Solutions of 10 Online Physics Brawl

Online Physics Brawl 10th year November 25, 2020
Solutions of 10 th Online Physics Brawl
1
Problem FoL.1 . . . (hopefully just) easy pulleys 3 points

Suppose that we have a pulley and some weights hanging from a cord as
shown in the figure. Assume that both the cord and the pulley are ideal
and massless, M = 2.0 kg and m = 1.0 kg. Find the acceleration of the
weight M (in the downwards direction). Lego wants
to discover the simplest pulley problem that half of all teams still fail.
The cord as well as the pulley are massless (assumption from the problem
statement), therefore the forces that the cord exerts on both weights are
equal. Let us denote each by T . To describe this, we can write a system m
of two equations
M
a1 M = M g − T ,
a2 m = mg − T ,
where both accelerations are oriented downwards.

We subtract the equations and set a1 = −a2 , because the cord is ideal, which means that
it does not change its length. We get
M −m .
a1 = g = 3.3 m·s−2 .
M +m
Šimon Pajger
legolas@fykos.cz
Problem FoL.2 . . . two circuits 3 points

Legolas found several old identical resistors and connected them in series. Such a circuit had
a total resistance RS = 10.0 kΩ. After that, he decided to connect the resistors in parallel and
measured a resistance RP = 1.0 Ω. What was the resistance of a single resistor?
Lego made himself a lego out of resistors.
Let R denote the desired resistance of one resistor and n the total number of the resistors. Then
we get
RS = nR ,
R
RP = .
n
2
By multiplying these two equations, we
√ get rid of n and the remaining equation is R = RS RP ,
from which we obtain the result R = RS RP = 100 Ω. We need n = 100 such resistors.
Šimon Pajger
legolas@fykos.cz
2
Problem FoL.3 . . . felling trees 3 points

A coniferous tree could be (from a mechanical point of view) approximated by a homogeneous
right circular cone with height h = 40 m and radius at the base r = 1.0 m. Find the maximal
angle by which its axis may be displaced from the vertical axis before it starts to fall due to its
weight. Dodo was procrastinating on Youtube.
A rigid body begins to fall when it is displaced in such a way that its centre of mass is no
longer straight above its base. The centre of mass of a cone is located at height h/4 above the
base. The maximal angle at which the tree does not fall yet is the same as the angle between
the vertical and the line connecting the centre of mass and the edge of the base. Its magnitude
.
r
is calculated using the formula tan Φ = h/4 , therefore Φ = arctan 4r
h
= 5.7◦ . If the tree is tilted
more, it will fall.
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.4 . . . the nearest asteroid 3 points

On 16th August 2020, an asteroid (later called 2020 QG) has been recorded as the closest
asteroid (spotted so far) that flew by the Earth without colliding with it. At the nearest point
of approach, it was only 2 950 km above Earth’s surface and it had a velocity v = 12.3 km·s−1 .
How much higher was its velocity compared to the escape velocity at that height above Earth’s
surface? Find the ratio v/vesc . Karel made a problem out of news from the website astro.cz.
The escape velocity is given by the condition that total energy of the object in the gravitational
field shall equal zero. Therefore
1 Mm
mv 2 − G = 0,
2 r
where M denotes the mass of the Earth, G is gravitational constant and v and r is the velocity
of the object and its distance from the centre of Earth, respectively. By expressing the velocity
we get √ √
2GM 2GM
vesc = = ,
r R+h
where r was substituted with the sum of radius of Earth and height of the flyby (measured
from the ground). If we divide these velocities, we get
√
v R+h .
=v = 1.33 .
vesc 2GM
Jozef Lipták
liptak.j@fykos.cz
3
Problem FoL.5 . . . unstable 3 points

Suppose that we have a cuboid with dimensions a = 20 cm, b = 30 cm, c = 50 cm and density ρ =
= 620 kg·m−3 . One of its faces with dimensions a and c is lying on a horizontal surface in
a homogeneous gravitational field. How stable is it with respect to rotation around one of
its bottom edges with length c? Find the smallest amount of energy needed to turn it over.
Assume g = 9.81 m·s−2 . Danka’s stuff was falling down.
To overturn the cuboid, we only have to get its center of gravity above the c edge and just barely
behind it. This change of the cuboid’s position requires an increase in the cuboid’s potential
energy. This can be counted as the difference in the potential energy of the cuboid’s center of
gravity. Let’s place the zero energy level on the horizontal plane. In the beginning the cuboid
has a potential energy of
b
Ep 1 = ρabcg .
2
When its center of gravity is above the c edge, it will be in h above the plane, where
1√ 2
h= a + b2 .
2
Then it will have a potential energy of
Ep 2 = ρabcgh .
The stability is the difference of these two energies, therefore
(√ )
1 .
∆E = Ep 2 − Ep 1 = ∆E = ρabcg a2 + b2 − b = 5.5 J .
2
The stability of the cuboid is therefore 5.5 J.
Daniela Pittnerová
daniela@fykos.cz
Problem FoL.6 . . . safe power 3 points

There are two appliances with resistances R1 = 500 Ω and R2 = Imax
= 2 000 Ω respectively, connected in parallel to the terminals of
a battery through a fuse with a maximum allowed current Imax =
U R1 R2
= 500 mA. What is the maximum power we could get from the
circuit? Dodo must pay attention at the dormitory.
The relation between the voltage and the current is described by the Ohm’s law U = RI,
where R is the total resistance of the connected appliances. The total resistance of the appliances
connected in parallel is
R1 R2
R= = 400 Ω .
R1 + R2
We can calculate the power of an appliance from the voltage on them and the current that
flows through them as P = U I = RI 2 . Maximum power implies maximum current, which is
the current that blows the fuse. Plugging in the numerical values, we get P = 100 W.
Jozef Lipták
liptak.j@fykos.cz
4
Problem FoL.7 . . . the ink-blooded prince 3 points

When Harry Potter did not behave, professor Umbridge would punish him by forcing him to
write with his own blood. He needed 1.21 μl of blood on average to write one lowercase letter.
A book of school rules has 259 standard pages, each of which contains 1 488 letters on average.
However, each thirty-seventh letter is a capital letter, which means it consumes three times
more ink than a lowercase letter. Assuming that Harry has 5 l of blood, how many copies of the
rules could be made from him? Jáchym found the whole book series considerably illogical.
For each 37 letters of the school rules the blood consumption equals writing 39 lower case
letters. Let k = 39/37 denote that ratio. Let Va , np , nl and VH denote the remaining quantities
respectively. The volume of the blood required to write on set of school rules is
Vb = np nl kVa .
Therefore the number of copies equals
VH .
nc = = 10.2 .
np nl kVa
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.8 . . . slowed down train 4 points

While arriving at a station, a train decelerates evenly. Its braking distance is s = 75 m and
during the penultimate (second to last) second before stopping, it drives a distance l = 2.25 m.
What is its initial velocity v0 before it begins to brake?
Verča took advantage of a train delay to think of new problems.
The train moves with evenly decelerated motion and during the next to last second it drives
the same distance as during the second second of a train accelerating from rest with the same
magnitude of acceleration. To make this solution more illustrative, let us mark the beginning
and the end of the next to last second as t1 and t2 respectively. The distance l could be written
as
1 1 1 ( )
l = at22 − at21 = a t22 − t21 .
2 2 2
We can express acceleration a from this equation as
2l
a= 2 .
t2 − t21
We will plug it into the formula for total distance
1 2 1 v02
s= at = ,
2 2 a
from which we can already express the initial velocity as
√
√ 4ls
v0 = 2as = .
t22 − t21
Numerical evaluation for values given in the task results in v0 = 15 m · s−1 .
Veronika Hendrychová
vercah@fykos.cz
5
Problem FoL.9 . . . we pump oil 4 points

Find the efficiency of a pump, if the electric current flowing through it has an effective value I =
= 125 mA when attached to a standard power supply of voltage U = 230 V. Our pump pumps
oil with density ρ = 870 kg·m−3 through pressure difference ∆p = 120 hPa with volumetric flow
rate Q = 0.83 ml·s−1 . Dodo wanted to get a rotary vane pump.
The efficiency η is defined as the ratio between the useful work done (in our case mechanical
work against pressure force) and energy supplied (by electric power). The electric network
supplies the pump with energy
W1 = U It .
in time t. The pump uses it to push the liquid through the given pressure difference, where the
useful work is given by the product of the pressure force Fp and the distance s, along which
the force is exerted
W2 = Fp s = S∆ps = ∆pV ,
where S is the cross-section of the pump. The distance s is, in fact, the length of the part of the
liquid which flowed through the pump in time t, so V = Ss is the volume that flowed through.
Plugging it in the equation we obtain
∆pV ∆pQ .
η= = = 0.000 35 ,
U It UI
where we used the formula for the flow rate V = Qt. The second part of the problem is solvable
by dimensional analysis as well.
In this task, we made a numerical error. We decided that the fairest resolution is to not
count any points gained or lost because of this problem. We apologize for the troubles.
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.10 . . . homogeneous air 4 points

From the point of view of classical optics, every substance is homogeneous. However, today we
know that everything consists of particles. Determine the number of particles (molecules) of air
that are, under standard conditions, contained in a cube, where the length of one edge of this
cube corresponds to the wavelength of the yellow D-line of sodium.
Dodo couldn’t sleep again.
The wavelength of the D-line is approximately λ = 590 nm. Let us first determine the mass of
the air contained in our cube
m = ρV = ρλ3 ,
where ρ = 1.29 kg·m−3 is the density of the air. We will determine the number of particles
using the average molar mass of air Mm = 29.0 g·mol−1 and the Avogadro constant NA =
= 6.022 · 1023 mol−1 as
mNA ρλ3 NA .
N= = = 5 500 000 .
Mm Mm
6
Therefore, in the cube where the edge corresponds to the wavelength of the sodium D-line there
are millions of molecules.
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.11 . . . our old clock 4 points

Our old clock is battery-powered and the capacity of the battery is E = 1.2 Wh. The hour
hand is rh = 5 cm long and the minute hand is rm = 8 cm long. The linear density of each hand
is τ = 10 g·m−1 . The efficiency of the clockwork is η = 7 %. How long does it take for the clock
to stop? The hands rotate continuously and any potential energy released during descending is
lost. Michal was late due to a delayed clock.
When the watch hands rotate continuously the clockwork does not have to accelerate them.
However, it needs to compensate the gravitational force during their movement up. Energy
which the clockwork has to spend for one cycle of its hand can be calculated as the difference
of the potential energy between the highest and the lowest point. For the minute hand we have
( )
rm −rm
∆Em = mg∆h = rm τ g − 2
= rm τg .
2 2
and for the hour hand
∆Eh = rh2 τ g .
The task does not provide information about the starting position of watch hands so we assume
consistent power as follows
∆Em ∆Eh .
P = Pm + Ph = + = 1.7 ·10−7 W + 5.7 ·10−9 W = 1.8 ·10−7 W .
3 600 s 12 · 3 600 s
The battery could power the watch for
Eη .
t= = 47 ·104 h .
P
Now we can see that it really does not matter that we assumed stable power output of
watch, because the period of 12 h is much smaller than the expected battery lifetime. Even
with such a low efficiency the watch could work for over 50 years.
Šimon Pajger
legolas@fykos.cz
Problem FoL.12 . . . cough cough 4 points

A deadly virus with a long incubation period has spread around the world and already infected
one ten-thousandth of the total population. Hard-working researchers haven’t found any cure
yet, but managed to develop a test which can decide whether a given person is infectious or not.
It returns the positive result in 99.99 % of infected cases. However, it returns the false positive
result in 0.03 % healthy cases. It may seem that the test is quite reliable. We have chosen and
tested a random person. The test returned a positive value. Estimate the probability that the
chosen person is infected. Matěj felt sick during statistical physics practice.
7
Let pP = 0.000 1 denote the probability that a random person is infected and pN = 0.999 9
the probability that he is healthy. Furthermore let pPP = 0.999 9 denote the probability that
an infectious person will have positive test and pNP = 0.000 3 that a healthy person will have
positive test.
The desired probability can be calculated as the ratio between the number of all cases in
which is infectious person correctly tested (this equals the product pP pPP )to the number of all
people with positive test, which leads to the formula
pP pPP 1
= .
pP pPP + pN pNP 4
On the other hand, even though the test seemed to be quite reliable, we obtained the result, that
a person with a positive test is infectious only with 25 % probability. Do not give up hope. . .
Matěj Mezera
m.mezera@fykos.cz
Problem FoL.13 . . . uneven illumination 4 points

Danka was sitting at a round table with a radius R = 1.0 m and noticed that the edge of the
table was illuminated much less than the center. What is the difference between illuminance in
the center and at the edge of the table if the only light source in the room is a light bulb with
luminous intensity I = 120 cd hanging h = 1.5 m above the center of the table? The light bulb
is an isotropic source of light and the ceiling is black. Danka couldn’t see her books well.
Illuminance E is a photometric quantity defined as the luminous flux incident on a unit
area. If we have a point source of light with luminous intensity I at the distance r away from
a surface, we can calculate the illuminance as
I
E= cos α ,
r2
where α is the angle between the normal of the surface and the direction of the incident light
rays. In the centre of the table, the rays arrive perpendicularly, that is cos α = 1. At the edge
of the table however, the rays arrive at an angle α and
h
cos α = √ ,.
h2 + R2
The distance between the edge of the table and the light source is
√
l= R 2 + h2 .
The difference of illuminance between the centre and the edge of the table is therefore
I I h
δE = − 2 √ ,
h2 R + h2 h2 + R 2
( )
1 h .
δE = I − 3 = 22.6 lx .
h2 (R2 + h2 ) 2
The difference of illuminance between the centre and the edge of the table is 22.6 lx.
Daniela Pittnerová
daniela@fykos.cz
8
Problem FoL.14 . . . autumn in a train 4 points

