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AP® Physics C: Mechanics 2007 Scoring Guidelines

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 AP ® PHYSICS C: MECHANICS

2007 SCORING GUIDELINES General General Notes About 2007 2007 AP Physics Scoring Guidelines 1.

The solutions contain the most common method of solving the free-response questions and the allocation of points points for this solution. Some also contain a common alternate solution. solution. Other methods of solution also receive appropriate credit for correct work.

2. Generally, double penalty for for errors is avoided. For example, if an incorrect answer to 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 concepts normally receive credit. For example, if use of the equation equation expressing a  particular concept is worth 1 point, point, and a student’s solution solution contains the the application of that equation equation to the problem but the student does not write the basic equation, the point is still awarded. 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 such as those given on the AP Physics Physics exam equation sheet. sheet. See pages 21–22 of the  AP Physics Course Description  for a description of the use of such terms as “derive” and “calculate” on the exams, and what is expected for each. 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 usually show numerical answers using both values when they are significantly different. 5.

Strict rules regarding significant digits 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 the published answer due to to differences in rounding throughout the question question typically receive full credit. Exceptions to these guidelines guidelines usually occur when rounding makes a difference difference in obtaining a reasonable reasonable answer. For example, suppose a solution 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 determine the difference in the numbers, and some credit may be lost.

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

2

®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 1 15 points total

(a)

Distribution of points

4 points

For each of the forces shown above with arrow correctly drawn and labeled, 1 point was awarded For each incorrect or extraneous vector, such as acceleration or velocity, a point was deducted, with the minimum possible score being 0.

(b)

2 points

 F  y =

0

For the correct  y component of F 1  N + F1 sin q  - mg =

1 point

0

For the correct answer  N = mg - F 1 sin q 

(c)

4 points

1 point

3 points

 F  x  = ma For show showing ing corr correct ect expr express ession ionss for the the horiz horizont ontal al force forcess and sett setting ing them them equa equall to ma1

1 point

F1 cos q - m N  = ma1

For substituting the expression for  N from part (b) F1 cos q - m (mg - F1 si n q ) = ma1

1 point

For the correct answer F cos q  - ma1  m = 1 mg - F 1 sin q 

1 point

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 3

®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 1 (continued) Distribution of points

(d)

3 points

For a linear relationship on the u  versus t graph with positive slope and with u = 0 at t  = 0 For a parabolic relationship on the  x versus t  graph  graph that is concave upward, with  x  = 0 at t  = 0 For the two graphs being consistent with each other

(e)

1 point 1 point 1 point

3 points For indicating that  N  =  = 0 is the condition for the maximum acceleration of the block before it loses contact For indicating that the friction force is zero  f =  m N  = 0

1 point 1 point

 F  x  = Fmax cos q  = mamax F max cos q  m  F  y = Fmax sin q  - mg =

amax =

0

mg sin q  Substituting F max into the expression for amax above F max =

amax =

mg sin q 

cos q 

m

For the correct answer amax = g cot q 

1 point

 Note: Since F 1 is a variable quantity in this problem and since the initial directions included F 1 as a quantity that could be used in expressions for the answers, the expression amax =

F 1 cos q  was also acceptable for the answer point. m

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®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 2 15 points total

(a)

Distribution of points

2 points For a correct expression of the relationship among T , R, and 2 p R T 

 R

 R

=

=

=

u

u uT 

2 p 

(3.40

)(

103 m s 7.08

¥

¥

103 s

)

2 p  For the correct numerical answer  R

=

(b)

1 point

1 point

3.83 ¥ 106 m

2 points For correctly equating centripetal force and gravitational force ms u

2

 M  M 

Gms MM  

=

 R =

 R

u

2

2  R

G

For substituting the value of  R from (a) into either the original equation or the simplified expression for  M  M   above  M  M   M  M 

=

=

1 point

(3.40

¥

103 m s

6.67 ¥ 10 6.64

¥

-

2

) (3.83

11

3

m

¥

106 m

kg s i

1 point

)

2

10 23 kg

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®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 2 (continued) Distribution of points

(c)

4 points For a correct expression that equates

 E tot 

to the sum of kinetic and gravitational potential

1 point

energies  Etot 

=



+



For a negative sign on a correct expression for

1 point

U G

Gms M M  1 2 ms u  R 2 For explicit substitution of the value of  R from (a) and the value of  Etot

=

 M  M 

 from (b) in the

1 point

equation above or for correct numerical answer if worked as follows: From (b),

 M  M 

=

u

2  R

G

Gms u 2 R 1 1 2 2 ms u m u =  R G 2 2 s 2 1 3  E tot  = - (930 kg ) 3.40 ¥ 10 m s 2 For a negative sign on the final answer  Etot

