Impulse & Collision Evidence — Print · Grade 12 Science
Chapter: Work, Energy & MomentumImpulse & Collision Evidence
Learning goals
- I can use a force-time area to calculate signed momentum change.
- I can explain collision time, average force, and system conservation.
Readings for this unit
Before this unit
Momentum, velocity signs, graph areas, Newton’s laws.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
How can cushioning change force without eliminating the stopping impulse?
Illustrative evidence case. Two ideal pulse shapes stop the same momentum change over different times.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
How can cushioning change force without eliminating the stopping impulse?
Your first model
Shade force–time areas and label the chosen body.
Build the language from the idea
- impulse
- Time integral of force, equal to momentum change.
- external impulse
- Impulse from forces exerted by objects outside the chosen system.
Momentum p=mv is a vector quantity. Impulse J=∫Fdt changes momentum; for a known average force, J=FavgΔt=Δp. A force-time graph’s signed area gives impulse. It is the change in momentum rather than change in speed alone that matters when direction reverses.
Choose signs before subtracting: Δp=m(vfinal−vinitial). A ball moving right and bouncing left has a larger momentum change magnitude than an otherwise similar ball merely stopping. A negative result identifies the impulse direction under the selected convention.
A longer stopping time can reduce average force for the same momentum change. Padding or a crumple zone increases interaction time and distance; it does not remove the need for momentum change. Peak force can differ greatly from average force, so an average-force calculation does not describe every instant.
Momentum conservation applies to a system when net external impulse is negligible during the interval. Internal forces cancel between interacting bodies by Newton’s third law. Individual object momentum changes while total system momentum can stay the same. Define the system before claiming conservation.
Kinetic energy need not be conserved in an inelastic collision; some energy becomes deformation, vibration, or thermal energy. Total energy remains accounted for in an appropriately inclusive system. In an elastic ideal collision, both total momentum and total kinetic energy are conserved.
Use supplied force-time data or gentle tabletop models. Collision experiments with people, fast projectiles, or fragile objects are unnecessary. Evaluate whether friction or an external support had enough impulse to undermine the isolated-system approximation.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Relate impulse to a momentum change
Impulse J is the time integral of force and equals momentum change. A 0.20 kg ball reversing from+10 to−5 m/s changes momentum by−3.0 kg·m/s, equivalently−3.0 N·s. Subtract signed velocities before multiplying by mass. Using the difference of speed magnitudes would miss part of the reversal. The momentum theorem concerns net force on the selected object; one sensor’s force may not be the entire net force.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What is the average net force if this change takes 0.020 s?
Visual reasoning — after your attempt
Favg=Δp/Δt=−3.0/0.020=−150 N.
Impulse is the time integral of net external force and equals the momentum change of the selected system. For a constant or average net force, J=FavgΔt. It is a vector account, so reversal can produce a larger change than merely stopping from the same initial speed.
A ball rebounding at the original speed changes momentum direction and has a change magnitude twice its initial momentum.
Pause and explain
Why can a rebound have greater impulse than stopping?
Check your reasoning after answering
The momentum must first decrease to zero and then change to the opposite direction, giving a larger vector difference.
Part 2 · Read a force-time trace using area
A force–time trace shows how force varies during contact. If its direction is fixed, the signed area gives impulse; the displayed positive triangle has area 3.0 N·s and average force 150 N over 0.020 s. Peak force 300 N is twice that average for this particular shape. A different pulse with the same area can have a different peak, so neither peak alone nor duration alone determines the momentum change.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What constant force over 0.040 s has the same impulse?
Visual reasoning — after your attempt
75 N gives 75×0.040=3.0 N·s in the same direction.
Signed area under a force-time graph gives impulse. The largest recorded force is a peak, not necessarily the average. A triangular pulse and rectangular pulse with the same peak and duration can have different areas. Include the full pulse and a suitable baseline.
A triangular 100 N pulse lasting 0.020 s has impulse 1.0 N·s, compared with 2.0 N·s for a rectangular pulse.
Pause and explain
Does equal peak force imply equal impulse?
Check your reasoning after answering
No. Pulse duration and shape matter because impulse depends on the complete force-time area.
Part 3 · Separate average-force reduction from guarantees
For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why must rebound be compared when testing two cushions?
Visual reasoning — after your attempt
Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
For the same momentum change, extending stopping time reduces average force in the simple model. Peak force and forces on particular tissues or parts can still depend on the pulse shape and contact geometry. An average-force calculation alone cannot certify protective performance.
A longer stopping interval can explain a design mechanism, but comparable force traces and repeated testing are needed to evaluate the device.
Pause and explain
What does J/Δt establish?
