The frictional force can often be approximated as being proportional to the velocity times the cross-sectional area of the falling object.. Suppose that the frictional force of air resis
Trang 1PHYSICS TOPICAL:
Translational Motion
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 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 1–5)
It is a fundamental fact of life that ants cannot be as
large as elephants, and whales cannot be as small as
goldfish For every type of animal there is a most
convenient size Each animal has a characteristic length L
From this length one can define a characteristic area which
is proportional to L2, and a characteristic volume which is
proportional to L3 The characteristic mass is also taken to
be proportional to L3.
The maximum size of an animal is constrained by the
material out of which it is made For example, consider
whether a giant human is a viable creature If we rescaled a
human by a factor of 10, we would necessarily increase the
radius of the legs by a factor of 10 However, the mass of
the human would increase by a factor of L3, or 1000.
Therefore, legs with 100 times the cross-sectional area
would have to support 1000 times the weight
Size also explains why an ant that falls off of the
roof of a four story building will not be seriously injured,
whereas a human most certainly would be injured A small
animal can survive such a fall because of its relatively large
surface area to volume ratio An object in free-fall within
the Earth’s atmosphere will only accelerate up to a certain
velocity because of friction caused by the air resistance
The frictional force can often be approximated as being
proportional to the velocity times the cross-sectional area
of the falling object The maximum velocity that an object
reaches in free-fall is called the terminal velocity
1 Which of the following statements must be assumed if
the characteristic mass of an animal is taken to be
proportional to L3?
I Every animal can be assigned a characteristic
length L.
II The characteristic volume of the animal is
proportional to L3
III The density of the animal is constant
throughout that particular animal
A I only
B I and III only
C II and III only
D I, II, and III
2 Suppose that the frictional force of air resistance is
only proportional to an object’s cross-sectional area and not its velocity A falling object would then: [Note: Assume that the frictional force is much smaller than the weight of the object.]
A take more time to reach terminal velocity.
B take less time to reach terminal velocity.
C reach terminal velocity in the same amount of
time
D never reach terminal velocity.
3 The terminal velocity of a human in the horizontal
spread eagle position would be:
A smaller than if he/she is rolled up in a ball
because the cross-sectional area is greater
B greater than if he/she is rolled up in a ball because
the cross-sectional area is also greater
C the same as if he/she is rolled up in a ball because
the surface area to volume ratio is constant
D the same as if he/she is rolled up in a ball because
the mass is the same in each case
4 The characteristic length of an ant is 0.5 cm, while
that of a human is 1 m If q is the surface area divided
by the volume, what is the ratio of q-ant to q-human?
A 200:1
B 400:1
C 2000:1
D 1:20
5 Cells can only absorb nutrients through their surfaces.
In order for a cell to absorb nutrients most efficiently,
it will grow to a maximum size and then divide If the total volume remains constant when a cell divides, then the ratio of the total surface area of the daughter cells to that of the original cell is:
A 22:1
B 2:1
C 21/3:1
D 1:22
GO ON TO THE NEXT PAGE.
Trang 4Passage II (Questions 6–11)
A circus wishes to develop a new clown act Figure 1
shows a diagram of the proposed setup A clown will be
shot out of a cannon with velocity v0 at a trajectory that
makes an angle θ = 45° with the ground At this angle, the
clown will travel a maximum horizontal distance The
cannon will accelerate the clown by applying a constant
force of 10,000 N over a very short time of0.24 sec The
height above the ground at which the clown begins his
trajectory is 10 m
45
Figure 1
A large hoop is to be suspended from the ceiling by a
massless cable at just the right place so that the clown will
be able to dive through it when he reaches a maximum
height above the ground After passing through the hoop
he will then continue on his trajectory until arriving at the
safety net Figure 2 shows a graph of the vertical
component of the clown’s velocity as a function of time
between the cannon and the hoop Since the velocity
depends on the mass of the particular clown performing the
act, the graph shows data for several different masses
[Note: The impulse J, or change in momentum p
generated by the force F is given by J = F∆t = p = mv0,
where m is the mass of the clown and t is the time over
which the force acts on the clown The acceleration of
gravity is g = 10 m/s2 sin 45° = cos 45° = 1
2 ]
14 17 21 28
42
vy (m/s)
t (s)
40 kg
60 kg
80 kg
100 kg
Figure 2
6 If the angle the cannon makes with the horizontal is
increased from 45°, the hoop will have to be:
A moved farther away from the cannon and lowered.
B moved farther away from the cannon and raised.
