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Unformatted text preview: AP® Physics C: Mechanics 2010 Scoring Guidelines The College Board
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General Notes
1. The solutions contain the most common method of solving the freeresponse questions and the allocation of points for the 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 — for example, 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 1 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 sheets. 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 Sections ⎯ Student 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. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 1
15 points total Distribution of points (a) 2 points Starting with Newton’s second law: Fnet = mg  Cu 2 = ma
For correctly indicating that at terminal velocity ( u = uT ), the net force and acceleration are zero
2 mg  CuT = 0 1 point For a correct relationship between uT and m
2 uT = 1 point g m C (b) (i) 4 points © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (b) (i) (continued) For recording in the table a row of calculated data points that involve only uT or a combination of uT and g
2 For graphing uT (or the equivalent) on the vertical axis to obtain a linear graph For including an appropriate linear scale on both axes For drawing a reasonable bestfit straight line Note: Correct graphs of uT versus m are given credit provided the quantities being plotted are clearly indicated. 1 point 1 point 1 point 1 point (ii) 3 points 1 point For a correct calculation of the slope, using points on the student’s bestfit line (not points from the data table) Using the example graph shown: (0.95  0.30 ) m 2 s2 slope = = 217 m 2 kg i s2 È( 4.4  1.4) ¥ 10 3 ˘ kg Î ˚ For a correct expression relating the slope to C g slope = 217m 2 kg i s2 = C 9.8 m s2 g C= = Slope 217 m 2 kgi s2 For correct units on C C = 0.045 kg m (0.046 kg m using g = 10 m s2 ) (c) (i) 3 points 1 point 1 point For a graph that starts at the origin with an initial positive slope For a graph that is concave down throughout 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 1 (continued)
For labeling uT and T at a point where the slope of the graph approaches zero (c) (ii) (continued) 1 point 1 point Distribution of points 1 point For a verbal statement that clearly indicates the distance Y is the area under the curve between times t = 0 and t = T Or an equivalent statement referring to the area under the curve when the u versus t graph ends at T
T Or an equivalent mathematical expression with limits: Y = Ú u ( t ) dt
0 (d) 2 points For a correct indication that the mechanical energy dissipated is the change in mechanical energy during the time the stack falls a distance y y Ê ˆ 2 DE = U final + K final  (Uinitial + Kinitial ) Á or DE = Cu dy ˜ Á ˜ Ë ¯ 0 DE = DU + DK For correct substitutions of y, m, and uT DU =  mgy 1 2 DK = muT 2 1 2 DE = muT  mgy (or any equation with an additional negative sign that has the 2 correct relative signs for potential and kinetic energies) 1 point ( ) Ú 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 2
15 points total Distribution of points (a) 3 points For each correct force for which the vector is drawn with the correct direction, label and point of application, 1 point was awarded. One earned point was deducted if any components or extraneous forces are present. (b) 5 points Starting with Newton’s second law (linear form): Fnet = Ma For expressing Fnet in terms of gravitational and frictional forces For including the correct component of the weight Mg sin q  Ff = Ma
t = Ia For correct substitution of torque into Newton’s second law (angular form) RFf = I a 3 points 1 point 1 point 1 point RFf = ( 2 5) MR 2 a
For the correct relationship between angular and linear acceleration (either explicitly stated or used in the calculation) a=aR 1 point RFf = ( 2 5) MR 2 ( a R )
Solving for Ma Ma = ( 5 2 ) F f Substituting into the linear equation above Mg sin q  Ff = ( 5 2 ) Ff Ff = ( 2 7) Mg sin q = (2 7) (6.0 kg)(9.8 m s 2 ) (sin 30∞)
For the correct value of Ff 1 point Ff = 8.4 N
