DO NOT REMOVE THIS QUESTION PAPER FROM THE EXAM VENUE
SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
PHYS201 : MECHANICS & MODERN PHYSICS
JUNE 2015 MAIN EXAMINATION
DURATION: 3 HOURS
TOTAL MARKS: 180
External Examiner:
Prof T. Konrad (UKZN)
Internal Examiners:
Dr M. Moodley (UKZN)
Dr A. P. Matthews (UKZN)
Instructions:
• Check that you have 12 pages including this cover page
• Answer Section A (Mechanics and Oscillations) in a different answer book to Section B
(Special Relativity and Thermal Physics)
• Information and formulae for Section B can be found in pages 11 and 12 of this exam
Students are requested, in their own interest to write legibly
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, PHYS201 Mechanics and Modern Physics
Section A: Mechanics and Oscillations (90 marks)
Question 1 [15 marks]
A particle of mass m moves
along the x-axis under the influence of a velocity dependent retard√
ing force, F (v) = −α v, where α is a positive constant. Assume that at t = 0, its position is
x = x0 and velocity v = v0 .
(a). Show that the displacement of the particle is given by
x(t) = x0 + v0 t −
√
(8)
1
α
v0 βt2 + β 2 t3 where β =
.
3
2m
(b). By first determining the duration, T , before the particle comes to rest, show that the range
R (maximum displacement) of the particle is
(7)
R = x0 +
1 3/2
v
3β o
Question 2 [15 marks]
The Cartesian cordinates and unit vectors in terms of cylindrical coordinates and their unit
vectors are given by
x = ρ cos ϕ, y = ρ sin ϕ, z = z
and
x̂ = cos ϕ ρ̂ − sin ϕ ϕ̂, ŷ = sin ϕ ρ̂ + cos ϕ ϕ̂, ẑ = ẑ.
(a). Show that the cylindrical unit vectors ρ̂ and ϕ̂ in terms of x̂ and ŷ are given by
ρ̂ = cos ϕ x̂ + sin ϕ ŷ
(5)
and ϕ̂ = − sin ϕ x̂ + cos ϕ ŷ.
(b). If the position vector in Cartesian coordinates is given by r = x x̂ + y ŷ + z ẑ, show that
the velocity vector in cylindrical coordinates is
(10)
v = ρ̇ ρ̂ + ρϕ̇ ϕ̂ + ż ẑ.
[Hint: Rewrite r in cylindrical coordinates and then take its time derivative.]
page 2 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, PHYS201 Mechanics and Modern Physics
Question 3 [30 marks]
In plane polar coordinates, the equation of motion in a central force field F (r) of a particle of
mass m in the radial direction is given by
F (r) = mr̈ − mrθ̇2 .
(a). Derive the orbital differential equation
(10)
m
d2 u
+ u = − 2 2 F (u−1 )
2
dθ
Lu
where u ≡ 1/r and L = mr2 θ̇ is the angular momentum.
(b). A particle moves along an exponential spiral orbit
r(θ) = r0 eθ
where r0 is a constant, under the influence of a central potential V (r).
(i) Show that the potential is V (r) = −L2 /mr2 .
(8)
(ii) Show that the total energy E of this orbit is zero.
(6)
(iii) Determine the angular motion in time θ(t) of the particle in this orbit.
(6)
Question 4 [20 marks]
A cylinder of mass m and cross-sectional area A floats vertically in a liquid of density ρ, as
shown in the figure. The positive z-axis points vertically upward from the surface of the liquid.
W and B indicate the weight and buoyancy force.
(a). If the cylinder is pushed down by a distance z from its equlibrium immersion depth z0 ,
show that the buoyancy force is B = (mg − ρAgz)ẑ.
(8)
(b). Show that the equation of motion for the cylinder is
(5)
√
z̈ +
ω02 z
= 0 with ω0 =
ρgA
.
m
page 3 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, PHYS201 Mechanics and Modern Physics
(c). If the cylinder is now subject to a vertical driving force Fd = 2 cos ωt Newtons, show that
the vertical displacement of the cylinder as a function of time is,
(7)
z(t) =
2 cos ωt
.
