Fluids — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Core fluid statics and dynamics concepts including density and pressure, Pascal's principle, Archimedes' buoyancy principle, the continuity equation, and Bernoulli's equation for AP Physics 2 exam success.

You should already know: AP Physics 1 or equivalent.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Are Fluids?

Fluids are substances that deform continuously under applied shear stress, including both liquids and gases. Unlike rigid solids covered in AP Physics 1, fluids do not retain a fixed shape, so their behavior is described using bulk properties rather than individual particle motion for most exam contexts. The AP Physics 2 CED splits fluid content into statics (fluids at rest) and dynamics (fluids in motion), with the topic making up 10-15% of your total exam score across multiple-choice and free-response sections.

2. Density and pressure

Density is an intensive bulk property of a fluid, defined as mass per unit volume, with the formula: $$\rho =\frac{m}{V}$$ Where $\rho$ = density (units: ${\text{kg/m}}^3$), $m$ = mass, $V$ = volume. Density is independent of sample size, so 1 L and 100 L of pure water both have a density of $1000{\text{ kg/m}}^3$, a value you should memorize for the exam.

Pressure is defined as the perpendicular force applied per unit area of a surface: $$P=\frac{F}{A}$$ Where $P$ = pressure (units: Pascals, $1\text{ Pa}=1{\text{ N/m}}^2$), $F$ = perpendicular force, $A$ = area. For static fluids, pressure increases with depth due to the weight of the fluid above a given point, a relationship called hydrostatic pressure: $$P=P_0+\rho gh$$ Where $P_0$ = atmospheric pressure ($\approx 1.013\times 10^5\text{ Pa}$ at sea level), $h$ = depth below the fluid surface, $g=9.8{\text{ m/s}}^2$. Gauge pressure, the pressure relative to atmospheric pressure, is equal to $\rho gh$, while absolute pressure includes the $P_0$ term.

Worked example

What is the gauge pressure 12 m below the surface of a freshwater swimming pool?

1. Identify known values: $\rho =1000{\text{ kg/m}}^3$, $g=9.8{\text{ m/s}}^2$, $h=12\text{ m}$
2. Use the gauge pressure formula: $P_{gauge}=1000\times 9.8\times 12=1.176\times 10^5\text{ Pa}$, or ~1.16 times atmospheric pressure.

3. Pascal's principle

Pascal's principle states that a change in pressure applied to an enclosed, incompressible fluid is transmitted undiminished to all portions of the fluid and the walls of its container. This principle is the basis for hydraulic systems, which use differences in piston area to multiply force for applications like car lifts and brake systems.

The core formula for Pascal's principle comes from the definition of pressure: if pressure change is uniform across the system, $$\Delta P=\frac{F_1}{A_1}=\frac{F_2}{A_2}$$ Rearranged to solve for output force: $$F_2=F_1\times \frac{A_2}{A_1}$$ A key tradeoff of hydraulic systems is that the smaller piston must move a larger distance to lift the larger piston a small distance, since the volume of fluid displaced is equal on both sides: $A_1{d}_1=A_2{d}_2$.

Worked example

A hydraulic car lift has a small input piston of area $0.008{\text{ m}}^2$ and a large output piston of area $0.4{\text{ m}}^2$. What is the minimum input force required to lift a 1500 kg car?

1. Calculate weight of the car: $F_2=mg=1500\times 9.8=14700\text{ N}$
2. Rearrange Pascal's formula for $F_1$: $F_1=F_2\times \frac{A_1}{A_2}=14700\times \frac{0.008}{0.4}=294\text{ N}$, or the equivalent of lifting a ~30 kg mass.

4. Buoyancy — Archimedes' principle

Archimedes' principle states that any object fully or partially submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object. The formula for buoyant force is: $$F_b={\rho }_{fluid}{V}_{displaced}g$$ Where $V_{displaced}$ is the volume of fluid pushed out of the way by the object (equal to the volume of the submerged portion of the object).

