Bernoulli's Principle

AP Physics 1· AP Physics 1 CED — Fluids and Thermal Physics· 14 min read

1. Core Concept of Bernoulli's Principle★☆☆☆☆⏱ 3 min

📘 Definition

Bernoulli's Principle

$P$ = absolute pressure, $ρ$ = fluid density, $v$ = average flow speed, $g$ = acceleration due to gravity, $h$ = height above reference

A fundamental energy conservation relationship for steady, incompressible, non-viscous fluid flow that links pressure, flow speed, and gravitational height.

Example:

Predicts lower pressure when fluid flow speed increases at constant height in a narrowing pipe.

Bernoulli's principle is derived directly from conservation of mechanical energy applied to moving fluid systems. For AP Physics 1, this topic makes up ~1-2% of the total exam score, appearing in both multiple-choice and short free-response questions. Notation used in this guide matches standard AP exam conventions.

2. The Continuity Equation (Conservation of Mass)★★☆☆☆⏱ 4 min

Before applying Bernoulli's principle to most AP problems, you first need the continuity equation, which comes from conservation of mass for incompressible steady flow. For flow through a closed pipe, mass entering per unit time equals mass leaving, so for constant density, volume flow rate is constant.

📘 Definition

Continuity Equation

$Q$ = volume flow rate, $A$ = cross-sectional area, $v$ = average flow speed

For steady incompressible flow, volume flow rate is constant along a closed pipe.

Q = A_1 v_1 = A_2 v_2

For circular pipes, $A = π d^{2} /4$ , so the constant terms cancel when taking the ratio of areas, giving the simplified relation $d_{1}^{2} v_{1} = d_{2}^{2} v_{2}$ , which saves significant time on AP 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:
2
$A_1 v_1 = A_2 v_2$
3
Substitute the area of a circular pipe. The constant $π /4$ terms cancel:
4
$d_1^2 v_1 = d_2^2 v_2$
5
Rearrange to solve for unknown output speed $v_{2}$ :
6
$v_2 = v_1 \left(\frac{d_1}{d_2}\right)^2$
7
Plug in given values:
8
$v_2 = (1.0 \text{ m/s}) \left(\frac{6.0 \text{ cm}}{2.0 \text{ cm}}\right)^2 = 9.0 \text{ m/s}$

Exam tip:

Always check intuition: if the pipe narrows, output speed must be higher than input. If you get a lower speed, you flipped the diameter ratio.

3. Bernoulli's Equation (Quantitative Principle)★★★☆☆⏱ 5 min

Bernoulli's equation is the quantitative expression of the principle, representing conservation of mechanical energy per unit volume of fluid. For any two points along a streamline in valid flow, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume is constant.

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 pressure (Pa, equivalent to J/m³). The key conceptual takeaway for same height ( $h_{1} = h_{2}$ ) is that an increase in flow speed always corresponds to a decrease in pressure, and vice versa. AP Physics 1 expects you to remember default values: $ρ_{water} = 1000$ kg/m³ and $P_{a t m} = 1.01 \times 1 0^{5}$ Pa.

📐 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 $2.5 \times 1 0^{5}$ Pa, and the fluid is water. What is the output gauge pressure?

1
We already know $v_{1} = 1.0$ m/s, $v_{2} = 9.0$ m/s from continuity. Since the pipe is horizontal, $h_{1} = h_{2}$ , so the $ρ g h$ terms cancel.
2
Write Bernoulli's equation for gauge pressure (atmospheric pressure cancels out):
3
$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$
4
Rearrange to solve for $P_{2}$ :
5
$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)$
6
Substitute values, using $ρ_{water} = 1000$ kg/m³:
7
$P_2 = 2.5 \times 10^5 + 0.5(1000)(1^2 - 9^2) = 2.1 \times 10^5 \text{ Pa}$

Exam tip:

When working with gauge pressure, you can ignore atmospheric pressure entirely if both points are open to the atmosphere, since it cancels out and simplifies arithmetic.

4. Torricelli's Law (Special Case for Tank Drainage)★★★☆☆⏱ 4 min

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. Two key approximations apply here: 1) the tank area is much larger than the hole area, so the surface speed is negligible ( $v_{surface} \approx 0$ ), and 2) both the surface and hole are open to the atmosphere, so pressures are equal.

🔬 Derivation

Goal:

Derive Torricelli's Law from Bernoulli's equation

Starting from:

Full Bernoulli's equation for a large open tank with a small hole

1
Set reference height $h = 0$ at the hole, so surface height equals the depth of the hole $h$ below the surface. We know $P_{surface} = P_{hole} = P_{a t m}$ and $v_{surface} \approx 0$ .
2
$P_{atm} + 0 + \rho g h = P_{atm} + \frac{1}{2} \rho v^2 + 0$
3
Cancel $P_{a t m}$ and $ρ$ from both sides and rearrange:
4
$v = \sqrt{2 g h}$

Result:

The exit speed equals the speed of an object in free fall dropped from height $h$ , matching energy conservation expectations.

