Fluids in AP Physics 1: Pressure, Buoyancy, Bernoulli

Fluids is Unit 8 of AP Physics 1, worth 10 to 15% of the multiple-choice section. You need density, pressure vs depth, buoyant force (fluid density times displaced volume times g), continuity, and Bernoulli. Prep materials that put fluids only in Physics 2 are out of date.

AP Physics: Unit 8 (topics 8.1 Internal Structure and Density, 8.2 Pressure, 8.3 Fluids and Newton's Laws, 8.4 Fluids and Conservation Laws). Covers Unit 8 of AP Physics 1 (10-15% of the multiple-choice section) in the course effective fall 2024, which moved fluids in from AP Physics 2.

Yes, Fluids Is on the AP Physics 1 Exam

Fluids is Unit 8 of AP Physics 1 and carries 10 to 15% of the multiple-choice section weight, the same range as kinematics. The unit has four topics: 8.1 Internal Structure and Density, 8.2 Pressure, 8.3 Fluids and Newton's Laws, and 8.4 Fluids and Conservation Laws.

This is a recent change. The course description effective fall 2024 moved fluids out of AP Physics 2 and into Physics 1, so any prep book, playlist, or review packet that files fluids under Physics 2 is describing the old course. Work from the current Unit 8 page and the official equation sheet instead.

The redesigned 8-unit course also changed outcomes: 67.3% of students earned a 3 or higher on the May 2025 exam, up from 47.3% in 2024 under the old structure. Skipping Unit 8 because an old book told you to is an unforced error.

Density and Pressure vs Depth

Density is mass per volume, ρ=m/V\rho = m/V, and fresh water sits at about 1000 kg/m31000 \text{ kg/m}^3. Pressure is force per area, P=F/AP = F_{\perp}/A, measured in pascals (1 Pa=1 N/m21 \text{ Pa} = 1 \text{ N/m}^2).

The pressure at depth hh in a fluid open to the surface is

P=P0+ρghP = P_0 + \rho g h

where P0P_0 is the pressure at the surface, usually atmospheric. Only depth and fluid density matter. The shape of the container and the total amount of water do not; a narrow tube and a lake produce the same pressure at the same depth.

A useful calibration: every 10 m of water adds ρgh=1000×9.8×10=9.8×104 Pa\rho g h = 1000 \times 9.8 \times 10 = 9.8 \times 10^4 \text{ Pa}, roughly one atmosphere. Also keep gauge vs absolute straight: ρgh\rho g h alone is the gauge pressure, and adding P0P_0 gives absolute. All of these equations appear on the AP Physics 1 formula sheet.

Buoyant Force: Archimedes' Principle

The buoyant force on an object in a fluid is

Fb=ρVgF_b = \rho V g

and the two classic traps live inside those symbols. First, ρ\rho is the density of the fluid, never the object. Second, VV is the volume of fluid displaced, which equals the object's volume only when the object is fully submerged. A boat displaces far less than its hull volume; a sunken anchor displaces exactly its own volume.

Physically, buoyancy exists because pressure grows with depth: the fluid pushes harder on the bottom of the object than on the top, and the difference is a net upward force.

Topic 8.3 is Fluids and Newton's Laws, so expect to put FbF_b on a free-body diagram alongside weight, tension, or a normal force and apply Fnet=maF_{net} = ma like any other dynamics problem. If your diagrams are shaky, review how to draw a free-body diagram first.

Float or Sink: The Density Logic

Whether an object floats comes down to comparing densities. If the object's average density is less than the fluid's, it floats; if greater, it sinks; if equal, it hovers at any depth.

For a floating object, equilibrium gives you a clean shortcut. Since Fb=mgF_b = mg,

ρfluidVsubg=ρobjVg\rho_{fluid} V_{sub} g = \rho_{obj} V g

so the submerged fraction is Vsub/V=ρobj/ρfluidV_{sub}/V = \rho_{obj}/\rho_{fluid}. That single line answers a whole family of exam questions. Ice floats with roughly 90% of its volume underwater because its density is about 90% of water's. A block half submerged has half the fluid's density.

For a sinking object, the buoyant force does not vanish; it just loses to gravity. The apparent weight of a fully submerged object is WFbW - F_b, which is why objects feel lighter underwater and why a scale reading changes when you lower a mass into a beaker.

