11Physics discretes test w solutions

Physics Discretes Test Time: 30 Minutes Number of Questions: 30 This test consists of 30 discrete questions—questions that are NOT based on a descriptive passage These discretes comprise 15 of the 77 questions on the Physical Sciences and Biological Sciences sections of the MCAT MCAT PHYSICS DISCRETES TEST DIRECTIONS: The following questions are not based on a descriptive passage; you must select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Unq (261) 105 Unp (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 as developed by Physics Discretes Test An ideal experiences: gas compressed adiabatically A a decrease in temperature and an increase internal energy B a decrease in temperature and a decrease internal energy C an increase in temperature and an increase internal energy D no change in either the temperature internal energy in in An object is moving counterclockwise in a circle at constant speed Which of the following diagrams correctly indicates both its velocity vector and its acceleration vector? A in or at v Two identical rods having temperatures 100°C and 50°C, respectively, are brought into contact If the system is isolated from the environment, its entropy: a v a B A B C D C v increases decreases remains the same cannot be determined ar D v a An ambulance originally at rest has its siren going The ambulance starts accelerating away from a boy standing nearby He will hear: A B C D a decrease in the pitch of the siren no change in the pitch of the siren an increase in the pitch of the siren an increase, and then a decrease in the pitch of the siren A 1.0 kg object moving east with a velocity of 10 m/s has a head-on collision with a 0.5 kg object which is at rest Neglecting friction, what is the momentum of the system after collision? A B C D 15 kg • m/s; east 15 kg • m/s; west 10 kg • m/s; east 10 kg • m/s; west GO ON TO THE NEXT PAGE KAPLAN The diagram shows energy levels E1, E 2, E 3, and E4 in an atom If the transition from energy E3 to E2 gives rise to a photon of radiation of a particular wavelength, which of the following transitions could give rise to a photon of radiation of longer wavelength? An object (O) is placed within the focal length f of a converging lens The image (I) is located at which of the following positions? O A f E4 f' Energy (eV) E3 I I B O E2 f C f' O f' E1 A B C D E4 to E1 E3 to E1 E2 to El E4 to E3 f D I O f f' The activity of a radioactive source falls to onesixteenth of its original value in 32 minutes What is the half-life, in minutes, of this decay process? A B C D 16 min min I When the 2/3 Ω resistor is replaced by an 8/3 Ω resistor, the voltage across the resistor changes from: 9V A parallel plate air capacitor is connected across a battery of constant emf If the separation of the plates is decreased, which of the following increases? I Charge on the plates of the capacitor II Potential difference between the plates of the capacitor III Capacitance of the capacitor A B C D III only I and II only I and III only II and III only 2/3 Ω 1/3 Ω A B C D 4 6 V V V V to to to to 8 V V V V Physics Discretes Test 1 From the data shown below, estimate the halflife of element X Number of Undecayed Nucleii 10 x 105 In the diagram below a stream of electrons leaves the electron gun at G and strikes the fluorescent screen at P When the current is switched on at S, it flows through the wire coils in an anticlockwise direction as seen by the observer at O; the observer sees the spot of light at P: Element X 7.5 x 105 G x 105 P O 2.5 x 105 1.5 3.0 4.5 6.0 S Years elapsed A B C D 1.5 years 3.0 years 4.5 years 6.0 years A planet has a diameter one half of the Earth’s diameter and a mass that is one half of the Earth’s mass What would be the acceleration due to gravity on the planet? A B C D 2.45 m/s2 4.90 m/s2 9.80 m/s2 19.6 m/s2 A B C D move downwards move to the left move to the right remain still Object A of mass M is released from height H while object B of mass 0.5 M is released from height 2H What is the ratio of the velocity of object A to the velocity of object B immediately before they hit the ground? (Note: Assume that air resistance is negligible.) A B C D 1:1 1: 1:2 1:4 GO ON TO THE NEXT PAGE KAPLAN 5 In the hydraulic lift shown below, piston A has a cross-sectional diameter of m and piston B a cross-sectional diameter of m If the force exerted by piston A on the liquid is doubled, the force lifting up piston B is: A A B C D B A fluid flows through a pipe such that the velocity at point A is 1/4 that at point B What is the ratio of the radius of the pipe at point A to that at point B? A B C D 1:4 2:1 4:1 1:2 A cube of wood whose sides are each 10 cm weighs 16 N in air When half submerged in an unknown liquid it weighs only 10 N What is the density of the liquid? (Note: Assume g = 10 m/s2.) decreased by a factor of increased by a factor of increased by a factor of increased by a factor of A 600 kg/m3 B 800 kg/m3 C 1,000 kg/m3 D 1,200 kg/m3 Which of the following is true of the adiabatic expansion from point Q to R? P T Q S R Two fixed charges of +Q and +3Q repel each other with a force, F If an additional charge of –2Q is now added to each of the two charges, what is the new force between them? A B C D Zero F/6 attraction F/3 repulsion F/3 attraction V A The internal energy at Q is greater than at R B The internal energy at R is greater than at Q C Heat is released by the gas in moving from Q to R D Heat is absorbed by the gas in moving from Q to R If the binding energy of a nucleus is 186.2 MeV, what is its mass defect? (Note: amu yields 931 MeV of energy, amu is equivalent to 1.66 × 10–27 kg.) A B C D 3.3 6.6 