Momentum & Collisions — Print · Grade 12 Science
Chapter: Work, Energy & MomentumMomentum & Collisions
Learning goals
- I can calculate momentum with p = mv and apply the impulse-momentum relationship J = FΔt = Δp.
- I can use conservation of momentum to analyze elastic and inelastic collisions.
Readings for this unit
Before this unit
Momentum vectors and external impulse.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why can colliding bodies conserve momentum while losing kinetic energy?
Illustrative evidence case. Two carts stick together after a short collision.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why can colliding bodies conserve momentum while losing kinetic energy?
Your first model
Choose both carts as the system and name external impulses.
Build the language from the idea
- momentum
- The 'quantity of motion' of an object, p = mv; a vector measured in kilogram-metres per second (kg·m/s).
- impulse
- The product of force and the time it acts, J = FΔt; it equals the change in momentum, Δp.
- conservation of momentum
- When no net external force acts, the total momentum of a system is the same before and after an interaction.
- elastic collision
- A collision in which both momentum and kinetic energy are conserved.
- inelastic collision
- A collision in which momentum is conserved but some kinetic energy is lost; in a perfectly inelastic collision the objects stick together.
Momentum is an object's 'quantity of motion,' defined simply as p = mv — mass times velocity. Because it multiplies the two, a slow but massive object and a fast but light one can carry the same momentum: a loaded truck rolling at walking pace and a bullet in flight are both hard to stop. Momentum is a *vector*, so direction counts; something moving left has momentum opposite in sign to something moving right, and that bookkeeping of signs is what makes momentum such a precise tool. Its unit is the kilogram-metre per second (kg·m/s).
To change an object's momentum you must apply a force for some time, and that combination is called impulse: J = FΔt. The impulse delivered exactly equals the change in momentum, J = Δp — the impulse-momentum theorem. This relationship has a life-saving twist. The momentum change in a car crash is fixed by how fast you were going, but F = Δp/Δt shows that stretching the collision over a longer time Δt shrinks the force. That is precisely how airbags, crumple zones, padded dashboards, and even bending your knees when you land all protect you: they lengthen the stopping time and soften the force.
The crown jewel of this lesson is conservation of momentum: if no net external force acts on a system, its total momentum is identical before and after any interaction. This follows straight from Newton's third law — in a collision the two objects push on each other with equal and opposite forces for the same time, so the impulses cancel and whatever momentum one gains, the other loses. It holds whether objects bounce apart, stick together, or explode outward, which is why it works for colliding billiard balls, coupling train cars, and a rocket hurling exhaust one way to drive itself the other.
Collisions come in two flavours, and telling them apart hinges on energy. In an elastic collision, both momentum *and* kinetic energy are conserved; the objects bounce off with no energy lost to heat, the way hardened steel balls or gas molecules nearly do. In an inelastic collision, momentum is still conserved but some kinetic energy is converted into heat, sound, and permanent deformation. The extreme case is a perfectly inelastic collision, in which the objects stick and move off together — like a dart burying itself in a board — losing the most kinetic energy possible while still, always, conserving momentum.
A neat special case makes the ideas concrete: in a head-on elastic collision between two *equal* masses, they simply exchange velocities. A moving ball striking an identical stationary one stops dead while the target shoots off at the incoming speed — exactly what you see click across a Newton's cradle. Analyzing any collision then becomes a two-part checklist: momentum is *always* conserved, so write p_before = p_after; kinetic energy is conserved *only if* the collision is elastic. Keeping those two rules straight lets you predict the outcome of impacts from the subatomic to the astronomical.
Momentum conservation is one of the deepest principles in physics, holding even where Newton's laws break down, in the quantum and relativistic worlds. For everyday engineering it is indispensable: it tells crash-safety teams how to design a survivable impact, lets rocket scientists compute thrust in the vacuum of space where there is nothing to push against, and allows physicists to reconstruct particle collisions they never directly see — all by insisting that the total momentum going in must equal the total coming out.
Model check: Momentum conservation does not automatically imply kinetic-energy conservation. Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Keep momentum vector information
Momentum is p=mv, a vector with units kg·m/s. Opposite velocities produce opposite momentum signs along a chosen axis. A 2 kg body moving at+3 m/s and a 1 kg body at−4 m/s have momenta+6 and−4 kg·m/s, totaling+2. Their kinetic energies are both positive and add differently. Keep component signs through the calculation rather than adding speed magnitudes as though all bodies moved together.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What is the total kinetic energy of these two bodies?
Visual reasoning — after your attempt
½×2×3²+½×1×4²=9+8=17 J.
