PHYSICS TOPICAL: Work, Energy and Momentum Test 1

Transcription

1 PHYSICS TOPICAL: Work, Energy and Momentum Test 1 Time: 22 Minutes* Number of Questions: 17 * The timing restrictions for the science topical tests are optional. If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit.

2 MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, 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 1 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (98) 44 Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra Ac Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce Pr Nd Pm (145) 62 Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa (231) 92 U 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. 2 as developed by

3 Work, Energy and Momentum Test 1 Passage I (Questions 1 4) Two stars bound together by gravity orbit each other because of their mutual attraction. Such a pair of stars is referred to as a binary star system. One type of binary system is that of a black hole and a companion star. The black hole is a star that has collapsed on itself and is so massive that not even light rays can escape its gravitational pull. Therefore, when describing the relative motion of a black hole and a companion star, the motion of the black hole can be assumed negligible compared to that of the companion. The orbit of the companion star is either elliptical with the black hole at one of the foci or circular with the black hole at the center. The gravitational potential energy is given by U = GmM/r, where G is the universal gravitational constant, m is the mass of the companion star, M is the mass of the black hole, and r is the distance between the center of the companion star and the center of the black hole. Since the gravitational force is conservative, the companion star s total mechanical energy is a constant of the motion. Because of the periodic nature of the orbit, there is a simple relation between the average kinetic energy <K> of the companion star and its average potential energy <U>. In particular, <K> = <U/2>. Two special points along the orbit are singled out by astronomers. Perigee is the point at which the companion star is closest to the black hole, and apogee is the point at which it is farthest from the black hole. 3. Which of the following prevents the companion star from leaving its orbit and falling into the black hole? A. The centripetal force B. The gravitational force C. The companion star s potential energy D. The companion star s kinetic energy 4. The work done on the companion star in one complete orbit by the gravitational force of the black hole equals: A. the difference in the kinetic energy of the companion star between apogee and perigee. B. the total mechanical energy of the companion star. C. zero. D. the gravitational force on the companion star times the distance that it travels in one orbit. 5. For a circular orbit, which of the following gives the correct expression for the total energy? A. 1 2 B. mv 2 2 mv C. GmM r 1. At which point in an elliptical orbit does the companion star attain its maximum kinetic energy? D. GmM 2r A. Apogee B. Perigee C. The point midway from apogee to perigee D. All points in the orbit, since the kinetic energy is a constant of the motion 2. For circular orbits, the potential energy of the companion star is constant throughout the orbit. If the radius of the orbit doubles, what is the new value of the velocity of the companion star? 6. What is the ratio of the acceleration of the black hole to that of the companion star? A. M/m B. m/m C. mm/r D. 1/1 A. It is 1/2 of the old value. B. It is 1/ 2 of the old value. C. It is the same as the old value. D. It is double the old value. GO ON TO THE NEXT PAGE. KAPLAN 3

4 MCAT Questions 7 through 12 are NOT based on a descriptive passage. 7. What is the power required to accelerate a 10 kg mass from a velocity of 5 m/s to a velocity of 15 m/s in 3 seconds? A. 33W B. 167W C. 225 W D. 333W 8. A block of mass m slides down a plane inclined at an angle θ. Which of the following will NOT increase the energy lost by the block due to friction? A. Increasing the angle of inclination B. Increasing the distance that the block travels C. Increasing the acceleration due to gravity D. Increasing the mass of the block 9. A block of mass m starts from rest and slides down a frictionless semi-circular track from a height h as shown below. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is: 10. A boy hits a baseball with a bat and imparts an impulse J to the ball. The boy hits the ball again with the same force, except that the ball and the bat are in contact for twice the amount of time as in the first hit. The new impulse equals: A. half the original impulse. B. the original impulse. C. twice the original impulse. D. four times the original impulse. 11. The power of solar radiation incident on a solar panel is 40 kw. If the efficiency of the solar panel is 10%, how much energy does the solar panel generate in 300 minutes? A. 1.2 MJ B MJ C. 120 MJ D. 720 MJ 12. Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed v towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v/3 in the same direction. What type of collision has occurred? A. inelastic B. elastic C. completely inelastic D. Cannot be determined from the information given. h h A. 4. h B. 2. C. h. D. independent of h. GO ON TO THE NEXT PAGE. 4 as developed by

