Hold up a single sheet of printer paper by two corners so it drapes down in front of your mouth. Ask students: 'If I blow HARD across the TOP of this paper, which way will the paper move — down, stay, or up?' Take a quick hand-vote and tally on the board. Most students predict down (they think fast air pushes the paper). Blow steadily across the top — the paper lifts UP. Do it twice so everyone sees it. Do NOT explain yet. Say: 'By the end of class you will be able to tell me exactly why that happened, and you will calculate pressure and speed changes in a real pipe.' Connect to real world: this is the same physics behind airplane lift, why shower curtains billow inward when the water runs, and why a semi-truck passing you at highway speed briefly pulls your car toward it.

1. 7m

   Continuity: why fluid speeds up in a narrow pipe

   Content

   For an incompressible fluid in steady flow, whatever volume of water enters one end of a pipe per second must exit the other end per second — mass is conserved and density is constant, so volume flow rate Q = Av is the same at every cross-section. That gives the continuity equation A₁v₁ = A₂v₂. When the pipe narrows, A shrinks, so v must grow in exact proportion: cut the area in half and the flow speed doubles. Worked example: water flows at v₁ = 2.0 m/s in a pipe of cross-sectional area A₁ = 4.0 × 10⁻⁴ m². The pipe narrows to A₂ = 1.0 × 10⁻⁴ m². Then v₂ = (A₁v₁)/A₂ = (4.0 × 10⁻⁴)(2.0)/(1.0 × 10⁻⁴) = 8.0 m/s. The volume flow rate stays Q = 8.0 × 10⁻⁴ m³/s at both points. The assumption of incompressibility matters: liquids are nearly perfect for this, but a real gas being compressed would violate the equation because ρ changes.

   Delivery

   Anchor the reasoning in 'nothing piles up, nothing disappears' — molecules can't compress and can't leak, so the same volume per second passes every cross-section. Head off the biggest misconception directly: many students think 'narrower pipe = less room = slower flow.' Ask, 'If the same water per second has to squeeze through a smaller opening, does it go faster or slower?' Guide them to see the water HAS to speed up. On the worked example, have students do the arithmetic silently on whiteboards before you reveal 8.0 m/s. Emphasize the proportional reasoning shortcut: area went down by a factor of 4, so speed goes up by a factor of 4 — no calculator needed. Finally flag the incompressible-fluid caveat so students don't misapply it to compressed air in an AP FRQ.

2. 8m

   Bernoulli's principle: pressure, speed, and height trade off

   Content

   Bernoulli's equation is conservation of energy per unit volume along a streamline of an ideal fluid: P + ½ρv² + ρgh = constant. The three terms are pressure energy density, kinetic energy density, and gravitational potential energy density. If a parcel of fluid speeds up along a streamline, its kinetic term grows, so something else must shrink — usually the pressure term. That is Bernoulli's principle: along a streamline, faster flow means lower pressure. For a horizontal pipe h doesn't change, so P₁ + ½ρv₁² = P₂ + ½ρv₂². Combined with continuity, a constriction has both higher speed AND lower pressure. Worked example: horizontal water pipe, ρ = 1000 kg/m³. At point 1 (wide): v₁ = 2.0 m/s, P₁ = 150 000 Pa. At point 2 (narrow, from prior example): v₂ = 8.0 m/s. Then P₂ = P₁ + ½ρ(v₁² − v₂²) = 150 000 + ½(1000)(4.0 − 64) = 150 000 − 30 000 = 120 000 Pa. Pressure dropped by 30 kPa where flow sped up. The Venturi tube visual makes this concrete: vertical pressure-tap tubes stand shorter above the narrow section because pressure there is lower.

   Delivery

   Frame Bernoulli as 'conservation of energy for a moving fluid — pressure, motion, and height share a fixed energy budget.' Slow down on the biggest misconception: students think 'faster air must push harder, so higher pressure.' Attack this head-on — the fast region has lower pressure precisely because energy was spent speeding the fluid up. Return to the hook: air blown across the top moves faster than the still air below, so pressure above drops, and the higher pressure below pushes the paper UP. On the worked example, ask students to predict the SIGN of ΔP before computing — they should reason 'v went up, so P went down.' Point out the Venturi visual: shorter water columns at the narrow section = lower static pressure. Flag the height term for a later problem: raise the pipe and ρgh grows, which changes the P and v balance.

