Solutions to the problems of the 5-th
International Physics Olympiad, 1971, Sofia, Bulgaria
The problems and the solutions are adapted by
Victor Ivanov
Sofia State University, Faculty of Physics, 5 James Bourcier Blvd., 1164 Sofia, Bulgaria
Reference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High
School Students”, eds. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).
Theoretical problems
Question 1.
The blocks slide relative to the prism with accelerations a
1
and a
2
, which are
parallel to its sides and have the same magnitude a (see Fig. 1.1). The blocks move
relative to the earth with accelerations:
(1.1) w
1
= a
1
+ a
0
;
(1.2) w
2
= a
2
+ a
0
.
Now we project w

Fig. 1.1
The equations of motion for the blocks and for the prism have the following vector
forms (see Fig. 1.2):
(1.7)
11111
TRgw
++=
mm
;
(1.8)
22222
TRgw
++=
mm
;
(1.9)
21210
TTRRRga
−−+−−=
MM
.
Fig. 1.2
The forces of tension T
1
and T
2
at the ends of the thread are of the same magnitude T
since the masses of the thread and that of the pulley are negligible. Note that in equation
(1.9) we account for the net force –(T
1

2
R
1
T
1
R
Mg
m
1
g
m
2
g
x
y
(1.11)
gmRTam
111111
cossinsin
−α+α=α
;
(1.12)
2220222
sincoscos
α+α−=−α
RTamam
;
(1.13)
gmRTam
222222

.
The straightforward elimination of the unknown forces gives the final answer for a
0
:
(1.17)
2
22112121
22112211
0
)coscos())((
)coscos)(sinsin(
α+α−+++
α+αα−α
=
mmmmMmm
mmmm
a
.
It follows from equation (1.17) that the prism will be in equilibrium (a
0
= 0) if:
(1.18)
1
2
2
1
sin
sin
α
α

= 0 °C, or T
0
= 273 K, and β its coefficient of expansion. The volume of the
hydrogen is given by:
(2.3) V
i
= S(H – h
i
).
Now we can write down the equations of state for hydrogen at points 0, 1, 2, and
3 of the PV diagram (see Fig. 2):
(2.4)
00000
)()( RT
M
m
hHSghP
=−ρ−
;
(2.5)
01101
)()( RT
M
m
hHSghP
=−ρ−
;
(2.6)
22212
)()( RT

33322
)()( RT
M
m
hHSghP
=−ρ−
where
[ ]
)(1
0302
TT
−β−ρ≈ρ
,
2
3
2
2
3
23
hH
hH
T
V
V
TT
−
−
==
for the isobaric process 2–3.
P

≈ 364 K;
P
2
≈ 1.067×10
5
Pa;
T
3
≈ 546 K;
P
2
≈ 4.8×10
4
Pa.
Question 3.
A circuit equivalent to the given one is shown in Fig. 3. In a steady state (the
capacitors are completely charged already) the same current I flows through all the
resistors in the closed circuit ABFGHDA. From the Kirchhoff’s second rule we obtain:
(3.1)
R
EE
I
4
14
−
=
.
Next we apply this rule for the circuit ABCDA:
(3.2)
121

5
4
14
242
=
−
−−=
EE
EEV
V,
(3.5)
1
4
14
344
=
−
−−=
EE
EEV
V.
Finally, the voltage V
3
across C
3
is found by applying the Kirchhoff’s rule for the
outermost circuit EHDAH:
(3.6)
5
4

(3.8)
R
E
I
2
4
=
′
.
The new steady-state voltage on C
2
is found by considering the BFGCB circuit:
(3.9)
242
EERIV
−=
′
+
′
or finally:
(3.10)
0
2
2
4
2
=−=
′
E
E

≈ R α;
(4.2) d

≈ R γ;
(4.3) α + γ = 2β;
(4.4) α ≈ nβ.
From equations (4.1) - (4.4) we find the vertical displacement of the first image in terms
of d:
(4.5)
d
n
n
d
1
2
=
−
,
and respectively its velocity v
1
in terms of v:
(4.6)
v
n
n
v
1
2
2
=

R
A
B
A
1
B
1
a
b
1
d
1
α
β
β
γ
α
d
Fig. 4.1
The rays, which are first reflected by the mirror, and then are refracted twice at
the walls of the aquarium form the second, real image (see Fig. 4.2). It can be considered
as originating from the mirror image of the fish, which move along the line A’B’ at
exactly the same distance d as the fish do.
Fig. 4.2
The vertical displacement A
2
B
2
of the second image is equal to the distance d
2

d
109
2
−
=
,
and the velocity v
2
of the second image in terms of v:
(4.12)
vv
n
n
v
3
2
109
2
=
−
=
.
The relative velocity of the two images is:
(4.13) v
rel
= v
1
– v
2
in a vector form. Since vectors v

4R
a
b
2