Physics 221 2005S Final Exam Solutions
PHYSICS 221
Spring 2005
Final Exam Solutions May 3 2005 2:15pm—4:15pm
Part A (18 Questions/Lectures 1-42)
[56]
In the figure below there are two vectors, 1 and 2. There exists a cross-product (“vector
product”), C, of the vectors (i.e., C = 1 × 2). Calculate the direction of C.
Note:
• represents a vector pointing out of the page.
× represents a vector pointing into the page.
1
×
2
Which of the following boxes best describes the direction of C ?
Box B
Box A
Box C
Both
×
•
and
correctly describe
the direction of C
Box D
Both Box A and
Box B correctly
describe the
direction of C
Box E
The cross-product
has a magnitude
equal to zero.
Answer [B]: Using the right hand rule, if you put your right thumb in direction 1 and your index finger
in direction 2, then your middle finger naturally points in the direction shown in box B.
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Physics 221 2005S Final Exam Solutions
[57] A particle of mass m = 10 g moves along the x-axis with acceleration
ax = (3m / s 3 ) t
The velocity of the particle at t = 0 is 3.0 m/s in the +x direction. What is the velocity of
the particle at t = 4 s?
(A) 12 m/s
(B) 27 m/s
(C) 37 m/s
(D) 41 m/s
(E) 51 m/s
Answer [B]: The velocity at t=4 is given by
4
v(t ) = v(0) + ∫ a dt
0
[ ]
= 3m / s + (3m / s 3 ) 12 t 2
t =4s
t =0
= 27m / s
The following information applies to questions [58] and [59]: A 50-g dry sock rotates
stuck to the inner wall of a dryer drum. As a first approximation, you can think of the
drum as made of a 2-kg hollow cylinder of radius R = 30 cm with a 1-kg disk of radius R
= 30 cm attached to it on one side (see figure). Both the disk and the hollow cylinder
have uniform density. The drum rotates at a constant angular of 100 rpm.
Disk (cap)
(1 kg, radius 30 cm)
Hollow cylinder
(2 kg, radius 30 cm)
Open front (no cap)
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Physics 221 2005S Final Exam Solutions
[58] Determine the force on the sock by the drum wall when the sock is at the top of the
drum.
(A) 0.5 N pointing down
(B) 0.5 N pointing up
(C) 1.2 N pointing down
(D) 1.2 N pointing up
(E) 1.6 N pointing down
Answer [C]:
The angular momentum of the drum is ω = (100)(2π radians ) /(60s ) = 10.47 radians / s
The acceleration of the sock is centripetal a=v²/r so the net force is Fnet=(mv²/r)(-j). The
two forces on the sock are gravity and the normal force so
Fnet = N − j mg
Thus N = ˆjm g − rω 2
(
)
(
)
Putting in the numbers N = ˆj (.05kg ) (9.8m / s 2 ) − (0.3m)(10.46s −1 ) 2 = −1.15 ˆj
It is therefore downwards and indeed must be downwards because the normal force can
only push not pull.
[59] What is the moment of inertia of the drum?
(A) 0.135 kg m²
(B) 0.180 kg m²
(C) 0.270 kg m²
(D) 0.225 kg m²
(E)0.0675 kg m²
Answer [D]: The back plate is a solid cylinder so the moment of inertia is
I back = 12 mr 2 = 12 (1kg )(0.3m) 2 = 0.045 kg m 2 . The curved cylinder is a hollow cylinder so
the moment of inertia is I curvedk = mr 2 = (2kg )(0.3m) 2 = 0.180 kg m 2 adding these two
together I = I curved + I back = (.045 + .180)kg m 2 = .225kg m 2
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Physics 221 2005S Final Exam Solutions
[60] The horizontal beam shown has uniform
density and weighs 250N. It is connected to the
wall with an ideal massless string as shown and
supports a 500N weight from the end. What is the
magnitude of the net force that the wall exerts on
the beam if the system is in equilibrium?