Three key equilibrium cases:

1. If $F_b>F_{g,object}$: The object accelerates upward and floats
2. If $F_b=F_{g,object}$: The object is neutrally buoyant, remaining stationary in the fluid
3. If $F_b<F_{g,object}$: The object accelerates downward and sinks

For floating objects, the buoyant force equals the weight of the object, so we can derive the fraction of the object submerged: $$\frac{V_{submerged}}{V_{total}}=\frac{{\rho }_{object}}{{\rho }_{fluid}}$$

Worked example

A block of wood with density $720{\text{ kg/m}}^3$ floats in freshwater ($\rho =1000{\text{ kg/m}}^3$). What percentage of the wood's volume is above the water surface?

1. Calculate the fraction submerged: $\frac{720}{1000}=0.72$, or 72%
2. Subtract from 1 to find the fraction above water: $1-0.72=0.28$, or 28%

5. Continuity equation

The continuity equation is a statement of conservation of mass for incompressible, laminar (smooth, non-turbulent) fluid flow in a closed system with no leaks or added sources of fluid. For these systems, the mass flow rate into a section of pipe equals the mass flow rate out. Since density is constant for incompressible fluids, we can simplify to conservation of volume flow rate $Q$: $$Q=Av=\text{constant}$$ Or for two points in the flow path: $$A_1{v}_1=A_2{v}_2$$ Where $A$ is the cross-sectional area of the pipe, $v$ is the average flow speed of the fluid, and $Q$ has units of ${\text{m}}^3/\text{s}$. This equation explains why water sprays faster when you pinch the end of a hose: reducing the cross-sectional area increases flow speed to keep volume flow rate constant.

Worked example

Water flows through a pipe of diameter 10 cm at a speed of 1.5 m/s. The pipe narrows to a diameter of 5 cm. What is the flow speed in the narrow section?

1. Area of a circle is proportional to the square of diameter, so $\frac{A_1}{A_2}={\left(\frac{d_1}{d_2}\right)}^2={\left(\frac{10}{5}\right)}^2=4$
2. Rearrange the continuity equation: $v_2=v_1\times \frac{A_1}{A_2}=1.5\times 4=6\text{ m/s}$

6. Bernoulli's equation

Bernoulli's equation is a statement of conservation of energy per unit volume for incompressible, laminar, non-viscous (no friction) fluid flow. It relates pressure, flow speed, and height at two points in the flow path: $$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$$ Each term represents a form of energy per unit volume: $P$ is pressure potential energy, $\frac{1}{2}\rho v^2$ is kinetic energy, and $\rho gh$ is gravitational potential energy.

A common application is the Venturi effect: for horizontal flow (where $h_1=h_2$), faster moving fluid has lower pressure. This effect is the source of lift for airplane wings, the operation of carburetors, and the force that pulls shower curtains inward when you run hot water. A special case of Bernoulli's equation is Torricelli's law, which gives the speed of fluid exiting a hole in a large open tank: $v=\sqrt{2gh}$, where $h$ is the depth of the hole below the fluid surface.

Worked example

Water flows through a horizontal pipe at a speed of 3 m/s with a pressure of $2.5\times 10^5\text{ Pa}$. The pipe narrows, increasing flow speed to 9 m/s. What is the pressure in the narrow section? Use ${\rho }_{water}=1000{\text{ kg/m}}^3$.

1. Cancel the $\rho gh$ terms since height is constant, rearrange Bernoulli's equation: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)$
2. Calculate the kinetic energy difference: $\frac{1}{2}\times 1000\times (3^2-9^2)=500\times (9-81)=-36000\text{ Pa}$
3. Solve for $P_2$: $2.5\times 10^5-3.6\times 10^4=2.14\times 10^5\text{ Pa}$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using the object's density instead of the fluid's density when calculating buoyant force. Why you might do it: You mix up the mass of the object and the mass of displaced fluid. Correct move: Always use ${\rho }_{fluid}$ in the buoyant force formula; only use the object's density to calculate its weight for equilibrium problems.
• Wrong move: Forgetting to distinguish between absolute and gauge pressure. Why you might do it: You default to adding atmospheric pressure even when the question asks for gauge pressure. Correct move: Read the question carefully: gauge pressure = $P_{absolute}-P_0$, so omit the $P_0$ term if gauge pressure is requested.
• Wrong move: Applying Bernoulli's equation to turbulent or compressible flow. Why you might do it: You forget the formula's assumptions. Correct move: Bernoulli's only applies to laminar, incompressible, non-viscous flow, which is explicitly stated for all AP Physics 2 fluid dynamics problems unless noted otherwise.
• Wrong move: Using diameter instead of cross-sectional area in continuity or hydraulic lift calculations. Why you might do it: You skip converting diameter to area to save time. Correct move: Area scales with the square of diameter, so always calculate $A=\frac{\pi d^2}{4}$ or use the area ratio ${\left(\frac{d_1}{d_2}\right)}^2$ to avoid linear scaling errors.
• Wrong move: Assuming volume flow rate decreases when a pipe narrows. Why you might do it: You confuse flow speed and volume flow rate. Correct move: For closed systems with no leaks, volume flow rate $Q$ is constant; only flow speed changes when pipe area changes.

