AP Physics 1 · Topic 8.1

Topic 8.1: Internal Structure and Density

Unit 8: Fluids10-15% of the multiple-choice section

Density is mass divided by volume, in kilograms per cubic meter. It is a property of the material, so cutting an object in half leaves it unchanged. A fluid is a substance with no fixed shape, so liquids and gases both count, and an ideal fluid is incompressible and has no viscosity.

AP Physics: Unit 8 (topics 8.1 Internal Structure and Density). AP Physics 1 Unit 8, Topic 8.1. The single learning objective, 8.1.A, asks students to describe the properties of a fluid. Four essential knowledge statements support it: 8.1.A.1 states that distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules; 8.1.A.2 states that a fluid is a substance that has no fixed shape; 8.1.A.3 states that fluids can be characterized by their density, defines density as a ratio of mass to volume, and gives rho = m/V; and 8.1.A.4 states that an ideal fluid is incompressible and has no viscosity. The topic prints no boundary statement. The CED's suggested skills here are 1.B, 2.B, 2.C, 3.A, and 3.C. Unit 8 carries 10 to 15 percent of the multiple-choice section and a suggested 12 to 17 class periods.

What Topic 8.1 requires

Topic 8.1 carries one learning objective, 8.1.A: describe the properties of a fluid. Four essential knowledge statements sit under it, and none of them has sub-statements.

8.1.A.1 states that distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules.
8.1.A.2 states that a fluid is a substance that has no fixed shape.
8.1.A.3 states that fluids can be characterized by their density, that density is defined as a ratio of mass to volume, and prints the relevant equation $\rho = \dfrac{m}{V}$ .
8.1.A.4 states that an ideal fluid is incompressible and has no viscosity.

That is the entire required content of the topic. Many topics in this course print a boundary statement to fence off what the exam will not ask; Topic 8.1 prints none, so there is no exception clause to hunt for and the four statements above have to be read literally rather than expanded.

The CED lists five suggested skills here, the largest set of the four topics in the unit: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Two of those five, 1.B and 3.A, are about producing data and procedures rather than about solving for a number, which tells you where this topic lands on a free-response section.

Unit 8 is weighted at 10 to 15 percent of the multiple-choice section, and the CED suggests roughly 12 to 17 class periods for the unit.

One piece of positioning worth checking before you trust an outside resource. Unit 8 belongs to the AP Physics 1 course framework effective fall 2024. In the AP Physics 2 CED of the same date, the units run from 9, Thermodynamics, to 15, Modern Physics, and none of the seven is fluids; that document refers in passing to torque, rotation, and fluids as AP Physics 1 topics. A study guide that files fluids under Physics 2 is not describing the current pair of courses. The fluids guide covers the whole unit in one pass.

What "internal structure" means here, and what it does not

The topic title promises internal structure, and essential knowledge 8.1.A.1 is the only statement that delivers it: distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules. Read that sentence carefully, because it is doing less work than a chemistry course would ask of it.

What it gives you is a causal direction. The difference between a steel bar, the water in a glass, and the air above the water is not a difference in the kind of matter at some abstract level; it is a difference in how strongly and how persistently the particles interact. Strong, fixed interactions hold particles in place and give a solid a shape of its own. Weaker or more mobile interactions let particles slide past one another, and the substance takes the shape of whatever holds it. That is the whole argument, and it is the reason 8.1.A.2 can define a fluid by shape rather than by chemistry.

What 8.1.A.1 does not give you is a list. It does not name a single type of intermolecular force, does not mention bonding, polarity, hydrogen bonding, phase diagrams, melting points, or the mole. None of those appear anywhere in the AP Physics 1 course framework. The AP Physics 1 equation sheet prints no gas law and no chemical constants: Avogadro's number, the universal gas constant, and Boltzmann's constant are on the AP Physics 2 sheet, not this one. If a review resource has you memorizing intermolecular force types for AP Physics 1, it is teaching a different subject.

