1Translational motion test w solutions

PHYSICS 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.


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)

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2

as developed by


Translational Motion Test 1
Passage 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 L 3 , 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.
B.
C.
D.

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.
B.
C.
D.

22:1
2:1
21/3:1
1:22

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

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KAPLAN

3


MCAT
Passage II (Questions 6–11)

vy (m/s)
28
40
kg

21
60
kg

17
14

kg
80 kg
0 kg
10 20
1

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 v 0 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 of 0.24 sec. The
height above the ground at which the clown begins his
trajectory is 10 m.

42

1.4 1.7 2.1

2.8

4.2

t (s)
Figure 2
6 . If the angle the cannon makes with the horizontal is
increased from 45°, the hoop will have to be:

45

A.
B.
C.
D.

moved farther away from the cannon and lowered.
moved farther away from the cannon and raised.
moved closer to the cannon and lowered.
moved closer to the cannon and raised.

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 = mv 0 ,
where m is the mass of the clown and t is the time over
which the force acts on the clown. The acceleration of
1
gravity is g = 10 m/s2. sin 45° = cos 45° =
]
2

7 . If the clown’s mass is 80 kg, what initial velocity v 0
will he have as he leaves the cannon?
A.
3
B.
15
C . 30
D . 300

m/s

m/s
m/s
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.
B.
C.
D.

–g
v0
y–y0
sin θ

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4

as developed by


Translational Motion Test 1
9 . 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.

B.
C.
D.

1.7
3.4
3.9
4.4

s
s
s
s

1 0 . If the mass of a clown doubles, his initial kinetic
energy, mv02/2, will:
A.
B.
C.
D.

remain the same.
be reduced in half.
double.
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 . v 0 ∆t M
v02
g
v02
C.
2g
v02
D.
4g
B.

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KAPLAN

5


MCAT
Questions 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 5 . Two blocks, whose masses are m 1 and m 2 , 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 m l is greater
than m 2 and θ 1 is greater than θ 2 , 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 . m 1 slides down faster than m 2 because it
experiences a greater gravitational force.
C . m 1 slides down faster because of the larger angle
of inclination.

D . m 2 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.

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.

END OF TEST

6

as developed by

Translational Motion Test 1

THE ANSWER KEY IS ON THE NEXT PAGE

KAPLAN

7


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

8

11.
12.
13.
14.

15.

D
A
B
B
C

16. B

as developed by


Translational Motion Test 1
EXPLANATIONS
Passage 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 + …
= ρ 1V 1 + ρ 2V 2 + ρ 3V 3 + ρ 4V 4 + …
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 L 3 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

KAPLAN

9


MCAT
The 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 1
= ( )
k’L3 k’ L

where k and k’ are the respective proportionality constants. The ratio q-ant to q-human is therefore:
q-ant : q-human =

k 1
k
1

1
1
1
1
(
) : (
) =
:
=
:
= 2 : 0.01 = 200 : 1
k’ Lant k’ Lhuman
Lant Lhuman 0.5 100

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:
Lhuman 100
1
1
200
:
=
=
= 200 =
= 200 : 1
Lant Lhuman
Lant
0.5
1
5.

C
Let L be the characteristic length of the original cell. It will have a surface area proportional to L 2 , 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, L 2 , in terms of λ. λ 2 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:
2λ 2 : L 2 = 2λ 2 : (22/3)λ 2 = 2:22/3
Again, recalling the equivalence between a ratio and a quotient, we can simplify this to:
2:22/3 =

2
2
2 –2/3 2 1/3
= 2/3 × –2/3 =
= 21/3:1
2 2/3
2
2
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:

10

as developed by


Translational Motion Test 1
1
mvy02 = mghmax
2
vy02
hmax =
2g
7.

C
This question calls for a straightforward application of the equation for the impulse:
Ft = mv0
The force is given in the passage as 10000 N and t as 0.24 s, while m is 80 kg from the question. Solving for v0:

v0 =

Ft
10000•0.24 2400
=
=
= 30 m/s
m
80
80

Alternatively, for a mass of 80 kg, Figure 2 shows that the intitial velocity in the y direction is 21 m/s. This is related
to the magnitude of initial velocity by
v0y = v0 sin 45°
21
= 21 2 . Without actually performing any calculations, we should be able
sin45°
to decide that choice C, 30 m/s, is the only reasonable answer.
The initial velocity is therefore v0 =

8.

