At this point, it is found to have kinetic energy equal to Mgh, where g is the acceleration due to gravity and h is the initial height at point A.. When the cart reaches point C, it is f
Trang 1PHYSICS TOPICAL:
Force, Motion, Gravitation and
Equilibrium Test 1
Time: 21 Minutes*
Number of Questions: 16
* 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 2
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.
Trang 3Passage I (Questions 1–5)
An amusement park wishes to build a roller coaster
with a single loop as shown in Figure 1
A
h
C
r
ϕ
Figure 1
For the test run, an empty cart having mass M is
released from point A with an initial velocity v a The cart
then increases in speed at a non-constant acceleration until
it reaches a velocity of v b at point B At this point, it is
found to have kinetic energy equal to Mgh, where g is the
acceleration due to gravity and h is the initial height at
point A The loop portion of the track closely
approximates a perfectly symmetric circle of radius r, and
the distance from point B to point C equals the distance
from point C to point D When the cart reaches point C, it
is found to have a velocity of v c, which is the minimum
velocity required to keep gravity from pulling it off the
track It then accelerates down the back side of the loop to
point D, where it reaches its final velocity, v d
Figure 2 shows a graph of the magnitude of the
normal force N that the track exerts on the cart as a
function of distance s between points B and D In the
graph, the energy lost due to friction has not been included
D
6 Mg
Mg B
C
s
Figure 2
1 The amount of energy lost between point A and point
B is what percentage of the cart’s initial kinetic energy?
A 0%
B 25%
C 50%
D 100%
2 Examination of Figure 2 reveals that the normal force
on the cart reaches a maximum value of 6Mg If we
measure the cart’s position in the loop by the angle ϕ, shown in Figure 1, at what angle does this maximum normal force occur?
A ϕ = 0°
B ϕ = 30°
C ϕ = 180°
D ϕ = 210°
3 If the effects of friction were included in Figure 2, the normal force on the cart in the loop would:
A remain the same everywhere in the loop.
B increase everywhere in the loop.
C decrease everywhere in the loop.
D increase in some places and decrease in other
places
4 According to Figure 2, what is the correct value for the
cart’s kinetic energy at point C?
A Mgr/2
B Mgr
C 3Mgr/2
D 5Mgr/2
5 When passengers are riding in the cart, its mass
approximately doubles The minimum velocity v c at point C will then:
A be cut in half.
B remain the same.
C be doubled.
D be quadrupled.
GO ON TO THE NEXT PAGE.
Trang 4
Passage II (Questions 6–11)
The general motion of a massive object under the
influence of the Earth’s gravitational force can be solved in
closed form using Newton’s second law The gravitational
force is given by:
F = GmMe/R2 where G is the universal gravitational constant, m is the
mass of the object, Me is the mass of the Earth, and R is
the distance of the object from the center of the Earth One
result of this analysis is that Kepler’s laws can be derived
for satellites orbiting the Earth These laws were originally
discovered by Johannes Kepler (1571–1630) based on
observations of the motion of the planets about the Sun
They are as follows:
I All satellites move in elliptical orbits with the
Earth at one focus
II A line that connects a satellite to the center of the
Earth sweeps out equal areas in equal times
III The square of the period of any satellite is
proportional to the cube of the semimajor axis of
its orbit
For objects near the surface of the Earth, a great
simplification in the mathematics occurs by approximating
the gravitational force to be a constant Let R = Re + h,
where Re is the radius of the Earth, and h is the distance of
the object from the surface of the Earth The gravitational
force can now be written as:
F = GmMe/(Re+ h)2,
which is approximately equal to GmMe/Re2 when h <<Re
This constant force is usually set equal to mg, where g is
defined to be the free-fall acceleration (Note: The universal
gravitational constant is G = 6.67 × 10–11 N•m2/kg2 The
mass of the Earth is Me = 5.98 × 1024 kg, and its radius is
Re = 6.37 × 106 m.)
6 Two balls of unequal mass are dropped from the same
height If the gravitational force is assumed to be
constant, then the theory predicts that they will hit the
Earth’s surface at the same time If the actual force is
used instead, then the theory predicts:
A that the ball with the larger mass will hit first.
B that the ball with the smaller mass will hit first.
C that they will hit at the same time.
