from the subatomic to the astronomical.

Momentum conservation is one of the deepest principles in physics, holding even where Newton's laws break down, in the quantum and relativistic worlds. For everyday engineering it is indispensable: it tells crash-safety teams how to design a survivable impact, lets rocket scientists compute thrust in the vacuum of space where there is nothing to push against, and allows physicists to reconstruct particle collisions they never directly see — all by insisting that the total momentum going in must equal the total coming out.

Model check: Momentum conservation does not automatically imply kinetic-energy conservation. Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Keep momentum vector information

Momentum is p=mv, a vector with units kg·m/s. Opposite velocities produce opposite momentum signs along a chosen axis. A 2 kg body moving at+3 m/s and a 1 kg body at−4 m/s have momenta+6 and−4 kg·m/s, totaling+2. Their kinetic energies are both positive and add differently. Keep component signs through the calculation rather than adding speed magnitudes as though all bodies moved together.

Momentum retains direction that kinetic energy does notMomentum retains direction that kinetic energy does notA:2 kg at+3 m/s → p=+6 kg·m/sB:1 kg at−4 m/s → p=−4 kg·m/sTotal p=+2 kg·m/s, not 10 kg·m/s.
Original two-body initial-state model; arrow lengths represent momentum magnitude.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What is the total kinetic energy of these two bodies?

Visual reasoning — after your attempt

½×2×3²+½×1×4²=9+8=17 J.

Momentum p=mv carries velocity direction and has units kg·m/s. In one dimension, choose a positive axis before adding momenta. Two moving objects can have zero total momentum with positive total kinetic energy, so a momentum sum does not determine all energy.

Equal masses moving at +2 and −2 m/s have opposite momenta that cancel; both still have kinetic energy.

Pause and explain

Why does zero total momentum not mean all objects are at rest?

Check your reasoning after answering

Opposite momenta can sum to zero while individual objects continue moving; their kinetic energies do not cancel by direction.

Part 2 · Choose the collision system and interval

For a selected system, total momentum change equals total external impulse. During a short two-body collision, the bodies’ mutual forces give opposite internal impulses that cancel in the combined account. Total momentum is conserved only when external impulse is negligible over that interval. Gravity, supports or friction may still exist; the approximation concerns their net impulse, not a claim that every external force vanishes.

Conservation requires negligible external impulse over the intervalConservation requires negligible external impulse over the intervalSystem A+B during a short collisionInternal impulses cancel; external impulse must be checked.
Original collision boundary and internal interaction arrows; no particular collision outcome is implied.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Can momentum be approximately conserved even when gravity acts?

Visual reasoning — after your attempt

Yes, if the relevant external impulse is negligible or cancels over the chosen short interval and components.

Total momentum is conserved for a selected system when net external impulse is negligible over the interval. Collision forces between included objects are internal and exchange momentum. A short collision can make some external effects small, but that assumption must be justified rather than automatic.

Two carts form one useful system during contact. Floor friction can still exert external impulse, especially over a long interval.

Pause and explain

What matters more than simply calling a collision isolated?

Check your reasoning after answering

The external impulse relative to the internal exchange and measurement precision must be assessed for the chosen interval.

Part 3 · Distinguish collision categories by energy

An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.

Collision names classify kinetic energy and final motionCollision names classify kinetic energy and final motionCollision modelMomentumKinetic energyElastic, isolatedConservedConservedInelastic, isolatedConservedMay decreasePerfectly inelasticConservedBodies share finalvelocityKinetic decrease becomes other forms, not destroyed energy.
Original classification table assumes an isolated collision system for its momentum column.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

For the initial A+B case sticking together, what final velocity follows?

Visual reasoning — after your attempt

vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.

Elastic collisions conserve total kinetic energy as well as momentum in the ideal system. Inelastic collisions do not conserve total kinetic energy; perfectly inelastic objects stick together. Total energy still has an account through deformation, thermal effects, and other stores.

Sticking is not evidence of momentum loss. It changes how the common final velocity is constrained and where energy is stored.

Optional guided lab · Compare momentum with kinetic energy

Optional guided lab · Compare momentum with kinetic energy

Optional external exploration — WeduLab. Complete the local explanation first. The task and practice record here are independently written for this course. Choose this in place of comparable practice or allow extra time; it does not add a required session to the year plan.

