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Unformatted text preview: AP® Physics B 2010 Scoring Guidelines The College Board
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General Notes
1. The solutions contain the most common method of solving the freeresponse 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 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
Question 1
15 points total Distribution of points (a) 3 points For a correct relationship between the vertical distance and time 1 h = gt 2 2 For substitution of the vertical height and the acceleration due to gravity 2 ( 0.80 m ) 2h t= = g 9.8 m s2 For the correct answer t = 0.40 s
2 Note: Credit was awarded for an alternate solution using u y = 2 gh with appropriate 1 point 1 point 1 point substitutions to find the vertical velocity when the block lands, followed by substitution of this velocity into u y = gt (or equivalent) to find the time. (b) 2 points For a correct relationship between the horizontal distance and time x = ut For a consistent substitution of time from part (a) into the correct equation x 1.2 m u= = t 0.40 s u = 3.0 m s 1 point 1 point (c) 3 points For any statement of conservation of energy For correct use of appropriate energy equations 12 1 kx = mu 2 2 2 For a consistent substitution of velocity from part (b) into the correct equation 4 kg m x= u= (3.0 m s) 650 N m k x = 0.24 m 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
Question 1 (continued)
Distribution of points (d) 4 points For any statement of conservation of momentum m Aui = ( m A + mB ) u f For substitution of both masses into the equation For substitution of the velocity from part (b) into the equation 4 kg Ê mA ˆ Ê ˆ uf = Á ˜ ui = Ë 4 kg + 4 kg ¯ 3.0 m s = 1.5 m s Ë m A + mB ¯ For substitution of time from part (a) into a correct relationship between the horizontal distance and time d = u f t = (1.5 m s) ( 0.40 s) 1 point 1 point 1 point 1 point d = 0.60 m
(e) 2 points For indicating that E2 < E1 For a correct justification stating one of the following: • the kinetic energy (or energy) is transformed into other forms of energy during the collision (e.g., by reference to heat, internal energy, sound) • the kinetic energy is not conserved in an inelastic collision • a numerical calculation of the relevant energies 1 point 1 point Units 1 point For correct units on all completed answers 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 2
15 points total (a) 2 points Distribution of points For a single upward force, appropriately labeled, representing the buoyant force For downward gravitational force (or forces), appropriately labeled, representing the cup and the sample One earned point was deducted if any extraneous forces were present. 1 point 1 point (b) 3 points For any statement of equilibrium FB = Fg For a correct substitution including both masses mC and mS
rOVO g = ( mC + mS ) g 1 point 1 point 1 point For correct statement of the overflow volume, VO VO = mC + mS rO (c) 4 points © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (c) (continued) For data plotted correctly For correct units on both axes For numerical scales that are linear and allow the plotted data to extend over about half the grid area For a reasonable single straight bestfit line that does not go through (0,0) 1 point 1 point 1 point 1 point (d) 4 points mC 1 + m rO rO S For properly calculating a slope using points on the straight line drawn, including data points only if they are on that line Example: Using the two points (0.060 kg, 75 ¥ 10 6 m 3 ) and (0.025 kg, 35 ¥ 10 6 m 3 ) that are on the line in the graph above For calculating a reasonable value of slope (75  35) ¥ 10 6 m 3 1 slope = = = 1.14 ¥ 10 3 m3 kg r0 (.060  0.025) kg For an explicit or implicit indication of inverting the slope For calculating a reasonable value for the oil’s density (including units and four significant figures or less) rO = 8.8 ¥ 10 2 kg m3
From part (b), VO = 1 point 1 point 1 point 1 point (e) 2 points For a complete statement that the y intercept is the volume of the oil displaced by the empty cup Note: 1 point is given for a partially correct answer 2 points © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 3
10 points total (a) 1 point For an indication that q1 is negative and q2 is positive 1 point Distribution of points (b) 2 points For force F1 drawn and labeled correctly For force F2 drawn and labeled correctly Notes: The force vectors must either originate or terminate on q3 . Forces on other particles are ignored. 1 point 1 point (c) 3 points For a correct statement or use of Coulomb’s law Applying Coulomb’s law to determine the magnitude of the forces F1 and F2 : 1 point F1 = F2 = kq1q3 r13
