Physics section test (12)

MCAT Section Tests
Dear Future Doctor,
The following Section Test and explanations should be used to practice and to assess
your mastery of critical thinking in each of the section areas. Topics are confluent and
are not necessarily in any specific order or fixed proportion. This is the level of
integration in your preparation that collects what you have learned in the Kaplan
classroom and synthesizes your knowledge with your critical thinking.
Simply
completing the tests is inadequate; a solid understanding of your performance through
your Score Reports and the explanations is necessary to diagnose your specific
weaknesses and address them before Test Day.
All rights are reserved pursuant to the copyright laws and the contract clause in your
enrollment agreement and as printed below. Misdemeanor and felony infractions can
severely limit your ability to be accepted to a medical program and a conviction can
result in the removal of a medical license. We offer this material for your practice in your
own home as a courtesy and privilege. Practice today so that you can perform on test
day; this material was designed to give you every advantage on the MCAT and we wish
you the best of luck in your preparation.
Sincerely,

Albert Chen
Executive Director, Pre-Health Research and Development
Kaplan Test Prep

© 2003 Kaplan, Inc.
All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm,
xerography or any other means, or incorporated into any information retrieval system, electronic
or mechanical without the written permission of Kaplan, Inc. This book may not be duplicated,
distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement.

PHYSICAL SCIENCES TEST 6 EXPLANATIONS
Passage I (Questions 1–7)
1.

C
This question deals with the stoichiometry of Reaction 1. Looking at the balanced reaction given, we see
that when 2 mol AgBr react with 1 mol hydroquinone, two moles of metallic silver are generated. 1 mol AgBr
would therefore be expected to react with 0.5 mol hydroquinone to produce 1 mol Ag. In this case, however,
hydroquinone is a limiting reagent. The 0.25 moles of hydroquinone we have will react with 0.5 mol AgBr to
produce 0.5 moles of metallic silver.
2.

A
The quickest way to get the answer is to remember that the passage mentions that hydroquinone is a mild
reducing agent at the beginning of the second paragraph. That means that the hydroquinone itself is oxidized in
the reaction. It loses electrons in the process. The only choice that shows hydroquinone as the reactant of an
oxidative half-reaction is choice A.
Choice B shows the other half-reaction from Reaction 1. It shows Ag+ gaining an electron and going to
solid silver. Since electrons are added in this half-reaction, it is a reduction.
Choice C can also be eliminated because it is a reduction half-reaction also, but since we know that we are
looking for a half-reaction of reaction 1, the half-reaction must conform to the direction of reaction 1. In choice C,
we see hydroquinone as a product, so we know that this choice cannot be right.
Looking at choice D, we see that while it is an oxidative half-reaction that releases one electron, it is the
oxidation of solid silver to form silver ion. We can eliminate this choice because the product of reaction 1 is solid
silver, not silver ion. So, if you forgot the definition of a reducing agent, you could have gotten the answer anyway
by eliminating choices B and C as reductions and choice D since it wasn't a part of reaction 1.
3.

D
Question 3 is really a reading comprehension question. To answer it, you have to think through what the

passage has told you about photographic negatives and the process involved in developing them, especially the
fixer. The third paragraph describes what a negative really is and the second describes how it is made. The second
paragraph also describes the role of the fixer solution in the development of film.
If a developing piece of film were left long enough, the unactivated silver bromide would eventually react
with the hydroquinone and deposit silver over the entire film. The fixer solution prevents this by stopping the
reaction. The fixer removes the unreacted silver bromide by ion-exchange solvation. That means that any activated
silver halide that hadn't reacted yet would also be removed from the film. If the film is never treated with the fixer
and we assume an excess of hydroquinone, we know that Reaction 1 would proceed until all the silver bromide on
the film had reacted, leaving metallic silver all over the film. The film would then appear black because the
metallic silver is black. That lets us know that choices A and B are wrong because they say that the negative would
be completely white. If we look at the rest of the answers, we see that C says that the negative is black because the
unactivated silver bromide would bond with the hydroquinone. Be careful! We just finished saying that all the
silver bromide will be reduced to metallic silver, but in the reaction, the silver bromide and the hydroquinone do
not bind together. The unreacted silver bromide will reduce to metallic silver as in choice D. So choice D is right
because it describes the correct appearance of the negative and also states the correct reasons why. Notice that
choice A also explains that silver will be deposited all over the film. It is imperative in this question, therefore, to
know not only that the entire film is covered in silver but how this makes the film look. The only choice with both
parts correct is choice D.
4.

C
In AgBr, silver has an oxidation number of +1: the oxidation numbers of Ag and Br have to add up to
zero since AgBr is neutral, and Br, as a halogen, will typically have an oxidation number of –1. On the product
side, silver is associated with the thiosulfate ion. Note that since the thiosulfate ion is already present on the
reactant side, and so instead of assigning oxidation numbers to sulfur and oxygen separately, we can just look at
the ion as a whole which has an oxidation number of –2, since that is its charge. In silver thiosulfate, there are two

1

Kaplan MCAT Physical Sciences Test 6 Explanations

thiosulfate ions, which together will give an oxidation state of -4. This, added to the oxidation number of silver,
must equal –3, the charge of the complex ion. The oxidation number of silver must therefore be +1.
5.

A
Essentially, all this question asks is which halogen has the greatest electronegativity. Electronegativity
measures the attractive force that an element has for electrons in a bond and is often used to help determine
whether a bond between two atoms will be covalent or ionic. It increases as we move right across a period (except
for the noble gases) and as we move up a column. That means that the halogens are the most electronegative
elements in their periods and that fluorine, at the top of the halogen group, has the strongest electronegativity of
the halogens.
6.

C
From paragraph one, we know that what the question really asks is where the most light struck the film.
The most light strikes the film where the image of the brightest object is on the print. That would mean the white
ball, or answer choice C. Remember that we were given a print to work backward from. The white ball on the print
would appear black on the negative, which means that the most metallic silver was deposited on the film there. The
black background, choice A, would have appeared white on the negative meaning that hardly any silver was
deposited on the area that showed its image. The gray ball, choice D, would be an intermediate between these two,
having more light than the black background but not as much as the white ball. Choice D, the boundary between
the white ball and the background, is more of a distraction than anything else. The boundary defines a part of the
film between two concentrations of activated silver bromide and doesn't represent any particular value.
7.

