PHYSICS TOPICAL: Translational Motion 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 Translational Motion Test Passage I (Questions 1–5) It is a fundamental fact of life that ants cannot be as large as elephants, and whales cannot be as small as goldfish For every type of animal there is a most convenient size Each animal has a characteristic length L From this length one can define a characteristic area which is proportional to L2, and a characteristic volume which is proportional to L3 The characteristic mass is also taken to be proportional to L3 The maximum size of an animal is constrained by the material out of which it is made For example, consider whether a giant human is a viable creature If we rescaled a human by a factor of 10, we would necessarily increase the radius of the legs by a factor of 10 However, the mass of the human would increase by a factor of L , or 1000 Therefore, legs with 100 times the cross-sectional area would have to support 1000 times the weight Size also explains why an ant that falls off of the roof of a four story building will not be seriously injured, whereas a human most certainly would be injured A small animal can survive such a fall because of its relatively large surface area to volume ratio An object in free-fall within the Earth’s atmosphere will only accelerate up to a certain velocity because of friction caused by the air resistance The frictional force can often be approximated as being proportional to the velocity times the cross-sectional area of the falling object The maximum velocity that an object reaches in free-fall is called the terminal velocity Which of the following statements must be assumed if the characteristic mass of an animal is taken to be proportional to L3? I Every animal can be assigned a characteristic length L II The characteristic volume of the animal is proportional to L3 III The density of the animal is constant throughout that particular animal A B C D Suppose that the frictional force of air resistance is only proportional to an object’s cross-sectional area and not its velocity A falling object would then: [Note: Assume that the frictional force is much smaller than the weight of the object.] A take more time to reach terminal velocity B take less time to reach terminal velocity C reach terminal velocity in the same amount of time D never reach terminal velocity The terminal velocity of a human in the horizontal spread eagle position would be: A smaller than if he/she is rolled up in a ball because the cross-sectional area is greater B greater than if he/she is rolled up in a ball because the cross-sectional area is also greater C the same as if he/she is rolled up in a ball because the surface area to volume ratio is constant D the same as if he/she is rolled up in a ball because the mass is the same in each case The characteristic length of an ant is 0.5 cm, while that of a human is m If q is the surface area divided by the volume, what is the ratio of q-ant to q-human? A 200:1 B 400:1 C 2000:1 D 1:20 Cells can only absorb nutrients through their surfaces In order for a cell to absorb nutrients most efficiently, it will grow to a maximum size and then divide If the total volume remains constant when a cell divides, then the ratio of the total surface area of the daughter cells to that of the original cell is: A B C D 22:1 2:1 21/3:1 1:22 I only I and III only II and III only I, II, and III GO ON TO THE NEXT PAGE KAPLAN MCAT Passage II (Questions 6–11) vy (m/s) 28 40 kg 21 60 kg 17 14 kg 80 kg kg 10 20 A circus wishes to develop a new clown act Figure shows a diagram of the proposed setup A clown will be shot out of a cannon with velocity v at a trajectory that makes an angle θ = 45° with the ground At this angle, the clown will travel a maximum horizontal distance The cannon will accelerate the clown by applying a constant force of 10,000 N over a very short time of 0.24 sec The height above the ground at which the clown begins his trajectory is 10 m 42 1.4 1.7 2.1 2.8 4.2 t (s) Figure If the angle the cannon makes with the horizontal is increased from 45°, the hoop will have to be: 45 A B C D moved farther away from the cannon and lowered moved farther away from the cannon and raised moved closer to the cannon and lowered moved closer to the cannon and raised Figure A large hoop is to be suspended from the ceiling by a massless cable at just the right place so that the clown will be able to dive through it when he reaches a maximum height above the ground After passing through the hoop he will then continue on his trajectory until arriving at the safety net Figure shows a graph of the vertical component of the clown’s velocity as a function of time between the cannon and the hoop Since the velocity depends on the mass of the particular clown performing the act, the graph shows data for several different masses [Note: The impulse J, or change in momentum p generated by the force F is given by J = F∆t = p = mv , where m is the mass of the clown and t is the time over which the force acts on the clown The acceleration of gravity is g = 10 m/s2 sin 45° = cos 45° = ] If the clown’s mass is 80 kg, what initial velocity v will he have as he leaves the cannon? A B 15 C 30 D 300 m/s m/s m/s m/s The slope of the line segments plotted in Figure is a constant Which one of the following physical quantities does this slope represent? A B C D –g v0 y–y0 sin θ GO ON TO THE NEXT PAGE as developed by Translational Motion Test From Figure 2, approximately how much time will it take