Physics section test (2)

MCAT Section Tests
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PHYSICAL SCIENCES TEST 1 EXPLANATIONS
Passage I (Questions 1–8)
Passage I is about a set of experiments that were performed on the gemstone malachite. The questions require you to
understand stoichiometry, calculations of yield, and the nature of acids and bases. Take note of how each experiment
was performed and to the methods used to analyze the products.
1.
The correct answer is C. This question asks you to examine Experiments 1 and 2 and determine what gas was
collected. To answer it, you must look carefully at the starting material, malachite, and recognize the hints given to
you in the descriptions of the experiments. The first choice that can be eliminated is choice B, water vapor. Since the
gas was collected by water displacement, it seems unlikely that water vapor, which would probably condense as it
bubbled through the water, could be collected in this way. Next, you can eliminate oxygen and hydrogen since both
will support combustion. We were told that a burning splint inserted into the gas was extinguished almost
immediately, so the gas has to be something that will stop the burning process. Since carbon dioxide does this, it
seems like the best answer, and it is indeed the correct answer. So, the gas contained in the collection bottles was
carbon dioxide, choice C.
Another way to arrive at the answer, or at least would make an intelligent guess, is to realize from the
molecular formula of malachite that there are four elements that need to be accounted for: copper, oxygen, hydrogen,
and carbon. The copper and (part of) the oxygen appear in the black solid copper oxide. The condensation is
presumably water, which takes care of the hydrogen and (more) oxygen. Where does the carbon go? A natural guess
would be that it would be in the gas.
2.
The correct answer for is choice B. This question asks you to calculate the percent composition of oxygen in
malachite. This sort of calculation should be pretty routine. The percent composition of an element in a compound is
determined by dividing the total mass of that element in the compound by the molecular mass of the whole
compound, and then converting that value into a percent. The experiment outline tells us that the molecular formula
for malachite is Cu 2(OH)2CO3. That lets us know that there are 2 moles of copper, 5 moles of oxygen, 2 moles of
hydrogen, and 1 mole of carbon in every mole of malachite. When we multiply these numbers by the molar masses of
their respective elements and add the results together, we find that the molecular mass of malachite is 221 g/mol. 80
of those 221 grams can be attributed to oxygen. 80 is one-third of 240, and so is slightly more than one-third of 221.
Inspection of the answer choices tells us that only choice B could be correct.

3.
The correct answer to question 3 is D. This question asks you to classify the reaction that is occurring in
Experiment 1, in which malachite is heated vigorously. This heating produces a gas, condensation, and a black solid.
From the information given in the passage and interpreted above in the discussion for number 1, we know that the
reaction is of the form:
CuCO3•Cu(OH)2 ? 2 CuO + H2O + CO2.
Because this reaction involves only one reactant, which breaks down into several components, it can be classified as a
decomposition reaction, answer choice D. Neither acid nor base were added to the reaction vessel, choice A is
incorrect. Choice B, a double displacement reaction, is wrong as well, since it requires two reactants. Finally, choice
C says that the reaction outlined in Experiment 1 is a redox reaction. However, examination of the products reveals
that none of the atoms has changed oxidation state, so choice C is incorrect.
4.
Choice D is the correct answer to question 4. This problem asks you to calculate how many 100 mL
collecting bottles will be filled with gas at the completion of Experiment 1. 22 grams of malachite were added to the
reaction vessel, and we know from question 1, that the molecular weight of malachite is 221 g/mol. That means that
0.1 mole of malachite were reacted. We know that the gas produced was carbon dioxide. So, how many moles of
carbon dioxide are produced by the decomposition of one mole of malachite? From the balanced reaction written for
number 3 above, we know that the stoichiometric ratio is 1:1. So 0.1 mole of malachite produced 0.1 mole of CO2.
Now, we know that at STP, the conditions presented in the question, a mole of any gas takes up a volume of 22.4
liters. So, 0.1 moles of CO2 will occupy one tenth of that, or 2.24 liters. Since each of the collection flasks holds 100
mL, or 0.1 liters, the total number of bottles we'll need is 2.24 liters divided by 0.1 liters, which is 22.4 bottles.
However, since the question asks us how many bottles we'll need to collect the gas and we can't have only four tenths
of a bottle, it will take 23 bottles to hold all the gas. That is choice D.

