PHYSICS 221
Spring 2004
Final Exam: May 5 2004 12:00pm—2:00pm
Solutions
Part A: Questions 76-95 (20 questions):
[Based on Lectures 1-41]
G
G
[76] Vectors A and B start at a corner of a cube of edge
G
length 1. Vector A lies along the edge of the cube. Vector
G
B goes to the opposite corner of the cube. What is the
G G
value of A ⋅ B ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 2
y
1
x
Take the origin at the base of the two vectors and define the x axis in the direction of the
G
G
a vector using the co-ordinate system shown. Thus A = iˆ, B = iˆ + ˆj + kˆ . Calculating
G G
the dot product: A ⋅ B = (iˆ) ⋅ (iˆ + ˆj + kˆ) = 1
[77] A rock is thrown upwards from a bridge 20.0m above level ground. The initial
upwards velocity of the rock is 10.0m/s. How fast is the rock moving when it hits the
ground? Neglect air resistance:
(A) 17.2 m/s
(B) 19.8 m/s
(C) 22.2 m/s
(D) 25.0 m/s
(E) 27.2 m/s
Using the v squared formula for constant acceleration:
v 2f = vi2 + 2 g∆x
v f = vi2 + 2 g∆x = (10.0m / s ) 2 + 2(−9.8m / s 2 )(−20m)
= 22.2m / s
Physics 221 2004 S Final Exam Solutions
z
Page 1 of 23
[78] The Figure at right shows the
trajectories of three cannon balls fired
simultaneously. In which order do they
strike the ground? Neglect air resistance.
(A) P Q R
(B) R Q P
(C) R P Q
(D) Q P R
(E) Cannot be determined without
more information.
The time it takes for the cannon ball to hit the ground along any given trajectory is equal
to twice the time it takes to fall from a height of the peak of the trajectory. The order will
thus be the same as the order of the peaks of the parabolas: R P Q
Physics 221 2004 S Final Exam Solutions
Page 2 of 23
[79] Four blocks of mass mP=4kg, mQ=3kg, mR=2kg, mS=1kg are on a frictionless
horizontal surface as shown on the figure below. The blocks are connected by ideal
massless strings. A force FL=30N is applied to the left block and is directed to the left.
Another force FR=50N is applied to the right block, and is directed to the right. What is
the magnitude of the tension T in the string between mQ and mR.
T=?
FL=30N
mP=4kg
mQ=3kg
mR=2kg
mS=1kg
FR=50N
(A) T=14N
(B) T=20N
(C) T=30N
(D) T=36N
(E) T=44N
Take right as positive. The acceleration of the whole train is (20N)/(10kg)=2m/s². This is
the acceleration of the first two blocks which has a mass of 7kg so the net force on those
blocks is 14N. The applied force is –30N therefore the force from the center string is
+44N. The tension in the string is therefore 44N.
[80] The body that is suspended by a massless rope has weight of 75N. The rope is
pulling the body up at decreasing speed. Is the tension in the rope is
(A) 75 N
(B) greater then 75 N
(C) less than 75 N
(D) there is not enough information to distinguish between the answers A, B and C
(E) 0 N
The block is accelerating downwards so the net force on the block must be downwards. If
the weight is 75N the upwards force due to the tension must be less than 75N.
Physics 221 2004 S Final Exam Solutions
Page 3 of 23
[81] A block of mass 10kg lies at rest on a floor. The coefficients of static friction
between the block and the floor is µS=0.4. The coefficient of kinetic friction between the
block and the floor is µK=0.3. What is the magnitude of the frictional force on the block?
(A) 40N
(B) 30N
(C) 4N
(D) 3N
(E) 0
By Newton’s 1st law, if the block is stationary and thus not accelerating then the net force
is 0. The only possible force in the horizontal direction is friction which is therefore 0.
[82] A simple pendulum has a string of length L=2m and a bob of mass 4kg. The bob
is pulled back through an angle of 30º and released from rest. What is
•
the kinetic energy of the bob when the string is vertical?
(A) 78.4J
2m
(B) 39.2J
30º
(C) 67.7J
(D) 10.5J
(E) 21.0J
4 kg
K=?