(A)1200N
(D)1505N
(B)1375N
(E)1625N
13m
5m
(C)1400N
250N θ
12m
500N
Answer [D]: First we need to find the tension T in
the 13m string. Let q be the angle between the
string and the beam. We use the condition that the net torque about the hinge must be 0:
τ = 0 = T (12m) sin θ − (12m)(500 N ) − (250 N )(6m)
since sinθ=5/13 and cosθ=12/13 we can solve for T:
T=1625N
Now we can figure out the net force on the hinge. This can be calculated using the fact
that the net force on the beam must be 0.
x-component: 0 = Fnet x = Fhinge x − T cosθ
y-component: 0 = Fnet y = Fhinge y + T sin θ − (500 N ) − (250 N )
Solving these two equations for the force of the hinge:
Fhinge x = (1625 N )(12 / 13) = 1500 N
Fhinge y = (750 N ) − (1625 N )(5 / 13) = 125 N
The magnitude of the vector is thus Fhinge = (1500 N ) 2 + (125 N ) 2 = 1505 N
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Physics 221 2005S Final Exam Solutions
[61] A flywheel has moment of inertia I=8kg m2. At t=0 it rotates with a period of
T=1.5s. A motor gradually speeds up the flywheel such that the period, T, shortens at a
rate of dT/dt= −2.0x10−3. What is the rate of change of the kinetic energy of the flywheel
at t=0?
(A)94mW
(B)140mW
(C)187mW
(D)376mW
(E)423mW
Answer [C]: Following the method in Problem 9.98 on Written Assignment 5, the total
rotational energy is:
I
T2
The rate of change in kinetic energy is obtained using the chain rule:
K = 12 Iω 2 = 2π 2
4π 2  dT 
P = K = 2π I d (1 / T ) / dt = − 3 

T  dt 
Plugging in the numbers from the question
d
dt
P=−
2
(
2
)
4π 2  dT 
4π 2 (8kg m 2 )
=
−
(−2.0 × 10− 3 ) = 187mW

3 
3
T  dt 
(1.5s)
[62] A girl stands in an elevator that has a constant upwards acceleration over a distance
of 20m. During the acceleration the normal force exerted by the floor of the elevator does
10.0kJ of work on the girl while gravity does –8.0kJ of work on her. What is the
acceleration of the elevator?
(A)2.45m/s²
(B)4.90 m/s²
(C)7.28 m/s²
(E)Cannot be determined from the information given.
(D)12.27 m/s²
Answer [A]: Following question 6.60 from Written Assignment 4, Gravity does –8kJ of
work over 20m so the girl’s weight is W=(8kJ/20m)=400N. The girl’s mass is
m=W/g=40.8kg.
The work done by the normal force is 10kJ so that the normal force is
(10kJ)/(20m)=500N. The net force on the girl is thus Fnet=500N-400N=100N. The
acceleration is thus given by Newton’s second law a=Fnet/m=2.45m/s².
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Physics 221 2005S Final Exam Solutions
[63] A cannon, located 60m from the base of a vertical 25m tall cliff, shoots a 15kg shell
at an angle of 45° above the horizontal toward the cliff at a speed of 20m/s. What is the
speed of the shell when it lands on the top of the cliff? Neglect air resistance.
(A) 0m/s
(B) 12.1m/s
(C) 13.7m/s
(E) The shell does not land on the top of the cliff
(D) 20.0m/s
Answer [E]: The kinetic energy of the shell is K=(1/2)(15kg)(20m/s)²=3kJ. The potential
energy (taking U=0 at the bottom of the cliff) of the shell at the top of the cliff, 25m
above the bottom is U=mgh=(15kg)(9.8m/s²)(25m)=3.675kJ>3kJ. The shell can therefore
never make it to the top of the cliff.
G
G
G
G
[64] If A = iˆ + 2 ˆj + 4 kˆ and B = 4iˆ − 4 ˆj + kˆ , what is the angle between A and B
in radians?