C
Molar solubility is the number of moles of a compound that will dissolve in a solution. That means that
we're looking for the number of moles of silver bromide that will dissolve in a liter of water that already has 0.10

moles of silver nitrate in it. From the definition of the solubility product constant, Ksp, we know that:
[Ag+][Br –] = 5.0 ∞ 10–13 (in the saturated solution)
So, the maximum concentrations of silver and bromide ions dissolved cannot have a product greater than
5.0 ∞ 10−13. x moles of silver halide produces x moles of Ag+ and x moles of Br − in a solution which means the
product is x2. We have, however, already got 0.1 moles of silver in the solution from silver nitrate, so the amount
of silver ions will be x + 0.10, the amount dissolved from AgBr plus the amount already in solution. So the
relationship that needs to be satisfied is:
5.0 ∞ 10–13 = x (x + 0.10)
Since the molar solubility of silver bromide is very small compared to the 0.10 moles of silver already in
the solution, whether we actually add the x to the 0.10 in our product or not won't make much difference. So, if we
just treat our product as 0.10x, all we have to do to find the molar solubility is divide 5.0 ∞ 10–13 by 0.10. The
answer is 5.0 ∞ 10-12, which is choice C.
Passage II (Questions 8–12)
8.

A
The last paragraph of the passage gives the definition of mass defect:
mass defect = sum of masses of nucleons – actual mass of nucleus

What is the sum of the masses of the constituent nucleons? There are two types of nucleons: protons and
neutrons. Their masses are given as mp and mn respectively. The nucleus under consideration has atomic number Z
and mass number A: the nucleus has Z protons and (A - Z) neutrons. The sum of the masses of its constituent
nucleons, then, is:


Kaplan MCAT Physical Sciences Test 6 Explanations

Z ∞ mp + (A – Z) ∞ mn
This is what one would expect the mass of the nucleus to be. It actually has a smaller mass, measured to
be M. The difference between the two is the mass defect; i.e.:

mass defect = Z ∞ mp + (A Z) ∞ mn – M
9.

C
The mass condition referred to in the question is simply a requirement of energy conservation. A decay
reaction is special in that it will happen spontaneously. That is, the initial particle will decay even if it is not given
any energy. So the energy condition for a spontaneous decay dictates that the rest energy of the reactants must be
greater than the rest energy of the products. The difference is converted into kinetic energy for the emitted electron,
for example.
We know that rest energies are related to masses by E = mc2, and, since c is a constant, we may conclude,
that the mass of the initial particle must be greater than the mass of the decay products. Some of the mass of the
initial particle is converted to energy to make the reaction occur, and therefore the remaining mass, which is the
mass of the product, must be smaller than the mass of the initial particle.
The initial particle in beta decay is called the parent nucleus, and the decay products are the daughter
nucleus and an electron. (Remember that the electron emitted in ordinary beta-decay comes from the nucleus itself,
not from the cloud of electrons around the nucleus.) So, we conclude that choice C is correct. The mass of the
parent nucleus must be greater than the sum of the masses of the daughter nucleus and an electron, in order for
beta decay to occur.
Choice A can be ruled out because the mass defect is related to the binding energy of the nucleus, which
in turn is an indication of its stability: the higher the binding energy, the more stable the nucleus since the more
energy needs to be supplied to break it apart. If the daughter nucleus has a lower mass defect than the parent
nucleus, that would mean it is less stable than the parent nucleus. Together with the kinetic energy of the emitted
beta particle, this would imply that energy needs to be supplied, which contradicts our knowledge of the
spontaneity of such reactions.
Answer choice B is wrong as well; although the statement it makes is true, it is not pertinent. The mass of
the parent particle plus the mass of an electron will be greater than the mass of the daughter particle, but this fact
is not enough to ensure that there is enough energy for the decay to occur.
Answer choice D is again true, but it is a statement that has no bearing on the beta decay of the parent
nucleus. It simply restates the concept of mass defect and is true of every nucleus.
10.

C
The second paragraph identifies the two forces acting within the nucleus. There is the familiar
electrostatic force which acts to repel protons from one another because they are similarly charged. There is also
the strong force which is stronger than electrostatic repulsion and works to hold the nucleus together. Both forces
decrease as distance increases—the electrostatic force falls off as the distance squared:
F=k

q1q2
r2

The passage does not tell us precisely how the strong force depends on the distance.
Since a large binding energy means a stable (and hence tightly bound) nucleus, and the strong force is
what keeps the nucleons together, stronger strong forces means a higher binding energy. The electrostatic force, on
the other hand, works to destabilize the nucleus: the stronger the electrostatic forces, the smaller the binding
energy. If the binding energy per nucleon is decreasing, then, either the strong force is decreasing or the
electrostatic force is increasing (per nucleon). This alone helps us eliminate choice A.
Choice B proposes that larger nuclei have stronger electrostatic forces because of their larger size.
Although stronger electrostatic forces would decrease binding energy, electrostatic forces decrease with larger
distances, not increase.

3


Kaplan MCAT Physical Sciences Test 6 Explanations

Choice C states that the strong nuclear forces decrease in magnitude as the size of the nucleus increases.
This is consistent with what we know about the strong force and also successfully accounts for the trend. Choice C
is the correct answer.
Choice D is incorrect because greater total charges increase the electrostatic repulsion between protons.

11.

C
This is the only question based on the graph and it requires the ability to interpret information presented
in a graphical format. This is a difficult question. How do we decide which reactions give out the most energy? The
reaction that gives off the most energy is that one with the greatest energy difference between the reactant particles
and the product particles.
We know from the passage that the total binding energy of a nucleus is the amount of energy needed to
completely break it apart into its constituents. Look at each reaction given in the answer choices. The number of
protons and neutrons on each side of the reaction balances, which means that the sum of the energies of the
nucleons on each side of the reaction is the same. Therefore:
Ereactants – Eproducts = binding E (products) – binding E (reactants)
So, to chose the reaction that gives out the most energy, we need to chose the reaction for which the product
particles have the highest total binding energy compared to the reactant particles.
The total binding energy of a nucleus is the binding energy per nucleon, shown on the graph, multiplied
by the number of nucleons, which is just the mass number, A. It looks like we are going to have to do a lot of math.
To make our calculations as short as possible, let's make integer estimates of the binding energies per nucleon from
the graph. This estimation should allow us to differentiate answer choices from each other.
Take answer choice A. The binding energy of N-14 is approximately 8 MeV (from the graph) multiplied
by 14, which is the number of nucleons. The binding energy of C-13 is approximately 8 MeV ∞ 13. The binding
energy of H-1 is zero. Therefore, the energy released for a single reaction is approximately:
(14 ∞ 8 – 13 ∞ 8) = (14 – 13) ∞ 8 = 8 MeV
Now look at answer choice B. The binding energy of He-4 is about 7 MeV ∞ 4. The binding energy of C12 is about 8 MeV ∞ 12. The binding energy of N-15 is about 8 MeV ∞ 15. The binding energy of H-1 is zero.
Therefore the energy released for a single reaction is approximately
7 ∞ 4 + 8 ∞ 12 – 8 ∞ 15 = 7 ∞ 4 – 8 ∞ 3 = 4 MeV
Next consider answer choice C. Following the reasoning we used for the first two answer choices, the
energy given off by one occurrence of this reaction is given by: about 7 MeV ∞ 4 ∞ 2 – about 1 MeV ∞ 2 – about 5
MeV ∞ 6, which is 24 MeV. Thus, the energy released in this reaction is much greater than that released in answer
choices A and B. In fact, answer C turns out to be the correct answer, as we can check by considering final
reaction, given in answer choice D.

