How to find the momentum modulus. Law of conservation of momentum, kinetic and potential energies, power of force

The momentum of a body is a vector physical quantity, which is equal to the product of the speed of the body and its mass. Also, the momentum of the body has a second name - the amount of motion. The direction of the momentum of the body coincides with the direction of the velocity vector. The momentum of a body in the SI system does not have its own unit of measure. Therefore, it is measured in units included in its composition is a kilogram meter per second kgm / s.

Formula 1 - Impulse of the body.

m - body weight.

v is the speed of the body.

The momentum of the body, in fact, is a new interpretation of Newton's second law. In which they simply decomposed the acceleration. In this case, the value Ft was called the momentum of the force, and mv the momentum of the body.

The impulse of a force is a physical quantity of a vector character, which determines the degree of action of a force over a period of time during which it acts.

Formula 2 - Newton's second law, momentum of a body.

m - body weight.

v1 - initial speed of the body.

v2 - final speed of the body.

a - acceleration of the body.

p is the momentum of the body.

t1 - start time

t2 - end time.

This was done in order to be able to calculate the tasks associated with the movement of bodies of variable mass and at speeds comparable to the speed of light.

The new interpretation of Newton's second law should be understood as follows. As a result of the action of force F during time t on a body of mass m, its speed will become equal to V.

In a closed system, the magnitude of the momentum is constant, this is how the law of conservation of momentum sounds. Recall that a closed system is a system that is not acted upon by external forces. An example of such a system is two dissimilar balls moving along a rectilinear trajectory towards each other, with the same speed. The balls have the same diameter. There are no friction forces during movement. Since the balls are made of different materials, they have different masses. But at the same time, the material provides absolute elasticity of the bodies.

As a result of the collision of the balls, the lighter one will rebound with greater speed. And the heavier one will roll back more slowly. Since the momentum of the body, reported by a heavier ball to a lighter one, is greater than the momentum given by a light ball to a heavy one.

Figure 1 - Law of conservation of momentum.

Thanks to the law of conservation of momentum, it is possible to describe reactive motion. Unlike other types of motion, reactive motion does not require interaction with other bodies. For example, a car moves due to the force of friction, which contributes to its repulsion from the surface of the earth. In jet motion, interaction with other bodies does not occur. Its cause is the separation from the body of part of its mass at a certain speed. That is, part of the fuel is separated from the engine in the form of expanding gases, while they move at great speed. Accordingly, the engine itself at the same time acquires a certain impulse that tells it the speed.

Often in physics they talk about the momentum of a body, implying the amount of motion. In fact, this concept is closely connected with a completely different quantity - with force. The impulse of force - what is it, how is it introduced into physics, and what is its meaning: all these issues are covered in detail in the article.

Number of movement

The impulse of the body and the impulse of the force are two interrelated quantities, moreover, they practically mean the same thing. First, let's look at the concept of momentum.

The amount of motion as a physical quantity first appeared in the scientific works of modern scientists, in particular in the 17th century. It is important to note two figures here: Galileo Galilei, the famous Italian, who called the quantity under discussion impeto (impulse), and Isaac Newton, the great Englishman, who, in addition to the magnitude of motus (movement), also used the concept of vis motrix (driving force).

So, the above-named scientists understood the product of the mass of an object and the speed of its linear movement in space as the amount of motion. This definition in the language of mathematics is written as follows:

Note that we are talking about the vector value (p¯), directed in the direction of body movement, which is proportional to the speed modulus, and the body mass plays the role of the proportionality coefficient.

Relation between the momentum of force and the change in p¯

As mentioned above, in addition to the momentum, Newton also introduced the concept of driving force. He defined this as follows:

This is the familiar law of the appearance of acceleration a¯ on a body as a result of some external force F¯ acting on it. This important formula allows us to derive the law of momentum of force. Note that a¯ is the time derivative of the rate (the rate of change of v¯), which means:

F¯ = m*dv¯/dt or F¯*dt = m*dv¯ =>
F¯*dt = dp¯, where dp¯ = m*dv¯

The first formula in the second line is the impulse of the force, that is, the value equal to the product of the force and the time interval during which it acts on the body. It is measured in newtons per second.

