Physics 221 – Spring 2007 – Final Exam
PHYSICS 221
Spring 2007
FINAL EXAM: May 1st 2007 4:30pm—6:30pm
Name (printed): ____________________________________________
ID Number: ______________________________________________
Section Number: __________________________________________
INSTRUCTIONS:
Each question is of equal weight, answer all questions. All questions are multiple choice. Choose
the best answer to each question.
Before turning over this page, put away all materials except for pens, pencils, erasers, rulers, your
calculator and “aid sheet”. An “aid sheet” is one two sided 8½×11 page of notes prepared by the
student. There is also a list of possibly useful equations at the end of the exam.
"In general, any calculator, including calculators that perform graphing numerical analysis
functions, is permitted. Electronic devices that can store large amounts of text, data or equations
are NOT permitted." If you are unsure whether or not your calculator is allowed for the exam ask
your TA.
Examples of allowed calculators: Texas Instruments TI-30XII/83/83+/89, 92+
Casio FX115/250HCS/260/7400G/FX7400GPlus/FX9750 Sharp EL9900C.
Examples of electronic devices that are not permitted: Any laptop, palmtop, pocket computer,
PDA or e-book reader.
In marking the multiple-choice bubble sheet use a number 2 pencil. Do NOT use ink. If you did
not bring a pencil, ask for one. Fill in your last name, middle initial, and first name. Your ID is
the middle 9 digits on your ISU card. Special codes K to L are your recitation section; for the
Honors sections please encode your section number as follows: H1⇒02; H2⇒13 and H3⇒25.
If you need to change any entry, you must completely erase your previous entry. Also, circle your
answers on this exam. Before handing in your exam, be sure that your answers on your bubble
sheet are what you intend them to be.
It is strongly suggested that you circle your choices on the question sheet. You may also copy
down your answers on a piece of paper to take with you and compare with the posted answers.
You may use the table at the end of the exam for this.
When you are finished with the exam, place all exam materials, including the bubble sheet, and
the exam itself, in your folder and return the folder to your recitation instructor. No cell phone
calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone
must hand in their work; their exam is over.
Total number of questions is 28 but score will be out of 27 (so one
question is “extra credit”). Questions are numbered 56-83.
Best of luck, David Atwood and Paula Herrera-Siklody
Physics 221 – Spring 2007 – Final Exam
Neglect air resistance in all problems.
56. The acceleration of a particle constrained to move along the x-axis
is ax = 4.0 m/s 4 t 2 . At t = 0, the particle is at x = 0 and its velocity is −3.0 m/s.
(
)
Where is the particle at t = 3.0 s?
(A)
(B)
(C)
(D)
(E)
x = 18 m
x = 30 m
x = 96 m
x = 150 m
x = 320 m
We must use integrals. First, v(t) (dropping units for simplicity):
t
v = v(0) + ∫ adt
0
t
v = −3.0 + ∫ ( 4.0 ) t 2 dt
0
t3
v = −3.0 + ( 4.0 )
3
Then, x(t):
t
x = x(0) + ∫ vdt
0
⎡
t3 ⎤
3.0+
4.0
−
(
)
⎥ dt
∫0 ⎢⎣
3⎦
t4
x = − ( 3.0 ) t +
3
x = 0+
t
At t = 3.0 s:
x(3.0 s) = − ( 3.0 ) (3.0) +
(3.0) 4
= 18 m
3
Physics 221 – Spring 2007 – Final Exam
57. You want to kick a ball into your neighbor’s yard, over the 2.0-m high fence. The
ball leaves the ground at 8.0 m/s and just barely clears the fence when it is at the
top of its trajectory. At what angle was the ball moving relative to the horizontal
right after the kick?
vy = 0
(A)
(B)
(C)
(D)
(E)
22°
38°
44°
52°
63°
h
v0
θ
Using conservation of energy (or the v2 projectile motion formula):
v 2 − v02 = 2 gh
But vx = v0 x , so
v y2 − v02y = 2 gh
0 − v02 sin 2 θ = 2 gh
sin θ =
θ = sin −1
2 gh
v0
2 gh
2(9.8)(2)
= sin −1
= 52°
v0
8
Physics 221 – Spring 2007 – Final Exam
58. A coin is placed near the edge of a turntable with radius 30 cm. The turntable
makes 20 turns in one minute. What is the minimum coefficient of static friction
needed to ensure that the coin moves along with the turntable?
