General Chemistry
1
FOR INSTRUCTORS ONLY
MCAT
®
In-Class Compendium
PHYSICS
TECHNIQUES & SOLUTIONS
FOR INSTRUCTORS ONLY
2016
2016
MCAT Physics Techniques and
MCAT In-Class Compendium Physics Solutions
FOR INSTRUCTORS ONLY
Physics Solutions written by:
Jonathan Fowler, M.A.
Steven Leduc, M.S.
Chris Pentzell, M.S.
Carolyn Shiau, M.D.
Edited for Production by:
Judene Wright, M.S., M.A.Ed.
National Content Director, MCAT Program
The Princeton Review
Copyright © 2016, 2015, 2009, 2008, 2005, 2004, 2003, 2002, 2001, 2000, 1999 by Princeton Review, Inc.
All rights reserved.
MCAT is a registered trademark of the Association of American Medical Colleges (AAMC).
The Princeton Review is not affiliated with Princeton University or AAMC.
www.PrincetonReview.com
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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MCAT Physics Information and Passage Techniques
The MCAT has changed! Effective Spring 2015, the MCAT included additional Biochemistry content as
well as an entirely new science section testing psychology and sociology concepts. For all the information
you would ever need about the MCAT, go to:
www.AAMC.org
OR
www.PrincetonReview.com/MCAT
There are four sections on the MCAT: Chemical and Physical Foundations of Biological Systems
(Chem/Phys), Critical Analysis and Reasoning Skills (CARS), Biological and Biochemical Foundations of
Living Systems (Bio/Biochem), and Psychological, Social, and Biological Foundations of Behavior
(Psych/Soc). All sections consist of multiple-choice questions.
Section
Chemical and Physical
Foundations of Biological
Systems
Critical Analysis and
Reasoning Skills
Biological and
Biochemical Foundations
of Living Systems
Psychological, Social, and
Biological Foundations of
Behavior
Concepts Tested
Basic concepts in chemical and physical
sciences, scientific inquiry, reasoning,
research and statistics skills
Critical analysis of information drawn
from a wide range of social science and
humanities disciplines
Basic concepts in biology and
biochemistry, scientific inquiry,
reasoning, research and statistics
Basic concepts in psychology, sociology,
and biology, scientific inquiry, reasoning,
research methods and statistics.
# of Questions and Timing
59 multiple-choice (freestanding and passage-based)
questions in 95 minutes
53 multiple-choice (passagebased) questions in 90
minutes
59 multiple-choice (freestanding and passage-based)
questions in 95 minutes
59 multiple-choice (freestanding and passage-based)
questions in 95 minutes
Note that the best way to become familiar with the new test yourself is to take a few practice tests. You
can access these through your online content.
MCAT Tools
There are a number of tools available on the test, including highlighting, strike-outs, the Mark button, the
Review button, a periodic table, and scratch paper. The following is a brief description of each tool.
Highlighting: This is done in passage text (including table entries and some equations, but excluding
figures and molecular structures) AND in the question stem by left-clicking and dragging the cursor across
the words you wish to highlight; the selected words will then be highlighted in blue. When you release the
mouse button, a highlighting icon will appear; clicking on the icon will highlight the selected text in yellow.
To remove the highlighting left click on the highlighted text.
Strike-outs: This is done on the various answer choices by right-clicking on the answer choice that you
wish to eliminate. As a result, the entire set of text associated with that answer choice is crossed out. The
strike-out can be removed by right-clicking again. Left-clicking selects an answer choice; note that an
answer choice that is selected cannot be struck out. When you strike out a figure or molecular structure,
instead of being crossed out, the image turns grey.
Mark button: This is available for each question and allows you to flag the question as one you would
like to review later if time permits. When clicked, the “Mark” button turns red and says “Marked.”
Review button: This button is found near the bottom of the screen, and when clicked, brings up a new
screen showing all questions and their status (either “completed,” “incomplete,” or “marked”). You can
then choose one of three options: “review all,” “review incomplete,” or “review marked.” You can only
review questions in that section of the MCAT, but this button can be clicked at any time during the allotted
time for that section; you do NOT have to wait until the end of the section to click it.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Periodic Table button: Clicking this button will open a periodic table. Note that the periodic table is large,
covering most of the screen. However, this window can be resized to see the questions and a portion of
the periodic table at the same time. The table text will not decrease, but scroll bars will appear on the
window so you can center the section of the table of interest in the window.
Scratch Paper: You will be given four blank pages (8 faces) of scratch paper at the start of the test.
While you may ask for more at any point during the test, your first set of paper will be collected before you
receive fresh paper. Scratch paper is only useful if it is kept organized; do not give in to the tendency to
write on the first available open space! Good organization will be very helpful when/if you wish to review a
question. Indicate the passage number in a box near the top of your scratch work, and indicate which
question you are working on in a circle to the left of the notes for that question. Draw a line under your
scratch work when you change passages to keep the work separate. Do not erase or scribble over any
previous work. If you do not think it is correct, draw one line through the work and start again. You may
have already done some useful work without realizing it.
The following is an outline of suggested techniques to modify the in-classroom experience to match the
expectations of a computer-based test. Be sure to remind your students to bring a highlighter to class!
Mapping the Passage
1) Read Chapter 2 of the Physics and Math Review text closely: it is both your and your students’
best guide to passage strategies.
2) Time is of the essence! Students should not spend more than 2 minutes mapping out the
passage. The MCAT gives points for answering questions correctly, not for reading thoroughly.
These passages should not be read like textbook material, with the intent of learning something
from every sentence (science majors especially will be tempted to read this way). Passages
should be skimmed to get a feel for the type of questions that will follow, and to prepare to answer
the anticipated questions as efficiently as possible.
3) Students should read the first sentence of the passage and look for figures. The topic of the
passage is often found within the first few words, and is certainly reflected in any pictures (which
can be worth a thousand words, though that conversion is suspect). This will allow the
identification of the passage as physics or something else. Many students find it more efficient to
tackle all of the passages from their stronger subjects, leaving the remaining time to work on all
passages from the remaining subjects. The type of passage is also quickly identifiable: an
experiment-based passage will have tables of data, an information presentation passage with a
technical bent will have several equations set apart from the text, a conceptual persuasive
reasoning passage will have a lot of text, few numbers and equations, and a lot of unfamiliar
terminology. The type of information management and problem-solving techniques required are
often the best ways for students to choose whether to do a passage now or save it for later.
Remember again, the MCAT gives points for answering questions correctly, regardless of their
difficulty level or subject association.
4) The Princeton Review teaching policy still enforces that no overhead projectors and no computer
PowerPoint presentations should be used during lecture time. You need to demonstrate how you
would highlight the passage on the computer and what notes you would put on your scratch
paper. Thus, when you prepare your passage on the board, you should draw a general
schematic or map of the passage outline. It does not have to be very fancy—boxes to represent
the paragraphs and the necessary formulae, numbers, and table headings. Then, on this map,
you should use a different color to write the few words that you would highlight: definitions of
italicized phrases, for example (see “Highlighting” below). Use the colored pens sparingly as
students need to focus on answering the questions instead of covering their passages in yellow
highlighter. Tell your students to use scratch paper (see “Scratch Paper” below) when they do the
passages: they cannot write directly on the computer screen, so they shouldn’t practice writing
directly on the passage.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Anything that you need to use in order to solve a question in the passage should appear on your
map (or should be basic outside knowledge students should know). This will help you to prove to
your students that a well-mapped passage can be used to solve all of the questions without
having to read the entire passage word-for-word. It will also allow you to be more independent of
your book while discussing the passage. Remember, you are teaching the students how to map
the passages themselves, so you must both explain how you chose to map the passage as you
did (and not by knowing which problems were coming!) and demonstrate the utility of your map.
5) Highlighting –this tool should be used sparingly otherwise you will end up with a passage that is
completely covered in yellow highlighter! Highlighting in a physics passage should be used to
draw attention to 1-2 words that demonstrate one of the following:
•
•
•
The overall topic of that paragraph
A general physics principle that has been taught (i.e., familiar terms)
Numbers floating in the middle of text (piles of numbers are easy to find without
highlighting)
•
•
Extreme words like All, None, or Except
An unusual term that is defined specifically for that passage (e.g., something that is
italicized), or a term or phrase that keeps repeating
6) Scratch Paper – the fact that this is being allowed in the test is KEY in making it through the
physics section. However, scratch paper is only useful if it is kept organized! See below for more
details on how to maximize the functionality of scratch paper. Some figures should be quickly
redrawn (for force diagrams and the like) and equations should be copied with space to be
manipulated.
Using Scratch Paper
1) For each passage, there should be a heading that says “Passage X” in a box – clearly indicating
that the following scratch work is associated with a particular part of the test. The tendency is to
write on the first available corner of the scratch paper, but this makes it very difficult to re-trace
your previous work.
2) For each question within the passage, there should be a question number circled beside the work
that is written for that question. This will help you to make sure that you address all of the
questions in the passage, especially if you decide to tackle them out of order. It also keeps your
work organized for reference later.
