Popular science physics and mathematics
quantum
12
1970
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

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4. Sections of polyhedra

Now we will consider images encountered when studying stereometry.

Example 4. At figure 11, a the flat shape of a triangular feature is depicted.

It is “constructed” very simply: arbitrary points $A_3$, $B_3$, $C_3$ are taken on the edges of the feature and connected by segments of the feature. At figure 11, $\delta$ the same “method” is applied to the construction of a section of a quadrangular feature.

Figure 11 is correct, but on figure 11, $\delta$ there was a mistake (a very common one!). In fact, the plane is determined by the third point. Therefore, when depicting the plane shape of a triangular feature, we can take points on its edges arbitrarily. You cannot do this with a quadrangular sign. Noting the pro-

arbitrarily the points $A_3$, $B_3$, $C_3$, we have already defined the cutting plane, and the point of its intersection with the four edges of the recognition cannot be taken arbitrarily.

We will show you how to correctly find the point of intersection of a cutting plane with four edges.

The plane passing through the edges $^*$) $A'_1A'_2$ and $C'_1C'_2$ intersects the plane section $A'_3B'_3C'_3D'_3$ along the diagonal $A'_3C'_3$, and the plane passing through the edges $B'_1B'_2$ and $D'_1D'_2$, along the diagonal $B'_3D'_3$. The intersection of these two planes is parallel to the edge of the plane. Consequently, in all plane sections of the line the points of intersection of the diagonal line lie on a straight line parallel to the edge of the line. At figure11, $\delta$ the points $P_1$ and $P_2$ of intersection of the diagonal bases are marked; The yama $P_1P_2$ is parallel to the edge of the yam. Point $P_3$ does not lie on this line -

*) Do not forget that the elements of the original are indicated by letters with dashes.

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Fig. 13.
therefore figure is incorrect. This means that the quadrilateral $A_3'B_3'C_3'D_3'$ depicted on it is not flat. A person with a trained eye, for example, an artist, will notice this immediately, without any construction.

This property allows you not only to check the correctness of the drawing, but also to construct a planar view of a quadrangular feature. Let us arbitrarily mark the points $A_3$, $B_3$, $C_3$ (Fig. 12). We construct points $P_1$ and $P_2$ and draw a straight line $P_1P_2$ (it turns out to be parallel to the straight line $A_1A_2$). We carry out a straight line $A_3C_3$. We find the point $P_3$ of intersection $A_3C_3$ and $P_1P_2$. We are holding $B_3P_3$. The intersection point $B_3P_3$ with $D_1D_2$ is the desired one.

Depicting flat sections of polyhedra requires the use of only two rules:

1. If planes $\beta'$ and $\gamma'$ intersect along line $l'$, and plane $\alpha'$ intersects their corresponding lines $b'$ and $c'$, then $b'$ and $c'$ either intersect at a point lying on $l'$ or are parallel to $l'$.

2. If planes $\beta'$ and $\gamma'$ are parallel, and plane $\alpha'$ intersects their corresponding lines $b'$ and $c'$, then $b'$ and $c'$ are parallel.

Example 5. Image of a flat section of a parallelped.

We draw images of linear sections $RQ$ and $QP$. The point of their intersection must lie on the line $A_2B_2$, otherwise the lines $PQ$ and $QR$ are arbitrary. Next we carry out $RS \parallel QP$ and $PS \parallel QR$.

On figure 13 the lines $PQ$ and $QR$ are drawn so that the point $Q$ falls outside the segment $A_2B_2$. Mark the points $T \equiv PQ \times A_2D_2$ and $U \equiv QR \times A_1A_2$ and connect $U$ and $T^*$). We carry out $RS \parallel QP$ and $PS \parallel QR$. It is clear that the point $S$ must lie on the line $C_1D_1$. If this point falls on the continuation of the segment $C_1D_1$, then we connect the points $V \equiv RS \times B_1C_1$ and $W \equiv PS \times C_1C_2$. Analyze on your own the case when the point $Q$ lies inside the segment $A_2B_2$.

Example 6. Image of a plane section of a quadrangular paramid (Fig. 14).

We carry out links $PQ$ and $QR$. From point $R$ it is necessary to draw a section line along the rear edge. The planes of the front and rear faces intersect along a straight line $S'X'$. The image of this line can be easily obtained by constructing the point $X \equiv AB \times CD$. The traces of the cutting plane $\alpha'$ on the front and back faces must intersect at $S'X'$. We find point $T \equiv PQ \times SX$ and draw $TR$. We mark the segment of this straight line $RU$, which is the interior of the triangle $SCD$. And $AB$ represents the line of intersection of the front and bottom edges. We build point $Y \equiv AB \times PQ$ and find point $V \equiv AD \times UY$.

*) The notation $M \equiv AB \times CD$ means: “M is the point of intersection of the lines $AB$ and $CD$”, and $m \equiv A \times B$ - “m is the line defining the points $A$ and $B$”.

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Fig. 17.
Each ellipse shows two diameters: in the original they were redistributed, and in the image they are conjugate (see the article by I. N. Bronshtein “Ellipse”, Quantum No. 9).

The base of the cone is also depicted as an ellipse. In school practice, the following mistake is often encountered: the extreme generatrices are used to depict the main section, and it is assumed that they touch the ellipse at the ends of the major axis (Fig. 18, a). This is not laziness. If you draw tangents to the ellipse from point S and connect tangency points A and B, then straight line AB will not pass through the center of the ellipse. However, some

Fig. 19.
The “draftsmen” take her through the center of the violent. Figure 18, b - correct. One of the points of tangency, for example A, is connected to the center O, and point C is marked on the ellipse, diametrically opposite to A. Triangle SAC is an image of its own section.

To depict a master section, it is not at all necessary to use

Fig. 18, a.
Fig. 18, b.
Fig. 18, p.

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Fig. 1.

ma or group of atoms is transferred to neighboring…

When one body moves relative to another, touching it, the atoms of these bodies “cling” and swing each other. If you are not afraid of vulgarization, then this process can be imagined like this: a person runs away from the pendants hanging nearby and swings them. Or better yet: the pendants are interconnected. By swinging one or more, all the others are forced to move (Fig. 1).

Vibrations of atoms located on the surface of the body are transmitted to atoms located in the depths. The transfer of energy from the surface to the depths of the body is a complex process that has not yet been fully traced (although its general features are understood). The path of energy transfer depends significantly on what the solid is. For example, in a metal, electrons can be carried away. But, whatever the path of energy removal from the surface, ultimately

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The interaction of one body with another leads to an increase in temperature. In addition, during the friction process, a change in the structure of the rubbing surfaces occurs. They must be taken into account if you want to understand all the phenomena accompanying trenna *). However, as long as we are not trying to construct a quantitative theory, it is not very important for us to know exactly what the process is in a solid. It is important to understand: the energy of a moving body is partially spent on exciting the movement of atomic particles that make up solids.