A train is climbing an icy track uphill. The steepest section of track which the train is still able
to climb has a slope angle α = 1.75◦ . When the train gets over the slope, it reaches a station
where tracks are horizontal and still icy. What is the shortest distance at which the train can
stop from a speed v = 52 km·h−1 ? Dodo was waiting for a train.
During the ascent, it is necessary to have a condition on static friction Ff ≤ f Fn . If only
the gravity force FG affects the train, we use its components in the equations Ff = FG sin α
and Fn = FG cos α. This implies a relation for the slope and the coefficient of friction tan α ≤ f .
During the deceleration a, the maximum friction force allowed before the wheels start slip-
ping is given by condition Fn = FG and Ff = ma. Using the condition from above we get a ≤ f g
where g is gravitational acceleration. The shortest distance at which the train stops is when a =
= f g = g tan α and the distance is
1 2 v2 v2 .
s= at = = = 348 m .
2 2a 2g tan α
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.15 . . . the power of a waterfall 4 points

−1
A h = 30 m high waterfall has flow rate Q = 1.2 m ·s . Find the total force with which water
3
impacts the ground under the waterfall. Assume that the water quickly flows away from the
point of impact and the depth of water under the waterfall is negligible.
Dodo is reminiscing about National park Plitvička jezera.
To solve this problem, we will use Newton’s second law in the formulation with momentum
dp
F = ,
dt
which states, that the force is defined as as the time derivative of momentum. In time dt, the
river bed decelerates falling water with mass dm = ρ dV = ρQ dt from the impact velocity v
to zero. We will denote the impact velocity by comparing kinetic and potential energy in
a homogeneous gravitational field
1 √
mgh = mv 2 , v= 2gh ,
2
where g is gravitational acceleration. Plugging in, we obtain the force
dp v dm vρQ dt √ .
F = = = = ρQ 2hg = 29 kN .
dt dt dt
Jozef Lipták
liptak.j@fykos.cz
9
Problem FoL.16 . . . burning coal 4 points

We burn m = 213 mg of pure carbon in a closed vessel of volume V = 20 l filled with air. Once
the temperature equilibrium between the vessel and its surroundings is restored, we measure
the pressure in the vessel. Find the ratio of the pressure after burning to the pressure before
burning. Dodo wanted to be malicious.
A chemical reaction occurring while burning carbon in an environment with enough oxygen is
described by equation
C + O2 −−→ CO2 .
Since carbon is solid, one mole of gas transforms into one mole of other gas. If we assume that
the gas is ideal, the ideal gas law
pV = nRT ,
must hold before the reaction as well as after it once the equilibrium is reached again. Consider
that the volume V has not changed, nor temperature T (in balance with the surrounding) nor
the amount of substance n. Therefore the pressure hasn’t changed either. The solution to the
task above is p/p0 = 1. We should make sure there is enough oxygen in the vessel. In the
case of room conditions, a mole of gas has a volume of approximate 24 l. Therefore there are
approximately 0.2 mol of oxygen molecules. The amount of our carbon is n = m/Mm ≈ 0.02 mol,
which means there is enough oxygen.
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.17 . . . heat and phases 4 points

How much higher is the amount of heat required to let ice boil away (Qb ) compared to the
amount of heat necessary to only melt it (Qm )? Find k = Qb /Qm . Our ice is taken from
a fridge with inner temperature t = −18 ◦C. The specific heat capacity of liquid water is c =
= 4 180 J·kg−1 ·K−1 and that of ice is c0 = 2 090 J·kg−1 ·K−1 . The enthalpy of fusion of ice
is l1 = 334 kJ·kg−1 and the enthalpy of vaporization of water is l2 = 2.26 MJ·kg−1 . We are
interested in the points when the ice fully melts to water (at 0 ◦C) and when the water fully
vaporizes (at 100 ◦C). Karel was wondering about the possibility of scalding above a kettle.
Let us divide the process into four parts. Q1 is the heat necessary to warm the ice up to 0 ◦C (the
temperature difference is ∆t1 = 18 ◦C). To calculate it we will use the equation Q1 = mc0 ∆t1 ,
where m is the mass of the ice and c0 is the specific heat capacity of the ice.
Q2 is the heat necessary to change ice to water at 0 ◦C. We will calculate it using the
equation Q2 = ml1 , where we use the enthalpy of fusion (after all, it’s about melting).
Q3 is the heat necessary to warm the water from 0 ◦C up to 100 ◦C (the temperature differ-
ence is ∆t2 = 100 ◦C). To calculate it, we will use analogous equation as for Q1 , i.e. Q3 = mc∆t2 .
However, now we used the specific heat capacity of water and a different temperature difference.
Q4 is the heat necessary to change water to steam at 100 ◦C. The calculation is again
analogous to the calculation of Q2 , but now we use the enthalpy of vaporization (after all, this
is about vaporization). The equation is therefore Q4 = ml2 .
Now, if we look closely at the situation, we will find that Qb = Q1 + Q2 + Q3 + Q4 and
that Qm = Q1 + Q2 . In total, after canceling out m we have
Qb c0 ∆t1 + l1 + c∆t2 + l2
= .
Qm c0 ∆t1 + l1
10
After numerical substitution we will find out that we have to supply 8.21 times more heat to
the ice to boil it away than to melt it.
Kateřina Fatková
katka@fykos.cz
Problem FoL.18 . . . boiling 4 points

Danka put V = 1.5 l of water with a temperature T0 = 25 ◦C in a pot and set it to boil on
a stove with wattage (electric power consumption) P = 1 200 W. The efficiency of heating the
pot with water (i.e. how much heat is transferred to the pot with water) is η = 0.69 and the heat
capacity of the pot is C = 500 J·K−1 . How long does Danka need to wait till the water starts to
boil? Find the necessary constants in tables. Danka had to wait a long time while cooking.
We assume the water to have the following properties: density ρ = 1 000 kg·m−3 , specific heat
capacity cw = 4 180 J·kg−1 ·K−1 and boiling point of Tw = 100 ◦C.The heat needed to be
absorbed by the pot and the water is
Q = V ρcw (Tw − T0 ) + C (Tw − T0 ) .
The stove transmits this heat in time t and therefore
Q = ηP t .
Thus
ηP t = (Tw − T0 ) (V ρcw + C) .
Final formula for the time is
(Tw − T0 ) (V ρcw + C) .
t= = 613 s ≈ 10 min .
ηP
Danka will wait approximately 10 minutes for the water to start boiling.
Daniela Pittnerová
daniela@fykos.cz
Problem FoL.19 . . . internal resistance 4 points

Consider two identical resistors connected in series to a non-ideal voltage source. In this case,
the efficiency of the voltage source is 0.87. Find the efficiency of the voltage source, i.e. the
ratio between the power consumed by our resistors and the total power supplied by the source,
when the two resistors are connected in parallel instead. We assume model the non-ideal voltage
source like an ideal voltage source (not a current source) with an internal resistor connected in
series. Matěj wanted to save some money – so he tried saving electricity.
The efficiency of a power supply source can be calculated as the ratio of the total power to the
power on attached resistors. To calculate the power on a resistor, we will use the formula P =
= RI 2 , where I is the current flowing through a resistor with resistance R. Let Ri denote the
internal resistance of the power supply source and R the total resistance of two resistors in
series. Then, for the efficiency of the connection in series, the equation
RI 2 R
η1 = = ,
(R + Ri )I 2 R + Ri
11
holds. We used the fact that the current is the same on all devices connected in series. We
denote Ri from the equation as
1 − η1
Ri = R .
η1
Parallel connection of two resistors can be substitued by one resistor with half resistance. There-
fore, two identical resistors connected in parallel have four times smaller resistance compared
to serial connection. For the efficiency of the parallel connection we obtain
R 2 R
I R R η1 .
η2 = ( R 4
) = 4
= = = = 0.626 ,
4
+ Ri I2 R
4
+ Ri R + 4Ri 1−η1
R + 4R η1 4 − 3η1
where we used the formula for Ri .

Matěj Mezera
m.mezera@fykos.cz
Problem FoL.20 . . . raft voyage 4 points

A raft consists of twelve cylinder-shaped wooden logs. Each log is l = 8.00 m long and has
a radius r = 12.0 cm. When loaded with m = 70 kg freight, it floats in such a way that the logs
stick s = 3.0 cm above the water level. What is the density of the wood?
Jarda was wondering about possible improvements for a trip on water.
The force of gravity, caused by the raft and its freight, must be balanced by the force of
buoyancy, which, according to Archimedes’ principle, equals the force of gravity of the water
displaced by the immersed part of the body
Fbu = Vi ρ0 g ,
where ρ0 is the density of water, g is the gravitational acceleration and Vi is the volume of
the immersed part of the raft. For the force of gravity, using the formula for the volume of
a cylinder, we have
FG = mg + V ρg = mg + 12πr2 lρg ,
where ρ is the density of the wood. The remaining task is to estimate the immersed volume of
the raft. From the geometrical point of view, these are twelve bodies, each of which was made
by cutting off a cylinder perpendicularly to its base. The base of each body is a circular segment
with the surface S. The formula for the surface can be found in the literature as
r−h √
S = r2 arccos − (r − h) 2hr − h2 ,
r
where h = 2r − s. After plugging it into the formula for the volume and enumeration we obtain
.
the volume of the immersed part Vi = 12Sl = 4.03 m3 .
If we compare the forces mentioned above we can express the density of the wood as
ρ0 Vi − m .
ρ= = 910 kg·m−3 .
12πr2 l
Therefore it is wood with quite high density, but still, we can float on it with a sufficient margin.
.
For the conditions given, a single log with density ρ = 730 kg·m−3 would be enough to carry
a person.
Jozef Lipták
liptak.j@fykos.cz
12
Problem FoL.21 . . . mysterious ball 4 points

We have a ball with radius r = 2.0 cm, which has an unknown mass and is made from an un-
known material, hanging from a massless spring. When submerged in water with density ρ =
= 1 000 kg · m−3 , the elongation of the spring drops to 80% of its value before the ball is
submerged. What is the mass of the ball?
Verča was reminiscing about experiments in high school.
The difference in elongation of the string is caused by the force of buoyancy acting on the
submerged ball. The relevant forces can be expressed as
0.8FG = FG − Fb ,
where FG is the force of gravity and Fb the force of buoyancy. Now we need to plug the formulas
for forces in and express mass m as
0.8mg = mg − V ρg ,
20 3
m= πr ρ .
3
.
Using the numerical values given we obtain the mass of the ball as m = 168 g.
vercah@fykos.cz
Problem FoL.22 . . . illuminated 5 points

Danka was sitting in a futuristic train, which was traveling at a constant speed v = 1.0·105 km·s−1 .
The train track was perpendicular to a long straight road and along this road, lamps were placed
with constant spacing l = 5.0 m. Danka crossed the intersection of the track and the road. In
the landscape’s reference frame, all the lamps were turned on at once when the train was at
a distance d0 = 2.0 km from the intersection. One lamp is at the intersection, let’s give it
index 0. What time passes in Danka’s reference frame between the moments when light from
lamp 0 reaches her and when light from a lamp with index 100 reaches her? Use the exact value
of speed of light. Danka was traveling by train.
The light from the lamp with index 0 reaches Danka in time t0 . In that moment she is in the
distance d0 + vt0 from the crossing. For time t0
d0
t0 = ,
c−v
holds. Let t100 denote the time when light from lamp with index 100 reaches Danka and let r
be the distance between the lamp and the position of Danka at time t100 . It satisfies
√
r= n2 l2 + (d0 + vt100 )2 ,
where n = 100. Then √

r n2 l2 + (d0 + vt100 )2
t100 = = ,
c c
13
holds. This equation rewrites as a quadratic equation for t100 , with solutions
√
vd0 ± c2 d0 2 + (c2 − v 2 ) n2 l2
t100 = .
c2 − v 2
The physically correct one is the positive one. Furthermore, let us calculate the difference ∆t =
= t100 − t0 . We obtain ∆t = 2.06 · 10−7 s. However, this time difference is in the reference
frame connected with the countryside. Since Danka moves in that reference frame, we need to
transform it into the reference frame connected with the train (let us denote it with apostrophe),
in which is Danka stationary and her proper time therefore equals the coordinate time t′ . We
will begin with the transformation formula t′ → t
( )
v ′
t=γ x + t′ ,
c2
where v is the velocity of one reference frame towards the other, γ = √ 1
2
and if Danka has
1− v2
c
′
constant x (coordinate in reference frame connected with the train), for the time difference we
obtain
1 .
∆t′ = ∆t = 1.94 · 10−7 s .
γ
Daniela Pittnerová
daniela@fykos.cz
Problem FoL.23 . . . a ball 5 points

Anička is playing with a ball of yarn with a radius R = 5 cm, which is formed by a l = 100 m
long string. She stands on an inclined plane with a slope angle α = 3◦ . She kicks the ball along
the slope upwards, but one end of the string stays stuck at the point where she kicked it away,
so the string is unraveling as the ball travels up the slope until the whole ball unravels and only
the straight string remains on the slope. Find the smallest initial velocity of the ball needed to
reach the state described above. Matěj read a children’s book.
Let’s start with the law of conservation of energy. Once the ball is fully unraveled (i.e. there
is only a straight string), its centre of mass is exactly in one half of its length, which is at
height 2l sin α. In the beginning, the centre of mass
( of the ball is) at height R cos α. Therefore,
the difference in potential energy is ∆E = M g 2l sin α − R cos α . Let v denote the velocity of
the kick. The initial kinetic energy of translation is 21 M v 2 and the kinetic energy of rotation
v2 2
is 12 J R 2
2 , where J = 5 M R . Therefore, we obtain
1 1 7
∆E = M v2 + M v2 = M v2 ,
2 5 10
( )
l 7
Mg sin α − R cos α = M v2 ,
2 10
√
5g .
v= (l sin α − 2R cos α) = 6.00 m·s−1 .
7
Matěj Mezera
m.mezera@fykos.cz
14
Problem FoL.24 . . . a traveler’s problem 5 points

A traveler needs to catch his train and does not have much time. The train station is s = 1 000 m
far from him. Unfortunately, he finds himself in the fields, where he has to force his way out
with velocity vf = 3.0 km·h−1 . There is a road leading to the station as well, and on this road,
he could jog with an average velocity vr = 7.0 km·h−1 . However, the road is l = 600 m far from
him. Find the optimal angle α (measured with respect to a perpendicular to the road) such
that the traveler reaches the station as soon as possible if he starts walking towards the road
at this angle. Verča went hiking.
We will begin with calculation of the time of travel depending on the angle α and other param-
eters as √
l s2 − l2 − l tan α
t= + .
vf cos α vr
In order to reach minimal time possible we will differentiate the formula with respect to α and
put it equal zero
dt l sin α l 1
= · − · = 0.
dα vf cos2 α vr cos2 α
.
The desired angle expresses as α = arcsin (vf /vr ) = 25.4◦ .
vercah@fykos.cz
Problem FoL.25 . . . a different parallel-plate capacitor 5 points

