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Unformatted text preview: AP Physics B 2009 Scoring Guidelines The College Board
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General Notes About 2009 AP Physics Scoring Guidelines
1. The solutions contain the most common method of solving the freeresponse 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 FreeResponse SectionsStudent 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. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 1
15 points total (a) 3 points For a correct statement of conservation of mechanical energy (which might be implied in the next statement) For correct energy expressions set equal 1 mgh = kx 2 2 For solving for h, consistent with the energy equation
h= kx 2 mg
2 Distribution of points 1 point 1 point 1 point (b) (i) 2 points For a correct combination of ( 1 m and h) or ( 1 h and m), with or without constants Notes: If both 1 m and 1 h were chosen, only 1 point was earned. If part (b)(i) contained quantities that would not yield a correct graph for part (c), no credit was given for parts (b)(ii) through (d). But if part (b)(i) was left blank, parts (b)(ii) through (d) were examined for correct results. The example of h versus 1 m is used in the remainder of the scoring guideline. (ii) 2 points 2 points 1 m kg1
50 33 25 20 17 ( ) m (kg)
0.020 0.030 0.040 0.050 0.060 h (m)
0.49 0.34 0.28 0.19 0.18 1 point 1 point For correctly filling in the table with the appropriate data For including the correct units in the table 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (c) 4 points For correctly plotting appropriate data (all five data points plotted correctly, assuming the data represented an inverse relationship between m and h) For correctly drawing a bestfit straight line (a single straight line with data points reasonably scattered above and below the line) For correctly labeling both axes For correctly indicating the scale on both axes 1 point 1 point 1 point 1 point (d) 2 points For correctly calculating the slope from points on the line (0.42  0.10) m 0.32 m = = 1.07 10 2 m i kg For example: slope = 30 kg 1 ( 40  10 ) kg1 For a correct numerical value of the spring constant From (a), h = 1 point 1 point kx kx , so the slope of the line = 2 mg 2g 2 2 k = k = 2 g (slope ) x2 2 9.8 m s2 1.07 10 2 mi kg ( (0.02 m ) )( ) 2 k = 524 N m (535 N m using g = 10 m s2 ) Note: Values between 450 N m and 550 N m were accepted. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (e) 2 points For a correct, complete procedure Examples: Use a meter stick and a visual recorder, such as the eye, a video camera, or a camera. Use a sonic range finder and appropriate computer software. Measure the time (either to rise or for a round trip) and give the correct appropriate kinematics equations (the equations must be included in the answer). One point was awarded for a partially correct or incomplete procedure. Examples: Use a meter stick. Use a photogate. 2 points 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 2
10 points total (a) 3 points Distribution of points For the direction of all field lines away from the source charges For the correct shape, symmetry, and curvature of the field lines For a clear indication that the net field about point A, the point midway between the two charges, is zero; example: an absence of field lines in the area around point A 1 point 1 point 1 point (b) 2 points
For clearly showing the addition of both contributions to the potential at point A kq kq 1 Q Q V = 1 + 2 OR V = + OR equivalent r1 r2 d 4p 0 d For a correct expression in terms of the given quantities Q 2kQ 2Q V = OR OR OR equivalent L sin q 4p 0 L sin q 2p 0 L sin q Notes: The answer with no work shown earned 1 point. The following expression or its equivalents, with no other work shown, earned both points: kQ kQ V = + L sin q L sin q
(c) 2 points ( ) 1 point 1 point For all three vectors correctly drawn, with arrowheads, and no extraneous vectors For appropriate labeling of vectors (only if first point was awarded) 1 point 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (d) 3 points For an expression indicating that the xcomponent of tension is equal to FE
