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Unformatted text preview: AP® Physics C: Mechanics
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For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. 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.
 Spring '08
 Reich
 Physics, mechanics

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