Predict the final velocity when a 0.50 kg cart moving right at 2.00 m/s sticks to a stationary 0.50 kg cart. Define right as positive and take both carts as the system. Momentum is mass times velocity; kinetic energy is one half mass times speed squared. Equal total momentum does not require equal kinetic energy before and after.

Explore for about 15–20 minutes after learning collision categories. Use the conservation-of-momentum topic. Compare spring/elastic bumpers with the sticking/Velcro case, using the same masses and starting velocities when possible. Record both carts’ signed initial and final velocities and calculate system momentum and kinetic energy. If controls cannot reproduce the selected numbers, label and analyze your actual setup, or use the original ideal records below.

Negligible net external impulse is the reason the selected two-cart momentum is conserved. Real tracks, bumpers, alignment and sensors differ from ideal models. Kinetic energy missing from cart motion is transferred into deformation, thermal energy, sound and other internal motion; energy has not vanished. Sticking gives the minimum final translational kinetic energy for the stated isolated two-cart initial condition in this frame, not a universal ranking of every possible collision.

No-lab route: the original supplied record below contains everything needed for the question. It is not measured evidence collected from WeduLab. Label any separate observations you collect as simulation output, with settings and date.

Original ideal case: m1 = m2 = 0.50 kgu1 / m s⁻¹u2 / m s⁻¹v1 / m s⁻¹v2 / m s⁻¹
Elastic equal-mass collision+2.000.000.00+2.00
Perfectly inelastic sticking collision+2.000.00+1.00+1.00

Question 1

Calculate total momentum and kinetic energy before and after each case. Explain why the common final velocity is +1.00 m/s in the sticking case. What evidence would you need before treating real-track momentum differences as a failure of the conservation law?

Worked solution and marking guidance

Initial momentum is 0.50 × 2.00 = +1.00 kg m/s. Initial kinetic energy is 0.5 × 0.50 × 2.00² = 1.00 J.

The elastic row retains +1.00 kg m/s and 1.00 J. For sticking, momentum gives (0.50 + 0.50)v = 1.00, so v = +1.00 m/s. Its final kinetic energy is 0.50 J.

Check the system boundary, external impulse from friction or apparatus, timing, velocity signs and measurement uncertainty before interpreting a discrepancy.

Both ideal rows conserve momentum at +1.00 kg m/s. Only the elastic row conserves translational kinetic energy. The sticking row transfers 0.50 J out of the two carts’ translational kinetic energy while total energy is conserved.

  • Define the two-cart system and signed direction.
  • Calculate both quantities with units, keeping momentum and energy distinct.
  • Account for the kinetic-energy change and discuss external impulse and measurement limits.

Pause and explain

Is momentum conservation restricted to elastic collisions?

Check your reasoning after answering

No. It also applies to inelastic collisions when net external impulse is negligible; kinetic energy is the quantity that differs.

Part 4 · Check results in the selected frame

Under ordinary low-speed frame changes by speed u, each velocity becomes v−u. The combined momentum changes by−Mu for total mass M, so its numerical value need not match across frames. Conservation can hold in either inertial frame if the same frame is used before and after. Kinetic-energy values also change with frame; do not mix a laboratory initial energy with a moving-frame final energy.

Momentum values depend on frame, while a valid balance persistsMomentum values depend on frame, while a valid balance persistsFrameInitial total pFinal total pLaboratory+2 kg·m/s+2 kg·m/sMoving+1 m/s2−3×1=−1−1 kg·m/sChange frames consistently for every body and event.
Calculated frame transformation for the same 3 kg two-body system; nonrelativistic velocities.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why is laboratory momentum+2 but moving-frame momentum−1 here?

Visual reasoning — after your attempt

The 3 kg system has its momentum shifted by−3×1=−3 kg·m/s under the stated frame change.

Use signed velocities, solve the momentum equation, and separately calculate kinetic energies if requested. A common final velocity can lie between initial velocities in a simple sticking case. Direction and dimensional checks can expose accidental use of speeds or wrong masses.

A negative final velocity means the joined system moves toward the negative axis, not an impossible collision.

Pause and explain

Why calculate energy after solving momentum?

Check your reasoning after answering

It tests a different account and can classify the collision; conserving momentum alone does not guarantee equal initial and final kinetic energy.

Apply the whole unit · Analyze two carts that stick

Illustrative 2.00 kg cart moves at +3.00 m/s and 1.00 kg cart at −1.00 m/s. They stick; net external impulse during contact is negligible.