Consider a capacitor consisting of a square conductive plate with surface area S = 6.0 cm2
and a parallel, infinite and grounded conductive plane at a distance d = 1.1 mm. Find the
capacitance of such a capacitor, defined as the ratio of charge on the plate to potential on the
plate. Originally, Vašek prepared a problem with a capacitor with infinite capacitance.
Both conductive planes create equipotential surfaces. As usual, we choose electrostatic poten-
tial φ such that equals zero on the grounded conductor. Let Q denote the electric charge of the
square plate and φr the electrostatic potential on it. When looking for electrostatic potential
in half-space containing the square plate bounded by the infinite plane (let us denote it right
half-space), the problem reduces to solving Poisson’s equation
ρ
∆φ = − (1)
ε0
in given half-space with boundary condition φ = 0 on the infinite plane, where ρ is the charge
density and ε0 is vacuum permittivity. This kind of problems is often solved using the method
of images. We will consider the infinite plane to be a plane of symmetry. If all charges from
the right half-space were mirrored to the left half-space with the opposite charge, we would
get a new electrostatic problem which solution equals the solution of the original problem in
the right half-space. It should be noted that mirroring and change of signs cause the charge on
the plane to be zero. Overall it causes the charge density ρ to be antisymmetric when mirrored
through the plane of symmetry. Consider that the Poisson equation (1) is linear – therefore
potential φ exists and it is antisymmetric when mirrored and just like the charge density ρ is
zero on the plane of symmetry. These observations form the basis of the method of images.
15
In our particular problem, the method of images gives electrostatic problem with two iden-
tical square plates in the distance 2d, from which one has potential φr and charge Q and the
second one has potential −φr and charge −Q. These plates together form a parallel-plate√ca-
pacitor. Considering that their distance 2d is significantly less than their characteristic size S,
we may approximate their capacitance as
C ′ = ε0 S/2d .
Now we need to realise, that such capacitor has potential difference twice the difference between
plate and the infinite plane and so capacitance, as defined in task, is C = 2C ′ = 4.8 pF.
Václav Mikeska
v.mikeska@fykos.cz
Problem FoL.26 . . . oscillating hoop 5 points

Suppose that we have two identical hoops, each with a radius R =
= 1.0 m. The upper hoop is in a fixed position, while the lower one
is attached to the upper one by several massless cords; each of these
cords has the same length l = 2.0 m and hangs vertically. Let the
mass of the lower hoop be m = 1.0 kg. If we rotate it a bit around
the vertical axis and release, what is its period of small oscillations?
Lego modified a mathematical problem into a physics problem.
As usual, there are plenty of ways how to calculate the period of small
oscillations. Here we will use a less well known but a very effective
method.
The main idea is to express the energy of our system as a sum of kinetic and potential energy.
We will use the angle of displacement from the equilibrium φ as a variable to the relations.
The kinetic energy is simply
1 2 1
Ek = Iω = mR2 φ̇2 .
2 2
The potential energy is slightly more difficult. For sufficiently small φ we can assume
that the cord attachment points have moved from their initial positions√ by Rφ. Applying
the Pythagorean theorem we get the new distance between the hoops l2 − R2 φ2 . If we as-
sume Rφ ≪ l, we can approximate the relation as
√ ( )2 ( ( )2 )
Rφ 1 Rφ R2 φ2
l 1− ≈l 1− =l− .
l 2 l 2l
Therefore, compared to the initial position (zero displacement was at the distance l below the
upper hoop), the lower hoop is lifted by
R 2 φ2
∆h = .
2l
The difference in potential energy is
1 mgR2 2
Ep = mg∆h = φ .
2 l
16
Consider that kinetic and potential energy of simple harmonic oscillator is described by
equations
1
Ek = mẋ2 ,
2
1
Ep = kx2 .
2
√
The period is T = 2π m/k. Now we just need to notice that the formulas, which we have
derived for our system, have the same ”pattern” as these for SHO (the only difference is that
our variable is angle, not position, but that is not an issue). Let us denote the expressions
equivalent to m and k as mef and kef , respectively, and plug them into the equation for the
period of small oscillations. We obtain
√ √ √
mef mR2 l .
T = 2π = 2π mgR2
= 2π = 2.84 s .
kef g
l
We probably could have guessed the result straight away. . .

Šimon Pajger
legolas@fykos.cz
Problem FoL.27 . . . oscillating 5 points

Suppose that we have a homogeneous square plate with edge length 1 m. We drill a tiny hole in
it, hammer a nail into a wall, hang the plate from the nail through this hole and let the plate
oscillate in the vertical plane in which it lies. How far from the centre of the square do we have
to drill the hole to maximise the frequency of small oscillations?
Matěj had some spare metal sheet and didn’t know what to do with it.
Let us use the equation for physical (or compound) pendulum
√
mgl
ω= ,
J
where m is the mass of the plate, l is the distance from the axis to the centre of gravity of
the plate and J is the moment of inertia of a square with respect to the rotation axis. The
moment of inertia can be calculated using the parallel axis theorem, since we know the moment
of inertia of a square with respect to the centre of mass J0 = 16 ma2 , where a is the length of
its side. For the desired moment of inertia we can write
1
J = J0 + ml2 = ma2 + ml2 .
6
We obtain
gl
ω2 = 1 2
6
a + l2
and put the derivative with respect to l equal zero giving
( )
1 2
g a + l2 − 2gl2 = 0 ,
6
17
from which we can calculate the required distance

1 .
l = √ a = 0.408 m .
6
Matěj Mezera
m.mezera@fykos.cz
Problem FoL.28 . . . lego and dice 1.0 4 points

We have two small cuboids, one lying on the other – the lower cuboid has mass M = 2.0 kg,
the upper one m = 1.0 kg. The coefficient of friction between the lower cuboid and the surface
under it is 0, the coefficient of static friction between the cuboids is fs = 0.50 and the coefficient
of kinetic friction between them is fk = 0.20.
Find the magnitude of the force we must exert on the lower cuboid that causes it to move
with a constant acceleration a = 10 m·s−2 .
Lego was helping his friend with physics, so he at least took some ideas for new problems.
.
The maximum lateral force that the lower cuboid can impart on the upper is Fmax = mgfs = 5 N.
.
In this case, the upper cuboid would accelerate with the acceleration amax = Fmax /m = 5 m·s−2 ,
which is less than the acceleration required by the problem task. Therefore, we are interested
in dynamic friction.
The magnitude of the force of friction between those two cuboids is simply Fk = mgfk .
Therefore, if we denote F to be the force that we impart on the lower cuboid, the resulting
acceleration will be equal
F − mgfk
a= .
M
.
Now we only need to express F = aM + mgfk = 22 N.
Šimon Pajger
legolas@fykos.cz
Problem FoL.29 . . . lego and dice 2.0 5 points

We have two small cuboids, one lying on the other – the lower cuboid has mass M = 2.0 kg,
the upper one m = 1.0 kg. The coefficient of friction between the lower cuboid and the surface
under it is 0, the coefficient of static friction between the cuboids is fs = 0.50 and the coefficient
of kinetic friction between them is fk = 0.20.
Find the magnitude of the force we must exert on the lower cuboid that causes it to move
with a constant acceleration a = 1.0 m·s−2 .
Lego was helping his friend with physics, so he at least took some ideas for new problems.
Consider that the upper cuboid will not accelerate faster than the lower, which implies that the
maximal acceleration of the upper cuboid is a. To achieve that, force am = 1 N is necessary.
.
The maximal force that the lower cuboid can impact the upper with is Fmax = mgfs = 5 N,
which is more than is necessary, therefore the upper cuboid will accelerate with the accelera-
tion a as well.
18
Now we need to subtract the force that the lower cuboid imparts on the upper from the
force we impart on it. Alternatively, it is enough to realise that these cuboids behave as a single
object, which means
F
a= .
M +m
Now we will only need to express the force as F = (M + m) a = 3.0 N.
Šimon Pajger
legolas@fykos.cz
Problem FoL.30 . . . impact accelerator 5 points

We have a large number of point masses in a series in one line. The first point mass has
a mass M0 = 1 kg and every subsequent point mass has a mass equal to 70 % of the previous
one. The first point mass starts to move towards the second one with a kinetic energy E0 = 50 J.
All collisions are perfectly elastic. Which point mass will be the first to have a speed larger
than one percent of the speed of light? Neglect any relativistic effects.
Jarda wanted to improve upon CERN technology.
During perfectly elastic collisions, we can use the law of conservation of momentum
M u = M vM + mvm
and the law of conservation of energy
1 1 1
M u2 = M v M 2
+ mvm 2
,
2 2 2
where M is the mass of the more massive of two points, m = 0.7M is the mass of the less
massive point, u is the speed of the more massive point before the collision, vM is it’s speed
after the collision, and vm is the speed of the lighter point after the collision. From the equations
we get
2M 2
vm = u= u.
M +m 1.7
Apparently, the speed of a point after a collision is simply a multiple of the speed of the previous
point. After n collisions, the speed of the last (fastest) point is
( )n
2
vmax = u0 ,
1.7
where √
2E0
u0 = = 10 m·s−1
M0
is the speed of the first mass point before the first collision. Now we plug in vmax and get n as
( 0.01c )
ln .
n= ( u20 ) = 77.6 .
ln 1.7
For the index of the first point that surpasses one percent of lightspeed, we need to round up
and add one (we get to the nth point after (n − 1) collisions), that is 79.
Jaroslav Herman
jardah@fykos.cz
19
Problem FoL.31 . . . jump on! 5 points

Imagine a sufficiently large cuboid with mass M = 32.5 kg lying on the ground. We throw
a smaller cuboid with mass m = 11.7 kg on it in such a way that right before the impact,
the vertical component of its velocity is almost zero, while the horizontal component is v =
= 19.2 m·s−1 . The coefficient of friction between the larger cuboid and the ground is f1 = 0.13,
the coefficient of friction between the two cuboids is f2 = 0.69 . What is the total distance
covered by the larger cuboid? Jachym wondered wherever is possible to land with a plane.
The friction force f2 mg will affect the smaller cuboid until its velocity reaches the velocity of
the lower cuboid (let’s denote it v1 and the time when this happens t1 ). From the magnitude
of this force it follows that the acceleration is a2 = −f2 g.
The smaller cuboid affects the bigger one with the same friction force. The bigger cuboid is
also affected by the friction force between it and the ground, which is f1 (m + M ) g. This force
decelerates it, while the friction force between it and the smaller cuboid accelerates it. In total,
it’s acceleration is
f2 mg − f1 (m + M ) g
a1 = ,
M
therefore its velocity at t1 will be v1 = a1 t1 . For the small block, on the other hand, we
have v1 = v + a2 t1 ; from this equation we can express time as
v Mv
t1 = = .
a1 − a2 (f2 − f1 ) (m + M ) g
During this time, the bigger cuboid will travel the distance of
1
x1 = a1 t21 .
2
During the rest of the trajectory, the cuboids move as one body, so their acceleration is a3 =
= −f1 g. Deceleration from v1 to zero will take them
v1 a1 t1
t3 = − =− .
a3 a3
During this time they travel the distance of
1 1 a2 t2
x3 = v1 t3 + a3 t23 = − a3 t23 = − 1 1 .
2 2 2a3
Therefore, overall covered distance will be
( )
1 a2 t2 1 a1
x = x1 + x3 = a1 t21 − 1 1 = a1 t21 1 − =
2 2a3 2 a3
mv 2
f2 m − f1 (m + M ) .
= = 3.60 m .
2g (m + M ) f1 (f2 − f1 ) (m + M )
Jáchym Bártík
tuaki@fykos.cz
20
Problem FoL.32 . . . disgusting piston 5 points

2
In a thermally insulated cylinder with an internal cross section S = 500 cm and a height l =
= 50 cm, there is a resistor with resistance R = 120 Ω. The cylinder is otherwise filled with air
at a temperature T0 = 20 ◦C and pressure p0 = 101 kPa, and the same kind of air surrounds
the cylinder. A current I = 200 mA flows through the resistor. A base of the cylinder breaks
away when pushed with a force exceeding F = 500 N. After what time does that happen?
Jarda wanted to break something using air.
For the base to break, the pressure difference must reach
F
∆p = .
S
At the beginning, both outside and inside the piston, there is the pressure p0 . The pressure
inside the cylinder must then rise to p0 + ∆p, that is 1 + ∆p/p0 times. Since the process will
be isochronic, this is also the ratio of the increase in temperature.
The resistor heats the air with the power P = I 2 R. All of the energy will be converted to
the internal energy of the gas, giving
Q = ∆U ,
2
I Rt = mcV ∆T ,
where m ist the mass of the gas which we are heating up, cV is it’s specific heat capacity
at constant volume which, for air, is approximately 0.72 kJ·kg−1 ·K−1 , and ∆T is the differ-
ence in temperatures at the beginning and the end of the process. We already know that the
temperature must increase by the factor 1 + ∆p/p0 , that is by ∆T = T0 ∆p/p0 .
Now we only need to find the mass of gas in the cylinder, which is simply m = ρV = ρSl,
where ρ = 1.20 kg·m−3 (the density of air in normal conditions). Finally we have
ρSlcV T0 F ρlcV T0 F .
t= = = 130 s .
Sp0 I 2 R p0 I 2 R
Šimon Pajger
legolas@fykos.cz
Problem FoL.33 . . . 4D scouring pad 5 points

