Princeton WB phys

MCAT SciENCE WoRKBOOK TABLE OF CONTENTS PERIODIC TABLE OF THE ELEMENTS BIOLOGY QuESTIONS AND PAssAGEs • • • •• •.•.• • SOLUTIONS 139 PHYSICS QuESTIONS AND pAS SAGES •• • .• • • • • .• 223 SoLUTIONS 353 GENERAL CHEMISTRY QuESTIONS AND PASSAGES 431 SoLUTIONS .• • •.• • .• • • • • • •.• 569 ORGANIC CHEMISTRY QuESTIONS AND PASSAGES 639 SoLUTIONS 731 MCAT SciENCE WoRKBOOK PASSAGE GUIDE BIOLOGY Chapter Number 10 MCAT Biological Sciences Review MCAT Biology Chapter Title Corresponding MCAT Science Workbook Passage Numbers Molecular Biology Microbiology Generalized Eukaryotic Cells Genetics and Evolution Nervous and Endocrine Systems Circulatory, Lymphatic, and Immune Systems Digestive and Excretory Systems Muscle and Skeletal Systems Respiratory and Skin Systems Reproductive Systems and Development 5, 12, 16, 18, 23,25, 27-32,55 13, 17, 19, 20, 21, 26, 34,35 1,2, 4,6, 7,10,24 33,56,62, 63,64, 65,66 8,9, 15,22,38,39,58,59,67-74 40,42,43,44,45, 46,47,61, 75 14,57, 76, 77,78 41,48,49,50,51,60, 79 52,53,54, 80 3, 11, 36,37,81, 82,83 MCAT Physical Sciences Review MCAT Physics Chapter Title Correspond;ng MCAT Science Workbook Passage Numbers Kinematics Mechanics I Mechanics II Mechanics Ill Fluids and Elasticity of Solids Electrostatics Electricity and Magnetism Oscillations and Waves Sound Light and Geometrical Optics 11 PHYSICS Chapter Number 10 3-9, 15-18 1Q-13, 19-24 31,32 25-30,33-39 14,40,45,46,48-55 42,43,56,57 41,44,47,58-60 GENERAL CHEMISTRY Chapter Number 10 MCAT Physical Sciences Review MCAT General Chemistry Chapter Title Corresponding MCAT Sqience Workbook Passage Numbers Atomic Structure Periodic Trends and Bonding Phases Gases Solutions Kinetics Equilibrium Acids and Bases Thermodynamics Redox and Electrochemistry 1-4,6,9-18,70 5, 7,19-28,43,72 35,57,59, 60,62, 63,65,66 40,61,64, 67 31,41, 42, 45,68, 69 33,44, 71 29,37,39,46 30,32,34,36,38,47-55 58, 73-78 8,56, 79-91 ORGANIC CHEMISTRY Chapter Number MCAT Biological Sciences Review MCAT Organic Chemistry Chapter Title Corresponding MCAT Science Workboo}f Passage Numbers Structure and Bonding Substitution and Elimination Reactions Electrophilic Addition Reactions Nucleophilic Addition/Cycloaddition Reactions Lab Techniques and Spectroscopy Biologically-Important Organic Chemistry 1,2, 4,5, 6, 7,8,9, 10 11, 12, 13, 14, 15, 16, 17, 18 19,20,21,22,23,24,25,26,27-30 31,32,33,34,35 36,37,38, 39,40,41,42,43 INTRODUCTION PERIODIC TABLE OF THE ELEMENTS r r He 4.0 H 1.0 Be B c Ll N 6.9 9.0 10.8 16.0 12 13 12.0 14 14.0 11 16 Na Mg AI Sl 23.0 24.3 27.0 28.1 15 p 31.0 19 20 21 22 24 26 27 28 29 Ca Sc: Ti v 2S K Cr Mo Fe Co Nl Cu 39.1 37 40.1 4.5.0 47.9 50.9 55.8 40 41 43 44 58.9 4.5 58.7 46 Kb Sr Zr Nb Mo Tc: Ku Rh 85 •.5 87.6 39 y 88.9 52.0 42 54.9 38 91.2 92.9 95.9 (98) 101.1 5.5 56 57 72 73 74 75 76 Cs 132.9 23 Ba La* ur Ta w Ke Os 137.3 138.9 178.5 180.9 10.5 Db (2621 183.9 186.2 107 Bb (262) 190.2 87 88 89 104 Fr Ka Act Kf (223) 226.0 227.0 (261) • t SCIENCE WORKBOOK s, _(263) 108 Hs (265) Ar 32.1 3.5 39.9 32 33 34 3.5 36 As Se Br Kr 63.5 69.7 72.6 74.9 79.0 79.9 83.8 47 48 49 50 51 Pd Ag 52 Cd lo So Sb Te 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 53 I 126.9 54 Xe 131.3 77 lr 192.2 78 79 80 81 82 83 81 209.0 85 Po At (209) (210) 86 Rn (222) 175.0 103 Pt Au Hg Tl Pb 195.1 197.0 200.6 204.4 207.2 84 109 Mt (267) 61 62 63 64 Sm Eu Gd 140.1 144.2 (145) 150.4 152.0 157.3 65 Tb 158.9 90 91· 92 93 94 95 96 97 Tb Pa Np Pu Am Cm Bk (243) (241} (247) (237) Cl Ge Pm 238.0 s 31 60 (231) 17 10 Ne 20.2 18 Ga Nd u F 19.0 30 Zn 65.4 59 Pr 140:9 58 Ce 232.0 MCAT 106 (244) 66 67 68 69 Dy 162.5 98 Ho Er Tm 164.9 167.3 168.9 70 Yb 173.0 cr 99 100 101 Es 102 Fm Md No Lr (251) (252) (257) (258) (259) (260) 7l Lu MCAT PHYSICS PRACTICE QUESTIONS AND pASS AGES I i i !' I I ! I 223 ~ What is the magnitude ofthex-component of Vector P? Passage (Questions 1-7) A A vector is a quantity that incorporates both magnitude and direction A vector can be pictured as an arrow whose orientation indicates direction and whose length indicates magnitude A scalar quantity possesses magnitude only 6m B 10m C 17m D 20m Vectors can be added (or subtracted) using the "tip-totail" method and resolved into components using trigonometry What is the magnitude of they-component of Vector P? A Figure shows three vectors plotted on a pair of x-y coordinate axes Vectors P and Reach have a magnitude of 20 m, and Q has a magnitude of 40 m 6m B 10m C 17m D 20m y Which of the following vectors best illustrates the direction of the vector -R? A./ c.~ B.~ D./ R Which of the following vectors best illustrates the direction of the vector P + R? A Figure ) B / What are the horizontal and vertical components, respectively, of Vector P? A B C D 20 20 20 20 MCAT sin 30° and 20 cos 30° sin 30° and 20 tan 30° cos 30° and 20 tan 30° cos 30° and 20 sin 30° A B Velocity Displacement Speed Acceleration SciENCE WoRKBOOK D A N N N N 0N B 25 N C 75 N D 125 N A 5.0 m/s 8.7 m/s c 10.0 m/s D 11.3 m/s 19 A 1-kg mass is at the end of a vertical spring with a spring constant of 20 N/m When the mass comes to B rest, by how many meters will the spring have stretched? A B c D 0.05 0.5 22 The acceleration experienced by a block moving down a frictionless plane inclined at a 30° angle: 20 A B C D MCAT SCIENCE WORKBOOK 228 decreases as the block moves down the plane is constant increases as the block moves down the plane depends on the height of the plane 23 A block is moving down the slope of a frictionless inclined plane The force parallel to the surface of the plane experienced by the block is: A B C D 27 A block is being pulled by a rope up the surface of an inclined plane at constant velocity Which of the following is true of the tension in the rope r, the force due to friction f, and the component of the block's weight parallel to the plane w P? less than the weight of the block equal to the weight of the block greater than the weight of the