ipho -1971

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V International Physics Olympiad, 1971
Sofia, Bulgaria
The problems and the solutions are adapted by
Victor Ivanov
University of Sofia, Faculty of Physics, 5 James Bourchier 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.
A triangular prism of mass M is placed one side on a frictionless horizontal plane as
shown in Fig. 1. The other two sides are inclined with respect to the plane at angles 1 and 2
respectively. Two blocks of masses m1 and m2, connected by an inextensible thread, can slide
without friction on the surface of the prism. The mass of the pulley, which supports the
thread, is negligible.
 Express the acceleration a of the blocks relative to the prism in terms of the
acceleration a0 of the prism.
 Find the acceleration a0 of the prism in terms of quantities given and the acceleration g
due to gravity.
 At what ratio m1/m2 the prism will be in equilibrium?
Fig. 1
Question 2.
A vertical glass tube of cross section S = 1.0 cm2 contains unknown amount of
hydrogen. The upper end of the tube is closed. The other end is opened and is immersed in a
pan filled with mercury. The tube and the pan are placed in a sealed chamber containing air at
temperature T0 = 273 K and pressure P0 = 1.334105 Pa. Under these conditions the height of
mercury column in the tube above the mercury level in the pan is h0 = 0.70 m.
One of the walls of the chamber is a piston, which expands the air isothermally to a
pressure of P1 = 8.00104 Pa. As a result the height of the mercury column in the tube
decreases to h1 = 0.40 m. Then the chamber is heated up at a constant volume to some
temperature T2 until the mercury column rises to h2 = 0.50 m. Finally, the air in the chamber
is expanded at constant pressure and the mercury level in the tube settles at h3 = 0.45 m above
the mercury level in the pan.
1 2
m1 m2
Provided that the system is in mechanical and thermal equilibrium during all the
processes calculate the mass m of the hydrogen, the intermediate temperature T2, and the
pressure P in the final state.
The density of mercury at temperature T0 is 0 = 1.36104 kg/m3, the coefficient of
expansion for mercury  = 1.8410‚ 4 K‚ 1, and the gas constant R = 8.314 J/(molK). The
thermal expansion of the glass tube and the variations of the mercury level in the pan are not
considered.
Hint. If T is the interval of temperature variations of the system then T = x << 1 In
that case you can use the approximation: x
x
 11
1 .
Question 3.
Four batteries of EMF E1 = 4 V, E2 = 8 V, E3 = 12 V, and E4 = 16 V, four capacitors
with the same capacitance C1 = C2 = C3 = C4 = 1 F, and four equivalent resistors are
connected in the circuit shown in Fig. 3. The internal resistance of the batteries is negligible.
 Calculate the total energy W accumulated on the capacitors when a steady state of the
system is established.
 The points H and B are short connected. Find the charge on the capacitor C2 in the
new steady state.
Fig. 3
Question 4.
A spherical aquarium, filled with water, is placed in front of a flat vertical mirror. The
radius of the aquarium is R, and the distance between its center and the mirror is 3R. A small
fish, which is initially at the point nearest to the mirror, starts to move with velocity v along
the wall. An observer looks at the fish from a very large distance along a horizontal line
passing trough the center of the aquarium.
What is the relative velocity vrel at which the two images of the fish seen by the observer
will move apart? Express your answer in terms of v. Assume that:
 The wall of the aquarium is made of a very thin glass.
 The index of refraction of water is n = 4/3.
E1 E2
E3 E4
C1
C2
C3
C4
A B
CD
E F
GH
Experimental Problem
Apparatus: dc source, ammeter, voltmeter, rheostat (coil of high resistance wire with sliding
contact), and connecting wires.
Problem: Construct appropriate circuit and establish the dependence of the electric power P
dissipated in the rheostat as a function of the current I supplied by the dc source.
1. Make a plot of P versus I.
2. Find the internal resistance of the dc source.
3. Determine the electromotive force E of the source.
4. Make a graph of the electric power P versus resistance R of the rheostat.
5. Make a graph of the total power Ptot dissipated in the circuit as a function of R.
6. Make a graph of the efficiency  of the dc source versus R.
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 a1 and a2, 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) w1 = a1 + a0;
(1.2) w2 = a2 + a0.
Now we project w1 and w2 along the x- and y-axes:
(1.3) 011 cos aaw x  ;
(1.4) 11 sin aw y ;
(1.5) 022 cos aaw x  ;
(1.6) 22 sin aw y .
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 T1 and T2 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 ‚ (T1 + T2), which the bended thread exerts on the
1 2
x
y
a0
a1
a2w2
w1
R2
T2R1 T1
R
Mg
m1g
m2g
x
y
prism through the pulley. The equations of motion result in a system of six scalar
equations when projected along x and y:
(1.10) 1110111 sincoscos  RTamam ;
(1.11) gmRTam 111111 cossinsin  ;
(1.12) 2220222 sincoscos  RTamam ;
(1.13) gmRTam 222222 sinsinsin  ;
(1.14) 2122110 coscossinsin  TTRRMa ;
(1.15) MgRRR  2211 coscos0 .
By adding up equations (1.10), (1.12), and (1.14) all forces internal to the system cancel
each other. In this way we obtain the required relation between accelerations a and a0:
(1.16)
2211
21
0 coscos 