Find the resistance between two neighbouring vertices of a four-dimensional cube made of wire.
Each edge of the cube has a resistance R = 1 000 Ω.
Karel was wondering about multi-dimensional budgeting problems.
A 4D cube can be visualised as two 3D cubes where their relevant vertices are connected. Two
neighbouring vertices in the 4D cube are e.g. two vertices neighbouring in the 4D cube but from
different 3D cubes. Let us discuss, how could the current flow from one vertex (we will denote
it A) into the second one (Z).
The first option, of course, is through the edge connecting them, which has the resistance R.
One could see that this trajectory is parallel to the others.
If the current does not flow from A through the edge connecting two 3D cubes, it must flow
through the edge of the 3D cube to one vertex of this cube which is neighbour to A. There
21
are three such vertices, and they are interchangeable. Therefore, we will use the often-used
trick – imagine, that these vertices are connected perfectly conductively (but still no current
will flow through these connections) and let B denote one of them. From A to B there are three
resistors with the resistance R in parallel, therefore for the resistance the following will hold
(for any n resistors)
1 1 R
=n· ⇒ Rn = .
Rn R n
For n = 3 we obtain that the resistance between A and B equals R/3.
In vertex B the current has two possible paths again – through the edge connecting the
two 3D cubes or “further” in the original cube. In the first case, it always flows to a vertex
neighbouring with Z. We can connect these vertices into one (similarly as we did with B) and
denote it Y . Consider that the resistance between Y and Z, as well as the resistance between B
and Y , equals R/3. In the second case the current has two possible paths (since it can not flow
back to A) in each of the three vertices equivalent to B, which means six edges in total. But
it will always flow into one of three vertices of the first cube, which is in the distance of two
edges from A. As before, we will connect these three vertices into a single one (let us denote
it C) and the resistance between B and C is R/6.
There are two possibilities in the vertex C. The current may either flow into the second 3D
cube – through resistance R/3 and end in the vertex X, where the resistance between X and Y
equals R/6; or it can continue in the first 3D cube – through resistance R/3 and it will end in
the vertex which is opposite to A – let us denote it D.
From the vertex D the current can flow only into the vertex of the second 3D cube, which
is opposite to Z. Let us denote it W . Then the resistance between D and W is R and be-
tween W and X again R/3. Therefore we can calculate the resistance between C and X as the
′
“direct” (RCX ) path and the path through D and W (RCDW X ) in parallel
(R )
R′ RCDW X
R R
3 3
+R+ 3 5
RCX = ′ CX = = R.
RCX + RCDW X 3R
3
+R 18
After that we can similarly calculate the resistance between B and Y as

(R )
R′ RBCXY
R 5
3 6
+ 18
R+ R
6 11
RBY = ′ BY = R R 5
= R.
RBY + RBCXY 3
+ 6
+ 18 R+ R6
51
Finally we will obtain the resistance between A and Z as

( )
R′ RABY Z R R3
+ 11
51
R+ R
3 15 .
RAZ = ′ AZ = = R = 469 Ω .
RAZ + RABY Z R
R + 3 + 51 R + R
11
3
32
Šimon Pajger
legolas@fykos.cz
22
Problem FoL.34 . . . levitation by firearm 5 points

Consider a body with mass M = 12.1 kg, which is kept in the air by sustained fire from AK-47.
Calculate the rate of fire necessary for the body to hover and constantly oscillate between two
points low above the ground. All the impacts are elastic, the mass of each bullet is m = 7.93 g
and the bullets are flying straight up with speed v = 715 m·s−1 .
Don’t forget to take cover and move only when the enemy is reloading.
Any drop of velocity of the bullets associated with gravity can be neglected due to the low
heights involved. We can therefore place the lowest point of the body’s oscillatory motion to
a height 0. At this height the body is moving downwards with a velocity v0 . In a perfectly
elastic collision, both momentum and energy are conserved. If the velocity of the body after
a collision is v0′ (this time moving upwards), the laws of conservation can be written as
mv − M v0 = −mv ′ + M v0′ ,
1 1 1 1
mv 2 + M v02 = mv ′2 + M v0′2 ,
2 2 2 2
where v ′ is the downwards velocity of the bullet after the collision. However, we can notice
that the mechanical energy is conserved so the velocity v0 has the same magnitude as v0′ .
Substituting v0 = v0′ , the second equation becomes v 2 = v ′2 . The solution v = −v ′ does not
make any physical sense (it would be as if the collision never happened), so we must have v = v ′ .
From the first equation, we obtain
m
v0 = v.
M
Let the fall of the body down from the highest point take a time t, then v0 = gt. The time it
takes the body to move up is the same as the time of the fall so we are dealing with a period
2v0 2mv
T = 2t = = .
g Mg
Now we can just calculate the frequency (the rate of fire)
1 Mg .
f= = = 10.5 s−1 .
T 2mv
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.35 . . . burning phosphorus 5 points

Assume that we burn m = 213 mg of phosphorus dust in a closed vessel with volume V = 20 l
filled with air. Once the temperature equilibrium between the vessel and its surroundings is
restored, we measure the pressure in the vessel. Find the ratio of the pressure after burning to
the pressure before burning the phosphorus. Dodo wanted to be malicious, v. 2.
We start from the equation
4P + 5O2 −−→ 2P2 O5 ,
23
i.e. five moles of gas and four moles of phosphorus change into a solid product. The amount of
substance of oxygen used in the reaction is therefore
5 5m .
nO2 = nP = = 0.008 59 mol ,
4 4Mm
where Mm = 31.0 g·mol−1 is the molar mass of the phosphorus atoms. We will use the ideal
gas law in the form of
pV
= const ,
Tn
where the amount of substance of the gas and the pressure changes during the examined process.
.
At the beginning there was n1 = V /Vmol = 0.892 9 mol in the vessel. We obtain the ratio of the
pressures as
p2 n2 nO2 5mVmol .
= =1− =1− = 0.990 4 .
p1 n1 n1 4Mm V
Jozef Lipták
liptak.j@fykos.cz
Problem FoL.36 . . . mixed current 4 points

Consider an electronic component which behaves like a resistor with resistance R = 42 Ω. We
let electric current flow through it; this current has an alternating part and a direct part. The
alternating part is harmonic with a frequency f = 50.0 Hz. The minimum value of the current
is 12 mA and the maximum value is 42 mA. The current does not change direction. What is the
average power consumed by the component? Karel wanted to combine.
Let us begin with the discussion about what is the current like. The problem statement reads,
that it has both direct and harmonic components, which can be written as I = IDC +IAC sin (2πf t).
In general, the formula should contain the phase as well, but since we are interested in the av-
erage power, it does not matter at all. Furthermore the problem statement mentiones that the
current does not change direction, which means IDC > IAC . We can write the following system
of equations
IDC + IAC = Imax = 42 mA ,

IDC − IAC = Imin = 12 mA ,
from which we can express

Imax + Imin
IDC = = 27 mA ,
2
Imax − Imin
IAC = = 15 mA .
2
We now know what current flows through the appliance. The remaining task is to denote the
power being dissipated at time t. Since the power is the product of the current and the voltage,
we get
( )
P (t) = I(t)U (t) = RI(t)2 = R IDC
2 2
+ 2IDC IAC sin (2πf t) + IAC sin2 (2πf t) .
24
The power changes periodically with the period T = 1/f , therefore if we want to calculate the
average power, it is enough to integrate it through one period and divide it by the length of
one period
∫
1
T
(2 2
)
P = R IDC + 2IDC IAC sin (2πf t) + IAC sin2 (2πf t) dt =
T 0
[ ( )]1/f
1 t sin (4πf t)
2
= f R tIDC − 2
IDC IAC cos (2πf t) + IAC − =
πf 2 8πf
( 2
) 0
2 IAC .
= R IDC + = 35 mW .
2
The evaluation of the integral could be simplified using the fact that the integral of a sine
through the whole period is zero and the integral of the square of a sine through an interval
equal to a multiple of the half-period is one half of the interval length.
Šimon Pajger
legolas@fykos.cz
Problem FoL.37 . . . stopping a train 5 points

−1
There is a train with mass m = 500 t and speed v = 100 km·h heading towards a superhero
standing on the tracks. The train is at distance s = 1 300 m from the superhero, who wants the
train to stop just in front of him. However, the only thing available to him is a laser pointer.
What power must his laser have to achieve this feat? The front of the locomotive is a mirror.
Jarda thought up the plot for a new film.
Let’s first calculate what the deceleration of the train must be if it stops just shy of the
superhero. The velocity of the train is much smaller than the speed of light, so we can consider
this to be movement under a constant (negative) acceleration and write
1 2 1
s= at = vt ,
2 2
i.e. the superhero needs to stop the train in t = 2s/v. This means that the train needs to
decelerate at a rate of
v v2
a= = .
t 2s
The force that needs to act on the train is
v2
F = ma = m .
2s
Now for the laser. The momentum of a single photon can be expressed as p = h/λ, where h
is Planck’s constant and λ is the wavelength of the photon. The energy of the photon is E =
= ch/λ = cp, where c is the speed of light (in fact, the momentum is defined as E/c, which is a
direct consequence of the formula for relativistic energy and the fact that a photon has no rest
mass).
25
We therefore have a simple relationship between the momentum of photons traveling in the
laser and the energy needed for the train to stop. It remains to find the momentum of photons
that need to leave the laser per unit time. We can use the well known formula
∆p
F = ,
∆t
but here we need to take care. The problem statement says that there is a mirror at the
front of the locomotive, i.e. the photons are not absorbed, but rather reflected. The change of
momentum the photons undergo upon impact with the train is double the magnitude of their
momentum when traveling. Symbolically,
v2 ∆plaser
m =2 ,
2s ∆t
where we only substituted for the necessary force.
At the end, we only need to realize that power is energy per unit time
∆Elaser ∆plaser v2 .
P = =c = cm = 22.2 TW
∆t ∆t 4s
We can note that even though lasers in use today can reach powers up to several petawatts,
but they are only emitting pulses of light. For continuous lasers, the highest power used tends
to be around tens of kilowatts.
Šimon Pajger
legolas@fykos.cz
Problem FoL.38 . . . solar cube 5 points

Imagine that we have a satellite in the shape of a perfect cube with its surface covered by solar
panels. We position it in such a way that it orbits a star with a circular trajectory much larger
than the star. The satellite is able to rotate in any way we want. Find the ratio of the maximum
to the minimum radiant power absorbed by the satellite. Assume that all incident radiation is
absorbed. Karel saw a solar cube on the Internet.
At a long distance from the star, light rays are be parallel, i.e. the wavefronts are parallel planes
perpendicular to the rays. The power absorbed by the satellite is directly proportional to the
cross-sectional area of the satellite (from the point of view of the rays). Therefore, the solution
of the problem is the ratio between the areas of maximal and minimal projections of a cube
onto a plane.
Let z be a unit vector in the direction of the radiation. Then the area of the projection
satisfies ∫
S= |z · dS| ,
ω
where ω is the part of the surface of the cube which is exposed to the radiation and dS = n dS,
where n denotes the normal vector of an infinitesimal surface dS. We want to calculate that
integral for all possible positions of the cube with respect to z, so we can choose the highest
and lowest value.
26
Consider that no more than three faces of the cube can be exposed simultaneously. Also,
all exposed faces must have one vertex in common. Let us denote these faces by A, B and C
and their normal vectors by a, b and c respectively.
Furthermore, the dot product of z and a given normal (mentioned above) is zero for all
faces which are not exposed to the radiation. We may claim that there are always exactly three
faces exposed, but some of them have normals perpendicular to the rays and therefore do not
add anything to the total surface. The set ω from the integral therefore represents the union of
faces A, B and C. We obtain
∫ ∫ ∫
S= |z · a| dS + |z · b| dS + |z · c| dS .
A B C
The expression z · n is constant on each face. For a cube with unit edges, we get
S = |z · a| + |z · b| + |z · c| .
The vector of the radiation may be divided into three components – each in the direction of
one coordinate vector – let us denote them by za , zb and zc respectively. The expression for
the surface area simplifies to
S = |za | |a| + |zb | |b| + |zc | |c| = |za | + |zb | + |zc | ,
where we used the fact that normal vectors have unit length. The vector of radiation has unit
length as well and assuming orthogonality of za , zb and zc (which holds because of orthogonality
of a, b and c), we get
1 = |za |2 + |zb |2 + |zc |2 .
Notice that the equation resembles the geometric mean of magnitudes of vectors z. Similarly,
the formula for calculation of the surface area can be understood as their arithmetic mean. It
is possible to obtain the maximal estimate of the surface area from the inequality between the
geometric and arithmetic mean, the minimum can be (with luck) estimated as well.
However, if we do not know the inequality yet, we must continue our work. Let x = |za |
and y = |zb |. Plugging them both into the formula for area, we obtain
√
S =x+y+ 1 − x2 − y 2 .
⟨ √ ⟩
Let us examine extrema inside intervals of permissible values x ∈ ⟨0, 1⟩, y ∈ 0, 1 − x2 . The
partial derivatives of the function S(x, y), which are
∂S x ∂S y
=1− √ , =1− √ ,
∂x 1 − x2 − y 2 ∂y 1 − x2 − y 2
must both equal zero. Solving the system of equations results in x = y and then a quadratic
equation
1
3x2 = 1 ⇒ x = √
3
√
giving the only solution. The resulting surface area is S1 = 3.
The second option is to find extrema on the boundaries of the intervals. For y = 0 we have
√
S =x+ 1 − x2 ,
27
√
which is a simple function whose maximum (on the permissible interval for x) is S2 = 2
for x = √12 and whose minimum is S3 = 1 for x ∈ {0, 1}.
√
For the second boundary condition y = 1 − x2 , we receive
√
S =x+ 1 − x2 ,
which was already examined before.
We have examined all possibilities and obtained all possible candidates for extrema.√Since S1 > S2
we can claim that the maximum of the function S(x, y) on the given interval is 3, while
the minimum
√ is 1. The desired solution is the ratio of the maximum to the minimum, which
equals 3.
Finally, it should be pointed out that it is possible to find the solution much easier using
intuition – it was enough to imagine a cross-section of the cube when cut by some plane. The
minimal value corresponds to the situation when the wavefronts are parallel with one face – the
maximal value is reached if the cross-section is a hexagon. However, these are only estimates,
not mathematical proofs.
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.39 . . . optimal direction 6 points

We have a rigid body with mass M = 1.0 kg lying on a horizontal surface. The coefficient of
friction between the body and the surface below it is f = 0.4. What is the maximum acceleration
which the body can reach if we exert a force F = 5 N on it? Lego was moving boxes.
Assume that the force vector, together with vector in horizontal direction, form the angle φ.
Therefore, the resulting net force acting on the body will be horizontal (since the vertical
component is compensated by the surface below) and its magnitude will equal
Fr = F cos φ − f (M g − F sin φ) .
Our only freedom is in the choice of angle φ, so we find the derivative of Fv with respect to φ
and put the result equal 0.
−F sin φ + f F cos φ = 0
f = tan φ
.
We obtain φ = arctan f = 0.38. If we plug the angle into Fr and consider that ar = Fr /M , we
. −2
obtain amax = 1.5 m·s .
Šimon Pajger
legolas@fykos.cz
28
Problem FoL.40 . . . watch out, Danka is throwing 6 points

