- What the printed page says
-
Page 48, the stem: “Two satellites orbit the same planet in circular orbits. Satellite X orbits at radius r with orbital speed v, and Satellite Y orbits at radius 2r with orbital speed v/2…”
Page 55, the middle term of the Fast Path chain: TY = 2π(2r)/(v/2) = 4πr/(v/1) = 4·2πr/v.
Page 55, choice (C): “Assumes the periods are equal, perhaps reasoning that both satellites orbit the same planet.”
- What it should say
-
Page 48: “Two objects move in uniform circular motion. Object X moves in a circle of radius r with speed v, and Object Y moves in a circle of radius 2r with speed v/2…” — and, further down, “the period TX of Object X… the period TY of Object Y”.
Page 55, the chain: TY = 2π(2r)/(v/2) = 4πr/(v/2) = 8πr/v = 4·(2πr/v).
Page 55, choice (C): “Assumes the periods must be equal because both objects travel in circles. Period depends on both the path length and the speed: T = 2πr/v, and here both change.”
- Why it matters
- Two satellites of the same planet cannot have those radii and speeds together: a circular orbit forces v2r = GM, so doubling the radius requires the speed to fall to v/√2, not to v/2, and the period ratio would then be 23/2 ≈ 2.83 — which is not one of the four choices. A reader who applies the orbital-speed relation the book teaches elsewhere finds no answer that fits. Recast as plain uniform circular motion, with no gravity in the picture, the scenario is consistent and every number in it stays the same. The printed middle term is a separate slip: 4πr/v is half of 8πr/v, so the chain as printed asserts that a quantity equals twice itself. The key (B), TY = 4TX, and all four choices are unaffected.
Corrections to the Workbook
Last updated 9 September 2026
This page lists every confirmed error found in the first edition (2026) of AP Physics 1: Algebra-Based Practice Workbook. It covers both printings — Full-Color and Black & White — which carry identical page numbering, so every page number below applies to either one. Each entry gives the printed wording, the corrected wording, and one sentence on what goes wrong if you work from the printed version.
Every correction below is already fixed in the book's source files and goes into the next printing, so every copy in circulation today still carries the printed wording shown in each entry. Reports that turn out to be real errors are published here; reports that turn out to be correct as printed are not.
At a glance
Read this before you start pencilling anything in.
There are eleven corrections, in chapters 2, 3, 6, 7 and 8. No multiple-choice answer key changes — every correct letter printed in the book is still the correct letter — and no boxed final answer in any free-response solution changes. Five of the eleven correct an explanation rather than a result: the letter you were told to pick, or the answer you were told to get, was right, but the reason printed beside it was not (pages 94, 236, 268, 273 and 285).
Numbers move in exactly two places. In Problem 6.6 the whole measured data set is replaced, so the values you read off the graph change with it — though the quantity the question finally asks for, μk = 0.28, does not. In the closing sanity check of Solution 6.3 one arithmetic illustration changes; the answer it was checking, I ≈ 0.20 kg·m2, does not.
Two entries are marked critical. Those are the ones that will change a number you calculate, or a claim you write in a free-response answer, if you follow the printed text: the normal-force rule on page 94 and the angular-momentum argument on page 268. If you only have time for two, do those.
Separately, and independently of any error in the book, the College Board has changed the length of the exam from May 2027 onwards. That affects every copy and is described in Exam format from May 2027 below.
Does this affect my copy?
The book carries no printing number, so the cover and the copyright page will not tell you which version you have. One page will. Turn to page 94, the Exam Readiness page at the end of Chapter 2, and find the list headed Run Your Final Checks. Read the first item, the one that begins “Before computing friction”, and look at the last formula in it.
If it reads FN = mg + F sin θ
… for a force applied above the horizontal, and the item ends there, you have a copy printed before the corrections. Everything on this page applies to it.
Affected copy
If it reads FN = mg − F sin θ
… followed by a bracket noting that a push directed below the horizontal gives mg + F sin θ instead, you have a corrected copy and nothing in the list below applies to it.
Corrected copy
Corrections
In order of printed page number. Page numbers are the ones printed at the foot of the page, and they are the same in both editions. Entries marked critical change a result; the rest correct a statement, an explanation, or a diagram.
- What the printed page says
- “…with an angled push F above horizontal, FN = mg + F sin θ. An error in FN propagates straight into f.”
