7Work, energy, and momentum test w solutions

PHYSICS TOPICAL:
Work, Energy and Momentum
Test 1
Time: 22 Minutes*
Number of Questions: 17

* The timing restrictions for the science topical tests are optional. If you
are using this test for the sole purpose of content reinforcement, you
may want to disregard the time limit.


MCAT

DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage
preceding each group of questions. Study the passage,
then select the single best answer to each question in the
group. Some of the questions are not based on a
descriptive passage; you must also select the best answer
to these questions. If you are unsure of the best answer,
eliminate the choices that you know are incorrect, then
select an answer from the choices that remain. Indicate
your selection by blackening the corresponding circle on
your answer sheet. A periodic table is provided below for
your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1
H
1.0

2
He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8
O
16.0

9
F

19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S
32.1

17
Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23
V
50.9

24
Cr

52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga
69.7

32
Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38
Sr
87.6

39
Y

88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd
106.4

47
Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53
I
126.9

54
Xe

131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re
186.2

76
Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82
Pb
207.2

83
Bi

209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Rf
(261)

105
Ha
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr
140.9

60
Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66
Dy
162.5

67
Ho

164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91
Pa
(231)

92
U

238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es
(252)

100
Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

GO ON TO THE NEXT PAGE.

2

as developed by


Work, Energy and Momentum Test 1
Passage I (Questions 1–4)
Two stars bound together by gravity orbit each other
because of their mutual attraction. Such a pair of stars is
referred to as a binary star system. One type of binary
system is that of a black hole and a companion star. The
black hole is a star that has collapsed on itself and is so

massive that not even light rays can escape its gravitational
pull. Therefore, when describing the relative motion of a
black hole and a companion star, the motion of the black
hole can be assumed negligible compared to that of the
companion.
The orbit of the companion star is either elliptical
with the black hole at one of the foci or circular with the
black hole at the center. The gravitational potential energy
is given by U = –GmM/r, where G is the universal
gravitational constant, m is the mass of the companion
star, M is the mass of the black hole, and r is the distance
between the center of the companion star and the center of
the black hole. Since the gravitational force is
conservative, the companion star’s total mechanical energy
is a constant of the motion. Because of the periodic nature
of the orbit, there is a simple relation between the average
kinetic energy <K> of the companion star and its average
potential energy <U>. In particular, <K> = –<U/2>.
Two special points along the orbit are singled out by
astronomers. Perigee is the point at which the companion
star is closest to the black hole, and apogee is the point at
which it is farthest from the black hole.

3 . Which of the following prevents the companion star
from leaving its orbit and falling into the black hole?
A.
B.
C.
D.

The centripetal force
The gravitational force
The companion star’s potential energy
The companion star’s kinetic energy

4 . The work done on the companion star in one complete
orbit by the gravitational force of the black hole
equals:
A . the difference in the kinetic energy of the
companion star between apogee and perigee.
B . the total mechanical energy of the companion star.
C . zero.
D . the gravitational force on the companion star
times the distance that it travels in one orbit.

5 . For a circular orbit, which of the following gives the
correct expression for the total energy?
A. −
B.

mv 2

C. −
1 . At which point in an elliptical orbit does the
companion star attain its maximum kinetic energy?
A.
B.
C.
D.

Apogee
Perigee
The point midway from apogee to perigee
All points in the orbit, since the kinetic energy is
a constant of the motion

D.

1
mv 2
2

GmM
r

GmM
2r

6 . What is the ratio of the acceleration of the black hole
to that of the companion star?
A.
B.
C.
D.

M/m
m/M
mM/r
1/1

2 . For circular orbits, the potential energy of the
companion star is constant throughout the orbit. If the
radius of the orbit doubles, what is the new value of
the velocity of the companion star?
A . It is 1/2 of the old value.
B . It is 1/ 2 of the old value.
C . It is the same as the old value.
D . It is double the old value.

KAPLAN

GO ON TO THE NEXT PAGE.