3) When mapping the physics passage, copy ALL FORMULAE onto the scratch paper in the order
that they appear in the passage, under the boxed heading for “Passage X”, leaving space below
each to manipulate them algebraically. Putting the formula on paper allows you to see it and
work with it without having to look back at the computer screen while answering questions. Also,
with the formulae on paper, you can annotate them more effectively, labeling unknown variables
with info from the passage. This may seem like a waste of valuable time, but the technique will
allow you to familiarize yourself with the formulae and their proportionalities. It will also prevent
copying errors that might occur if you use the formula on-screen and have to look up and down
from computer screen to scratch paper to answer the question.
4) Many figures are worth copying and elaborating as well. Sometimes a passage will describe a
simple phenomenon and the forces involved without providing a picture: drawing what is
described is a useful way to prepare for questions.
5) Do not erase or scribble over any previous work. If you do not think it is correct, draw one line
through the work and start again. The reason for this is that you may have already done some
useful work without realizing it. Taking an eraser to the page only serves to waste precious time
during the test.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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General Approach to Physics Questions
1) Do the questions generally in order from easiest to most difficult, but don’t overthink the
decisions: in practice this typically means skipping one or two obviously different and difficult
questions and doing them last (what constitutes “difficult” is somewhat idiosyncratic, but for
example, Q6 on ICC 14 would certainly be one to do last if it weren’t already last). Students
receive points for questions answered correctly, regardless of the difficulty level. Doing the
questions in order of difficulty will help save time for the questions that may require those few
extra seconds.
2) Process of Elimination is still paramount! The strike-out tool on the CBT format allows you to
indicate which choices you have eliminated. Process of elimination improves your chances of
guessing the correct answer if you are not able to narrow it down to one choice.
If you are guessing between a couple of choices, use the CBT tools to your advantage:
a. Strike out any choices that you are sure are incorrect, improbable, or do not answer the issue
addressed in the question.
b. Jot down some notes on your scratch paper to help clarify your thoughts if you return to the
question.
c.
Click the “Mark” button so that the test will flag the question for review at a later time.
d. Do not leave it blank! If you are not sure and you have already spent more than 60 seconds
on that question, just pick one of the remaining choices. If you have time to review it at the
end, you can always debate the remaining choices based on your previous notes.
3) Predict the required formula. Often, the questions on the test are independent of the
accompanying passage. After reading the text of the question, try to predict the physics formula
that will be required to answer the question before looking at the answer choices. Write this
formula on your scratch paper beside the circled label for the question’s scratch work.
4) Units, units, units! If you cannot remember the formula, think about the units that are most
appropriate for the answer. A unit-analysis can help you regenerate or confirm the correct
formula. Also, it may help you to eliminate choices right away. Ensure that all calculations are
done with the “m.k.s.” unit system (meters, kilograms, seconds) unless otherwise requested by a
specific formula.
5) Do not substitute numbers when letters will do! Do not be tempted to substitute values into the
equations until you are at the very end of the algebra. It is easier to correct your mistake if you
leave the formula will all of its variables intact and solve the equation for the desired variable.
Once the algebra is done, it is a matter of arithmetic to complete the calculation.
6) Use approximations. It is not necessary to use the precise numbers in calculations. Simplify the
arithmetic by using estimates of the actual number. To decide between answer choices at the
end of the calculation, take into consideration whether your original estimate was greater or less
than the actual value. To evaluate square-roots and logarithms, decide on an upper bound and a
lower bound for the actual values (e.g., the square root of 12 is somewhere between the square
root of 9 and the square root of 16. Therefore, it must be 3-point-something.)
7) The majority of the concepts discussed in the TPR MCAT Physics course take place in the ideal
world. If the question is asking for an approximation, the ideal-world formulae are valid. If the
question is asking to take into consideration the real-world forces, look for a new formula in the
passage that addresses these issues.
8) Two of the same answer cannot be correct. This is commonly used in physics problems. Make
sure that you cannot already eliminate choices by simplifying the fractions or evaluating the
answers.
9) True vs. Explains. For long-text answers in physics questions, always review the last few words
in the prompt text of the question. Often, the question asks for a specific principle and gives four
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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statements – two of which are completely false, and two that are true. However, only one of
those remaining “true” choices actually answers the question. The other choice is meant to be a
distracter. Make sure that the final choice you make addresses the actual question.
REMEMBER: you must model these techniques using the In-Class Compendium passages during your
lectures. Students will want to see what you highlighted, what you struck out, and what notes you put on
your scratch paper.
A Final Note: You should use this document in three ways:
• First, you should read through the above guide to passage strategy and let it inform your teaching
of all the passages (in particular your introduction to Passage 1).
• Second, you should read the summary paragraphs at the beginning of each passage as you plan
to teach them: these paragraphs offer the best ways to map the passages and to explain the
problems so that students take generally applicable lessons from the passages and not merely
particular solutions.
• Third, you should use the solutions to each problem in each passage as a reference when
preparing to teach and to help answer student questions. You do not need to approach each
problem solution exactly the way we have, but you should take advantage of the insights of the
passages’ authors and editors. Remember: careful preparation is the key to excellent teaching!
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 1
Primary Topic: Mechanics
Passage Type: Experiment presentation
Subtopics:
Friction, pulleys
Techniques:
This is the classic MCAT Physics passage showing an experiment to demonstrate the
physics concepts of kinetic and static friction. Because this is the first passage, take extra
time to discuss mapping the passage (i.e., what to highlight, how long it should take,
etc.). It is also another opportunity to practice drawing free-body diagrams on the pulley
system. Q1 and Q4 are straight calculation questions, though you might note for your
students with math anxiety that the answer to Q4 need not be calculated so long as they
recognize that the answer must be less than 15 N. Q2 and Q5 refer conceptually to the
friction formula: which changes affect the variables in the formula and which do not (don’t
forget the I-II-III technique for Q5). Q3 requires a free body diagram and reminds
students that “none of the above” is a valid answer.
Before going over the solutions to Passage 1, quickly go over the following with the students.

The passage is a resource for reference, not something to spend a lot of time reading. Physics is not
like CARS; the student will not be asked for the theme of the passage or anything so general.
Students should begin by skimming to see what information is where and mapping the passage. They
should not try to master the material before looking at the questions. In fact, a few students may
prefer to skip the passage altogether and to begin directly with the questions. As this is a personal
choice, ask students to try both methods at home and to use the one that works better for them.

The conceptual and strategic challenges presented by each question are more important than the
informational content. Despite the fact that the passages are chosen according to the material
covered, your treatment of passages should never be limited to clarification of facts. It is absolutely
essential that you use passages as an opportunity to let students enter your mind and learn from your
style of attack. That is why you are here.

There are three basic challenges in questions: information management, careful reasoning, and
efficient use of time.
Information Management: There are three kinds of information to be managed (three sources for
information): explicit information (acquired by scanning for key words or graphics), information
inferred from the passage (requires scanning plus careful reading and thought), and information that
the student should know walking in the door (memory).
It is helpful for the student to identify in each question the type of information requested. This allows
the student to know how and where to search for the answer. If a question asks for explicit
information, a quick look back at the passage is all that is necessary, while memory questions do not
require the passage at all. (Keep in mind, though, that every student has a different knowledge base,
and what may be an explicit question for one student could be a memory question for another.)
When answering an implicit question, before looking back at the passage or at any answer choices,
students should spend a few seconds thinking about the answer. After formulating a basic answer in
their heads, they can look back at the passage for clarification. Only then, when the correct answer is
fixed in their minds, should they look at the answer choices. This simple exercise often makes the
correct choice stand out clearly. The MCAT has a way of offering similar-sounding answer choices
that can muddle students’ thinking if they have not crystallized a response before looking at the
answer choices. (This technique is actually helpful for all three types of questions but is most helpful
for implicit information questions.)
When going over the solutions to In-Class Compendium Passage 1, identify the types of information
asked for in the questions and encourage the students to do this at home. After working through a
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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few passages, they get the hang of this and don’t really need to stop to think about it. An additional
benefit is that they can then look back to see which type of question they most often get wrong. This
can help guide them in their studying. For example, if they frequently miss memory-style questions,
they should review basic material, if they miss more explicit-style questions, they need to work on
their passage mapping and reading comprehension skills, etc.
Careful Reasoning: Eliminating incorrect choices is the dominant problem-solving strategy on the test!
Even if the correct choice is discovered immediately, it is often a good idea to double-check the
answer by eliminating the others. Eliminating choices is more certain than choosing the correct choice
except in the simplest fact-oriented or calculation questions. One can usually be certain that a choice
is wrong based on a single solid piece of evidence. Conversely, the student is only sometimes able to
be absolutely certain of a correct choice, since the MCAT is designed to present false choices that
sound appetizing. Also, remain aware of the relevance of information. MCAT passages often distract
the student with details that sound important but are actually irrelevant. Show them how to assess
relevance.
Efficient Use of Time: Time is of the essence. One of the most valuable secrets for doing well on the
MCAT is this: do the easy questions and passages first, postpone the difficult ones. Make absolutely
sure that when time runs out and a few questions are left undone, they are difficult ones the student is
less likely to get right anyway. The single worst mistake is not looking at the last few questions in a
section just because of time. They may have been easily answered in 15 seconds each. The other
most important time-saving tip has already been covered: skim the passage. Do not read for detailed
understanding; skim selectively to map the passage with judicious highlighting and brief but clear
notes on the scratch paper.