Your task

Find common velocity, initial and final kinetic energy, and mechanical-energy redistribution. Explain why opposing momenta must carry signs.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Initial momentum=2×3+1×(−1)=5.00 kg·m/s.

  2. Final velocity=5.00/(2+1)=+1.67 m/s.

  3. Initial K=9.00+0.500=9.50 J; final K=½×3×(5/3)²≈4.17 J; difference≈5.33 J.

The final direction is positive, and the collision is perfectly inelastic in the supplied model. Momentum is conserved while about 5.33 J of translational kinetic energy enters other stores. That difference does not specify how much goes separately into deformation, sound, or heating without further measurements.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require signs before multiplication and separate momentum and energy totals. Ask where the interaction forces act when students question why sticking can conserve momentum. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Total momentum is conservedBeforeAvBat restAfterA+Bv′m₁v₁ + m₂v₂ = (m₁ + m₂)v′
Selected sticking collision with negligible external impulse. The common final velocity follows the signed momentum sum and need not universally be slower than every initial body.

The key idea

System momentum change equals external impulse. Negligible external impulse allows momentum conservation; kinetic energy is also conserved only in the elastic collision model.

Momentum

p = mv

m is mass in kg; v is velocity in m/s; p is a vector in kilogram-metres per second (kg·m/s).

Impulse-Momentum Theorem

J = FΔt = Δp

F is the average force; Δt is the contact time; the impulse equals the change in momentum, Δp = mv_f − mv_i.

Conservation of Momentum

m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′

with no net external force, the total momentum before an interaction equals the total after. For a perfectly inelastic collision the objects stick together: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.

Worked examples

Example 1

Find the momentum of a 1,500 kg car moving at 20 m/s.
  1. Use p = mv.
  2. p = 1,500 × 20.
  3. = 30,000.
Answer: 30,000 kg·m/s

Example 2 · Worked application

A 2 kg cart at 3 m/s collides with a stationary 1 kg cart and sticks. Find final velocity with negligible external impulse.
  1. Initial momentum=2×3+1×0=6 kg·m/s.
  2. Total mass after sticking=3 kg.
  3. v_final=6/3=2 m/s in the initial direction.
Answer: 2 m/s.
Check the reasoning

Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.

Example 3

A 2.0 kg cart moving at 3.0 m/s collides with and sticks to a 1.0 kg cart at rest. Find their common velocity afterward.
  1. Momentum is conserved: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.
  2. 2.0 × 3.0 + 1.0 × 0 = (2.0 + 1.0) × v′, so 6.0 = 3.0 × v′.
  3. v′ = 6.0 ÷ 3.0 = 2.0.
Answer: 2.0 m/s (in the original direction)

Example 4 · Find and fix the mistake

Context: A 2 kg cart at 3 m/s collides with a stationary 1 kg cart and sticks. Find final velocity with negligible external impulse. A student claims: “Momentum conservation implies kinetic energy is conserved in every collision.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.
  2. Kinetic energy can transfer to deformation, heat and sound; momentum conservation requires negligible net external impulse.
Answer: Kinetic energy can transfer to deformation, heat and sound; momentum conservation requires negligible net external impulse.
Check the reasoning

Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.

Example 5 · Reason from the visual model

Collision names classify kinetic energy and final motionCollision names classify kinetic energy and final motionCollision modelMomentumKinetic energyElastic, isolatedConservedConservedInelastic, isolatedConservedMay decreasePerfectly inelasticConservedBodies share finalvelocityKinetic decrease becomes other forms, not destroyed energy.
Original classification table assumes an isolated collision system for its momentum column.
For the initial A+B case sticking together, what final velocity follows?
  1. An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.
  2. vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
Answer: vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
Check the reasoning

Original classification table assumes an isolated collision system for its momentum column.

Avoid these mistakes

Check your understanding

2 kg at 3 m/s sticks to 1 kg atrest. Find final velocity.

Discuss the answer

Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.

Return to the phenomenon

Return to the initial case. Why can colliding bodies conserve momentum while losing kinetic energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Internal collision impulses cancel in the system account; kinetic energy can become deformation and thermal energy.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why can colliding bodies conserve momentum while losing kinetic energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Internal collision impulses cancel in the system account; kinetic energy can become deformation and thermal energy.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A 3 kg cart at 4 m/s sticks to a 1 kg cart initially at rest. Neglect external impulse. Find final velocity and kinetic-energy decrease.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Initial momentum=12 kg·m/s, so vf=3 m/s. Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.