8.3 3.1 × × × × 10–28 10–28 10–27 10–25 kg kg kg kg Physics Discretes Test A block starts from rest and slides down along one quadrant of a frictionless circular track whose radius is one meter Which of the following is true? In a frictionless horizontal mass spring system, as the mass moves from the point of maximum expansion through the equilibrium position to the point of maximum compression, the potential energy of the system: A B C D 1m A The speed of the block at the bottom will be the same as if it had fallen vertically B Mechanical energy will not be conserved C The acceleration of the block is constant throughout the descent D The total work done by gravity is greater than if the block had fallen vertically decreases continually increases then decreases decreases then increases remains the same Which of the following diagrams accurately depicts the electric field lines between charges of equal magnitude but opposite sign? A C B D 2 Which one of the following is a scalar quantity? A B C D Velocity Force Impulse Kinetic energy A fireboat is fighting a fire to its right but also has its hoses going full blast into the water to its left Which law of physics best explains the reason for this? A B C D Newton’s law of gravity Newton’s second law Newton’s third hew Conservation of energy GO ON TO THE NEXT PAGE KAPLAN MCAT Four positively charged spheres (+1 C each) and four negatively charged spheres (–1 C each) are distributed at equally spaced intervals around a circular hoop whose radius is m as shown below What is the electrostatic potential at the center of the hoop? (Note: k is the electrostatic constant.) In an attempt to restore the natural rhythm of a heart attack victim, electric shock plates are sometimes used The shock plates are run off a 10 V battery and draw current at a constant 30 A If each shock draws 300 J of electrical energy, what is the duration of the shock? A B C D + + + + – – – – A bat locates obstacles in its flight path by emitting sound in the same direction as it travels When the sound encounters an obstacle, it is reflected back to the bat Assuming that the bat’s velocity and the frequency of the emitted sound are constant, and assuming the obstacle is stationary, which of the following describes the frequency observed by the bat? A It is greater than the frequency emitted by the bat B It is less than the frequency emitted by the bat C It is the same as the frequency emitted by the bat D Its character cannot be determined A 0V B 4k V C 8k V D 16k V Which of the following quantities can be expressed dimensionally as M • L2/T2, where M is mass, L is length, and T is time? 0.1 sec 0.3 sec 1.0 sec 3.0 sec An object is moved from 2f towards f, where f is the focal length of this converging lens The image formed on the other side of the lens will move from: I Torque II Work III Energy A B C D I only I and II only II and III only I, II, and III 2f f A B C D f f to 2f 2f to f/2 2f to f 2f to infinity END OF TEST 2f Physics Discretes Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: C 11 B A 12 D C 13 D A 14 B A 15 B 10 10 D C C B C 16 17 18 19 20 A B D D A 21 22 23 24 25 A D C C A 26 27 28 29 30 A D C A D Physics Discretes Test EXPLANATIONS C This is a first law of thermodynamics question, which states that the change in internal energy, U, equals the heat, Q, minus the work done, W We are asked to find what happens to a gas that is compressed adiabatically To solvethis problem one must know that adiabatic means that no heat enters or leaves the system This means that Q is zero, and the first law simplifies to U = –W Because the system is compressed, work is done on the system, and not by the system Work is therefore negative Another way we could have reached this conclusion from the formula W = PV for an ideal gas Since in this case the final volume is less than its initial volume, V is negative and so is W Since W is negative, –W is positive and so the change in internal energy is positive Choices B and D can therefore be eliminated Now all that is left is to decide whether an increase in internal energy would increase or decrease the temperature Since temperature increases with the average kinetic energy of the gas molecules, as the internal energy increases, so would the temperature A In an isolated system, the entropy never decreases The only possibilities are entropy remains the same (if the process is reversible) or it increases (if the process is irreversible) Under what category does this process fall? When the two rods are put in thermal contact, heat will flow from the hot rod to the cold one until both rods are at the same temperature (in thermal equilibrium) Is it possible for the original situation (hot rod/cold rod) to redevelop spontaneously? No: the process is irreversible The answer is thus choice A Note that the rod that is hot originally will experience a decrease in entropy as it cools This rod by itself is however not an isolated system (energy is flowing out of it into the cold rod) and therefore it doesn’t violate what we have just said The increase in entropy of the cold hot upon warming is larger in magnitude than the decrease in entropy of the hot rod as it cools, thus leading to a net increase in entropy C The two concepts necessary to keep in mnd to solve this problem are (1) momentum is conserved in a collision and (2) momentum is the product of mass and velocity Conservation of momentum states that the momentum immediately before a collision equals the momentum immediately after the collision Before the collision, the 1.0-kg object has a momentum of 1.0 kg × 10 m/s = 10 kg•m/s This is the total initial momentum since the other object