Solutions to the problems of the 5th
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
1
and
w
2
along the
x
and
y
axes:
(1.3)
0
1
1
cos
a
a
w
x

α
=
;
(1.4)
1
1
sin
α
=
a
w
y
;
(1.5)
0
2
2
cos
a
a
w
x

α
=
;
(1.6)
2
2
sin
α

=
a
w
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)
1
1
1
1
1
T
R
g
w
+
+
=
m
m
;
(1.8)
2
2
2
2
2
T
R
g
w
+
+
=
m
m
;
(1.9)
2
1
2
1
0
T
T
R
R
R
g
a


+


=
M
M
.
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
+
T
2
), which the bended thread exerts
on the prism through the pulley
.
The equations of motion result in a system of six
scalar equations when projected along
x
and
y
:
α
1
α
2
x
y
a
0
a
1
a
2
w
2
w
1
R
2
T
2
R
1
T
1
R
M
g
m
1
g
m
2
g
x
y
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document(1.10)
1
1
1
0
1
1
1
sin
cos
cos
α

α
=

α
R
T
a
m
a
m
;
(1.11)
g
m
R
T
a
m
1
1
1
1
1
1
cos
sin
sin

α
+
α
=
α
;
(1.12)
2
2
2
0
2
2
2
sin
cos
cos
α
+
α

=

α
R
T
a
m
a
m
;
(1.13)
g
m
R
T
a
m
2
2
2
2
2
2
sin
sin
sin

α
+
α
=
α
;
(1.14)
2
1
2
2
1
1
0
cos
cos
sin
sin
α
+
α

α

α
=

T
T
R
R
Ma
;
(1.15)
Mg
R
R
R

α

α

=
2
2
1
1
cos
cos
0
.
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
a
0
:
(1.16)
2
2
1
1
2
1
0
cos
cos
α
+
α
+
+
=
m
m
m
m
M
a
a
.
The straightforward elimination of the unknown forces gives the final answer for
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 NA
 Physics, Equations, Velocity, Bulgaria, Sofia, DC source

Click to edit the document details