Download 2005 C Mechanics 1. (a) ____ increases

Mechanics
1. (a)
2005 C
____ increases ____ decreases ____
x remains the same
The force of gravity is a constant throughout the path and is always in the downward direction.
The force of air resistance, F, depends on the speed. The magnitude of the air resistance force is
directly proportional to the speed of the ball and in the opposite direction of the velocity. As the
ball moves upward its speed decreases, so, the magnitude of F is decreasing and its direction is
downward. The net force is the vector sum of these two downward forces (gravity and air
resistance) and this net force equals Ma according to Newton's Second Law of Motion. Since F is
decreasing as the ball moves upward, the net force decreases, thus, the acceleration decreases
(b) -Mg - kv = Ma
-g -
k
dv
v=
M
dt
(c) -Mg + kv = Ma = 0
v=
(d)
M
g
k
x
____longer to rise ____longer
to fall
The acceleration is greater on the way up because the forces due to gravity and air resistance are
in the same direction, thus, making a greater net force on the ball than on the way down where
these two forces act in opposite directions. The distance it rises is the same distance it falls.
Therefore, it will take longer to fall due to its smaller acceleration than it did to rise. Furthermore,
its average velocity will greater on the way up than on the way down due to this difference in the
accelerations.
(e)
Velocity
vt
O
Where vt is the terminal velocity of the ball if it reaches this value which depends v0.
tf
Time
Mechanics
2005 C
2. (a) F = G
M S mm
r2
MSm
v2
2pr
/m
, where v =
m
=
/
m
2
r/
T
r/
2
M
æ 2pr ö
G S =ç
÷
r
è T ø
(b) G
4p 2 3 2
T =
r
GM S
(c) T 2 vs. r 3 or r 3 vs. T 2
(d)
Orbital Period, T Orbital Radius, R
(seconds)
(meters)
4
8.14 x 10
1.85 x 108
(e)
(Orbital Period)2, T 2
[(seconds)2]
6.63 x 109
(Orbital Radius)3, R3
[(meters)3]
6.33 x 1024
1.18 x 105
2.38 x 108
1.39 x 1010
1.35 x 1025
1.63 x 105
2.95 x 108
2.66 x 1010
2.57 x 1025
2.37 x 105
3.77 x 108
5.62 x 1010
5.35 x 1025
T 2 (x 109 s2)
60
50
40
30
20
10
0
(f) m =
10
20
30
40
50
60
r3 (x 1024 m3)
4p 2
Dy DT 2
50 x 10 9 s 2 - 2 x 10 9 s 2
-15 s 2
10
10
=
=
=
.
x
=
, so M s = 592
. x 10 26 kg
3
24
3
3
24
3
m
GM s
Dx Dr
48 x 10 m - 2 x 10 m
Mechanics
3. (a) L = Iw
2005 C
1
L = M 1 d 2w
3
(b) L1 + L2 = L1 '+ L2 '
Iw + 0 = 0 + I 2 w 2
1
v
M 1 d 2w = M 2 d 2
3
d
v=
1 M1
dw
3 M2
(c) KE = KE'
TKE + RKE = TKE'+ RKE"
1
1
0 + Iw 2 = mv 2 + 0
2
2
2
æ 1 M1
ö
1æ1
1
2ö 2
dw ÷÷
ç M 1 d ÷w = M 2 çç
2è3
2
ø
è3 M2
ø
M1
=3
M2
(d) L1 + L2 = L1 '+ L2 '
Iw + 0 = 0 + I 2 w 2
1
v
M 1 d 2w = M 1 x 2
3
x
1 d2
v=
w
3 x
KE = KE'
TKE + RKE = TKE'+ RKE"
1
1
0 + Iw 2 = mv 2 + 0
2
2
2
æ1 d2 ö
1æ1
1
2ö 2
ç
÷
w÷
ç M 1 d ÷w = M 1 ç
2è3
2
ø
è3 x ø
1 d2
1=
3 x2
1
x2 = d2
3
x=
1
d
3
OR
since v = rw
1 M1
dw = dw
3 M2
1 M1
=1
3 M2
M1
=3
M2
E&M
2005 C
1. (a) i. The magnitude of the electric field is greatest at point C because that is where the electric field
lines are the most tightly spaced.
ii. The electric
r rpotential is greatest at point A. Electric potential is related to electric field by
V = -ò E × dl . Applying this relation to this field diagram shows the position of point A is at the
greatest potential. In other words, it is furthest from the apparent source of these field lines.
The source appears to be a negative charge to the right of the diagram.
(b) i. The electron will move to the left with an increasing speed and a decreasing acceleration.
ii. qV = 12 mv 2
16
. x 10 -19 C (10 V ) = 12 911
. x 10 -31 kg v 2
(
)
(
)
v = 187
. x 10 6 m / s
r r
r r
(c) Assuming the electric field is essentially constant over this short distance, V = -ò E × dl = -E ò dl
20 V = E(001
. m)
E = 2000 V/m
(d)
y (m)
0.1
0.08
0.06
A
B
C
0.04
D
0.02
0
0
0.02
0.04
0.06
0.08
0.1
x (m)
E&M
2005 C
2 (a) Immediately after the switch is closed, the inductor has a very large impedance (like an infinite
resistance) resulting in, essentially, no current flow in that branch (it is as if the branch is open),
so, the rest of the circuit is two resistors in series.
e
I0 =
RT
I0 =
e
R1 + R2
(b) V L = LdI / dt = IR2
dI / dt =
e
1
R2
L ( R1 + R2 )
dI / dt =
eR2
L( R1 + R2 )
(c) A long time after the switch has been closed, there is essentially no impedance in the inductor
since it opposes changes in current and the current will be essentially constant. Therefore, almost
all the current passing through R1 will flow through that branch (it is as if the branch is shorted
out) and almost no current will flow through resistor, R2.
I=
e
RT
I=
e
R1
(d)
Current
e
=I
RT
e
= I0
R1 + R2
O
æ e ö
(e) V 2 = I 0 R2 = çç ÷÷R2
è R1 ø
æR
V 2 = eçç 2
è R1
ö
÷÷
ø
Time
E&M
3. (a)
2005 C
Trial
1
Position of End Q Measured Magnetic Field (T)
(cm)
(directed from P to Q)
40
9.70 X 10-4
2
3
50
7.70 X 10-4
200
60
6.80 X 10
-4
167
4.90 X 10
-4
125
4.00 X 10
-4
100
80
4
100
5
n
(turns/m)
250
Sample Calculation for Trial 1: n =
N 100 turns
=
= 250 turns / m
l
0.40 m
(b)
10.0
9.0
8.0
B (x10-4 T)
7.0
6.0
5.0
4.0
3.0
2.0
1.0
O
50
100
150
200
250
300
n (turns/m)
B
. The current is a constant 3.0 A/turn in this experiment. Therefor, the slope
nI
divided by 3.0 A/turn will give m0.
. - 20
. ) x 10 -4 T
DB (100
Slope, m =
= 392
=
. x 10 -6 T × turns / m
Dn (254 - 50) turns / m
(c) B = m 0 nI , so, m 0 =
m0 =
m 392
. x 10 -6 T × turns / m
=
= 131
. x 10 -6 ( T × m) / A = m 0
I
30
. A / turn
(d) % error =
Experimental - Actual
Actual
. x 10
[(131
x 100% =
-6
- 4p x 10 -7
)( T × m) / A]
4p x 10 -7 ( T × m) / A
= 0.425%