Impulse after collision. Savelyev I.V.

Solution. The mass can be calculated using the formula. A force that is twice as strong imparts 4 times the acceleration to a body with mass .

Correct answer: 2.

A3. At what stage of flight in a spacecraft that becomes an Earth satellite in orbit will weightlessness be observed?

Solution. Weightlessness is observed in the absence of all external forces, with the exception of gravitational forces. These are the conditions in which a spacecraft finds itself during an orbital flight with the engine turned off.

Correct answer: 3.

A4. Two balls with masses m and 2 m move with speeds equal to 2, respectively v And v. The first ball moves after the second and, having caught up, sticks to it. What is the total momentum of the balls after impact?

Solution. According to the law of conservation, the total momentum of the balls after the collision is equal to the sum of the impulses of the balls before the collision: .

Correct answer: 4.

A5. Four identical sheets of plywood thickness L Each one, tied in a stack, floats in the water so that the water level corresponds to the boundary between the two middle sheets. If you add another sheet of the same type to the stack, the immersion depth of the stack of sheets will increase by

Solution. The immersion depth is half the height of the stack: for four sheets - 2 L, for five sheets - 2.5 L. The immersion depth will increase by .

Correct answer: 3.

A6. The figure shows a graph of the change over time in the kinetic energy of a child swinging on a swing. At the moment corresponding to the point A on the graph, its potential energy, measured from the equilibrium position of the swing, is equal to

Solution. It is known that in the equilibrium position a maximum of kinetic energy is observed, and the difference in potential energies in two states is equal in magnitude to the difference in kinetic energies. The graph shows that the maximum kinetic energy is 160 J, and for the point A it is equal to 120 J. Thus, the potential energy measured from the equilibrium position of the swing is equal to .

Correct answer: 1.

A7. Two material points move in circles with radii and equal velocities. Their periods of revolution in circles are related by the relation

Solution. The period of revolution around a circle is equal to . Because , then .

Correct answer: 4.

A8. In liquids, particles oscillate near an equilibrium position, colliding with neighboring particles. From time to time the particle makes a “jump” to another equilibrium position. What property of liquids can be explained by this nature of particle motion?

Solution. This nature of the movement of liquid particles explains its fluidity.

Correct answer: 2.

A9. Ice at a temperature of 0 °C was brought into a warm room. Temperature of ice before it melts

Solution. The temperature of the ice before it melts will not change, since all the energy received by the ice at this time is spent on destroying the crystal lattice.

Correct answer: 1.

A10. At what air humidity does a person tolerate high air temperatures more easily and why?

Solution. A person can more easily tolerate high air temperatures with low humidity, since sweat evaporates quickly.

Correct answer: 1.

A11. The absolute body temperature is 300 K. On the Celsius scale it is equal to

Solution. On the Celsius scale it is equal to .

Correct answer: 2.

A12. The figure shows a graph of the volume of an ideal monatomic gas versus pressure in process 1–2. The internal energy of the gas increased by 300 kJ. The amount of heat imparted to the gas in this process is equal to

Solution. The efficiency of a heat engine, the useful work it performs and the amount of heat received from the heater are related by the equality , whence .

Correct answer: 2.

A14. Two identical light balls, the charges of which are equal in magnitude, are suspended on silk threads. The charge of one of the balls is indicated in the figures. Which of the pictures corresponds to the situation when the charge of the 2nd ball is negative?

Solution. The indicated charge of the ball is negative. Like charges repel each other. The repulsion is observed in the figure A.

Correct answer: 1.

A15. An α particle moves in a uniform electrostatic field from a point A exactly B along trajectories I, II, III (see figure). Work of electrostatic field forces

Solution. The electrostatic field is potential. In it, the work of moving the charge does not depend on the trajectory, but depends on the position of the starting and ending points. For the drawn trajectories, the starting and ending points coincide, which means that the work of the electrostatic field forces is the same.

Correct answer: 4.

A16. The figure shows a graph of the current in a conductor versus the voltage at its ends. What is the resistance of the conductor?

Solution. In an aqueous salt solution, current is created only by ions.

Correct answer: 1.

A18. An electron flying into the gap between the poles of an electromagnet has a horizontally directed speed perpendicular to the magnetic field induction vector (see figure). Where is the Lorentz force acting on the electron directed?

