7Work, Energy, And Momentum Test w. Solutions

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PHYSICS TOPICAL:

Work, Energy and MomentumTest 1

Time: 22 Minutes*Number of Questions: 17

* 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.



Work, Energy and Momentum Tes

KAPLAN

Passage I (Questions 1–4)

Two stars bound together by gravity orbit each otherbecause of their mutual attraction. Such a pair of stars isreferred to as a binary star system. One type of binarysystem is that of a black hole and a companion star. Theblack hole is a star that has collapsed on itself and is somassive that not even light rays can escape its gravitational

pull. Therefore, when describing the relative motion of ablack hole and a companion star, the motion of the black hole can be assumed negligible compared to that of thecompanion.

The orbit of the companion star is either ellipticalwith the black hole at one of the foci or circular with theblack hole at the center. The gravitational potential energyis given by U = –GmM/r , where G is the universalgravitational constant, m is the mass of the companionstar, M is the mass of the black hole, and r is the distancebetween the center of the companion star and the center of the black hole. Since the gravitational force isconservative, the companion star’s total mechanical energy

is a constant of the motion. Because of the periodic natureof the orbit, there is a simple relation between the averagekinetic energy <K > of the companion star and its averagepotential energy <U >. In particular, <K > = –<U /2>.

Two special points along the orbit are singled out byastronomers. Perigee is the point at which the companionstar is closest to the black hole, and apogee is the point atwhich it is farthest from the black hole.

1 . At which point in an elliptical orbit does the

companion star attain its maximum kinetic energy?

A . ApogeeB . PerigeeC . The point midway from apogee to perigeeD . All points in the orbit, since the kinetic energy is

a constant of the motion

2 . For circular orbits, the potential energy of thecompanion star is constant throughout the orbit. If theradius of the orbit doubles, what is the new value of

the velocity of the companion star?

A . It is 1/2 of the old value.

B . It is 1/ 2 of the old value.C . It is the same as the old value.D . It is double the old value.

3 . Which of the following prevents the companion sfrom leaving its orbit and falling into the black hole

A . The centripetal forceB . The gravitational forceC . The companion star’s potential energyD . The companion star’s kinetic energy

4 . The work done on the companion star in one complorbit by the gravitational force of the black hoequals:

A . the difference in the kinetic energy of tcompanion star between apogee and perigee.

B . the total mechanical energy of the companion stC . zero.D . the gravitational force on the companion s

times the distance that it travels in one orbit.

5 . For a circular orbit, which of the following gives tcorrect expression for the total energy?

A . −1

2

2mv

B . mv2

C . −GmM

r

D .G

2

mM

r

6 . What is the ratio of the acceleration of the black hoto that of the companion star?

A . M/ mB . m/M C . mM/r D . 1/1

GO ON TO THE NEXT PAG



MCAT

4 as developed by

Questions 7 through 12 areNOT based on a descriptivepassage.

7 . What is the power required to accelerate a 10 kg massfrom a velocity of 5 m/s to a velocity of 15 m/s in 3 seconds?

A . 33WB . 167WC . 225 WD . 333W

8 . A block of mass m slides down a plane inclined at anangle θ . Which of the following will NOT increase theenergy lost by the block due to friction?

A . Increasing the angle of inclinationB . Increasing the distance that the block travelsC . Increasing the acceleration due to gravity

D . Increasing the mass of the block

9 . A block of mass m starts from rest and slides down africtionless semi-circular track from a height h asshown below. When it reaches the lowest point of thetrack, it collides with a stationary piece of putty alsohaving mass m . If the block and the putty stick together and continue to slide, the maximum heightthat the block-putty system could reach is:

h

A .h

4 .

B .h

2 .

C . h.

D . independent of h.

1 0 . A boy hits a baseball with a bat and imparts animpulse J to the ball. The boy hits the ball again withthe same force, except that the ball and the bat are incontact for twice the amount of time as in the first hit.The new impulse equals:

A . half the original impulse.B . the original impulse.

C . twice the original impulse.D . four times the original impulse.

1 1 . The power of solar radiation incident on a solar panelis 40 kW. If the efficiency of the solar panel is 10%,how much energy does the solar panel generate in 300minutes?

A . 1.2 MJB . 72.0 MJC . 120 MJD . 720 MJ

1 2 . Two billiard balls undergo a head-on collision. Ball 1is twice as heavy as ball 2. Initially, ball 1 moveswith a speed v towards ball 2 which is at rest.Immediately after the collision, ball 1 travels at aspeed of v /3 in the same direction. What type of collision has occurred?

