2010BAP 2 Solutions

2010BAP 2 Solutions - AP® Physics B 2010 Scoring...

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Unformatted text preview: AP® Physics B 2010 Scoring Guidelines Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the College Board is composed of more than 5,700 schools, colleges, universities and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,800 colleges through major programs and services in college readiness, college admission, guidance, assessment, financial aid and enrollment. Among its widely recognized programs are the SAT®, the PSAT/NMSQT®, the Advanced Placement Program® (AP®), SpringBoard® and ACCUPLACER®. The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities and concerns. © 2010 The College Board. College Board, ACCUPLACER, Advanced Placement Program, AP, AP Central, SAT, SpringBoard and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service is a trademark owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) General Notes 1. The solutions contain the most common method of solving the free-response questions and the allocation of points for the solution. Some also contain a common alternate solution. Other methods of solution also receive appropriate credit for correct work. 2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded. One exception to this may be cases when the numerical answer to a later part should be easily recognized as wrong — for example, a speed faster than the speed of light in vacuum. 3. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a particular concept is worth 1 point and a student’s solution contains the application of that equation to the problem but the student does not write the basic equation, the point is still awarded. However, when students are asked to derive an expression it is normally expected that they will begin by writing one or more fundamental equations, such as those given on the AP Physics Exams equation sheet. For a description of the use of such terms as “derive” and “calculate” on the exams and what is expected for each, see “The FreeResponse Sections ⎯ Student Presentation” in the AP Physics Course Description. 4. The scoring guidelines typically show numerical results using the value g = 9.8 m s 2 , but use of 10 m s 2 is of course also acceptable. Solutions usually show numerical answers using both values when they are significantly different. 5. Strict rules regarding significant digits are usually not applied to numerical answers. However, in some cases answers containing too many digits may be penalized. In general, two to four significant digits are acceptable. Numerical answers that differ from the published answer due to differences in rounding throughout the question typically receive full credit. Exceptions to these guidelines usually occur when rounding makes a difference in obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that should have five significant figures and that differ starting with the fourth digit (e.g., 20.295 and 20.278). Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and some credit may be lost. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 1 10 points total (a) 2 points For a correct conservation of energy equation for this situation 1 mghi = mgh f + mu 2 f 2 1 point Distribution of points u f = 2 g hi - h f u f = 2 9.8 m s2 ( 2.0 m - 0.50 m ) ( ( ) ) For the correct answer 1 point 2 u f = 5.4 m s (or 5.5 using g = 10 m s ) (b) 3 points For correctly drawing and appropriately labeling the weight of the block For correctly drawing and appropriately labeling the normal force For no extraneous forces 1 point 1 point 1 point (c) 2 points At the top of the track, the net force on the block is the centripetal force ma = mu 2 r = mg + N The condition for minimum speed is that the normal force is zero. For a correct equation that can be solved for the minimum speed 2 mumin r = mg umin = rg umin = 1 point (0.60 m ) (9.8 m s2 ) 1 point For the correct answer umin = 2.4 m s © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 1 (continued) Distribution of points (d) 3 points For a correct conservation of energy equation for this situation 12 mghmin = mg ( 2r ) + mumin 2 2 hmin = 2r + umin 2 g 1 point ( ) For correctly substituting the value of umin from part (c) hmin = 2 ( 0.60 m ) + ((2.4 m s) 2 (9.8 m s )) 2 2 1 point For the correct answer hmin = 1.5 m 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 2 15 points total (a) 3 points To determine the frequency, all that is needed is a stopwatch to measure the period. For choosing only the stopwatch OR choosing the stopwatch