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Viktor Chivilev1

Viktor Chivilev1

Associate Professor of Physics. Author of study materials for CPUE.

Professor Victor I. Chivilev ('70) teaches graduate courses in Physics at MIPT since 1977. In addition, every year he writes six to seven new study guides for MIPT's Distance Learning School for high school students. His problem sets and study guides are considered the best in the areas he covers. These include physics of gases and mechanics (dynamics and statics). Professor's speaks about importance of learning through solving physical problems:

"A physical problem may take anything from a few minutes to many years to solve. Solving a hard problem requires a subtle mix of a strong technical background and ability for non-standard thinking. The latter also includes imagination. And the subtle art of teachers of Physics requires composing the right blend of the technicalities and the stimuli for imagination for their students.

How do you measure the scope of knowledge and the quality of professional imagination of a prospective student? How do you estimate the ability to grasp the physical essence of the problem? Every summer we think about these questions when putting together the problems for the Admission exam at MIPT. Here is one of those problems, invented by me a few years ago and used in MIPT's Admission Exam in Physics in 1993:

A soccer player kicks a ball B trying to hit with it a point M on the vertical wall. The distance between the player and the wall is L=32m. After it's kicked, the ball flies with the initial speed of V0=25m/s at an angle a to the horizon. There is no wind before the player kicks the ball. However, right after the ball starts to fly, wind begins to blow at the speed of V1=10m/s. The direction of the wind is horizontal and parallel to the wall. The ball hits the wall. However, because of the wind, it deviates from the mark M by S=2m in horizontal direction and hits point D. Find the time ball was in flight. Assume cosa=0.8. The ball doesn't rotate in flight.

View from above: wind starts to blow as the ball starts to fly. Note: the ball was supposed to hit the wall at right angle BMD if viewed from above. Instead, the ball hits wall at point D. Side view: initial position
>>> Click here to see the solution

Now, you can try to find the answer by solving differential equations (this likely will not work!!). But such a method doesn't require a lot of imagination or in-depth understanding of how the wind affects the ball along its path. However, the one who is ready to stretch his or her imagination and use mental abilities may come up with a SHORT, ELEGANT and PRESICE (without any approximations!) solution, which doesn't require any knowledge of differential and integral calculus.

First of all, I offered this problem for the USSR's National Physical Olympiad, but I COULD NOT convince the Committee to accept it! They simply did not quite understand the logic of my "easy" solution. In about a year after that, I was able to explain the logic of this problem to another committee - this time to the MIPT's Admission Exam committee. And the problem was offered to our prospective students.

Every year in July these new candidates come to Moscow from virtually all regions of Russia to take this quite difficult Exam. Although, as we always stress, no knowledge of Physics beyond our quite limited high-school program is necessary to pass the Admission Exam in Physics, the ways to use even standard knowledge may be quite ingenious depending on the depth of understanding of physical phenomena. We are trying to gauge the ability of the candidates by offering them problems requiring non-standard thinking. As a rule, these problems are not difficult technically and do not take long calculations or overly sophisticated algebra. The ideal problem may be solved in one or two lines of formulas supported by a short explanation. Again, we focus on the physical phenomena as we see it instead of building an overly sophisticated problem structure around them. It takes sometimes a lot of efforts in Physics to enliven a concept, or a problem, or a solution for the problem. That's why I am so happy to work at MIPT: we have here some of the best people in the world, whose profession is to make this all happen."

Professor Chivilev became involved in the Olympiads for High Schools in 1965, on his second year of study at MIPT. Now he is a well-known author of many original problems in general Physics of various levels of difficulty. As far as the Olympiads are concerned, Professor Chivilev's interests include: Inventing new problems and refining the problems prepared by the other MIPT's Professors for three major Olympiads in Physics supported by MIPT annually. Editing of the series of annual collections of the problems recommended for the MIPT's Annual Outreach Olympiad in Physics. Participating on the Commissions in Methodology of All-Russian Olympiads of all levels from Regional to National. Training the Russian National Teams for International Physical Olympiads.


During the last second of its fall, a free falling body travels 8/9 of the length of its total path. What was the height the body fell from? Comment: This problem was offered on the MIPT's 35th Outreach Olympiad in Physics and Mathematics in 1996. It's an easy one. High up in the mountains the temperature is -20 Centigrade (corresponds to absolute temperature T0=253 K), the air pressure is P0 = 40,000 Pascal (1Pascal = 1Pa= 1N/m2). Find the ratio of the air density at normal conditions to the air density high up in the mountings. Normal conditions are defined as such that the temperature is 0 Centigrade (absolute T1 = 273 K) and the air pressure is P1 = 100,000 Pa (one standard atmosphere). Comment: This problem was offered in 1992 on MIPT's 31st Outreach Olympiad in Physics and Mathematics. It's also an easy one. The stew-pan has the form of a cylinder with the bottom being a circle. The area of thiscircle is S = 0.1 m2. The stew-pan is submerged into the water so that its bottom is horizontal and is 4 centimeters below the waterline. Then the bottom of the pan is carefully cut from the rest of the stew-pan. It does not separate from it because of the water pressure. What minimal weight would you put on the bottom to cause it separate from the stew-pan and sink? Where would you put that weight? The geometrical sizes of the weight are very small compared to the bottom of the pan. See the picture for details. Comment: Average difficulty. Distance Learning School Olympiad of MIPT, 1991. The Martian year is k = 1.9 times longer than Earth's. A collecting lens on Earth produces an image of Sun on Earth with diameter d1 = 4 mm. Find the diameter of Sun's image produced by the same lens on Mars. Assume planets' orbits are circular. Comment: Average difficulty. This problem was offered in 1999 on MIPT's Traditional Physical and Mathematical Olympiad. Magnetic field is uniform and has a magnitude B in the interior of a very long solenoid far from its ends. One of the ends of the solenoid is closed with a thin flat plastic cover. A single electrical loop of radius R is laid on the cover so that its center is on the axis of the solenoid. The electrical current flowing in the loop is I. Find the mechanical tension in the loop's wire. Comment: This rather difficult problem was offered on the XXIX Russian National Olympiad in Physics in 1995. The Olympiad took place in Cheliabinsk. You have 5 kg of kerosene (aviation fuel) at 0 Centigrade (absolute T0 = 273) and 2 kg of water at 100 Centigrade (T1 = 373 K). How would you organize the exchange of heat energy between these substances to increase the temperature of kerosene to the maximal possible one? You can not use energy drawn from any other sources. You can't burn kerosene. The specific heat of water is 4,200 J/(kg K), the specific heat of kerosene is 2,100 J/(kg K). Comment: This difficult problem was offered on the Distance Learning School Olympiad of MIPT in 1991. Only a few of several thousands participants solved it.


Approximately 11 meters. Approximately 2.3. The weight must be 2 kg, placed next to the pan's wall. d2 = k2/3 d1 F = BIR/2. 80 Centigrade (absolute T = 353 K).

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