Motion — Print · Physical Science: Matter, Forces & Energy, Grade 8

Describe movement with measurements and graphs, then use Newton’s laws to explain changes.

Printable chapter lessons and review.

Speed, Velocity & Acceleration

Learning goals

Readings for this unit

Before this unit

Divide distance by time and include direction when needed.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in speed, velocity & acceleration?

Illustrative evidence case. An illustrative walk covers 30 m east and then 10 m west in 20 s. The starting position is the chosen origin, with east positive.

Distance and displacement differDistance and displacement differ0 m: start20 m: finish30 m: turn30 m east10 m westDisplacement: 20 m eastIllustrative model • use the labels and stated assumptions
The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Notice and wonder

Over 20 s, what are average speed and velocity? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Find distance, displacement, average speed and average velocity. Explain why the two rates differ. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

speed
How fast an object moves, found by dividing distance traveled by the time it took.
velocity
Speed in a specific direction, like 60 km/h north.
acceleration
How quickly velocity changes over time, found by dividing the change in velocity by the time it took.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Speed answers the everyday question, 'how fast is something moving?' You calculate it by dividing the distance traveled by the time it took: speed = distance ÷ time. If a cyclist rides 24 km in 3 hours, her average speed is 24 ÷ 3 = 8 km/h. That word 'average' matters, because she probably sped up on the flats and slowed on the hills — average speed smooths all of that into a single number for the whole trip.

Velocity takes speed and adds one crucial extra piece of information: direction. '8 km/h' is a speed, but '8 km/h heading north' is a velocity. This might sound like a fussy distinction, but it's a big deal in the real world. Two cars can share the exact same speed yet have completely different velocities if they're pointed different ways — and that difference is everything when they're both approaching the same intersection. Anytime an object turns, its velocity is changing even if its speed stays constant.

Acceleration measures how quickly velocity is changing — how fast you're speeding up, slowing down, or turning. Its formula is acceleration = (final velocity − initial velocity) ÷ time. Because it describes a change in velocity spread over a stretch of time, acceleration carries units like m/s² (meters per second, per second). A car that goes from 10 m/s to 25 m/s in 5 seconds is accelerating at (25 − 10) ÷ 5 = 3 m/s².

Acceleration can be positive or negative once a direction has been chosen positive. Its sign gives direction, not whether speed increases. With right positive, a right-moving cart with negative acceleration initially slows, while a left-moving cart with negative acceleration speeds up. Compare velocity and acceleration directions. Changing direction around a curve also changes velocity even if speed stays constant.

Negative acceleration does not always mean slowingNegative acceleration does not always mean slowingRight is positive. Compare velocity AND acceleration.Rightward motion + rightward acceleration: speeds upRightward motion + leftward acceleration: slows downLeftward motion + leftward acceleration: speeds upBlue: velocity (m/s). Red: acceleration (m/s²). Separate arrow scales.
The arrows distinguish present motion from change in motion. When velocity and acceleration point in the same direction, speed increases. Opposite directions initially reduce speed. In the third row both signed quantities are negative, yet the object speeds up to the left. The sign depends on the chosen coordinate direction. This compares straight-line cases; turning also changes velocity even when speed stays constant.

Keeping these three ideas straight is the backbone of describing any kind of motion. Speed tells you the pace, velocity pins down the pace and the heading, and acceleration captures how that motion is changing from moment to moment. Engineers use exactly these quantities to time traffic lights, design safe highway on-ramps, and figure out how much room a car needs to stop — the same math a radar gun uses to clock a driver going over the limit.

Average speed uses total distance divided by elapsed time. Average velocity uses displacement divided by elapsed time and includes direction. A round trip can have a nonzero average speed but zero average velocity because the final position equals the starting position.

Distance and displacement differDistance and displacement differ0 m: start20 m: finish30 m: turn30 m east10 m westDisplacement: 20 m eastIllustrative model • use the labels and stated assumptions
The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Acceleration is change in velocity divided by elapsed time. Direction matters: a cart moving right and slowing has acceleration left. The sign of acceleration alone does not tell you whether speed increases; compare it with the velocity direction.

Keep units consistent before calculating. If position is in meters and time in seconds, speed is in m/s and acceleration in m/s². State an average when data cover an interval, since an average can hide different motions within it.

Build your understanding

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Choose a reference and describe the path

Motion is a change in position relative to a chosen reference. Distance measures total path length; displacement compares final and initial position and includes direction. A return trip can have substantial distance with zero displacement. State the reference and positive direction before using signed positions.

Distance and displacement differDistance and displacement differ0 m: start20 m: finish30 m: turn30 m east10 m westDisplacement: 20 m eastIllustrative model • use the labels and stated assumptions
The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Read the visual model

How can a trip have zero displacement without zero distance?

Visual reasoning — after your attempt

The final position can equal the initial position even after a nonzero path. Displacement uses endpoints; distance adds the traveled path.

A student walks 6 m east and 6 m west back to the start. Distance is 12 m; displacement is zero. Mark the route and endpoints separately so the calculation answers the intended question.

Pause and explain

How can a trip have zero displacement without zero distance?

Check your reasoning after answering

The final position can equal the initial position even after a nonzero path. Displacement uses endpoints; distance adds the traveled path.

Part 2 · Distinguish average speed from average velocity

Average speed is total distance divided by elapsed time. Average velocity is displacement divided by elapsed time and includes direction. Both describe an interval, not every instant within it. A stopwatch result alone cannot show whether the object moved steadily, stopped, or changed pace.

Distance against displacementSpeed needs a reference and an intervalstartendpath travelled — distancestraight line start to end — displacementAverage speed uses the path; average velocity uses the displacement and its direction
Two measurements of one journey that answer different questions. A runner finishing a lap has covered a distance and has zero displacement, which is why average velocity for the lap is zero and average speed is not.

Read the visual model

Does a 2 m/s average speed establish a constant 2 m/s at every moment?

Visual reasoning — after your attempt

No. Different speeds and stops can produce the same distance/time average. More detailed position or speed measurements are needed.

For the 12 m return trip lasting 6 s, average speed is 2 m/s and average velocity is zero. A zero average velocity does not imply the student stood still throughout the six seconds.

Pause and explain

Does a 2 m/s average speed establish a constant 2 m/s at every moment?

Check your reasoning after answering

No. Different speeds and stops can produce the same distance/time average. More detailed position or speed measurements are needed.

Part 3 · Recognize three ways velocity can change

Acceleration is change in velocity divided by time over a specified interval. It can describe speeding up, slowing down, or changing direction. Choose a sign convention; negative acceleration means the acceleration points in the negative direction, not automatically that speed is decreasing.

Negative acceleration does not always mean slowingNegative acceleration does not always mean slowingRight is positive. Compare velocity AND acceleration.Rightward motion + rightward acceleration: speeds upRightward motion + leftward acceleration: slows downLeftward motion + leftward acceleration: speeds upBlue: velocity (m/s). Red: acceleration (m/s²). Separate arrow scales.
The arrows distinguish present motion from change in motion. When velocity and acceleration point in the same direction, speed increases. Opposite directions initially reduce speed. In the third row both signed quantities are negative, yet the object speeds up to the left. The sign depends on the chosen coordinate direction. This compares straight-line cases; turning also changes velocity even when speed stays constant.

Read the visual model

Can motion at constant speed still involve acceleration?

Visual reasoning — after your attempt

Yes. Changing direction changes velocity, so curved motion can be accelerated even when the speed remains constant.

A car traveling around a bend at constant speed still changes velocity because direction changes. A right-moving cart with leftward acceleration initially slows; a left-moving cart with leftward acceleration speeds up. Label both arrows.

The blue and red arrows carry different units: velocity in m/s and acceleration in m/s². Compare directions rather than adding their lengths. A negative velocity means motion left in this reference; a negative acceleration means a change toward the left. If both are negative, the object can move left faster. A cart moving right with leftward acceleration may slow to rest and later move left; the description depends on the interval.

Pause and explain

Can motion at constant speed still involve acceleration?

Check your reasoning after answering

Yes. Changing direction changes velocity, so curved motion can be accelerated even when the speed remains constant.

Part 4 · Calculate an interval and check its meaning

Use average acceleration=(final velocity−initial velocity)/time. Include direction or signs and units m/s². The unit describes change in m/s per second, not a traveled distance. An average calculation does not establish constant acceleration during the entire interval.

Read acceleration from a velocity–time graphRead acceleration from a velocity–time graphRightward velocity (m/s)024680123Time (s)Run: 3 sRise: 6 m/sSlope = rise ÷ run = 6 m/s ÷ 3 s = 2 m/s².
For this modeled constant-acceleration cart, velocity rises from 2 to 8 m/s in 3 s. The graph slope is (8−2)/(3−0) = 2 m/s². The units mean velocity changes by 2 m/s each second. A velocity–time graph uses slope for acceleration, whereas a position–time graph uses slope for velocity. If only the endpoints were measured, this slope would establish an average rather than prove the straight line at every instant.

Read the visual model

Why is m/s² appropriate for acceleration?

Visual reasoning — after your attempt

Velocity changes in m/s and is divided by seconds. The resulting unit describes how much velocity changes per second.

A cart changes from 2 m/s right to 8 m/s right in 3 s. With right positive, average acceleration is (8−2)/3=2 m/s² right. Confirm that the velocity increased by 6 m/s over the stated three seconds.

To read the slope, choose two points on the velocity–time graph, subtract their velocities, then divide by their time difference. Here the rise is 6 m/s and the run is 3 s. The result 2 m/s² describes a rate of velocity change. The graph is a straight line only because constant acceleration was assumed in this illustration; endpoint records alone establish an average.

Pause and explain

Why is m/s² appropriate for acceleration?

Check your reasoning after answering

Velocity changes in m/s and is divided by seconds. The resulting unit describes how much velocity changes per second.

Apply the whole unit · Analyze a journey and a speeding cart

Illustrative journey: 8 m east then 2 m west in 5 s. Separate cart record: velocity changes from 1 m/s east to 7 m/s east in 3 s. Take east as positive.

Your task

Find journey distance, displacement, average speed, and average velocity; then find the cart’s average acceleration. Draw direction arrows and explain why the interval averages do not establish constant motion.

What your response should show

  • Use the supplied observations and measurements accurately; include units where needed.
  • Explain the science behind your answer, rather than giving only a number or label.
  • Show a labeled model, calculation, or evidence comparison as requested.
  • State an assumption or limitation and suggest a useful next measurement.
Response guidance — after your attempt

Distance=10 m; displacement=6 m east; average speed=2 m/s; average velocity=1.2 m/s east. Cart acceleration=(7−1)/3=2 m/s² east. More detailed readings are needed to determine within-interval changes. Keep journey and cart evidence separate; they are different records.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Begin with a drawn route before equations. Ask distance or displacement on every division-by-time task. Use a turning example to distinguish speed from velocity. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

Speed = distance ÷ time; velocity adds direction; acceleration = change in velocity ÷ time.

Speed

speed = distance ÷ time

distance in meters or km; time in seconds, minutes, or hours; speed in matching units like m/s or km/h.

Acceleration

a = (v final − v initial) ÷ t

v final and v initial are velocities in m/s; t is time in seconds; a comes out in m/s².

Worked examples

Example 1

A cyclist travels 24 km in 3 hours. What is her average speed?
  1. Use speed = distance ÷ time.
  2. 24 ÷ 3 = 8.
Answer: 8 km/h

Example 2 · Worked application

A runner travels 120 m east, then 120 m west in 60 s. Compare average speed and average velocity.
  1. Total distance=240 m, so average speed=240/60=4 m/s.
  2. Final position equals initial position, so displacement=0 m.
  3. Average velocity=displacement/time=0 m/s.
Answer: Average speed 4 m/s; average velocity 0 m/s.
Check the reasoning

Distance counts the whole path; displacement compares endpoints.