D nothing, since the equation of motion cannot be
solved without the constant force approximation
7 Consider a satellite in circular orbit around the Earth at
a distance R from the Earth’s center The area that the
satellite sweeps out as it moves through an angle θ is
given by A = R2θ/2 Kepler’s second law implies that
the time that it takes for the satellite to move through
2θ is:
A double the time that it takes to move through θ
B half the time that it takes to move through θ
C the square of the time that it takes to move
through θ
D equal to the time that it takes to move through θ
8 Which of the following graphs represents how the
magnitude of acceleration of a falling object changes with the distance through which it has fallen, if no approximation is made with regard to the distance?
height fallen
acceleration acceleration
A.
B.
C.
D.
height fallen
height fallen height fallen
9 Which of the following expressions shows how the
acceleration due to gravity depends on the height of an object above the surface of the Earth?
A G(Re + h)
B 2G(Re + h)2
C GMe/(Re + h)
D GMe/(Re + h)2
1 0 A satellite of mass m is in circular orbit around the
Earth at a distance R from the Earth’s center If R is
increased by a factor of 4, then the period of the orbit will be: (Note: The semimajor axis of a circle equals its radius.)
A unchanged.
B increased by a factor of 4.
C increased by a factor of 8.
D increased by a factor of 64.
Trang 5GO ON TO THE NEXT PAGE 1 1 The amount of energy needed to move a rocket having
mass m from the surface of the Earth to a distance R from the Earth’s center is E = GmMe(1/Re – 1/R).
What is the minimum velocity the rocket must have
in order to completely escape the Earth’s gravitational field?
A (2Gm/Re)1/2
B (2GMe/Re)1/2
C (2gh)l/2
D (2mGRe)l/2
Trang 6
Questions 12 through 16 are NOT
based on a descriptive passage
1 2 A 1-kg aluminum sphere and a 3-kg brass sphere, both
having the same diameter, are allowed to fall freely
from the same height above the ground Neglecting air
resistant and assuming both spheres are released at the
same instant, they will reach the ground at:
A the same time but with different speeds.
B the same time and with the same speed.
C different times and with different speeds.
D different times but with the same speed.
1 3 A 20-kg boy, starting from rest, slides down a slide
that makes a 30° angle with the ground The
coefficient of kinetic friction between the boy and the
slide is 0.20 What is the magnitude of the boy’s
acceleration as he slides? (Note: Assume that the
acceleration due to gravity, g, equals 10 m/s2 The
cosine of 30° is 0.866, and the sine of 30° is 0.5.)
A 1.7 m/s2
B 3.3 m/s2
C 5.0 m/s2
D 7.7 m/s2
1 4 A car traveling at 25 m/s requires 10 seconds to come
to rest, with an average braking force of
2,000 N What is the mass of the car?
A 200 kg
B 400 kg
C 800 kg
D 1,000 kg
GO ON TO THE NEXT PAGE.
1 5 A massless rod of 4 meters is placed on a fulcrum as
shown in the diagram below An object of mass m is attached to one end and an object of mass 2m is attached to the other end Where must the fulcrum be placed for the system to remain in equilibrium?
A 1.33 meters from the left
B 1.50 meters from the left
C 2.33 meters from the left
D 2.67 meters from the left
1 6 A 10 kg mass on a horizontal table top is being
swung counterclockwise in a circular path and is slowed down by a frictional force The net force on the mass when it lies in the position shown, points toward:
C
10 kg
D A
B
v
A A
B B
C C
D D
END OF TEST
Trang 7THE ANSWER KEY IS ON THE NEXT PAGE
Trang 8
ANSWER KEY:
Trang 9Passage I (Questions 1—5)
One should recognize that it would not be efficient to try to memorize all of the information presented in the passage
as one is reading it for the first time Instead, it is better to concentrate on the physical content of the situation First of all, we suspect that energy conservation is an important tool since the acceleration is not constant: most of the kinetics equations would therefore not apply We have not been given explicit information about friction, but we know from energy conservation and the information given about points A and B in Figure 1 that it must be present We are also told at the end of the passage that friction is not included in the graph of Figure 2 That is however something we will need to take into account
1 D
In the passage we are told that at point A, the cart has an initial velocity of va Its initial kinetic energy is therefore 1
2
Mva2 In addition to this, however, the cart also has gravitational potential energy equal to Mgh The sum of the two is its total initial energy At point B, the cart has no gravitational potential energy but only kinetic energy We are not given the velocity
of the cart at that point, but are told that the kinetic energy at B (and therefore the total energy at B) is equal to Mgh: the initial potential energy of the cart Therefore the amount of energy that has been lost to friction is 1
2 Mva
2, or 100% of the initial kinetic energy