3


MCAT
Questions 7 through 12 are
NOT based on a descriptive
passage.
7 . What is the power required to accelerate a 10 kg mass
from a velocity of 5 m/s to a velocity of
15 m/s in 3 seconds?
A . 33W
B . 167W
C . 225 W
D . 333W

8 . A block of mass m slides down a plane inclined at an
angle θ. Which of the following will NOT increase the

energy lost by the block due to friction?
A.
B.
C.
D.

Increasing the angle of inclination
Increasing the distance that the block travels
Increasing the acceleration due to gravity
Increasing the mass of the block

9 . A block of mass m starts from rest and slides down a
frictionless semi-circular track from a height h as
shown below. When it reaches the lowest point of the
track, it collides with a stationary piece of putty also
having mass m . If the block and the putty stick
together and continue to slide, the maximum height
that the block-putty system could reach is:

1 0 . A boy hits a baseball with a bat and imparts an
impulse J to the ball. The boy hits the ball again with
the same force, except that the ball and the bat are in
contact for twice the amount of time as in the first hit.
The new impulse equals:
A.
B.
C.
D.

half the original impulse.

the original impulse.
twice the original impulse.
four times the original impulse.

1 1 . The power of solar radiation incident on a solar panel
is 40 kW. If the efficiency of the solar panel is 10%,
how much energy does the solar panel generate in 300
minutes?
A.
1.2 MJ
B.
72.0 MJ
C . 120 MJ
D . 720 MJ

1 2 . Two billiard balls undergo a head-on collision. Ball 1
is twice as heavy as ball 2. Initially, ball 1 moves
with a speed v towards ball 2 which is at rest.
Immediately after the collision, ball 1 travels at a
speed of v/3 in the same direction. What type of
collision has occurred?
A.
B.
C.
D.

inelastic
elastic
completely inelastic
Cannot be determined from the information given.

h

h
.
4
h
B.
.
2
C . h.
D . independent of h.
A.

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4

as developed by


Work, Energy and Momentum Test 1
Passage II (Questions 13–17)
Particle storage rings facilitate collisions between
electrons and positrons (positively charged electrons).
When electrons and positrons collide at high energies, they
can annihilate each other and produce a variety of
elementary particles, including photons. In these reactions,
momentum is always conserved. Powerful magnets are
placed at various points along the ring to create a force

directed toward the center, thereby guiding the particles in a
circular motion. In a scattering experiment, the particle
beams circle in opposite directions and collide head-on at
the interaction points, which are surrounded by particle
detectors. The particles in the ring are accelerated at one or
more points along the ring by an external electric field.
When charged particles accelerate, they radiate
electromagnetic energy. Radio frequency power is
continually fed into the storage ring to compensate for the
energy loss.
The reaction rate, R , is the number of particles
scattered per second in the storage ring. It is given by the
formula: R = L σ , where L is the luminosity and σ is the
cross-section of the reaction. The cross-section is a quantity
that depends only on the particular type of reaction being
considered. The luminosity contains all of the information
about the initial conditions for a reaction and is given by:
L=

N e− N e+
A

f,

where N e– refers to the number of electrons, N e+ refers to
the number of positrons, A is the cross-sectional area of
the storage ring, and f is the number of revolutions per
second made by the particles. Electron-positron reaction
rates for modern particle storage rings are typically on the
order of l0–3 s–1.

1 3 . What percentage of the work done on the circulating
particles is done by the magnetic field?
A . 0%, because the direction of the magnetic force is
perpendicular to the direction in which the
particles travel
B . 0%, because the magnetic field doesn’t exert a
force on the particles
C . 50%, because the magnetic field and the electric
field provide equal amounts of energy to the
particles
D . 100%, because the magnetic field is the source for
the centripetal force that accelerates the particles

1 4 . Which of the following would increase the reaction
rate in an electron-positron storage ring?
I. Decreasing the cross-sectional area of the
storage ring
II. Increasing the energy of the particles
III. Increasing the number of positrons in the
storage ring
A.
B.
C.
D.

