1
Your Name:
PHYSICS 101 FINAL EXAM
January 23, 2006
3 hours
Please Circle your section
1
9 am
Galbiati
2
10 am
Wang
3
11 am
Hasan
4 12:30 pm Hasan
5 12:30 pm Olsen
Problem
1
2
3
4
5
6
7
8
9
Total
Score
/20
/15
/15
/15
/15
/20
/20
/20
/15
/155
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 23. The exam contains 9 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you
may want to tear off and keep handy.
Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the
Honor Code during this examination.
Signature
2
Problem 1
Miscellaneous Questions
1. (6 points) A bag of lead pellets (specific heat capacity C = 128 J kg−1 ◦ C−1 ) is
dropped from an height of 10 m. Upon impact, 50% of the kinetic energy of the
bag is turned into thermal energy dissipated throughout the lead bag. Calculate
the change in temperature of the lead.
The energy balance is:
1
× mgh = mC∆T
2
Therefore:
∆T =
gh
(9.8 m/s2 )(10 m)
=
= 0.38 ◦ C
2C
2(128 J kg−1 ◦ C−1 )
2. (5 points) For this question, you MUST box your answer AND offer your
arguments as to why you are choosing that answer.
Imagine you are floating on a lake in a boat in which you have a large rock.
You throw the rock overboard. Does the level of the lake:
• Stay the same?
• Increase?
• Decrease?
The level decreases. In the initial situation, the water level is raised by the amount
necessary to make the rock float on the boat; i.e., the hull of the boat displaces a
volume of water whose mass is equal to the mass of the rock. In the final situation
the rock (now on the bottom of the lake) displaces a volume of water equal to its own
volume. The volume displace in the initial situation is larger than the rock’s own volume
by a factor given by the ratio of the rock density to the water density.
3
3. Two disks of identical mass m but different radii (2r and r for disk 1 and 2
respectively) are spinning on frictionless bearings at the same angular speed
ω0 but in opposite directions. The two disks are brought slowly together.
The resulting frictional force between the surfaces eventually brings them to
a common angular velocity.
Disk 1 Disk 2
a.) (6 points) What is the magnitude of that final angular velocity in terms of
ω0 ?
The momenta of inertia of the two disks are:
I1 = 2mr2
I2 =
1 2
mr
2
The angular momenta are:
L1 = 2mr2 ω0
1
L2 = − mr2 ω0
2
The total angular momentum in the initial condition is:
Ltot = L1 + L2 =
3 2
mr ω0
2
The angular momentum is conserved. Therefore, in the final condition:
Ltot = (I1 + I2 )ωf
where ωf is the final angular velocity:
ωf =
Ltot
=
I1 + I2
3
2
2 mr ω0
5
2
2 mr
=
3
ωf
5
b.) (3 points) In the final state - when the two disks are moving together - which
is the direction in which they are spinning? Specify if it’s the same direction in
which disk 1 or 2 was originally moving.
In the direction of the disk carrying the larger momentum of inertia, and therefore the
larger initial angular momentum, i.e. disk 1.
4
Problem 2
The human voice is generated by vibration of the vocal cords and amplified by the
resonance characteristics of the vocal tract, which acts like a hollow tube closed at
one end.
a.) (5 points) What is the fundamental resonant frequency of Bob’s vocal tract,
which has a length of 170 mm? (Assume vsound = 340 m/s.)
For a tube closed at one end we have:
λ = 4L
So:
f=
v
v
340 m/s
=
=
= 500 Hz
λ
4L
4 × 0.17 m
b.) (5 points) Helium is a monatomic gas with an atomic mass of 4 u. Breathing
in helium changes the resonant frequency of the vocal tract by changing the
speed of sound in the airway. Assuming a temperature of 20◦ C for the helium
enhaled, what would be the fundamental frequency of Bob’s vocal tract after
breathing helium?
The speed of sound in He at 20◦ C is:
r
v=
γkT
m
where γ = Cp /Cv = 5/3 for a monatomic gas. Therefore:
s
(5/3)(1.38 × 10−23 J/K)(293 K)
v=
= 1007 m/s
(4)(1.66 × 10−27 kg)
The new fundamental frequency would be:
f=
1007 m/s
= 1482 Hz
4(0.17 m)
5
c.) (5 points) After inhaling way too much helium, Bob proposes the following
experiment: while screaming at his new resonant frequency, he will attempt
to run away from you fast enough so that you hear the resonant frequency of
his natural voice (before he inhaled the helium). Show how severely Bob has
been impaired by calculating how fast he would need to run to accomplish this
feat. (If you got stuck on the earlier parts, you can assume fa = 400 Hz and
fb = 1200 Hz for the the answers to parts a and b, respectively.)