ω02 − ω 2
[Hint: Without damping you only need to consider the particular solution, zp (t).]
Question 5 [20 marks]
Consider a location at northern latitude λ on the Earth’s surface. A particle of mass m starts
falling from rest at position r0 = (0, 0, h) in the local coordinate system with axes as shown in
the figure. The effective acceleration felt by the particle is a = g − 2(ω × v) where in the local
coordinate system,
g = (0, 0, −g), ω = (−ω cos λ, 0, ω sin λ), and v = (0, 0, −gt).
(a). Show that the position of the particle during the fall is,
(12)
1
1
r(t) = (0, ωgt3 cos λ, h − gt).
3
2
(b). If h = 100 m, g = 9.81 m/s2 and λ = 45◦ , what is the magnitude and direction of
horizontal deflection from the vertical line of the point where the particle hits the ground.
(The angular frequency of the Earth’s rotation is, ω = 7.29 × 10−5 rads/s.)
(8)
————– End of Section A ————–
page 4 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
Section B: Modern Physics (90 marks)
Question 1 [15 marks]
The diagram below shows a scenario viewed in frame S at time t = 0. A rod moves with
speed v = 4c/5 in the positive x-direction, and is at rest in frame Sʹ′. Point A marks the back
end of the rod and point B marks the front end. The rod’s rest length is L0 = 45 m. The
Lorentz factor γ = 5/3.
S
E1
E2
A
B
v
x
Event E1 is a light signal emitted at A at t = 0. Event E2 is B at t = 0. Events not shown on the
diagram are: E3: B at tʹ′ = 0, and E4: light signal reaches B. Coordinates of events are:
E1: in S: (x1 = 0, t1 = 0), in Sʹ′: (x1ʹ′ = 0, t1ʹ′ = 0).
E2: in S: (x2 = 27 m, t2 = 0), in Sʹ′: (x2ʹ′ = ?, t2ʹ′ = ?).
E3: in S: (x3 = ?, t3 = ?), in Sʹ′: (x3ʹ′ = 45 m, t3ʹ′ = 0).
E4: in S: (x4 = ?, t4 = ?), in Sʹ′: (x4ʹ′ = 45 m, t4ʹ′ = ?).
.
(a) Calculate t2ʹ′ (time coordinate of event E2 in Sʹ′) in units of ns.
(4)
(b) Calculate x3 (x-coordinate of event E3 in S) in units of m.
(4)
(c) Let event E4 be the arrival at B of light from E1. Measured in S, E4 has position x4.
Calculate x4, in m.
(3)
(d) The length of the rod may be measured as the difference in x-coordinates of two events.
Using this method, which two events (of E1 to E4) must be used to find the length Lʹ′ of
the rod as measured in Sʹ′? Explain your answer (explain your choice of both E1 and E2).
(4)
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page 5 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
Question 2 [6 marks]
This question refers to Question 1. Reproduce the the Minkowski diagram below in your
answer book, and on it show and label the following:
(a) The worldline of point B.
(1)
(b) Events E1, E2, E3 and E4.
(4)
(c) The coordinate t2ʹ′ of event E2 in Question 1, marked off on the tʹ′ axis.
(1)
Note: Lines that are meant to be parallel should be shown as close to parallel as possible. The
magnitudes of coordinate values are arbitrary, and units are chosen such that c = 1.
t
tʹ′
x = ct
xʹ′ = ctʹ′
xʹ′
x
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page 6 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
Question 3 [10 marks]
Frames S and Sʹ′ are in standard configuration. Viewed in frame Sʹ′, a projectile has velocity
𝑢′ = 𝑢′! , 𝑢′! where uʹ′x = uʹ′y. In frame S the velocity components are ux and uy, where
ux = 2uy. Show that:
𝑢!! =
𝛾𝑣
2−𝛾
(Note: Do not derive the velocity addition formulae.)
Question 4 [8 marks]
The space-time interval 4-vector dR ≡ (cdt, dx, dy, dz), where the index µ labels coordinates.