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Kaplan MCAT Physical Sciences Test 1 Explanations

5.

For question 5, choice B is the correct answer. Answering this question correctly relies on your
understanding of Experiment 2. Experiment 2 involved a reaction between aqueous sulfuric acid and malachite. As a
result of the reaction, gas was evolved and a neutral solution was left in the reaction vessel. Which of the choices is
consistent with all that? Choice A says that an excess of sulfuric acid was added to the reaction vessel. If an excess of
sulfuric acid had been added, the final solution in the reaction vessel would have been acidic, so it's pretty easy to
eliminate this choice. Choice B says that an aqueous solution of copper sulfate was formed. The reaction between
malachite and sulfuric acid is an acid-base reaction with H2SO4 as the acid and malachite as the base. In an acid-base
reaction, you know that water and salt are formed as products and, in this case, the extra atoms are released as a gas.
So carbon dioxide, a gas, water, and the salt copper sulfate are formed, so choice B is correct. Now let's look at the
last two choices. Choice C say that an aqueous solution of copper hydroxide was formed. Well, we already said that
this is an acid-base reaction that ended up neutral when acid was added. Any hydroxide ions produced would go into
the formation of water, not copper hydroxide. Furthermore, copper hydroxide would be basic. Choice D says that the
pH paper turned blue when placed in the reaction solution, which is not consistent with the description that the
solution is neutral. (pH paper turns blue in basic solutions.) Hence, the correct answer to this question is B.
6.
The correct answer to question 6 is A. This question asks for the pH of the milky white mixture that is
formed when stoichiometric amounts of calcium hydroxide solution are added to the gas filled collection bottle. We
know that the gas is carbon dioxide. In order to answer this question, we must first determine what reaction is taking
place in Experiment 3. When water was added to the carbon dioxide in the bottle, and the bottle was shaken, the final
solution was acidic. This means that carbon dioxide reacted with water to produce an acidic solution, carbonic acid to
be precise. Therefore, when stoichiometric amounts of the aqueous base, calcium hydroxide, are added to the carbon
dioxide, the carbonic acid formed and the base added should neutralize each other. The products of this neutralization
reaction are calcium carbonate and water; calcium carbonate is insoluble in water and thus precipitates, accounting
for the milky white color, and the final solution would be neutral, as stated in choice A. Choice B, which says that the
pH would be acidic, would be correct only if calcium hydroxide were added at less than the stoichiometric amount,
meaning that there was excess carbon dioxide. Choice C, which says that the final solution would be basic, would be
correct only if excess calcium hydroxide solution were added to the carbon dioxide, or if the carbonate ion were left
in solution and could therefore hydrolyze. (This latter would occur if a different base were used and a soluble
carbonate were formed in the neutralization; for instance, if sodium hydroxide were added, soluble sodium carbonate
would form and the solution would then be basic-but clear and colorless.)

7.
The correct choice is B. This question asks you to predict the products of a reaction similar to the one
performed in Experiment 2, but using hydrochloric acid rather than sulfuric acid. From answering other questions in
this problem set, we already know that the reaction between malachite and sulfuric acid is an acid-base reaction. The
products formed in this reaction were carbon dioxide, water, and a copper sulfate salt. So, since hydrochloric acid is a
strong acid, just like sulfuric acid, the same sort of reaction should take place between HCl and malachite. However,
since there is no longer a sulfate anion in the reaction, but a chloride anion, the identity of the copper salt will change
to copper chloride. So the products of this reaction are carbon dioxide gas, water, and aqueous copper chloride salt.
This is choice B.
8.
The correct answer is B. This question tells us that four grams of copper oxide are formed from the reaction
in Experiment 1, and asks us to calculate the percent yield of the experiment. Percent yield is defined as the amount of
product collected divided by the theoretical yield of the reaction. Therefore, the first step is to calculate the theoretical
yield of copper oxide from the decomposition of malachite. The easiest way to do this is to convert everything to
moles. We discussed earlier that the 22 grams of malachite used in Experiment 1 represents 0.1 mole of malachite. 1
mole of malachite contains 2 moles of copper, so 1 mole of malachite will theoretically decompose to yield 2 moles
of copper oxide. Using this ratio, we know that the theoretical yield of copper oxide from this experiment is 0.2 mole
of copper oxide. How many moles were actually produced? From the Periodic Table and the molecular formula of
copper oxide, CuO, we know that the molecular weight of copper oxide is about 80 grams. The question tells us we
have 4 grams of copper oxide produced. The number of moles is therefore 4 divided by 80, or about .05 mole of
copper oxide. We were expecting 0.2 mole and we got 0.05 mole. Converting that into a percentage, we get 0.05/0.2
times 100%, or 25%. That's choice B. The other answers come from calculation mistakes like forgetting that there
were 2 moles of copper in 1 mole of malachite or trying to find the percent yield by directly dividing the 4 grams of
copper oxide produced by the 22 grams of malachite used.
Passage II (Questions 9–13)

