Lesson 10 of 1915 minutes

Impulse, Collisions, and System Momentum

Start with the lesson question, connect the representations, and test the model with evidence.

momentumimpulsecollisionsconservationcenter of masssystems

Learning objectives

  • Relate impulse and momentum change.
  • Apply momentum conservation to collisions and explosions.
  • Evaluate when a momentum model is valid.
Lesson flowHook, model, explanationShow guidance

Inspect the opening phenomenon

Predict what changes, then name the evidence.

Apply in the lab

Name the evidence before reading the answer.

Read only what helps

Then use the lab and recall check.

More when needed

Transcript and resources stay available below.

Course progress

AP Physics 1 — Algebra-Based · Linear Momentum · Lesson 10

Impulse, Collisions, and System Momentum

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Decision challenge

Observe the phenomenon. Then connect the representations.

Use the opening example to make a prediction, identify evidence, and explain which model supports it.

Why Momentum Survives a Crash | AP Physics 1

Choose right as positive. Predict the sign of each cart's momentum and whether the stuck pair moves left or right.

Predict whether momentum and kinetic energy survive a sticking collision, then check the signed calculation.

Before

Choose right as positive. Predict the sign of each cart's momentum and whether the stuck pair moves left or right.

During

Pause after the two initial momenta appear. Add them with signs, then divide by the combined mass before the result is revealed.

After

Explain why system momentum is conserved while kinetic energy decreases in the sticking collision.

Reference drawerTranscript, source notes, scripts, and package status stay tucked away until you need them.7 files

Lesson reading

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15 min

Video script

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Transcript fallback

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courses/ap-physics-1/modules/04-linear-momentum/lessons/01-impulse-collisions-and-system-momentum/video-transcript.md

Momentum Before and After a Cart Collision

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1 hr 30 min

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7 questions / 15 min

Book section:courses/ap-physics-1/modules/04-linear-momentum/lessons/01-impulse-collisions-and-system-momentum/book-section.md
Transcript for accessibility and fallback

# Accessible transcript: Why Momentum Survives a Crash A crash can destroy kinetic energy—but not necessarily momentum. Put both carts inside one system. During the brief collision, their huge forces are internal and cancel in the system total. Point four kilograms moves right at three meters per second. Point six moves left at one. Signed momentum is one point two minus point six: positive point six kilogram-meter per second. They stick, so divide by their combined one kilogram. Final velocity: point six meters per second right. Quick check: was kinetic energy also conserved? Pause. No. Sticking converts some kinetic energy, while system momentum survives when external impulse is negligible. Learn it free at EduQuest AI. ## Visual description Two labeled carts approach, collide, and stick. A boundary encloses both carts. Blue and coral arrows encode opposing signed momenta. The joined cart moves slowly right. A kinetic-energy symbol dims while the total-momentum indicator remains unchanged. All numerical information is repeated in narration.

Reading lab

Core explanation

Connect the lesson's words, diagrams, graphs, evidence, and equations.

Driving question

What can we predict about an interaction when external impulse is limited?

Before learning: predict the collision

Two carts approach each other. One is twice as massive but moves half as fast. Before calculating, decide whether the system's momentum points left, right, or is zero. Then state what additional information you need to predict each cart's final velocity.

Hook: the interaction can be messy while the system stays predictable

A collision may bend bumpers, create sound, and turn kinetic energy into thermal energy. Yet the total momentum of a well-chosen system can remain nearly constant. Momentum lets us predict the before-to-after change without resolving every force during the brief interaction.

Momentum is a vector

For a particle of constant mass,

p=mv.\vec p=m\vec v.

Momentum has units kgm/s\text{kg}\cdot\text{m/s}. Its direction is the velocity direction. A larger object does not automatically have more momentum; mass and velocity both matter.

For a multi-object system,

psystem=imivi.\vec p_{system}=\sum_i m_i\vec v_i.

Choose axes first. In one dimension, right may be positive and left negative. Never add speeds when directions differ.

Momentum vectors before and after a sticking collision

The diagram is equivalent to the statement that the vector sum before equals the vector sum after when external impulse is negligible.

Impulse changes momentum

Impulse is the time accumulation of force:

J=titfFnetdt=Δp.\vec J=\int_{t_i}^{t_f}\vec F_{net}\,dt=\Delta\vec p.

For constant or average net force,

J=FavgΔt.\vec J=\vec F_{avg}\Delta t.