A
The easiest way to do this problem is through dimensional analysis: Since all the choices have different units, if we
know what units the slope is in, we can just select the one that matches. The graph in Figure 2 is a plot of velocity versus
time. Slope is in general the change in y over the change in x, and therefore in this case is the change in velocity over the
m/s
change in time, which will have units of
= m/s 2. These are the dimensions of acceleration, and, looking at the answer
s

choices, the slope must therefore be equal to the acceleration due to gravity, g, or choice A.
Choice B is the initial velocity and has units of m/s; choice C is vertical displacement which just has the units of
distance or m; choice D is dimensionless.
Alternatively, knowing that the slope is a constant, we must find the kinematics equation that has the form y = mx +
b, where y is the velocity in the y direction, and x is time. The equation we are looking for is
v = v0 + at
or, more specifically in this case,
v y = v0y – gt
where m, the slope, is –g.
9.

C
This question involves both interpretation of the graph in Figure 2 and general knowledge of projectile motion. The
graph enables us to determine the time the clown takes to reach the hoop. According to the graph, a 100-kg clown will exit the
cannon with an initial y-velocity of 17 m/s. (One can also obtain this value indirectly by finding the velocity using the formula
for impulse as in #7 and then calculating the y-component from trigonometry.) One can then find the time t taken for the trip
by the equation:
y = vy 0 t +

KAPLAN

1
2

at2

11


MCAT

In substituting values, we need to be careful in making sure that our signs are consistent. Defining the upward
direction to be positive, y would be the net vertical displacement which is –10 m, v y0 is 17 m/s, and a, the acceleration, is
–10 m/s2 due to gravity:
–10 = 17t – 5t2
Solving for t, however, means solving a quadratic equation that does not factor neatly. One can try the different answer
choices to see which would satisfy this equation, but this would still be time-consuming and it is easy to make numerical
mistakes. Let us also consider another approach that is more instructive in elucidating the kinetics behind the problem.
When the y-velocity falls to zero, the clown has reached the hoop. The time at that point (i.e. the x-intercept in the
graph), then, is the time it takes for the clown to reach the maximum height from the cannon: in this case it takes 1.7 s. On his
trip down, the clown will take this same amount of time to reach the same level as the cannon; i.e. he will have taken a total of
3.4 s to complete the projectile trip from the cannon back to the same level. That, however, is not enough. Additional time is
needed for the clown to fall from the cannon level to the safety net 10 m further down. (At this point, you should be able to
eliminate choices A and B. If you are pressed for time, therefore, a guess between C and D will at least maximize your chances
of getting the right answer.) A diagram should be helpful:

time = 3.4 s
vy = initial vy = 17 m/s

time = 1.7 s
vy = 0

Point B
10 m

How much time is needed for this last part of the trip, from point B to the safety net? One can avoid working with
cumbersome kinetics equations by examining the two remaining answer choices more closely. First of all, we should realize
that choice C essentially says that it takes 0.5 s (= 3.9 – 3.4), while choice D says that it takes 1 s (= 4.4 – 3.4). Which is
more reasonable? At point B, the clown already has a vertical velocity of 17 m/s. It would thus take slightly more than half a
distance
second to travel 10 m (time =

). Besides, 17 m/s is only an “initial” value; the clown is also accelerating from gravity.
speed
Therefore choice C is the only reasonable answer.
10.

B
This is a trick question in that if one merely looks at the formula for the kinetic energy, one might conclude that if
mass doubles the initial kinetic energy will double as well. This, however, is true only if the initial velocity stays the same.
But the first paragraph in the passage indicates that it is the impulse, not the initial velocity, that stays constant. If the mass
doubles, then, the initial velocity is also going to change. The equation given in the passage is:
Ft = mv0
v0 =

Ft
m

The numerator is the impulse that is constant (= 10000 N × 0.24 s). If the mass is doubled, then, the initial velocity would be
halved. Using a prime (’) to designate the new quantities, then, we have:
m’ = 2m
v0
v0’ =
2
The new initial kinetic energy would therefore be:

12

as developed by


Translational Motion Test 1

v0
1
1
1
1 1
1
m’v0’2 = 2m ( )2 = mv02 = ( mv02) = KEold
2
2
2
4
2 2
2
11.