The energy emitted in answer choice D is given by: about 3 MeV ∞ 3 – about 1 MeV ∞ 2, which is about
7 MeV. The energy given off in reaction C, over 20 MeV is much greater than the energy given off in the other
three reactions. Answer choice C is correct.
12.

C
This question requires some interpretation and a calculation using the formula given. To answer this
question, we must figure out what causes the neutron to lose mass in a γ decay. We are told in the question stem
that a photon is emitted. So the mass that the nucleus loses during its decay must correspond to the energy that it
loses in emitting the gamma photon. The energy taken away by the photon is related to its wavelength by the
hc
. Since we are told to assume that the nucleus does not recoil, we know that the nucleus does not
formula E =
λ
carry any kinetic energy. So, the difference between the rest energy of the nucleus before the decay, and the rest
energy of the nucleus after the decay, is equal to the energy of the photon.


Kaplan MCAT Physical Sciences Test 6 Explanations

We can eliminate choices A and B immediately because they are negative numbers. The initial rest mass
of the nucleus must be greater, not less, than the final mass of the nucleus. This must be so, since the emission of
the gamma photon takes away some of the energy, and therefore some of the mass, from the nucleus.
Now we are going to have to do some calculating to choose between answers C and D. We know that the
difference in rest energies of the initial and final nuclei, E, has an associated rest mass, m, where E = mc2.
Coupling this equation with the photon energy, we find that:
hc
λ
hc
h

m= 2=
λc
λc
E = mc2 =

m=

(6.63 ∞ 10–34)
(3.83 ∞ 10–12 ∞ 3 ∞ 108)

Since the remaining answer choices are pretty far apart, it is a good idea to save time by estimating the
answer:
m

7
∞ 10(–34 + 12 – 8)
4∞3
♠ 0.5 ∞ 10–30 = 5 ∞ 10–31 kg
♠

This is closest to choice C so it must be the correct answer.
Passage III (Questions 13–18)
13.

A
The final paragraph of the passage states that a sacrificial anode for iron is a metal that is more easily
oxidized than iron. In other words, a sacrificial anode for iron has a reduction potential that is more negative than
that of iron. Remember that a positive potential corresponds to a negative free energy, and so a more negative
reduction potential would mean a lower tendency to be reduced, or equivalently, a higher tendency to be oxidized.
Looking at table one, you can see that iron, with a reduction potential of –0.44 volts, would be located between zinc

and nickel. Since choices B and D, tin and palladium, have reductions potentials that are less negative than that of
iron, they can be eliminated. So, those metals that are more easily oxidized than iron are zinc, aluminum, and
magnesium. Looking at the answer choices you can see that you have two possibilities: magnesium, with a
reduction potential of –2.37 volts, and zinc, with a reduction potential of –0.76 volts. Both of these metals can be
used as sacrificial anodes for iron, but since magnesium has the more negative potential of the two, it is the best
choice.
14.

D
Corrosion is an electrochemical process, and in any electrochemical process, charged conductors, ions, are
needed in order for charge to flow. So, let's start eliminating some answer choices. Choice C, salt water, is out. As
is well known, salt water contains sodium and chlorine ions, and these ions definitely increase the rate of
corrosion. Choice A, dissolved carbon dioxide gas, also increases the rate of corrosion. Dissolved CO2 gas produces
carbonic acid and hence protons according to the following equilibria:
CO2 + H2O

H2CO3

H+ + HCO3–

You can see that H+ is one of the reactants in Reaction 1; increasing its concentration will drive this reaction
forward increasing the rate of corrosion. So, choice A is out. Choice B, air pollution, increases the rate of corrosion
as well. The sulfur oxides in air pollution react with water to form dissolved sulfuric acid; this speeds up corrosion
much in the same way as dissolved CO2 does. The only choice left is choice D, making it the correct response.
Nitrogen gas is relatively inert and in no way assists in the corrosion of iron.

5


Kaplan MCAT Physical Sciences Test 6 Explanations

15.

A
Recall that oxidation occurs at the anode, and upon oxidation iron will acquire a positive charge to
become cations. In a typical galvanic cell, because the anode and cathode are separated, the ions themselves cannot
migrate, and so charge is balanced by the movement of the ions in the salt bridge. In this case, however, no such
separation is performed, and so the positive Fe ions will be attracted to the cathode. Looking at Equation 2, which
is the reaction for the formation of rust, you can see that oxygen and water are required. Since the exposed areas of
iron are rich in oxygen and water, they serve as the cathodes and the unexposed areas serve as the anodes.
16.

C
To answer this question you need to know how to interpret cell schematics or cell diagrams. First, the
anode process is represented on the left with a vertical line drawn to represent phase boundaries. So in this case,
we have a zinc anode where zinc metal is oxidized to Zn 2+. Next, two vertical lines are drawn to represent the
separation between anode and cathode, and the cathode is represented on the right side of the schematic. In this
example we have an iron cathode that has Fe2+ being reduced to metallic iron. Now that we know how to construct
the cell schematic, all we have to do is determine the cell potential. The cell potential is determined by subtracting
the emf of the cathode from the emf of the anode when they are written as reductions:
E cell = E cathode – E anode
Again, it is important to realize that subtraction is valid only when both potentials are given as reduction
potentials. If the anode reaction is already written as an oxidation reaction, and the emf given accordingly, then we
can simply add the two half-cell potentials. Looking at Table 1 we can see that the reduction potential of zinc is –
0.76, and the passage tells us that the reduction potential of iron is –0.44. So the cell potential is (–0.44) – (–0.76)
= 0.32 volts.
17.

D
The standard hydrogen electrode is the common reference electrode to which all half-cell potentials are

measured. The standard hydrogen electrode is a hydrogen-gas electrode specifying hydrogen ion concentration at
one molar and the partial pressure of hydrogen gas at 1 atmosphere. By convention, the SHE is assigned a value of
zero volts. Choice A says the SHE “uses hydrogen gas at 1 atmosphere (760 mm Hg)” — this is a true statement,
so we can rule out this choice. Statement II says the SHE “potential is assigned a value of zero volts” — again this
is a true statement which eliminates choice B. Choice C is also a true statement: if a half-cell has a negative
reduction potential, it has a lower tendency to be reduced than the SHE, and so when the two are coupled it will
undergo oxidation instead, yielding a positive emf overall which corresponds to a spontaneous net reaction. The
statement in choice D is untrue and so this is the correct choice.
18.

