1Translational Motion Test w. Solutions

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PHYSICS TOPICAL:

Translational MotionTest 1

Time: 21 Minutes*Number of Questions: 16

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



MCAT

2 as developed by

DIRECTIONS: Most of the questions in the following testare organized into groups, with a descriptive passagepreceding each group of questions. Study the passage,then select the single best answer to each question in thegroup. Some of the questions are not based on adescriptive 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, thenselect an answer from the choices that remain. Indicateyour selection by blackening the corresponding circle onyour answer sheet. A periodic table is provided below foryour use with the questions.

PERIODIC TABLE OF THE ELEMENTS

1

H

1.0

2

He

4.0

3Li

6.9

4Be

9.0

5B

10.8

6C

12.0

7N

14.0

8O

16.0

9F

19.0

10Ne

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



Translational Motion Tes

KAPLAN

Passage I (Questions 1–5)

It is a fundamental fact of life that ants cannot be aslarge as elephants, and whales cannot be as small asgoldfish. For every type of animal there is a mostconvenient size. Each animal has a characteristic length L.From this length one can define a characteristic area whichis proportional to L2, and a characteristic volume which is

proportional to L3. The characteristic mass is also taken tobe proportional to L3.

The maximum size of an animal is constrained by thematerial out of which it is made. For example, considerwhether a giant human is a viable creature. If we rescaled ahuman by a factor of 10, we would necessarily increase theradius 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 areawould have to support 1000 times the weight.

Size also explains why an ant that falls off of theroof of a four story building will not be seriously injured,

whereas a human most certainly would be injured. A smallanimal can survive such a fall because of its relatively largesurface area to volume ratio. An object in free-fall withinthe Earth’s atmosphere will only accelerate up to a certainvelocity because of friction caused by the air resistance.The frictional force can often be approximated as beingproportional to the velocity times the cross-sectional areaof the falling object. The maximum velocity that an objectreaches 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 beproportional to L3?

I. Every animal can be assigned a characteristiclength L.

II. The characteristic volume of the animal isproportional to L3.

III. The density of the animal is constantthroughout that particular animal.

A . I onlyB . I and III onlyC . II and III onlyD . I, II, and III

2 . Suppose that the frictional force of air resistance only proportional to an object’s cross-sectional arand not its velocity. A falling object would the[Note: Assume that the frictional force is musmaller 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 time.

D . never reach terminal velocity.

3 . The terminal velocity of a human in the horizonspread eagle position would be:

A . smaller than if he/she is rolled up in a bbecause the cross-sectional area is greater.

B . greater than if he/she is rolled up in a ball becauthe cross-sectional area is also greater.

C . the same as if he/she is rolled up in a ball becauthe surface area to volume ratio is constant.

D . the same as if he/she is rolled up in a ball becau

the mass is the same in each case.

4 . The characteristic length of an ant is 0.5 cm, whthat of a human is 1 m. If q is the surface area dividby the volume, what is the ratio of q-ant to q-human

A . 200:1B . 400:1C . 2000:1D . 1:20

5 . Cells can only absorb nutrients through their surfacIn order for a cell to absorb nutrients most efficientit will grow to a maximum size and then divide. If ttotal volume remains constant when a cell dividthen the ratio of the total surface area of the daughcells to that of the original cell is:

A . 22:1B . 2:1C . 21/3:1D . 1:22

GO ON TO THE NEXT PAG



MCAT

4 as developed by

Passage II (Questions 6–11)

A circus wishes to develop a new clown act. Figure 1shows a diagram of the proposed setup. A clown will beshot out of a cannon with velocity v0 at a trajectory that

makes an angle θ = 45° with the ground. At this angle, theclown will travel a maximum horizontal distance. Thecannon will accelerate the clown by applying a constant

force of 10,000 N over a very short time of 0.24 sec. Theheight above the ground at which the clown begins histrajectory is 10 m.

45

Figure 1

A large hoop is to be suspended from the ceiling by amassless cable at just the right place so that the clown willbe able to dive through it when he reaches a maximumheight above the ground. After passing through the hoophe will then continue on his trajectory until arriving at thesafety net. Figure 2 shows a graph of the vertical

component of the clown’s velocity as a function of timebetween the cannon and the hoop. Since the velocitydepends on the mass of the particular clown performing theact, the graph shows data for several different masses.

[Note: The impulse J , or change in momentum pgenerated 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 overwhich the force acts on the clown. The acceleration of

gravity is g = 10 m/s2. sin 45° = cos 45° =1

2 ]

1.4 1.7 2.1 2.8 4.2

14

17

21

28

42

vy (m/s)

t (s)

4 0 k g

6 0 k g 8 0 k g 1 0 0 k g

1 2 0 k g

Figure 2

6 . If the angle the cannon makes with the horizontal isincreased 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/sB . 15 m/sC . 30 m/sD . 300 m/s

8 . The slope of the line segments plotted in Figure 2 is aconstant. Which one of the following physicalquantities does this slope represent?

A . –gB . v0

C . y–y0

D . sin θ

GO ON TO THE NEXT PAGE.