Take the lowest position of the bob as the zero for potential energy. If L
is the length of the string and θ is the angle it is pulled back to then the initial mechanical
energy is E = U i + K i = mgL(1 − cosθ ) + 0 = mgL(1 − cosθ ) . This stays the same
through the swing but at the bottom of the swing it is converted all to kinetic energy since
the potential energy is 0. The final kinetic energy is thus
K f = mgL(1 − cosθ ) = (4kg )(9.8m / s 2 )(2m)(1 − cos(30o )) = 10.5 J
Physics 221 2004 S Final Exam Solutions
Page 4 of 23
[83]
Particles Q, R, S, T have the masses and momenta given in the following table:
Particle
Q
R
S
T
Mass
m0
2m0
m0
2m0
Momentum
p0
p0
2p0
2p0
Particle Q has mass m and momentum p0. Particle
R has mass 2m0 and momentum p0. Particle S has
mass m0 and momentum 2p0. Particle T has mass
2m0 and momentum 2p0.
If the kinetic energy of these particles is KQ, KR, KS and KT respectively. Which of the
following statements concerning the relative size of the kinetic energy is true?
(A) KP<KQ<KR<KS
(B) KP<KR<KQ<KS
(C) KQ<KP<KS<KR
(D) KP=KQ=KS<KR
(E) KR<KQ<KT<KS
Applying the formula for kinetic energy K =
p2
to each of the particles we find that
2m
 p2 
KQ =  0 
 2m0 
1  p02 
KR = 

2  2m0 
 p2 
K S = 4 0 
 2m0 
 p2 
KT = 2  0 
 2m0 
The correct ordering is therefore KR<KQ<KT<KS
Physics 221 2004 S Final Exam Solutions
Page 5 of 23
[84] The first diagram below shows xcomponent of force as a function of
position ( Fx ) for a particle confined to
move along the x-axis. Which of the
following diagrams correctly indicates the
corresponding graph of potential energy U
as a function of position.
(A) Graph A
(B) Graph B
(C) Graph C
(D) Graph D
(E) Graph E
The force is always to the right so the potential must be always decreasing hence only E
or C could be correct. The first bump in the force graph is bigger and since U is the
negative integral of force, Graph C which has the big downslope to the left is the only
possibility.
Physics 221 2004 S Final Exam Solutions
Page 6 of 23
2m
[85] Consider a square slab of uniform
density and mass 10kg is resting on its
edge. How much work does it take to
rotate the slab 45º so that it is balanced on
its corner?
(A) 98 J
(B) 138 J
(C) 41 J
(D) 82 J
(E) 196 J
2m
The potential energy of a system is U=mgh where m is the mass and h is the height of the
center of gravity. In this case the center of gravity is the center of the square so initially
hi=1.00m while in the final state hf=1.41m. The work required is thus the difference in
potential energy W=(Uf-Ui)=mg(hf-hi)=41J.
[86] Consider a system that consists of four 2kg masses
connected by massless rods and arranged in a square with edge
length 4m. What is the moment of inertial of the system about the
axis going through the diagonal of the square?
(A) 4 kg m²
(B) 8 kg m²
(C) 16 kg m²
(D) 32 kg m²
(E) 64 kg m²
2kg
4m
4m
2kg
4m
4m
The two masses on the axis do not contribute to the moment of inertia. The other two are
distance 8 m from the axis and so the contribution of EACH of them is mr²=16 kg m².
There are two of these masses so the total is I=32 kg m²
Physics 221 2004 S Final Exam Solutions
Page 7 of 23
2kg
2kg
[87] A skater is spinning on a vertical axis with her arms extended. She pulls in her
arms and her angular velocity increases. Which of the following statements most
accurately describes the situation:
(A) The rotational kinetic energy of the skater remains the same as she pulls in her
arms.
(B) The angular momentum of the skater remains the same as she pulls in her arms.
(C) The period of the skater’s rotation remains the same as she pulls in her arms.
(D) Both the rotational kinetic energy and angular momentum of the skater remains
the same as she pulls in her arms.
(E) The rotational kinetic energy of the skater decreases as she pulls in her arms.
Angular momentum is preserved since pulling in arms exerts no external torque. The
rotational KE increases because the act of pulling in the arms does work. You can see this
form the relation K=L²/I for a rigid body that if I decreases due to pulling in arms, K must
increase if L is constant. Thus, only B is correct.
Physics 221 2004 S Final Exam Solutions
Page 8 of 23
[88] A 1.00m long rod of uniform mass distribution
weighing 200N is supported at its ends by wires A and B.
A 100N weight is attached to the rod 10cm from wire A.
What is the tension in wire A if the system is in
equilibrium.