block unrelated to the weight of the block A w p = T+f B T= wP + f C f T + wP D T + f+ wP = 24 A block is moving down the slope of an inclined plane at constant velocity The normal force exerted by the plane on the block: A B C D 28 An object is moving in a circle at constant speed Its acceleration vector must be directed: increases with increasing velocity decreases with increasing velocity is independent of velocity depends on the coefficient of kinetic friction between the plane and block A tangent to the circle and opposite the direction of motion B tangent to the circle and in the direction of motion C radially and toward the center of the circle D radially and away from the center of the circle 25 A block is sitting motionless on the surface of an inclined plane as the angle of elevation is gradually increased The normal force exerted by the plane on the block: A B C D 29 An object is traveling in a circular path If the velocity of the object is doubled without changing the path, the force required to maintain the object's motion is: A increases with increasing angle of elevation decreases with increasing angle of elevation is independent of angle of elevation depends on the coefficient of 1)tatic friction between the plane and block B c D halved unchanged doubled quadrupled 30 How far from the heavier end must the fulcrum of a massless 5-m seesaw be if an 800-N man on one side is to balance his 200-N daughter at the other end? 26 A block is sliding down the surface of an inclined plane while the angle of elevation is gradually decreased Which of the following is true about the results of this process? A The force due to friction decreases, and the weight of the block remains constant B The force due to friction decreases, and the weight of the block decreases C The force due to friction increases and the weight of the block decreases D • The force due to friction increases, and the weight of the block remains constant A 0.5 m B C m m m D 31 An object is traveling in a circular path If the radius of the circular path is doubled_,rithout changing the speed of the object the force required to maintain the object's motion is: A B c D 229 halved unchanged doubled quadrupled ( PHYSICS [You may wish to the calculation as an exercise you get kax from each of the two end springs and tk (2x) from the central spring (if you squeeze your left fist a distance x to the right and your right fist a distance x to the left, the central spring will be compressed a distance 2x) This gives (k + 2kb)x ] Passage 55 t [Note: The analysis of coupled oscillators is an advanced topic usually studied by physics majors in their second or third year So what would it be doing in an MCAT passage? The MCAT often has "advanced-level" passages that ask introductory-level questions That is, the topic of the passage may be upper-division and not something you would be expected to know walking in to the test, but all the questions are answerable using the information in the passage and your knowledge of introductory physics This passage is excellent practice for the real thing.] S B In the symmetric mode only the end springs are stretched or compressed so the answer must involve k only Choice B is the only possibility Note that V:hether you stretch or compress a spring a distance x, the stored potential energy will be the same A The third sentence in the third paragraph gives this one away In the symmetric mode, the central spring is never stretched and the two masses oscillate independently just as if the central spring were not there So, in the symmetric mode, the period of oscillation of each mass can be calculated using the equation for the period for a mass attached to a single spring given in the first paragraph: D In either normal mode, the initial displacements have the same magnitude, the masses are equal, and they are both attached to two springs of force constants k 11 and kb, so there is nothing to make one mass move faster than the other If you have visualized the motion in each mode, this may be obvious to you Anyway, the above reasoning eliminates choices A and B The velocities are opposite in direction in the anti-symmetric mode The only way to see this is to visualize the motionthis fact is not explicitly stated in the passage T = 21t~mfka This is the period of each mass and also the period for the whole system since the two masses oscillate in synch with each other A The last two sentences in the third paragraph are the key to this one Each of the masses feels forces from two springs in the anti-symmetric mode For example, when they are squeezed close together, the left mass is pushed to the left by the central spring and pulled to the left by the leftmost spring In any case, they oscillate faster, so the period (the number of seconds per oscillation) is shorter A You are expected to infer this answer from the last paragraph Different initial conditions imply a different mixture of normal modes which implies a different period Actually, as long as you realize that the period is different for' the two normal modes, you can infer that the period depends on initial conditions since that is what determines which mode you are in Choice B is a bit silly since whether or not friction is important depends how well you greased your surface, not what kind of oscillation you're looking at~ Choice C is something that happens in all kinds of oscillations with springs Choice D is true for uncoupled oscillations but the last three words of the question ask you to consider coupled oscillations A The central spring is never stretched in the symmetric mode Visualize the motion Make a fist with each hand Hold your fists about inches apart and move them back and forth together-that's symmetric motion The spring connecting your fists never has any stored energy since it never stretches (or compresses) B This is very similar to question except now you're asked for force