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Kaplan MCAT Physical Sciences Test 1 Explanations

9.
The correct answer is choice C, strontium 94. Question 9 gives a typical uranium fission reaction and asks
you to identify one of the products. To solve this problem all you need to do is balance the mass numbers and atomic
numbers on each side of the arrow.
The atomic number is the subscript which represents the number of protons in the nucleus. On the left side
of the arrow you have 92 protons from uranium, and on the right side you have 54 protons in xenon, which leaves 92
minus 54, or 38 protons for our unknown product. Looking at the answer choices, we can eliminate A and D.
To distinguish between answers B and C, we need to look at the mass numbers. The mass number is the
superscript which represents the total number of protons and neutrons in the nucleus. The total mass number on the
left is given by 235 from the uranium and 1 from the neutron, which adds up to a total of 236. Xenon has a mass
number of 139, so adding this to the mass numbers of the 3 neutrons gives a mass number of 142. Therefore, our
product must have a mass number equal to 236 minus 142, which is 94. Now we know our product must have an
atomic number of 38 and a mass number of 94. So the correct answer is C, strontium 94.
10.
The correct answer choice is C, the binding energy is equal to the mass defect of the nucleus times the speed
of light squared. This is an outside knowledge question that requires you to understand the concepts of mass defect
and binding energy and how the two are related through Einstein’s equation E = mc2. As nucleons (neutrons and
protons) come together to form a nucleus, some of the mass is converted into energy and released, analogous to the
way that energy is released when a common chemical bond is formed. Because of this conversion, the mass of the
nucleus formed is less than the sum of the masses of the constituent nucleons. The difference between the two mass
values is known as the mass defect, and is related to the energy released by the equation E = mc2. Because c is such a
large number, a small mass defect corresponds to a significant amount of energy. The energy released is called the
binding energy. Its name implies that this binding energy is the energy holding the nucleus together, which is a valid
description because this same amount of energy needs to be supplied to break apart the nucleus into individual
nucleons, again analogous to the fact that breaking a bond is an endothermic process that requires energy input. The
missing mass will be restored through this input of binding energy.
11.
The correct answer is choice D, not deflected at all. Because the neutron is electrically neutral, it does not
interact either with the magnetic field or the electric field. You can see this if you look at the equations for the forces

due to an electric field and a magnetic field. Both forces depend on the charge on the particle. So if the charge of the
particle is zero, the force must also be zero, and if no forces act on the particle it will be undeflected.
12.
The correct answer choice is B, 9.5 hours. This question can be answered with a basic understanding of what
a half-life is. If we can figure out how many half-lives have passed in 19 hours, then we can divide to get the value for
the half-life itself. After one half-life, 50% of the sample will remain by definition. After two half-lives have passed,
50% ∗ 50% or 25% of the sample will remain. We don't need to go further since if 25% of the sample remains then
75% of the sample has decayed. Since two half-lives have passed in 19 hours, the half-life must be 19 divided by 2
giving 9.5 hours.
13.
The correct answer choice is D. In order for fission to occur, energy must be supplied to the nucleus to
initiate the reaction. This energy comes from the kinetic energy of the particle that is used to bombard the nucleus. In
order for this kinetic energy to be transferred to the nucleus, the particle must be able to get within close proximity to
it. Protons would not work well because being positively-charged themselves, they would experience repulsion from
the positively-charged nucleus.
Passage III (Questions 14–18)
14.
For question 14 the correct answer is C. You are told in the first paragraph of the passage that enzymes
catalyze reactions. The passage also tells you that catalysts increase the rate of reactions by decreasing the amount of
energy needed to bring the reactants to the transition state. This energy is called the activation energy. Like all
catalysts, enzymes do nothing to change the equilibrium of a reaction. The only thing that it changes is the rate of
reaction by making the transition states more accessible.
15.
For question 15 the correct answer is A. If you know the equation for a line, which is required knowledge
for the MCAT, you shouldn't have had any problem with this one. Looking at Equation 1, since it's already in linear

