Physics Competitions Vol. 14 No 2 2012
The Final Round of the First World
Physics Olympiad
held in Lombok, West Nusa Tenggara Indonesia: Problems and Results
Herry J. Kwee and Yohanes Surya
Faculty of Physics Education Department, Surya College of Education, Banten, Indonesia
Abstract
A brief report of the final round of the first World Physics Olympiad (WoPhO) held in Lombok,
West Nusa Tenggara, Indonesia is presented. The theoretical and experimental problems
are presented and the mark distribution is discussed.
— 1. Introduction
The final round of the first WoPhO was held in Lombok, West Nusa Tenggara, Indonesia from December 28th, 2011 to January 3rd, 2012 and was organized by Surya
College of Education and Indonesian Society for the Promotion of Science under the
auspices of Surya Institute.
WoPhO is a secondary school level individual physics competition initiated by one of
the authors, Yohanes Surya. It is a unique competition that lasts for a full year and
consists of three rounds: selection, discussion and final round. The selection round is
meant to provide as many opportunities as possible to students to participate in an
Olympiad-level physics competition. Problems are presented online and any student
who is eligible 1 can participate. For the first WoPhO selection round, 345 students
from all over the world register to participate. The discussion round provides an opportunity for the participants and the general public to discuss the problems and solutions of the selection round and other physics Olympiad-level problems. And the final
round, which is organized very similar to the International and Asian Physics Olympiad (IPhO and APhO) provides an opportunity for every participants to challenge the
APhO and IPhO gold medalists. Just like in APhO and IPhO, there are 5 hours at the
students´ disposal to solve the three theoretical problems and another 5 hours for the
experimental tasks. There were 122 students from 13 countries who participated in
the first WoPhO final round, even though the distribution was not evenly spread. The
majority of students were from the host country, Indonesia.
1
Students who have not started college or reached 20 years of age by June 30th of competition year.
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Physics Competitions Vol. 14 No 2 2012
Another unique characteristic of WoPhO is the fact that the problems for the final
round are not provided by the host country but are selected from a competition (or
simply called WoPhO problem competition). Anyone can participate in this competition, in fact one of the winners of this first WoPhO problem competition was a high
school student and another one was a university student. Three theoretical problems
were chosen from thirty one entrants and two experimental problems were chosen
from four entrants. The number of entries for experimental problems was much lower
than for theoretical problems due to the amount of work needed to prepare a good
experiment and its apparatus.
In section 2, we present the theoretical problems.
In section 3, we present the experimental problems.
In section 4, we present the summary of the result of the final round.
— 2. Theoretical Examination
Theoretical Problem No. 1:
Motion of a Rolling Rod
In this problem, the motion of a uniform rod (stick) with length L , ended with casterwheels at both ends, will be investigated on a flat surface. The casters at each end of
the rod can spin freely and independently (see Figure 1) and have a negligible mass
compared to the rod. The friction between the rod and the caster-wheels is negligible.
The diameters of the caster-wheels are a bit larger than the diameter of the rod, but
both diameters are much smaller than the length of the rod. The gravitational acceleration is g .
Figure 1: Sketch of the rod with the caster-wheels.
The rod is placed on a horizontal flat surface and pushed such that each end of the
rod get different horizontal initial velocity ( v1 and v2 , pointing in the same direction)
perpendicular to the axis of the rod. The casters roll without slipping on the surface.
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Physics Competitions Vol. 14 No 2 2012
 Calculate the initial velocity v0 of the center of the rod and the initial angular velocity ω0 of the rod using v1 , v2 and L !
[0.8 points]
 Describe the motion of the center of mass of the rod! Determine the parameter(s)
of its orbit!
[0.8 points]
 What should be the minimum value of the coefficient of static friction µ for the
casters to not slip on the surface?
[0.6 points]
In the following sections the case of the inclined surface will be considered. The angle between the inclined surface and the horizontal plane is α .
If α is infinitesimally small, the motion of the rod slightly changes: the motion of the
center of mass is approximately the same as in the previous section but a constant
drift velocity vdrift added to the solution. Use a coordinate system as in Figure 2.
 Calculate the magnitude and the direction of vdrift as a function of the small α ,
the initial velocities of the two ends of the rod ( v1 and v2 , pointing in the same direction) and the gravitational acceleration g !
[1.9 points]
 Sketch the orbit of the center of mass of the rod!
[0.5 points]
If α is finite, the details of the motion of the rod changes. Place the rod on the inclined plane along the steepest line of the surface (so the rod is parallel with the inclined edges of the plane). Consider that the initial velocity v0 of the center of mass of
the rod is perpendicular to the axis of the rod and the initial angular velocity ω0 is
perpendicular to the surface as shown in Figure 2.
Figure 2: The initial conditions of the rod

Calculate the time evolution of the ve-
locity v (t ) = (vx (t ), v y (t )) of the center of mass
of the rod in the Cartesian coordinate system
shown in Figure 2.
[0.8 points]

Depending on the magnitude of the v0
and ω0 , it can occur, that the center of the rod
stops for a moment during its motion. Express the condition(s) for such a behavior
using the parameters v0 , ω0 , g , α and L !