Words to know

momentum
The 'quantity of motion' of an object, p = mv; a vector measured in kilogram-metres per second (kg·m/s).
impulse
The product of force and the time it acts, J = FΔt; it equals the change in momentum, Δp.
conservation of momentum
When no net external force acts, the total momentum of a system is the same before and after an interaction.
elastic collision
A collision in which both momentum and kinetic energy are conserved.
inelastic collision
A collision in which momentum is conserved but some kinetic energy is lost; in a perfectly inelastic collision the objects stick together.

Descartes, Huygens, and the birth of conservation of momentum

Long before the word 'energy' was well understood, René Descartes proposed in the 1600 s that the total 'quantity of motion' in the universe is fixed. His early version was flawed because he ignored direction, but the Dutch physicist Christiaan Huygens corrected it by treating motion as a directed (vector) quantity and worked out the rules for elastic collisions. When Newton later published his laws, conservation of momentum fell out naturally from his third law. It remains one of the most robust principles in all of physics, still holding in the quantum and relativistic realms where much of Newton's mechanics needs revision.

Misconception: momentum and kinetic energy are not the same 'amount of motion'

Both p = mv and KE = ½mv² measure something about motion, so students often blur them — but they behave very differently. Momentum is a vector (direction matters, and opposite motions can cancel) and it is conserved in every collision. Kinetic energy is a scalar (always positive) and is conserved only in elastic collisions. In a head-on crash where two identical cars moving at equal speeds stick together and stop, the total momentum was zero before and after (conserved), while all the kinetic energy vanished into crumpling metal and heat. Keeping the two quantities separate is the key to analyzing any collision.

Impulse and the physics of staying alive in a crash

Impulse, J = FΔt = Δp, is the single most important idea in crash safety. Your change in momentum in an accident is fixed the instant the crash begins — but the force you feel depends on how long the stop takes. Crumple zones let the front of a car collapse over a longer distance and time; airbags cushion your head over tens of milliseconds instead of one; even a boxer 'rolling with' a punch or a gymnast bending their knees on landing is stretching Δt to shrink F = Δp/Δt. The same equation explains why landing on concrete hurts far more than landing on a crash mat, even from the same height.

Rockets: pushing on nothing at all

A common myth is that rockets fly by 'pushing against the air.' They don't — they work best in the vacuum of space, and the reason is pure conservation of momentum. A rocket hurls a stream of exhaust gas backward at enormous speed; to keep the system's total momentum constant, the rocket must gain an equal and opposite momentum forward. No ground, no air, and nothing external to push against is required — only the exhaust being thrown one way and the rocket recoiling the other. The same principle lets an astronaut drift back to a spacecraft by throwing a tool in the opposite direction.

In the real world

Rockets work by conservation of momentum: they hurl exhaust gas backward at high speed, and the equal-and-opposite forward momentum drives the rocket ahead — which is why a rocket can accelerate in the vacuum of space with nothing to push against.

Exit check

2 kg at 3 m/s sticks to 1 kg atrest. Find final velocity.

Check your explanation after answering

Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Impulse & Collision Evidence

Learning goals

Readings for this unit

Before this unit

Momentum, velocity signs, graph areas, Newton’s laws.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

How can cushioning change force without eliminating the stopping impulse?

Illustrative evidence case. Two ideal pulse shapes stop the same momentum change over different times.

Impulse includes a reversal’s full signed velocity changeImpulse includes a reversal’s full signed velocity changem=0.20 kg; vi=+10 m/s; vf=−5 m/sΔp=m(vf−vi)=0.20(−5−10)=−3.0 N·s
Original signed reversal example; force direction is negative along the chosen axis.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

How can cushioning change force without eliminating the stopping impulse?

Your first model

Shade force–time areas and label the chosen body.

Build the language from the idea

impulse
Time integral of force, equal to momentum change.
external impulse
Impulse from forces exerted by objects outside the chosen system.

Momentum p=mv is a vector quantity. Impulse J=∫Fdt changes momentum; for a known average force, J=FavgΔt=Δp. A force-time graph’s signed area gives impulse. It is the change in momentum rather than change in speed alone that matters when direction reverses.

Choose signs before subtracting: Δp=m(vfinal−vinitial). A ball moving right and bouncing left has a larger momentum change magnitude than an otherwise similar ball merely stopping. A negative result identifies the impulse direction under the selected convention.