C moved closer to the cannon and lowered.
D moved closer to the cannon and raised.
7 If the clown’s mass is 80 kg, what initial velocity v0
will he have as he leaves the cannon?
A 3 m/s
B 15 m/s
C 30 m/s
D 300 m/s
8 The slope of the line segments plotted in Figure 2 is a
constant Which one of the following physical quantities does this slope represent?
A –g
B v0
C y–y0
D sin θ
GO ON TO THE NEXT PAGE.
Trang 59 From Figure 2, approximately how much time will it
take for a clown with a mass of 100 kg to reach the
safety net located 10 m below the height of the
cannon?
A 1.7 s
B 3.4 s
C 3.9 s
D 4.4 s
1 0 If the mass of a clown doubles, his initial kinetic
energy, mv02/2, will:
A remain the same.
B be reduced in half.
C double.
D quadruple.
1 1 If a clown holds on to the hoop instead of passing
through it, what is the expression for the minimum
length of the cable so that he doesn’t hit his head on
the ceiling as he swings upward?
A v0∆t M
B v0
2
g
C v0
2
2g
D v0
2
4g
GO ON TO THE NEXT PAGE.
Trang 6Questions 12 through 16 are NOT
based on a descriptive passage.
1 2 Which of the following is the LEAST amount of
information that could be used to determine how far a
cannonball lands from a cannon on Mars?
A Initial speed, angle of inclination, acceleration due
to gravity
B Initial speed, time of travel
C Initial speed, acceleration due to gravity
D Initial speed, angle of inclination
1 3 A boat can move with a speed of 20 m/s in still water.
It starts at the shore of a river that flows at a rate of 10
m/s In what direction must the boat move to reach the
exact same point on the opposite shore?
[Note: sin 30° = 0.5, cos 30° = 0.866, sin 60° = 0.866,
cos 60° = 0.5]
A The boat must be pointed at an angle of 90° to the
shore
B The boat must be pointed upstream at an angle of
60° to the shore
C The boat must be pointed upstream at an angle of
30° to the shore
D The boat will not be able to make it directly
across the river because the current will always
push it past that point
1 4 Which statement is true of a projectile at the highest
vertical point in its path?
A The projectile has its greatest kinetic energy.
B The projectile has its greatest potential energy.
C The vertical component of the projectile’s speed is
greater at this point than at any other point
D The horizontal component of the projectile speed
is greater at this point than at any other point
1 5 Two blocks, whose masses are m1 and m2, have the same coefficient of sliding friction Starting from rest, they slide down two planes of equal length inclined at angles θ1 and θ2 respectively Given that ml is greater than m2 and θ1 is greater than θ2, which of the following statements is consistent with the information?
A Both blocks slide down with the same speed
because they both have the same coefficient of sliding friction
B m1 slides down faster than m2 because it experiences a greater gravitational force
C m1 slides down faster because of the larger angle
of inclination
D m2 slides down faster because it experiences less friction
1 6 Two balls of equal mass are shot upward simultaneously from the same point on the ground with the same initial speed, but at different angles to the horizontal Which of the following statements must be true?
A The ball launched at the larger angle hits the
ground first
B The ball launched at the smaller angle hits the
ground first
C The two balls hit the ground at the same time.