So the honest scope of "internal structure" in Topic 8.1 is qualitative and one sentence deep. You should be able to explain, in words, why a liquid takes the shape of its container and a solid does not, and to attribute that difference to the interactions between the particles. You should not expect to calculate anything from it. Every number in this unit comes from the equations, and the first of those is density.

A fluid is a substance that has no fixed shape (8.1.A.2)

The CED's definition is short enough to quote whole: a fluid is a substance that has no fixed shape. Notice what the definition is built from. Not "a liquid." Not "something wet." Not "something that pours." Shape.

Run the test on three cases.

Water in a beaker. Pour it into a taller, narrower beaker and it takes the new shape. It is a fluid.
Air in a room. It fills whatever container holds it, with no shape of its own. It is a fluid, and the CED treats it as one: an essential question printed at the front of Unit 8 asks why we do not feel the miles of air above us pushing us down.
A steel block. Move it to a different container and it keeps its shape. It is not a fluid.

The practical consequence for the exam is that gases count. A question about atmospheric pressure, about a column of air, or about gas flowing through a duct is a Unit 8 question, and the same four equations of fluid statics and flow apply. Students who read "fluid" as "liquid" lose the second half of the unit.

One distinction the definition deliberately leaves alone is volume. A fluid has no fixed shape, but a given amount of an incompressible fluid does have a fixed volume: that is essential knowledge 8.2.A.3, over in Topic 8.2, which states that the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid. Shape is free; volume is not. Holding those two apart is what lets you say that squeezing a sealed syringe of water raises the pressure without changing the density inside it.

Density is a ratio, not an amount (8.1.A.3)

\rho = \frac{m}{V}

Kilograms divided by cubic meters gives kilograms per cubic meter, and the symbol is the Greek letter rho. The equation is printed on the AP Physics 1 equation sheet, and the sheet's symbol list defines $\rho$ as density, $m$ as mass, and $V$ as volume, so nothing here has to be recalled from memory.

The single idea that makes density behave differently from mass and volume is that it is a ratio. Mass and volume are amounts: take more of the substance and both go up. Density is what you get when you divide one by the other, so the increases cancel and the number does not move. Cut an aluminum block in half and each half has half the mass, half the volume, and exactly the original density. Melt a candle into a puddle and, ignoring any change of state effects, the same argument applies. This is why 8.1.A.3 says fluids can be characterized by their density: the density is a fingerprint of the substance, while the mass is a fingerprint of the sample.

Two unit habits save time.

Convert to kilograms per cubic meter before doing anything else. One gram per cubic centimeter equals 1000 kilograms per cubic meter, because a gram is $10^{-3}$ kg and a cubic centimeter is $10^{-6}$ m $^3$ . Fresh water sits at about $1000 \, \mathrm{kg/m^3}$ , which is 1.00 g/cm $^3$ in the other set of units.
Watch the cube when you convert a length. One centimeter is $10^{-2}$ m, so one cubic centimeter is $10^{-6}$ m $^3$ , not $10^{-2}$ m $^3$ . Getting this wrong scales a density by a factor of ten thousand and is the most common arithmetic failure in the topic.

For an object that is not made of one uniform material, or that is hollow, $\rho = m/V$ still applies, but with $V$ the total volume the object occupies. The result is an average density, and the average is what matters for whether the object floats, which is Topic 8.3. A steel ship and a steel bolt are made of the same material at the same material density, and only one of them floats, because only one of them has most of its volume filled with air. The third worked example below runs that calculation.

Ideal fluids: incompressible and no viscosity (8.1.A.4)

Essential knowledge 8.1.A.4 defines an ideal fluid with exactly two properties. It is incompressible, and it has no viscosity. Each one buys you a specific simplification later in the unit, and it is worth knowing which is which.