rather that potential energy The same trick is applicable You must realize that the central spring does not contribute, so the answer involves only the leftmost spring (remember, you're asked to consider the left mass) This narrows it down to either choice A or B Hooke's Law says that the magnitude of the force exerted by a spring with spring constant k stretched a distance x is just kx The answer is B [Hooke's Law may be given in the passage, but then again it may not It is worth memorizing because (1) it is so simple, and (2) it the equation for spring problems.] D Again, visualize the motion Hold your fists inches apart and move them alternately together and apart When they are at their closest the central spring is compressed, and the two end springs are stretched When they are at their greatest separation the central spring is stretched, and the two end springs are compressed These are the points of maximum potential energy and energy is stored in all three springs Since all three springs are involved the answer must involve both spring constants There is no need for calculation-the answer must be is D MCAT SCIENCE WORKBOOK 416 D This is similar to question The masses are either squeezed together or pulled apart The magnitude of the force on each mass is the same in either configuration The quick way to answer this question is to realize that both spring constants must be involved since all the springs are stretched (or compressed) obeyed only to the extent that the spring retains its shape perfectly You know that if you pull a spring too far you can ruin it Even small oscillations will eventually wear out a spring The answer is C The question is testing whether or not you know that gravitational forces are virtually zero except when one of the objects is planet-sized It is possible to observe the gravitational force between two ordinary objects but it requires extremely sophisticated equipment Or, you can use the reasoning shown in the answer to question and Hooke's Law to calculate the answer: For example, when the masses are squeezed together, the leftmost spring exerts a force on the left mass equal to kax to the left, and the central spring exerts a force on the left mass equal to kb(2x) to the left This gives a total force of (ka + 2kb)x Independent Questions B We compute the speed v' through the new material in terms of the speed v through Material #1: 10 D As in some of the previous questions, you must reason that the answer has to depend on both spring constants since both springs are involved in the oscillation That m~ans the answer is either C or D But choice C has the wrong units The equation for the period given in the first paragraph shows us that period must have the units the square root of mass over force constant Choice C has a units of force constant squared in the denominator inside the square root and therefore cannot have units of period = C Using the fundamental equation v A.f, we get A.= vff =340/20000 = 17/1000 =1.7/100 =1.7 em B Since the wave speed is independent of the frequency tlie speed is still v (Of course, the wave with frequency 4/ will have 114 the wavelength, but the wave speed will stay the same.) [As an exercise, let's the units more carefully Force constants have units of N/m as you can see from the equation F = kx, where F is in newtons and x is in meters A newton can also be written as kg·m/sl, as can be seen from the equation F = ma So the units of a force constant are kg/s 2• That means the units of mass over force constant are s2• The square root gives seconds, the correct units for period Obviously this doesn't work if you've got kg/s x kg/s in the denominator as in choice C.] A Since the wave speed v is constant, the equation v = A.f tells us that wavelength is inversely proportional to the frequency ~If the frequency increases, then the wavelength decreases S C The given equation tells us that the intensity is proportional to tqe square of the pressure amplitude Therefore, since Wave #1 has twice the pressure amplitude, it will have 2 = times the intensity of Wave #2 11 A Choices C and Dare the points of maximum potential energy Choice B only happens in the symmetric mode In the anti-symmetric mode, when the masses pass the equilibrium point, they are either moving toward each other or away from each other At the equilibrium point, none of the springs are stretched and therefore the stored potential energy is zero (The energy has not disappeared, it is in the form of kinetic energy.) B The round-trip distance for the chirp is 2d =2 x meters = 10 meters, so using the fact that time equals dist~nce divided by speed, we find t = 2dlv 10/340 = l/34 = 3/100 0.03 sec = 12 C Although air resistance (choice A) is usually ignored, it is certainly observable/measurable in a careful experiment Now, if you bend a paper clip back and forth for awhile and then touch the point of the bend to your cheek, you will feel heat Thus, choice B, although a small effect, is observable and counts for something As forD, Hooke's Law is = A Since the intensity is inversely proportional to the square of the distance r, if the distance increases by a factor of (which happens when moving from meters to 6), the intensity will decrease by a factor of 2 = Therefore; I' is of/, or; equivalently, I= 41' t 417 PHYSICS SOLUTIONS 14 B Because A/= v and v is fixed, A and fare inversely proportionaL Thus, if one is increased, the other must decrease This observation eliminates choices A and D Now in a closed-end pipe, we know that f, = n(v/4L), for odd n only, so the fundamental frequency is / = v/4L Thus, longer pipes resonate at lower frequencies (and longer wavelengths) than shorter pipes C Since f = ~ = 340 mfs == 340 = 3400 = 425 Hz A 77 m 0.8 the best choice is C {The decibel level is irrelevant.) C If beats are heard every 250 ms {which= 1/4 sec), then there must be beats per second Since • = Hz ' the frequencies of the two tuning forks Jbeat must differ by 4Hz If one fork has a frequency of 498 Hz, then the other must have a frequency of either 498 = 494 Hz or 498 + = 502 Hz 15 C A shout is 80- 20 = 60 dB louder than a whisper We know from the given equation that if I changes by a factor of 10, then /3 changes by adding (or subtracting) 10 Therefore, a 60-dB drop (i.e., drops of 10) corresponds to factors of 10, and 10 = one million 10 C Sincef11 = n(vi2L), it is clear that/11 = nf1• Thus, the fifth harmonic frequency is five times higher than the fundamental (This is a general fact which you will find worthwhile to simply know Whether the pipe is open r closed, the nth harmonic frequency is n times higher than the fundamental frequency.) 