5 Work, Energy and Momentum Test 1 Passage II (Questions 13 17) Particle storage rings facilitate collisions between electrons and positrons (positively charged electrons). When electrons and positrons collide at high energies, they can annihilate each other and produce a variety of elementary particles, including photons. In these reactions, momentum is always conserved. Powerful magnets are placed at various points along the ring to create a force directed toward the center, thereby guiding the particles in a circular motion. In a scattering experiment, the particle beams circle in opposite directions and collide head-on at the interaction points, which are surrounded by particle detectors. The particles in the ring are accelerated at one or more points along the ring by an external electric field. When charged particles accelerate, they radiate electromagnetic energy. Radio frequency power is continually fed into the storage ring to compensate for the energy loss. The reaction rate, R, is the number of particles scattered per second in the storage ring. It is given by the formula: R = Lσ, where L is the luminosity and σ is the cross-section of the reaction. The cross-section is a quantity that depends only on the particular type of reaction being considered. The luminosity contains all of the information about the initial conditions for a reaction and is given by: N L = N A + e e where N e refers to the number of electrons, N e + refers to the number of positrons, A is the cross-sectional area of the storage ring, and f is the number of revolutions per second made by the particles. Electron-positron reaction rates for modern particle storage rings are typically on the order of l0 3 s What percentage of the work done on the circulating particles is done by the magnetic field? A. 0%, because the direction of the magnetic force is perpendicular to the direction in which the particles travel B. 0%, because the magnetic field doesn t exert a force on the particles C. 50%, because the magnetic field and the electric field provide equal amounts of energy to the particles D. 100%, because the magnetic field is the source for the centripetal force that accelerates the particles f, 14. Which of the following would increase the reaction rate in an electron-positron storage ring? I. Decreasing the cross-sectional area of the storage ring II. Increasing the energy of the particles III. Increasing the number of positrons in the storage ring A. I only B. III only C. II and III only D. I, II, and III 15. The work done on the particles by the gravitational field of the Earth is not considered when figuring their energy loss per revolution. This is because: A. the gravitational force is perpendicular to the gravitational acceleration. B. the energy lost due to gravity is equal to the energy gained from the magnetic fields. C. the particles do not experience a significant gravitational force. D. the luminosity is not dependent on gravitational acceleration. 16. Two electrons with equal speed collide head-on at a total energy of 180 GeV. If the speed of the first electron after the collision is 0.9c, what is the speed of the second electron after collision? (Note: The speed of light in a vacuum is c = m/s.) A. 0.45c B. 0.7c C. 0.8c D. 0.9c 17. An electron and a positron are held in a storage ring, and they each lose 260 MeV of energy per revolution in the form of electromagnetic radiation. If the frequency of revolution is 10,000 Hz, how much power must be supplied to keep the total energy constant at 180 GeV? A MeV/s B MeV/s C MeV/s D MeV/s END OF TEST KAPLAN 5

6 MCAT ANSWER KEY: 1. B 6. B 11. B 16. D 2. B 7. D 12. B 17. C 3. D 8. A 13. A 4. C 9. A 14. D 5. A 10. C 15. C 6 as developed by

