Momentum (AQA 3.4.1.6) — Handout + Slides + Notes

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Momentum: conserved motion with direction

AQA A Level Physics (3.4.1.6)
Lower Sixth • 35 minutes

Learning objectives

  1. Define momentum: p=mv\mathbf{p}=m\mathbf{v} (vector; units).
  2. Apply conservation of momentum in 1D for an isolated/approximately isolated system.
  3. Use F=Δp/ΔtF=\Delta p/\Delta t and interpret impulse as area under an FF-tt graph.
  4. Distinguish momentum conservation from kinetic-energy conservation.

Key vocabulary

  • vector (magnitude + direction), scalar (magnitude only)
  • system, external force, internal force
  • impulse, elastic, inelastic

What you already know vs what is new today

Already covered (3.4.1.1—3.4.1.5 and 3.4.1.7—3.4.1.8)

  • Core mechanics language: vector/scalar, components, sign conventions, units.
  • Kinematics basics: vv, aa, and interpreting velocity–time graphs (including “area under graph” reasoning).
  • Rate-chain recap: Δx\Delta x over time gives vv, and Δv\Delta v over time gives aa.
    For interested students only: Δa\Delta a over time is jerk (not required for this lesson/spec point).
  • Newton’s laws, especially F=maF=ma for constant mass and free-body diagrams. ([aqa.org.uk][1])
  • Work/energy: Ek=12mv2E_k=\tfrac12 mv^2, W=FscosθW=Fs\cos\theta, power ideas. ([aqa.org.uk][1])

New in 3.4.1.6 (today)

  • Momentum defined and used as a conserved quantity.
  • Impulse and force as rate of change of momentum (F=Δ(mv)/Δt)\left(F=\Delta(mv)/\Delta t\right), and for constant FF, FΔt=ΔpF\Delta t=\Delta p.
  • Force–time graphs: area under the curve as impulse.
  • Applying conservation of momentum quantitatively in 1D; elastic vs inelastic; explosions; “impact time reduces force” contexts. ([aqa.org.uk][1])

One key judgement call for the interview

Prioritise (A) conservation in 1D and (B) impulse = area under FF-tt. These are the most “AQA-visible” outcomes for 3.4.1.6. ([aqa.org.uk][1])

Introduction: symmetry suggests which quantities matter

Reversing direction changes vvv \to -v.
But squaring removes direction: (v)2=v2(-v)^2=v^2.

So it is natural that:

  • momentum is linear in vv because it tracks direction;
  • kinetic energy is proportional to v2v^2 because it ignores direction.

Symmetry helps justify why energy uses v2v^2.
The factor 12\tfrac12 is not explained by symmetry; it comes from geometry of areas under graphs, later.

A debate about “what counts as motion”

For a long time (especially in the 18th–19th centuries) there were serious arguments about the “true measure of motion” in impacts: should it be something like mvmv (directional) or something like mv2mv^2 (non-directional)? Those debates mattered during the industrial revolution, when engines, collisions, and efficiency were practical concerns and “fundamental laws” could bring reputation, patronage, and influence. Engels’ Dialectics of Nature discusses “laws of motion” in this period—when it was not culturally obvious which quantities were the right foundations.

Textbook core 1: momentum and mass

Momentum

[ \mathbf{p}=m\mathbf{v} ]

  • Units: kgms1\mathrm{kg\,m\,s^{-1}}
  • Momentum is a vector (direction matters).

1D sign convention: choose a positive direction and use signs for velocities.

Mass as proportionality (and as “quantity of matter”)

In this topic we can treat mass as the constant of proportionality between pp and vv:

[ p \propto v \quad\Rightarrow\quad p = mv. ] Experiments with ordinary matter show something stronger: mass is also (very nearly) a conserved property of matter in everyday mechanics, used as a measure of “quantity of matter”.

Aside: directional motion is vector motion, whereas “amount” without direction is scalar motion.

Textbook core 2: conservation of momentum

Conservation law (1D)

If the resultant external force on a system is zero (or external impulse is negligible during the interaction),

[ \sum p_{\text{before}} = \sum p_{\text{after}}. ]

What “system” means

A system is the collection of objects you decide to include in the momentum balance (e.g., two trolleys together).

  • Internal forces act between objects inside the system (they come in equal and opposite pairs).
  • External forces come from outside the system (track friction, a hand pushing, gravity components along the track).

Momentum conservation is strongest when external forces are negligible during the short interaction.

Textbook core 3: force, impulse, and graphs

Newton’s second law in momentum form

[ F=\frac{dp}{dt}. ] Definition: Impulse is the change in momentum,

[ \text{Impulse} \equiv \Delta p. ] For (approximately) constant force over Δt\Delta t,

[ \Delta p = F\Delta t. ] More generally,

[ \Delta p=\int F,dt. ]

Graph meaning (exam gold)

  • On an FF-tt graph, area =Δp=\Delta p (impulse).
  • On a pp-tt graph, gradient =dpdt=F=\dfrac{dp}{dt}=F.

