AP® Physics B 2013 Scoring Guidelines

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AP® PHYSICS B 2013 SCORING GUIDELINES

Question 1 10 points total Distribution of points

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(a) 3 points

For showing the buoyant force in the correct direction and labeling it 1 pointFor showing the tension in the correct direction and labeling it 1 pointFor showing the gravitational force in the correct direction and labeling it 1 pointOne earned point was deducted for any extraneous forces.

(b) 2 points

For use of the correct expression for the buoyant force 1 point

BF Vgr

Substitute correct values

3 3 3 21000 kg m 6.25 10 m 9.8 m sBF

For a correct answer, with units 1 point

61.3 NBF (62.5 N if using 210 m sg )

(c) 3 points

For a correct expression of Newton’s second law when the anchor is at equilibrium 1 point

T BF F mg

For substituting a value consistent with the buoyant force from part (b)

1 point

For substituting correct values for calculation of the gravitational force 1 point

250 kg 9.8 m s 61.3 NTF

429 NTF (438 N if using 210 m sg )

AP® PHYSICS B 2013 SCORING GUIDELINES

Question 1 (continued)

Distribution of points

© 2013 The College Board. Visit the College Board on the Web: www.collegeboard.org.

(d) 2 points

For selecting 'd d 1 pointFor a correct justification 1 point

Example When the anchor is lifted onto the boat, the buoyant force on the boat must now support the weight of both the boat and the anchor. This will increase the buoyant force, which requires a greater volume of water, which means the boat will reach a greater depth into the water.

W KD=

( )2 2 22 1 1

1 1

2 2W m mu u u= - = -

( ) ( ) ( )21

20 kg 4.0 m s2

W = -

160 JW =

1 2K U=

2 21 2

1 1

2 2m kxu =

( ) ( ) ( )21

160 J 0.50 m2k=

1280 N mk =

max maxa F m kx m= =

( ) ( ) ( )1280 N m 0.50 m 20 kga =

232 m sa =

1 1

2

kf

T mp= =

( ) ( )( )

1280 N m1

2 20 kgf

p=

1.27 Hzf =

− +

−

slopen = 1 slope sin rq sin iq

( )( )1.00 0.20

0.70 0.14n

-=

-

1.43n =

2 1n =

cq 1 sin cn q=

21

2yiy t atuD = +

0yiu =( )2 y

ta

D=

( ) ( )

( )2

2 2.4 m0.70 s

9.8 m st = =

xx

tu

D=

( )( )1.8 m

2.6 m s0.70 sxu = =

2 20 2x adu u= +

2

2xad

u=

( )( ) ( )

222.6 m s

3.5 m s2 0.95 m

a = =

F ma=å

( )obj obj ballm g m M m a= + +

( ) ( )22.5 kg 9.8 m s 25 Nobjm g = =

( ) ( )(2.5 kg 0.3 kg ) 2.8 kgobj ballm M m a M a M a+ + = + + = +

( ) ( )225 N 2.8 kg 3.5 m sM= +

4.24 kgM =

( )launch falltu ´

U Q WD = +

( ) ( )3200 J 2100 JUD = +

5300 JUD =

W P VD= - VD

PV T

0UD =

U Q WD = +

0m

( ) ( )

( ) ( )

7

04 10 T m A 65 A

2 2 0.025 m

IB

r

pmp p

-´= =

45.2 10 TB -= ´

BW F=

mg BIL=

( ) gmI

L B=

( )( )

( )

2

3

4

9.8 m s5.6 10 kg m

5.2 10 TI -

-= ´

´

106 AI =210 m sg =

OR B Lt

fue e D

= = -D

AB B L

tue D

= - = -D

( )4 415.2 10 T 2.6 10 T

2B - -= ´ = ´

( ) ( )

( ) ( )

7

404 10 T m A 65 A

2.6 10 T2 2 0.050 m

IB

r

pmp p

-

-´

= = = ´

u

( ) ( )4 4(2.6 10 T) 3.0 m s 1.2 m 9.4 10 Ve - -= ´ = ´

3n = 2n =

3n = 1n =

2n = 1n =

E hf hc l= =

( ) ( )( )1240 eV nm 3.0 0.75 eVhc El = = -

7551 nm or 5.51 10 ml -= ´

12 eVE = 181.9 10 J-´

−−