A . inelasticB . elasticC . completely inelasticD . Cannot be determined from the information given.

GO ON TO THE NEXT PAGE.




KAPLAN

Passage II (Questions 13–17)

Particle storage rings facilitate collisions betweenelectrons and positrons (positively charged electrons).When electrons and positrons collide at high energies, theycan annihilate each other and produce a variety of elementary particles, including photons. In these reactions,momentum is always conserved. Powerful magnets are

placed at various points along the ring to create a forcedirected toward the center, thereby guiding the particles in acircular motion. In a scattering experiment, the particlebeams circle in opposite directions and collide head-on atthe interaction points, which are surrounded by particledetectors. The particles in the ring are accelerated at one ormore points along the ring by an external electric field.When charged particles accelerate, they radiateelectromagnetic energy. Radio frequency power iscontinually fed into the storage ring to compensate for theenergy loss.

The reaction rate, R , is the number of particlesscattered per second in the storage ring. It is given by the

formula: R = Lσ , where L is the luminosity and σ is thecross-section of the reaction. The cross-section is a quantitythat depends only on the particular type of reaction beingconsidered. The luminosity contains all of the informationabout the initial conditions for a reaction and is given by:

L N N

A f e e=

− +

,

where N e– refers to the number of electrons, N e+ refers to

the number of positrons, A is the cross-sectional area of

the storage ring, and f is the number of revolutions per

second made by the particles. Electron-positron reaction

rates for modern particle storage rings are typically on theorder of l0–3 s–1.

1 3 . What percentage of the work done on the circulatingparticles is done by the magnetic field?

A . 0%, because the direction of the magnetic force isperpendicular to the direction in which theparticles travel

B . 0%, because the magnetic field doesn’t exert aforce on the particles

C . 50%, because the magnetic field and the electricfield provide equal amounts of energy to theparticles

D . 100%, because the magnetic field is the source forthe centripetal force that accelerates the particles

1 4 . Which of the following would increase the reactirate in an electron-positron storage ring?

I. Decreasing the cross-sectional area of thestorage ring

II. Increasing the energy of the particlesIII. Increasing the number of positrons in the

storage ring

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

1 5 . The work done on the particles by the gravitationfield of the Earth is not considered when figuring thenergy loss per revolution. This is because:

A . the gravitational force is perpendicular to gravitational acceleration.

B . the energy lost due to gravity is equal to tenergy gained from the magnetic fields.

C . the particles do not experience a significgravitational force.

D . the luminosity is not dependent on gravitationacceleration.

1 6 . Two electrons with equal speed collide head-on attotal energy of 180 GeV. If the speed of the fielectron after the collision is 0.9c, what is the speedthe second electron after collision? (Note: The speedlight in a vacuum is c = 3.0 × 108 m/s.)

A . 0.45cB . 0.7c

C . 0.8cD . 0.9c

1 7 . An electron and a positron are held in a storage rinand they each lose 260 MeV of energy per revolutiin the form of electromagnetic radiation. If frequency of revolution is 10,000 Hz, how mupower must be supplied to keep the total enerconstant at 180 GeV?

A . 1.8 × 106 MeV/sB . 2.6 × 106 MeV/sC . 5.2 × 106 MeV/s

D . 9.0 × 108 MeV/s

END OF TES



MCAT

6 as developed by

ANSWER KEY:1. B 6. B 11. B 16. D2. B 7. D 12. B 17. C3. D 8. A 13. A4. C 9. A 14. D5. A 10. C 15. C




KAPLAN

EXPLANATIONS

Passage I (Questions 1—6)

1 . BIn the passage we are told that the total mechanical energy of the companion star is constant as it orbits around t

black hole. The total mechanical energy is the sum of the gravitational potential energy and the kinetic energy. We do not reaknow anything directly about the kinetic energy in this case, except that it equals mv2 /2. We are, however, given a formula

the gravitational potential energy of the companion star. Since the mechanical energy is constant, the point at which the kineenergy is maximum must also be the point at which the potential energy is minimum. From the more familiar case of droppa ball on Earth, we know that the potential energy is lower when the height is smaller. We may thus expect that the potentenergy is minimum when the star is closest to the black hole. We can confirm this by using the formula given in the passagU = – GmM/r. Since G, m, M, and r are all positive quantities, U is always going to be negative and has a maximum valuezero when the two are infinitely far apart. As r decreases, U becomes more and more negative. It follows, then, that the pointwhich the potential energy is minimum is the point of closest approach to the black hole, i.e. the perigee. The perigee, then,also the point where the kinetic energy is the greatest.

