PHY315
Data Provided:
A formula sheet and table of physical constants is attached to this paper.
DEPARTMENT OF PHYSICS AND ASTRONOMY
Spring Semester 2015
TECHNIQUES OF PROBLEM SOLVING IN PHYSICS
2 hours
Answer EIGHT questions.
All questions are marked out of 10.
If a personal formula sheet has been brought to the examination, it must be attached
to the answer booklet.
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1
This question consists of parts A to E. If you attempt this question, answer all
parts.
A) A battle ship fires two shells at enemy ships at the same time. Both shells follow
parabolic paths. Which of the following statements is correct?
i.
Ship A is hit first.
ii.
Ship B is hit first.
iii. Both ships are hit at the same time.
iv.
More information is needed.
B) A student views a distant object through a telescope. If the lower half of the
objective lens is covered up, which one of the following statements is correct?
i.
The image will disappear.
ii.
The lower half of the image will disappear.
iii. The upper part of the image will disappear.
iv.
The image will appear dimmer.
v. There will be no effect.
C) A student sits on a swing and the swing oscillates at its natural frequency. If the
student now stands up on the swing which of the following is correct?
i.
The new frequency is greater than the original frequency.
ii.
The new frequency is smaller than the original frequency.
iii. The new frequency is equal to the original frequency.
D) A body falls from rest through air so that there is air resistance. If the body gains
20J of kinetic energy, how much gravitational potential energy did the body lose?
i.
20 J.
ii.
More than 20 J.
iii. Less than 20 J.
iv. It is not possible to say.
E) A car travels round an unbanked curve. (An unbanked track is one which is
entirely in a horizontal plane.) What force causes the car to be held in the curve?
i.
The horizontal component of the normal force.
ii.
The vertical component of the normal force.
iii. The weight of the car.
iv.
The frictional force.
2
Four identical cells, of EMF  and internal resistance r, are connected in series to a bulb
of resistance 3.00 Ω, and the voltage across the bulb is measured as 5.35 V. One of the
cells is then reversed, and the voltage across the bulb is recorded as only 2.61 V. As
the filament in the bulb reached a lower temperature, its resistance dropped to 2.55 Ω.
Assuming the internal resistance of the reversed cell remained constant, find the EMF
and internal resistance of a single cell.
PHY315
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3
Comet 67P has a mass of 10 billion (1010) tonnes, and an estimated density of
400 kg m–3. When the Philae probe attempted to land on the comet, it initially bounced
to a height above the surface of 1 km. Estimate the speed with which it left the surface
after the first contact, making clear any assumptions involved.
4
The filament of a 100 W light bulb has a length of 5 cm. Estimate its radius given
Wien’s Displacement Law which relates the temperature of an emitting body to the
hf
frequency for which maximum power is emitted
 2.8 .
kT
5
The speed of light in a vacuum, c, depends on the relative permittivity 0 and
permeability 0 of free space. Use dimensional analysis to find the relationship
between c, 0, and 0.
6
The π0 is a meson of rest mass 135 MeV/c2 which decays into two photons. It should
therefore be possible to produce it through the interaction of two photons. Roughly
what energy gamma ray would be required to interact with a thermal photon (in an
environment at room temperature) to produce a π0?
7
A horizontal disk of radius r has two loudspeakers attached at its edges, diametrically
opposite each other. Each loudspeaker is fed with the same amplitude sinusoidal signal
of frequency f0. The radius of the disk is small compared to the wavelengths of the
sound produced. If the disk rotates around a vertical axis with angular velocity  show
that an observer situated a distance much greater than the size of the disk away from the
disk hears the frequency f0 modulated by a beat frequency which varies between zero
and a maximum value. Derive an expression for the maximum beat frequency.

v
The formula for the Doppler effect for sound is f  1   f 0 .
 vs 
8
How large would a helium balloon need to be in order to lift an adult from the ground?
9
A hoop, a uniform cylinder and a solid ball, all of radius R and mass M, roll down a
slope inclined at an angle θ to the horizontal. All three roll without slipping. If the
cylinder takes 30.0 s to travel a distance s, what are the times thoop and tball taken by the
other two objects?
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10
Polonium is the only element that crystallises in a simple cubic structure. In a
diffraction experiment using X-rays of wavelength 0.1789 nm, a first order diffraction
peak associated with the (110) crystal plane is deflected through an angle of 44.51° (see
diagram).
a) What is the lattice constant of polonium?
b) At what angle would you expect to find the first order peak corresponding to the
(111) plane?
c) At what angle would you expect to find the second order peak corresponding to the
(111) plane?