On a net-force-versus-time graph, signed area is impulse. Peak force alone does not determine momentum change.

Force-time graph with a triangular area representing impulse

Worked example: catching safely

A 0.150 kg0.150\text{ kg} ball moving at +20.0 m/s+20.0\text{ m/s} is caught and brought to rest. Its momentum change is

Δp=0(0.150)(20.0)=3.00 kgm/s.\Delta p=0-(0.150)(20.0)=-3.00\text{ kg}\cdot\text{m/s}.

If the catch lasts 0.060 s0.060\text{ s}, the average net force is

Favg=3.000.060=50.0 N.F_{avg}=\frac{-3.00}{0.060}=-50.0\text{ N}.

If the catcher moves their hands so the same momentum change takes longer, the average force magnitude decreases. The impulse remains 3.00 Ns-3.00\text{ N}\cdot\text{s}.

Momentum conservation is a system statement

For a chosen system,

Δpsystem=Jexternal.\Delta\vec p_{system}=\vec J_{external}.

If external impulse is zero or negligible during the interval,

psystem,ipsystem,f.\vec p_{system,i}\approx\vec p_{system,f}.

Internal interaction forces occur in Newton's-third-law pairs. Their impulses cancel in the total system momentum, although each object can undergo a large momentum change.

Collision categories

  • Elastic collision: momentum and kinetic energy are conserved for the system.
  • Inelastic collision: momentum is conserved when external impulse is negligible, but kinetic energy is not.
  • Perfectly inelastic collision: objects stick and share a final velocity.

Momentum conservation alone generally cannot determine two unknown final velocities. An elastic collision needs the additional kinetic-energy condition; a sticking collision supplies a shared-velocity condition.

Worked example: carts stick

A 0.40 kg0.40\text{ kg} cart moving right at +3.0 m/s+3.0\text{ m/s} sticks to a 0.60 kg0.60\text{ kg} cart moving left at 1.0 m/s-1.0\text{ m/s}. Neglect external impulse during impact.

(0.40)(3.0)+(0.60)(1.0)=(0.40+0.60)vf,(0.40)(3.0)+(0.60)(-1.0)=(0.40+0.60)v_f, vf=+0.60 m/s.v_f=+0.60\text{ m/s}.

The positive result means the joined carts move right. Initial kinetic energy is 2.10 J2.10\text{ J} and final kinetic energy is 0.18 J0.18\text{ J}; the difference becomes other forms of energy. Momentum conservation does not imply kinetic-energy conservation.

Explosions and recoil

If a system initially at rest separates under internal forces and external impulse is negligible,

\vec p_1+ ec p_2=0.

The pieces have equal-magnitude, opposite momenta—not necessarily equal speeds. The less massive piece moves faster.

Two recoil objects with equal and opposite momentum arrows

Center-of-mass connection

System momentum relates to center-of-mass velocity:

psystem=Mvcm.\vec p_{system}=M\vec v_{cm}.

Internal collisions can radically change individual velocities while the center-of-mass motion changes only through external impulse.

During learning: model check

For each scenario, identify the system and interaction interval before choosing “momentum conserved.” Ask:

  1. Which forces are external to the system?
  2. Over the selected interval, is their total impulse negligible compared with object momenta?
  3. Is the problem one-dimensional, or must momentum be conserved separately in xx and yy?
  4. What collision condition provides any additional equation?

Two-dimensional collisions

Momentum conservation is vector conservation:

px,i=px,f,py,i=py,f.\sum p_{x,i}=\sum p_{x,f},\qquad \sum p_{y,i}=\sum p_{y,f}.

Choose axes, resolve momentum vectors into components, and keep signs consistent. Kinetic energy is a scalar and does not have xx and yy components.

Misconception clinic

“Momentum is conserved for each object.” Interaction impulses usually change each object's momentum; the total system momentum is the conserved quantity.

“Kinetic energy is conserved in every collision.” It is conserved only in elastic collisions, while total energy is always accounted for in all forms.

“Equal and opposite forces mean equal accelerations.” Third-law forces are equal, but different masses can have different accelerations.

“A larger object always has more momentum.” Momentum depends on both mass and velocity.

