Bernoulli's Principle — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: the continuity equation for steady incompressible flow, full statement of Bernoulli's principle, Bernoulli's energy equation, problem-solving for unknown pressure, velocity, and height changes, and analysis of common AP-style real-world applications.

You should already know: Conservation of mass and energy for closed mechanical systems. Basic definitions of fluid pressure, density, and incompressibility. Flow rate concepts for dynamic fluids.

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 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Bernoulli's Principle?

Bernoulli's principle is a fundamental relationship linking pressure, flow velocity, and gravitational potential energy per unit volume for steady, incompressible, non-viscous fluid flow. It is derived directly from the conservation of mechanical energy applied to fluid flow, so it is fundamentally an energy conservation statement for moving fluids. For the AP Physics 1 CED, this topic accounts for approximately 1-2% of the total exam weight, and it appears in both multiple-choice (MCQ) and short free-response (FRQ) questions, usually as a conceptual or calculation-based item connected to fluid properties. Synonyms sometimes used include Bernoulli's relation or Bernoulli's equation, though the principle refers to the conceptual statement while the equation is the quantitative expression of that statement. Notation conventions used throughout this guide are consistent with AP Physics 1 expectations: $P$ is absolute pressure, $\rho$ is fluid density, $v$ is average flow speed, $g$ is acceleration due to gravity, and $h$ is height relative to an arbitrary reference level. AP exam questions almost always use these same symbols, so you do not need to memorize alternative notation for this topic.

2. The Continuity Equation (Conservation of Mass for Fluids)

Before applying Bernoulli's principle, we first need the continuity equation, which is derived from conservation of mass for incompressible flow. For steady flow through a closed pipe or tube, the mass of fluid entering any cross-section per unit time must equal the mass of fluid leaving per unit time, because mass cannot accumulate inside the pipe. For incompressible fluids, density $\rho$ is constant, so we can cancel $\rho$ from both sides of the mass balance, leading to the continuity equation in terms of volume flow rate $Q$, defined as the volume of fluid passing a point per unit time. $$Q=A_1{v}_1=A_2{v}_2$$ Where $A_1$ and $A_2$ are the cross-sectional areas at the two points, and $v_1$, $v_2$ are the average flow speeds at those points. Intuitively, this means when the pipe narrows (smaller area), flow speed increases, and when it widens, flow speed decreases. This is a non-negotiable prerequisite for almost all AP Bernoulli problems, because most problems involve changing pipe diameter, so you will always need continuity to find an unknown flow speed before plugging into Bernoulli's equation. For circular pipes, cross-sectional area is $A=\pi r^2=\pi d^2/4$, so the $\pi /4$ terms cancel when taking the ratio of areas, leaving $d_1^2{v}_1=d_2^2{v}_2$, a simplification that saves time on exams.

Worked Example

A circular water input pipe with diameter 6.0 cm narrows to an output pipe with diameter 2.0 cm. If the input flow speed is 1.0 m/s, what is the output flow speed?

1. Write the continuity equation for the two points: $A_1{v}_1=A_2{v}_2$
2. Substitute the area of a circular pipe: $\pi (d_1^2/4)v_1=\pi (d_2^2/4)v_2$. The $\pi /4$ terms cancel, leaving $d_1^2{v}_1=d_2^2{v}_2$
3. Rearrange to solve for the unknown output speed $v_2$: $v_2=v_1{\left(\frac{d_1}{d_2}\right)}^2$
4. Plug in the given values: $v_2=(1.0\text{ m/s}){\left(\frac{6.0\text{ cm}}{2.0\text{ cm}}\right)}^2=1.0\times (3{)}^2=9.0\text{ m/s}$

Exam tip: Always check that your output speed makes intuitive sense: if the pipe narrows, the speed must be higher than the input speed. If you get a lower speed, you flipped the ratio, so correct it before proceeding to avoid losing points on later steps.

3. Bernoulli's Equation (Quantitative Principle)

Bernoulli's equation is the quantitative expression of Bernoulli's principle, derived directly from conservation of mechanical energy per unit volume of fluid. For any two points along a streamline in steady, incompressible, non-viscous flow, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume is constant. This gives the full equation: $$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$$ Each term has units of energy per unit volume (Pa, which is equivalent to J/m³). $P$ is the pressure energy per unit volume, $\frac{1}{2}\rho v^2$ is kinetic energy per unit volume, and $\rho gh$ is gravitational potential energy per unit volume. The core conceptual takeaway is that for two points at the same height ($h_1=h_2$), an increase in flow speed $v$ is always accompanied by a decrease in pressure $P$, and vice versa. When a point is open to the atmosphere, $P=P_{atm}$ (atmospheric pressure), which is a common value to plug into problems. AP Physics 1 expects you to remember ${\rho }_{\text{water}}=1000$ kg/m³ and $P_{atm}=1.01\times 10^5$ Pa for most problems.