[0.8 points]
Determine the maximum displacement of the center of the rod in the direction
of the steepest line ( y -direction) as function of v0 and ω0 !
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[1.2 points]
Physics Competitions Vol. 14 No 2 2012
Investigate another situation where the rod is placed horizontally on the inclined surface. Consider that the initial angular velocity ω0 of the rod is perpendicular to the
surface but the initial velocity of the center of the rod is zero (see Figure 3).
Figure 3: The initial conditions of the rod

Describe the motion of the center
of mass of the rod! Determine the parameter(s) of its orbit!
[1.6 points]

What should be the minimum value of the coefficient of static friction µ in
this case for the casters to not slip on the
surface?
[1.0 points]
— Theoretical Problem No. 2:
Why Maglev Trains Levitate
Maglev is a technology for magnetic suspension (levitation) and propulsion of trains
or other vehicles. Since there is no friction force between the train and the rails, Maglev trains reach record velocities approaching 600 km / h .
There are three types of Maglev systems -- EMS (Electromagnetic Suspension), EDS
(Electro-dynamic Suspension), and the experimental Inductrack technology. In this
problem you are going to explore the physical principles of Inductrack suspension on
a simplified model of a Maglev train. The principle of magnetic propulsion will not be
considered here because it has a lot in common with the physics of magnetic levitation. Shown in Figure 4 is a schematic side view of an Inductrack train-car. When at
rest or moving at a low speed, the car lies with its wheels on the rails like any ordinary train. The car, however, detaches from the rails at a specific takeoff velocity 𝑣𝑡
due to the system described below: two long parallel arrays of permanent cubicshape magnets of size a = 5 cm, each is located at the bottom of the car. The magnetic dipole moments of the neighboring magnets are tilted at an angle of 45° relative
to each other (the so called Halbach array). As a result a static magnetic field is produced below each array with components of the magnetic field given by the equations:
=
Bx B0 exp(− ky ) sin(kx)
(2.1)
=
By B0 exp(−ky ) cos(kx)
(2.2)
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Physics Competitions Vol. 14 No 2 2012
where the x-coordinate is measured from the rear end of the car in the direction of
motion and y-coordinate from the bottom of the car in a downward direction. The parameter k is the wavenumber of the magnetic field. The amplitude of the magnetic
field is B0 = 1.4 T .
Figure 4: A schematic side view of an Inductrack train-car. The arrows show the directions of magnetic
moments of the cubic magnets in the Halbach array. The wheels and the rails of the train are not shown
for convenience.
Two inductive arrays of horizontal rectangular wire frames of the same width a made
of permanent magnets are arranged along the rail as seen from the top view
Figure 5. The length of the inductive frames as well as the distance between them is
b. Each inductive array is located below the corresponding Halbach array at a distance corresponding to the y-coordinate of the inductive array (see Figure 4). Assume that the equations (2.1) and (2.2) for the components of the magnetic field hold
only for the inductive frames, which are located below the car, while the magnetic
field outside the car area is strictly zero.
Figure 5: A schematic top-view of an Inductrack train-car. The wheels and the rails of the train are not
shown for convenience.
Note: In the real Inductrack trains, the frames of the inductive array are electrically
connected and form a continuous ladder-like array. In order to simplify the theoretical
consideration, we use a simplified model shown in
Figure 5.
Helpful mathematics formulas:
x− y
x+ y
sin( x) − sin( y ) =
2sin(
) cos(
)
2
2
(2.3)
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Physics Competitions Vol. 14 No 2 2012
y−x
x+ y
cos( x) − cos( y ) =
2sin(
) sin(
)
2
2
(2.4)
Assume that the train car is infinitely long (i.e. equations (2.1) and (2.2) hold for all values of
the x-coordinate.
a) Electromotive force in the inductive wires
 Derive a relationship between the wavenumber of the magnetic field k and the
lateral size of the cubic magnets in the Halbach array a. Calculate k numerically.
[0.3 points]
 Suppose that the train moves with a constant velocity v in the positive direction of
the x-axis. Consider an inductive frame whose center in the initial moment t = 0
is located below the car at a point of coordinate xc relative to the car. Derive an
expression for the electromotive force (EMF)
ε induced in the frame as a func-
tion of the time t and the other parameters already defined.
 What is the angular frequency ω of the induced EMF.
[1.5 points]
[0.4 points]
b) Current in the inductive wires
Each frame in the inductive array is characterized by a self-inductance L and a resistance R. The mutual inductance between different frames is negligible compared
to L. The current I induced in the frame varies with time according to the equation:
=
I (t ) I 0 sin(ωt − kxc +ψ )