Example 3

A car speeds up from 10 m/s to 25 m/s in 5 seconds. What is its acceleration?
  1. Use acceleration = (final velocity − initial velocity) ÷ time.
  2. (25 − 10) ÷ 5 = 15 ÷ 5.
  3. = 3.
Answer: 3 m/s²

Example 4

A student walks 30 m east then 30 m west in 40 s. Find average speed and average velocity.
  1. Total distance=60 m; speed=60/40=1.5 m/s.
  2. Displacement=0 m; average velocity=0/40=0 m/s.
Answer: Average speed 1.5 m/s; average velocity 0 m/s.

Example 5 · Find and fix the mistake

Context: A runner travels 120 m east, then 120 m west in 60 s. Compare average speed and average velocity. A student claims: “Negative acceleration always means slowing down.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Distance counts the whole path; displacement compares endpoints.
  2. The sign depends on the chosen axis. An object moving in the negative direction speeds up if acceleration is also negative.
Answer: The sign depends on the chosen axis. An object moving in the negative direction speeds up if acceleration is also negative.
Check the reasoning

Distance counts the whole path; displacement compares endpoints.

Example 6 · Interpret the visual model

Distance and displacement differDistance and displacement differ0 m: start20 m: finish30 m: turn30 m east10 m westDisplacement: 20 m eastIllustrative model • use the labels and stated assumptions
The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.
How can a trip have zero displacement without zero distance?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A student walks 6 m east and 6 m west back to the start. Distance is 12 m; displacement is zero. Mark the route and endpoints separately so the calculation answers the intended question.
  3. The final position can equal the initial position even after a nonzero path. Displacement uses endpoints; distance adds the traveled path.
Answer: The final position can equal the initial position even after a nonzero path. Displacement uses endpoints; distance adds the traveled path.
Check the reasoning

Keep the model’s stated limits explicit: The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Avoid these mistakes

Check your understanding

A student walks 30 m east then 30 m west in 40 s. Find average speed and average velocity.

Discuss the answer

Average speed 1.5 m/s; average velocity 0 m/s.

Can an object accelerate while its speed remains constant?

Discuss the answer

Average speed uses total distance divided by elapsed time. Average velocity uses displacement divided by elapsed time and includes direction. A round trip can have a nonzero average speed but zero average velocity because the final position equals the starting position. Acceleration is change in velocity divided by elapsed time. Direction matters: a cart moving right and slowing has acceleration left. The sign of acceleration alone does not tell you whether speed increases; compare it with the velocity direction.

Return to the phenomenon

An illustrative walk covers 30 m east and then 10 m west in 20 s. The starting position is the chosen origin, with east positive. Find distance, displacement, average speed and average velocity. Explain why the two rates differ. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

Distance is 40 m; displacement is +20 m east. Average speed is 40/20=2 m/s; average velocity is +20/20=+1 m/s east. Distance counts the full path, while displacement compares final and initial positions with direction. These averages do not establish a constant speed at every instant.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

An illustrative walk covers 30 m east and then 10 m west in 20 s. The starting position is the chosen origin, with east positive. Find distance, displacement, average speed and average velocity. Explain why the two rates differ. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

Distance is 40 m; displacement is +20 m east. Average speed is 40/20=2 m/s; average velocity is +20/20=+1 m/s east. Distance counts the full path, while displacement compares final and initial positions with direction. These averages do not establish a constant speed at every instant.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Over 20 s, what are average speed and velocity? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Response guidance — discuss after your attempt

Average speed is 40/20=2 m/s; average velocity is 20/20=1 m/s east. The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

Words to know

speed
How fast an object moves, found by dividing distance traveled by the time it took.
velocity
Speed in a specific direction, like 60 km/h north.
acceleration
How quickly velocity changes over time, found by dividing the change in velocity by the time it took.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Average speed vs. the number on your speedometer

The speed = distance ÷ time formula gives you average speed over a whole trip, smoothing out every stoplight and highway stretch into one number. But your car's speedometer shows instantaneous speed — how fast you're going at that exact moment, which can be very different from the average. A trip with an average speed of 40 km/h might have included stretches at 0 km/h (stopped at lights) and 70 km/h (open highway); average speed just cannot capture that up-and-down detail, which is exactly why scientists use graphs to show motion changing moment to moment.

Reading motion straight off a graph

If you plot an object's distance against time, the slope (steepness) of that line is literally its speed — a steeper line means faster motion, and a flat line means the object has stopped. Plot velocity against time instead, and the slope becomes acceleration: a rising line means speeding up, a falling line means slowing down, and a flat line means constant velocity (zero acceleration). This is one of the most useful shortcuts in physics: instead of doing arithmetic, you can often just look at how steep — and which way — a graph tilts.

Galileo's ramps: clocking acceleration before stopwatches existed

In the early 1600s, Galileo Galilei wanted to study how falling objects accelerate, but free fall happens too fast to time accurately with the water clocks and pulse-counting available then. His clever workaround: roll a ball down a gently sloped ramp, which 'dilutes' gravity's pull and slows everything down enough to measure. By timing balls rolling different distances, Galileo showed that acceleration due to gravity is constant and doesn't depend on an object's weight — centuries before anyone had a precise clock, let alone a computer.

Acceleration is sneakier than 'speeding up' — turning counts too

Because velocity includes direction, any change in direction is technically acceleration — even if speed never changes at all. A car rounding a curve at a perfectly steady 50 km/h is still accelerating, because its direction is constantly changing; this is called centripetal acceleration. Pilots and astronauts describe strong accelerations in multiples of Earth's gravity, or 'g-forces' — a fighter jet pulling a tight turn might subject its pilot to 6–9 g's, meaning their body effectively feels 6 to 9 times heavier than normal, even though their speed might barely change.

In the real world

Traffic radar guns calculate a car's speed to catch drivers going faster than the posted limit.

Exit check

Over 20 s, what are average speed and velocity? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

Average speed is 40/20=2 m/s; average velocity is 20/20=1 m/s east. The path covers 30 m east and 10 m west, for 40 m distance and 20 m east displacement. Arrows separate the journey from the start-to-finish comparison. The schematic horizontal positions are not a precision scale and do not establish a changing speed without elapsed-time data.

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

Reading Motion Graphs

Learning goals

Readings for this unit

Before this unit

Read axes, intervals and units on a graph.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in reading motion graphs?

Illustrative evidence case. Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points.

Read the position–time axesRead the position–time axesTime (s)Position (m)001.52344.5668Illustrative model • use the labels and stated assumptions
Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Notice and wonder

Does the final downward line prove decreasing speed? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

position
Location relative to a chosen origin.
slope
Change in vertical value divided by change in horizontal value.
average velocity
Displacement divided by elapsed time, including direction.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

A position–time graph shows where an object is relative to a chosen origin. Put time on the horizontal axis and position on the vertical axis. The slope tells average velocity over an interval: change in position divided by elapsed time. A graph shows a relationship, not necessarily the physical path the object traveled.

Read the position–time axesRead the position–time axesTime (s)Position (m)001.52344.5668Illustrative model • use the labels and stated assumptions
Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

A horizontal segment means position stays the same: the object is at rest during that interval. A straight rising segment means constant positive velocity. A falling segment means motion in the negative direction, not automatically slowing down. Greater steepness means greater speed; the sign of the slope indicates direction.

Choose two data points within the interval you are studying. Subtract both the positions and times. On a curved position–time graph the velocity is changing. Do not confuse position–time with speed–time: a horizontal speed–time line means constant speed, which may be nonzero. Always read the axes first.

Velocity–time area: displacement and distance differVelocity–time area: displacement and distance differSigned velocity (m/s)-202401234Time (s)+6 m−4 mDisplacement: +6 − 4 = +2 m. Distance: 6 + 4 = 10 m.
Positive is right. The object travels right at 3 m/s for 2 s, then left at 2 m/s for 2 s. A rectangle’s time width times velocity height gives displacement: +6 m followed by −4 m. Signed areas add to +2 m displacement; their magnitudes add to 10 m distance. The instantaneous switch is an idealization; an actual reversal takes time. This velocity graph is a motion record, not a map of the object’s path.

The slope of a position–time graph represents velocity. A horizontal segment means unchanged position, and a straight rising segment means constant positive velocity. A steep segment indicates a larger magnitude of velocity only when the axes and scales are comparable.

On a velocity–time graph, slope represents acceleration. A horizontal segment can mean moving at constant velocity, not standing still. The area between the graph and time axis represents displacement; a section below the axis contributes negative displacement.

Always identify which quantity is plotted before interpreting the shape. Use two points separated by a useful interval to calculate slope as vertical change divided by horizontal change. A line on a graph is not a picture of a hill or the actual path of the object.

Build your understanding

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Read axes before interpreting a line

A position-time graph shows position on the vertical axis and time on the horizontal axis. Its line is not a drawing of a hill or the object’s route. Position is measured relative to a stated reference, and negative position can simply mean the other side of that reference.

Read the position–time axesRead the position–time axesTime (s)Position (m)001.52344.5668Illustrative model • use the labels and stated assumptions
Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Read the visual model

What does a horizontal position-time section show?

Visual reasoning — after your attempt

Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

A graph rising from position 0 m to 6 m over 3 s describes increasing position. It does not mean an object physically climbed a diagonal track. Write a sentence explaining each axis before describing motion.

Pause and explain

What does a horizontal position-time section show?

Check your reasoning after answering

Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

Part 2 · Use position-time slope to find velocity

Slope compares change in position with change in time. A straight segment has constant velocity in the model. A steeper magnitude means greater speed; a negative slope indicates motion in the negative direction. Two different starting positions can have the same slope and therefore the same velocity.

Position rises from 0 to 4 meters in 2 seconds, stays at 4 through 4 seconds, then returns to zero at 6 seconds.Time (s)Position (m)024604+2 m/sRest−2 m/s
Position-time graph for the supplied data, with straight segments assuming constant interval velocity. It is a graph of measurements and a model between them, not the shape of the route.

Read the visual model

Why must both coordinate changes be used for an interval slope?

Visual reasoning — after your attempt

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

A line changing from 2 m to 8 m between 1 s and 4 s has slope (8−2)/(4−1)=2 m/s. Using total position divided by total clock time would answer a different question unless both starting values are zero.

Pause and explain

Why must both coordinate changes be used for an interval slope?

Check your reasoning after answering

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

Part 3 · Read velocity-time graphs differently

On a velocity-time graph, a horizontal line means constant velocity, not necessarily rest. Slope represents acceleration. The signed area under the graph gives displacement; distance requires adding the magnitudes of traveled portions. A negative velocity means motion in the negative direction under the chosen convention.

Velocity–time area: displacement and distance differVelocity–time area: displacement and distance differSigned velocity (m/s)-202401234Time (s)+6 m−4 mDisplacement: +6 − 4 = +2 m. Distance: 6 + 4 = 10 m.
Positive is right. The object travels right at 3 m/s for 2 s, then left at 2 m/s for 2 s. A rectangle’s time width times velocity height gives displacement: +6 m followed by −4 m. Signed areas add to +2 m displacement; their magnitudes add to 10 m distance. The instantaneous switch is an idealization; an actual reversal takes time. This velocity graph is a motion record, not a map of the object’s path.

Read the visual model

Why is a flat velocity-time line not always a stopped object?

Visual reasoning — after your attempt

Its vertical value can be nonzero. A flat line at 3 m/s describes constant motion; only a line at zero velocity describes rest.

A velocity of −2 m/s maintained for 3 s gives displacement −6 m. The graph is flat but the object moves. Always name the graph type before applying a horizontal-line rule learned from position-time graphs.

For the supplied graph choose right positive. From 0–2 s, velocity is +3 m/s; the rectangular area is 2 s×3 m/s = +6 m. From 2–4 s, velocity is −2 m/s; its area is −4 m. Net displacement is +2 m, but total distance is 10 m. Area units are seconds multiplied by metres per second, leaving metres. The vertical switch is an idealized reversal; a real object needs time to reverse.