1. B Before the block starts to slide in Experiment 1, the tension in the cord, T, must equal fs max. That
maximum force of static friction on M is fs max = µsN = µsMg = (0.4)(1 kg)(10 m/s2) = 4 N. However,
because the tension force on M is just T = mg = (0.2 kg)(10 m/s2) = 2 N to the right, static friction
exerts only a force of 2 N to the left (as per Newton’s First Law).
2. B The block in Trial 4 would have the same volume (and therefore the same mass) as the block in
Trial 3. Hence, the threshold mass for Trial 4 would be the same as that in Trial 3: 1210 g. (The
macroscopic area of contact would be doubled, but for MCAT purposes the frictional force is
independent of macroscopic contact area.)
3. D The force exerted by the table on the block has a component perpendicular to the tabletop (the
normal force) and a component parallel to the tabletop (the frictional force). Therefore, the total
force exerted by the table, which is the vector sum of these components, is neither perpendicular
nor parallel to the tabletop.
4. A Look at the hanging mass m. The net force on it is mg – T, and this gives it an acceleration a = g/6.
Therefore, mg – T = m(g/6), so T = (5/6)mg = (5/6)(1.5 kg)(10 m/s2) = 12.5 N.
5. A Neither Item II nor III would make any difference, so the answer must be choice A. (If a layer of oil
were placed on the bottom of the block, this would reduce the coefficient of static friction with the
tabletop and thus lower the threshold masses.)
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 2
Primary Topic: Projectile motion,
Passage Type: Situation presentation, technical (expect explicit questions)
Subtopics:
Non-ideal motion, conservation of energy
Techniques:
This passage requires an efficient map to focus on the information relevant to solving the
questions. Draw a quick diagram of the projectile under the influence of gravity, drag, and
lift at some point along the trajectory: this shows students how to translate a few
sentences into a picture that for most will be easier to interpret than those sentences at
the end of the first paragraph. Q1 is the classic MCAT trap that asks if the object is still
moving when it reaches the peak of the parabolic trajectory. Q2 is another classic
question about the concepts of perpendicular forces. Q3 requires the common exercise
of extracting a proportion from an equation given in the passage. Q4, Q5, and Q6 are
variations on the “two-by-two” question, where one fact eliminates 2 choices, and a
second fact eliminates a different set of 2 choices, leaving the correct answer. Emphasize
process of elimination as a powerful tool on the MCAT. When looking at graphs, make
sure students know to check the extremes of the graph (e.g., t = 0 and t = total time).
1. B The kinetic energy at any point is equal to ½mv2 = ½m(vx2+ vy2). Because vx always equals v0x and
vy = 0 at the top of the projectile’s trajectory, the kinetic energy there is simply ½mv0x2.
2. D The passage states that L is perpendicular to D and, therefore, perpendicular to v. Thus, the
acceleration due to L is also perpendicular to v, so it cannot affect the magnitude of v (only the
direction of v).
3. A According to Equation 2, aD is proportional to v2. Therefore, if v decreases by a factor of 2, then aD
decreases by a factor of 22 = 4.
4. D There are two relevant proportionalities to be derived from the Universal Gravitational Law (prior
knowledge) and the Prandtl Equation (given in the passage). The low-altitude shell will feel a
greater gravitational force because it is closer to the earth, and more drag because air density is
greater at lower altitudes.
5. B First, the passage states that drag decreases a projectile’s range; this eliminates choices A and D.
Now, because the drag force always acts to reduce the projectile’s speed (because it always acts
opposite to the velocity), the vertical velocity will drop to zero sooner. Therefore, the object’s
maximum height will be reduced from the ideal case (eliminating choice C).
6. B First, eliminate choices A and C; PE first increases as the projectile rises then decreases as it falls.
Now, because PE = mgy = mg(y0+ v0yt – ½gt2), the graph of PE vs. t must be a parabola (because
of the t2 term). Thus, the correct graph must be the one in choice B.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 3
Primary Topic: Momentum – perfectly inelastic collisions
Passage Type: Straightforward situation presentation (expect memory questions)
Subtopics:
Kinematics, work-energy theorem
Techniques:
There’s so little information in the passage that the questions have to be memory
questions. The biggest challenge for students is to remember that momentum is a vector.
Students will generally solve (Q1) the kinetic energy question correctly, and proceed to
Q2 forgetting that the car has negative velocity. Q2 is a good chance to remind students
that one error in logic will give a completely different answer, which is often listed as a
distracting answer choice. (The student who chooses D has forgotten that momentum is
a vector.) Q6 is a straight calculation question, but there are two ways to solve it. As the
instructor, you should be prepared to demonstrate a solution using “work done by friction”
(hence the coefficient of kinetic friction provided in the last line of the passage) and
“conservation of momentum” (remember that it’s a vector!). Q3 provides a good
conceptual example of Newton’s Third Law. Remember the graph technique when doing
Q4 (the truck is the dotted line, starting at t = 0).
1. C Let v be the speed of both the car and the truck just before the collision. (We know their speeds are
the same by the information given in the last paragraph of the passage.) If m is the mass of the car,
then K = ½mv2. The kinetic energy of the truck is ½(2m)v2, which is equal to 2K. Therefore, the total
kinetic energy of the vehicles just before the collision is K + 2K = 3K.
2. B If v is the velocity of the truck just before the collision, then the velocity of the car just before the
collision is –v. Therefore, the momentum of the truck is pT= Mv, and the momentum of the car is pC
= (½M)(–v) = –½Mv. So, the total momentum of the vehicles before the collision is pT+ pC= ½Mv,
which is equal to ½pT, half the momentum of the truck. Because total momentum is conserved, the
total momentum after the collision is the same, equal to half the momentum of the truck before the
collision.
3. D By Newton’s Third Law, the force that the truck exerts on the car is equal but opposite to the force
that the car exerts on the truck. Because the forces point in opposite directions, so do the resulting
accelerations. Furthermore, because the car has half the mass of the truck, it will experience twice
the acceleration. [Note: After the collision, the car and truck—as one combined mass—experience
the same acceleration as they slide, Fnet/(M + m), where Fnet is Ffric = µk(M + m)g.]
4. B Because the truck has a greater momentum than the car, the vehicles will slide to the right after the
collision. That is, the position x of the vehicles will increase after the collision. The answer must
therefore be choice B.
5. A The last paragraph tells us that the vehicles skid for 7 m before coming to rest. The work done by
friction is therefore –Ffricd = –µk(M + ½M)gd = –(0.35)(3000 kg)(10 m/s2)(7 m) = –73,500 J. By the
work-energy theorem, this must equal the change in kinetic energy of the vehicles between the time
they collided and the time they came to rest. Therefore, the total kinetic energy of the vehicles just
after the collision must have been 73,500 J. Alternatively, we recall from problem 3 that the total
momentum of the car and truck immediately after the collision is half the momentum of the truck
just before the collision, or ½(2000 kg)(21 m/s) = 21,000 kgm/s. This implies that the velocity of the
truck and car combined immediately after the collision is (M + ½M)v = (3000 kg)v = 21,000 kgm/s,
so v = 7 m/s. The kinetic energy of the truck and car combined must therefore be ½(M + ½M)v2 =
½(3000 kg)(49 m2/s2) ≈ 75,000 J.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 4
Primary Topic: Conservation of energy
Passage Type: Straightforward situation presentation (expect memory questions)
Subtopics:
Torque, impulse
Techniques:
When creating a map of the passage on the board, ask students which figures they might
redraw and which ignore—they will have an immediate sense for this. Then draw a
version of Figure 2 with all of the numerical labels (including those given in the text but
not on the figure, like the mass of the chaser); you should incorporate Figure 3 into this
drawing as well. Aside from a few numbers given in the text and the definition of marker
(or chaser), most of the text is not necessary for answering the questions. Q3 is a
completely free-standing question. Q1 and Q2 are straight calculation questions.
Students often find Q4 difficult to approach. There are two equally valid ways to solve the
problem: many might find it easiest simply to locate the center of mass at the fulcrum;
others might prefer to note that a system that in static equilibrium has zero net force and
zero net torque. If they start with this latter concept, they can solve the question using
torque. Q5 is another calculation question that reveals the inefficiency of energy transfer
(and is an entertaining concept for students who have previously attempted this game at
a carnival).
1. B Use Big Five #3 with v0 = 0 and a = g. If y is the distance the chaser falls in time t, then y = ½gt2.
Solving for t, we get t = √(2y/g). So, if y = 4 m, then t = √(0.8) ≈ √(0.81) = 0.9 s.
2. C When the head of the mallet drops to the rubber mat, the decrease in its gravitational potential
energy becomes kinetic energy: mgh = ½mv2. Solving for v, we get v = (2gh)½. Because the length
of the mallet is 1 m, we have h = 1 m, so v = 20½ ≈ 4.5 m/s (as another useful estimation technique,
note that 16 < v2 < 25 ⇒ 4 < v < 5, eliminating all but one possible answer).
3. A The area under the F vs. t graph is the impulse, by definition. We need to estimate this area. The
figure below shows that the area under the given curve is a little less than the area of the indicated
triangle. The area of this triangle is ½ × base × height = ½(20 × 10–3 s)(2000 N) = 20 N-s, choice A
is best.