D
A galvanic cell is an electrochemical cell where a chemical reaction produces electrical energy. An
electrolytic cell is an electrochemical cell that requires an input of electrical energy to produce a chemical change.
Since the corrosion of iron does not require an input of electrical energy, but rather occurs spontaneously, it is
considered a galvanic cell. Since this is the case, choice A and B can be eliminated. You are told in the passage
that Fe2+ is further oxidized to Fe3+ to form rust. As discussed above in the explanation to #15, the positive ions
migrate toward the cathode. This piece of information doesn’t really help us in this question since the remaining
two choices both contain refer to cathodes. What you should realize, however, is that these iron cations go on to
react with oxygen and water to form rust. Where is there an abundance of these two reactants? In the atmosphere.
Choice D is the correct response to question 18.
Discrete Questions
19.

B
The acceleration of gravity near the surface of the earth is the constant g = 9.8 m/s2. The fact that this is a
constant means that all objects, regardless of their mass, are accelerated towards the earth at the same rate. If their
acceleration is the same, their velocity changes the same amount in any given time. So, if they fall from the same
height with similar initial velocities, in this case zero, they will hit the ground with the same speed. Since distance

Kaplan MCAT Physical Sciences Test 6 Explanations

= (1/2)at2 and they fall the same distance with the same acceleration, the time it takes to reach the ground is also
the same.
20.

B
A Brønsted acid is defined as a substance that is capable of donating a proton, and a Brønsted base is a
substance capable of accepting a proton. In this question HCl donates a proton to form Cl–: HCl is the Brønsted
acid and Cl– is the Brønsted base. In addition, water accepts a proton to form the hydronium ion--water accepts a
proton making it a Brønsted base and H3O+ the Brønsted acid. Two substances that differ from each other only by a
proton are called an acid-base conjugate pair. This question is asking for the conjugate base of the hydronium ion.
If we remove a proton from H3O+ we have water, making choice B the correct response.
21.

C
Radiation is the transfer of energy by electromagnetic waves. In this manner energy is transferred from
the sun to the Earth. Convection and conduction both require the presence of a material substance. Conduction is
the transfer of heat energy from molecule to molecule in a material substance, and convection is the transfer of heat
energy as a result of the large scale motion of a material substance. In a vacuum neither of these can occur.
22.

D
A reducing agent loses electrons and gets oxidized and an oxidizing agent gains electrons and gets
reduced. Since a reducing agent loses electrons, its oxidation number increases, and since an oxidizing agent gains
electrons, its oxidation number decreases.
23.

A
The original weight is 180 newtons. Weight is the force on a mass due to gravity, and is given by the

equation W = mg, where g is the acceleration due to gravity. Since the mass, m, doesn't change, the acceleration
due to gravity, g, is what changes with W. Now the weight force is also given by Newton's law of gravity which is
given by the equation:
F=

GM1M2
r2

where G is the universal gravitational constant, M1 and M2 are the two masses, and r is the distance between the
centers of the two masses. If we let the mass of the Earth be Me, the mass of an object be m, and the distance that
separates them be r, then we get that F = GMem/r2.
Here r is the distance from the center of the Earth, which at the Earth's surface is the radius of the Earth.
Now, when the object's distance from the center of the Earth is tripled, then
F=

GMem GMem
=
(3r)2
9r2

The force is hence only 1/9 that of its magnitude on the surface. At the surface of the Earth the object
weighs 180 newtons. So at the distance under consideration W = 180/9 = 20 newtons, answer choice A.
Passage IV (Questions 24–29)
24.

A
In the second paragraph we are told that the pressure in the chamber is lowered. This has the effect of
lowering the temperature at which the hydrogen in the chamber boils. So by lowering the pressure in the chamber,
the hydrogen may remain a liquid even though its actual temperature is above the boiling temperature of the liquid
at the new lower pressure. By quickly looking over the answer choices, we see there is only one choice involving

pressure. Answer choice A states that the pressure is decreased in an isothermal process. An isothermal process is
one in which the temperature remains constant. We are told in the passage that the temperature of the liquid

7


Kaplan MCAT Physical Sciences Test 6 Explanations

remains constant. So answer choice A corresponds to what happens when the piston is lowered, and is probably the
correct answer.
However let's look over the other possibilities. Choice B states that the temperature of the liquid is
increased in an adiabatic process. We don't even need to know that an adiabatic process is one in which no heat
flow occurs. We are told in the passage that the temperature of the liquid in the chamber remains constant so this
choice must be incorrect. Choice C states that the volume of the liquid is decreased in an isothermal process.
Again, just by looking at the first part of this answer choice, we can determine that this is the opposite of what
happens. In the first paragraph we are told that the piston is lowered. This creates a bigger space inside the
chamber and increases the volume. So choice C is also incorrect. Choice D states that the density of the liquid is
increased in an adiabatic process. You should know that liquids are not very compressible so the variation in
density of a liquid is very small. We know that the volume of the chamber containing the liquid increases. If
anything this might cause a small decrease in the density of the liquid, so answer choice D is also incorrect.
25.

B
The passage states that when the pressure drops slightly below the liquid-gas coexistence line and no
phase change occurs, it is considered an unstable superheated liquid. Looking at the answer choices we see they all
involve pressure and temperature so we need to determine what the defining pressure and temperature is of a
superheated liquid. In the passage we are told that if we drop the pressure below the liquid-gas coexistence line
while keeping the temperature constant the temperature of the liquid will be above the boiling temperature of the
liquid at the new lower pressure. Now let's look over the answer choices to see if any match this condition.
Choice A states that a superheated liquid is a liquid at a temperature above its boiling temperature at

atmospheric pressure. Boiling points are defined by specific pressures and temperatures. So the temperature at
which the liquid boils at atmospheric pressure will be different than the temperature at which the liquid boils at its
present pressure. Therefore, choice A is incorrect. We are looking for a liquid that is above the boiling temperature
that corresponds with its present pressure. Choice B states that a superheated liquid is a liquid that is at a
temperature above its boiling point at its present pressure. Therefore it is the correct answer. Let's quickly review
the other answer choices to make sure we are not missing anything. Look at choice C. It states that a superheated
liquid is normally a gas at room temperature and atmospheric pressure. This isn't mentioned anywhere in the
passage and it doesn't really make sense that the term superheated liquid would be used to refer to a substance
which is a gas at room temperature. Choice D states that a superheated liquid is a liquid which remains a liquid at
1000 °C and atmospheric pressure. Nowhere in the passage is this quantitative description of a superheated liquid
given.
26.

