2009BAP 2 Solutions

2009BAP 2 Solutions - AP Physics B 2009 Scoring Guidelines...

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Unformatted text preview: AP Physics B 2009 Scoring Guidelines Form B 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 association is composed of more than 5,600 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 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. 2009 The College Board. College Board, Advanced Placement Program, AP, AP Central, 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. 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 AP teachers: apcentral.collegeboard.com. AP PHYSICS 2009 SCORING GUIDELINES General Notes About 2009 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 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 (Form B) Question 1 15 points total (a) 4 points For any indication that the centripetal force on the small disk equals the weight of the hanging objects For using the correct expression for the centripetal acceleration m1u 2 = m2 g r For the correct expression for the speed of the disk in terms of the period u = 2 pr P m1 2 pr 2 = m2 g r P For a correct expression relating m2 and P in terms of the specified quantities Distribution of points 1 point 1 point 1 point ( ) 1 point 4 p m1r P2 P = 2p 2 = m2 g m1r m2 g (b) 2 points For listing two quantities, one of which is correct, consistent with equation obtained in part (a) For having both quantities correct, consistent with equation obtained in part (a) For example: 1 P 2 and m2 , P 2 and 1 m2 , P and 1 m2 1 point 1 point (c) 4 points Using the example of 1 P 2 and m2 m2 (kg) 0.020 1.40 0.51 0.040 1.05 0.91 0.060 0.80 1.6 0.080 0.75 1.8 P (s) 1 P2 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES (Form B) Question 1 (continued) Distribution of points (c) (continued) For correctly labeling both axes For correctly scaling both axes For a reasonably correct plotting of the data For a reasonably correct best-fit line 1 point 1 point 1 point 1 point (d) 5 points For associating the slope of the line with the correct coefficient for the quantities graphed m2 g g 1 , so slope = For the example graph given, 2 = 2 2 P 4 p m1r 4 p m1r For a correct method of calculating the slope, including substitution of points on the line For a correct substitution of given values (i.e., all values except the points from the line) For example: 1 point 1 point 1 point g = (slope) 4p 2 m1r = ( ) (0.084 kg - 0.016 kg) ( (2 s-2 - 0.4 s-2 ) 4 p 2 ( 0.012 kg)( 0.80 m) 1 point 1 point ) For a reasonable numerical value For correct units g = 8.9 m s 2 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES (Form B) Question 2 15 points total (a) 6 points The solution starts with the expression for the force between point charges. kq q F = 12 2 r For having equal contributions to the force from each charge on the x axis (e.g., by including a factor of 2 in the calculation) For using the correct distance between charges 2 r 2 = y2 + xB A For including only the y component of the forces, since the x components cancel (e.g., by including sin q in the equation, where q is the angle between the x axis and the line connecting particle A and either of the other two particles) For correctly determining the angle q from the geometry For example: q = tan ( y A x B ) These elements combine to yield the following expression. kq q F = 2 2 A B2 sin tan -1 ( y A x B ) y +x -1 Distribution of points 1 point 1 point 1 point 1 point F =2 ( A B 9.0 109 Ni m 2 C2 0.20 10 -9 C 0.30 10 -9 C (0.030 m ) + (0.040 m ) 2 )( )( 2 ) sin tan-1 0.030 m ( 0.040 m ) 1 point 1 point For the correct numerical answer F = 2.59 10 -7 N For correctly indicating that the force is in the negative y direction (b) 2 points For any indication that particle A will move toward the origin after release For any indication that the particle goes past the origin, then reverses direction For example: The particle oscillates between y = 0.030 m 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 (Form B) Question 2 (continued) Distribution of points (c) 2 points For any path that (1) is an arc of a circle whose center lies on the line defining the side of the field where the particle enters and (2) goes completely through the field region For the path curving to the left 1 point 1 point (d) 2 points For the correct expression for the force on a particle moving perpendicular to a magnetic field F = qu B sin q = qu B F = 0.20 10 -9 C (6000 m s)( 0.50 T ) 1 point ( ) For the correct answer F = 6.0 10 -7 N ( ) 1 point (e) 3 points For equating the expression for the electric and magnetic forces on the particle qE = qu B E = uB E = (6000 m s)(0.50 T ) For the correct numerical answer E = 3000 N C For correctly indicating that the field is in the negative x direction 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 (Form B) Question 3 15 points total (a) 3 points For a meaningful use of the correct expression for flow rate, Au (e.g., an attempt to substitute for volume and speed) For relating the flow rates at points A and B AAu A = AB u B u A = ABu B AA = 0.50 10 -4 m 2 (8.2 m s) 1.0 10 -4 m 2 Distribution of points 1 point 1 point ( ) ( ) For the correct answer u A = 4.1 m s 1 point (b) 5 points For applying Bernoulli's equation to this situation 1 2 1 2 PA + rgy A + ru A = PB + rgyB + ru B 2 2 1 2 2 PA = PB + rg ( yB - y A ) + r uB - u A 2 1 point ( ) For using atmospheric pressure (1.0 10 5 Pa ) for the pressure at point B For correctly substituting values for y A and y B For correctly substituting values for u A and u B consistent with the work in part (a) 1 point 1 point 1 point PA = (1.0 105 Pa ) + (1000 kg m3 )(9.8 m s2 ) (0.50 m - 0) 1 2 2 + (1000 kg m3 ) ([8.2 m s] - [ 4.1 m s] ) 2 For correct units for the answer PA = 1.3 10 5 N m 2 (or Pa) 1 point (c) 2 points For correctly relating the initial speed and the maximum height 1 mu 