Obtain expressions for the amplitude I 0 and the phase-shift ψ of the alternat-
ing current induced in the frame.
[2.0 points]
Note: The positive direction of circulation of the induced current in the frame is related to the
positive direction of the y-axis according to the right-hand rule.
c) The dynamics of the train
 Derive formulas for the time-averaged components Fx and Fy acting on a single
inductive frame in terms of velocity v, the distance between the Halbach array
and the inductive array y, and the others parameters already defined. Sketch
qualitative graphs of Fx and Fy versus train velocity v for a fixed value of the distance y.
[3.0 points]
 Consider a train-car of a large but finite length (l and b). Derive expressions for
the magnitude of the total lift (vertical) and drag (horizontal) forces FL and FD
acting on the car.
[1.2 points]
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Physics Competitions Vol. 14 No 2 2012
 For a given value of a, what is the minimum aspect ratio b / a for which the lift
force is maximum for given values of the distance y, velocity v, resistance R, and
the inductance L.
[1.0 points]
 Consider a train-car of length l = 10 m and mass m = 10000 kg . Assume that the
aspect ratio b / a corresponds to the optimal value found in 3(c). The inductance
and the resistance of the inductive frames are =
L 1.0 ×10−7 H and R =1.0 ×10−5 Ω
respectively. The acceleration due to gravity is g = 9.8 /
m s 2 . Assume that the distance between the Halbach array and the inductive array is y = 0 m , when the
train is at rest.
Obtain an expression and calculate the takeoff velocity vt , the velocity at which
the train detaches from the rails.
[0.3 points]
 What is the distance y between the train and the rails at an operating velocity of
360 km/h?
— Theoretical Problem No. 3: Mirage
The refractive index of the air varies with temperature. Cold air is denser than warm
air and has therefore a greater refractive index. Thus a temperature gradient in the
atmosphere is always associated with a gradient of the refractive index. Under certain conditions, this gradient of the refractive index could be strong enough so that
light rays are bent to produce a displaced image of distant objects. This amazing
phenomenon is called a mirage.
Mirages can be categorized as "inferior" and "superior". Inferior mirages can be seen
on deserts and highways, and superior mirages occur over the sea. To describe in
detail the phenomenon of the mirage, we need to analyze the path of light rays in
media with a refractive index gradient.
Figure 6:
Left: A superior image.
Right: An inferior image
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Physics Competitions Vol. 14 No 2 2012
The path of a light ray and the trajectory of a mass point
Media with a refractive index gradient can be treated as being formed from thin homogeneous layers with different refractive indices. The path of a light ray can be determined then by analyzing refractions of the light ray at interfaces between these
thin homogeneous layers. But there exists a more convenient way for determining the
path of a light ray in media with a refractive index gradient. Instead of analyzing the
propagation of the light, one may study the motion of a mass point that moves along
the path of the light ray, driven by a conservative force. The potential energy of the
conservative force field depends on the distribution of the refractive index. Once the
potential energy is established, one may study the motion of the mass point by using
well developed tools in classical mechanics, and find the trajectory of a mass point
which is also the path of the light ray.
Refraction of a light ray at the interface between a medium with a refractive index n1
and a medium with a refractive index n2 is shown in Figure 7a. The angles i1 and i2
obey the Snell's law:
n1 sin i1 = n2 sin i2 .
(2.1)
The path of a mass point in a conservative force field is illustrated in Figure 7b. The
speed of the mass point is v1 in the region with the potential energy E p1 , and v2 in the
region with the potential energy E p 2 .
 Find expressions for v1 and v2 so that the relation n1 sin i1 = n2 sin i2 holds also for
the trajectory of the mass point. You may use an arbitrary constant expression
[0.2 points]
v0 with the dimension of speed in your expressions.
Figure 7:
(a) Left: Refraction of a light ray.
(b) Right: Deviation of a mass point in a conservative force field
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Physics Competitions Vol. 14 No 2 2012
 Assume that the mass of the mass point is m, and the total energy of the mass
point equals to zero. Find the potential energies E p1 and E p 2 in v0 , n1 and n2
[0.2 points]
 The trajectory of a mass point with a mass m in a conservative force field is the

same as the path of a light ray in a medium with a refractive index n(r ) which is
a function of the position. The total energy of this mass point is zero. Find ex
pressions for the potential energy E p (r ) of the conservative force field and the

[0.4 points]
speed of the mass point v(r )
 To describe the motion of a mass point, one expresses the position of the mass

point as a function of time: r (t ) . To describe the trajectory of a mass point, one
needs to express the position of the mass point as a function of the distance s
traveled by the mass point from the start point of the trajectory to the current po

sition: r ( s ) . Derive a differential equation for the trajectory r ( s ) of a mass point
with a mass m and a total energy E that moves in a conservative force field.
[0.6 points]

Hint: in a conservative force field, r (t ) satisfies Newtons´ second law:


d r
m 2 = −∇E p ,
dt
2
(2.2) where
ds

= v ( r ),
dt

∂
∂
∂
=
∇ xˆ + yˆ + zˆ .
∂x
∂y
∂z
(2.3)
 Derive the light ray equation (a differential equation for the path of a light ray) by
using the results obtained in (c) and (d).
[0.6 points]