for a clown with a mass of 100 kg to reach the safety net located 10 m below the height of the cannon? A B C D 1.7 3.4 3.9 4.4 s s s s If the mass of a clown doubles, his initial kinetic energy, mv02/2, will: A B C D remain the same be reduced in half double quadruple 1 If a clown holds on to the hoop instead of passing through it, what is the expression for the minimum length of the cable so that he doesn’t hit his head on the ceiling as he swings upward? A v ∆t M v02 g v02 C 2g v02 D 4g B GO ON TO THE NEXT PAGE KAPLAN MCAT Questions 12 through 16 are NOT based on a descriptive passage Which of the following is the LEAST amount of information that could be used to determine how far a cannonball lands from a cannon on Mars? A Initial speed, angle of inclination, acceleration due to gravity B Initial speed, time of travel C Initial speed, acceleration due to gravity D Initial speed, angle of inclination A boat can move with a speed of 20 m/s in still water It starts at the shore of a river that flows at a rate of 10 m/s In what direction must the boat move to reach the exact same point on the opposite shore? [Note: sin 30° = 0.5, cos 30° = 0.866, sin 60° = 0.866, cos 60° = 0.5] A The boat must be pointed at an angle of 90° to the shore B The boat must be pointed upstream at an angle of 60° to the shore C The boat must be pointed upstream at an angle of 30° to the shore D The boat will not be able to make it directly across the river because the current will always push it past that point Two blocks, whose masses are m and m , have the same coefficient of sliding friction Starting from rest, they slide down two planes of equal length inclined at angles θ1 and θ2 respectively Given that m l is greater than m and θ is greater than θ , which of the following statements is consistent with the information? A Both blocks slide down with the same speed because they both have the same coefficient of sliding friction B m slides down faster than m because it experiences a greater gravitational force C m slides down faster because of the larger angle of inclination D m slides down faster because it experiences less friction Two balls of equal mass are shot upward simultaneously from the same point on the ground with the same initial speed, but at different angles to the horizontal Which of the following statements must be true? A The ball launched at the larger angle hits the ground first B The ball launched at the smaller angle hits the ground first C The two balls hit the ground at the same time D The ball launched at the larger angle always has more total mechanical energy Which statement is true of a projectile at the highest vertical point in its path? A The projectile has its greatest kinetic energy B The projectile has its greatest potential energy C The vertical component of the projectile’s speed is greater at this point than at any other point D The horizontal component of the projectile speed is greater at this point than at any other point END OF TEST as developed by Translational Motion Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: D D D C A A A C C 10 B 11 12 13 14 15 D A B B C 16 B as developed by Translational Motion Test EXPLANATIONS Passage I (Questions 1–5) D This question asks you to identify the underlying assumptions made in drawing certain conclusions from statements presented (This kind of reasoning is also required in the verbal reasoning section.) Let us examine each statement in turn Statement I says that every animal can be assigned a characteristic length L This is certainly a necessary assumption if we were to define the characteristic mass to be proportional to L3 This is, in fact, one of the main points of the first paragraph of the passage Statement II says that the characteristic volume of the animal is proportional to L3 The passage does make this claim Whether this statement is a necessary assumption depends on the relationship between mass and volume The concept of density, then, will play a role Since statement III explicitly addresses this concept, we can consider these two statements together The density ρ of an object is defined as the mass per unit volume, and one often rearranges this definition to obtain the formula M = ρV where M is the mass and V the volume If, however, the object does not have a constant density throughout, the total mass is found by adding the individual masses of smaller regions in which the density is constant: Mtot = M1 + M2 + M3 + M4 + … = ρ 1V + ρ 2V + ρ 3V + ρ 4V + … where 1, 2, etc., are labels for the regions of the object within which the density is constant In this case then, as one can see, the mass of the object is not simply proportional to its volume, i.e.: Mtot kV for some constant k (V = V1 + V2 + V3 + V4 + …) Going back to the statements, then, only if the density is constant and if the volume is proportional to L can we express the mass of an animal to be: M = ρV = ρ(kL3) or M ∝ L3 where k is the proportionality constant between the volume and L3 In sum, then, all three statements are necessary to make the claim D An object that has reached terminal velocity is falling at a constant velocity, i.e it is no longer accelerating or picking up speed From Newton’s second law, we know that this would have to imply that the net force acting on the object is zero In particular, the gravitational force on the object is exactly balanced by the frictional force of air resistance If the frictional force is only proportional to the cross-sectional area and not the velocity of the falling