Impulse includes a reversal’s full signed velocity changeImpulse includes a reversal’s full signed velocity changem=0.20 kg; vi=+10 m/s; vf=−5 m/sΔp=m(vf−vi)=0.20(−5−10)=−3.0 N·s
Original signed reversal example; force direction is negative along the chosen axis.

A longer stopping time can reduce average force for the same momentum change. Padding or a crumple zone increases interaction time and distance; it does not remove the need for momentum change. Peak force can differ greatly from average force, so an average-force calculation does not describe every instant.

A longer stop can reduce average force for a fixed changeA longer stop can reduce average force for a fixed changeForce magnitude (N)010020030000.010.020.030.04Time (s)Short pulse:3 N·sLong pulse:3 N·s
Calculated equal-impulse pulses, not medical or crash-test evidence.

Momentum conservation applies to a system when net external impulse is negligible during the interval. Internal forces cancel between interacting bodies by Newton’s third law. Individual object momentum changes while total system momentum can stay the same. Define the system before claiming conservation.

Kinetic energy need not be conserved in an inelastic collision; some energy becomes deformation, vibration, or thermal energy. Total energy remains accounted for in an appropriately inclusive system. In an elastic ideal collision, both total momentum and total kinetic energy are conserved.

Use supplied force-time data or gentle tabletop models. Collision experiments with people, fast projectiles, or fragile objects are unnecessary. Evaluate whether friction or an external support had enough impulse to undermine the isolated-system approximation.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Relate impulse to a momentum change

Impulse J is the time integral of force and equals momentum change. A 0.20 kg ball reversing from+10 to−5 m/s changes momentum by−3.0 kg·m/s, equivalently−3.0 N·s. Subtract signed velocities before multiplying by mass. Using the difference of speed magnitudes would miss part of the reversal. The momentum theorem concerns net force on the selected object; one sensor’s force may not be the entire net force.

Impulse includes a reversal’s full signed velocity changeImpulse includes a reversal’s full signed velocity changem=0.20 kg; vi=+10 m/s; vf=−5 m/sΔp=m(vf−vi)=0.20(−5−10)=−3.0 N·s
Original signed reversal example; force direction is negative along the chosen axis.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What is the average net force if this change takes 0.020 s?

Visual reasoning — after your attempt

Favg=Δp/Δt=−3.0/0.020=−150 N.

Impulse is the time integral of net external force and equals the momentum change of the selected system. For a constant or average net force, J=FavgΔt. It is a vector account, so reversal can produce a larger change than merely stopping from the same initial speed.

A ball rebounding at the original speed changes momentum direction and has a change magnitude twice its initial momentum.

Pause and explain

Why can a rebound have greater impulse than stopping?

Check your reasoning after answering

The momentum must first decrease to zero and then change to the opposite direction, giving a larger vector difference.

Part 2 · Read a force-time trace using area

A force–time trace shows how force varies during contact. If its direction is fixed, the signed area gives impulse; the displayed positive triangle has area 3.0 N·s and average force 150 N over 0.020 s. Peak force 300 N is twice that average for this particular shape. A different pulse with the same area can have a different peak, so neither peak alone nor duration alone determines the momentum change.

Force–time area determines the impulse of a pulseForce–time area determines the impulse of a pulseForce magnitude for one-direction pulse (N)010020030000.010.02Contact time (s)Triangle area=½×0.020×300=3.0 N·s.
Calculated illustrative triangular pulse; positive force direction is chosen independently of the preceding reversal example.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What constant force over 0.040 s has the same impulse?

Visual reasoning — after your attempt

75 N gives 75×0.040=3.0 N·s in the same direction.

Signed area under a force-time graph gives impulse. The largest recorded force is a peak, not necessarily the average. A triangular pulse and rectangular pulse with the same peak and duration can have different areas. Include the full pulse and a suitable baseline.

A triangular 100 N pulse lasting 0.020 s has impulse 1.0 N·s, compared with 2.0 N·s for a rectangular pulse.

Pause and explain

Does equal peak force imply equal impulse?

Check your reasoning after answering

No. Pulse duration and shape matter because impulse depends on the complete force-time area.

Part 3 · Separate average-force reduction from guarantees

For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.

A longer stop can reduce average force for a fixed changeA longer stop can reduce average force for a fixed changeForce magnitude (N)010020030000.010.020.030.04Time (s)Short pulse:3 N·sLong pulse:3 N·s
Calculated equal-impulse pulses, not medical or crash-test evidence.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why must rebound be compared when testing two cushions?