(A)
(B)
(C)
(D)
(E)
0.13
0.42
0.51
0.70
0.82
The friction is the force that produces the required centripetal acceleration:
f = mRω 2
Since static friction cannot be larger than its maximum value,
f ≤ μ s N = μ s mg
⇓
μ s mg ≥ mRω 2
μs ≥
Rω 2
g
The angular speed needs to be in rad/s:
ω = 20
turns 2π rad 1 minute
2π
rad/s
=
minute 1 turn 60 seconds
3
2
⎛ 2π ⎞
0.3) ⎜
(
2
⎟
Rω
3 ⎠
⎝
μs ≥
=
= 0.13
g
9.8
Physics 221 – Spring 2007 – Final Exam
59. Two blocks 1 and 2 (m1 = 4m2) on an incline are connected through an ideal,
massless string that goes through an ideal, massless pulley as shown in the figure.
The incline makes an angle of 30° with the horizontal. The blocks are released
from rest and there is no friction between the blocks or between block 1 and the
incline.
T
T
2
m2g sinθ
1
m1g sinθ
30°
Find the acceleration of block 1 right after it is released.
(A)
(B)
(C)
(D)
(E)
0
2.9 m/s2
5.9 m/s2
7.4 m/s2
9.8 m/s2
In the free-body diagram shown above, only the forces along the incline have been drawn
(the forces perpendicular to the incline –normal forces and the rest of the weights- are
irrelevant for this problem).
Newton’s second law for each block, taking positive down for 1 and up for 2 (it’s
intuitively how the system will move):
m1 :
m1 g sin θ − T = m1a
m2 :
T − m2 g sin θ = m2 a
Adding both equations:
m1 g sin θ − m2 g sin θ = ( m1 + m2 ) a
a=
m1 − m2
g sin θ
m1 + m2
Since m1 = 4m2, a =
3
g sin θ = 2.9 m/s 2
5
Physics 221 – Spring 2007 – Final Exam
60. A 100g ball is launched vertically using a spring as shown below. The spring is
compressed by 10cm from its relaxed length so that the ball is at ground level.
The ball rises to a maximum height of 1m. What is the force constant, k, of the
spring?
1m
100g
10 cm
(A) 49 N/m
(B) 98N/m
(C) 196 N/m
(D) 392 N/m
(E) 588 N/m
Energy is conserved. Since the kinetic energy is zero both at the bottom and at the top,
1 2
kx = mgh
2
2mgh 2(0.1 kg)(9.8 m/s 2 )(1.0 m)
=
= 196 N/m
k=
x2
(0.1 m) 2
Physics 221 – Spring 2007 – Final Exam
61. Consider the two situations depicted below where a block is attached to an ideal
string which winds around the given pulley without slipping. In both cases the
mass of the pulley is m, but the blocks have different masses m and M (see
figure). The blocks accelerate downwards at the same rate in both cases. What is
the ratio M:m?
Solid uniform disk of
Hoop of mass m (all mass
mass m
concentrated on rim)
(A) 3:1
(B) 2:1
(C) 1:1
(D) 1:2
(E) 1:3
T
T
T
T
m
M
mg
Mg
The free-body diagram is formally the same in both cases. The only differences
between the two cases is the mass of the block and the moment of inertia of the
pulley:
1
I = γ mR 2
with γ =
in case 1
2
γ = 1 in case 2
We’ll save time if we solve the situation in the more general case where the mass
of the block is not equal to the mass of the pulley:
τ net = Iα with I = γ mR 2
Fnet = ma
a = Rα
Rotational part (torque):
TR = γ mR 2α
TR = γ mR 2
T = γ ma
a
R
[1]
Physics 221 – Spring 2007 – Final Exam
Translation of the block:
Mg − T = Ma
[2]
Putting [1] and [2] together:
Mg − γ ma = Ma
a=g
M
M +γm
For the solid disk, M = m and γ = ½, so a =
For the hoop, γ =1, so a = g
M
.
M +m
Both accelerations should be the same:
M
2
=
M +m 3
⇒
M
=2
m
2
g.
3
Physics 221 – Spring 2007 – Final Exam
62. Two particles of masses m1 and m2 = 4m1 head toward each other along the x axis
with velocities v1x = 30 m/s and v2x = –10 m/s. The particles collide elastically.
What is the speed of particle 1 right after the collision?