What we will do in 35 minutes (lesson spine)

  1. Warm-up diagnostic (3–4 min): quick questions to activate prior knowledge.
  2. Define momentum (5 min): p=mvp=mv, vector and units; why it’s different from energy.
  3. Conservation in 1D (12–15 min): one model example + pupil practice (stick then rebound).
  4. Impulse (8–10 min): F=Δp/ΔtF=\Delta p/\Delta t; force–time area; “increase impact time, reduce force”.
  5. Exit check (2 min): one conservation question + one impulse-from-graph question.

Why this ordering works

It aligns to the spec and uses the demo as “evidence” rather than a fragile dependency.

Warm-up diagnostic questions (3—4 minutes)

  1. A 2kg2\,\text{kg} object at +3m s1+3\,\text{m s}^{-1}: what is EkE_k?
  2. If the resultant force is zero, what can you say about velocity?
  3. Sketch what a large force over a short time might look like on an FF-tt graph.

Aim of the warm-up

Elicit (i) Ekv2E_k\propto v^2, (ii) Newton 1, (iii) readiness for “area under graph” reasoning.

Board plan (teacher script, left → right boards)

BOARD 1 (0:00–0:04) — Title + Do Now

Write:

  • Momentum (AQA 3.4.1.6)
    Aim: define momentum, use conservation in 1D, link to impulse.
  • Do Now:
    1. m=2.0kg, v=3.0m s1m=2.0\,\text{kg},\ v=3.0\,\text{m s}^{-1}. Find EkE_k.
    2. If F=0\sum F=0, what happens to velocity?
    3. State Newton 2 in symbols.

Script: “Start Do Now.” Circulate; cold call.
Feedback: Ek=12mv2=9 JE_k=\tfrac12mv^2=9\text{ J}; velocity constant; F=maF=ma (tease F=dp/dtF=dp/dt).

BOARD 2 (0:04–0:09) — Definition, units, sign convention

Write:

  • p=mv\mathbf{p}=m\mathbf{v}; units kgms1\mathrm{kg\,m\,s^{-1}}; vector
  • Choose ++ direction; signs on velocities
  • Quick check: m=0.50m=0.50 kg, v=4.0v=-4.0 m/s → p=2.0p=-2.0

Script: prediction questions: double vv or mm doubles pp. Add: compare pp vs EkE_k: same EkE_k can have different pp, and vice versa.

BOARD 3 (0:09–0:14) — Demo + conservation condition

Write:

  • pbefore=m1u1+m2u2p_\text{before}=m_1u_1+m_2u_2
  • pafter=(m1+m2)vp_\text{after}=(m_1+m_2)v
  • If u20u_2\approx0: vpred=m1u1m1+m2v_\text{pred}=\dfrac{m_1u_1}{m_1+m_2}
  • Condition: short collision ⇒ external impulse small ⇒ total momentum \approx constant

Demo: stick collision; read u1u_1 and vv. Round; emphasise “close supports model”.

BOARD 4 (0:14–0:24) — Conservation engine + worked example + practice

Write:

  • If Fext0\sum F_\text{ext}\approx0: pbefore=pafter\sum p_\text{before}=\sum p_\text{after}
  • Method: direction → totals → solve
  • Worked: m1=0.40,u1=+3.0;m2=0.60,u2=0v=1.2m_1=0.40,u_1=+3.0; m_2=0.60,u_2=0\Rightarrow v=1.2
  • Practice A: u2=1.0u_2=-1.0 stick → v=0.6v=0.6
  • Practice B: rebound v1=1.0v2=2.672.7v_1=-1.0\Rightarrow v_2=2.67\approx2.7

Optional extension (if flying): explosion-from-rest: momenta equal and opposite.

BOARD 5 (0:24–0:31) — Impulse and FF-tt

Write: F=Δp/ΔtF=\Delta p/\Delta t; Δp=\Delta p= area under FF-tt; triangle Δp=12FmaxΔt\Delta p=\tfrac12F_\text{max}\Delta t Script: crumple zones/catching with “give”.

BOARD 6 (0:31–0:35) — Energy link + exit + summary

Write: Ek=12mv2=12(mv)v=12pvE_k=\tfrac12mv^2=\tfrac12(mv)v=\tfrac12pv.
Exit: p=1.2p=1.2, m=1.5v=0.80m=1.5\Rightarrow v=0.80.
Summary bullets.

