Unformatted text preview: AP Physics B 2008 Scoring Guidelines Form B The College Board: Connecting Students to College Success
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General Notes About 2008 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 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 B 2008 SCORING GUIDELINES (Form B)
Question 1
10 points total Distribution of points (a) 2 points For a correct expression of conservation of momentum m w u w = ms u s 1 point mwuw (70 kg)(0.55 m s) = 35 kg ms For the correct final answer us = 1.1 m s us = 1 point (b) 2 points For correct use of the impulsemomentum relationship Favg Dt = ms Dus 1 point ms Dus (35 kg )(1.1 m s  0 m s) = Dt 0.60 s For the correct final answer Favg = 64 N Favg = 1 point Alternate solution For correct calculation of the average acceleration of the son during the push Dus as = Dt 1.1 m s as = = 1.83 m s2 0.6 s For the correct final answer using Newton's second law Alternate points 1 point 1 point Favg = ms as = (35 kg)(1.83 m s )
2 Favg = 64 N (c) 3 points For indicating that the average force exerted on the mother by the son is equal in magnitude to that exerted on the son by the mother For indicating that the average force exerted on the mother by the son is opposite in direction to that exerted on the son by the mother For a correct justification invoking Newton's third law 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 B 2008 SCORING GUIDELINES (Form B)
Question 1 (continued)
Distribution of points (d) 3 points For correctly finding the acceleration of the mother In the following solution, the positive direction for all quantities is to the right relative to the figure shown in the question. 1 point uwf 2 = uwi 2 + 2aw Dxw
0 = ( 0.55 m s) + 2aw ( 7 m )
2 aw = 0.022 m s2 For recognizing that the accelerations of both the mother and the son are the same magnitude, since the acceleration is caused by friction Ffric = m N = mmg = ma a = mg aw = as
as = 0.022 m s2 For correctly finding the displacement of the son, with units 1 point 1 point usf 2 = usi + 2as Dxs
2
2 0 = (1.1 m s) + 2 0.022 m s2 Dxs
Dxs = 28 m ( ) 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 2
15 points total (a) 2 points For correct use of a kinematic relationship to find acceleration, and correct substitution of values 1 x = x0 + u0t  at 2 2 2 [( x  x0 )  u0t ] a=t2 2 [55 m  ( 25 m s)(3.0 s)] a= (3.0 s)2 For the correct final answer, regardless of sign or units 1 point Distribution of points 1 point a = 4.4 m s
(b) 3 points 2 For each correct force for which the vector was correctly drawn and labeled, 1 point was awarded. For each extraneous or incorrect force vector, 1 point was deducted with the minimum possible score being zero. (c) (i) 3 points For equating the frictional force to ma SF = Ffric = ma For a correct expression for the frictional force Ffric = m N = mmg Substituting the expression for Ffric into Newton's second law mmg = ma 3 points 1 point 1 point 4.4 m s2 a = g 9.8 m s2 For the correct final answer, without units m=
m = 0.45 ( m = 0.44 for g = 10 m s )
2 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 2 (continued)
Distribution of points (ii) 1 point 1 point For indicating that the friction is static (d) 3 points For correct use of a kinematic relationship to find acceleration u = u0 + at 1 point u x  u0 x 25 m s  0 m s = = 2.5 m s2 t 10 s For correct substitutions into Newton's second law SF = ma = kx kx = max ax = 1 point (900 kg ) 2.5 m s2 ma x = = 9200 N m k For the correct answer with units x = 0.24 m
(e) 3 points For indicating that the extension of the spring is less than in part (d) For a correct justification For example: When the truck is moving at a constant speed, the crate is also moving at the same constant speed with zero acceleration. This means the net force on the crate must be zero; since the bed of the truck is frictionless, the force of the spring on the crate must also zero, and so the spring is not extended at all. Notes: A single point could be awarded for partial justification, e.g., for either of the statements above given in the absence of the other. The justification points could only be earned if the point was awarded for the proper extension of the spring. ( ) 1 point 1 point 2 points 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 3
15 points total (a) 2 points Distribution of points For each new point being above one of the points shown on the graph For each new point being about the same distance from the old point, a distance equal to onehalf the distance between the horizontal grid lines shown (b) 3 points For an attempted application of a correct relationship to find the current I m I B= 0 2p r 2 prB I = m0 For correct substitutions using one of the new data points plotted in part (a), for example, using the point (0.01 m, 10.0 10 4 T)