A
This question requires some outside physics knowledge because it deals with a situation that is not
discussed in the passage. Let's examine what is going on when the liquid hydrogen boils. The liquid becomes a gas
and expands. Therefore, the vapor pressure inside the chamber increases. In the question stem we are told we need
to keep the pressure of the system constant. So to compensate for the increasing pressure inside the chamber, the
volume must expand. The hydrogen gas will exert a force on the piston moving it farther away from the interior of
the chamber. Now let's look over the answer choices to see if any agree with our analysis of the situation.
In looking over the answer choices, we realize they have two parts. Both parts of the correct answer choice
must be correct. So if we can rule out one part of an answer choice, we can eliminate that choice. Choice A states
that the piston moves away from the interior. We determined earlier that this is true. The second part of this choice
states that work is done by the system in order to move the piston away. We decided that the piston moves out
because the expanding gas exerts a force on it, so the gas is in fact doing work. Both parts of choice A seem to be
correct so A is probably the correct answer. Let's go over the other answer choices to make sure.
The first part of answer choice B is correct, the piston does move away from the interior of the chamber.
What about the second part of choice B? It states that work is done on the system. We decided that this is not true,
and indeed the piston does not do work on the gas when the gas expands. The work is done by the liquid-gas
system when the hydrogen absorbs heat and becomes a gas. So choice B is incorrect. The first part of answer

choice C states that the piston moves toward the interior of the chamber; this is incorrect. So choice C must not be
the right choice, because both parts of the correct answer must be correct. Answer choice D is incorrect for the
same reason; the first part states that the piston moves toward the interior of the chamber. So answer choice A is
the correct answer.


Kaplan MCAT Physical Sciences Test 6 Explanations

27.

B
The answer to this question depends on the information given in Figure 2 and requires a bit of common
sense. Let's look at the diagram and try to figure out what is going on in the tank and what interaction is indicated
by the tracks labeled “a”. We note that the particles enter the tank from the left and exit from the right, so the
initial particle or particles must be indicated by the tracks on the left side of the tank. So looking at the track
labeled “a”, we see that there is one line which starts from the left and splits into two lines as we move towards the
right side of the tank. So this indicates that one moving particle experiences some event in which the single line
becomes two lines, which represent two particles. Now we have to look over the answer choices to decide which
one best describes the situation.
Answer choice A suggests a particle decays into three charged particles. If this were the case, we would
observe one line which starts from the left and then splits into three lines at some point in the chamber which then
exit on the right. Well, this is not the case so answer choice A must be incorrect.
Choice B states that an elastic collision between two particles occurs. You should know that in an elastic
collision two particles collide and then go off in different directions. If this type of collision occurred between two
moving particles, we would observe two tracks which intersect and then split again. We don't observe this.
However, one of the particles in the elastic collision can initially be at rest somewhere in the chamber. In this case
we would not see a track for it before the collision. We would only observe one line branching into two lines as we
look from left to right. The particle at rest is sitting at the point of intersection and only starts to move after the
other particle collides into it. So choice B is correct if we consider this case. There is nothing in the question stem
or answer choice excluding the case where one particle is at rest so choice B is probably correct. Let's look over the

other answer choices for completeness.
Choice C states that a completely inelastic collision between two particles occurs. In a completely inelastic
collision two particles collide and then stick together. So we would have either two tracks intersecting and merging
into one track or one track if one of the particles in the collision were initially at rest. Clearly, we don't have one
track, but let's make sure the tracks labeled “a” don't indicate an inelastic collision between two moving particles.
We must be careful and pay close attention to the direction from which the particles are entering the chamber. In
the fourth paragraph and in Figure 2, we are told that the particles enter from the left and exit from the right. So an
inelastic collision between two moving particles would be observed as two lines starting from the left and becoming
one line as we move toward the right side of the chamber. This is the opposite of the case shown, and therefore
choice C is incorrect.
Answer choice D suggests that an inelastic collision of three particles occurs. If the particles were all
moving initially, this would show up as three lines starting from the left which intersect, merge into one line, and
leave the chamber on the right. Clearly the tracks labeled “a” do not show three lines intersecting and forming a
single line. An inelastic collision between three particles might also be represented by two tracks merging into one
if one particle were initially at rest, but we've already seen that this is not what the tracks labeled “a” indicate. So
choice D must be incorrect. Therefore answer choice B best describes the diagram and is the correct choice.
Note that, of course, the tracks may also indicate the decay of one particle into two particles. That,
however, is not one of the answer choices.
28.

B
To answer this question we need to figure out what feature of the particle tracks in a bubble chamber
indicate its charge. We are told in the passage that charged particles curve due to the magnetic field in the chamber
and that the direction in which they curve indicates their charge. The direction of the magnetic force can be figured
out by applying the right hand rule. The magnetic field points into the page, and that is the direction our fingers
should point. For track “b,” the particle is experiencing a force pointing upwards. Our palm should therefore be
facing upwards. Our thumb is now pointing to the right, and this is the direction of qv. Since all particles enter the
chamber from the left and exit at the right, our thumb is pointing in the same direction as the velocity of the
particle. This implies that q must be positive for the particle that left track “b.”
As for track “d,” our fingers still point into the page, but now our palm faces down. Our thumb thus points

opposite to the direction of the paricle’s velocity, and so q must be negative.

9


Kaplan MCAT Physical Sciences Test 6 Explanations

F

v

qv

qv
(also direction of v)

B X

B X

F
d

b

29.

C
To answer this question we need to know what feature of a particle track relates to the momentum of the
particle responsible for that track. The logical place to go for this information is the passage. Sure enough in the

third paragraph we are told that the curve of the track is related to the momentum of the particle that left it. We are
given a formula involving the radius of the circle that the particle travels in, the momentum of the particle, the
charge of the particle, and the magnetic field. In the question stem we are told to assume that all particles in the
chamber have the same magnitude of charge. From the equation we see that the radius of the circle is directly
proportional to the momentum of the particle. So we can gauge the momentum of the particles in this bubble
chamber solely by the curve of their tracks. Therefore, a particle traveling in a small circle will have a small
momentum. Now we need to relate the size of the circle in which a particle travels to the curved paths we see in the
chamber. The curved paths are just sections of a circle:
track
track
R

larger R: higher momentum

R

smaller R, lower momentum

So the track that is the most curved must come from the smallest circle. If we are looking for the particle
with the smallest momentum, we must look for the particle that left the most curved track. Looking at the diagram,
the upper branch of track c is definitely the most curved. Therefore it must be the track of the particle with the
lowest momentum in the chamber. The spiral occurs because the particle is losing energy, and therefore the
momentum and radius are getting smaller.
Passage V (Questions 30–35)
30.

C
The kinetic theory of gases attempts to explain the macroscopic properties of gases from a microscopic
perspective. The basic assumptions are:
(1) the gas particles are separated by distances much greater than their size;

(2) the only intermolecular forces occur during collisions, which are totally elastic;
(3) the gas particles are moving in a random fashion with a distribution of speeds; and
(4) the average kinetic energy of a gas is directly proportional to the absolute temperature.
You should be aware that since the AVERAGE kinetic energy is constant at a particular temperature so is
the AVERAGE speed. This does not imply that the speed of a particular molecule remains constant, it changes
with each elastic collision. It is the average speed that remains constant. Just one more aside about elastic
collisions, an elastic collision is one in which the sum of the kinetic energies of the colliding particles does not
change. Anyway, it is the third and fourth assumptions that allow us to pick C as the correct response to this


Kaplan MCAT Physical Sciences Test 6 Explanations

question. As just stated, it is the average speed of the collection of gas molecules that remains constant, not the
speed of the individual particles.
31.