Incompressible means the density does not change when the pressure does. Squeeze it, take it deeper, push it through a constriction, and $\rho$ stays put. That single property is what lets you carry one value of $\rho$ through an entire problem: through $P = P_0 + \rho g h$ in Topic 8.2, through the continuity equation, and through both sides of Bernoulli's equation in Topic 8.4. If the density could change with depth, the pressure equation would not be linear in $h$ and none of the rest would follow.

No viscosity means no internal friction between neighboring layers of the fluid, and none between the fluid and the pipe wall. Nothing dissipates mechanical energy as the fluid moves. That is the property that makes Bernoulli's equation a conservation statement rather than an approximation, and it is why a Unit 8 problem never asks you to account for energy lost along a length of pipe. Real water in a real hose does lose energy that way; the ideal fluid does not.

The exam does not make you guess when these apply. The conventions printed with the AP Physics 1 Table of Information state that fluids are assumed to be ideal, and pipes are assumed to be completely filled by fluid, unless otherwise stated. Note both halves. The default is ideal, and the default is a full pipe. The phrase "unless otherwise stated" is the part that keeps this from being a blanket rule, and the CED's own materials use it: the sample instructional activity listed for Topic 8.2 has students download a graph of air pressure against elevation and explain why its slope decreases with height, and the explanation is that the air gets less dense higher up. That is a compressible fluid, described in the CED, one topic later. When a question hands you a situation where the density plainly varies, the question has otherwise stated.

Mass, volume, weight, size, density: the CED's own vocabulary warning

The Unit 8 overview in the CED singles out language as a scoring risk. It notes that students may believe objects float in water because the objects are "lighter" than water, or sink because they are "heavier" than water, and it says students should know the difference between the meanings of "mass," "volume," "weight," "size," and "density." The overview ties this to the free-response section, where written expression and justification carry many of the available points.

The fix is to keep five quantities apart on paper.

Quantity	Symbol	Unit	What it describes
Mass	$m$	kg	How much matter is in this sample
Volume	$V$	m $^3$	How much space this sample occupies
Weight	$F_g = mg$	N	The gravitational force on this sample
Size	none	none	An informal word; not a physics quantity
Density	$\rho = m/V$	kg/m $^3$	A property of the material, independent of sample

Read the misconception again with that table in view. "Lighter than water" is not a claim you can even check until you say lighter in what sense. A cruise ship weighs far more than a bucket of water and floats; a steel bolt weighs far less than a lake and sinks. Weight alone predicts nothing. The comparison that works is between densities, because density is the only entry in the table that does not depend on how much of the stuff you happen to have, so it is the only one that lets you compare two samples of wildly different size.

When you write a justification, name the quantity you are comparing and say it is a comparison of densities, not of weights or sizes. "The block's density is less than the fluid's density, so it floats" earns the point. "The block is lighter, so it floats" does not, and it is wrong in general, which is exactly why the CED flags it. The same discipline pays off in the free-body diagram language you will need in Topic 8.3, where weight and buoyant force are two separate forces in newtons and density is not a force at all.

Measuring density: the graph and the procedure (skills 1.B and 3.A)

Two of the five suggested skills for this topic are about generating evidence rather than consuming it. Skill 1.B is creating quantitative graphs with appropriate scales and units, including plotting data, and skill 3.A is creating experimental procedures appropriate for a given scientific question. Density is the natural place to practice both, because it is a slope.

Rewrite the definition as $m = \rho V$ . Plotted with volume on the horizontal axis and mass on the vertical axis, a set of samples of the same material gives a straight line through the origin whose slope is the density. That framing is worth more than the equation itself for three reasons.

The slope uses every data point. One sample gives you one ratio and one shot at a mistake. Five samples give you a line, and a line averages the random error out.
The intercept is a diagnostic. A best-fit line that misses the origin by a consistent amount is telling you about a systematic error, most often an unsubtracted container mass. The slope survives that error; a single-point ratio does not. The first worked example below puts numbers on how badly.
Curvature is information. If the plot of mass against volume bends, the samples are not all the same material, or the volume measurements are wrong, or the substance is not behaving as an incompressible one.