2 = (9 ¥ 109 N ∑ m 2 C2 4.0 ¥ 10 6 C 1.0 ¥ 10 6 C )( kq2 q3 r232 (9 ¥ 10 = ( 4.0 m ) )( 2 ) = 2.25 ¥ 103 N 9 N∑m 2 C 2 ) (1.7 ¥ 106 C) (1.0 ¥ 106 C) = 1.70 ¥ 103 N
(3.0 m )2
1 point For any indication that F is the vector sum of the two forces: F = F1 + F2 Since F1 and F2 are at right angles to each other, the magnitude can be found using the Pythagorean theorem. F = F12 + F2 2 = (2.25 ¥ 103 N) + (1.70 ¥ 103 N)
2 2 Alternate solution: The y components cancel, so the magnitude of F is the sum of the x components. F = F1x + F2 x = F1 cos37 + F2 cos 53
F = 2.25 ¥ 10 3 N cos37 + 1.70 ¥ 10 3 N cos 53 ( ) ( ) F = 1.8 ¥ 10 3 N + 1.0 ¥ 10 3 N For the correct answer with units F = 2.8 ¥ 10 3 N 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (d) 2 points For substituting the value of F from part (c) and using the correct value for q3 1 point F 2.8 ¥ 10 3 N = q3 1.0 ¥ 10 6 C This point could also be earned for substituting F1 and F2 from part (c) into E = F q3 E=
and then calculating the magnitude of the vector sum, or calculating E1 and E2 from E = kq r 2 with correct q’s and r’s and then calculating the magnitude of the vector sum. For a calculated answer with correct units E = 2.8 ¥ 103 N C (e) 2 points 1 point For an ¥ in the correct position as shown above For a correct justification that refers to forces For example: Positive charges repel and a force to the right would cancel force F . 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 4
10 points total (a) 3 points
W tˆ Ê e = Pout Pin Á i.e. out ˜ Ë Qin t ¯ For a correct substitution of the efficiency into a correct equation 0.12 = Wout Qin 0.12 = Pout Pin or For a correct recognition of the relationship between power, energy and time Examples of exhibiting that relationship include: starting with e = Pout Pin , using Distribution of points e = Wout Qin or 1 point 1 point P = W t or Q t in the efficiency equation or referring to the power as a rate. Pin = Pout e = 4.5 ¥ 106 W 0.12 For a correct answer with correct units Pin = 3.8 ¥ 107 W
(b) 2 points 1 point F = Pout (u cos q )
For a correct substitution of Pout into a correct expression The resistive force acts opposite to the velocity, so q = 180∞ .
F = 4.5 ¥ 10 6 W [( 7.0 m s)( cos180∞)] For an answer consistent with the value of Pout substituted, with correct units F = 6.4 ¥ 10 5 N Pout = Fu cos q 1 point 1 point (c) (i) 1 point For an answer that uses the word “work” to represent the area (ii) 2 points 1 point 1 point For a correct calculation of the work (either on the gas or by the gas) represented by the rectangular path W = base ¥ height = (VD  VA ) ( PB  PA )
W = 0.60 m 3  0.20 m 3 3.0 ¥ 10 5 N m 2  1.0 ¥ 10 5 N m 2 = 8.0 ¥ 10 4 J ( )( ) There are four cycles per second, so the time for one cycle is 0.25 s. Pout = Wout Dt = 8.0 ¥ 10 4 J 0.25 s For a consistent calculation of the power output from the work calculated, with units Pout = 3.2 ¥ 105 W 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 4 (continued)
Distribution of points (d) 2 points For indicating AB as a correct process For indicating BC as a correct process One point is deducted for each incorrect process indicated, up to the number of points earned for correct processes. 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 5
10 points total (a) 2 points Distribution of points n1 sin q1 = n2 sin q2
n1 sin q1 n2 For correct substitutions into Snell’s Law (1.0)sin 40∞ sin q2 = = 0.390 1.65 For the correct answer q2 = 22.9∞ or 23∞ sin q2 = 1 point 1 point (b) 3 points In order for total internal reflection to occur, q3 must increase until it is greater than qcritical . For this to occur, q2 must decrease. Finally, to decrease q2 there must be
an increase in q1 . For stating that q3 must increase to become greater than qcritical For stating that q2 must decrease For stating that q1 must decrease Alternate solution For calculating the minimum value of q3 that will result in total internal reflection
qcritical = sin 1 1 (1.65 ) = 37.3∞ 1 point 1 point 1 point Alternate points 1 point For calculating the corresponding value of q2 1 point 1 point q2 = 60∞  q3 = 22.7∞
For calculating the corresponding value of q1 q1 = sin 1 [1.65 (sin 22.7∞)] = 39.5∞ , therefore q1 must be decreased.
