ap physics c: mechanics 2004 free-response …® physics c: mechanics 2004 free-response questions...

AP Physics C: Mechanics2004 Free-Response Questions

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 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

For further information, visit www.collegeboard.com

Copyright 2004 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central,

AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the

College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.

Other products and services may be trademarks of their respective owners.

For the College Boards online home for AP professionals, visit AP Central at apcentral.collegeboard.com.

The materials included in these files are intended for noncommercial 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. This permission does not apply to any

third-party copyrights contained herein. These materials and any copies made of them may not be resold, and the copyright notices

must be retained as they appear here.

2

TABLE OF INFORMATION FOR 2004 and 2005

CONSTANTS AND CONVERSION FACTORS

1 unified atomic mass unit, 1 1 66 1027

u kg

931 MeV/2

=

=

-

.

c

Proton mass, mp = 1 67 10 27. kg

Neutron mass, mn = 1 67 10 27. kg

Electron mass, me = 9 11 10 31. kg

Magnitude of the electron charge, e = 1 60 10 19. C

Avogadros number, N023 16 02 10= . mol

Universal gas constant, R = 8 31. / ( )J mol K

Boltzmanns constant, kB = 138 10 23. J / K

Speed of light, c = 3 00 108. /m s

Plancks constant,

h

hc

=

=

=

=

6 63 10

414 10

1 99 10

1 24 10

34

15

25

3

.

.

.

.

J s

eV s

J m

eV nm

Vacuum permittivity, 012 2 28 85 10= . /C N m

Coulombs law constant, k = = 1 4 9 0 100 9 2 2/ . / N m C Vacuum permeability, 0

74 10= ( ) /T m A Magnetic constant, k' = T m A0 / ( ) /4 10

7= Universal gravitational constant, G =

-

6 67 1011 3 2

. /m kg s

Acceleration due to gravity at the Earths surface, g = 9 8 2. m/ s

1 atmosphere pressure,

1 1 0 10

1 0 10

5 2

5

atm N / m

Pa

=

=

.

.

1 electron volt, 1 1 60 10 19 eV J= .

UNITS

Name Symbol meter m

kilogram kg

second s

ampere A

kelvin K

mole mol

hertz Hz

newton N

pascal Pa

joule J

watt W

coulomb C

volt V

ohm

henry H

farad F

tesla T

degree Celsius C

electron- volt eV

PREFIXES

Factor Prefix Symbol 109 giga G

106 mega M

103 kilo k

10 2 centi c

10 3 milli m

10 6 micro

10 9 nano n

10 12 pico p

VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES

sin cos tan

0 0 1 0

30 1/2 3 2/ 3 3/

37 3/5 4/5 3/4

45 2 2/ 2 2/ 1

53 4/5 3/5 4/3

60 3 2/ 1/2 3

90 1 0

The following conventions are used in this examination. I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial. II. The direction of any electric current is the direction of flow of positive charge (conventional current). III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge.

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3

ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2004 and 2005

MECHANICS ELECTRICITY AND MAGNETISM

( )

2

2 2

2

22

2 2

2

2

0

0 0

0 0

0

0 0

12

2

12

12

12

12

fric

net

g

c

net

cm

s

s

at

x x t at

a x x

m

ddt

dt

m

F N

W d

K m

dWP

dtP

U mgh

a rr

I

I r dm mr

m m

r

I

K I

t

t t

k

U

u u

u

u u

m

u

uw

u w

w

w w a

q q w a

D

D

= +

= + +

= + -

= =

=

= =

=

=

=

=

=

=

= =

=

= =

= =

=

=

= =

=

= +

= + +

= -

=

F F a

pF

J F p

p v

F r

F v

r F

r r

L r p

F x

t

t t a

w

2

1 22

1 2

2 1

2

2

G

G

s

p

kx

Tf

mT

k

Tg

Gm m

rGm m

Ur

p

w

p

p

= =

=

=

= -

= -

F r

a = acceleration F = force f = frequency h = height I = rotational inertia J = impulse K = kinetic energy k = spring constant = length L = angular momentum m = mass N = normal force P = power p = momentum r = radius or distance r = position vector T = period t = time U = potential energy u = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque w = angular speed a = angular acceleration

Fq q

r

q

dQ

EdV

dr

Vq

r

U qVq q

r

CQ

V

CA

d

C C

C C

IdQ

dt

U QV CV

RA

V IR

R R

R R

P IV

q

d I

I d

B nI

d

d

dt

LdI

i

ii

E

i

i

M

p ii

s ii

c

s i

p i

s

m

m

=

=

=

=

=

= =

=

=

=

=

=

= =

=

==

=

==

=

= z

=

= z

=

=

1

4

1

4

1

4

1 1

1

2

1

2

1 1

0

1 2

0

0

0

1 2

0

2

0

0

2p

p

p

k

r

m

m

f

fe

e

EF

E A

F v B

B

F B

B A

dt

U LIL =1

22

A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current L = inductance = length n = number of loops of wire

per unit length P = power Q = charge q = point charge R = resistance r = distance t = time U = potential or stored energy V = electric potential u = velocity or speed r = resistivity fm = magnetic flux k = dielectric constant

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4


GEOMETRY AND TRIGONOMETRY

CALCULUS

Rectangle

Triangle

Circle

Parallelepiped

Cylinder

Sphere

Right Triangle

A bh

A bh

A rC r

V wh

V rS r r

V r

S r

a b cac

bc

ab

=

=

==

=

== +

=

=

+ =

=

=

=

12

2

2 2

434

2

2

2

3

2

2 2 2

p

p

p

p p

p

p

sin

cos

tan

q

q

q

A = area C = circumference V = volume S = surface area b = base h = height = length w = width r = radius

c a

b90q

( )

( )

( )

( )

( )

1

1

11n

sin cos

cos sin

1 , 11

ln

cos sin

sin cos

n n

x x

n n

x x

d f d f dudx du dxd

x nxdxd

e edxd

xdx xd

x xdxd

x xdx

x dx x nn

e dx e

dxx

x

xdx x

xdx x

-

+

=

=

=

=

=

= -

= -+

=

=

=

= -

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2004 AP PHYSICS C: MECHANICS FREE-RESPONSE QUESTIONS

Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

GO ON TO THE NEXT PAGE.

