2001 answers - AP Physics C: Mechanics 2001 Scoring...

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Unformatted text preview: AP Physics C: Mechanics 2001 Scoring Guidelines The materials included in these files are intended for non-commercial use by AP teachers for course and exam preparation; permission for any other use must be sought from the Advanced Placement Program. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein. These materials were produced by Educational Testing Service (ETS), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed of more than 3,900 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3,500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT™, the Advanced Placement Program® (AP®), and Pacesetter®. The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES General Notes about 2001 AP Physics Solutions 1. The solutions contain the most common method(s) of solving the free-response questions, and the allocation of points for these solutions. 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. 3. An 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. 4. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a particular concept is worth one point, and a solution contains the application of the equation to the problem but does not separately list the basic equation, the point is still awarded. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 2 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 1 15 points total 1. (a) 6 points Distribution of Points The average acceleration is the change in velocity divided by the time interval For correct subtraction to find the time interval ∆t = t f − ti = 0.37 − 0.33 = 0.04 s 1 point From graph: υi = 0.22 m s For getting υi in the range 0.2 < υi ≤ 0.25 m s 1 point From graph: υ f = −0.18 m s For getting υ f in the range −0.15 ≥ υ f > −0.20 m s 1 point For getting ∆υ consistent with the student’s values of υi and υ f , including subtracting in 1 point the correct direction ∆υ = υ f − υi = −0.18 m s − 0.22 m s = −0.40 m s a= ∆υ −0.4 m s = ∆t 0.04 s For correct substitution of values in the above equation a = −10 m s 1 point 2 For showing deceleration (e.g., with a minus sign) 1 point Note: There were three alternate methods for solving parts (b) and (c) that could receive full credit. Method 1. 1. (b) 3 points For any indication of the concept of finding the area under the curve in the second graph 1 point ∆p = ò F dt or ∆ p = the area under the F vs. t curve ∆p = 0.6 N • s or ∆p = 0.6 kg • m s For correct numerical value of 0.6 For correct units 1. (c) 1 point 1 point 2 points For any statement of the correct equation for the change in momentum 1 point For correct substitution of values consistent with those obtained above 1 point ∆p = m∆υ m= ∆p 0.6 N s = = 1.5 kg ∆υ 0.4 m s • Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 3 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 1 (cont.) Method 2. 1. (b) 3 points Distribution of Points Expressing the change in momentum in terms of the average force: ∆p = F ∆t For some method of using the graph to find the average force for the four non-zero intervals such as indicating that the area is equivalent to 6 boxes each with a height 1 point of 10 N, so that F = 60 4 = 15 N ∆p = (15 N )( 0.04 s ) =0.6 N • s or ∆p = 0.6 kg • m s For correct numerical value of 0.6 For correct units 1. (c) 1 point 1 point 2 points Expressing the average force in terms of the average acceleration: F = ma For correct equation ( F = ma also accepted) For correct substitution of values consistent with those obtained above m= 1 point 1 point 15 N F = = 1.5 kg a 10 m s 2 Method 3. Student solved part (c) first and went back to part (b) 1. (c) 3 points For some method of using the graph to find the average force for the four non-zero intervals such as indicating that the area is equivalent to 6 boxes each with a height of 10 N, so that F = 60 4 = 15 N For a correct expression for Newton’s second law F = ma For correct substitution of values consistent with those obtained above m= 1 point 1 point 1 point 15 N F = = 1.5 kg a 10 m s 2 Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 4 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 1 (cont.) Method 3. (continued) 1. (b) 2 points Distribution of Points ∆p = m∆υ = (1.5 kg ) ( 0.4 m s ) =0.6 N • s or ∆p = 0.6 kg • m s For correct substitution of values consistent with those obtained above For correct units 1. (d) 1 point 1 point 4 points For a correct statement of energy change 1 point ∆E = E f − Ei For a kinetic energy equation 1 point 1 E = mυ 2 2 For correct substitution of values consistent with those obtained above including the squared velocities 1 1 2 2 (1.5 kg ) ( 0.22 m s ) − (1.5 kg ) ( −0.18 m s ) 2 2 ∆E = 0.012 J 1 point ∆E = For correct units 1 point Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 5 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 2 15 points total 2. (a) i. 3 points Distribution of Points There were two methods generally used to solve this part. Method 1. F = mac GM J m F= R2 For a statement of at least one of Newton’s laws above Equating the two equations above and substituting expression for centripetal force: mac = 1 point GM J m R2 For substituting the expression for centripetal force 1 point mυ GM J m = R R2 2 For a solution for υ that follows algebraically from previous work 1 point GM J R υ= Method 2. a= GM J R2 or g= GM J R2 For statement of either of the above, which are derived from Newton’s laws For a correct statement of centripetal acceleration ac = 1 point 1 point υ2 R Equating the two expressions above for ac and solving for υ : υ= GM J R For a solution for υ that follows algebraically from previous work. Two points were awarded for an approach that started with K = 1 point 1 GM J m mυ 2 = and 2 2R solved for υ as long as there was no sign error in the equation and there were no incorrect statements regarding energy prior to the equation. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 6 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 2 (cont.) 2. (a) ii. 3 points Distribution of Points There were three methods generally used to solve this part. Method 1. d 2π R = t T υ= For an expression for υ in terms of the period T For substitution of 2π R for the length d of the orbital path Solving for T gives T = 2π R υ For correct substitution for υ from (a) i. T= 2π R 4π R = GM J GM J R 2 1 point 1 point 1 point 3 Method 2. T= 2π ω = 2π R υ For the equation for T in terms of ω For substitution of υ R for ω in the equation For correct substitution for υ from (a) i. T= 1 point 1 point 1 point 2π R 4π 2 R3 = GM J GM J R Method 3. 2 points 2 GmM J mυ 2 æ 2π ö = mω 2 R = m ç F= ÷ R= R R2 èTø 4π 2 mR GmM J = T2 R2 4π 2 R3 2 For T = GM J 1 point 4π 2 R3 Note: Direct use of T = was awarded 1 point only, if it was defined as GM J 2 Kepler’s law or Law of Orbits. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 7 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 2 (cont.) 2. (b) 3 points Distribution of Points For use of correct equation for T from (a) ii. or derivation of this relationship T= 4π R GM J 2 For correct solution for R or R 3 , numerical or symbolic, from above equation R3 = 1 point 3 GM J T 2 4π 2 1 point 1/ 3 or æ GM J T 2 ö R=ç ÷ 2 è 4π ø or R = é( 6.67 × 10−11 N • m 2 kg 2 )(1.9 × 10 27 kg )( 3.55 × 104 s ) ê ë For a correct answer 2 4π 2 ù ú û 1/ 3 1 point R = 1.59 × 108 m Note: If RJ was subtracted from R the answer point was only awarded if the difference was clearly indicated to be the height of the orbit above the surface. 2. (c) i. 3 points For stating that the orbit is an ellipse For diagram with orbit drawn completely outside the circle with point of contact only at point P and major axis along PJ. Partial credit of 1 point awarded for any path or orbit completely outside the circle. No points were awarded in any part of path or orbit was inside the circle. 1 point 2 points Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 8 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 2 (cont.) 2. (c) ii. 3 points Distribution of Points For stating that the orbit is an ellipse For diagram with orbit drawn completely inside the circle with point of contact only at point P and major axis along PJ. Partial credit of 1 point awarded for any path or orbit completely inside the circle. No points were awarded if any part of path or orbit was outside the circle. 1 point 2 points Note: Three points may also be awarded in this part for a path in which the satellite “crashes” into Jupiter only if there is specific reference to the scale of the orbit from part (b) and the given radius of Jupiter. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 9 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 3 15 points total 3. (a) 3 points Distribution of Points 1 point For a correct formula for the rotational inertia I = å mr 2 For a sum containing a term of the form mL2 (may include extra incorrect terms, but the point was not awarded if the expression does not contain an mL2 term) I = mL2 + mL2 For the correct answer 1 point 1 point I = 2 mL 2 3. (b) 6 points For a correct expression of Newton’s 2nd law 1 point For correct substitutions into Newton’s law 1 point For a correct formula for torque τ = Iα or Tr Iα = Tr Iα 1 point F = ma 4mg − T = 4ma T= r From Newton’s 2nd law equation above: T = 4mg − 4ma Substituting into the torque equation: Iα = 4mg − 4ma r For substituting the expression for I from part (a) into Newton’s law 2mL2α 1 point = 4mg − 4ma r For the expression α = a r Substituting this expression into the previous equation: 1 point 2mL2 a = 4mg − 4ma r2 For the correct answer a= 1 point 2 2 gr L + 2r 2 2 Note: For the solution a = 4mg − T , obtained by solving 4mg − T = 4ma for a directly, 4m a maximum of 3 points was awarded for part (b) as follows; 1 point for Newton’s law, 1 point for substitutions, and 1 point for answer. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 10 AP® PHYSICS C: MECHANICS 2001 SCORING GUIDELINES Question 3 (cont.) 3. (c) 3 points Distribution of Points For correctly checking the space in front of “Equal to 4mgD” For correct justification, such as “The kinetic energy gained by the two smaller blocks comes from the decrease in the potential energy of the 4m block.” OR “Total energy is conserved.” 1 point 2 points Note: No points awarded for part (c) if wrong box was checked. 3. (d) 3 points For correctly checking the space in front of “Less” For correct justification, such as “The small blocks rise and gain potential energy. The total energy available is still 4mgD. Therefore the kinetic energy must be less than in part (c).” 1 point 2 points Note: No points awarded for part (d) if wrong box was checked. Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 11 ...
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This note was uploaded on 08/05/2009 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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