Formula Analysis

The expression for the impulse of force in the previous paragraph also reveals the physical meaning of this quantity: it shows how much the amount of motion changes over a period of time dt. Note that this change (dp¯) is completely independent of the total momentum of the body. The impulse of force is the cause of a change in the momentum, which can lead to both an increase in the latter (when the angle between the force F¯ and the speed v¯ is less than 90 o) and its decrease (the angle between F¯ and v¯ is greater than 90 o).

An important conclusion follows from the analysis of the formula: the units of measurement of the impulse of force are the same as those for p¯ (newton per second and kilogram per meter per second), moreover, the first value is equal to the change in the second, therefore, instead of the impulse of force, the phrase "body impulse" is often used, although it is more correct to say "change in momentum".

Forces that depend and do not depend on time

Above, the law of force impulse was presented in differential form. To calculate the value of this quantity, it is necessary to carry out integration over the action time. Then we get the formula:

∫ t1 t2 F¯(t)*dt = Δp¯

Here, the force F¯(t) acts on the body during the time Δt = t2-t1, which leads to a change in the momentum by Δp¯. As you can see, the impulse of a force is a quantity determined by a force that depends on time.

Now let's consider a simpler situation, which is realized in a number of experimental cases: we assume that the force does not depend on time, then we can easily take the integral and obtain a simple formula:

F¯*∫ t1 t2 dt = Δp¯ => F¯*(t2-t1) = Δp¯

When solving real problems on changing the momentum, despite the fact that the force generally depends on the time of action, it is assumed to be constant and some effective average value F¯ is calculated.

Examples of manifestation in practice of an impulse of force

What role this value plays is easiest to understand with specific examples from practice. Before we give them, we write out the corresponding formula once again:

Note that if Δp¯ is a constant value, then the momentum modulus of the force is also a constant, so the larger Δt, the smaller F¯, and vice versa.

Now let's give concrete examples of the momentum of force in action:

A person who jumps from any height to the ground tries to bend his knees when landing, thereby increasing the time Δt of the impact of the ground surface (the reaction force of the support F¯), thereby reducing its strength.
The boxer, deflecting his head from the blow, prolongs the contact time Δt of the opponent's glove with his face, reducing the impact force.
Modern cars are trying to design in such a way that in the event of a collision, their body is deformed as much as possible (deformation is a process that develops over time, which leads to a significant decrease in the force of a collision and, as a result, a decrease in the risk of injury to passengers).

The concept of the moment of force and its momentum

And the impulse of this moment is other quantities different from those considered above, since they no longer relate to linear, but to rotational motion. So, the moment of force M¯ is defined as the vector product of the shoulder (the distance from the axis of rotation to the point of action of the force) and the force itself, that is, the formula is valid:

The moment of force reflects the ability of the latter to perform torsion of the system around the axis. For example, if you hold the wrench away from the nut (large lever d¯), you can create a large moment M¯, which will allow you to unscrew the nut.

By analogy with the linear case, the momentum M¯ can be obtained by multiplying it by the time interval during which it acts on a rotating system, that is:

The quantity ΔL¯ is called the change in angular momentum, or angular momentum. The last equation is important for considering systems with an axis of rotation, because it shows that the angular momentum of the system will be conserved if there are no external forces that create the moment M¯, which is written mathematically as follows:

If M¯= 0 then L¯ = const

Thus, both momentum equations (for linear and circular motion) turn out to be similar in terms of their physical meaning and mathematical consequences.

Bird and Airplane Collision Challenge

This problem is not something fantastic. Such collisions do occur quite often. Thus, according to some data, in 1972, about 2.5 thousand bird collisions with combat and transport aircraft, as well as with helicopters, were recorded in Israeli airspace (the zone of the densest bird migration).

The task is as follows: it is necessary to approximately calculate what impact force falls on a bird if an airplane flying at a speed of v = 800 km / h is encountered on its path.

Before proceeding to the solution, let's assume that the length of the bird in flight is l = 0.5 meters, and its mass is m = 4 kg (it can be, for example, a drake or a goose).