Notes: Credit is awarded for solutions that use the value of u calculated in (c) to calculate acceleration and, from there, the value of the frictional force. If Mg cos q is used, the point was awarded for a value of 14.5 N. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (c) 3 points For an expression of conservation of energy For including gravitational potential energy, translational kinetic energy and rotational kinetic energy in a correct energy equation or statement 1 1 MgDh = M u 2 + I w 2 2 2 1 12 2 Mgd sin q = M u + MR 2 w 2 2 25 w=uR 1 1 7 Mgd sin q = M u 2 + M u 2 = Mu 2 2 5 10 1 point 1 point ( ) u= (10 7) gd sin q = (10 7) ( 9.8 m s2 ) ( 4.0 m )(sin 30∞) For a correct numerical answer u = 5.3 m s Alternate solution (kinematics method) For determination of the linear acceleration in terms of mass and frictional force a = 5Ff 2 M from work shown in part (b)
a = 5 (8.4 N ) 2 (6.0 kg ) = 3.5 m s2 For a correct substitution into an appropriate kinematics equation using u0 = 0
2 u 2 = u0 + 2ad = 2ad 1 point Alternate points 1 point 1 point u = 2ad = (2) (3.5 m s2 ) ( 4.0 m)
1 point For a correct numerical answer u = 5.3 m s (d) 3 points For a correct statement of conservation of momentum Miui = M f u f 1 point u f = Mi M f ui
For correctly equating ui with the horizontal component of the ball as it leaves the roof For setting M f equal to the total mass of the ball and the wagon/box
u f = Mi M f u cos q = (
( ) 1 point 1 point ) [(6.0 kg ) (18.0 kg )] ( 5.3 m s2 cos 30 ) u f = 1.5 m s
Units 1 point For correct units in at least two of the parts (b), (c) and (d) 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 3
15 points total Distribution of points (a) 4 points For a correct relationship between velocity and acceleration 1 point u = Ú a(t ) dt OR u = Ú a(t ) dt
0 t OR du =a dt
1 point For a correct substitution of the expression for acceleration into the integral relationship
u= Ú (amax sin T ) dt
pt OR u = Ú (amax sin T ) dt
t pt (0 < t < T )
1 point 0 For a correct evaluation of the integral, with an integration constant or correct limits aT aT pt t pt u =  max cos + C OR u =  max cos (0 < t < T ) p T0 p T For a correct determination of the integration constant or evaluation between the limits aT aT aT pt 1 OR u =  max cos u ( 0 ) =  max + C = 0 ﬁ C = max (0 < t < T ) p T p p aT pt u = max 1  cos (0 < t < T ) p T ( ) ( ) 1 point (b) 2 points For indicating that the work done by the net force is equal to the change in kinetic energy 1 W = m u 2  ui2 f 2 For a correct substitution of velocity from (a) into the workenergy expression aT 2a T u f = uT = max (1  cosp ) = max p p amaxT ui = u0 = (1  cos 0 ) = 0 p 1 point ( ) 1 point W=
W= 2 1 Ê 2amaxT ˆ mÁ ˜ 2Ë p¯
2 2 mamaxT 2 p2 Alternate solution (integral form) W = F ∑ dx Alternate points 1 point Ú For a correct substitution of the expression for force into the integral πt W = ∫ mamax sin dx T For a correct expression for dx in terms of time 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (b) (continued)
T W= W= Ú mamax sin T
0 T 0 p t Ê amaxT pt ˆ Á p 1  cos T ˜ dt Ë ¯ pt  sin ( ) 2 mamaxT p Ú (sin T
pt ( pt pt cos T T )) dt ma 2 T W = max p W=
2 mamaxT 2 T Ú (sin T
0 1 2 pt  sin dt 2 T ) W=
(c) p2 2 2 mamaxT 2 p2 (  cos pt 1 2 pt  cos T T 4 ) T 0 1 point Starting with Newton’s second law: Fnet = Frope  mg sin q = ma At terminal velocity, the net force and acceleration are zero: Frope  mg sin q = 0 For a correct expression for the force Frope = mg sin q 1 point (d) 2 points J = Ú F dt
For a correct substitution of force into the impulsetime relationship 1 point J = mamax Ú sin
0 T pt dt T T mamaxT pt  cos p T0 For evaluation at the limits of integration ma T J = max [  cos p + cos 0 ] p 2 mamaxT J= p Alternate solution (impulsemomentum) J= ( ) 1 point Alternate points J = Dp = muT © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: MECHANICS 2010 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (d) (continued) For a correct substitution of the velocity mamaxT pt J= 1  cos T p For setting t = T mamaxT J= (1  cos p ) p 2 mamaxT J= p ( ) 1 point 1 point (e) 6 points
F1 = mg sin q + mamax sin ( pTt )
2T (0 < t < T ) F2 = mg sin q + mamax e  pt For a graph labeled F1 : for starting at mg sin q for half a sine wave with a maximum at ∼ T 2 for returning to original starting point at t = T for a horizontal line at the original starting point for t > T For a graph labeled F2 : for starting on the vertical axis at a point above the starting point of F1 (if there is no F1 graph, this point was awarded if the F2 graph starts above mg sin q ) for an exponential decay graph 1 point 1 point 1 point 1 point 1 point 1 point © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. ...
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This note was uploaded on 05/05/2011 for the course CHEM 504 taught by Professor John during the Fall '11 term at American College of Computer & Information Sciences.
 Fall '11
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