I will start with a couple of definitions, without knowing 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:
- The greater the mass of the body, the more it resists the forces that are trying to bring it out of rest.
- The greater the mass of the body, the more it resists the forces that try to change its speed if the body moves uniformly.
Summarizing, we can say that the inertia of the body counteracts attempts to give the body acceleration. And the mass serves as an indicator of the level of inertia. The greater the mass, the greater the force must be applied to influence the body in order to give it acceleration.
Closed system (isolated)- a system of bodies that is not influenced by other bodies that are 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. Imagine that there was an interaction between the points, as a result of which the speeds of the points changed. Denote by and the increments of these velocities during the time of interaction between the points . We will assume that the increments have opposite directions and are related by the relation 
. We know that the coefficients and do not depend on the nature of the interaction of material points - this is confirmed by many experiments. The coefficients and 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 of the velocities of these material points as a result of the interaction between them.
The above relation can be presented in another form. Let us denote the speeds of the bodies before the interaction as and respectively, and after the interaction - and . In this case, the speed increments can be represented in this form - and . Therefore, the ratio can be written as -.
Impulse (the amount of energy of a material point) is a vector equal to the product of the mass of a material point and its velocity vector —
Impulse of the system (amount of motion of the system of material points) is the vector sum of the impulses of material points that this system consists of - .
It can be concluded that in the case of a closed system, the momentum before and after the interaction of material points must remain the same - , where and . It is possible to formulate the law of conservation of momentum.
The momentum of an isolated system remains constant in time, regardless of the interaction between them.
Required definition:
Conservative forces - forces, the work of which does not depend on the trajectory, but is due only to 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 transformations of potential energy into kinetic energy and vice versa are possible.
The potential energy of a material point is a function of only the coordinates of this point. Those. potential energy depends on the position of the point in the system. Thus, the forces acting on a point can be defined as follows: can be defined as: . is the potential energy of a material point. 
Multiply both sides by and we get 
. We transform and obtain 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 are connected and then move as one.
Two balls , s and experience a perfectly inelastic gift with each other. According to the law of conservation of momentum. From here we can express the speed of two balls moving as a whole after the collision - 
. Kinetic energies before and after impact: 
And 
. Let's find the difference
,
Where - reduced mass of balls . This shows that in the case of an absolutely inelastic collision of two balls, the kinetic energy of the macroscopic motion is lost. This loss is equal to half the product of the reduced mass times the square of the relative velocity.
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The laws of conservation of momentum are the fundamental laws of nature. An example of the application of these laws can be the phenomenon of collision. Absolutely elastic and inelastic impacts - a change in the state of bodies as a result of a short-term interaction during their collision.
Interaction mechanism
The simplest type of interaction of physical bodies is the central collision of balls that have an ideal geometric shape. The contact time of these objects is within hundredths of a second.
According to the definition, a center shot is one in which the collision line crosses the centers of the balls. In this case, the trajectory of interaction is a straight line drawn exactly to the element of the contact surface at the moment of contact. In mechanics, absolutely elastic and inelastic impacts are distinguished.
Interaction types
An absolutely inelastic impact occurs when two bodies made of plastic materials or a plastic and elastic bodies collide. After it is completed, the speeds of the colliding objects become the same.
An absolutely elastic impact is observed when objects made of elastic materials interact (for example, two balls made of hard steel or balls made of some types of plastic, etc.).
Stages
The process of elastic collision occurs in two stages:
- Stage I - the moment after the start of the collision. The forces acting on the balls increase with increasing strain. An increase in deformation is accompanied by a change in the speed of objects. Bodies whose speed was greater slow down their movement, and bodies with a lower speed accelerate. When the deformation becomes maximum, the speed of the balls after an absolutely elastic impact becomes equilibrium.
- II stage. From the moment that characterizes the beginning of the second stage of elastic impact, the value of deformations decreases. In this case, the deformation forces push the balls apart. After the deformation disappears, the balls are removed and completely restore their original shape and move at different speeds. Thus, at the end of the second stage, the central absolutely elastic impact converts the entire supply of potential energy of elastically deformed bodies into kinetic energy.
Isolated systems
In practice, no impact is absolute (elastic or inelastic). The system in any case interacts with the surrounding matter, exchanges energy and information with the environment. But for theoretical studies, the existence of isolated systems is allowed, in which only the objects of research interact. For example, both absolutely inelastic and absolutely elastic ball impacts are possible.