For the procedure half, the question a Topic 8.1 experiment answers is usually "what is the density of this material?" A workable procedure names the instruments, says what is varied and what is measured, says how many trials, and says how the data will be analyzed. Mass comes from a balance. Volume comes from geometry when the shape is regular, and the sheet's geometry table prints the formulas you would need, including $V = \ell w h$ for a rectangular solid, $V = \pi r^2 \ell$ for a cylinder, and $V = \frac{4}{3}\pi r^3$ for a sphere. Volume comes from water displacement when the shape is irregular. The CED's own sample activity for the unit does the irregular case with buoyancy rather than with a graduated cylinder: it has students use a spring scale and a deep sink to find the volume and density of an irregularly shaped metal object. That activity is filed under Topic 8.3, because it needs the buoyant force to work, which is a good reminder that the measurement techniques in this unit build on each other.

How Topic 8.1 is tested, and where it leads

Topic 8.1 rarely fills a question by itself. It supplies the quantity every other Unit 8 question runs on, and the ways it shows up are narrow enough to rehearse.

Compute a density from a mass and a set of dimensions, or the reverse, with a unit conversion buried in the given values (skill 2.B).
Compare two samples and decide what changed. Same material, different sizes: the densities match. Same mass, different volumes: the denser one is the smaller. This is skill 2.C, and it is usually a multiple-choice question with no arithmetic in it at all.
Read a density off the slope of a mass-versus-volume graph, or sketch what that graph looks like for two materials of different density (skill 1.B).
Design a procedure that would determine an unknown density, and say what you would plot (skill 3.A).
Justify a claim about floating, sinking, or ranking materials using density as the evidence rather than weight (skill 3.C).
Decide whether the ideal-fluid assumption is doing any work in a described situation.

From here the unit builds in one direction and never comes back. Topic 8.2 puts density into the pressure at a depth, $P = P_0 + \rho g h$ , where it is the fluid's density that appears. Topic 8.3 puts it into the buoyant force, $F_b = \rho V g$ , where the density is again the fluid's and the volume is the volume displaced, and where comparing the object's average density with the fluid's decides whether it floats. Topic 8.4 puts it into the continuity and Bernoulli equations, where it is the constancy of $\rho$ that lets the terms be written per unit volume at all.

That is four appearances of one symbol, and in three of them the trap is the same: the $\rho$ in the equation is the density of the fluid, not the density of the object sitting in it. Fix that association now, while there is only one density in the problem. The Unit 8 overview shows how the four topics fit together, and the AP Physics 1 hub puts the unit back in the context of the seven that come before it.

Density from a mass-versus-volume graph, and what a tray does to it

A student measures four samples of the same metal. The volumes are $1.00 \times 10^{-5}$ , $2.00 \times 10^{-5}$ , $3.00 \times 10^{-5}$ , and $4.00 \times 10^{-5} \, \mathrm{m^3}$ , and the corresponding masses are 0.0890, 0.1780, 0.2670, and 0.3560 kg. (a) Find the density from the graph. (b) A second student repeats the measurements but leaves each sample in a 0.0250 kg tray on the balance and forgets to subtract the tray. What density does that student get from a single sample, and what density from the slope?