(A) 0
(B) 6.0 m/s
(C) 10 m/s
(D) 25 m/s
(E) 34 m/s
This is a 1D elastic collision. Total linear momentum is conserved, and total
kinetic energy is also conserved. Instead of using conservation of kinetic energy,
though, we can use the relation between relative velocities before and after the
collision, which will make the algebra easier:
m1v1i + m2 v2i = m1v1 f + m2 v2 f
v1i − v2i = − ( v1 f − v2 f
)
m1 (30) + 4m1 (−10) = m1v1 f + 4m1v2 f
30 − (−10) = − ( v1 f − v2 f
)
−10 = v1 f + 4v2 f
40 = −v1 f + v2 f
Solving this system yields:
v1 f = 34 m/s
v2 f = 6.0 m/s
Physics 221 – Spring 2007 – Final Exam
63. John wants to ride a jet ski across a river that is 1.0-km wide and flows from
north to south at a rate of 5.0 km/h. Starting at point A on the west bank he
would like to drive his jet ski in a straight line to point B due east of point A.
If the jet ski can move at a speed of 10 km/h with respect to the water, how
long does it take to cross the river?
(A)
(B)
(C)
(D)
(E)
0.013 hours
0.089 hours
0.10 hours
0.12 hours
0.20 hours
5.0 km/h
A
John's
trajectory
Let us draw a diagram of the velocities:
vski relative to shore
vwater relative to shore
(5 km/h)
vski relative to water
(10 km/h)
From the diagram, using Pythagoras’s theorem:
2
2
vski , shore = vski
, water − vwater , shore
vski , shore =
(10 km/h ) − ( 5 km/h )
2
Thus the time to cross the river is:
t=
d
vski , shore
=
1 km
= 0.12 h
8.7 km/h
2
= 8.7 km/h
B
Physics 221 – Spring 2007 – Final Exam
64. A uniform sphere of mass 5M and radius R is glued to a uniform rod of mass 3M
and length 2R. The rod is perpendicular to the surface of the sphere. The system
rotates about axis P, which is perpendicular to the page through a point in the
surface of the sphere (see figure). What is the moment of inertia of this rigid body
for this axis?
(A) 3MR 2
(B) 6MR 2
(C) 8MR 2
(D) 11MR 2
(E) 35MR 2
P
R
5M
2R
3M
We need the moment of inertia about an axis through the center of mass for each
object (sphere and rod) and the corresponding parallel axis theorem correction
for each.
2
(5M ) R 2 + ( 5M ) R 2 = 7 MR 2
5
1
2
2
= (3M ) ( 2 R ) + 3M ( 3R ) = 28MR 2
12
I sphere =
I rod
I total = 35MR 2
Physics 221 – Spring 2007 – Final Exam
65. The graph below shows a position-dependent net force acting on a 1.0-kg block
moving along the x-axis. Initially the block is at x = 0 m, moving in the +x
direction. When it reaches 8.0 m, it has a kinetic energy of 30 J. What was the
initial kinetic energy of the block?
(A) 12 J
(B) 18 J
(C) 24 J
(D) 36 J
(E) 48 J
Net Force
3N
2N
1N
1m
2m
3m
4m
5m
6m
7m
8m
Work-kinetic-energy theorem: W=Kf− Ki
The work is the area under the F-x curve between x = 0 and x = 8 m.. Breaking
the figure into two triangles and one rectangle:
W=
1
1
(2 m)(3 N) + (2 m)(3 N) + (2 m)(3 N) = 12 J
2
2
The initial kinetic energy is thus:
K i = K f − W = 30 J − 12 J = 18 J
x
Physics 221 – Spring 2007 – Final Exam
66. A uniform plank of length 4.0 m and mass 4.0 kg is placed on the edge of a table
with 75% of its length on the table. We want to place a 6.0 kg cube of side 1.0 m
on the free end of the plank without tipping over the system. What is the
maximum allowed value of d?
1.0 m
x=0
4.0 kg
CM
4.0 m
(A)
(B)
(C)
(D)
(E)
6.0 kg
d?
CM
x
1.0 m
0.052 m
0.17
0.33
0.67
The system tips over for any (positive) value of d.
In order to have equilibrium, the center of mass of the whole system (plank +
block) must be on the table. If we take +x to the right and x=0 at the edge of the table
(see figure), the condition is xCM ≤ 0.