object, the magnitude of the force would not change Since the question stem states that this force is much smaller than the weight of the object (i.e the gravitational force on it), it will never be able to balance the downward force: There will always be a net downward force acting on it that causes it to accelerate downward throughout its entire fall The object therefore never reaches a constant terminal velocity A The friction force is proportional to the velocity and the cross-sectional area, as stated towards the end of the passage So when terminal velocity is reached, the magnitudes of the two forces are related by: kvtermA = mg where k is the proportionality constant in the friction force, A is the area, and the velocity is indicated as terminal since only at that point are the two forces equal (see #2 above) Rearranging this, one obtains: vterm = mg kA A human in the horizontal spread eagle position presents a greater cross-sectional area to the air than if he is rolled up into a ball Therefore the terminal velocity would hence be smaller in the spread eagle case A KAPLAN MCAT The surface area is proportional to L2, while the volume is proportional to L3 The parameter q is defined as the surface area divided by the volume, i.e.: q= kL2 k = ( ) k’L3 k’ L where k and k’ are the respective proportionality constants The ratio q-ant to q-human is therefore: q-ant : q-human = k k 1 1 ( ) : ( ) = : = : = : 0.01 = 200 : k’ Lant k’ Lhuman Lant Lhuman 0.5 100 Note that we need to convert the two length dimensions into the same units to calculate the ratio As a side note, you should also recognize the equivalence between a ratio and a quotient: Lhuman 100 1 200 : = = = 200 = = 200 : Lant Lhuman Lant 0.5 C Let L be the characteristic length of the original cell It will have a surface area proportional to L , and volume proportional to L3 Since the total volume remains constant, the daughter cells, each taken to have characteristic length λ, will have a volume λ3 that satisfies: L3 = 2λ3 where we have dropped the proportionality constant The factor of two appears because the original cell gives rise to two daughter cells To find the relationship between the surface areas, we take the cube root of each side and square: L2 = (22/3)λ2 and this is the surface area of the original cell, L , in terms of λ λ is the surface area of one daughter cell, and so the total surface area of the daughter cells is 2λ2 The surface area ratio is therefore: 2λ : L = 2λ : (22/3)λ = 2:22/3 Again, recalling the equivalence between a ratio and a quotient, we can simplify this to: 2:22/3 = 2 –2/3 1/3 = 2/3 × –2/3 = = 21/3:1 2/3 2 Notice that 21/3 is greater than one, and so the total surface area has increased upon division Passage II (Questions 6–11) D This question can be answered using the equations of projectile motion, but since the answer choices are not quantitative, we should use a more intuitive and time-saving approach For the horizontal component, all one has to is read the first paragraph of the passage which states that when the cannon is at an angle of 45°, the clown will travel the maximum horizontal distance If the angle deviates from 45°, then, the horizontal distance traveled would decrease The hoop is situated (roughly) halfway between the cannon and the landing point (roughly because the two may be at different heights so it is not exactly symmetric) A smaller horizontal distance, then, would mean that the hoop needs to be closer to the cannon For the vertical component, if the angle is increased from 45°, the vertical component of the initial velocity of the clown would increase The kinetic energy associated with this initial vertical motion would therefore be greater The hoop is situated at the point of maximum height, i.e when all the vertical kinetic energy is transformed into gravitational potential energy The higher its initial vertical kinetic energy, the greater the gravitational potential energy it can ultimately gain, and the higher the hoop will have to be placed: 10 as developed by Translational Motion Test 1 mvy02 = mghmax vy02 hmax = 2g C This question calls for a straightforward application of the equation for the impulse: Ft = mv0 The force is given in the passage as 10000 N and t as 0.24 s, while m is 80 kg from the question Solving for v0: v0 = Ft 10000•0.24 2400 = = = 30 m/s m 80 80 Alternatively, for a mass of 80 kg, Figure shows that the intitial velocity in the y direction is 21 m/s This is related to the magnitude of initial velocity by v0y = v0 sin 45° 21 = 21 Without actually performing any calculations, we should be able sin45° to decide that choice C, 30 m/s, is the only reasonable answer The initial velocity is therefore v0 = A The easiest way to this problem is through dimensional analysis: Since all the choices have different units, if we know what units the slope is in, we can just select the one that matches The graph in Figure is a plot of velocity versus time Slope is in general the change in y over the change in x, and therefore in this case is the change in velocity over the m/s change in time, which will have units of = m/s These are the dimensions of acceleration, and, looking at the answer s choices, the slope must therefore be equal to the acceleration due to gravity, g, or choice A Choice B is the initial velocity and has units of m/s; choice C is vertical displacement which just has the units of distance or m; choice D is dimensionless Alternatively, knowing that the slope is a constant, we must find the kinematics