AP® Physics B
2013 Scoring Guidelines
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AP® PHYSICS B
2013 SCORING GUIDELINES
Question 1
10 points total
(a)
Distribution
of points
3 points
For showing the buoyant force in the correct direction and labeling it
For showing the tension in the correct direction and labeling it
For showing the gravitational force in the correct direction and labeling it
One earned point was deducted for any extraneous forces.
(b)
1 point
1 point
1 point
2 points
For use of the correct expression for the buoyant force
FB  rVg
Substitute correct values



FB  1000 kg m3 6.25  103 m3 9.8 m s2
1 point

For a correct answer, with units
1 point
2
FB  61.3 N (62.5 N if using g  10 m s )
(c)
3 points
For a correct expression of Newton’s second law when the anchor is at equilibrium
FT  FB  mg
For substituting a value consistent with the buoyant force from
part (b)
For substituting correct values for calculation of the gravitational force

FT   50 kg 9.8 m s
2
   61.3 N
FT  429 N (438 N if using g  10 m s2 )
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1 point
1 point
1 point
AP® PHYSICS B
2013 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points
(d)
2 points
For selecting d '  d
For a correct justification
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.
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1 point
1 point


W = DK
1
1
W = m u22 - u12 = - mu12
2
2
(
W =-
)
( 12 ) ( 20 kg)( 4.0 m s)
2
W = 160 J
K1 = U2
1 2 1 2
mu = kx2
2 1
2
( 2)
(160 J) = 1 k ( 0.50 m ) 2
k = 1280 N m
a = Fmax m = kxmax m
a = (1280 N m)( 0.50 m) ( 20 kg)
a = 32 m s2
1
1 k
=
T
2p m
f =
f =
( 21p )
(1280 N m )
( 20 kg)
f = 1.27 Hz
−
+
−
n = slope
1 slope
sin qr
n=
n = 1.43
(1.00 - 0.20)
( 0.70 - 0.14)
sin qi
n2 = 1
qc
n = 1 sin qc
Dy = uyi t +
1 2
at
2
uyi = 0
t =
( 2)( 2.4 m )
t =
( 9.8 m s2 )
2 ( Dy )
a
= 0.70 s
Dx
t
(1.8 m )
ux =
= 2.6 m s
( 0.70 s)
ux =
ux2 = u02 + 2ad
a=
ux2
2d
a=
( 2.6 m s) 2
= 3.5 m s2
( 2) ( 0.95 m )
åF
= ma
mobj g = ( mobj + M + mball ) a
(
)
mobj g = ( 2.5 kg) 9.8 m s2 = 25 N
( mobj + M + mball ) a = (2.5 kg + 0.3 kg + M)a = ( 2.8 kg + M ) a
(
25 N = ( 2.8 kg + M ) 3.5 m s2
)
M = 4.24 kg
( ulaunch ´ t fall )
DU = Q + W
DU = ( 3200 J) + ( 2100 J)
DU = 5300 J
W = -PDV
PV T
DV
DU = 0
DU = Q + W
m0
B=
(
)
4 p ´ 10 -7 T m A ( 65 A)
m0 I
=
2 pr
( 2 p )( 0.025 m )
B = 5.2 ´ 10-4 T
W = FB
mg = BIL
I =
( mL ) Bg
(
I = 5.6 ´ 10
I = 106 A
-3
kg m
)
( 9.8 m s2 )
( 5.2 ´ 10-4 T)
g = 10 m s2



e = BuL
OR
B=
e = - DDft
(
e = - B DDAt = - BuL
)
1
5.2 ´ 10 -4 T = 2.6 ´ 10 -4 T
2
(
)
4 p ´ 10 -7 T m A ( 65 A )
m0 I
B=
=
= 2.6 ´ 10 -4 T
2 pr
( 2 p )( 0.050 m )
u
e = (2.6 ´ 10-4 T)( 3.0 m s)(1.2 m) = 9.4 ´ 10-4 V
n=3
n=3
n=2
n=2
n =1
n =1
E = hf = hc l
l = hc E = (1240 eV nm)
( ( 3.0 - 0.75) eV)
l = 551 nm or 5.51 ´ 10 -7 m
E = 12 eV
1.9 ´ 10-18 J
−
−