2 . BThe question stem states that the gravitational potential energy for circular orbits is constant. This implies that

kinetic energy is also constant (from the conservation of total mechanical energy). In the last line of the second paragraph of passage, we are given a relationship between the average value of the potential energy and the average value of the kineenergy. Since in this case the two are constants, their average values are identical to their actual values. Therefore, for a circu

orbit, the kinetic energy, K, is : K = – U/2. The potential energy, U, is equal to – GmM/r. If the distance r is doubled, then, absolute value of the potential energy is reduced in half, and thus the kinetic energy will also be cut in half. (Recall from odiscussion for the question above that the potential energy is negative; a reduction of its absolute value thus leads to a lnegative number or a higher potential energy, as is reasonable when one separates two objects that are attracted to each otheSince the kinetic energy is proportional to the velocity squared, as the kinetic energy becomes 1/2 its original value, t

velocity must become 1/ 2 its original value to satisfy the definition of kinetic energy mv2 /2.

3 . DLet us examine each choice and try to eliminate those that seem incorrect. Choice A states that the companion s

does not fall into the black hole because of the centripetal force. The centripetal force on the star is an attractive force directowards the black hole. It is provided by the gravitational attractive force between the two objects and hence, if anything, itwhat leads us to expect the two to come together.

Choice B, the gravitational force, is, as just mentioned, what is providing the centripetal force. This is one way th

we could have eliminated these two choices: since they are essentially the same thing, neither of them can be correct!Choice C states that the companion star’s potential energy keeps it in orbit. Does this make sense? Think about a b

that you hold in your hand. It has some potential energy by virtue of the fact that you are holding it above the ground, but tpotential energy does not keep the ball from falling: it is your hand that does this. If you let go, the potential energy converted to kinetic energy as the ball goes speeding towards the ground. Analogously, then, we would not expect potenenergy to be the agent responsible for keeping the star in orbit here.

Choice D states that the companion star does not fall into the black hole because of its kinetic energy. This maksense because its tangential velocity is what keeps it in orbit and kinetic energy is certainly related to tangential velocity. Tgravitational force acts as a centripetal force which keeps redirecting the star so it doesn’t fly off into space. Why would it off into space? Because it has kinetic energy from the velocity in the tangential direction. A stable orbit occurs because of balance between the kinetic energy and the gravitational attraction. If the kinetic energy were not large enough, the star wofall or spiral into the black hole.

4 . C

There are several ways that one could have obtained the answer. One is by the work-energy theorem: The net work doon an object is the change in its kinetic energy. In the case of a complete orbit, this is the kinetic energy the star has at the eof one round of the orbit minus the kinetic energy it had at the beginning (at the same point in space). Since potential energydependent only on position, we know that the star would have the same potential energy after one orbit that brings it back to starting point. Since total mechanical energy is conserved, the fact that the potential energy is the same before and after implthat the kinetic energy is also the same before and after. The change in kinetic energy after one orbit is therefore zero. From work-energy theorem, then, we know that the work done on the star has to be zero as well.



MCAT

8 as developed by

In addition, the passage tells us that the gravitational force is conservative (which is something you might alreadyknow anyway). One of the properties of a conservative force is that the work it does is path-independent, which in turn impliesthat the work it does after one loop (or orbit) that starts and ends at the same place is zero. Since this force is the only oneacting on the star, the work done on the star is zero.

Beware of choice D, which states that the work done equals the gravitational force of the black hole on the companionstar times the distance traveled. This is very close to the definition of work and looks like an obvious choice. However, whatthis formulation leaves out is the angle between the force and the direction of travel. The formula for work done by a force is: W= Fd cosθ. In this case the force is directed towards the center (or focus of an ellipse) and the direction of travel is tangent to the

orbit. Therefore, for an elliptical orbit, the angle between the two is never zero (the cosine is never 1):

black hole

star

gravitational force

direction of travel

The work done over any segment of the orbit, then, is never simply the product of the gravitational force and thedistance traveled. The symmetry of the problem, in fact, leads to net cancellation of the work done after one complete orbit.Also, note that in the special case of a circular orbit, the two are always perpendicular, and thus the cosine of the angle, andthe work done, are zero all throughout the orbit.