We can say that corpses “work”, only this work is done in vain - energy is wasted on heating the body, which most often comes to be artificially cooled (a Stone Age man, accustomed to adding fire to the morning, would not agree with us).

The greater the mutual speed of the bodies, the more energy is spent on rocking the atoms, which means the more work the bodies produce. On the other hand, it is known that work per unit time (power) is equal to force multiplied by speed. This force is the force of trenya.

We spring formally “introduced” the power of trenya. True, they tried to explain its nature.

You can do it a little differently. Find out how much momentum (amount of motion) is expended by a moving body per unit time on the excitation of atomic vibrations. According to Newton’s law, the change in the momentum of a body per unit time is equal to the force acting on it. This gives us the opportunity to determine the strength of trenya. Of course, both definitions lead to the same meaning of force.

The force of trenya characterizes the interaction between macrosoppe bodies, and although trenna is the result*

*) Teams of scientists are occupied with the problem of trenya, using the most modern research methods and the latest ideas about the movement of atomic particles in solids.

Because of the interaction of individual atomic particles, the frictional force is strikingly different from the forces that act between atoms and molecules. Indeed, the forces acting between atoms and molecules depend only on the distance between them, and the force of friction depends on the relative speed of the body. However, this is not the only reason why the frictional force differs from the forces of interaction between microsoptic bodies. It has properties called by different words, the meaning of which is the same:

Irreversibility, dissipation, loss

Under the influence of the force of training, the body always loses energy and is always thermalized. This is due to the fact that the energy of movement of bodies during friction is spent on exciting the movement of atomic particles. The atoms are unimaginable (10^23 particles per 1 cm³), they move randomly. The energy transferred to it is simply dissipated (dispersed, from the French word “dissiparer” - to dissipate) between countless atoms. The new owners of energy continuously exchange it with each other and are not at all inclined to repay the debt to the macrosoptic body. The described process is, of course, a transition from integral energy to warm energy. We see that this process is irreversible.

A fruitful method of generalization is that one of the main properties inherent in a given concept is taken as its new definition. Let us use this method to generalize the concept of trenya. The most important property of trenya, as we have seen, is irreversibility associated with the transition of the energy of ordered motion into heat. The acceptance and rejection of this property will be called tren any irreversible process that accompanies the transition of the energy of ordered motion into warm energy. Terrier, we can talk about the friction of two counter flows of gas - they are mutually thermalized and heated, about

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friction of a solid body against a gas, etc.

What about the friction force? It is still accompanied by a friction process. In fact, during the process of friction, the energy of ordered motion decreases, which means its speed also decreases. In other words, acceleration is directed in the direction opposite to the speed of the body.

Let’s remember Newton’s second law. Where there is acceleration, there is force. This force is the friction force. It is directed in the direction opposite to the speed, it always slows down…

Simple case

There are cases when the calculation of siliprene is relatively simple and does not require a deep penetration into the theory of solids.

Let, for example, a solid body move through a fairly rarefied gas. Then silipren is caused by collisions of gas atoms with the surface of the body.

Before proceeding to the calculation of silipren, we will consider a separate act of collision. A particle with momentum $P$ falls on the surface of the body. The collision of an atom with a surface is a complex event. Let’s think about what could happen. Atom can remain on the surface (float), can penetrate deeply into the body, or can be reflected from the surface of the body. With reflection (we will be interested in this particular silipran, we will explain how much later) various situations are also possible. For example, an Atom can innise or be knocked out of an atom from a surface, a molecule can disintegrate into its constituent atoms, and so on. In short, a lot will appear, but we will assume that atom was simply reflected, like a ball from a wall. Not for simplification, although this situation is indeed the simplest, but because in the overwhelming majority of silipevs, when a body moves through a gas not too quickly, this is exactly what happens - the atoms are simply reflected (bounced off) from the surface of the body ). Large complex situations (the introduction of atoms into a solid body, nucleation, etc.), of course, are also observed and studied in detail. Each of these processes occurs with most probability at a certain particle speed. For example, the higher the particle speed, the greater the likelihood that an atom will penetrate to a relatively greater depth into a solid body. To imagine what speeds we are talking about, we note: an oxygen atom incident on the surface of a solid body at a speed of $10^7$ cm/sec*) penetrates approximately 10-12 atomic distances into the depth of the solid body. The average speed of $v$ atoms of a gas with a mass $m$, as shown by the molecular-kinetic theory of gases, is equal to $\sqrt{\frac{kT}{m}}$, where $T$ is the temperature in absolute units, and $k$ is the so-called Boltzmann constant, equal to $1,4 \cdot 10^{-23}$ J/sec. At $T = 300^\circ$ K ($=27^\circ$ C) the average speed of an oxygen atom is less than $10^5$ cm/sec. If a body accumulates in a gas, then the average speed of gas molecules with the surface of the body approximately coincides with the average body speed; if the body moves, it differs from it. However, in the case when the speed of a moving body is small compared to the average solid speed (you should never go through the minimum partial silipevs, in which the question under study looks simpler), the speed of the moving body is almost no different from the solid body, and the energy of the atom is simply not enough to penetrate into the depths of the solid body.

Although we called the essence of reflection of atoms from the surface of a solid body the simplest, it also needs to appear…

*) Fortunately, very rarely water molecules from the air float to the flying person. Otherwise, the description of glaciation would have increased many times.
**) The kinetic energy of oxygen atom is equal to 10 keV (see below).

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Fig. 3.
Now we have really moved away from the subject that interests us.

What is the friction force?

So, let’s return to the calculation of siliprene acting on a body moving at a relatively low speed $u$ through a rarefied gas, the temperature of which is $T$. It is convenient to consider the body to be at rest and the gas to be moving at speed $-u$. Let us select an infinitesimal area $^*$ $\Delta S$ (Fig. 3) on the surface of the body. The momentum of a gas molecule, if it collides elastically with a body at a point on the visible area, will change to $2P_n$, where $P_n$ is the projection of the molecule’s momentum onto the normal to the surface of the body. This means that a gas molecule, when colliding with a body, gives it an impulse equal to $2P_n$. Now our task is to calculate the total impulse transmitted by the gas to the body per unit time. The change in momentum per unit time is the force (in this case, acting from the gas on the body). To calculate the total transmitted momentum, it is necessary, firstly, to take into account all the molecules falling per unit time onto a given area, and secondly, to sum over all areas.