(A) 95N
(B) 100N
(C) 109N
(D) 190N
(E) 200N
1.00m
200N
100N
Take the attachment of wire B as the center of torque. Since the bar is in equilibrium, the
net torque is 0. The attached weight and the weight of the bar give positive torques while
the tension in the wire gives a negative torque. Recalling that the weight of the bar acts at
its center of mass which in this case is at its center, we write down the balance:
τ net = 0 = τ A + τ weight + τ bar
= −T (1.00m) + (100 N )(.90m) + (200 N )(0.5m)
⇒ T = (90 Nm + 100 Nm) /(1.00m) = 190 N
Physics 221 2004 S Final Exam Solutions
Page 9 of 23
[89] Consider an asteroid which orbits the sun in
the elliptical orbit shown. The closest approach to the
sun at point A is 2AU while the farthest the asteroid
moves from the sun is 8AU at point B. What is the
ratio between the velocity of the asteroid at point A,
vA and the velocity of the asteroid at point B, vB?
(A) v A : v B = 1 : 1
(B) v A : v B = 2 : 1
(C) v A : v B = 4 : 1
(D) v A : v B = 8 : 1
(E) v A : v B = 16 : 1
Sun
•
2AU
•
8AU
From Kepler’s second law, the asteroid’s angular momentum about the sun is constant.
At both point A and B the velocity is perpendicular to the radius vector. Thus the angular
momentum is given by L = rAv A = rB vB thus v A : vB = rB : rA = 4 : 1 .
[90] What is the angular frequency of small oscillation of a square picture with side
length 1m which is hung by the corner. Assume that the square has uniform density.
Suspension Point
(A) ω=3.22 s-1
(B) ω=4.56 s-1
(C) ω=6.45 s-1
(D) ω=2.90 s-1
(E) ω=5.80 s-1
1m
d=L/ 2
L
1m
The moment of inertia of the picture about its center is Icm=(mL²)/6. We need the moment
of inertia about the suspension point which can be obtained via the parallel axis theorem:
I = I CM + md 2 =(mL²)(1/6 + 1/2)= (2/3)(mL²).
For a physical pendulum, the angular frequency is given by:
ω=
mgd
=
I
mgL / 2
=
(2 / 3)mL2
3g
=
2 2L
3(9.8m / s 2 )
= 3.22s −1
2 2 (1m)
Physics 221 2004 S Final Exam Solutions
Page 10 of 23
[91] Three uniformly charged rods with their axes parallel to the y-axis A, B and C are
equally spaced as shown below where the spacing is small compared to the length of the
rods. The charge on rod A is Q, the charge on rod B is 2Q
and the charge on rod C is 3Q. What is the ratio between
y
the x component of the electrostatic force on rod A, FAx
to the electrostatic force on rod B, FBx.
(A) FAx:FBx= +7:8
2
(B) FAx:FBx= +11:16
2
(C) FAx:FBx= +1:1
3/2
(D) FAx:FBx= −11:16
(E) FAx:FBx= −7:8
6
x
Q
2Q 3Q
The force between two charged rods is F ∝ q1q2 / r . The
x components of the forces is therefore given by:
FAx = FABx + FACs ∝ −2 − 3 / 2 = −7 / 2
FBx = FBAx + FBCs ∝ +2 − 6 = −4
therefore FAx:FBx= +7:8
[92] Consider three particles of charge Q arranged in an equilateral
triangle. How much net work does it take to move one of the particles
to a point exactly half way in between the other two while keeping
those two charges fixed?
(A) 0
(B) kQ²/L
(C) (3/2) kQ²/L
(D) 2kQ²/L
(E) 4kQ²/L
U =
The electric potential energy (zero at infinity) is given by
∑
+Q
L
L
+Q
W=?
L
k
qi q j
rij
. Thus,
the initial potential energy is U i = 3k E Q 2 / L . The final potential energy is
U f = k E Q 2 / L + 2k E Q 2 /( L / 2) = 5k E Q 2 / L . The work done by the external force is thus
U f − U i = 2k E Q 2 / L .
Physics 221 2004 S Final Exam Solutions
Page 11 of 23
+Q
[93] A capacitor consists of two circular disks of radius 10cm which are separated by a
distance of 1mm. What is the capacitance?
(A) 28 pF
(B) 140 pF
(C) 28 nF
(D) 70 pF
(E) 278 pF
For a parallel plate capacitor,
C = ε 0 A / d = ε 0πr 2 / d = (8.85 pF / m)π (0.1m) 2 /(.001m) = 278 pF
Physics 221 2004 S Final Exam Solutions
Page 12 of 23
[94] A parallel plate capacitor consists of two plates separated by a distance d. It is
attached to a battery and an amount of energy U is stored in the capacitor. While it is
connected to the battery, the distance between the two plates is increased to 2d. What is
the energy stored in the capacitor after the distance is increased?
(A) U/4
(B) U/2
(C) U
(D) 2U
(E) 4U
The energy stored in a capacitor is given by U=CV²/2. In this case V is held constant and
by doubling the separation, the capacitance is reduced by a factor of 2 therefore U is
reduced by a factor of 2 to U/2.
4Ω+2Ω=6Ω
[95] How much current will be supplied by the
battery in the circuit shown?
(A)
(B)
(C)
(D)
(E)
1A
2A
3A
4A
6A
4Ω
12V
The 3Ω and 6Ω resistor combined in parallel gives
an equivalent resistance of 2Ω. This, added to the
4Ω resistor in series gives a total resistance of
6Ω.From Ohm’s law, the current is thus
12V/6Ω=2A.
Physics 221 2004 S Final Exam Solutions
3Ω
I=?
1/3Ω+1/6Ω=1/2Ω
Req=2Ω
Page 13 of 23
6Ω
Part B: Questions 96-100 (5 questions) [Based on Lectures 33-41]
[96]
A parallel plate capacitor consists of two plates separated by a distance d. It is
attached to a battery and an amount of energy U is stored in the capacitor. It his then
disconnected from the battery and the plates are insulated so the charge on the plates
remains constant. The distance between the two plates is then increased to 2d. What is the
energy stored in the capacitor after the distance is increased?
(A) U/4
(B) U/2
(C) U
(D) 2U
(E) 4U
The energy stored in a capacitor is given by U=I²/(2C). In this case I is held constant and
by doubling the separation, the capacitance is reduced by a factor of 2 therefore U is
increased by a factor of 2 to 2U.
Physics 221 2004 S Final Exam Solutions
Page 14 of 23
[97] Consider the two capacitors depicted. Both
are parallel plate capacitors with the same area of
plates and the same separation between the plates.
Capacitor #1 has vacuum between the plates and the
capacitance of this capacitor is C1. Capacitor #2 has
the lower half of the space filled with a dielectric of
dielectric constant κA and the upper half of the
space is filled with a dielectric of dielectric constant
κB. Assuming the spacing of the gap is much
smaller than the dimensions of the plate, which of
the following is the best estimate of the capacitance
of #2, C2?
2
C1
(A) C 2 =
1/ κ A + 1/ κ B
1
C1
(B) C 2 =
1/ κ A + 1/ κ B
(C) C 2 = (κ A + κ B )C1
(D) C 2 = 12 (κ A + κ B )C1
C1
Capacitor #1
Vacuum
C2
κA
Capacitor #2
κB
(E) C 2 = κ Aκ B C1
If the area of the plates is A and the separation is d then the capacitance of capacitor #1 is
C1 = ε 0 A / d . Capacitor #2 may be regarded as two capacitors based on half the area in
parallel. The upper half has capacitance C A = κ Aε 0 ( A / 2) / d and the lower half has
capacitance CB = κ Bε 0 ( A / 2) / d . The capacitance of these two parts in parallel are
C2 = C A + CB = (κ A + κ B )ε 0 A /(2d ) = 12 (κ A + κ B )C1 .
Physics 221 2004 S Final Exam Solutions
Page 15 of 23
Right Loop
Left Loop
I2
[98] What is the magnitude of the current through the 4Ω
resistor in the circuit diagram shown?
(A) 0.25 A
(B) 0.50 A
(C) 1.00 A
(D) 2.00 A
(E) 4.00 A
I1
I3
I1
First, I will label the currents as shown therefore the current we want to determine is I1.
Use the Kirchhoff’s loop rule on the right loop:
-4V+1V+2V-(4Ω)I1=0
Therefore I1=0.25A
Physics 221 2004 S Final Exam Solutions
Page 16 of 23
[99] In the circuit diagram shown at right what is the
charge on the 1F capacitor?
(A) 4C
(B) 8C
(C) 12C
(D) 15C
(E) 36C
1/(2F)+1/(1F)=1/(2/3 F)
Ceq=2/3 F
The two capacitors on the right are equivalent to a 2/3 F capacitance. The charge on the
equivalent capacitor is the same as the actual charge on each of the capacitors. That
charge is therefore Q=CV=(2/3 F)(12 V)=8C.