Visual reasoning — after your attempt

Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.

For the same momentum change, extending stopping time reduces average force in the simple model. Peak force and forces on particular tissues or parts can still depend on the pulse shape and contact geometry. An average-force calculation alone cannot certify protective performance.

A longer stopping interval can explain a design mechanism, but comparable force traces and repeated testing are needed to evaluate the device.

Pause and explain

What does J/Δt establish?

Check your reasoning after answering

It gives the average net force for that interval. It does not uniquely determine peak force or local stresses.

Part 4 · Check system boundaries and sensor meaning

A sensor records a particular contact force, with finite calibration, sampling and response time. The momentum theorem uses net external force on the object over the same interval. If weight matters along the measured component, include its impulse rather than silently equating contact force with net force. Consistent sign conventions, time alignment and measured velocities are essential when comparing a force-trace area with mΔv. Disagreement can reveal a boundary or measurement problem.

Match sensor force with the selected momentum accountMatch sensor force with the selected momentum accountRecordWhat it measuresNeeded checkContact sensorOne interaction forceOther forces / calibrationMotion recordVelocity before/afterFrame and time intervalComparisonArea versus mΔvSame object and componentA matching area supports an account within uncertainty.
Original evidence-comparison checklist for force and motion records.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

When might weight’s impulse be negligible?

Visual reasoning — after your attempt

For a sufficiently short interval or a horizontal component with no gravity projection, justified against the size of other impulses.

A sensor records one interaction, while momentum change depends on net external force. Weight or other contacts may need inclusion over the interval. Calibration, sampling rate, timing, and rebound velocity affect whether a recorded pulse matches the momentum account.

For a short horizontal cart impact, other horizontal impulses may be negligible. For a longer vertical event, gravity may require explicit inclusion.

Pause and explain

When can one sensor force be treated as the net force?

Check your reasoning after answering

Only when other relevant force contributions are negligible or already accounted for in the stated direction and interval.

Apply the whole unit · Compare rebound and stop intervals

Illustrative 0.200 kg ball has incoming velocity +10.0 m/s and outgoing velocity −5.00 m/s. Compare contact intervals 0.0100 s and 0.0300 s; neglect other impulse along the axis.

Your task

Find momentum change and average contact force for each interval. Explain what the comparison establishes and what it cannot establish about peak force or a protective design.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Δp=0.200×(−5−10)=−3.00 kg·m/s.

  2. For 0.0100 s, Favg=−300 N; for 0.0300 s, Favg=−100 N.

  3. The impulse stays −3.00 N·s despite the different average-force magnitudes.

The longer interval lowers average force for the same specified momentum change. A peak force cannot be reconstructed from average and duration alone, and real designs may also alter rebound velocity. Compare calibrated full traces and consistent initial conditions before drawing a design conclusion.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask the student to subtract signed velocities and distinguish average from peak. Avoid presenting this supplied calculation as an injury or equipment-safety rating. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Equal impulse, different force and duration20 N × 0.10 s10 N × 0.20 sBoth areas:2 N·sTime (s) →F(N)
Two rectangular force-time pulses have equal 2 N·s area. A 20 N pulse over 0.10 s and 10 N pulse over 0.20 s give the same momentum change.

The key idea

Impulse is signed momentum change; the relevant system boundary determines conservation.

Worked examples

Example 1

A .20 kg ball goes from +6 to −4 m/s. Find impulse.
  1. Δp=.20(−4−6).
  2. Track the negative direction.
Answer: −2.0N·s

Example 2 · Worked application

A 0.20 kg ball's velocity changes from +10 to −5 m/s in 0.10 s. Find impulse and average net force.
  1. Δp=m(v_f−v_i)=0.20(−5−10)=−3.0 kg·m/s.
  2. Impulse=Δp=−3.0 N·s.
  3. F_avg=Δp/Δt=−3.0/0.10=−30 N.
Answer: Impulse −3.0 N·s; average net force −30 N.
Check the reasoning

A reversal requires signed velocities, not just a speed difference.

Example 3

A 2N·s impulse acts over .010 s versus .100 s. Compare average forces.
  1. Favg=J/Δt.
  2. 2/.010=200N and 2/.100=20N.
Answer: A tenfold longer time gives one-tenth the average force.