is at rest (v = 0, hence p = mv = 0) This initial momentum of the system points in the same direction as the velocity of the moving object, i.e towards the east This must then also be the momentum of the system after collision A The object is undergoing uniform circular motion as its trajectory is circular and it is moving at a constant speed The velocity of the object is always tangential to the circle On this criterion alone, choices B and C can be eliminated For uniform circular motion, there is no tangential component to the acceleration The acceleration is therefore entirely radial and is called the centripetal acceleration, with magnitude equal to v2/r It points towards the center of the circle A This is a Doppler effect question where we have a source moving away from a stationary observer The exact form of the equation that describes this situation is: f’ = fv (v + v s ) where f’ is the observed frequency, f is the actual frequency (from the point of view of the source), v the velocity of sound in the medium, and vs the velocity of the source We not need to remember this equation specifically though to answer the question (In fact it is very unlikely that the MCAT will expect you to have memorized this.) We should, however, be able to reason out that if the source is constantly moving away from the observer, successive wave crests would take longer and longer to reach the observer, leading to a perception of a lower frequency or pitch The faster the source is moving away, the lower the perceived pitch would be Here the ambulance is accelerating from the boy; its speed is therefore increasing The pitch would then be decreasing D KAPLAN 11 MCAT We are asked for a transition that would give rise to a photon of a longer wavelength When a particle (for example an electron) relaxes from a higher energy state to a lower energy state, the excess energy is released in the form of a photon The energy of a photon is directly proportional to its frequency via E = hf, but is inversely proportional to the wavelength: E = hc/λ We are therefore looking for the transition that corresponds to a smaller energy difference between the initial and final states than between E3 and E2 The energy difference between levels is easily read off from the figure as the vertical distance between the energy levels involved Out of all the choices, only choice D, a transition from level to level 3, represents a transition that releases less energy than the transition from level to level C When one half-life has passed there is one-half of the original sample left When two half-lives have passed, one quarter of the original sample is left Continuing in the same way, we find that when four half-lives have passed there is one sixteenth of the sample left This means that in 32 minutes half-lives have passed This corresponds to minutes per half-life C The capacitance of a parallel-plate capacitor is approximately given by: C = κε0 A d where κ is the dielectric constant of the medium between the plates, ε0 is the permittivity of free space, A is the area of overlap between the two plates, and d is the separation of the two plates From this equation we can see that if the separation of the plate decreases, the capacitance will increase A more general relationship involving capacitance, which applies to capacitors that are not necessarily of the parallel-plate variety, is C = Q/V, where Q is the charge stored on one of the pair of conductors making up the capacitor, and V the voltage across the two We are told in the question stem that the capacitor is connected across a battery of constant emf Applying Kirchhoff’s law, this means that the voltage across the plates will have to constant as well—in fact, its magnitude will be the same as the emf of the battery Since C increases while V has to remain constant, Q, the charge, will have to increase as well B This is one of those questions where you may already know the answer, but if you don't you can reason it out For any object that is placed inside the focal point of a converging lens, the image is on the same side of the lens, erect and enlarged The converging lens is then a magnifying lens If you didn't know this, you could have 1 selected some sensible numbers and plugged them into the equation = + If we assume that the focal f i o length is centimeters and the object distance is centimeters, we can calculate an approximate value for the image distance Rearranging the equation and putting in the numbers, we get: 1 1 = – =– i The image distance is then –6 centimeters Since the value is negative, the image is virtual For a lens this means that the image is located on the same side of the lens as the object There is only one answer choice where this is the case, and that is answer choice B 10 D In both cases the two resistors are in series Case has a 2/3-Ω resistor and a 1/3-Ω resistor in series, while case has an 8/3-Ω resistor and a 1/3-Ω resistor in series Let's first look at case The total resistance, since they are in series, is Ω, i.e 2/3 + 1/3 and, since the potential difference in this circuit is volts, the current from I = V/R is volts/1 Ω or amps It follows then that the potential drop across the 2/3-Ω resistor is A x 2/3 Ω or volts from V = IR For case the current in the circuit is volts/3 Ω or amps The potential drop across the 8/3-Ω resistor is IR, or amps x 8/3 Ω , which is volts The change in voltage drop across that resistor is from volts to volts, and that is choice D 11 12 B Physics Discretes Test This question requires you to interpret a graph On the x-axis you are given the time in years, and on the yaxis you are given the number of undecayed nuclei