16 B If f3 is the sound level at intensity I, then the sound level at intensity 51 is 1logC:J 1{logs+ log: J P' = o = I == 10(0.7)+ 10log- = + f3 11 A If you did the preceding question, you know that Io f, = nf1• Therefore, since Anfn = v (and v is fixed), if the nth harmonic frequency is n times higher than the fundamental, then the nth harmonic wavelength is n times shorter than the fundamental: 17 A If f3 is the sound level at intensity I, then the sound level at intensity I is /3' = Therefore, A3 = (1/3)A 1, regardless of the exact length of the pipe IOio{ ~:) = IO[Iog2+ log:.] I = 10{0.3)+ lOlog- Io 12 B The Doppler Effect tells us that if the source moves toward the detector, then the perceived =3+{3 = 3+50=53 dB frequency will be higher than that of the source But is the source •s speed is constant, this higher pitch will remain constant as well Finally, it's clear that if the source of sound gets closer, the intensity increases 18 D The first harmonic (or fundamental) frequency is found by first sketching the first harmonic wavelength in the pipe (of length L) A sketch would reveal that L = A/2 for the first harmonic {or recall the equation for harmonic wavelengths) Then using the equation v = 'A/, we get 13 B First of all, it is important to note that when a wave passes from one medium to another (or is reflected at a boundary between two media), the frequency of the transmitted or reflected wave is no different from that of the original wave This eliminates choices A and C Now, since sound travels more slowly through air than it does through metal, the equation A = vlf tells us that the wavelength in air will be shorter than in metal MCAT SCIENCE WORKBOOK lv 340 340 L=-A = = =-=3.4m 2 J; 2·50 100 418 19 C The first harmonic (or fundamental) frequency is found by first sketching the first harmonic wavelength in the pipe (of length L) A sketch would reveal that L = A/4 for the first harmonic Then using the fundamental equation v = A/ we get 23 B Make a sketch to find that A3 = 2Lf3 (or recall the equation for the harmonic wavelengths) Then with L = 1.5 meters, JNe get ~ = 2Lf3 = 2(1.5)/3 = meter 24 C Using the given equation for the harmonic wavelengths (L = nA/2 => A= 2Lfn) together with the equation v = Af, we find the following expression for the harmonic frequencies: In = nvf2L Therefore, the desired ratio is A1 v 340 340 L=-=-= =-=1.7 m 4ft 4·50 200 20 B Solving the given equation for A, we find An = 2Lfn Since higher harmonic numbers give shorter wavelengths, harmonic number n gives the 2.5-meter wave, and harmonic number n + (i.e., the next consecutive one) gives the 2-meter wave: An = 2nL = 2.5) => J; 3vj2L -= =3 ft vj2L Note that the precise length of the pipe (in this case it was given to be m) is irrelevant In general, the nth resonant frequency is n times the fundamental: In = nft · 2L=2.5n {2L = 2(n+1) 2L } a= =2 "+ n+1 25 B The third-lowest frequency corresponds to the third-longest wavelength (since frequency and wavelength are inversely proportional) Writing the given equation in the form An = 4L/n we see that the longest wavelength is found by letting n = (the fundamental), and the third-longest wavelength corresponds ton = (since n = 1, 3, 5, ) The ratio of these two harmonic wavelengths is => 2.5n = 2(n + 1) => n = => L = m where the last step (finding L = m) was done by substituting n = back into either one of the original equations 21 B Solving the given equation for A, we find An= 4Lfn Since higher harmonic numbers give shorter wavelengths, harmonic number n gives the 7-meter wave, and harmonic number n + (the next consecutive one is now found by adding (not 1) since only odd harmonic numbers are allowed) gives the 5-meter wave: A.,.= ~L =7 4L n+2 A = =5 "+ ) 26 A Since the given equation can be rewritten as A= 2Lfn and since v = Af, we can quickly derive the following equation for the harmonic frequencies: In = nvf2L Therefore, the difference between consecutive harmonic frequencies is 4L=7n => {4L= 5(n+2) v 340 340 100 =-= = =-=25Hz 2L · (6.8) · (2)(3.4) ~ · 35 => 1n = 5(n + 2) => n = => L = - m where the last step (finding L = 35/4 m) was done by substituting n = back into either one of the original equations 22 C Simply use the equation v 27 A Since beats of frequency Hz are heard, the violin string must be vibrating at either 330 - = 327 Hz or 330 + = 333 Hz But since we are told that the string is known to be too taut, we expect that its frequency is too high; therefore, the string vibrates at 333 Hz Since the period is the reciprocal of the frequency, T = } If= 1/333 second = A/: f =!! _ = 340 = 340 = 340 ·1 00 = 8500 Hz } 0.04 4/100 419 PHYSICS SoLUTIONS 33 C The time interval between successive pressure compressions (adjacent maxima) is equal to the period of the wave Using v /.f, we can solve for the period, T: 28 C Call the four tuning forks W, X, Y, and Z Then we could have beats with W and X, W and Y, W and = Z, X andY, X and Z, or Y and Z Since we have six possible pairs, we have the possibility of six different beat frequencies ~ v=A/ 29 B Since X and Y give beats of Hz, we know that X is either less or more than Y Similarly, Z is v=A·T ~ ') 0.68 68 T=-= = = sec=2ms v 340 34000 1000 either more or less than Y Therefore, X differs from Z by either 4- = Hz or + =8Hz 34 B Period is the reciprocal of the frequency, regardless of the wave speed: 30 A Solve the given equation for the vehicle speed v, then substitute (using V = l f) and simplify: 2vf V !lf=- VA/ 2f (l f)llf 2f T= f = - - - = - - sec=0.005 sec 200Hz 1000 W v=-= =- ~ 35 A We first find the detected frequency Since the speed of the listener is zero, and the source is moving toward the listener, we expect a detected frequency higher than the source frequency: = (0.1 m)(200 Hz) _ mfs 10 v =-!