3


Kaplan MCAT Physical Sciences Test 1 Explanations

form, you should be able to see that the slope of the line is equal to KM over Vmax and that the y-intercept is equal to
1/Vmax. Now, if you examine the plot for competitive inhibition in Figure 2--the top one--you can see that the yintercept, 1/Vmax, is the same for both the inhibited and the uninhibited reactions. So, in this case, the slope is directly
proportional to the KM for both the uninhibited and the inhibited reactions. Since the slope is greater for the inhibited
reaction, it has the greater KM and choice A is the correct response. Choices C and D can be eliminated because you
know that the Vmax,'s for both the inhibited and the uninhibited reactions are the same. Choice B can be eliminated
because, as you just determined, the KM increases, not decreases. Again, choice A is the correct response.
16.
For question 16 the correct answer is D. According to the passage, Table 1 contains data from experiments
run with and without a noncompetitive inhibitor. The first thing you need to do is look at the plot in Figure 2 that
deals with noncompetitive inhibition, the lower one. From this plot you should be able to see that the intercept with
the y-axis is equal to 1/Vmax, and that the reaction with inhibitor--the dashed line--has a smaller Vmax, than the one
without inhibitor. We now know that the Vmax, decreases in the presence of a noncompetitive inhibitor, but what
happens to KM ? Solving Equation 1 for 1/[S] by setting y = 1/V = 0, you can see that the intercept with the x-axis is
equal to -1/KM , and since both lines in figure two have the same x-intercept, KM is the same with and without
inhibitor. So, with inhibitor, Vmax, decreases, but KM stays the same. Choice D is the correct response.
17.
For question 17 the correct answer is C. To answer this question, you didn't need to know much science
since the important information is in the passage. The passage describes the two types of inhibition, telling you that
competitive inhibitors bind to the same site on the enzyme that the substrate does and can be overcome by increasing
the substrate concentration. In contrast, noncompetitive inhibitors bind to different sites on the enzyme and cannot be
overcome by increasing the substrate concentration. Noncompetitive inhibitors work by binding to the enzyme and
changing its configuration, making it unable to recognize its substrate. It may be interesting to note that the reverse
has also been observed, there are some chemicals, called enzyme cofactors, that bind to a site on the enzyme and
change its configuration so that the enzyme can recognize its substrate. Anyway, the question tells you that the
inhibition on enzyme a caused by substance B is overcome by increasing the concentration of substance A, the
substrate of enzyme a. Therefore B is a competitive inhibitor, and molecules of substance B must bind at the catalytic
site. That information lets you eliminate choices B and D. However, the inhibition by substance D cannot be
overcome by increasing the concentration of substance A. Therefore, substance D must be a noncompetitive inhibitor
that binds at a different regulatory site. That makes choice C the correct answer.