equation that has the form y = mx + b, where y is the velocity in the y direction, and x is time The equation we are looking for is v = v0 + at or, more specifically in this case, v y = v0y – gt where m, the slope, is –g C This question involves both interpretation of the graph in Figure and general knowledge of projectile motion The graph enables us to determine the time the clown takes to reach the hoop According to the graph, a 100-kg clown will exit the cannon with an initial y-velocity of 17 m/s (One can also obtain this value indirectly by finding the velocity using the formula for impulse as in #7 and then calculating the y-component from trigonometry.) One can then find the time t taken for the trip by the equation: y = vy t + KAPLAN at2 11 MCAT In substituting values, we need to be careful in making sure that our signs are consistent Defining the upward direction to be positive, y would be the net vertical displacement which is –10 m, v y0 is 17 m/s, and a, the acceleration, is –10 m/s2 due to gravity: –10 = 17t – 5t2 Solving for t, however, means solving a quadratic equation that does not factor neatly One can try the different answer choices to see which would satisfy this equation, but this would still be time-consuming and it is easy to make numerical mistakes Let us also consider another approach that is more instructive in elucidating the kinetics behind the problem When the y-velocity falls to zero, the clown has reached the hoop The time at that point (i.e the x-intercept in the graph), then, is the time it takes for the clown to reach the maximum height from the cannon: in this case it takes 1.7 s On his trip down, the clown will take this same amount of time to reach the same level as the cannon; i.e he will have taken a total of 3.4 s to complete the projectile trip from the cannon back to the same level That, however, is not enough Additional time is needed for the clown to fall from the cannon level to the safety net 10 m further down (At this point, you should be able to eliminate choices A and B If you are pressed for time, therefore, a guess between C and D will at least maximize your chances of getting the right answer.) A diagram should be helpful: time = 3.4 s vy = initial vy = 17 m/s time = 1.7 s vy = Point B 10 m How much time is needed for this last part of the trip, from point B to the safety net? One can avoid working with cumbersome kinetics equations by examining the two remaining answer choices more closely First of all, we should realize that choice C essentially says that it takes 0.5 s (= 3.9 – 3.4), while choice D says that it takes s (= 4.4 – 3.4) Which is more reasonable? At point B, the clown already has a vertical velocity of 17 m/s It would thus take slightly more than half a distance second to travel 10 m (time = ) Besides, 17 m/s is only an “initial” value; the clown is also accelerating from gravity speed Therefore choice C is the only reasonable answer 10 B This is a trick question in that if one merely looks at the formula for the kinetic energy, one might conclude that if mass doubles the initial kinetic energy will double as well This, however, is true only if the initial velocity stays the same But the first paragraph in the passage indicates that it is the impulse, not the initial velocity, that stays constant If the mass doubles, then, the initial velocity is also going to change The equation given in the passage is: Ft = mv0 v0 = Ft m The numerator is the impulse that is constant (= 10000 N × 0.24 s) If the mass is doubled, then, the initial velocity would be halved Using a prime (’) to designate the new quantities, then, we have: m’ = 2m v0 v0’ = The new initial kinetic energy would therefore be: 12 as developed by Translational Motion Test v0 1 1 1 m’v0’2 = 2m ( )2 = mv02 = ( mv02) = KEold 2 2 11 D This question is most easily answered using the principle of conservation of energy Since the clown reaches the hoop at the maximum height of his trajectory, the vertical component of his velocity is zero There are no forces acting on him in the v0 horizontal direction, and so his horizontal velocity is the same as its initial value, that is, v0cos 45° or The clown’s v0 1 kinetic energy as he reaches the hoop is therefore m ( ) = mv After grabbing the hoop, the clown will swing upward until all of this kinetic energy is converted into potential energy, mgy, where y is the vertical distance above the relaxed position of the hoop that the clown swings upward This is also the minimum length of the cable: y = minimum length of cable The condition that y needs to satisfy is therefore: mv v02 y= 4g mgy = Independent Questions (Questions 12–16) 12 A All of the answer choices include the initial speed What else is needed to determine how far a cannonball lands from a cannon? Remember that speed is a scalar: it has a magnitude but no direction In order to resolve the cannonball’s motion into vertical and horizontal components (which we need to do), we need the direction of travel—or in terms of the choices in this question—the angle of inclination In addition, we need to know how large the force is that is pulling the cannonball down: The acceleration due to gravity on Mars, which will differ from the familiar 9.8 m/s2, will affect the amount of time the cannon spends in the air and thus the horizontal distance traveled Let us also consider a slightly more quantitative approach that would also reveal to us why the other combinations are not sufficient There is no net force in the horizontal direction, and