I only
III only
II and III only
I, II, and III

1 5 . The work done on the particles by the gravitational
field of the Earth is not considered when figuring their
energy loss per revolution. This is because:
A . the gravitational force is perpendicular to the
gravitational acceleration.
B . the energy lost due to gravity is equal to the
energy gained from the magnetic fields.
C . the particles do not experience a significant
gravitational force.
D . the luminosity is not dependent on gravitational
acceleration.
1 6 . Two electrons with equal speed collide head-on at a
total energy of 180 GeV. If the speed of the first
electron after the collision is 0.9c, what is the speed of
the second electron after collision? (Note: The speed of
light in a vacuum is c = 3.0 × 108 m/s.)
A.
B.
C.
D.

0.45c
0.7c
0.8c
0.9c

1 7 . An electron and a positron are held in a storage ring,
and they each lose 260 MeV of energy per revolution
in the form of electromagnetic radiation. If the
frequency of revolution is 10,000 Hz, how much

power must be supplied to keep the total energy
constant at 180 GeV?
A.
B.
C.
D.

1.8 × 106 MeV/s
2.6 × 106 MeV/s
5.2 × 106 MeV/s
9.0 × 108 MeV/s

END OF TEST

KAPLAN

5


MCAT
ANSWER KEY:
1. B
6.
2. B
7.
3. D 8.
4. C 9.
5. A 10.

6

B
D
A
A
C

11.
12.
13.
14.
15.

B
B
A
D
C

16.
17.

D
C

as developed by


Work, Energy and Momentum Test 1
EXPLANATIONS

Passage I (Questions 1—6)
1.

B
In the passage we are told that the total mechanical energy of the companion star is constant as it orbits around the
black hole. The total mechanical energy is the sum of the gravitational potential energy and the kinetic energy. We do not really
know anything directly about the kinetic energy in this case, except that it equals mv2/2. We are, however, given a formula for
the gravitational potential energy of the companion star. Since the mechanical energy is constant, the point at which the kinetic
energy is maximum must also be the point at which the potential energy is minimum. From the more familiar case of dropping
a ball on Earth, we know that the potential energy is lower when the height is smaller. We may thus expect that the potential
energy is minimum when the star is closest to the black hole. We can confirm this by using the formula given in the passage:
U = – GmM/r. Since G, m, M, and r are all positive quantities, U is always going to be negative and has a maximum value of
zero when the two are infinitely far apart. As r decreases, U becomes more and more negative. It follows, then, that the point at
which the potential energy is minimum is the point of closest approach to the black hole, i.e. the perigee. The perigee, then, is
also the point where the kinetic energy is the greatest.
2.

B
The question stem states that the gravitational potential energy for circular orbits is constant. This implies that the
kinetic energy is also constant (from the conservation of total mechanical energy). In the last line of the second paragraph of the
passage, we are given a relationship between the average value of the potential energy and the average value of the kinetic
energy. Since in this case the two are constants, their average values are identical to their actual values. Therefore, for a circular
orbit, the kinetic energy, K, is : K = – U/2. The potential energy, U, is equal to – GmM/r. If the distance r is doubled, then, the
absolute value of the potential energy is reduced in half, and thus the kinetic energy will also be cut in half. (Recall from our
discussion for the question above that the potential energy is negative; a reduction of its absolute value thus leads to a less
negative number or a higher potential energy, as is reasonable when one separates two objects that are attracted to each other.)
Since the kinetic energy is proportional to the velocity squared, as the kinetic energy becomes 1/2 its original value, the
velocity must become 1/ 2 its original value to satisfy the definition of kinetic energy mv2/2.
3.

D
Let us examine each choice and try to eliminate those that seem incorrect. Choice A states that the companion star
does not fall into the black hole because of the centripetal force. The centripetal force on the star is an attractive force directed
towards the black hole. It is provided by the gravitational attractive force between the two objects and hence, if anything, it is
what leads us to expect the two to come together.
Choice B, the gravitational force, is, as just mentioned, what is providing the centripetal force. This is one way that
we could have eliminated these two choices: since they are essentially the same thing, neither of them can be correct!
Choice C states that the companion star’s potential energy keeps it in orbit. Does this make sense? Think about a ball
that you hold in your hand. It has some potential energy by virtue of the fact that you are holding it above the ground, but that
potential energy does not keep the ball from falling: it is your hand that does this. If you let go, the potential energy is
converted to kinetic energy as the ball goes speeding towards the ground. Analogously, then, we would not expect potential
energy to be the agent responsible for keeping the star in orbit here.
Choice D states that the companion star does not fall into the black hole because of its kinetic energy. This makes
sense because its tangential velocity is what keeps it in orbit and kinetic energy is certainly related to tangential velocity. The
gravitational force acts as a centripetal force which keeps redirecting the star so it doesn’t fly off into space. Why would it fly
off into space? Because it has kinetic energy from the velocity in the tangential direction. A stable orbit occurs because of the
balance between the kinetic energy and the gravitational attraction. If the kinetic energy were not large enough, the star would
fall or spiral into the black hole.
4.

