Unit 8: Fluids · First Principles Physics

AP Physics 1 · Unit 08 of 08

Fluids.

A swimming submarine. An airplane wing. A drinking straw. A weather front. Five miles of atmosphere pressing on your shoulders right now. Liquids and gases obey their own beautiful physics — built on the same conservation laws you've used all year.

Exam Weight · 10–15% Class Periods · ~12–17 CED Topics · 8.1–8.4 Difficulty · Moderate

The lesson

Why ships float, planes fly, and there's an ocean of air on your shoulders

A 100,000-ton steel ship floats on the ocean. A 1-gram steel paperclip sinks. The steel is identical. So why does one float and the other sink? The answer Archimedes worked out in 250 BCE, in a moment of legendary inspiration in his bathtub, is also the answer to questions like "how does a hot air balloon rise?", "why does ice float on water?", and "how does a fish control its depth?" Buoyancy is one of those rare laws that explains an enormous range of phenomena from a single, simple principle.

Unit 8 is the capstone of AP Physics 1. Almost everything in it is built from concepts you already know — Newton's laws (Unit 2), conservation of energy (Unit 3), and the idea of forces and pressures from Unit 2. What's new is applying those tools to something continuous: fluids that flow, fill containers, and exert distributed pressures. Fluids are simply collections of particles too numerous to track individually — so we model them by their bulk properties: density, pressure, and flow rate.

The single most important sentence in this unit: a fluid pushes outward in all directions with pressure that depends only on depth — and any object immersed in it feels a buoyant force equal to the weight of the fluid it displaces.

Density: how much stuff is packed in

Density is mass per unit volume:

$$\rho = \dfrac{m}{V}$$

Units: kg/m³. Water is the canonical reference: $\rho_{\text{water}} = 1000$ kg/m³ (1 g/cm³). Air is about 1.2 kg/m³ — almost a thousand times less dense than water. Steel is around 7800 kg/m³. Mercury (metallic liquid) is 13,600 kg/m³ — the densest common liquid.

For an "incompressible" fluid (which AP Physics 1 assumes for liquids), density is constant — applying pressure doesn't change $\rho$. For a gas, density depends strongly on pressure and temperature, but for the limited gas problems on AP Physics 1, we'll usually treat air as having a fixed density too.

Pressure: force per unit area

Pressure is the perpendicular force per area exerted on (or by) a surface:

$$P = \dfrac{F_\perp}{A}$$

Units: pascals (Pa = N/m²). Pressure is a scalar — at any point in a fluid, the pressure has the same value regardless of which direction you measure it. (This is famously different from force, which is a vector.) Atmospheric pressure at sea level is about $P_{\text{atm}} = 1.013 \times 10^5$ Pa ≈ 100 kPa. This is the weight of a column of air about 10 meters tall pressing on every square centimeter of you.

Pressure in a static fluid: it grows with depth

Imagine a column of fluid of height $h$ and cross-sectional area $A$. The fluid above any point pushes down with weight $\rho gh \cdot A$. Per unit area, that's pressure $\rho gh$. So the pressure inside a static fluid increases linearly with depth:

$$P_{\text{gauge}} = \rho g h$$

Here $P_{\text{gauge}}$ is the pressure due to the fluid above — what a "gauge" would read if the surface pressure (atmospheric) were zero'd out. The absolute pressure includes atmospheric pressure on top:

$$P_{\text{abs}} = P_0 + \rho g h$$

where $P_0$ is whatever pressure exists at the top of the fluid (usually atmospheric). For a swimming pool, the gauge pressure at 3 m depth is $\rho gh = (1000)(10)(3) = 30{,}000$ Pa = 30 kPa. The absolute pressure is $101{,}000 + 30{,}000 \approx 131$ kPa — atmospheric pressure plus the water above.

Pressure depends only on depth, not shape

Two open containers, one tall and narrow, the other short and wide, are both filled with water to the same height $h$. The pressure at the bottom is the same in both: $P_0 + \rho gh$. The shape of the container doesn't matter — only the vertical depth. This is sometimes called the "hydrostatic paradox" because it's counterintuitive (the wider container has way more water!). The reason: the water above any specific point on the bottom only consists of the column directly above it; the rest of the water in the container is supported by the sides.

Worked example 1 · pressure at depth

A submarine's hull

A submarine dives to 200 m below the ocean surface. Find (a) the gauge pressure at this depth, (b) the absolute pressure, (c) the inward force on a 0.5 m × 0.5 m hatch on the submarine. Use $\rho_{\text{seawater}} = 1025$ kg/m³, $g = 10$ m/s², and $P_{\text{atm}} = 101$ kPa.

(a) Gauge pressure

$P_{\text{gauge}} = \rho gh = (1025)(10)(200) = 2{,}050{,}000$ Pa = 2.05 MPa.

(b) Absolute pressure

$P_{\text{abs}} = 101{,}000 + 2{,}050{,}000 \approx 2.15$ MPa.

If the inside of the submarine is at atmospheric pressure, the net inward force on the hatch is the gauge pressure × area: $F = P_{\text{gauge}} A = (2.05 \times 10^6)(0.25) \approx 513{,}000$ N. About 500 kN — over 50 metric tons of force!

Net inward force on hatch ≈ 513 kN.

Buoyancy: the upward push of a fluid

An object immersed in a fluid experiences pressure on all its surfaces. Pressure is greater at the bottom of the object than at the top (because the fluid is deeper there). The result: a net upward force on the object, called the buoyant force. Archimedes' principle tells us its magnitude:

$$F_b = \rho_{\text{fluid}} V_{\text{displaced}} g$$

The buoyant force equals the weight of fluid displaced by the object. Notice that this depends on the fluid's density (not the object's) and the volume submerged (not the total volume of the object). A boat floating on water displaces just enough water for the displaced water's weight to equal the boat's weight — even though only a small part of the boat is below the surface.

Why ships float and paperclips sink

The maximum buoyant force on an object is the weight of fluid that would fill its entire volume. If this max buoyant force is greater than the object's weight, it floats; if less, it sinks. For a steel paperclip, max buoyant force = weight of water with volume of the paperclip = (1000)($V$)(g). Paperclip's weight: (7800)($V$)(g). Weight wins, paperclip sinks. For a steel ship, the volume includes all the air inside the hull — so the displaced water is the volume of the entire hull, including the air. That total water weight exceeds the ship's actual weight (which is mostly hull metal + cargo + air-empty cargo holds), so it floats.

Worked example 2 · floating object

An iceberg

An iceberg has density 920 kg/m³. It floats in seawater of density 1025 kg/m³. What fraction of the iceberg's volume is above water?