KAPLAN

9 . From Figure 2, approximately how much time will ittake for a clown with a mass of 100 kg to reach thesafety net located 10 m below the height of thecannon?

A . 1.7 sB . 3.4 sC . 3.9 s

D . 4.4 s

1 0 . If the mass of a clown doubles, his initial kineticenergy, mv0

2 /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 passingthrough it, what is the expression for the minimum

length of the cable so that he doesn’t hit his head onthe ceiling as he swings upward?

A . v0∆tM

B .v0

2

g

C .v0

2

2g

D .v0

2

4g

GO ON TO THE NEXT PAG



MCAT

6 as developed by

Questions 12 through 16 are NOTbased on a descriptive passage.

1 2 . Which of the following is the LEAST amount of information that could be used to determine how far acannonball lands from a cannon on Mars?

A . Initial speed, angle of inclination, acceleration due

to gravityB . Initial speed, time of travelC . Initial speed, acceleration due to gravityD . 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 10m/s. In what direction must the boat move to reach theexact 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 theshore.

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 directlyacross the river because the current will alwayspush it past that point.

1 4 . Which statement is true of a projectile at the highestvertical 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 atangles θ 1 and θ 2 respectively. Given that ml is greater

than m2 and θ 1 is greater than θ 2, which of the

following statements is consistent with theinformation?

A . Both blocks slide down with the same speedbecause 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 groundwith the same initial speed, but at different angles tothe horizontal. Which of the following statementsmust be true?

A . The ball launched at the larger angle hits theground first.

B . The ball launched at the smaller angle hits theground 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




KAPLAN

THE ANSWER KEY IS ON THE NEXT PAGE



MCAT

8 as developed by

ANSWER KEY:1. D 6. D 11. D 16. B2. D 7. C 12. A3. A 8. A 13. B4. A 9. C 14. B5. C 10. B 15. C





MCAT

10 as developed by

The surface area is proportional to L2, while the volume is proportional to L3. The parameter q is defined as the surfacearea divided by the volume, i.e.:

q =k L2

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

= 10.5

: 1100

= 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 shouldalso recognize the equivalence between a ratio and a quotient:

1

Lant :

1

Lhuman =

Lhuman

Lant =

100

0.5 = 200 =

200

1 = 200 : 1

5 . CLet 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, willhave 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 twodaughter 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 totalsurface 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 =21/3

1 = 21/3:1

Notice that 21/3 is greater than one, and so the total surface area has increased upon division.

Passage II (Questions 6–11)

6 . DThis 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 readthe first paragraph of the passage which states that when the cannon is at an angle of 45°, the clown will travel the maximumhorizontal 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 issituated at the point of maximum height, i.e. when all the vertical kinetic energy is transformed into gravitational potentialenergy. The higher its initial vertical kinetic energy, the greater the gravitational potential energy it can ultimately gain, and thehigher the hoop will have to be placed:




KAPLAN

1

2 mvy0

2 = mghmax

hmax =vy0

2

2g

7 . CThis 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

m =

10000•0.24

80 =

2400

80 = 30 m/s

Alternatively, for a mass of 80 kg, Figure 2 shows that the intitial velocity in the y direction is 21 m/s. This is relato the magnitude of initial velocity by

v0y = v0 sin 45°

The initial velocity is therefore v0 =21

sin45° = 21 2 . Without actually performing any calculations, we should be a

to decide that choice C, 30 m/s, is the only reasonable answer.

8 . AThe easiest way to do this problem is through dimensional analysis: Since all the choices have different units, if

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 verstime. Slope is in general the change in y over the change in x, and therefore in this case is the change in velocity over

change in time, which will have units ofm/s

s = m/s2. These are the dimensions of acceleration, and, looking at the answ

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

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,

vy = v0y – gt

where m, the slope, is –g.

9 . CThis question involves both interpretation of the graph in Figure 2 and general knowledge of projectile motion. T

graph enables us to determine the time the clown takes to reach the hoop. According to the graph, a 100-kg clown will exit cannon with an initial y-velocity of 17 m/s. (One can also obtain this value indirectly by finding the velocity using the form

for impulse as in #7 and then calculating the y-component from trigonometry.) One can then find the time t taken for the tby the equation:

y = vy0t +1

2 at2



MCAT

12 as developed by

In substituting values, we need to be careful in making sure that our signs are consistent. Defining the upwarddirection to be positive, y would be the net vertical displacement which is –10 m, vy0 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 answerchoices 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 histrip 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 isneeded for the clown to fall from the cannon level to the safety net 10 m further down. (At this point, you should be able toeliminate choices A and B. If you are pressed for time, therefore, a guess between C and D will at least maximize your chancesof getting the right answer.) A diagram should be helpful:

time = 1.7 svy = 0

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

10 m

Point B

How much time is needed for this last part of the trip, from point B to the safety net? One can avoid working withcumbersome kinetics equations by examining the two remaining answer choices more closely. First of all, we should realizethat 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 ismore reasonable? At point B, the clown already has a vertical velocity of 17 m/s. It would thus take slightly more than half a

second to travel 10 m (time =distance

speed ). Besides, 17 m/s is only an “initial” value; the clown is also accelerating from gravity.