Example 4 · Find and fix the mistake

Context: A 0.20 kg ball's velocity changes from +10 to −5 m/s in 0.10 s. Find impulse and average net force. A student claims: “A longer stopping time always increases average stopping force for the same momentum change.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: A reversal requires signed velocities, not just a speed difference.
  2. F_avg=Δp/Δt; increasing time reduces average magnitude for fixed Δp.
Answer: F_avg=Δp/Δt; increasing time reduces average magnitude for fixed Δp.
Check the reasoning

A reversal requires signed velocities, not just a speed difference.

Example 5 · Reason from the visual model

A longer stop can reduce average force for a fixed changeA longer stop can reduce average force for a fixed changeForce magnitude (N)010020030000.010.020.030.04Time (s)Short pulse:3 N·sLong pulse:3 N·s
Calculated equal-impulse pulses, not medical or crash-test evidence.
Why must rebound be compared when testing two cushions?
  1. For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.
  2. Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
Answer: Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
Check the reasoning

Calculated equal-impulse pulses, not medical or crash-test evidence.

Avoid these mistakes

Check your understanding

Why distinguish average from peak collision force?

Discuss the answer

The force varies through the collision; impulse fixes its area, not a single maximum value.

Return to the phenomenon

Return to the initial case. How can cushioning change force without eliminating the stopping impulse? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Equal impulse permits different average and peak forces; rebound and actual pulse shapes must also be checked.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. How can cushioning change force without eliminating the stopping impulse? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Equal impulse permits different average and peak forces; rebound and actual pulse shapes must also be checked.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A 0.50 kg object slows from 6 m/s to 0 over 0.15 s. Find impulse and average net force. If it rebounds at−2 m/s instead, find the new impulse.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N. With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.

Words to know

impulse
Time integral of force, equal to momentum change.
external impulse
Impulse from forces exerted by objects outside the chosen system.

In the real world

Use the chapter investigation to connect this unit to measured evidence and a practical decision.

Exit check

Why distinguish average from peak collision force?

Check your explanation after answering

The force varies through the collision; impulse fixes its area, not a single maximum value.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A 25 N force pushes a cart 8.0 m in the direction of the force. How much work is done?
Answer and explanation

200 J

W = Fd = 25 × 8.0 = 200 J.

Problem 2

A crane lifts a 700 N crate 3.0 m upward in 5.0 s. Find the work done and the power output.
Answer and explanation

Work = 2,100 J; power = 420 W

Work against gravity: W = Fd = 700 × 3.0 = 2,100 J. Power: P = W/t = 2,100 ÷ 5.0 = 420 W.

Problem 3

A car cruises at a constant 30 m/s while its engine supplies 12,000 W of power to overcome resistance. Using P = Fv, find the total resistive force.
Answer and explanation

400 N

P = Fv, so F = P/v = 12,000 ÷ 30 = 400 N. (At constant speed the driving force equals the resistive force.)

Problem 4

Find the kinetic energy of a 5.0 kg object moving at 4.0 m/s.
Answer and explanation

40 J

KE = ½mv² = ½ × 5.0 × 4.0² = ½ × 5.0 × 16 = 40 J.

Problem 5

A spring with spring constant k = 300 N/m is compressed 0.10 m. How much elastic potential energy is stored?
Answer and explanation

1.5 J

PE = ½kx² = ½ × 300 × 0.10² = ½ × 300 × 0.01 = 1.5 J.

Problem 6

A 500 kg roller-coaster car starts from rest at the top of a 40 m hill. Ignoring friction, find (a) its gravitational potential energy at the top and (b) its speed at the bottom. (g = 9.8 m/s²)
Answer and explanation

PE = 196,000 J; speed = 28 m/s

PE = mgh = 500 × 9.8 × 40 = 196,000 J. All of it becomes kinetic energy: 196,000 = ½ × 500 × v² = 250v², so v² = 784 and v = 28 m/s.

Problem 7

A device is supplied 1,200 J of energy and produces 900 J of useful output. Find its efficiency and the energy lost as waste heat.
Answer and explanation

75% efficient; 300 J lost

Efficiency = (900 ÷ 1,200) × 100% = 75%. Energy lost = 1,200 − 900 = 300 J, dissipated as heat.

Problem 8

The net force on a 5.0 kg cart, initially at rest, does 250 J of work on it. Using the work-energy theorem, find its final speed.
Answer and explanation

10 m/s

W_net = ΔKE, so 250 = ½ × 5.0 × v² = 2.5v². Then v² = 100 and v = 10 m/s.