Apply equilibrium

For a floating iceberg, the buoyant force equals its weight: $\rho_{\text{water}} V_{\text{sub}} g = \rho_{\text{ice}} V_{\text{total}} g$. Cancel $g$: $\rho_w V_{\text{sub}} = \rho_i V_{\text{total}}$.

So $V_{\text{sub}}/V_{\text{total}} = \rho_i/\rho_w = 920/1025 \approx 0.898$. About 89.8% of the iceberg is below the surface.

Above water: 1 − 0.898 ≈ 10.2%.

Bonus observation

The famous "tip of the iceberg" idiom is approximately right — only ~10% of an iceberg is visible above water, with the bulk lurking below. This made icebergs especially dangerous to early shipping.

Fully submerged objects: buoyancy and apparent weight

For an object fully submerged in a fluid, the buoyant force is $\rho_{\text{fluid}} V g$ where $V$ is the object's full volume. The net upward force on a submerged object is:

$$F_{\text{net}} = F_b - mg = (\rho_{\text{fluid}} - \rho_{\text{object}}) V g$$

If the fluid is denser than the object, this is positive — the object accelerates upward (and rises until partially out of the water, where it floats). If the fluid is less dense, the object sinks. If they're equal density, the object is neutrally buoyant — floats stationary in the middle.

The 2025 AP Physics 1 exam Q4 was exactly this scenario: a swimmer holds a block at rest underwater. When released, the block accelerates upward. The acceleration depends on $(\rho_{\text{fluid}} - \rho_{\text{block}})/\rho_{\text{block}} \cdot g$. In denser saltwater, the block accelerates upward faster than in fresh water — denser fluid means more buoyant force on the same volume.

Continuity: mass flow in, mass flow out

For a fluid flowing through a pipe or constriction, mass conservation requires that whatever flows in must flow out. If the pipe doesn't leak, the rate of mass crossing any cross-section is the same. For an incompressible fluid (constant density), this becomes a constraint on the volume flow rate:

$$A_1 v_1 = A_2 v_2$$

where $A$ is the cross-sectional area of the pipe and $v$ is the fluid speed. This is the continuity equation. Squeeze a hose and water comes out faster; widen the river and the current slows. The product of area and speed is constant.

Why doesn't a slow-moving river have more water?

Visit a wide, slow river and a narrow, fast stream — the volume of water passing per second can be exactly the same. The wide one has more cross-sectional area, but slower speed. The narrow one has less area, but faster speed. $Av$ (volume per second) is conserved as the same water flows down to the sea.

Bernoulli's equation: energy conservation for fluids

For a flowing incompressible fluid, energy conservation can be written as:

$$P_1 + \rho g y_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \tfrac{1}{2}\rho v_2^2$$

This is Bernoulli's equation. The three terms have direct interpretations:

$P$ = pressure energy per unit volume
$\rho g y$ = gravitational PE per unit volume (analog of $mgh$ divided by volume)
$\tfrac{1}{2}\rho v^2$ = kinetic energy per unit volume (analog of $\tfrac{1}{2}mv^2$ divided by volume)

The equation says: along a streamline, the sum of pressure, gravity-PE-per-volume, and KE-per-volume is constant. It's the same energy conservation you used in Unit 3, just expressed in a way appropriate for a continuously flowing fluid.

Bernoulli explains why airplanes fly (sort of)

Air flowing over the curved top of an airplane wing speeds up. Higher $v$ at the same height means lower $P$ (Bernoulli). Lower pressure on top, normal pressure on bottom: net upward force = lift. (The full explanation involves angle of attack and Newton's third law — air pushed downward exerts equal upward force on wing — but Bernoulli captures part of it.) The same principle explains how a curveball curves, why prairie dog burrows have ventilation, and why a roof can be peeled off in high winds (low pressure above the roof, normal pressure inside).

Torricelli's theorem: a tank draining

A classic Bernoulli application: a fluid tank with a small hole at depth $h$ below the surface. Apply Bernoulli between the top (where pressure is atmospheric and velocity is essentially zero) and the hole (where pressure is also atmospheric, since fluid emerges into open air, but velocity is some value $v$):

$$P_{\text{atm}} + \rho g h + 0 = P_{\text{atm}} + 0 + \tfrac{1}{2}\rho v^2$$

Solving: $v = \sqrt{2gh}$ — the same as if a particle had fallen from height $h$ in free fall. This is Torricelli's theorem: water emerging from a hole at depth $h$ below the surface moves at $\sqrt{2gh}$, which is exactly the free-fall speed it would have falling that height. Energy conservation in disguise.

Worked example 3 · Bernoulli + continuity

Squeezing a horizontal pipe

Water flows through a horizontal pipe. At point 1, the cross-sectional area is 0.04 m² and the speed is 2 m/s; at the constriction (point 2), the area is 0.01 m². If pressure at point 1 is 200 kPa, find the pressure at point 2.

Use continuity to find v₂

$A_1 v_1 = A_2 v_2 \Rightarrow v_2 = (A_1/A_2)v_1 = (0.04/0.01)(2) = 8$ m/s.

Apply Bernoulli (horizontal, so y₁ = y₂)

$P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2 \Rightarrow P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 200{,}000 + \tfrac{1}{2}(1000)(4 - 64) = 200{,}000 - 30{,}000 = 170{,}000$ Pa.

$P_2 = 170$ kPa

Sanity check

The water sped up at the constriction, so it gained KE. By energy conservation, this had to come from somewhere — pressure dropped to provide it. Faster fluid → lower pressure. This is the core insight Bernoulli's equation captures.

Where this unit fits in the bigger picture

Unit 8 ties together everything you've learned. Pressure relates to forces (Unit 2). Buoyancy is gravity + Newton's laws applied to extended objects. Continuity is mass conservation. Bernoulli is energy conservation. All four major themes of the course — forces, energy, momentum, and conservation laws — show up in fluid contexts. Master Unit 8 and you've shown you can apply your physics tools to any continuous medium.

In AP Physics 2 (if you take it), fluids are revisited and extended. You'll learn about viscosity, surface tension, and gas laws. You'll also study electric fields and circuits — which use the same conservation/flow ideas in different language.

Conceptual summary

Density is mass per volume; pressure is force per area. Inside a static fluid, pressure increases with depth as $P_{\text{gauge}} = \rho g h$. Any object in a fluid experiences a buoyant force equal to the weight of fluid displaced (Archimedes' principle). Whether an object floats or sinks depends on whether the maximum buoyant force exceeds its weight — equivalently, whether its density is less than or greater than the fluid's. For flowing fluids, mass conservation gives the continuity equation $A_1 v_1 = A_2 v_2$; energy conservation gives Bernoulli's equation $P + \rho g y + \tfrac{1}{2}\rho v^2$ is constant along streamlines. Torricelli's theorem ($v = \sqrt{2gh}$ for a draining tank) is a special case of Bernoulli.