and other equipment, using the stopwatch correctly in part (b), and indicating there a plausible use of the other equipment that does not interfere with correctly using the stopwatch (e.g., using the protractor to set the initial angle) Partial credit was awarded if part (b) did not use the stopwatch or, in addition to the stopwatch, used other equipment incorrectly. For choosing one item in addition to the stopwatch, 2 points were awarded; for choosing two additional items, 1 point was awarded. Distribution of points 3 points (b) 3 points For a reasonable and complete procedure that correctly measures time and allows determination of the frequency, either by calculating the period and indicating the correct relationship between the period and frequency, or by directly determining the frequency from the measurements. Some mention of error reduction (e.g., measuring over multiple cycles) was expected as part of a complete experimental procedure. 3 points (c) 3 points For indicating that one of the following is a parameter that can be varied: mass of bob, length of string, angle of release or height of release For a reasonable and complete experimental procedure and description of data analysis 1 point 2 points (d) 3 points For correctly indicating that the temperature of the room would slightly increase For stating that the pendulum loses kinetic energy For stating that the lost kinetic energy is converted to heat energy 1 point 1 point 1 point (e) 3 points For correctly indicating that the period of the pendulum would increase For indicating that the length of the rod increases For using the relationship between pendulum length and period, T = 2 p that the increase in length leads to an increase in period 1 point 1 point 1 point g , to show © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 3 15 points total (a) 2 points Applying Newton’s second law to one of the objects Distribution of points ma = T - kq2 r 2 = 0 For a correct expression for the tension T T = kq2 r 2 T = 9.0 ¥ 10 9 N im 2 C 2 -4.0 ¥ 10 -9 C 1 point ( )( ) 2 (0.020 m )2 1 point For the correct answer, with units T = 3.6 ¥ 10 -4 N (b) 4 points For a full representation of the field in the vicinity of the objects For field lines that begin or end on the objects For having curved or bent field lines in the region between the objects that indicate an asymptotic approach to a vertical line midway between the objects For showing the direction of the field as inward (toward the objects) 1 point 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 3 (continued) Distribution of points (c) 3 points The acceleration is caused by the electrostatic force, and initially that force has the same magnitude calculated in part (a). ma = kq2 r = T For a correct force equation ma = T a=T m For correctly substituting the value of T from part (a), and mass m1 a1 = 3.6 ¥ 10 -4 N a1 = 3.6 ¥ 10 -4 N 1 point 1 point ( ) (0.030 kg ) = 1.2 ¥ 10 -2 m s2 1 point For correctly substituting the value of T from part (a), and mass m2 ( ) (0.060 kg ) = 6.0 ¥ 10 -3 m s2 (d) 3 points For beginning the graph at a value d that is clearly greater than zero For a concave upward curve For acceleration approaching zero as d approaches infinity 1 point 1 point 1 point (e) 3 points For indicating that the speeds increase For indicating that as the objects move apart, their speeds increase at a slower rate (i.e., the acceleration decreases) For indicating that the speeds approach a constant value as d approaches infinity 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 4 10 points total (a) 2 points For correctly determining the equivalent resistance of the two parallel resistors 1 1 1 2 = + = Rp R2 R2 R2 1 point Distribution of points 1 R 22 For correctly determining the total equivalent resistance of the circuit RT = R1 + R1 + R p Rp = 1 point RT = 2 R1 + 1 R 22 (b) 2 points For a correct expression for the power in terms of emf and resistance P = e2 R T 1 point 1 point For correctly substituting the value of total resistance from part (a) 1 P = e 2 2 R1 + R2 2 ( ) (c) 3 points For correctly indicating that the field is directed out of the plane of the page For using the right-hand rule to determine the direction of the field at point P from each wire (into the page from the top wire, out of the page from the bottom wire) For indicating that the magnitude of the field at point P from the bottom wire is greater because it is closer to point P 1 point 1 point 1 point (d) 3 points For correctly indicating that the force is directed toward the bottom of the page For indicating that the magnetic