*) See regarding the “infinitesimal platform” the article by N. Ya. Vilenkin “Secrets of Infinity” («Kvant» No. 3, 1970).

A similar operation was carried out in the article by Ya. A. Smorodinsky “Ideal Gas” («Kvant» No. 10). Here we will use ready-made results. The gas pressure on a wall at rest is equal to $1/3$ nmv², where $n$ is the number of gas molecules per unit volume, $m$ is the mass of the molecule and $v^2$ is the arithmetic mean of the squares of the velocities of the molecules. If the body ends ($u = 0$), the number of blows to the right and left is on average the same, and therefore no impulse is transferred to the body. At $u \neq 0$ this is not the case. Let the body move. In this case, the speed of molecules hitting the front is $u$ greater, and the speed of molecules hitting the rear is $u$ less. Therefore, the force acts on section $\Delta S$ of the moving plane

$$F = (P_{\text{спереди}} - P_{\text{задди}}) \Delta S = \frac{1}{3} nm \Delta S \left[ (v + u)^2 - (v - u)^2 \right] = \frac{2}{3} nmvu \Delta S.$$

Recalling that $v = \sqrt{\frac{3kT}{m}}$ (see the article “Ideal Gas”), we obtain

$$F = \frac{2}{3} n \sqrt{3kTm} u \Delta S.$$

For a body of arbitrary shape we must obtain the formula

$$F = \beta Sn \sqrt{mkT} u.$$

Here $S$ is the surface area of ​​the body, and $\beta$ is a numerical coefficient, depending on the shape of the body. It takes into account that molecules enter one area perpendicular to the surface, and others at an angle.

If we use Clapeyron’s law $PV = NkT$ ($N = nV$ is the total number of the gas part in the volume $V$), then we can express the friction force through the pressure $P$ and temperature $T$ of the gas:

$$F_{\text{Tp}} = \beta SP \sqrt{\frac{m}{kT}} u.$$

The result was natural: at fixed temperatures and

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speed of the body, the greater the friction force, the greater the gas pressure.

Gas viscosity

We have described in some detail why, when a body moves through a gas, a force acts on it - the force of friction - the result of the impacts of individual molecules on the surface of the moving body. At the same time, we were not at all interested in what happens to the gas, to its molecules after the collision. Let us think for a moment about the fate of the molecules that come into contact with the body. It is clear that if we had the opportunity to change the speed of all molecules interacting with a body, we would be convinced that their average speed is not equal to zero. Molecures, albeit slightly, are carried away by a moving body. And if you change the speed of molecules far from the body, then (in the absence of finite wind, their average speed is zero). In slightly different words: near the surface of a moving body there is an ordered movement of gas, but far from the body there is none. There is a certain impulse associated with ordered movement. A layer of gas in the surface layer collides more often with a moving body than other gas molecules. Because of this, their average momentum would have to increase if there were no process of momentum removal from the boundary layer.

This process is easy to understand. Gas molecules that escape from the boundary layer simply carry away excess momentum with them. Far from the body, they “disperse” it in multiple collisions with other molecules or in collisions with bodies at rest.

The ability of a gas to transfer impulse from one place to another and, most importantly, to waste it irreversibly is called viscousness. Viscosity and friction are the same. Viscosity is an interesting phenomenon in which collisions between molecules play a major role. We neglected when calculating the friction force acting on the moving body from the gas side.

We were wrong about this. But how? Did you exaggerate or underestimate the friction force? It’s easy to understand that they exaggerated. Indeed, we believed that a molecule, once in contact with a body, would never return to it and, therefore, would not be able to “return” to the body the impulse it received. In fact, if we carefully monitor many molecules colliding with the body, we will be convinced that some of them return to the body (after colliding with any of the gas molecules), others give the excess implus acquired from the body to their collide partners, and they, in turn, can collide with the body. Strictly speaking, as a result of such new molecules interacting with each other, a partial return of the body’s impulse occurs, which means that the body loses less impulse than we thought.

The formula derived here is applicable only to very rarefied gases, in which the average stretch between stolnov molecules is much greater than the size of the body. If we calculate the friction force acting on dust particles with dimensions not exceeding several microns, our air may also fit into the category of rarefied gases.

ABOUT THE ORDIBAL PERIOD OF SATELLITES

In September 1970, the automatic station “Luna 16” was launched. On September 17, it entered a lunar orbit with a period of 1 hour 59 minutes (119 minutes). What can be said about the Moon based on these data?

Let us remember that when a satellite is launched around the Earth, its period is usually about one and a half hours (90 minutes). What can you say about the Earth?

It turns out that from these data it is possible to estimate the densities of the Moon and Earth.

If the height of the satellite is small (compared to the radius of the celestial body around which it rotates), then we can approximately assume that the satellite is moving near the surface. Let us equate the acceleration of gravity to the centripetal acceleration:

$$G \frac{M}{R^2} = \frac{v^2}{R} = \left(\frac{2\pi}{T}\right)^2 R,$$

where $G$ is the gravitational constant, $M$ is the mass of the celestial body, $R$ is its radius, $v$ is the speed of the satellite, $T$ is its period of revolution. Next we write the mass in terms of the radius:

$$M = \frac{4}{3} \pi R^3 \rho,$$

where $\rho$ is the average density.

Substituting this expression into the previous formula, we get a remarkable result:

$$T = \sqrt{\frac{3\pi}{\rho G}}.$$

It follows that the orbital period of a low satellite depends only on the size of the celestial body and does not depend on its size. Therefore, knowing the period, you can determine the density of the Moon and Earth. The table shows the period and density values ​​for different planets of the solar system.

| Central body | Density ($\rho/cm^3$) | Period |
| :— | :— | :— |
| Sun | 1,42 | 166 |
| Mercury | 5,61 | 83 |
| Venus | 5,16 | 87 |
| Earth | 5,52 | 84 |
| Moon | 3,36 | 108 |
| Mars | 3,95 | 100 |
| Jupiter | 1,34 | 171 |
| Saturn | 0,64 | 238 |
| Uranus | 1,36 | 170 |
| Neptune | 1,30 | 174 |

For Earth, the period is 84 minutes. In gases, a slightly larger overall value is reported. This means that the satellite is not flying very low. If we denote the height of the satellite by $h$, then it is not difficult to obtain the exact formula for a circular orbit (try to derive it yourself):

$$T = \sqrt{\frac{3\pi}{\rho G}} \left(1 + \frac{h}{R}\right)^{\frac{3}{2}} \approx T_0 \left(1 + \frac{3h}{2R}\right),$$

where $T_0$ is the period of the low satellite (given in the table), and the last equality is an approximate one - true if $h \ll R$. You can also obtain a formula for an elliptical orbit if you know Kepler’s laws. You will read about them in «Kvant» No. 1 for 1971.