In this reference frame, the centers of mass of the plank and block are respectively
x plank = −1.0 m
xblock = d + 0.5 m
The CM of the entire system:
xCM =
x plank m plank + xblock mblock
xCM ≤ 0
m plank + mblock
⇒
x plank m plank + xblock mblock ≤ 0
(−1.0 m)(4.0 kg) + (d + 0.5 m)(6.0 kg) ≤ 0
d ≤ 0.17 m
Physics 221 – Spring 2007 – Final Exam
67. A 1.0-kg particle attached to a spring with k = 400 N/m oscillates with simple
harmonic motion along the x axis. As it passes the equilibrium position, the
velocity of the particle is vx = 2.0 m/s. What is the position of the particle 2.0 s
after?
(A)
(B)
(C)
(D)
(E)
x = –6.7 cm
x = –3.2 cm
x = +1.2 cm
x = +7.5 cm
x = +10 cm
As a function of time, the velocity of the particle is:
v = − Aω sin (ωt + ϕ )
We know Aω because this is the speed of the particle at the equilibrium position
(and the maximum possible speed), vmax = Aω = 2.0 m/s
k
(400 N/m)
=
= 20 s −1
m
1.0 kg
v
2.0 m/s
Therefore, the amplitude is A = max =
= 0.10 m = 10 cm
ω
20 s −1
But we need the phase angle φ. For this, we use the initial conditions on position
and velocity. At t=0,
We can find ω =
0 = A cos ( 0 + ϕ )
cos ϕ = 0
2.0 = − Aω sin ( 0 + ϕ )
sinϕ < 0
The solution to this system is ϕ = −
π
2
We can then go back to the position equation:
x = A cos (ωt + ϕ )
π⎞
⎛
x = (10 cm ) cos ⎜ 20 s −1 t − ⎟
2⎠
⎝
x(2.0 s) = 7.5 cm
(
)
Physics 221 – Spring 2007 – Final Exam
The solar system of the Ugly Star (mass 1030 kg) has two planets, 1 and 2, that
have circular orbits of radii r1 = 1011 m and r2 = 2 r1. A 1000-kg spacecraft is to
be sent from planet 1 to planet 2. The spacecraft leaves planet 1 when this is in the
position shown in the figure, and reaches planet 2 on the other side of the star,
traveling along the trajectory shown in the figure below. Note that the velocity
vector of the spacecraft is parallel to the velocity vector of planet 1 at launch and
parallel to the velocity vector of planet 2 on arrival.
Planet 1 (position upon Ugly Star
launching)
Planet 2 (position upon
spacecraft arrival)
spacecraft
68. How long will the spacecraft trip take?
(A)
(B)
(C)
(D)
(E)
700 hours
2650 hours
6200 hours
11200 hours
23900 hours
The path of the satellite is half of an elliptical orbit about the Ugly Star with semir +r
major axis a = 1 2 = 1.5 ×1011 m . Thus the time needed for the trip is half of
2
the period of the entire orbit,
ttrip =
=
T
2
2π
2
a3
GM Ugly Star
(1.5 ×10
11
=π
( 6.7 ×10
−11
m
Nm 2 /kg 2
)
)(10
3
30
kg
)
= 6200 h
Physics 221 – Spring 2007 – Final Exam
The situation below refers to the next two questions:
Two charges are fixed at the positions shown below.
y
Q1 = 3.0μC
Q2 = −3.0μC
Q1
Q2
x
x = 3.0 cm
69. Find the force on Q1 by Q2.
(A)
(B)
(C)
(D)
(E)
3.0 ×107 i N
−3.0 × 107 i N
90i N
−90i N
None of the above
The interaction is attractive, so the force on Q1 must point to the right (ie,
direction +x). So (B) and (D) cannot be the right answer.
The magnitude of the force is given by Coulomb’s law:
F = ke
Q1Q2
r2
= (9.0 × 109 Nm 2 /C2 )
= 90 N
Thus the answer is (C).
