4Force, Motion, Gravitation, And Equilibrium Test w. Solutions

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PHYSICS TOPICAL:

Force, Motion, Gravitation andEquilibrium Test 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.





MCAT

4 as developed by

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 theEarth 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 isproportional to the cube of the semimajor axis of its orbit.

For objects near the surface of the Earth, a greatsimplification in the mathematics occurs by approximatingthe 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 gravitationalforce 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 isdefined to be the free-fall acceleration. (Note: The universalgravitational constant is G = 6.67 × 10–11 N•m2 /kg2. Themass 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 sameheight. If the gravitational force is assumed to beconstant, then the theory predicts that they will hit the

Earth’s surface at the same time. If the actual force isused 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

a c c e l e r a t i o n

a c c e l e r a t i o

n



A.

B.

C.

D.

height fallen

height fallen height fallen

9 . Which of the following expressions shows how theacceleration due to gravity depends on the height of anobject 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 theEarth at a distance R from the Earth’s center. If R isincreased by a factor of 4, then the period of the orbit

will be: (Note: The semimajor axis of a circle equalsits radius.)

A . unchanged.B . increased by a factor of 4.C . increased by a factor of 8.D . increased by a factor of 64.



Force, Motion, Gravitation and Equilibrium Tes

KAPLAN

GO ON TO THE NEXT PAGE. 1 1 . The amount of energy needed to move a rocket havimass m from the surface of the Earth to a distancefrom the Earth’s center is E = GmMe(1/Re – 1/ R

What is the minimum velocity the rocket must hain order to completely escape the Earth’s gravitatiofield?

A . (2Gm /Re)1/2

B . (2GMe /Re)1/2

C . (2gh)l/2

D . (2mGRe)l/2



MCAT

6 as developed by

Questions 12 through 16 are NOTbased on a descriptive passage.

1 2 . A 1-kg aluminum sphere and a 3-kg brass sphere, bothhaving the same diameter, are allowed to fall freelyfrom the same height above the ground. Neglecting airresistant and assuming both spheres are released at thesame 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 slidethat makes a 30° angle with the ground. Thecoefficient of kinetic friction between the boy and theslide is 0.20. What is the magnitude of the boy’sacceleration as he slides? (Note: Assume that the

acceleration due to gravity, g, equals 10 m/s2. Thecosine 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 kgB . 400 kgC . 800 kgD . 1,000 kg

GO ON TO THE NEXT PAGE.

1 5 . A massless rod of 4 meters is placed on a fulcrum asshown in the diagram below. An object of mass m isattached to one end and an object of mass 2m isattached to the other end. Where must the fulcrum beplaced for the system to remain in equilibrium?

2m m

A . 1.33 meters from the leftB . 1.50 meters from the leftC . 2.33 meters from the leftD . 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 isslowed down by a frictional force. The net force on themass when it lies in the position shown, pointstoward:

C

10 kg

D

A

B

v

A . AB . BC . CD . D

END OF TEST




KAPLAN

THE ANSWER KEY IS ON THE NEXT PAGE



MCAT

8 as developed by

ANSWER KEY:1. D 6. C 11. B 16. D2. A 7. A 12. B3. C 8. A 13. B4. A 9. D 14. C5. B 10. C 15. A




KAPLAN

EXPLANATIONS

Passage I (Questions 1—5)

One should recognize that it would not be efficient to try to memorize all of the information presented in the passaas one is reading it for the first time. Instead, it is better to concentrate on the physical content of the situation. First of all, suspect that energy conservation is an important tool since the acceleration is not constant: most of the kinetics equatiowould therefore not apply. We have not been given explicit information about friction, but we know from energy conservat

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 passathat 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 v a. Its initial kinetic energy is therefore

Mva2. In addition to this, however, the cart also has gravitational potential energy equal to Mgh. The sum of the two is its to

initial energy. At point B, the cart has no gravitational potential energy but only kinetic energy. We are not given the velocof 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 ini

potential energy of the cart. Therefore the amount of energy that has been lost to friction is1

2 Mva

2, or 100% of the ini

kinetic energy.

2 . AChoice C, 180°, can be eliminated immediately since that corresponds to point C, at which the normal force is zaccording to Figure 2. Within the loop the cart is undergoing circular motion (not uniform since the velocity is changing). any point the instantaneous centripetal acceleration is v2 /r, where v is the velocity at that point. This acceleration needs to provided by a net force perpendicular to the track. The normal force, by definition, is perpendicular to the track, but there walso be a component of gravity that acts in the same or opposite direction. The component of gravity pointing in the opposdirection as the normal force is given by Mgcosϕ:

Mg

normal force

j

Mgcos j

Thus the equation is:

N – Mgcosϕ = Mv2

r

N = M (v2

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 cowould be positive, leading to a larger value for the normal force: the normal force needs to be greater because it needscounteract gravity and provide the centripetal acceleration. Conversely, at the top half of the loop, ϕ is between 90° and 27and cosϕ is negative, leading to a smaller value for the normal force. Physically, there is a component of gravity that is para



MCAT

10 as developed by

(rather than antiparallel) to the normal force and therefore it “reduces the burden” of the normal force to provide the centripetalacceleration, 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 amaximum 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 amaximum.