Y. A. Smorodinsky

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by the cosine theorem from triangle ABD
$$BD^2 = 4x_0^2 + 64 - 16x_0;$$
on the other hand, the triangle inequality gives
$$BD < BC + CD = 6x_0 - 8,$$
where
$$4x^2 + 64 - 16x < (6x - 8)^2, 2x^2 - 5x > 0,$$
and taking into account $2 < x_0 < 8$ we obtain $x_0 > \frac{5}{2}$. A more accurate value of the distance $AB + BC$, according to the author of this letter, is: 11.044 versts.

Some readers misunderstood the conditions of the problem and believed that $\angle D = 90^\circ$, that is, that triangle ADC is equilateral and right-angled. This is, of course, not true.

M18. a) Prove that for any point $M$ of the circle circumscribed about a regular triangle ABC, one of the three segments MA, MB, MC is equal to the sum of the other two.
6) When equal, the circles $\gamma_1, \gamma_2, \gamma_3$ touch each other in pairs, and a circle $\gamma$ is described around them, which touches all three: $\gamma_1, \gamma_2$ and $\gamma_3$. Prove that for any point $M$ of circle $\gamma$, the tangent drawn from point $M$ to one of the three circles $\gamma_1, \gamma_2, \gamma_3$ is equal to the sum of tangents drawn from point $M$ to the other two circles.

Solution a) Let the point $M$ lie on the arc $AB$ of the circumscribed circle. Let’s rotate triangle AMB around point $A$ by $60^\circ$ so that it takes position AKC (Fig. 3). Since angles MBA and MCA are equal (they rest on the same arc AM), point $K$ lies on segment MC. Since AM = AK and $\angle MAK = 60^\circ$ (by construction), the triangle AMK is equilateral, therefore MK = AM. So,
$$MA + MB = MK + KC = MC.$$

Another solution is obtained immediately from Ptolemy’s theorem, according to which in any cyclic quadrilateral the product of the diagonals is equal to the sum of the product of the opposite sides: if in the equality $MC \times AB = MA \cdot BC + MB \cdot AC$ we replace $BC$ and $AC$ by $AB$, we obtain $MC \times AB = (MA + MB) AB$, whence $MC = MA + MB$.

Our readers have solved this problem a lot. A. Braverman from Perm, A. Veselov from the Kalinin region, V. Metlitsky from Angarsk, N. Mirolyubova from Aktyubinsk, O. Khudaverdyan from Evrev even added several solutions.

6. Let us prove a more general fact. Let the centers of three circles of radius $r$ be the vertices of an equilateral triangle $C_0C_1C_2$ and $O$ the center of a circle of radius $a$ circumscribed around it. Then for any point $M$

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of a circle of radius $R = a + r$ with center $O$, the tangent drawn from the point $M$ to one of the two circles of radius $r$ is equal to the sum of the tangents drawn from the point $M$ to the other two circles. Note that for $r = 0$ the result of problem a) is obtained.

Let the radius $OM$ form angles $\alpha_k = \alpha_0 + k \cdot 120^\circ$ (Fig. 4) with the radii $OC_k$ ($k = 0, 1, 2$); then $MC_k^2 = R^2 + a^2 - 2Ra \cos \alpha_k^*$), and the square of the tangent drawn from the point $M$ to the corresponding circle is $MC_k^2 - r^2 = R^2 + a^2 - 2Ra \cos \alpha_k - (R - a)^2 = 2Ra (1 - \cos \alpha_k)$.

Therefore the tangents are equal to $2\sqrt{Ra} \sin \frac{\alpha_k}{2} \left( \sin \frac{\alpha_k}{2} \geq 0, \text{ поскольку } 0 \leq \alpha_k < 2\pi \right)$. It remains to check the equality $\sin \frac{\alpha_0}{2} + \sin \frac{\alpha_2}{2} = \sin \frac{\alpha_1}{2}$. It’s already simple: $\sin \frac{\alpha_0}{2} + \sin \frac{\alpha_2}{2} = 2 \cos 60^\circ \cdot \sin \frac{\alpha_0 + \alpha_2}{4} = \sin \frac{\alpha_1}{2}$.

M. Preezer sent a solution to this problem using analytical geometry. A. Braverman sent us an interesting proof (using Ptolemy’s theorem) of the following generalization of problem a): let $M$ be a point on the arc $A_1A_{2n+1}$ of a circle circumscribed about a regular ($2n + 1$)-gon $A_1A_2A_3 \dots A_{2n}A_{2n+1}$; Then

$$MA_1 + MA_3 + \dots + MA_{2n+1} = MA_2 + MA_4 + \dots + MA_{2n}.$$

(The solution is to write the Ptolemaic equations for the quadrilaterals $MA_1A_2A_3, MA_2A_3A_4, \dots, MA_{2n-1}A_{2n}A_{2n+1}, MA_{2n+1}A_{2n}A_1, MA_{2n+1}A_2A_1$, placing all terms containing $MA_{2k}$ on the left side of the equality, and all terms containing $MA_{2k+1}$ on the right side, and then adding all the equalities.) Consider how to prove this statement and a similar generalization of problem 6) using trigonometric functions.

M19. In an endless chain of faithful cells, each cell can be in one of two states: “rest” and “excitement”. If at a given moment a cell is excited, then it sends a signal that, after a unit of time (say, one millisecond), reaches both cells neighboring it. Each cell excites if and only if it receives a signal from one of its neighboring cells; if the signal arrives simultaneously from both sides, then they are extinguished, and the cell does not excite.

Let only one cell be excited at the initial time $t = 0$. How many cells will be in an excited state after 15 months? in 65 months? in 1000 months? in $t$ months?

What will happen if the chain is not infinite, but contains only $N$ cells connected in a circle - will the excitation remain infinite for a long time or will it die out?

Solution. It is enough to trace how the excitation from a single cell spreads above ten to fifteen clock cycles (Fig. 5) to notice the following patterns.

1. At time $t = 2^k$, where $k = 0, 1, 2, \dots$, only two cells are excited: $x = -2^k$ and $x = 2^k$.

2. At time $t = 2^k - 1$, $2^k$ cells from $x = -2^k + 1$ to $x = 2^k - 1$ are excited, with $x$ odd.

3. Let $0 \leq t < 2^k$. then at time $2^k + t$ there are twice as many excited cells as at time $t$.

To prove these regularities (by induction on $k$), we can use the fact that the “excitation waves” generated by each of the two cells active at time $2^k$ do not overlap until time $t = 2^{k+1} - 1$ and therefore each of them is structured exactly the same as the excitation wave from one cell at $t < 2^k$.