mm
mmMaa .
The straightforward elimination of the unknown forces gives the final answer for a0:
(1.17) 2
22112121
22112211
0 )coscos())((
)coscos)(sinsin(


mmmmMmm
mmmma .
It follows from equation (1.17) that the prism will be in equilibrium (a0 = 0) if:
(1.18)
1
2
2
1
sin
sin


m
m .
Question 2.
We will denote by H (H = const) the height of the tube above the mercury level
in the pan, and the height of the mercury column in the tube by hi. Under conditions of
mechanical equilibrium the hydrogen pressure in the tube is:
(2.1) iairH ghPP 2 ,
where  is the density of mercury at temperature ti:
(2.2)  t 10
The index i enumerates different stages undergone by the system, 0 is the density of
mercury at t0 = 0 C, or T0 = 273 K, and  its coefficient of expansion. The volume of
the hydrogen is given by:
(2.3) Vi = S(H hi).
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 )()( RTM
mhHSghP  ;
(2.5) 01101 )()( RTM
mhHSghP  ;
(2.6) 22212 )()( RTM
mhHSghP  ,
where
0
21
2 T
TPP  ,  )(1
)(1 02002
0
1 TTTT

 since the process 1 3 is
isochoric,and:
(2.7) 33322 )()( RTM
mhHSghP 
where  )(1 0302 TT  ,
2
3
2
2
3
23 hH
hHT
V
VTT 
 for the isobaric process 2 3.
Fig. 2
After a good deal of algebra the above system of equations can be solved for the
unknown quantities, an exercise, which is left to the reader. The numerical answers,
however, will be given for reference:
H  1.3 m;
m  2.1110 6 kg;
T2  364 K;
P2  1.067105 Pa;
T3  546 K;
P2  4.8104 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
(3.1)
R
EEI
4
14  .
Next we apply this rule for the circuit ABCDA:
(3.2) 121 EEIRV  ,
where V1 is the potential difference across the capacitor C1. By using the expression
(3.1) for I, and the equation (3.2) we obtain:
(3.3) 1
4
14
121  EEEEV V.
Similarly, we obtain the potential differences V2 and V4 across the capacitors C2 and C4
by considering circuits BFGCB and FGHEF:
(3.4) 5
4
14
242  EEEEV V,
P0
P2
P1
P
V0 V1= V2 V3 V
1
2 3
0
(3.5) 1
4
14
344  EEEEV V.
Finally, the voltage V3 across C3
outermost circuit EHDAH:
(3.6) 5
4
14
133  EEEEV V.
The total energy of the capacitors is expressed by the formula:
(3.7)   26
2
2
4
2
3
2
2
2
1  VVVVCW J.
Fig. 3
When points B and H are short connected the same electric current flows
through the resistors in the BFGH circuit. It can be calculated, again by means of the
(3.8)
R
EI
2
4 .
The new steady-state voltage on C2 is found by considering the BFGCB circuit:
(3.9) 242 EERIV 
or finally:
(3.10) 0
2 2
4
2  EEV V.
Therefore the charge 2q on C2 in the new steady state is zero.
Question 4.
In a small time interval t the fish moves upward, from point A to point B, at a
small distance d = vt. Since the glass wall is very thin we can assume that the rays
leaving the aquarium refract as if there was water air interface. The divergent rays
undergoing one single refraction, as show in Fig. 4.1, form the first, virtual, image of the
fish. The corresponding vertical displacement A1B1 of that image is equal to the distance