Using the Doppler formula we have:
1±
fo = fs
1∓
vo
v
vs
v
= fs
1
1 + vvs
Rearranging gives:
vs = v
fs
−1
fo
= (340 m/s)
1482 Hz
−1
500 Hz
= 668 m/s
6
Problem 3
Two blocks of mass mA = 20 kg and mB = 5 kg are connected by a massless cable
wound around a massless pulley. A force F is applied to block A causing it to slide
over a horizontal surface with kinetic coefficient of friction µk1 = 0.25, while block B
slides over block A with µk2 = 0.15.
B
μ k2 = 0.15
A
F
μ k1 = 0.25
a.) (8 points) What force is needed to move block A with constant speed?
The frictional force between block A and the surface is:
fA = µk1 (mA + mB )g
The frictional force between blocks A and B is:
fAB = µk1 mB g
The force diagram for block A gives:
F − T − fAB − fA = 0
where fAB is the friction force between the two blocks and fA is the friction force
between block A and the surface. The force diagram for block B gives:
T − fAB = 0
Substituting for T in the first equation gives:
F = 2fAB + fA = 2µk1 mB g + µk1 (mA + mB )g
2
2
F = 2(0.15)(5 kg)(9.8 m/s ) + (0.25)(25 kg)(9.8 m/s ) = 76 N
7
b.) (7 points) If the cable can sustain a maximum tension of 100 N, what is the
maximum force Fmax allowable without breaking the cable?
The maximum force will occur when T = 100 N. From:
T − fAB = mB a
we get:
a = 18.5 m/s
2
Substituting this into the equation for block A, which is:
F − T − fAB − fA = mA a
gives:
2
2
2
F = (100 N)+(0.15)(5 kg)(9.8 m/s )+(0.25)(25 kg)(9.8 m/s )+(20 kg)(18.5 m/s ) = 539 N
8
Problem 4
As long as the air temperature is less than the temperature of your skin (36 ◦ C),
your body will radiate a net amount of heat per unit time corresponding to the
Stefan-Boltzman law. A minimum amount of fuel (food) is needed each day to
replenish this heat energy and maintain a constant body temperature. (Some data for
water that might be useful in this problem: c = 4186 J/◦ C/ kg; Lf = 33.5 × 104 J/kg;
Lv = 24.2 × 105 J/kg.)
a.) (5 points) Assuming a body surface area of 1.3 m2 , an emissivity of 0.75, and
an environmental temperature of 22 ◦ C, calculate how much net energy is lost
to the environment in one day (24 hours). Give your answer in units of “food
Calories” (1 Calorie = 4186 J).
Q = eσAt(T 4 − T04 ) =
= (0.75)(5.67×10−8 J/(s · m2 · K4 ))(1.3 m2 )(24 hr)(3600 s/hr) (309 K)4 − (295 K)4
= 7.37 × 106 J = 1761 Calories
9
b.) (5 points) In contrast, if the environmental temperature is higher than your skin
temperature, you will absorb a net amount of heat per unit time. Assuming
an outside temperature of 40 ◦ C, how much net radiant energy (in Joules) is
absorbed by the body in one hour?
Q = eσAt(T 4 − T04 ) =
= (0.75)(5.67 × 10−8 J/(s · m2 · K4 ))(1.3 m2 )(3600 s/hr) (313 K)4 − (309 K)4
= 9.6 × 104 J
c.) (5 points) By sweating, the body dissipates excess heat through evaporation of
moisture from the skin. Assuming the net radiant heat gain in (b), how much
water would need to be evaporated every hour to carry away the excess heat?
(Assume Qnet = 104 J if you could not get part b.)
Since:
Q = mLv
We get:
m = Q/Lv = (9.6 × 104 J)/(24.2 × 105 J/kg) = 0.040 kg (also about 40 mL)
10
Problem 5
A solid cylinder of mass 10 kg and radius 0.5 m is placed on the edge of a 30◦
incline of height 2 m. The coefficient of static friction of the incline’s surface µ is just
high enough such that the cylinder barely rolls without slipping down the incline.
2m
θ = 30o
a.) (7 points) What is µ?
Friction creates torque
mgcosθµR = Iα,
where α is the angular acceleration, and I the moment of inertia of the cylinder. Force
creates acceleration
mgsinθ − mgcosθµ = mαR
solve for µ,
µ = 0.19
11
b.) (4 points) How long does it take the cylinder to roll down to the bottom?
Solve for α from the above equations,
α=
2gsinθ
= 6.53 rad/s2
3R
the cylinder’s acceleration in the direction along the incline surface is
a = αR = 3.27 m/s2
1 2
at = 4 m
2
t = 1.56 s
x=
c.) (4 points) What is the angular speed of the cylinder at the bottom of the incline?
w = αt = 6.53 rad/s2 × 1.56 s = 10.18 rad/s
12
Problem 6
Two thermally insulated tanks of equal volume V1 = V2 = 1 m3 , are connected by
a thermally insulated partition. The two tanks communicate through a valve which
initially closed. The two tanks contain two samples of the same ideal and monoatomic
gas.
Tank 1
Tank 2
P1 = 2 x 105 Pa
n1 = 500 mol
P2 = 1 x 105 Pa
n1 = 200 mol
a.) (4 points) What are the temperatures T1 and T2 of the two tanks?
P V = nRT
T1 =
(2 × 105 Pa)(1 m3 )
P1 V 1
=
= 48.1 K
n1 R
(500 mol)(8.3 J mol−1 K−1 )
T2 =
P2 V 2
(1 × 105 Pa)(1 m3 )
=
= 60.2 K
n2 R
(200 mol)(8.3 J mol−1 K−1 )
13
b.) (4 points) What are the internal energies U1 and U2 of the gas in the two tanks?
U=
3
nRT
2
3
3
3
n1 RT1 = P1 V1 = (2 × 105 Pa)(1 m3 ) = 3 × 105 J
2
2
2
3
3
3
U2 = n2 RT2 = P2 V2 = (1 × 105 Pa)(1 m3 ) = 1.5 × 105 J
2
2
2
U1 =
c.) (4 points) The valve is opened and the two gasses mix. What is the final
temperature Tf ?
There is no work done by the gas, and there is no exchange of heat from the
surroundings, thus the total internal energy of the two tanks remains the same after
the valve opens.
Uf = U1 + U2 =
Tf =
3
(n1 + n2 )RTf = 4.5 × 105 J
2
2(4.5 × 105 J)
= 51.6 K
3(700 mol)(8.3 J mol−1 K−1 )
14
d.) (4 points) What is the final pressure of the mixed gas?
Pf Vf = (n1 + n2 )RTf
Pf =
(700 mol)(8.3 J mol−1 K−1 )(51.6 K)
= 1.5 × 105 Pa
2 m3
e.) (4 points) What is the change in entropy of the system?
You can think that the mixing of the gas is done in two steps. In the first step, each of
the two samples of gas expands adiabatically to occupy the entire volume (with little or
no interaction with the other gas). In a subsequent step, the two gases mix and come
to the same temperature. In the first step, there is no change of entropy because no
heat is exchanged. What matters is the heat exchanged between the two samples of
gas during the second step.
A quantity of heat Q flows from the hotter to the cooler gas. Q is given by the change
in internal energy of the sample 1 of gas (or the opposite of the change in internal
energy of the sample 2 of gas):
Q=
3
3
n1 R(Tf − T1 ) = (500 mol)(8.315 J/K/mol)(51.6 K − 48.1 K) = 21830 J
2
2
Therefore:
Q
Q
∆S =
−
= (21830 J)
T1
T2
1
1
−
48.1 K 60.2 K
= 91.2 J/K
15
Problem 7
A plastic tube with radius R = 1 cm is placed at the bottom of a well on a hill slope.
The second end of the tube is located close to a large bucket (capacity 200 liters =
0.2 m3 ). Once the syphone is primed (the tube is filled with water), the second end of
the tube is inserted into a large bucket located downhill. The distance between the
level of the water in the well and the level of the water in the bucket is y = 2 m. The
distance between the level of the water in the well and the top of the tube is h = 7 m.
Throughout the problem, you can use the approximation patm = 105 Pa.
(top point
of water tube)
C
Water Tube
h
A
(water level
in the well)
y
(water level
in bucket)
B
a.) (5 points) What is the velocity of the water at the end of the pipe?
Let’s apply Bernoulli equation between point A (surface of the water in the well) and
point B (exit of the water tube):
1 2
1 2
pA + ρgy + ρvA
= pB + ρvB
2
2
Taking into account that pA = pB = patm and vA = 0, we get:
p
p
v = 2gy = 2(9.8 m s−2 )(2 m) = 6.3 m/s
16
b.) (5 points) How much time is needed to fill the bucket (you can neglect the
variation of y during the time the bucket is filled)?
Let’s call with Q = AvB the volume flow rate at the exit of the pipe; A is the area of
the pipe. Then:
V = Qt = (AvB )t = πR2 vB t
Hence:
t=
V
0.2 m3
=
= 101 s
2
πR vB
π(0.01 m)2 (6.3 m)
c.) (5 points) What is the water pressure inside the tube at the top point (marked
as C in the associated figure)?
Let’s apply Bernoulli equation between point C and point B (exit of the water tube):
1 2
1 2
pC + ρg(h + y) + ρvC
= pB + ρvB
2
2
Taking into account that pB = patm and vC = vB , we get:
pC = pB − ρg(h + y) = 105 Pa − (1000 kg/m3 )(9.8 m s−2 )(9 m) = 11800 Pa
17
d.) (5 points) If the distance h gets increased (for example, by raising the top of
the tube C) at some point the syphone will stop working. For what value of h
does the syphon stop working?
The syphone stops working when the pressure in point C becomes equal to zero (can’t
have negative pressure). Applying the Bernoulli equation between points B and C, the
condition is:
ρg(h + y) = patm
Or:
h=
patm
105 Pa
−y =
− 2 m = 8.20 m
ρg
(1000 kg/m3 )(9.8 m s−2 )
18
Problem 8
Consider the thermodynamic cycle shown in the figure. It is composed by an isocoric
(1 → 2), an adiabatic (2 → 3), and an isobaric transformation (3 → 1). The
substance enclosed in the system is 1 mole of ideal monoatomic gas (γ = 5/3). The
initial pressure and temperature are P1 = 105 Pa and T1 = 200 K. During the isocoric
transformation, the gas is heated at constant volume until it reaches the temperature
T2 = 800 K. It is then expanded adiabatically until its temperature reaches again the
initial value, P3 = P1 = 105 Pa. Finally, the gas is compressed at constant pressure
back to its original state.
P2
P1,P3
2
1
V1,V2
3
V3
19
a.) (6 points) Find the temperature T3 at the end of the adiabatic expansion.
Since the points 1 and 2 are connected through an isocoric transformation:
P2 = P1
Also:
V1 = V2 =
800 K
T2
= P1
= 4P1 = 4(105 Pa) = 4 × 105 Pa
T1
200 K
nRT2
(1 mol)(8.315 J/K/mol)(800 K)
=
= 0.017 m3
P2
4 × 105 Pa
Let’s first find the volume V3 in terms of V2 . Points 2 and 3 are connected through an
adiabatic transformation, therefore:
P2 V2γ = P3 V3γ
V3 = V2
P2
P3
1/γ
= V2
P2
P3
1/γ
= V2 ×43/5 = 2.3×V2 = 2.3×0.017 m3 = 0.038 m3
Therefore, since points 1 and 3 are connected through an isobaric transformation and
V1 = V2 :
V3
V3
= T1
= 2.3 × T1 = 2.3 × 200 K = 460 K
T3 = T1
V1
V2
b.) (6 points) Find the heat entering or leaving the system during one cycle.
Along the entire cycle ∆U = 0. Therefore Q = W . The net work is the sum of the
work along the transformations (2 → 3) and (3 → 1):
Q = W = W2→3 + W3→1 =
=
=
3
nR(T2 − T3 ) − P3 (V3 − V1 ) =
2
3
(1 mol)(8.315 J/K/mol)(800 K − 460 K) − (105 Pa)(0.038 m3 − 0.017 m3 ) =
2
= 4200 J − 2100 J = 2100 J
One could also solve by calculating directly the heat exchanged in each of the three
transformations:
Q1→2 = cV ∆T =
3
3
nR(T2 −T1 ) = (1 mol)(8.315 J/K/mol)(800 K−200 K) = 7500 J
2
2
Along the adiabatic there is no exchange of heat: Q2→3 = 0.