11
Calculate the total resistance of an infinite “ladder” of resistors, each of resistance R, as
illustrated.
R
R
R
Hint: consider the effect of adding an additional “stage” on the left of the ladder.
12
Particles emit Cherenkov radiation when they pass through a medium with a speed
greater than the speed of light in that medium. Air has a refractive index of 1.00027.
Which, if any, of pions, kaons and protons, each with an energy of 15 GeV, will emit
Cherenkov radiation in the atmosphere? The rest masses of pions, kaons and protons
are 139.6 Mev/c2, 493.7 Mev/c2 and 938.3 Mev/c2 respectively.
13
An elastic rope of natural length L, negligible mass and elastic constant k is attached
between two supports of equal height and a distance L apart. If a mass m is attached to
the centre of the rope what is the vertical displacement, y, of the centre of the rope
assuming the acceleration due to gravity is g and that L>>y?
14
You stand in a pool, fishing with a spear. The tip of your spear touches the surface of
the pool. A fish appears to float in the pool under an angle of 45° to the normal of the
water surface, taken at the point where your spear touches the water. The refractive
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CONTINUED
index of water is 1.333, assume n = 1 exactly for air. At what angle to the normal
should you thrust the spear into water to spear the fish?
15
The official specifications for a tennis ball state that its mass should be between 56.0
and 59.4 g and that its diameter should be between 6.54 and 6.86 cm. The drag
coefficient Cd of a tennis ball has been measured to be 0.507±0.024. The drag force on
a projectile is proportional to the drag coefficient (a pure number) and depends on the
density of air, the cross-sectional area of the object and the speed of the object.
During a championship, the service speed of a particular player is measured to be
58 m s−1. A sports scientist employed by this player wishes to determine the effect of
drag on his tennis serves. If the error on the service speed measurement is negligible,
calculate the fractional variation in the acceleration due to drag caused by the tolerance
in the ball specifications and the uncertainty in the drag coefficient (which is also
caused by ball-to-ball variations), stating any assumptions made.
END OF EXAMINATION PAPER
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PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE
Physical Constants
electron charge
electron mass
proton mass
neutron mass
Planck’s constant
Dirac’s constant (~ = h/2π)
Boltzmann’s constant
speed of light in free space
permittivity of free space
permeability of free space
Avogadro’s constant
gas constant
ideal gas volume (STP)
gravitational constant
Rydberg constant
Rydberg energy of hydrogen
Bohr radius
Bohr magneton
fine structure constant
Wien displacement law constant
Stefan’s constant
radiation density constant
mass of the Sun
radius of the Sun
luminosity of the Sun
mass of the Earth
radius of the Earth
e = 1.60×10−19 C
me = 9.11×10−31 kg = 0.511 MeV c−2
mp = 1.673×10−27 kg = 938.3 MeV c−2
mn = 1.675×10−27 kg = 939.6 MeV c−2
h = 6.63×10−34 J s
~ = 1.05×10−34 J s
kB = 1.38×10−23 J K−1 = 8.62×10−5 eV K−1
c = 299 792 458 m s−1 ≈ 3.00×108 m s−1
ε0 = 8.85×10−12 F m−1
µ0 = 4π×10−7 H m−1
NA = 6.02×1023 mol−1
R = 8.314 J mol−1 K−1
V0 = 22.4 l mol−1
G = 6.67×10−11 N m2 kg−2
R∞ = 1.10×107 m−1
RH = 13.6 eV
a0 = 0.529×10−10 m
µB = 9.27×10−24 J T−1
α ≈ 1/137
b = 2.898×10−3 m K
σ = 5.67×10−8 W m−2 K−4
a = 7.55×10−16 J m−3 K−4
M = 1.99×1030 kg
R = 6.96×108 m
L = 3.85×1026 W
M⊕ = 6.0×1024 kg
R⊕ = 6.4×106 m
Conversion Factors
1 u (atomic mass unit) = 1.66×10−27 kg = 931.5 MeV c−2
1 astronomical unit = 1.50×1011 m
1 eV = 1.60×10−19 J
1 atmosphere = 1.01×105 Pa
1 Å (angstrom) = 10−10 m
1 g (gravity) = 9.81 m s−2
1 parsec = 3.08×1016 m
1 year = 3.16×107 s
Polar Coordinates
x = r cos θ
y = r sin θ
∂
1 ∂2
1 ∂
2
r
+ 2 2
∇ =
r ∂r
∂r
r ∂θ
dA = r dr dθ
Spherical Coordinates
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
dV = r2 sin θ dr dθ dφ
1
∂
1
∂2
1 ∂
∂
2
2 ∂
∇ = 2
r
+ 2
sin θ
+ 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
Calculus
f (x)
f 0 (x)
f (x)
f 0 (x)
xn
ex
nxn−1
ex
tan x
sin−1
ln x = loge x
1
x
sin x
cos x
cos x
− sin x
cosh x
sinh x
sinh x
cosh x
a
cos−1 xa
tan−1 xa
sinh−1 xa
cosh−1 xa
tanh−1 xa
cosec x
−cosec x cot x
uv
sec x
sec x tan x
u/v
sec2 x
x
√ 1
a2 −x2
− √a21−x2
a
a2 +x2
√ 1
x2 +a2
√ 1
x2 −a2
a
a2 −x2
0
0
u v + uv
u0 v−uv 0
v2
Definite Integrals
Z
∞
xn e−ax dx =
0
Z
+∞
n!
