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Fluids Dynamics Speed Review

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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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Unit 8 Fluids we just added to the 2025 AP Physics 1 exam. This post will cover everything you need to know in 10 minutes of less.

🚀 AP Physics 1: The Ultimate Fluids Speed Review

So, fluids made the cut for AP Physics 1. Great. Another topic to cram.

But don’t worry—this guide will get you up to speed with everything you need to know about fluids, pressure, buoyancy, and flow, minus the fluff. Let’s get into it.

💡 First Things First: What Even is a Fluid?

Fluids are anything that flows—liquids, gases, and whatever weird in-between states physics hasn’t classified yet. Unlike solids, fluids don’t hold their shape; they take the form of whatever container they’re in.

The key quantity to track? Density, which is just mass packed into a given volume:

\[ \rho = \frac{m}{V} \]

Think about it: oil floats on water because it’s less dense. Helium rises in air for the same reason. Simple.

🔧 Pressure: The Invisible Force You Never Notice

Pressure is how much force a fluid applies over an area. Mathematically, that’s:

\[ P = \frac{F_{\perp}}{A} \]

More force? More pressure. Smaller area? Even more pressure. That’s why a sharp knife cuts better than a dull one.

📝 A Note on Pascal’s Law

Normally, this is where we’d talk about Pascal’s Law—the principle that explains how a change in pressure is transmitted equally throughout a fluid.

This is the reason hydraulic systems work, like the braking system in your car. A small force applied at one end can create a much larger force at the other, making it possible to stop a car with just the pressure from your foot.

The equation for Pascal’s Law is:

\[ F_1 = \frac{A_1}{A_2} F_2 \]

Although Nerd Notes includes Pascal’s Law in our practice question bank under fluids, it’s important to note that this is not covered on the AP Physics 1 exam. The exam only tests concepts related to fluid pressure, which we’ll focus on next.

⏬ How Fluid Pressure Changes with Depth

Go deeper underwater, and pressure increases. The equation for absolute pressure is:

\[ P = P_0 + \rho gh \]

Where \(P_0\) is atmospheric pressure at the surface, and \(\rho gh\) is the extra pressure from the fluid above you.

That’s why deep-sea fish look like nightmares—high pressure squashes anything that isn’t built for it.

⚖️ Buoyancy: Why Some Things Float and Others Sink

Ever tried to push a beach ball underwater? The harder you push, the stronger the water pushes back. That’s buoyancy in action.

Archimedes figured out that the buoyant force equals the weight of the displaced fluid:

\[ F_b = \rho V g \]

If \(F_b\) is greater than the object’s weight, it floats. If not, down it goes.

This also explains why ships, despite being made of heavy metal, don’t sink—because they displace enough water to generate a buoyant force greater than their own weight.

🚰 Fluid Flow: The Traffic Laws of Physics

Fluids follow a strict rule: what goes in must come out. That’s where the continuity equation comes in:

\[ A_1 v_1 = A_2 v_2 \]

Here’s what it really means: the volume of fluid flowing per second stays the same, no matter how the pipe’s shape changes. This is called the flow rate (\( Q \)), defined as:

\[ Q = A v \]

Where:

  • \(Q\) = flow rate (m³/s)
  • \(A\) = cross-sectional area (m²)
  • \(v\) = velocity of the fluid (m/s)

If a pipe gets narrower, the velocity increases to maintain the same flow rate. If it widens, the velocity decreases. That’s why river rapids form when a wide river suddenly narrows—water speeds up to keep the same volume moving through.

💡 Key takeaway: The flow rate always stays constant for an incompressible fluid. If area changes, velocity adjusts to compensate.

🛫 Bernoulli’s Equation: Why Airplanes Fly

Bernoulli’s principle is just conservation of energy applied to fluids. It states that a fluid’s total energy (pressure, gravitational potential, and kinetic) stays constant:

\[ P_1 + \rho g y_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2} \rho v_2^2 \]

This explains:

  • How airplanes generate lift (faster air = lower pressure)
  • Why a shower curtain gets sucked inward when you turn on the water
  • How a perfume atomizer sprays liquid

🚀 Torricelli’s Theorem: Fluid Speed Without the Drama

When a fluid drains out of a hole, it acts like it’s in free fall. Torricelli’s equation gives its speed:

\[ v = \sqrt{2g\Delta y} \]

The greater the height difference, the faster the fluid moves. That’s why water from a tall tower shoots out faster than from a short one.


📌 Final Takeaways (a.k.a. Your Cheat Sheet)

  • Pressure increases with depth: \( P = P_0 + \rho gh \)
  • Buoyant force = weight of displaced fluid: \( F_b = \rho V g \)
  • Fluid flow speeds up in narrow spaces: \( A_1 v_1 = A_2 v_2 \)
  • Fast-moving fluids have low pressure: Bernoulli’s equation
  • Fluid draining from a hole follows \( v = \sqrt{2g\Delta y} \)

Master these, and you’ll crush any fluids question the AP throws at you.

If you’re looking for free fluid practice questions in AP style, check out UBQ, our collection of over 1000+ Physics questions, with explanations, AI powered assistance, and automatic FRQ grading. Over 200,000 students use it daily to accelerate this success! Try it here.

Practice Questions for Mastery

Question 1
Difficulty - Advanced
Solve Type - Mathematical
A spherical balloon has a radius of \(7.15\) \(\text{m}\) and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of \(930\) \(\text{kg}\)? Take the density of helium and air to be \(0.18\) \(\text{kg/m}^3\) and \(1.24\) \(\text{kg/m}^3\), respectively.
View Full Question and Explanation
Question 2
Difficulty - Intermediate
Solve Type - Proportional Analysis
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
  1. \[ \frac{1}{4} \]

  2. \[ \frac{1}{2} \]

  3. \[ 2 \]

  4. \[ 4 \]

View Full Question and Explanation
Question 3
Difficulty - Intermediate
Solve Type - Mathematical
Water at a gauge pressure of \( 3.8 \) \( \text{atm} \) at street level flows into an office building at a speed of \( 0.78 \) \( \text{m/s} \) through a pipe \( 5.0 \) \( \text{cm} \) in diameter. The pipe tapers down to \( 2.8 \) \( \text{cm} \) in diameter by the top floor, \( 16 \) \(\text{m} \) above, where the faucet has been left open. Calculate the flow velocity AND the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.
View Full Question and Explanation
Question 4
Difficulty - Intermediate
Solve Type - Proportional Analysis
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
  1. \[ \frac{3}{4} F_0 \]

  2. \[ F_0 \]

  3. \[ \frac{4}{3} F_0 \]

  4. The answer cannot be determined without knowing the rock’s volume.

View Full Question and Explanation
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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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