Q3→1 = cp ∆T =
5
5
nR(T1 −T3 ) = (1 mol)(8.315 J/K/mol)(200 K−460 K) = −5400 J
2
2
Then:
Q = Q1→2 + Q2→3 + Q3→1 = 7500 J + 0 J − 5400 J = 2100 J
20
c.) (5 points) Find the change in entropy of the gas along the adiabatic
transformation (2 → 3).
There is no exchange of heat along an adiabatic trasnformation.
∆S = 0
d.) (3 points) Find the efficiency of an hypothetic Carnot cycle which would be
designed to operate between the two extremes temperatures experienced during
this cycle (make a reasonable assumption if you could not derive earlier the
extremes values for the temperature in the cycle).
η =1−
TL
200
=1−
= 0.75
TH
800
21
Problem 9
One block of mass 1 kg is connected on two sides by two springs of different stiffnesses.
The block rests on a frictionless surface. The block is displaced by 5 cm from its
equilibrium position and released from rest.
k2= 2 N/m
k1= 1 N/m
a.) (5 points) What is the frequency of ensuing oscillation?
r
w=
k
m
k = k1 + k2 = 3 N/m
s
3 N/m
w=
= 1.73 rad/s
1 kg
2πf = w
f = 0.28 Hz
22
b.) (5 points) What is the speed of the block when it passes the equilibrium
position?
1
1
mv 2 = kA2
2
2
s
v=
(3 N/m)(0.05 m)2
= 0.087 m/s
1 kg
c.) (5 points) When the block is passing through the equilibrium point where the
springs are unstrained, another block of equal mass and speed as the first block
is attached to it. What is the new amplitude of the oscillation?
1
1
(m + m)v 2 = kA2f
2
2
√
Af = 2A = 7 cm
23
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
x = x0 + v0 t + 12 at2
F = µFN
s = rθ
p = mv
ω = ω0 + αt
KE = 21 Iω 2
W = F s cos θ
2
ac = vr
τ = F ` sin θ
Ifull cylinder = 21 mr2
Ifull sphere = 52 mr2
F = Y A ∆L
L0
P V = nRT
Q = mL
V
Wisotherm = nRT ln Vfi
Q
= kA
∆T
t
L
Q = σT 4 At
v = λf
v = v0 + at
F = −kx
v = rω
F ∆t = ∆p
P E = 12 kx2
KE = 12 mv 2
L = Iω
ω 2 =p
ω02 + 2α∆θ
ω = k/m√
2πr3/2 = T GM
Ihollow cylinder = mr2
∆L = αL0 ∆T
P1 V1γ = P2 V2γ
P V = nkT
Q = cm∆T
γ =q
CP /CV
v=
v=
√
γkT
m
v2 = v02 + 2ax
F = −G Mr2m
a = rα
P
xcm = M1tot i xi mi
P E = mgh
Wnc = ∆KE + ∆P E
P
τ = Iα
∆θ =pω0 t + 12 αt2
ω = g/l
P
I = mi ri2
Q = πR4 ∆P/8ηL
Q = Av
P + ρgh + 21 ρv 2 = const
∆S = ∆Q
T
∆U = Q − W
U = 32 nRT
q
F
v = m/L
√
v = Yρ
β = (10 dB) log II0
f0 = fs / 1 ∓ vvs
sin θ = Dλ
Bad ρ v
1± 0
f0 = fs 1∓ vvs
v
v
fn = n 2L
sin θ = 1.22 λd
Monatomic:
Diatomic:
CV = 3R/2
CV = 5R/2
CP = 5R/2
CP = 7R/2
R = 8.315 J/K/mol
u = 1.66 × 10−27 kg
σ = 5.67 × 10−8 W/m2 /K4
NA = 6.022 × 1023 mol−1
k = 1.38 × 10−23 J/K
Mearth = 6.0 × 1024 kg
Mmoon = 7.4 × 1022 kg
0◦ C = 273.15 K
Rearth = 6.4 × 106 m
Rmoon = 1.7 × 106 m
1 kcal = 4186 J
GNewton = 6.67 × 10−11 Nm2 /kg2
f0 = fs (1 ± vv0 )
v
fn = n 4L
vsound = 343 m/s