18.
Choice B is the correct answer. As stated in the passage, KM represents the substrate concentration that lets
the reaction proceed at half its maximum rate. Anything that will affect the reaction rate or the amount of substrate
the enzymes can bind to at any one time will affect the KM of a given reaction. We can go through the choices and see
which could do that. The pH of the reaction will affect this because enzymes are proteins and the pH of a solution can
interfere with the secondary and tertiary structure of a protein, a change in pH can change the conformation of the
enzyme, making the enzyme less likely to recognize the substrate. Since the catalyst is thus less effective, the rate of
reaction will change, which also changes the KM. Anyway, since roman numeral I appears in all the answer choices,
we knew it had to be right. Roman Numeral II also appears in all the answer choices, so it must also be right. The
temperature also affects the enzyme conformation and the rate of reaction. Since the rate of reaction changes with
temperature, KM changes, too. So that leaves roman numerals III and IV. The KM is also affected by the concentration
of enzyme because the greater the number of enzyme molecules in the reaction mixture, the greater the degree of
catalysis. The greater the catalysis, the faster the reaction, so the KM will also change. Since III is correct, we know
that the answer is either B or D. So, looking at statement IV, we know that the answer is choice B. KM is a constant
that describes a substrate concentration at a particular pH, temperature, and enzyme concentration. No matter what the
substrate concentration in the reaction mixture is, the KM will be the same if the other factors are held constant. So
the correct answers are I, II, and III, which corresponds to answer choice B.
Discrete Questions
19.
The correct answer choice is C. Since the change in temperature can be calculated from the equation E =
mc? T, where m and c are constant, the rate at which the temperature increases will be dependent on the rate at which
heat is supplied to the system. The immersion heater is rated at 2000W, which means that it delivers 2000 J every
second. This rate is constant, and therefore the temperature will be increasing at a constant rate, i.e., it will increase
linearly. The choice is then between A and C, and we can see that C is correct because the initial temperature is given
as 30 °C in the question.

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Kaplan MCAT Physical Sciences Test 1 Explanations

It may be worth pointing out that since we are not given the specific heat of water, there is in principle the
possibility that the heat supplied may have been sufficient to cause the water to undergo a phase change (to boil),
which would show up as a flat portion of the temperature versus time graph. Since this is not given as a choice, we do
not have to worry about this scenario. And in fact, the specific heat of water is high enough (about 4000 J/kg.K) such
that this is not the case; given the numbers in the problem, the temperature after 100 seconds can be verified to be 50
°C.
20.
For question 20 the correct answer is D. Elements with partially filled d subshells are located in the block of
transition elements of the Periodic Table. Boron and silicon, choices A and B, are both p-block elements. Their
valence electrons are in the p subshell. Since the valence shell is the only one that is partially filled in ground state
elements, these two choices cannot be correct. Magnesium, choice C, is an s-block element on the left of the periodic
table. Its valence electrons are in the s subshell, so this too is an incorrect answer. That leaves chromium, choice D.
Chromium is element 24, the fourth element in the fourth period transition elements. It has four unpaired electrons in
its 3d subshell, so this is indeed the correct answer. If you did not remember how to find this information from the
structure of the Periodic Table, you could have written out the electron configurations of all the choices. It would
have been a bit more time consuming, but equally effective. For the record, you can quickly determine that any
element with an atomic number lower or equal to 18 will not even begin to fill the 3d subshell.
21.
The correct answer choice is D, 9F over 8. This question requires you to have a good understanding of
Coulomb's law, and the flow of charge. Originally the two charges of Q and 2Q are separated by a distance d, and
exert a force of F on one another. In terms of Coulomb's law, F is equal to Coulomb's constant k times Q times 2Q
2kQ2
. Now the charges are brought together and made to touch one another.
all over d squared which is equal to
d2
What happens to the charge on the two spheres? The total charge on the two spheres was Q plus 2Q or 3Q. When the
spheres, which have the same size, touch, the charge becomes evenly distributed. Each sphere now carries a total
charge of 3Q over 2. The spheres are then separated again by a distance d, but because they are now carrying the new
charges the force between them has changed. We'll call this new force F 1. From Coulomb's law, F 1 is equal to k, times