External forces do not act on such a system or their influence is compensated. In an isolated system, the law of conservation of momentum works in full measure - the total momentum between colliding bodies is conserved:
∑=m i v i = const.
Here "m" and "v" are the mass of a certain particle ("i") of the isolated system and its velocity vector, respectively.
To conserve mechanical energy (a special case of the general law of energies), there is a need for the forces that act in the system to be conservative (potential).
Conservative forces
Conservative forces are those that do not convert mechanical energy into other types of energy. These forces are always potential - that is, the work that such forces perform in a closed loop is zero. Otherwise, the forces are called dissipative or non-conservative.
In conservative isolated systems, mechanical energy between colliding bodies is also conserved:
W=Wk+Wp=∑(mv 2 /2)+Wp=const.
Here Wk and Wp are kinetic (k) and potential (p) energies respectively.
To check the relevance of the laws of conservation of energy (the above formulas), if absolutely elastic bodies strike, provided that one of the balls does not move before the collision (the speed of the stationary body v 2 = 0), scientists have derived the following pattern:
m 1 v 1 Ki \u003d m 1 U 1 +m 2 U 2
(m 1 v 1 2)/2×Ke=(m 1 U 1 2)/2+(m 2 U 2 2)/2.
Here m 1 and m 2 are the masses of the first (impact) and second (fixed) balls. Ki and Ke are coefficients showing how many times the momentum of two balls (Ki) and energy (Ke) increased at the moment when an absolutely elastic impact is made. v 1 - the speed of the moving ball.
Since the total momentum of the system must be conserved under any collision conditions, it should be expected that the momentum recovery factor will be equal to unity.
Impact force calculation
The speed of a shock (deflected on a thread) ball that hits a stationary (freely suspended on a thread) ball is determined by the formula of the law of conservation of energy:
m 1 gh=(m 1 v 1 2)/2
h=l-lcosα=2lsin 2 (α/2).
Here h is the deviation of the plane of the impact ball relative to the plane of the stationary ball. l - the length of the threads (absolutely the same) on which the balls are suspended. α is the angle of deflection of the impact ball.
Accordingly, an absolutely elastic impact in the collision of a shock (deflected on a thread) and a motionless (freely hanging on a thread) ball is calculated by the formula:
v 1 =2sin(α/2)√gl.
Research facility
In practice, a simple setup is used to calculate the interaction forces. It is designed to study the types of impacts of two balls. The installation is a tripod with three screws that allow you to set it horizontally. On the tripod there is a central rack, to the upper end of which special suspensions for balls are attached. An electromagnet is fixed on the rod, which attracts and holds one of the balls (shock ball) in the deflected state at the beginning of the experiment.
The value of the initial deflection angle of this ball (coefficient α) can be determined from the arc-shaped scale diverging in both directions. The magnitude of its curvature corresponds to the trajectory of the movement of the interacting balls.
Research Process
First, a pair of balls is prepared: depending on the tasks, elastic, inelastic, or two diverse balls are taken. The masses of the balls are recorded in a special table.
Then the shock element is docked to the electromagnet. The scale determines the angle of deflection of the thread. Then the electromagnet is turned off, it loses its attractive properties, and the ball rushes down in an arc, colliding with a second, free, motionless ball, which, as a result of an impulse (impact), deviates to a certain angle. The deviation value is fixed on the second scale.
Absolutely elastic impact is calculated on the basis of experimental data. To confirm the veracity of the laws of conservation of momentum and energy during elastic and inelastic impacts of two balls, their velocities are determined before and after the collision. It is based on the ballistic method of measuring the speed of the balls by the magnitude of their deviation. This value is measured on scales made in the form of arcs of a circle.
Features of calculations
When calculating impact in classical mechanics, a number of indicators are not taken into account:
- impact time;
- degree of deformation of interacting objects;
- heterogeneity of materials;
- the rate of deformation (transfer of momentum, energy) inside the ball.
The collision of billiard balls is a good example of an elastic impact.
Ball speeds before impact
Ball speeds after impact
Let us write down the equations according to the law of conservation of momentum and the law of conservation of energy.
By solving the system of these two equations, we can obtain the following formulas for the speeds of the balls after impact
Let's consider special cases.
Collision of identical balls, m 1 =m 2 .
That is, when the balls collide, they exchange speeds.
If one of the balls is stationary, for example v 20 =0, then after the impact it will move at a speed equal to the speed of the first ball (and in the same direction), and the first ball will stop.
2). Ball hitting a massive wall, m 2 >>m 1 .