After learning: retrieval and transfer

  1. What does signed area under a net-force-time graph represent?
  2. Why can internal forces change each cart's momentum without changing total system momentum?
  3. Which extra condition defines a perfectly inelastic collision?
  4. A firework initially at rest separates into unequal masses. Compare their momenta and speeds.
  5. Describe a situation in which momentum conservation is only approximate and defend the chosen interval.

AP-style evidence routine

  1. Define system, interval, axes, and initial/final states.
  2. Draw a before/after momentum representation.
  3. Evaluate external impulse explicitly.
  4. Write vector momentum conservation symbolically.
  5. Add only justified collision constraints.
  6. Solve, interpret signs, check units, and compare limiting cases.

Key takeaway

Momentum conservation is not a slogan about collisions. It is a consequence of negligible external impulse on a clearly defined system over a stated interval.

Further learning and alignment

Practice labMomentum Before and After a Cart CollisionOpen this when you are ready to apply the model, collect evidence, and check your explanation.1 hr 30 min

Lab: Momentum Before and After a Cart Collision

Objective

How closely is measured system momentum conserved in elastic-like and sticking cart collisions, and how does kinetic-energy behavior differ?

Safety

Work under teacher or responsible-adult supervision. Secure and level the track, use low-speed carts and small masses, keep faces and fingers away from bumpers and magnets, install end stops, and keep the travel lane clear. Do not alter spring bumpers or use projectiles. Inspect carts and track before use.

Materials

  • two low-friction carts with hook-and-loop and elastic or magnetic bumpers;
  • track with soft end stops;
  • balance;
  • two motion sensors, photogates, or fixed slow-motion video with scale markers;
  • spreadsheet or graph paper.

Low-cost alternative: two toy cars on a smooth floor, removable modeling-clay coupler, meterstick markers, and slow-motion video.

Simulation alternative: a teacher-approved collision simulation with exported raw data. Explain which real losses and measurement uncertainties it omits.

Steps

  1. Measure each cart's mass and instrument resolution.
  2. Define the two-cart system, positive direction, and collision interval.
  3. Draw before/during/after diagrams and predict signs of both momentum changes.
  4. Record at least three elastic-like trials with different initial conditions. Keep speeds low.
  5. Record at least three sticking trials using the approved coupler.
  6. Determine velocities from fitted position-time slopes immediately before and after collision; do not use a single noisy frame difference.
  7. Preserve raw position-time data and record rejected trials with reasons.
  8. Calculate total momentum and total kinetic energy before and after each trial.

Expected Result

Total two-cart momentum should agree before and after within experimental uncertainty when external impulse during the collision is small. Kinetic energy should be closer to conserved for elastic-like trials and decrease more substantially for sticking collisions.

Analysis

For each trial calculate

pi=m1v1i+m2v2i,pf=m1v1f+m2v2fp_i=m_1v_{1i}+m_2v_{2i},\qquad p_f=m_1v_{1f}+m_2v_{2f}

and

Ki=12m1v1i2+12m2v2i2,Kf=12m1v1f2+12m2v2f2.K_i=\frac12m_1v_{1i}^2+\frac12m_2v_{2i}^2,\qquad K_f=\frac12m_1v_{1f}^2+\frac12m_2v_{2f}^2.
  • Report pfpip_f-p_i with propagated or bounding uncertainty.
  • Graph pfp_f versus pip_i with a reference line of slope one.
  • Compare fractional momentum discrepancy and fractional kinetic-energy change.
  • Discuss track tilt, rolling resistance, sensor alignment, frame rate, rotational energy, and collision-duration selection.

Reflection Questions

  1. Which system choice makes the collision forces internal?
  2. Did a small momentum discrepancy prove momentum failed to be conserved? Explain using uncertainty.
  3. Why can kinetic energy decrease while total energy remains accounted for?
  4. Which measured velocity contributes most to uncertainty in system momentum?

Claim-evidence-reasoning conclusion

Make separate claims about momentum and kinetic energy for each collision category. Cite numerical comparisons and uncertainties, then connect them to external impulse and energy transformation.

Accessibility

Offer roles for apparatus safety, release, measurement, data recording, graphing, uncertainty analysis, and oral reporting. Use high-contrast cart markers, tactile track orientation, screen-reader-friendly tables, and verbal graph descriptions. All conclusions can be completed from shared class data without physically operating carts.

Extension Challenge

Predict the shared final velocity for a new safe sticking-collision condition using only measured initial velocities and masses. Test it once, compare prediction with measurement and uncertainty, and explain any discrepancy.