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Reference sheet

Every equation in Unit 8

★ Density

$$\rho = \dfrac{m}{V}$$

$\rho$: density (kg/m³)
$m$: mass (kg)
$V$: volume (m³)

Use when finding density from mass and volume, or vice versa. Water = 1000 kg/m³.

★ Pressure

$$P = \dfrac{F_\perp}{A}$$

$F_\perp$: perpendicular force (N)
$A$: area (m²)
Units: pascals (Pa)

Use when converting between force and pressure. Pressure is a scalar.

★ Hydrostatic gauge pressure

$$P_{\text{gauge}} = \rho g h$$

$h$: depth below surface

Use when finding pressure inside a static fluid at depth $h$. Above atmospheric.

Absolute pressure

$$P_{\text{abs}} = P_0 + \rho g h$$

$P_0$: reference pressure (often atmospheric)

Use when the surface is open to the atmosphere or another reference pressure.

★ Buoyant force (Archimedes)

$$F_b = \rho_{\text{fluid}} V_{\text{disp}} g$$

$V_{\text{disp}}$: volume of fluid displaced
$\rho_{\text{fluid}}$: fluid's density

Use when object is partially or fully submerged. Always upward.

Floating object equilibrium

$$\rho_{\text{fluid}} V_{\text{sub}} = \rho_{\text{obj}} V_{\text{total}}$$

$V_{\text{sub}}$: volume below surface

Use when object floats. Gives the fraction submerged: $V_{\text{sub}}/V_{\text{tot}} = \rho_{\text{obj}}/\rho_{\text{fluid}}$.

★ Continuity equation

$$A_1 v_1 = A_2 v_2$$

$A$: cross-sectional area
$v$: flow speed

Use when incompressible fluid flows through a varying pipe. Mass conservation.

★ Bernoulli's equation

$$P + \rho gy + \tfrac{1}{2}\rho v^2 = \text{const}$$

$P$: pressure
$y$: height above reference
$v$: flow speed

Use when ideal fluid flows along a streamline. Energy conservation.

Torricelli's theorem

$$v = \sqrt{2gh}$$

$h$: depth below fluid surface

Use when fluid drains from a hole at depth $h$. Special case of Bernoulli.

Volume flow rate

$$Q = Av$$

$Q$: volume per second (m³/s)

Use when a problem mentions flow rate. Same value at all points by continuity.

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Study guide

One page. Night before the test.

The 6 concepts that matter most

Pressure increases linearly with depth: $P = \rho g h$ (gauge). Doesn't depend on container shape — only on depth.
Buoyant force = weight of displaced fluid: $F_b = \rho_{\text{fluid}} V_{\text{disp}} g$. Always upward. Depends on fluid density (not object's).
An object floats if its density is less than the fluid's. Fraction submerged = ratio of densities.
Continuity: $A_1 v_1 = A_2 v_2$. Squeeze pipe → fluid speeds up. Conservation of mass for incompressible flow.
Bernoulli: $P + \rho gy + \tfrac{1}{2}\rho v^2$ is constant along a streamline. Faster flow → lower pressure (at same height).
Pressure is a scalar; force is a vector. Pressure exists "in the fluid" with no preferred direction.

Key equations

$\rho = m/V$
$P = F_\perp/A$
$P_{\text{gauge}} = \rho gh$, $P_{\text{abs}} = P_0 + \rho gh$
$F_b = \rho_{\text{fluid}} V_{\text{disp}} g$ (Archimedes)
$A_1 v_1 = A_2 v_2$ (continuity)
$P_1 + \rho g y_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \tfrac{1}{2}\rho v_2^2$ (Bernoulli)
$v = \sqrt{2gh}$ (Torricelli)

5 traps students fall for

Using object's density (instead of fluid's) in buoyancy formula.
Forgetting that buoyancy depends on volume submerged, not total volume.
Thinking pressure depends on container shape or width (it doesn't — only depth).
Not converting between gauge and absolute pressure when problem requires it.
Forgetting the ½ in the kinetic energy term of Bernoulli.

Float or sink? The density rule

Condition	Result
$\rho_{\text{obj}} < \rho_{\text{fluid}}$	Floats. Fraction below surface = $\rho_{\text{obj}}/\rho_{\text{fluid}}$.
$\rho_{\text{obj}} = \rho_{\text{fluid}}$	Neutrally buoyant. Hovers wherever placed.
$\rho_{\text{obj}} > \rho_{\text{fluid}}$	Sinks. Net force $(\rho_{\text{obj}} - \rho_{\text{fluid}}) V g$ downward.

Problem-solving checklist

Static fluid problem? Use $P = \rho gh$ for pressure. Use Archimedes for buoyancy.
Floating? Buoyant force = weight. Find the submerged fraction or volume.
Submerged? Find net force from buoyancy minus weight, then $F = ma$.
Flowing fluid? Use continuity for speed changes. Use Bernoulli for pressure changes.
Tank with hole? Torricelli: $v = \sqrt{2gh}$ at the hole.
Sanity check: Faster flow → lower pressure. Deeper → higher pressure. Less dense object → higher fraction above water.

If you only remember one thing Buoyancy is just gravity acting on the displaced fluid, redirected upward onto the immersed object. $F_b = \rho_{\text{fluid}} V_{\text{displaced}} g$. Float or sink depends on density comparison. Pressure increases with depth. Bernoulli is energy conservation in disguise. Master those four ideas — density, pressure with depth, buoyancy, and Bernoulli — and Unit 8 is essentially done.

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Worksheet A · Foundation

Ten problems building from basic to moderate

Use $g = 10$ m/s², $\rho_{\text{water}} = 1000$ kg/m³, $P_{\text{atm}} = 10^5$ Pa unless otherwise noted.

Foundation

A block has mass 4 kg and volume $2 \times 10^{-3}$ m³. Find its density. Will it float in water?

Foundation

A 2000 N force is applied perpendicular to a 0.04 m² surface. Find the pressure.

Foundation

Find the gauge pressure at a depth of 8 m below the surface of a freshwater lake.

Moderate

A wooden block of density 600 kg/m³ floats in water. What fraction of its volume is submerged? What fraction is above the water?

Moderate

A solid metal cube of side 0.1 m and density 8000 kg/m³ is fully submerged in water. Find (a) the buoyant force on it, (b) the weight of the cube, (c) the apparent weight (the tension in a string holding it submerged at rest).

Moderate

Water flows through a horizontal pipe. At point 1, the radius is 5 cm and the speed is 2 m/s. At point 2, the radius is 2 cm. Find the speed of the water at point 2.

Moderate

A water tank is open at the top. A small hole is punched in the side of the tank, 1.8 m below the water surface. Find the speed at which water exits the hole.