remaining You are asked to determine the half-life The half-life is the time it takes for one-half of a sample to decay We start off at time zero with 10 x 105 nuclei When one half-life elapses, one-half of the sample decays, and that implies that there is one-half of the sample remaining One-half of 10 x 105 is x 105 At what time we have x 105 nuclei remaining? From the graph, you see that it is years, and therefore years is the half-life If you wanted to make sure, you could check to see when the next half-life occurs After years we have x 105 nuclei left In another half-life, only half of x 105 or 2.5 x 105 are left When does that occur? Looking at the graph, that occurs in years So another years have passed Therefore, the half-life is years, answer choice B 12 D Any two objects with mass attract each other with a particular force given by Newton's law of gravity which says F=Gm1m2/R2, where G is the universal gravitational constant, m1 and m are the two masses, and R is the distance between their centers Let's look at the case when an object on the Earth's surface is attracted to the Earth Under these conditions, we'll let m1 be the Earth's mass, m2 be the mass of the object on the Earth's surface, and R be the distance between the centers of the two masses, which is equal to the Earth's radius However, we also know that this gravitational force on m is equal to m 2g, where g is the acceleration due to gravity at the Earth's surface.If we equate these two expressions for the force of attraction on m2 and cancel m which appears in both, we see that the acceleration due to gravity g is equal to Gm 1/R Moving to another planet, the analysis would be the same but m1 and R would be different for the new planet In this problem the new m1 and R are one-half of the Earth values So the mass of the new planet equals m1/2, and the radius of the new planet equals R/2 This gives us that the new acceleration a is a=G m 1/2 m 1/2 m1 =G = 2G 2 (R/2) R /4 R Now Gm 1/R2 is the acceleration due to gravity on Earth so substituting in gives us that a = 2g The gravitational acceleration due to gravity on the Earth is 9.8 m/s which would make the acceleration on the other planet two times that or 19.6 m/s2 13 D This question asks you to determine how an electron beam is deflected when current flows in the wire coils When current flows in the wire coils, it creates a magnetic field This magnetic field can exert a force on the electron beam, causing the beam to be deflected In order to find the direction of the magnetic force, we first need to know the direction of the magnetic field To find the direction of the field, we put the thumb of our right hand in the direction of the current Our remaining fingers curl around the wire until we get to the point in question The direction of the field is the direction in which our fingers point We are told that the current travels counter-clockwise with respect to the observer So if you imagine yourself at O looking at the tube, the current travels from your right to your left at the top of the tube Therefore as seen from O, the thumb of your right hand should point towards the left across the top of the tube Now curl your fingers around Inside the tube, where the electron beam is, your fingers go along the dashed line from G to P In terms of the electron beam, your fingers and therefore the magnetic field are going in the same direction as the electron beam Now whenever the magnetic field is in the same direction or in the opposite direction to the motion of the charged particle, the magnetic force on the particle is zero This can be seen from the equation F=qvBsinθ If the magnetic field is in the same, or opposite direction, then θ is either 0° or 180° Therefore the sine of θ is 0, so the magnetic force is also zero When no magnetic force is present, there is no deflection of the beam 14 B We are asked to find the ratio of the velocity between two objects just before they hit the ground Object A has a mass of M and is released from a height of H Object B has a mass of M/2 and dropped from a height of 2H We want to find the ratio of the velocity of object A to object B just before they hit the ground How can we relate final velocity to height? Let's use the kinematic equation: v2 = v o + 2as where v is the final velocity, vo is the initial velocity, as is the acceleration in this case, it is the acceleration due to gravity, g, -and s is the distance traveled In our problem, the two objects are released from a given height This implies that they had no initial velocity, so v o = This simplifies our equation to v2 = 2gs, or v = (2gs)1/2 Notice that this KAPLAN 13 MCAT equation says that velocity is independent of mass That is true Specifying the mass in the question is superfluous, because all objects free-fall at the same rate regardless of their mass So the velocity of object A is given by va = For object B, vb = 2g2h = 2gh 4gh Now taking the ratio we find that va = vb 2gh = 4gh 1/2 = 1/ 15 B The hydraulic lever is based on Pascal's principle, which states that an applied pressure is transmitted undiminished to all portions of a fluid Therefore, the pressure under piston A equals the pressure under piston B Since pressure equals force divided by area, we can say F1/A1=F2/A2 In the question stem we are told that piston A has a diameter of meters, and that the diameter of piston B is meters But notice that the areas of the pistons not change The information that the question stem gives you about the diameters of the two pistons is not needed to answer the problem The point is that if the areas remain constant, the force on piston A is directly proportional to the force on