5 . AAs discussed in the explanation to #2, the kinetic and potential energies are both constants of the motion for a circular

orbit because r is constant. We can thus take out the average signs <> and write the equation given in the passage as K = – U/2.Since the passage also gives us an expression for the gravitational potential energy, we are tempted to substitute that in and get:

E = K + U = – U

2 + U = – (–

GmM

2r ) + (–

GmM

r ) =

GmM

2r –

GmM

r =

GmM

2r –

2GmM

2r = –

GmM

2r

While this is certainly correct, this is not one of our answer choices: choice D does not have the right sign. Wetherefore must find another way of expressing the quantity by using the fact that K = mv2 /2. Since K = – U/2, U = – 2K:

E = K + U = K + (– 2K) = K – 2K = –K = – 1

2

mv2

You may be surprised that the total energy is negative, but remember that we can define the zero of potential energyany way we want and in this case, as has been discussed, the potential energy is always negative. This is purely a matter of convention (and convenience); since the kinetic energy has a magnitude half of that of the potential energy, it is not sufficient tooverwhelm the potential energy term and thus the total energy remains negative.

6 . BFrom Newton’s second law, the acceleration an object undergoes is equal to the force it experiences divided by its mass.

In this particular case, M is the mass of the black hole while m is the mass of the star. The force each experiences is the samein magnitude from Newton’s third law: The gravitational force the star exerts on the black hole is the same as the gravitationalforce the black hole exerts on the star. The magnitude of the force is GmM/r2. The acceleration that the black hole undergoes istherefore this force divided by M, or Gm/r2, while the acceleration the star undergoes is the same force divided by m, or GM/r2.The ratio of the former to the latter is therefore Gm/r2:GM/r2 or m:M.

Note that we need not even have come up with the expression for the force explicitly: all we need is to be able torecognize that the two forces will have equal magnitude from Newton’s third law. We can then write the ratio as:

ablack hole:astar =F

M :

F

m =

1

M :

1

m = m:M

where the last step can be made if one recognizes that a ratio is itself just a quotient.




KAPLAN

Independent Questions (Questions 7–12)

7 . DTo do this problem one needs to understand the relation between power, work and kinetic energy. Power is defined

the amount of work done per unit time. The time in this case is given as three seconds, but we need to determine the amountwork done. This can be obtained using the work-energy theorem, which states that the work done is equal to the changekinetic energy:

W = ∆KE = ∆ (12

mv2) = 12

m ∆(v2) = 12

m (vf 2 – vi

2)

where vf is the final velocity and vi is the initial velocity. (Note: ∆(v2) is not the same as (∆v)2! I.e. we cannot write

change in kinetic energy as1

2 m(vf – vi)

2.) Substituting in numbers supplied by the question, we have the change in energy

be equal to1

2 × 10 kg × (225 – 25) m2 /s2 = 1000 J which is the work done. This divided by 3 seconds gives 333 J/s or 333 W

8 . AThe energy lost to, or dissipated by, friction is the work done by friction, which is equal to the magnitude of the fo

times the distance traveled. Anything that increases either of these two factors will increase the energy lost to friction. We thus looking for something that does not increase either of these two. Choice B can be immediately ruled out since it explic

suggests increasing the distance traveled. To arrive at the correct answer, we can almost rely on intuition alone. Increasing acceleration due to gravity (choice C) or the mass of the block (choice D) will increase the weight, i.e. the gravitational forcethe block. It is therefore “held more tightly” to the plane, and so one would expect a higher friction force. Increasing the anof inclination, on the other hand, decreases the component of gravity that is perpendicular to the plane, and so the block is held so tightly to the plane. This would lead us to expect that the friction force would be less, and so less energy will dissipated.

For a fuller understanding of the scenario, we can derive an expression for the energy lost because of friction in terof the quantities that appear in the answer choices. First, we notice that the force and the motion are in opposite directions so:

W = Fdcos180° = – Fd = ∆E due to friction

The negative simply implies that the work done by the friction force causes the block to lose rather than gain energSince we are only interested in the magnitude of the energy lost, the negative sign can be ignored in our calculation. At tpoint it is helpful to refer to a diagram with which you should already be familiar:

θ

mg θ

friction = µN = µmgcosθ

normal force = mgcosθ

If we let d be the distance traveled by the block as it slides down the inclined plane, then the energy dissipated friction is µmgdcosθ. With this expression in hand, we can evaluate the answer choices. Choice A states that increasing angle of inclination would not increase the amount of energy lost. The cosine of angle decreases if the angle increases within range of 0° to 90° (cos 0° = 1; cos 90° = 0). Since the angle of inclination is within this range, increasing the angle inclination would decrease the cosine term in the expression above, thus decreasing the amount of energy lost. It is therefotrue that increasing the angle of inclination would not increase the amount of energy lost, making choice A the correct respon

9 . A



MCAT

10 as developed by

The most important thing to realize in this problem is that energy is no t conserved in the collision which is totallyinelastic since the two objects stick together. What is conserved is momentum, with which one can determine the velocities of the bodies. Energy conservation, however, does come into play as potential energy is converted into kinetic energy up to thepoint right before the collision, and vice versa after the collision. We shall see exactly how all these come together.