7 Work, Energy and Momentum Test 1 EXPLANATIONS Passage I (Questions 1 6) 1. B In the passage we are told that the total mechanical energy of the companion star is constant as it orbits around the black hole. The total mechanical energy is the sum of the gravitational potential energy and the kinetic energy. We do not really know anything directly about the kinetic energy in this case, except that it equals mv 2 /2. We are, however, given a formula for the gravitational potential energy of the companion star. Since the mechanical energy is constant, the point at which the kinetic energy is maximum must also be the point at which the potential energy is minimum. From the more familiar case of dropping a ball on Earth, we know that the potential energy is lower when the height is smaller. We may thus expect that the potential energy is minimum when the star is closest to the black hole. We can confirm this by using the formula given in the passage: U = GmM/r. Since G, m, M, and r are all positive quantities, U is always going to be negative and has a maximum value of zero when the two are infinitely far apart. As r decreases, U becomes more and more negative. It follows, then, that the point at which the potential energy is minimum is the point of closest approach to the black hole, i.e. the perigee. The perigee, then, is also the point where the kinetic energy is the greatest. 2. B The question stem states that the gravitational potential energy for circular orbits is constant. This implies that the kinetic energy is also constant (from the conservation of total mechanical energy). In the last line of the second paragraph of the passage, we are given a relationship between the average value of the potential energy and the average value of the kinetic energy. Since in this case the two are constants, their average values are identical to their actual values. Therefore, for a circular orbit, the kinetic energy, K, is : K = U/2. The potential energy, U, is equal to GmM/r. If the distance r is doubled, then, the absolute value of the potential energy is reduced in half, and thus the kinetic energy will also be cut in half. (Recall from our discussion for the question above that the potential energy is negative; a reduction of its absolute value thus leads to a less negative number or a higher potential energy, as is reasonable when one separates two objects that are attracted to each other.) Since the kinetic energy is proportional to the velocity squared, as the kinetic energy becomes 1/2 its original value, the velocity must become 1/ 2 its original value to satisfy the definition of kinetic energy mv 2 /2. 3. D Let us examine each choice and try to eliminate those that seem incorrect. Choice A states that the companion star does not fall into the black hole because of the centripetal force. The centripetal force on the star is an attractive force directed towards the black hole. It is provided by the gravitational attractive force between the two objects and hence, if anything, it is what leads us to expect the two to come together. Choice B, the gravitational force, is, as just mentioned, what is providing the centripetal force. This is one way that we could have eliminated these two choices: since they are essentially the same thing, neither of them can be correct! Choice C states that the companion star s potential energy keeps it in orbit. Does this make sense? Think about a ball that you hold in your hand. It has some potential energy by virtue of the fact that you are holding it above the ground, but that potential energy does not keep the ball from falling: it is your hand that does this. If you let go, the potential energy is converted to kinetic energy as the ball goes speeding towards the ground. Analogously, then, we would not expect potential energy to be the agent responsible for keeping the star in orbit here. Choice D states that the companion star does not fall into the black hole because of its kinetic energy. This makes sense because its tangential velocity is what keeps it in orbit and kinetic energy is certainly related to tangential velocity. The gravitational force acts as a centripetal force which keeps redirecting the star so it doesn t fly off into space. Why would it fly off into space? Because it has kinetic energy from the velocity in the tangential direction. A stable orbit occurs because of the balance between the kinetic energy and the gravitational attraction. If the kinetic energy were not large enough, the star would fall or spiral into the black hole. 4. C There are several ways that one could have obtained the answer. One is by the work-energy theorem: The net work done on an object is the change in its kinetic energy. In the case of a complete orbit, this is the kinetic energy the star has at the end of one round of the orbit minus the kinetic energy it had at the beginning (at the same point in space). Since potential energy is dependent only on position, we know that the star would have the same potential energy after one orbit that brings it back to its starting point. Since total mechanical energy is conserved, the fact that the potential energy is the same before and after implies that the kinetic energy is also the same before and after. The change in kinetic energy after one orbit is therefore zero. From the work-energy theorem, then, we know that the work done on the star has to be zero as well. KAPLAN 7