C
There are several ways that one could have obtained the answer. One is by the work-energy theorem: The net work done
on an object is the change in its kinetic energy. In the case of a complete orbit, this is the kinetic energy the star has at the end
of one round of the orbit minus the kinetic energy it had at the beginning (at the same point in space). Since potential energy is
dependent only on position, we know that the star would have the same potential energy after one orbit that brings it back to its
starting point. Since total mechanical energy is conserved, the fact that the potential energy is the same before and after implies
that the kinetic energy is also the same before and after. The change in kinetic energy after one orbit is therefore zero. From the
work-energy theorem, then, we know that the work done on the star has to be zero as well.

KAPLAN

7


MCAT
In addition, the passage tells us that the gravitational force is conservative (which is something you might already
know anyway). One of the properties of a conservative force is that the work it does is path-independent, which in turn implies
that the work it does after one loop (or orbit) that starts and ends at the same place is zero. Since this force is the only one
acting on the star, the work done on the star is zero.
Beware of choice D, which states that the work done equals the gravitational force of the black hole on the companion
star times the distance traveled. This is very close to the definition of work and looks like an obvious choice. However, what
this formulation leaves out is the angle between the force and the direction of travel. The formula for work done by a force is: W
= Fd cosθ. In this case the force is directed towards the center (or focus of an ellipse) and the direction of travel is tangent to the
orbit. Therefore, for an elliptical orbit, the angle between the two is never zero (the cosine is never 1):

direction of
star
travel
gravitational force
black hole

The work done over any segment of the orbit, then, is never simply the product of the gravitational force and the
distance traveled. The symmetry of the problem, in fact, leads to net cancellation of the work done after one complete orbit.
Also, note that in the special case of a circular orbit, the two are always perpendicular, and thus the cosine of the angle, and
the work done, are zero all throughout the orbit.
5.

A
As discussed in the explanation to #2, the kinetic and potential energies are both constants of the motion for a circular
orbit because r is constant. We can thus take out the average signs <> and write the equation given in the passage as K = – U/2.

Since the passage also gives us an expression for the gravitational potential energy, we are tempted to substitute that in and get:
E=K+U=–

U
GmM
GmM
GmM GmM GmM 2GmM
GmM
+ U = – (–
) + (–
)=
–
=
–
=–
2
2r
r
2r
r
2r
2r
2r

While this is certainly correct, this is not one of our answer choices: choice D does not have the right sign. We
therefore must find another way of expressing the quantity by using the fact that K = mv2/2. Since K = – U/2, U = – 2K:
E = K + U = K + (– 2K) = K – 2K = –K = –

1
mv2

2

You may be surprised that the total energy is negative, but remember that we can define the zero of potential energy
any way we want and in this case, as has been discussed, the potential energy is always negative. This is purely a matter of
convention (and convenience); since the kinetic energy has a magnitude half of that of the potential energy, it is not sufficient to
overwhelm the potential energy term and thus the total energy remains negative.
6.