Check your reasoning after answering
It gives the average net force for that interval. It does not uniquely determine peak force or local stresses.
Part 4 · Check system boundaries and sensor meaning
A sensor records a particular contact force, with finite calibration, sampling and response time. The momentum theorem uses net external force on the object over the same interval. If weight matters along the measured component, include its impulse rather than silently equating contact force with net force. Consistent sign conventions, time alignment and measured velocities are essential when comparing a force-trace area with mΔv. Disagreement can reveal a boundary or measurement problem.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
When might weight’s impulse be negligible?
Visual reasoning — after your attempt
For a sufficiently short interval or a horizontal component with no gravity projection, justified against the size of other impulses.
A sensor records one interaction, while momentum change depends on net external force. Weight or other contacts may need inclusion over the interval. Calibration, sampling rate, timing, and rebound velocity affect whether a recorded pulse matches the momentum account.
For a short horizontal cart impact, other horizontal impulses may be negligible. For a longer vertical event, gravity may require explicit inclusion.
Pause and explain
When can one sensor force be treated as the net force?
Check your reasoning after answering
Only when other relevant force contributions are negligible or already accounted for in the stated direction and interval.
Apply the whole unit · Compare rebound and stop intervals
Illustrative 0.200 kg ball has incoming velocity +10.0 m/s and outgoing velocity −5.00 m/s. Compare contact intervals 0.0100 s and 0.0300 s; neglect other impulse along the axis.
Your task
Find momentum change and average contact force for each interval. Explain what the comparison establishes and what it cannot establish about peak force or a protective design.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Δp=0.200×(−5−10)=−3.00 kg·m/s.
For 0.0100 s, Favg=−300 N; for 0.0300 s, Favg=−100 N.
The impulse stays −3.00 N·s despite the different average-force magnitudes.
The longer interval lowers average force for the same specified momentum change. A peak force cannot be reconstructed from average and duration alone, and real designs may also alter rebound velocity. Compare calibrated full traces and consistent initial conditions before drawing a design conclusion.
Scientific references and source-model foundations
Parent / teacher teaching notes
Ask the student to subtract signed velocities and distinguish average from peak. Avoid presenting this supplied calculation as an injury or equipment-safety rating. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
Impulse is signed momentum change; the relevant system boundary determines conservation.
Worked examples
Example 1
- Δp=.20(−4−6).
- Track the negative direction.
Example 2 · Worked application
- Δp=m(v_f−v_i)=0.20(−5−10)=−3.0 kg·m/s.
- Impulse=Δp=−3.0 N·s.
- F_avg=Δp/Δt=−3.0/0.10=−30 N.
A reversal requires signed velocities, not just a speed difference.
Example 3
- Favg=J/Δt.
- 2/.010=200N and 2/.100=20N.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: A reversal requires signed velocities, not just a speed difference.
- F_avg=Δp/Δt; increasing time reduces average magnitude for fixed Δp.
A reversal requires signed velocities, not just a speed difference.
Example 5 · Reason from the visual model
- For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.
- Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
Calculated equal-impulse pulses, not medical or crash-test evidence.
Avoid these mistakes
- A bounce requires direction signs; do not subtract speed magnitudes only.
Check your understanding
Why distinguish average from peak collision force?
Discuss the answer
The force varies through the collision; impulse fixes its area, not a single maximum value.
Return to the phenomenon
Return to the initial case. How can cushioning change force without eliminating the stopping impulse? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Equal impulse permits different average and peak forces; rebound and actual pulse shapes must also be checked.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. How can cushioning change force without eliminating the stopping impulse? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Equal impulse permits different average and peak forces; rebound and actual pulse shapes must also be checked.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A 0.50 kg object slows from 6 m/s to 0 over 0.15 s. Find impulse and average net force. If it rebounds at−2 m/s instead, find the new impulse.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N. With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.
Words to know
- impulse
- Time integral of force, equal to momentum change.
- external impulse
- Impulse from forces exerted by objects outside the chosen system.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
Why distinguish average from peak collision force?
Check your explanation after answering
The force varies through the collision; impulse fixes its area, not a single maximum value.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
-15 N·s
3(0−5)=−15.
Problem 2
Answer and explanation
No; it changes momentum over a longer time and can reduce average force.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 3
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
8 N·s
2(1−(−3))=8.
Problem 4
Answer and explanation
External forces can transfer momentum into or out of the selected system.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 5
Hints
A reversal requires signed velocities, not just a speed difference.
Answer and explanation
12/0.30=40 N.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
It extends the time over which the same impulse is delivered; the force–time area remains the momentum change.
Problem 7
Answer and explanation
Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.
Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 8
Answer and explanation
Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N. With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.
Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N.
With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.