Figures (caption immediately after each image)

Experimental results (demo table)

Trial m1m_1 (kg) m2m_2 (kg) u1u_1 (m s1^{-1}) u2u_2 (m s1^{-1}) vv (m s1^{-1}) pbeforep_\text{before} (kg m s1^{-1}) pafterp_\text{after} (kg m s1^{-1})
1 0.50 0.50 +0.80 0.00 +0.41 0.50(0.80)+0=0.400.50(0.80)+0=0.40 1.00(0.41)=0.411.00(0.41)=0.41
2 0.50 0.50 +1.10 0.00 +0.56 0.50(1.10)=0.550.50(1.10)=0.55 1.00(0.56)=0.561.00(0.56)=0.56

Interpretation: agreement won’t be perfect (friction, alignment, sensor noise). “Close” is evidence for the conservation model during the short interaction.

Textbook fill: standard collision outcomes

Inelastic vs elastic

  • Inelastic collision: momentum conserved (if isolated), kinetic energy decreases (converted to heat/sound/deformation).
  • Perfectly inelastic: objects stick together (maximum kinetic energy loss consistent with momentum conservation).
  • Elastic collision: momentum conserved and kinetic energy conserved (idealised; approximately true in some interactions).

Important: momentum conservation depends on external forces, not on whether the collision is elastic. ([aqa.org.uk][1])

Explosions (useful extension)

If a system starts at rest and explodes into two parts, total momentum is still zero:

[ p_1 + p_2 = 0 \quad\Rightarrow\quad p_1 = -p_2. ] This is a clean “direction matters” example.

Textbook fill: common mistakes (and how to avoid them)

  1. Forgetting direction: always set a positive direction and keep signs.
  2. Conserving momentum for the wrong system: include all interacting objects; exclude the hand pushing.
  3. Mixing up graphs:
    • FF-tt: area is Δp\Delta p.
    • pp-tt: gradient is FF.
  4. Assuming kinetic energy is conserved: only in elastic collisions.

Questions to test understanding

Quick checks

  1. Why is momentum a vector? Give a 1D example where the sign matters.
  2. Show dimensionally: N=kgms2\mathrm{N}=\mathrm{kg\,m\,s^{-2}} and dp/dtdp/dt has the same units.
  3. What assumption are you making when you apply momentum conservation to two trolleys on a track?
  4. On an FF-tt graph, what does the area represent?

Short problems

  1. Stick collision: m1=0.50m_1=0.50 kg at +1.2+1.2 m/s hits m2=0.50m_2=0.50 kg at rest and they stick. Find vv.
  2. Rebound: m1=0.40m_1=0.40 kg at +3.0+3.0 m/s hits m2=0.60m_2=0.60 kg at rest. After, v1=1.0v_1=-1.0 m/s. Find v2v_2.
  3. Impulse: approximate a triangular force pulse with Fmax=120 NF_{\max}=120\ \mathrm{N} lasting 0.040 s0.040\ \mathrm{s}. Estimate Δp\Delta p.

Conceptual

  1. Two pulses have the same area but different peaks. What is the same physically? What differs?
  2. Why does increasing collision time reduce injury risk if Δp\Delta p is fixed?

Practical/apparatus notes (low risk, high payoff)

If you want apparatus, the most interview-efficient is something that makes the collision story concrete without setup risk:

  • two low-friction trolleys with Velcro/magnets for “stick together” collisions; and/or
  • a motion sensor/light gates to measure before/after velocities.

If the lesson is fully runnable without apparatus, the demo becomes an enrichment/evidence moment rather than a dependency.

Interactive trolley simulation

Further thoughts (beyond syllabus): curved pp-vv and “effective mass”

What could it represent? (suggestions)

  • A system gaining/losing mass as it moves (e.g. collecting material).
  • A situation where the simple model “one constant mm” isn’t adequate across the whole range (you would test what extra physics is missing).

Thinking exercises

  1. If the curve bends upward (slope increasing with vv), what does that suggest about m(v)m(v)?
  2. How would you estimate the local slope near v=v0v=v_0 using a small Δp\Delta pΔv\Delta v triangle?
  3. Why is it risky to label the shaded area “kinetic energy” in the variable-mass case?
  4. What extra measurements would you want to decide whether curvature is truly “mass changing” or a different effect?

A patent clerk and a deeper unification

A famous patent clerk, Albert Einstein, pursued a single consistent framework in which energy and momentum fit together at very high speeds. Special relativity governs high-velocity particles far beyond Newton’s everyday regime, while reproducing Newtonian mechanics as an excellent approximation at low speeds.

Plenary (syllabus)

  • p=mv\mathbf{p}=m\mathbf{v} (vector; sign convention in 1D).
  • If external impulse is negligible: pbefore=pafter\sum p_{\text{before}}=\sum p_{\text{after}}.
  • Δp=Fdt\Delta p=\int F\,dt (area under FF-tt).
  • Ek=12mv2E_k=\tfrac12 mv^2 is not always conserved even when momentum is.

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