Solution. Let’s use the “left hand” rule: point four fingers in the direction of the electron’s movement (away from ourselves), and turn the palm so that the magnetic field lines enter it (to the left). Then the protruding thumb will show the direction of the acting force (it will be directed downward) if the particle were positively charged. The electron charge is negative, which means the Lorentz force will be directed in the opposite direction: vertically upward.

Correct answer: 2.

A19. The figure shows a demonstration of an experiment to verify Lenz's rule. The experiment is carried out with a solid ring, not a cut one, because

Solution. The experiment is carried out with a solid ring, because an induced current arises in a solid ring, but not in a cut ring.

Correct answer: 3.

A20. The decomposition of white light into a spectrum when passing through a prism is due to:

Solution. Using the formula for the lens, we determine the position of the image of the object:

If you place the film plane at this distance, you will get a clear image. It can be seen that 50 mm

Correct answer: 3.

A22. Speed ​​of light in all inertial frames of reference

Solution. According to the postulate of the special theory of relativity, the speed of light in all inertial frames of reference is the same and does not depend either on the speed of the light receiver or on the speed of the light source.

Correct answer: 1.

A23. Beta radiation is

Solution. Beta radiation is a stream of electrons.

Correct answer: 3.

A24. The thermonuclear fusion reaction releases energy, and:

A. The sum of the charges of the particles - the reaction products - is exactly equal to the sum of the charges of the original nuclei.

B. The sum of the masses of the particles - the reaction products - is exactly equal to the sum of the masses of the original nuclei.

Are the above statements true?

Solution. The charge is always maintained. Since the reaction occurs with the release of energy, the total mass of the reaction products is less than the total mass of the original nuclei. Only A is correct.

Correct answer: 1.

A25. A load weighing 10 kg is applied to a moving vertical wall. The coefficient of friction between the load and the wall is 0.4. With what minimum acceleration must the wall be moved to the left so that the load does not slide down?

Solution. To prevent the load from sliding down, it is necessary that the friction force between the load and the wall balances the force of gravity: . For a load that is motionless relative to the wall, the following relation is true, where μ is the friction coefficient, N- the support reaction force, which, according to Newton’s second law, is related to the acceleration of the wall by the equality . As a result we get:

Correct answer: 3.

A26. A plasticine ball weighing 0.1 kg flies horizontally at a speed of 1 m/s (see figure). It hits a stationary cart of mass 0.1 kg attached to a light spring and sticks to the cart. What is the maximum kinetic energy of the system during its further oscillations? Ignore friction. The blow is considered instantaneous.

Solution. According to the law of conservation of momentum, the speed of a cart with a plasticine ball stuck to it is equal to

Correct answer: 4.

A27. Experimenters pump air into a glass vessel, simultaneously cooling it. At the same time, the air temperature in the vessel decreased by 2 times, and its pressure increased by 3 times. How many times has the mass of air in the container increased?

Solution. Using the Mendeleev-Clapeyron equation, you can calculate the mass of air in the vessel:

.

If the temperature dropped by 2 times and its pressure increased by 3 times, then the mass of air increased by 6 times.

Correct answer: 3.

A28. A rheostat is connected to a current source with an internal resistance of 0.5 Ohm. The figure shows a graph of the dependence of the current in the rheostat on its resistance. What is the emf of the current source?

Solution. According to Ohm's law for a complete circuit:

.

When the external resistance is equal to zero, the emf of the current source is found by the formula:

Correct answer: 2.

A29. A capacitor, inductor and resistor are connected in series. If, with a constant frequency and voltage amplitude at the ends of the circuit, the capacitance of the capacitor is increased from 0 to , then the amplitude of the current in the circuit will be

Solution. The AC resistance of the circuit is . The current amplitude in the circuit is equal to

.

This dependency as a function WITH on the interval has a maximum at . The amplitude of the current in the circuit will first increase and then decrease.

Correct answer: 3.

A30. How many α- and β-decays must occur during the radioactive decay of a uranium nucleus and its eventual transformation into a lead nucleus?