Danka wants to throw a cricket ball on flat ground. Find the launch angle (measured with
respect to the horizontal plane) which she should choose in order to throw the ball as far as
possible. Danka is h = 1.6 m tall and throws with a velocity v = 4.5 m·s−1 .
Throwing a cricket ball has never been Danka’s favourite sport.
Let us choose a coordinate system, where x denotes the horizontal axis and y is the vertical
one. Let φ denote the angle under which Danka throws the ball, measured from the horizontal
axis. The force, which acts on the ball, acts in −y direction only, therefore the horizontal
velocity vx = v cos φ is constant. The horizontal distance satisfies
x = v cos φ t .
The vertical position will change according to the formula for motion with constant acceleration
giving
1
y = h + v sin φ t − gt2 .
2
By elimination of time from these two equations we obtain height y as a function of x
gx2
y = h + x tan φ − .
2v 2 cos2 φ
We are interested in the moment when the ball hits the ground, which means y = 0. Let d denote
the horizontal coordinate of the point of impact. We will use the formula cos−2 φ = tan2 φ + 1,
plug it to the equation and obtain
gd2 gd2 tan2 φ

0 = h + d tan φ − 2
− .
2v 2v 2
This is quadratic equation not only in d, but also in tan φ. Let us write down the variant
where tan φ is the desired unknown variable
gd2 gd2
− tan2 φ 2
+ d tan φ − 2 + h = 0 .
2v 2v
Now a number of tiny and simple physics ideas. For most of possible values of d we will find
two angles that lead to the distance given. If we increase d, the difference between these two
angles approaches zero – for d = dmax there is only one angle which leads to such distance.
In terms of mathematics, the discriminant equals zero. Let us calculate it from the quadratic
equation for tan φ and put it equal 0. This leads to
( )
gd2 gd2
2
dmax + 4 max h − max = 0,
2v 2 2v 2
from which we express dmax

v√
dmax = 2gh + v 2 .
g
Now we continue solving the quadratic equation for tan φ, which has reduced to
v2
tan φ = .
gd
29
Plugging in d = dmax we obtain

v
tan φ = √ = 0.63 .
2gh + v 2
The desired optimal angle is 32◦ . Consider that for h > 0 the optimal angle is always smaller
than 45◦ and with increasing h it approaches zero.
Šimon Pajger Jaroslav Herman
legolas@fykos.cz jardah@fykos.cz
Problem FoL.41 . . . laser at a dormitory 4 points

Jáchym shines a laser from a dormitory window towards the ground. Matěj measures the radia-
tion of the laser on the ground, ∆h = 42.0 m below Jáchym. The laser shines with a frequency f0
but Matěj measures a different frequency f . What is the ratio |f − f0 | /f0 ?
Karel repeatedly heard about a black hole being observed.
A photon with frequency f has energy E = hf , where h is the Planck’s constant. As the photon
moves down in the gravitational field, it loses its potential energy and ∆E = mg∆h, where m
is the “mass” of the photon. While it may seem strange to talk about mass of a photon (which
has zero rest-mass), we know from relativity that E = mc2 and therefore
hf hf0
m= ≈ 2 .
c2 c
A change in energy therefore leads to a change in frequency. The change of potentional energy
is
hg∆h
E − E0 = ∆E ⇒ h (f − f0 ) = mg∆h ≈ f0 .
c2
Finally we obtain the required ratio as
|f − f0 | g∆h .
k= ≈ 2 = 4.58 · 10−15 .
f0 c
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.42 . . . pyramid 6 points

−3
Imagine a wooden (density ρw = 600 kg·m ) symmetrical square-base pyramid floating on the
surface of a calm lake. A part of the pyramid with height h = 6.0 cm is under the water surface.
We impact the pyramid in the vertical direction. Find the frequency of small oscillations of the
pyramid. The density of the water is ρl = 1 000 kg·m−3 . Assume that these oscillations do not
affect the height of water level in the lake. The pyramid has an ideal shape, it is homogeneous
and pointing downwards (i.e. the apex is lower than the base).
Vítek wondered about global warming.
Firstly we will choose the coordinate system such that the origin is in the apex of the pyra-
mid when the pyramid remains at rest. Next, before we make the impact, the system is in
equilibrium, which means that buoyant force equals the weight. If s denotes the surface of
30
a cross-section at the height h and S denotes the surface of the base at the height H (where H
is the height of the whole pyramid) and if we consider that the weight of the pyramid equals M =
= 1/3ρw SH, we get
shρl = SHρw . (2)
If we look at a vertical cross-section along the pyramid altitude, looking the angles between the
vertical altitude and the horizontal base we get
h H s h2
= , = 2,
a A S H
where A and a denote the length of the base at given heights. Using this relation and (2) we
can find the equation
h3 ρ l = ρ w H 3 .
The motion of the pyramid after that small vertical impact could be described using equation
M z̈ = −ρl g dV = −ρl gsz ,
where dV ≈ sz for sufficiently small z is infinitesimal volume difference of the part which is
under water in addition (compared to the system in equilibrium). Plugging in for the mass we
get
1
ρw SH z̈ = −ρl gsz ,
3
3gρl h2
z̈ + z = 0.
ρw H 3
The equation above is the equation of a simple harmonic oscillator, therefore for the frequency
of small oscillations, we can write
3gρl h2 3g
Ω2 = = ,
ρw H 3 h
√
1 3g
f= .
2π h
.
Using numerical values from the problem statement, we obtain the desired frequency f = 3.52 Hz.
Vít Beran Šimon Pajger
vit.beran@fykos.cz legolas@fykos.cz
Problem FoL.43 . . . pole dance 6 points

We have two coaxial cylinders – the smaller one has an outer radius r = 1.0 cm and the bigger
one has an inner radius R = 10 cm. The cylinders are connected by a rod positioned in the
radial direction. On the rod, next to the surface of the smaller cylinder, there is a small bead.
There is one more small ball on the surface of the smaller cylinder. We spin the whole system
as a rigid body around the common axis with an angular velocity ω = 2.0 rad·s−1 and then
release the bead and the ball simultaneously. The bead can only move on the rod; the ball can
move freely. What is the absolute difference between the times when the ball and the bead hit
the inner surface of the larger cylinder? Jarda got a bit flummoxed from this.
31
We will calculate the motion of the ball (the one which is not on the rod) first. If we assume
both cylinders rotate with the angular velocity ω, the ball will move with the velocity rω in
the direction tangential to the smaller cylinder. When we release the ball, it will not change its
velocity and move in a straight line until it hits the larger cylinder. The question is√how long
is the straight line. We obtain its length using the Pythagorean theorem as s1 = R2 − r2 .
Therefore, the time between the release of the ball and the moment when it hits the larger
cylinder is √
R2 − r 2 .
t1 = = 5.0 s .
rω
Let us investigate the bead on the rod now. Since it is on the rod, it moves with the constant
angular velocity ω even after its release from the cylinder. We will be using the reference frame
connected to the rotating cylinders (and most importantly, with the rod). In this reference
frame, the bead will be acted on by the centrifugal force, which is given as Fcen = mω 2 r(t),
where r(t) denotes the position of the bead on the rod in time t. By dividing the formula by m
we get the acceleration of the bead as a function of time t. Consider that the acceleration is
the second derivative of r(t), so we get the differential equation
r̈(t) = ω 2 r(t) .
Similar equations are most easily solved using the so-called characteristic polynomial. This
means that we assume that the solution is exponential function r(t) = r0 eλt . We plug it into
the equation and get
λ2 r0 eλt = ω 2 r0 eλt .
First of all, consider that r0 = 0 would satisfy the equation. This makes sense – if the bead’s
initial position was on the axis, there would not be any force to push it. However, its initial
position is on the smaller cylinder and therefore such a solution is not interesting for us. We
also know, that an exponential is never zero, therefore we can cancel it out. We cancel out r0 as
well. From the remaining equation, we obtain λ = ±ω. If we solve the differential equation of
order n, we get n solutions. The next step is to write the solution as sum of individual solutions.
In our case
r(t) = r1 eωt + r2 e−ωt .
This is general solution of the motion of the bead on the rod rotating with the angular
velocity ω. The remaining task is to investigate how will it move in our case. We know that the
position at time t = 0 is r and the velocity at time t = 0 is zero. We get a system of equations
r = r1 + r2
0 = ωr1 − ωr2 ,
which holds for r1 = r2 = r/2. Plugging this to the general solution, we get
r ( ωt )
r(t) = e + e−ωt = r cosh ωt .
2
If we had not noticed the cosh, it would not have been a mistake, but then we would have to
find the resulting time somehow else – using the substitution x = eωt – we would get a quadratic
equation or using a numerical calculation.
32
The remaining task is to find the time when the bead hits the larger cylinder. Mathematically
said, for which t2 the formula R = r(t2 ) holds? From the formula above we obtain
1 R .
t2 = argcosh = 1.5 s .
ω r
Subtracting the times we get the desired time difference between the impacts of the ball and
.
the bead ∆t = t1 − t2 = 3.5 s.
Šimon Pajger
legolas@fykos.cz
Problem FoL.44 . . . a fall into the unknown 7 points

We have an infinite grounded conductive plane in free space without any gravitational fields.
At a distance d = 1.00 m from the plane, there is a small ball with mass m = 2.00 g and electric
charge with magnitude q = 4.00 μC. We release the ball. How long does it take for the charge
to drop on the plane?
Jarda knew that it would fall eventually, but he wanted to know when exactly.
Firstly, we want to find the force that the plane exerts on the charge. A charge with the opposite
sign and the same magnitude is electrostatically induced on the plane. In the equilibrium state,
the charges do not move inside the plane, therefore no force is exerted on them, the potential
on the plane is constant and since we are free to add a constant to potential, we choose the
potential to be zero on the plane. The net potential is the sum of the potential φp due to
charges of the plane and the potential φq due to the charge q. Therefore
φq + φp = 0 ,
φp = −φq .
The potential on the plane is the same as if there was an opposite charge −q on the opposite
side of the plane at the same distance d. Now we can substitute all induced charges on the
plane by a single charge −q. The force that the charge q exerts on −q is
1 q2 1 q2
F = 2
= .
4πε (d + d) 16πε d2
Because the charges have opposite signs, the charge q is attracted to the plane. This calculation
is based on the method of image charges. We can see that the force is inversely proportional to
the square of the distance from the plane to the charge. When we have utilized the plane as a
mirror, why not utilise motion of planets too? Gravitational force is, similarly to our force F ,
proportional to r−2 . The motion of the charge, therefore, satisfies Kepler’s laws of planetary
motion with changed constants. Kepler’s third law is
a3 MG
= .
T2 4π2
We need to use our constant instead of a constant derived from Newton’s gravitational law, the
physical behaviour of the system remains unchanged otherwise. The equation becomes
a3 q2
2
= .
T 64π3 εm
33
What is this equation good for? The charge’s trajectory will not be an ellipse, but a straight
line perpendicular to the plane. However, if we assume that it is an infinitely thin ellipse, with
its foci almost in the original and final point of the trajectory, then such an ellipse approximates
the straight line segment on which the charge moves. This ellipse has a semi-major axis a with
length d2 . The period equals
√ ( d )3 √
64π3 εm 2 8π3 εmd3
T = = .
q2 q2
The move from the initial point to the plane lasts only half a period, therefore the time, which
is the final solution to this problem, equals
√
T 2π3 εmd3 .
t= = = 0.262 s .
2 q2
Jaroslav Herman
jardah@fykos.cz
Problem FoL.45 . . . spring of knowledge 7 points

Jáchym owns a well of knowledge, which has depth h = 32 m and a constant circular cross-
section with radius r = 1.5 m. At the bottom of the well, in the middle, there is a point
which radiates knowledge uniformly into the whole space around it with power P0 . Knowledge
propagates in space similarly to light rays. The reflectance of the sides of the well is k = 0.42.
The bottom of the well of knowledge does not reflect anything, it absorbs knowledge perfectly.
Let P denote the total knowledge power radiated out of the well. Find the ratio P/P0 .
Wells are inexhaustible sources of ideas.
The problem has radial symmetry. Imagine one of the possible two-dimensional cross-sections –
the well is projected into a rectangle with dimensions 2r × h, while the source of knowledge
is in the middle of the bottom side. Let φ denote the angle between a knowledge ray and the
vertical. Then all rays with angles from φ0 = 0 to
r
φ1 = arctan
h
exit the well without any reflection and therefore with their original intensity (power). In
general, rays with the angle between φi and φi+1 , where
(2i − 1) r
φi = arctan ,
h
reflect i-times in total (let us call them rays in the i-th zone). Their intensity is ki times the
original intensity.
In order to solve the problem, we need to find out what power corresponds to each zone.
Imagine a ball with a radius R and with its center in the source of knowledge. The flux of
knowledge is uniform on its surface and has the magnitude
P0
I0 = ,
4πR2
34
which is the total power divided by the surface area. The power in each zone equals I0 Si ,
where Si is the part of the ball’s surface coresponding to the zone between angles φi and φi+1 .
We can calculate it as
∫ φi+1
Si = 2πR sin φR dφ = 2πR2 [− cos φ]φi+1
φi = 2πR2 (cos φi − cos φi+1 ) .
φi
The total power leaving the well in the i-th zone is

ki
pi = ki I0 Si = P0 (cos φi − cos φi+1 ) ,
2
while the desired ratio of powers is the sum
( )
P ∑ pi
∞
1 (1 − k) ∑ i
∞
η= = = 1− k cos φi .
P0 P0 2 k
i=0 i=1
Using the trigonometric identity

( )− 21
cos arctan x = 1 + x2
we can modify the formula to
 ( 
( )2 )− 12
(1 − k) ∑ i
∞
1 (2i − 1) r
η = 1 − k 1+ .
2 k h
i=1
The sum must be calculated numerically; it is approximately 0.716. The ratio of power radiated
.
out of the well to total radiated power is η = 5.53 · 10−3 .
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.46 . . . an incredible space battle 8 points