5

PHYSICS C Section II, MECHANICS

Time45 minutes 3 Questions

Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the booklet in the spaces provided after each part, NOT in this green insert.

Mech. 1.

A rope of length L is attached to a support at point C. A person of mass m1 sits on a ledge at position A holding the other end of the rope so that it is horizontal and taut, as shown above. The person then drops off the ledge and swings down on the rope toward position B on a lower ledge where an object of mass m2 is at rest. At position B the person grabs hold of the object and simultaneously lets go of the rope. The person and object then land together in the lake at point D, which is a vertical distance L below position B. Air resistance and the mass of the rope are negligible. Derive expressions for each of the following in terms of m1, m2, L, and g.

(a) The speed of the person just before the collision with the object

(b) The tension in the rope just before the collision with the object

(c) The speed of the person and object just after the collision

(d) The ratio of the kinetic energy of the person-object system before the collision to the kinetic energy after the collision

(e) The total horizontal displacement x of the person from position A until the person and object land in the water at point D.

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6

Mech. 2.

A solid disk of unknown mass and known radius R is used as a pulley in a lab experiment, as shown above. A small block of mass m is attached to a string, the other end of which is attached to the pulley and wrapped around it several times. The block of mass m is released from rest and takes a time t to fall the distance D to the floor.

(a) Calculate the linear acceleration a of the falling block in terms of the given quantities.

(b) The time t is measured for various heights D and the data are recorded in the following table.

D (m) t (s)

0.5 0.68

1 1.02

1.5 1.19

2 1.38

i. What quantities should be graphed in order to best determine the acceleration of the block? Explain

your reasoning.

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7

ii. On the grid below, plot the quantities determined in (b)i., label the axes, and draw the best-fit line to the data.

iii. Use your graph to calculate the magnitude of the acceleration.

(c) Calculate the rotational inertia of the pulley in terms of m, R, a, and fundamental constants.

(d) The value of acceleration found in (b)iii, along with numerical values for the given quantities and your answer to (c), can be used to determine the rotational inertia of the pulley. The pulley is removed from its support and its rotational inertia is found to be greater than this value. Give one explanation for this discrepancy.

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8

Mech. 3.

A uniform rod of mass M and length L is attached to a pivot of negligible friction as shown above. The pivot is

located at a distance 3L from the left end of the rod. Express all answers in terms of the given quantities and

fundamental constants.

(a) Calculate the rotational inertia of the rod about the pivot.

(b) The rod is then released from rest from the horizontal position shown above. Calculate the linear speed of the bottom end of the rod when the rod passes through the vertical.

(c) The rod is brought to rest in the vertical position shown above and hangs freely. It is then displaced slightly from this position. Calculate the period of oscillation as it swings.

END OF SECTION II, MECHANICS

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AP Physics C: Mechanics2004 Scoring Guidelines

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 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

For further information, visit www.collegeboard.com

Copyright 2004 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central,

AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the

College Entrance Examination Board and National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.

Other products and services may be trademarks of their respective owners.

For the College Boards online home for AP professionals, visit AP Central at apcentral.collegeboard.com.

The materials included in these files are intended for noncommercial 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. This permission does not apply to any

third-party copyrights contained herein. These materials and any copies made of them may not be resold, and the copyright notices

must be retained as they appear here.

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AP PHYSICS C MECHANICS 2004 SCORING GUIDELINES

General Notes about 2004 AP Physics Scoring Guidelines

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. 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 students solution contains the application of that equation to the problem but the student does not write the basic equation, the point is still awarded.

4. The scoring guidelines typically show numerical results using the value 29.8 m sg = , but use of

210 m s is of course also acceptable. 5. Numerical answers that differ from the published answer due to differences in rounding throughout the

question typically receive full credit. The exception is usually 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.


2

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Question 1

15 points total Distribution of points (a) 2 points

For any statement of conservation of energy 1 point Taking the zero of potential to be at the height of point B, and setting the kinetic energy at point B equal to the potential energy at point A:

21 1

12 Bm mu = gL

For the correct answer 1 point 2B gLu =

(b) 4 points

For any indication that there are two forces acting on the person 1 point For an indication that the acceleration of the person is centripetal, i.e. equal to ru2 or L2u

1 point

For a correct application of Newtons second law that includes the two forces (tension T and weight) and a non-zero acceleration

1 point

21

1BmT m g r

u- =

21

1BmT mr

u= + g

For substitution of the expression for Bu from part (a) and L for the radius 1 point ( )1

1 12

2m gL

T m g m gL= + = + 1m g

g

after

13T m= (c) 3 points

For any statement of conservation of momentum 1 point For a correct application of conservation of momentum, with a clear indication that both masses have a common final velocity

1 point

( )1 1 2Bm m mu u= +

( )1

after1 2

Bm

m mu u=

+

For an answer in terms of the required quantities and of the form:

( ) ( )1

1 2answer from part(a)

mm m+

1 point

( )1

after1 2

2m

gLm m

u =+


3

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Question 1 (continued)

Distribution of points (d) 2 points

For correct expressions for the kinetic energy before and after the collision, using the answers to parts (a) and (c)

1 point

( )2before 1 1 11 1 22 2BK m m gL mu= = = gL

( ) ( )( ) ( )

2 22 1 1

after 1 2 after 1 2 21 21 2

1 1 22 2m m

K m m m m gLm mm m

u= + = + =++

gL

For constructing the ratio before afterK K from valid expressions for kinetic energy, in terms of the required quantities. The ratio does not need to be simplified, but if it is the algebra needs to be correct.

1 point

( )

121

1 2

b

a

K m gLK m gL

m m

= +

( )1 21

b

a

K m mK m

+=

(e) 4 points

For a correct expression relating the distance fallen, L, to the time it takes to fall from point B to the water:

1 point

212

L g= t

For indicating that the horizontal displacement from B to D is the answer to part (c) multiplied by the time

1 point

afterBDx tu= For correctly solving the first equation for t and substituting two quantities into the second equation (this must yield an expression in terms of the required given quantities)

1 point

2t L= g

1 1after

1 2 1 2

22 2BD

m mx t gL L gm m m mu

= = = + +L

For indicating that the total horizontal displacement from A to D is BDx plus L 1 point

1tot

1 2

2BD

m Lx x L m m= + = ++ L

( )1 2tot

1 2

3m m Lx m m

+= +


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Question 2


Using the kinematic equation: 2

012x t au= + t

For applying this equation with x D= and u 0 0= 1 point 21

2D at=

For the correct answer 1 point

22Dat

=

(b) i. 2 points

For indicating a correct pair of quantities to graph 1 point For a correct explanation 1 point Method 1: Graph quantities such as D versus t or u versus t. Such graphs are linear and the slope will yield the acceleration.