4 p 10 ( Tim ) A For the correct answer I = 50 A
7 1 point 1 point 1 point 1 point I = 2 p ( 0.01 m ) 10.0 10 4 T ( ) 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 3 (continued)
Distribution of points (c) 2 points For the direction that the needle points being toward the northwest quadrant Note: If the needle was drawn pointing directly to the west, 1 point was awarded. If the needle was drawn pointing directly to the north, no points were awarded. (d) 4 points For a correct relationship to find the westward component of B, and a correct substitution of the current determined in part (b) m I 4 p 10 7 ( Tim ) A ^ 35 A Bw = 0 = ~ 0.040 m 2p r 2p 2 points 1 point ( ) For the correct answer for Bw 1 point 1 point Bw = 17.5 10 5 T For a correct relationship relating the northward and westward components to the angle B 17.5 10 5 T tan q = w = = 3.5 Bn 5.0 105 T For the correct answer q = 74 or 1.3 rad
(e) 2 points For correct substitution of both values into Ohm's law e = IR e = 120 V R= I 35 A For the correct answer including units R = 3.4 W 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 B 2008 SCORING GUIDELINES (Form B)
Question 3 (continued)
Distribution of points (f) 2 points For correct substitution of values into a correct expression for power 1 point P = IV (or P = I R or P = V R ) P = (35 A )(120 V ) For the correct answer including units P = 4200 W
2
2 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 4
10 points total Distribution of points (a) 3 points For a correct expression for the volume flow rate IV
IV = Au For a correct substitution for area (only if the first point was awarded) A = pr
2 1 point 1 point IV = pr 2u = (3.14 )( 0.015 m ) (6.0 m s ) For the correct answer IV = 4.2 10 3 m 3 s
2 1 point (b) 3 points For a correct application of Bernoulli's equation to point 1 in the feeder pipe below ground and point 2 at the surface where the water emerges 1 2 1 P + rgy1 + ru1 = P2 + rgy2 + ru2 2 1 2 2 P = Pabs and P2 = Patm 1 1 point 1 2 r u2 2  u1 2 For the correct use of the equation of continuity to find u1 Pabs = Patm + rg ( y2  y1 ) +
A1u1 = A2u2 ( ) 1 point u A u pr2 2 (6.0 m s)(0.015 m ) u1 = 2 2 = 2 2 = = 2.2 m s A1 pr1 (0.025 m )2
2 u2 2  u12 = (6.0 m s )  ( 2.2 m s) = 31 m 2 s 2
2 2 Pabs = 1.0 10 5 Pa + 103 kg m 3 9.8 m s2 ( 2.5 m ) +
For the correct answer with units Pabs = 1.4 10 5 Pa ( )( ) 1 103 kg m3 31 m 2 s2 2 ( )( )
1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 4 (continued)
Distribution of points (c) 4 points For correct use of kinematics or conservation of energy or Torricelli's theorem to find the exit speed, with correct substitution of values 2 u2 = 2gh
u2 = 2 gh = 2 9.8 m s2 ( 4.0 m ) 1 point ( ) For the correct answer for the exit speed u2 = 8.9 m s For correct use of the equation of continuity to find rnew , with correct substitution of values Au = Anewunew
2 pr 2u = prnewunew
2 rnew = r 2 1 point 1 point u unew For the correct answer with units, consistent with the new exit speed found above r2 = 1.2 10 2 m = ( 0.015 m ) 2 (6.0 m s) (8.9 m s) 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 5
10 points total (a) 3 points Example: Distribution of points For each correct ray from the object used to locate the image, 1 point was awarded, to a maximum of 2 points For the correct size and orientation of the image 2 points 1 point (b) (i) 1 point For an indication that the image is virtual (ii) 1 point 1 point 1 point For a correct justification Example: The rays emerging from the lens did not actually converge at the image but only appear to have done so. This point was only awarded if the point for part (b)(i) was awarded. (c) 2 points For correct use of the equation relating image distance to object distance and focal length, with correct substitutions 1 1 1 + = si so f 1 point si = 6.0 cm  10.0 cm For the correct answer with the correct sign si = 15 cm si = (10.0 cm )(6.0 cm) fso so  f 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 5 (continued)