Worked Example

The pipe system from the previous worked example lies horizontally, so input and output are at the same height. The input gauge pressure is $2.5\times 10^5$ Pa, the fluid is water. What is the output gauge pressure?

1. We already know from continuity: $v_1=1.0$ m/s, $v_2=9.0$ m/s. Since the pipe is horizontal, $h_1=h_2$, so the $\rho gh$ terms cancel out.
2. Write Bernoulli's equation for gauge pressure (atmospheric pressure cancels out for gauge pressure calculations): $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
3. Rearrange to solve for $P_2$: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)$
4. Plug in values: $P_2=2.5\times 10^5+0.5(1000)(1^2-9^2)=250000+500(-80)=250000-40000=2.1\times 10^5$ Pa. The result makes sense: higher speed gives lower pressure.

Exam tip: When working with gauge pressure, you can ignore atmospheric pressure entirely if both points are open to the atmosphere, because it cancels out of the equation, which simplifies your arithmetic.

4. Torricelli's Law (Special Case for Tank Drainage)

One of the most common AP applications of Bernoulli's principle is Torricelli's Law, which describes the exit speed of fluid from a small hole in a large open tank. For a large tank, the cross-sectional area of the tank is much larger than the area of the hole, so from continuity, the speed at which the water surface drops is negligible ($v_{\text{surface}}\approx 0$). Both the water surface and the hole are open to the atmosphere, so $P_{\text{surface}}=P_{\text{hole}}=P_{atm}$. If we set our reference height $h=0$ at the hole, then the height of the surface is $h$, which is equal to the depth of the hole below the surface. Substituting all these values into Bernoulli's equation cancels both the pressure terms and the kinetic energy term for the surface, giving the simple result: $$v=\sqrt{2gh}$$ This result makes sense from energy conservation: the exit speed equals the speed of an object in free fall dropped from height $h$, which matches the expectation that gravitational potential energy is converted to kinetic energy of the exiting fluid. This special case comes up constantly in AP MCQ and FRQ problems.

Worked Example

A large open tank is filled with water to a total height of 6.0 m. A small hole is punctured 2.0 m above the bottom of the tank. What is the exit speed of water from the hole?

1. Confirm the assumptions for Torricelli's Law: the tank is large, so $v_{\text{surface}}\approx 0$, and both the surface and hole are open to the atmosphere, so pressures are equal.
2. Calculate the depth of the hole below the surface: $h=\text{total water height}-\text{height of hole above bottom}=6.0\text{ m}-2.0\text{ m}=4.0\text{ m}$.
3. Substitute into Torricelli's Law: $v=\sqrt{2gh}=\sqrt{2(9.8{\text{ m/s}}^2)(4.0\text{ m})}=\sqrt{78.4}\approx 8.9\text{ m/s}$.
4. Check for consistency: this matches the free fall speed from 4 m, so the result is reasonable.

Exam tip: When explaining Torricelli's Law on a conceptual FRQ, always state the two key approximations (large tank gives negligible surface speed, both points open to atmosphere) to earn full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Flipping the diameter/area ratio in continuity, calculating $v_2=v_1(d_2/d_1{)}^2$ and getting a lower speed for a narrower pipe. Why: Students mix up which area corresponds to which speed, and forget the inverse relationship between area and speed. Correct move: Always check your final speed after solving for continuity; if the pipe narrows, your output speed must be larger than the input, so flip the ratio if it is not.
• Wrong move: Using absolute pressure for one point and gauge pressure for another point when plugging into Bernoulli's equation. Why: Problems often give gauge pressure, and students forget all pressure terms must use the same reference. Correct move: Convert all pressures to the same type at the start of the problem; gauge pressure is simpler for problems with multiple open ends, since $P_{atm}$ cancels automatically.
• Wrong move: Automatically canceling $\rho gh$ terms even when the pipe is not horizontal. Why: Students get used to horizontal pipe problems and forget to check for height differences. Correct move: Always explicitly write the $\rho gh$ term for both points before canceling any terms, so you do not miss height changes.
• Wrong move: Claiming higher speed always causes lower pressure, regardless of height differences. Why: Students memorize the "higher speed = lower pressure" rule and forget it only applies at the same height. Correct move: Always use the full Bernoulli equation to compare pressures, accounting for height changes; pressure can increase with speed if the height drop offsets the kinetic energy increase.
• Wrong move: Using the full $v_{\text{surface}}^2$ term for Torricelli's Law when the tank is much larger than the hole, leading to an incorrectly low exit speed. Why: Students forget the continuity approximation that surface speed is negligible for large tanks. Correct move: If the tank diameter is more than 10 times the hole diameter, approximate $v_{\text{surface}}=0$, which introduces an error of less than 1%, well within AP exam tolerance.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A large open tank is filled with water to a height of 7.0 m above the tank's base. A small hole is 2.0 m above the base of the tank and is open to the atmosphere. What is the approximate speed of water exiting the hole? (A) 6.3 m/s (B) 9.9 m/s (C) 11.7 m/s (D) 14 m/s