2

Method 2: Graph quantities such as a versus t or a versus x, using the equation from part (a) to determine the accelerations. Such graphs are linear and the a intercept will yield the acceleration.

Method 3: Graph D versus t, which is parabolic. The equation that describes the best fit curve is quadratic, and the second derivative is the acceleration.


5

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Distribution of points (b) (continued)

ii. 3 points One example of a correct graph:

For using data consistent with the answer to (b) i. 1 point For plotting the data correctly, including labeling the axes and their scales 1 point For constructing a best fit line or curve, depending on the data graphed 1 point

iii. 2 points

For using points from the best fit line, not data points 1 point For correctly determining the acceleration, including correct units and explicitly showing how the points yield the answer

1 point

For example, using the graph shown:

( ) 22 2 22.0 m 0.5 m 1.5 m2 slope 2 2 2.04 m s

1.97 s 0.5 s 1.47 sa - = = = = -


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Distribution of points (c) 4 points

For a correct torque equation that includes the tension T in the string 1 point TR Ia= For the correct relationship between and a a 1 point

a Ra = Substituting for and solving for I: a

aTR I R=

2I TR= a

For correctly applying Newtons second law for the block 1 point ma mg T= - Solving for T:

( )T m g a= - For the correct answer, obtained for substituting T into the equation for I 1 point

( ) 2m g a RI a-= or 2 1gI mR a

= -

Alternate solution Alternate points For a correct application of conservation of energy for the system 1 point The change in potential energy of the block equals the change in linear kinetic energy of the block plus the change in rotational kinetic energy of the pulley

2 21 12 2mgD m Iu w= +

For the correct relationship between the rotational and linear speeds 1 point

Ru

w =

For a correct relationship between the linear speed, the acceleration, and the distance D 1 point 2 2aDu =

Substituting the last two expressions into the energy equation: 2

2 22 2

1 1 12 2 2

ImgD m I mR Ru

u u = + = +

( ) 21 22

ImgD aD mR

= +

For the correct answer 1 point ( ) 2m g a RI a

-= or 2 1gI mR a = -


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For a reasonable analysis of the answer to part (c) that explicitly indicates either the effective mass or radius is greater than given in the initial paragraph of the question, or the experimental acceleration is obtained in (b) iii is greater than it would be without experimental error.

1 point

For a possible physical reason for the discrepant value 1 point Examples: The string was wrapped around the pulley several times, causing the effective radius at which the torque acted to be larger than the radius of the pulley used in the calculation. The string slipped on the pulley, allowing the block to accelerate faster than it would have otherwise, resulting in a smaller experimental moment of inertia.

Note: Friction is not a correct answer, since the presence of friction would make the experimental value of the moment of inertia too large.


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Question 3

15 points total Distribution of points

m

(a) 4 points

Using the integral expression for the moment of inertia: 2I r d=

For a correct change of variables 1 point ( )dm M L dr=

For correct limits of integration 1 point 2 3

2

3

L

L

MI rL-

= dr ( )or other appropriate combination of limits and distance expression, such as replacing by 3 and integrating from zero to r x L L

-

For correctly integrating 1 point 2 33

33

L

L

M rI L-

=

( ) ( )3 323 3 3M L LI L = - -

For the correct answer 1 point 2

9MLI =

Alternate Solution Alternate pointsFor any statement of the parallel axis theorem 1 point

2cmI I mr= + , where r is the distance from the center of mass to the pivot point

For a correct value of the center of mass inertia (calculated or remembered) 1 point 21

12cmI M= L

For indicating that 6L=r 1 point

( )22112 6LI ML M= + For the correct answer 1 point

2

9MLI =

Appropriate credit was also awarded for adding inertias for the parts of the rod on either side of the pivot. Credit was given for either calculating the inertias or remembering an appropriate expression for inertia.


9

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Distribution of points (b) 7 points

For any indication of conservation of energy 2 points For correctly calculating the change in potential energy of the rod (or the work done on it)

2 points

For example: ( )6cm LU Mgh MgD = = For writing a conservation equation that includes a rotational kinetic energy (regardless of whether the potential energy is correct)

1 point

212 6

MgLIw =

For any indication that is linear speed divided by a distance (regardless of whether the correct distance is used)

w 1 point

Substituting and solving for u :

( )2212 9 6MgLML ru =

2

2 2318

6MgL g

Lr MLu = =

( )22 23 3 2 43 3g g Lr gL Lu = = = L For the correct answer 1 point

2 3gL

u =

(c) 4 points

For an equation for the period of a physical pendulum 1 point

2 IT mgdp=

For substitution of the inertia from part (a) 1 point For indicating that the distance d is the distance from the pivot to the center of

mass, i.e. 2 3 6L L L= - =d

1 point

2 92 6MLT MgLp=

For the correct answer 1 point 22 3

LT gp=

Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents)..

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Distribution of points (c) (continued)

Alternate solution Alternate points For an equation relating the angular acceleration to the torque and inertia 1 point

2

2d

Idtq t

a = =

For substituting the inertia from part (a) and the torque as a function of q 1point ( )2

2 26 sin 3 sin29

Mg L gdLdt ML

qqq

-= = -

For using the approximation sinq q 1 point 2

232

gdLdt

qq= -

Taking , the second derivative is sinkq = tw2

22 sin

d k tdtq

w w= -

Substituting into the differential equation and solving for : w 2 3sin sin2

gk t kLw w- = - tw

32

gLw =

Using the relationship between T and : w 2T p w=

For the correct answer 1 point 22 3

LT gp=

No credit was awarded for the incorrect approach of using the period of a simple pendulum, 2p=T , and the length g 2 3L , which happens to give the same result as the correct method

Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents)..

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AP Physics C: Mechanics 2005 Free-Response Questions

The College Board: Connecting Students to College Success

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 4,700 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three and a half million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

Copyright 2005 by College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. Admitted Class Evaluation Service, CollegeEd, Connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: http://www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program and Pre-AP: apcentral.collegeboard.com.

