E&M - Scoring 2008 - AP® Physics C: Electricity...

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Unformatted text preview: AP® Physics C: Electricity and Magnetism 2008 Scoring Guidelines The College Board: Connecting Students to College Success 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 association is composed of more than 5,400 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). 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. © 2008 The College Board. All rights reserved. College Board, AP Central, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of 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 online home for AP teachers: apcentral.collegeboard.com. AP® PHYSICS 2008 SCORING GUIDELINES General Notes About 2008 AP Physics Scoring Guidelines 1. The solutions contain the most common method of solving the free-response questions and the allocation of points for this 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, e.g., 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 one 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 exam 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 Free-Response Sections⎯Student Presentation” in the AP Physics Course Description. 4. The scoring guidelines typically show numerical results using the value g = 9.8 m s2 , 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. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 1 15 points total Distribution of points (a) (i) 2 points For indicating that the charge on the inner surface of the shell is -Q For a correct explanation with no incorrect statements Examples: • Applying Gauss’s law to a Gaussian surface within the shell gives Qenclosed = 0 , since the field within a conductor is zero. Therefore the charge on the inner surface of the shell is -Q . • The +Q on the sphere attracts an equal and opposite charge onto the inner surface of the shell. (Equal magnitude could be implied by a statement that earned the first point.) (ii) 2 points 1 point 1 point 1 point 1 point For indicating that the charge on the inner surface of the shell is +Q For a correct explanation with no incorrect statements Examples: • Applying Gauss’s law to a Gaussian surface outside the shell gives Qenclosed = +Q , therefore the sum of the charges on the inner and outer surfaces of the shell must be 0. • The net charge on the shell is zero. Therefore the charge on the outer surface must be the opposite of the charge on the inner surface because of conservation of charge. Note: If the correct sign of the charge is given in part i or ii without the magnitude (Q), a correct explanation could receive 1 point. (b) (i) 1 point Since the sphere is a conductor all the charge lies on the outside surface. Applying Gauss’s law to any Gaussian surface inside the sphere gives Qenclosed = 0 . For a correct answer E=0 (ii) 1 point 1 point For any surface between the sphere and the shell the net enclosed charge is +Q . Applying Gauss’s law E 4 pr 2 = Q 0 For a correct answer E = Q 4 p 0r 2 1 point or E = kQ r 2 © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 1 (continued) Distribution of points (b) (continued) (iii) 1 point For any surface inside the shell the net enclosed charge is zero. For a correct answer E=0 (iv) 1 point 1 point For any surface outside the shell the net enclosed charge is +Q . Applying Gauss’s law E 4 pr 2 = Q 0 For a correct answer E = Q 4 p 0r 2 1 point or E = kQ r 2 (c) 4 points For drawing E = 0 for 0 £ r £ a (must have a line drawn on the axis) For drawing a positive, decreasing, concave up function with E ( b) π 0 for a £ r £ b For drawing E = 0 for b £ r £ c (must have a line drawn on the axis) For drawing a positive, decreasing, concave up function with E ( c ) < E ( b ) for r ≥ c 1 point 1 point 1 point 1 point © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 1 (continued) Distribution of points (d) 3 points Note: The intent of the problem was to ask for the speed at a distance of 10c instead of the stated 10r. Very few students did anything other than treat 10r as if it were outside the shell. For a correct statement of conservation of energy For example, K + U = 0 For correct substitution of all variables into a correct relationship (including integration limits of • and 10r if integration is used) Q ( - e) 1 =0 For example, meu 2 + 2 4 p 0 (10r ) For a correct solution for the speed Qe 1 meu 2 = 2 4 p 0 (10r ) 1 point 1 point 1 point u2 = 2Qe Qe 1 = 4 p 0 5mer 4 p 0 me (10r ) Qe 1 or equivalent 4 p 0 5mer Note: If a student noted that 10r did not define a definite radius, points were awarded as appropriate for attempting a calculation in any region or several regions, or for a clear indication that the solution is region-dependent. u= © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 2 15 points total Distribution of points (a) (i) 4 points For calculating the equivalent resistance of the parallel branch 1 1 1 3 = + = RresP (100 + 50 ) W 300 W 300 W RresP = 100 W 1 point For calculating the total resistance of the circuit RresT = R1 + RresP = 200 W + 100 W = 300 W For correctly using the total resistance to compute the current through the battery IresT = e RresT = 1500 V 300 W = 5 A For correctly using the total current to calculate the voltage across R2 V1res = IresT R1 = ( 5 A ) ( 200 W ) = 1000 V V2 res = 1 point 1 point 1 point e - V1res = 1500 V - 1000 V V2 res = 500 V Alternate solution For the correct Kirchhoff junction equation I1 = I 2 + I3 For one correct loop equation ( SV = 0 ) For a second correct loop equation 1500 V - ( 200 W ) I1 - (300 W ) I 2 = 0 1500 V - ( 200 W ) I1 - (150 W ) I 3 = 0 Alternate points 1 point 1 point 1 point Using these three equations to solve for I 2 1500 V - ( 200 W ) I1 - (300 W ) I 2 = 1500 V - ( 200 W ) I1 - (150 W ) I3 (300 W) I 2 = (150 