(a) Plot mass on the vertical axis against volume on the horizontal axis. Take two widely separated points and find the slope: $\rho = \dfrac{0.3560 \, \mathrm{kg} - 0.0890 \, \mathrm{kg}}{4.00 \times 10^{-5} \, \mathrm{m^3} - 1.00 \times 10^{-5} \, \mathrm{m^3}} = \dfrac{0.2670 \, \mathrm{kg}}{3.00 \times 10^{-5} \, \mathrm{m^3}} = 8900 \, \mathrm{kg/m^3}$ .
Check that the line really is straight through the origin by taking each ratio separately: $0.0890/1.00 \times 10^{-5} = 8900$ , and the same for the other three. All four give $8900 \, \mathrm{kg/m^3}$ , so the data are consistent with a single material.
(b) With the tray included, the first sample reads $0.0890 + 0.0250 = 0.1140 \, \mathrm{kg}$ . Treating that as the sample mass gives $\rho = 0.1140 \, \mathrm{kg} / (1.00 \times 10^{-5} \, \mathrm{m^3}) = 11400 \, \mathrm{kg/m^3}$ .
That is wrong by $(11400 - 8900)/8900 = 0.281$ , about 28 percent high, and nothing in the single measurement reveals the error.
Now take the slope of the second student's data instead. Every mass is shifted up by the same 0.0250 kg, so the differences are unchanged: $\dfrac{(0.3560 + 0.0250) - (0.0890 + 0.0250)}{4.00 \times 10^{-5} - 1.00 \times 10^{-5}} = \dfrac{0.2670}{3.00 \times 10^{-5}} = 8900 \, \mathrm{kg/m^3}$ .
The tray shows up as a vertical intercept of 0.0250 kg, which is the diagnostic: a mass-versus-volume line that does not pass through the origin is reporting a constant offset in the mass measurement.

(a) $\rho = 8900 \, \mathrm{kg/m^3}$ , or 8.90 g/cm $^3$ . (b) A single sample gives $11400 \, \mathrm{kg/m^3}$ , about 28 percent too high, while the slope still gives $8900 \, \mathrm{kg/m^3}$ and the 0.0250 kg intercept exposes the mistake. This is the argument for plotting a line rather than dividing once.

Unit conversion, and why cutting the block changes nothing

A rectangular block measures 0.080 m by 0.050 m by 0.025 m and has a mass of 0.250 kg. (a) Find its density in kg/m $^3$ and in g/cm $^3$ . (b) The block is cut exactly in half. Find the density of one half. (c) Compare the block with fresh water at $1000 \, \mathrm{kg/m^3}$ .

(a) Volume of a rectangular solid, from the geometry table on the equation sheet: $V = \ell w h = (0.080 \, \mathrm{m})(0.050 \, \mathrm{m})(0.025 \, \mathrm{m}) = 1.0 \times 10^{-4} \, \mathrm{m^3}$ .
Density: $\rho = \dfrac{m}{V} = \dfrac{0.250 \, \mathrm{kg}}{1.0 \times 10^{-4} \, \mathrm{m^3}} = 2500 \, \mathrm{kg/m^3}$ .
Convert: $1 \, \mathrm{g/cm^3} = 10^{-3} \, \mathrm{kg} / 10^{-6} \, \mathrm{m^3} = 1000 \, \mathrm{kg/m^3}$ , so $2500 \, \mathrm{kg/m^3} = 2.5 \, \mathrm{g/cm^3}$ .
(b) Half the block has mass $0.125 \, \mathrm{kg}$ and volume $5.0 \times 10^{-5} \, \mathrm{m^3}$ . Its density is $0.125 / (5.0 \times 10^{-5}) = 2500 \, \mathrm{kg/m^3}$ , unchanged. Both numerator and denominator halved, so the ratio held.
(c) $2500 > 1000$ , so the block is 2.5 times as dense as fresh water. Its mass and volume tell you nothing about that comparison on their own; only the ratio does.

(a) $\rho = 2500 \, \mathrm{kg/m^3} = 2.5 \, \mathrm{g/cm^3}$ . (b) Still $2500 \, \mathrm{kg/m^3}$ : density is a property of the material, not of the amount. (c) The block is 2.5 times as dense as fresh water.

Average density of a hollow object

A sealed hollow ball has an outer radius of 0.050 m and a mass of 0.30 kg. (a) Find its average density. (b) If the shell is made of a material of density $2500 \, \mathrm{kg/m^3}$ , what fraction of the ball's outer volume is shell?