Imagine us in the middle of a fight in space between civilization 1 and civilization 2. A battle
cruiser of civilization 1 has just launched a rocket on a battle cruiser of civilization 2. The
cruisers are s = 5.00 km from each other and do not move with respect to each other. The
properties of the rocket are: initial mass m0 = 5.00 t, engine thrust T = 1.50 · 105 N, specific
impulse (exhaust velocity with respect to the rocket) u = 3.00 km·s−1 . The engines are set to
maximum thrust from launch till the moment of impact. Find the velocity of the rocket right
before it crashes into the cruiser of civilization 2. Jindra was watching Star Wars.
The velocity of the rocket is described by the Tsiolkovsky equation
m0
v(t) − v0 = u ln . (3)
m(t)
The initial velocity v0 is zero and the rocket’s mass depends on time according to the for-
mula m(t) = m0 − Rt, where R is the fuel mass flow rate
T
R= .
u
35
−1
Numerically, it is R = 50.0 kg·s . Substituting it into the equation (3), we get
m0
v(t) = u ln . (4)
m0 − Rt
By integrating the equation (4) we obtain the formula for the distance covered at a given time.
The initial condition is s(t = 0) = 0, therefore
m0 m0 u m0
s(t) = ut + ut ln − ln (5)
m0 − Rt R m0 − Rt
After numerically solving the equation (5), we get the time when the rocket reaches the dis-
tance s = 5.00 km. Let’s plug it into the equation (4) and calculate the rocket’s velocity at the
. .
point of collision. We get t = 17.7 s and v = 584 m·s−1 . We can tell that the warring civilizations
are not very advanced, since their rockets move very slowly compared to the speed of light.
Therefore, we could calculate the solution safely without using the special theory of relativity.
Jindřich Jelínek
jjelinek@fykos.cz
Problem FoL.47 . . . rotational pump 8 points

Vítek would like to get some water from his well, but he does not want to keep pulling the
bucket up. Therefore, he gradually stirred the water around faster and faster until, at an angular
velocity ω = 11 rad·s−1 , the water started flowing out of the well all by itself. Vítek knows the
depth of the well (from the top edge to the ground at the bottom) h = 47 m. The well has
a circular cross-section with a radius r0 = 1.6 m. What was the height of the water column
(from the bottom of the well to the water surface) before Vítek started spinning the water
around? Jáchym heard that problems about wells were popular in Fyziklání 2020.
Let’s introduce cylindrical coordinates in which we label the horizontal radial distance from
the axis of the well as r, an angle of rotation with respect to the axis of the well as φ and the
height above the bottom as z.
The forces acting on a small volume of rotating water with mass m and at distance r from
the axis are centrifugal force Fo = mω 2 r and force of gravity Fg = mg, plus some buoyant
hydrostatic forces. The surface is nothing else than a region with a constant potential energy.
The potential energy due to the centrifugal force can be calculated as
∫ r ∫ r
1
Ec (r) = −Fc (x) dx = − mω 2 x dx = − mω 2 r2 .
0 0
2
The minus sign comes from the fact that we are integrating against the force which we would
need to exert to counteract the centrifugal force. The potential energy due to the force of gravity
is ∫ z
Eg (z) = Fg dx = mgz .
0
As we said, the water surface is a surface with constant potential. (There are also no buoyant
forces on the surface.) For every point on the surface with coordinates rs , zs , the following holds
Ec (rs ) + Eg (zs ) = const ,
36
From this condition, we obtain the height of the surface as a function of radial distance
ω2 2
zs (r) = r + z0 ,
2g
where z0 is the height of the water surface in the center of the well. Of course, this only holds
if the surface is always above the bottom of the well. If there wasn’t enough water in the well,
the parabola we obtained could intersect the bottom of the well, but for now, let’s assume that
this is not the case. We can later check whether this assumption holds.
The volume of the water does not change when it is spinning. THe initial volume was
V = πr02 zv , where zv is the original height of the water column. After spinning, we have
∫ r0 ∫ 2π ∫ r0 ∫ r0 ( ) ( )
ω2 3 ω2 4
V = zs (r)r dφ dr = 2πzs (r)r dr = 2π r + z0 r dr = π r0 + z0 r02 .
0 0 0 0
2g 4g
From this, we can express the height of water in the center
ω2 2
z0 = zw − r0 .
4g
We know from the problem statement that at the given angular velocity, the water just started
to flow out of the well. We can express that as zs (r0 ) = h. From this condition, we get
ω2 2 ω2 2 ω2 2
h= r 0 + z0 = r 0 + zw − r0 .
2g 2g 4g
The original height of the water in the well that we are looking for is then
ω2 2 .
zw = h −
r0 = 39 m .
4g
.
At the end, we just check that in this case z0 = 31 m, which means our assumption was correct.
Jáchym Bártík
tuaki@fykos.cz
Problem FoL.48 . . . ideal centrifuge 9 points

Imagine a tube filled with ideal gas with molar mass Mm = 36 g·mol−1 . The length of the tube
is r0 = 1.00 m, its thickness is negligible compared to its length. We spin the tube around the
axis perpendicular to the tube and passing through one of its bases, with an angular velocity ω =
= 451.00 s−1 . The gas inside will settle in equilibrium, at a temperature T = 300 K along with
the tube. Find the distance between the axis of rotation and the centre of mass of the air in
the tube. The tube is rotating in a horizontal plane. Jáchym wanted to separate air.
Let us choose a coordinate describing distance from the axis of rotation up to the end of
the tube, i.e. from 0 to r0 , denoted by r. In some section dr, we have gas with mass dm =
= λ(r) dr. When the equilibrium is reached, the temperature in the tube must be the same as
at the beginning (since the tube is thermally isolated, the process must be adiabatic). From
the equation of state, we express the pressure in a given section
dnRT dnRT dmRT RT
p= = = = λ,
dV S dr Mm S dr Mm S
37
where S is the cross-sectional area of the tube. From this formula, we get
RT ′
dp = λ dr
Mm S
There must be a centripetal force
dFd = dmω 2 r ,
exerted on a given section of the tube. This force must be caused by difference in pressure.
Therefore S dp = dFd holds. Putting it together, we get
RT ′
λ dr = dmω 2 r ,
Mm
which is a differential equation
ω 2 Mm
λ′ = λr = 2k2 λr ,
RT
where k is a wisely-defined constant
√
ω 2 Mm .
k= = 1.212 m−1 .
2RT
The solution of the equation is
2 2
λ = Aek r
,
where A is a constant. We calculate the center of mass
∫ ∫ r0 ∫r 2 2
1
r0
rλ dr A 0 0 rek r dr
Tr = r dm = ∫ r0
0
= ∫ r0 2 2
m 0 λ dr k r A e dr
0 0
and substituting x = kr, we get the result

[ ]kr0
∫ kr0 x2 ex
2
2 2 2 2
1 0 2xe dx 1 1 ek r0 − 1 1 ek r0 − 1 .
Tr = ∫ kr0 2 = ∫ kr0 02 = ∫ kr0 2 = √ = 0.629 m ,
2k ex dx 2k ex dx 2k ex dx k πerfi(kr0 )
0 0 0
where erfi is the imaginary error function.
Statistical solution
It is possible to solve the problem using statistical physics as well, but the ability to work with
statistical ensembles is required.
The one-particle Hamiltonian is
p2 p2 p2 − 2mωp · n3 × x
H1 (X1 ) = −ω·L= − ωn3 · x × p = =
2m 2m 2m
′2
(p − mωn3 × x)2
mω 2
p 2
mω 2
= − (n3 × x)2 = − r ,
2m 2 2m 2
where X1 is microstate of one particle, m is its mass and vectors x = (x1 , x2 , x3 ) and p =
= (p1 , p2 , p3 ) represent its coordinates and momentum. The vector n3 is the unit vector in the
38
direction of the third coordinate axis, which we choose to be identical with the axis of rotation.
We also define a new coordinate p′ = p − mωn3 × x, which is shifted momentum. As we will
see later, the shift has no effect during integration. Compared to the Hamiltonian of a free
particle, there is also a potential term which decreases with increasing distance from the axis.
This potential represents the centrifugal force exerted on the particles in a corotating system.
The variable r is the distance from the axis of rotation and it satisfies r2 = x21 + x22 .
Since this is a system with constant temperature, we describe it using a canonical statistical
ensemble. We divide the tube into small layers with thickness ∆r, perpendicular to the axis of
the tube. We will focus on one layer, which is at a distance r from the axis of rotation. The
one-particle partition function is
∫ (∫ ∞ )3
1 1 βp′2 βmω 2 r 2 V − βmω22 r2
Z1 (r) = e−βH1 (X1 ) dX1 = e− 2m dp′ S∆re− 2 = e ,
h3 h3 −∞
κ3
1
where the integral is over all possible states X1 of one particle, β = kB T
and h is Planck’s
constant. We used an auxiliary constant
√
2πm
κ=h
β
and denoted the volume of a layer by V = S∆r. Since we are working with an ideal gas, the
particles do not interact with each other and therefore we can write the total partition function
as
Z1 (r)N (r)
Z(r) = ,
N (r)!
where N (r) is the number of particles in the given layer. From statistical physics, we know
that the result is connected to Helmholtz free energy F = − β1 ln Z and we also know that the
chemical potential µ(r) can be calculated as the derivative of F with respect to the number of
particles N . We can write
( ) ( )
∂F 1 ∂ ln Z 1 eZ1 eZ1 1 N 1 N κ3 − βmω22 r2
µ(r) = =− =− ln − ≈ ln = ln e ,
∂N β ∂N β N N β eZ1 β eV
where we used the approximation

( )N
N
N! ≈
e
and neglected the term eZ N
1
, because it is zero in the thermodynamic limit. Now, we use the fact
that the equilibrium between individual layers requires µ(r) = µ0 to be constant. This means
that the whole formula inside the logarithm must be constant as well. If we denote the volume
density of particles n(r) = N V
, we obtain
N κ3 − βmω22 r2
e = const ,
eV
βmω 2 r 2
n(r) = n0 e 2 ,
where n0 is the volume density of particles at the axis of rotation.
39
The remaining task is to calculate the position of the center of mass

∫ r0 ∫ r0 k2 r2
rn(r) dr re dr
Tr = ∫ r0
0
= ∫0 r0 k2 r2 ,
0
n(r) dr 0
e dr
√
βmω 2
where k = 2
. We obtain the same result as in the case above.
Jáchym Bártík Matěj Mezera

tuaki@fykos.cz m.mezera@fykos.cz
Problem FoL.49 . . . falling impulse 9 points

There is a thread with a length l = 1.00 m and a total mass mv = 1.00 g hanging from the
ceiling. On its bottom end, a small weight with unknown mass m is hung. We send an impulse
down the thread from the top and simultaneously drop a point mass from the ceiling. We notice
two things:
1. the impulse is traveling along the thread with a constant speed,
2. the impulse reaches the bottom end of the thread at the same time as the point mass.
What is the mass of the little weight? Everything is taking place in homogeneous gravity with
acceleration g = 9.81 m·s−2 . Jirka was unable to figure out how this idea struck his mind.
Let’s use x as a coordinate of distance measured from the top of the thread. The bottom end
of the thread is at x = l. The impulse is traveling along the thread with speed
√
F (x)
v(x) = = const,
λ(x)
where F (x) is the force of tension within the thread at a distance x and λ(x) is the linear
density of the thread at the same point. We know that the speed of the impulse is constant
along the whole length of the thread. We can find the force F (x) by integrating the infinitesimal
contributions from pieces of thread hanging between x and l
( ∫ l )
F (x) = m+ λ(t) dt g .
x
We express the force from the first equation and substitute into the second one
∫ l
v2
λ(x) = m + λ(t) dt = m + L(l) − L(x) , (6)
g x
where L is the indefinite integral of λ. We differentiate this equation by x and obtain the
differential equation
v2 ′
λ = −λ ,
g
solvable by the method of separation of variables. We get
g
− 2x
λ = λ0 e v ,
40
where λ0 is an unknown constant which we need to infer from boundary conditions. We can
see that the linear density of the thread decreases exponentially with length. The total mass of
the thread is
∫ l ( )
− 2t
g λ0 v 2 gl
− 2
mt = λ0 ev dt = 1−e v ,
0
g
mt g
λ0 = ( gl
).
− 2
v2 1−e v
In the equation (6), we set x = 0 and express the mass of the little weight
∫ l
v2
m= λ0 − λ dx .
g 0
We’ve already calculated this integral once, the result is mt . We substitute for λ0 and get
( )
v2 mg 1 mt
m= · ( t ) − mt = mt gl
−1 = gl
.
g v 2 1 − e− vgl2 1−
−
e v2 e v2 −1
Now we can use the information that the impulse reaches the end in the same time t as the
point mass falling from height l. We have
l = vt ,
1
l = gt2 ,
2
but the velocity v is not the final velocity of the point mass, but the speed of the impulse,
i.e. the mean velocity of the point mass during fall. From that follows
gl
v2 = .
2
Putting everything together, we get
mt .
m= = 0.157 g.
e2 − 1
The mass of the little weight is roughly 0.157 g.
Jindřich Jelínek
jjelinek@fykos.cz
Problem FoL.50 . . . a problem with potential 3 points

Lego took two hemispherical shells with the same radii and connected them U1
in a non-conducting way such that together they formed a sphere. Then he
set a potential U1 = 100 V on one hemisphere and a potential U2 = −100 V
on the other hemisphere. What is the potential in the middle of the sphere?
Lego knew that he has a potential. . .
U2
41
The problem can be solved using Laplace’s equation, of course. However,

let’s wait with that for now. . .
Let’s denote the desired potential in the center of the sphere by φ. If we exchange U1 and U2 ,
the signs of all potentials given in the problem change. Therefore, the signs of all potentials in
the solution change as well and φ changes to −φ (we can also look at it as if there were aliens
who define the signs of electric charge in the exactly opposite way; this would be the problem
they would need to solve).
However, if we exchange the potentials on the spherical shell, it is also the same as if we
only turned the sphere over (or if we looked at it from the opposite side). And that cannot
have any impact on the potential in its center. We get the equation φ = −φ, which has only
one solution, 0.
Therefore, the potential in the center of the sphere is φ = 0 V.
Šimon Pajger
legolas@fykos.cz
Problem FoL.51 . . . another problem with potential 9 points