From formulas (11) and (12) we obtain in this case:
The wall speed remains unchanged. If the wall is stationary, (v 20 = 0), that is, the ball hitting the wall will bounce back with almost the same speed.
Table 1 Study of elastic collision
v 10 and v 1 calculated by the formulas - where =0.1 m - the length of the plates inserted into the carts.
Table 2 Measurements at different bogie weights
Table 3
Conclusion: In an absolutely elastic impact, the kinetic energy of the colliding bodies first transforms into the potential energy of elastic deformation. Then the bodies return to their original shape, repelling each other. As a result, the potential energy of elastic deformation again turns into kinetic energy, and the bodies fly apart with velocities, the magnitude and direction of which are determined by two laws - the law of conservation of energy and the law of conservation of momentum.
Table 4 Study of inelastic collision
Table 5
since we are considering a special case when the hit body (m 2) is motionless (v 20 \u003d 0) and the mass of the hit body is large, (m 2 >> m 1), then
Table 6
Conclusion: in an absolutely inelastic impact, the kinetic energy is completely or partially converted into internal energy, leading to an increase in the temperature of the bodies. After the impact, the colliding bodies either move together at the same speed or are at rest. In this case, after the impact, the bodies move together. In a perfectly inelastic impact, only the law of conservation of momentum is satisfied.
As an example of the practical application of a new form of Newton's second law, consider the problem of an absolutely elastic impact of a ball with a mass on a fixed wall (Fig. 4.11).
Let us assume that the ball before the impact has a speed and moves perpendicular to the wall. You need to find the speed with which it will move after the impact, and the momentum that the wall will receive during the impact.
Let us consider separately the successive stages of impact.
From the moment of contact, deformations begin to develop in the ball and the wall. Together with them, gradually increasing elastic forces will arise, acting on the wall and on the ball and slowing down the motion of the ball. The growth of deformations and forces will stop at the moment when the speed of the ball becomes zero:
Thus, for this stage of the impact, we know the initial and final values of the momentum of the ball, and from them we can determine the momentum received by the ball from the wall during this time. The force at this time changes its value from zero to some maximum
magnitude, so expressing momentum directly in terms of force is quite difficult. Let's introduce the so-called average force: we will call the average force a constant force imparting to the body the same impulse as the variable force imparts to it in the same time.
For the impulse of the average force that acted on the ball during its deformation, now we can write the equation of Newton's second law: So we finally get:
The change in the momentum of the ball during the first half of the impact and the momentum received by the ball turn out to be equal to the initial momentum, taken with the opposite sign.
During the second half of the impact, after the ball comes to a complete stop, the elastic forces will cause it to move in the opposite direction. Deformations, and with them elastic forces, will begin to decrease. In this case, all values of deformations and forces will be repeated in the reverse order for the same time. Consequently, during the second stage of the impact, the ball will additionally receive the same momentum from the wall as in the first stage. Now let's substitute into the equation of Newton's second law the found values of momentum and velocities corresponding to the second half of the impact. Since we will get
Equating the left parts of the expressions written for the first and second halves of the beat, we find:
After an elastic impact on the wall along the normal, the ball will have a velocity equal in absolute value to the initial velocity and directed oppositely to it. The total momentum received by the ball during the entire time of impact, and the total change in the momentum will be equal
According to Newton's third law, the wall will receive the same momentum from the ball, but directed in the opposite direction.
Let us assume that the wall experiences such impacts in one second. During each impact, the wall will receive an impulse. In just one second, the wall will receive an impulse. Knowing this impulse, it is possible to calculate the average force that acts on the wall and is created by the impacts of the balls. The total momentum received by the wall will be
where is the time during which the strikes occurred. Substituting, we find that in one second the average force will act on the wall
The considered example is especially important because it is in this way that the forces of gas pressure on the walls of the vessel are calculated. As you will learn in a molecular physics course, the pressure of a gas on the walls of a vessel arises due to the impulses that fast-moving gas molecules impart to the wall during impacts. In this case, it is assumed that each impact of the molecule is absolutely elastic. Our calculations are fully applicable to this case. The whole difficulty in calculating the gas pressure lies in the correct calculation of the number of impacts of molecules on the walls of the vessel per unit of time. We also note that the coincidence of the modulus of force with the modulus of momentum imparted by this force per unit time is often used in solving many practical problems.
Finally, we note that in our reasoning there is one unsaid assumption that the time spent on creating deformations during impact is equal to the time for removing deformations. A little later we will prove its validity.