Initially we have a block of mass m at a height of h. The potential energy, which is also the total energy, is mgh.This will be entirely converted into kinetic energy of the block at the point immediately before collision, and we can thus solvefor the velocity of the block before collision via energy conservation:

mgh = 12

mv2

v = 2gh

The momentum before the collision is thus p = mv = m 2gh . Momentum is conserved in the collision, so the

momentum of the block-putty system after the collision will also be m 2gh . We can use this relation to determine thevelocity of the system after collision, v’:

(2m)v’ = m 2gh

where the mass on the left hand side is 2m since the putty also has a mass of m and so the total mass of the system is m + m =2m after collision. Solving for v’ gives us:

v’ =2gh

2

Immediately after the collision, the block-putty system has no potential energy. The kinetic energy is thus the totalenergy, and this is equal to:

1

2 (2m) v’2 =

1

2 (2m)

2gh

4 =

mgh

2

Note that this is less than the initial energy of the block, mgh. As we mentioned, energy is not conserved in aninelastic collision. In this particular case, the energy has been cut in half. The rest has been dissipated as heat and/or sound asthe collision occurs. This reduced total energy, which is entirely in the form of kinetic energy right after the collision, is

converted to potential energy as the block-putty system moves up the track. At maximum height h’, we can write the equality:

(2m)gh’ = total energy right after collision =1

2 (2m) v’2 =

mgh

2

2mgh’ =mgh

2

h’ =h

4

Among the other answer choices, choice B, h/2, is what one would erroneously obtain if one assumes that energy is conserved.

1 0 . CThis question requires you to remember the definition of impulse, J:

J = ∆p = F∆t

where ∆p is the change in momentum, F is the force and ∆t the time over which the force acts. You should recognize the lastequality as simply Newton’s second law rearranged:

F = ma = m∆v

∆t (from the definition of acceleration)




KAPLAN

Multiplying both sides by ∆t, we get:

F ∆t = m ∆v

= ∆ (mv) (assuming mass is constant)

= ∆p

The larger the force, or the longer the time over which the force acts, the more the velocity (and hence momentumchanged. In this problem, the time that the baseball and the bat are in contact is doubled, while the force remains constant. Timpulse is therefore doubled.

1 1 . BSince the efficiency of the solar panel is only 10%, only 10% of the incident solar radiation is converted to use

energy. The output power of the solar panel is therefore 40 kW × 0.1 = 4 kW = 4000 W = 4000 J/s.Power is energy per unit time, and from the calculation above, we know that the solar panel delivers 4000 J of ene

every second. In 300 minutes, then, the energy generated is:

4000J

s × 300 min ×

60 s

1 min = 4000

J

s × 18000 s = ((4 × 103) × (1.8 × 104)) J = 7.2 × 107 J = 72 MJ

where 1 MJ = 106 J.One could also have gotten the answer by first calculating the energy delivered in 300 minutes and then multiplying

the efficiency factor.

1 2 . BLet the masses of ball 1 and ball 2 be 2m and m respectively. The momentum of ball 1 before the collision is (2m)v

2mv. Since ball 2 is at rest, it has no momentum and so the momentum of ball 1 is also the total momentum of the systeRegardless of the nature of the collision, momentum is conserved (kinetic energy may or may not be conserved). The tomomentum of the system after the collision would therefore also be 2mv. Ball 1 moves with a speed of v/3 after the collisioits momentum is therefore (2m)(v/3) = (2/3) mv. Since it’s moving in the same direction as before, its momentum vecpoints in the same direction. Ball 2 must then either move in the same or the opposite direction; otherwise the vector for ttotal momentum will have acquired a component along a new coordinate. The velocity of ball 2 after the collision, v’, msatisfy the equation:

mv’ +2

3 mv = 2mv

v’ = 43 v

At this point we could eliminate the possibility that the collision is completely inelastic. If that were the case, the tballs would have moved as one, i.e. with the same velocity, after the collision. Choice C is therefore incorrect. Since we hthe velocities both before and after the collision for both balls, we could determine whether the collision is elastic or inelasby computing the kinetic energies; choice D is also incorrect.