  
Hint: ∇f 2 (r ) = 2 f (r )∇f (r )
The inferior mirage
When an inferior mirage appears, an image of a distant object can be seen under the
real object. A direct image of that object is seen because some of the light rays enter
the eye in a straight line without being refracted. The double images seem to be that
of the object and its upside-down reflection in water. For exhausted travelers in the
desert it seems like that there is a lake of water in front of them.
An inferior mirage occurs when a strong positive gradient of refractive index is present near the ground. We use the following model to describe the variation of the
refractive index of the air with elevation:
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Physics Competitions Vol. 14 No 2 2012
n02 + α z for h ≥ z ≥ 0
,
n ( z) = { 2
n0 + α h
for z > h
2
(2.4)
with n02 − 1  1 and α h  1 .
Figure 8: The geometry for analyzing an inferior mirage
 Find the path of a light ray (i.e. z as a function of x) that enters in an observer's
eye at an angle θ (see Figure 8). The height of the observer eye is H.
[1.9
points]
 Due to the inferior mirage, the observer can see an inverted image of the upper
part of a camel at a large distance. The parameters for the refractive index of the
air are =
α 3.0 ×10−5 m −1 , h = 0.50 m .
The height of the observers eye is
H = 1.5 m , and the height of the camel is l = 2.2 m . Find the minimum values for
the distance Dm between the observer and the camel so that the observer still
can see the inverted image of the upper part of the camel. You may use the approximation n02 ≈ 1 .
[0.7 points]
 As a result of the light refraction in the region with refractive index gradient, the
observer cannot see the lower part of the camels legs. Find the height of the lowest point ( lm ) on the camel at the distance Dm that can still be seen by the observer.
[0.7 points]
 Find the distance d between the observer and the imaginary lake of water.
[0.3 points]
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Physics Competitions Vol. 14 No 2 2012
The superior mirage
When a superior mirage occurs, light rays that were originally directed above the line
of sight of the observer will reach the observer's eyes. Thus, an object ordinarily below the horizon will be apparently above the horizon.
A superior mirage occurs when a negative gradient of refractive index is present over
a body of water or over large sheets of ice. We use the following model to describe
the refractive index of the atmosphere (see Figure 9):
n 2 [1 − β (r − r0 ) ] for b ≥ (r − r0 ) ≥ 0
n 2 (r ) = { 0 2
,
n0 [1 − β b ]
for (r − r0 ) > b
(2.5)
where r0 is the radius of the earth, b  r0 and β b  1 .
Figure 9: The geometry for analyzing a superior mirage}
 Find the path of a light ray (i.e. r as a function of φ ) within the range
b ≥ (r − r0 ) ≥ 0 . Use γ , the angle between the light ray and the vertical direction at
the sea level as the parameter of the path.
[1.4 points]
Hint: The trajectory of a mass point with mass m, angular momentum L and total
energy E in a conservative force field with
mA
Ep =
−
+ E0 (2.6)
r
Where
ρ=
r=
is
L2
,
m2 A
and
ρ
,
1 − ε cos φ
=
ε
1+
We also have
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2( E − E0 ) L2
.
m 3 A2
(2.7)
(2.8)
Physics Competitions Vol. 14 No 2 2012
1 1 r − r0
≈ − 2
r r0
r0
if
| r − r0 | r0 .
(2.9)
 Find the minimum value of β ( β m ) at which the superior mirage occurs. Use the
following values for n0 and r0 : n0 ≈ 1 , =
r0 6.4 ×106 m .
[0.8 points]
 Under a certain atmospheric condition, with b = 100 m and =
β 6.0 ×10−7 m −1 , calculate the largest distance DM at which the surface of the sea can be seen by an
observer at an altitude y = 10 m (y is the altitude of the observer's eye).
[1.4 points]
1
Useful formula: cos φ ≈ 1 − φ 2 for φ  1 .
2
 For comparison, calculate the largest distance D′ M at which the surface of the
sea can be seen by an observer at the same altitude y = 10 m , when the refractive index of the air is constant.
[0.3 points]
 Calculate the angular difference ∆ϑ between the apparent horizon when a superior mirage occurs as and the apparent horizon in a normal day, seen at the
[0.5 points]
same altitude y = 10 m .
— 3. Experimental Examination
Experimental Problem No.1: Granular Material
The vast majority of mechanics experiments normally conducted in high school physics laboratories are those that have to do with solid objects such as spheres, springs,
carts, and rods. Moreover, these solid objects are additionally assumed to be ideal
rigid objects such as the ones usually dealt with in any classroom discussion involving dynamics. This approach unfortunately omits the exploration of another class of
solid objects that are rather ubiquitous in real life, granular materials. These are materials consisting of many small grains or particles; while each grain or particle is a
solid object, the collective behavior of any group of those grains or particles is usually
nothing like their more rigid counterparts. The following experiments depart from the
common high school mechanics experiments in that they are designed to explore
many aspects of the behaviors of granular materials.
The experiments consist of four independent parts. In the first part, we will investigate
the static and dynamic properties of a granular material (dry sand) by measuring the
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Physics Competitions Vol. 14 No 2 2012
angle of repose and the maximum angle of stability. This experiment accordingly
gives us the information about coefficient of static friction of the sand grains. The
mass flow rate of the dry sand through an orifice is investigated in the second experiment. In the third part, we will observe how the sand settles in a fluid. Having obtained the settling velocity of the sand sediment, we estimate the average size of a