Moderate

A balloon is filled with helium ($\rho_{\text{He}} = 0.18$ kg/m³) and has a total volume of 0.5 m³. The balloon material itself has mass 50 g. Air density is 1.2 kg/m³. Find the net upward force on the balloon (or downward, if applicable).

Moderate

Water flows horizontally through a pipe that narrows from area 0.02 m² to 0.005 m². At the wide section, the speed is 1 m/s and the gauge pressure is 50 kPa. Find (a) the speed at the narrow section, (b) the gauge pressure at the narrow section.

Moderate

A submerged 3 kg object has volume $5 \times 10^{-4}$ m³. When fully submerged in a fluid, its apparent weight (force needed to hold it stationary) is 25 N. Find the density of the fluid.

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Worksheet B · AP exam style

Ten problems in AP exam format

Part I — Multiple Choice

AP MC

An ideal fluid flows through two sections of cylindrical pipe. The narrow section has radius $R$; the wide section has radius $2R$. The ratio of the fluid's speed in the wide section to its speed in the narrow section is:

AP MC

Two identical objects are submerged in two different fluids. Object 1 is in water (density $\rho_w$); Object 2 is in oil (density $\rho_o < \rho_w$). Compared to Object 1, the buoyant force on Object 2 is:

Greater, because oil is less dense.
Less, because oil is less dense.
The same, because the objects are identical.
Cannot be determined without knowing the volumes.

AP MC

Two open-top containers, one tall and narrow and the other short and wide, are filled with water to the same height $h$. Comparing the pressure at the bottom of each:

Pressure is greater at the bottom of the wider container.
Pressure is greater at the bottom of the taller container.
Pressure is the same in both — depends only on depth.
Cannot be determined without knowing volumes.

AP MC

A solid block of density $\rho_b$ floats in a fluid of density $\rho_f$, with 30% of the block's volume submerged. The ratio $\rho_b/\rho_f$ is:

0.30
0.70
1.00
3.33

AP MC

Air flows over the curved top of an airplane wing at speed $v_t$ and along the flatter bottom at $v_b$, with $v_t > v_b$. According to Bernoulli's equation, the pressure on the top of the wing is:

Greater than the pressure on the bottom (gives downward lift).
Less than the pressure on the bottom (gives upward lift).
Equal to the pressure on the bottom.
Bernoulli's equation does not apply to airplane wings.

Part II — Short Free Response

AP SFR

(~7 min) A solid block has density 800 kg/m³ and volume $2 \times 10^{-3}$ m³. It floats in water.

Draw a free-body diagram for the floating block at equilibrium.
Calculate the volume of the block submerged below the water surface.
The block is now pushed entirely below the surface and held there. Calculate the additional downward force needed to hold it submerged.

AP SFR

(~7 min) Water flows through a horizontal pipe that has area $A_1$ at a wide section and $A_2$ at a narrow section. The pressure in the wide section is $P_1$ and the pressure in the narrow section is $P_2$.

Starting from continuity and Bernoulli's equation, derive a symbolic expression for $P_2$ in terms of $P_1$, $\rho$, $v_1$, $A_1$, and $A_2$.
If $A_1 = 2 A_2$ and $v_1 = 1$ m/s with $P_1 = 200$ kPa, calculate $P_2$.
Will the pressure in the narrow section be greater than, less than, or equal to that in the wide section? Justify.

AP SFR

(~7 min) A cylindrical tank of cross-sectional area $A_T$ is filled with water to height $H$. A small circular hole of area $A_h \ll A_T$ is punched in the side at height $h$ above the ground (so the depth of water above the hole is $H - h$).

Starting from Bernoulli's equation, derive the speed at which water exits the hole.
The water exits horizontally and lands on the ground. Derive a symbolic expression for the horizontal distance $d$ it travels.
Optimize: at what value of $h$ does $d$ become maximum?

Part III — Extended Free Response

AP EFR

(~18 min — mirrors 2025 AP Physics 1 FRQ Q4) In Scenario 1, a swimmer holds a block of mass $m$ at rest in a tank of freshwater with density $\rho_1$. The block is released from rest and accelerates upward with initial acceleration $a_1$. All frictional forces are negligible.

In Scenario 2, the same swimmer holds the same block at rest in a tank of saltwater with density $\rho_2$, where $\rho_2 > \rho_1$. The block is again released from rest and accelerates upward with initial acceleration $a_2$.

Indicate whether $a_1$ is greater than, less than, or equal to $a_2$. Justify your answer in terms of all forces exerted on the block in each scenario, using qualitative reasoning beyond just citing equations.
Consider the general case where a block of mass $m$ and volume $V$ is fully submerged in a fluid of density $\rho$. Starting with Newton's second law, derive a symbolic expression for the initial upward acceleration $a$ of the block when released from rest. Express your answer in terms of $m$, $V$, $\rho$, and physical constants.
Indicate whether the expression for $a$ you derived in part (b) is or is not consistent with the claim made in part (a). Briefly justify by referencing your derivation.
Suppose the block has density $\rho_{\text{block}} > \rho_2 > \rho_1$. Will the block in either scenario sink or rise? Justify using your derived expression.

AP EFR

(~18 min — experimental design) A student wants to determine the density of an unknown irregularly-shaped solid object. The student has a spring scale, a beaker of water, and a string.

Describe a procedure that will let the student determine the density. Specify what to measure and any steps to reduce uncertainty.
Derive a symbolic expression for the object's density in terms of the spring scale readings in air ($W_{\text{air}}$) and submerged in water ($W_{\text{sub}}$), and the density of water $\rho_w$.
If the object reads 5.0 N in air and 3.0 N submerged, calculate the object's density.
Suggest one reason why the object might appear to weigh more or less than its true value when measured in this way, and how to correct for it.
Could this method be used to determine the density of a substance less dense than water (e.g., an oily wood)? Explain how to modify the procedure if so, or why it wouldn't work if not.

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Worksheet C · Challenge

Five problems at olympiad level

Challenge

Iceberg melting. An iceberg of density 920 kg/m³ floats in seawater of density 1025 kg/m³. The iceberg melts entirely. Does the water level in the sea rise, fall, or stay the same? Justify with calculation. (Hint: compare the volume of seawater displaced by the floating iceberg to the volume of water produced by melting the same mass of ice.)

Challenge

Pascal's hydraulic lift. A hydraulic lift consists of two vertical cylinders connected by a horizontal pipe at the bottom, both filled with oil. The narrow cylinder has area $A_1$; the wide one has area $A_2 \gg A_1$. A force $F_1$ is applied to a piston in the narrow cylinder. (a) Derive an expression for the force $F_2$ on the piston in the wide cylinder, due to pressure transmitted through the oil. (b) Show that the work done by the operator equals the work done on the load (energy conservation). (c) If the operator pushes the narrow piston down 1 m, by how much does the wide piston rise? Take $A_2/A_1 = 100$.