Force (N)
2000
1000
4
12
20
Time (ms)
4. D The passage states that the “the fulcrum is positioned so that the plank is horizontal between
attempts to ring the bell.” Therefore, the torques exerted by the weights of the rubber mat and the
chaser must balance:
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
12
Thus, we have r1(m1g) = r2(m2g), so m = (r2/r1)m2 = (0.80/0.40)(1 kg) = 2 kg. Alternatively, we can
set the center of mass at the fulcrum and calculate m1: xCM = (–40m1 + 80m2)/(m1 + m2) = 0 ⇒ 40m1
= 80m2 ⇒ m1 = 2(1 kg) = 2 kg.
5. D If the chaser reaches the 3-meter level, then the increase in its gravitational potential energy is mgh
= (1 kg)(10 m/s2) × (3 m) = 30 J. So, if the input energy was 180 J and the output energy was 30 J,
that means 180 – 30 = 150 J of energy was lost in the mat-plank mechanism.
Physics Passage 5
Primary Topic: Mechanics
Passage Type: Information presentation, technical and conceptual (expect all Q types, but mostly implicit
and memory given how the information is and isn’t presented—no equations or figures,
but some new terminology)
Subtopics:
Kinematics, forces, torque, conservation of energy
Techniques:
This is an intimidating passage for students and teachers because it takes all of the
mechanics topics of the first four lectures and puts them together in a biomechanical
context. Note that the majority of the information is presented verbally in the first three
paragraphs without any equations or figures. A careful map of the passage that highlights
the familiar terms to ground the passage in physics instead of biology. Q1 is a memory
question relying almost entirely on recall of Newton’s First Law. Q2 is a memory question
using just some basic data from the passage. Q3 is an implicit question requiring a bit
more subtlety to determine the mathematical approach to take given the variety of
numbers in the passage. Q4-Q6 are implicit questions requiring a synthesis of the
explanatory text in the passage with a basic understanding of Newton’s Laws and
energy. Approach these questions using process of elimination and be very clear on why
every answer choice is right or wrong: this is particularly important for Q5, where the
choices all have a believable amount of physics jargon.
1. A Because the runner is accelerating to the right, Fnet must point the same way. There are two forces,
Fground and w, which add to give Fnet. Since the force of gravity, w, is straight down, the force
exerted by the ground must be up (the normal force) and to the right (the static frictional force).
2. A The acceleration is a = (8 – 0)/3 = 8/3 m/s2. The net force, therefore, is Fnet = ma = (60)(8/3) = 160
N.
3. B At each leap the body is pushed up as it moves forward, so that it has a given initial horizontal and
vertical velocity. This is like a cannonball problem in which the cannonball reaches its greatest
height when the vertical velocity becomes zero. Thus we can use Big Five #5 to write
2
v=
v 02y + 2gh . Setting vy = 0 and v0y = 1.5 m/s yields h = 0.11 m.
y
4. A Energy starts as chemical energy in the muscles. It gets converted into kinetic and potential energy
in the push and rise of the body off the ground. It gets converted to heat in the breaking of the fall.
5. C Nothing in the passage supports choices A, B, or D. The biped must bring forward a large limb to
catch themselves in each step, compared with the small limbs used by the quadruped. This takes
more force or more time in each step.
6. C Concerning choice A, although the forelimb may convert some muscle energy into kinetic energy in
a forward thrust, this is not what happens in breaking the fall. The passage does not support choice
B. Concerning choice C, the elbow indeed increases the distance over which the fall is broken
(because it bends during the impact process), and the mention of force and distance hints that we
could write Fd = W, where W is the work performed in breaking the fall, which is constant (all that
kinetic energy must become something). Increasing d will decrease F and make for a less jarring
run. As for choice D, the elbow does not decrease the time of the break, rather increases it.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 6
Primary Topic: Thermodynamics
Passage Type: Information / situation presentation, technical (expect explicit questions)
Subtopics:
thermodynamics processes, P-V diagrams
Techniques:
This passage requires an understanding of the First Law of Thermodynamics, the names
of the thermodynamic processes, and the proper interpretation of P-V diagrams. The
mapping is not a challenge, however, as there is a single equation to copy and two
obvious definitions to highlight. Q1 is the First Law question, and can be done either by
memory or explicitly. Emphasize that students understand the physics definitions of Q
and W so they’re not confused about sign. Q2 is a straight definition question from
memory, and Q4 implicitly relies on the definitions from the passage. Q5 explicitly
requires the equation from the passage; and Q3 explicitly requires estimating the area
enclosed by the set of process curves, so show your students how best to do with the
basic figures they recognize, triangles and rectangles.
1. A For any closed thermodynamic cycle, ΔE = Q – W, where ΔE is the change in internal energy of the
system, Q is the heat added to the system and W is the work done by the system. The passage
states that a thermodynamic process has a net zero change in internal energy over a cycle, so Q =
W. In other words, net heat input equals net work output. Since heat and work will have the same
magnitude, answer choices C and D are eliminated. Note that the total heat input will never be
100% converted to work. The theoretical maximum work output is equal to the net heat added to
the system. This depends on the difference between TH and TC.
2. D The passage states that the second stage of the Rankine cycle occurs at a constant pressure.
Therefore it is by definition an isobaric process.
3. B Since the loop goes in a clockwise direction, it is a heat engine, eliminating choices C and D. The
work is the area contained within the pressure-volume loop. It can be approximated by adding the
area of two triangles where area = (1/2)(base)(height). The top triangle has a base of 2vo, height of
3/2po, and area = (1/2)(2vo)(3/2po) = (3/2)(vopo). The bottom triangle has a height of 2/3po, base of
2vo, and area = (1/2)(2vo)(2/3po) = (2/3)(vopo). The magnitude of total work = (3/2)(vopo) + (2/3)(vopo)
= (13/6)(vopo).
4. B The passage states that the expanding steam does the work in a steam turbine, so water is the
working body, eliminating choice A. In a steam engine, the coal is burned to provide the heat for the
working body, eliminating C and D, and making choice B correct.
5. C The passage states that the cold reservoir is limited to ambient temperature, which is approximately
25 °C or 298 K. Given the hot temperature in the question of 550 °C = 823 K, theoretical efficiency
can be solved using Equation 1:
T
η= 1 − C
TH
η = 1−
298 K
300 5
≈ 1−
=
823 K
800 8
This fraction corresponds to a percentage closest to choice C.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 7
Primary Topic: Fluid statics
Passage Type: Situation presentation, technical (expect explicit question with all of those new
equations!)
Subtopics:
Buoyancy, viscosity, drag force
Techniques:
This algebra-heavy passage requires little outside information other than Newton’s
Second Law and the fact that that the buoyant force points up. Everything else is
included: show them how to map it by writing down the four equations with room to
manipulate them, and render the paragraph about terminal speed as a free-body diagram
of a sphere falling through a fluid (you will use this diagram for Q1). Q1 is the Newton’s
Law question. You can explain that at terminal speed the forces balance, so if you reduce
one of the upward forces (drag, which depends on speed), the downward force (gravity)
wins. It is also worth noting that Choice A is a true statement (since gravity is already the
largest of the three forces), but the question asks, “Which BEST describes the
relationship...?” Q2 and Q4 are classic “2 x 2” questions—you can eliminate two choices
first. In Q3, don’t derive anything, just use POE.
1. D When an object begins descending through a fluid, its speed is too low for the upward drag force
(which depends on v) to help the buoyant force balance the downward gravitational force. That is,
when v < vT, the sum of the drag force and the buoyant force is less than the gravitational force.
2. D Equation 3 shows that vT is proportional to r2, so a large ball would have a faster terminal speed
than a small ball. Equation 3 also shows that the greater the value of ρ, the greater the value of vT.
Because gold has a greater density than steel, the answer is D.
3. B Because ρ0 < ρf, the answer cannot be A, because this expression would give a negative value for
the speed vT. Because the air bubble feels both the buoyant force (which involves ρf) and the
gravitational force (which involves ρ0), neither C nor D can be correct.
4. A Equation 4 shows that η is proportional to √T, so the viscosity of the air increases with increasing
temperature. Therefore, competitors who jump when the air is cooler (when the air viscosity is
lower) would have an advantage.
Physics Passage 8
Primary Topic: Fluid dynamics
Passage Type: Situation presentation, technical (expect explicit questions drawing on the equations)
Subtopics:
Flow rate, viscosity, turbulence, the Bernoulli (Venturi) Effect
Techniques:
This is another algebra-heavy passage, so map it by writing down the equations with
room to work. It may also be useful to sketch the figure and to include r, so that the
velocity gradient across the artery cross section given by equation 2 can be visualized.
Most of the questions can be answered using these facts and a few numbers from the
passage, and knowing that high speed means low pressure and f = Av. Q1 and Q6 are
straightforward calculation problems. Q2 is a proportion problem using a given equation:
you should explain the difference between Q1 and Q2 as the first relies on prior
knowledge and the second on a new equation (attempting to rely on f = Av for Q2 yields
a wrong answer—new information always trumps prior knowledge). Q3 is another
example of functional dependence, i.e., knowing how changing certain variable in an
equation affects the overall value. In Q4 it should be noted that the values for the
Reynolds number, R, given in the passage contain no units, but you should also show
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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them the long way. Q5 involves graphing a given equation: remember to emphasize the
technique of testing the extremes or endpoints.