[100] An infinitely long charge wire has charge density of λ=+1 µC/m. How much
work must be preformed by an external force on a +2 µC charge to move it from a point
10cm from the wire to a point 1cm from the wire?
(A) +0.083J
(B) +0.041J
(C) +0.021J
(D) –0.021J
(E) –0.041J
The electric field of the line charge is E = (k E / 2)
λ
r
outwards from the wire. The work
done by the external force is
rB
rA
G G rA
r 
dr
− ∫ qE ⋅ dl = ∫ qEdr = ∫ (k E / 2)qλ
=2k E qλ log A 
r
 rB 
rA
rB
rB
= 2(8.99 × 109 Nm 2 / C 2 )(1 × 10− 6 C / m)(2 × 10− 6 C ) log(10cm / 1cm)
= 0.0828 J
Physics 221 2004 S Final Exam Solutions
Page 17 of 23
Part C: Questions 101-105 (5 questions) [Lab Questions]
Consider a cart whose position is measured with an
[101]
ultrasonic transducer (a “motion detector”), as you did in lab.
(Assume the sensor gives positions relative to the X axis illustrated in
the figure).
Which of the following represents a graph of position vs. time that
might result when the cart is accelerating at a fixed, positive rate?
A
X
X
B
t
E)
X
C
t
X
t
X
0
D
Uniform positive
acceleration gives
a upwards
opening parabola
t
None of the above graphs.
Physics 221 2004 S Final Exam Solutions
Page 18 of 23
L
two cells, chain,
seat and subject
R
Y
W
X
W
X component
0
L
-56.2
R
+57.1
[102] In the "Forces and Vectors" lab, you accumulated data similar to that shown
above (for simplicity, we consider only the X components). Assume that, done with care,
the procedure and apparatus which you used yields results (for each of the components of
L, R, and W) that have an uncertainty (for each) that may be as large as ± 1%.
With this in mind, are the experimental results shown in the table above consistent with
theoretical expectations? Which of the following is the most suitable response?
(A) The data are not consistent with theoretical expectations, and the uncertainties typical of
this apparatus have little to do with this conclusion.
(B) The data are not consistent with theoretical expectations, given the uncertainties expected
for this apparatus.
(C) The data are consistent with theoretical expectations, given the uncertainties expected for
this apparatus
(D) The data are not consistent with the theoretical predictions; this indicates that the
particular apparatus used needs repair.
(E) One cannot say anything based upon the information given.
The theoretical expectation is that the left and right x-components of the forces sum to 0.
From the data, this sum is 0.9N. Each of the two forces has an error of 1% which is about
±0.5N so if we combine the errors linearly, the sum has a maximum error of ±1N. The
experimentally observed sum is within that range of the theoretical value (0) and C is
correct.
Physics 221 2004 S Final Exam Solutions
Page 19 of 23
[103] 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 0.5 I0 , 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
t
C
ω
t
D
ω
E
t
The moment of inertia increases by a factor of 1.5 when the disk is dropped. Angular momentum
L=Iω is conserved so ω is reduced by a factor of 0.67. Graph D must be the correct graph because
it appears to have the appropriate drop in ω. Graph B shows a reduction which is in excess of the
expectation and the other graphs do not even have the correct qualitative behavior.
Physics 221 2004 S Final Exam Solutions
Page 20 of 23
t
[104] 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.)
The initial state seems to have a net x-component of momentum which is close to 0 because the
initial tracks have the same spacing of dots hence are moving at about the same velocity. After the
collision the left puck has 0 x-component of momentum but the right puck has a considerable x
component of momentum. This means that the net x-momentum has changed during the collision
which is impossible without a large external force.
Physics 221 2004 S Final Exam Solutions
Page 21 of 23
[105] 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
surface of sphere
surface of sphere
center 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
Charge leaks out due to breakdown of the air and this breakdown happens where the electric field
is largest, the surface of the sphere. Note that inside the sphere the electric field is zero since it is
a conducting sphere.
Physics 221 2004 S Final Exam Solutions
Page 22 of 23
Physics 221 Final Exam Answer Key
51
81
E
91
A
101
D
52
82
D
92
D
102
C
83
E
93
E
103
D
74
84
C
94
B
104
B
75
85
C
95
B
105
A
76
B
86
D
96
D
77
C
87
B
97
D
78
C
88
D
98
A
79
E
89
C
99
B
80
C
90
A
100
A
Physics 221 2004 S Final Exam Solutions
Page 23 of 23