2 A
Choice C, 180°, can be eliminated immediately since that corresponds to point C, at which the normal force is zero according to Figure 2 Within the loop the cart is undergoing circular motion (not uniform since the velocity is changing) At any point the instantaneous centripetal acceleration is v2/r, where v is the velocity at that point This acceleration needs to be provided by a net force perpendicular to the track The normal force, by definition, is perpendicular to the track, but there will also be a component of gravity that acts in the same or opposite direction The component of gravity pointing in the opposite direction as the normal force is given by Mgcosϕ:
Mg
normal force j
Mgcosj
Thus the equation is:
N – Mgcosϕ = Mv
2
r
N = M (v
2
r + gcosϕ) Notice that at the bottom half of the loop, ϕ is either between 0° and 90°, or between 270° and 360°, and so cosϕ
would be positive, leading to a larger value for the normal force: the normal force needs to be greater because it needs to counteract gravity and provide the centripetal acceleration Conversely, at the top half of the loop, ϕ is between 90° and 270°, and cosϕ is negative, leading to a smaller value for the normal force Physically, there is a component of gravity that is parallel
Trang 10
(rather than antiparallel) to the normal force and therefore it “reduces the burden” of the normal force to provide the centripetal acceleration, allowing it to have a smaller value
Going back to the question, the value of cosϕ is at a maximum when ϕ = 0° At that same point, v is also at a maximum since the cart loses kinetic energy as it rises in the loop Thus, ϕ = 0° is the point at which the normal force is at a maximum
3 C
Friction between the track and the cart would cause the cart to slow down This has the effect of lowering the value of
v in the right-hand side of the equation given above (in the explanation to #2) for the magnitude of the normal force The equation is otherwise unaffected since the friction force is always directed opposite in direction to that of travel, and is thus perpendicular to the normal force: the friction force does not enter into the equation directly since it has no component in the normal direction
4 A
According to Figure 2, the normal force is zero at point C The centripetal acceleration is therefore provided by the gravitational force entirely At point C, the direction perpendicular to the track is parallel to gravity, which means that all of gravity (rather than just a component) is providing the centripetal acceleration:
Mg = Mvc
2
r
vc2 = gr where vc is the velocity of the cart at point C, and is the minimum velocity required for the cart to stay on the track rather than fall off The kinetic energy at that point is 1
2 Mvc2 Substituting in the relation we just obtained, we have:
K.E = 1
2 Mgr.
5 B
The minimum velocity vc is always that corresponding to a zero normal force Its value can be determined using the equations given above in the explanation to #4 As you can see, it is unaffected by the mass since M cancels out Again, friction force does not appear explicitly since it is perpendicular to the normal direction
Passage II (Questions 6—11)
6 C
The question asks us to determine how our more exact theory of gravity would predict the relative times it will take for two balls of unequal mass dropped from the same height to reach the ground The approximate theory neglected the difference in height of various objects since it was small compared to the radius of the Earth; the approximation lies not in the masses involved In the scenario given, the balls are dropped from the same height, and thus whatever correction is made to the falling time of one will also be made to the other The two will therefore still hit the Earth at the same time
You may want to note, for example, that in the usual approximate case, F = mg = ma A larger mass would mean a larger gravitational force, but does not mean a larger acceleration since the m’s cancel What we have here is a more accurate expression for g: one that is not constant but varies with distance The mass, however, still does not affect the acceleration
7 A
A close reading of Kepler’s laws is necessary These laws, originally derived for planets orbiting around a star, applies also to satellites revolving around the Earth The second statement in the laws, when applied to satellites, tells us that a line connecting the satellite to the center of the Earth sweeps out equal areas in equal times This means that if the satellite sweeps out an area A in time t, it will sweep out an area 2A in time 2t, etc
In the question stem, we are told that the satellite travels in a circular orbit We are also given a formula relating the area and the angle θ that the satellite sweeps out: the area is directly proportional to θ Therefore, if the satellite sweeps out an angle 2θ, the area swept out will be 2A The time is therefore twice that of sweeping out θ, from the reasoning in the paragraph above