8. Practice Questions (AP Physics 2 Style)

Question 1

A rectangular research buoy of length 8 m, width 3 m, floats in saltwater of density $1025{\text{ kg/m}}^3$. When a 1200 kg scientific instrument is mounted to the top of the buoy, how far does the buoy sink into the water?

Solution

1. The additional buoyant force equals the weight of the instrument, per Archimedes' principle: ${\rho }_{fluid}{V}_{displaced}g=m_{instrument}g$
2. Cancel $g$, rearrange to solve for displaced volume: $V_{displaced}=\frac{m}{\rho }=\frac{1200}{1025}\approx 1.17{\text{ m}}^3$
3. Displaced volume = length × width × depth sunk: $d=\frac{V}{l\times w}=\frac{1.17}{8\times 3}\approx 0.049\text{ m}$, or ~5 cm.

Question 2

A fire hose of inner diameter 12 cm is connected to a nozzle with an exit diameter of 2 cm. If the volume flow rate through the hose is $0.012{\text{ m}}^3/\text{s}$, what is the flow speed of water exiting the nozzle?

Solution

1. Use the volume flow rate formula $Q=Av$, rearranged to solve for exit speed: $v=\frac{Q}{A_{exit}}$
2. Calculate area of the nozzle: $A_{exit}=\frac{\pi d^2}{4}=\frac{\pi (0.02{)}^2}{4}\approx 3.14\times 10^{-4}{\text{ m}}^2$
3. Solve for speed: $v=\frac{0.012}{3.14\times 10^{-4}}\approx 38.2\text{ m/s}$

Question 3

A large open water tank has a small hole 3.5 m below the water surface. What is the speed of water flowing out of the hole? Use $g=9.8{\text{ m/s}}^2$.

Solution

1. Apply Bernoulli's equation, comparing the top of the water (point 1) to the hole (point 2). $P_1=P_2=P_0$ (both open to atmosphere), $v_1\approx 0$ (tank is large, so water level drops slowly), $h_1=3.5\text{ m}$, $h_2=0$.
2. Cancel $P$ terms and the $v_1$ term: $\rho g{h}_1=\frac{1}{2}\rho v_2^2$
3. Cancel density, rearrange for $v_2$: $v_2=\sqrt{2g{h}_1}=\sqrt{2\times 9.8\times 3.5}=\sqrt{68.6}\approx 8.3\text{ m/s}$

9. Quick Reference Cheatsheet

10. What's Next

The fluid concepts you learned here directly connect to later AP Physics 2 topics, including thermodynamics (where fluid pressure changes and expansion drive heat engine cycles) and electromagnetism (where the continuity equation for electric charge is analogous to the fluid continuity equation, and Bernoulli's principle parallels energy conservation in DC circuits). Fluids are also frequently tested in interdisciplinary free-response questions that combine mechanics, energy, and lab design skills, such as designing an experiment to measure unknown fluid density or flow speed.

To reinforce your understanding of fluids, practise with official College Board AP Physics 2 past papers, and use the practice questions in this guide to test your problem-solving speed and accuracy. If you get stuck on any concept or question, you can ask Ollie for step-by-step explanations, additional practice problems, or targeted review of specific subtopics at any time via the homepage.