Other correct methods also received appropriate credit. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 5 (continued)
Distribution of points (c) (i) 2 points For a correct relationship between the wavelength in air and the wavelength in the film, which can be derived from l = u f and n = c u l lfilm = air nfilm 1 point 6.65 ¥ 10 7 m 1.38 For the correct answer with units lfilm =
7 1 point lfilm = 4.82 ¥ 10 m Alternate solution Alternate points For the correct calculation of either the velocity of light in the medium or the frequency 1 point of the light 3.00 ¥ 108 m s c ufilm = = = 2.17 ¥ 108 m s nfilm 1.38
OR f = 3.00 ¥ 108 m s c = = 4.51 ¥ 1014 Hz 7 lair 6.65 ¥ 10 m ufilm 2.17 ¥ 108 m s = f 4.51 ¥ 1014 Hz For the correct answer with units lfilm = 4.81 ¥ 10 7 m lfilm =
(ii) 3 points 1 point The light that enters the film and reflects off the prism travels a total distance 2t through the film. At both interfaces, there is a 180∞ phase change when the light is reflected, so the relative phase change of the interfering rays is zero. For destructive interference, the minimum path length in the film must equal lfilm 2 . Therefore, we have the relationship 2t = lfilm 2 . For the relationship between the thickness of the film and the wavelength of light in the film lair l t = film or t = 4 4 nfilm For the correct substitution of the wavelength of light in the film l l 4.82 ¥ 10 7 m 6.65 ¥ 10 7 m t = film = OR t = air = 4 4 4 nfilm 4 (1.38) For the correct answer with units 1 point 1 point 1 point t = 1.20 ¥ 10 7 m © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 6
10 points total (a) 3 points Distribution of points For an unambiguous indication that the induced current in the loop is in the counterclockwise direction For a justification that includes two correct and relevant principles, such as the following: The flux is changing (or increasing into the page). The induced current will oppose the change in flux. A counterclockwise current will produce flux out of the page. The magnetic forces on charges in the righthand wire will drive a counterclockwise current. Velocity is to the right, and B into the page, so qv ¥ B points toward the top of the page. The induced current must produce a magnetic drag force (opposite the motion). A single relevant principle earns 1 point. 1 point 2 points (b) (i) 2 points For writing relevant algebraic expressions for both the current and the emf somewhere in the part (b) answer space f e and I= e =  DDtm or e = B u R I = B u R = ( 2.0 T ) ( 0.10 m ) (3.0 m s ) 4.0 W For the correct magnitude of the current I = 0.15 A (ii) 1 point 1 point 1 point FB = BI sin q , where q = 90∞ because the field is perpendicular to the direction of the current. For an unambiguous substitution of current and wire length consistent with part (i) and the correct magnetic field into a correct expression for force FB = ( 2.0 T) ( 0.15 A ) (0.10 m ) sin 90∞ FB = 0.030 N
Units 1 point For correct units in the answers to both parts (i) and (ii) 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 6 (continued)
Distribution of points (c) 3 points For an unambiguous indication that the net force is zero For stating that the current is zero For either correctly explaining why the current is zero (such as “there is no change in magnetic flux” or “magnetic forces on charges in the two sides of the loop push the charges in opposite directions”) or explaining how zero current results in zero magnetic force on the wire loop 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
Question 7
10 points total Distribution of points (a) 2 points u = fl For substitution of the appropriate values of the speed of light and the wavelength into the correct expression 1 point 3.0 ¥ 108 m s u c = = l l 400 ¥ 109 m For the correct answer f=
f = 7.5 ¥ 10
14 1 point Hz (b) 2 points K max = hf  f f = hf  K max For consistent substitution of the maximum kinetic energy into the correct expression For consistent substitution of the frequency into the correct expression f = 6.63 ¥ 10 34 Jis 7.5 ¥ 1014 Hz  1.1 ¥ 10 19 J ( )( ) 1 point 1 point f = 3.9 ¥ 10 19 J (c) 2 points eV = K max For substitution of the appropriate values of the maximum kinetic energy and the charge of the electron into the correct expression K 1.1 ¥ 10 19 J V = max = e 1.6 ¥ 10 19 C For the correct magnitude of the stopping potential V = 0.69 V 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS B 2010 SCORING GUIDELINES
Question 7 (continued)
Distribution of points (d) 3 points 1 mu 2 2 For substitution of the appropriate values of the maximum kinetic energy and the mass of the electron into the correct expression K= 9.11 ¥ 10 kg For consistent substitution of velocity and mass of the electron into the correct expression p = mu = 9.11 ¥ 10 31 kg 4.91 ¥ 10 5 m s u= 2 K max = m 2 1.1 ¥ 10 19 J
31 1 point ( ) = 4.91 ¥ 10 5 m s 1 point ( )( ) For the correct answer p = 4.5 ¥ 10 25 kgim s 1 point Units 1 point For using correct units in completed answers 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. ...
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This note was uploaded on 02/09/2011 for the course PHYS 10 taught by Professor Davidnewton during the Spring '11 term at DeAnza College.
 Spring '11
 DavidNewton
 Physics

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