9kQ2
9kQ2
2kQ2
F
3Q over 2, times 3Q over 2, all over d squared, which equals
. Dividing F 1 =
by F = 2 , we find that 1
2
2
4d
4d
d
F
9
9F
= . Therefore, F 1 = , which is answer choice D.
8
8
22.
The correct answer to this question is choice D. When you are asked to match up electronic configurations,
the easiest thing to do is count the number of electrons. After all, since orbital filling is the same, a certain number of
electrons will always have the same electronic configuration around a nucleus. So how many electrons are there in the
configuration described in the question? Adding the superscripted numbers which show the number of electrons in
each subshell, we see that there are 18 electrons here. So let's look at the choices. Choice A, argon, has an atomic
number of 18, so in its ground state, there are 18 electrons and it will have the indicated configuration. Chlorine has
an atomic number of 17, but we need to add one for its negative 1 charge. That brings the number up to 18 and choice
B is also wrong. Choice C, K+, is also wrong, since its atomic number minus one electron for its +1 charge also
equals 18. So all three of these species have the described electronic configuration. That leaves Ga3+ as the answer
we're looking for. It's atomic number of 31, minus 3 for its +3 charge gives 28, so there are 28 electrons in Ga3+. In
fact, its electronic configuration is the same as the one in the question, except that it also has a full 3d subshell,

adding a 3d10 to the end of the question's configuration.
23.
The correct answer choice A, 5 ∗ 106. To do this question, we must first find the period of the wave, which
is the time it takes for a wave to complete one cycle. However, in order to find the period we must know the
frequency of the wave since the period is one over the frequency. In the question we are given the wavelength and the
speed of light, and the equation that relates these two values to the frequency is f = Error!. So putting in values
gives 3 ∗ 108 m/s divided by 600 ∗ 10–9 m, which gives a frequency of 5 ∗ 1014 Hertz. This is equivalent to a period
of 2 ∗ 10–15 seconds. If one wavelength has a period of 2 ∗ 10–15 seconds, there will be 0.01 ∗ 10–6 over 2 ∗ 10–15
complete waves in 0.01 µs, or 5 ∗ 106, which is answer choice A.
Passage IV (Questions 24–31)

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Kaplan MCAT Physical Sciences Test 1 Explanations

The correct answer is choice B, 1:3. To find the distance traveled in the first four seconds, we can use the
1
equation x - x0 = v0t + at2, where the left-hand side is the distance traveled. Since the car starts from rest, its initial
2
1
velocity, v0, is zero. The distance traveled in the first four seconds is therefore a (4)2, or 8a. There are two ways to
2
find the distance traveled in the next four seconds: one can determine the velocity at the end of the first four seconds
1
via the equation v = v0 + at, and use this value as the new v0 in the equation x - x0 = v0t + at2 again with t = 4 s; or,
2
alternatively, one can just determine the total distance traveled at the end of 8 seconds, and subtract 8a from that. We
shall try this latter approach. Using the same equation, again with v0 = 0, but this time with t = 8 s, we find that the
1

distance traveled is a (8)2, or 32a. The distance traveled in the last four seconds is therefore 32a - 8a = 24a, and the
2
ratio of the distances is 8a:24a = 1:3.

24.

25.
The correct answer choice is B. This problem requires two formulas. The first is F = ma. Since the mass of
the car remains constant, if the net force doubles then so will the acceleration. The second formula states when
acceleration is constant, the change in velocity equals acceleration times time: v = v0 + at. The change in velocity of
28 m/s is the same in both cases. Therefore, if the acceleration doubles, the time will only be 1/2 the old value.
26.
The correct answer choice is C, increase if the mass of the car increases. The maximum force that an object
can sustain without sliding is equal to µs times the normal force. In the present case the normal force is just the weight
of the car, mg. Therefore, we can immediately rule out answer choices A and B since this force does not depend on
velocity. Since the force is directly proportional to the mass of the car, it will increase if the mass increases. Once
again, C is the correct choice.
mg
. This question requires you to know that pressure is force divided by the
4A
area over which the force is applied. In this case, the force is the weight of the car, W = mg, and the area over which it
is applied is the contact patch of the tires. It is important to note that this weight is equally distributed between the
four tires; so the force on each of the contact patches is one quarter of the total weight of the car, mg/4. Since the area
mg
of the contact patch for each tire is A, the pressure is .
4A
27.