Tx  FE = 0 OR T sin q = FE OR equivalent For an expression indicating that the ycomponent of tension is equal to mg Ty  mg = 0 OR T cos q = mg OR equivalent For stating the two equations in terms of the given quantities kQ 2 Q2 Q2 1 OR OR OR equivalent T sin q = 4p 0 4L2 sin 2 q 4L2 sin 2 q 16p 0 L2 sin 2 q T cos q = mg Notes: Correct statement of both of the final two equations, in the absence of any other expressions, earned full credit. One point was deducted if sine and cosine were interchanged at any point. 1 point
1 point 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 3
15 points total Distribution of points (a) 3 points For a correct statement of Ohm's law (symbolic or numeric) For a correct statement of Faraday's law (symbolic or numeric) e=Bu I = R R (0.80 T )(0.52 m )(1.8 m s) I = 3.0 W For the correct answer I = 0.25 A 1 point 1 point 1 point (b) 4 points For a correct expression for the friction force (symbolic or numeric) Ff = mk mg 1 point Ff = (0.20)(0.22 kg ) 9.8 m s2 ( ) Ff = 0.43 N (0.44 N using g = 10 m s2 ) For a correct expression for the magnetic force (symbolic or numeric) FB = BI
FB = (0.80 T )(0.25 A )(0.52 m ) 1 point FB = 0.10 N For any indication of multiple forces (explicit, implied, freebody diagram, etc.) F = Ff + FB
For a correct answer consistent with the current from part (a) F = 0.43 N + 0.10 N = 0.53 N (0.54 N using g = 10 m s2 ) 1 point 1 point (c) 3 points For a correct expression for energy (including time) Ediss = Pt For a correct expression for power (individually or in another expression) P = I 2 R OR P = IV OR P = V 2 R For consistent substitution of current or voltage from part (a) into a correct equation Ediss = I 2 Rt OR Ediss = IVt OR Ediss = V 2 t R For example: 2 Ediss = (0.25 A ) (3.0 W)( 2.0 s) 1 point 1 point 1 point Ediss = 0.38 J 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (d) 2 points For a correct expression for distance d = ut d = (1.8 m s)( 2.0 s) d = 3.6 m For a consistent substitution of force from part (b) W = Fd W = (0.54 N )(3.6 m ) W = 1 .9 J Full credit could also be given for equivalent solutions (e.g., combining the expression P = Fu for power with the expression W = Pt for work to get W = Fut , followed by consistent substitutions). 1 point 1 point (e) 2 points For mentioning the force of friction but not including the concept of work For including a statement that friction does work or dissipates energy Note: If the answer for part (d) is less than or equal to the answer for part (c), no credit was given. 1 point 1 point Units point For correct units on all numerical answers 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 4
10 points total Distribution of points (a) 3 points For any attempt to use the ideal gas equation of state, i.e., by attempting to substitute values into the equation (simply writing the equation did not earn this point) PV = nRT T = PV nR For correctly calculating the number of moles n = 2.2 kg 18 103 kg mol = 122 mol For substituting the correct values for pressure, volume, and the universal gas constant into the ideal gas equation TA = 3.0 105 Pa 2.0 m3 (122 mol )(8.31 J moliK ) 1 point 1 point 1 point ( )( ) TA = 590 K (b) 1 point T = PV nR TC = 4.0 105 Pa 2.5 m3 ( )( ) (122 mol)(8.31 J moliK )
1 point For the correct answer, with the correct unit TC = 980 K Note: An alternate solution uses the ratios PAVA TA = PCVC TC to obtain TC = 5TA 3 . The point could be earned if the answers in parts (b) and (a) were in the ratio 5 3 and the temperatures were in Kelvin. (c) 3 points For indicating that the internal energy increases For a correct justification, with no incorrect statements For example: internal energy is proportional to the temperature, which increases as the gas is taken from A to C. One point could be earned for an incomplete justification or one that would be complete and correct but for an incorrect physical statement. (d) 3 points
W = area under the curve =  P V (the minus sign is not needed for this point) 1 point 2 points W =  4.0 105 Pa 0.5 m3 ( )( ) 1 point 1 point 1 point For an answer having the correct magnitude, including the correct unit For the negative sign on the answer W = 2.0 105 J 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 5