Challenge

Block on spring underwater. A block of mass 0.5 kg and volume $2 \times 10^{-4}$ m³ is attached to the bottom of a tank of water by a spring (k = 100 N/m) of natural length 0.2 m. At equilibrium underwater, find the length of the spring. (Hint: balance buoyancy, gravity, and spring force.) Then find the period of small oscillations.

Challenge

Atmospheric pressure decrease with altitude. The atmosphere is well-modeled by an "isothermal exponential" with $P(h) = P_0 e^{-h/H}$ where $H \approx 8500$ m is the scale height. (a) Show qualitatively that this comes from $dP/dh = -\rho g$ combined with the ideal gas law (constant temperature). (b) Find the altitude at which atmospheric pressure is half its sea-level value. (c) Mt. Everest's summit is at 8848 m. Estimate the air pressure there as a fraction of sea-level pressure.

Challenge

Water hammer / column oscillation. A long horizontal pipe of area $A$ and length $L$ is full of water. A pump pushes the water with constant pressure $\Delta P$ on one end while the other end is open. (a) Find the steady-state speed of water through the pipe (assuming Bernoulli applies). (b) The pump suddenly stops. Treating the column of water in the pipe as a single mass under pressure-driven motion, find the deceleration of the water just after the pump stops. (c) If a valve at the open end is suddenly closed while water is still flowing, the moving column compresses against the closed valve, creating very high pressures (the "water hammer effect"). Explain qualitatively why this happens, using momentum/impulse arguments.

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Complete worked solutions

Every problem, every step

Worksheet A — Foundation

· density check

$\rho = m/V = 4/(2 \times 10^{-3}) = 2000$ kg/m³. Density > water (1000 kg/m³), so it sinks.

$\rho = 2000$ kg/m³; sinks.

· pressure from force

$P = F/A = 2000/0.04 = 50{,}000$ Pa = 50 kPa.

$P = 50$ kPa

· pressure at depth

$P = \rho gh = (1000)(10)(8) = 80{,}000$ Pa = 80 kPa (gauge).

$P_{\text{gauge}} = 80$ kPa

· floating block

Fraction submerged = $\rho_{\text{block}}/\rho_{\text{water}} = 600/1000 = 0.60$ (60%). Above water: 40%.

60% submerged, 40% above.

Quick check: a wooden block of density 600 kg/m³ has 60% submerged. Density less than water → floats. Fraction below = density ratio. Memorize this.

· submerged dense cube

Volume: $V = 0.001$ m³. Mass: $m = \rho V = 8000(0.001) = 8$ kg. Weight: 80 N.

(a) $F_b = \rho_w V g = (1000)(0.001)(10) = 10$ N.

(b) Weight = 80 N.

(c) Tension = weight − buoyancy = 80 − 10 = 70 N.

$F_b = 10$ N; W = 80 N; tension = 70 N.

· pipe constriction continuity

$A \propto r^2$, so $A_1/A_2 = (5/2)^2 = 6.25$. Continuity: $v_2 = v_1(A_1/A_2) = 2(6.25) = 12.5$ m/s.

$v_2 = 12.5$ m/s

· Torricelli — water from hole

$v = \sqrt{2gh} = \sqrt{2(10)(1.8)} = \sqrt{36} = 6$ m/s.

$v = 6$ m/s

· helium balloon

Buoyant force from displaced air: $F_b = \rho_{\text{air}} V g = (1.2)(0.5)(10) = 6$ N.

Weight of balloon system: helium + balloon material = $\rho_{\text{He}} V + m_{\text{balloon}} = (0.18)(0.5) + 0.05 = 0.14$ kg. Weight = 1.4 N.

Net upward = $F_b - W = 6 - 1.4 = 4.6$ N upward.

Net force: 4.6 N upward.

A balloon rises because it (with its lighter-than-air gas) displaces air heavier than itself. Buoyancy in air works exactly like in water — just smaller densities and forces.

· Bernoulli with continuity

(a) Continuity: $v_2 = v_1 A_1/A_2 = 1(0.02/0.005) = 4$ m/s.

(b) Bernoulli (horizontal): $P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$. $P_2 = 50000 + \tfrac{1}{2}(1000)(1 - 16) = 50000 - 7500 = 42{,}500$ Pa.

$v_2 = 4$ m/s; $P_2 = 42.5$ kPa

· density of unknown fluid from apparent weight

Object's weight = mg = 30 N. Apparent weight 25 N means buoyancy = 30 − 25 = 5 N. Buoyancy: $F_b = \rho V g \Rightarrow \rho = F_b/(Vg) = 5/(5 \times 10^{-4} \cdot 10) = 1000$ kg/m³. (It's water.)

$\rho = 1000$ kg/m³

Worksheet B — AP Exam Style

· pipe radius ratio

Wide section has area $\pi(2R)^2 = 4\pi R^2$, narrow has $\pi R^2$. Ratio of areas wide/narrow = 4. Continuity: $v_{\text{wide}} = v_{\text{narrow}}/4$. So $v_{\text{wide}}/v_{\text{narrow}} = 1/4$.

(A) 1/4

· buoyancy in water vs oil

$F_b = \rho_{\text{fluid}} V g$. Same volume. Less dense fluid → less buoyancy.

(B) Less.

· pressure depends only on depth

$P = \rho gh$. Depends only on depth, not container shape or width.

(C) The same.

The "hydrostatic paradox" — but it's not really a paradox. The pressure at the bottom comes from the weight of the column of water directly above that specific point, not the total weight of water in the container.

· 30% submerged → density ratio

Floating: fraction submerged = density ratio. $V_{\text{sub}}/V = \rho_b/\rho_f = 0.30$.

(A) 0.30

· airplane wing lift

Bernoulli: faster flow → lower pressure (at same height). Top of wing has faster flow → lower pressure → upward lift.

(B) Less than the bottom (gives upward lift).

· floating block analysis

(a) FBD: weight (down, mg) and buoyant force (up, magnitude equal to mg for equilibrium).

(b) Submerged fraction = $\rho_b/\rho_w = 800/1000 = 0.8$. Submerged volume = $0.8(2 \times 10^{-3}) = 1.6 \times 10^{-3}$ m³.

(c) Fully submerged: max buoyant force = $\rho_w V g = (1000)(0.002)(10) = 20$ N. Weight = $\rho_b V g = 800(0.002)(10) = 16$ N. Net upward (without external force): 4 N. To hold submerged: apply 4 N downward.

$V_{\text{sub}} = 1.6 \times 10^{-3}$ m³; need 4 N downward to hold submerged.