- What it should say
- “…with a force F applied at angle θ above the horizontal, FN = mg − F sin θ (a push directed θ below the horizontal instead gives FN = mg + F sin θ). An error in FN propagates straight into f.”
- Why it matters
- A force tilted above the horizontal has an upward component, which unloads the surface and lowers the normal force: with +y up, ΣFy = FN + F sin θ − mg = 0, so FN = mg − F sin θ. The printed sign makes the normal force too large, and since friction is computed from it, every friction force that follows is too large as well — for a 5 kg block pulled at 30° by 20 N with μk = 0.40, that is 24 N instead of 16 N. The checklist also contradicts the book's own Problem 2.2(b), which derives FN = Mg − F sin θ correctly. No worked solution in the book uses the printed form; the error is confined to this one checklist item.
- What the printed page says
- “…The data are shown in the table. Use g = 10 m/s2.”
- What it should say
- “…The data are shown in the table. Do not assume a value for g; the purpose of Parts (c) and (d) is to determine g from the data.”
- Why it matters
- Parts (c) and (d) ask you to determine g experimentally from those measurements, and the measurements give about 9.8 m/s2 — which is what the printed solution reports. The instruction as printed hands you the answer before you measure it, and then the measurement disagrees with what you were told to assume. No part of the problem needs a numerical g as an input, so the sentence simply comes out. The answer, g ≈ 9.8 m/s2, is unchanged.
- What the printed page says
- “(A) Omits one factor of ½, either from the solid cylinder's rotational inertia or from Krot = ½Iω2.”
- What it should say
- “(A) Applies the factor ½ one time too many — for example writing I = ¼mR2, or halving the result of Krot = ½Iω2 a second time. That halves the correct 9.0 J.”
- Why it matters
- Dropping one factor of ½ — from either place — gives 18 J, which is choice (C), not choice (A). Choice (A), 4.5 J, is exactly half the correct 9.0 J, so it comes from applying the ½ once too often. As printed, the note trains you to recognise the wrong trap, and it duplicates the diagnosis already given three lines below for (C). The key (B), 9.0 J, is unaffected.
- What the printed page says
-
Page 251, the data table reads t = 0.0, 0.5, 1.0, 1.5, 2.0 s with vcm = 8.0, 6.5, 5.2, 3.7, 2.4 m/s and Rω = 0.0, 1.4, 2.9, 4.2, 5.5 m/s. The blank grid on page 252 runs from 0 to 2.5 s.
Page 270, under the worked graph: “The two best-fit lines intersect at approximately t ≈ 1.4 s…” In the legend of that graph, the sample line drawn beside Rω is the solid one, which is the style used for the vcm series.
Page 271, Part (d) reads the slope from the points (0.25 s, 7.3 m/s) and (1.75 s, 3.1 m/s).
- What it should say
-
Page 251, the data table:
t (s) vcm (m/s) Rω (m/s) 0.00 8.0 0.0 0.20 7.4 1.5 0.40 6.9 2.8 0.60 6.4 4.1 0.80 5.7 5.6 The blank grid on page 252 should run from 0 to 1.0 s, tick-marked every 0.2 s. On the printed grid, the corrected data and their crossing point all fall in the left-hand third, so either plot them there or relabel the horizontal axis before you start.
Page 270: the two best-fit lines intersect at about t = 0.82 s and v = 5.7 m/s, which is where the ball starts rolling without slipping — and 5.7 m/s ≈ (5/7)v0, the standard result for a solid sphere launched with no spin. In the legend, each series should be identified by its line style and its marker: vcm is the solid line with round markers, Rω the dashed line with square markers.
Page 271: read the slope from (0.10 s, 7.7 m/s) and (0.60 s, 6.3 m/s), which gives (6.3 − 7.7)/(0.60 − 0.10) = −2.8 m/s2 — the same slope, and therefore the same μk = 2.8/10 = 0.28, as before.
- Why it matters
-
The problem states that the ball is a solid sphere, I = (2/5)MR2. In the printed table the rim speed Rω climbs at 2.76 m/s² while the centre slows at 2.80 m/s², and those two rates together imply I = MR2 — a hoop, not a sphere. The last two rows are worse than inconsistent: they put Rω above vcm, which cannot happen while the ball is still slipping forward. Plotted, the printed data cross at 1.4 s and half the launch speed, instead of the (5/7)v0 the same solution invokes a page later.