an+1
r
(n ≥ 0 and a > 0)
π
a
−∞
r
Z +∞
1 π
2 −ax2
xe
dx =
2 a3
−∞
Z b
b Z b du(x)
dv(x)
Integration by Parts:
u(x)
dx = u(x)v(x) −
v(x) dx
dx
dx
a
a
a
−ax2
e
dx =
Series Expansions
(x − a) 0
(x − a)2 00
(x − a)3 000
f (a) +
f (a) +
f (a) + · · ·
1!
2!
3!
n X
n n−k k
n
n!
n
Binomial expansion: (x + y) =
x y
and
=
(n − k)!k!
k
k
k=0
Taylor series: f (x) = f (a) +
(1 + x)n = 1 + nx +
ex = 1 + x +
n(n − 1) 2
x + ···
2!
x 2 x3
+ +· · · ,
2! 3!
sin x = x −
ln(1 + x) = loge (1 + x) = x −
Geometric series:
n
X
rk =
k=0
Stirling’s formula:
(|x| < 1)
x3 x5
+ −· · ·
3! 5!
x2 x3
+
− ···
2
3
and
cos x = 1 −
x2 x4
+ −· · ·
2! 4!
(|x| < 1)
1 − rn+1
1−r
loge N ! = N loge N − N
or
ln N ! = N ln N − N
Trigonometry
sin(a ± b) = sin a cos b ± cos a sin b
cos(a ± b) = cos a cos b ∓ sin a sin b
tan a ± tan b
1 ∓ tan a tan b
sin 2a = 2 sin a cos a
tan(a ± b) =
cos 2a = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a
sin a + sin b = 2 sin 21 (a + b) cos 12 (a − b)
sin a − sin b = 2 cos 12 (a + b) sin 12 (a − b)
cos a + cos b = 2 cos 12 (a + b) cos 12 (a − b)
cos a − cos b = −2 sin 12 (a + b) sin 21 (a − b)
eiθ = cos θ + i sin θ
1 iθ
1 iθ
cos θ =
e + e−iθ
and
sin θ =
e − e−iθ
2
2i
1 θ
1 θ
cosh θ =
e + e−θ
and
sinh θ =
e − e−θ
2
2
sin a
sin b
sin c
Spherical geometry:
=
=
and cos a = cos b cos c+sin b sin c cos A
sin A
sin B
sin C
Vector Calculus
A · B = Ax Bx + Ay By + Az Bz = Aj Bj
A×B = (Ay Bz − Az By ) î + (Az Bx − Ax Bz ) ĵ + (Ax By − Ay Bx ) k̂ = ijk Aj Bk
A×(B×C) = (A · C)B − (A · B)C
A · (B×C) = B · (C×A) = C · (A×B)
grad φ = ∇φ = ∂ j φ =
∂φ
∂φ
∂φ
î +
ĵ +
k̂
∂x
∂y
∂z
∂Ax ∂Ay ∂Az
+
+
∂x
∂y
∂z
∂Ax ∂Az
∂Ay ∂Ax
∂Az ∂Ay
−
î +
−
ĵ +
−
k̂
curl A = ∇×A = ijk ∂ j Ak =
∂y
∂z
∂z
∂x
∂x
∂y
div A = ∇ · A = ∂ j Aj =
∇ · ∇φ = ∇2 φ =
∂ 2φ ∂ 2φ ∂ 2φ
+
+ 2
∂x2 ∂y 2
∂z
∇×(∇φ) = 0
and
∇ · (∇×A) = 0
∇×(∇×A) = ∇(∇ · A) − ∇2 A