Momentum p=mv carries velocity direction and has units kg·m/s. In one dimension, choose a positive axis before adding momenta. Two moving objects can have zero total momentum with positive total kinetic energy, so a momentum sum does not determine all energy.
Equal masses moving at +2 and −2 m/s have opposite momenta that cancel; both still have kinetic energy.
Pause and explain
Why does zero total momentum not mean all objects are at rest?
Check your reasoning after answering
Opposite momenta can sum to zero while individual objects continue moving; their kinetic energies do not cancel by direction.
Part 2 · Choose the collision system and interval
For a selected system, total momentum change equals total external impulse. During a short two-body collision, the bodies’ mutual forces give opposite internal impulses that cancel in the combined account. Total momentum is conserved only when external impulse is negligible over that interval. Gravity, supports or friction may still exist; the approximation concerns their net impulse, not a claim that every external force vanishes.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Can momentum be approximately conserved even when gravity acts?
Visual reasoning — after your attempt
Yes, if the relevant external impulse is negligible or cancels over the chosen short interval and components.
Total momentum is conserved for a selected system when net external impulse is negligible over the interval. Collision forces between included objects are internal and exchange momentum. A short collision can make some external effects small, but that assumption must be justified rather than automatic.
Two carts form one useful system during contact. Floor friction can still exert external impulse, especially over a long interval.
Pause and explain
What matters more than simply calling a collision isolated?
Check your reasoning after answering
The external impulse relative to the internal exchange and measurement precision must be assessed for the chosen interval.
Part 3 · Distinguish collision categories by energy
An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
For the initial A+B case sticking together, what final velocity follows?
Visual reasoning — after your attempt
vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
Elastic collisions conserve total kinetic energy as well as momentum in the ideal system. Inelastic collisions do not conserve total kinetic energy; perfectly inelastic objects stick together. Total energy still has an account through deformation, thermal effects, and other stores.
Sticking is not evidence of momentum loss. It changes how the common final velocity is constrained and where energy is stored.
Optional guided lab · Compare momentum with kinetic energy
Optional guided lab · Compare momentum with kinetic energy
Optional external exploration — WeduLab. Complete the local explanation first. The task and practice record here are independently written for this course. Choose this in place of comparable practice or allow extra time; it does not add a required session to the year plan.
Predict the final velocity when a 0.50 kg cart moving right at 2.00 m/s sticks to a stationary 0.50 kg cart. Define right as positive and take both carts as the system. Momentum is mass times velocity; kinetic energy is one half mass times speed squared. Equal total momentum does not require equal kinetic energy before and after.
Explore for about 15–20 minutes after learning collision categories. Use the conservation-of-momentum topic. Compare spring/elastic bumpers with the sticking/Velcro case, using the same masses and starting velocities when possible. Record both carts’ signed initial and final velocities and calculate system momentum and kinetic energy. If controls cannot reproduce the selected numbers, label and analyze your actual setup, or use the original ideal records below.
Negligible net external impulse is the reason the selected two-cart momentum is conserved. Real tracks, bumpers, alignment and sensors differ from ideal models. Kinetic energy missing from cart motion is transferred into deformation, thermal energy, sound and other internal motion; energy has not vanished. Sticking gives the minimum final translational kinetic energy for the stated isolated two-cart initial condition in this frame, not a universal ranking of every possible collision.
No-lab route: the original supplied record below contains everything needed for the question. It is not measured evidence collected from WeduLab. Label any separate observations you collect as simulation output, with settings and date.
| Original ideal case: m1 = m2 = 0.50 kg | u1 / m s⁻¹ | u2 / m s⁻¹ | v1 / m s⁻¹ | v2 / m s⁻¹ |
|---|---|---|---|---|
| Elastic equal-mass collision | +2.00 | 0.00 | 0.00 | +2.00 |
| Perfectly inelastic sticking collision | +2.00 | 0.00 | +1.00 | +1.00 |
Question 1
Calculate total momentum and kinetic energy before and after each case. Explain why the common final velocity is +1.00 m/s in the sticking case. What evidence would you need before treating real-track momentum differences as a failure of the conservation law?
Worked solution and marking guidance
Initial momentum is 0.50 × 2.00 = +1.00 kg m/s. Initial kinetic energy is 0.5 × 0.50 × 2.00² = 1.00 J.
The elastic row retains +1.00 kg m/s and 1.00 J. For sticking, momentum gives (0.50 + 0.50)v = 1.00, so v = +1.00 m/s. Its final kinetic energy is 0.50 J.
Check the system boundary, external impulse from friction or apparatus, timing, velocity signs and measurement uncertainty before interpreting a discrepancy.