In the replacement set the rim speed climbs at 6.9 m/s², which implies I ≈ 0.41 MR2 — a solid sphere, as stated — and every row keeps vcm > Rω. The legend is a separate defect of the same figure: the graph draws four lines (data and best fit for each series) but carries only two legend entries, so the label Rω attached itself to the best-fit line through the vcm points. A reader matching series by line style would read the two curves the wrong way round.
The answer the question asks for, μk = 0.28, is the same for both data sets, because the slope of vcm versus t was never the faulty part.
- What the printed page says
- “The measured I ≈ 0.20 kg·m2 is comparable to a disk of mass ~2 kg and radius ~0.45 m — a sensible laboratory turntable.”
- What it should say
- “The turntable's rim radius is given as R = 0.15 m, so a uniform disk with I ≈ 0.20 kg·m2 would need M = 2I/R2 ≈ 18 kg — a heavy massive-platter turntable. That is exactly the kind of platter whose large rotational inertia makes the approximation I ≫ mR2 safe for the small hanging masses used here.”
- Why it matters
- The problem gives the rim radius as R = 0.15 m — it is the radius the string is wrapped around and the one every earlier line of the solution uses. A reader who runs the sanity check as printed is checking a turntable three times the size of the one in the problem, and the check appears to pass for the wrong reason. With the stated radius, M = 2(0.20)/(0.15)2 ≈ 18 kg, and that heavy platter is precisely why the later approximation I ≫ mR2 holds. The result being checked, I ≈ 0.20 kg·m2, is unaffected.
- What the printed page says
-
“…Similarly, the fixed axle exerts external forces on the flywheel. Because of these external torques from the support structure and axle, angular momentum is not conserved for System 2 either.”
And in the Common Mistake box: “…the support structure and axle that keep the brake and flywheel in place are external — and they exert nonzero torques.”
- What it should say
-
The axle force acts through the rotation axis, so its lever arm about that axis is zero and, by τ = rF sin θ, it contributes no torque about that axis no matter how large the force is. The decisive external torque is the one exerted by the brake's mounting, which does have a nonzero lever arm; because that torque is nonzero, angular momentum is not conserved for System 2.
The Common Mistake box should list blaming the axle as a second common error, not as the explanation: naming the axle as a torque source earns no credit, because conservation is decided by torques with nonzero lever arms about the chosen axis.
- Why it matters
- Torque about an axis is τ = rF sin θ, with r measured from that same axis. An ideal axle force is applied at the axis, so r = 0 and its torque about that axis vanishes — the force is real and large, and its torque is still zero. On the exam, writing “the axle exerts an external torque” is a standard non-credited answer, so a reader who learns the printed argument arrives at the right conclusion by reasoning that would score nothing. Both boxed answers — System 1 and System 2 both not conserved — are correct as printed and do not change.
- What the printed page says
- “When mass moves closer to the axis, update I with mr2. Halving r quarters that contribution, so the compensating change in ω is a factor of four.”
- What it should say
- “When mass moves closer to the axis, update I with mr2. Halving r quarters that contribution — but ω rises by the ratio Ii/If of the total rotational inertia, so the factor is four only when the moving mass carries all of I.”
- Why it matters
- L = Iω gives ωf/ωi = Ii/If for the total rotational inertia, and the mass that moves is usually only part of it. The book's own Problem 6.1 halves exactly this radius and gets a factor of 1.6, not 4, because the platform carries most of I. Four appears only in the limiting case where the moving mass is all of I — which is why MCQ 6.9 has to state “neglect the rotational inertia of the student and stool” before the answer comes out as 4. As printed, the bullet turns a limiting case into a general rule, and a reader who applies it to a problem with a massive platform overestimates the new angular speed — in the book's own Problem 6.1, by a factor of 2.5.
- What the printed page says
- “(C) Finds the correct kinetic-energy fraction but then forgets the factor of ½ in K = ½mv2, making the speed too large by a factor of √2.”
- What it should say
- “(C) Finds the correct kinetic-energy fraction but drops the factor of ½ on the spring side, using E = kA2 instead of E = ½kA2. That inflates the speed by a factor of √2.”
- Why it matters
- The error named in the printed text makes the speed √2 times too small, not too large: dropping the ½ from K = ½mv2 gives v = A√(3k/8m), which is not among the four choices at all. Choice (C), A√(3k/2m), is √2 times too large, and the only way to land on it is to write the spring energy as E = kA2. The distinction is the whole point of the question, so getting it backwards teaches the wrong habit twice over. The key (A), A√(3k/4m), is unaffected.
