Download 1. (a) GPE = KE m gL = ½m

Mechanics
1. (a) GPE1 = KE2
m1gL = ½m1v2
v =
2004 C
2gL
(b) SF = FT - Fg = ma
FT - m1g = m1(0 m/s2)
FT = m1g
(c) p1 + p2 = p1' + p2'
m1v1 + m2v2 = m1v1' + m2v2'
m1v1 + m2v2 = (m1 + m2)v' Inelastic Collision
m1 2gL + 0 = ( m1 + m 2 )v'
ö
æ
m1
÷÷ 2gL
v 2 ' = çç
è m1 + m 2 ø
KE sys
2
1
2
m1 v1
(d) ratio =
= 1
=
2
KE sys '
m
+
m
v'
(
)
1
2
2
ratio =
1
2
m1
(
)
2
2gL
ìïé
üï
ù
m1
1
2gL
m
+
m
(
)
í
ý
1
2
ê
ú
2
ïþ
îïë( m1 + m 2 ) û
2
m1 + m 2
m1
(e) From position A to position B the horizontal distance is L
The system experiences projectile motion from B to D in which the initial velocity is all horizontal.
x
y
v xo = 2gL
v yo = 0 m / s
x=?
g = -9.8 m/s2
yo = L
y=0m
t=?
y = yo + vyot + ½at2
0 m = L + (0 m/s)t + ½gt2
½gt2 = L
L
t = 1
2g
x = v xo t =
x = 2L
so the total horizontal distance from A to D is given by:
xTotal = xA-B + xB-D = L + 2L
xTotal = 3L
(
æ
2gL × ç
ç
è
)
L ö÷
1
÷
2g ø
Mechanics
2. (a) y = yo + vot + ½at2
0 = D + 0 + ½at2
a =
2004 C
2D
t2
(b) i. D should be graphed as a function of t2 to best determine the acceleration of the block because the
relationship between D and t is D a t2, so graphing D vs. t would give a parabola which would be
difficult to analyze for acceleration. Graphing D vs. t2 would yield a straight line (direct
proportion). The slope of this graph would determine the proportionality constant for the relation
between D and t2 (D a t2). From the kinematic equation y = yo + vot + ½at2 which simplifies to D =
½at2 in this circumstance, the proportionality constant is one-half of the acceleration. Therefore, the
slope of this graph (D vs. t2) would equal one-half of the acceleration.
ii. 2
2.0
D (m)
1.5
1.0
0.5
0.0
0.0
0.5
1.0
t (s2)
1.5
2.0
2
iii. slope, m =
Dy
Dx
=
DD
Dt 2
=
(2.0 m -
(1.91 s
2
0.00 m)
- 0.00 s
2
)
= 1.05 m / s 2 , but m = ½a
1.05 m/s2 = ½a
a = 2.10 m/s2
(c) SF = FT - Fg = ma
FT - mg = ma
FT = ma - mg = m(a - g)
ST = Ia
RFT = I
a
R
a
R
a
g
gö
æ
ö
2æ
I = mR 2 ç
÷ = mR ç1 - ÷
aø
è a ø
è
Rm(a - g) = I
(d) The weight of the string was ignored which would reduce the experimental value of the rotational
inertia. As the string unravels, the mass of the falling weight would increase, thus, increasing the
measured acceleration, thereby, reducing the experimental rotational inertia.
Mechanics
2004 C
3. (a) dI = dmr2 = ldxrx2
I =
ò dI
2L
3
-L
3
ò
=
l =
x3
lx 2dx = l
3
2L
3
-L
3
3
éæ 8L3
éæ 2L ö 3
æ -L ö ù
ç ÷ ú
êçç
êç ÷
M
3
3
è
ø
è
ø
êè 27
ú =
I = lê
3 ú
Lê 3
ê 3
ê
ê
ú
ë
û
ë
I =
1
9
ö
÷÷
ø -
( )
g =
v =
æ -L3
çç
è 27
3
öù
÷÷ ú
3
3
3
ø ú = M é8L + L ù = M é9L ù
ú
L êë 81
81 úû
L êë 81 úû
ú
û
ML2
(b) GPE1 = RKE2
mgy1 = ½Iw2
1
1
Mg L = Iw 2
6
2
2
1æ1
1
2 öæ v ö
Mg L = ç ML ÷ç ÷
2è9
6
øè r ø
2
1 2 æ v ö
gL =
L çç 2 ÷÷
3
è 3 Lø
gL =
m
dm
=
L
dx
( )
1 2
L
3
v2
( 23 L)2
1 2 æ 9 v2
L çç
2
3
è4 L
( )
3
4
Note: Use the lowest point of center of mass (cm)
as 0 height, therefore, y1, the initial height of
the cm (located in the center of the rod),
would be the distance the center of mass is
from the pivot point. The cm is at 12 L while
the pivot point is at 23 L. The distance
between these two points,y1, is given by:
y1 = 12 L - 13 L = 36 L - 26 L = 16 L
Note: r is the distance from the pivot point to the
bottom of the rod when it is in the vertical
position. This distance is 23 L.