Both ideal rows conserve momentum at +1.00 kg m/s. Only the elastic row conserves translational kinetic energy. The sticking row transfers 0.50 J out of the two carts’ translational kinetic energy while total energy is conserved.
- Define the two-cart system and signed direction.
- Calculate both quantities with units, keeping momentum and energy distinct.
- Account for the kinetic-energy change and discuss external impulse and measurement limits.
Pause and explain
Is momentum conservation restricted to elastic collisions?
Check your reasoning after answering
No. It also applies to inelastic collisions when net external impulse is negligible; kinetic energy is the quantity that differs.
Part 4 · Check results in the selected frame
Under ordinary low-speed frame changes by speed u, each velocity becomes v−u. The combined momentum changes by−Mu for total mass M, so its numerical value need not match across frames. Conservation can hold in either inertial frame if the same frame is used before and after. Kinetic-energy values also change with frame; do not mix a laboratory initial energy with a moving-frame final energy.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is laboratory momentum+2 but moving-frame momentum−1 here?
Visual reasoning — after your attempt
The 3 kg system has its momentum shifted by−3×1=−3 kg·m/s under the stated frame change.
Use signed velocities, solve the momentum equation, and separately calculate kinetic energies if requested. A common final velocity can lie between initial velocities in a simple sticking case. Direction and dimensional checks can expose accidental use of speeds or wrong masses.
A negative final velocity means the joined system moves toward the negative axis, not an impossible collision.
Pause and explain
Why calculate energy after solving momentum?
Check your reasoning after answering
It tests a different account and can classify the collision; conserving momentum alone does not guarantee equal initial and final kinetic energy.
Apply the whole unit · Analyze two carts that stick
Illustrative 2.00 kg cart moves at +3.00 m/s and 1.00 kg cart at −1.00 m/s. They stick; net external impulse during contact is negligible.
Your task
Find common velocity, initial and final kinetic energy, and mechanical-energy redistribution. Explain why opposing momenta must carry signs.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Initial momentum=2×3+1×(−1)=5.00 kg·m/s.
Final velocity=5.00/(2+1)=+1.67 m/s.
Initial K=9.00+0.500=9.50 J; final K=½×3×(5/3)²≈4.17 J; difference≈5.33 J.
The final direction is positive, and the collision is perfectly inelastic in the supplied model. Momentum is conserved while about 5.33 J of translational kinetic energy enters other stores. That difference does not specify how much goes separately into deformation, sound, or heating without further measurements.
Scientific references and source-model foundations
Parent / teacher teaching notes
Require signs before multiplication and separate momentum and energy totals. Ask where the interaction forces act when students question why sticking can conserve momentum. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
System momentum change equals external impulse. Negligible external impulse allows momentum conservation; kinetic energy is also conserved only in the elastic collision model.
Momentum
p = mv
m is mass in kg; v is velocity in m/s; p is a vector in kilogram-metres per second (kg·m/s).
Impulse-Momentum Theorem
J = FΔt = Δp
F is the average force; Δt is the contact time; the impulse equals the change in momentum, Δp = mv_f − mv_i.
Conservation of Momentum
m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
with no net external force, the total momentum before an interaction equals the total after. For a perfectly inelastic collision the objects stick together: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.
Worked examples
Example 1
- Use p = mv.
- p = 1,500 × 20.
- = 30,000.
Example 2 · Worked application
- Initial momentum=2×3+1×0=6 kg·m/s.
- Total mass after sticking=3 kg.
- v_final=6/3=2 m/s in the initial direction.
Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.
Example 3
- Momentum is conserved: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.
- 2.0 × 3.0 + 1.0 × 0 = (2.0 + 1.0) × v′, so 6.0 = 3.0 × v′.
- v′ = 6.0 ÷ 3.0 = 2.0.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.
- Kinetic energy can transfer to deformation, heat and sound; momentum conservation requires negligible net external impulse.
Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.
Example 5 · Reason from the visual model
- An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.
- vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
Original classification table assumes an isolated collision system for its momentum column.
Avoid these mistakes
- Momentum conservation does not automatically imply kinetic-energy conservation.
Check your understanding
2 kg at 3 m/s sticks to 1 kg atrest. Find final velocity.
Discuss the answer
Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.
Return to the phenomenon
Return to the initial case. Why can colliding bodies conserve momentum while losing kinetic energy? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Internal collision impulses cancel in the system account; kinetic energy can become deformation and thermal energy.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why can colliding bodies conserve momentum while losing kinetic energy? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Internal collision impulses cancel in the system account; kinetic energy can become deformation and thermal energy.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A 3 kg cart at 4 m/s sticks to a 1 kg cart initially at rest. Neglect external impulse. Find final velocity and kinetic-energy decrease.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Initial momentum=12 kg·m/s, so vf=3 m/s. Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.