(3.0 ×10−6 C)(3.0 × 10−6 C)
(3.0 × 10−2 m) 2
Physics 221 – Spring 2007 – Final Exam
70. Determine the y-component of the total electric field at point (x, y) = (0, 3.0 cm).
(A)
(B)
(C)
(D)
(E)
Ey = 0
Ey = 1.2 × 107 N/C
Ey = 1.5 × 107 N/C
Ey = 1.7 × 107 N/C
Ey = 1.9 × 107 N/C
We need to add the E fields by Q1 and by Q2:
E1
y
E2
Q1 = 3.0μC
Q2 = −3.0μC
a 2
a
Q1
Q2
x
a = 3.0 cm
The magnitude of each of these electric fields is:
G
Q
(3.0 × 10−6 C)
E1 = ke 21 = (9.0 ×109 Nm 2 /C2 )
= 3.0 ×107 N/C
−2
2
a
(3.0 × 10 m)
G
Q1
(3.0 × 10−6 C)
9
2
2
E2 = ke
=
(9.0
×
10
Nm
/C
)
= 1.5 × 107 N/C
2
−2
2
2(3.0 × 10 m)
a 2
(
)
We only need the y components:
E y = E1 − E2 cos 45° = 1.9 × 107 N/C
Physics 221 – Spring 2007 – Final Exam
71. Shown below are four conducting spheres, each with a total charge Q. Point P is
in each case at the same distance from the center of the sphere. In cases 1 and 2,
point P is outside the spheres. In cases 3 and 4, it is inside the spheres. (Each
sphere is to be considered an isolated system, very far from any other charged
object)
4
3
2
1
P
P
P
P
Rank the magnitude of the electric field at point P in each case.
(A)
(B)
(C)
(D)
(E)
E1 = E2 = E3 = E4
E1 < E2 = E3 = E4
E1 < E2 < E3 < E4
E1 = E2 > E4 > E3
E1 = E2 > E3 = E4
E3 = E4 = 0 because these points are inside a conductor in equilibrium.
For 1 and 2, we can use the spherical shell theorem (the electric field produced
by a spherically symmetric charge distribution, at points outside the charged
object, is the same as if all the charge was concentrated at the center of the
system). Thus for both 1 and 2, the electric field is the same (same total charge Q,
same distance from the center of the sphere to point P), but not zero (unless Q =
0).
Physics 221 – Spring 2007 – Final Exam
72. A sphere of radius R has a charge Q uniformly distributed throughout its volume.
Consider an imaginary sphere of radius R/2 with center located at distance R/2
from the real sphere (see figure). What is the electric flux through the imaginary
sphere?
(A) 0
Q
(B)
ε0
R/2
Q
2ε 0
Q
(D)
4ε 0
Q
(E)
8ε 0
(C)
R
R/2
This is a straightforward application of Gauss’s law. The charge
enclosed by the little, imaginary sphere is
qenclosed = (charge density)(volume sphere)
4 ⎛R⎞
π⎜ ⎟
=
4
π R3 3 ⎝ 2 ⎠
3
Q
=
8
Q
Thus Φ =
qenclosed
ε0
=
Q
8ε 0
3
Physics 221 – Spring 2007 – Final Exam
73. Two very thin surfaces with equal uniform charge density σ = +8.85 nC/m2 are
placed parallel to the yz axes, at x1 = 0 and at x2 = 2.0 cm, respectively. If we take
the electric potential to be zero at x = 0, what is the electric potential at x = 6.0
cm?
(A)
(B)
(C)
(D)
(E)
0V
−60 V
60 V
−40 V
40 V
σ
σ
x = 2.0 cm
●
x
x = 6.0 cm
The electric field is the superposition of the contribution from each sheet (
σ
in the
2ε 0
appropriate direction). Overall:
⎧ σ
⎪− ε
⎪⎪ 0
Ex = ⎨ 0
⎪σ
⎪
⎪⎩ ε 0
x<0
0 < x < 2 cm
x > 2 cm
Since E = 0 between the plates, the potential is constant in that region. Thus the
potential difference between 0 and 6 cm is the same as the potential difference
between 2 cm and 6 cm. For x > 2 cm, the electric field is uniform, so
ΔV = Ex Δx =
σ
8.85 ×10−9 C/m 2
Δx =
(6 − 2) ×10−2 m = 40 V
−12
2
2
ε0
8.85 ×10 Nm /C
This is just the magnitude of the change in potential. Since the electric field points in
the positive x direction, the potential must be decreasing in that direction, so V(6 cm)
< V(2 cm). Since V(2 cm ) = V(0) = 0, we conclude V(6 cm) = -40 V.
Physics 221 – Spring 2007 – Final Exam
74. The figure below shows the electric field lines produced by an electric dipole.
Compare the electric potential at points A, B and C.
B
A
(A)
(B)
(C)
(D)
(E)
C
VB < VA < VC
VA = VB < VC
VB < VC < VA
VA > VB = VC
VA < VB = VC
All the points on the dotted line have the same potential because the electric field
is perpendicular to the line (no electric field component along the line means no
changes in potential). So VB = VC.