*) $\cos(2\pi - \alpha) = \cos \alpha$, so the formula is also valid for $\alpha > \pi$.

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bones. To do this we can use the law of conservation of energy
$$\frac{mv^2}{2} + mgh = \frac{mu^2}{2}.$$
Hence $u = \sqrt{v^2 + 2gh}$.

Now you can find the angle at which the block will move on the bottom
half-plane:
$$\sin \beta = \frac{v_1^2}{u} = \frac{v}{\sqrt{v^2 + 2gh}} \sin \alpha.$$
This solution was sent by G. Zaitsev, a 9th grade student from Gagra, GSSR.

F27. The hemispherical bell, lying tightly on the table, is poured through the hole.
at the top of the water Fig. 11. When the water reaches the hole, it lifts the bell and begins to flow out. Find the mass of the bell if its radius is $R$ and the density of water is $\rho$.

Solved and e. The force of pressure on the table is equal to the sum of the forces of gravity acting on the bell and water: $F = Mg + \frac{4}{3} \pi R^3 \rho g$. But at the moment when the water comes
lifts the bell and begins to leak, the bell itself does not press on the table. Therefore, the force of pressure on the table is equal to the water pressure multiplied by the area
base of the vessel. The water pressure on the table is the same at all points and
equals $\rho g R$, so $F = \rho g R \cdot \pi R^2$.

This means $\pi R^3 \rho g = Mg + \frac{4}{3} \pi R^3 \rho g$. Hence $M = \frac{1}{3} \pi R^3 \rho$.

An interesting idea for a solution was proposed by Anatoly Vorovlyansky from Gorky.

Suppose the bell is placed in a cylindrical vessel of radius and height. Let’s fill this vessel with water and assume that the mass of the bell is negligible.
It is easy to report that since the pressure on the bell equipment and from the inside at all points
equal to each other (this follows from Pascal’s law and from the conditions that the bell is very thin),
then the balance of water will not be disturbed, if the bell is at all. Water pressure on the table
In this case it won’t change either. But this means that the force of water pressure that we put into a cylindrical vessel acts in the same way as a bell, which means that the mass of the bell is exactly equal to the mass of the added water

$$M = (V \text{ цилиндра} - V \text{ полушара}) \rho = \left( R \pi R^2 - \frac{2}{3} \pi R^3 \right) \rho = \frac{1}{3} \pi R^3 \rho.$$

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ONCE AGAIN ABOUT EQUATIONS
WITH PARAMETERS

You have already come across equations and inequalities that contain one parameter more than once. What if there are more than one parameters? It is natural to expect that the number of possible cases to be considered will increase. None of the possible options should be missed. How to do this? There is no single method suitable for any equations and inequalities with parameters. In this article we will show with examples how equations with two parameters are solved.

The general form of the equation for $x$ with two parameters $a, b$ is:

$$F(x; a; b) = 0.$$

(1)

Equation (1) for a given function $F$ is, in fact, an infinite set of equations, each of which is determined by an ordered pair of numbers $(a; b)$.

As is known, such an ordered pair can be interpreted as a point on the coordinate plane. This makes it possible to establish a one-to-one correspondence between equations of the form (1) for a given function $F$ and points of the plane.

To do this, we introduce the coordinate system $aOb$ and agree on how to assign an equation of the form (1) to a point in the plane $aOb$ with coordinates $(a; b)$, and vice versa. Let’s turn this plane into a plane of parameters. Such a correspondence is somewhat unusual (the equation corresponds to a point), but it will help us in each specific case to find a set of permissible parameter values ​​and not to miss a single case: to which point of the plane of parameters an equation of the form (1) corresponds and a certain answer must correspond.

Example 1: Solve the equation

$$\frac{2a + b}{a + x} - \frac{2a - b}{a - x} = \frac{2a}{b}$$

(2)

Solution. Let us immediately note that for $b = 0$ equation (2) does not make sense, that is, for any pair of numbers of the form $(a; 0)$, where $a$ is any real number, there are no solutions (red line on figure 1).

Let $b \neq 0$. Let’s find the ODZ: $x \neq \pm a$, get rid of the denominator and bring similar terms. We get

$$a(x^2 - 2bx + b^2 - a^2) = 0.$$

(3)

If $a = 0$, then equation (3) has its solutions for any values ​​of $x$. Of these, for equation (2), the values ​​$x = \pm a$, that is, $x = 0$, are prohibited.

Thus, for any pair of numbers of the form $(0; b)$, where $b \neq 0$, equation (2) has its solutions to any numbers except 0 (green line). Let now $a \neq 0$. Solving the equation $x^2 - 2bx + b^2 - a^2 = 0$, we obtain $x_1 = b + a$, $x_2 = b - a$.

*) In the form (1) this equation is: $$\frac{2a + b}{a + x} - \frac{2a - b}{a - x} - \frac{2a}{b} = 0.$$

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Can we assume that for any $a, b$ different from zero, the found values ​​$x_1$ and $x_2$ are the roots of equation (2)? No, or it may turn out that for some values ​​of the parameters $x_1$ or $x_2$ will coincide with the forbidden values ​​$\pm a$.

So, it remains to find out when $x_1 = \pm a$ or $x_2 = \pm a$. If $x_1 = a$ or $x_2 = -a$ then $b = 0$, but we are considering the case $b \neq 0$. So, $x_1 \neq a$, $x_2 \neq -a$.

Let $x_1 = -a$, that is, $b + a = -a$, whence $b = -2a$. But then $x_2 = -3a$. Thus, if the parameters $a$ and $b$ are such that $b = -2a$, then equation (2) has a single root $x = -3a$ (blue line at figure 1). Let $x_2 = a$, that is $b - a = a$. This is possible with $b = 2a$.

But then $x_1 = 3a$. Thus, if the parameters $a$ and $b$ are such that $b = 2a$, then equation (2) has a single root $x = 3a$ (yellow line).

To summarize, we get the following answer:

if $b = 0$, $a$ are any real number, then there are no solutions;
if $b \neq 0$, $a = 0$, then $x$ is any real number other than zero;
if $b = -2a$, $b \neq 0$, then $x = -3a$;
if $b = 2a$, $b \neq 0$, then $x = 3a$;
if $a, b \neq 0$, $b \neq \pm 2a$, then $x_1 = b + a$, $x_2 = b - a$.

Example 2*). Solve the equation

$$\sqrt{a^2 - x} + \sqrt{b^2 - x} = a + b.$$

(4)

Solution. ODZ is determined by a system of inequalities

$$x \leq a^2, \quad x \leq b^2, \quad \text{или } x \leq \min\{a^2; b^2\}.$$

The left side of equation (4) is non-negative, and therefore for $a + b < 0$ this equation cannot have solutions. On figure 2 this is marked by the fact that the result under the straight line $b = -a$ is colored pink.