Solution. During α decay, the mass of the nucleus decreases by 4 a. e.m., and during β-decay the mass does not change. In a series of decays, the mass of the nucleus decreased by 238 – 198 = 40 a. e.m. For such a decrease in mass, 10 α decays are required. With α-decay, the charge of the nucleus decreases by 2, and with β-decay, it increases by 1. In a series of decays, the charge of the nucleus decreased by 10. For such a decrease in charge, in addition to 10 α-decays, 10 β-decays are required.

Correct answer: 1.

Part B

IN 1. A small stone thrown from a flat horizontal surface of the earth at an angle to the horizon fell back to the ground after 2 s, 20 m from the point of throw. What is the minimum speed of the stone during flight?

Solution. In 2 s, the stone covered 20 m horizontally; therefore, the component of its velocity directed along the horizon is 10 m/s. The speed of the stone is minimal at the highest point of flight. At the top point, the total speed coincides with its horizontal projection and, therefore, is equal to 10 m/s.

AT 2. To determine the specific heat of melting of ice, pieces of melting ice were thrown into a vessel with water with continuous stirring. Initially, the vessel contained 300 g of water at a temperature of 20 °C. By the time the ice stopped melting, the mass of water had increased by 84 g. Based on the experimental data, determine the specific heat of melting of ice. Express your answer in kJ/kg. Neglect the heat capacity of the vessel.

Solution. The water gave off heat. This amount of heat was used to melt 84 g of ice. The specific heat of melting of ice is .

Answer: 300.

AT 3. When treating with an electrostatic shower, a potential difference is applied to the electrodes. What charge passes between the electrodes during the procedure, if it is known that the electric field does work equal to 1800 J? Express your answer in mC.

Solution. The work done by the electric field to move a charge is equal to . Where can we express the charge:

.

AT 4. A diffraction grating with a period is located parallel to the screen at a distance of 1.8 m from it. What order of magnitude maximum in the spectrum will be observed on the screen at a distance of 21 cm from the center of the diffraction pattern when the grating is illuminated by a normally incident parallel beam of light with a wavelength of 580 nm? Count .

Solution. The deflection angle is related to the lattice constant and the wavelength of light by the equality . The deviation on the screen is . Thus, the order of the maximum in the spectrum is equal to

Part C

C1. The mass of Mars is 0.1 of the mass of the Earth, the diameter of Mars is half that of the Earth. What is the ratio of the orbital periods of artificial satellites of Mars and Earth moving in circular orbits at low altitude?

Solution. The orbital period of an artificial satellite moving around the planet in a circular orbit at low altitude is equal to

Where D- diameter of the planet, v- the speed of the satellite, which is related to the centripetal acceleration ratio.

Momentum is a physical quantity that, under certain conditions, remains constant for a system of interacting bodies. The modulus of momentum is equal to the product of mass and velocity (p = mv). The law of conservation of momentum is formulated as follows:

In a closed system of bodies, the vector sum of the bodies’ momenta remains constant, i.e., does not change. By closed we mean a system where bodies interact only with each other. For example, if friction and gravity can be neglected. Friction can be small, and the force of gravity is balanced by the force of the normal reaction of the support.

Let's say one moving body collides with another body of the same mass, but motionless. What will happen? Firstly, a collision can be elastic or inelastic. In an inelastic collision, the bodies stick together into one whole. Let's consider just such a collision.

Since the masses of the bodies are the same, we denote their masses by the same letter without an index: m. The momentum of the first body before the collision is equal to mv 1, and the second is equal to mv 2. But since the second body is not moving, then v 2 = 0, therefore, the momentum of the second body is 0.

After an inelastic collision, the system of two bodies will continue to move in the direction where the first body was moving (the momentum vector coincides with the velocity vector), but the speed will become 2 times less. That is, the mass will increase by 2 times, and the speed will decrease by 2 times. Thus, the product of mass and speed will remain the same. The only difference is that before the collision the speed was 2 times greater, but the mass was equal to m. After the collision, the mass became 2m, and the speed was 2 times less.

Let us imagine that two bodies moving towards each other inelastically collide. The vectors of their velocities (as well as impulses) are directed in opposite directions. This means that the pulse modules must be subtracted. After the collision, the system of two bodies will continue to move in the direction in which the body with greater momentum was moving before the collision.