=tf 31 C As the train (with speed vT and whistle frequency f) approaches the observer, the Doppler effect predicts that the frequency f' heard will be higher than f, in fact, iv Now, since the frequency has increased by 817, the wavelength must decrease by 817, i.e., A.'= A t f'=-v-f v-vT f" heard by the f', we write As the train recedes, the frequency observer will be lower than f: 36 D The distance between the source (the plane) and the listener, 500 m, never changes (see the diagram below) Since there is no relative motion between the source and the detector, the detected frequency will be no different than that of the source v !"=-! v+vT Since we are told that f" is l/2 of 32 A In the first case (stationary source, moving detector), the Doppler effect predicts that , v+w v+tv fv v v f =-!=-!=-!=-! y 37 C The distance from S to Q is meters (it's the hypotenuse of a 3-4-5 right triangle), and the distance from S to Q is meters Thus, the difference between the path lengths to point Q is meter, which is half a wavelength Therefore, the sound waves are out of phase when they meet at Point Q, causing destructive interference there (It'll be pretty quiet at Point Q.) However, since Poin~P is the same distance from S and S 2, the waves will arrive in phase at Point P, and interfere constructively This tells us that AP > AQ In the second case (stationary detector, moving source), we get v v v !"=-! J=-/=2/ v-w v-tv tv Thus we see that MCAT f' < f" SciENCE WoRKBOOK 429 38 A The speed of the wave is v =A/= (0.5)(700) = 350 m/s Using distance = rate x time, we find that the distance that could be covered in half a second is d = vt = (350)(0.5) = 175 m Passage 56 C The truck is considered the detector in this question (since we are asked at what frequency the waves strike the truck) Since the source (the police car) is not moving, but the truck is moving toward the source, the Doppler effect predicts that the frequency received will be 39 A Using v = Aj, Trial #1 tells us that the speed of sound through the material under study is v = (20)( 100) = 2000 m/s Since the sound wave is Trial #2 also travels at 2000 m/s (since it's traveling through the same medium as the one in Trial #1), the frequency times the wavelength in Trial #2 must also equal 2000 m/s Since the frequency is 400 Hz, the wavelength must be v/f = 2000/400 = m f' = v + vtruck f = v + nr v f = -M v f v v v 11 = -(1 MHz) = 1.1 MHz 10 40 D Since / > / 1, it is clear that {32 will be greater than {3 1, so the difference {32 - {3 is positive; eliminate choices A and B Since the ratio of / to / is 10 ( = factors of 10), the difference between {32 and {3 is 10 + 10 + 10 = 30 (dB) B Now we consider the truck the source and the car to be the detector (since the waves are bounced off the truck and head back to the police car) Since the source is moving toward the stationary detector, v !"= v-vtruck 41 C All that is required is to combine the given equation with c = 1{: v v v-lb-v -fov !'= -!'=-!' 10 = g-(1.1 MHz) !'=(1±%)! ~ ;, =(1±%)i ~ v = 1.2MHz A -=(1±!1.)- ~ A=A' c A 1±~ I B We now combine the two different situations In the first part, the police car is the source and the truck is the detector Then the waves bounce off the truck and head back to the police car, so for the second part of the sound wave's journey, the truck is the source and the car is the detector Letf be the frequency emitted by the car, let f' be the frequency at the truck, and let f" be the frequency received back at the police car Then 42 D The Doppler Effect is qualitatively similar for light waves and for sound waves: if the source and observer are moving toward each oth.er, the detected frequency will be higher than the source frequency, and if the source and observer are moving away from each other, the detected frequency will be lower than the source frequency Since the galaxy is moving away from the Earth, the detected frequency will be lower; this eliminates choices A and B Since the speed of light through space is constant(= c), and c = Aj, a lower detected frequency means a longer detected wavelength Since the detected wavelength is longer than the emitted wavelength, we'd say that the wavelength has been shifted toward the red end of the visible spectrum (since the visible light with the longest wavelength is red) police car ~ truck : /' = x f v (o-v (source , /) • (detectoP• /') V I ~ police car : f" ;= _v_ x f' truck (source a/') (detector '"r> v +to- v t· ~ 43 B Choices A and D can be eliminated immediately: when a wave is transmitted into a new medium (in this case a warmer region of air), the frequency does not change The equation given for v makes it clear that at higher temperatures, the wave speed is higher Since v = Aj, and the frequency f doesn't change, the wavelength A will increase when the wave speed increases f , ~ v kv =-X-j=-j f-M-v v 11 Sincef= MHz, we have /" = 9/11 = 0.82 MHz 421 PHYSICS SOLUTIONS Passage 57 C The third paragraph of the passage implies that the energy coupling in a megaphone is most efficient when the wavelength of the sound is smaller than the transition length (which is the length of the megaphone) Since a higher frequency has a shorter wavelength, the megaphone is most efficient for a high voice B Choices A and D are properties of all waves Choice Cis true of both sound and string waves, since