D The ball launched at the larger angle always has
more total mechanical energy
END OF TEST
Trang 7THE ANSWER KEY IS ON THE NEXT PAGE
Trang 8ANSWER KEY:
Trang 9Passage I (Questions 1–5)
1 D
This question asks you to identify the underlying assumptions made in drawing certain conclusions from statements presented (This kind of reasoning is also required in the verbal reasoning section.) Let us examine each statement in turn
Statement I says that every animal can be assigned a characteristic length L This is certainly a necessary assumption if
we were to define the characteristic mass to be proportional to L3 This is, in fact, one of the main points of the first paragraph
of the passage
Statement II says that the characteristic volume of the animal is proportional to L3 The passage does make this claim Whether this statement is a necessary assumption depends on the relationship between mass and volume The concept of density, then, will play a role Since statement III explicitly addresses this concept, we can consider these two statements together
The density ρ of an object is defined as the mass per unit volume, and one often rearranges this definition to obtain the formula M = ρV where M is the mass and V the volume If, however, the object does not have a constant density throughout, the total mass is found by adding the individual masses of smaller regions in which the density is constant:
Mtot = M1 + M2 + M3 + M4 + …
= ρ1V1 + ρ2V2 + ρ3V3 + ρ4V4 + … where 1, 2, etc., are labels for the regions of the object within which the density is constant In this case then, as one can see,
the mass of the object is not simply proportional to its volume, i.e.: Mtot kV for some constant k (V = V1 + V2 + V3 + V4
+ …) Going back to the statements, then, only if the density is constant and if the volume is proportional to L3 can we express the mass of an animal to be:
M = ρV = ρ(kL3) or M ∝ L3
where k is the proportionality constant between the volume and L3
In sum, then, all three statements are necessary to make the claim
2 D
An object that has reached terminal velocity is falling at a constant velocity, i.e it is no longer accelerating or picking
up speed From Newton’s second law, we know that this would have to imply that the net force acting on the object is zero In particular, the gravitational force on the object is exactly balanced by the frictional force of air resistance If the frictional force
is only proportional to the cross-sectional area and not the velocity of the falling object, the magnitude of the force would not change Since the question stem states that this force is much smaller than the weight of the object (i.e the gravitational force
on it), it will never be able to balance the downward force: There will always be a net downward force acting on it that causes it
to accelerate downward throughout its entire fall The object therefore never reaches a constant terminal velocity
3 A
The friction force is proportional to the velocity and the cross-sectional area, as stated towards the end of the passage
So when terminal velocity is reached, the magnitudes of the two forces are related by:
kvtermA = mg where k is the proportionality constant in the friction force, A is the area, and the velocity is indicated as terminal since only at that point are the two forces equal (see #2 above) Rearranging this, one obtains:
vterm = mg kA
A human in the horizontal spread eagle position presents a greater cross-sectional area to the air than if he is rolled up into a ball Therefore the terminal velocity would hence be smaller in the spread eagle case
4 A
Trang 10The surface area is proportional to L2, while the volume is proportional to L3 The parameter q is defined as the surface area divided by the volume, i.e.:
q = kL2
k’L3 = k k’ (
1
L)
where k and k’ are the respective proportionality constants The ratio q-ant to q-human is therefore:
q-ant : q-human = k
k’ (
1
Lant) :
k k’ (
1
Lhuman) =
1
Lant :
1
Lhuman =
1 0.5 :
1
100 = 2 : 0.01 = 200 : 1 Note that we need to convert the two length dimensions into the same units to calculate the ratio As a side note, you should also recognize the equivalence between a ratio and a quotient:
1
Lant :
1
Lhuman =
Lhuman
Lant =
100 0.5 = 200 =
200
1 = 200 : 1
5 C
Let L be the characteristic length of the original cell It will have a surface area proportional to L2, and volume
proportional to L3 Since the total volume remains constant, the daughter cells, each taken to have characteristic length λ, will have a volume λ3 that satisfies:
L3 = 2λ3
where we have dropped the proportionality constant The factor of two appears because the original cell gives rise to two daughter cells To find the relationship between the surface areas, we take the cube root of each side and square:
L2 = (22/3)λ2
and this is the surface area of the original cell, L2, in terms of λ λ2 is the surface area of one daughter cell, and so the total surface area of the daughter cells is 2λ2 The surface area ratio is therefore:
2λ2 : L2 = 2λ2 : (22/3)λ2 = 2:22/3
Again, recalling the equivalence between a ratio and a quotient, we can simplify this to:
2:22/3 = 2
22/3 = 2
22/3 × 2–2/3
2–2/3 = 2
1/3
1 = 2
1/3:1 Notice that 21/3 is greater than one, and so the total surface area has increased upon division
Passage II (Questions 6–11)
6 D
This question can be answered using the equations of projectile motion, but since the answer choices are not quantitative, we should use a more intuitive and time-saving approach For the horizontal component, all one has to do is read the first paragraph of the passage which states that when the cannon is at an angle of 45°, the clown will travel the maximum horizontal distance If the angle deviates from 45°, then, the horizontal distance traveled would decrease The hoop is situated (roughly) halfway between the cannon and the landing point (roughly because the two may be at different heights so it is not exactly symmetric) A smaller horizontal distance, then, would mean that the hoop needs to be closer to the cannon
For the vertical component, if the angle is increased from 45°, the vertical component of the initial velocity of the clown would increase The kinetic energy associated with this initial vertical motion would therefore be greater The hoop is situated at the point of maximum height, i.e when all the vertical kinetic energy is transformed into gravitational potential energy The higher its initial vertical kinetic energy, the greater the gravitational potential energy it can ultimately gain, and the higher the hoop will have to be placed: