physical sciences: second-round sample tasks

DEMO VERSION OF THE SECOND ROUND TASKS

1

Physical Sciences: second-round sample tasks

Dear participants of Open Doors Olympiad!

We are glad to present you a demonstrational version of problems in physics, which is

intended to help you to prepare yourself for a competitive part of the Olympiad 2021-2022. The

content of these problems correlates with the tasks and topics of the physics course which we have

prepared for the competition.

We hope that some ideas and well-known mathematical formulations of physics laws from

the solutions of this version will help you to cope with the Olympics tasks. Remember that it is

not enough only to read the tips and solutions that we prepared. To succeed you are also expected

to have a physical erudition, fantasy and the ability to find non-traditional approaches to the

solutions. These skills are necessary to continue learning on the master and doctoral programmes

of the universities of Russia and successfully starting your scientific work.

To reap better benefits out of this demo-version, we strongly recommend you to follow the

advice:

1. Carefully read the problem and try to solve it by yourself.

2. If the answer you got coincides with the given one, be pleased with your chances of a

successful performance at the Olympiad. But, before proceeding to the next task, read quickly

through the given solution. There is a chance to pay attention to some ideas, which can be useful

in solving more complicated problems.

3. If you don’t manage to solve the task, read the given solution, realize it and then try to repeat

it by yourself.


2

Problem 1

A ball of mass M = 100 g falls from rest and has an elastic collision with the horizontal plate.

The initial height of the ball above the plate is H = 2.5 m. Assuming that the mass of the plate

is much greater than the mass of the ball, find the momentum, obtained by the plate.

Solution:

2пp p mv= − =

2v gh=

2 2 1.4 kg /sпp m gh m= =

Answer: 1.4 N·s.

Problem 2.

After pouring water into the pot, air bubbles were formed on the

bottom of the pot. Determine the acceleration of the bubble with a

radius R = 1 mm immediately after the separation from the bottom.

Give the answer in m/s2 and round it to the first decimal figure.

Solution:

The second Newton’s law for the bubble which starts floating from the rest:

( )gV mg m M a − = + .

Here: ρ – the density of water, m = ρaV – the mass of air in the bubble, V – the volume of the

bubble, M – the added mass of water, where the bubble accelerates.

The added mass of water when the bubble accelerates through it is equal to:

2

VM

= .

When the bubble accelerates through water, it must move some surrounding layers of water.

This process needs the energy. Therefore we have to take into account the added mass.

After substitution:


3

(𝜌 − 𝜌𝑎)𝑔 = (𝜌𝑎 +𝜌

2)𝑎.

The density of air in the bubble can be determined with the use of the equation of state of an

ideal gas:

𝑝𝑉 =𝑚

𝜇 𝑅𝑇 ⇒ 𝜌𝑎 =

𝑃𝜇

𝑅𝑇.

Here p – atmospheric pressure, 29 = g/mole – molar mass of air, T – absolute temperature.

1.2aP

RT

= kg/m3 << 1000 kg/m3 = ρ.

The obtained density of air in the bubble is 3 order less than the density of water. According to

this result, the density of air in the bubble can be considered as negligible. Then we get an

easy but a little unexpected answer:

𝑎 =𝜌−𝜌𝑎

𝜌𝑎+𝜌

2

𝑔 ≈ 2𝑔.

Answer: 19.6a = (m/s2)

Problem 3.

A dynamometer consists of a body of mass dm = 60 g and a metal spring of mass sm

= 40 g. A weight of mass wm = 100 g is hung to the dynamometer’s hook. The

dynamometer is pulled up by the force F = 4 N, which is acted on the body of the

dynamometer. What measurement indicates the dynamometer? The mass of the

connecting wire between the spring and the body is negligible. Give the answer in N

and round it to the first decimal figure.

Solution:

Let us divide the spring into N equal parts and give them numbers starting from the bottom.