Pause and explain

Why is a flat velocity-time line not always a stopped object?

Check your reasoning after answering

Its vertical value can be nonzero. A flat line at 3 m/s describes constant motion; only a line at zero velocity describes rest.

Part 4 · Separate data from an interpolated model

A few measured points do not reveal every detail between readings. Joining points with straight lines is a modeling choice. Label measured values and explain any assumption of constant velocity between them. Avoid interpreting a smooth curve as more precise than the measurement supports.

Two endpoints do not reveal every momentTwo endpoints do not reveal every momentPosition (m)01234012Time (s)Straight: steady 2 m/sCurve: changing velocityTwo measured endpoints fit both models; intermediate readings test them.
The only supplied measurements are position 0 m at 0 s and position 4 m at 2 s (gold dots). The blue straight-line model has constant 2 m/s velocity. The red model, position = time² in these units, has increasing velocity and reaches 1 m at 1 s instead of 2 m. Both fit the endpoints and have the same average velocity. More measurements between the endpoints are needed to decide which account fits the actual motion. Intermediate curves are proposed models, not measured records.

Read the visual model

What would test a claim of constant velocity between two measurements?

Visual reasoning — after your attempt

Collect additional position-time readings within the interval and compare their slopes. Endpoint data alone establish an interval average, not every moment.

Positions measured at 0, 2, and 4 s can support average velocities for two intervals. If no intermediate readings exist, several different within-interval motions might fit the same points. More frequent measurements can test the assumed straight segments.

The gold endpoints in the new figure are the only actual supplied position records. The blue straight interpolation and red curved proposal both reach 4 m after 2 s, so they share average velocity 2 m/s. At 1 s they predict different positions. Measuring an intermediate position tests the model. Connecting records with a straight line should not silently turn missing data into proof of constant velocity.

Pause and explain

What would test a claim of constant velocity between two measurements?

Check your reasoning after answering

Collect additional position-time readings within the interval and compare their slopes. Endpoint data alone establish an interval average, not every moment.

Apply the whole unit · Make and interpret a motion graph

Illustrative positions: at 0, 2, 4, and 6 s a cart is at 0, 4, 4, and 0 m. Model each intervening interval as constant velocity, with positive position to the right.

Your task

Draw a labeled position-time graph, calculate velocity for each interval, and describe the motion in words. Find total distance and displacement, then identify how a new measurement could test the straight-segment assumption.

What your response should show

  • Use the supplied observations and measurements accurately; include units where needed.
  • Explain the science behind your answer, rather than giving only a number or label.
  • Show a labeled model, calculation, or evidence comparison as requested.
  • State an assumption or limitation and suggest a useful next measurement.
Response guidance — after your attempt

Velocities are +2, 0, and −2 m/s. The cart moves right 4 m, pauses, and returns left 4 m. Distance=8 m and displacement=0 m. Label time in seconds and position in meters. Intermediate readings would test constant velocity; the connected segments are a stated model rather than additional observations.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Use graph-reading sentences before calculations. Contrast a flat section on position-time and velocity-time axes. Ask the learner to separate measured dots from assumed connections. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

On a position–time graph, slope describes velocity; a horizontal segment means rest.

Worked examples

Example 1

An object is at 2 m at 1 s and 10 m at 5 s. Find its average velocity.
  1. Change in position = 10 − 2 = 8 m.
  2. Elapsed time = 5 − 1 = 4 s.
  3. Average velocity = 8 ÷ 4.
Answer: 2 m/s in the positive direction.

Example 2 · Worked application

Position is 0 m at 0 s, 8 m at 2 s and 8 m at 4 s. Interpret the two graph segments.
  1. From 0 to 2 s, position changes 8 m: slope=8/2=4 m/s.
  2. From 2 to 4 s, position stays 8 m: slope=0.
  3. Assuming straight segments, the object first moves at constant positive velocity, then stops.
Answer: 4 m/s for the first segment and 0 m/s for the second.
Check the reasoning

Read slope, not the vertical height, as velocity on a position–time graph.

Example 3

Position changes from 4 m at 2 s to 16 m at 6 s on a straight segment. Find its velocity.
  1. Vertical change=16−4=12 m; time change=6−2=4 s.
  2. Slope=12/4=3 m/s.
Answer: Constant velocity of +3 m/s over this segment.

Example 4 · Find and fix the mistake

Context: Position is 0 m at 0 s, 8 m at 2 s and 8 m at 4 s. Interpret the two graph segments. A student claims: “A horizontal position–time line means fast constant motion.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Read slope, not the vertical height, as velocity on a position–time graph.
  2. Horizontal means position is constant, so velocity is zero. A horizontal velocity–time line has a different meaning.
Answer: Horizontal means position is constant, so velocity is zero. A horizontal velocity–time line has a different meaning.
Check the reasoning

Read slope, not the vertical height, as velocity on a position–time graph.

Example 5 · Displacement from signed areas

For the supplied velocity–time graph, motion is +3 m/s for 2 s then −2 m/s for 2 s. Find displacement, distance and average velocity.
  1. First displacement = +3×2 = +6 m. Second displacement = −2×2 = −4 m.
  2. Displacement = +2 m; distance = 6+4 = 10 m.
  3. Average velocity = +2 m / 4 s = +0.5 m/s.
Answer: Displacement 2 m right; distance 10 m; average velocity 0.5 m/s right.
Check the reasoning

Average speed would be 10/4 = 2.5 m/s. It answers a different question.

Example 6 · Interpret the visual model

Read the position–time axesRead the position–time axesTime (s)Position (m)001.52344.5668Illustrative model • use the labels and stated assumptions
Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.
What does a horizontal position-time section show?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A graph rising from position 0 m to 6 m over 3 s describes increasing position. It does not mean an object physically climbed a diagonal track. Write a sentence explaining each axis before describing motion.
  3. Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.
Answer: Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.
Check the reasoning

Keep the model’s stated limits explicit: Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Avoid these mistakes

Check your understanding

Position changes from 4 m at 2 s to 16 m at 6 s on a straight segment. Find its velocity.

Discuss the answer

Constant velocity of +3 m/s over this segment.

How does a horizontal velocity–time line differ from a horizontal position–time line?

Discuss the answer

The slope of a position–time graph represents velocity. A horizontal segment means unchanged position, and a straight rising segment means constant positive velocity. A steep segment indicates a larger magnitude of velocity only when the axes and scales are comparable. On a velocity–time graph, slope represents acceleration. A horizontal segment can mean moving at constant velocity, not standing still. The area between the graph and time axis represents displacement; a section below the axis contributes negative displacement.

Return to the phenomenon

Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

The interval velocities are +3, 0 and −2 m/s. A level position graph means unchanged location; a level nonzero velocity graph means steady motion. The axes determine the interpretation. Straight segments describe the simplified interval model rather than measurements at every possible instant.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

The interval velocities are +3, 0 and −2 m/s. A level position graph means unchanged location; a level nonzero velocity graph means steady motion. The axes determine the interpretation. Straight segments describe the simplified interval model rather than measurements at every possible instant.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Does the final downward line prove decreasing speed? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Response guidance — discuss after your attempt

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude. Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Words to know

position
Location relative to a chosen origin.
slope
Change in vertical value divided by change in horizontal value.
average velocity
Displacement divided by elapsed time, including direction.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Exit check

Does the final downward line prove decreasing speed? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude. Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

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

Newton's Three Laws of Motion

Learning goals

Readings for this unit

Before this unit

Calculate net force and distinguish mass from weight.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in newton's three laws of motion?

Illustrative evidence case. Illustrative carts of 2 kg and 4 kg each experience a 6 N horizontal net force. Other conditions are comparable.

Equal net force, different accelerationEqual net force, different accelerationQuantity (m/s²)2 kg cart: a34 kg cart: a1.5Each cart has the same horizontal net force: 6 N.a = Fnet / m; compare mass alone.Illustrative model • use the labels and stated assumptions
For the same 6 N net force, doubling mass halves acceleration from 3 to 1.5 m/s². The bars represent rates of velocity change, not current velocity or distance. Equal net force is a condition of this comparison; interaction pairs would require named separate receivers.

Notice and wonder

What happens if mass and force both double? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Calculate acceleration magnitudes, identify an equal-and-opposite interaction pair and explain why zero net force does not require rest. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

inertia
The tendency of an object to resist a change in its motion.
Newton's first law
An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by an unbalanced force.
Newton's second law
Force equals mass times acceleration: F = ma.
Newton's third law
For every action force, there is an equal and opposite reaction force.
newton (N)
The unit used to measure force; 1 newton = 1 kg·m/s².
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Isaac Newton tied together how forces and motion relate in just three elegant laws, and together they explain almost everything you see moving around you. Newton's first law is the law of inertia: an object at rest stays at rest, and an object already moving keeps moving at the same speed and in the same straight-line direction, unless an unbalanced force acts on it. Inertia is simply an object's built-in resistance to any change in its motion, and the more mass something has, the more inertia it has — that's why it's easy to stop a rolling tennis ball but nearly impossible to stop a rolling truck.

You feel the first law every time you ride in a car. When the car suddenly brakes, your body lurches forward — not because something pushed you, but because your body was already moving and 'wants' to keep moving until a force stops it. That force is your seatbelt, which is exactly why it can save your life. The same law explains why a stack of coins stays put when you flick the bottom one out, and why you slide sideways when a car turns sharply.

This first law also corrects one of the most common misunderstandings in all of physics: the belief that keeping something moving requires a constant push. It seems obvious from daily life — stop pedaling and your bike coasts to a halt. But the truth is the opposite: a moving object only slows down because hidden forces like friction and air resistance are pushing back on it. Remove those forces, as in the near-frictionless emptiness of deep space, and an object will glide onward forever with no engine at all.

Zero net force allows rest OR constant velocityZero net force allows rest OR constant velocityNet force = 0: cart at rest0 s1 s2 sSame position each secondNet force = 0: cart at constant rightward velocity0 s1 s2 s
Equal-time snapshots can show no position change or equal changes in position. Both are consistent with zero net force in an inertial reference frame. The moving cart does not need a forward net force to keep its constant velocity. On a real surface, rolling resistance can slow it unless balanced. The rest snapshots overlap intentionally; no extra force called inertia should be drawn.

Newton's second law puts numbers to all of this with one compact equation: F = ma, meaning force equals mass times acceleration. It says two intuitive things at once. First, the harder you push something, the more it accelerates. Second, the more mass an object has, the harder it resists that acceleration — which is why the same kick sends a soccer ball flying but barely budges a bowling ball. Force is measured in a unit called the newton (N), where 1 newton is the force needed to accelerate 1 kilogram at 1 m/s².

Use the net force, not just the applied pushUse the net force, not just the applied push2 kg cartPush: 10 N →Resistance: 4 N ←Net force: 6 N →Acceleration = net force ÷ mass = 6 N ÷ 2 kg = 3 m/s² right.Acceleration direction is known; current velocity is not supplied.
Both red forces act on the cart and their lengths share one force scale. Subtract the opposing forces: 10−4 = 6 N right. For a 2 kg cart, acceleration is 6/2 = 3 m/s² right. The green arrow summarizes that net force, not a third force to add again. Vertical weight and support balance in the stated level-surface model and are omitted. A rightward acceleration alone does not tell you the cart’s current direction of travel.

Newton's third law delivers the famous line: for every action force, there is an equal and opposite reaction force — and, crucially, those two forces act on different objects. When a swimmer's hands push water backward, the water pushes the swimmer forward with equal strength. When you jump, your legs push down on the Earth and the Earth pushes back up on you. The forces always come in matched pairs, pointing in opposite directions.

This third law is the secret behind how a rocket can fly through the vacuum of space, where there's nothing to push against. The rocket doesn't push on the air — it pushes a huge amount of hot gas downward out of its engines, and by Newton's third law that gas pushes the rocket upward with equal force. From walking (your feet push the ground backward so it pushes you forward) to a squid jetting through the ocean, these three laws together let you predict how any push or pull will change the way things move.