Lego took two hemispherical shells with the same radii and connected them U1
in a non-conducting way such that together they formed a sphere. Then he
set a potential U1 = 100 V on one hemisphere and a potential U2 = −100 V
on the other hemisphere. Find the potential in the middle of a straight line
connecting the centre of the sphere and the apex of the positively charged
hemisphere (depicted in the figure)?
. . . and he also thought that it is greater than 0.
U2
Numerical solution
The electric potential everywhere except for the regions with non-zero charge (therefore every-
where inside of the sphere) satisfies Laplace’s equation
∆Φ = 0 .
For some basic idea, it is enough to realise that the Laplace operator is a sum of second deriva-
tives in all (three) directions. In theory, we could discretize the sphere as a 3D grid. Then, using
differentiation of the Laplace equation, we would obtain that the potential at each point is equal
to the arithmetic mean of potentials at the neighbouring 6 points, and could let the computer
iterate the given formula (with fixed potential on the border) until it converges. However, to
obtain the result with the required precision, the calculation would be very time-consuming
(without an external server, it could even last longer than our competition). Therefore, as
usual in electrostatics problems, we use symmetry. The Laplace operator in spherical coordi-
nates equals
( )
1 ∂2 1 ∂ ∂Φ 1 ∂2Φ
(rΦ) + sin θ + 2
.
r ∂r2 r2 sin θ ∂θ ∂θ r2 sin θ ∂φ2
If we choose the z-axis as identical with the symmetry axis of the problem, the potential
does not depend on the φ coordinate (written as Φ = Φ(r, θ)), i.e. the third term is zero.
Most importantly, we can then discretize the problem using two variables only. Therefore, we
42
have a 2D grid where each point is described with two indices i, j and its potential Φ(r, θ) =
= Φ(ihr , jhθ ) = f (i, j), where we defined a function f for clarity; hr , hθ are constants denoting
the length of one step in the direction of r or θ. If we index our points starting from 0,
then hr = R/imax and hθ = π/2jmax , where R is the radius of the sphere, imax is the largest
index i and jmax is the largest index j, because we know from the solution of the previous
problem that the potential in the plane connecting both hemispheres is everywhere equal to 0.
Then we only need to find the potential inside the hemisphere which is interesting for us,
i.e. where θ ∈ ⟨0, π/2⟩.
In the direction of increasing r, we have discrete points with spacing hr . The derivative
of f r at a point i, j is
f (i + 1, j)r(i + 1) − f (i, j)r(i)
[f (i, j)r(i)]′ = ,
hr
where we used only the definition of a derivative and omitted the detail that h shall be infinitely
small. To be precise, we have written the so-called forward derivative, which somewhat describes
the derivative of f in the region between i and i+1. It is useful to keep this in mind. The second
derivative is the derivative of the derivative of f . Let us think about it for a while. Instead of
using two forward derivatives, which would mean comparing the derivative between i and i + 1
with the one between i + 1 and i + 2 (this does not seem to be the correct way to express
the second derivative in the point i), we use the backward derivative (symbolically described
as f ′ (i) = (f (i) − f (i − 1))/h, which is, of course, only another first derivative) instead. Next,
we plug in r(i) = ihr and calculate
1 ∂2 1 [f (i, j)r(i)]′ − [(f (i − 1, j)r(i − 1)]′
2
(f r) = =
r ∂r r(i) hr
f (i + 1, j)(i + 1) − 2f (i, j) + f (i − 1, j)(i − 1)
= .
ih2r
Similarly (with slightly more difficulty), we differentiate the second term
( )
1 ∂ ∂f
sin θ(j) =
r2 (i) sin θ(j) ∂θ ∂θ(j)
[( ) ] [( ) ]
sin j+ 1
2
hθ (f (i, j + 1) − f (i, j)) − sin j− 1
2
hθ [f (i, j) − f (i, j − 1)]
= .
sin (jhθ ) i2 h2r h2θ
[( ) ]
The expression sin j + 21 hθ may look quite weird, but it results from our desire to get the
value of the second derivative “exactly” between j and j + 1.
The remaining task is to use the fact that the sum of these two terms equals 0 and express
sin (jhθ ) ih2θ [f (i − 1, j) (i − 1) + f (i + 1, j) (i + 1)]
f (i, j) = [( ) ] [( ) ]+
2i2 h2θ sin (jhθ ) + sin j + 12 hθ + sin j − 12 hθ
[( ) ] [( ) ]
f (i, j − 1) sin j− 1
hθ + sin j + 12 hθ f (i, j + 1)
+ 2
[( ) ] [( ) ] .
2i2 h2θ sin (jhθ ) + sin j+ 1
2
hθ + sin 1
j− 2
hθ
This holds for the potential at all points except these with non-zero charge. Such points are
on the hemispheres. However, the potential on the hemispheres is given by the task, so these
are our boundary conditions.
43
We discretize the space into a 2D grid, pass through all its points and calculate the formula
given above for each of them. We repeat this over and over again until f (i, j) converges. The
code in Python is attached.
import numpy as np
Fi=np.zeros((101,101))
Fi2=np.zeros((101,101))
h=np.pi/200
b=True
eps=10**(-3)
F=100
while b:
b=False
for i in range(101):
for j in range(101):
if (i==100):
Fi2[i,j]=F
elif (i==0):
Fi2[i,j]=0
elif (j==100):
Fi2[i,j]=0
elif (j==0):
Fi2[i,j]=(np.sin(j*h)*i*h*h*(Fi[i-1,j]*(i-1)+Fi[i+1,j]*(i+1))
+2*np.sin((j+0.5)*h)*Fi[i,j+1])/(2*i*i*h*h*np.sin(j*h)+2*np.sin((j+0.5)*h))
if abs(Fi2[i,j]-Fi[i,j])>eps:
b=True
else:
Fi2[i,j]=(np.sin(j*h)*i*h*h*(Fi[i-1,j]*(i-1)+Fi[i+1,j]*(i+1))+
Fi[i,j-1]*np.sin((j-0.5)*h)+np.sin((j+0.5)*h)*Fi[i,j+1])/
(2*i*i*h*h*np.sin(j*h)+np.sin((j+0.5)*h)+np.sin((j-0.5)*h))
if abs(Fi2[i,j]-Fi[i,j])>eps:
b=True
for x in range(101):
for y in range(101):
Fi[x,y]=Fi2[x,y]
print(Fi[50,100])
The result after rounding is 66 V.
Analytical solution
Let us use the general solution of Laplace’s equation ∆Φ = 0 in spherical coordinates
∑
∞
∑
l
( )
Φ(r, θ, φ) = Aml rl + Blm r−l−1 Ylm (θ, φ) ,
l=0 m=−l
44
where Ylm (θ, φ) are spherical harmonics and Aml and Bml are constants determined from the
initial conditions. We will choose the coordinates in such a way that the ray θ = 0 intersects the
apex of one hemisphere. In such a case, the coordinate φ characterizes the azimuthal symmetry
of the problem and therefore the potential is independent from it, which means that only
components with m = 0 remain in the sum and the solution with azimuthal symmetry satisfies
∑
∞
( )
Φ(r, θ) = Al rl + Bl r−l−1 Pl (cos θ) ,
l=0
where Pl (x) are Legendre polynomials. We look for potential inside the ball with radius R,
which does not diverge in the centre (for r = 0), therefore Bl = 0 for all l. The problem has
simplified into
∑
∞
Φ(r, θ) = Al rl Pl (cos θ) . (7)

l=0
The potential V is specified on the sphere r = R, which leads to

∑
∞ { ⟨ )
U0 θ ∈ 0, π2
V (θ) = l
Al R Pl (cos θ) = ( ⟩ (8)
−U0 θ ∈ π2 , π .
l=0
Our objective is to find constants Al such that the equation holds for each θ ∈ ⟨0, π⟩. We use
orthogonality of Legendre polynomials given by equation
∫ 1
2
Pl (x)Pk (x) dx = δlk .
−1
2l + 1
We multiply the equation (8) by Pk (cos θ) and integrate, which gives

∫ 1
2Al Rl
= V (θ)Pl (cos θ) d cos θ ,
2l + 1 −1
∫ π
2l + 1
Al = V (θ)Pl (cos θ) sin θ dθ . (9)
2Rl 0
In the next step, we calculate the integral

∫ π ∫ 0 ∫ 1
V (θ)Pl (cos θ) sin θ dθ = (−U0 ) Pl (x) dx + U0 Pl (x) dx =
0 −1 0
( ∫ ∫ ) ∫
1 1
( ) 1
= U0 − Pl (−x) dx + Pl (x) dx = U0 1 − (−1)l Pl (x) dx . (10)
0 0 0
For even l, Al = 0 holds, and for odd l we need to calculate the last integral. To do this, we use
several properties of Legendre polynomials, namely the knowledge of the values on the borders
l ( )
of the interval of integration Pl (1) = 1, Pl (0) = (−1) l 2
2l l for even l and the integral formula
2
for Legendre polynomials ∫
Pl+1 (x) − Pl−1 (x)
Pl (x) = .
2l + 1
45
Using these properties we get

∫ 1
Pl+1 (1) − Pl−1 (1) Pl+1 (0) − Pl−1 (0)
Pl (x) dx = − =
2l + 1 2l + 1
0
( l+1 ( ) l−1 ( ))
1 (−1) 2 l+1 (−1) 2 l−1
=− l+1 − l−1 =
2l + 1 2l+1 2
2l−1 2
l−1 ( ( ) ( ))
(−1) 2 1 l+1 l−1
= l+1 + l−1 =
(2l + 1)2l−1 22 2 2
l−1 l−1 ( )
(−1) 2 (l − 1)(l + 1)(l − 2)! (−1) 2 l+1
= (( l+1 ) )2 = l+1 .
2l+1 ! 2l+1 l 2
2
We plug this result into (10) and then into (9) and (7), which lets us explicitly express the
potential inside the sphere using Legendre polynomials as
∑
∞ ( )
( ) (−1)
l−1
2l + 1 2 l+1
Φ(r, θ) = U0 1 − (−1) l
l+1 rl Pl (cos θ)
2Rl 2l+1 l 2
l=0
∑ (−1) l−1 ( )
2 (2l + 1) l + 1 rl
= U0 l+1 Pl (cos θ)
2l+1 l 2
Rl
lodd
∑ (−1)n 4n + 3 (2n + 2 ) r2n+1
∞
= U0 P2n+1 (cos θ)
22n+2 2n + 1 n+1 R2n+1
n=0
∑ (−1)n (4n + 3) (2n ) r2n+1
∞
= U0 P2n+1 (cos θ) .
22n+1 (n + 1) n R2n+1
n=0
Now we plug in the coordinates of the point where we want to calculate the potential, which
are θ = 0, r = R
2
, and get
( )
∑ (−1)n 4n + 3 2n
∞ ( )
R 1
Φ , 0 = U0
2 22n+1 n + 1 n 22n+1
n=0
∑
∞ ( )
(−1)n 4n + 3 2n
= U0
42n+1 n + 1 n
n=0
( )
3 .
= U0 2− √ = 0.658 U0 .
5
Halfway from the centre to the positively charged apex of the hemisphere, the potential
.
is Φ = 66 V. It is enough to evaluate the first three terms of the series to obtain the result
with sufficient precision, i.e. the first two significant figures.
Šimon Pajger Matěj Mezera
legolas@fykos.cz m.mezera@fykos.cz
46
Potential
0
−1
Fig. 1: Visualization of the potential between the hemispheres (normalized).
Problem M.1 . . . at the train station 3 points

−1
Legolas was trying to catch a train. He was running with a speed v = 3.0 m·s when he noticed
that one of his shoelaces was untied. Therefore, he stopped and spent t = 10.0 s tying it. He
then jumped on an escalator, which was moving with a speed u = 1.0 m·s−1 (on which he ran
with speed v as well). Suddenly, he facepalmed, realizing he could have made one more step,
tied the shoelace on the escalator and saved some time. But how much exactly?
You do not need to know the exact length of the escalator – it is sufficient to know that it
was long enough for Legolas to manage to tie his shoelace on it. Lego missed the train.
We’re interested in the difference of time between these two scenarios. In the first one, Legolas
stops moving for a time t and then travels on the escalator with velocity u + v. In the second
scenario, he first travels on the escalator for the time t with velocity u and then he traverses
the rest of the escalator with velocity u + v (the problem statement clarifies that there is some
distance left).
The velocity of traversing the second part is the same in both scenarios. This means we are
only interested in the time difference after traversing the first part of the escalator, which is
the distance Legolas covered while tying his shoelaces in the second scenario. The distance of
this point from the entrance to the escalator is l = ut.
In the second scenario, he reaches this point in time t. In the first one, he’s first stationary
for time t and then covers the distance l in time
l u
tl = =t .
u+v u+v
47
Hence, the time difference is

u
∆t = (t + tl ) − t = tl = t = 2.5 s .
u+v
Šimon Pajger
legolas@fykos.cz
Problem M.2 . . . at the airport 3 points

Dano and Danka are in a hurry to catch a plane to Mexico, so they both run on a moving
walkway (in the direction of its movement). Both of them have the same step length, which
is exactly 1 m. Dano runs twice as fast as Danka and he has to take 28 steps to run from one
end of the walkway to the other end, while Danka needs to take just 21 steps. How long is the
walkway? Legolas wanted to catch a plane.
We denote the velocity of the walkway by v and the frequency of Danka’s steps by f (the
step length is the same, so Dano’s frequency is 2f ). From Dano’s point of view, the length of
the walkway is sDano + sww , where sDano is the distance Dano walked on the walkway, that is
sDano = 28 m, and sww is the distance the walkway traveled while Dano was standing on it.
Dano spent the time 28/(2f ) on the belt, so the length of the walkway is sDano + 14v/f . We can
do the same calculation for Danka, to obtain the length of the walkway as sDanka +21v/f , where
sDanka = 21 m. Comparing these two relationships, we obtain v/f = 1 m, and substituting into
(either of) the formulas for the length of the walkway, we determine the length of the walkway
as 42.0 m.
Šimon Pajger
legolas@fykos.cz
Problem M.3 . . . under the ground 3 points

