Download [ ] ( )

Mechanics
1. (a)
2006 C
Block
Slab
FNB
FNS
FfB
FfS
FgB
FBS
(b) M B v B + M S v S = ( M B + M S )v f
(0.50 kg) × ( 4.0 m / s) + 0 = (0.50 kg + 3.0 kg)v f
Note: An alternative (but much longer)
solution would be to use Newton's
second law to get relative accelerations
and calculate relative velocities. See
last page for this solution.
vf = 0.57 m/s
(c) W NET = DKE
2
F f × d = 12 M s v f - 12 M s v 2
(
M (v
)
-v )
mFNB × d = 12 M S v f - v
mM B g × d =
1
2
S
2
2
f
2
åF
FNB
FgS
= FNB - FgB = M B a
- MBg = 0
y
FNB = M B g
2
(0.20)(0.50 kg)(9.8 m / s 2 )d = 12 (3.0 kg)[(0.57 m / s) 2 - 0]
d = 0.500 m
Note: An alternative solution would be to use kinematic equations. See last page for this solution.
(d) W NET = DKE
2
W NET = 12 M S v f - v 2
(
W NET =
1
2
)
(3.0 kg)[(0.57 m / s) 2 - 0]
WNET = 0.49 J
Note: An alternative solution would be to use the definition of work. See last page for this solution.
Mechanics
2. (a) Plot F vs. x2
2006 C
(b)
x (m)
F (N)
x2 (m2)
0.05
4
0.0025
0.10
17
0.0100
0.15
38
0.0225
0.20
68
0.04
0.25
106
0.0625
(c)
Force vs. Distance Compressed Squared
F (N)
100
50
0
0
0.0250
0.0500
x2 (m2)
(d) A = m =
(106 - 4) N
Dy DF
= 1700 N / m 2 = A
=
=
2
2
Dx Dx
(0.0625 - 0.0025) m
(
)
(e) W = F × d cos q = Ax 2 × ( x ) cos 0 =
(f) W = DKE = 12 mv 2 - 12 mv 0
1.7 J =
1
2
[(1700 N / m )(0.10 m) ]× (0.10 m) = 1.7 J = W
2
(0.50 kg)v 2 - 12 (0.50 kg)(0 m / s) 2
v = 2.61 m/s
2
2
Mechanics
3. (a) SF = Fgx - F f = ma
2006 C
St = t f = Ia
æa ö
RF f sinf = mR 2 ç T ÷ Note: aT = a, f = 90, sin 90 = 1
è R ø
F f = ma
(
mg sinq - ma = ma
mg sinq = 2ma
)
a = 12 g sinq
(b) GPE1 = TKE 2 + RKE 2
2
2
mgh = 12 mv 2 + 12 Iw 2
mgL sinq = mv 2 +
1
2
gL sinq = 12 v 2 + 12 v 2
2
2
1
2
2
(
æv ö
mR ç 2 ÷
èRø
2
)
2
v 2 = gL sinq
(c)
x
x=?
vx0 = v2
y
y0 = H
y=0
vy0 = 0
ay = g
t=?
y = y 0 + v 0 t + 12 a y t 2
H = 0 + 0 + 12 gt 2
t = 2gH
x = vx0t = v2t
x = gL sinq
(
)(
2gH
)
x = g 2HL sinq
(d)
____ Less than
____ The same as
X Greater than
____
A disk has half the moment of inertia of the thin hoop resulting in a greater acceleration down the ramp
and, thus, a greater velocity at the bottom of the ramp. Since the disk leaves the tabletop at a greater
speed it will land a greater distance from the edge of the table than the thin hoop did.
E&M
1. (a)
2006 C
P
(b) i. The electric field due to the charges in the upper left and lower right corner will cancel each
other because they are of the same magnitude at point P and in opposite directions. The electric
field due to the charges in the lower left (LL) and upper right (UR) corner are in the same
direction (toward the upper right corner) at point P, therefore, adding to each other; and of the
same magnitude.
Q
Q
2
Note: r =
E NET = E LL + EUR = k 2 + k 2
a
2
r
r
Q
Q
E NET = k
+k
2
2
æ 2 ö
æ 2 ö
ç
÷
ç
÷
ç 2 a÷
ç 2 a÷
è
ø
è
ø
E NET = 4k
Q
a2
ii. V NET = V LL + VUL + VUR + V LR = k
V NET = (c)
Q
-Q
-Q
-Q
Q
Q
+k
+k
+k
= -2k = -2k
r
r
r
r
r
2
a
2
4 Q
k
2 a
X Positive
____
____ Negative
____ Zero
The positive charge at point P is being moved into a more positive region of space (closer to the
lower left corner) and, therefore, against an increasingly stronger electric field. Thus, the
movement of this charge will require an applied force (that will need to increase against the
increasing electric field) to counter this electric field. At least a component of this force is in the
direction of the displacement of the charge as it moves from point P to point R, so the work done
(W = FD cosq ) will be positive.