8 MCAT In addition, the passage tells us that the gravitational force is conservative (which is something you might already know anyway). One of the properties of a conservative force is that the work it does is path-independent, which in turn implies that the work it does after one loop (or orbit) that starts and ends at the same place is zero. Since this force is the only one acting on the star, the work done on the star is zero. Beware of choice D, which states that the work done equals the gravitational force of the black hole on the companion star times the distance traveled. This is very close to the definition of work and looks like an obvious choice. However, what this formulation leaves out is the angle between the force and the direction of travel. The formula for work done by a force is: W = Fd cosθ. In this case the force is directed towards the center (or focus of an ellipse) and the direction of travel is tangent to the orbit. Therefore, for an elliptical orbit, the angle between the two is never zero (the cosine is never 1): direction of travel star gravitational force black hole The work done over any segment of the orbit, then, is never simply the product of the gravitational force and the distance traveled. The symmetry of the problem, in fact, leads to net cancellation of the work done after one complete orbit. Also, note that in the special case of a circular orbit, the two are always perpendicular, and thus the cosine of the angle, and the work done, are zero all throughout the orbit. 5. A As discussed in the explanation to #2, the kinetic and potential energies are both constants of the motion for a circular orbit because r is constant. We can thus take out the average signs <> and write the equation given in the passage as K = U/2. Since the passage also gives us an expression for the gravitational potential energy, we are tempted to substitute that in and get: E = K + U = U 2 + U = ( GmM 2r ) + ( GmM r ) = GmM 2r GmM r = GmM 2r 2GmM 2r = GmM 2r While this is certainly correct, this is not one of our answer choices: choice D does not have the right sign. We therefore must find another way of expressing the quantity by using the fact that K = mv 2 /2. Since K = U/2, U = 2K: E = K + U = K + ( 2K) = K 2K = K = 1 2 mv2 You may be surprised that the total energy is negative, but remember that we can define the zero of potential energy any way we want and in this case, as has been discussed, the potential energy is always negative. This is purely a matter of convention (and convenience); since the kinetic energy has a magnitude half of that of the potential energy, it is not sufficient to overwhelm the potential energy term and thus the total energy remains negative. 6. B From Newton s second law, the acceleration an object undergoes is equal to the force it experiences divided by its mass. In this particular case, M is the mass of the black hole while m is the mass of the star. The force each experiences is the same in magnitude from Newton s third law: The gravitational force the star exerts on the black hole is the same as the gravitational force the black hole exerts on the star. The magnitude of the force is GmM/r 2. The acceleration that the black hole undergoes is therefore this force divided by M, or Gm/r 2, while the acceleration the star undergoes is the same force divided by m, or GM/r 2. The ratio of the former to the latter is therefore Gm/r 2 :GM/r 2 or m:m. Note that we need not even have come up with the expression for the force explicitly: all we need is to be able to recognize that the two forces will have equal magnitude from Newton s third law. We can then write the ratio as: a black hole :a star = F M : F m = 1 M : 1 m = m:m where the last step can be made if one recognizes that a ratio is itself just a quotient. 8 as developed by

9 Work, Energy and Momentum Test 1 Independent Questions (Questions 7 12) 7. D To do this problem one needs to understand the relation between power, work and kinetic energy. Power is defined as the amount of work done per unit time. The time in this case is given as three seconds, but we need to determine the amount of work done. This can be obtained using the work-energy theorem, which states that the work done is equal to the change in kinetic energy: W = KE = ( 1 2 mv2 ) = 1 2 m (v2 ) = 1 2 m (v f 2 v i 2 ) where v f is the final velocity and v i is the initial velocity. (Note: (v 2 ) is not the same as ( v) 2! I.e. we cannot write the change in kinetic energy as 1 2 m(v f v i ) 2.) Substituting in numbers supplied by the question, we have the change in energy to be equal to kg (225 25) m2 /s 2 = 1000 J which is the work done. This divided by 3 seconds gives 333 J/s or 333 W. 8. A The energy lost to, or dissipated by, friction is the work done by friction, which is equal to the magnitude of the force times the distance traveled. Anything that increases either of these two factors will increase the energy lost to friction. We are thus looking for something that does not increase either of these two. Choice B can be immediately ruled out since it explicitly suggests increasing the distance traveled. To arrive at the correct answer, we can almost rely on intuition alone. Increasing the acceleration due to gravity (choice C) or the mass of the block (choice D) will increase the weight, i.e. the gravitational force on the block. It is therefore held more tightly to the plane, and so one would expect a higher friction force. Increasing the angle of inclination, on the other hand, decreases the component of gravity that is perpendicular to the plane, and so the block is not held so tightly to the plane. This would lead us to expect that the friction force would be less, and so less energy will be dissipated. For a fuller understanding of the scenario, we can derive an expression for the energy lost because of friction in terms of the quantities that appear in the answer choices. First, we notice that the force and the motion are in opposite directions and so: W = Fdcos180 = Fd = E due to friction The negative simply implies that the work done by the friction force causes the block to lose rather than gain energy. Since we are only interested in the magnitude of the energy lost, the negative sign can be ignored in our calculation. At this point it is helpful to refer to a diagram with which you should already be familiar: normal force = mgcosθ friction = µn = µmgcosθ mg θ θ If we let d be the distance traveled by the block as it slides down the inclined plane, then the energy dissipated by friction is µmgdcosθ. With this expression in hand, we can evaluate the answer choices. Choice A states that increasing the angle of inclination would not increase the amount of energy lost. The cosine of angle decreases if the angle increases within the range of 0 to 90 (cos 0 = 1; cos 90 = 0). Since the angle of inclination is within this range, increasing the angle of inclination would decrease the cosine term in the expression above, thus decreasing the amount of energy lost. It is therefore true that increasing the angle of inclination would not increase the amount of energy lost, making choice A the correct response. 9. A KAPLAN 9