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2

TABLE OF INFORMATION FOR 2005


1 unified atomic mass unit, 1 1 66 1027

u kg

931 MeV/2

=

=

-

.

c

Proton mass, mp = 1 67 10 27. kg

Neutron mass, mn = 1 67 10 27. kg

Electron mass, me = 9 11 10 31. kg

Magnitude of the electron charge, e = 1 60 10 19. C

Avogadros number, N023 16 02 10= . mol

Universal gas constant, R = 8 31. / ( )J mol K

Boltzmanns constant, kB = 138 10 23. J / K

Speed of light, c = 3 00 108. /m s

Plancks constant,

h

hc

=

=

=

=

6 63 10

414 10

1 99 10

1 24 10

34

15

25

3

.

.

.

.

J s

eV s

J m

eV nm

Vacuum permittivity, 012 2 28 85 10= . /C N m

Coulombs law constant, k = = 1 4 9 0 100 9 2 2/ . / N m C Vacuum permeability, 0

74 10= ( ) /T m A Magnetic constant, k' = T m A0 / ( ) /4 10

7= Universal gravitational constant, G =

-

6 67 1011 3 2

. /m kg s

Acceleration due to gravity at the Earths surface, g = 9 8 2. m/ s

1 atmosphere pressure,

1 1 0 10

1 0 10

5 2

5

atm N / m

Pa

=

=

.

.

1 electron volt, 1 1 60 10 19 eV J= .

UNITS

Name Symbol meter m

kilogram kg

second s

ampere A

kelvin K

mole mol

hertz Hz

newton N

pascal Pa

joule J

watt W

coulomb C

volt V

ohm

henry H

farad F

tesla T

degree Celsius C

electron- volt eV

PREFIXES

Factor Prefix Symbol 109 giga G

106 mega M

103 kilo k

10 2 centi c

10 3 milli m

10 6 micro

10 9 nano n

10 12 pico p

VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES

sin cos tan

0 0 1 0

30 1/2 3 2/ 3 3/

37 3/5 4/5 3/4

45 2 2/ 2 2/ 1

53 4/5 3/5 4/3

60 3 2/ 1/2 3

90 1 0


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3



( )

2

2 2

2

22

2 2

2

2

0

0 0

0 0

0

0 0

12

2

12

12

12

12

fric

net

g

c

net

cm

s

s

at

x x t at

a x x

m

ddt

dt

m

F N

W d

K m

dWP

dtP

U mgh

a rr

I

I r dm mr

m m

r

I

K I

t

t t

k

U

u u

u

u u

m

u

uw

u w

w

w w a

q q w a

D

D

= +

= + +

= + -

= =

=

= =

=

=

=

=

=

=

= =

=

= =

= =

=

=

= =

=

= +

= + +

= -

=

F F a

pF

J F p

p v

F r

F v

r F

r r

L r p

F x

t

t t a

w

2

1 22

1 2

2 1

2

2

G

G

s

p

kx

Tf

mT

k

Tg

Gm m

rGm m

Ur

p

w

p

p

= =

=

=

= -

= -

F r

a = acceleration F = force f = frequency h = height I = rotational inertia J = impulse K = kinetic energy k = spring constant = length L = angular momentum m = mass N = normal force P = power p = momentum r = radius or distance r = position vector T = period t = time U = potential energy u = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque w = angular speed a = angular acceleration

Fq q

r

q

dQ

EdV

dr

Vq

r

U qVq q

r

CQ

V

CA

d

C C

C C

IdQ

dt

U QV CV

RA

V IR

R R

R R

P IV

q

d I

I d

B nI

d

d

dt

LdI

i

ii

E

i

i

M

p ii

s ii

c

s i

p i

s

m

m

=

=

=

=

=

= =

=

=

=

=

=

= =

=

==

=

==

=

= z

=

= z

=

=

1

4

1

4

1

4

1 1

1

2

1

2

1 1

0

1 2

0

0

0

1 2

0

2

0

0

2p

p

p

k

r

m

m

f

fe

e

EF

E A

F v B

B

F B

B A

dt

U LIL =1

22

A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current L = inductance = length n = number of loops of wire

per unit length P = power Q = charge q = point charge R = resistance r = distance t = time U = potential or stored energy V = electric potential u = velocity or speed r = resistivity fm = magnetic flux k = dielectric constant

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4



CALCULUS

Rectangle

Triangle

Circle

Parallelepiped

Cylinder

Sphere

Right Triangle

A bh

A bh

A rC r

V wh

V rS r r

V r

S r

a b cac

bc

ab

=

=

==

=

== +

=

=

+ =

=

=

=

12

2

2 2

434

2

2

2

3

2

2 2 2

p

p

p

p p

p

p

sin

cos

tan

q

q

q

A = area C = circumference V = volume S = surface area b = base h = height = length w = width r = radius

c a

b90q

( )

( )

( )

( )

( )

1

1

11n

sin cos

cos sin

1 , 11

ln

cos sin

sin cos

n n

x x

n n

x x

d f d f dudx du dxd

x nxdxd

e edxd

xdx xd

x xdxd

x xdx

x dx x nn

e dx e

dxx

x

xdx x

xdx x

-

+

=

=

=

=

=

= -

= -+

=

=

=

= -

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GO ON TO THE NEXT PAGE. 5

PHYSICS C Section II, MECHANICS


Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces provided after each part, NOT in this green insert. Mech. 1.

A ball of mass M is thrown vertically upward with an initial speed of 0u . It experiences a force of air resistance

given by k= -F v , where k is a positive constant. The positive direction for all vector quantities is upward. Express all algebraic answers in terms of M, k, 0u , and fundamental constants.

(a) Does the magnitude of the acceleration of the ball increase, decrease, or remain the same as the ball moves upward?

____ increases ____ decreases ____ remains the same

Justify your answer.

(b) Write, but do NOT solve, a differential equation for the instantaneous speed u of the ball in terms of time t as the ball moves upward.

(c) Determine the terminal speed of the ball as it moves downward.

(d) Does it take longer for the ball to rise to its maximum height or to fall from its maximum height back to the height from which it was thrown?

____longer to rise ____longer to fall

Justify your answer.