piston B, so if the force on A is doubled, then the force on B must also double 16 A An adiabatic expansion means that there is no heat flow in or out of the system That fact alone immediately eliminates choices C and D When a gas expands, it does work on the environment The magnitude of the work done by the gas is expressed as PV P is the pressure that it expands against and V is the increase in volume Now since under adiabatic conditions no heat flows into the system, the energy needed to this work of expansion must come from the internal energy of the gas itself So the internal energy of the gas after adiabatic expansion will have to be less than the internal energy before the expansion From points Q to R the volume of the gas has expanded and work has been done by the gas so that the internal energy of the gas must be greater at point Q than it is at R 17 B To answer this question we must use the continuity equation, which states that the product of the crosssectional area and the velocity is a constant This means that we equate the product of the cross-sectional area and the velocity at point A with that at point B, so we have that AAvA=ABvB, where A is the area, and v is the velocity Now, in the question stem we are told that the velocity at point A is one quarter that at point B, so vA = v B /4, or vB = 4vA If we let the radius at point A be rA, and the radius at point B be rB , we can write the cross-sectional area at point A as being r A 2, and the cross sectional-area at point B as being r B Substituting into the continuity equation, we get that rA2vA equals rB2vA Canceling out the 's and the vA's and rearranging the terms, we are left with rA2/rB2 = Taking the square root of this we get that rA/rB = Therefore, the ratio of the radius at point A to that at point B is to 1, which is answer choice B 18 D To solve this question properly you need a good understanding of Archimedes' Principle, which states that a body wholly or partially immersed in a fluid will be buoyed up by a force equal to the weight of the fluid that it displaces In other words, the buoyant force, Fb, equals the weight of the displaced fluid, which is the mass of the displaced fluid times g, the acceleration due to gravity But we're looking for the density of the fluid, so we need to define the buoyant force in terms of density The equation for density is ρ = m/V, where ρ is the density, m is the mass, and V is the volume We can rewrite this equation to get an equation for m, and we get that m = ρV So now we have the buoyant force Fb = ρVg Solving for the density of the liquid, we find ρ1 = Fb/V1g, where the subscript denotes liquid The problem tells you to assume that g equals 10 m/s2 Now we have to calculate the buoyant force and the volume of liquid displaced The buoyant force is the difference between an object's weight in air and its weight in the fluid We're told that the cube weighs 16 newtons in air and only 10 newtons in the liquid So, the buoyant force is equal to the difference, or 16 – 10 = N We're told that the block is only half submerged, so, the volume of liquid displaced V1 equals half of the volume of the cube Since the volume of any cube is its side cubed and each side is 10 cm or 0.1 m, we find that the volume of the block is 10–3 m3 And so the volume of the water displaced is one-half the volume of the block, or 10–3/2, or x 10–4 m3 14 Physics Discretes Test In sum, we found the buoyant force, Fb = newtons, the volume of the liquid displaced, V1 = x 10 –5 m , and the acceleration due to gravity, g = 10 m/s Now we just have to substitute into the equation, ρ1 = F b/V 1g Doing the math we find that the density of the liquid equals 1,200 kg/m3, which is answer choice D 19 D This is an electrostatics question that requires you to know Coulomb's law, which gives the magnitude of the electrostatic force The force F = kQ1Q2/r2, where k is the electrostatic constant, Q1 and Q2 are the charges of the two particles and r is the distance separating them Now, in this question we originally have two particles with charges of +Q and +3Q From Coulomb's law, the force, which we are given as F, = k(Q)(3Q)/r2 So we can say that F = 3kQ2/r2 Now the situation changes A charge of –2Q is added to each of the two charges So the +Q becomes –Q, and the +3Q becomes +Q So now we have two charged particles at +Q and –Q Since the question specifies that the charges are fixed, we can assume that r has not changed Let's plug these values into Coulomb's law The force equals kQ(–Q)/r2 which equals –kQ2/r2 Remember that the original force F was equal to 3kQ2/r2 So this new force is actually one third the original force, and this eliminates answer choices A and B Now we have to determine if the force is an attractive one or a repulsive one Oppositely charged particles attract, and similarly charged particles repel Initially we have charges of +Q and +3Q Since they are both positive, they repel each other However, when we add the –2Q to each charge, one of the charges becomes +Q and the other becomes –Q Since they have opposite signs, the force is attractive Therefore, answer choice D, F/3 attraction is correct 20 A The binding energy, E, equals the mass defect, m, times c2 c is the speed of light in vacuum, and has the value x 108 m/s However, you not need to know this, and in fact, if you try to plug this value in and calculations, you are more likely to make mistakes We are given the fact in the question stem that amu yields 931 MeV of energy, and so we have: E = mc2 931 MeV = (1 amu)c2 c2 term is therefore equal to 931 MeV/amu This may be a weird unit to report the square of a velocity, but all we are after is some proportionality