D
This question is most easily answered using the principle of conservation of energy. Since the clown reaches the hoop
at the maximum height of his trajectory, the vertical component of his velocity is zero. There are no forces acting on him in the
v0
horizontal direction, and so his horizontal velocity is the same as its initial value, that is, v0cos 45° or
. The clown’s
2
v0 2 1
1
kinetic energy as he reaches the hoop is therefore m (
) = mv 0 2 .
2
4
2

After grabbing the hoop, the clown will swing upward until all of this kinetic energy is converted into potential
energy, mgy, where y is the vertical distance above the relaxed position of the hoop that the clown swings upward. This is also
the minimum length of the cable:

y = minimum
length of cable

The condition that y needs to satisfy is therefore:
1
mv 0 2
4
v02
y=
4g

mgy =

Independent Questions (Questions 12–16)
12.

A
All of the answer choices include the initial speed. What else is needed to determine how far a cannonball lands from a
cannon? Remember that speed is a scalar: it has a magnitude but no direction. In order to resolve the cannonball’s motion into
vertical and horizontal components (which we need to do), we need the direction of travel—or in terms of the choices in this
question—the angle of inclination. In addition, we need to know how large the force is that is pulling the cannonball down: The
acceleration due to gravity on Mars, which will differ from the familiar 9.8 m/s2, will affect the amount of time the cannon
spends in the air and thus the horizontal distance traveled.
Let us also consider a slightly more quantitative approach that would also reveal to us why the other combinations are
not sufficient. There is no net force in the horizontal direction, and so the horizontal velocity is constant until the ball hits the
ground. The horizontal distance traveled is then simply vxt, where vx is the horizontal component of the initial velocity, i.e.

vx = v0cosθ, where v0 is the initial speed and θ is the angle of inclination. t is the time the ball stays in the air, and is dictated
by the kinetics in the vertical component. Specifically, it can be found using the kinetics equation:
y = v0 y t +

1
2

at2

where in this particular case y will be zero (assuming that the cannonball is launched and hits the ground at the same level) and
care has to be taken to make sure that v0y and a have opposite signs since they point in opposite directions. v 0y is v0sinθ, and
thus no additional information is needed to determine this that was not already required to find vx (i.e. v 0 and θ). The only other
piece of information, then, is the acceleration due to gravity. Let us now also consider why the other answer choices are not
correct.
Choice B gives us the time of travel, and thus it would save us the calculation to find t. However, initial speed by
itself is not enough to let us find vx, for which we also need the angle of inclination: what we need is the initial speed in the

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MCAT
horizontal direction, which we cannot determine with the quantities in choice B. (Incidentally, we can use the kinetics equation
above to solve for the angle of inclination if we know the initial speed, which we do, the time, which we do, and the
acceleration due to gravity, which we don’t. In other words, no matter how one chooses to look at it, choice B is insufficient.)
Choice C is likewise lacking in the angle of inclination or time of travel. Choice D does not enable us to solve for the
time spent traveling without the value of the acceleration due to gravity on Mars.
13.

B
This is essentially a vector addition problem. Since the current would carry the boat downstream, the boat needs to aim
upstream if it were to “cancel” the effect of the downstream current and arrive at the same spot on the opposite shore. We can
thus eliminate choice A immediately. The two velocity vectors, the one of the downstream current and the one of the boat in
still water, need to add to give a resultant vector that points directly to the opposite shore (i.e. at an angle of 90° to the shore).

current: 10 m/s

resultant

θ

boat: 20 m/s

The question, in other words, is: What should θ be so that the resultant vector would point directly towards the
opposite shore? (The angle with the shore that the boat should make would then be 90° – θ.) The sine of θ is 10/20 or 0.5. In
the question stem we are told that sin 30° = 0.5, and so θ = 30°, and the boat should make an angle of 60° with the shore.
Since the boat has a speed of 20 m/s that is stronger than the current (10 m/s), it is able to combat the current and get
to the opposite shore, making choice D incorrect in this case. If the current is, say, 25 m/s instead, then the boat would not
have been able to make it to the opposite shore at the same point.
14.