1. C Using the equation f = Av, we find that v = 1×10–4/[π × (1×10–2)2] = 1/π ≈ 0.3 m/s = 30 cm/s.
2. B According to Poiseuille’s Law (Equation 3), f is proportional to R4. So, if R is decreased by a factor
of 2, then f will decrease by a factor of 24 = 16. The flow rate equation (f = Av) would also work
here, but one would have to consider the area of a circle (A = πr2) and the non-ideal flow speed
given by Equation 2 (which includes a factor of r2).
3. A By Equation 1, the Reynolds number, R, would be increased—and thus possibly signal a laminar
flow becoming turbulent—if ρ, vavg, or R were increased; this eliminates choices B, C, and D.
However, because R is inversely proportional to η, an increase in the viscosity coefficient (choice
A) would cause R to decrease.
4. D Using Equation 1, [R] = [ρ][vavg][ R]/[η] = (kg/m3)(m/s)(m)/(kg/m·s) = 1 (note that 1 Pa = 1 N/m2 =
1 kg/m·s2), so R is dimensionless. It is worth noting both that the Reynolds number is called a
“number” (i.e., something without units) and that, unlike all other physical values in the passage, it
is presented without units: both of these facts point to its unitlessness.
5. D The question is asking for the graph of Equation 2. This equation says that r = 0 gives the
maximum v, and r = R gives v = 0. Only the graph in choice D illustrates these two points.
6. C Take Equation 3 and solve it for ∆P/L, the pressure drop per unit length: ∆P/L = f(8η)/πR4 =
8(1 × 10–4)(2 × 10–3)/π (10–8) = 160/π ≈ 50 Pa/m.
Physics Passage 9
Primary Topic: Fluids
Passage Type: Straightforward situation presentation (expect memory questions with so little info to draw
upon)
Subtopics:
Flow rate, Bernoulli’s Equation, gauge pressure, buoyancy, projectile motion
Techniques:
This short and (deceptively) simple passage combines various fluid concepts with a little
kinematics and Newton’s Second Law thrown in. It shouldn’t take more than 30 seconds
to map (a quick sketch of the figure is a good place to put all the given values, or they
could just jot down the numbers not already in the drawing), so emphasize that simplicity
to students: this gives more time for the questions! Q1 is a straightforward calculation
problem using f = Av. Remind them of the conversion between liters and cubic meters.
Q2 involves using Bernoulli’s Equation conceptually—pressure increases with depth and
decreases with speed. You can turn this into a “2 x 2” question by eliminating choices A
and B based on the first fact and then choice C using the second fact. Q3 is all POE. Q4
requires reading the question carefully—the problem gives heights whereas we need
depths (this is where the drawing can help). Q5 is free standing but provides a good
conceptual test of students’ understanding of buoyancy: it tends to be counterintuitive for
many so be sure to explain carefully how to reason through the problem by referring to
the equation (equations aren’t subject to bad intuition!).
1. D The flow rate is equal to the cross-sectional area of the pipe multiplied by the flow speed. Because
we know the flow speed at the ejection point, we multiply this by the cross-sectional area of the
horizontal arm: f = Av = πr2v = π (1/4 m)2(20 m/s) = π (1/16 m2)(20 m/s) ≈ 4 m3/s = 4000 L/s
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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2. D According to Bernoulli’s Equation, fluid pressure is lowest when the elevation is highest and the
flow speed is fastest. So, we’d certainly expect the water pressure to be lowest within the horizontal
arm before ejection. (We can eliminate choices A, B, and C because the pressure must be greater
at the base and within the vertical arm of the pump to push the water up into the horizontal arm.)
3. D Choice A is false; the buoyant force is equal to the weight of the water displaced by the stone,
whether it’s sitting on the bottom of the lake or not. Choice B is false, because if the buoyant force
were greater than the weight of the stone, the stone would rise. And choice C is false because the
hydrostatic pressure depends only on the density of the water and the depth (and the value of g
and Patm), but not on any properties of the stone itself.
4. C The gauge pressure is proportional to the depth, not to the distance from the bottom of the lake.
Because the passage gives the depth of the lake as 12 m, the depth of Fish #1 is 12 – 6 = 6 m.
Because Fish #2 is 3 m from the bottom, its depth is 12 – 3 = 9 m. So, the hydrostatic gauge
pressure on Fish #1 is 6/9 = 2/3 of the gauge pressure on Fish #2.
5. A The boat is floating both before and after the anchor is thrown, so the buoyant force must be equal
to the weight of the boat (and its contents) in either case. Because the weight of the boat and its
contents is less after the anchor has been tossed overboard, the buoyant force on the boat must be
less as well.
Physics Passage 10
Primary Topic: Electrostatics
Passage Type: Information presentation, conceptual (expect memory and some implicit questions with
no given equations but some conceptual definitions)
Subtopics:
Electric field, electric potential and potential energy, current, capacitors
Techniques:
This is a short passage presenting relatively little information; use your map to integrate
the numbers given in the passage directly onto the picture. You might also list the basic
equations called to mind by the phenomenon described: F = qE, ∆PE = q∆ϕ, Q = CV,
V = IR. This illustrates that in a passage with no new equations, basic knowledge is
paramount, and it’s not a bad idea to write down a few reminders before confronting
possibly confusing questions. Q1 is definitional. Q2 is a subtle problem, in that it requires
students to imagine the air, which we normally consider to be the same as vacuum in a
capacitor, as a dielectric present between the cloud and the ground. Take time to explain
the reasoning here. Q3 is best treated as a POE problem, as the three wrong answers
are nonphysical. Q4 is a straight calculation problem. Q5 tests the interpretation of ∆PE =
q∆ϕ: make sure students realize that negative charges tend toward higher potential.
1. D Electric field lines point from positive charges toward negative charges. Because the top of the
cloud is positively charged and the bottom is negatively charged, the electric field lines point from
the top of the cloud to the bottom.
2. A Before the lightning strikes, air is acting as an insulator, that is, as a dielectric. The presence of a
dielectric between two isolated conductors carrying equal but opposite charges cannot change the
magnitude of the separated charge, but it does increase the capacitance. Because Q = CV, an
increase in C for the same Q implies a decrease in V. Therefore, the presence of the air reduces
the potential difference (slightly) from what its value would be if the intervening space were vacuum.
3. B Electrons are free to move in a conductor. They will repel each other, causing them to come to rest
on the outer (metallic) surface of the car. The persons inside will remain safe if there is no electrical
contact with the surface. Choices A, C, and D are false statements.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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4. C The passage states that a typical lightning bolt transfers –30 C of charge to the Earth. If this is done
in 1 ms, then the magnitude of the average current is, by definition, I = Q/t = (30 C)/(10–3 s) =
3 × 104 A.
5. B Because the electron, which is negatively charged, is moving from a negative region to a positive
region, the electric field is doing positive work in pulling it down. Therefore, the electron experiences
a decrease in electrical potential energy (because, intuitively, it’s doing what it naturally wants to
do). This eliminates choices C and D. Because ΔPE = qΔϕ, the fact that both ΔPE and q are
negative implies that Δϕ is positive; that is, the change in electric potential is positive. The electron
is moving toward a region which is at a higher electric potential. (All negatively charged particles
are naturally accelerated toward points at higher electric potential.)
6. C Because the unit “V/m” is equivalent to the unit “N/C,” the equation F = qE yields
F = (2 × 10–9 C)(6 × 105 N/C) = 12 × 10–4 N = 1.2 × 10–3 N.
Physics Passage 11
Primary Topic: Electric Circuits
Passage Type: Experiment/data presentation (expect explicit and implicit questions requiring data
interpretation and use of the new terminology)
Subtopics:
Resistance and current
Techniques:
This passage isn’t exactly an experiment, but given the tabular information and the graph
carefully labeled with values, it’s legitimate to think of this passage as requiring the same
reasoning as an experiment presentation. Q1 is an implicit question that is simply
memory of Ohm’s law combined with the ability to read the table. Q2 is an implicit
question that requires getting a general sense for the trends shown in the table and
interpreting it in light of basic understanding of circuit terms as well and chemistry
(electrolytes). Q3 and Q5 are explicit, “read the graph” questions. Q4 is deceptively a
memory question, essentially freestanding, requiring reasoning from known principles of
electrodynamics.
1. D Since the voltage required to produce 1 mA of current through dry skin is 10 V—from the first line of
data in the table—the resistance of dry skin must be R = V/I = (10 V)/(10–3 A) = 10,000 Ω. You can
use any other table entry and get the same answer.
2. C According to the data in Table 1, wet skin provides ten times less resistance to current than dry skin
does. The wetness of the skin must increase the conductance, due to dissolved electrolytes. As
with a wire, the resistance decreases with area and increases with length or thickness. This means
that that conductivity does the opposite—it increases with area and decreases with thickness. This
eliminates choices A and B. The skin cannot, in most cases, be bypassed whether wet or dry,
which eliminates choice D.
3. C According to the graph in Figure 1, the threshold of sensation for current alternating at a frequency
of 10,000 Hz is approximately 5 mA.
4. C Current flows only when there is a conducting pathway connecting points at different electric
potentials. The feet of a bird which lands on a single wire are both at the same potential, and
consequently, no current is established.
5. B The greatest danger would come from exposure to “can’t-let-go” current. Of the given frequency
choices, only 100 Hz places 10 mA in the “can’t-let-go” region of Figure 1.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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Physics Passage 12
Primary Topic: Magnetic fields and forces
Passage Type: Situation presentation, mostly straightforward (expect memory and implicit questions
drawing on recalled equations and definitions to supplement the passage)
Subtopics:
Electric fields and forces, right-hand rules, dipoles, Newton’s Second Law, equilibrium
Techniques:
This is a good passage from which to teach visualization skills (though its phenomenon is
not typically as difficult for students to visualize as a few on the diagnostic exams): use
your passage map carefully to illustrate just what is directed where and how the students
should discover and record this on their own maps (draw the figure elaborated with the
charges distributed and E induced). Beyond working out the ramifications of the diagram,
there is little new information in the passage. Q1 and Q2 are implicit questions that
require an accurate diagram and use of the correct right-hand rule: be sure to
demonstrate slowly and more than once the correct hand orientation and motion. Q2
adds the step of finding the torque on a dipole, which you should demonstrate physically
with a pen or ruler (a good teaching strategy is to show your students how they can “do”
mechanics even at their desks with only pencil and paper). Q3 reminds us that F still
equals ma, even when we’re discussing ions and magnetic fields. Q4 uses the definition
of equilibrium, which students should know, and simple algebra. For Q5, simply ask the
class, “how much work do magnetic fields do?” Students need to know which are the 10second memory questions.
1. D First, apply the right-hand rule to determine the direction of the magnetic force on the positively
charged sodium ion:
FB
Because FB is downward, the answer must be D. Also, notice that as a result of FB on the Na+ ions
being downward, these ions will drift toward the bottom of the artery and the negatively charged
chloride ions will drift upward. Because electric field vectors always point away from positive source
charges, the induced electric field within the artery will point upward.
2. B Because the molecule is neutral, it will feel no net force due to the magnetic field; this eliminates
choice D. If the molecule’s dipole moment is as depicted in choice B, the magnetic forces on the
two oppositely charged ends will create a net counterclockwise torque:
FB on δ–
FB on δ+
However, in either case A or C, there will be no net torque:
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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FB on δ–
FB on δ+
FB on δ–
FB on δ+
3. B Because both ions carry the same magnitude of charge (namely e), both ions feel the same
magnitude of magnetic force. Because FNa = FCl, we have mNaaNa = mClaCl, so 23aNa ≈ 35aCl (use
the periodic table to find these atomic masses), which gives aCl ≈ (2/3)aNa.
4. C The last sentence of the passage states that the blood’s flow speed can be determined when
equilibrium between the magnetic and electric forces is achieved; that is, when FE = FB, or,
equivalently, when eE = evB. Solving this for v, we get v = E/B. But E = Veq/d, so v = Veq/Bd.
5. A Remember that one of the fundamental facts about magnetic forces is that they can do no work, for
example on a moving charged particle. (This is because the magnetic force is always perpendicular
to the particle’s velocity.)
Physics Passage 13
Primary Topic: Magnetic fields and forces
Passage Type: Persuasive/Scientific Reasoning, conceptual (expect all three Q types with a challenge
distinguishing when outside knowledge is a benefit or hindrance)
Subtopics:
Uniform circular motion, radioactive decay
Techniques:
This conceptual passage can seem difficult, but the questions can be answered by
knowing our rules and equations for magnetic fields and forces and by using the passage
and process of elimination. Q1 involves noticing from the figure that the magnetic field
would point vertically downward and then using the right-hand rule and the equation for
magnetic force. Q2 seems to involve passage information, but choice D must be the
answer, given the definition of half-life. Choice C may be the only choice that they can
definitively eliminate, so be sure to explain how common sense about physical
phenomena allows you to cut choices A and B. Q3 is a free standing memory question: it
involves knowing the right-hand (and left-hand, if one uses it) rule and that the magnetic
force provides the centripetal force and solving for r. Q4 may seem to require outside
information. However, it is explicit, and choices B, C, and D can all be eliminated based
on referring to the theories mentioned in the passage.
1. A Charged particles moving perpendicularly to a uniform magnetic field experience a force |FB|=|q|vB.
The magnetic field strength in Siberia is given in the passage as 60 µT. Therefore,
FB = (2 × 10–3 C)(100 m/s)(60 × 10–6 T) = 1.2 × 10–5 N. This eliminates choices C and D. To find the
direction, we need to use the right-hand rule. According to the Figure in the passage, the Earth’s
magnetic field points inward toward the earth in the northern hemisphere, in other words, straight
down. The thumb of the right hand points in the direction of the particle’s velocity (north), and the
fingers point in the direction of the magnetic field (down). The force would be directed outward from
the palm (west) if the charge were positive. However, the force a negative charge feels is in the
opposite direction. Therefore, the correct answer is choice A. (Note: for negatively charged
particles, a left-hand rule may be used).
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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2. D Knowledge of nuclear physics is not required for this problem—we can answer this from the
passage and from common sense. Choices A and B may or may not be true, so we’ll put them
aside for the moment. Choice C is false because 232U has a half-life and must therefore be
radioactive. Choice D must be true because the half-life of 232U is only 69 years. Thus little to no
traces of 232U would be left from the creation of the Earth.
3. C Charged particles moving in the presence of a uniform magnetic field travel in circular paths,
eliminating choice D. The centripetal force is provided by the magnetic force: |q|vB = mv2/r which
means that r = mv/|q|B. Given the same v,|q|, and B, then r is proportional to m. Ions are more
massive then electrons, therefore choice B is incorrect. Because the ion and the electron are
oppositely charged, the forces they feel will be in opposite direction, thus causing rotation in
opposite directions. The correct answer is choice C.
4. A There is nothing in the passage that incorporates pole reversals into either theory, and indeed the
last two paragraphs emphasize that the theories about the Earth’s magnetic field do not predict pole
reversals, which eliminates B and D. Choice C can be eliminated because the Free Electron Theory
mentions heavy elements, not light ones. The correct answer is choice A.
Physics Passage 14
Primary Topic: Simple harmonic motion
Passage Type: Experiment presentation (expect explicit and implicit questions requiring data
interpretation and use of the given equation)
Subtopics:
Spring-block oscillator, Hooke’s Law, elastic potential energy, energy conservation, static
friction, experimental design.
Techniques:
This is another experiment-based passage (the tabular data is a dead giveaway). As for
passage 1, the pictures tell almost the entire story, so drawing and annotating one of
them (probably figure 2, because the free-body diagram is more interesting with gravity
included) carefully is a good way to map this passage. Q1 is a straightforward calculation
problem relying on basic formulas. Q2 and Q4 are examples of functional dependence,
i.e., knowing how changing certain variables in an equation affects the overall value, but
Q2 is implicit (you need to provide the equations yourself) whereas Q4 is explicit. Q5 is
one of the more difficult implicit problems to solve among all these passages, but it is
worth pointing out that POE gets students to choices A and C quickly. And, for those
who don’t see clearly how to proceed with the solution, a 50% chance in 30 seconds or
so is better than spending several minutes trying to figure out the problem.
1. B When the block comes to rest, the upward force due to the stretched spring balances the weight of
the block, so ks = Mg. Solving for s gives s = Mg/k. Because the mass of the block is 0.5 kg (the
block used in Trial 3 of Experiment 1) and the spring constant is 100 N/m (the spring used in Trial 1
of Experiment 1), we find that s = (0.5 kg)(10 m/s2)/(100 N/m) = 5 cm.
2. B When the block is pulled a distance A from its equilibrium position and released from rest, the total
energy of the spring-block system is elastic potential energy, ½kA2. As it passes through
equilibrium, all this energy is converted to kinetic energy, ½Mvmax2. Therefore, ½Mvmax2 = ½kA2, so
vmax = A(k/M)½; thus, vmax is proportional to A.
3. B The amount of work done to initiate the oscillations is equal to the maximum potential energy of the
spring-block system, which is given by ½kA2. Because we’re only asked to compare the values and
determine the smallest, we can ignore the factor of ½ and simply compare the values of kA2. For
Trials 2 through 5, the values of kA2 are as follows: Trial 2: kA2 = (50 N/m)(2 cm)2,
Trial 3: kA2 = (40 N/m)(2 cm)2, Trial 4: kA2 = (25 N/m)(4 cm)2, and Trial 5: kA2 = (50 N/m)(2 cm)2.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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We can immediately eliminate choices A and D (Trials 2 and 5) since their values are the same.
The value of 40 × 22 is smaller than 25 × 42, so the correct answer is choice B.
4. D While the period of a simple pendulum that oscillates in a vertical plane depends on the local
gravitational acceleration, T = 2π(L/g)½, the period of a spring-block oscillator, even one oscillating
vertically, is given by T = 2π(M/k)½. Because T is independent of g, a change in g will not affect the
period of the oscillations.
5. A First, the small block is most likely to slide off the big block when the force on the big block is the
greatest. (Static friction would have to withstand the force of the spring pulling the big block in order
to keep the small block resting on top.) Because the force on the big block is greatest when the
block is at its maximum displacement from equilibrium (which follows from Hooke’s Law, F = –kx),
choices B and D can be eliminated. To decide between choices A and C, we can note that if C were
correct, then a very massive block M would require a very large μ, but intuitively, this is backwards.