Consider N>>1. Then mass of a part of the spring can be found as m N , where m – mass of the

spring. When N is great enough, this mass of a part is negligible in compare with mass of all the

parts which are located below.

In case of series of N equal springs (when springs are connected end to end) the equivalent spring

constant is 1k k N= . Thus, the spring constant for one part can be found as 1k kN= .


4

On the part with number ( 1i + ) acted gravity force F mg i N= . The corresponding extension is

1 21

iF mg i

lk kN

+ = = .

The total extension of the spring can be determined by summing the extensions of the parts.

Using the formula for arithmetic progression sum:

( )1

2 21 0

1

2

N N

ii i

mg i mg N Nl l

kN kN

−

= =

− = = = .

In the limit N →+ we have

2

mgl

k = .

Thus, the extension of a massive spring is twice less than the extension of a weightless spring

with the point weight in the bottom.

Total mass of the system is d s wm m m m= + + . Acceleration is a F m= . Let us go to non-

inertial reference frame connected with the dynamometer.

effg g a= + .

Without a weight under the action of the spring's own weight, the dynamometer shows 0,

therefore

02

ss

m gl

k = .

The extension of a massive spring when moving with acceleration arises from the extension due

to the weight w w effl m g k = and the extension due to the mass of the spring

( )2s s effl m g k = .

02 2 2 2

eff s s s ss w s w w

g m m g m m gg al l l l m m

k k k k

+ = + − = + − = + −

,

2 2 2 2

s s s sw w

m m g m m gg F m g Fl m m

k k km k

+ − = + − = + −

.

Dynamometer scale graduates in units of force. According to Hooke’s law


5

2 2

s sd w

d s w

m m gFF k l m

m m m

= = + − + +

.

2.2dF = N

Answer: 2.2dF = N.

Problem 4.

A gas enclosed in a cylinder at a temperature t1=100 °C. To which temperature the gas must be

heated to double the pressure?

Solution:

1 2 1 22

1 2 1

373 2 746 КP P T P

TT T P

= = = = .

Answer: 2 473 Ct =

Problem 5.

Hydrogen of mass m = 0,2 kg is heated at constant pressure from the initial temperature t1=0

°С to the final temperature t2=100 °С. Find the heat absorbed by hydrogen, the work done by

the gas and the change in its internal energy.

Solution:

The heat absorbing in isobaric process is determined by the formula

2 2291kJ

2 2p

i i mQ C T R T R T

M

+ += = = = .

Ответ: 291kJ

Problem 6.

Solution:


6

One mole of an ideal diatomic gas undergoes the closed cycle in a

form of a circle presented in the figure. Find the work done by the

gas, if V1 =10 l, V2=2V1, P1 =1 bar, P2=2P1.

( )( )22 1 2 1 785

4 4

V V P PdA PdV

− −= = = = J.

Answer: A=785 J.

Problem 7.

A point charge Q = 10 nC is located in a center of a sphere of radius R = 20 cm. Find the

electric flux

through the part of the spherical surface with square S = 20 cm2.

According to the Gauss’s law the electric flux through the sphere is

00

4q

kq = =

.

The flux through the part of the sphere

02 2

0

4

4

kqS kqSS

S R R

= = =

.

Answer: 4.5 V m= .

Problem 8.

Two batteries with emfs ε1 = 1 V and ε2 = 3 V and the same internal

resistances r = 0.4 Ω are connected as shown in the figure. Find the

potential difference U of point A relative to point B.

V

P

V1 V2

P1

P2

ε1

ε2

A B


7

Solution:

1 2 2rI − = , 1 2 1 21 1

2 2ABU Ir r

r

− + = − = − =

Answer: 2 VABU = .

Problem 9.

Find the magnitude of the magnetic field at a distance r = 5 mm from the axis of symmetry in

the middle region between the round capacitor’s plates with radius R = 25 mm. The capacitor

is charging by the current Ic = 500 mA from the source of a constant voltage. Give the answer

in μT and round it to the second decimal figure.