(e) On the axes below, sketch a graph of velocity versus time for the upward and downward parts of the balls flight, where ft is the time at which the ball returns to the height from which it was thrown.

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Mech. 2.

A student is given the set of orbital data for some of the moons of Saturn shown below and is asked to use the data to determine the mass SM of Saturn. Assume the orbits of these moons are circular.

Orbital Period, T (seconds)

Orbital Radius, R (meters)

48.14 10 81.85 10 51.18 10 82.38 10 51.63 10 82.95 10 52.37 10 83.77 10

(a) Write an algebraic expression for the gravitational force between Saturn and one of its moons.

(b) Use your expression from part (a) and the assumption of circular orbits to derive an equation for the orbital period T of a moon as a function of its orbital radius R.

(c) Which quantities should be graphed to yield a straight line whose slope could be used to determine Saturns mass?

(d) Complete the data table by calculating the two quantities to be graphed. Label the top of each column, including units.

(e) Plot the graph on the axes below. Label the axes with the variables used and appropriate numbers to indicate the scale.

(f) Using the graph, calculate a value for the mass of Saturn.

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7

TOP VIEWS

Mech. 3.

A system consists of a ball of mass 2M and a uniform rod of mass 1M and length d. The rod is attached to a horizontal frictionless table by a pivot at point P and initially rotates at an angular speed w , as shown above

left. The rotational inertia of the rod about point P is 2113

M d . The rod strikes the ball, which is initially at rest.

As a result of this collision, the rod is stopped and the ball moves in the direction shown above right. Express all

answers in terms of 1M , 2M , w , d, and fundamental constants.

(a) Derive an expression for the angular momentum of the rod about point P before the collision.

(b) Derive an expression for the speed u of the ball after the collision.

(c) Assuming that this collision is elastic, calculate the numerical value of the ratio 1 2M M .

(d) A new ball with the same mass 1M as the rod is now placed a distance x from the pivot, as shown above. Again assuming the collision is elastic, for what value of x will the rod stop moving after hitting the ball?

END OF SECTION II, MECHANICS

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AP Physics C: Mechanics 2005 Scoring Guidelines


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 4,700 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three and a half million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

Copyright 2005 by College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. Admitted Class Evaluation Service, CollegeEd, Connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: http://www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program and Pre-AP: apcentral.collegeboard.com.

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General Notes About 2005 AP Physics Scoring Guidelines

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.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 students solution contains the application of that equation to the problem but the student does not write the basic equation, the point is still awarded.

4. The scoring guidelines typically show numerical results using the value 29.8 m sg , but use of210 m s is of course also acceptable.

5. Numerical answers that differ from the published answer due to differences in rounding throughout the question typically receive full credit. The exception is usually 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 and20.278). Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and some credit may be lost.

Copyright 2005 by College Entrance Examination Board. All rights reserved.Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

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Question 115 points total Distribution

of points


(a) 2 points

For indicating that the magnitude of the acceleration decreases as the ball moves upward 1 point For a correct, reasonable justification For example: Since velocity is upward, air resistance is downward, in the same direction

as gravity. The velocity will decrease, causing the force of air resistance to decrease.Therefore, the net force and thus the total acceleration both decrease.

1 point

(b) 3 points

For showing the expression dadtu 1 point

For any clear indication that the forces of air resistance and gravity are in the samedirection, such as by showing an equation or a free-body diagram 1 point

netF Mg kuFor a correct differential equation with the correct signs 1 point dM Mg kdtu

u

(c) 3 points

For recognizing that at terminal speed net 0F 1 point For any clear indication that the forces of air resistance and gravity are now in opposite

directions, such as by showing an equation or a free-body diagram 1 point netF Mg ku

T0 Mg kuFor a correct expression for the terminal speed 1 point

T Mg ku

(d) 2 points

For indicating that it takes longer for the ball to fall 1 point For a correct, reasonable justification For example: The ball loses mechanical energy on the way up and on the way down.

This means the average speed is greater on the way up than on the way down.Since the distance traveled is the same, the time must be longer on the way down.

1 point

3

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Question 1 (continued) Distribution

of points

(e) 5 points

For an exponentially decreasing curve beginning with positive initial velocity , crossing the time axis at t less than

0u

2ft , and having the final speed less than the initial speed 5 points One point partial credit was awarded for each of the following curve characteristics.

For showing that when , the curve is differentiable (i.e., no discontinuity in slope) and has a negative slope

0u

For showing the curve to be concave upward both for when the ball is rising and when the ball is falling

For showing time intervals for when the ball is rising and when the ball is falling that are consistent with the answer to part (d)

For showing that the final velocity is negative and that the speed at ft is less than 0u


4 www.oneplusone.cn



of points(a) 1 point

For a correct expression of Newtons law of gravity containing the mass of Saturn, withor without the minus sign

1 point

2SGM mFR

, where m is the mass of a moon

(b) 3 points

Circular orbit means the force is centripetal. For equating Newtons law of gravity to the centripetal force 1 point

2

2SGM m m

RRu OR g cF F

For correctly relating the orbital speed and the period 1 point 2 R Tu p OR and 2r Tu w w p

2

22SGM m m R

R TRp

For a correct relationship between T and R 1 point

2 24SGM T Rp3 OR

23 24

ST R

GMp

(c) 1 point

For any correct pair of quantities that would yield a straight line when graphed 1 point For example, versus 2T 3R OR T versus 3 2R

(d) 3 points

For appropriate and correct table headings, i.e., powers of T and R consistent withanswer to part (b) 1 point

For correct units in both columns 1 point For correct numerical values in at least 75 percent of the table entries, provided that

the table will yield a linear graph 1 point Full credit was awarded for completing only one column containing 3 2R .(Example data on next page.)


5 www.oneplusone.cn



of points(d) (continued)

Example data Orbital Period, T

(seconds)Orbital Radius, R

(meters)2 2sT 3 3 mR

48.14 10 81.85 10 100.663 10 250.633 1051.18 10 82.38 10 101.39 10 251.35 1051.63 10 82.95 10 102.66 10 252.57 1052.37 10 83.77 10 105.62 10 255.36 10

(e) 4 points

For axis labels consistent with the table and having correct units 1 point For appropriate numbering of axes 1 point For correctly plotting data from table (only awarded if numbering is correct) 1 point For drawing a best-fit linear graph (only awarded if data define a linear graph) 1 point

Example response


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of points

(f) 3 points

For correctly indicating a calculation of the slope of the graph (only awarded if graphis drawn) 1 point

From the graph shown here 10 10 3 10 3

25 25 2 25 2

5.6 10 1.4 10 m 4.2 10 mslope5.4 10 1.4 10 s 4.0 10 s

For relating the slope to the equation obtained in part (b) 1 point

In the example shown here,2

2 34

ST R

GMp .