both involve the displacement of matter (as opposed to light waves) Note that in the case of sound, the spring-like forces are created by pressure differences A The small wavelength of light implies the diffraction angle is quite small Concerning B, all waves diffract As for C, the only apparatus needed are the ear and eye D is simply not true A Although the full chain of energy is given in D, the megaphone facilitates the conversion of the mechanical energy of the vocal cords to sound energy near the mouth to sound energy at the end of the horn Independent Questions C Light passing through a pinhole spreads due to diffraction The spreading angle is given by B Snell's Law immediately gives us A8 = Nd = 0.65 J.l.m I J.l.m = 1/6 radian The size of the spot can now be obtained from the following diagram: B The angle of reflection equals the angle of incidence ( =60°), so the angle that the reflected ray makes with the boundary is 30° Since the refracted ray is perpendicular to the reflected ray, the angle that the refracted ray makes with the boundary is 90° - 30° 60°, so the angle of refraction is 30° Snell's Law then gives D = Applying the definition of radian measure, we find lspot = D&8 =(3 m)(l/6 rad) =0.5 m C First, eliminate A: the sound is not passing through a hole, so diffraction does not apply B is a true statement but does not explain the observation Concerning C, a poor impedance match would result in good reflection and poor transmission of the wave, in analogy with the ropes in the passage Nothing in the passage supports D and, in fact, the statement is false incident ray reflected ray "' , : question says the S B In order to have a better defined direction (i.e., to minimize the spreading angle ll8), we need to reduce diffraction by decreasing A or increasing d The megaphone does not change the wavelength of the sound, which depends only on sound speed and frequency, but it does increase the effective size of the hole MCAT SCIENCE WoRKBOOK ,~~~ 422 """' are perpendiculru D Since TIR occurs, we know that the angle of incidence(= 45°) must be greater than the critical angle for TIR If n is the index of the plastic (and, of course, the index for air is 1), then B Because the index for polystyrene(= 1.55) is greater than the index for air ( = 1), the incident medium is optically denser than the refracting medium, so TIR can occur But the angle of incidence must be large enough: TIR ~ so TIR ~ 81 > sin-1(1/n) TIR ::) >ecru 81 >sin- (n2 fn ), ::) sin45° > lfn ~ ~ 45° >sin-1(1/n) n>- ::) n > fi sin45° D The angle of incidence is 60° [Not 30°! Remember: Even if we are given information about the angles made with the surface, we must turn that into information about the angles made with the normal.] Snell's Law then says n1 sin9 =n2 sin8 ~ l·sin60°=-J3·sin8 ~ sin82 = 1/2 ~ e2 = 30°, so the angle the refracted ray makes with the surface is 60° B Since the prism has a higher index of refraction than the air, the beam of light bends toward the normal upon entering the prism, then back away upon leaving Fr:om the figure below, we clearly see that the ray is directed toward the base upon leaving the prism S D The angle of refraction is greater than the angle of incidence since the angle the refracted ray makes with the boundary is less than the angle that the jncident ray makes with the boundary [Remember: Even if we are given information about the angles made with the boundary, we must turn that into information about the angles made with the normal.] Recall that the refracted ray bends away from the normal when the refracting medium is optically less dense than the incident medium Since this is the case here, we must have n > nr (See the figure at the top of the next page.) Notice that choices A and B can be ruled out immediately, since the index of refraction for a medium can never be less than normal ', air air incident bea ~ Or - ·This angle's smaller B Since the index of the glass is greater than the index of air, we know that the beam of light must bend toward the normal upon entering the glass, then· back away from the normal upon leaving Only choice B depicts this so "'"' "'· this angle's bigger B Since the angle of incidence at the surface of the liquid is 0° (not 90°!), Snell's law immediately tells us (see the calculation below) that the angle of refraction will be 0° as well: the refracted ray continues straight downward Then the angle of reflection at the first mirrored surface equals the angle of incidence(= 45°) there, so the beam travels normal 423 PHYSICS SoLUTlONS 12 A From the mirror equation we get directly to the right and hits the other mirrored surface at 45° It bounces off at 45°, which means it heads straight up toward the surface of the liquid As before, since the angle of incidence is 0° again, the refracted ray will be at 0°, so the beam just emerges straight upward If = 0°, then 1 i =0 sin92 => e2 f fo => t= o-f To make i negative (to produce a virtual image), we must have o-f< 0; that is, o n1 sin0° = n2 sin8 => -+-=- => =0° 13 C Since the image is upright and times the size of the object, m =+4 Therefore -i/o= +4, which implies that i = -4o Then 111 1 +-=- => -+ =- => i f o -4o f => - = - => o=3JJ4 4o f 411 =4o 4o f 14 D The object's distance from the mirror is 2/, so the mirror equation gives 1 1 1 i f 2/ i f 1 1 => -= -= =- => i=2f i f 2f 2f 2/ 2f -+-=- => -+-=- 10 C The object's distance from the mirror isf/2, so the mirror equation tells us that 1 i f f/2 i f -+-=- => -+-=- !