Newton's first law describes motion when net force is zero. The second law connects net force to acceleration: Fnet=ma. The third law identifies paired forces between two interacting objects. Use the law that addresses the question rather than memorizing isolated slogans.

A force diagram belongs to one object. If a hand pushes a cart, include the hand's force on the cart, not the cart's force on the hand. The latter belongs on the hand's diagram. Pair forces have equal magnitude, but different masses can have different accelerations.

For a fixed net force, a larger mass has a smaller acceleration. For a fixed mass, doubling net force doubles acceleration. These are model predictions that can be tested with repeated measurements; friction and measurement uncertainty can alter an ideal classroom result.

Build your understanding

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Use inertia to explain persistence of motion

Newton’s first law describes constant velocity when the net external force is zero. Inertia is resistance to changes in velocity and increases with mass. A forward force is not required merely to maintain steady motion in the ideal case; everyday objects slow because resistance supplies an unbalanced force.

Zero net force allows rest OR constant velocityZero net force allows rest OR constant velocityNet force = 0: cart at rest0 s · 1 s · 2 sSame position each secondNet force = 0: cart at constant rightward velocity0 s1 s2 s
Equal-time snapshots can show no position change or equal changes in position. Both are consistent with zero net force in an inertial reference frame. The moving cart does not need a forward net force to keep its constant velocity. On a real surface, rolling resistance can slow it unless balanced. The rest snapshots overlap intentionally; no extra force called inertia should be drawn.

Read the visual model

Why is inertia not an extra force on a force diagram?

Visual reasoning — after your attempt

Inertia describes how mass resists acceleration. It is not an interaction from another object and should not be added as a force arrow.

A released toy cart slows due to interactions such as rolling resistance, not because motion uses up a stored force. A passenger’s motion changes when a seat or restraint exerts force; connect the law to interactions rather than calling inertia a pushing force.

Balanced and unbalanced forcesNet force decides the change in motionBalancedno change in velocityUnbalancedaccelerates rightOne forceaccelerates rightBalanced forces keep velocity constant — which includes staying still
Constant velocity and being at rest are the same case as far as the forces are concerned. Something moving steadily needs no net force, which is the part of the first law that contradicts intuition.

Read the equal-time snapshots. The rest case overlaps at one position; the constant-velocity case moves the same distance each second. Neither needs a net forward force. Inertia describes resistance to a velocity change; it is not another force acting forward. Identify real interactions such as rolling resistance before explaining why an ordinary cart slows.

Pause and explain

Why is inertia not an extra force on a force diagram?

Check your reasoning after answering

Inertia describes how mass resists acceleration. It is not an interaction from another object and should not be added as a force arrow.

Part 2 · Use the net force in the second law

For the introductory constant-mass model, net force=mass×acceleration. Acceleration points in the net-force direction. A larger applied push does not automatically give larger acceleration if opposing forces or mass also change. Find the net force on one chosen object before using the equation.

Use the net force, not just the applied pushUse the net force, not just the applied push2 kg cartPush: 10 N →Resistance: 4 N ←Net force: 6 N →Acceleration = net force ÷ mass = 6 N ÷ 2 kg = 3 m/s² right.Acceleration direction is known; current velocity is not supplied.
Both red forces act on the cart and their lengths share one force scale. Subtract the opposing forces: 10−4 = 6 N right. For a 2 kg cart, acceleration is 6/2 = 3 m/s² right. The green arrow summarizes that net force, not a third force to add again. Vertical weight and support balance in the stated level-surface model and are omitted. A rightward acceleration alone does not tell you the cart’s current direction of travel.

Read the visual model

Which force belongs in F=ma for this calculation?

Visual reasoning — after your attempt

Use the net external force on the chosen cart, not a single applied force when other forces also act.

A 2 kg cart pushed right with 10 N against 4 N resistance has net force 6 N right and acceleration 3 m/s² right. Dividing 10 N by mass would ignore resistance and overestimate acceleration.

Read the red arrows on the cart, subtract opposite forces and then use the resulting net force. The green arrow is their summary and must not be counted as an additional interaction. A rightward acceleration could mean a right-moving cart speeds up or a left-moving cart slows; no present velocity was supplied by the force diagram.

Pause and explain

Which force belongs in F=ma for this calculation?

Check your reasoning after answering

Use the net external force on the chosen cart, not a single applied force when other forces also act.

Part 3 · Place third-law pairs on different objects

When object A exerts a force on B, B exerts an equal opposite force on A. These two forces act on different objects, so they do not cancel within one object’s force diagram. Balanced forces on a single object are a different kind of pair.

A third-law pairThird-law pairs act on different objectsskater Askater BA pushes BB pushes AThese two never cancel, because they act on different objects.Only forces on the same object can be added to find a net force.
The commonest error in this topic is adding an action and reaction together and concluding nothing can ever accelerate. The pair acts on two different bodies, so neither appears in the other's net force.

Read the visual model

Why do equal action-reaction forces not prevent the cart accelerating?

Visual reasoning — after your attempt

One force acts on the cart and the other on the hand. They are not both forces on the cart, so they cannot be canceled in its net-force calculation.

A hand pushes a cart and the cart pushes the hand equally in the opposite direction. The cart can accelerate because its own net force need not be zero. The two objects can have different accelerations because their masses and other interactions differ.

OpenSciEd sketch of a wire coil facing a separate magnet across a gap; the diagram is a template for adding the two interaction-force arrows.
The OpenSciEd coil and magnet are two separate objects. Under a specified interaction their force pair is equal and opposite, although their masses and other forces can give different accelerations. Add named source-on-receiver arrows; do not infer motion merely from this uncompleted force template. Coil and magnet before force arrows are added · OpenSciEd · 8.3 Forces at a Distance, Lesson 2, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.

Pause and explain

Why do equal action-reaction forces not prevent the cart accelerating?

Check your reasoning after answering

One force acts on the cart and the other on the hand. They are not both forces on the cart, so they cannot be canceled in its net-force calculation.

Part 4 · Combine the laws in one explanation

A complete explanation names the object, its interacting partners, the net force, and the change in velocity. Use the first law when net force is zero, the second law for acceleration, and the third law to describe interaction partners. The laws answer related but different questions.

Two second-law graphs answer different questionsTwo second-law graphs answer different questionsAcceleration (m/s²)01.534.50369Net force (N)Mass fixed: 2 kgAcceleration (m/s²)024612345Mass (kg)Net force fixed: 6 NChange one factor at a time. These are model predictions, not collected results.
The left graph holds mass at 2 kg and predicts acceleration = force/2: double net force gives double acceleration. The right graph holds net force at 6 N and predicts acceleration = 6/mass: double mass gives half the acceleration. The curved relationship on the right is not a straight decrease. The mass graph begins at 1 kg, not zero; dividing by zero mass is outside this model. Compare actual repeated cart data with these predictions and account for resistance and uncertainty.

Read the visual model

Do equal interaction forces require equal accelerations?

Visual reasoning — after your attempt

No. Acceleration depends on each object’s mass and all forces acting on it. Equal opposite partner forces act on different objects.

For two carts with equal net force but different masses, the heavier cart has smaller acceleration. Each push still has an equal opposite partner force on the pusher. These facts are consistent because interaction-force equality is not acceleration equality.

Equal net force, different accelerationEqual net force, different accelerationQuantity (m/s²)2 kg cart: a34 kg cart: a1.5Each cart has the same horizontal net force: 6 N.a = Fnet / m; compare mass alone.Illustrative model • use the labels and stated assumptions
For the same 6 N net force, doubling mass halves acceleration from 3 to 1.5 m/s². The bars represent rates of velocity change, not current velocity or distance. Equal net force is a condition of this comparison; interaction pairs would require named separate receivers.

The two graphs hold different quantities fixed. Left: mass is 2 kg, so a = Fnet/2 and acceleration increases in direct proportion to net force. Right: net force is 6 N, so a = 6/m and acceleration decreases with increasing mass along a curve. Do not change mass and force together and attribute the result to one factor. Repeated measurements can be compared with these model predictions without pretending the model curves are collected data.

Pause and explain

Do equal interaction forces require equal accelerations?

Check your reasoning after answering

No. Acceleration depends on each object’s mass and all forces acting on it. Equal opposite partner forces act on different objects.

Investigate force and mass separately

A change in a cart’s motion is not established by a force arrow alone. A planned comparison needs motion records, controls and a receiver.

Acceleration is the rate of velocity change, not the object’s current speed. To investigate its relation to net force, keep mass fixed and compare different controlled net forces. To investigate mass, keep net force fixed and compare masses. A horizontal force estimate must include opposing resistance; an applied pull during acceleration is not automatically the net force. An inertial reference is a frame where the introductory laws apply without additional apparent forces.

Investigate force and mass separatelyInvestigate force and mass separatelyQuantity (acceleration, m/s²)Fixed 1 kg: Fnet 1 N1Fixed 1 kg: Fnet 2 N2Fixed 2 N: mass 4 kg0.5Illustrative model • read the labels and caption
The bars select three supplied cases rather than claim every hand push is a controlled force. At fixed mass, greater net force gives greater acceleration; at fixed net force, greater mass gives less acceleration. Read which condition is held fixed before drawing a causal conclusion.

Define motion measurement before testing: specify the start, positive direction, time interval, and recorded positions or velocities. Equal-interval records can establish average velocities and their changes. Repeating a comparison reveals variation; it does not repair a method that changed both intended factors together. Use safe toy observations to investigate qualitative patterns only, or the supplied quantitative model data; do not invent a constant force from an unmeasured hand push.

Braking illustrates a net force opposite an existing motion direction. A right-moving cart under a leftward net force can slow before reversing. Friction, a brake pad or another interaction transfers energy into deformation and thermal changes. If force becomes balanced while the cart still moves, constant velocity can continue in the ideal model. The three laws connect persistence, net-force/mass response and separate interaction partners rather than provide three unrelated slogans.

Evidence to interpret

Mass (kg)Net force (N)Acceleration (m/s²)
111
122
221
420.5

Investigate or evaluate the supplied records

  1. State one intended variable and choose matching record pairs.
  2. Specify initial state, time measurements and a fixed positive direction.
  3. Use supplied values or an approved safe simulation; ordinary hand-push observations do not provide calibrated net-force values.
  4. Plot acceleration against the changed factor while identifying fixed conditions.
  5. Plan repeated trials and a result that would challenge the explanation.

Explain and apply

Plan two comparisons from the records: one isolating net force, the other mass. Specify motion measurements, repeats, controls and what outcome would challenge the predicted relation. Explain braking separately from velocity direction.

  • Design two separately controlled comparisons.
  • Specify recorded motion, units and repeats.
  • Explain a challenge result and braking mechanism.
Reasoning and marking guidance — after your attempt

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Apply the whole unit · Compare two carts

Illustrative trials: cart A has mass 2 kg and cart B mass 4 kg. Each receives a 10 N push right and 2 N resistance left. Both begin at rest; assume constant forces over the comparison interval.

Your task

Draw a force diagram for each cart, calculate acceleration, and describe how the first, second, and third laws apply. Label the partner force exerted on the pusher without adding it to the cart diagram.

What your response should show

  • Use the supplied observations and measurements accurately; include units where needed.
  • Explain the science behind your answer, rather than giving only a number or label.
  • Show a labeled model, calculation, or evidence comparison as requested.
  • State an assumption or limitation and suggest a useful next measurement.
Response guidance — after your attempt

Each cart has net force 8 N right. A accelerates at 4 m/s²; B at 2 m/s². Their velocities change because net force is nonzero. Each cart exerts a 10 N opposite force on its pusher, acting on the pusher rather than the cart. Equal applied forces and resistance do not imply equal acceleration when masses differ.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Check each force with the sentence source acts on object. Require net-force subtraction before division by mass. Use a zero-net-force comparison to connect all three laws without suggesting they are separate unrelated rules. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation. The illustrated study “Investigate force and mass separately” is required curriculum content. Teach it before the associated application or distribute it over the assigned unit sessions; it is not an optional enrichment. Its practice/review/quiz response provides evidence of understanding. Model/data work must not be recorded as an actual physical experiment.