10 MCAT The most important thing to realize in this problem is that energy is not conserved in the collision which is totally inelastic since the two objects stick together. What is conserved is momentum, with which one can determine the velocities of the bodies. Energy conservation, however, does come into play as potential energy is converted into kinetic energy up to the point right before the collision, and vice versa after the collision. We shall see exactly how all these come together. Initially we have a block of mass m at a height of h. The potential energy, which is also the total energy, is mgh. This will be entirely converted into kinetic energy of the block at the point immediately before collision, and we can thus solve for the velocity of the block before collision via energy conservation: mgh = 1 2 mv2 v = 2gh The momentum before the collision is thus p = mv = m 2gh. Momentum is conserved in the collision, so the momentum of the block-putty system after the collision will also be m 2gh. We can use this relation to determine the velocity of the system after collision, v : (2m)v = m 2gh where the mass on the left hand side is 2m since the putty also has a mass of m and so the total mass of the system is m + m = 2m after collision. Solving for v gives us: v = 2gh 2 Immediately after the collision, the block-putty system has no potential energy. The kinetic energy is thus the total energy, and this is equal to: 1 2 (2m) v 2 = 1 2gh (2m) 2 4 = mgh 2 Note that this is less than the initial energy of the block, mgh. As we mentioned, energy is not conserved in an inelastic collision. In this particular case, the energy has been cut in half. The rest has been dissipated as heat and/or sound as the collision occurs. This reduced total energy, which is entirely in the form of kinetic energy right after the collision, is converted to potential energy as the block-putty system moves up the track. At maximum height h, we can write the equality: (2m)gh = total energy right after collision = 1 2 (2m) v 2 = mgh 2 2mgh = mgh 2 h = h 4 Among the other answer choices, choice B, h/2, is what one would erroneously obtain if one assumes that energy is conserved. 10. C This question requires you to remember the definition of impulse, J: J = p = F t where p is the change in momentum, F is the force and t the time over which the force acts. You should recognize the last equality as simply Newton s second law rearranged: F = ma = m v t (from the definition of acceleration) 10 as developed by

11 Work, Energy and Momentum Test 1 Multiplying both sides by t, we get: F t = m v = (mv) (assuming mass is constant) = p The larger the force, or the longer the time over which the force acts, the more the velocity (and hence momentum) is changed. In this problem, the time that the baseball and the bat are in contact is doubled, while the force remains constant. The impulse is therefore doubled. 11. B Since the efficiency of the solar panel is only 10%, only 10% of the incident solar radiation is converted to useful energy. The output power of the solar panel is therefore 40 kw 0.1 = 4 kw = 4000 W = 4000 J/s. Power is energy per unit time, and from the calculation above, we know that the solar panel delivers 4000 J of energy every second. In 300 minutes, then, the energy generated is: 4000 J 60 s 300 min s 1 min = 4000 J s s = ((4 103 ) ( )) J = J = 72 MJ where 1 MJ = 10 6 J. One could also have gotten the answer by first calculating the energy delivered in 300 minutes and then multiplying by the efficiency factor. 12. B Let the masses of ball 1 and ball 2 be 2m and m respectively. The momentum of ball 1 before the collision is (2m)v = 2mv. Since ball 2 is at rest, it has no momentum and so the momentum of ball 1 is also the total momentum of the system. Regardless of the nature of the collision, momentum is conserved (kinetic energy may or may not be conserved). The total momentum of the system after the collision would therefore also be 2mv. Ball 1 moves with a speed of v/3 after the collision; its momentum is therefore (2m)(v/3) = (2/3) mv. Since it s moving in the same direction as before, its momentum vector points in the same direction. Ball 2 must then either move in the same or the opposite direction; otherwise the vector for the total momentum will have acquired a component along a new coordinate. The velocity of ball 2 after the collision, v, must satisfy the equation: mv mv = 2mv v = 4 3 v At this point we could eliminate the possibility that the collision is completely inelastic. If that were the case, the two balls would have moved as one, i.e. with the same velocity, after the collision. Choice C is therefore incorrect. Since we have the velocities both before and after the collision for both balls, we could determine whether the collision is elastic or inelastic by computing the kinetic energies; choice D is also incorrect. The kinetic energy before the collision is: The kinetic energy after the collision is: 1 2 (2m) v m(0)2 = mv (2m) (v 3 ) (m) (4v 2 3 ) 2 = mv mv2 9 = mv 2 Kinetic energy is therefore conserved during the collision and thus the collision is elastic. KAPLAN 11