We will neglect the speed of the bird (it is small compared to that of the aircraft), and we will also consider the mass of the aircraft to be much greater than that of the birds. These approximations allow us to say that the change in the momentum of the bird is equal to:

To calculate the impact force F, you need to know the duration of this incident, it is approximately equal to:

Combining these two formulas, we get the desired expression:

F \u003d Δp / Δt \u003d m * v 2 / l.

Substituting the numbers from the condition of the problem into it, we get F = 395062 N.

It will be more visual to translate this figure into an equivalent mass using the formula for body weight. Then we get: F = 395062/9.81 ≈ 40 tons! In other words, a bird perceives a collision with an airplane as if 40 tons of cargo had fallen on it.

Momentum is one of the most fundamental characteristics of a physical system. The momentum of a closed system is conserved for any processes occurring in it.

Let's start with the simplest case. The momentum of a material point of a mass moving at a speed is called the product

Law of change of momentum. From this definition, using Newton's second law, you can find the law of change in the momentum of a particle as a result of the action of a certain force on it. Changing the speed of a particle, the force also changes its momentum: . In the case of a constant acting force, therefore

The rate of change of momentum of a material point is equal to the resultant of all forces acting on it. At a constant force, the time interval in (2) can be taken by anyone. Therefore, for the change in the momentum of the particle over this interval, it is true

In the case of a force that changes with time, the entire period of time should be divided into small intervals during each of which the force can be considered constant. The change in the momentum of a particle for a separate interval is calculated by formula (3):

The total change in momentum over the entire time interval under consideration is equal to the vector sum of momentum changes over all intervals

If we use the concept of a derivative, then instead of (2), obviously, the law of change in the momentum of a particle is written as

Force impulse. The change in momentum over a finite period of time from 0 to is expressed by the integral

The value on the right side of (3) or (5) is called the impulse of the force. Thus, the change in momentum Dr of a material point over a period of time is equal to the momentum of the force acting on it during this period of time.

Equalities (2) and (4) are essentially another formulation of Newton's second law. It was in this form that this law was formulated by Newton himself.

The physical meaning of the concept of momentum is closely related to the intuitive or everyday experience that each of us has about whether it is easy to stop a moving body. What matters here is not the speed or mass of the stopped body, but both together, that is, precisely its momentum.

system momentum. The concept of momentum becomes especially meaningful when it is applied to a system of interacting material points. The total momentum P of a system of particles is the vector sum of the momenta of individual particles at the same time:

Here the summation is performed over all the particles in the system, so that the number of terms is equal to the number of particles in the system.

Internal and external forces. It is easy to arrive at the law of conservation of momentum for a system of interacting particles directly from Newton's second and third laws. The forces acting on each of the particles included in the system will be divided into two groups: internal and external. The internal force is the force with which the particle acts on the external force is the force with which all bodies that are not part of the system under consideration act on the particle.

The law of particle momentum change in accordance with (2) or (4) has the form

We add term by term equations (7) for all particles of the system. Then, on the left side, as follows from (6), we obtain the rate of change

total momentum of the system Since the internal forces of interaction between particles satisfy Newton's third law:

then when adding equations (7) on the right side, where the internal forces occur only in pairs, their sum will turn to zero. As a result, we get

The rate of change of total momentum is equal to the sum of external forces acting on all particles.

Let us pay attention to the fact that equality (9) has the same form as the law of change in the momentum of one material point, and only external forces enter the right side. In a closed system, where there are no external forces, the total momentum P of the system does not change, regardless of what internal forces act between the particles.

The total momentum does not change even in the case when the external forces acting on the system are summed to zero. It may turn out that the sum of external forces is equal to zero only along some direction. Although the physical system in this case is not closed, the component of the total momentum along this direction, as follows from formula (9), remains unchanged.

Equation (9) characterizes the system of material points as a whole, but refers to a certain point in time. It is easy to obtain from it the law of change in the momentum of the system over a finite period of time. If the acting external forces are unchanged during this period, then from (9) it follows

If the external forces change with time, then the right side of (10) will contain the sum of integrals over time from each of the external forces:

Thus, the change in the total momentum of a system of interacting particles over a certain period of time is equal to the vector sum of the impulses of external forces over this period.