The key idea

First law: objects resist changes in motion (inertia). Second law: F = ma. Third law: every action has an equal, opposite reaction.

Newton's Second Law

F = ma

F is net force in newtons (N); m is mass in kg; a is acceleration in m/s². Rearranges to a = F ÷ m or m = F ÷ a.

Worked examples

Example 1

A 5 kg cart is pushed with a net force of 20 N. Find its acceleration.
  1. Rearrange F = ma to solve for acceleration: a = F ÷ m.
  2. a = 20 ÷ 5.
  3. = 4.
Answer: 4 m/s²

Example 2 · Worked application

A 2 kg cart has net force 6 N right. Find acceleration and identify the force partner when a hand pushes the cart.
  1. Apply a=Fnet/m: 6/2=3 m/s² right.
  2. The hand's push on the cart has a partner: the cart's push on the hand.
  3. The two partner forces act on different objects, so they do not cancel within the cart's force sum.
Answer: Acceleration is 3 m/s² right; the interaction partner acts on the hand.
Check the reasoning

Use net force on one object in Newton's second law.

Example 3

A soccer ball with a mass of 0.4 kg accelerates at 25 m/s² when kicked. What net force was applied?
  1. Use F = ma.
  2. F = 0.4 × 25.
  3. = 10.
Answer: 10 N

Example 4

A net horizontal force of 12 N acts on a 4 kg cart. Calculate acceleration and identify the reaction to a hand's push.
  1. Acceleration=Fnet/m=12/4=3 m/s².
  2. The reaction to the hand pushing the cart is the cart pushing the hand, on a different object.
Answer: 3 m/s² in the net-force direction; the cart exerts an equal opposite force on the hand.

Example 5 · Find and fix the mistake

Context: A 2 kg cart has net force 6 N right. Find acceleration and identify the force partner when a hand pushes the cart. A student claims: “The action and reaction forces cancel, so no object can accelerate.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Use net force on one object in Newton's second law.
  2. They act on different objects. Acceleration depends on the net force on each individual object.
Answer: They act on different objects. Acceleration depends on the net force on each individual object.
Check the reasoning

Use net force on one object in Newton's second law.

Example 6 · Read both second-law graphs

Use the two model graphs. What is acceleration for net force 9 N at mass 2 kg? What is acceleration for mass 4 kg at net force 6 N?
  1. Left graph: a = 9/2 = 4.5 m/s².
  2. Right graph: a = 6/4 = 1.5 m/s².
  3. State what was fixed before comparing results.
Answer: 4.5 m/s² and 1.5 m/s² respectively.
Check the reasoning

The examples change different variables; neither graph predicts current velocity from force alone.

Example 7 · Interpret the visual model

Zero net force allows rest OR constant velocityZero net force allows rest OR constant velocityNet force = 0: cart at rest0 s · 1 s · 2 sSame position each secondNet force = 0: cart at constant rightward velocity0 s1 s2 s
Equal-time snapshots can show no position change or equal changes in position. Both are consistent with zero net force in an inertial reference frame. The moving cart does not need a forward net force to keep its constant velocity. On a real surface, rolling resistance can slow it unless balanced. The rest snapshots overlap intentionally; no extra force called inertia should be drawn.
Why is inertia not an extra force on a force diagram?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A released toy cart slows due to interactions such as rolling resistance, not because motion uses up a stored force. A passenger’s motion changes when a seat or restraint exerts force; connect the law to interactions rather than calling inertia a pushing force.
  3. Inertia describes how mass resists acceleration. It is not an interaction from another object and should not be added as a force arrow.
Answer: Inertia describes how mass resists acceleration. It is not an interaction from another object and should not be added as a force arrow.
Check the reasoning

Keep the model’s stated limits explicit: Equal-time snapshots can show no position change or equal changes in position. Both are consistent with zero net force in an inertial reference frame. The moving cart does not need a forward net force to keep its constant velocity. On a real surface, rolling resistance can slow it unless balanced. The rest snapshots overlap intentionally; no extra force called inertia should be drawn.

Avoid these mistakes

Check your understanding

A net horizontal force of 12 N acts on a 4 kg cart. Calculate acceleration and identify the reaction to a hand's push.

Discuss the answer

3 m/s² in the net-force direction; the cart exerts an equal opposite force on the hand.

Why can a truck and a small car exert equal collision forces but have different accelerations?

Discuss the answer

Newton's first law describes motion when net force is zero. The second law connects net force to acceleration: Fnet=ma. The third law identifies paired forces between two interacting objects. Use the law that addresses the question rather than memorizing isolated slogans. A force diagram belongs to one object. If a hand pushes a cart, include the hand's force on the cart, not the cart's force on the hand. The latter belongs on the hand's diagram. Pair forces have equal magnitude, but different masses can have different accelerations.

Return to the phenomenon

Illustrative carts of 2 kg and 4 kg each experience a 6 N horizontal net force. Other conditions are comparable. Calculate acceleration magnitudes, identify an equal-and-opposite interaction pair and explain why zero net force does not require rest. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

Accelerations are 3 and 1.5 m/s² because a=Fnet/m. Equal interaction forces act on different objects, such as hand-on-cart and cart-on-hand. A zero net force means no acceleration in the simple inertial-frame model and permits constant nonzero velocity. Mass affects response, not an extra permanent force needed just to keep moving.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Illustrative carts of 2 kg and 4 kg each experience a 6 N horizontal net force. Other conditions are comparable. Calculate acceleration magnitudes, identify an equal-and-opposite interaction pair and explain why zero net force does not require rest. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

Accelerations are 3 and 1.5 m/s² because a=Fnet/m. Equal interaction forces act on different objects, such as hand-on-cart and cart-on-hand. A zero net force means no acceleration in the simple inertial-frame model and permits constant nonzero velocity. Mass affects response, not an extra permanent force needed just to keep moving.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

What happens if mass and force both double? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

For the same 6 N net force, doubling mass halves acceleration from 3 to 1.5 m/s². The bars represent rates of velocity change, not current velocity or distance. Equal net force is a condition of this comparison; interaction pairs would require named separate receivers.

Response guidance — discuss after your attempt

The ratio Fnet/m, and thus acceleration, remains unchanged in this simple model. For the same 6 N net force, doubling mass halves acceleration from 3 to 1.5 m/s². The bars represent rates of velocity change, not current velocity or distance. Equal net force is a condition of this comparison; interaction pairs would require named separate receivers.

Words to know

inertia
The tendency of an object to resist a change in its motion.
Newton's first law
An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by an unbalanced force.
Newton's second law
Force equals mass times acceleration: F = ma.
Newton's third law
For every action force, there is an equal and opposite reaction force.
newton (N)
The unit used to measure force; 1 newton = 1 kg·m/s².
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Newton stood on giants' shoulders — and corrected a 2,000-year-old mistake

The ancient Greek philosopher Aristotle taught that a constant force was needed to keep anything moving, and that objects naturally wanted to come to rest — an idea that felt intuitive and went essentially unquestioned for almost 2,000 years. Galileo's experiments with rolling balls suggested otherwise: on a perfectly frictionless surface, a moving object would just keep going forever. In 1687, Isaac Newton formalized this insight as his first law in a book called the Principia Mathematica, one of the most influential science books ever written, building directly on Galileo's and Johannes Kepler's earlier work.

Inertia depends on mass, not speed

It's tempting to think a fast-moving object has 'more inertia' than a slow one, but inertia only depends on mass — how much matter something has — not on how fast it's currently moving. A parked truck and a truck cruising at highway speed have exactly the same inertia (resistance to a change in motion), which is why it takes the same enormous force to start a heavy truck rolling as it does to bring it to a sudden stop. This is exactly why seatbelts and headrests exist: in a sudden stop, your body's inertia makes it want to keep moving forward (or your head want to stay still) even after the car has stopped.

Momentum and impulse: the hidden link between the second and third laws

Multiply an object's mass by its velocity and you get its momentum (p = mv) — a kind of 'quantity of motion.' It turns out force is really just how fast momentum changes: force × time = change in momentum. That's why airbags work: by stretching a car crash's stopping time from a fraction of a second into a slightly longer fraction of a second, the same change in momentum happens with much less force on your body. This connects directly to Newton's third law, too — the passenger pushes back on the airbag exactly as hard as the airbag pushes on the passenger.

Real rockets, real numbers: the third law at full throttle

The Saturn V rocket that carried astronauts to the Moon weighed nearly 3 million kilograms fully fueled, and its engines pushed exhaust gases downward with a combined thrust of over 34 million newtons at liftoff. By Newton's third law, that exhaust pushed back on the rocket with just as much force in the opposite direction — enough to overcome the rocket's enormous weight and accelerate it off the launch pad. No engine 'pushes against the air' the way a swimmer pushes against water; rockets work identically in the vacuum of space, since the action-reaction pair is between the rocket and its own exhaust, not the surrounding air.

In the real world

Newton's third law explains exactly how rockets launch into space — by pushing exhaust gas downward, they get pushed upward in return.

Exit check

What happens if mass and force both double? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

The ratio Fnet/m, and thus acceleration, remains unchanged in this simple model. For the same 6 N net force, doubling mass halves acceleration from 3 to 1.5 m/s². The bars represent rates of velocity change, not current velocity or distance. Equal net force is a condition of this comparison; interaction pairs would require named separate receivers.

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

Collisions, Deformation & Changes in Motion

Learning goals

Readings for this unit

Before this unit

Review the earlier motion graphs and Newton’s laws units. The visible language notes explain this unit’s quantities; then connect them to its concrete records and figure.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in collisions, deformation & changes in motion?

Illustrative evidence case. Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop.

Deformation can happen and recoverDeformation can happen and recoverBefore contactBumper: 15 mmCart movingDuring contactBumper: 8 mmInteraction forcesAfter contactBumper: 15 mmShape recoveredIllustrative model • use the labels and stated assumptions
The three snapshots show temporary elastic deformation in a simplified bumper record. A photograph taken only after contact would miss the compression. The model does not assign a force from thickness alone or certify real protective equipment, and it omits the full time-dependent motion.

Notice and wonder

Does recovered thickness establish that no energy transferred? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Explain the temporary deformation and compare peak values without declaring all collisions safe. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

collision
Short interaction in which objects exert forces on each other.
deformation
Change in an object’s shape.
elastic deformation
Shape change that substantially reverses after the force is removed.
permanent deformation
Shape change remaining after the interaction.
peak force
Largest force recorded during a specified interval.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

A toy cart can hit soft packing material and stop without a permanent dent, while an otherwise similar collision against a hard stop can damage it. A collision is an interaction over a short interval when objects exert forces on each other.

Separate motion before contact from forces during contactSeparate motion before contact from forces during contactBefore contactA: 1 kgB: 1 kgAt restDuring contactA: 1 kgB: 1 kgAfter ideal elastic collisionA: 1 kgB: 1 kgAt restRed: paired contact forces
These equal-mass carts illustrate an ideal one-dimensional elastic collision: the initially moving cart can stop while the initially stationary cart moves. Interaction-force arrows apply during contact, while velocity arrows describe motion before or after. The snapshots omit detailed deformation and other forces and do not claim that all real collisions exchange velocities or conserve mechanical kinetic energy.

Deformation means a change in shape. Elastic deformation is substantially reversed when the force is removed; permanent deformation remains.

A collision changes energy stores rather than deleting energyA collision changes energy stores rather than deleting energyInitial kinetic energy: 4 JIllustrative after-contact accountMotion: 2 JThermal: 1 JDeformation: 1 JTotal stays 4 J when the relevant objects and changes are included.
In this explicitly illustrative account, 4 J of initial kinetic energy becomes 2 J of motion plus 1 J of thermal increase and 1 J of retained deformation energy. The bar widths share an energy scale. A real event can require additional objects and pathways, and visible damage is not a complete energy meter or a universal indicator of peak force.