sand particle. Lastly, we model an asteroid impact crater using craters formed by
balls dropped into dry, non-cohesive, granular media.
The following physical quantities are given:
Gravitational acceleration at surface of Earth:
Average density of Earth's continental crust:
Average radius of Earth:
Density of iron at room temperature:
The explosive power of 1000 tons of TNT:
Viscosity of water at room temperature:
g = 9.81 m/s2
ρE = 2.7 g/cm3
RE = 6400 km
ρiron = 7.874 g/cm3
1 kiloton = 4.184x109 J
η = 8.94x10-4 Pa.s
Table 1: List of available apparatus and materials for experiment
Name
Quantity
Name
Quantity
Inclined plane with acrylic box +
stand
1
Graduated Cylinder
1
Sand in the container with pink cover
1
Colored (red/blue) sand
1
Containers with different orifice size
6
A glass of water
1
Empty container without orifice
2
Ruler
1
Vernier caliper
1
Flashlight
1
Digital weight scale
1
Graph paper
1
Digital stopwatch
1
Tripod
1
Metal stand with release mechanism
1
Glass ball
1
Pile of sand in plastic basin
1
Spoon
1
ATTENTION:
Do not mix different types of sands (sand in the container with pink cover, colored
sand, and pile of sand in plastic basin are different.)
Use the correct sand for each experiment.
Angle of Repose
The hourglass is a fascinating and antique timekeeping device. Its distinctive geometrical shape makes it immediately recognizable, and, as some people claim, it has a
certain elegance which modern timekeeping devices lack. Nevertheless, the geometric shape of the common hourglass must exist for a reason, since one does not find
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Physics Competitions Vol. 14 No 2 2012
cubic hourglasses. The reason, as it turns out, is that granular materials have a certain parameter called the angle of repose, θ r which is a measure of the final angle
formed by the surface of a pile of granular material when it is poured onto a horizontal surface. The angle of repose θ r is measured with respect to horizontal surface.
In this problem, we will investigate the angle of repose of dry sand in a simple experiment.
Initially, a flat rectangular container is filled with sand and the top surface of the sand
is flattened out horizontally. The expanse of sand can be stable when its container is
tilted as long as its slope is less than the angle of maximal stability, θ m . As the slope
is increased above θ m , some of the sand grains begin to flow and an avalanche occurs.
Note: Use Sand in the container with pink cover
Figure 10: Configuration to determine maximum angle of stability.
 Determine the angle of maximal stability, θ m .
[1.0 points]
The angle you have just calculated, however, is not the angle of repose. If the sand is
too compact it can affect your measurement result.
 You may perform another experiment to determine the angle of repose of the
sand, θ r . Draw a schematic diagram of your experiment.
[1.0 points]
 Determine the intergranular friction coefficient, µ s , which is the coefficient of static friction between a sand grain and other sand grains.
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[0.5 points]
Physics Competitions Vol. 14 No 2 2012
Flow Rate of Granular Matter
The mass flow rate of granular materials through an orifice due to gravitational force
is generally independent on the geometry of the container. This statement holds as
long as the dimensions of the container are sufficiently large compared to the size of
the grains and the diameter of the outlet orifice. Note that it was this fact which,
though perhaps unknown to the people of that time, enabled hourglasses to be relatively good timekeeping devices. In this case we will determine parameters that can
affect the flow rate of sand through a circular orifice.
Note: Use Sand in the container with pink cover
Let us suppose that the mass flow rate of sand, W, through a circular orifice depends
on the density of sand ρ , the acceleration due to gravity g, the orifice diameter D,
and the intergranular friction coefficient µ s . Based on this assumption, the flow rate
fulfills the equation W = C ( µ ) ρ g D athr ,
(2.1)
where C ( µ ) is a dimensionless empirical factor which depends on the internal friction
coefficient of the sand and athr is a constant.
 Find the theoretical value of athr .
[0.5 points]
 Measure the flow rate of sand using various orifice diameters and plot the dependence of lnW with respect to ln D . Using the same assumption used in deriving eq. (2.1) that W ∝ D
aexp
, determine the value of aexp from the experiment.
[1.2 points]
The most widely accepted law that predicts the flow rate of grains through an orifice
was proposed by Beverloo, which is modified from eq. (2.1) and has the form
=
W C ρ g ( D − kd ) athr ,
(2.2)
where d is the grain diameter and k is a dimensionless fitting parameter. According
to the Beverloo law, the grains in fact do not flow through the whole orifice, but
through an effective exit aperture whose diameter is given by ( D − kd ) .
 Based on this law, determine the value of kd using linear regression from the data you obtained.
[0.8 points]
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Settling Rate of Sand
It is obvious that sand is a porous medium with different shapes and sizes of grains.
In order to estimate the size of a sand grain, we can assume that the shape of the
grain is spherical; however, the size of any individual grain might differ from that of
the other grains. Despite this fact, we can still estimate the average size of sand particles. In this experiment, you will observe the settling rate of sand particles in water
and hence estimate the average radius of a sand grain.
Note: Use colored sand for this section. Plan your experiment very carefully before
