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
Trang 1PHYSICS 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
Trang 2DIRECTIONS: 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.
Trang 3Passage 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
1 At which point in an elliptical orbit does the
companion star attain its maximum kinetic energy?
A Apogee
B Perigee
C The point midway from apogee to perigee
D All points in the orbit, since the kinetic energy is
a constant of the motion
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.
3 Which of the following prevents the companion star
from leaving its orbit and falling into the black hole?
A The centripetal force
B The gravitational force
C The companion star’s potential energy
D 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 −1
2
2
mv
B mv2
C −GmM
r
D G
2
mM r
6 What is the ratio of the acceleration of the black hole
to that of the companion star?
A M/m
B m/M
C mM/r
D 1/1
GO ON TO THE NEXT PAGE.
Trang 4Questions 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 Increasing the angle of inclination
B Increasing the distance that the block travels
C Increasing the acceleration due to gravity
D 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:
h
A h
4
B h
2
C h.
D independent of h.
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 half the original impulse.
B the original impulse.
C twice the original impulse.
D 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?
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 inelastic
B elastic
C completely inelastic
D Cannot be determined from the information given.
GO ON TO THE NEXT PAGE.
Trang 5Passage 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 = Lσ, 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:
e e
= − + ,
where Ne– refers to the number of electrons, Ne+ 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 I only
B III only
C II and III only
D 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 0.45c
B 0.7c
C 0.8c
D 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 1.8 × 106 MeV/s
B 2.6 × 106 MeV/s
C 5.2 × 106 MeV/s
D 9.0 × 108 MeV/s
END OF TEST
Trang 6ANSWER KEY:
Trang 7Passage I (Questions 1—6)
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
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
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
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
Trang 8In 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):
black hole
star gravitational force
direction of travel
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.
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
2 + U = – (–
GmM 2r ) + (–
GmM
r ) =
GmM 2r –
GmM
r =
GmM 2r –
2GmM 2r = –
GmM 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
2 mv2 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
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
M :
F
m =
1
M :
1
m = m:M where the last step can be made if one recognizes that a ratio is itself just a quotient
Trang 9Independent Questions (Questions 7–12)
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
2 mv
2) = 1
2 m ∆(v2) = 1
2 m (vf
2 – vi2)
where vf 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 change in kinetic energy as 1
2 m(vf – vi)2.) Substituting in numbers supplied by the question, we have the change in energy to
be equal to 1
2 × 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
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:
θ
mg θ
friction = µ N = µ mgcos θ
normal force = mgcos 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
Trang 10The 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 = 1
2 mv
2
v = 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
2 (2m) v’2 =
1
2 (2m)
2gh
4 =
mgh 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
2 (2m) v’
2 = mgh 2 2mgh’ = mgh
2 h’ = h 4 Among the other answer choices, choice B, h/2, is what one would erroneously obtain if one assumes that energy is conserved
1 0 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∆v
∆t (from the definition of acceleration)