Distribution of points (d) 3 points Example: For an indication that the height of the new image is larger For a correct justification (only when the previous point has been awarded) Justification approach 1: The rays passing through the lens become less diverging from each other due to the geometry. Their extensions meet further away from the lens making the image larger. Justification approach 2: Construct a new ray diagram For each correct ray from the object used to locate the image, 1 point was awarded, to a maximum of 2 points. The student must explicitly refer to the diagram as the justification for the larger height in order for the diagram to be considered. Justification approach 3: Calculate the change in magnification For a correct calculation of the image location after the object is moved 1 1 1 + = si so f 1 point 2 points 2 points 2 points 1 point fso (10.0 cm)(9.0 cm ) = = 90 cm 9.0 cm  10.0 cm so  f For a correct calculation of the magnification both prior to moving the object and after the object is moved, leading to the conclusion that the new image is larger s 15 cm M1 =  i = = +2.5 6 cm so si = M2 =  1 point si 90 cm == +10 9 cm so The magnification increases so the height of the new image is larger. 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES (Form B)
Question 6
10 points total (a) 4 points Apply the ideal gas law to pairs of points, recognizing that nR is the same at all three points. For example, comparing points 1 and 2 and points 1 and 3 For correct use of the ideal gas law at points 1 and 2, with correct substation of values PV1 PV 1 = 2 2 , with V1 = V2 T1 T2 Distribution of points 1 point (300 K ) 500 103 Pa T1P2 T2 = = P 100 103 Pa 1 For the correct temperature at point 2 T2 = 1500 K For correct use of the ideal gas law at points 1 and 3, with correct substitution of values PV PV1 1 = 3 3 , with P = P3 1 T1 T3 (300 K ) 6.0 10 4 m3 T1V3 T3 = = V1 1.0 10 4 m3 For the correct temperature at point 3 T3 = 1800 K
(b) 2 points Approach 1: Calculate the area enclosed by the triangular path For recognition that the magnitude of the work done on the gas in one cycle is equal to the area enclosed by the triangular path 1 Wtot = (V3  V1 ) ( P2  P ) 1 2 For the correct answer 1 Wtot = (6.0  1.0 ) 10 4 m3 ( 500  100 ) 103 Pa 2 Wtot = 100 J ( ) 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 B 2008 SCORING GUIDELINES (Form B)
Question 6 (continued)
Distribution of points (b) (continued) Approach 2: Calculate the work done on the gas in each of the three processes and take the sum For recognition that the work done on the gas is equal to the sum of the work done in each of the three processes Wtot = W1 2 + W2 3 + W31 For a correct computation of the numerical value W1 2 =  P (V2  V1 ) , but V2 = V1 , so W1 2 = 0 1 1 point 1 point ( 500 + 100 ) 103 Pa (6.0  1.0 ) 10 4 m3 = 150 J W23 =  2 W31 =  P3 (V1  V3 ) =  100 103 Pa (1.0  6.0) 10 4 m3 = +50 J W31 = +50 J ( ) Wtot = W1 2 + W2 3 + W31 = 0  150 J + 50 J = 100 J Note: The minus sign was not necessary since the question asks only for the amount of work; the sign is asked for in part (c). (c) 1 point For indicating that the work done on the gas in one complete cycle is negative 1 point (d) 3 points For an application of the first law of thermodynamics to process 1 2 , recognizing that the work done from point 1 to point 2 is zero DU = Q + W Q = DU  W Q = DU For a correct expression for Q from point 1 to point 2, with correct substitution of values 3 3 Q = nRDT = ( 0.0040 mol )(8.31 J moliK )(1500 K  300 K ) 2 2 For the correct answer Q = 60 J 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 B 2008 SCORING GUIDELINES (Form B)
Question 7
10 points total (a) 3 points For a correct equation relating energy and wavelength hc Eg = hf = l For correct substitutions (and conversions if necessary) 6.63 10 34 Jis 3.00 108 m s hc OR = l = Eg 1.02 10 6 eV 1.60 10 19 J eV 1 point Distribution of points ( ( l= 1.24 103 eVi nm 10 9 m nm hc = Eg 1.02 106 eV ( )( )( )( ) ) 1 point )
1 point For the correct answer with units l = 1.22 10 12 m (b) 2 points For correct use of the equation relating wavelength and momentum (or energy and momentum) and correct substitution of values h l= (or E = pc ) pg
h 6.63 10 34 J is OR pg = = l 1.22 10 12 m For the correct answer with units 1 point 1.02 10 6 eV 1.6 10 19 J eV E p= = c 3.0 108 m/s ( )( )
1 point pg = 5.43 10
(c) 3 points 22 kgi m s For an indication that momentum is conserved pg = pnuc For a correct expression for the momentum of the nucleus pnuc = mnucunuc For either a correct substitution from part (b) or the correct answer pg p 5.43 10 22 kgim s unuc = nuc = = = 1.21 10 4 m s mnuc mnuc 4.48 10 26 kg 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 B 2008 SCORING GUIDELINES (Form B)
Question 7 (continued)
Distribution of points (d) 2 points For correct use of the expression for kinetic energy 1 2 K nuc = munuc 2 For a correct substitution from part (c) 2 1 4.48 10 26 kg 1.21 10 4 m s = 3.28 10 18 J K nuc = 2 1 point ( )( ) 1 point 2008 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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