Even a material that looks rigid can change shape slightly during contact. Study soft toy objects or supplied records, never a real phone, glass object, helmet or a person.

Temporary changes that escape normal-speed viewing can still matter to the explanation.

An observation is what a person or instrument records. An inference is a proposed interpretation of those records. Keep the two separate: a graph can show a pattern while more than one mechanism might explain it. State the chosen system, or part of the world being studied, and its boundary before accounting for material or energy transfers.

Use each figure as a reasoning tool. Read its caption and labels, describe the pattern, then explain a cause. If the picture is schematic, use the stated numbers rather than measuring its printed size. Check whether its simplified conditions match the problem before applying a formula.

Build your understanding

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Compare before, during and after contact

Observations before contact establish motion and shape; observations during contact can reveal bending or compression; afterward they show rebound or permanent change. A system model includes both objects and their surroundings. A cartoon with three frames is a model, not proof of unseen changes.

Separate motion before contact from forces during contactSeparate motion before contact from forces during contactBefore contactA: 1 kgB: 1 kgAt restDuring contactA: 1 kgB: 1 kgAfter ideal elastic collisionA: 1 kgB: 1 kgAt restRed: paired contact forces
These equal-mass carts illustrate an ideal one-dimensional elastic collision: the initially moving cart can stop while the initially stationary cart moves. Interaction-force arrows apply during contact, while velocity arrows describe motion before or after. The snapshots omit detailed deformation and other forces and do not claim that all real collisions exchange velocities or conserve mechanical kinetic energy.

Read the visual model

A timeline is important because different interactions exist at different stages. An elastic collision is an ideal case where the system’s kinetic energy is retained after separation. A damaged or thermally dissipative collision can have a different outcome.

Why should the contact-force arrows not be carried into the separated after-collision snapshot?

Visual reasoning — after your attempt

The contact interaction has ended once the carts separate. A remaining velocity is not evidence of a continuing contact push; other real interactions must be named separately.

Illustrative foam bumper images show 20 mm thickness before contact, 12 mm during contact and 19 mm afterward. The shape mostly recovers but not exactly. A slow record permits an observation that a single after-collision photograph would miss.

Deformation can happen and recoverDeformation can happen and recoverBefore contactBumper: 15 mmCart movingDuring contactBumper: 8 mmInteraction forcesAfter contactBumper: 15 mmShape recoveredIllustrative model • use the labels and stated assumptions
The three snapshots show temporary elastic deformation in a simplified bumper record. A photograph taken only after contact would miss the compression. The model does not assign a force from thickness alone or certify real protective equipment, and it omits the full time-dependent motion.

Pause and explain

Why is an afterward photograph insufficient to detect temporary deformation?

Check your reasoning after answering

An object can deform during contact and then recover. The after-image does not show the complete time sequence.

Part 2 · Connect equal forces with different motion changes

During contact, each object exerts an equal-size opposite-direction force on the other. Different masses can undergo different accelerations under these equal interaction forces. For each object, include other forces before calculating a net force.

Equal paired forces can produce unequal accelerationsEqual paired forces can produce unequal accelerationsA: 2 kgB: 1 kgContact force 10 N leftContact force 10 N rightA: acceleration 10/2 = 5 m/s² leftB: acceleration 10/1 = 10 m/s² rightEqual interaction force ≠ equal acceleration; compare mass and other forces.
The two interaction forces have equal magnitude and opposite directions on different receivers. In the stated horizontal model, these are the net horizontal forces: the 2 kg cart accelerates at 5 m/s² left and the 1 kg cart at 10 m/s² right. Additional horizontal resistance would have to be included before using the force as the net value.

Read the visual model

Force equality belongs to the interaction pair. Acceleration also depends on the receiver’s mass and full force account. Calculate a change in velocity separately from the object’s already existing velocity.

Over an interval of 0.1 s at these average forces, what are the velocity changes?

Visual reasoning — after your attempt

A changes by 0.5 m/s left and B by 1 m/s right. Their acceleration and velocity changes differ even though the paired interaction forces are equal.

If the contact force dominates, a 1 kg cart and a 2 kg cart under equal 4 N contact forces have acceleration magnitudes 4 and 2 m/s². Metres per second squared describes the rate of velocity change, not the distance travelled.

Pause and explain

Does greater acceleration of the lighter cart imply a larger collision force on it?

Check your reasoning after answering

No. Equal interaction forces can cause different acceleration because masses differ, and net-force calculations must account for other forces.

Part 3 · Track energy and deformation without a damage shortcut

Kinetic energy is energy associated with motion. In the same simplified situation, more mass at the same speed means more kinetic energy, and doubling speed gives four times the kinetic energy. During stopping, energy can transfer into temporary deformation, thermal changes, sound and surroundings.

A collision changes energy stores rather than deleting energyA collision changes energy stores rather than deleting energyInitial kinetic energy: 4 JIllustrative after-contact accountMotion: 2 JThermal: 1 JDeformation: 1 JTotal stays 4 J when the relevant objects and changes are included.
In this explicitly illustrative account, 4 J of initial kinetic energy becomes 2 J of motion plus 1 J of thermal increase and 1 J of retained deformation energy. The bar widths share an energy scale. A real event can require additional objects and pathways, and visible damage is not a complete energy meter or a universal indicator of peak force.

Read the visual model

A system boundary determines which energy changes are included. Temporary deformation may later release stored energy; permanent change and warming can remain. Specify the stage of the collision before deciding what the bars represent.

How much kinetic energy is no longer in motion here, and why is calling it destroyed incorrect?

Visual reasoning — after your attempt

2 J is no longer kinetic in this account. It is represented by thermal and deformation changes inside the stated full system, so total energy remains 4 J.

A supplied cart model has 1 J at one speed and 4 J at twice that speed. The energy ratio is useful, but damage also depends on contact area, material, duration and breaking limits. Equal energy does not guarantee equal damage.

Pause and explain

Can speed alone predict every dent?

Check your reasoning after answering

No. It influences motion energy, but material response, geometry and how the object stops also affect damage.

Part 4 · Evaluate a cushioning mechanism

Cushioning can allow a moving object to stop through a longer distance and time, often reducing typical and peak stopping forces for comparable starting motion. A peak force is the largest recorded force; it differs from the average over contact. A cushion that bottoms out can lose that advantage.

Cushioning can spread a motion change over more timeCushioning can spread a motion change over more timeContact force (N)01020304000.020.040.060.08Time during contact (s)Shorter stopLonger cushioned stopBoth ideal pulse areas = 0.4 N·s; same 0.2 kg object changes velocity by 2 m/s.
The two modeled triangular force pulses stop the same 0.2 kg object from 2 m/s in the stated horizontal comparison. The short pulse lasts 0.02 s with peak 40 N; the longer lasts 0.08 s with peak 10 N. Both have area 0.4 N·s. These are ideal illustrative pulse shapes, not measured OpenSciEd results; a real protector must be tested for its force profile, bottoming out and all design criteria.

Read the visual model

Start with the time width and force height. Average force can be found from the measured velocity change and stopping time in this model. The force–time area is an optional further connection; it does not have units of distance or energy.

Compare average forces using mass×velocity change/time. Why is “soft-looking always protects” too strong?

Visual reasoning — after your attempt

The averages are 0.2×2/0.02 = 20 N and 0.2×2/0.08 = 5 N. Longer stopping time can reduce force for the same motion change, but a material may bottom out, change the stopping conditions or fail another requirement.

Illustrative tests for matched carts at the same speed show peak readings of 18, 19, 20 N without padding and 9, 10, 11 N with padding. Check repeated trials, measuring method and pad thickness before generalizing to other objects.

OpenSciEd cushioning design drawing with three rings side by side between a loaded plate and a table, for comparing material shape and load sharing.
OpenSciEd offers three side-by-side rings as a shape to compare with one ring. Arrangement can change deformation and how a load is shared. This unloaded illustration does not itself establish a stopping time or peak-force reduction; use the separate force–time model and appropriate repeated records to test the mechanism. Three rings sharing a load · OpenSciEd · 8.1 Contact Forces, Lesson 13, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.

Pause and explain

What do these peak records support?

Check your reasoning after answering

Lower measured peak force for this padded setup under the stated matched conditions. They do not certify every pad or establish identical average forces or injury protection.

Apply the whole unit · Temporary shape and a peak reading

Illustrative case for analysis, not measurements collected by the learner. Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop.

Your task

Explain the temporary deformation and compare peak values without declaring all collisions safe. Use the supplied conditions, show intermediate reasoning and distinguish the supported conclusion from a broader claim.

What your response should show

  • Use the specified quantities and units.
  • Explain an intermediate step or comparison.
  • Connect the pattern to the physical mechanism.
  • State an assumption or limitation.
Response guidance — after your attempt
  1. Compare thickness at three times: 15→8→15 mm shows compression and recovery in the record.

  2. Classify the modeled shape change as elastic because its thickness recovers.

  3. Compare peaks: 12/24=0.5, so the padded peak is half the rigid-stop peak under matched conditions.

  4. Relate cushioning to stopping time/distance and account for redistributed energy, while limiting the claim to these records.

The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Ask the learner to describe the phenomenon before naming a formula. Discuss each checkpoint after an attempt. Use the supplied records when equipment is unavailable and label them illustrative rather than actual measurements. The final explanation should identify a mechanism and a limit, with different defensible wording accepted. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

It may have changed temporarily during contact; use a before–during–after record. The bumper undergoes elastic deformation in this simplified record because it regains its thickness.

Worked examples

Example 1 · Build the first reasoning step

Why is an afterward photograph insufficient to detect temporary deformation?
  1. Identify what the question asks and what the stated observation establishes.
  2. An object can deform during contact and then recover. The after-image does not show the complete time sequence.
Answer: An object can deform during contact and then recover. The after-image does not show the complete time sequence.

Example 2 · Temporary shape and a peak reading

Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop. Explain the temporary deformation and compare peak values without declaring all collisions safe.
  1. Compare thickness at three times: 15→8→15 mm shows compression and recovery in the record.
  2. Classify the modeled shape change as elastic because its thickness recovers.
  3. Compare peaks: 12/24=0.5, so the padded peak is half the rigid-stop peak under matched conditions.
  4. Relate cushioning to stopping time/distance and account for redistributed energy, while limiting the claim to these records.
Answer: The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.
Check the reasoning

Explain why each stated condition matters.

Example 3 · Transfer to a different condition

Two collisions have the same initial kinetic energy but different damage. Is energy conservation violated?
  1. Identify the changed condition.
  2. No. Different materials and stopping conditions can distribute the energy differently. Conservation does not require the same peak force, deformation or damage.
Answer: No. Different materials and stopping conditions can distribute the energy differently. Conservation does not require the same peak force, deformation or damage.

Example 4 · Find and fix the mistake

A student claims: “A recovered bumper never changed shape.” Evaluate the claim.
  1. Distinguish what is observed from what is inferred.
  2. It may have changed temporarily during contact; use a before–during–after record.
Answer: It may have changed temporarily during contact; use a before–during–after record.
Check the reasoning

Compare the claim with the stated observations and model assumptions.