Let the proper time interval be dτ, and the rest mass be m0. Recall that 𝑑𝜏 = 𝑑𝑡 𝛾.
µ
(a) Show that the 4-momentum P ≡ 𝑚!
µ
!!!
!"
evaluates to 𝛾 𝑢 𝑚! 𝑐, 𝑢 .
! !
𝑚!! 𝑐 !
(b) Use the invariant interval for P to show that 𝑚 𝑐 =
relativistic mass and 𝑝 = 𝑚𝑢 is relativistic momentum.
µ
(4)
+ 𝑝 𝑐 , where 𝑚 = 𝛾𝑚! is
! !
(4)
Question 5 [5 marks]
A particle has rest mass m0 = 5.0 × 10-27 kg and speed v = 0.95 c.
(a) Calculate γ.
(1)
(b) Calculate the kinetic energy of the particle, in J.
(4)
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page 7 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
Question 6 [28 marks]
The diagrams below show three different states of a system enclosed in a container with rigid,
adiabatic walls. Initially (panel (a)) the container is divided into two compartments of equal
volume separated by an impermeable separating wall. In the left compartment is an ideal gas
in equilibrium with temperature T1 = 290 K, pressure P1 = 200 kPa and volume V1 = 6 litres.
In the right compartment is a vacuum. The separating wall is then quickly removed, allowing
the gas to expand freely to fill the container. Panel (b) shows the system immediately after
the separating wall has been removed. Panel (c) shows the final equilibrium state of the
system when the gas has occupied the container and has T2 = 290 K, P2 = 100 kPa and V2 =
12 litres. Let process A be the process that takes the gas from the state in panel (b) to the state
in panel (c).
vacuum
gas
T1 = 290 K
P1 = 200 kPa
V1 = 6 litres
separating wall
(a) Gas is in equilibrium in the left
side of the container when it is
separated from the vacuum by
a separating wall.
adiabatic wall
gas starts to expand
freely into vacuum
Gas now occupies all volume
T2 = 290 K
P1 = 100 kPa
V1 = 12 litres
(b) The separating wall is
removed, and the gas is free to
expand to fill the container. In
the state shown, the gas has
just started to expand.
(c) The gas is in its final
equilibrium state in which it
has filled all the volume of the
container.
(a) Is process A reversible or irreversible? Explain your answer.
(3)
(b) In panel (b), is the system in equilibrium? Explain your answer.
(3)
(c) Does the internal energy change during process A? Explain your answer.
(3)
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page 8 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
(d) A mercury thermometer is used to measure the temperature of the gas. The thermometer’s
thermometric property is the height h of the mercury column, and uses a two-point (icesteam) scale, with h = 2.4 cm at 0 °C and h = 8.6 cm at 100 °C. What is the value of h
when T2 is measured? (Let the ice point be at 273 K).
(6)
(e) Consider a different process B that takes the gas from the state in panel (b) to the state in
panel (c). In process B, the exterior adiabatic wall is replaced by a diathermal wall, and
then the internal separating wall is moved very slowly to the right until it is against the
right-hand interior wall of the container and the gas has filled the container. Note that
process B, which is a quasi-static isothermal expansion, is reversible.
(i) Explain what is meant by “adiabatic” and “diathermal”.
(4)
(ii) Explain why it is necessary for the adiabatic wall to be replaced by a diathermal wall.
(3)
(iii) Calculate the entropy change ΔS of the gas by integrating over process B.
(6)
Question 7 [10 marks]
A fixed quantity of an ideal gas undergoes a cycle.
State variables P (pressure), V (volume), T
(temperature) and U (internal energy) are shown in
the table. Process AB is isochoric, process BC is
adiabatic, and process CA is isobaric.
State
A
B
C
P (Pa)
100
400
100
V(m3)
2
2
4.59
T (K)
200
400
459
U (J)
300
1200
689
P
B
C
A
V
Calculate the following:
(a) Heat QAB added to the gas during process AB (note that Q is positive if absorbed and
negative if rejected).
(2)
(b) Heat QBC added to the gas during process BC.
(2)
(c) Heat QCA added to the gas during process CA.