(

 E tot 

(d)

msu

=

2

= -

1 2 mu 2 s

)

1 point

9

5.38 ¥ 10 J

= -

3 points For correct selection of “Less than” check space For a correct justification Example 1: From Kepler’s third law ( r decrease Example 2: Use relationships among  R, following: From (b), T 

=

2 p R u

u

2  R

3

u

2



=

1 point 2 points  decreases, then T  must  must also constant ), if r  decreases,

, and  T  T with no incorrect physics such as the

, so as R decreases,

u

 must increase. From (a),

, so both a decrease in  R and an increase in

u

 contribute to a decrease in T .

 M  M 

=

G

 Note: 1 point partial partial credit was awarded awarded for using only that

u



=

2 p R u

 , unless it was stated

 was constant, in which case no credit was awarded.

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®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 2 (continued) Distribution of points

(e)

2 points For a correct expression of conservation of angular momentum equivalent such as  I1w1 = I 2 w2  or u1r1 = u2 r 2 ms u1r1 = ms u 2r2  or   u

2

= u 1

r 1 r2

= u 1

 RC

+

RM 

RF

+

RM 

, where

approaches, respectively, and

 RC   and  RF   are  R M   is

the distances of closest and farthest

the radius of Mars

For explicit substitution of radii (not altitudes) into the equation or for the correct numerical answer 2 =

(3.40

2 =

3.34

u

u

¥

¥

3

10 m s

3.71 ¥ 10 5 m

) 4.36

¥

5

10 m

1 point

10 5 m

+

34.3

+

34.3 ¥ 10 m

¥

1 point

5

103 m s

 Alternatively, if the longer longer approach using conservation conservation of energy was was taken, 1 point was awarded for a correct statement of conservation of energy if explicitly written as Gms M M Gms M M   1 1 2 2   ms u1 msu2 =  , and 1 point was awarded awarded for the explicit r 1 r 2 2 2 substitution of radii radii (not altitudes) or for a correct numerical numerical answer.

Units point For including correct units on at least three numerical answers Significant figures point For including less than five significant digits on at least three numerical answers for which a calculation was shown

1 point

1 point

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®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 3 15 points total

(a)

Distribution of points

2 points For a correct equation using conservation of energy 1 1 2 2 mu = kx  2 2 For a correct substitution of the numerical value of k  in  in part (a) or in a subsequent part of the  problem 1 1 2 2 mu = ( 40) x  2 2

1 point

1 point

(b) and (c)

(b)

3 points For correct axis labels and units on both axes For correct linear scales on both axes For plotting at least 4 of 5 points in the correct location  Note: Full credit was was awarded if both axes axes were reversed from the graph graph shown above and everything else was correct.

1 point 1 point 1 point

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®

AP  PHYSICS C: MECHANICS 2007 SCORING GUIDELINES Question 3 (continued) Distribution of points

(c) (i)

1 point

For a reasonable best-fit straight line  Note: This point was was awarded only if the axes had linear scales. scales. (ii)

3 points

1 2 mu 2

=

1 2 kx  2

k  2  x  , so k m is the slope of the graph of u 2 versus  x 2 m For use of a slope derived from the data For using two points in the calculation of the slope that are clearly on the best-fit line (Students using data points not on the line could not receive this second point.) u

1 point

2

=

Example: Selecting the points ( 2.4 shown in the graph Slope =

m

=

( 4.8 - 1.0) m2 s2 (2.4 - 0.5) ¥ 10 2 m2

k  slope

-

=

=

¥

2.0

1 point 1 point

2 2 10 - , 4.8) and ( 0.5 ¥ 10 - , 1.0), which are on the line

¥

10 2 s-2

=

k  m

40 N m

2.0 ¥ 102 s-2 For a numerical answer in the range 0.18 kg to 0.22 kg m = 0.20 kg

1 point

(d) (i)

4 points

For use of the correct energy types ( K , U g , and U s ) in a single equation

1 point

For recognition that the difference in height is greater than h in the figure For a correct expression for U g

1 point 1 point

For substitution of U g into a correct equation

1 point

1 1 2 2 mu = kx + mg ( h + x sin   q ) 2 2  Note: Third and fourth fourth points were awarded awarded only if the first first two points were awarded. awarded. (ii)

2 points

For checking “No” For a clear justification explaining that

u

2

2

varies with both  x  and  x.  x.

1 point 1 point

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