3. 5m

   Putting them together: ranking speed and pressure

   Content

   On any horizontal pipe with wide-narrow-wide sections and points 1, 2, 3 labeled, continuity fixes the speed ranking and Bernoulli fixes the pressure ranking. If A₁ = A₃ > A₂, then v₂ > v₁ = v₃, and therefore P₂ < P₁ = P₃. The two laws point opposite directions: where speed is highest, pressure is lowest. This is the exact pattern AP FRQs use — students must justify BOTH rankings, one with continuity (conservation of mass), one with Bernoulli (conservation of energy). Real-world tie-in: a roof lifts off a house in high wind because air races over the roof (high v, low P outside) while interior air is nearly still (high P inside); the pressure difference pushes the roof upward.

   Delivery

   Do this beat as rapid call-and-response. Point to each labeled region and have the class chorus 'higher speed, lower pressure' or 'lower speed, higher pressure.' This builds the AP-FRQ verbal template. Emphasize that on the AP exam, a correct RANKING is not enough — students must cite the LAW (continuity or Bernoulli) as justification. Close by returning explicitly to the paper demo from the hook: fast air on top, slow air on bottom, higher pressure below pushes the paper up. Students should now be able to explain the demo in one sentence using Bernoulli.

1. 25m

   Bernoulli demos + Venturi analysis labLab

   Groups of 3. Distribute the handout below. Part 1 is three quick Bernoulli demos where students predict FIRST, then test, then explain using P + ½ρv² = constant. Part 2 is a Venturi calculation using continuity and Bernoulli. Circulate and push groups to CITE the law in their explanations — this mirrors AP FRQ scoring. Targets SP 1 Creating Representations (streamline sketches), SP 2 Mathematical Routines (continuity + Bernoulli calculations), and SP 3 Scientific Questioning and Argumentation (predict-test-justify). Timing: 3 min setup and read handout, 12 min Part 1 demos, 8 min Part 2 problem, 2 min share-out. Student handout — Fluids in Motion Lab Part 1 — Predict, Test, Explain (3 demos) For each demo: (1) write your PREDICTION before doing anything, (2) perform the demo, (3) write a one-sentence EXPLANATION citing either the continuity equation or Bernoulli's principle by name. Predict before you blow. No exceptions. - Demo A — Single sheet. Hold one sheet of paper by two top corners so it hangs in front of your face. Blow HARD and steadily across the TOP surface. - Prediction: the paper will move ______ (up / down / stay). - Observation: the paper moved ______. - Explanation (name the principle): ____________________ - Demo B — Two hanging sheets. Tape two sheets of paper to the edge of the table so they hang parallel, about 5 cm apart. Blow steadily straight down BETWEEN them. - Prediction: the sheets will ______ (spread apart / come together / stay). - Observation: ______ - Explanation: ____________________ - Demo C — Two ping-pong balls on strings. Tape two 20 cm strings to the underside of the table, hang a ping-pong ball from each so they dangle about 3 cm apart at the same height. Blow steadily through a straw straight down BETWEEN the two balls. - Prediction: the balls will ______ (spread apart / come together / stay). - Observation: ______ - Explanation: ____________________ Part 2 — Venturi Calculation A horizontal pipe carrying water (ρ = 1000 kg/m³) has a wide section (point 1) and a narrow section (point 2). At point 1: A₁ = 6.0 × 10⁻⁴ m², v₁ = 1.5 m/s, P₁ = 200 000 Pa. At point 2: A₂ = 2.0 × 10⁻⁴ m². 1. Sketch the pipe. Label points 1 and 2. Draw velocity arrows — make the arrow at point 2 longer than at point 1 to show the speed increase. 2. Use continuity to find v₂. Show the equation and your algebra. - v₂ = ______ m/s 3. Use Bernoulli's equation (horizontal, so no height term) to find P₂. Show your work. - P₂ = ______ Pa 4. Rank and justify: - Higher speed at point ______ — justified by ______________. - Higher pressure at point ______ — justified by ______________. 5. Argue from evidence: If a Venturi tube had vertical open pressure-tap columns rising out of the pipe at points 1 and 2, at which point would the water column stand HIGHER? Explain in one sentence.