For example, if one body had a mass of 2 kg and moved with a speed of 3 m/s, and the other had a mass of 1 kg and a speed of 4 m/s, then the impulse of the first is 6 kg m/s, and the impulse of the second is 4 kg m /With. This means that the velocity vector after the collision will be codirectional with the velocity vector of the first body. But the speed value can be calculated like this. The total impulse before the collision was equal to 2 kg m/s, since the vectors are opposite directions, and we must subtract the values. It should remain the same after the collision. But after the collision, the body mass increased to 3 kg (1 kg + 2 kg), which means that from the formula p = mv it follows that v = p/m = 2/3 = 1.6(6) (m/s). We see that as a result of the collision the speed decreased, which is consistent with our everyday experience.

If two bodies are moving in one direction and one of them catches up with the second, pushes it, engaging with it, then how will the speed of this system of bodies change after the collision? Let's say a body weighing 1 kg moved at a speed of 2 m/s. A body weighing 0.5 kg, moving at a speed of 3 m/s, caught up with him and grappled with him.

Since the bodies move in one direction, the impulse of the system of these two bodies is equal to the sum of the impulses of each body: 1 2 = 2 (kg m/s) and 0.5 3 = 1.5 (kg m/s). The total impulse is 3.5 kg m/s. It should remain the same after the collision, but the body mass here will already be 1.5 kg (1 kg + 0.5 kg). Then the speed will be equal to 3.5/1.5 = 2.3(3) (m/s). This speed is greater than the speed of the first body and less than the speed of the second. This is understandable, the first body was pushed, and the second, one might say, encountered an obstacle.

Now imagine that two bodies are initially coupled. Some equal force pushes them in different directions. What will be the speed of the bodies? Since equal force is applied to each body, the modulus of the impulse of one must be equal to the modulus of the impulse of the other. However, the vectors are oppositely directed, so when their sum will be equal to zero. This is correct, because before the bodies moved apart, their momentum was equal to zero, because the bodies were at rest. Since momentum is equal to the product of mass and speed, in this case it is clear that the more massive the body, the lower its speed will be. The lighter the body, the greater its speed will be.

I’ll start with a couple of definitions, without knowledge of which further consideration of the issue will be meaningless.

The resistance that a body exerts when trying to set it in motion or change its speed is called inertia.

Measure of inertia – weight.

Thus, the following conclusions can be drawn:

1. The greater the mass of a body, the greater its resistance to the forces that try to bring it out of rest.
2. The greater the mass of a body, the more it resists the forces that try to change its speed if the body moves uniformly.

To summarize, we can say that the inertia of the body counteracts attempts to give the body acceleration. And mass serves as an indicator of the level of inertia. The greater the mass, the greater the force that must be applied to the body in order to give it acceleration.

Closed system (isolated)- a system of bodies that is not influenced by other bodies not included in this system. Bodies in such a system interact only with each other.

If at least one of the two conditions above is not met, then the system cannot be called closed. Let there be a system consisting of two material points with velocities and, respectively. Let's imagine that there was an interaction between the points, as a result of which the velocities of the points changed. Let us denote by and the increments of these speeds during the interaction between points. We will assume that the increments have opposite directions and are related by the relation . We know that the coefficients do not depend on the nature of the interaction of material points - this has been confirmed by many experiments. The coefficients are characteristics of the points themselves. These coefficients are called masses (inertial masses). The given relationship for the increment of velocities and masses can be described as follows.

The ratio of the masses of two material points is equal to the ratio of the increments in the velocities of these material points as a result of the interaction between them.

The above relationship can be presented in another form. Let us denote the velocities of the bodies before the interaction as and , respectively, and after the interaction as and . In this case, the speed increments can be presented in the following form - and . Therefore, the relationship can be written as follows - .

Momentum (amount of energy of a material point)– a vector equal to the product of the mass of a material point and its velocity vector –

Momentum of the system (amount of motion of the system of material points)– vector sum of the momenta of the material points of which this system consists - .

We can conclude that in the case of a closed system, the momentum before and after the interaction of material points should remain the same - , where and . We can formulate the law of conservation of momentum.

The momentum of an isolated system remains constant over time, regardless of the interaction between them.

Required definition:

Conservative forces – forces whose work does not depend on the trajectory, but is determined only by the initial and final coordinates of the point.

Formulation of the law of conservation of energy:

In a system in which only conservative forces act, the total energy of the system remains unchanged. Only the transformation of potential energy into kinetic energy and vice versa is possible.