so the horizontal velocity is constant until the ball hits the ground The horizontal distance traveled is then simply vxt, where vx is the horizontal component of the initial velocity, i.e vx = v0cosθ, where v0 is the initial speed and θ is the angle of inclination t is the time the ball stays in the air, and is dictated by the kinetics in the vertical component Specifically, it can be found using the kinetics equation: y = v0 y t + at2 where in this particular case y will be zero (assuming that the cannonball is launched and hits the ground at the same level) and care has to be taken to make sure that v0y and a have opposite signs since they point in opposite directions v 0y is v0sinθ, and thus no additional information is needed to determine this that was not already required to find vx (i.e v and θ) The only other piece of information, then, is the acceleration due to gravity Let us now also consider why the other answer choices are not correct Choice B gives us the time of travel, and thus it would save us the calculation to find t However, initial speed by itself is not enough to let us find vx, for which we also need the angle of inclination: what we need is the initial speed in the KAPLAN 13 MCAT horizontal direction, which we cannot determine with the quantities in choice B (Incidentally, we can use the kinetics equation above to solve for the angle of inclination if we know the initial speed, which we do, the time, which we do, and the acceleration due to gravity, which we don’t In other words, no matter how one chooses to look at it, choice B is insufficient.) Choice C is likewise lacking in the angle of inclination or time of travel Choice D does not enable us to solve for the time spent traveling without the value of the acceleration due to gravity on Mars 13 B This is essentially a vector addition problem Since the current would carry the boat downstream, the boat needs to aim upstream if it were to “cancel” the effect of the downstream current and arrive at the same spot on the opposite shore We can thus eliminate choice A immediately The two velocity vectors, the one of the downstream current and the one of the boat in still water, need to add to give a resultant vector that points directly to the opposite shore (i.e at an angle of 90° to the shore) current: 10 m/s resultant θ boat: 20 m/s The question, in other words, is: What should θ be so that the resultant vector would point directly towards the opposite shore? (The angle with the shore that the boat should make would then be 90° – θ.) The sine of θ is 10/20 or 0.5 In the question stem we are told that sin 30° = 0.5, and so θ = 30°, and the boat should make an angle of 60° with the shore Since the boat has a speed of 20 m/s that is stronger than the current (10 m/s), it is able to combat the current and get to the opposite shore, making choice D incorrect in this case If the current is, say, 25 m/s instead, then the boat would not have been able to make it to the opposite shore at the same point 14 B The potential energy of a projectile is mgh, and clearly increases as the height increases At the highest vertical point in its path, then, its potential energy must be at a maximum As has been discussed in earlier problems in this topical test, the kinetic energy associated with the projectile’s motion in the vertical direction is completely transformed into gravitational potential energy at its highest point, and so choice A is incorrect At the highest point, the projectile is turning around, and so its instantaneous vertical velocity is zero The horizontal component, however, remains constant throughout its flight (assuming no air resistance) and does not change until impact with the ground causes it to stop 15 C We need to derive an equation for the acceleration of a black down the inclined plane A diagram would be most helpful: friction = µN = µmgcosθ mgsinθ θ mg θ The component of the weight that is parallel to the plane is the force that pulls the block down As can be seen from the diagram, this force has a magnitude of mgsinθ, where m is the mass of the block The friction force acts in the opposite direction of motion and has a magnitude given by the general formula F = µN, where µ is the coefficient of friction and N the normal force If the plane is level, the normal force would simply be the weight of the object, but in the case of the inclined 14 as developed by Translational Motion Test plane, it is the component of the weight that is perpendicular to (or normal to, hence the name) the plane: mgcosθ The friction force is therefore µmgcosθ The net force on the object in the direction parallel to the plane is therefore: F = mgsinθ – µmgcosθ This force will provide the acceleration down the inclined plane: mgsinθ – µmgcosθ = ma gsinθ – µgcosθ = a Notice that the masses on each side cancel So, in the scenario given by the question, the fact that m is greater than m is irrelevant They both start from rest, and so whichever experiences a larger acceleration will slide down faster, and the acceleration depends only on the angle of inclination, since the coefficient of friction is the same in both cases The question, then, is, does increasing the angle increase or decrease acceleration? The value of sine is increasing on going from 0° to 90° (from to 1), while the value of cosine is decreasing on going from 0° to 90° (from to 0) Presumably θ1 and θ2 will both be in between the limits of 0° and 90°, and so the fact that θ1 is greater than θ2 means that sinθ1>sinθ2, and that cosθ1