Let us now consider the case $a + b \geq 0$. After estimated transformations we obtain

$$x(a + b)^2 = 0$$

(5)

Let’s consider two cases:

I case: $a + b > 0$. Then $x = 0$. In this case, conditions $x \leq a^2$ and $x \leq b^2$ are satisfied automatically, which cannot be said about condition $x \geq -ab$. If $ab > 0$, then this condition is very much satisfied. If $ab < 0$, then $-ab > 0$ and $x = 0 < -ab$. Therefore, at figure 2 in zones II and III (where $a + b > 0$, but $ab < 0$) there are no solutions. So $x = 0$ is the root

*) This example is taken from the book by N.P. Antonov et al. “Collection of problems in elementary mathematics” (Moscow, “Nauka”, 1969), where it was solved incorrectly.

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Fig. 2.

Equation (4), only if $a + b > 0$ and $ab \geq 0$, that is, for the points of the first four plane of parameters, including the points of its boundary (zone $I$ to figure 2).

P case: $a + b = 0$. In this case, equation (5) becomes the identity $x \cdot 0 = 0$, but this does not mean at all that any real number is the root of the original equation (4) (explain yourself why this is so).

For $a + b = 0$ we successively find $b = -a$, $b^2 = a^2$, and equation (4) takes the form $2\sqrt{a^2 - x} = 0$, whence $x = a^2 = b^2$, and all conditions are obviously satisfied. At figure 2 this answer is indicated near the blue line $b = -a$.

Combining all the results obtained, we get the answer:
if $a + b > 0$ and $ab \geq 0$, then $x = 0$;
if $a + b = 0$, then $x = a^2$;
if $a + b < 0$ or $a + b > 0$ and $ab < 0$, then there are no solutions.

Example 3. Solve a system of equations

$$tg x + tg y = a, \quad x + y = b.$$

Solution. Let’s note the ODZ: $x, y \neq \frac{\pi}{2} + n\pi$ (hereinafter $k, l, m, n = 0; \pm 1; \ldots$). Let us transform the first equation of the system to the form $\frac{\sin(x + y)}{\cos x \cos y} = a$ and then, taking into account the fact that $x + y = b$, we obtain

$$a \cos x \cos y = \sin b.$$

I case: $a = 0$. It naturally breaks down into two options: I, a) $a = 0$, $\sin b = 0$; I, 6) $a = 0$, $\sin b \neq 0$.

If $sin b = 0$, that is, $b = m\pi$, then equation (7) satisfies any pairs $(x; y)$, and system (6) satisfies any pairs $(x; y)$ satisfying the following conditions:

$$x \neq \frac{\pi}{2} + n\pi; \quad y \neq \frac{\pi}{2} + n\pi; \quad x + y = b.$$

If $sin b \neq 0$, that is, $b \neq m\pi$, then equation (7), and with it system (6), has no solution.

II case: $a \neq 0$. Then equation (7) is transformed to the form

$$\cos x \cos y = \frac{\sin b}{a} \text{ и, далее,}$$

$$\cos (x + y) + \cos (x - y) = \frac{2 \sin b}{a}.$$

If we again take into account that $x + y = b$, then equation (8) can be rewritten as follows:

$$\cos (x - y) = \frac{2 \sin b}{a} - \cos b.$$

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Further alienation depends on the assessment of the right side of equation (9):
if $\left| \frac{2 \sin b}{a} - \cos b \right| > 1$, then equation (9), and with it system (6), has no solution; if $\left| \frac{2 \sin b}{a} - \cos b \right| \leq 1$, then
$$x - y = \pm \arccos \left( \frac{2 \sin b}{a} - \cos b \right) + 2k\pi.$$
Solving the last equation in the system with equation $x + y = b$, we obtain
$$\begin{cases} x = \frac{b}{2} \pm \frac{1}{2} \arccos \left( \frac{2 \sin b}{a} - \cos b \right) + k\pi, \\ y = \frac{b}{2} \pm \frac{1}{2} \arccos \left( \frac{2 \sin b}{a} - \cos b \right) - k\pi. \end{cases}$$
We just have to find out whether it could happen that for some parameter values ​​the found values ​​$x, y$ will coincide with the forbidden values ​​$\frac{\pi}{2} + n\pi$. It is easy to see that this will happen in the case when $b \pm \arccos \left( \frac{2 \sin b}{a} - \cos b \right) = \pi + 2l\pi$. Let’s rewrite the last equality as follows: $\pm \arccos \left( \frac{2 \sin b}{a} - \cos b \right) = \pi - b + 2l\pi$, and take the cosine from both sides. We get
$$\frac{2 \sin b}{a} - \cos b = -\cos b, \quad \sin b = 0, \quad b = m\pi.$$
So, if $b = m\pi$ and $a \neq 0$, then system (6) will not have a solution. Combining all the results obtained, we write down the answer:
if $a = 0, b = m\pi$, then the solution to system (6) is any pair $(x, y)$ as well as $x \neq \frac{\pi}{2} + n\pi, \quad y \neq \frac{\pi}{2} + n\pi, \quad x + y = b$;
if $a = 0, b \neq m\pi$, or $a \neq 0, b = m\pi$, then there are no solutions;
if $a \neq 0, b \neq m\pi$ but $\left| \frac{2 \sin b}{a} - \cos b \right| \leq 1$, then there are no solutions.

We limited ourselves to considering examples with only two parameters. Further notification of the number of parameters, without giving anything fundamentally new, only leads to the notification of technical difficulties.

Exercises

1. Solve the following equations:
2. Solve the system of equations:

a) $a^2 - \frac{a^2 - b^2}{2x - x^2} = \frac{b^2 (x + 2)}{x - 2}$;
b) $\frac{\sqrt{a - x} - \sqrt{x - b}}{\sqrt{a - x} + \sqrt{x - b}} = \frac{a - b}{a + b - 2x}$;
b) $\frac{\mathrm{tg} ax}{\sin bx} = 0$.

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HOW TO CHECK ANSWER

Probably, after a seemingly exhaustive solution to a problem, everyone sometimes gets the mistake that something is still wrong here. This state of mind is especially natural if the problem is fundamentally new or if the solution was cumbersome. What should you do in this case? How can you at least make sure that the answer is not completely incorrect? The most natural way to self-test in such a situation is to examine the answer. Let us note right away that there is no universal scheme for this. We can only highlight a few relatively common cases.