An electric field line flows from B to A. Thus for a trajectory along this line, the
electric potential is constantly decreasing (the electric field always points in the
direction of decreasing potential). Thus VB > VA.
Physics 221 – Spring 2007 – Final Exam
75. Three capacitors are connected to an ideal battery as shown below. When the
capacitors are all fully charged, a dielectric with κ > 1 is inserted between the
plates of capacitor C2. Due to this, and compared with the values before the
dielectric was inserted, the charge in C1 __________________ and the charge in
C3 __________________
.
C2
V0
C1
C3
(A) decreases, decreases
(B) decreases, remains the same
(C) increases, remains the same
(D) remains the same, increases
(E) remains the same, decreases
No matter what we do with C3, the potential difference between the plates of C1 is
fixed by the battery, and thus remains the same. Therefore the charge on C1
remains the same.
The charge on C3 must be the same as the charge on C2, and the same as that of
the combined equivalent capacitor C2-3 (series connection). Since the capacitance
of C2 increases due to the dielectric, C2-3 also increases. But again, the potential
difference across C2-3 is fixed by the battery. So the charge must increase, Q2-3 =
C2-3 V0.
Physics 221 – Spring 2007 – Final Exam
76. An electric dipole made of two charges ± 3.0 ×10−18 C separated a distance
1.0 ×10−6 m is placed between the plates of a parallel plate capacitor that is
connected to a 9.0-V battery. The plates of the capacitor have an area of 400 m2
and are separated by an air-filled gap 0.10 mm wide. The dipole moment makes
an angle of 60° with the plates of the capacitor. Determine the magnitude of the
torque on the dipole.
(A) 0
(B) 4.1×10−20 Nm
(C) 8.2×10−20 Nm
(D) 1.4×10−19 Nm
(E) 2.7×10−19 Nm
60° p
E
d
The dipole moment is
(
)(
)
p = Qd between charges = 3.0 × 10−18 C 1.0 × 10−6 m = 3.0 ×10−24 Cm
The electric field between the plates is dictated by the potential difference
between its plates and the distance between them:
E=
V
9.0 V
=
= 9.0 ×104 N/C
d 0.10 ×10−3 m
The torque on the dipole is
G
τ = Ep sin ( angle between E and p )
(
)(
G
)
= 9.0 ×104 N/C 3.0 ×10−24 C sin 30°
= 1.4 × 10−19 N
Physics 221 – Spring 2007 – Final Exam
The situation below refers to the next two questions:
An ideal battery and four resistors are connected as shown below. All wires are ideal.
The ideal ammeter reads 0.40 A.
R1 = 10 Ω
R2 = 20 Ω
R3 = 30 Ω
R4 = 40 Ω
R2
R1
R3
A
R4
V0
77. Find the equivalent resistance of the circuit.
(A)
(B)
(C)
(D)
(E)
6.7 Ω
17 Ω
26 Ω
55 Ω
100 Ω
Resistors 1 and 2 are in series:
R1− 2 = R1 + R2 = 30 Ω
Resistor R1-2 is in parallel to R3:
1
1
1
=
+
R1− 2−3 R1− 2 R3
R1− 2−3 = 15 Ω
Resistor R1-2-3 is in series with R4:
Rall = R1− 2−3 + R4 = 55 Ω
Physics 221 – Spring 2007 – Final Exam
78. Determine the power dissipated in resistor R3.
(A)
(B)
(C)
(D)
(E)
1.2 W
3.1 W
4.8 W
8.8 W
16 W
Since both R1-2 and R3 are 30 Ω each, the current splits evenly between the
two branches. The current through R3 is thus 0.40/2 = 0.20 A. The power
dissipated in R3 is thus:
P3 = I 32 R3 = ( 0.20 A ) ( 30 Ω ) = 1.2 W
2
Physics 221 – Spring 2007 – Final Exam
79. In the circuit below, all resistors are 10 Ω and all batteries are 10 V. The wires
and the battery are ideal. Determine the current through the central wire.
I2
I1
I3
(A)
(B)
(C)
(D)
(E)
0.33 A
0.67 A
1.0 A
1.3 A
2.0 A
Due to the left-right symmetry of the circuit, I1 = I2, so in fact we only have
two unknowns. We need to equations using Kirchhoff’s laws.