If M were very large, then the acceleration it (and its accompanying “passenger” block m) would
feel would be small, and it would be easy to keep the small block on top. Thus, the minimum μ
needed to keep m from sliding off would be lower, and the correct answer must be choice A. To see
this worked out in more detail, note that at x = A, the magnitude of the force exerted by the spring is
kA and it acts on a total mass of (m + M), so the force acting on m (due to the spring) is
ma = mkA/(m + M). The maximum force of static friction is equal to μmg, the coefficient of static
friction times the normal force on m. The small block will stay put if μmg > mkA/(m + M). Dividing
both sides by mg, we get μ > kA/[(m + M)g].
Physics Passage 15
Primary Topic: Standing waves
Passage Type: Information presentation, conceptual (expect all three Q types with a challenge
distinguishing when outside knowledge is a benefit or hindrance)
Subtopics:
Wave-speed equation, Big Rules for Waves, electromagnetic waves,
Techniques:
Like Passage 13, this is a largely conceptual passage, though it does include a new
equation. The greatest difficulty in passages like these (at least one of which students
should anticipate on the MCAT) can be recognizing the fact that basic physics principles
and common sense remain relevant even when the novel ideas are complicated—
indeed, especially so. The passage is also long, so in your map be sure to focus on those
elements of the text such as definitions of italicized terms and numerical values in
paragraphs that you would either highlight or jot down on your scratch paper. You might
also reduce Figure 2 to an annotation like “l = # of vertical nodal lines, m = # of horizontal
nodal lines.” Finally, again to emphasize what students do know over what seems weird
and threatening, you might write T = 1/f and v = λf on the map. Q1 is an application of
the first Big Rule for Waves: recognizing that this is a memory question and not an
explicit question about the provided equation is key. Q3 is a straightforward calculation; it
is worth pointing out the common answer pattern where all that is needed is the correct
order of magnitude. Q2 and Q4 are the more difficult conceptual questions (as indicated
partly by the long, purely verbal questions and answer choices): for these be sure to
focus on POE and the way that common sense and a little basic physics eliminates most
of the incorrect answers (microwaves don’t burn you when you open them, milk isn’t
boiling after we’ve driven it back from the store, etc.).
1. A The speed of a wave in a medium is a property of the medium, not of the wave frequency. So for a
given membrane, the wave speed will be the same for all normal modes on the membrane.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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2. D The passage tells us that it is the frictional effect of rotating molecules on adjacent molecules that
produces heating. Each molecule of water in the liquid phase has many adjacent neighbors, but
each molecule of water vapor has none (or very few). The frictional effects would be absent, and
the molecules would heat up less. (This is why the air in a microwave oven does not reach the
same temperature as the food that is being cooked.)
3. C We use the basic equation for waves, λf = v. For electromagnetic waves, v is c = 3 × 108 m/s,
so λ = c/f. We then find that λ = (3 × 108 m/s)/(2450 × 106 Hz) ≈ 0.12 m = 12 cm.
4. D Choice A is wrong because, for one thing, centripetal forces do not push anything out from the
center of rotation. Choice B can be eliminated because any additional kinetic energy that might be
absorbed by the food would be negligible and would not significantly shorten cooking time. There is
no reason to believe that the half of the oven that is closer to the microwave source (the
magnetron) contains higher intensity microwaves; in fact, because the interference of the reflected
waves creates a standing wave pattern (as the passage indicates), we know that the wave
amplitude, and therefore intensity, must not be greater on one side than another. The correct
answer is choice D. Because of nodal points and planes within the oven’s interior, if the food
remained motionless, those regions of the food at and near those nodal points would heat up much
more slowly than the other regions of the food, and the cooking would be non-uniform.
Physics Passage 16
Primary Topic: Waves
Passage Type: Information presentation, technical (expect explicit questions with the definitions and
equation, and implicit questions applying basic knowledge about sound)
Subtopics:
Sound waves, Doppler shift, intensity attenuation
Techniques:
Though the phenomenon described in the passage can seem complex, this is a fairly
straightforward passage. Because it is relatively short, one can spend the time to read it
fully and jot down the numbers, Equation 1, and a quick synopsis of each paragraph on
his or her map (along the lines of “ultrasounds generated by crystal vibrations” and
“detects I, f, t of reflection of wave pulses—chirps—off of tissues”). Q1 is an explicit
recall/comprehension question: one statement in the passage supports the answer and
other choices can be eliminated or are unsupported. Q2 and Q4 are memory questions,
testing understanding of basic wave phenomena. Q3 and Q5 are relatively
straightforward algebra/calculation problems; Q6 is as well, though it proves difficult for
many students, so you should explain carefully how to interpret the language of the
question mathematically (the analogy to half-life problems is useful here).
1. B In paragraph 3, the passage states that “ultrasonic waves are entirely reflected at air/tissue
boundaries, but they transmit well through liquids.” We can conclude that the liquid gel helps to
eliminate the air layer between the scanner head and the skin, thus increasing the intensity of the
ultrasound waves entering the body (choice B). Nothing in the passage supports choices A, C, or D.
Choice A is false (the waves would move faster through liquid), and choices C and D don’t make
sense in context (though you should not dwell on explaining why unless asked).
2. C Ultrasound waves do experience reflection, so Item II must be true; this eliminates choice B.
Because sound waves are longitudinal, they cannot be polarized (only transverse waves can be
polarized), so Item I cannot be true; this eliminates choices A and D. Refraction (Item III) is
experienced whenever a wave passes (at a nonzero angle of incidence) into a new medium.
3. D Frequency times wavelength gives wave speed, so v = XY. Using distance = rate × time, the
distance the wave can travel in time t = Z is vt = (XY)Z.
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4. D Statement D correctly describes the result of the Doppler Effect in this situation.
5. C Because a typical ultrasound scan uses waves with a frequency of 3 MHz (paragraph 2), which
travel at a speed of 1500 m/s through the body (paragraph 3), the wavelength is:
λ = v/f = (1500 m/s)/(3 × 106 Hz) = 5 × 10–4 m = 0.5 mm.
6. B Equation 1 gives the intensity as a function of depth. Because the question tells us that I = kA2
(where k is a constant of proportionality), we have kA2 = I0e–zd, so A = A0e–zd/2. Therefore, A will be
equal to A0e–1 when zd/2 = 1; that is, when d = 2/z.
Physics Passage 17
Primary Topic: Sound waves
Passage Type: Experiment/data presentation (expect explicit questions interpreting all these crazy data
and using the definitions)
Subtopics:
Sound level, impedance, resonance, loudness as a function of frequency
Techniques:
This is a daunting passage for many students, with a couple of subtle definitions
(impedance and phons) and a difficult graph. It can be seen as a hybrid between a
technical passage like Passages 8 or 11 and a more conceptual passage like Passages
13 or 15 (of course, these categories themselves are fluid). As usual, your map should
pay attention to definitions of new terms and phenomena and the provided equations.
Obviously students would not themselves redraw the graph on the exam, but you may
want to draw a legible approximation with a couple curves of equal loudness in order to
explain how to read it. Q1 is an explicit, straight calculation. Q2 and Q4 explicitly rely on
correct reading of the graph; Q4 can also be solved using Equation 3. Q3 and Q6 require
facts from the passage about impedance, along with a given equation for Q3 (Q6 is
POE). Q5 is a POE functional-dependence problem.
1. B We use Equation 2 given in the passage: I = p/(4πr2) ⇒ rmin = (p/4πImax)½ = (500/40π)½ ≈ 4½ = 2 m.
2. D A 200 Hz, 60 dB sound is located on the sixth curve in Figure 3 (counting from the bottom), but a
4 kHz, 60 dB sound is located on the seventh curve from the bottom. Thus, the 4 kHz, 60 dB sound
would be perceived as louder than the 200 Hz, 60 dB sound (choice A is wrong). As for choice B,
both of these sounds (300 Hz, 30 dB and 7 kHz, 30 dB) are on the same loudness curve (the third
one from the bottom), so they would be perceived as having the same loudness (choice B is
wrong). Like choice A, choice C is backwards; each loudness curve slopes downward from 100 Hz
to 1000 Hz, so a 1000 Hz sound generally requires a lower decibel level to have the same loudness
as a 100 Hz sound. Thus, the average human would perceive a 1000 Hz sound to be louder than a
100 Hz sound of the same sound level (choice C is wrong). The correct answer is choice D. Each
loudness curve slopes upward from 4 kHz to 10 kHz, so a 10 kHz sound generally requires a higher
decibel level to have the same loudness as a 4 kHz sound; thus, the average human would
perceive a 4 kHz sound to be louder than a 10 kHz sound of the same sound level.
3. B In the paragraph below Equation 2, the passage tells us that “there is a 99.9% loss of intensity
when sound travels from air to water.” So, if cochlear fluid has the same impedance as water,
sound should experience a 99.9% decrease in intensity in passing from the middle ear to the inner
ear. If I is the intensity in air, then the intensity in the cochlear fluid (if the impedance-matching
mechanism of the middle ear were absent) would be just 0.1%, or 1/1000, of I. If I decreased by a
factor of 1000, then according to Equation 1, the sound level would decrease by 10 log 1000 = 30
dB. Thus, the impedance matching of the middle ear must overcome this 30 dB drop.