(a) Outer volume, from the sphere formula on the equation sheet's geometry table: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.050 \, \mathrm{m})^3 = 5.24 \times 10^{-4} \, \mathrm{m^3}$ .
Average density: $\rho_{\text{avg}} = \dfrac{0.30 \, \mathrm{kg}}{5.24 \times 10^{-4} \, \mathrm{m^3}} = 5.7 \times 10^{2} \, \mathrm{kg/m^3}$ (573 kg/m $^3$ before rounding to two significant figures).
(b) Almost all of the ball's mass is in the shell, so the shell's volume is $V_{\text{shell}} = \dfrac{m}{\rho_{\text{shell}}} = \dfrac{0.30 \, \mathrm{kg}}{2500 \, \mathrm{kg/m^3}} = 1.2 \times 10^{-4} \, \mathrm{m^3}$ .
Fraction: $\dfrac{1.2 \times 10^{-4}}{5.24 \times 10^{-4}} = 0.23$ , so about 23 percent of the ball is shell material and the other 77 percent is enclosed air.
Read the two densities against each other. The material is 2500 kg/m $^3$ and the object averages 570 kg/m $^3$ , a factor of more than four apart, and the only difference between them is which volume went in the denominator.

(a) $\rho_{\text{avg}} \approx 5.7 \times 10^{2} \, \mathrm{kg/m^3}$ . (b) About 23 percent of the outer volume is shell. The same object has a material density and an average density, and questions about floating always mean the average one.

Frequently asked questions

What is a fluid in AP Physics 1?

A fluid is a substance that has no fixed shape, which is the definition given in essential knowledge 8.1.A.2 of the AP Physics 1 course framework. That covers both liquids and gases, since a gas also takes the shape of its container, so questions about air and atmospheric pressure are fluids questions. A solid keeps its own shape and is not a fluid.

What is the formula for density in AP Physics 1?

Density is mass divided by volume, rho = m/V, printed on the AP Physics 1 equation sheet and given in essential knowledge 8.1.A.3. The SI unit is kilograms per cubic meter. One gram per cubic centimeter equals 1000 kilograms per cubic meter, so fresh water at about 1.00 g/cm^3 is about 1000 kg/m^3.

Does density change if you cut an object in half?

No. Density is a ratio of mass to volume, and cutting an object in half halves both, so the ratio is unchanged. A 0.250 kg block of volume 1.0 x 10^-4 m^3 has a density of 2500 kg/m^3, and each half has a mass of 0.125 kg, a volume of 5.0 x 10^-5 m^3, and the same density of 2500 kg/m^3. Density describes the material; mass and volume describe the sample.

What is an ideal fluid in AP Physics 1?

An ideal fluid is incompressible and has no viscosity, per essential knowledge 8.1.A.4. Incompressible means its density does not change when the pressure changes, which is what lets one value of rho be used throughout a problem. No viscosity means no internal friction and no energy lost to flow. The conventions printed with the AP Physics 1 Table of Information state that fluids are assumed to be ideal, and pipes are assumed to be completely filled by fluid, unless otherwise stated.

Do I need to know intermolecular forces for AP Physics 1 fluids?

No. Essential knowledge 8.1.A.1 says only that the distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules. It names no specific force, and the AP Physics 1 course framework contains no bonding, no phase diagrams, and no gas law. You need the qualitative idea that particle interactions explain why a fluid takes the shape of its container, not any chemistry.

How do you measure the density of an irregular object?

Measure the mass on a balance, then measure the volume by displacement rather than by geometry: submerge the object and record the volume of fluid it pushes aside, then divide mass by volume. The AP Physics 1 CED's sample activity for Unit 8 does this with buoyancy, using a spring scale and a deep sink to get the volume and density of an irregularly shaped metal object, and files that activity under Topic 8.3 because it needs the buoyant force.

Why is density better than weight for predicting whether something floats?

Because weight depends on how much of the substance you have and density does not. A ship weighs far more than a bucket of water and floats; a small steel bolt weighs far less than a lake and sinks. Comparing average densities removes the size of the sample from the question. The AP Physics 1 CED flags the belief that objects float because they are lighter than water as a misconception to correct in Unit 8.