B
From Newton’s second law, the acceleration an object undergoes is equal to the force it experiences divided by its mass.
In this particular case, M is the mass of the black hole while m is the mass of the star. The force each experiences is the same
in magnitude from Newton’s third law: The gravitational force the star exerts on the black hole is the same as the gravitational
force the black hole exerts on the star. The magnitude of the force is GmM/r2. The acceleration that the black hole undergoes is
therefore this force divided by M, or Gm/r2, while the acceleration the star undergoes is the same force divided by m, or GM/r2.
The ratio of the former to the latter is therefore Gm/r2:GM/r2 or m:M.
Note that we need not even have come up with the expression for the force explicitly: all we need is to be able to
recognize that the two forces will have equal magnitude from Newton’s third law. We can then write the ratio as:
ablack hole:astar =

F
F
1
1
:
=
:
= m:M
M m M m

where the last step can be made if one recognizes that a ratio is itself just a quotient.

8

as developed by


Work, Energy and Momentum Test 1
Independent Questions (Questions 7–12)
7.

D
To do this problem one needs to understand the relation between power, work and kinetic energy. Power is defined as
the amount of work done per unit time. The time in this case is given as three seconds, but we need to determine the amount of
work done. This can be obtained using the work-energy theorem, which states that the work done is equal to the change in
kinetic energy:
W = ∆KE = ∆ (

1
1
1
mv2) = m ∆(v2) = m (vf2 – vi2)
2
2
2

where v f is the final velocity and vi is the initial velocity. (Note: ∆(v2) is not the same as (∆v)2! I.e. we cannot write the
1
change in kinetic energy as m(vf – vi)2.) Substituting in numbers supplied by the question, we have the change in energy to
2
1

be equal to × 10 kg × (225 – 25) m2/s2 = 1000 J which is the work done. This divided by 3 seconds gives 333 J/s or 333 W.
2
8.

A
The energy lost to, or dissipated by, friction is the work done by friction, which is equal to the magnitude of the force
times the distance traveled. Anything that increases either of these two factors will increase the energy lost to friction. We are
thus looking for something that does not increase either of these two. Choice B can be immediately ruled out since it explicitly
suggests increasing the distance traveled. To arrive at the correct answer, we can almost rely on intuition alone. Increasing the
acceleration due to gravity (choice C) or the mass of the block (choice D) will increase the weight, i.e. the gravitational force on
the block. It is therefore “held more tightly” to the plane, and so one would expect a higher friction force. Increasing the angle
of inclination, on the other hand, decreases the component of gravity that is perpendicular to the plane, and so the block is not
held so tightly to the plane. This would lead us to expect that the friction force would be less, and so less energy will be
dissipated.
For a fuller understanding of the scenario, we can derive an expression for the energy lost because of friction in terms
of the quantities that appear in the answer choices. First, we notice that the force and the motion are in opposite directions and
so:
W = Fdcos180° = – Fd = ∆E due to friction
The negative simply implies that the work done by the friction force causes the block to lose rather than gain energy.
Since we are only interested in the magnitude of the energy lost, the negative sign can be ignored in our calculation. At this
point it is helpful to refer to a diagram with which you should already be familiar:

normal force = mgcosθ
friction = µN = µmgcosθ

mg θ
θ
If we let d be the distance traveled by the block as it slides down the inclined plane, then the energy dissipated by
friction is µmgdcosθ. With this expression in hand, we can evaluate the answer choices. Choice A states that increasing the
angle of inclination would not increase the amount of energy lost. The cosine of angle decreases if the angle increases within the

range of 0° to 90° (cos 0° = 1; cos 90° = 0). Since the angle of inclination is within this range, increasing the angle of
inclination would decrease the cosine term in the expression above, thus decreasing the amount of energy lost. It is therefore
true that increasing the angle of inclination would not increase the amount of energy lost, making choice A the correct response.
9.

A

KAPLAN

9


MCAT
The most important thing to realize in this problem is that energy is not conserved in the collision which is totally
inelastic since the two objects stick together. What is conserved is momentum, with which one can determine the velocities of
the bodies. Energy conservation, however, does come into play as potential energy is converted into kinetic energy up to the
point right before the collision, and vice versa after the collision. We shall see exactly how all these come together.
Initially we have a block of mass m at a height of h. The potential energy, which is also the total energy, is mgh.
This will be entirely converted into kinetic energy of the block at the point immediately before collision, and we can thus solve
for the velocity of the block before collision via energy conservation:
mgh =
v=