Example 5 · Interpret the visual model

Separate motion before contact from forces during contactSeparate motion before contact from forces during contactBefore contactA: 1 kgB: 1 kgAt restDuring contactA: 1 kgB: 1 kgAfter ideal elastic collisionA: 1 kgB: 1 kgAt restRed: paired contact forces
These equal-mass carts illustrate an ideal one-dimensional elastic collision: the initially moving cart can stop while the initially stationary cart moves. Interaction-force arrows apply during contact, while velocity arrows describe motion before or after. The snapshots omit detailed deformation and other forces and do not claim that all real collisions exchange velocities or conserve mechanical kinetic energy.
Why should the contact-force arrows not be carried into the separated after-collision snapshot?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A timeline is important because different interactions exist at different stages. An elastic collision is an ideal case where the system’s kinetic energy is retained after separation. A damaged or thermally dissipative collision can have a different outcome.
  3. The contact interaction has ended once the carts separate. A remaining velocity is not evidence of a continuing contact push; other real interactions must be named separately.
Answer: The contact interaction has ended once the carts separate. A remaining velocity is not evidence of a continuing contact push; other real interactions must be named separately.
Check the reasoning

Keep the model’s stated limits explicit: These equal-mass carts illustrate an ideal one-dimensional elastic collision: the initially moving cart can stop while the initially stationary cart moves. Interaction-force arrows apply during contact, while velocity arrows describe motion before or after. The snapshots omit detailed deformation and other forces and do not claim that all real collisions exchange velocities or conserve mechanical kinetic energy.

Avoid these mistakes

Check your understanding

Why is an afterward photograph insufficient to detect temporary deformation?

Discuss the answer

An object can deform during contact and then recover. The after-image does not show the complete time sequence.

Does greater acceleration of the lighter cart imply a larger collision force on it?

Discuss the answer

No. Equal interaction forces can cause different acceleration because masses differ, and net-force calculations must account for other forces.

Can speed alone predict every dent?

Discuss the answer

No. It influences motion energy, but material response, geometry and how the object stops also affect damage.

What do these peak records support?

Discuss the answer

Lower measured peak force for this padded setup under the stated matched conditions. They do not certify every pad or establish identical average forces or injury protection.

Return to the phenomenon

Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop. Explain the temporary deformation and compare peak values without declaring all collisions safe. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop. Explain the temporary deformation and compare peak values without declaring all collisions safe. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Does recovered thickness establish that no energy transferred? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

The three snapshots show temporary elastic deformation in a simplified bumper record. A photograph taken only after contact would miss the compression. The model does not assign a force from thickness alone or certify real protective equipment, and it omits the full time-dependent motion.

Response guidance — discuss after your attempt

No. Energy can transfer through deformation, motion, sound and thermal pathways even if the original shape returns. The three snapshots show temporary elastic deformation in a simplified bumper record. A photograph taken only after contact would miss the compression. The model does not assign a force from thickness alone or certify real protective equipment, and it omits the full time-dependent motion.

Words to know

collision
Short interaction in which objects exert forces on each other.
deformation
Change in an object’s shape.
elastic deformation
Shape change that substantially reverses after the force is removed.
permanent deformation
Shape change remaining after the interaction.
peak force
Largest force recorded during a specified interval.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop.

Exit check

Does recovered thickness establish that no energy transferred? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

No. Energy can transfer through deformation, motion, sound and thermal pathways even if the original shape returns. The three snapshots show temporary elastic deformation in a simplified bumper record. A photograph taken only after contact would miss the compression. The model does not assign a force from thickness alone or certify real protective equipment, and it omits the full time-dependent motion.

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

Designing & Revising a Collision Protector

Learning goals

Readings for this unit

Before this unit

Review the earlier collisions/deformation unit. The visible language notes explain this unit’s quantities; then connect them to its concrete records and figure.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in designing & revising a collision protector?

Illustrative evidence case. Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm.

Compare means and constraints togetherCompare means and constraints togetherQuantity (N)Design A mean peak13Design B mean peak9A thickness: 4 cm • B thickness: 7 cmHard limit: 5 cm • B exceeds the allowed thickness.Illustrative model • use the labels and stated assumptions
The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

Notice and wonder

Which is a qualifying starting design under the stated limit? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Choose a qualifying starting design and propose an evidence-based revision. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

criterion
Feature used to evaluate success.
constraint
Limit a design must meet.
stakeholder
Person or group affected by the design.
operational definition
Stated rule for measuring or judging a property.
iteration
Another test-and-revision cycle informed by evidence.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Engineering uses scientific evidence to solve a defined problem. A criterion is a feature used to judge success, such as keeping a model payload undamaged.

Compare means and constraints togetherCompare means and constraints togetherQuantity (N)Design A mean peak13Design B mean peak9A thickness: 4 cm • B thickness: 7 cmHard limit: 5 cm • B exceeds the allowed thickness.Illustrative model • use the labels and stated assumptions
The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

A constraint is a limit, such as maximum size or cost. A stakeholder is someone affected by a design.

A good protector must satisfy the actual task, not merely contain the most padding. Use a soft toy model and supplied records; do not use people, animals, real electronics, fragile glass or certification claims about safety equipment.

The design cycle is iterative: testing informs a revision, and the revised solution must be tested again.

Transfer a mechanism without assuming the result transfersTransfer a mechanism without assuming the result transfersDesign BWorks in original testNew orientation / fragile objectNew prediction → new testKeep the original records; a new use may change contact area or stopping path.A revision is a hypothesis until its actual outcome is tested.
A protector that meets the original tests is proposed for a different orientation or object. Changes in contact, mass, shape or stopping travel may change performance. The arrows show a design-reasoning sequence, not confirmed data. Retain the successful original version and state which new records could support or challenge the revised prediction.

An observation is what a person or instrument records. An inference is a proposed interpretation of those records. Keep the two separate: a graph can show a pattern while more than one mechanism might explain it. State the chosen system, or part of the world being studied, and its boundary before accounting for material or energy transfers.

Use each figure as a reasoning tool. Read its caption and labels, describe the pattern, then explain a cause. If the picture is schematic, use the stated numbers rather than measuring its printed size. Check whether its simplified conditions match the problem before applying a formula.

Build your understanding

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Define success and the limits before testing

Specify the protected object, collision conditions, damage measure and permitted resources. Operational definition means exactly how a property will be judged in the test. A visible mark on a paper target can be a model damage indicator, but it is not a measurement of real injury.

Compare means and constraints togetherCompare means and constraints togetherQuantity (N)Design A mean peak13Design B mean peak9A thickness: 4 cm • B thickness: 7 cmHard limit: 5 cm • B exceeds the allowed thickness.Illustrative model • use the labels and stated assumptions
The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

Read the visual model

Why define the dent measure before seeing results?

Visual reasoning — after your attempt

A fixed rule allows consistent judgments and avoids changing success criteria to favor a preferred design.

A toy protector may need to leave no permanent target dent, fit within 5 cm thickness and use at most 20 g of padding. A design that exceeds a hard limit does not qualify even if its force reading is lower.

Pause and explain

Why define the dent measure before seeing results?

Check your reasoning after answering

A fixed rule allows consistent judgments and avoids changing success criteria to favor a preferred design.

Part 2 · Predict how a feature could work

An initial design model should connect a physical feature to a mechanism. Compressible padding can increase stopping distance; a broad face can distribute contact over greater area. These are different mechanisms and need different observations. More padding can add mass or run out of compression space.

A cushioning feature needs a mechanism and a testA cushioning feature needs a mechanism and a testFlat pad: little travelRings: room to compressSame test loadA predicted mechanism: geometry changes deformation and stopping travel.A drawing predicts a feature; repeated measurements decide whether it helps.
The comparison proposes that a compressible ring arrangement provides more deformation travel than a very thin pad under a matched test load. Geometry is shown schematically, and no measured force reduction is supplied by this drawing. A protector must still be checked for stability, maximum compression, size and repeat performance; adding material without a mechanism is not enough.

Read the visual model

A mechanism explains how the feature changes the interaction. OpenSciEd’s ring sketches below give concrete shapes to compare, but an unloaded drawing cannot establish their performance. Predict, test, revise and retest.

Which measurement could distinguish “more travel” from “a prettier design”?

Visual reasoning — after your attempt

Measure deformation or stopping distance/time under matched loads and then evaluate peak-force/damage criteria using an appropriate safe test or supplied records.

Sketch two designs with the same total padding amount but different arrangements. Predict the shape changes during contact and identify the measurement that could distinguish the ideas. Drawing a thick case alone is not an explanation.

OpenSciEd drawing of six cushioning rings arranged in a triangle with a broad base on the table and one top ring touching the loaded plate.
This OpenSciEd six-ring triangle has a broad base at the table and a narrow contact at the top plate. It is a proposed geometry for testing deformation and load distribution, not proof of successful impact protection. Predict the mechanism and which criterion it could change before examining supplied test records. Triangular stack of cushioning rings · OpenSciEd · 8.1 Contact Forces, Lesson 13, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
OpenSciEd drawing of six cushioning rings arranged as an inverted triangle, with three top rings under a loaded plate and a single bottom ring touching the table.
OpenSciEd also shows the same six rings with the triangle inverted. Comparing the two arrangements helps isolate shape while keeping the ring count fixed. Contact distribution and stability can differ; a wider top does not automatically establish lower peak force. Keep actual trial records and evaluate the stated criteria. Inverted triangular stack of cushioning rings · OpenSciEd · 8.1 Contact Forces, Lesson 13, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.

Pause and explain

Does a larger contact face automatically mean a smaller total force?

Check your reasoning after answering

No. It can lower pressure for the same force by spreading contact area, but total force depends on the stopping interaction.

Part 3 · Compare repeated tests under matched conditions

Hold starting speed, payload mass, collision direction, surface and measuring procedure comparable. Repeat with fresh or restored samples when previous deformation affects results. Keep all planned measurements, including inconvenient failures. A mean is the sum divided by the number of readings; a range is maximum minus minimum.

Averages do not hide a design’s failing trialAverages do not hide a design’s failing trialPeak-force limit: every stated trial must remain below 35 N.01020304050ABCDots = trials; bars = means. C’s occasional high force still matters.
The supplied peak forces are A: 40, 44, 42 N; B: 24, 28, 26 N; and C: 30, 31, 48 N under comparable tests. The bars are means and dots preserve individual readings. With a criterion that every stated trial stay below 35 N, only B passes these force records. Other criteria such as cost and size must also be checked before choosing a design.

Read the visual model

State whether a requirement concerns a mean, a maximum or every trial. Different definitions lead to different decisions. Graphing both the mean and trials helps a parent or reviewer see the evidence instead of relying on one favorable result.

Which design meets the stated force limit in all three trials? Why must C’s 48 N reading remain in the report?

Visual reasoning — after your attempt

Only B does. C fails one trial even though two are below the limit; omitting the 48 N record would misrepresent reliability and the actual criterion.

Illustrative peaks for A are 12, 14, 13 N and for B 10, 11, 15 N. A averages 13 N with range 2 N; B averages 12 N with range 5 N. B has a lower mean but greater variability.

Pause and explain

Why is B’s single 10 N reading an incomplete comparison?

Check your reasoning after answering

It omits its other readings and variability. Evaluate all planned trials along with constraints and damage observations.

Part 4 · Revise and transfer the explanation

Choose a revision from evidence, then predict what should change and retest. A design can be strongest for one criterion and weaker for another. A transfer assessment asks whether the mechanism still applies in a new setting, rather than whether the student remembers a brand or drawing.

Transfer a mechanism without assuming the result transfersTransfer a mechanism without assuming the result transfersDesign BWorks in original testNew orientation / fragile objectNew prediction → new testKeep the original records; a new use may change contact area or stopping path.A revision is a hypothesis until its actual outcome is tested.
A protector that meets the original tests is proposed for a different orientation or object. Changes in contact, mass, shape or stopping travel may change performance. The arrows show a design-reasoning sequence, not confirmed data. Retain the successful original version and state which new records could support or challenge the revised prediction.

Read the visual model

Transfer means applying an explanation to a new situation. Keep the mechanism explicit: which part should deform, what motion change should occur, and which constraint might fail? A prediction is not an already completed retest.

Does passing three horizontal tests guarantee a successful corner impact? Name one changed condition to test.

Visual reasoning — after your attempt

No. Orientation/contact area or the stopping path changes. Define a safe comparable corner-impact test or analyze supplied records and evaluate the same stated criteria.