=]:-~=-.! => f f f Since i is positive, the image is real and therefore inverted (Note that choices B and C couldn't possibly be correct.) => i=-f Therefore, the magnification is = 15 B Since the image is virtual, i is negative, so m +3 This gives -i/o= => i -3o Since o = 10 em, we have i =-30 em, and therefore i -! m= = =2 f/2 1 1 1 1 -+-=- => + =- => - - - = 10 -30 f i f 30 30 f 11 D The objecfs distance from the mirror is 3//2 (since halfway between f and 2f is 3f /2 ) Therefore, the mirror equation gives 1 1 ~ i=y- 3f= 3f- 3f= 3f -+-=- => +-=0 i f 3ff2 i f MCAT 1 SCIENCE WORKBOOK => i=3f .424 16 A The focal length of a convex mirror must be specified as negative when using the mirror equation Since f is the magnitude of the focal length, we must write -!as the actual focal length The mirror equation then tells us 1 i 1 foeallength 1 -+-= - -= =-i f f f 1 f i -! -+-= ==} ::::::) 20 C Since convex mirrors can only form virtual images (see #17), the magnification must be positive (virtual images are always upright) Therefore, m = 1/2, so -i/o = 1/2, and therefore i =-o/2 Since the focal length is half the radius of curvature, f -24/2 = -12 em Thus, = ::::::) 1 i f -+-=- f t= 2' 12 = ::::::) so the image is at distance the virtual side) f /2 -12 ::::::) 12 -=-0 o= 12 em., :::::) from the mirror (on 21 B Eliminate A and D The lens equation tells us that i is positive, so the image is real and therefore inverted: 17 D Since the object distance o is positive, and the focal length f of a convex mirror is negative, the mirror equation tells us that i could not possibly be positive, because a positive plus a positive could not equal a negative Since i cannot be positive, the image cannot be real positive\ -o/2 -+ = :::::) 1 1 1 => -+-= 40 i 20 i f 1 1 => - = - - - = - - - = - => i=40cm 20 40 40 40 40 -+-= /negative 1 i f 22 D Since the image is real, it is inverted, so the magnification is -2 (not +2) This tells us that -i/o = -2, which implies o = i/2 Since the power is diopters, the focal length must be 114 meter = 25 em The lens equation then gives us i: -+-=- t ••• cannot' be positive 1 18 B The mirror equation and the magnification equation give us 1 i f 1 i 1 ==} 1 20 10 -20 20 = = => z= em 20 20 20 i -20/3 em :.m=-;= 10em 10 -+-=- => -+-= => -= -0 6 ==} 1 if f 7, => = -=25 -+-= 1 -+-=- :::::) -+-=0 i f -6 -3 1 1 => - = + = +- = => o = -6 em 1 1 i=75 em 23 D Remember that for a lens, the side where the object sits defines the virtual side for the image Therefore, if the' image is to appear at the same position as the object, the magnitudes of i and o will be the same, but i must be negative Therefore i = -o Then 19 D We want the image distance to be i = r = 2f = 2( -3 em) =-6 em, so the mirror equation says 1 -+-=- => -+-=- => -+-=i i 25 i/2 i 25 o i 1 1 :::::) -+-=0 -o f which is impossible which is impossible (o can't be negative) 425 PHYSICS SOLUTIONS 27 C Since the image is real, i > 0, so m = -2 Uust like in #22) So, -i/o =-2, which gives i = 2o = 2( 10 em) = 20 em, and therefore, 24 A We will use the lens equation after choosing some numbers for the focal length and object distance Let's take f = 10 em, and then, because we want to place the object just inside the focus, let's choose o = em With these choices (which are arbitrary except for the fact that o is a little less than f), the lens equation tells us that 1 i 1 1 1 i 10 i f 1 1 10 => - = - - - = - - - = - - => i=-90 em 90 10 90 90 -+-=- => -+-=- Since we got a negative value for i, we know that the image is virtual The magnification equation then gives m = -i/o = -( -90)/9 = 10, so clearly the image is enlarged Note that although the choices for f and o were arbitrary (but with o o = -2i = -2(-10) = 20 29 A Since the image is upright, m = +112 so o =-2i = -2(-10 em)= 20 em Then I 1 1 => -= = => t= m 4 10 _!_ =1- _!_ = _2._ => i =119 em· il 10 10 }- -+-=- -+-=0 i f -+-=- => -+-=- f ~ 28 B The lens equation tells us 25 D As we did above in #24, let's just pick some numbers! Choose f = em For the object positions, first take o = 10 em then take o = em (which is closer to the lens) We now compute the two image distances: 01 1 1 ~ -+-=20 20 f 10 20 f f 20 ~ -=- ~ !=-em 20 f 20 1m :.f=- emx -meter 100 em 15 => P = _! = = 15 D f 1/15 m -:-+-=- 1 -+-=- => -+-=i2 1 1 1 -+-=- => - + - = - =-= 11 em 4 12 Since i > i > f, the image moves away from the focus, away from the lens f i ~ f=-20 em 20 => P=-= f 26 D Since o is positive, andfis negative (it's a diverging lens), there is no way i could be positive, so a real image cannot be formed (see the solution to #17 above) => -1/5 m -10 f meter f=-=-5 D 30 D Since the focal length is negative, and o is positive, i can never be positive (see question 17) MCAT SCIENCE WORKBOOK 426 = s 31 B The power of the combination is P P + Pr Since power is the reciprocal of the focal length (in meters), we have ft = 10 em= 1/10 meter h =20 em =1/5 meter :::::} P1 = 1/.ft = 10 D, :::::} P2 =1/h = D, A If we look only at the spreading of the wave, then our diagram looks like the following: so the power of the combination is P = P + P = 10 + B Changing the characteristics of the cornea certainly changes the index of refraction of the light, so C and D are out Changing the wavelength of the light (choice A) changes the index of refraction for reasons given in #4 above 15 D Passage 58 A If we replace ') by v/f, the equation for the spreading angle becomes 6.8 = vlfd Thus, if both d and f increase by a factor of 2, 6.8 will decrease by a factor of x = Once again applying the definition of radian measure, we find l sp1oleh = l oyo6.8 B Again we use the equation given for 6.0 The lower limit for the uncertainty in direction in the human eye is given by 6.8 =lre /1 pup 11 = (0.65 x lO-ti m) I (2 x I0-3 m) = x 10-4 radian Passage 59 D The decay scheme mentioned for the radioactive C We draw the following diagram: hydrogen isotope does not produce a positron, and without a positron there will be no pair annihilation and therefore no gamma rays to detect This would be a problem since the passage states (last paragraph) that the PET image is formed by the detection of gamma rays Nothing in the passage supports choices A, 8, or C, and, in fact, they are false (and/ or irrelevant) r and then calculate (by definition ofradian measure) D = s = r8ro• = (10 m)(I0-2 rad) = 0.1 m C Choices A anfl