The potential energy of a material point is a function only of the coordinates of this point. Those. potential energy depends on the position of a point in the system. Thus, the forces acting on a point can be defined as follows: can be defined as follows: . – potential energy of a material point. Multiply both sides by and get . Let's transform and get an expression proving law of energy conservation .

Elastic and inelastic collisions

Absolutely inelastic impact - a collision of two bodies, as a result of which they connect and then move as one.

Two balls, with and experience a completely inelastic gift with each other. According to the law of conservation of momentum. From here we can express the speed of two balls moving after a collision as a single whole - . Kinetic energies before and after impact: And . Let's find the difference

,

Where - reduced mass of balls . From this it can be seen that during an absolutely inelastic collision of two balls there is a loss of kinetic energy of macroscopic motion. This loss is equal to half the product of the reduced mass and the square of the relative velocity.

In this lesson we continue to study the laws of conservation and consider the various possible impacts of bodies. From your own experience, you know that an inflated basketball bounces well off the floor, while a deflated one barely bounces at all. From this you could conclude that the impacts of different bodies can be different. In order to characterize impacts, the abstract concepts of absolutely elastic and absolutely inelastic impacts are introduced. In this lesson we will study different strokes.

Topic: Conservation laws in mechanics

Lesson: Colliding bodies. Absolutely elastic and absolutely inelastic impacts

To study the structure of matter, one way or another, various collisions are used. For example, in order to examine an object, it is irradiated with light, or a stream of electrons, and by scattering this light or a stream of electrons, a photograph, or an X-ray, or an image of this object in some physical device is obtained. Thus, the collision of particles is something that surrounds us in everyday life, in science, in technology, and in nature.

For example, a single collision of lead nuclei in the ALICE detector of the Large Hadron Collider produces tens of thousands of particles, from the movement and distribution of which one can learn about the deepest properties of matter. Considering collision processes using the conservation laws we are talking about allows us to obtain results regardless of what happens at the moment of collision. We don't know what happens when two lead nuclei collide, but we do know what the energy and momentum of the particles that fly apart after these collisions will be.

Today we will look at the interaction of bodies during a collision, in other words, the movement of non-interacting bodies that change their state only upon contact, which we call a collision, or impact.

When bodies collide, in the general case, the kinetic energy of the colliding bodies does not have to be equal to the kinetic energy of the flying bodies. Indeed, during a collision, bodies interact with each other, influencing each other and doing work. This work can lead to a change in the kinetic energy of each body. In addition, the work that the first body does on the second may not be equal to the work that the second body does on the first. This can cause mechanical energy to turn into heat, electromagnetic radiation, or even create new particles.

Collisions in which the kinetic energy of the colliding bodies is not conserved are called inelastic.

Among all possible inelastic collisions, there is one exceptional case when the colliding bodies stick together as a result of the collision and then move as one. This inelastic impact is called absolutely inelastic (Fig. 1).

A) b)

Rice. 1. Absolute inelastic collision

Let's consider an example of a completely inelastic impact. Let a bullet of mass fly in a horizontal direction with speed and collide with a stationary box of sand of mass , suspended on a thread. The bullet got stuck in the sand, and then the box with the bullet began to move. During the impact of the bullet and the box, the external forces acting on this system are the force of gravity, directed vertically downward, and the tension force of the thread, directed vertically upward, if the time of impact of the bullet was so short that the thread did not have time to deflect. Thus, we can assume that the momentum of the forces acting on the body during the impact was equal to zero, which means that the law of conservation of momentum is valid:

.

The condition that the bullet is stuck in the box is a sign of a completely inelastic impact. Let's check what happened to the kinetic energy as a result of this impact. Initial kinetic energy of the bullet:

final kinetic energy of bullet and box:

simple algebra shows us that during the impact the kinetic energy changed:

So, the initial kinetic energy of the bullet is less than the final one by some positive value. How did this happen? During the impact, resistance forces acted between the sand and the bullet. The difference in the kinetic energies of the bullet before and after the collision is exactly equal to the work of the resistance forces. In other words, the kinetic energy of the bullet went to heat the bullet and the sand.

If, as a result of the collision of two bodies, kinetic energy is conserved, such a collision is called absolutely elastic.