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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4. C There are two ways to answer this question. First, Equation 3 gives the resonant frequency of the
external auditory canal. Substituting in the given values for vs and L, we find that
f = (340 m/s)/[4(0.025 m)] = 3400 Hz. Alternatively, look at Figure 3. Each loudness curve has a
minimum around 3-4 kHz, implying that the ear is most sensitive to frequencies in this range.
5. D If the maximum gauge pressure P increases by a factor of 10, then according to Equation 1, the
sound level increases by 20 log 100 = 20 dB. Impedance and wave speed are properties of the
medium, not the source (choices A and B are wrong), and there is no information in the passage
relating gauge pressure to source power (eliminating choice C).
6. A The passage mentions “the impedance of the medium” in which sound waves travel. Thus, we can
infer that impedance is a physical property of the conducting medium, not a property of the source
of the sound. Choices B, C, and D are all physical properties of a material medium; choice A,
frequency, is a property of a sound source.
Physics Passage 18
Primary Topic: Mirrors and Lenses
Passage Type: Situation Presentation, relatively straightforward (expect mostly memory questions, with
some explicit once the “equation” of paragraph 3 is written down)
Subtopics:
Refraction, wave properties of light
Techniques:
This is a medium-length passage with some information that is redundant: the figure,
correctly interpreted, allows one to skip over the subsequent paragraph. A good map
should translate the first sentence of the third paragraph directly into an equation,
because that sentence is the only “equation” provided by the passage! Highlight the
sentence describing the image formed by the telescope (“virtual and still inverted”), as
students need to know when to focus on new information that trumps remembered basic
information (in this case, that virtual images are always upright, which applies to single
mirror or lens optics). Q1 can be difficult if students don’t know how to rely on the figure.
Emphasize that the given D in the figure indicates what is meant by diameter of the
objective lens. Q2 explicitly requires that translated equation in the mapping. Q3 is a freestanding memory question. Q4 implicitly requires understanding that increasing of index
of refraction means the light incident on a lens will bend more, thus increasing its power
and decreasing its focal length: this is a challenging connection between physical and
geometric optics, so sketch comparative diagrams quickly if you didn’t discuss lenses in
these terms in lecture. Q5 is another implicit question, but mainly relies on prior
knowledge of dispersion. Show students how POE could get them to the right answer
even if they had forgotten this wave property.
1. B The diameter D of the objective lens indicates how wide it is, not anything about its radius of
curvature. This eliminates choice A. A larger lens with the same focal length—and thus radius of
curvature—as a smaller lens would be thicker in the middle, not thinner, thus eliminating choice C.
A careful consideration of Figure 1 should show students that the larger diameter lens would
intersect a greater number of equally-spaced incident rays, thus collecting more light. The
implication of focusing more light is that fainter objects will be visible, as indicated by choice B.
2. C The passage’s last paragraph states that magnification is equal to the fraction: F (focal length of
objective lens) / f (focal length of eyepiece). Magnification is greatest when that fraction is greatest.
Fractions, of course, increase with an increasing numerator and a decreasing denominator. Choice
C represents a fraction with a large numerator and a small denominator.
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3. C This is a straightforward computation. Apply the lens equation:
1/f=1/o+1/i
1 / (2 cm) = 1 / (1.5cm) + 1 / i
i = –6 cm
4. A Increasing the refractive index would decrease the focal length. Decreasing the focal length of the
eyepiece lens causes the magnification of the telescope to increase since it is proportional to 1 / f,
as indicated by the equation extracted from the third paragraph.
5. B Choices A, C, and D can each be eliminated: any positive focusing at all would cause the inversion
of the image of a distance object, and just because the index of refraction might be different, a
converging lens is still a converging lens and still focuses incident light. For the telescope to be in
focus, the focal lengths of the objective and eyepiece lenses must coincide at a point inside the
telescope. If the index of refraction were not constant, then this could only happen for one
wavelength of light at a time, thus leaving all other wavelengths out of focus to varying degrees.
Physics Passage 19
Primary Topic: Quantum physics
Passage Type: Situation presentation, technical (expect all three Q types with a challenge distinguishing
when outside knowledge is a benefit or hindrance)
Subtopics:
Photons, nuclear physics and chemistry
Techniques:
This is a long passage with a lot of information: it’s a combination of a technical situation
passage and an experiment-based (though the table is as simple as possible). There are
a few terms to highlight with their definitions, a reaction and a figure to interpret (but not
redraw: it’s too detailed and it doesn’t yield enough). Q1 is a free-standing memory
question. Q2 is also a memory question, basically, given that the salient fact from the
passage is restated in the stem. Note the difference between “true” and “explains.” Q3
and Q4 are relatively straightforward implicit questions. Q5 and Q6 are more subtle
implicit questions, with Q5 relying on the table and recall of kinematics and Q6 a
combination of more obscure outside knowledge and careful reliance on one’s passage
map.
1. C We need to write a balanced nuclear reaction to arrive at the correct daughter nucleus. The
radioactive parent nucleus is 13N from 13N-ammonia. According to the passage, it emits a positron
upon decay. The atomic number of nitrogen is 7 and positrons are written as +10ß to balance nuclear
reactions. Therefore, the reaction is:
0
13
7N→ +1ß
+ AZX
Since the question is only asking about the identity of the daughter nucleus, we only need to
balance the nuclear charges (the subscripts). This yields: 7 = 1 + Z, so Z = 6. Using the periodic
table we arrive at carbon (answer choice C) as the correct answer.
2. A Even though I, II, and III all occur during the process of annihilation, only I explains the emission of
the gamma photons in opposite directions. The question is looking for an explanation for the 180
degree angle, and that directionality suggests vector quantities. Mass, energy, and electric charge
are scalars and cannot explain a vector phenomenon such as the 180 degree angle of emitted
photons (II and III are therefore incorrect so B, C, and D are wrong). Momentum, however, is a
vector quantity, and conservation of momentum dictates the 180 degree angle between the emitted
photons.
3. A The first paragraph of the passage states that upon annihilation, conversion of mass to energy
occurs. So the total mass of a positron and electron becomes converted to energy in the form of
MCAT ICC Physics Passage Techniques and Solutions – FOR INSTRUCTORS ONLY
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two photons. The mass of a positron is equal to the mass of an electron (me = 9.11 x 10−31 kg).
Using Einstein's mass-energy equivalence formula, E = mc2, we can calculate the total energy
created upon annihilation. Using SI units, we have E=2mec2 = (2)(9.11 x 10−31 kg)(3 x 108 m/s)2 ≈
162 x 10−15 J. The answer choices are all in units of eV, so (162 x 10−15 J)(1 eV / 1.6 x 10−19 J) ≈
106 eV = 1000 keV. At this point you may be tempted by answer choice B, but since there are two
photons generated per annihilation event, each will have half of that amount of energy (̴ 500 keV).
4. B The last paragraph of the passage mentions that the typical administered dosage of 18F-FDG is 10
mCi. We need to determine the number of half lives that must elapse before the activity drops to 1
mCi. Starting with 10 mCi and splitting the activity in half until 1 mCi is achieved:
10 → 5 → 2.5 → 1.25 → 0.625
It appears that this is accomplished between 3 and 4 half lives. From table 1, the half life of 18F is
110 minutes, so the amount of time needed to reach 1 mCi is between (3 x 110 min = 330 min) and
(4 x 110 min =440 min). The only answer choice between 330 min and 440 min is answer choice B
(6 hours).
5. C This is a "2 x 2" question, where one fact eliminates 2 answer choices, and a second fact eliminates
a different set of 2 answer choices. According to table 1, the maximum kinetic energy (KE) of the
positrons emitted from 15O is greater than the maximum KE of the positrons emitted by 13N. Since
1
KE = mv 2 , and the mass of the positron is the same regardless of its parent nucleus, we can
2
conclude that KE is proportional to v2. Therefore, the initial speed of the positron emitted from 15O
will be greater than the initial speed of the positron emitted from 13N. This eliminates answer
choices A and D. The passage states that positrons lose most of their kinetic energy before they
are annihilated. This means that their final speed can be considered to be zero upon annihilation (or
some other small nonzero value). An object with a higher initial speed will travel more distance
before coming to rest (consider v2 = 2ad). Therefore the positron with faster initial speed will have
the longer range. This eliminates answer choice D.
6. C The emitted gamma-ray photons may be absorbed by atoms of the surrounding tissue and release
photoelectrons from these atoms. This is the photoelectric effect which will cause the atoms to be
ionized (lose electrons). Answer choice A is a true statement and therefore eliminated. From the
electromagnetic spectrum, we know that gamma -rays have higher energy (E), higher frequency (f),
and lower wavelengths (λ) than X-rays (E = hf = hc/λ). This makes answer choice B a true
statement and it is also eliminated. Increasing the concentration or dosage of the radioactive
material will result in more radioactive decay which means more positrons and more annihilation
events. However, similar to the photoelectric effect, there is only one photon per one electron, or
one photon per one positron. This means that no matter how many positrons are converted to
energy, the energy of each created photon will be the same. More positrons will result in more
photons created (higher intensity) but the energy of each photon will stay the same. Since the
energy of one photon is given by E = hf, the frequency of the emitted photons will not change.
Answer choice D is therefore a true statement and eliminated.