Words to know
- momentum
- The 'quantity of motion' of an object, p = mv; a vector measured in kilogram-metres per second (kg·m/s).
- impulse
- The product of force and the time it acts, J = FΔt; it equals the change in momentum, Δp.
- conservation of momentum
- When no net external force acts, the total momentum of a system is the same before and after an interaction.
- elastic collision
- A collision in which both momentum and kinetic energy are conserved.
- inelastic collision
- A collision in which momentum is conserved but some kinetic energy is lost; in a perfectly inelastic collision the objects stick together.
Descartes, Huygens, and the birth of conservation of momentum
Long before the word 'energy' was well understood, René Descartes proposed in the 1600 s that the total 'quantity of motion' in the universe is fixed. His early version was flawed because he ignored direction, but the Dutch physicist Christiaan Huygens corrected it by treating motion as a directed (vector) quantity and worked out the rules for elastic collisions. When Newton later published his laws, conservation of momentum fell out naturally from his third law. It remains one of the most robust principles in all of physics, still holding in the quantum and relativistic realms where much of Newton's mechanics needs revision.
Misconception: momentum and kinetic energy are not the same 'amount of motion'
Both p = mv and KE = ½mv² measure something about motion, so students often blur them — but they behave very differently. Momentum is a vector (direction matters, and opposite motions can cancel) and it is conserved in every collision. Kinetic energy is a scalar (always positive) and is conserved only in elastic collisions. In a head-on crash where two identical cars moving at equal speeds stick together and stop, the total momentum was zero before and after (conserved), while all the kinetic energy vanished into crumpling metal and heat. Keeping the two quantities separate is the key to analyzing any collision.
Impulse and the physics of staying alive in a crash
Impulse, J = FΔt = Δp, is the single most important idea in crash safety. Your change in momentum in an accident is fixed the instant the crash begins — but the force you feel depends on how long the stop takes. Crumple zones let the front of a car collapse over a longer distance and time; airbags cushion your head over tens of milliseconds instead of one; even a boxer 'rolling with' a punch or a gymnast bending their knees on landing is stretching Δt to shrink F = Δp/Δt. The same equation explains why landing on concrete hurts far more than landing on a crash mat, even from the same height.
Rockets: pushing on nothing at all
A common myth is that rockets fly by 'pushing against the air.' They don't — they work best in the vacuum of space, and the reason is pure conservation of momentum. A rocket hurls a stream of exhaust gas backward at enormous speed; to keep the system's total momentum constant, the rocket must gain an equal and opposite momentum forward. No ground, no air, and nothing external to push against is required — only the exhaust being thrown one way and the rocket recoiling the other. The same principle lets an astronaut drift back to a spacecraft by throwing a tool in the opposite direction.
In the real world
Rockets work by conservation of momentum: they hurl exhaust gas backward at high speed, and the equal-and-opposite forward momentum drives the rocket ahead — which is why a rocket can accelerate in the vacuum of space with nothing to push against.
Exit check
2 kg at 3 m/s sticks to 1 kg atrest. Find final velocity.
Check your explanation after answering
Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
2 kg·m/s
p = mv = 0.50 × 4.0 = 2.0 kg·m/s.
Problem 2
Answer and explanation
10 kg·m/s
Impulse equals the change in momentum: J = Δp = m(v_f − v_i) = 0.50 × (20 − 0) = 10 kg·m/s.
Problem 3
Answer and explanation
3 m/s
Conservation of momentum: 3.0 × 4.0 + 1.0 × 0 = (3.0 + 1.0)v′, so 12 = 4.0v′ and v′ = 3.0 m/s.
Problem 4
- Momentum only
- Kinetic energy only
- Both momentum and kinetic energy
- Neither momentum nor kinetic energy
Answer and explanation
Momentum only
Momentum is conserved in every collision with no external force. In an inelastic collision, some kinetic energy is converted to heat, sound, and deformation, so only momentum is conserved.
Problem 5
Answer and explanation
0.6 m/s
Total momentum starts at 0, so 50 × v + 2.0 × 15 = 0, giving 50v = −30 and v = −0.6 m/s. The astronaut recoils at 0.6 m/s in the opposite direction.
Problem 6
Hints
Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.
Answer and explanation
Initial momentum=4; final mass=4; velocity=1 m/s.
Problem 7
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Momentum is a vector; opposite directions contribute opposite signed components.
Problem 8
Answer and explanation
vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.
vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.
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 9
Answer and explanation
Initial momentum=12 kg·m/s, so vf=3 m/s. Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.
Initial momentum=12 kg·m/s, so vf=3 m/s.
Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.