4Force, motion, gravitation, and equilibrium test w solutions

PHYSICS TOPICAL: Force, Motion, Gravitation and Equilibrium Test Time: 21 Minutes* Number of Questions: 16 * 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 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 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 Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Force, Motion, Gravitation and Equilibrium Test Passage I (Questions 1–5) An amusement park wishes to build a roller coaster with a single loop as shown in Figure 1 The amount of energy lost between point A and point B is what percentage of the cart’s initial kinetic energy? A 0% B 25% C 50% D 100% A C Examination of Figure reveals that the normal force on the cart reaches a maximum value of 6Mg If we measure the cart’s position in the loop by the angle ϕ , shown in Figure 1, at what angle does this maximum normal force occur? h r ϕ B D A B C D Figure For the test run, an empty cart having mass M is released from point A with an initial velocity v a The cart then increases in speed at a non-constant acceleration until it reaches a velocity of v b at point B At this point, it is found to have kinetic energy equal to Mgh, where g is the acceleration due to gravity and h is the initial height at point A The loop portion of the track closely approximates a perfectly symmetric circle of radius r, and the distance from point B to point C equals the distance from point C to point D When the cart reaches point C, it is found to have a velocity of v c , which is the minimum velocity required to keep gravity from pulling it off the track It then accelerates down the back side of the loop to point D, where it reaches its final velocity, vd Figure shows a graph of the magnitude of the normal force N that the track exerts on the cart as a function of distance s between points B and D In the graph, the energy lost due to friction has not been included If the effects of friction were included in Figure 2, the normal force on the cart in the loop would: A B C D remain the same everywhere in the loop increase everywhere in the loop decrease everywhere in the loop increase in some places and decrease in other places According to Figure 2, what is the correct value for the cart’s kinetic energy at point C? A B C D Mgr/2 Mgr 3Mgr/2 5Mgr/2 When passengers are riding in the cart, its mass approximately doubles The minimum velocity v c at point C will then: Mg A B C D N Mg ϕ = 0° ϕ = 30° ϕ = 180° ϕ = 210° B C be cut in half remain the same be doubled be quadrupled D s Figure GO ON TO THE NEXT PAGE KAPLAN MCAT where G is the universal gravitational constant, m is the mass of the object, M e is the mass of the Earth, and R is the distance of the object from the center of the Earth One result of this analysis is that Kepler’s laws can be derived for satellites orbiting the Earth These laws were originally discovered by Johannes Kepler (1571–1630) based on observations of the motion of the planets about the Sun They are as follows: I All satellites move in elliptical orbits with the Earth at one focus II A line that connects a satellite to the center of the Earth sweeps out equal areas in equal times A double the time that it takes to move through θ B half the time that it takes to move through θ C the square of the time that it takes to move through θ D equal to the time that it takes to move through θ Which of the following graphs represents how the magnitude of acceleration of a falling object changes with the distance through which it has fallen, if no approximation is made with regard to the distance? A III The square of the period of any satellite is proportional to the cube of the semimajor axis of its orbit height fallen height fallen B D acceleration For objects near the surface of the Earth, a great simplification in the mathematics occurs by approximating the gravitational force to be a constant Let R = R e + h, where Re is the radius of the Earth, and h is the distance of the object from the surface of the Earth The gravitational force can now be written as: C acceleration F = GmM e/R2 F = GmMe/(Re + h)2, acceleration The general motion of a massive object under the influence of the Earth’s gravitational force can be solved in closed form using Newton’s second law The gravitational force is given by: Consider a satellite in circular orbit around the Earth at a distance R from the Earth’s center The area that the satellite sweeps out as it moves through an angle θ is given by A = R2θ/2 Kepler’s second law implies that the time that it takes for the satellite to move through 2θ is: acceleration Passage II (Questions 6–11) height fallen height fallen which is approximately equal to GmMe/Re2 when h