1
mv2
2
2gh

The momentum before the collision is thus p = mv = m 2gh . Momentum is conserved in the collision, so the
momentum of the block-putty system after the collision will also be m 2gh . We can use this relation to determine the

velocity of the system after collision, v’:
(2m)v’ = m 2gh
where the mass on the left hand side is 2m since the putty also has a mass of m and so the total mass of the system is m + m =
2m after collision. Solving for v’ gives us:
v’ =

2gh
2

Immediately after the collision, the block-putty system has no potential energy. The kinetic energy is thus the total
energy, and this is equal to:
1
1
2gh mgh
(2m) v’2 = (2m)
=
2
2
4
2
Note that this is less than the initial energy of the block, mgh. As we mentioned, energy is not conserved in an
inelastic collision. In this particular case, the energy has been cut in half. The rest has been dissipated as heat and/or sound as
the collision occurs. This reduced total energy, which is entirely in the form of kinetic energy right after the collision, is
converted to potential energy as the block-putty system moves up the track. At maximum height h’, we can write the equality:
(2m)gh’ = total energy right after collision =

1
mgh
(2m) v’2 =
2

2

mgh
2
h
h’ =
4

2mgh’ =

Among the other answer choices, choice B, h/2, is what one would erroneously obtain if one assumes that energy is conserved.
10.

C
This question requires you to remember the definition of impulse, J:
J = ∆p = F∆t

where ∆p is the change in momentum, F is the force and ∆t the time over which the force acts. You should recognize the last
equality as simply Newton’s second law rearranged:
F = ma = m

10

∆v
(from the definition of acceleration)
∆t

as developed by

Work, Energy and Momentum Test 1
Multiplying both sides by ∆t, we get:
F ∆t = m ∆v
= ∆ (mv) (assuming mass is constant)
= ∆p
The larger the force, or the longer the time over which the force acts, the more the velocity (and hence momentum) is
changed. In this problem, the time that the baseball and the bat are in contact is doubled, while the force remains constant. The
impulse is therefore doubled.
11.

B
Since the efficiency of the solar panel is only 10%, only 10% of the incident solar radiation is converted to useful
energy. The output power of the solar panel is therefore 40 kW × 0.1 = 4 kW = 4000 W = 4000 J/s.
Power is energy per unit time, and from the calculation above, we know that the solar panel delivers 4000 J of energy
every second. In 300 minutes, then, the energy generated is:
J
60 s
J
4000 × 300 min ×
= 4000 × 18000 s = ((4 × 103) × (1.8 × 104)) J = 7.2 × 107 J = 72 MJ
s
1 min
s
6
where 1 MJ = 10 J.
One could also have gotten the answer by first calculating the energy delivered in 300 minutes and then multiplying by
the efficiency factor.
12.

B

Let the masses of ball 1 and ball 2 be 2m and m respectively. The momentum of ball 1 before the collision is (2m)v =
2mv. Since ball 2 is at rest, it has no momentum and so the momentum of ball 1 is also the total momentum of the system.
Regardless of the nature of the collision, momentum is conserved (kinetic energy may or may not be conserved). The total
momentum of the system after the collision would therefore also be 2mv. Ball 1 moves with a speed of v/3 after the collision;
its momentum is therefore (2m)(v/3) = (2/3) mv. Since it’s moving in the same direction as before, its momentum vector
points in the same direction. Ball 2 must then either move in the same or the opposite direction; otherwise the vector for the
total momentum will have acquired a component along a new coordinate. The velocity of ball 2 after the collision, v’, must
satisfy the equation:
2
mv = 2mv
3
4
v’ = v
3

mv’ +

At this point we could eliminate the possibility that the collision is completely inelastic. If that were the case, the two
balls would have moved as one, i.e. with the same velocity, after the collision. Choice C is therefore incorrect. Since we have
the velocities both before and after the collision for both balls, we could determine whether the collision is elastic or inelastic
by computing the kinetic energies; choice D is also incorrect.
The kinetic energy before the collision is:
1
1
(2m) v2 + m(0)2 = mv2
2
2
The kinetic energy after the collision is:
1
v

1
4v
mv2 8mv 2
(2m) ( ) 2 + (m) ( ) 2 =
+
= mv2
2
3
2
3
9
9
Kinetic energy is therefore conserved during the collision and thus the collision is elastic.

