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ipho_Solutions71

# ipho_Solutions71 - Solutions to the problems of the 5-th...

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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 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

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(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
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ipho_Solutions71 - Solutions to the problems of the 5-th...

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