B
The potential energy of a projectile is mgh, and clearly increases as the height increases. At the highest vertical point
in its path, then, its potential energy must be at a maximum. As has been discussed in earlier problems in this topical test, the
kinetic energy associated with the projectile’s motion in the vertical direction is completely transformed into gravitational
potential energy at its highest point, and so choice A is incorrect. At the highest point, the projectile is turning around, and so
its instantaneous vertical velocity is zero. The horizontal component, however, remains constant throughout its flight
(assuming no air resistance) and does not change until impact with the ground causes it to stop.
15.

C
We need to derive an equation for the acceleration of a black down the inclined plane. A diagram would be most

helpful:

friction = µN = µmgcosθ
mgsinθ
θ mg
θ
The component of the weight that is parallel to the plane is the force that pulls the block down. As can be seen from
the diagram, this force has a magnitude of mgsinθ, where m is the mass of the block. The friction force acts in the opposite
direction of motion and has a magnitude given by the general formula F = µN, where µ is the coefficient of friction and N the
normal force. If the plane is level, the normal force would simply be the weight of the object, but in the case of the inclined

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as developed by


Translational Motion Test 1
plane, it is the component of the weight that is perpendicular to (or normal to, hence the name) the plane: mgcosθ. The friction
force is therefore µmgcosθ. The net force on the object in the direction parallel to the plane is therefore:
F = mgsinθ – µmgcosθ
This force will provide the acceleration down the inclined plane:
mgsinθ – µmgcosθ = ma
gsinθ – µgcosθ = a
Notice that the masses on each side cancel. So, in the scenario given by the question, the fact that m 1 is greater than
m 2 is irrelevant. They both start from rest, and so whichever experiences a larger acceleration will slide down faster, and the
acceleration depends only on the angle of inclination, since the coefficient of friction is the same in both cases. The question,

then, is, does increasing the angle increase or decrease acceleration?
The value of sine is increasing on going from 0° to 90° (from 0 to 1), while the value of cosine is decreasing on going
from 0° to 90° (from 1 to 0). Presumably θ1 and θ2 will both be in between the limits of 0° and 90°, and so the fact that θ1 is
greater than θ2 means that sinθ1>sinθ2, and that cosθ1block 1, we would have a greater number minus a smaller number compared to block 2. The acceleration for block 1 is therefore
greater.
16.

B
As has been discussed above for question #12, the time a projectile spends in air can be found by solving for t in the
equation:
y = 0 = v0 y t –

1
2

gt2

where we have substituted in –g for the acceleration. Since this is a quadratic equation, there are two solutions that one can
obtain through straightforward factoring:
1
gt) = 0
2
2v0y
t = 0 or t =
g

t (v0y –

The t = 0 solution corresponds to the launch point and thus is not what we are interested in. The time spent in flight is

2v0y
2v0sinθ
hence
, or
. The ball launched at a smaller angle would have a smaller value of sinθ (see discussion for previous
g
g
question), and thus would spend less time in air and hit the ground first.
It is important to realize that less time spent in air does not necessarily mean a smaller horizontal
distance traveled. As is explicitly stated in passage 2, the horizontal distance is maximized if the angle is 45°. There is a
trade-off between the time spent in air and the horizontal velocity, both of which determine the horizontal distance traveled (see
discussion to #12 above): by increasing the time of flight, we have to increase the angle which reduces the horizontal velocity.
Choice D is wrong because the total mechanical energy is equal to the sum of kinetic and gravitational potential
energies. From the principle of conservation of energy, this sum has to be equal to the initial kinetic energy that the ball
possesses as it exits the cannon. Since the two balls are launched with the same initial speed and have the same mass, their
initial kinetic energies, and thus total mechanical energy, will be equal and have a value of mv02/2.

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