(A) 0
(B) 3π/4
(C) π/4
(D) π
(E) π/2
G G
Answer[E]: Taking the dot product: A ⋅ B = (iˆ + 2 ˆj + 4kˆ) ⋅ (4iˆ − 4 ˆj + kˆ) = 0 . The two
vectors are therefore perpendicular and the angle between them is thus π/2.
[65] Consider a physical pendulum consisting of a thin rod of uniform density, length
L=2m and mass m=4kg. The rod is suspended from one end. What is the period of this
pendulum?
(A) 1.64s
(B) 2.32s
(C) 2.84s
(D) 3.28s
Answer [B]: The period of a physical pendulum is given by T = 2π
d=L/2 and I=(1/3)mL² so
T = 2π
I
(1 / 3)mL2
2L
= 2π
= 2π
= 2.32s
mgd
(1 / 2)mgL
3g
Page 6 of 19
(E) 4.02s
I
. In this case
mgd
Physics 221 2005S Final Exam Solutions
[66] A spaceship travels through empty space at 300 km/s in the positive x direction and
splits into two pieces. One of the pieces has twice the mass of the other one. They both
come out at 20° with the initial direction (see figure). Find the velocity of the center of
mass of the system after the separation.
(A)+150 km/s
(D)+460 km/s
iˆ
iˆ
(B)+282 km/s
(E)+564 km/s
iˆ
iˆ
(C)+300 km/s iˆ
y
Part 1
spaceship
20°
x
20°
Part 2
Answer [C]: there is no external force on the system so the center of mass velocity does
not change. Initially the center of mass velocity is +300 km/s iˆ so after the separation
the center of mass velocity is also +300 km/s iˆ .
[67] A 1kg car and a 2 kg car are constrained to move along a one dimensional air track.
Initially the 1kg car has a velocity of + 3m / s while the 2kg car has a velocity of − 3m / s .
The two cars collide and the collision is fully elastic. What is the velocity of the 1kg car
after the collision?
(A) −5m/s
(B) −4m/s
(C) −3m/s
(D) −2m/s
Answer [A]: The velocity of the center of mass is
(1kg )(3m / s ) + (2kg )(−3m / s )
vcm =
= −1m / s
1kg + 2kg
The final velocity after a 1D elastic collision is given by
vf=2vcm−vi=2(−1m/s)−3m/s=−5m/s.
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(E) −1m/s
Physics 221 2005S Final Exam Solutions
[68]The point charges in the figures below are placed at distance s from points A and B.
Compare the magnitude of the net electric field and the electric potential at points A and
B. Assume that each system is very far from any other charges. Take the potential to be
zero at infinity.
Q
+
Q +
s
s
2Q
+
•A
s
•B
(A) EA = EB ; VA = VB
(B) EA > EB ; VA = VB
(C) EA = EB ; VA > VB
(D) EA < EB ; VA = VB
(E) EA = EB ; VA < VB
Answer[D]: Think of the 2Q object as two Q charges on top of each other. At point B
both of the field vectors from the two Q charges are aligned so the net field is just the
sum. At point A they are not aligned so the electric field is less. The contribution to the
potential depends only on the distance so the potentials in both cases are the same.
[69] A parallel plate capacitor is connected to a battery with EMF=V. If the space
between the plates is filled with air the energy stored in the capacitor is Uair. If the space
is filled with a dielectric having dielectric constant κ, the energy stored in the capacitor is
Udielectric. What is the ratio Udielectric/Uair?
(A) κ
(B) κ2
(C) 1/κ
(D) 1/κ2
(E) 1 (same energy in both cases)
Answer[A]: If C is the capacitance with air then the capacitance with dielectric is κC.
Here the voltage in the capacitor is the same in both cases so the energy stored in the air
capacitor is Uair=(1/2)CV² while the energy stored in the dielectric capacitor is
Udielectric=(1/2)(κC)V². The ratio Udielectric/Uair=κ.
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Physics 221 2005S Final Exam Solutions
[70] Determine the escape speed from the surface of a spherical asteroid of constant
density with a radius of 700 m and a mass of 4×1012 kg.
(A) 0.44m/s
(B) 0.62m/s
(C) 0.87m/s
(D) 1.1m/s
(E)1.2m/s
Answer[C]: The escape speed is given by
ve = 2GM / R
= 2(6.7 × 10−11 N m 2 / kg 2 )(4 × 1012 kg ) /(700m) = 0.873m / s
[71] Consider the following three systems:
: Spherical Gaussian surface inside the shells
Q
Q
Q
A
Empty conducting
spherical shell with
charge Q > 0
B
A conducting
spherical shell,
uncharged, with a
point-charge Q > 0
in its center.
C
Empty insulating spherical
shell with charge Q > 0
uniformly distributed
throughout the volume of
the insulator
Let ΦA, ΦB and ΦC be the net electric flux through a Gaussian spherical surface inside the
shells (dotted lines in the figure). Which of the following is true?
(A) ΦA = ΦB = ΦC = 0
(B) ΦA = ΦB = 0; ΦC > 0
(C) ΦA = 0; ΦB > 0; ΦC > 0
(D) ΦA > 0; ΦB = 0; ΦC > 0
(E) ΦA > 0; ΦB >0; ΦC > 0
Answer[B]: Within a conductor the electric field is 0 so the flux in case A
and B are 0. In case C the surface encloses a positive finite amount of charge
and therefore ΦA = ΦB = 0; ΦC > 0.
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Physics 221 2005S Final Exam Solutions
[72] In the circuit shown, what is the power dissipated by the 2Ω resistor?
(A) 16W
(B)32W
(C)64W
(D)128W
(E)256W
2Ω
20V
1Ω
0.5Ω
1Ω
Answer[D]: The two 1Ω resistors are in parallel and are equivalent to 0.5Ω. The
equivalent resistance to the whole circuit is 2.5Ω. By ohms law, the current through the
circuit is 8A=(20V)/(2.5Ω). All that current goes through the 2Ω resistor so that the
power dissipated is P=I²R=(8A)²(2Ω)=128W.
[73][Extra Credit] Two identical springs have spring constant k=100N/m. These
springs are connected end to end as shown below, with the left end connected to a fixed
wall. How much force does it take to extend the system of two springs 1cm beyond their
unstretched length?
(A)5N
(B)2N
(C)1N
(D)0.5N
k=100N/m
(E)0.25N
k=100N/m
Force=?
Unstreched Length
1cm
Answer[D]: Following the logic of 13.96 from written assignment 6, the force which
each spring exerts is the same. By symmetry, each of the springs stretches 0.5cm. Since
the spring constant is 100N/m the force exerted by each spring is
(100N/m)(.005cm)=0.5N.
Page 10 of 19
Physics 221 2005S Final Exam Solutions
Part B (6 Questions/Lectures 29-42)
[74] In Figure 1, two infinite parallel conducting plates carrying charge densities σ and
−σ are separated by a distance d. Let E1 be the magnitude of the electric field half way
between the two plates. In Figure 2 the two identical plates are separated by a distance
2d and the electric field half way between the plates is E2. What is the ratio between E1
and E2?
(A)E1:E2=4:1
−σ
(B)E1:E2=2:1
E1
(C)E1:E2=1:1
(D)E1:E2=1:2
+σ
−σ
d
(E)E1:E2=1:4
E2
2d
Figure 1
Figure 2
Answer[C]: Just as in problem 23.38a on written assignment 8, note that the electric field
of due to an infinite sheet of charge is independent of distance so E1=E2.
[75] A particle with charge 1mC and mass 1kg is in a uniform electric field directed to
the left. It is released from rest and moves to the left; after it has moved 50cm, the
velocity of the particle is 2m/s. What is the magnitude of the electric field?
(A) 250 V/m
(B)500 V/m
(C)1000 V/m
(D)2000V/m
(E)4000V/m
Answer[E]: Just as in Problem 23.18 we see that the particle has gained a kinetic energy
of K=(1/2)mv²=2J which is the work W that the electric field. The electric force is
therefore F=W/d=(2J)/(0.5m)=4N. In terms of the electric field,
E=F/q=(4N)/(1mC)=4000 V/m
Page 11 of 19
+σ
Physics 221 2005S Final Exam Solutions
[76] A satellite of mass m orbits in a circular, near earth orbit (radius of orbit≈RE=radius
of Earth). What is the minimum amount of work that must be done on the satellite in
order to boost it to a circular orbit of radius 2 RE ? (ME=mass of Earth)
1
1
1
(A) 0
(B) GmME/RE (C) GmME/RE
(D) GmME/RE
(E) GmME/RE
2
4
8
Answer[D]: From the result of problem 12.66 on Assignment 7, the mechanical energy of
a object in circular orbit is exactly half the potential energy. The mechanical energy of
1
the satellite in low earth orbit is thus − GmME/RE while the mechanical energy of the
2
1
satellite in the 2R orbit is − GmME/RE. The amount of work which must be done to
4
1
raise the satellite is thus GmME/RE.
4
Page 12 of 19
Physics 221 2005S Final Exam Solutions
[77] In the circuit below, R1 = 10 Ω, R2 = 20 Ω, R3 = 30
Ω and V0 = 100 V. Determine the magnitude of the
current through R1.
(A) 1.4 A
(D) 2.4 A
(B) 1.8 A
(E) 2.9 A
(C) 2.0 A
Answer[B]: R1 and R2 are in parallel with
equivalent resistance R12=1/(1/10Ω+1/20Ω)=6.67Ω. R3 is in series with that
system so the equivalent resistance of the three resistors is R12+R3=36.67Ω.
The current is therefore (100V)/(36.67Ω)=2.73A giving a voltage drop across R3 of
(2.73A)(30Ω)=81.82V. The voltage across R1 is thus 18.18V=100V-81.82V hence the
current through it is (18.18V)/(10Ω)=1.82A.
Page 13 of 19
Physics 221 2005S Final Exam Solutions
[78] Consider the two circuits depicted below. In circuit A the three capacitors are all
1pF and are arranged in a delta network between terminals P, Q and R. In circuit B three
capacitors, each of which has capacitance CY , are arranged in a Y-network between
terminals P, Q and R. For what value of CY are the two capacitor networks equivalent
circuits?
(A) CY=4pF
(B) CY=3pF
(C) CY=2pF
(D) CY=1pF
(E) CY=0.33pF
Q
P
1pF
P
Q
CY
CY
1pF
1pF
CY
R
R
Circuit B
Circuit A
Answer[B]: Following the method of problem 24.75 on assignment 8, if we connect P
and Q in circuit A to an external EMF and disconnect R the equivalent capacitance is
Ceq=1pF+1/(1/1pF+1/1pF)=1.5pF. If we hook up B the same way we must get the same
equivalent capacitance. In circuit B we the capacitor attached to R is irrelevant so it is
just two CY capacitors in series. Thus Ceq=CY/2. CY/2=1.5pF so CY=3pF.
[79] [Extra Credit] In this problem, consider the electric field produced by two separate
systems depicted in the figures below
System A is a very long, solid insulating cylinder with radius R that has a cylindrical hole
with radius R/2 bored along its entire length. The axis of the hole is parallel to the axis of
the insulating cylinder and is at distance R/2 from the axis of the cylinder. The solid
material of the cylinder has a uniform charge density ρ. Let X be the point at the center of
the hole and EX is the magnitude of the electric field at point X.
System B is a very long, solid insulating cylinder with radius R with no hole. The
substance of system B also has uniform charge density ρ. In system B the point Y is
located at distance R/2 from the center of the axis of the cylinder and EY is the magnitude
of the electric field at point Y.
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Physics 221 2005S Final Exam Solutions
What is the correct relation between EX and EY?
(A) EX :EY=1 : 1
(B) EX:EY=0 : 1
(C) EX:EY=1 : 2
(D) EX:EY=1 : 4
(E) EX:EY=2 : 1
R/2
R/2
R
R
R/2
X
Y
System A:
End view of cylinder with hole
System B:
End view of cylinder without hole
Answer[A]: Using the same method as problem 22.62, we can think of the charge
distribution in system A as being a superposition of system B with a cylinder of charge
density −ρ centered at X with radius R/2. By symmetry, the electric field produced by the
negative cylinder at its center is 0 so the electric field at X is the same as Y.
Page 15 of 19
Physics 221 2005S Final Exam Solutions
Part C (4 Questions/Lab Final)
[80]An electrophorus is equipped with an insulating plate that
becomes positively charged when rubbed with tissue. Which metal
of the following best describes the change in the charge on
plate
the insulating plate, when the metal plate gains 1 µC of
charge as the electrophorus is operated normally.
insulating
handle
(A)
The insulating plate gains approximately 1 µC of
positive charge.
(B)
The insulating plate gains approximately 1 µC of
negative charge.
(C)
The insulating plate loses approximately 1 µC of
positive charge.
(D)
The insulating plate loses approximately 1 µC of negative charge.
(E)
The charge on the insulating plate will be relatively unchanged, compared to 1
µC.
insulating plate
ELECTROPHORUS
Answer[E]: The bulk of the charge on the metal plate is placed there by electrostatic
induction. When the metal plate is placed on the charged insulating plate and grounded, the
1µC flows into the metal from the ground. The insulating plate is only responsible for the
separation of charges between the metal plate and the ground but no significant charge is
transferred directly from the insulating plate to the metal plate.
Page 16 of 19
Physics 221 2005S Final Exam Solutions
[81] As done in lab, a student, seat, chain, and two load cells are at rest and supported by
cables attached to the upper (sensitive) fittings of the two load cells. The cell readings
and those of the large protractors are given in the
protractors
table below.
load cells
LEFT
RIGHT
Cell reading
531 N
600 N
Protractor reading
60°
50°
0
Y
seat
Use this information to estimate the X component
of the force exerted by the left cell upon the end
of the left chain.
(A) 0 (The force must be zero, since nothing is accelerating.)
(B) +266 N
(C) −266 N
(D) +460 N
(E) −460 N
Answer [E]: From the figure it should be clear that the force which the cell exerts on the
chain is generally up and to the left so the answer should be negative. The magnitude of
the total force is F=531N and paying attention to which way the protractor reads, the x
component of that force is Fx=−Fsin(60º)=−460N.
Page 17 of 19
0
chain
X
Physics 221 2005S Final Exam Solutions
[82] 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).
Which of the following graphs best illustrates the velocity, VX , of the cart versus time
after it leaves your hand?
Vx
Vx
t
0
A
Vx
Vx
t
B
Vx
t
C
t
D
t
E
Answer[E]: Since it is initially moving to the left, the velocity must start at a negative
value. The drag decreases the magnitude of the velocity so the curve has a positive slope
and is linear since the acceleration due to friction should be independent of velocity. Only
E looks like that.
Page 18 of 19
Physics 221 2005S Final Exam Solutions
[83] In the rotational motion experiment, the following graph was obtained when the
wheel of the apparatus which you used was rotated by hand. Provide the most likely
explanation for this particular graph.
A) The wheel was rotated slowly, and the angular encoder on the wheel axis has
a resolution of 1/200 of a revolution.
B) The wheel was rotated at different velocities, increasing in steps as time went
on.
C) The wheel was rotated slowly, and the angular encoder on the wheel axis has
a resolution of 1/20 of a revolution.
D) The wheel was rotated rapidly then stopped for a second or two, then rotated
rapidly, then stopped again, and so on.
E) The apparatus likely was malfunctioning.
Answer[A]:The jumps in the graph are artifacts of the way the angular encoder works,
the output of the encoder is advanced only at discrete angles. Looking at the graph we see
that 10 jumps in angular position correspond to about 0.31 radians so each jump
corresponds to about .031radians. A full circle would thus give about (2π)/(0.031)=203
jumps. Choice A is thus the most plausible possibility.
Page 19 of 19