ö
÷÷
ø
gL
(c) Period, T, for a physical pendulum is given by T = 2p
I
where D is the distance from the
MgD
pivot point to the cm. For this rod, D is 16 L.
1
ML
I
T = 2p
= 2p 9 1
MgD
Mg 6 L
2L
3g
T = 2p
D
q
q
Fg
Note: A simple pendulum consists of a mass (bob) suspended from a string. A
physical pendulum consists of a rigid body suspended body suspended by
some point other than its cm such as the rod in this problem. When the
body is displaced from its equilibrium position it will oscillate. When the
angle of the displacement is small, the physical pendulum's oscillation
approximates simple harmonic motion just as a simple pendulum does.
The torque created by the physical pendulum is given by:
T = -rFsinq = Ia
let r = D which represents the distance from the pivot
d 2q
point to the cm. For small angles, sin q » q
-Dmgq = I 2
dt
MgD
2p
d 2q
MgD
and the period is given by T =
.
= q so w 2 =
2
I
w
I
dt
E&M
1. (a) i.
ii.
2004 C
+
+
-
+
+
-
+
-
-
+
-
+
-
-
+
+
+
(b)
_____
1 Va
_____
3 Vb
_____
5 Vc
_____
1 Vd
_____
3 Ve
(c) For a Gaussian surface, use a cylindrical shell with its center at the center of the infinite line charge for each
of the following (i., ii., iii.). The electric field through the end faces of the cylindrical shell is 0. The only
differences among i., ii., and iii. is the size of the Gaussian surface (shown as dotted lines below).
r r
ldl
Q
i. ò E × dA =
=
eo
eo
l/l
E2prl/ =
r2
eo
E =
1 l
2pe o r
r1
r r
ll + rV
ll + rAl
Q
ii. ò E × dA =
=
=
eo
eo
eo
E2prl/ =
(
l/l + r pr 2 - pr1
2
eo
)/l
i.
(
2
1 l + r pr - pr1
E =
2pe o
r
2
)
ii.
r r
ldl + rdV
Q
iii. ò E × dA =
=
eo
eo
E2prl/ =
(
l/l + r pr2 - pr1
l +
1
E =
2pe o
E =
2
2
eo
(pr
1 l + lC
2pe o
r
2
)/l
lC
2
- pr1
2
r
)
iii.
Note: r =
(pr
2
2
- pr1
2
)
Q
=
V
(pr
2
Q2
2
- pr1
2
)l
=
(pr
2
lC
2
- pr1
2
)
because l C =
Q
l
E&M
2. (a) 20 V
(b)
8V
2004 C
Voltage across the capacitor, C, is zero initially. The branch with the capacitor acts as a short
circuit since charges will flow directly onto the top plate unimpeded (there is no charge,
initially, on this plate so no repulsion to charges flowing onto this plate) and away from the
bottom plate. Since the sum ov the voltages in any loop must be zero (Kirchoff's loop
rule), there must be a voltage drop of 20 V (equivalent to EMF of battery) across R 2.
A long time after the switch is closed, the voltage across the C is 12 V which is the voltage
across R1 (parallel). Since the voltage drops in the circuit must add to the EMF of the
battery (Kirchoff's loop rule), the voltage across R2 would be 8 V (20 V - 12 V).