B
Ideal gases obey the ideal gas law at all temperatures and pressures--ideal gases exist only in theory. Real
gases, on the other hand, follow the ideal gas law only at high temperatures and low pressures. At low
temperatures real gases exhibit a lower volume than Charles' law would predict. This is because attractive forces
exist between the molecules of a real gas; whereas, it is assumed that the molecules of an ideal gas do not interact
in any way. We know that these attractive forces exist because if we continue to lower the temperature of a gas, the
molecules associate with each other, which eventually leads to condensation. Ideal gases never condense; they just
continue to follow Charles' law. Another way in which real gases differ from ideal gases, is that at high pressures,
real gases occupy a larger volume than that predicted by Boyle’s law. This is because the ideal gas law does not
take into consideration the volume occupied by the molecules themselves. It assumes that the volume predicted is
the volume that the gas would occupy if the molecules themselves occupied no space.
You should remember from the passage that the Van der Waals equation of state attempts to account for
these deviations. Anyway, looking at the answer choices we can see that choice B, attractive forces, is the correct
response. But what if you didn't remember this--is there any way that the passage could help? The answer is yes.

The answer to this question can be found directly in the passage. Equation 2, the Van der Waals equation, predicts
the behavior of nonideal, or real, gases. How does this equation differ from the ideal gas equation? It has two
additional constants that account for the intermolecular forces of the gas molecules and for the "excluded volume"
of the gaseous molecules. Looking at the answer choices you can see that choice B, attractive forces, is the correct
response.
32.

D
The Gay-Lussac law states that at constant volume the pressure of an ideal gas is directly related to its
temperature. (This has been subsumed into the ideal gas law so there is no reason to memorize these separately.)
So, since pressure is directly related to temperature, whenever the temperature increases so goes the temperature.
With this in mind you can eliminate choices A and B since they state that the pressure decreases. Answer choices
C and D both state that the pressure increases, but choice C says that the density increases as well. If the volume is
constant, this can't be true! Choice D is, therefore, the correct response. If you didn't realize this, you could have
used your knowledge of the kinetic theory in arriving at choice D as the correct response. Choice D states that the
pressure increases because the frequency of collisions between gas particles and the container increase. This is
absolutely true.
33.

B
This has been discussed above in question 31.

34.

D
This question is testing you on the physical meaning of the constant b in the van der Waals equation,
which is Equation 2 in the passage. The passage tells you that the purpose of this constant is to account for the
volume that the gas molecules occupy, something that the ideal gas doesn't take into consideration. How can you
approximate the magnitude of this constant for a particular molecule? If you need to figure out how much space the
molecules take up in the gas phase, a good place to look is at the molar volume of the liquid state. In this state the

molecules are in extremely close contact. The molar volume can be found from dividing the molecular weight by
the density:
molecular weight ÷ density =

mass
mass volume volume
mass
÷
=
∞
=
mass
mole
mole volume mole

You can see from the table that the densities are all pretty close to each other so it is the molecular weight
that predominates. Ethanol, choice D, is by far the heaviest so it is the correct response.
35.

C

11


Kaplan MCAT Physical Sciences Test 6 Explanations

In a nutshell, the kinetic energy of a gas depends only on the absolute temperature, and the average speed
of a gas is inversely proportional to the square root of its molecular weight:
vave =

k
mw

According to the kinetic molecular theory, since there are no intermolecular interactions, the energy of a
gas is all kinetic. The higher the temperature, the more energetic is the gas, and at a specific temperature, the
lighter the gas, the faster is its average speed. The question stem tells you that gases X and Y are in a closed
container at a constant temperature. You are also told the gas Y is heavier than gas X. Let's look for the true
statement in the answer choices. Choice A states that both gases have the same average speed. This isn't true: since
Y is heavier than X, X has the greater average speed. Choice B states that gas X has a greater average kinetic
energy than Y. Not true, according to the kinetic molecular theory: all gases at the same temperature have the same
average kinetic energy. Choice C states that gas X has a greater average speed than that of gas Y. This is a true
statement. If two gases are maintained at the same temperature, the lighter of the gases will have the higher
average speed. Choice C is the correct response. Choice D is another choice that does not equate the kinetic
energies, so it is also incorrect.
Passage VI (Question 36–41)
36.

B
The question stem indicates that we only need to look at one experiment. First we have to figure out how
power is related to the parameters described in the passage. We are told that a constant force is applied to the
spaceship for the first 200 kilometers in Experiment 1. The formula power = force ∞ velocity may have escaped
your mind, but you should be able to see why it makes sense and ideally be able to drive it on your own. Power is
defined as energy per unit time, and energy in this case is the change in kinetic energy which is also the work done
by the engine. Work = force ∞ distance, and so:
power = (force ∞ distance) ÷ time = force ∞

distance
= force ∞ velocity
time

So for the first 100 km of Experiment 1, in which the ship experiences a constant force, the power is
directly proportional to the velocity. Also, a constant force results in a constant acceleration, and thus the velocity
is directly proportional to the time as seen in the kinematic equation v = at. Therefore, the power is also directly
proportional to the time. So the power also must increase linearly with time. Now we must look for a graph with a
line that tilts upward as we go farther away from zero on the positive side of the x-axis. Therefore choice B is the
correct answer choice.
37.

D
In this question we are asked to determine in which of the three experiments the spaceship achieves the
greatest maximum velocity. We can assume that the greatest maximum velocity in Experiments 1 and 2 will occur
before the spaceship hits the wall because the collision only serves to slow the spaceship down. So let's figure out
what determines the velocity. We are told that in each experiment there are stretches over which the spaceship is
accelerated. The velocity must increase during these stretches. We are also told that there are stretches over which
the engines remain off and that the velocity is constant over these stretches. So the stretches we should be
concerned with for this question are only those over which the ship is accelerating. Now to relate acceleration,
velocity, and distance, we can use the equation
vf2 – vi2 = 2ad
So to find the maximum velocity in each experiment, we need to consider the velocity at the beginning
and end of each stretch over which the spaceship is accelerated for each experiment. Since the spaceship starts at