The correct answer choice is C,

28.
The correct answer is choice A. There are two forces acting on the car along the direction of the plane: the
force acting down the slope due to the weight of the car, and the force of friction acting in the opposite direction to
the motion of the car. First, let us consider the force due to friction. This equals the normal force on the car exerted by
the surface of the slope multiplied by the coefficient of friction, where the normal force is equal to the weight of the
car times cosθ:

normal force

friction

θ

mg

θ

This friction force remains constant since the angle of the slope is constant and the coefficient of friction remains
constant. Now let us consider the force down the slope. This is equal to the weight of the car times sinθ, and this
value also remains constant since the angle of the slope is constant. We have already said that the total force acting on
the car on the slope is equal to the force due to the weight of the car minus the force due to friction, and since both of
these values remain constant the force acting on the car must be constant. This implies that the acceleration of the car
must be constant, and the only graph where this is the case is graph A. So the correct answer choice is A.

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Kaplan MCAT Physical Sciences Test 1 Explanations

29.

The correct answer is 4.1 seconds, choice B. As long as the tires do not start to slide or skid, the force of
friction is static friction. The question asks for the least time, which means that the static force of friction is at its
maximum value, which is µs times N, where µs is the coefficient of static friction and N is the normal force. The
normal force N is equal to the weight of the car, which is m ∗ g or 1,000 kg ∗ 10 m/s2 which is 10,000 N. Note that
we approximated g to be 10 m/s2, to make the calculations more straightforward. The coefficient of static friction, µs,
is given as 0.6, so the maximum static friction force is 0.6 ∗ 10,000 N which equals 6,000 N. From Newton's second
law, the acceleration is the force, 6,000 N, divided by the mass, 1,000 kilograms, or 6 m/s2. Because the car is
stopping, the acceleration is opposite to the direction of the motion. If we take the direction of motion to be in the
positive direction, then the velocity vectors are in the positive direction and the acceleration vector is in the negative
direction. The equation that we need to use to find the stopping time is v = v0 + at. Rearranging this we get that the
time equals the final velocity minus the initial velocity divided by the acceleration. Substituting our values into this
(0 – 24)
= 4 seconds. Since we made an approximation earlier, we must pick the
equation, we get that the time equals
–6
closest answer and this is answer choice B, 4.1 seconds.
30.
The correct answer here is choice A. This question requires an understanding of the principle behind the
hydraulic lever—namely, Pascal’s principle. This principle states that the pressure exerted on a fluid is transmitted
throughout the entire fluid. Since pressure is force/area, if the pressure were to remain constant over the entire fluid,
this would have to imply that a small force exerted over a small area would create a large force over a larger area.
This is how one could exert a large force somewhere else by exerting a small one locally. The force required to stop
the wheel from turning is expected to be quite large; this is how the mechanism described in the passage is helpful: the
driver exerts a force f on the brake pedal piston which has an area a. The pressure is thus f/a. This pressure is
transmitted to the brake shoe piston which has an area A. The force at the brake shoe piston F thus satisfies:
F f
= , or
A a
A
F=f∗ .

a
If we want to magnify the force (have F > f), we would want A to be greater than a, i.e. to have the brake
shoe piston larger than the brake pedal piston.
Choice B is incorrect because there is no evidence that the coefficient of friction is dependent on the area of
the brake shoe piston. Choice C is incorrect because a larger brake pedal piston would actually decrease the pressure
given the same force. Choice D is incorrect because the brake pedal piston is not actually in direct contact with the
brake shoe piston.
31.
The correct answer choice is B, 4900 N. This question is based on Archimedes' principle which states that if
a body is partially or totally immersed in water, it will be buoyed up by a force equal to the weight of water that it
displaces. In this problem we are asked to find the apparent weight of the car given that it displaces 0.5 m3 of water
when it rolls into a lake. The apparent weight of the car is the weight of the car in air minus the buoyant force due to
the weight of water displaced. To calculate the buoyant force, we need to find the mass of the water displaced, which
we can work out from the equation: ρ = m/V, where ρ is the density, m the mass, and V the volume. We know that the
density of water is 1000 kg/m3 and that the volume of water displaced is 0.5 m3. So putting these numbers into the
equation and rearranging, we get that the mass of water displaced is 500 kg. This is equivalent to a buoyant force of
4900 N since the buoyant force is the mass of water displaced times the acceleration due to gravity. The weight of the
car in air is 1,000 kg times 9.8 m/s2, which equals 9,800 N. Therefore, the apparent weight of the car is 9800 – 4900,
or 4900 N which is answer choice B.

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