(c) A long time after the switch is closed, when the capacitor is fully charged, the branch with the capacitor
acts as an open circuit, so no current flows into that branch. At this time, the current through R 2 is the
same as that through R1. Since the voltage across R1 is 12 V, the current can be determined from Ohm's
V
12 V
law: I =
=
= 8.0 x 10 -4 A = 0.80 mA. Since this is the same current through R2
R
15,000 W
which has a voltage of 8 V across it, the value of the resistance R 2 can be determined from Ohm's law.
V2 = IR2
8 V = (8 x 10-4 A)R2
R2 = 10,000 W = 10 kW
(d) E = ½QV, but Q = CV, so E = ½CV2
E = ½(20 mF)(12 V)2
E = 1.44 x 10-3 J = 1.44 mJ
(e)
2.0
Current
in R2
(mA)
1.5
1.0
0.5
0
0
5
10
Time (s)
(f)
Initially, with no charge on
the capacitor, the branch
with the capacitor acts as a
short circuit with no
resistance. So current
through R2 can be
determined by using Ohm's
law.
V
20 V
Io =
=
R
10,000 W
I o = 2.0 x 10 -3 A
I o = 2.0 mA
It then would follow an
exponential decay with a
time constant of 1.5 s (the
same as the voltage graph)
15 to 0.80 mA (when the
capacitor branch acts as an
open circuit).
_____ Greater than _____ Less than
_____
x The same as
The energy stored depends on the physical nature of the capacitor and the voltage across the capacitor
which is the same as the battery after a long time. Decreasing the resistance of R 2 will only reduce the
time needed to fully charge this capacitor and, thereby, reach full energy storage.
E&M
2004 C
mo I
dr
2p r
l
l m I
mo Iù
m æ1
mo æ 4
mo æ 3 ö
1ö
1ö
o
B = ò
dr =
= o Iç Iç
Iç ÷
÷ =
÷ =
ú
4 l 2p r
2p r û 4 l
2p è l
4l ø
2p è 4l
4l ø
2p è 4l ø
é m æ 3 öù
f B = BAcosq = ê o Iç ÷ú 12l 2 cos(0)
ë 2 p è 4 l øû
3. (a) db =
(
fB =
)
mo
I(9l )
2p
(b)
_____
x Counterclockwise
_____ Clockwise
The current is decreasing exponentially with time causing the flux within the rectangular loop
which is out of the page to decrease with time. Lenz's law states that the induced current will be
in a direction such that the magnetic field created by this induced current will oppose the change.
Since the current in the long straight wire is decreasing (exponentially) resulting in a decrease in
flux out of the page, the induced current will travel counterclockwise producing a magnetic field
out of the page to oppose the decrease caused by the decreasing current in the long straight wire.
(
)
d I oe -kt
dI ( t )
(c)
=
= I oe -kt (-k ) = - kI oe -kt
dt
dt
df B
m
m
= o -kI oe -kt (9l ) = (-9kl ) o I oe -kt
e =
dt
2p
2p
mo
-kt
(-9kl ) I oe
e
2p
I =
=
R
R
(
)
æ -9kl ö m o
-kt
I = ç
÷ I oe
è R ø 2p
2
æ 81m o 2 k 2 l 2 ö 2 -2kt
æ 81k 2 l 2 öæ m o ö 2 -2kt
éæ -9kl ö m o
-kt ù
÷I o e
÷÷ç ÷ I o e
(d) P = I R = êç
= çç
÷ I oe ú R = çç
2
÷
2
R
2
R
p
p
4
R
p
è
ø
è
ø
ë
û
è
ø
ø
è
æ 81m o 2 k 2 l 2 ö 2 -2kt
dE
÷I o e dt
dP =
so dE = dP × dt = çç
2
÷
dt
è 4p R ø
¥
2 2 2
¥ æ 81m k l ö 2
æ 81m o 2 kl 2 ö 2 -2kt ù
æ 81m o 2 kl 2 ö 2 -2k( ¥ )
-2k 0
-2kt
o
ç
÷
ç
÷I o e
÷
e
dt
=
E = ò çç
I
e
=
- e ()
I
ú
o
o
2
2
2
ç
÷
ç
÷
÷
0
úû 0
è 8p R ø
è 8p R ø
è 4p R ø
æ 81m o 2 kl 2 ö 2
÷I o (0 - 1)
E = çç
2
÷
è 8p R ø
2
2
[
æ -81m o 2 kl 2
E = çç
2
è 8p R
ö 2
÷I o
÷
ø
]