24slope of lineSGM

p

24 slopeS

MG

p

2 2

11 3 10 34 4.0 10 s

6.67 10 m kg s 4.2 10 mSM

p 5 2

For the correct numerical answer 1 point 265.64 10 kgSM

Solutions for which the slope yields Saturns mass directly (e.g., table columns of 2 3 24 versusR GTp ) received full credit.


7 www.oneplusone.cn



of points


(a) 1 point

rL Iw , where 2

11 3I M dFor the correct expression for angular momentum before the collision 1 point

211 3rL M d w

Note: An expression without an equation was acceptable. This point was not awarded if the expression was not given in terms of the given quantities.

(b) 4 points

For any indication of conservation of angular momentum 1 point b rL L

For the correct expression for bL 1 point For substitution for consistent with part (a) rL 1 point

22 1

13

M d M du w

For the correct final expression for u 1 point1

2

13M

dM

u w

(c) 4 points

For any indication of conservation of kinetic energyNo points were awarded for conservation of mechanical energy.

1 point

b rK KFor the correct expressions for both kinetic energies 1 point

2 22

1 12 2M Iu w

2 22M Iu w

For correct substitutions for I and for consistent with part (b) u 1 point 2

2 212 1

2

1 13 3M

M d M dM

w w

22 2 2 21

2 12

1 19 3M

M d MM

w wd

21

12

1 19 3M

MM

For the correct final expression for the ratio 1 point 1 2 3M M (3:1 was also acceptable)

8

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of points(d) 6 points

For the correct equation for conservation of angular momentum 2 points 2

1 113

M x M du w

For solving this equation for u 1 point21

3dx

u w

For the correct equation for conservation of kinetic energy 1 point 2 2

11 12 2M Iu w

2 21 1

13

M M du w2

2 213du w2

For the correct substitution of the above expression for u from momentum conservation into the equation for conservation of kinetic energy

1 point

222 21 1

3 3d dxw w

42 2

21 19 3d d 2x

w w

2 21 9 1

3x dFor the correct final answer 1 point

3dx


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AP Physics C: Mechanics 2006 Free-Response Questions


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 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. 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 the AP Program: apcentral.collegeboard.com.

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2

TABLE OF INFORMATION FOR 2006 and 2007


1 unified atomic mass unit, 27

2

1 u 1.66 10 kg

931 MeV c

=

=

Proton mass, 271.67 10 kgpm=

Neutron mass, 271.67 10 kgnm=

Electron mass, 319.11 10 kgem=

Electron charge magnitude, 191.60 10 Ce =

Avogadros number, 23 10 6.02 10 molN=

Universal gas constant, 8.31 J (mol K)R i=

Boltzmanns constant, 231.38 10 J/KBk=

Speed of light, 83.00 10 m/sc =

Plancks constant,

34

15

6.63 10 J s

4.14 10 eV s

h i

i

=

=

25

3

1.99 10 J m

1.24 10 eV nm

hc i

i

=

=

Vacuum permittivity, 12 2 20 8.85 10 C N mi=

Coulombs law constant, 9 2 201 4 9.0 10 N m Ck i= = Vacuum permeability, 70 4 10 (T m) Ai

=

Magnetic constant, 70 /4 10 (T m) Ak = =

Universal gravitational constant, 11 3 26.67 10 m kg sG i= Acceleration due to gravity at Earths surface, 29.8 m sg =

1 atmosphere pressure, 5 2

5

1 atm 1.0 10 N m

1.0 10 Pa

=

=

1 electron volt, 191 eV 1.60 10 J=

UNITS

Name Symbol meter m

kilogram kg

second s

ampere A

kelvin K

mole mol

hertz Hz

newton N

pascal Pa

joule J

watt W

coulomb C

volt V

ohm W

henry H

farad F

tesla T

degree Celsius C

electron- volt eV

PREFIXES

Factor Prefix Symbol

910 giga G

610 mega M

310 kilo k

210- centi c

310- milli m

610- micro m

910- nano n

1210- pico p

VALUES OF TRIGONOMETRIC

FUNCTIONS FOR COMMON ANGLES

sin cos tan

0 0 1 0

30 1/2 3 2/ 3 3/

37 3/5 4/5 3/4

45 2 2/ 2 2/ 1

53 4/5 3/5 4/3

60 3 2/ 1/2 3

90 1 0


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3



0 atu u= +

20 0

12

x x t atu= + +

( )2 20 02a x xu u= + -

net m = =F F a

ddt

= pF

dt D= =J F p m=p v

fricF Nm

W d= F r

212

K mu=

dWPdt

=

P = F v

gU mghD =

22

ca rru

w= =

= r Ft

net I = =t t a

2 2I r dm mr= =

cm m m= r r

ru w=

I= =L r p w

212

K Iw=

0 tw w a= +

20 0

12

t tq q w a= + +

a = acceleration F = force f = frequency h = height I = rotational inertia J = impulse K = kinetic energy k = spring constant

= length L = angular momentum m = mass N = normal force P = power p = momentum r = radius or distance r = position vector T = period t = time U = potential energy u = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque w = angular speed a = angular acceleration

s k= -F x

212s

U kx=

2 1Tf

pw

= =

2smTk

p=

2pT gp=

1 22

GGm m

r= -F r

1 2G

Gm mU

r= -

21 2

0

14

q qF

rp=

q= FE

0

Qd = E A

dVEdr

= -

0

14

i

ii

qV

rp=

1 2

0

14E

q qU qV

rp= =

QC

V=

0 ACd

k=

p ii

C C=

1 1

s iiC C=

dQI

dt=

21 12 2c

U QV CV= =

RAr=

r=E J

dI Ne Au=

V IR=

iis

R R=

1 1

i ipR R

=

P IV=

M q= F v B

A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current J = current density L = inductance