KAPLAN

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MCAT
Passage II (Questions 13—17)
13.

A
The work done by a force is the product of the distance and that component of the force that is in the same direction:
that is what the cosθ term means in the formula W = Fdcosθ. If the two are perpendicular, the angle between them is 90°, and
the cosine term is zero. The work done will be zero in such a case. In the particle storage ring, the particles are moving in a
circle. Their instantaneous velocity, and therefore, their direction of motion at any time, is tangent to the circle. The direction of
the magnetic force, on the other hand, is toward the center of the ring and is acting as a centripetal force. The force and the
direction of motion are perpendicular, and thus the work done by the magnetic force is zero.

14.

D
The first and most natural step in approaching this problem is to identify the factors that affect the reaction rate from
the information given in the passage and determine how they affect the rate. The formula for the rate is given in the passage:
rate = luminosity × cross section. Since the cross section is fixed for a given reaction, the only way one can control the rate is
by changing the luminosity. (Do not confuse cross section with cross-sectional area!) The expression for luminosity tells us
that it is increased by increasing the number either of electrons or of positrons, by increasing the frequency of revolution, and by
decreasing the cross-sectional area of the storage ring. Since luminosity is directly proportional to the reaction rate, any of the
above that increases luminosity would also increase the reaction rate. Statements I and III are exactly two ways we have
identified to increase the rate. Statement II, increasing the energy of the particles, would increase the velocity of the particles.
They would thus move faster, completing more revolutions per second in the storage ring. This is also one of the ways we have
identified to increase the rate. Thus all three statements are correct. Note also that once we have established statements I and III
as being correct, choice D has to be the correct response.
15.

C
The mass of the positron or electron is on the order of 10–31 kg. Multiplying this by the acceleration due to gravity
results in a number that is still negligibly small compared to the magnitude of the electric interactions. Choice A is incorrect
because gravity points in the same direction as the acceleration due to gravity. This is a consequence of Newton’s second law F
= ma, where F and a are vectors and m is a scalar. In order for this vector equality to hold, F and a must point in the same
direction. (One must, in general, of course be aware of the distinction between force and net force.) Choice B states that the
energy lost due to gravity is equal to that gained from the magnetic fields. The passage tells us that magnets are placed at
various points in the storage ring. Their purpose, however, is to keep the particles moving in a circular path. Since the
magnetic force is perpendicular to the particle’s velocity, the magnetic force does no work, and thus causes no energy gain or
loss. Choice D states that the luminosity is not dependent on the gravitational acceleration. This is a true statement, but is not
relevant to the issue we are trying to address here.
16.

D

This is a conservation of momentum question in disguise. The energy value given is extraneous information. In the
question stem, we are told that the two electrons with equal speed collide head-on. There is really a lot of information that can
be garnered from this statement. Since both particles are electrons, they have the same mass. Furthermore, since they collide
with equal speed head-on, we know that their velocities are equal and opposite. So the total initial momentum of the system is
zero: The initial momentum of one electron is a vector that is equal and opposite to that of the other electron.
Momentum conservation tells us that the total momentum after the collision must in this case also be equal to zero.
After the collision, therefore, the two electrons must again move with equal speed in opposite directions. The second electron
will thus also have a speed of 0.9 c.
17.

C
Again, the energy value of 180 GeV is irrelevant here: the only important point is that the total energy needs to be
kept constant. In order to maintain this, the power we need to supply has to be the same as the power lost. The question tells us
that 260 MeV of energy is lost per revolution by each particle, and that the frequency of revolution is 10000 Hz. In one second,
then, the number of revolutions made is 10000, and the amount of energy lost in one second by one particle is 260 × 10000
MeV = 2.6 × 102 × 104 MeV = 2.6 × 106 MeV. The total amount of energy lost in one second is this value multiplied by two,
since there are two particles: 5.2 × 106 MeV. This is the amount of energy lost per second, and so energy needs to be supplied
at a rate of 5.2 × 106 MeV per second for it to remain constant.

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