- What the printed page says
-
Page 330, the paragraph before the figure ends: “…so the string is taut and the system is in static equilibrium. The string has negligible mass.”
Page 332, the paragraph before the figure ends: “The pressure sensor can be lowered into the liquid so that the tip of the sensor is at a depth h below the surface, as shown.”
Neither problem states a value for g.
- What it should say
- Each of those two sentences should be followed by: “Use g ≈ 10 m/s2.”
- Why it matters
- Both worked solutions are computed with g = 10 m/s2, but neither problem says so. A reader who takes 9.8 m/s2 from the equation sheet — the reasonable default when a problem is silent — gets T = 6.9 N instead of 7.0 N in Problem 8.1, and ρ ≈ 1.19 × 103 kg/m3 instead of 1.17 × 103 in Problem 8.3, with no way to tell whether the book or their own work is at fault. Both problems were designed around g = 10 m/s2, so the fix is the missing sentence, not new arithmetic: the printed answers T = 7.0 N and ρ ≈ 1.17 × 103 kg/m3 are correct and stay as they are.
- What the printed page says
-
In the diagram, the upward buoyant-force arrow is drawn shorter than the tension and weight arrows added together.
Below it: “The upward buoyant-force arrow must be longer than either downward arrow individually because static equilibrium requires FB = Mg + T.”
- What it should say
-
The buoyant arrow should be drawn as long as the two downward arrows combined. With the printed arrow lengths, tension 1.5 and weight 1.8 sum to 3.3, so the buoyant arrow should be 3.3 rather than the 2.4 shown.
And the sentence: “The upward buoyant-force arrow must be drawn as long as the two downward arrows combined, because static equilibrium requires FB = Mg + T.”
- Why it matters
- The question explicitly asks for arrow lengths that reflect the relative magnitudes of the forces, and the model diagram fails the very equilibrium condition printed three lines below it — being longer than each downward arrow is not enough when the two of them add. On the exam this is the diagram's second scoring point, so the model answer as printed would lose it. The rest of the diagram — buoyancy up, weight and tension both down — is correct, as is every number in the parts that follow.
Exam format from May 2027
This one is not an error in the book — it is a change the College Board made after the book went to press, and it applies to every copy, corrected or not.
What changed
From the May 2027 administration onwards, Section I is 42 multiple-choice questions in 85 minutes and Section II is 4 free-response questions in 95 minutes. The book was printed with the previous figures: 40 questions in 80 minutes, and 100 minutes for the free-response section. The exam is still three hours long overall, and the two sections are still worth 50% each.
The old figures are printed in four places, and read the old way in every copy printed so far: the exam-format table on page xi, the free-response summary table on page xiii, the pacing plan on pages 363–364, and the one-line Legend at the head of each chapter's free-response section (pages 16, 56, 110, 159, 200, 243, 289 and 329).
Two consequences worth pencilling in. The per-question time targets for the four free-response questions become 22 minutes (MR), 27 (TBR), 23 (LAB) and 18 (QQT) — 90 minutes, leaving about five in reserve, and replacing the printed ranges whose upper ends added up to more than the time available. The book's multiple-choice heuristic survives untouched: 85 minutes over 42 questions is 2.0 minutes each, the same “about two minutes a question” the appendix already teaches.
What did not change
Everything the book is actually about. Checked item by item against the current course and exam description:
- the eight units and their exam weightings;
- the Science Practices and the share of the exam each skill carries;
- the four free-response types, their fixed order — MR, TBR, LAB, QQT — and their 10 / 12 / 10 / 8 point values, adding to 40;
- the Table of Information, reproduced in the book's appendix;
- the calculator policy: a four-function, scientific or graphing calculator on both sections.
In other words, no problem, solution, diagram or strategy in the book goes out of date because of this change — only the clock does.
Source: AP Physics 1 Course and Exam Description — Clarifications and Corrections, under the heading “To be implemented for Fall 2026”. The 2027 exam is on Wednesday, 5 May 2027, session 2; the section figures and the date were both re-checked against the College Board's own course page on 9 September 2026.
Found something else?
Write to ap.physics.feedback@gmail.com
The address printed in the book's preface. A page number and a sentence about what looks wrong is enough — a photograph of the page is better still.
Every report is worked through from first principles before anything happens to it, including a re-derivation of whatever the page claims. Real errors are corrected in the source, published here, and acknowledged. Suspected errors that turn out to be unclear explanations rather than mistakes count too — those are rewritten for the next printing, even though they never appear on this page.