constant between E and m and as long as we keep our units consistent we are fine In the problem we are given the binding energy as 186.2 MeV hence, the mass defect, expressed in amu, is m= E amu = 186.2 MeV x = 0.2 amu c2 931 MeV But we want the answer =in kilograms We are also told that in amu there are 1.66 x 10 –27 kilograms Multiplying 0.2 amu by 1.66 x 10-27 kg/amu, we find the mass defect is 0/33 x 10–27 or 3.3 x 10–28 kilograms 21 A Here we have a block sliding down one quadrant of a frictionless circular track whose radius is one meter, and we are asked which of the statements A, B, C, or D is true Let's go through each of the answer choices in turn Answer choice A states that the speed of the block at the bottom will be the same as if it had fallen vertically Initially the block has a potential energy of mgh, where m is the mass, g is the acceleration due to gravity, and h is the height Now we are told that the radius of the track is meter So the initial potential energy of the block is mg Now this potential energy converts to kinetic energy as the block moves down the track, and since the track is frictionless, when the block reaches the bottom all the potential energy will have been converted to kinetic energy Kinetic energy is given by the equation K = (1/2)mv2, and this is equal to the potential energy which we found to be mg So we get that (1/2)mv2 = mg, or v2 = 2g For an object in free fall the equationn that relates the final velocity v, to the initial velocity vo, and the height h, is v2 = vo2 + 2ah, where a is the acceleration, which in this case is g, the acceleration due to gravity Now we know that the initial velocity is zero since it starts from rest, and we also know that the height is meter since it starts from the same height as the block that slides down the track So our equation becomes v2 = 2g This is exactly the same as the result that we obtained from the block sliding down the track, which tells us that the final velocity in both cases is equal Therefore answer choice A is correct KAPLAN 15 MCAT Let's go through the other answer choices and see why they are incorrect Answer choice B states that mechanical energy will not be conserved This statement is false since we have already said that because the track is frictionless there are no non-conservative forces acting on the block, and so the sum of the potential and kinetic energy at any point is always constant In other words, mechanical energy is conserved Answer choice C states that the acceleration of the block is constant throughout the descent You may have thought that this answer was correct since the acceleration due to gravity remains constant However, you must remember that the block is not in free fall but is following a circular path, and so the direction of the acceleration vector is continually changing In order for a vector to be constant, both the magnitude and the direction must remain constant, so the fact that the direction changes along the circular track is sufficient to eliminate choice C Answer choice D states that the total work done by gravity is greater than if the block had fallen vertically This is untrue in this case since, as we said before, when the block moves down the track there are no nonconservative forces acting on the block, so no extra work has to be done by gravity Thus answer choice D is incorrect, and once again the correct answer is choice A 22 D A scalar is a quantity that has a magnitude but no dierection as opposed to a vector which has both magnitude and direction Velocity, choice A, has both a magnitude and direction An example of a velocity vector is meters per second, south The magnitude in this case is meters per second and the direction is south Answer choice B suggests force We draw force diagrams all the time For instance, a girl pushes a box to the left with a force of newtons Now, the magnitude of the force is newtons and the direction is to the left So force is also a vector Impulse is defined as the force vector times the time Time is a scalar, but when you multiply a scalar by a vector you get a vector, so impulse is a a vector Impulse is also equal to the change in momentum and changes in momentum are vectors So again, impulse must be a vector Answer choice D suggests kinetic energy.Energy is a scalar quantity "But wait a minute," you say, "Doesn't kinetic energy also depend on velocity?" No, kinetic energy depends on the square of the speed, and the speed, which is the magnitude of the velocity vector, is a scalar Kinetic energy is a scalar, and answer choice D is correct 23 C Here we have a fireboat fighting a fire to its right, but it also has its hoses throwing water to the left We are asked to find which of the given physical laws best explains why it does this The answer is choice C -Newton's third law, which states that to every action there must be an equal and opposite reaction When the fireboat is fighting the fire, it has its hoses going to the right, and it is exerting a force to the right on that water Now, by Newton's third law, the water provides a reactive force that is equal in magnitude but opposite in direction towards the left, and this causes the fireboat to be pushed towards the left In order to prevent the boat from moving towards the left, the fireboat must also have its hoses pushing water to the left, creating a second reactive force on it which is to the right The other answer choices don't imply that the boat would have to blast water in the opposite direction to stay in place Newton's law of gravitation, choice A, states that there exists a force of mutual attraction between all matter which is proportiomal to the