(d) i. Replace the charge at the lower left corner with charge of -Q as the electric field from this
charge will cancel the electric field at the center from the charge in the upper right corner. The
other two charges would also cancel each other for similar reasons as described above in 1. (b) i.
ii. Replace the charge at the upper left corner OR the lower right corner with a charge of +Q as this
will create two positive charges and two negative charges all equidistance to the center of the
square, so the net electric potential will be zero. However, the individual electric fields created
by the four charges (each of which will be equal in magnitude) will not cancel because the
direction of these electric fields are not in exactly opposite directions.
E&M
2. (a) Apply Kirchoff's loop rule
Q
e - IR1 - = 0
C
e-
(b)
2006 C
dQ
Q
R1 =
dt
C
dQ
Q
R1 = e dt
C
dQ
dt
Now, Integrate both sides. Let U = Ce - Q, then dU = -dQ.
=
Ce - Q R1C
-dQ
-dt
ò Ce - Q = ò R1C
t
ln(Ce - Q ) + L = + M , where L and M are constants of integration. Let N = M - L.
R1C
t
ln(Ce - Q ) = +N
R1C
Ce - Q = e
-
t
+N
R1C
-
, rearranging Ce - Q = e
-
t
R1C
× e , or Ce - Q = Be
N
-
t
R1C
, where B = e N is a constant.
t
R1C
To determine B, use the information that initially (t = 0), the charge, Q = 0.
Q = Ce - Be
0 = Ce - B, or B = Ce because e 0 = 1. Substituting this into the original expression for Q yields:
t
t
é
ù
æ
ö
( 4700 W)( 0. 060 F )
R1C ÷
ç
, for the values given, this expression becomesQ = (0060
Q = Ce 1 - e
.
F)(12 V )ê1 - e
ú
ç
÷
ê
úû
è
ø
ë
t
é
ù
( 282 s )
Q = (0.72 C )ê1 - e
ú
êë
úû
Q
Q
for the capacitor, so 4.0 V =
, or Q = 0.24 C. Substituting this value into the expression
C
0.060 F
t
é
ù
( 282 s )
above yields: 024
. C = (0.72 C )ê1 - e
ú. Now solve for t to get t = 310 s.
êë
úû
Current
(c) V =
(d)
mA
2.55
I2
1.28
I1
Time
E&M
3. (a)
2006 C
(b) å F = FS - FB = ma = 0
kx = IwB0
B0 =
kx
Iw
(c) i.
Df B0 wDl
Dl
=
= B0 w
Dt
Dt
Dt
e = B0 wv 0
e
I=
R
ii. e =
I=
B0 wv 0
R
2
æ B wv ö
(d) P = I R = ç 0 0 ÷ R OR
è R ø
2
2
B0 w 2 v 0
P=
R
(e)
2
X Increases
____
W Fd cosq
where å F = FA - FB = ma = 0
=
t
t
d
where FA = FB = IwB0
P = F cos0 = Fv
t
2
2
B0 w 2 v 0
æ B0 wv 0 ö
P = IwB0 v 0 = ç
÷B0 wv 0 =
R
è R ø
P=
____Decreases
____ Remains the same
As shown in the boxed derivation of part (d), the external force applied to the loop is directly
promotional to the magnetic field strength.
Mechanics
1. (b) Alternate Solution:
2006 C
Slab
Block
åF
= F fB = M B a å Fy = FNB - FgB = M B a
FNB - M B g = 0
-mFNB = M B a
FNB = M B g
-mM B g = M B a
a = -mg = -(0.20) × 9.8 m / s 2 = -1.96 m / s 2
x
(
)
v = v 0 + at
v f = 40
. m / s + -1.96 m / s 2 t
(
åF
= F fS = M S a
mFNB = M S a
mM B g = M S a
M
a=m B g
MS
æ 0.50 kg ö
a = (0.20)ç
÷ 9.8 m / s 2 = 0.327 m / s 2
è 3.0 kg ø
x
(
)
)
v = v 0 + at
v f = 0 + 0.327 m / s 2 t
(
Set the two vf expressions equal to each other and solve for time.
4.0 m/s - (1.96 m/s2)t = (0.327 m/s2)t
4.0 m/s = (2.287 m/s2)t
t = 1.75 s
Now substitute this value of t in either expression of vf to determine vf.
v f = 0 + 0.327 m / s 2 t
(
(
v f = 0.327 m / s
)
2
)(1.75 s)
vf = 0.57 m/s
(c) x = x 0 + v 0 t + 12 at 2
x = 0 + 0 + ½(0.327 m/s2)(1.75 m/s2)
x = 0.50 m
(d) W NET = FNET × d = F f × d = (mM B g ) × d
W NET = (020
. )(0.50 kg) 9.8 m / s 2 × (0.50 m)
[
WNET = 0.49 J
(
)]
)