(2)
(d) Net work done by the gas in one cycle.
(2)
(e) Efficiency of one cycle.
(2)
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page 9 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
Question 8 [10 marks]
Answer ONE of (a) or (b):
(a) Show, using a diagram, that if the Kelvin-Planck statement (applied to a cycle) is false,
then the Clausius statement (applied to a cycle) is false. Do this by constructing a
composite device that includes a Carnot cycle.
(b) Show, using a diagram, that if the Clausius statement (applied to a cycle) is false, then the
Kelvin-Planck statement (applied to a cycle) is false. Do this by constructing a composite
device that includes a Carnot cycle.
Note: Answer only one of (a) or (b). If you answer both, only your first answer will be
marked.
--------------------- End of Section B -------------------
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page 10 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
PHYS201 Special Relativity: Information and Formulae
Postulates:
1. All inertial frames are equivalent for all physical experiments.
2. The speed of light c in vacuo is constant in all inertial frames.
c = 3.00 × 108 m s-1
Lorentz Transformation
Let inertial frames S(x,y,z,t) and Sʹ′(xʹ′,yʹ′,zʹ′,tʹ′) in standard configuration be Cartesian frames
such that their origins and spatial axes coincide (i.e. in particular xʹ′ = x = 0) at t = tʹ′ = 0. Sʹ′
has constant speed v in the x-direction, relative to S. The coordinates of any event E(x,y,z,t) in
S are observed in Sʹ′ to have coordinates:
⎛ vx ⎞
t ʹ′ = γ ⎜ t − 2 ⎟
z ʹ′ = z
xʹ′ = γ (x − vt )
y ʹ′ = y
⎝ c ⎠
1
where
Lorentz factor γ =
and c is the speed of light.
2
v
1−
c2
Velocity Transformation
u ʹ′x =
ux − v
uv
1 − x2
c
u ʹ′y =
uy
⎛ u v ⎞
γ ⎜1 − x2 ⎟
c ⎠
⎝
u ʹ′z =
uz
⎛ u v ⎞
γ ⎜1 − x2 ⎟
c ⎠
⎝
Note that the inverse transformations from Sʹ′ to S are obtained by changing the sign of v in
the formulae above.
(Recall that 1 ns (nanosecond) = 10-9 s)
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page 11 of 12
University of KwaZulu-Natal, School of Chemistry and Physics, Westville Campus
June 2015 Main Examination Session, Phys201 Mechanics and Modern Physics
PHYS201
Classical Thermodynamics
Gas constant
Atmospheric pressure
Avogadro’s number
Triple point temperature of water
Data and Formulae
R = 8.314 J K-1 mol-1
Patm = 101 kPa
NA = 6.022 × 1023 mol-1
TTP = 273.16 K
Δ𝑈 = 𝑄 + 𝑊
ð𝑄 = 𝑑𝑈 + 𝑃𝑑𝑉
Ideal Gas
Equation of state
Work done on gas in a reversible
isothermal process
Reversible adiabatic process
𝑃𝑉 = 𝑛𝑅𝑇
𝐶! − 𝐶! = 𝑛𝑅
𝛾 = 𝐶! 𝐶!
𝑊 = 𝑛𝑅𝑇 ln 𝑉! 𝑉!
𝑇𝑉 !!! = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑃𝑉 ! = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Second Law
1. Kelvin-Planck statement: “No process is possible whose sole result is the complete
conversion of heat into work”
2. Clausius statement: “No process is possible whose sole result is the transfer of heat from
a lower-temperature to a higher-temperature body”
Carnot engine efficiency
𝜂! = 1 − 𝑇! 𝑇!
(T2 is lower temperature, T1 is higher temperature)
ð!!"#
Clausius inequality
=0
!
(T0 is temperature of surroundings)
Entropy change in a reversible process
Central Equation of Thermodynamics
Enthalpy
ð!
!!
<0,
(reversible cycle)
(irreversible cycle)
! ð!
Δ𝑆 = ! !!"#
𝑇𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉
𝐻 = 𝑈 + 𝑃𝑉
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