Therefore choice C is the only reasonable answer.

1 0 . BThis 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 massdoubles, 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 behalved. Using a prime (’) to designate the new quantities, then, we have:

m’ = 2m

v0’ =v0

2

The new initial kinetic energy would therefore be:




KAPLAN

1

2 m’v0’2 =

1

2 2m (

v0

2 )2 =

1

4 mv0

2 =1

2 (

1

2 mv0

2) =1

2 KEold

1 1 . DThis question is most easily answered using the principle of conservation of energy. Since the clown reaches the ho

at the maximum height of his trajectory, the vertical component of his velocity is zero. There are no forces acting on him in

horizontal direction, and so his horizontal velocity is the same as its initial value, that is, v0cos 45° orv0

2 . The clow

kinetic energy as he reaches the hoop is therefore12

m (v0

2 )2 =

14

mv02.

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

y = minimum

length of cable

The condition that y needs to satisfy is therefore:

mgy =1

4 mv0

2

y =v0

2

4g

Independent Questions (Questions 12–16)

1 2 . AAll of the answer choices include the initial speed. What else is needed to determine how far a cannonball lands from

cannon? Remember that speed is a scalar: it has a magnitude but no direction. In order to resolve the cannonball’s motion ivertical and horizontal components (which we need to do), we need the direction of travel —or in terms of the choices in tquestion—the angle of inclination. In addition, we need to know how large the force is that is pulling the cannonball down: Tacceleration due to gravity on Mars, which will differ from the familiar 9.8 m/s2, will affect the amount of time the cannspends 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 not sufficient. There is no net force in the horizontal direction, and so the horizontal velocity is constant until the ball hits 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 dictat

by the kinetics in the vertical component. Specifically, it can be found using the kinetics equation:

y = v0yt +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) care has to be taken to make sure that v0y and a have opposite signs since they point in opposite directions. v0y is v0sinθ, a

thus no additional information is needed to determine this that was not already required to find vx (i.e. v0 and θ). The only ot

piece of information, then, is the acceleration due to gravity. Let us now also consider why the other answer choices are ncorrect.

Choice B gives us the time of travel, and thus it would save us the calculation to find t. However, initial speed 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



MCAT

14 as developed by

horizontal direction, which we cannot determine with the quantities in choice B. (Incidentally, we can use the kinetics equationabove to solve for the angle of inclination if we know the initial speed, which we do, the time, which we do, and theacceleration 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 thetime spent traveling without the value of the acceleration due to gravity on Mars.

1 3 . BThis 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 canthus eliminate choice A immediately. The two velocity vectors, the one of the downstream current and the one of the boat instill 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).

boat: 20 m/s

current: 10 m/s

resultantθ

The question, in other words, is: What should θ be so that the resultant vector would point directly towards theopposite 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. Inthe 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 getto the opposite shore, making choice D incorrect in this case. If the current is, say, 25 m/s instead, then the boat would nothave been able to make it to the opposite shore at the same point.

1 4 . BThe 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, thekinetic energy associated with the projectile’s motion in the vertical direction is completely transformed into gravitationalpotential 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.

1 5 . CWe need to derive an equation for the acceleration of a black down the inclined plane. A diagram would be most

helpful:

θ

mgθ

friction = µN = µmgcosθ

mgsinθ

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




KAPLAN

plane, it is the component of the weight that is perpendicular to (or normal to, hence the name) the plane: mgcosθ. The frictforce 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 m1 is greater th

m2 is irrelevant. They both start from rest, and so whichever experiences a larger acceleration will slide down faster, and

acceleration depends only on the angle of inclination, since the coefficient of friction is the same in both cases. The questiothen, 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 gofrom 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

greater than θ2 means that sinθ1>sinθ2, and that cosθ1<cosθ2. Looking at the equation for a derived above, one sees that

block 1, we would have a greater number minus a smaller number compared to block 2. The acceleration for block 1 is therefgreater.

1 6 . B

As has been discussed above for question #12, the time a projectile spends in air can be found by solving for t in tequation:

y = 0 = v0yt – 1

2 gt2

where we have substituted in –g for the acceleration. Since this is a quadratic equation, there are two solutions that one obtain through straightforward factoring:

t (v0y – 1

2 gt) = 0

t = 0 or t =2v0y

g

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

hence2v0y

g , or

2v0sinθ

g . The ball launched at a smaller angle would have a smaller value of sinθ (see discussion for previo

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 horizon

distance traveled. As is explicitly stated in passage 2, the horizontal distance is maximized if the angle is 45 °. There itrade-off between the time spent in air and the horizontal velocity, both of which determine the horizontal distance traveled (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 potentenergies. From the principle of conservation of energy, this sum has to be equal to the initial kinetic energy that the bpossesses as it exits the cannon. Since the two balls are launched with the same initial speed and have the same mass, thinitial kinetic energies, and thus total mechanical energy, will be equal and have a value of mv0

2 /2.

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1Translational Motion Test w. Solutions