Kaplan MCAT Physical Sciences Test 6 Explanations

rest, the velocity at the beginning of the first stretch is zero. The velocity at the end of the first stretch can be
determined from the equation
v12 = 2ad1
where v1 is the velocity at the end of the first stretch, a is the acceleration of the spaceship over the stretch and d1 is
the length of the first stretch.
Now we need to find the velocity at the end of the second stretch. What is the velocity at the beginning of

the second stretch over which the spaceship is accelerated? It must be the final velocity at the end of the first
stretch because we are told in the passage that between the two accelerated stretches in each experiment the
spaceship travels at a constant velocity. So we can use v1 for the velocity at the beginning of the second stretch. So
the velocity at the end of the second stretch v2 can be determined by the following formula
v22 = v12 + 2ad2
where d2 is the length of the second stretch. If we plug in the relation for v12 which we determined earlier we get
v22 = 2a(d1 + d2). Now v22 is the velocity at the end of the second stretch so it is also the maximum velocity. d1 + d2
is the total distance over which the spaceship was accelerated in each experiment. We know that the acceleration a
is the same in each experiment because F = ma, the same force F is applied in each experiment, and the same
spaceship is used so m is the same in each experiment. Therefore the maximum velocity the spaceship reaches in
each experiment will be the same if the total distance over which the spaceship is accelerated is the same in each
experiment. In Experiment 1 the spaceship is accelerated over two 200-kilometer stretches, so the total distance
over which the spaceship is accelerated is 400 kilometers. In Experiment 2 the spaceship is accelerated over a
stretch of 100 kilometers and then 300 kilometers, so the total distance over which the spaceship is accelerated is
also 400 kilometers. In the third experiment the distance which the spaceship travels with its engines shut off is
increased and the wall is removed. Removing the wall just increases the distance the spaceship can travel without
meeting any resistance. The distance over which the spaceship is accelerated is still 400 kilometers. So the
maximum velocity of the spaceship in all three cases is the same. Therefore, answer choice D is correct.
38.

B
The question asks us how we can determine if the fuel company charged the correct amount for the fuel
used in the experiments. So we need to find out from the passage when the fuel tank was filled. We also need to
figure out when fuel is used in each experiment. Looking to the passage, we see that the tank is refilled at the end
of each experiment. So we need to calculate how much fuel was used in each experiment. Now ask yourself when
fuel is used by the engines. Fuel is only consumed when the engines are on. So we only need to consider the
stretches in each experiment when the engines are on. Now we know which part of each experiment to consider,
but we still need to determine which parameters of each experiment to consider. From looking over the answer
choices, we see they involve time, force times distance, and mass times average speed. We know that fuel provides
the energy the engines need to propel the spaceship. Also, we are told in the passage that the fuel used is

proportional to the square root of the work done. This information is crucial. Now we can determine how much
fuel was used by considering the work done in each experiment. The work done in each case is the force times the
distance over which the force is applied. So by multiplying the force applied and the distance over which the
spaceship was accelerated, we can determine the energy used in the form of fuel. Thus, the correct answer is choice
B. Let's look over the other choices. Choice A suggest that we should consider the entire distance traveled. The
engines aren't running over the entire distance, and fuel is not used when the engines aren't running. So choice A
is incorrect because we are only concerned with the distances over which fuel is being consumed. Choice C claims
all we need to know is the time each experiment took. But we can't figure out the work done from the time alone,
we must know the magnitude of the applied force in addition to the time. Choice D is incorrect because the work
done cannot be determined from the mass times the average speed over the entire distance traveled. Again we are
only concerned with the distance over which the spaceship is accelerated, and we are looking for a quantity which
has units of work. Mass times average speed has units of momentum.
39.

C

13


Kaplan MCAT Physical Sciences Test 6 Explanations

We are asked to compare the time for which the spaceship was in contact with the wall in each
experiment. The fact that we are given average force and that we are looking for the time for which the wall and
spaceship are in contact should bring to mind impulse, because impulse equals the average force times time. The
impulse is the change in momentum the spaceship experiences after a collision with the wall. So the change in
momentum of the ship due to its collision with the wall equals the average force multiplied by the time. In the
question stem we are told that the average force imparted by the wall is the same in both experiments. So the
change in momentum of the spaceship must be directly proportional to the time of contact of the spaceship and
wall in the collision. So we need to compare the change in momentum of the spaceship in each experiment.
Momentum is mass times velocity. It is a vector quantity. Since mass has no direction associated with it,

the momentum of an object will have the same direction as the velocity of the object. To figure out the initial
momentum, we must first determine the velocity of the spaceship right before it hits the wall in each experiment.
We know the mass is the same in each experiment because the same ship is used. To find the velocity before the
ship hits the wall, we can use the formula vf2 = vi2 + 2ad, where a is the acceleration and d is the distance over
which the ship is accelerated. The spaceship experiences the same constant acceleration in each experiment
because acceleration equals force over mass, and the force and mass associated with the spaceship are the same in
both experiments. Let d1 be the first distance over which the spaceship is accelerated in either experiment. In each
experiment the spaceship starts from rest, so the initial velocity is zero. Therefore the speed at the end of the first
distance is given by v12 = 2ad1. Let d2 be the second distance over which the spaceship is accelerated in either
experiment. The speed at the end of the second distance is given by v22 = v12 + 2ad2. Now plugging in the
previously determined value of 2ad1 for v12, we determine that v22 = 2ad1 + 2ad2, or v22 = 2a (d1 + d2). In
Experiment 1, d1 + d2 = 200 + 200 or 400 kilometers. In Experiment 2, d1 + d2 = 100 + 300 or 400 kilometers. So
d1 + d2 = 400 kilometers in both experiments. Therefore the velocity of the spaceship when it reaches the wall must
be the same in each experiment. So the momentum of the spaceship before it hits the wall is the same in each
experiment.
Now we must consider the momentum of the spaceship after the collision with the wall. So we need to
determine the velocity of the spaceship after the collision with the wall in each experiment. Well looking back to
the passage, we find that we are given information about the spaceship after the collision. We are told that it takes
less time for the spaceship to coast back to the starting point in the second experiment. Since the distance is the
same in each experiment, this implies that the velocity and therefore the speed with which the spaceship travels
after the collision is greater in the second experiment. So the momentum of the spaceship after the collision is
greater in the second experiment. Now to figure out the change in momentum we subtract the momentum of the
spaceship before it hits the wall from the momentum of the spaceship after it hits the wall. Don't forget to take into
consideration the direction of the velocity. It changes after the ship hits the wall. So if we take the direction the
ship travels when it is moving towards the wall as positive, then the direction it travels in after it hits the wall is
negative. But when we subtract this negative vector, it becomes positive. So the change in momentum is the mass
times the initial speed plus the mass times the final speed. Therefore the change in momentum is greater in the
second experiment because the mass times the speed after the collision in this experiment is greater than in the
first experiment. Since the change in momentum equals average force times time and the average force is the same
in each experiment, the change in momentum is proportional to time. Therefore the spaceship must be in contact

with the wall for a greater amount of time in Experiment 2 because the change in momentum due to the collision is
greater in Experiment 2.
40.