10 points total Distribution of points (a) 3 points For checking "No" For a complete and correct justification, with no incorrect statements For example: the tension in each string depends on the weight and the buoyant force. The buoyant force depends on the volume of the object. Objects of identical mass have the same weight but need not have identical volumes. One point could be earned for a partial justification, or one that contained an incorrect statement. 1 point 2 points (b) 3 points For any correct expression of the equilibrium of the three forces on object A T = mg  B , where B is the buoyant force For correctly using the density and volume of object A to calculate its weight and labeling as such mg = Vg = 1300 kg m3 1.0 105 m3 9.8 m s2 = 0.13 N 1 point 1 point B = mg  T = 0.13 N  0.0098 N For an answer consistent with preceding work B = 0.12 N
(c) 2 points ( )( )( ) 1 point The buoyant force equals the weight of the displaced liquid, which depends on its density. B = Vg = B Vg For substituting the buoyant force from part (b) and the correct values for volume and g into a correct expression = (0.12 N ) 1.0 105 m3 9.8 m s2 1 point ( )( ) For an answer consistent with the above substitutions, including correct units = 1200 kg m3 An alternate solution using the ratio of forces FB Fg = Vg AVg = A earned similar substitution and answer points. (d) 2 points For indicating that the tension increases For a correct justification For example: less of the object submerged means less liquid displaced, which means less buoyant force. By the equation in part (b), the tension is greater. 1 point 1 point 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 6
10 points total Distribution of points (a) 2 points For a meaningful attempt to use the equation f = c f = 3.00 10 m s
8 ( ) (550 10 9 m ) 1 point For the correct answer, with correct units f = 5.5 1014 Hz 1 point (b) 3 points For writing one or more equations that can be used to solve the problem x m L d OR d sin = m and sin x L (or tan x L ) For indicating that the calculation is for the spacing between fringes, rather than the position of a single fringe x = m L d Adjacent fringes means m = 1 x = L d For correct substitutions into the correct expression x = 550 109 m ( 2.2 m ) 1.8 105 m 1 point 1 point x = 0.067 m ( ) ( ) 1 point (c) 2 points For indicating that the frequency is the same as in part (a) f = 5.5 1014 Hz Note: If there was no answer in part (a) and a calculation was done in this part, the scoring guideline for part (a) was applied. 2 points (d) 3 points For correctly indicating that the fringe spacing decreases For indicating in the explanation that the wavelength decreases For indicating in the explanation that x is proportional to the wavelength 1 point 1 point 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 7
10 points total Distribution of points (a) 3 points = h p = h mu u = h m
For substituting the correct value of the electron wavelength into a correct expression For substituting a correct value of Planck's constant into a correct expression u = 6.63 1034 Jis 9.11 1031 kg 0.85 109 m ( )( )( ) 1 point 1 point u = 8.56 105 m s For substituting the correct value of the electron mass into a correct expression for the kinetic energy K = mu 2 2
K = 9.11 1031 kg 8.56 105 m s K = 3.3 1019 J (or 2.1 eV) 1 point ( )( ) 2 2 (b) 3 points For any indication that the student used the equation for the photoelectric effect K max = hf  1 point = hf  K max = (hc )  K max For substituting the correct value of the photon wavelength into the correct expression For substituting for hc in appropriate energy units into the correct expression = 6.63 1034 Jis 3.00 108 m/s 250 109 m  3.3 1019 J ( )( )( ) 1 point 1 point = 4.7 1019 J (or 2.9 eV) Units point For correct units in both parts (a) and (b) 1 point 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES
Question 7 (continued)
Distribution of points (c) 3 points For choosing the correct transition For indicating that the emission of a photon requires the atom to go to a lower energy level For indicating that E = hc for a photon, so a smaller wavelength means a larger energy difference Notes: Two points could be earned by choosing transition a and stating that the emission of a photon requires a drop to a lower energy level. Two points could be earned by choosing transition e and stating that the shorter photon wavelength implies a larger difference between energy levels. 1 point 1 point 1 point 2009 The College Board. All rights reserved. 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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