2 = mgh u 2 = u0 2 + 2a ( x - x 0 ) OR 2 2 u B = 2 gh 1 point 2 h = u B 2 g = (8.2 m s) 2 2 9.8 m s2 ( ) 1 point For the correct answer with correct units h = 3.4 m 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES (Form B) Question 3 (continued) Distribution of points (d) 5 points For correctly analyzing the vertical motion with constant acceleration, to find the time for the water to reach maximum height u = u0 + at 1 point 0 = uBy + gt uBy = gt For the correct vertical component of speed uBy = uB sin q 1 point t = u By g = u B sin q g For correctly analyzing the horizontal motion with constant speed x = u Bx ( 2 t ) For the correct horizontal component of speed u Bx = u B cos q 2 x = u B cos q (2u B sin q g ) = 2u B cos q sin q g 1 point 1 point x = 2 (8.2 m s) cos 60 sin 60 9.8 m s 2 2 ( ) For the correct answer with units x = 5.9 m (or 5.8 m using g = 10 m s2 ) 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 (Form B) Question 4 15 points total (a) 2 points For a correct expression for the pressure due to the weight of the piston For including atmospheric pressure P = Patm + ( mg A ) Notes: The second point was also awarded for stating the assumption that the cylinder is in a vacuum. A numeric value for atmospheric pressure instead of a variable in the equation was acceptable. 1 point 1 point Distribution of points (b) 2 points For combining resistors in parallel 1 1 = Rp Ri i 1 1 1 1 3 = + + = Req R0 R0 R0 R0 1 point For the correct answer Req = R0 3 1 point (c) 3 points For correctly relating the change in internal energy of the gas to the energy provided by the circuit DU = Pcir t For a correct relationship for the power delivered by the circuit in terms of voltage difference and resistance Pcir = e 2 Req R ^ DU = e 2 0 ~ t 3 For the correct answer DU = 3e 2 t R0 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 (Form B) Question 4 (continued) Distribution of points (d) 3 points For indicating that the temperature decreases Justifying using the first law of thermodynamics: DU = Q + W For indicating that Q = 0 (the definition of an adiabatic process) For indicating that negative work is done on the piston, so applying the first law gives a net loss of energy in the gas and thus the internal energy decreases Since the internal energy is proportional to temperature, the temperature must also decrease. 1 point 1 point 1 point (e) 5 points For a vertical straight segment One point for each curved segment from one end of the straight line to a point at a lower pressure and higher volume (Only the segments were scored. Presence or correctness of arrowheads was not considered.) For correctly labeling Vi For correctly labeling V f 1 point 2 points 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 (Form B) Question 5 10 points total (a) 2 points Using the relationship for the speed of light inside a material: n = c u For correct substitutions uoil = c n = 3.00 108 m s 1.7 Distribution of points ( ) 1 point 1 point For the correct answer uoil = 1.8 108 m s (b) 2 points For using the relationship for the wavelength of light inside a material, which can be derived as follows: foil = fair , and f = u l uoil loil = uair lair 1 point (c n ) loil = c lair loil = lair n For correct substitutions loil = 5.2 10 -7 m 1.7 ( ) 1 point loil = 3.1 10 -7 m (c) 3 points There is a 180 phase shift when light is reflected at the lower interface, so for constructive interference the path length in the film must be an odd multiple of a half wavelength. For any indication of the phase shift (e.g., a description as above or a loil 2 correction in an equation such as 2t + loil 2 = loil , where t is the thickness of the film) For work that correctly accounts for constructive interference (e.g., a description as in the first statement above for the net path length, or correct factors of l in an equation as in the second statement above) 2t = loil 2 For substituting the correct value for the wavelength t = loil 4 = 3.1 10 -7 m 4 1 point 1 point ( ) 1 point t = 7.8 10 -8 m 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2009 SCORING GUIDELINES (Form B) Question 5 (continued) Distribution of points (d) 3 points For a ray at the first interface that bends toward the normal For a ray at the second interface that bends away from the normal For a ray at the third interface that bends away from the normal (This ray does not have to be parallel to the initial ray.) 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 (Form B) Question 6 10 points total (a) 3 points Distribution of points The given equation, E n = n 2 E1 , is used to calculate the other energies One point for each correctly labeled energy level 3 points (b) 2 points From the relationship between frequency and energy, DE = hf , it is seen that the smallest frequency corresponds to the smallest energy difference. For a correct determination of the energy difference DE = 4 E1 - E1 = 3 E1 For a correct relationship between the minimum energy difference and the frequency hf = 3 E1 f = 3E1 h 1 point 1 point (c) 2 points For the transition from 9E1 to E1 For the pair of transitions from 9E1 to 4E1 and 4E1 to E1 One point earned for this part was deducted for one extraneous line, and two or more extraneous lines resulted in no credit for this part. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 1 point 1 point AP PHYSICS B 2009 SCORING GUIDELINES (Form B) Question 6 (continued) Distribution of points (d) 3 points For a correct relationship between the maximum energy difference and the frequency hf = 9 E1 - E1 = 8 E1 f = 8 E1 h 1 point For a correct relationship between the wavelength and frequency of the light f = c l A single equation that directly relates energy and wavelength could earn both points. c l = 8 E1 h For the correct answer l = hc 8 E1 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.

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