d1 between the optical axis a and the ray b1, which leaves the aquarium parallel to a.
Since distances d and d1 are small compared to R we can use the small-angle
approximation: sin  tan   (rad). Thus we obtain:
(4.1) d1  R ;
(4.2) d  R ;
(4.3)  +  = 2;
(4.4)   n.
E1 E2 E3E4
C1 C2
C3
C4
A B
CD
EF
G H
R R
R
R
From equations (4.1) - (4.4) we find the vertical displacement of the first image in terms
of d:
(4.5) d n
n
d1 2
  ,
and respectively its velocity v1 in terms of v:
(4.6) v n
n
v1 2
2   .
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
at exactly the same distance d as the fish do.
Fig. 4.2
The vertical displacement A2B2 of the second image is equal to the distance d2 between
the optical axis a and the ray b2, which is parallel to a. Again, using the small-angle
approximation we have:
(4.7)  4R - d,
(4.8) d2  R
Following the derivation of equation (4.5) we obtain:
(4.9) d n
n
d2 2
   .
Now using the exact geometric relations:
A
B
A1
B1
a
b1
d1 
  

d
    

d'
dd
d2
B2
A2 A
B 4R
a
b2
(4.10)  = 2 2
-angle limit, we finally express d2 in terms of d:
(4.11) d
n
nd
1092  ,
and the velocity v2 of the second image in terms of v:
(4.12) vv
n
nv
3
2
1092
 .
The relative velocity of the two images is:
(4.13) vrel = v1 v2
in a vector form. Since vectors v1 and v2 are oppositely directed (one of the images
moves upward, the other, downward) the magnitude of the relative velocity is:
(4.14) vvvv
3
8
21rel  .
Experimental problem
The circuit is given in the figure below:
Sliding the contact along the rheostat sets the current I supplied by the source. For each
value of I the voltage U across the source terminals is recorded by the voltmeter. The
power dissipated in the rheostat is:
P = UI
provided that the heat losses in the internal resistance of the ammeter are negligible.
1. A typical P I curve is shown below:
I
P
Pmax
I0
A
V
R
E
If the current varies in a sufficiently large interval a maximum power Pmax can be
detected at a certain value, I0, of I. Theoretically, the P(I) dependence is given by:
(5.1) rIEIP 2 ,
where E and r are the EMF and the internal resistance of the dc source respectively. The
maxim value of P therefore is:
(5.2)
r
EP
4
2
max  ,
and corresponds to a current:
(5.3)
r
EI
20
 .
2. The internal resistance is determined trough (5.2) and (5.3) by recording Pmax and I0
from the experimental plot:
2
0
max
I
Pr  .
3. Similarly, EMF is calculated as:
0
max2
I
PE  .
4. The current depends on the resistance of the rheostat as:
rR
EI  .
Therefore a value of R can be calculated for each value of I:
(5.4) r
I
ER  .
The power dissipated in the rheostat is given in terms of R respectively by:
(5.5) 2
2
)( rR
REP  .
The P R plot is given below:
Its maximum is obtained at R = r.
5. The total power supplied by the dc source is:
(5.6)
rR
EPtot 
2
.
R
Ptot
E2/r
R
P
R = r
E2/(4r)
6. The efficiency respectively is:
(5.7)
rR
R
P
P
tot 
 .
R

1
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