The kinetic energy before the collision is:

1

2 (2m) v2 +

1

2 m(0)2 = mv2

The kinetic energy after the collision is:

12

(2m) (v3) 2 + 1

2 (m) (4v

3) 2 = mv2

9 + 8mv2

9 = mv2

Kinetic energy is therefore conserved during the collision and thus the collision is elastic.



MCAT

12 as developed by

Passage II (Questions 13—17)

1 3 . AThe work done by a force is the product of the distance and that component of the force that is in the same direction:

that is what the cosθ term means in the formula W = Fdcosθ. If the two are perpendicular, the angle between them is 90°, andthe cosine term is zero. The work done will be zero in such a case. In the particle storage ring, the particles are moving in acircle. Their instantaneous velocity, and therefore, their direction of motion at any time, is tangent to the circle. The direction of the magnetic force, on the other hand, is toward the center of the ring and is acting as a centripetal force. The force and the

direction of motion are perpendicular, and thus the work done by the magnetic force is zero.

1 4 . DThe first and most natural step in approaching this problem is to identify the factors that affect the reaction rate from

the information given in the passage and determine how they affect the rate. The formula for the rate is given in the passage:rate = luminosity × cross section. Since the cross section is fixed for a given reaction, the only way one can control the rate isby changing the luminosity. (Do not confuse cross section with cross-sectional area!) The expression for luminosity tells usthat it is increased by increasing the number either of electrons or of positrons, by increasing the frequency of revolution, and bydecreasing the cross-sectional area of the storage ring. Since luminosity is directly proportional to the reaction rate, any of theabove that increases luminosity would also increase the reaction rate. Statements I and III are exactly two ways we haveidentified to increase the rate. Statement II, increasing the energy of the particles, would increase the velocity of the particles.They would thus move faster, completing more revolutions per second in the storage ring. This is also one of the ways we haveidentified to increase the rate. Thus all three statements are correct. Note also that once we have established statements I and IIIas being correct, choice D has to be the correct response.

1 5 . CThe mass of the positron or electron is on the order of 10–31 kg. Multiplying this by the acceleration due to gravity

results in a number that is still negligibly small compared to the magnitude of the electric interactions. Choice A is incorrectbecause gravity points in the same direction as the acceleration due to gravity. This is a consequence of Newton’s second law F= ma, where F and a are vectors and m is a scalar. In order for this vector equality to hold, F and a must point in the samedirection. (One must, in general, of course be aware of the distinction between force and net force.) Choice B states that theenergy lost due to gravity is equal to that gained from the magnetic fields. The passage tells us that magnets are placed atvarious points in the storage ring. Their purpose, however, is to keep the particles moving in a circular path. Since themagnetic force is perpendicular to the particle’s velocity, the magnetic force does no work, and thus causes no energy gain orloss. Choice D states that the luminosity is not dependent on the gravitational acceleration. This is a true statement, but is notrelevant to the issue we are trying to address here.

1 6 . D

This is a conservation of momentum question in disguise. The energy value given is extraneous information. In thequestion stem, we are told that the two electrons with equal speed collide head-on. There is really a lot of information that canbe garnered from this statement. Since both particles are electrons, they have the same mass. Furthermore, since they collidewith equal speed head-on, we know that their velocities are equal and opposite. So the total initial momentum of the system iszero: The initial momentum of one electron is a vector that is equal and opposite to that of the other electron.

Momentum conservation tells us that the total momentum after the collision must in this case also be equal to zero.After the collision, therefore, the two electrons must again move with equal speed in opposite directions. The second electronwill thus also have a speed of 0.9 c.

1 7 . CAgain, the energy value of 180 GeV is irrelevant here: the only important point is that the total energy needs to be

kept constant. In order to maintain this, the power we need to supply has to be the same as the power lost. The question tells usthat 260 MeV of energy is lost per revolution by each particle, and that the frequency of revolution is 10000 Hz. In one second,then, the number of revolutions made is 10000, and the amount of energy lost in one second by one particle is 260 × 10000

MeV = 2.6 × 102 × 104 MeV = 2.6 × 106 MeV. The total amount of energy lost in one second is this value multiplied by two,since there are two particles: 5.2 × 106 MeV. This is the amount of energy lost per second, and so energy needs to be suppliedat a rate of 5.2 × 106 MeV per second for it to remain constant.

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7Work, Energy, And Momentum Test w. Solutions