· pipe constriction with derivation

(a) Continuity: $v_2 = v_1 A_1/A_2$. Bernoulli (horizontal): $P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$. Substitute $v_2$: $P_2 = P_1 + \tfrac{1}{2}\rho v_1^2(1 - (A_1/A_2)^2)$. For $A_1 > A_2$, the term in parentheses is negative, so $P_2 < P_1$.

(b) $A_1/A_2 = 2$. $v_1 = 1$ m/s. $P_2 = 200000 + \tfrac{1}{2}(1000)(1)(1 - 4) = 200000 - 1500 = 198{,}500$ Pa.

(c) Less. Faster flow at narrow section means more KE per volume, which by energy conservation came from pressure energy. Pressure is lower in narrow section.

$P_2 \approx 198.5$ kPa, less than $P_1$.

· tank with hole optimization

(a) Bernoulli from top to hole: $P_{\text{atm}} + \rho g(H - h) + 0 \approx P_{\text{atm}} + 0 + \tfrac{1}{2}\rho v^2 \Rightarrow v = \sqrt{2g(H-h)}$. (We use $A_h \ll A_T$ to ignore velocity at the top.)

(b) Water exits horizontally at speed $v$ from height $h$. Time to fall: $t = \sqrt{2h/g}$. Distance: $d = vt = \sqrt{2g(H-h)}\sqrt{2h/g} = 2\sqrt{h(H-h)}$.

(c) Maximize $h(H-h)$. Derivative wrt $h$: $H - 2h = 0$ ⇒ $h = H/2$ gives max distance. Plug back: $d_{\max} = 2\sqrt{(H/2)(H/2)} = H$. So with the hole at the middle, the water lands a distance H from the tank base.

$v = \sqrt{2g(H-h)}$; $d = 2\sqrt{h(H-h)}$; maximum at $h = H/2$.

Symmetric optimum at $h = H/2$: hole too low gives high speed but no time to fly; too high gives long fall but slow exit speed. The middle is the sweet spot.

· 2025 AP-style: block in fresh vs salt water

(a) Forces on block in each scenario: gravity (mg, down) and buoyancy ($\rho_{\text{fluid}} V g$, up). Net force up = buoyancy − weight = $\rho_{\text{fluid}} V g - mg$. Since $\rho_2 > \rho_1$, buoyancy is bigger in scenario 2 (saltwater), so net upward force is bigger, so initial acceleration is bigger. $a_2 > a_1$, i.e., $a_1 < a_2$.

(b) Newton's law: $F_b - mg = ma$ → $\rho V g - mg = ma$ → $a = g(\rho V - m)/m = g(\rho V/m - 1) = g\rho V/m - g$.

(c) Yes, consistent. Plug in $\rho_2 > \rho_1$: $a$ increases with $\rho$, confirming $a_2 > a_1$. Same answer as part (a) reasoning.

(d) If $\rho_{\text{block}} > \rho_2$: $\rho V > m$ requires $\rho > m/V = \rho_{\text{block}}$, which fails. So $\rho V < m$, making $a$ negative — block accelerates downward (sinks) in both scenarios.

$a = g(\rho V - m)/m$; $a_2 > a_1$ when $\rho_2 > \rho_1$; block sinks if denser than fluid.

· density via apparent weight experiment

(a) Procedure: weigh object in air to get $W_{\text{air}}$. Submerge object fully in water and weigh again to get $W_{\text{sub}}$ (difference is the buoyant force). Use buoyant force = weight of displaced water → volume → density. To reduce uncertainty: ensure object is fully submerged but not touching beaker bottom; take multiple readings.

(b) Buoyant force: $F_b = W_{\text{air}} - W_{\text{sub}} = \rho_w V g$, so $V = (W_{\text{air}} - W_{\text{sub}})/(\rho_w g)$. Mass: $m = W_{\text{air}}/g$. Density: $\rho = m/V = W_{\text{air}}/(W_{\text{air}} - W_{\text{sub}}) \cdot \rho_w$.

(c) $\rho = (5.0/(5.0 - 3.0))(1000) = (2.5)(1000) = 2500$ kg/m³.

(d) Source of error: surface tension of water can pull object slightly downward. Correct by using a string attached to object, lowered into water carefully.

(e) Yes, but for objects less dense than water (which would float), tie a heavier sinker to the wood and submerge both. Measure $W_{\text{sub, both}}$ and compare to known sinker submerged alone. Subtract to find wood's contribution.

$\rho = W_{\text{air}}\rho_w/(W_{\text{air}} - W_{\text{sub}})$; object's density 2500 kg/m³.

This is the same procedure Archimedes used (with a balance instead of a spring scale). It works for any solid object you can submerge — clean, fast, and accurate.

Worksheet C — Challenge

· iceberg melting

Iceberg has mass $m_i$, volume $V_i = m_i/\rho_{\text{ice}}$. Submerged volume = $V_{\text{sub}} = m_i/\rho_{\text{seawater}}$. When ice melts, water produced has mass $m_i$, and this water has some volume in the sea. Density of melted water (fresh) = $\rho_w = 1000$ kg/m³, but the surrounding seawater is denser at 1025 kg/m³. So the water from melted ice has volume $V_{\text{melted}} = m_i/\rho_w$, but only takes up volume $V_{\text{filled}} = m_i/\rho_{\text{seawater}}$ in the sea (because it mixes/displaces equivalent salt water).

Wait — let me reconsider. Displaced water by the iceberg = $m_i/\rho_{\text{seawater}}$. Volume occupied by melted water of same mass in the sea = ?? Well, melted water has density of fresh water $\rho_w$ but it's now in sea — assume it mixes away. The total mass of water in the ocean increased by $m_i$. Sea level rises by $m_i/(\rho_{\text{seawater}} \cdot A_{\text{sea}})$. Iceberg displaced $m_i/\rho_{\text{seawater}}$, then disappeared. Net change: mass added $m_i$ → volume added (at seawater density) $m_i/\rho_{\text{seawater}}$. But it WAS already displacing exactly that volume! So sea level stays the same.

Wait — iceberg fresh water mixed with salty water means lower density of overall sea, complications. AP-level answer: an iceberg floating displaces exactly its own mass of water; when it melts, the water produced has exactly that mass; if the water densities matched, sea level wouldn't change. Since meltwater is fresh and less dense, it occupies slightly more volume, raising sea level a small amount. But for a freshwater iceberg in a freshwater lake, sea level wouldn't change at all.

In freshwater lake: no change. In salt sea: tiny rise (because freshwater meltwater is less dense than the saltwater it displaced).

This is why Arctic sea ice melting (which floats already) doesn't directly raise sea levels — but Antarctic ice sheets melting (which were on land, not floating) does, and significantly. The "floating ice melts" scenario only barely changes sea level via salinity effects.

· Pascal's hydraulic lift

(a) Pressure transmitted through fluid: $P = F_1/A_1 = F_2/A_2 \Rightarrow F_2 = F_1 A_2/A_1$. Mechanical advantage = $A_2/A_1$.