Pozny our story is structured as follows. First of all, four tasks are formulated. For each problem three answers are given, two of which are incorrect. We will try, without solving the problem, to determine which answer is correct. Solutions are provided at the end of the article.

Problem 1. The car passed the first level of the track at a speed of $v_1$, and the second - at a speed of $v_2$. What is the average speed automatically?

Omyems:

1. $v_{cp} = \frac{v_1 + v_2}{2v_1v_2}$.
2. $v_{cp} = v_1 + \frac{v_1v_2}{2(v_1 + v_2)}$.
3. $v_{cp} = 2\frac{v_1v_2}{v_1 + v_2}$.

The first test an answer must pass is its dimensionality. Obviously, in case 1) the situation is unfavorable in this regard. The answer has the dimension of inverse velocity $\left(\frac{cek}{cm}\right)$. Let us show that the second answer is also incorrect. Let’s try to understand what’s going on. For this purpose, the list is

Fig. 1.

Let’s look at the conditions of the problem again. It is easy to see that in relation to the question posed, the speeds $v_1$ and $v_2$ are equal, that is, it is not clear which part of the path is automatically with the speed $v_1$, and which one with the speed $v_2$. This means that replacing $v_1$ with $v_2$ and $v_2$ with $v_1$ should not change the answer. In this case, they say that it is symmetrical with respect to $v_1$ and $v_2$. Having made the indicated replacement in the second answer, we get

$v_{cp} = v_2 + \frac{v_1v_2}{2(v_1 + v_2)} \neq v_1 + \frac{v_1v_2}{2(v_1 + v_2)}$.

Thus, the third answer is correct.

Problem 2. Two weights with masses $m_1$ and $m_2$ are suspended on a thread and thrown over a block (Fig. 1). Counting

51

threads and block weightless and neglecting friction, determine the acceleration of the loads $a_1$ and $a_2$ and the tension of the thread $T$.

Answers:

1. $a_1 = \frac{m_1 - m_2}{m_1 + m_2} g$;
   $a_2 = \frac{m_2 - m_1}{m_1 + m_2} g$; $T = 2 \frac{m_1 m_2}{m_1 + m_2} g$.
2. $a_1 = \frac{m_1 - m_2}{m_1} g$; $a_2 = \frac{m_1 - m_2}{m_1} g$;
   $T = 2 \frac{m_1 m_2}{m_1 + m_2} g$.
3. $a_1 = \frac{m_1 - m_2}{m_1 + m_2} g$; $a_2 = \frac{m_2 - m_1}{m_1 + m_2} g$;
   $T = (m_1 + m_2) g$.

We consider the direction indicated by the arrow at figure to be positive.

As in the previous problem, first of all we check the dimension. It is easy to verify that in all three cases the right and left sides of each formula have the same dimension. After that you can move on. Please note that if the weights $m_1$ and $m_2$ are swapped, then the tension of the thread will not change, that is, the expression for the tension force must be symmetrical with respect to $m_1$ and $m_2$. All answers pass this test as well.

It is easy to see that in the second case a mistake was made. Indeed, the weights $m_1$ and $m_2$ hang on the same thread. This means that if mass $m_1$ moves with acceleration $a_1$, then mass $m_2$ must move with acceleration $-a_1$ (kinematic connection), which is not true in the second answer.

So, we just have to make a choice between the first and third results. They differ only in the formula for the amount of thread tension. We will reason like this: let some mass, for example $m_1$, be equal to zero. It is quite obvious that in this case the second weight falls freely and the thread is not tensioned. The same result, of course, should also be obtained from

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3. $h = \cos^2 \alpha \frac{(Mv)^2}{2g(m+M)^2}$;

$Q = \frac{mv^2}{2} \frac{m \cos^2 \alpha + M}{M}$.

The first check, as always, is the dimension. However, let’s not rush. How to do this quickly and easily? Let’s reveal a little secret: it is much easier to see the dimension of the correspondence if you first transform it as follows: from the parameters included in the formula, you need to see dimensionless relationships, and do this in such a way that the remaining dimensional combination is as simple or familiar as possible.

This is exactly what the first answer looks like. Let’s do a similar transformation for the second one:

$$h = \frac{(mv)^2}{2(m+M)} = \frac{mv^2}{2} \frac{m}{m+M};$$

$$Q = \frac{m^2}{2(m+M)} \left[ \left( \frac{M}{m} \right)^2 + v^2 \cos^2 \alpha \right].$$

Now we can see that both correspondences are incorrect. In the first case, the right part has the dimension of energy, and the left part has the dimension of length. In the second case, the sum of the dimensional $v^2 \cos^2 \alpha$ and dimensionless $\left( \frac{M}{m} \right)^2$ quantities is in parentheses, which should not be the case either. Using a similar transformation, make sure that the dimension of the third answer is correct. Unfortunately, unlike the two previous problems, there is no symmetry in the conditions. Let us therefore select a value of some parameter at which the answer is very clear, and check whether it can be obtained from our formulas. Let, for example, $\alpha = 0$. Then it is clear that the ball will not collapse, that is, the lifting height $h$ is zero, and all the kinetic energy will turn out to be $\frac{mv^2}{2}$ transferred to the body. Substituting the value $\alpha = 0$ into the first and third results, we get:

Fig. 3.

1. $h = 0$; $Q = \frac{mv^2}{2}$.

2. $h = \frac{v^2}{2g} \left( \frac{M}{m+M} \right)^2 \neq 0$;

$Q = \frac{mv^2}{2} \frac{m+M}{M} > \frac{mv^2}{2}$.

Thus the correct answer.

Let us also note that for self-test we can use the situation at $m = 0$ or $M \to \infty$. Try it yourself.

Task 4. A wooden cube on side $a$, whose density is $p$, floats in water. What will be the change in the depth of immersion of the cube $\Delta h$ if a layer of oil with a thickness of $l$ (Fig. 3) is poured onto the water? The densities of water and oil are $\rho_B$ and $\rho_M$ respectively.

Answers:

1. $\Delta h = \frac{\rho_M - \rho_B}{\rho} (a - l)$.

2. $\Delta h = \frac{l (\rho_B - \rho_M)}{\rho_B}$.

3. $\Delta h = l \frac{\rho_B + \rho_M}{\rho_M}$.

It is immediately clear that the dimensionality of all three answers is correct, and there is no symmetry in the problem.

Let’s try to check our answers on some very unique case. If $l = 0$ (that is, there is no oil), then the immersion depth should not change ($\Delta h = 0$). It is clear that the first answer, from which

53

case $l=0$ turns out to be $\Delta h = \frac{\rho_M - \rho_B}{\rho} a$, incorrect.