   Materials

   • Printer paper (2 sheets per pair)
   • 2 ping-pong balls per group
   • String (~40 cm per group) and tape
   • Straws (2 per group)
   • Small beaker of water with a few drops of food coloring
   • Stopwatch or phone timer
   • Metric ruler
   • Whiteboard markers

   Example outputs

   • Demo A: Paper moves UP. Faster air across the top has lower pressure by Bernoulli's principle; the slower (higher-pressure) air underneath pushes the sheet upward.
   • Part 2 Q2: A₁v₁ = A₂v₂ → v₂ = (6.0 × 10⁻⁴)(1.5)/(2.0 × 10⁻⁴) = 4.5 m/s.
   • Part 2 Q3: P₂ = P₁ + ½ρ(v₁² − v₂²) = 200 000 + ½(1000)(2.25 − 20.25) = 200 000 − 9000 = 191 000 Pa.
   • Part 2 Q4: Higher speed at point 2, by continuity (A₂ < A₁ forces v₂ > v₁). Higher pressure at point 1, by Bernoulli (slower flow has more of the energy in pressure).
   • Part 2 Q5: The column at point 1 stands higher because the static pressure there is greater; the narrow, fast region at point 2 has lower pressure and therefore a shorter column.

1. Water flows through a horizontal pipe. At point 1 the cross-sectional area is A₁ = 8.0 × 10⁻⁴ m² and the flow speed is v₁ = 3.0 m/s. At point 2 the pipe narrows to A₂ = 2.0 × 10⁻⁴ m². Calculate v₂. (Targets SP 2 Mathematical Routines.)

   calculationBy continuity A₁v₁ = A₂v₂, so v₂ = (A₁v₁)/A₂ = (8.0 × 10⁻⁴)(3.0)/(2.0 × 10⁻⁴) = 12 m/s. The area dropped by a factor of 4, so the speed rose by a factor of 4.

2. Using the values from the previous problem, if the water pressure at point 1 is P₁ = 180 000 Pa and ρ = 1000 kg/m³, what is the pressure P₂ at the narrow section? (Targets SP 2 Mathematical Routines.)

   calculationBernoulli (horizontal): P₁ + ½ρv₁² = P₂ + ½ρv₂². P₂ = P₁ + ½ρ(v₁² − v₂²) = 180 000 + ½(1000)(9 − 144) = 180 000 − 67 500 = 112 500 Pa. Pressure dropped 67.5 kPa where the fluid sped up — consistent with Bernoulli's principle.

3. During a hurricane, wind blows very fast across the top of a flat roof while the air inside the house is nearly still. Explain, in 2–3 sentences using Bernoulli's principle, why the roof can be lifted off the house. (Targets SP 3 Scientific Questioning and Argumentation.)

   short answerBy Bernoulli's principle, along a streamline higher flow speed corresponds to lower pressure. The fast wind over the roof has low pressure above, while the still air inside the house maintains higher (near-atmospheric) pressure below. The pressure difference produces a net upward force on the roof — if it exceeds the roof's weight and the strength of the fasteners, the roof lifts off.

4. A student claims: 'Fluid moving through a narrower pipe must slow down because there is less room for it to move.' Is this correct? Justify your answer using a conservation law. (Targets SP 3 Scientific Questioning and Argumentation.)

   short answerIncorrect. By conservation of mass (continuity, A₁v₁ = A₂v₂) an incompressible fluid must have the same volume flow rate at every cross-section. If the area A decreases, the flow speed v must INCREASE in inverse proportion so that Av stays constant. So fluid speeds up — not down — where the pipe narrows.

5. A horizontal pipe has three labeled points: point 1 (wide), point 2 (narrow), point 3 (wide again, same area as point 1). Which of the following correctly ranks the pressures? (Targets SP 3 Scientific Questioning and Argumentation.)

   multiple choiceCorrect answer: P₂ < P₁ = P₃. By continuity v₂ > v₁ = v₃ (smaller area means larger speed). By Bernoulli's principle along a horizontal streamline, higher speed corresponds to lower pressure, so P₂ is the lowest and P₁ equals P₃. (Distractors students might pick: P₂ > P₁ = P₃ — the 'faster means more forceful' misconception; P₁ > P₂ > P₃ — treating the pipe as if energy is being lost linearly.)

• ☐Printer paper — 2 sheets per group of 3
• ☐Ping-pong balls — 2 per group
• ☐String — ~40 cm per group
• ☐Masking tape
• ☐Drinking straws — 2 per group
• ☐Small beaker of water with food coloring (optional visual)
• ☐Metric rulers
• ☐Stopwatches (or phone timers)
• ☐Whiteboards and markers for group work
• ☐Handout printed for each student (Parts 1 and 2)
• ☐Calculators