Junction equation: I1 + I 2 = I 3
⇒
2 I1 = I 3
Loop equation (see loop in the figure):
10 V − (10 Ω ) I1 + 10 V − (10 Ω ) I 3 − (10 Ω ) I1 = 0
20 − 20 I1 − 10 I 3 = 0
2 − 2 I1 − I 3 = 0
Using [1] and [2],
2 − 2 I3 = 0
I 3 = 1.0 A
[2]
[1]
Physics 221 – Spring 2007 – Final Exam
Laboratory Final
80. Consider the motion of a cart like the one you used in lab,
and which is adjusted to have significant frictional drag.
Assume that the friction is well described by the "laws" of
(dry) friction that you studied this semester. Also assume
that the cart is given an initial velocity toward the left on
the horizontal track and then released. (Assume the sensor
gives positions relative to the X axis illustrated in the figure).
X
0
Which of the following best illustrates the acceleration, aX, versus time of the cart
during the same time interval?
ax
ax
ax
t
0
A
ax
t
B
ax
t
C
t
D
t
E
If the cart is moving to the left, the frictional force points to the right (i.e., is
positive). Thus the acceleration will also be positive. Since the magnitude of
the frictional force is constant, so is the acceleration.
Physics 221 – Spring 2007 – Final Exam
81. Consider a hollow metal sphere mounted on a thin insulating rod. Using
standard apparatus (e.g., an electrophorus), the largest possible electric
charge is placed on the sphere.
Where is the electric field largest, and for a sphere of a given size, what
factor determines the magnitude of the maximum charge?
location where electric
field is largest
A
B
C
D
E
center of sphere
surface of sphere
surface of sphere
center of sphere
none of the above
factor that determines the
magnitude of the maximum
charge
dielectric strength of air
type of metal
dielectric strength of air
type of metal
The electric field is strongest on the surface. It decreases as you move away
from the sphere, and it is zero anywhere inside the sphere because it is a
conductor.
The maximum charge is dictated by the maximum electric field that the
insulator around the sphere (air) can withstand without becoming a conductor
(dielectric breakdown). This maximum electric field is what is called the
dielectric strength of air.
Physics 221 – Spring 2007 – Final Exam
82. Using the rotating wheel apparatus such as you used in lab, a disk (which is not
spinning) is dropped concentrically upon the wheel as it is rotating freely. Assume
that the disk has a moment of inertia of 0.5I0 , where I0 is the moment of inertia of
the rotating wheel. Which of the following graphs best represents the angular
velocity, ω, as a function of time before, during, and after the disk is dropped?
ω
A
ω
t
B
ω
C
t
ω
t
D
ω
E
t
When the disk is dropped on the wheel, no external torque is exerted, so angular
momentum is conserved:
I wheelω0 = ( I wheel + I disk ) ω f
ω f = ω0
I wheel
I0
2
= ω0
= ω0
I wheel + I disk
I 0 + 0.5 I 0 3
So the answer is D. Note that B is qualitatively correct, too, but the angular speed
drops to about 1/3 of its initial value instead of 2/3.
The value of ω is decreasing all the time (slight slope in the lines) because of
friction and air resistance.
t
Physics 221 – Spring 2007 – Final Exam
83. Consider two air pucks (made of some unknown
material) which can slide upon a smooth
horizontal surface with little friction, leaving
trails of spark marks, with marks produced at a
fixed frequency. The pucks are of equal mass,
and are pushed (and released) toward one
another, and then collide. Assume that the pucks
rotate little before and after the collision.
For the record shown above, select the comment listed below which is most
appropriate.
(A) The data looks O.K., although the collision clearly is not elastic.
(B) The data must be invalid since clearly momentum is not conserved.
(C) The data must be invalid since the collision clearly is not elastic.
(D) The data looks O.K.; both momentum and mechanical energy appear to be
conserved.
(E) One must know the time between sparks to make a definitive statement.
NOTE: By the data being invalid is meant that, for the situation described, such
data is impossible. To obtain such data, then some large extraneous factor must be
at work (such as hidden magnets, angels, etc.)
Both pucks seem to be moving at approximately equal speed (the spacing
of the dots is the same for both) before and after the collision. From the
trajectories before the collision, the net linear momentum should be zero.
However, after the collision, the total linear momentum seems to be nonzero vector pointing down and to the left. Thus something went wrong in
our experiment, because linear momentum appears to not be conserved.