W) I3 2 I 2 = I3 I1 = I 2 + 2 I 2 = 3I 2 I 2 = 1.67 A 1500 V - ( 200 W ) 3I 2 - (300 W ) I 2 = 0 For correctly using the current I 2 to calculate the voltage across R2 V2 res = (1.67 A ) (300 W ) V2 res = 500 V 1 point © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 2 (continued) Distribution of points (a) (continued) (ii) 2 points 1 point For indicating that the current in branch 3 is zero immediately after the switch is closed, either explicitly or by correctly calculating the total resistance at this instant RindT = R1 + R2 = 200 W + 300 W = 500 W For correctly using the total resistance to calculate the voltage across resistor R2 V2 ind = (3 A ) (300 W ) = 900 V IindT = 1 point e RindT = 1500 V 500 W = 3 A Alternate solution For one correct Kirchhoff equation indicating knowledge that there is no current through resistor R3 1500 V - ( 200 W ) IindT - (300 W ) IindT = 0 IindT = 3 A Alternate points 1 point For correctly using the current to calculate the voltage across resistor R2 V2 ind = (3 A ) (300 W ) V2 ind = 900 V 1 point (iii) 3 points 1 point For indicating that the voltage across the capacitor is zero immediately after the switch is closed, either explicitly or by correctly calculating the total resistance 1 1 1 4 = + = RcapP 100 W 300 W 300 W RcapP = 75 W RcapT = R1 + RcapP = 200 W + 75 W = 275 W For correctly using the total resistance to compute the current through the battery I capT = e RcapT = 1500 V 275 W = 5.45 A For correctly using the total current to compute the voltage across R2 1 point 1 point V2 cap = I capT RcapP = ( 5.45 A ) (75 W) V2 cap = 410 V (rounded to two significant digits) © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 2 (continued) Distribution of points (a) (continued) (iii) (continued) Alternate points 1 point 1 point Alternate solution For one correct Kirchhoff equation indicating the current flowing in R3 For a second correct Kirchhoff equation 1500 V - ( 200 W ) I1 - (100 W ) I 3 = 0 1500 V - ( 200 W ) I1 - (300 W ) I 2 = 0 I1 = I 2 + I3 Using a Kirchhoff junction equation and solving the three equations for I 2 1500 V - ( 200 W ) I1 - (100 W ) I3 = 1500 V - (200 W ) I1 - (300 W ) I 2 (100 W) I3 = (300 W) I 2 I3 = 3I 2 I1 = I 2 + 3I 2 = 4 I 2 I 2 = 1.36 A 1500 V - ( 200 W ) 4 I 2 - (300 W ) I 2 = 0 For correctly using I 2 to calculate the voltage across R2 V2 = I 2 R2 = (1.36 A ) (300 W ) V2 = 410 V (rounded to two significant digits) 1 point © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 2 (continued) Distribution of points (b) 6 points Resistor graph: The current is constant with the resistor placed between points A and B. The resistance of that branch is more than when the capacitor and inductor are placed there, so the current will be less. For drawing a horizontal line, indicating a constant current For having the value of the resistor graph less than the initial value of the capacitor graph or the steady state value of the inductor graph Inductor graph: The inductor initially opposes the flow of current, so the initial current in that branch is zero. Eventually, the inductor acts like a wire and does not impede the flow of charge, as the rate of change of current decreases to zero. For starting the graph at I 3 = 0 at time t = 0 For a graph that is concave down and asymptotic to the initial current in the capacitor case Capacitor graph: Initially, the capacitor is uncharged and current is a maximum in the branch containing R3 . As the capacitor charges the current in that branch decreases to zero. For a finite, nonzero initial value for the current I3 at t = 0 For a graph that is concave up and asymptotic to I = 0 1 point 1 point 1 point 1 point 1 point 1 point © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 3 15 points total Distribution of points (a) (i) 1 point For indicating that the magnetic field B1 at point P is toward the top of the page (ii) 6 points dB r 1 point d (In plane of loop and tangent to it.) 1 point For using the Biot-Savart law m Id ¥r dB = 0 4p r3 For correctly indicating or applying the fact that d is perpendicular to r m I d r sin q m Id dB = 0 =02 3 4p 4p r r For recognizing that the net horizontal component will be zero, so that just the vertical components are summed R B1 = dBvert = dB cos a = dB , where a is the elevation angle of the vector r r For the correct expression for r 1 point 1 point Ú Ú Ú 1 point r = R2 + B1 = Ú R 5 = R 4 2 R R m0 I d dB = Ú r r 4p r 2 2 = m0 IR Ê5ˆ R 4p Á Ë2 ˜ ¯ 3 Úd 1 point For the correct calculation of Úd is the angle around the loop Úd 2p = Ú R d b = 2 p R , where b 0 For a correct final expression m0 IR m0 IR 4 m0 I B1 = d= 2p R = 3 3 55 R Ê5ˆ Ê5ˆ 4p Á 4p Á R R Ë2 ˜ ¯ Ë2 ˜ ¯ 1 point Ú © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2008 SCORING GUIDELINES Question 3 (continued) Distribution of points (b) 2 points For recognizing that Bnet is the vector sum of the field generated by the first loop and the field generated by the second loop For recognizing that the B field from top loop is in the same direction and has the same magnitude as that from the bottom loop 8 m0 I Bnet = 2 B1 = 55 R 1 point 1 point (c) 2 points For identifying B as Bnet in a correct expression for magnetic flux f= 1 point Ú BidA = Ú Bnet dA = Bnet A 1 point For correctly substituting the area as s 2 f = Bnet s 2 (d) 4 points For using Faraday’s law with f identified as magnetic flux 1 point e = - df dt with some work showing understanding of f 1 point 1 point 1 point For recognizing that there is an angular dependence f = Bnet s 2 cos q For correctly relating the angle to the angular velocity f = Bnet s 2 cos wt For the correct final expression d e = - Bnet dt s 2 cos wt = Bnet s 2 w sin wt ( ) © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ...
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