Matěj likes to travel by metro. One day, he got off at a station which was h = 50 m deep under
the surface. He walked upwards with a speed v = 1 m·s−1 on an escalator, which itself moves
upwards with a speed u = 1.5 m·s−1 . What work has Matěj done by walking upwards? Assume
that his mass is m = 60 kg. Matěj likes escalators and the metro.
The total work done was W = mgh. Part of it was done by Matěj by walking upwards, part
was done by the escalator. Because the work is directly proportional to the traveled height h,
it will be split in the same ratio as the heights covered by each of them. The time of motion is
the same for both Matěj and the escalator and they move in the same direction, so the height is
directly proportional to the speeds of movement. The height covered by the effort of the person
is
v
hM = h ,
v+u
whilst the escalator lifts him by the height
u
he = h .
v+u
48
From this we can calculate the work done by Matěj as

v .
WM = mghM = mgh = 11 800 J .
v+u
Matěj Mezera
m.mezera@fykos.cz
Problem M.4 . . . in the shopping mall 3 points

You have traveled by train, by metro and by plane to visit a shopping mall, which is, of course,
equipped with many escalators. On a switched-on escalator moving upwards, the time it takes
you to walk up and down again is ton = 50 s. On a switched-off escalator, it takes toff = 30 s.
What is the speed of the stairs of a switched-on escalator, if you walk with a speed vu =
= 1.0 m·s−1 upwards and vd = 2.0 m·s−1 downwards?
Legolas runs up and down escalators in metro stations.
We can find the length of the escalator from the case when it is switched off as
l l toff vu vd
toff = + ⇒ l= .
vu vd vu + vd
For the switched-on escalator, we label its speed as u and we have
l l ton
ton = + ⇒ l= (vu + u) (vd − u) .
vu + u vd − u vu + vd
We can therefore construct an equation
ton toff vu vd
(vu + u) (vd − u) = ,
vu + vd vu + vd
toff vu vd
vu vd + u (vd − vu ) − u2 = ,
ton
( )
toff
u2 − (vd − vu ) u + − 1 vu vd = 0 .
ton
We’ve obtained a quadratic formula for u, the solutions of which are
√
vd − vu ± (vd + vu )2 − 4vd vu ttoff
on
u= ,
2
and we are interested in the solution with the plus sign (the one with the minus sign is negative),
.
which is approximately u = 1.5 m·s−1 .
Šimon Pajger
legolas@fykos.cz
49
Problem E.1 . . . statistical 1 - intro 3 points

In statistical physics, we can describe a system of particles enclosed in an unchanging volume.
We call this a cannonical system and the probability that we find the system in a certain state
n follows Boltzmann’s Law
1 − En
pn = e k B T ,
Z
where En is the energy of the system in the state n, T is the temperature of the system and kB =
= 1.381·10−23 m2 ·kg·s−2 ·K−1 is the Boltzmann constant. The value of Z is chosen in such a way
that the sum of pn over all the states equals one. Typically, there is a large number of states
a system can reside in, but in this simple problem, let’s work with a system with only three
distinguishable states (n = 1, 2, 3) with energies E1 = 1.00 · 10−20 J, E2 = 2E1 and E3 = 3E1 .
Determine the value of Z at a temperature T0 = 275 K.
Matěj missed statistical physics among FYKOS problems.
Let’s start with the fact that the sum of all probabilities is equal to 1
( E E E )
1 − k 1T − k 2T − k 3T
1 = p1 + p2 + p3 = e B +e B +e B ,
Z
E E E
− k 1T − k 2T − k 3T .
Z=e B +e B +e B = 0.077 4 .
The specific value of Z usually does not have any physical meaning. It is simply a normalizing
constant. Only its derivatives will be physically relevant, as we shall see in the subsequent
problems.
Matěj Mezera
m.mezera@fykos.cz
Problem E.2 . . . statistical 2 - basics 3 points

Now, let’s examine a canonical system which has only two possible states. The difference of
energies between these two states is exactly ∆E = 10−20 J. For example, you can imagine a
molecule that exists either in its ground state or an excited state, but this description works for
any general two-state system. What is the probability of finding the system in the state with
higher energy if the temperature of the system is T0 = 275 K?
Note: use knowledge from the previous problem. Because the intro was not enough.
According to the problem statement, we have only two available states with energies E1 and
E2 = E1 + ∆E. In this case, we can easily plug in all (i.e. both) cases into the condition that
the sum of all probabilities equals 1, so we get
p1 + p2 = 1 ,
1 − kE1T 1 − E1 +∆E
e B + e kB T = 1 ,
Z Z
E1
− k∆E
1+e TB = Ze kB T .
50
Substituting into Boltzmann’s law we obtain

− k∆ET
1 − k∆ET e B 1 .
p2 = E1
e B = = = 0.067 0 .
− k∆E ∆E
Ze kB T 1+e TB 1+e kB T
Matěj Mezera
m.mezera@fykos.cz
Problem E.3 . . . statistical 3 - the quantum way 4 points

Earlier, you had to express Z, which you can calculate using
∑ E
− k nT
Z= e B ,
n
where the sum goes over all possible states (such as the two in the previous problem). Then
you had to use Boltzmann’s Law to calculate probabilities. The sum Z is called the partition
function.
Now let’s consider a more complicated system - the quantum harmonic
( oscillator,
) which can
be found in many different states. The energy of the n-th state is n + 12 ℏω, where n ∈ N0
and ℏω = 10−21 J is the parameter of the oscillator. Determine the probability that the system
is in the ground state (n = 0). The temperature of the system is T0 = 275 K.
∑
∞
Hint qn = 1
1−q
for q ∈ (0, 1). This is where the fun begins.
n=0
We calculate the partition function
∑
∞ (n+ 21 )ℏω ∞ (
∑ )n − 2kℏωT
− − 2kℏωT − kℏωT e B
Z= e kB T
=e B e B =
− kℏωT
n=0 n=0 1−e B
and substitute it into Boltzmann’s law

− kℏωT
1−e B − 2kℏωT − kℏωT .
p0 = e B =1−e B = 0.231 5 .
− 2kℏωT
e B
Matěj Mezera
m.mezera@fykos.cz
Problem E.4 . . . statistical 4 - the second quantum one 4 points

In the previous( problem,
) you calculated the partition function of a quantum harmonic oscillator
with energies n + 12 ℏω, where ℏω = 10−21 J. It’s actually possible to obtain all the different
thermodynamic properties of the system just using the partition function. For example, the
internal energy of the system E (the expected value of energy) can be calculated as
∂
E = kB T 2 ln (Z(T )) .
∂T
51
Determine the internal energy of the quantum harmonic oscillator from the previous problem
in the units of ℏω. The temperature of the system is T0 = 275 K. This
was supposed to be a problem with a second partial derivative, but it would probably be too hard.
We differentiate the partition function and get
∂ 1 ∂Z
E = kB T 2 ln (Z(T )) = kB T 2 =
∂T ( Z ∂T )
ℏω − ℏω − kℏωT ℏω − ℏω − 2kℏωT
2kB T 2
e 2kB T 1−e B + kB T 2
e kB T e B
1
( )2
2
= kB T =
Z − kℏωT
1−e B
( )
− kℏωT − kℏωT
1
1−e B +e B − ℏω
2 1 1 + e kB T 1 ℏω .
= ℏω ( ) = ℏω = ℏω coth = 3.819 7 ℏω .
1−e
− kℏωT
B
2 1 − e− kℏω
B T 2 2k BT
Matěj Mezera
m.mezera@fykos.cz
Problem X.1 . . . the strongman and the weight 3 points

Imagine a m = 10 kg weight attached to the end of a d = 1.0 m long (and very light) rod. How
many times bigger is the force our biceps needs to exert to hold the rod with the weight steady,
compared to simply holding it in hand? Assume that our forearm is l = 30 cm long. We hold
the rod in such a way that it forms a “straight-line extension” of our forearm.
Dodo was carrying a pan full of water.
The force that the biceps must exert can be calculated from an equlibrium of forces on a lever,
with the elbow joint serving as a pivot. By changing the position of the weight, we extend the
length of the lever arm. We obtain the ratio of forces
F2 l+d .
= = 4.3 .
F1 l
Jozef Lipták
liptak.j@fykos.cz
Problem X.2 . . . (not) breathing 3 points

In a single day, approximately V = 11 000 l of air pass through human lungs. Find the average
speed of air flowing through the larynx, assuming that its diameter is d = 40 mm.
Dod and his treacherous factors of two.
To solve this, we use the relation for volumetric flow rate of fluids Q
Q = Sv
where S is the cross-sectional area and v is the flow speed. The flow rate can be determined
from the volume of air exchanged. The volume V has to flow both into the lungs and out of
52
them in a day, so the average flow rate is Q = 2V /1 day. The cross-section of the larynx has an
area ( )2
d
S=π .
2
By expressing the flow speed and substituting for S and Q, we get
Q 8V .
v= = 2 = 0.20 m·s−1 .
S πd · 1 day
Jozef Lipták
liptak.j@fykos.cz
Problem X.3 . . . the pressure of a ballerina 3 points

A ballerina that weighs m = 50 kg is landing on a big toe after jumping into the height h =
= 0.5 m. At its narrowest point, the last phalanx bone of the big toe has a diameter d =
= 1.3 cm. Find the highest stress across this bone during the landing. Assume that the ballerina
decelerates upon impact with a constant force for t = 0.5 s. Dodo stood on one toe.
The stress can be calculated from its definition as a force across a cross-sectional area
F 4F
p= = 2.
S πd
The force transferred through the bone is given by the sum of the weight of the ballerina and
the force used for deceleration, which can be obtained from Newton’s 2nd law as the rate of
change of the momentum √
mv m 2gh
Fd = = ,
t t
into which we substituted the free fall velocity right before impact.
Substituting the expression for the force into the equation for stress, we obtain
( √ )
m 2gh
4 mg + t .
p= = 6.06 MPa .
πd 2
Jozef Lipták
liptak.j@fykos.cz
Problem X.4 . . . saliva 3 points

Fykos organizers are very skillful and managed to create 100 g a substance such that 99 % of its
mass is water. Unfortunately, after a while, some water evaporated away and now only 98 % of
the sample is water. What is the current mass of the sample? Lego likes Vsauce2.
The key is to realize that the mass of the non-water part of the sample is conserved. In the
beginning, it is 100 % − 99 % = 1 % of the total mass. Since the sample weighed 100 g, the
non-water part weighs 100 g · 1 % = 1 g.
53
After drying out, the non-water part makes up 100 % − 98 % = 2 % of the total mass. The
whole sample must then weigh 1 g/2 % = 50 g.
Šimon Pajger
legolas@fykos.cz
FYKOS
UK, Matematicko-fyzikální fakulta
Ústav teoretické fyziky
V Holešovičkách 2
180 00 Praha 8
www: http://fykos.cz
e-mail: fykos@fykos.cz
FYKOS is also on Facebook
http://www.facebook.com/Fykos
FYKOS is organized by students of Faculty of Mathematics and Physics of Charles University.

It’s part of Public Relations Office activities and is supported by Institute of Theoretical
Physics, MFF UK, his employees and The Union of Czech Mathematicians and Physicists.
This work is licensed under Creative Commons Attribution-Share Alike 3.0 Unported.
To view a copy of the license, visit http://creativecommons.org/licenses/by-sa/3.0/.
54

Solutions of 10 Online Physics Brawl

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How is force equilibrium applied to solve pulley problems?

How is force equilibrium applied to solve pulley problems?

How can the resistance of a single resistor be determined from measurements of a circuit with resistors in series and parallel?

How can the resistance of a single resistor be determined from measurements of a circuit with resistors in series and parallel?

Online Physics Brawl 10th year November 25, 2020

Solutions of 10 th Online Physics Brawl

Problem FoL.1 . . . (hopefully just) easy pulleys 3 points

where both accelerations are oriented downwards.

Problem FoL.2 . . . two circuits 3 points

Problem FoL.3 . . . felling trees 3 points

Problem FoL.4 . . . the nearest asteroid 3 points

Problem FoL.5 . . . unstable 3 points

Problem FoL.6 . . . safe power 3 points

Problem FoL.7 . . . the ink-blooded prince 3 points

Problem FoL.8 . . . slowed down train 4 points

Problem FoL.9 . . . we pump oil 4 points

Problem FoL.10 . . . homogeneous air 4 points

Problem FoL.11 . . . our old clock 4 points

Problem FoL.12 . . . cough cough 4 points

Problem FoL.13 . . . uneven illumination 4 points

Problem FoL.14 . . . autumn in a train 4 points

Problem FoL.15 . . . the power of a waterfall 4 points

Problem FoL.16 . . . burning coal 4 points

Problem FoL.17 . . . heat and phases 4 points

Problem FoL.18 . . . boiling 4 points

Problem FoL.19 . . . internal resistance 4 points

where we used the formula for Ri .

Problem FoL.20 . . . raft voyage 4 points

Problem FoL.21 . . . mysterious ball 4 points

Problem FoL.22 . . . illuminated 5 points

where n = 100. Then √

Problem FoL.23 . . . a ball 5 points

Problem FoL.24 . . . a traveler’s problem 5 points

Problem FoL.25 . . . a different parallel-plate capacitor 5 points

Problem FoL.26 . . . oscillating hoop 5 points

We probably could have guessed the result straight away. . .

Problem FoL.27 . . . oscillating 5 points

from which we can calculate the required distance

Problem FoL.28 . . . lego and dice 1.0 4 points

Problem FoL.29 . . . lego and dice 2.0 5 points

Problem FoL.30 . . . impact accelerator 5 points

Problem FoL.31 . . . jump on! 5 points

Problem FoL.32 . . . disgusting piston 5 points

Problem FoL.33 . . . 4D scouring pad 5 points

After that we can similarly calculate the resistance between B and Y as

Finally we will obtain the resistance between A and Z as

Problem FoL.34 . . . levitation by firearm 5 points

Problem FoL.35 . . . burning phosphorus 5 points

Problem FoL.36 . . . mixed current 4 points

IDC + IAC = Imax = 42 mA ,

from which we can express

Problem FoL.37 . . . stopping a train 5 points

Problem FoL.38 . . . solar cube 5 points

S = |za | |a| + |zb | |b| + |zc | |c| = |za | + |zb | + |zc | ,

which was already examined before.

Problem FoL.39 . . . optimal direction 6 points

Problem FoL.40 . . . watch out, Danka is throwing 6 points

gd2 gd2 tan2 φ