(b) Volume conservation: oil pushed down on narrow side equals oil rising on wide side. $A_1 d_1 = A_2 d_2 \Rightarrow d_2 = d_1 A_1/A_2$. Work check: $W_{\text{op}} = F_1 d_1$; $W_{\text{load}} = F_2 d_2 = (F_1 A_2/A_1)(d_1 A_1/A_2) = F_1 d_1$. Same work.

(c) If $A_2/A_1 = 100$ and $d_1 = 1$ m, then $d_2 = 1/100 = 0.01$ m = 1 cm.

$F_2 = F_1 (A_2/A_1)$; load rises by $d_1 (A_1/A_2)$; same work either side.

· block on spring underwater

Forces on block: weight $mg = 5$ N down; buoyant force $\rho_w V g = (1000)(2 \times 10^{-4})(10) = 2$ N up; spring force on block depends on stretch. Equilibrium: spring force balances net of weight − buoyancy = 3 N down. So spring pulls block UP with 3 N (i.e., spring is stretched). Stretch = 3 N / 100 N/m = 0.03 m. New length: $0.2 + 0.03 = 0.23$ m.

Period of small oscillations: same as ordinary spring (gravity and buoyancy just shift equilibrium): $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/100} = 2\pi(0.0707) \approx 0.444$ s.

Spring length 0.23 m; period ≈ 0.44 s.

· atmospheric scale height

(a) $dP/dh = -\rho g$. Ideal gas: $\rho = PM/(RT)$. So $dP/dh = -(M g/RT)P$, giving $P = P_0 e^{-h/H}$ with $H = RT/(Mg)$.

(b) $P = P_0/2$ when $e^{-h/H} = 1/2$, i.e., $h = H \ln 2 \approx 8500(0.693) \approx 5900$ m. So pressure halves at about 5.9 km — roughly the elevation of high-altitude airplane cruising.

(c) Mt. Everest: $h = 8848$ m. $P/P_0 = e^{-8848/8500} \approx e^{-1.04} \approx 0.35$. So about 35% of sea level pressure — which is why climbers need supplemental oxygen above ~7500 m.

Halves at ≈ 5.9 km; Everest summit ≈ 35% of sea level.

· water hammer

(a) Bernoulli: $\Delta P + 0 = 0 + \tfrac{1}{2}\rho v^2 \Rightarrow v = \sqrt{2\Delta P/\rho}$.

(b) Pump stops, pressure on pumping end drops to atmospheric (assume). Net force on water column = 0 (pressure even), so deceleration is initially zero — but friction and gravity (if pipe slopes) will eventually slow it.

(c) Water column has momentum $p = mv = (\rho A L)v$. When valve slams shut, this momentum must go to zero in a very short time (essentially the time for sound to travel to closed valve and back). Force = $\Delta p/\Delta t$ — small $\Delta t$ gives very large force, hence very high pressure spike. This is the water hammer — pipes can crack from this. Real plumbing systems use air chambers ("hammer arrestors") to provide a compressible buffer.

Steady-state $v = \sqrt{2\Delta P/\rho}$; water hammer = momentum decreased over very short time → huge force.

· · ·

Common mistakes

Seven errors that cost students points on Unit 8

Using the object's density instead of the fluid's in buoyancy

A 5 kg metal cube with density 5000 kg/m³ is fully submerged in water. Find the buoyant force.

$F_b = \rho_{\text{object}} V g = 5000 \cdot 0.001 \cdot 10 = 50$ N. (This is just the weight!)

$F_b = \rho_{\text{fluid}} V g = 1000 \cdot 0.001 \cdot 10 = 10$ N. Buoyancy depends on the fluid the object is in, not the object itself.

Archimedes' principle: buoyancy = weight of fluid displaced. The displaced fluid has the fluid's density, not the object's. Mixing these up gives the object's weight, which is wrong.

A 2 kg block of wood ($\rho = 800$ kg/m³) is held submerged underwater. Find the buoyant force on it. (Hint: don't use 800.)

Using total volume instead of submerged volume for floating objects

A boat of volume 50 m³ has 20 m³ submerged. Find the buoyant force on the boat.

$F_b = \rho_w (50)(10) = 500{,}000$ N (using total volume).

$F_b = \rho_w V_{\text{sub}} g = 1000(20)(10) = 200{,}000$ N. For a floating object, only the submerged part displaces water.

Buoyancy depends on volume of displaced fluid. For a floating object, that's only the submerged part — the air above the waterline doesn't displace water. For a fully submerged object, use full volume.

A wooden block of volume $2 \times 10^{-3}$ m³ floats with 60% of its volume below water. Find the buoyant force on it.

Thinking pressure depends on container shape or amount of fluid

Two open containers, one tall narrow and one short wide, both filled with water to height 1 m. Compare pressures at the bottom.

Wider container has more water → more pressure.

Same pressure: $P = \rho gh$. Depends only on depth, not container shape or volume.

The pressure at the bottom is from the column of fluid directly above that specific point. Width and shape don't matter — only how deep you are.

A swimming pool and a small fish tank are both filled with water to a depth of 0.5 m. Compare the pressure at the bottom of each.

Forgetting that pressure at top of fluid isn't always atmospheric

A sealed tank with gas at 200 kPa above water. Find the absolute pressure 5 m below the water surface inside the tank.

$P = P_{\text{atm}} + \rho gh = 100 + 50 = 150$ kPa.

$P = P_{\text{top}} + \rho gh = 200 + 50 = 250$ kPa. Top pressure is whatever the gas pressure is, not necessarily atmospheric.

In sealed containers, the pressure on top is the gas pressure (or vacuum, if evacuated), not atmospheric. Only when the surface is open to the atmosphere does $P_0 = P_{\text{atm}}$.

A cylinder is sealed at the top with vacuum ($P = 0$) above the water. Find the pressure at a depth of 5 m below the water surface.

Forgetting the ½ in Bernoulli's KE term

Water flows horizontally. At point 1: $P = 100$ kPa, $v = 2$ m/s. At point 2: $v = 6$ m/s. Find $P_2$.

$P_2 = P_1 + \rho(v_1^2 - v_2^2) = 100000 + 1000(4 - 36) = 68$ kPa.

$P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 100000 + 500(4 - 36) = 84$ kPa. The factor of ½ is critical.

Bernoulli's KE term is $\tfrac{1}{2}\rho v^2$, not $\rho v^2$ — same factor of ½ as the regular kinetic energy formula. Skipping it gives twice the correct pressure change.

Water flows horizontally through a pipe at 1 m/s with 200 kPa pressure. The pipe narrows so the speed becomes 4 m/s. Find the pressure at the narrow section.