Comparison with dynamic approach. Let us compare approaches to solving mechanical problems based on the equations of dynamics and based on the law of conservation of momentum using the following simple example.

A railway wagon of a mass moving at a constant speed collides with a stationary wagon of a mass and is coupled with it. How fast are the coupled wagons moving?

We do not know anything about the forces with which the cars interact during a collision, except for the fact that, based on Newton's third law, they are at every moment equal in absolute value and opposite in direction. With a dynamic approach, it is necessary to set up some kind of model for the interaction of cars. The simplest possible assumption is that the interaction forces are constant during the entire time that the coupling occurs. In this case, using Newton's second law for the speeds of each of the cars, after a time after the start of the coupling, we can write

Obviously, the coupling process ends when the speeds of the cars become the same. Assuming that this happens after time x, we have

From this we can express the momentum of the force

Substituting this value into any of the formulas (11), for example, into the second one, we find the expression for the final speed of the cars:

Of course, the assumption made about the constancy of the force of interaction of cars in the process of their coupling is very artificial. The use of more realistic models leads to more cumbersome calculations. However, in reality, the result for the final speed of the cars does not depend on the pattern of interaction (of course, provided that at the end of the process the cars are coupled and move at the same speed). The easiest way to verify this is using the law of conservation of momentum.

Since no external forces act on the cars in the horizontal direction, the total momentum of the system remains unchanged. Before the collision, it is equal to the momentum of the first car After coupling, the momentum of the cars is Equating these values, we immediately find

which naturally coincides with the answer obtained on the basis of the dynamic approach. The use of the law of conservation of momentum made it possible to find an answer to the question posed with the help of less cumbersome mathematical calculations, and this answer has greater generality, since no particular model of interaction was used to obtain it.

Let us illustrate the application of the law of conservation of momentum of the system by the example of a more complex problem, where the choice of a model for a dynamic solution is already difficult.

Task

Projectile burst. The projectile breaks at the top of the trajectory, which is at a height above the ground, into two identical fragments. One of them falls to the ground exactly below the break point after a time.

Solution First of all, let's write an expression for the distance over which an unexploded projectile would fly. Since the speed of the projectile at the top point (let's denote it as is directed horizontally, then the distance is equal to the product and times the time of falling from a height without an initial speed, equal to which an unexploded projectile would have flown. Since the speed of the projectile at the top point (let's denote it as directed horizontally, then the distance is equal to the product of the time of falling from a height without an initial velocity, equal to the body considered as a system of material points:

The rupture of the projectile into fragments occurs almost instantly, i.e., the internal forces that tear it apart act for a very short period of time. Obviously, the change in the speed of fragments under the action of gravity over such a short period of time can be neglected in comparison with the change in their speed under the action of these internal forces. Therefore, although the system under consideration, strictly speaking, is not closed, we can assume that its total momentum remains unchanged when the projectile breaks.

From the law of conservation of momentum, one can immediately reveal some features of the motion of fragments. Momentum is a vector quantity. Before the break, he lay in the plane of the projectile trajectory. Since, as stated in the condition, the velocity of one of the fragments is vertical, i.e., its momentum remains in the same plane, then the momentum of the second fragment also lies in this plane. This means that the trajectory of the second fragment will remain in the same plane.

Further, from the law of conservation of the horizontal component of the total momentum, it follows that the horizontal component of the velocity of the second fragment is equal to because its mass is equal to half the mass of the projectile, and the horizontal component of the momentum of the first fragment is equal to zero by condition. Therefore, the horizontal flight range of the second fragment from

the break point is equal to the product by the time of its flight. How to find this time?

To do this, we recall that the vertical components of the momenta (and, consequently, the velocities) of the fragments must be equal in absolute value and directed in opposite directions. The flight time of the second fragment of interest to us obviously depends on whether the vertical component of its velocity is directed upwards or downwards at the moment the projectile bursts (Fig. 108).