An example of perfectly elastic impacts is the collision of billiard balls. We will consider the simplest case of such a collision - a central collision.

A collision in which the velocity of one ball passes through the center of mass of the other ball is called a central collision. (Fig. 2.)

Rice. 2. Center ball strike

Let one ball be at rest, and the second fly at it with some speed, which, according to our definition, passes through the center of the second ball. If the collision is central and elastic, then the collision produces elastic forces acting along the line of collision. This leads to a change in the horizontal component of the momentum of the first ball, and to the appearance of a horizontal component of the momentum of the second ball. After the impact, the second ball will receive an impulse directed to the right, and the first ball can move both to the right and to the left - this will depend on the ratio between the masses of the balls. In the general case, consider a situation where the masses of the balls are different.

The law of conservation of momentum is satisfied for any collision of balls:

In the case of an absolutely elastic impact, the law of conservation of energy is also satisfied:

We obtain a system of two equations with two unknown quantities. Having solved it, we will get the answer.

The speed of the first ball after impact is

,

Note that this speed can be either positive or negative, depending on which of the balls has more mass. In addition, we can distinguish the case when the balls are identical. In this case, after hitting the first ball will stop. The speed of the second ball, as we noted earlier, turned out to be positive for any ratio of the masses of the balls:

Finally, let's consider the case of an off-center impact in a simplified form - when the masses of the balls are equal. Then, from the law of conservation of momentum we can write:

And from the fact that kinetic energy is conserved:

An off-central impact will be in which the speed of the oncoming ball will not pass through the center of the stationary ball (Fig. 3). From the law of conservation of momentum, it is clear that the velocities of the balls will form a parallelogram. And from the fact that kinetic energy is conserved, it is clear that it will not be a parallelogram, but a square.

Rice. 3. Off-center impact with equal masses

Thus, with an absolutely elastic off-center impact, when the masses of the balls are equal, they always fly apart at right angles to each other.

Bibliography

1. G. Ya. Myakishev, B. B. Bukhovtsev, N. N. Sotsky. Physics 10. - M.: Education, 2008.
2. A.P. Rymkevich. Physics. Problem book 10-11. - M.: Bustard, 2006.
3. O.Ya. Savchenko. Problems in physics - M.: Nauka, 1988.
4. A. V. Peryshkin, V. V. Krauklis. Physics course vol. 1. - M.: State. teacher ed. min. education of the RSFSR, 1957.

Answer: Yes, such impacts really exist in nature. For example, if the ball hits the net of a football goal, or a piece of plasticine slips out of your hands and sticks to the floor, or an arrow that gets stuck in a target suspended on strings, or a projectile hits a ballistic pendulum.

Question: Give more examples of a perfectly elastic impact. Do they exist in nature?

Answer: Absolutely elastic impacts do not exist in nature, since with any impact, part of the kinetic energy of the bodies is spent on doing work by some external forces. However, sometimes we can consider certain impacts to be absolutely elastic. We have the right to do this when the change in the kinetic energy of the body upon impact is insignificant compared to this energy. Examples of such impacts include a basketball bouncing off the pavement or metal balls colliding. Collisions of ideal gas molecules are also considered elastic.

Question: What to do when the impact is partially elastic?

Answer: It is necessary to estimate how much energy was spent on the work of dissipative forces, that is, forces such as friction or resistance. Next, you need to use the laws of conservation of momentum and find out the kinetic energy of bodies after a collision.

Question: How should one solve the problem of off-center impact of balls having different masses?

Answer: It is worth writing down the law of conservation of momentum in vector form, and that kinetic energy is conserved. Next, you will have a system of two equations and two unknowns, by solving which you can find the speeds of the balls after the collision. However, it should be noted that this is a rather complex and time-consuming process that goes beyond the scope of the school curriculum.

When bodies collide with each other, they undergo deformations. In this case, the kinetic energy that the bodies possessed before the impact is partially or completely converted into the potential energy of elastic deformation and into the so-called internal energy of the bodies. An increase in the internal energy of bodies is accompanied by an increase in their temperature.