Thinking faster fluid has higher pressure

Air flows over an airplane wing. The top has faster flow than the bottom. Compare pressures.

Faster flow = more force = higher pressure on top.

Faster flow = lower pressure (Bernoulli). The top of the wing has lower pressure than the bottom — this is what generates lift.

Counterintuitive but correct. Bernoulli says $P + \tfrac{1}{2}\rho v^2$ is constant. If $v$ is bigger, $P$ must be smaller to compensate. Higher speed means kinetic energy "comes from" pressure energy, leaving less pressure.

Wind blows over the top of a flat roof at 30 m/s. Inside the house, air is still at atmospheric pressure. Will the pressure on top of the roof be more or less than atmospheric? What does this mean for the roof in a strong wind?

Confusing area-area ratio with radius-radius ratio in continuity

Pipe with radius 4 cm narrows to radius 2 cm. If speed in wide section is 1 m/s, find speed in narrow section.

$v_2 = v_1 (r_1/r_2) = 1(2) = 2$ m/s.

$v_2 = v_1 (A_1/A_2) = v_1 (r_1/r_2)^2 = 1(4) = 4$ m/s. Continuity uses areas, which scale as radius squared.

Continuity equation is $A_1 v_1 = A_2 v_2$. For circular pipes, $A = \pi r^2$, so the ratio of areas is the ratio of radii squared. Forgetting the square gives wrong speeds by a factor of the radius ratio.

A garden hose has inner radius 1 cm and the nozzle has inner radius 0.5 cm. If water flows through the hose at 0.5 m/s, find the speed at the nozzle.

· · ·

Tutor's corner

What I learned going through this unit — and from helping others through it

What every student struggles with (in order)

The #1 sticking point in Unit 8 is the buoyancy formula's dependence on the fluid's density rather than the object's. Students keep writing $F_b = \rho_{\text{object}} V g$, which is wrong — that's just the object's weight. The correct version is $F_b = \rho_{\text{fluid}} V_{\text{disp}} g$. When something feels off in a buoyancy problem, double-check that you're using the fluid's density, not the object's.

The #2 struggle is "submerged vs total volume." For floating objects, only the submerged part displaces water. Students often plug the total volume into the buoyancy formula and get answers that are way off. Read the problem carefully: is the object floating? Then $V_{\text{disp}}$ is just the submerged part. Is it fully submerged? Then $V_{\text{disp}} = V_{\text{total}}$.

The #3 struggle is the counterintuitive Bernoulli relationship: faster fluid → lower pressure. It feels backward (shouldn't faster mean more force?). The reason: kinetic energy and pressure energy trade off. If the fluid speeds up, that KE came from somewhere — the pressure dropped. Drill problems on this until the relationship is reflexive.

What separates a 4 from a 5

Recognizing when to use which conservation law. Students who score 5 read a fluids problem and immediately classify: static pressure (use $P = \rho gh$), buoyancy (Archimedes), flow with area change (continuity), flow with pressure change (Bernoulli). Students who score 4 try one approach, hit a dead end, then guess at another. Build the classification reflex by drilling Worksheet B problems.

The question that reveals true understanding

"A boat full of rocks floats in a small pond. The rocks are dumped overboard, sinking to the bottom of the pond. Does the water level rise, fall, or stay the same?"

The deep answer: it falls. While in the boat, the rocks displaced their weight in water (since they were part of the boat, which was floating in equilibrium). Once on the bottom, they only displace their volume. Since rocks are denser than water, weight of rocks > volume of rocks (in water), so they displaced more water as cargo than they do as sunken objects. Water level falls. Students who reason through this carefully — distinguishing weight-of-displaced vs volume-of-object — have grasped buoyancy at the level the AP exam tests.

Exam-day strategy

Static fluid problem? Use $P = \rho gh$ for pressure at depth, and Archimedes for buoyant force. Identify whether the object is floating or fully submerged — different formulas for displaced volume.

Flowing fluid problem? Continuity $A_1 v_1 = A_2 v_2$ for speed changes. Bernoulli $P + \rho gy + \tfrac{1}{2}\rho v^2 = \text{const}$ for pressure changes. Together they handle most flow problems.

Buoyancy problem? Always check: is it floating or submerged? Use $\rho_{\text{fluid}}$ (not the object's). Get $V_{\text{disp}}$ right.

Comparison problem? Reason through which density appears where. The 2025 AP exam Q4 was exactly this — comparing acceleration in two different fluid densities.

A surprising real-world connection

A submarine controls its depth by adjusting buoyancy, not propulsion. To dive: it floods ballast tanks with seawater, increasing the submarine's effective density above seawater's, making it sink. To surface: it pumps the water out and replaces it with compressed air, decreasing density below seawater's, making it rise. To hover at constant depth: it adjusts ballast until its density exactly matches seawater. Modern submarines do this with millimeter precision, allowing them to "park" silently at any depth. Same physics: scuba divers use a buoyancy compensator vest that they inflate or deflate to neutral buoyancy. Fish do this with a swim bladder — an internal gas-filled organ they inflate or deflate to control depth without effort. Buoyancy is a tool that life and engineering exploit constantly.

One piece of advice

Drill the buoyancy + Newton's second law combination until it's automatic. The 2025 AP Physics 1 exam ended with exactly this pattern: a block of mass $m$ and volume $V$ submerged in fluid of density $\rho$, released from rest, accelerates at $a = g(\rho V - m)/m$. This single formula handles every "block in fluid" problem on the exam. Memorize the derivation: $F_{\text{net}} = F_b - mg$, $F_{\text{net}} = ma$, so $a = (\rho V g - mg)/m = g(\rho V/m - 1) = g(\rho/\rho_{\text{obj}} - 1)$. The result is positive (rises) if $\rho > \rho_{\text{obj}}$, negative (sinks) if not. Worksheet B problem 19 in this unit drills exactly the 2025 exam pattern. Master that one and you've covered the most likely Unit 8 free-response pattern for any future AP exam.

Congratulations — if you've worked through all 8 units of First Principles Physics, you're ready for the AP Physics 1 exam. Trust your problem-solving instincts, draw clear diagrams, write conservation laws explicitly, and remember: every problem decomposes into the patterns you've drilled. Good luck.

End of Unit 8 · Fluids · First Principles Physics · End of Course

Condition	Result
\(\rho_{\text{obj}} < \rho_{\text{fluid}}\)	Floats. Fraction below surface = \(\rho_{\text{obj}}/\rho_{\text{fluid}}\).
\(\rho_{\text{obj}} = \rho_{\text{fluid}}\)	Neutrally buoyant. Hovers wherever placed.
\(\rho_{\text{obj}} > \rho_{\text{fluid}}\)	Sinks. Net force \((\rho_{\text{obj}} - \rho_{\text{fluid}}) V g\) downward.