Continuity: Narrow Pipe, Faster Flow

For an incompressible fluid moving through a pipe, the volume flow rate AvAv is the same everywhere, which gives the continuity equation

A1v1=A2v2A_1 v_1 = A_2 v_2

The logic is bookkeeping, not force analysis: fluid cannot pile up or vanish, so whatever volume enters one end each second must leave the other end each second. Where the cross-sectional area shrinks, the speed must rise in exact proportion, which is why covering part of a garden hose opening with your thumb makes the stream shoot out faster.

Watch the units. AA is in m2\text{m}^2 and vv in m/s, so AvAv comes out in m3/s\text{m}^3/\text{s}, a volume per time. If a pipe's radius halves, the area drops by a factor of 4 (area scales with radius squared), so the speed quadruples. Ratio questions like that are quick multiple-choice points once you trust the proportionality.

Bernoulli Is Conservation of Energy

Bernoulli's equation, as printed on the formula sheet, is

P1+ρgy1+12ρv12=P2+ρgy2+12ρv22P_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

Every term is an energy per unit volume. 12ρv2\frac{1}{2}\rho v^2 mirrors kinetic energy 12mv2\frac{1}{2}mv^2, ρgy\rho g y mirrors gravitational potential energy mgymgy, and the pressure term accounts for work done by the surrounding fluid. That is why fluids landed in topic 8.4, Fluids and Conservation Laws: it is the same reasoning you built in conservation of energy, applied per cubic meter.

The qualitative rule the exam loves: along a horizontal flow, faster fluid means lower pressure. Pair Bernoulli with continuity and you can reason through most setups without a calculator. A pipe narrows, so continuity says the fluid speeds up, so Bernoulli says the pressure there drops. The worked example below runs those numbers.

How Unit 8 Shows Up on Exam Day

The exam is 3 hours: 42 multiple-choice questions in 85 minutes, then 4 free-response questions in 95 minutes, each section worth 50%. A four-function, scientific, or graphing calculator is allowed on both sections, so arithmetic like 12(1000)(8.0)2\frac{1}{2}(1000)(8.0)^2 costs you seconds, not minutes.

The four FRQ types are Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. Fluids fits all four, and buoyancy is a natural experimental-design setting (think measuring density with a scale and a beaker of water).

Every fluids equation you need is printed on the sheet: ρ=m/V\rho = m/V, P=F/AP = F_{\perp}/A, P=P0+ρghP = P_0 + \rho g h, Fb=ρVgF_b = \rho V g, continuity, and Bernoulli. Memorization is not the job; picking the right relation and defending it is. Keep the formula sheet open while you practice the Unit 8 problem sets.

Apparent Weight of a Submerged Block

An aluminum block (ρAl=2700 kg/m3\rho_{Al} = 2700 \text{ kg/m}^3, V=2.0×104 m3V = 2.0 \times 10^{-4} \text{ m}^3) hangs from a string, fully submerged in fresh water (ρw=1000 kg/m3\rho_w = 1000 \text{ kg/m}^3). Find the buoyant force and the string tension.

  1. Find the block's weight. Mass: m=ρAlV=2700 kg/m3×2.0×104 m3=0.54 kgm = \rho_{Al} V = 2700 \text{ kg/m}^3 \times 2.0 \times 10^{-4} \text{ m}^3 = 0.54 \text{ kg}. Weight: W=mg=0.54 kg×9.8 m/s2=5.29 NW = mg = 0.54 \text{ kg} \times 9.8 \text{ m/s}^2 = 5.29 \text{ N}.

  2. Find the buoyant force using the fluid density and the displaced volume (the full block volume, since it is fully submerged): Fb=ρwVg=1000×2.0×104×9.8=1.96 NF_b = \rho_w V g = 1000 \times 2.0 \times 10^{-4} \times 9.8 = 1.96 \text{ N}.

  3. Apply Newton's second law with a=0a = 0: tension plus buoyant force balances weight, so T=WFb=5.29 N1.96 N=3.33 NT = W - F_b = 5.29 \text{ N} - 1.96 \text{ N} = 3.33 \text{ N}.

Fb=1.96 NF_b = 1.96 \text{ N} and T=3.33 NT = 3.33 \text{ N}. The block still sinks without the string (2700 is greater than 1000 kg/m^3), but it feels about 37% lighter underwater.

Fraction of a Floating Block Below the Surface

A block of mass 0.75 kg and volume 1.25×103 m31.25 \times 10^{-3} \text{ m}^3 is placed in fresh water. Does it float, and if so, what fraction of its volume is submerged?

  1. Compute the block's density: ρ=m/V=0.75 kg/(1.25×103 m3)=600 kg/m3\rho = m/V = 0.75 \text{ kg} / (1.25 \times 10^{-3} \text{ m}^3) = 600 \text{ kg/m}^3. That is less than 1000 kg/m31000 \text{ kg/m}^3, so it floats.

