2003 answers - AP® Physics C: Mechanics 2003 Scoring...

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Unformatted text preview: AP® Physics C: Mechanics 2003 Scoring Guidelines The materials included in these files are intended for 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 noncommercial, face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained herein. This material may not be mass distributed, 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. These materials were produced by Educational Testing Service® (ETS®), which develops and administers the examinations of the Advanced Placement Program for the College Board. 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AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 1 15 points total (a) Distribution of points 3 points For indicating speed as the time derivative of position dx K= dt For taking the correct derivative K = 1.5t 2 + 2 For finding the correct initial speed at t = 0 K0 = 2 m/s (b) 1 point 1 point 1 point 6 points i. (1 point) 1 K = mK 2 2 For correctly substituting for the mass and the expression for K found in (a) 1 K = (100)(1.5t 2 + 2)2 = 50(1.5t 2 + 2) 2 2 ii. (3 points) Fnet = ma For indicating acceleration as the time derivative of the velocity dK a= dt For taking the correct derivative a = 3t For the correct expression Fnet = (100)(3t ) = 300t Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 2 1 point 1 point 1 point 1 point AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 1 (cont’d.) Distribution of points iii. (2 points) For the correct equation relating power to force and velocity P = FK For substituting expressions for F and K found in previous parts P = (300t )(1.5t 2 + 2) = 450t 3 + 600t Alternate method points For indicating power as the time derivative of kinetic energy dK P= dt For substituting the expression for kinetic energy from (b)i. d P= 112.5t 4 + 300t 2 + 200 = 450t 3 + 600t dt ( (c) 1 point 1 point Alternate 1 point 1 point ) 4 points For a statement that the work done on the box is equal to the change in its kinetic energy W = ∆K For finding K at 2 seconds K = (1.5)(2)2 + 2 = 8 m/s Substituting, using the value of υ0 from part (a): 1 2 2 W = (100 kg) ( 8 m/s ) − ( 2 m/s ) 2 For the correct answer with correct unit W = 3000 J ( 2 point 1 point ) Alternate method points For a statement that the work done by the box is equal to the integral of power over time W = ò Pdt For substituting the expression for power found in (b)iii 1 point Alternate 1 point 2 points 2 W = ò (450t 3 + 600t ) dt 0 450 4 2 600 2 2 ( t ) 0 + 2 (t ) 0 4 W = 1800 J + 1200 J For the correct answer with unit W = 3000 J W= 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 3 AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 1 (cont’d.) Distribution of points (d) 2 points For checking the box that the work done by the student is greater than in part (c) For a reasonable justification recognizing that the student had to perform work against friction, such as Wstudent = ∆KE + W friction Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 4 1 point 1 point AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 2 15 points total (a) Distribution of points 2 points For a statement of conservation of energy 1 MgH = M Kc 2 2 For the correct answer Kc = 2 g H 1 point 1 point Alternate solution points For use of correct kinematics equation Kc 2 = K0 2 + 2 gH , 1 OR the combination of a = g ; Kc = gt ; and H = gt 2 , 2 For the correct answer Kc = 2 g H (b) Alternate 1 point 1 point 3 points For recognition that momentum is conserved in the inelastic collision For use of the correct equation expressing conservation of momentum M Kc = 2 M K p For the correct answer 1 Kp = 2 gH 2 (c) 1 point 1 point 1 point 4 points For use of the correct equation for the period of a mass on a spring m T = 2π k For recognition that m = 2M For correct calculation of k using the force equation for the initial stretching of the spring Mg Mg = kD , giving k = D For the correct answer after substituting for m and k 2D T = 2π g Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 1 point 1 point 1 point 1 point AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 2 (cont’d.) Distribution of points (d) 3 points For recognition that the speed K is a maximum at the equilibrium point, which can be correctly described by one or more of the following statements: equilibrium point, F = 0, a = 0, kinetic energy is a maximum, midpoint of the oscillation, etc. For recognition that there is a new equilibrium point given by the following equation kx = 2 Mg , where x is the distance the spring is stretched from its initial unstretched length Substituting the value of k found in part (c) æ Mg ö ç ÷ x = 2Mg èDø For the correct answer x = 2D 1 point 1 point 1 point For a correct answer, x = 2 D , but with no justification, only 1 point was awarded (e) 3 points For a check in the “Less than” space in part (c) For a correct justification that states that in the second case there is less mass oscillating than m . in part (c) and that the period decreases with decreasing mass, T = 2π k (The formula is not necessary for full credit.) Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 1 point 2 points AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 3 15 points total (a) Distribution of points 2 points i. (1 point) For a smooth concave downward curve in the region between the points, that passes near all the points. ii. (1 point) For a reasonable interpolation of the value of x when M = 250 kg, based on the graph that was drawn For example, using the above graph x ≈ 33 m (b) 1 point 1 point 10 points i. (2 points) For using correct kinematic equation(s) For example, the equation y = (1/ 2) gt 2 can be used to directly solve for the time 1 point t = 2 y g = 2 (15 m ) 9.8 m s 2 For the correct answer t = 1.75 s (or 1.73 s using g = 10 m s 2 ) 1 point ii. (3 points) For determining the potential energy of both the load in the counterweight bucket and the projectile For the correct value of the potential energy of the bucket load U b = mgh = M ( 9.8 ) 3 = 29.4 M For the correct value of the potential energy of the projectile U p = mgh = (10 )( 9.8) 3 = 294 U init = U b + U p = 29.4 M + 294 (or 30M + 300 using g = 10 m s 2 ) Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 7 1 point 1 point 1 point AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 3 (cont’d.) Distribution of points (b) (continued) iii. (5 points) For a valid statement or equation indicating conservation of energy U init = U final + K 1 point For the correct final potential energy of the bucket load For the correct final potential energy of the projectile U final = M ( 9.8 )(1) + (10 )( 9.8 )(15 ) = 9.8M + 1470 1 point 1 point For having terms for the final kinetic energy of both the bucket load and the projectile 2 2 K p = (1/ 2) 10K x and K b = (1/ 2) M Kb OR K p = (1/ 2)(1440) M 2 and K b = (1/ 2)( 4 M ) M 2 1 point For using one of the following relationships to write all expressions in terms of Kx 1 point Kb = (1/ 6) K x OR M = Kx / 12 Substituting into the conservation of energy equation above and solving for Kx : 2 2 29.4M + 294 = 9.8M + 1470 + 5Kx + ( M 72) Kx Kx = (19.6M − 1176) ( 5 + ( M (or 72) ) ( 20M − 1200) (5 + ( M 72) ) using g = 10 m s 2 ) Alternate solution For an application of the equation for torque å J = I = = I ( d M dt ) For determining the torque applied by the bucket load J b = Fr = Mgr sin G = 19.6 M sin G For determining the torque applied by the projectile J p = 10 ( 9.8) sin G (12) = 2940sin G For determining the total rotational inertia I = 10(12) 2 + M (2) 2 = 1440 + 4 M For using the proper relationship to change from a rotational to a linear solution M = (1/12) Kx Substituting: 19.6M sin G − 2940sin G = (1 12) (1440 + 4M ) ( d Kx dt ) This equation can then be solved to obtain the expression for Kx Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8 Alternate points 1 point 1 point 1 point 1 point 1 point AP PHYSICS C: MECHANICS 2003 SCORING GUIDELINES Question 3 (cont’d.) Distribution of points (c) 3 points i. (1 point) Using the given relationship for x x = Kx t For substituting the answers answer from parts (b) iii. and (b) i. 1 point x = 1.75 (19.6M − 1176 ) ( 5 + ( M 72 ) ) (or 1.73 ( 20M − 1200 ) ( 5 + ( M 72 ) ) using g = 10 m s 2 ) ii. (2 points) For using the equation from part (c) i. to predict xtheor 1 point xtheo = 1.75 (19.6 ( 300 ) − 1176 ) ( 5 + ( 300 72 ) ) = 39.7 m (or 40.0 m using g = 10 m s 2 ) For a reasonable explanation for the fact that xexp < xtheor Examples: friction at the pivot, air resistance, neglected masses are not really negligible One point was awarded if no equation is available from (c) i. to make a theoretical prediction but the student developed the reasonable explanation for xexp < xtheor One point was awarded for a reasonable explanation if evaluation of equation in part (c) resulted in xexp > xtheor . Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 9 1 point ...
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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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