Problem 9

A pump lifts 200 kg of water to a tank 15 m higher, using 40,000 J of electrical energy. Find (a) the useful work done on the water and (b) the pump's efficiency. (g = 9.8 m/s²)
Answer and explanation

Useful work = 29,400 J; efficiency = 73.5%

Useful work = gain in gravitational PE = mgh = 200 × 9.8 × 15 = 29,400 J. Efficiency = (29,400 ÷ 40,000) × 100% = 73.5%.

Problem 10

Find the momentum of a 2.0 kg ball moving at 8.0 m/s.
Answer and explanation

16 kg·m/s

p = mv = 2.0 × 8.0 = 16 kg·m/s.

Problem 11

A 1.0 kg ball moving at 6.0 m/s strikes an identical 1.0 kg ball at rest in a head-on elastic collision. Find the velocity of each ball afterward.
Answer and explanation

The incoming ball stops (0 m/s); the struck ball moves off at 6.0 m/s.

For a head-on elastic collision between equal masses, the objects exchange velocities. Check: momentum 1.0 × 6.0 = 1.0 × 6.0 ✓, and kinetic energy ½ × 1.0 × 6.0² = ½ × 1.0 × 6.0² ✓. This is exactly the behavior seen in a Newton's cradle.

Problem 12

A 2.0 kg cart moving at 5.0 m/s catches up to a 3.0 kg cart moving at 1.0 m/s in the same direction and couples to it. Find their common velocity.
Answer and explanation

2.6 m/s

Conservation of momentum: 2.0 × 5.0 + 3.0 × 1.0 = (2.0 + 3.0)v′, so 10 + 3 = 5.0v′, giving 13 = 5.0v′ and v′ = 2.6 m/s.

Problem 13

A 0.20 kg ball hits a wall at 5.0 m/s and bounces straight back at 5.0 m/s. Find the magnitude of the impulse the wall gives the ball.
Answer and explanation

2.0 kg·m/s

Taking the outgoing direction as positive, Δp = m(v_f − v_i) = 0.20 × (5.0 − (−5.0)) = 0.20 × 10 = 2.0 kg·m/s. (Bouncing back doubles the impulse compared with simply stopping.)

Problem 14

A10N parallel force moves a box 3 m in 2 s. Find work and average power.
Answer and explanation

W=30 J; P=W/t=15W, if that force is constant.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 15

Chapter review: A motor raises a 20 kg load by 3 m in 6 s at 70% efficiency. Find useful work, output power and input energy.
Answer and explanation

Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total252 J under the stated boundary.

Useful work=mgh=588 J.

Output power=98 W.

Input energy=588/0.70=840 J; other transfers/storage total252 J under the stated boundary.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 16

A2 kg mass at 3 m has what U usingg=9.8 andU=0 atground?
Answer and explanation

U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 17

Chapter review: A1.5 kg cart has speed4 m/s. It compresses a loss-free spring with k=600 N/m on a level track. Find initial K and maximum compression.
Answer and explanation

K=½×1.5×16=12 J. Set12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.

K=½×1.5×16=12 J.

Set12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 18

A cart loses 20 J mechanical energy due to friction. Where is it?
Answer and explanation

Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 19

Chapter review: A system begins with200 J mechanical energy. It ends with120 J mechanical,50 J thermal increase and20 J transferred outward as sound/heat. Find the unaccounted energy and a useful next check.
Answer and explanation

The accounted final/output total is190 J, leaving10 J. Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.

The accounted final/output total is190 J, leaving10 J.

Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 20

2 kg at 3 m/s sticks to 1 kg atrest. Find final velocity.
Answer and explanation

Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 21

Chapter review: A3 kg cart at4 m/s sticks to a1 kg cart initially at rest. Neglect external impulse. Find final velocity and kinetic-energy decrease.
Answer and explanation

Initial momentum=12 kg·m/s, so vf=3 m/s. Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.

Initial momentum=12 kg·m/s, so vf=3 m/s.

Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 22

A triangular force pulse has base .020 s and height 100N. Find impulse.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

1 N·s

.5×.020×100=1.

Problem 23

Why define the system for collision conservation?
Answer and explanation

External forces can transfer momentum into or out of the selected system.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 24

Chapter review: A0.50 kg object slows from6 m/s to0 over0.15 s. Find impulse and average net force. If it rebounds at−2 m/s instead, find the new impulse.
Answer and explanation

Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N. With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.

Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N.

With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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