can be eliminated because ( 1) pair annihilation gives off two gamma rays, and neither answer has two or because (2) the total charge of the reaction is not conserved The total momentum of the "before" state is zero since the momentum vectors are equal in magnitude and opposite in direction Since total momentum is conserved, we must exclude choice D, because both momentum vectors are pointing in the same direction and therefore would not cancel Only in choice C are the momentum vectors of the two gamma ray photons equal and opposite D Since the index of refraction is greater for blue light, it bends more easily as it goes through a lens The following diagram shows, consequently, that blue light has a different focus than red light: Although A is true, it is not a consequence of the refractive index of the cornea but of the diffraction of the eye 427 PHYSICS SOLUTIONS Passage 60 B The total energy released in the pair annihilation is shared equally by the two emitted gamma rays, so each gamma ray has an energy E = MeV = x 106 eV Solving E = hjfor fwe find !=~= h 2x106 eV :::::]_x1021 Hz=S.Ox102o Hz 4.1x10-15 eV·s 2 C Choice A is wrong because the index of refraction of copper is irrelevant: light does not travel through a copper cable While B and D may be true statements, there is no evidence to support them as reasons for the difficulty of tapping into optical fibers The passage mentions, however, that light will be lost at imprecise connections This implies that C is the best choice C Although we are told that each reaction conserves charge, energy, and momentum, the only one of these three properties that is evident from the available choices is the charge before and after the reaction Choices A, B, and D all have the same net charge before and after the reaction: reaction A conserves charge +e, while reactions B and D conserve net charge Reaction C, however, has charge zero before but charge +e + e = +2e after, so it does not conserve charge and therefore cannot be a beta decay reaction D Choice A is untrue Choice B is also untrue: infrared light contains less energy than visible does Choice C is untrue: infrared light has a lower index of refraction than visible light (In general, the longer the wavelength, the lower the index of refraction; see the discussion of dispersion in the text.) The answer is D In general, electromagnetic radiation with a short wavelength is more easily absorbed, bent, scattered, etc C The speed of electromagnetic radiation through air is virtually the same as through free space: c Therefore, '\ -~- 3x10 mjs _ 11.- f - 2x10 21 Hz - 15 • x 10 _14 m B After traveling 20 km, the attenuation of the ray is (20 km),x (1 dB/km) = 20 dB Because the decibel level is logarithmic, f:J = 10 log (lll0 ), a decrease of 20 dB = ( 10 + 10) dB translates into a decrease in intensity by a factor of 10 x 10 = 100 A Curve C shows the neutron having positive charge throughout, which would imply the overall charge of the neutron is positive CurveD shows the neutron having negative charge throughout, implying that the neutron is negative Clearly, both of these are incorrect, since the neutron has zero overall charge Choice B is eliminated since it shows nonzero charge density for values of r beyond R, which is impossible since the neutron only extends to r = R Curve A must be the correct one It shows the neutron with a region of positive charge density (r < R/2) and a region of negative charge density (r > R/2), which would account for the overall neutrality of the particle D If the light originates in the core (the medium with the higher index of refraction), then the critical angle for total intern.al r:eflection from the cladding is -1 -n2 =sm • -1 ncladding ecrit =sm nl The positron is ejected, and the nucleus that remains (the daughter nucleus) is 15 N SCIENCE WORKBOOK ncore • -1 1.73 -1 J3= 600 =sm :::::sm 2.0 C As a result of the reaction p -+ n + e+ + n, a proton is lost (so the atomic number drops by 1) and a neutron is gained (so the mass number-the total number of protons and neutrons-stays the same) That is, the reaction is ~0 -+ ~N + +~e+ + n MCAT A The theoretical bandwidth of a fiber-optic cable is given to be 50 x 10 12 Hz A telephone conversation needs only kHz = x 10 Hz Therefore, there could be a maximum of (50 x 10 12 )/(3 x 103 ) = 17 x 109 = 1.7 x 10 10 conversations 428 A Choice B is untrue, and there is nothing in the passage to support choice C Choice D is wrong because electromagnetic radiation would not be a factor with optical fibers, since only charged particles would feel its effect Since light bounces back and forth between the boundaries of the core, there could be a number of different possible light paths within the core, all striking at different angles These different paths have different path lengths, and therefore will take different amounts of time to reach the end As a result, the light rays will interfere with each other The narrower the core, the fewer different possible paths the light can travel, and hence, the smaller the distortion B As the light ray enters a more optically dense medium, its speed decreases (by definition of the index of refraction) Since the frequency of the wave does not change, the wavelength must decrease (since J f= v) 429 PHYSICS SOLUTIONS MCAT SCIENCE WORKBOOK 430 ... 14,57, 76, 77,78 41,48,49,50,51,60, 79 52,53,54, 80 3, 11, 36,37,81, 82,83 MCAT Physical Sciences Review MCAT Physics Chapter Title Correspond;ng MCAT Science Workbook Passage Numbers Kinematics... Geometrical Optics 11 PHYSICS Chapter Number 10 3-9, 15-18 1Q-13, 19-24 31,32 25-30,33-39 14,40,45,46,48-55 42,43,56,57 41,44,47,58-60 GENERAL CHEMISTRY Chapter Number 10 MCAT Physical Sciences Review... 168.9 70 Yb 173.0 cr 99 100 101 Es 102 Fm Md No Lr (251) (252) (257) (258) (259) (260) 7l Lu MCAT PHYSICS PRACTICE QUESTIONS AND pASS AGES I i i !' I I ! I 223 ~ 3 What is the magnitude ofthex-component