Let us now consider another possible situation. Let us assume that the density of oil is equal to the density of water. Obviously, in this case, the immersion depth of the cube will not change. The third answer, as is easy to see, does not satisfy this condition. Thus, the second answer is correct.

Now, perhaps, it’s time to draw some conclusions. How to research the answer?

We have already seen that first of all we need to make sure that the dimension is correct. It is best to do this in letter form, highlighting the corresponding dimensionless and dimensional components.

The next step is to study the symmetry of the problem. If any parameters are included in the condition equally, then they must be included in the response equally (symmetrically). Otherwise, the symmetry of the problem is easier to detect by the appearance of the answer itself. What does it mean? Imagine that the response under study is symmetrical with respect to some parameters of the same type (for example, two masses, two velocities, and so on). Then you need to check whether the problem conditions have the same symmetry. If not, then the answer is incorrect.

Another convenient way to check is as follows: the result obtained depends on the parameters of the problem (for example, mass, initial speed, angle, capacity, …). We will try, by changing any parameter $x$, to reduce the problem to a known one or to one whose answer is very visible. Let’s assume that we succeeded and we know the correct answer for $x = x_0$. Then the check consists of comparing this result with ours at $x = x_0$. Note that most often you have to resort to this method of verification. But in order to successfully apply it, you need to have a good understanding of the physical side of the problem,

Fig. 4.

Let us now present a detailed solution to the proposed problems.

Task 1. The average speed $v_{cp}$ is equal to $\frac{S}{t}$, where $S$ is the path, and $t$ is the time it took to cover this path.

$$t = \frac{S}{2v_1} + \frac{S}{2v_2} = \frac{S(v_1 + v_2)}{2v_1v_2}.$$

Substituting this expression for $t$ into the formula for $v_{cp}$, we get

$$v_{cp} = \frac{S \cdot 2v_1v_2}{S(v_1 + v_2)} = 2 \frac{v_1v_2}{v_1 + v_2}.$$

Task 2. Due to the fact that the thread is inextensible and weightless, $a_2 = -a_1 = a$ (kinematic connection) and the tension of the thread are the same everywhere (Fig. 4). Let us write down the equation of motion of the group:

$$-m_1a = m_1g - T$$ and $$m_2a = m_2g - T.$$

Solving this system for $a$ and $T$, we find

$$a = \frac{m_2 - m_1}{m_1 + m_2} g,$$

The law of temperament in the ball-bullet system is preserved.

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Fig. 5.

of impulse (amount of movement), we obtain $mv \sin \alpha = (M+m) v'$, where $v'$ is the speed of the bullet and the ball immediately after the bullet hits the ball (Fig. 5).

From the law of conservation of mechanical energy it follows that

$$\frac{(M+m) v'^2}{2} = (M+m) gh,$$

where

$$h = \frac{v^2}{2g} \sin^2 \alpha \left( \frac{m}{M+m} \right)^2.$$

The amount of heat released is equal to the difference between the initial kinetic energy of the bullet and the kinetic energy of the ball with the bullet after the bullet hits the ball:

$$Q = \frac{mv^2}{2} - \frac{(m+M) v'^2}{2} = \frac{mv^2}{2} \frac{m \cos^2 \alpha + M}{m+M}.$$

Task 4. A cube immersed in water and oil is subject to a buoyant force $F = \rho_M g l a^2 + \rho_B g x a^2$ (Fig. 6). This force must be equal to the force of gravity acting on the cube: $P = \rho g a^3$. Equating these two relations, we obtain $\rho_a = \rho_M l + \rho_B x$, whence the immersion depth is $h_1 = x + l = \frac{\rho a}{\rho_B} - \frac{l \rho_M}{\rho_B} + l$. If there is no oil

Fig. 6.

$(l=0)$, then the immersion depth is $h_0 = a \frac{\rho}{\rho_B}$. Thus, the change in immersion depth is equal to

$$\Delta h = h_1 - h_0 = l - l \frac{\rho_M}{\rho_B} = \frac{\rho_B - \rho_M}{\rho_B} l.$$

Now try to figure out on your own which of the answers is correct in the following problems.

1. A man of mass $m$ is on the bow of a boat that is floating on the water. Boat length $L$, weight $M$. The man moves to the stern. How far will $l$ and $b$ move the boat and the person relative to the shore?

Answers:

1. $l = \frac{L}{m+M}$, $b = \frac{LM}{m(M+m)}$;

2. $l = \frac{m}{m+M} L$, $b = \frac{M}{M+m} L$;

3. $l = \frac{M}{m+M} L$, $b = \frac{m}{m+M} L$.

2. A ball of radius $R$ and mass $m$ stands in front of the height step $h$ (Fig. 7). What is the minimum horizontal force that must be applied to the center of the ball for it to rise up the step?

Answers:

1. $F \geqslant mg \frac{\sqrt{h(2R-h)}}{R-h}$;

2. $F \geqslant mg \frac{\sqrt{h(2R-h)}}{R}$;

3. $F \geqslant mg \frac{2R-h}{R-h}$.

Fig. 7.

3. A body of mass $m$ moves vertically with acceleration $a$. Determine what force $F$ acts on the body.

Answers:

1. $F = m(g + a)$;
2. $F = m(g - a)$;
3. $F = m \frac{ga}{g + a}$.

4. Two lenses with focal lengths $F_1$ and $F_2$ are placed side by side (Fig. 8). Determine the focal length of such a system.

Answers:

1. $F = F_1 + F_2$;

2. $F = \frac{F_1 F_2}{F_1 + F_2}$;

3. $F = \frac{F_1 F_2}{F_1 - F_2}$.

5. Two charged balls of radii $R_1$ and $R_2$ are located at a distance from each other much greater than their radii. The charges of the balls are $q_1$ and $q_2$ respectively. Find the charges $q'_1$, $q'_2$ and the potential f of the balls after they are connected with a wire.

Answers:

1. $q'_1 = q_1 \frac{R_1}{R_2}$; $q'_2 = q_2 \frac{R_2}{R_1}$;
   $\varphi = \frac{(q_1 - q_2)}{R_1 + R_2}$.
2. $q'_1 = \frac{(q_1 + q_2) R_1}{R_1 - R_2}$;
   $q_2 = -\frac{(q_1 + q_2) R_2}{R_1 - R_2}$;
   $\varphi = \frac{q_1 + q_2}{R_1 - R_2}$.
3. $q'_1 = \frac{(q_1 + q_2) R_1}{R_1 + R_2}$;
   $q'_2 = \frac{(q_1 + q_2) R_2}{R_1 + R_2}$;
   $\varphi = \frac{q_1 + q_2}{R_1 + R_2}$.