Worked Solution: First, calculate the depth of the hole below the water surface: total water height is 7.0 m, hole is 2.0 m above the base, so $h=7.0-2.0=5.0$ m. For Torricelli's Law, we use the approximation that the surface speed is negligible and pressures are equal, so $v=\sqrt{2gh}$. Plugging in $g=9.8$ m/s² and $h=5.0$ m gives $v=\sqrt{2(9.8)(5)}=\sqrt{98}\approx 9.9$ m/s. The most common wrong answer is C, which comes from using the total height of 7 m instead of the depth below the surface. The correct answer is B.

Question 2 (Free Response)

A horizontal factory pipe carries water from a wide inlet to a narrow outlet open to the atmosphere. The inlet has a diameter of 12 cm, flow speed of 0.5 m/s, and an absolute pressure of $2.0\times 10^5$ Pa. The outlet has a diameter of 4 cm. (a) Calculate the flow speed at the outlet. (b) Calculate the absolute pressure at the outlet. (c) Explain the pressure difference between inlet and outlet using conservation of energy reasoning.

Worked Solution: (a) Use the continuity equation for circular pipes: $v_2=v_1(d_1/d_2{)}^2=0.5\text{ m/s}(12/4{)}^2=0.5\times 9=4.5\text{ m/s}$. (b) The pipe is horizontal, so $h_1=h_2$ and potential energy terms cancel. Bernoulli's equation: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)=2.0\times 10^5+0.5(1000)(0.5^2-4.5^2)=200000+500(0.25-20.25)=200000-10000=1.9\times 10^5$ Pa absolute. (c) Total mechanical energy per unit volume is constant along the flow, per conservation of energy. The flow speed increases at the outlet, so kinetic energy per unit volume increases. Height (and thus gravitational potential energy per unit volume) is unchanged, so pressure energy per unit volume must decrease to keep total energy constant, leading to lower outlet pressure.

Question 3 (Application / Real-World Style)

An airplane wing has a flow speed of 260 m/s over the top surface and 220 m/s over the bottom surface. The density of air is 1.2 kg/m³, and the total surface area of the wing is 20 m². Assuming the top and bottom of the wing are at approximately the same height, calculate the net lift force on the wing from the pressure difference.

Worked Solution: For same height, Bernoulli's principle gives $\Delta P=P_{\text{bottom}}-P_{\text{top}}=\frac{1}{2}\rho (v_{\text{top}}^2-v_{\text{bottom}}^2)$. Plugging in values: $\Delta P=0.5(1.2)(260^2-220^2)=0.6(67600-48400)=0.6(19200)=11520$ Pa. Net lift force is $F=\Delta P\times A=11520\times 20=230400$ N ≈ 2.3 × 10⁵ N. In context, this lift force is large enough to support a small commercial airplane, which matches real-world expectations for this wing size and flow speed.

7. Quick Reference Cheatsheet

8. What's Next

Bernoulli's principle is the capstone of the fluid mechanics portion of AP Physics 1 Unit 8, and it reinforces the core course theme of using conservation laws to solve dynamic problems. Next in Unit 8, you will move to thermal physics topics including temperature, heat transfer, and the ideal gas law, where the energy conservation reasoning you practiced with Bernoulli will be applied again to macroscopic thermal systems. Understanding how conservation laws apply to fluid systems also builds intuition for energy problems in mechanics and thermodynamics that make up a large portion of the AP Physics 1 exam. Without mastering the continuity equation and Bernoulli's problem-solving method, you will struggle with energy-based reasoning in the thermal physics topics that follow. The next topics you will study are: Thermal Energy and Temperature Heat Transfer and Thermal Equilibrium Ideal Gas Law