This preview shows page 1. Sign up to view the full content.
Unformatted text preview: AP® Physics C: Electricity and Magnetism 2010 Scoring Guidelines 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 2010 SCORING GUIDELINES
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 Exam equation sheets. 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 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 C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
15 points total Distribution of points (a) 3 points For indicating that the potential at point B ranks 1 (has the highest potential) For indicating that the potentials at points A and C are equal and rank 2 For a correct justification Example: Compared to points A and C, point B is closer to most, and possibly all, points along the charge distribution. Since potential varies inversely with distance, point B has the highest potential. Points A and C have the same potential by symmetry. 1 point 1 point 1 point (b) 2 points For any indication of correct qualitative reasoning about the potential for this particular geometry Example: All points on the arc are a distance R from point P. Since potential is a scalar quantity, the potential will be the same as that of a point charge with charge Q located a distance R away. For a correct answer V = kQ R 1 point 1 point Alternate solution Alternate points For indicating the potential is obtained by integrating the contributions from each part 1 point of the charge distribution kdq Q 2Q V=Ú r dq = where dq = lr dq = dq r p r ( p 2) Noting that r = R for the entire distribution, the integral becomes: 2 kQ V= pR p2 Ú
1 point For a correct answer V = kQ R (c) 4 points For an indication that mechanical energy is conserved Ui + K i = U f + K f For correct substitution of potential energies Uf = 0 1 point 1 point Ui = qV For substituting the potential at P from part (b) q ( kQ R ) = K f
For substituting correctly for the kinetic energy and solving for the velocity q ( kQ R ) = (1 2 ) mu 2 1 point 1 point u = 2 kqQ mR © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (d) 1 point For a vector drawn in the correct direction — horizontally to the right 1 point (e) 5 points For any indication that the net electric field is the integral of all of the horizontal components from each part of the charge 1 point E = Ex = Ú dEx
1 point For correctly using cosine in computing the x-component k dQ E = dE cos q = cos q R2 For changing variables to integrate with respect to q 2Q dq = dq p For correct limits of integration Ú Ú 1 point 1 point p4 E= -p Ú 2 kQ 3p 5p and ) cos q dq (or equivalent limits such as 2 4 4 4 pR
p4 2 kQ sin q p R2 -p 4 For a correct answer 2 2 kQ E= p R2 E= 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
15 points total Distribution of points (a) 2 points For correctly stating that there is no current in the steady state I =0A 2 points (b) 2 points Q = CV For correct substitution of capacitance into the equation above Q = (10 mF ) (30 V ) For a correct numerical answer with units Q = 300 mC 1 point 1 point (c) 3 points 1 CV 2 2 For substitution of the correct capacitance (in units of mF or F) into the correct expression for energy For substitution of the battery voltage into the correct expression for energy 1 2 U = ( 5 mF )(30 V ) 2 For correct units on a numerical answer U = 2250 mJ U= 1 point 1 point 1 point (d) 2 points For recognizing the two resistors are now in series and correctly calculating the equivalent resistance RT = 20 W + 40 W = 60 W V = IR I =V R For substitution of the correct voltage and the calculated equivalent resistance into Ohm’s law I = 30 V 60 W I = 0.5 A 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (e) 3 points For recognizing that the voltage across the 5.0 mF capacitor is the same as that for the 40 W resistor and calculating that voltage V40 W = ( 0.5 A ) ( 40 W) = 20 V Q = CV For correctly substituting the value of the capacitance into the above equation Q = ( 5.0 mF ) ( 20 V ) For the correct answer Q = 100 mC 1 point 1 point 1 point Alternate solution Alternate points 1 point For correctly applying the loop equation to the loop that includes the 5.0 mF capacitor, the 20 W resistor and the 30 V battery Q V - IR =0 C Q = (C ) (V - IR ) For correctly substituting the current and the resistances 1 point Q = ( 5.0 mF ) [(30 V ) - ( 0.5 A ) ( 20 W )] For the correct answer 1 point Q = 100 mC (f) 3 points
P = I 2R For correct substitution of the total current from part (d) and the resistance into the above equation for power (or substituting correct values into P = V 2 R or P = IV ) 1 point P = ( 0.5 A ) ( 40 W) = 10 W For correctly substituting the power into an equation for energy For correctly substituting the time in seconds into an equation for energy E = Pt E = (10 W ) (60 s ) E = 600 J Note: If time is not substituted using seconds, the units on the final answer must be consistent with the substitution.
2 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
15 points total Distribution of points (a) 4 points For a correct indication that the current in the loop is in the counterclockwise direction For indicating that the magnetic field through the loop is directed out of the page (which can be done on the diagram) For indicating that the current in the wire is decreasing, either explicitly or by indicating a decrease in field or flux through the loop For indicating that the direction of the induced current is such as to oppose the change in flux Example: The flux due to current I is out of the plane of the page (by the right-hand rule) and is decreasing with time. The induced current will be in the direction that will produce a compensating flux (by Lenz’s law). Again using a right-hand rule, the current must be counterclockwise. 1 point 1 point 1 point 1 point (b) 2 points For a correct indication that the brightness of the lightbulb remains the same For a correct justification Example: The field and the flux both vary linearly with time. The emf, which is the time derivative of the flux, must then be constant. Since the power output of the lightbulb depends only on the emf and resistance (which are both constant), the power must be constant. 1 point 1 point (c) 2 points B2 pr = m0 I mI B= 0 2 pr mI B= 0 2 pr t = 0
For the correct answer mI B= 00 2 pr One point partial credit could be earned for either correctly applying Ampere’s law and leaving the result in terms of I, or obtaining an incorrect expression for B in terms of I0 . 2 points Ú Bid = m0 I © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2010 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (d) 4 points f= ÚB ∑ dA
1 point For correctly substituting the expression for B as a function of r into the flux equation Ê m Iˆ f = Ú Á 0 ˜ dA Ë 2 pr ¯ For correctly recognizing that dA = b dr Ê m Iˆ f = Ú Á 0 ˜ b dr Ë 2 pr ¯ For correctly integrating with respect to r 1 point 1 point f= f= m0 Ib 2p d +a Ú
d dr r
d +a m0 Ib ln (r ) 2p d m0 Ib d+a ln 2p d For correctly substituting the current as a function of time m ( I - Kt ) b d+a f= 0 0 ln 2p d f= () ( ) 1 point (e) 3 points
P = V 2 R (which can be derived from P = I 2 R and V = IR ) For recognizing that the voltage across the bulb is the induced emf in the loop, and using that emf in the above expression for power P = e 2 R , where e = - d f dt For correctly substituting the flux from part (d) into the above equation for emf dÈ +˘ e = - dt Í m0b ( I20p- Kt ) ln d d a ˙ Î ˚ For correctly taking the derivative (with respect to time) of the flux d + e = - m0pb ln d d a È dt ( I0 - Kt )˘ Í ˙ 2 Î ˚ m0 b + e = - 2 p ln d d a ( - K ) 1 point () 1 point () () 1 point 2 1 È m0 bK d+a ˘ ln R R Í 2p d˙ Î ˚ Note: If P = I e is used with the expression for the current in the long wire (rather than the loop) being substituted for I, the last 2 points for correctly determining the emf could still be earned. P= e2 = ( ) © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. ...
View Full Document
This note was uploaded on 05/05/2011 for the course CHEM 504 taught by Professor John during the Fall '11 term at American College of Computer & Information Sciences.
- Fall '11