📐 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 assumptions for Torricelli's Law: tank is large so $v_{surface} \approx 0$ , both points open to atmosphere so pressures are equal.
2
Calculate the depth of the hole below the surface:
3
$h = 6.0 \text{ m} - 2.0 \text{ m} = 4.0 \text{ m}$
4
Substitute into Torricelli's Law:
5
$v = \sqrt{2gh} = \sqrt{2(9.8)(4.0)} \approx 8.9 \text{ m/s}$

Exam tip:

On conceptual FRQs, always state the two key approximations (negligible surface speed, equal atmospheric pressure) to earn full credit.

5. Common AP Applications★★★★☆⏱ 5 min

Bernoulli's principle explains many common real-world phenomena tested on AP Physics 1, including airplane lift, which arises from pressure difference between the top and bottom of the wing.

📐 Worked Example

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

1
For same height, Bernoulli's principle gives pressure difference $Δ P = P_{bottom} - P_{top}$ :
2
$\Delta P = \frac{1}{2} \rho (v_{\text{top}}^2 - v_{\text{bottom}}^2)$
3
Substitute values:
4
$\Delta P = 0.5(1.2)(260^2 - 220^2) = 11520 \text{ Pa}$
5
Net lift force is pressure difference multiplied by wing area:
6
$F = \Delta P \times A = 11520 \times 20 = 2.3 \times 10^5 \text{ N}$

✓ Quick check

Test your understanding of Torricelli's Law:

A large open tank is filled with water to a height of 7.0 m above the base. A small hole is 2.0 m above the base, open to the atmosphere. What is the approximate exit speed of water?
- 6.3 m/s
- 9.9 m/s
- 11.7 m/s
- 14 m/s
Reveal answer
9.9 m/s —
Correct: $h = 7.0 - 2.0 = 5.0$ m, so $v = 2 (9.8) (5) \approx 9.9$ m/s. The most common error uses total tank height instead of depth below the surface.

6. Common Pitfalls

Wrong move:

Flipping the diameter/area ratio in continuity, calculating $v_{2} = v_{1} (d_{2} / d_{1})^{2}$ for a narrowing pipe

Why:

Students forget the inverse relationship between area and speed, mixing up which area corresponds to which speed.

Correct move:

Always check your result after solving continuity: if the pipe narrows, output speed must be larger than input. Flip the ratio if it is not.

Wrong move:

Using absolute pressure for one point and gauge pressure for another in Bernoulli's equation

Why:

Problems often give gauge pressure, and students forget all pressure terms must use the same reference frame.

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, as $P_{a t m}$ cancels automatically.

Wrong move:

Automatically canceling $ρ g h$ terms even when the pipe is not horizontal

Why:

Students get used to horizontal pipe problems and forget to check for height differences between points.

Correct move:

Always explicitly write the $ρ g h$ 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 when points are 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 a height drop offsets the kinetic energy increase.

Wrong move:

Retaining the full $v_{surface}^{2}$ term for Torricelli's Law when the tank is much larger than the hole

Why:

Students forget the continuity approximation that surface speed is negligible for large tanks, leading to incorrectly low exit speed.

Correct move:

If the tank diameter is more than 10 times the hole diameter, approximate $v_{surface} = 0$ , which introduces <1% error, well within AP exam tolerance.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Continuity Equation	$A_{1} v_{1} = A_{2} v_{2}$	Conservation of mass; for steady incompressible flow
Full Bernoulli's Equation	$P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$	Only for steady, incompressible, non-viscous flow; use same pressure type
Bernoulli (Same Height)	$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$	Higher flow speed → lower pressure; most common AP form
Torricelli's Law	$v = 2 g h$	$h$ = depth of hole below surface; assumes large tank, open to atmosphere
Volume Flow Rate	$Q = A v$	Constant for incompressible flow; units = m³/s
Circular Pipe Speed Ratio	$v_{2} = v_{1} (\frac{d _{1}}{d _{2}})^{2}$	$π$ cancels, use diameter ratio directly to save time
Standard AP Values	$ρ_{water} = 1000 kg/m^{3}$ , $P_{a t m} = 1.01 \times 1 0^{5} Pa$	Default values for most AP problems
Core Conceptual Rule	Higher flow speed at same height = lower pressure	Required for full credit on conceptual FRQ explanations

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Torricelli's law exit speed calculation
2021 · FRQ
Pipe pressure change explanation

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. The energy conservation reasoning you practice here will be applied again in the thermal physics topics that follow, which make up the rest of Unit 8. 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. Mastering the continuity equation and Bernoulli's problem-solving method will prepare you for the energy-based reasoning required for upcoming thermal physics topics.