A
The stem of this question might seem a little vague. So it is important to quickly read over the answer
choices to get a sense of what factors must be considered. From reading the answer choices, one can determine that
getting the spaceship to stop at the starting line will involve the application of a force. So to answer this question
we must go back to the passage and find out what is the last thing mentioned about Experiment 1. The last thing
we are told about experiment one is that the spaceship collides into the wall and drifts back to the starting line. We
need to have the spaceship come to a stop at the starting line. This will happen if the spaceship experiences a
deceleration on its way back. A deceleration must be caused by some external force.
Now we have to figure out if there is any force that the spaceship experiences on its way back to the
starting point. Since it is in outer space, we know there is no air resistance. The term outer space also implies that
there are no planets or stars nearby, therefore there will be no gravity due to nearby stars or planets. In the first
paragraph of the passage, we are told that the force of gravity between the spaceship and wall is negligible. So the


Kaplan MCAT Physical Sciences Test 6 Explanations

spaceship will not experience an external force when it coasts back to the starting line. But, what about the
collision with the wall? Didn't that cause the spaceship to decelerate? Initially it did, but once the spaceship left the
wall, it was traveling at a velocity which was no longer affected by the wall. Therefore the spaceship must be
coasting back toward the starting line at a constant velocity. This means that the spaceship will drift back right
past the starting line. So in order to have the spaceship come to rest at the starting line, we must apply a force. In
order to slow the spaceship down, a force must be applied opposite to the direction in which the spaceship is
traveling. Therefore the force must be applied in the direction pointing from the starting line toward the wall.
Now that we've analyzed the situation, let's look over the answer choices to see if any concur with our
reasoning. Answer choice A states that a force must be applied in the direction pointing from the starting line to
the wall. This agrees with our analysis so it is probably the correct answer. But to make sure we are not

overlooking anything, let's look at the other choices. Choice B states that a force must be applied in the direction
pointing from the wall toward the starting line. Well this means a force is applied in the same direction in which
the spaceship is traveling. This will accelerate the spaceship causing it to go past the starting line at a greater
speed. So choice B is incorrect. Answer choice C states that no force need be applied. Well, we figured out that the
spaceship is moving at a constant velocity, and according to Newton's first law it will keep moving until an outside
force is applied. So this choice is also incorrect. Choice D states that a force perpendicular to the direction the
spaceship is traveling in must be applied. Well this will only serve to change the direction in which the spaceship
is traveling. A force must be applied in the opposite direction to that in which an object is traveling in order to
slow it down. So answer choice D is incorrect. Therefore, choice A is the correct answer choice.

41.

A
All you need to do to answer this question is rearrange the given equation solving for Mi/Mf and relate it
to one of the graphs in the answer choices. First divide each side by u to obtain:
ln

Mi (vf – vi)
=
u
Mf

Now to get rid of the ln, you must raise each side to be a power of e. (Remember elnx = x.) So we get Mi/Mf = e to
(v – v )
the [(vf – vi)/u] power. This implies that Mi/Mf increases exponentially as the quantity f i increases. Now look
u
at the answer choices to determine the axes of the graphs. We see that all of the choices have Mi/Mf as the y(vf – vi)
as the x-coordinate. So we are looking for a graph that shows exponential growth of the ycoordinate and
u
coordinate as the x-coordinate increases. The correct answer choice is therefore choice A. Choice B implies an

exponential decay of the y-coordinate as the x-coordinate increases. Choice C implies a linear increase of the ycoordinate with respect to the x-coordinate, and choice D implies that the y-coordinate increases and then
decreases as the x-coordinate increases.
Discrete Questions
42.

D
The key word here is “melting”. When ice is melting the temperature remains constant until all of the ice
has melted, after which the temperature of the melted water starts to rise. The question stem does not tell us how
much ice there is and whether the heat is enough to melt all of it. Yet choice D is the only possibility of the four
choices: it shows that the temperature is constant as the ice is undergoing the phase change. Choices A and B
cannot be correct since it shows the temperature falling in at least one portion of the graph. Choice C is also
incorrect because at least at the beginning, the temperature should be flat as the ice finishes melting.
43.

B
The capacitance of a parallel-plate capacitor is given by

15


Kaplan MCAT Physical Sciences Test 6 Explanations

C=

ε0A
d

where ε0 is the permittivity of free space and is a constant, A is the area of overlap of the two plates, and d is the
separation of the two plates. From this we can see that the capacitance is proportional to the area of the plates and
inversely proportional to the separation of the plates. Now in the question we are told that the separation of the

plates increases which means that the capacitance must decrease assuming that all the other variables remain
constant. This eliminates Roman numeral I and therefore answer choices A and C. That leaves us choices B and D.
So we know that Roman numeral III must be part of the answer, and we only need to look at Roman numeral II.
Now, we are also told in the question stem that the capacitor is isolated. Since it is isolated, charge cannot
flow or be generated. This tells us that the charge on the plates must remain constant. This eliminates Roman
numeral II, and therefore answer choice D. This leaves us with answer choice B which is indeed the correct
answer. Let's go on and see why Roman numeral III is part of the answer. The capacitance of a capacitor is related
to the voltage across the plates and the charge stored on the plates by the equation C = Q/V, where C is the
capacitance, Q is the charge stored, and V is the voltage across the plates. We have already established that the
charge on the capacitor remains constant, and we can therefore say that the capacitance of the capacitor is
inversely proportional to the voltage across the plates. In other words as the capacitance decreases, the voltage
across the plates increases. So Roman numeral III is correct, and choice B is the answer.
44.

A
The force of a spring is a conservative force which means that the total energy inherent in a spring system
is equal to the sum of its potential and kinetic energies. Remember, potential energy is the energy that's stored,
while kinetic energy is the energy of motion. Now, the potential energy of a spring can be expressed by the formula
U=

1 2
kx
2

where U is the potential energy, k is the spring constant specific for that spring, and x is the distance that the
spring is either stretched or compressed from its equilibrium length. As the distance of compression or stretching
increases, the x2 term increases so the potential energy must increase. Let's say we have a compressed spring with a
certain amount of stored potential energy. When this spring is released, its potential energy will be converted to
kinetic energy until the spring has reached its equilibrium length. It will then continue to stretch, converting the
kinetic energy back to potential energy. The spring will stretch until its displacement from the equilibrium length

is equal to the compressed spring's displacement from the equilibrium length.
Looking at the energies involved, it's easy to see that the potential energy would be a maximum at the
maximum compression, or the maximum stretch, because at either point the displacement x would be at a
maximum. Furthermore, potential energy would be a minimum at the equilibrium length because at this point x
would be equal to 0. Since we're dealing in a conservative system, it follows that the sum of the potential and
kinetic energies will be constant. Therefore, the maximum kinetic energy would have to occur at the x = 0 point, or
at the equilibrium length, and the minimum kinetic energy will occur at the maximum stretch and maximum
compression points. In this question we're told that the spring is compressed to its minimum length and that it is
not allowed to spring back, so this would be a point of maximum potential energy and minimum kinetic energy (=
0), and choice A is correct.
45.

A
This question tests how well you understand vector addition. If the two forces act in the same direction,
they will reinforce each other and then the resultant force will be 12 + 5 or 17 newtons. If they act in opposite
directions, they will detract from each other in which case the resultant force is 12 – 5 or 7 newtons. These
represent the maximum and minimum resultant forces that can occur. But any other value between 7 and 17 is also
possible -- these will occur when the forces are at an angle to each other. So we are looking for an answer between
7 and 17 newtons, inclusive. The only Roman numeral that satisfies this condition is Roman numeral II, 13
newtons.