There are two limiting types of impact: absolutely elastic and absolutely inelastic. Absolutely elastic is an impact in which the mechanical energy of bodies does not transform into other, non-mechanical, types of energy. With such an impact, kinetic energy is converted completely or partially into potential energy of elastic deformation. Then the bodies return to their original shape by repelling each other. As a result, the potential energy of elastic deformation again turns into kinetic energy and the bodies fly apart at speeds, the magnitude and direction of which are determined by two conditions - conservation of total energy and conservation of total momentum of the system of bodies.

A completely inelastic impact is characterized by the fact that no potential strain energy arises; the kinetic energy of bodies is completely or partially converted into internal energy; After the impact, the colliding bodies either move at the same speed or are at rest. With an absolutely inelastic impact, only the law of conservation of momentum is satisfied, but the law of conservation of Mechanical energy is not observed - there is a law of conservation of the total energy of various types - mechanical and internal.

We will limit ourselves to considering the central impact of two balls. A hit is called central if the balls before the hit move along a straight line passing through their centers. With a central impact, an impact can occur if; 1) the balls are moving towards each other (Fig. 70, a) and 2) one of the balls is catching up with the other (Fig. 70.6).

We will assume that the balls form a closed system or that the external forces applied to the balls balance each other.

Let us first consider a completely inelastic impact. Let the masses of the balls be equal to m 1 and m 2, and the velocities before the impact V 10 and V 20. By virtue of the conservation law, the total momentum of the balls after the impact must be the same as before the impact:

Since the vectors v 10 and v 20 are directed along the same line, the vector v also has a direction coinciding with this line. In case b) (see Fig. 70) it is directed in the same direction as the vectors v 10 and v 20. In case a) the vector v is directed towards that of the vectors v i0 for which the product m i v i0 is greater.

The magnitude of the vector v can be calculated using the following formula:

where υ 10 and υ 20 are the modules of the vectors v 10 and v 20; the “-” sign corresponds to case a), the “+” sign to case b).

Now consider a perfectly elastic impact. With such an impact, two conservation laws are satisfied: the law of conservation of momentum and the law of conservation of mechanical energy.

Let us denote the masses of the balls as m 1 and m 2, the velocities of the balls before the impact as v 10 and v 20, and, finally, the velocities of the balls after the impact as v 1 and v 2. Let us write the conservation equations for momentum and energy;

Taking into account that , let us reduce (30.5) to the form

Multiplying (30.8) by m 2 and subtracting the result from (30.6), and then multiplying (30.8) by m 1 and adding the result with (30.6), we obtain the velocity vectors of the balls after impact:

For numerical calculations, let's project (30.9) onto the direction of the vector v 10 ;

In these formulas, υ 10 and υ 20 are modules, and υ 1 and υ 2 are projections of the corresponding vectors. The upper “-” sign corresponds to the case of balls moving towards each other, the lower “+” sign to the case when the first ball overtakes the second.

Note that the velocities of the balls after an absolutely elastic impact cannot be the same. In fact, by equating expressions (30.9) for v 1 and v 2 to each other and making transformations, we obtain:

Consequently, in order for the velocities of the balls to be the same after the impact, it is necessary that they be the same before the impact, but in this case the collision cannot occur. It follows that the condition of equal velocities of the balls after impact is incompatible with the law of conservation of energy. So, during an inelastic impact, mechanical energy is not conserved - it partially transforms into the internal energy of the colliding bodies, which leads to their heating.

Let's consider the case when the masses of the colliding balls are equal: m 1 =m 2 . From (30.9) it follows that under this condition

i.e., when the balls collide, they exchange speed. In particular, if one of the balls of the same mass, for example the second, is at rest before the collision, then after the impact it moves with the same speed as the first ball initially used; The first ball after impact turns out to be motionless.

Using formulas (30.9), you can determine the speed of the ball after an elastic impact on a stationary, non-moving wall (which can be considered as a ball of infinitely large mass m2 and infinitely large radius). Dividing the numerator and denominator of expressions (30.9) by m 2 and neglecting terms containing the factor m 1 / m 2 we obtain:

As follows from the results obtained, soon the walls remain unchanged. The speed of the ball, if the wall is stationary (v 20 = 0), changes the opposite direction; in the case of a moving wall, the speed of the ball also changes (increases to 2υ 20 if the wall moves towards the ball, and decreases 2υ 20 if the wall “moves away” from the ball catching up with it)