product of the masses and inversely proportiomal to the square of the distance separating them The constant of proportionality is G, the universal gravitational constant Newton's second law, choice B, states that a force exerted on an object equals the mass of the object times the acceleration And finally conservation of energy, choice D, states that in the absence of non-conservative forces like friction or air resistance, the total energy of a system must remain constant None of the other answer choices directly affects the situation 24 C This question tests how well you understand energy exchanges in a mass-spring system There are two types of energy: energy of motion, or kinetic energy, and energy of position, or potential energy Kinetic energy in any system, including a mass-spring system, is defined as mv2/2 Potential energy does not have such a universal formula that applies for every case; the potential energy function for gravity is different from the potential energy for the mass-spring system which is different from the electrostatic potential energy, etc For a mass-spring system, potential energy is one-half times the spring constant times the distance from the equilibrium point squared: (1/2)kx2 The other important piece of information you should know about energy in a mass-spring system is that, provided there are no frictional forces acting, energy is conserved This means that the total mechanical energy, which is the sum of the kinetic and potential energies, stays constant Here we have a spring traveling from the point of maximum extension to the point of maximum compression Let's consider the energy at maximum extension The words maximum expansion should tell you that 16 Physics Discretes Test the spring will not extend any more At this point the speed should be zero If it were not, then the spring would continue to expand If the speed is zero at this point, the kinetic energy must also be zero Now we know from conservation of energy that the total mechanical energy is a constant So when the kinetic energy equals, zero, the potential energy must be a maximum If potential energy is a maximum at this point, it cannot increase This eliminates answer choice B Now we are told that the spring moves from this position through the equilibrium point At the equilibrium point there is no expansion or compression Here, the speed of the mass is a maximum, and therefore the kinetic energy is a maximum Now from conservation of energy if kinetic energy is a maximum, potential energy must be a minimum So we can eliminate answer choice D since the potential energy does change Now the mass moves to the point of maximum compression The story at this point is much the same as at the point of maximum expansion At the point of maximum compression the speed is zero, otherwise the spring would continue to compress If the speed is zero, the kinetic energy must be also By conservation of energy when the kinetic energy is a minimum, potential energy must be a maximum So the potential energy has gone from its maximum value at maximum extension, to its minimum value at the equilibrium point, to its maximum value at maximum compression In other words, the potential energy has decreased and then increased Therefore, answer choice C is the correct answer If you know that the potential energy was given by the formula U = kx2/2, where U is the potential energy, k is the spring constant, and x is the distance from the equilibrium point, you could have also reached the same conclusion At maximum compression or expansion, x is a maximum, and therefore the potential energy is a maximum At the equilibrium point, x equals zero so the potential energy is zero Therefore, the potential energy goes from maximum, to zero to maximum This means that it decreases and then increases 25 A An electric field line shows how a positively charged test particle would be accelerated in an electric field Keep in mind that a positive test charge will be attracted to a negative particle and will be repelled from a positive particle We know that because opposite charges attract and like charges repel in electrostatics So we are looking for an answer choice in which the lines point away from the positive charge and toward the negative charge Now in answer choice A, we see field lines directed away from the positive charge and toward the negative charge Answer choice B shows the opposite situation So it is wrong Answer choices C and D are the kind of field patterns you would expect from two similarly charged particles They repel each other Therefore, the lines don't meet and, in fact, they diverge from each other If you didn't recognize the pattern, you could have eliminated answer choice C because it depicts field lines going away from, not toward, a negative particle Similar reasoning could be used for answer choice D for a positive charge the field lines point away from, not toward, the charge 26 A We are asked to find the electrostatic potential at a specific point given a distribution of charged spheres We have four spheres charged to +1 coulomb and another four spheres charged to –1 coulomb These spheres are arranged in a circle of radius meter What is the electrostatic potential? The electrostatic potential is defined as the work per unit charge required to move a test charge from infinity to a particular point Mathematically, we can say V = W/q, where V is the potential, W is the work, and q is the charge of the test charge Now, work is defined as the force times the distance So we can also say that V = Fr/q Now the force is the electrostatic force of Coulomb's law: F = kq1q2/r2 Plugging that into the formula V = Fr/q, we find that