field at the bottom wire due to the top wire is directed into the page For using the right-hand rule to determine the direction of the force on the bottom wire due to the magnetic field 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 5 10 points total (a) 1 point Distribution of points For drawing the lens between the source and the card, and indicating that the lens is one focal length from the source 1 point (b) 2 points For drawing diverging rays from the source to the lens For drawing parallel rays from the lens to the card One earned point was deducted if a diverging lens was drawn, if it was obvious that a mirror was being used, or if the lens was set right next to the card. 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 5 (continued) Distribution of points (c) 4 points For using the correct equation b sin q = ml For the correct approximation for sin q sin q ª y3 D For correctly substituting m = 3 b ( y3 D ) = 3 l For the correct answer l = by3 3 D Notes: The first 2 points could also be earned by starting directly with the equation m lL xm ª from the equation table. d The second point was also earned for use of either of the exact relationships 1 point 1 point 1 point 1 point sin q = y3 2 y3 + D2 or q = tan -1 ( y3 D ) . (d) 3 points For indicating that the fringe spacing would decrease For a clear, correct justification For example: If the index of refraction increases, the wavelength in that region decreases. From the relationship in part (c), one can see that that means a decrease in fringe spacing. No credit was awarded when multiple choices were marked unless they were affirmative and negative marks (e.g., a checkmark for the intended choice and an x for the others). 1 point 2 points © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 6 10 points total (a) 2 points For indicating that the buoyant force is the difference between the scale readings Fb = 17.8 N - 16.2 N For the correct answer with units Fb = 1.6 N 1 point 1 point Distribution of points (b) 3 points For indicating that the buoyant force equals the weight of the displaced water Fb = mw g For indicating the correct relationship between mass and volume m = rV The volume of displaced water equals the volume of the object. Fb = rwVw g = rwVo g Vo = Fb rw g For substituting the value of buoyant force from part (a) 1 point 1 point Vo = (1.6 N ) 1000 kg m Vo = 1.6 ¥ 10 -4 m3 ( 3 ) (9.8 m s ) 2 1 point (c) 2 points For using the relationship between mass and density ro = mo Vo 1 point wo g = wo gVo = (17.8 N ) 9.8 m s2 1.6 ¥ 10 -4 m3 Vo For the correct answer ro = 1.1 ¥ 10 4 kg m 3 ro = ( )( ) 1 point (d) 3 points For indicating that the pressure would decrease For correctly indicating an intermediate effect: the height of the water will decrease or the total force at the bottom of the water is less. For a correct relationship between the intermediate effect and the pressure: P μ rgh or P=F A 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES (Form B) Question 7 10 points total (a) 2 points K max = hf - f Distribution of points The work function can be determined when K max = 0 . For a correct relationship for determining the work function f = hf0 For correctly substituting f0 , determined from the graph 1 point 1 point f = 4.14 ¥ 10 -15 eVis 4.5 ¥ 1014 Hz f = 1.9 eV One earned point was deducted if the answer was not expressed in eVs. (b) 3 points For using the correct equation K max = hf - f For correctly substituting the value of f from part (a) For correctly substituting the value of f from the graph 1 point 1 point 1 point ( )( ) K max = 4.14 ¥ 10 -15 eVis 8 ¥ 1014 Hz - 1.9 eV K max = 1.4 eV One earned point was deducted if the answer was not expressed in eVs. Alternate solution: Alternate points For indicating a direct connection between the 1.5 V needed to stop the highest-energy 1 point electrons (as read from the graph) and the 1.5 eV maximum initial kinetic energy For justifying that connection 2 points ( )( ) (c) 2 points For a correct relationship to calculate the wavelength of light l=c f 1 point l = 3.00 ¥ 108 m s ( ) (8 ¥ 1014 Hz) For the correct answer l = 3.75 ¥ 10 -7 m Alternatively, the equation l = hc E can be used with E equal to the sum of the answers to parts (a) and (b). (d) 3 points For correctly indicating that the required wavelength would be longer For indicating that a lower K max means the light must have a lower frequency For indicating that a lower frequency corresponds to a longer wavelength 1 point 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. ...
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