3 . CFriction 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. Theequation is otherwise unaffected since the friction force is always directed opposite in direction to that of travel, and is thusperpendicular to the normal force: the friction force does not enter into the equation directly since it has no component in thenormal direction.

4 . AAccording 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 is1

2 Mvc

2. Substituting in the relation we just obtained, we have:

K.E. =1

2 Mgr.

5 . BThe 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 . CThe 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 inheight of various objects since it was small compared to the radius of the Earth; the approximation lies not in the massesinvolved. In the scenario given, the balls are dropped from the same height, and thus whatever correction is made to the fallingtime 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 alarger gravitational force, but does not mean a larger acceleration since the m’s cancel. What we have here is a more accurateexpression for g: one that is not constant but varies with distance. The mass, however, still does not affect the acceleration.

7 . AA 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 lineconnecting the satellite to the center of the Earth sweeps out equal areas in equal times. This means that if the satellite sweepsout 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 thearea and the angle θ that the satellite sweeps out: the area is directly proportional to θ. Therefore, if the satellite sweeps out anangle 2θ, the area swept out will be 2A. The time is therefore twice that of sweeping out θ, from the reasoning in the paragraphabove.




KAPLAN

8 . AWe are used to thinking of acceleration due to gravity as being constant (9.8 m/s2), and so may have been tempted

choose choice D. The passage however points out that this is only an approximation. In reality, the acceleration is changibecause the gravitational force on an object is greater as it gets closer and closer to the Earth. The force is inversely proportioto the square of the distance; and the acceleration is directly proportional to the force (ma = F g). As the distance between

object and the Earth gets smaller and smaller, then, the force and the acceleration is getting larger and larger.We then need to decide between choices A and B. Would the acceleration increase linearly with time, as in B,

nonlinearly, as in A? The gravitational force depends on the inverse distance squared. This would lead us to expect that

increase would be nonlinear.

9 . DThe passage indicates that the exact gravitational force is:

F =GMem

(R e + h) 2

This is the force that provides the acceleration as the object falls, i.e. the expression is also equal to ma. Solving fthe acceleration, then, one notices that the mass of the object, m, cancels to result in:

a =GMe

(R e + h)2

Notice that in the case of the approximate theory, a =GMe

Re2 and this is the constant g with which we are so familiar

1 0 . CThis question requires us to find a relationship between the period and the radius of orbit for a satellite. Kepler ’s th

law is precisely what we need: It says that the square of the period of a satellite is proportional to the cube of the semimaaxis of its orbit. The term “semimajor axis” is used in reference to an elliptical orbit, but we need not know its exact definitsince, in the question stem, we are told that this is equal to the radius of the circle in the case of a circular orbit. Trelationship is therefore:

T2 = kR3

where T is the period, or the time it takes for the satellite to complete one round trip, R is the radius (or semimajor axis), anis the proportionality constant. If the radius increased by a factor of four, i.e. R’ = 4R, then, the new period T’ would satisfy:

(T’)2 = k(4R)3

= 64kR3

= 64 T2

= (8T)2

The last two lines of the equation tell us that the new period squared is 64 times the old period squared, or equivalenthat the new period is eight times the old period.

1 1 . BThis question can be handled by an energy conservation approach. The question stem gives us the energy required

move a rocket having mass m from the surface of the Earth to a distance R from the Earth’s center. In order to find the enerrequired to escape the Earth’s gravitational field completely, we need to move the rocket from the surface of the Earth to infinite distance from the Earth’s center. Setting R equal to infinity and using the fact that 1/ = 0, we find that the eneneeded to escape the Earth’s gravitational field is:

E = GmMe (1

Re –

1 ) =

GmMe

Re



MCAT

12 as developed by

Where does this energy come from? The initial kinetic energy of the rocket. If the initial kinetic energy is greater thanthis amount, it will be able to escape the Earth’s gravitational field, and the difference will be the leftover kinetic energy. Theminimum velocity the rocket must have therefore corresponds to a kinetic energy that is just equal to the expression above:

1

2 mvmin

2 =GmMe

Re

Canceling m from both sides and solving for vmin gives:

vmin =2GMe

Re

Independent Questions (Questions 12–16)

1 2 . BEven though the force of gravity is dependent on the mass of the dropped object, it does not enter into the acceleration

since the mass cancels when one equates the gravitational force to ma. In other words, all objects, regardless of their mass, areaccelerated towards the Earth at the same rate (assuming there is no air resistance, as we are instructed to do so in the stem). If the initial height of the object is small compared to the radius of the Earth, this rate is a constant g, equal to 9.8 m/s 2. The timeit takes for the spheres to fall can be determined from the kinetics equation:

h = v0t +1

2 gt2

where h is the initial height of the object, and v0 the initial velocity. This is a quadratic equation in t, but if the spheres are

dropped (rather than thrown down), the initial velocity is zero and thus the time for each sphere to reach the ground is2h

g .