one of the r's cancel and one of the q's cancel We are then left with V = kq/r Let's calculate the potential for one negatively charged sphere The charge q equals –1 coulomb, and the distance from the charge to the point in question, which is the center of the circle, is meter From the equation V = kq/r, the potential due to each negatively charged sphere is –k Now to calculate the potential caused by each positively charged sphere, we use q = +1 coulomb, and r = meter We find that each positively charged sphere produces a potential of +k Now since electrostatic potential is scalar, we can add the effect from each charge directly In other words, there is no directional component to worry about So, Vtotal = 4(+k) + 4(–k) which equals 4k – 4k = 27 D In this Roman numeral question we are being asked which of the given quantities torque, work, and energy can be expressed dimensionally as ML2/T2 The first thing to note is that work and energy have the same units So you can immediately eliminate any answer choice that includes II but not III or that includes III but not II This enables us to eliminate choice B You can also consider the remaining choices strategically If the formula is KAPLAN 17 MCAT applicable to work and energy, then the answer is either choice C or choice D; and if it's wrong for work and energy, then the answer is choice A So let's look at the units of work Work is force times distance; let's see how those units compare to the formula ML2/T2 Distance has units of length, L What about force? From Newton's second law, force is mass which is M and acceleration, which has units of meters over seconds-squared or L/T2 So force is ML/T2, and force times distance is ML2/T2 So work and therefore energy can be expressed dimensionally as ML2/T2 Now, torque is also force times distance Even though torque has a physical meaning different from work or energy, it has the same units of ML 2/T2 Therefore all three Roman numeral statements are correct, which means that the correct answer is choice D 28 C In this medical application of physics we want to find the duration of a shock given to restore natural rhythm to a heart attack victim We are told that the shock plates are run off a 10 volt battery with a current of 30 amps, and the shock draws 300 joules of energy We want to relate work and time here, so we can use the equation P = W/t, where P is the power, W is the work or energy, and t is the time We don't directly have a value for power, but we can figure it out In any electrical system the power is the product of the current, i, and the voltage, V In other words, P = iV We have values for both of those quantities, so we can equate our two expressions for power, and find that W/t = iV We want to solve this for the time, t, so we find that t = W/iV Now it's just a case of plugging in the numbers We are told that the work or energy drawn is 300 joules, the current is 30 amps, and the voltage is 10 volts, so the time that the shock lasts is 300/(10 x 30), or 300/300 and that's just second, answer choice C 29 A The bat emits sound that is reflected off a stationary obstacle then travels back to the bat We want to know about the frequency heard by the bat relative to the frequency that it emitted First the bat emits the sound, then the sound bounces off a stationary object, and then the sound returns to the bat The Doppler effect says that when an observer and a source are moving with respect to each other, the frequency perceived by the detector will be different than the frequency emitted, and when they are approaching each other the perceived frequency will be higher There is a formula associated with this that allows us to calculate the actual frequency perceived by the detector, but we can get the answer without having to use the formula Since the bat and the stationary object are approaching each other, the bat intercepts more wavelengths per second than it would intercept if it were stationary Since frequency is the number of wavelengths per second, more wavelengths mean a higher frequency, answer choice A 30 D To find the image distance we use the formula 1/o + 1/i = 1/f, where o is the object distance, i is the image distance, and f is the focal length We're interested in the image distance, so we can rewrite the equation to get 1/i = 1/f – 1/o Since it's a converging lens, f is positive; and since it's a single-lens system, o will also be positive Now looking at the answer choices you might notice that the answers for the image distance when the object is at f are all different So we can just find that distance; that will be enough to locate the correct answer So when the object distance is f, our equation for i becomes 1/i = 1/f – 1/f = To find i we must take the inverse 1/0 is infinity, so answer choice D is correct For the sake of completeness, let's figure out what the image distance is when the object is at a distance of 2f Our equation becomes 1/i = 1/f – 1/(2f), which equals 1/(2f) And taking the reciprocal we find that i = 2f So, as an object is moved from 2f to f, the image formed will move from 2f to infinity, answer choice D 18 ... only I, II, and III 2f f A B C D f f to 2f 2f to f/2 2f to f 2f to infinity END OF TEST 2f Physics Discretes Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: C 11 B A 12 D C 13... (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Physics Discretes Test An ideal experiences: gas compressed adiabatically A a decrease in temperature and an... and III only II and III only 2/3 Ω 1/3 Ω A B C D 4 6 V V V V to to to to 8 V V V V Physics Discretes Test 1 From the data shown below, estimate the halflife of element X Number of Undecayed Nucleii