Solution:

According to Maxwell’s equation:

0 0 0d d dt

= +

E

B l j S SС .

As there is no conduction current, we get:

20 02 0

EB r r

t

= +

,

r R

Ic


8

0 02

E rB

t

=

.

The electric field in vacuum capacitor:

0

cQE

S=

.

Then a displacement current density:

0 2

1 1.c c

cdQ IE

It S dt S R

= = =

.

For the magnetic field we get:

00 0 2 2

02 2

c cI I rrB

R R

= =

,

Answer: В = 0.8 μT.

Problem 10.

A point source of light is located between two parallel plane mirrors in the center. The first

virtual images of the source in the mirrors get closer with the speed 5 m/s. Find the speed of

the mirrors. Both mirrors move with the same speed and remain parallel to each other.

Solution:

A virtual image is located behind a mirror at the same distance as a source. When the mirror

moves towards the source, the virtual image moves with the doubled speed. Thus, the speed

with which images get closer to each other is 4 times more than the speed of the mirrors.

Answer: v = 1.25 m/s.


9

Problem 11.

A beam of light falls on a system of two Nicol prisms whose transmission planes are turned

through an angle 400. Assuming that the absorption coefficient of each prism is 0.15, find how

many times will the intensity of light decrease after its passing through the system.

Solution:

( ) ( )22 22 1 0

11 cos 1 cos

2I I k I k= − = − ,

( )

02 2

2

2

1 cos

I

I k=

−

Answer: 0

2

I

I= 4.7.

Problem 12.

The distance between the lines of a diffraction grating is d = 4 μm. Light with wavelength λ =

0,58 μm falls normally on the grating. What is the highest order maximum that can be

observed?

Solution:

sind k = , sin 1k

d

= ,

dk

, max 6.9

dk

= =

Answer: 6k = .

Problem 13.

A photon emitted in electron-positron annihilation collides with ion 92238U which has only one

single electron. Photon liberated this electron. Find the velocity of the liberated electron.

Assume that the velocities of electron and positron prior the annihilation are small, all particles

move along the same straight. The ionization energy of hydrogen atom (Rydberg’s constant) is

HE =13.6 eV. Give the answer in Mm/s and round it to the integer.

Solution:

Under the circumstances in electron-positron annihilation two photons are emitted in opposite

directions. Each photon has the energy 20 0.511 MeVeE m c= = that is equal to electron rest


10

energy. Ion 92238U with single electron is hydrogen-like atom. The ionization energy of this

hydrogen-like atom is 2i HE Z E= where Z is atomic number (Z=92).

Kinetic energy of the liberated electron is

0 iE E E= − .

Here we do not take into account kinetic energy of the nucleus because of its mass.

On the other side

( ) ( )20 01 1E m c E= − = − , 0

0

2 iE E

E

− =

0

2 0 0 0

11 1 2

1

i iE E EE

E E E

− = = + = + = −

−

,

2 2

0

1 11 1

1E

E

= − = −

+

,

v c= = 248 Мm/s

Answer: v =248 Мм/с

Problem 14.

Illuminating a photocell with light of hydrogen lamp

which corresponds to the transition from fifth energy

level (n =5) to ground state, it was found that the

stopping potential is U = – 3.53 V. Illuminating the same

photocell with visible light from the same lamp it was

found that the change of polarity of the applied voltage

doesn’t affect on the current through the photocell. What


11

is the maximum number of the spectral lines could be

present in applied visible radiation?