From kinetics equations, one can also determine the speed of the sphere right before it hits the ground. All thesecalculations are obviously not necessary for this question; the most important thing is to realize that the accelerationexperienced by the two spheres would be equal, which in turn implies that the time taken and the final speed are the same for thetwo as well.

1 3 . BThis question is a straightforward application of Newton’s second law which states that the vector sum of all the forces

acting on a body is equal to its mass times its acceleration vector. In this case there are three forces acting on the boy:

30°

mg

normal force

friction

The first is his weight mg acting in the vertical direction downwards. The second is the normal force actingperpendicular to the slide upwards. The third is the force of friction parallel to the slide but opposite to the direction of motion.What is the magnitude of the acceleration vector? The boy accelerates downward along the surface of the slide. The acceleration

vector is therefore parallel to the slide. Its magnitude is the difference between the magnitude of the component of gravityparallel to the slide and the magnitude of the friction force, divided by the mass of the boy. The magnitude of the friction forceis dependent on the normal force N via the equation:

Ffr = µN




KAPLAN

where µ is the coefficient of kinetic friction. The magnitude of the normal force N is in turn dependent on the weight and tangle of the slide: it must exactly balance the component of gravity perpendicular to the slide; otherwise the boy wouaccelerate in that direction. I.e.:

N = mg cos30°

Ffr = µN = µmg cos30°

The component of gravity parallel to the slide is mg sin30°, and so the acceleration a satisfies:

mg sin30° – µmg cos30° = ma

m cancels on both sides since it appears in all three terms. Approximating g by 10 m/s 2 and substituting in the valufor the coefficient of friction, we get:

10 sin30° – 0.2 × 10 cos30° = a

a = 10 × 0.5 – 0.2 × 10 × 0.866 = 5 – 1.74 = 3.26 m/s2

Choice A is the deceleration caused by friction. Choice C is the answer obtained if friction were ignored. If yreversed the sine and the cosine, you would have gotten choice D. It is important to understand the trigonometry involved:

30°

mgx

y

The angle between the vertically downward force and the component perpendicular to the plane (between mg and xthe diagram above) is always the same as the angle that the inclined plane makes with the horizontal, in this case 30°. Tlengths x and y can then be found by simple trigonometry (make sure you can rotate the triangle in your mind to see how oobtains the diagram below):

mgy

x

30°

y

mg = sin30° or y = mg sin30°

x

mg = cos30° or x = mg cos30°

1 4 . CThe car goes from an initial velocity of 25 m/s to 0 m/s in 10 s. The average deceleration can be found by:

a =v

t =

(0 – 25)

10 = –2.5 m/s2

The mass of the car can then be determined by using Newton’s second law F = ma, where F in this case is the averaforce, equal to 2000 N.

m =F

a =

2000

2.5 = 800 kg



MCAT

14 as developed by

Alternatively, we know that the change in momentum is equal to the impulse:

p = (mv) = mv = Ft

m =Ft

v =

2000 (10)

25 = 800 kg

1 5 . AFor the rod to remain in equilibrium, the torque acting in the clockwise direction must equal the torque acting in the

counterclockwise direction. Torque is equal to force × (distance from fulcrum). In this case the force is the weight of the objects.x 4 – x

F = mgF = 2mg

If we let the distance between the fulcrum and the left end of the rod be x meters, then the torque caused by the mass onthe left is 2mgx in the counterclockwise direction. The torque generated by the mass on the right will then be mg(4 – x) in theclockwise direction. Setting the two torques equal and solving for x:

2mgx = mg (4 – x)2x = 4 – x

3x = 4x =

4

3 = 1.33 m

1 6 . DFor any object traveling in a circle, there is an acceleration directed towards the center of the circle that captures the fact

that the object is constantly changing direction. From Newton’s second law, there must be a force (or a component) that pointstowards the center of the circle since the force and acceleration vectors are parallel. In the case of uniform circular motion(constant speed), then, the net force and the net acceleration would both point to the center. In this question, however, inaddition to this centripetal acceleration, we also have a change in speed brought about by friction. Friction force acts in adirection opposite to the tangential motion of the object. Since the block is moving counterclockwise, the friction force mustthen act in a clockwise direction. When the object is in the position shown in the diagram, the friction force points downward,parallel to C. This, added vectorially to the force sustaining centripetal acceleration, would give a vector that points in the samedirection as D. This can be seen using the pictorial method of vector addition:

centripetal force

friction forcevector sum

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4Force, Motion, Gravitation, And Equilibrium Test w. Solutions