Combine A’s stable arrangement with B’s more compressible layer if the size and mass limits allow it. Predict reduced typical peak while retaining consistency, then collect new data. The combination is a proposal until tested.

Pause and explain

Does combining useful features prove the new protector is better?

Check your reasoning after answering

No. Features can interact and violate limits. The revised design requires new tests under the same evaluation rules.

Apply the whole unit · A design decision with a hard constraint

Illustrative case for analysis, not measurements collected by the learner. Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm.

Your task

Choose a qualifying starting design and propose an evidence-based revision. Use the supplied conditions, show intermediate reasoning and distinguish the supported conclusion from a broader claim.

What your response should show

  • Use the specified quantities and units.
  • Explain an intermediate step or comparison.
  • Connect the pattern to the physical mechanism.
  • State an assumption or limitation.
Response guidance — after your attempt
  1. Read the hard thickness limit before ranking peak force.

  2. A at 4 cm satisfies the 5 cm limit; B at 7 cm does not.

  3. Choose A as a qualifying starting design, not as proof of universally best protection.

  4. Propose a changed padding arrangement within 5 cm and retest with matched motion, repeated peaks and deformation records.

A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Ask the learner to describe the phenomenon before naming a formula. Discuss each checkpoint after an attempt. Use the supplied records when equipment is unavailable and label them illustrative rather than actual measurements. The final explanation should identify a mechanism and a limit, with different defensible wording accepted. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

Judge repeated evidence, variability, damage measures and hard constraints, rather than one selected reading. A qualifies under the thickness rule; B does not despite its lower mean peak.

Worked examples

Example 1 · Build the first reasoning step

Why define the dent measure before seeing results?
  1. Identify what the question asks and what the stated observation establishes.
  2. A fixed rule allows consistent judgments and avoids changing success criteria to favor a preferred design.
Answer: A fixed rule allows consistent judgments and avoids changing success criteria to favor a preferred design.

Example 2 · A design decision with a hard constraint

Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm. Choose a qualifying starting design and propose an evidence-based revision.
  1. Read the hard thickness limit before ranking peak force.
  2. A at 4 cm satisfies the 5 cm limit; B at 7 cm does not.
  3. Choose A as a qualifying starting design, not as proof of universally best protection.
  4. Propose a changed padding arrangement within 5 cm and retest with matched motion, repeated peaks and deformation records.
Answer: A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.
Check the reasoning

Explain why each stated condition matters.

Example 3 · Transfer to a different condition

A revised case protects one trial but fails two. Can its successful trial justify calling it reliable?
  1. Identify the changed condition.
  2. No. All planned trials matter. Report the failures, investigate their causes, and revise the design or the limited claim rather than selecting only the success.
Answer: No. All planned trials matter. Report the failures, investigate their causes, and revise the design or the limited claim rather than selecting only the success.

Example 4 · Find and fix the mistake

A student claims: “The design with the lowest one-trial reading is always best.” Evaluate the claim.
  1. Distinguish what is observed from what is inferred.
  2. Judge repeated evidence, variability, damage measures and hard constraints, rather than one selected reading.
Answer: Judge repeated evidence, variability, damage measures and hard constraints, rather than one selected reading.
Check the reasoning

Compare the claim with the stated observations and model assumptions.

Example 5 · Interpret the visual model

A cushioning feature needs a mechanism and a testA cushioning feature needs a mechanism and a testFlat pad: little travelRings: room to compressSame test loadA predicted mechanism: geometry changes deformation and stopping travel.A drawing predicts a feature; repeated measurements decide whether it helps.
The comparison proposes that a compressible ring arrangement provides more deformation travel than a very thin pad under a matched test load. Geometry is shown schematically, and no measured force reduction is supplied by this drawing. A protector must still be checked for stability, maximum compression, size and repeat performance; adding material without a mechanism is not enough.
Which measurement could distinguish “more travel” from “a prettier design”?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A mechanism explains how the feature changes the interaction. OpenSciEd’s ring sketches below give concrete shapes to compare, but an unloaded drawing cannot establish their performance. Predict, test, revise and retest.
  3. Measure deformation or stopping distance/time under matched loads and then evaluate peak-force/damage criteria using an appropriate safe test or supplied records.
Answer: Measure deformation or stopping distance/time under matched loads and then evaluate peak-force/damage criteria using an appropriate safe test or supplied records.
Check the reasoning

Keep the model’s stated limits explicit: The comparison proposes that a compressible ring arrangement provides more deformation travel than a very thin pad under a matched test load. Geometry is shown schematically, and no measured force reduction is supplied by this drawing. A protector must still be checked for stability, maximum compression, size and repeat performance; adding material without a mechanism is not enough.

Avoid these mistakes

Check your understanding

Why define the dent measure before seeing results?

Discuss the answer

A fixed rule allows consistent judgments and avoids changing success criteria to favor a preferred design.

Does a larger contact face automatically mean a smaller total force?

Discuss the answer

No. It can lower pressure for the same force by spreading contact area, but total force depends on the stopping interaction.

Why is B’s single 10 N reading an incomplete comparison?

Discuss the answer

It omits its other readings and variability. Evaluate all planned trials along with constraints and damage observations.

Does combining useful features prove the new protector is better?

Discuss the answer

No. Features can interact and violate limits. The revised design requires new tests under the same evaluation rules.

Return to the phenomenon

Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm. Choose a qualifying starting design and propose an evidence-based revision. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm. Choose a qualifying starting design and propose an evidence-based revision. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Which is a qualifying starting design under the stated limit? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

Response guidance — discuss after your attempt

A qualifies at 4 cm. B at 7 cm does not, despite its lower recorded mean peak. A revision needs retesting. The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

Words to know

criterion
Feature used to evaluate success.
constraint
Limit a design must meet.
stakeholder
Person or group affected by the design.
operational definition
Stated rule for measuring or judging a property.
iteration
Another test-and-revision cycle informed by evidence.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm.

Exit check

Which is a qualifying starting design under the stated limit? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

A qualifies at 4 cm. B at 7 cm does not, despite its lower recorded mean peak. A revision needs retesting. The bar comparison favors B for lower mean peak in the supplied tests, but B fails the hard size constraint. A decision needs both criteria and constraints and should check repeated variability and damage outcomes. These are illustrative model-protector records, not safety certification.

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

Problem 1

Does a 2 m/s average speed establish a constant 2 m/s at every moment?
Answer and explanation

No. Different speeds and stops can produce the same distance/time average. More detailed position or speed measurements are needed.

No. Different speeds and stops can produce the same distance/time average. More detailed position or speed measurements are needed.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 2

Over 20 s, what are average speed and velocity?
Answer and explanation

Average speed is 40/20=2 m/s; average velocity is 20/20=1 m/s east.

Average speed is 40/20=2 m/s; average velocity is 20/20=1 m/s east.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 3

Right is positive. A cart changes velocity from −2 to −8 m/s in 3 s. Calculate average acceleration and explain whether it speeds up or slows down.
Answer and explanation

(−8−(−2))/3 = −2 m/s². It speeds up toward the left because speed rises from 2 to 8 m/s.

Subtract signed velocities before dividing.

Compare velocity direction with acceleration direction.

Response criteria

  • Calculate −2 m/s².
  • Explain leftward motion.
  • Reject negative-always-slows.

Problem 4

Visual-model chapter review: Right is positive. A cart’s velocity changes from −3 to −9 m/s in 2 s. Find average acceleration and explain its speed change.
Answer and explanation

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

Subtract signed velocities.

Compare the magnitudes and directions instead of using a sign slogan.

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 5

Why must both coordinate changes be used for an interval slope?
Answer and explanation

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 6

Does the final downward line prove decreasing speed?
Answer and explanation

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude.

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 7

Read the supplied +3 m/s then −2 m/s graph, each interval lasting 2 s. Calculate distance, displacement and average speed.
Answer and explanation

Distance 10 m; displacement +2 m; average speed 2.5 m/s.

Compute signed areas +6 and −4 m.

Add magnitudes for distance and divide by 4 s for average speed.

Response criteria

  • Use signed area for displacement.
  • Use magnitudes for distance.
  • State average-speed units.

Problem 8

Visual-model chapter review: A velocity–time record is +4 m/s for 3 s followed by −1 m/s for 2 s. Calculate distance, displacement and average velocity over the full five seconds.
Answer and explanation

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Use signed rectangle areas for displacement.

Add path magnitudes for distance and divide displacement by total time for average velocity.

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 9

Which force belongs in F=ma for this calculation?
Answer and explanation

Use the net external force on the chosen cart, not a single applied force when other forces also act.

Use the net external force on the chosen cart, not a single applied force when other forces also act.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 10

What happens if mass and force both double?
Answer and explanation

The ratio Fnet/m, and thus acceleration, remains unchanged in this simple model.

The ratio Fnet/m, and thus acceleration, remains unchanged in this simple model.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 11

Plan two comparisons from the records: one isolating net force, the other mass. Specify motion measurements, repeats, controls and what outcome would challenge the predicted relation. Explain braking separately from velocity direction.
Hints

Use the illustrated study titled Investigate force and mass separately. Keep the observations, model and conditions separate.

Answer and explanation

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Response criteria

  • Design two separately controlled comparisons.
  • Specify recorded motion, units and repeats.
  • Explain a challenge result and braking mechanism.

Problem 12

The cart diagram shows 10 N right and 4 N left on a 2 kg cart. A student calculates 5 m/s² using the push alone. Correct the method and say whether the cart must currently move right.
Answer and explanation

Net force is 6 N right; acceleration is 3 m/s² right. The current velocity is not given: the cart could be moving left and slowing.

Subtract resistance from the applied push.

Divide net force by mass.

Keep acceleration distinct from current motion.

Response criteria

  • Calculate 6 N and 3 m/s².
  • Do not count the green net arrow again.
  • Avoid inferring current velocity.

Problem 13

Visual-model chapter review: A 2.5 kg cart has a 15 N applied push and 5 N resistance. Find acceleration, and explain why its equal force on the pusher does not cancel that acceleration.
Answer and explanation

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

Compute net force before dividing by mass.

Name the receiver of each interaction partner.

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 14

Does greater acceleration of the lighter cart imply a larger collision force on it?
Answer and explanation

No. Equal interaction forces can cause different acceleration because masses differ, and net-force calculations must account for other forces.

No. Equal interaction forces can cause different acceleration because masses differ, and net-force calculations must account for other forces.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 15

Does recovered thickness establish that no energy transferred?
Answer and explanation

No. Energy can transfer through deformation, motion, sound and thermal pathways even if the original shape returns.

No. Energy can transfer through deformation, motion, sound and thermal pathways even if the original shape returns.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 16

Visual-model chapter review: Two carts receive equal opposite 12 N contact forces in a stated horizontal interval. Their masses are 3 and 1.5 kg. Compare accelerations and explain whether equal force means equal velocity change over the same duration.
Answer and explanation

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

Use each receiver’s mass in Fnet/m under the supplied net-horizontal-force assumption.

Multiply acceleration by the common interval to compare velocity changes.

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 17

Does a larger contact face automatically mean a smaller total force?
Answer and explanation

No. It can lower pressure for the same force by spreading contact area, but total force depends on the stopping interaction.

No. It can lower pressure for the same force by spreading contact area, but total force depends on the stopping interaction.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 18

Which is a qualifying starting design under the stated limit?
Answer and explanation

A qualifies at 4 cm. B at 7 cm does not, despite its lower recorded mean peak. A revision needs retesting.

A qualifies at 4 cm. B at 7 cm does not, despite its lower recorded mean peak. A revision needs retesting.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 19

Visual-model chapter review: A design must stay below 30 N in every trial. Its supplied results are 22, 24 and 35 N. Does a mean of 27 N establish that it meets the criterion? Explain a proper next step.
Answer and explanation

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

Read the criterion as every trial, not the mean.

Use all records before recommending a tested revision.

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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