= length n = number of loops of wire

per unit length N = number of charge carriers

per unit volume P = power Q = charge q = point charge R = resistance r = distance t = time U = potential or stored energy V = electric potential u = velocity or speed r = resistivity fm = magnetic flux k = dielectric constant

0d Im = B

034

I dd

r

m

p= rB

I d= F B

0sB nIm=

m df = B A md

dtfe = -

dILdt

e = -

212L

U LI=

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4



CALCULUS

Rectangle

A bh= Triangle

12

A bh=

Circle

2A rp= 2C rp= Parallelepiped

V wh= Cylinder

2V rp=

22 2S r rp p= +

Sphere

343

V rp=

24S rp= Right Triangle

2 2 2a b c+ =

sin ac

q =

cos bc

q =

tan ab

q =

A = area C = circumference V = volume S = surface area b = base h = height

= length w = width r = radius

c a

b90q

d f d f dudx du dx

=

( ) 1n nd x nxdx-=

( )x xd e edx =

( ) 11n d xdx x

=

( )sin cosd x xdx

=

( )cos sind x xdx

= -

11 , 11

n nx dx x nn

+= -+

x xe dx e=

lndx xx

= cos sinx dx x= sin cosx dx x= -

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2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).


PHYSICS C: MECHANICS SECTION II


Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces provided after each part, NOT in this green insert.

Mech 1.

A small block of mass 0.50 kgBM = is placed on a long slab of mass 3.0 kgSM = as shown above. Initially, the slab is at rest and the block has a speed 0u of 4.0 m/s to the right. The coefficient of kinetic friction between the

block and the slab is 0.20, and there is no friction between the slab and the horizontal surface on which it moves. (a) On the dots below that represent the block and the slab, draw and label vectors to represent the forces acting on

each as the block slides on the slab.

At some moment later, before the block reaches the right end of the slab, both the block and the slab attain identical speeds fu .

(b) Calculate fu .

(c) Calculate the distance the slab has traveled at the moment it reaches fu .

(d) Calculate the work done by friction on the slab from the beginning of its motion until it reaches fu .

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Mech 2.

A nonlinear spring is compressed various distances x, and the force F required to compress it is measured for each distance. The data are shown in the table below.

x (m) F (N)

0.05 4

0.10 17

0.15 38

0.20 68

0.25 106

Assume that the magnitude of the force applied by the spring is of the form 2( )F x Ax= . (a) Which quantities should be graphed in order to yield a straight line whose slope could be used to calculate a

numerical value for A ?

(b) Calculate values for any of the quantities identified in (a) that are not given in the data, and record these values in the table above. Label the top of the column, including units.

(c) On the axes below, plot the quantities you indicated in (a) . Label the axes with the variables and appropriate numbers to indicate the scale.

(d) Using your graph, calculate A.

The spring is then placed horizontally on the floor. One end of the spring is fixed to a wall. A cart of mass 0.50 kg moves on the floor with negligible friction and collides head-on with the free end of the spring, compressing it a maximum distance of 0.10 m.

(e) Calculate the work done by the cart in compressing the spring 0.10 m from its equilibrium length.

(f) Calculate the speed of the cart just before it strikes the spring.

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Mech 3.

A thin hoop of mass M, radius R, and rotational inertia 2MR is released from rest from the top of the ramp of length L above. The ramp makes an angle q with respect to a horizontal tabletop to which the ramp is fixed. The table is a height H above the floor. Assume that the hoop rolls without slipping down the ramp and across the table. Express all algebraic answers in terms of given quantities and fundamental constants. (a) Derive an expression for the acceleration of the center of mass of the hoop as it rolls down the ramp.

(b) Derive an expression for the speed of the center of mass of the hoop when it reaches the bottom of the ramp.

(c) Derive an expression for the horizontal distance from the edge of the table to where the hoop lands on the floor.

(d) Suppose that the hoop is now replaced by a disk having the same mass M and radius R. How will the distance from the edge of the table to where the disk lands on the floor compare with the distance determined in part (c) for the hoop?

____ Less than ____ The same as ____ Greater than Briefly justify your response.

END OF EXAM

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AP Physics C: Mechanics 2006 Scoring Guidelines


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 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in 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.

2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. 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 the AP Program: apcentral.collegeboard.com.

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General Notes About 2006 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 1 point, and a students 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. See pages 2122 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 29.8 m sg = , but use of

210 m s 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.

2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

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Question 1


For the block force diagram: For correctly labeled horizontal force (friction to the left, no other forces or vectors) 1 point For correctly labeled vertical forces (normal up and weight down; gravity alone not

accepted) 1 point

For the slab force diagram: For correctly labeled horizontal force (friction to the right, no other forces or vectors) 1 point For correctly labeled vertical forces (normal up and combined weight down; combined

weight can be shown with two arrows or identifying weight as or S BW W++S BM g M g ; gravity alone not accepted)

1 point

(b) and (c) These two parts were scored together because of the different approaches that could be used to answer them. Momentum approach to part (b); Newtons second law and kinematics approach to part (c) (b) 3 points

For any statement of conservation of momentum 1 point No net external forces act on the two-block system, so linear momentum is conserved. For a correct momentum equation 1 point

( )0u u= +B B SM M M f

00.50 kg

4.0 m s0.50 kg + 3.0 kg

u u= =+

Bf

B S

MM M

For the correct answer 1 point 0.57 m su =f


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Distribution of points Momentum approach (continued) (c) 6 points

For a correct expression for the friction force (awarded if found in the solution to any of parts (a) through (d))

1 point

m=f mg or m=f N For correct substitution of = Bm M for the friction force on the block 1 point

m= Bf M g For recognizing that the friction force on the slab is equal in magnitude to the friction force

on the block and for an equation relating this force to the acceleration of the slab 1 point

= S Sf M a For a correct expression for the acceleration of the slab or its numerical value 1 point

20.33 m sBSS

M ga

Mm

= =

For a correct kinematic equation for the slab 1 point 2 2

0 2u u= +f Sa x , where 0 0u =

2 2

2 2

u u

m= =f f S

S B

Mx

a M g

For correct substitutions consistent with earlier values 1 point

( )( )( )

2

2

0.57 m s 3.0 kg2 0.20 0.50 kg 9.8 m s

=x

x = 0.49 m or 0.50 m, depending on use of 29.8 or 10 m sg = and where substitution and rounding took place