you begin your experiment, because you will not be able to recover the sand once it
gets wet.
Determine the following quantities:
 Density of sand in air, mass of sand grains divided by the total volume (including
the volume of air inside the pores),
Average density of sand grain, mass of sand grains divided by the total volume of
sand grains.
Density of sand in water, mass of sand grains divided by the total volume (including the volume of water inside the pores)
[1.0 points]
Pour all the remaining colored sand into the graduated cylinder and fill it up with water. Close the cylinder using the plastic bag and the rubber band of the coloured
sand, shake it well, and stand it in a vertical position. The sand will settle on the bottom with certain settling rate, which is the rate of the sediment height rises.
 Measure how fast the sand settles at various heights of the sediment and determine the settling rate of the sand within the region where the sand size is nearly
homogenous.
[1.0 points]
It is recommended to take measurements over the entire range of sediment volume
in order to obtain the region where the sand size distribution is nearly uniform/homogeneous.
 Derive an equation relating the sand settling rate and the density/ies of sand you
calculated before, taking note of the gravitational, Archimedean, and viscous
forces. Give your estimation of the average size of the sand particles. [0.5 points]
The sand particle is assumed to be a spherical and any turbulent effect is neglected.
You may also assume that a good shake causes the sand to be distributed uniformly.
The formula for the viscous force is Fviscous = 6π rη vterminal ,
where r is the radius of the grain and η is the viscosity of water.
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(2.3)
Physics Competitions Vol. 14 No 2 2012
Asteroid Impacts on Earths Surface
When an asteroid impacts the surface of the Earth, it creates a large crater with circular cross-section. The exact relation between the crater's dimensions, the physical
properties of the impacting body, and the physical properties of the general area of
impact is complicated and beyond the scope of this experiment. Nevertheless, there
exists an empirical formula to estimate the diameter of an impact crater on the surface of the Earth based on experimental nuclear explosions at Yucca Flat, Nevada.
From the experimental nuclear crater, the following empirical formula is obtained
1/σ
 ρ 
D = c f Kn W a 
 ρe 
(2.4)
,
where D is the diameter of the crater, c f is the so-called crater collapse factor
(whose value is 1 for craters  3 km in diameter and 1.3 for craters  4 km ),
K n = 0.074 km kilotons −1/σ is an empirical constant, W is the kinetic energy of the as-
teroid just prior to impact, ρ a is the mass density of the asteroid, ρe is the mean density of the target rocks at the point of impact and σ is a constant exponent. The diameter of the crater is defined by the peak of the "sand hill", not the outer most visible boundary. The empirical formula above can be verified by modeling the collision
using sand and glass balls.
Note: Use the pile of sand in plastic basin
 Using the equipment provided, devise an experiment to determine the numerical
value of the constant exponent σ . After you make your measurement, you may
need to loosen and flatten the surface of the sand again. If the sand is too compact it can affect your measurement result.
[2.0 points]
 Actually the σ of the typical asteroid is 25% bigger than the σ that we get from
previous question. Estimate the diameter of a crater that would be caused by an
iron meteorite of mass 107 kg drifting into Earth's path from very far away.
[0.5 points]
The students got detailed information about the use of the stopwatch in an appendix.
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Physics Competitions Vol. 14 No 2 2012
Experimental Problem No. 2:
A Rotary Magnetic Drag System for
Conductivity Measurement
Attention:
Use the answer sheet provided to summarize your answer.
Provide diagrams to help explaining your answer or your experiments, especially if
your answer is not in English.
Apparatus
Rotary magnet assembly (x 1 pc)
Copper plate (x 1 pc)
"Black box" containing unknown metal X (x
1pc)
Thin cardboard spacers (5 pcs)
Thick cardboard spacers (2 pcs)
Power supply unit (x 1pc)
Digital multimeter (DMM) – generic (x2 pcs).
Digital multimeter (DMM) with frequency
measurement (x1 pc). See Appendix A.
Ruler (x 1 pc)
Figure 11: Experimental setup
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Physics Competitions Vol. 14 No 2 2012
Important Experimental Data:
Copper plate thickness:
t = 0.6 mm
Copper plate conductivity:
σ = 6.0 x 107 (Ohm-m)-1
Metal X thickness:
tX = 1.05 mm
— Introduction
Electrical conductivity measurement of a material is important for many applications such
as metallurgy and semiconductor industries.
Usually one measures conductivity (σ)
(1/resistivity) by simple resistance measurement that requires making electrical contacts.
This is troublesome and sometimes impossible due to the presence of insulating layer.
Thus a contactless conductivity measurement technique is desired. In this problem we
will explore a simple and fascinating system to perform contactless conductivity measurement utilizing magnetic drag or braking effect that occurs between fast moving magnets and a metal sheet.
If a magnet moves with velocity v parallel to the plane of a non-magnetic, conducting material with conductivity σ and thickness t, it will experience a magnetic braking effect also
known as eddy current braking effect as shown below:
Figure 12: The magnetic braking effect of a moving magnet near a metal sheet
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Physics Competitions Vol. 14 No 2 2012