12 MCAT Passage II (Questions 13 17) 13. A The work done by a force is the product of the distance and that component of the force that is in the same direction: that is what the cosθ term means in the formula W = Fdcosθ. If the two are perpendicular, the angle between them is 90, and the cosine term is zero. The work done will be zero in such a case. In the particle storage ring, the particles are moving in a circle. Their instantaneous velocity, and therefore, their direction of motion at any time, is tangent to the circle. The direction of the magnetic force, on the other hand, is toward the center of the ring and is acting as a centripetal force. The force and the direction of motion are perpendicular, and thus the work done by the magnetic force is zero. 14. D The first and most natural step in approaching this problem is to identify the factors that affect the reaction rate from the information given in the passage and determine how they affect the rate. The formula for the rate is given in the passage: rate = luminosity cross section. Since the cross section is fixed for a given reaction, the only way one can control the rate is by changing the luminosity. (Do not confuse cross section with cross-sectional area!) The expression for luminosity tells us that it is increased by increasing the number either of electrons or of positrons, by increasing the frequency of revolution, and by decreasing the cross-sectional area of the storage ring. Since luminosity is directly proportional to the reaction rate, any of the above that increases luminosity would also increase the reaction rate. Statements I and III are exactly two ways we have identified to increase the rate. Statement II, increasing the energy of the particles, would increase the velocity of the particles. They would thus move faster, completing more revolutions per second in the storage ring. This is also one of the ways we have identified to increase the rate. Thus all three statements are correct. Note also that once we have established statements I and III as being correct, choice D has to be the correct response. 15. C The mass of the positron or electron is on the order of kg. Multiplying this by the acceleration due to gravity results in a number that is still negligibly small compared to the magnitude of the electric interactions. Choice A is incorrect because gravity points in the same direction as the acceleration due to gravity. This is a consequence of Newton s second law F = ma, where F and a are vectors and m is a scalar. In order for this vector equality to hold, F and a must point in the same direction. (One must, in general, of course be aware of the distinction between force and net force.) Choice B states that the energy lost due to gravity is equal to that gained from the magnetic fields. The passage tells us that magnets are placed at various points in the storage ring. Their purpose, however, is to keep the particles moving in a circular path. Since the magnetic force is perpendicular to the particle s velocity, the magnetic force does no work, and thus causes no energy gain or loss. Choice D states that the luminosity is not dependent on the gravitational acceleration. This is a true statement, but is not relevant to the issue we are trying to address here. 16. D This is a conservation of momentum question in disguise. The energy value given is extraneous information. In the question stem, we are told that the two electrons with equal speed collide head-on. There is really a lot of information that can be garnered from this statement. Since both particles are electrons, they have the same mass. Furthermore, since they collide with equal speed head-on, we know that their velocities are equal and opposite. So the total initial momentum of the system is zero: The initial momentum of one electron is a vector that is equal and opposite to that of the other electron. Momentum conservation tells us that the total momentum after the collision must in this case also be equal to zero. After the collision, therefore, the two electrons must again move with equal speed in opposite directions. The second electron will thus also have a speed of 0.9 c. 17. C Again, the energy value of 180 GeV is irrelevant here: the only important point is that the total energy needs to be kept constant. In order to maintain this, the power we need to supply has to be the same as the power lost. The question tells us that 260 MeV of energy is lost per revolution by each particle, and that the frequency of revolution is Hz. In one second, then, the number of revolutions made is 10000, and the amount of energy lost in one second by one particle is MeV = MeV = MeV. The total amount of energy lost in one second is this value multiplied by two, since there are two particles: MeV. This is the amount of energy lost per second, and so energy needs to be supplied at a rate of MeV per second for it to remain constant. 12 as developed by