  2. Floating means equilibrium: Fb=mgF_b = mg, so ρwVsubg=mg\rho_w V_{sub} g = mg. Solve for the submerged volume: Vsub=m/ρw=0.75 kg/(1000 kg/m3)=7.5×104 m3V_{sub} = m/\rho_w = 0.75 \text{ kg} / (1000 \text{ kg/m}^3) = 7.5 \times 10^{-4} \text{ m}^3. Notice gg cancels before you ever plug in 9.8.

  3. Take the ratio: Vsub/V=(7.5×104)/(1.25×103)=0.60V_{sub}/V = (7.5 \times 10^{-4}) / (1.25 \times 10^{-3}) = 0.60.

The block floats with 60% of its volume submerged, matching the shortcut ρobj/ρfluid=600/1000=0.60\rho_{obj}/\rho_{fluid} = 600/1000 = 0.60.

Bernoulli in a Narrowing Horizontal Pipe

Water (ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3) flows through a horizontal pipe. In the wide section, A1=8.0×104 m2A_1 = 8.0 \times 10^{-4} \text{ m}^2, v1=2.0 m/sv_1 = 2.0 \text{ m/s}, and P1=1.50×105 PaP_1 = 1.50 \times 10^5 \text{ Pa}. The pipe narrows to A2=2.0×104 m2A_2 = 2.0 \times 10^{-4} \text{ m}^2. Find v2v_2 and P2P_2.

  1. Continuity first: v2=A1v1/A2=(8.0×104 m2)(2.0 m/s)/(2.0×104 m2)=8.0 m/sv_2 = A_1 v_1 / A_2 = (8.0 \times 10^{-4} \text{ m}^2)(2.0 \text{ m/s}) / (2.0 \times 10^{-4} \text{ m}^2) = 8.0 \text{ m/s}. The area dropped by 4x, so the speed rose by 4x.

  2. The pipe is horizontal, so y1=y2y_1 = y_2 and the ρgy\rho g y terms cancel in Bernoulli's equation, leaving P2=P1+12ρv1212ρv22P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2.

  3. Evaluate each kinetic term: 12ρv12=0.5×1000×(2.0)2=2.0×103 Pa\frac{1}{2}\rho v_1^2 = 0.5 \times 1000 \times (2.0)^2 = 2.0 \times 10^3 \text{ Pa} and 12ρv22=0.5×1000×(8.0)2=3.2×104 Pa\frac{1}{2}\rho v_2^2 = 0.5 \times 1000 \times (8.0)^2 = 3.2 \times 10^4 \text{ Pa}.

  4. Combine: P2=1.50×105+2.0×1033.2×104=1.20×105 PaP_2 = 1.50 \times 10^5 + 2.0 \times 10^3 - 3.2 \times 10^4 = 1.20 \times 10^5 \text{ Pa}.

v2=8.0 m/sv_2 = 8.0 \text{ m/s} and P2=1.20×105 PaP_2 = 1.20 \times 10^5 \text{ Pa} (120 kPa). The pressure is lower where the water moves faster, exactly what the qualitative rule predicts.

Frequently asked questions

Is fluids in AP Physics 1 or AP Physics 2?

AP Physics 1. Since the course revision effective fall 2024, fluids is Unit 8 of Physics 1, covering density, pressure, buoyancy, continuity, and Bernoulli. AP Physics 2 now begins at Unit 9, Thermodynamics. Older prep materials that cover fluids only in Physics 2 describe the pre-2024 course.

How much of the AP Physics 1 exam is fluids?

Unit 8 accounts for 10 to 15% of the multiple-choice section, the same weighting range as kinematics, momentum, and rotation. Fluids can also appear in any of the four free-response questions.

Which density goes in the buoyant force equation?

The fluid's density, not the object's. Buoyant force equals fluid density times displaced volume times g. The displaced volume equals the object's full volume only when the object is completely submerged; a floating object displaces less than its own volume.

Do I need to memorize Bernoulli's equation?

No. Bernoulli's equation, continuity, pressure vs depth, and the buoyant force equation are all printed on the AP Physics 1 equation sheet, which you get on both sections of the exam. The skill being tested is choosing the right equation and setting it up, not recall.

Can I use a calculator on fluids problems?

Yes. A four-function, scientific, or graphing calculator is allowed on both the multiple-choice and free-response sections of AP Physics 1, so quantitative Bernoulli and buoyancy problems are fair game anywhere on the exam.