The magnetic braking force (using a magnetic dipole model and a thin metal sheet approximation) is given as: FMB = −α σ t
v
,
dm
(2.1)
where α is the magnetic braking coefficient of the system which depends on magnetic
moment of the magnet and magnetic permeability of the metal sheet, σ and t are the
conductivity and thickness of the metal, v is the velocity of the moving magnet, d is the
distance between the center of the magnet and the metal and m is the distance power
law factor to be determined in this experiment. The negative sign indicates a force that
opposes the velocity thus it is called "magnetic braking".
In this experiment we mount two strong magnetic pucks on a rotating disc driven by motor as shown below. When the disc rotates and the metal plate is inserted underneath,
the disc will slow down due to the magnetic braking effect. This effect can be exploited to
measure the conductivity (or thickness) of a metal sheet.
Figure 13: Connection diagram of the rotary magnet setup
System Information:
The motor is driven by a variable voltage power supply with coarse and fine control.
DMM (Digital Multimeter) #1 measures the motor voltage VM. DMM#2 measures the motor current IM as a voltage across 1 Ω resistor.
The Hall sensor serves as speed sensor. It provides a voltage pulse each time a magnet
passes by. When the disc rotates, the Hall sensor frequency fH can be measured using
DMM#3.
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Physics Competitions Vol. 14 No 2 2012
To demonstrate the magnetic braking effect, a metal plate is placed at a distance d from
the rotating disc. Note that d is measured from the center of the magnets (see Figure 13).
Thick and thin card board spacers are provided to vary the distance d.
There are two metals provided: a copper plate (with a known conductivity σ and thickness
t) and an unknown metal X inside a "black box" (with a known thickness tX). See Section
1: “Important Experimental Data”
Attention: The disc can rotate up to fH = 200 Hz. In general, useful data can be obtained approximately < 120 Hz. Do not run the motor at maximum frequency (around 200
Hz) for too long.
To save battery power, don't run the motor unnecessarily. Turn off the power if not in use.
— Experiment and Questions
This experiment is divided into four small sections:
A. Introduction: Speed Sensor and Basic Operation
B. Basis System Characterization
C. Magnetic Braking Effect Characterization
D. Conductivity Measurement of an Unknown Metal X
[1.5 points]
[2.0 points]
[4.5 points]
[2.0 points]
--
A: Introduction: Speed Sensor and Basic Operation
[1.5 points]
This section will help to familiarize yourself with the experimental setup. First we will try
to understand the operation of the Hall effect sensor that measures the rotation frequency. We will observe the voltage signal coming from the Hall sensor. Set the DMM #3 to
DC Voltage mode. Turn the rotating disc (mounted on the motor) manually so that the
magnet passes through the Hall sensor.
 Sketch the voltage waveform for two full rotations, mark how the voltage changes
with or without magnet near the sensor. Indicate the period of the waveform.
[0.25 points]
Now switch DMM#3 to Frequency mode (Hz) to measure the Hall sensor frequency fH.
No metal plate underneath. Turn on the motor Power Supply and increase the voltage
gradually to speed up the rotation. Operate the motor < 150 Hz. Observe how the frequency reading fH increases with speed.
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Physics Competitions Vol. 14 No 2 2012
 Express the disc angular frequency ω (in rad/sec) in terms of Hall sensor frequency
fH!
[0.25 points]
 Now insert the copper plate. Notice how the rotation slows down due to the magnetic
braking effect. How does the magnetic braking effect work? To help answer this
question, consider only interaction between the moving magnet and an "elemental
ring" from the metal sheet as shown below. Draw all the necessary electromagnet
fields involved in this diagram in the answer sheet.
[1.0 points]
B. Basic System Characterization
[2 points]
The motor has an internal series resistance RM which is not negligible in this experiment.
RM is the sum of resistance of the rotor coil inside the motor. Therefore when the motor is
driven by a voltage source, not all of the power is converted to kinetic or rotational energy. A fraction of the energy is turned into heat due to RM. Thus the real motor can be
modeled as an ideal motor (whose coil has no resistance) plus a series resistance RM as
shown below.
Figure 14. The equivalent circuit of the motor
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Physics Competitions Vol. 14 No 2 2012
 Determine the internal series resistance of the motor RM to at least two significant
figures!
Note: The series resistance is very small. Direct resistance measurement using ohm
meter mode of the DMMs (Digital Multimeters) does not have enough accuracy. Try
other method. You don't have to change the experimental setup.
[0.7 points]
Now we will explore how the system behaves at various ranges of rotation speeds with
no metal plate inserted. This is important to pick the operating range for further experiment. Under normal condition the motor should run smoothly, however at some frequencies the system vibrates strongly and become noisy and we want to avoid this. Do not
insert the copper plate.
 Increase the motor voltage gradually from 0 to near 200 Hz, record the Hall sensor
frequency fH and the motor current IM (DMM#2). Plot both data with respect to voltage
VM. Mark the range of frequencies where the system vibrates strongly and become
noisy. Note that there is one dominant noisy region.
[0.8 points]
 What may cause this strong vibration/noise at certain frequencies? What anomaly