Решение:

1

{𝑚𝑣2

𝑅= 𝑘

𝑞2

𝑅2

𝑚𝑣𝑅 = 𝑛ℏ

⇒ 𝑅𝑛 =ℏ2

𝑘𝑞2𝑚𝑛2 = 𝑛2 𝑎0

Calculation of the stationary orbit of

electron in state with principal quantum

number n in Coulomb’s field of the point

nucleus with charge q in the framework of

the simplest planetary Bohr-Rutherford

model (a0 = 0.5Å – the first Bohr radius)

2 𝑊𝑛 =

𝑚𝑣2

2− 𝑘

𝑞2

𝑅= −𝑘

𝑞2

2𝑅= −

𝑘

2

𝑞2

𝑎0

1

𝑛2

= −𝑊0

1

𝑛2

Electron energy in the state with principal

quantum number n in Coulomb’s field of the

point nucleus with charge q (W0 = 13.6 eV)

3.

ℏ𝜔 = 𝐴 +𝑚𝑣2

2

The photoelectric effect equation proposed

by A. Einstein

4. 𝐼 = 0 ⇒ |𝑞𝑢| ≥

𝑚𝑣2

2= ℏ𝜔 − 𝐴

Cancellation condition for the current

through the photocell by external stopping

potential

5 ℏ𝜔 = 𝑊0 (1 −

1

𝑛2)

Photon’s energy in experiment on the

determination of the work function

6 𝐴 = ℏ𝜔 − |𝑞𝑢| = 𝑊0 (1 −

1

𝑛2) − |𝑞𝑢|

The determination of work function with the

use of equations (2-4)


12

7 𝜆 =

2𝜋с

𝜔=

2𝜋сℏ

𝑊0 (1 −1

𝑛2)

Calculation of the wavelength of the

radiation used for the determination of work

function

8

𝜆 =2 ∙ 3.14 ∙ 3 ∙ 108 ∙ 1.05 ∙ 10−34

13.6 ∙ 1.6 ∙ 10−19= 91(нм)

Numerical calculation of the wavelength of

the spectral line that corresponds to the

transition 5-1 gives that the radiation

belongs to the UV-region. Energies of

optical photons are less, therefore they can’t

provide the current through the photocell

cancelled by the stopping potential U.

According to the condition of the task in the

experiment with the visible light the current

through the element is 0 for arbitrary

polarity of the applied potential.

9

ℏ𝜔𝑚𝑎𝑥 ≤ 𝐴

The condition for frequency of optical

radiation satisfying the requirement of

absence of the current for arbitrary polarity

of the applied potential.

1

0

𝑊0 (1

22−

1

𝑛𝑚𝑎𝑥2

) ≤ 𝐴 = 𝑊0 (1 −1

𝑛2) − |𝑞𝑢|

The condition for spectral line of visible

radiation to be out of threshold of

photoelectric effect (visible radiation in

hydrogen spectra corresponds to the

transitions on the second energy level)

1

1

|𝑞𝑢|

𝑊0+

1

𝑛2−

1

22≤

1

𝑛𝑚𝑎𝑥2

⇒

𝑛𝑚𝑎𝑥 ≤ (

|𝑞𝑢|

𝑊0+

1

𝑛2−

1

22)

−1/2

The limitation on the number of the initial

level of the optical transition arising from

(10)


13

1

2 𝑛𝑚𝑎𝑥 ≤ (

3.53

13.6+

1

52−

1

22)

−12

= 4.49

Numerical calculation of the upper threshold

of principal quantum number of the initial

level of the optical transition

1

3 [

(𝑛 = 3) → (𝑛 = 2 ) (+)(𝑛 = 4) → (𝑛 = 2 ) (+)(𝑛 = 5) → (𝑛 = 2 ) (−)

⇒ 𝑁 = 2

Optical spectra of visible radiation should

contain only two lines (the red one and the

blue-green line)

Answer: 2

Problem 15.

A photon with wavelength λ1 = 15 pm is scattered by a free electron. The wavelength of the

scattered photon is λ2 = 16 pm. Find the scattering angle.

Solution:

( )( )

2 12 1

1 cos cos 1h mc

mc h − = − = −

−

Answer: 54 = .

physical sciences: second-round sample tasks

Documents