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Distribution of points Newtons second law and kinematics approach to part (b); kinematics approach to part (c) (b) 7 points

For a correct expression for the friction force (awarded if found in the solution to any of parts (a) through (d))

1 point

m=f mg or m=f N For correct substitution of = Bm M 1 point

m= Bf M g For recognizing that the friction force on the slab is equal in magnitude to the friction force

on the block and for an equation relating this force to the acceleration of the slab 1 point

= S Sf M a For a correct expression for the acceleration of the slab or its numerical value 1 point

20.33 m sm

= =BSS

M ga

M

For a correct expression for the acceleration of the block or its numerical value 1 point

22.0m sm

m= = =BBB

M ga g

M

For a solution of the following simultaneous kinematic equations for the block and the slab, such as by setting the times equal and solving for u f

1 point

0u u= -f a tB for the block

u =f Sa t for the slab

( )( )202 2

0.33 m s 4.0 m s

0.33 m s + 2.0 m sS

fS B

aa a

uu = =

+

For the correct answer 1 point 0.57 m sfu =

(c) 2 points

For a correct kinematic equation for the slab 1 point 2 2

0 2u u= +f Sa x , where 0 0u =

2 2

2 2

u u

m= =f f S

S B

Mx

a M g

For correct substitutions consistent with earlier values 1 point

( )( )( )

2

2

0.57 m s 3.0 kg2 0.20 0.50 kg 9.8 m s

=x

x = 0.49 m or 0.50 m, depending on use of 29.8 or 10 m sg = and where substitution and rounding took place


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Distribution of points

x

(d) 2 points

For a correct expression for the work done 1 point

m= = BW Fd M g OR 21

2uD= = S fW K M

For consistent substitution from parts (b) and (c) 1 point

( )( )(20.20 0.5 kg 9.8 m s 0.50 m=W ) OR ( )( )21 3.0 kg 0.57 m s2=W

W = 0.49 J (or W = 0.50 J using 210 m sg = )


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Question 2

15 points total Distribution of points (a) 1 point

For indicating that F vs. 2x or F vs. x should be graphed, or other equivalent correct response (Must clearly specify two variables in order to earn this point.)

1 point

(b) 2 points

For a correct column label, including units 1 point For calculated values that match what is indicated in (a) 1 point Note: If answer to (a) was incorrect or incomplete, (b) received no credit.

Example using F vs. 2x

x (m) F (N) ( )2 2mx 0.05 4 0.0025

0.10 17 0.010

0.15 38 0.023

0.20 68 0.040

0.25 106 0.063

(c) 3 points

For appropriate linear axes scales 1 point For correct axes labels 1 point For plotting the points 1 point Note: Axes and scales must match answer in (a). However, if (a) was incorrect or

incomplete, points were awarded in (c) if graph was executed correctly. If (a) was blank or didnt include any variables, no credit was awarded for (b) or (c).

Example using data above


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For indication of a correct relationship between the coefficient A and the slope for the values graphed in (c)

1 point

For correct units and no more than four significant figures on value of A 1 point Example using data in the table and two points on the line in the graph

2=F Ax , so A is equal to the slope of the F vs. 2x line.

3 22 2 2

100 N 50 N slope 1.7 10 N m0.060 m 0.030 m

D

D

-= = = = -

FAx

Notes: This part stated to calculate, so an answer with correct units and significant figures but

with no work shown earned 1 point.

Since all the data points are on the best-fine line, additional credit was not awarded for a correctly drawn best-fine line or for use of points on the line instead of data points.

(e) 4 points

Using the definition of work

= W F dx For correct substitution of F(x) into the integral for work 1 point For correct limits on the integral 1 point For correct evaluation of the integral 1 point

( ) ( )(0.10 m

32 3 2

0

1 10.10 m 1.7 10 N/m 1.0 10 m3 3

W Ax dx A -= = = )3 3

For the correct answer with correct units 1 point 0.57 J=W

Note: This part stated to calculate, so a correct answer with correct units, but with no work shown, earned 1 point.

(f) 3 points

For an appropriate expression of conservation of energy or the work-energy theorem 1 point For a correct expression for K and substitution of W from part (e), expressed algebraically or

numerically 1 point

212

uD= =W K m

( )2 0.57 J20.5 kg

u = =Wm

For a value of u consistent with the value of W in (e), with correct units 1 point 1.5 m su =

Note: This part stated to calculate, so an answer consistent with (e) and with correct units, but with no work shown, earned 1 point.


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Question 3

15 points total Distribution of points (a) and (b) These two parts were scored together because of the different approaches that could be used to answer them. The parts could be answered in either order. Approach using translational and rotational dynamics (a) 5 points

For use of Newtons 2nd law in both translational and rotational forms 1 point = cmF ma and t a= cmI

For a correct equation applying Newtons second law in translational form 1 point sinq - = cmMg f Ma

For a correct equation applying Newtons second law in rotational form 1 point a= cmf R I

For a correct relationship between linear and angular acceleration for rolling without slipping

1 point

a = cmcmaR

Substituting for I and into the rotational equation above acm

2= cma

f R MRR

= cmf Ma Substituting this expression for f into the equation for translational motion above

sinq - =cm cmMg Ma Ma For the correct answer 1 point

sin2

q=cmg

a

(b) 3 points

For a correct kinematic equation containing a and u 1 point 2 2

0 2u u D= + a x , 0 0u =

For correct substitution of the expression for acceleration from part (a) 1 point For correct substitution of the distance traveled 1 point

2 2 sin2

u q = g

L

sinu q= gL


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Distribution of points Approach using torque about point of contact between hoop and ramp and parallel axis theorem (a) 5 points

For use of Newtons 2nd law in rotational form and the parallel axis theorem 1 point

t a= cmI and 2= +cmI I Mh For a correct rotational inertia about the point of contact using the parallel axis theorem 1 point

2 2 2= + =I MR MR MR2 For a correct torque about the point of contact 1 point

sint q= RMg For a correct relationship between linear and angular acceleration for rolling without

slipping 1 point

a = cmcmaR

Substituting for , I, and a into the rotational equation above t cm 2sin 2 cm

aRMg MR

Rq =

For the correct answer 1 point

sin2

q=cmg

ap physics c: mechanics 2004 free-response …® physics c: mechanics 2004 free-response questions...

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