do you observe in the motor current (IM) vs. voltage (VM) plot in this noisy region?
[0.5 points]
C. Magnetic Braking Effect Characterization
[4.5 points]
Now we will study the characteristics of the magnetic braking effect with the copper plate
provided. Here in section C and D, you can operate at low frequency (approximately fH <
120 Hz) and outside the dominant noisy region that you have identified in section B.
 Derive expression for power dissipation PMB due to the magnetic braking force of the
two magnets and the metal plate. Use (2.1) and express PMB in terms of the Hall
sensor frequency fH.
[0.5 points]
 Perform experiments to verify the relationship of PMB and fH from your previous answer!
[1.5 points]
 Perform experiment to determine the magnetic braking force coefficient α (note that
α is associated with one magnet) and distance power law factor m!
[2.5 points]
Note: To maintain the validity of the distance power law in Eq. (2.1) (due to magnetic
dipole approximation) don't place the metal too close. Maintain sufficient distance,
approximately d >8 mm.
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Physics Competitions Vol. 14 No 2 2012
D. Conductivity Measurement of an Unknown Metal X
[2.0 points]
To obtain conductivity (assuming thickness is known), the distance between the magnet
and the metal plate d needs to be determined accurately because the magnetic braking
force falls off strongly with distance due to large power factor m. In actual industrial application, it is difficult or even impossible to obtain an accurate distance d without making
a contact to the metal (remember this is a contactless method). However it is possible to
perform the conductivity measurement without knowing the exact distance between the
magnet and the metal plate d. This problem simulates this situation. We have an unknown metal X inside a "black box" as shown below.
tX = 1.05 mm.
The thickness is known:
However its exact location from the surface (∆d0) is not known. In fact, you do not need
to know it to obtain the conductivity σ.
Figure 15: The cross section of the "black box" containing the unknown metal X.
 Perform an experiment to determine the conductivity of the metal X!
Note:
[2.0 points]
Please don't open the "black box". It will invalidate your answer.
The students got detailed information about the use of the multimeter in an appendix.
— 4. Result
WoPhO is an individual competition, that is why we will present only students’ results.
There will be no compilation of results by countries. The best overall participant is Kexin
Yi from China with a score of 35.1 out of a total maximum of 50.0. The best result for a
theoretical examination is also achieved by Kexin Yi with a score of 22.75 out of a total
maximum of 30.0. The best experimental result is achieved by Eugen Hruska from Germany with a score of 13.5 out of a total maximum of 20.0. The average score for each
problem is very low due to many guest participants from the host country. We will present
here only the average score of the medalists. The overall average score of medalists is
17.32, the theoretical average score is 9.59 and the experimental average score is 7.73.
The lowest average for a theoretical problem is for problem no. 3, “Mirage”, with an average of 1.5, followed by problem no. 1, “Motion of a Rolling Rod”, with an average of 2.44
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Physics Competitions Vol. 14 No 2 2012
and the highest average is for problem no. 2, “Why Maglev Trains Levitate”, with an average of 5.66. For the experimental problems, the average of problem no. 2 is 4.28, followed by problem no. 1 with an average of 3.46.
In the following, we present the result of the 10 best students for their overall examination, both the theoretical and the experimental examination.
Table 1: The best 10 participants overall
1
Yi, Kexin
China
35.10
2
Ants Remm
Estonia
27.63
3
Eugen Hruska
Germany
27.48
4
Attila Szabo
Hungary
25.35
5
Danila Parinov
Russia
23.80
5
Christian George Emor
Indonesia
23.78
7
Lin Sen
Singapore
22.63
8
Jan Pulmann
Slovakia
22.45
9
Li Kewei
Singapore
20.75
Indonesia
20.15
10 Evan Laksono
Table 2: The best 10 participants for the theoretical examination
1
Yi, Kexin
China
22.75
2
Ants Remm
Estonia
18.68
3
Attila Szabo
Hungary
14.40
4
Eugen Hruska
Germany
13.98
5
Danila Parinov
Russia
13.90
5
Christian George Emor Indonesia
13.63
7
Evan Laksono
Indonesia
12.90
8
Jan Pulmann
Slovakia
12.50
9
Ivan Ivashkovskiy
Russia
11.98
Indonesia
11.83
10 Samuel Wirajaya
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Physics Competitions Vol. 14 No 2 2012
Table 3: The best 10 participants for the experimental examination
1
Eugen Hruska
Germany
13.50
2
Yi, Kexin
China
12.35
3
Lin Sen
Singapore
11.40
4
Li Kewei
Singapore
11.15
5
Attila Szabo
Hungary
10.95
5
Christian George Emor Indonesia
10.15
7
Jan Pulmann
Slovakia
9.95
8
Danila Parinov
Russia
9.90
9
Wojciech Tarnowski
Poland
9.60
Bulgaria
9.55
10 Momchil Molnar
The authors would like to make two observations.
First, WoPhO has been promoted as an opportunity for any student to challenge the winners of APhO and IPhO. However, all the gold medalists at the first WoPhO were still
gold medalists of either the competition year APhO or IPhO, with only 2 exceptions. One
of the two students, however, was a former gold medalist from the previous year IPhO
and another one will be the absolute winner of the next IPhO.
Second, the average score even for the medalists of the first final round of WoPhO was
quite low. The problems were difficult and both the academic committee and the students acknowledged this fact. However, WoPhO is also a competition for the winners of
the previous Olympiads and the level of difficulty of the problems in that regard can be
justified.
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