Rice. 108. The trajectory of the fragments after the burst of the projectile

It is easy to find out by comparing the time given in the condition for the vertical fall of the first fragment with the time for free fall from height A. If then the initial velocity of the first fragment is directed downward, and the vertical component of the velocity of the second is upward, and vice versa (cases a and in Fig. 108). At an angle a to the vertical, a bullet flies into the box with a speed u and almost instantly gets stuck in the sand. The box starts moving and then stops. How long did the box move? The ratio of the mass of the bullet to the mass of the box is y. Under what conditions will the box not move at all?

2. During the radioactive decay of the initially resting neutron, a proton, an electron and an antineutrino are formed. The momenta of a proton and an electron are equal and the angle between them is a. Determine the momentum of the antineutrino.

What is called the momentum of one particle and the momentum of a system of material points?

Formulate the law of change of momentum of one particle and system of material points.

Rice. 109. To determine the impulse of force from the graph

Why are internal forces not explicitly included in the law of change in the momentum of the system?

In what cases can the law of conservation of momentum of a system be used in the presence of external forces?

What are the advantages of using the law of conservation of momentum over the dynamic approach?

When a variable force acts on a body, its momentum is determined by the right side of formula (5) - the integral of over the time interval during which it acts. Let us be given a dependency graph (Fig. 109). How to determine the impulse of force for each of the cases a and

In some cases, it is possible to study the interaction of bodies without using expressions for the forces acting between the bodies. This is possible due to the fact that there are physical quantities that remain unchanged (conserved) during the interaction of bodies. In this chapter, we will consider two such quantities - momentum and mechanical energy.
Let's start with momentum.

The physical quantity, equal to the product of the mass of the body m and its speed, is called the momentum of the body (or simply the momentum):

Momentum is a vector quantity. The momentum modulus p = mv, and the direction of the momentum coincides with the direction of the body's velocity. The unit of momentum is 1 (kg * m)/s.

1. A truck with a mass of 3 tons is driving along a highway in the north direction at a speed of 40 km/h. In what direction and at what speed should a passenger car with a mass of 1 ton drive so that its momentum is equal to the momentum of the truck?

2. A ball weighing 400 g falls freely without initial velocity from a height of 5 m. After the impact, the ball bounces upwards, and the modulus of the ball's velocity does not change as a result of the impact.
a) What is the momentum of the ball just before the impact and what is its direction?
b) What is the momentum of the ball immediately after the impact and what is its direction?
c) What is the change in the momentum of the ball as a result of the impact and how is it directed? Find the momentum change graphically.
Clue. If the momentum of the body was equal to 1, and became equal to 2, then the change in momentum ∆ \u003d 2 - 1.

2. Law of conservation of momentum

The most important property of momentum is that under certain conditions the total momentum of interacting bodies remains unchanged (conserved).

Let's put experience

Two identical trolleys can roll along the table in a straight line with virtually no friction. (This experiment can be done with modern equipment.) The absence of friction is an important condition for our experiment!

We install latches on the carts, thanks to which the carts move as one body after a collision. Let the right cart first be at rest, and with the left push we will report the speed 0 (Fig. 25.1, a).

After the collision, the carts move together. Measurements show that their total speed is 2 times less than the initial speed of the left cart (25.1, b).

Let us denote the mass of each cart m and compare the total impulses of the carts before and after the collision.

We see that the total momentum of the carts remained unchanged (preserved).

Maybe this is true only when the bodies after the interaction move as a whole?

Let's put experience
Let's replace the latches with an elastic spring and repeat the experiment (Fig. 25.2).

This time the left cart stopped, and the right one acquired a speed equal to the initial speed of the left cart.

3. Prove that in this case the total momentum of the carts is also preserved.

Perhaps this is true only when the masses of the interacting bodies are equal?

Let's put experience
Let's fix another similar cart on the right cart and repeat the experiment (Fig. 25.3).

Now, after the collision, the left cart started moving in the opposite direction (that is, to the left) at a speed equal to -/3, and the double cart started moving to the right at a speed of 2/3.

4. Prove that in this experiment the total momentum of the carts is also preserved.

To determine under what conditions the total momentum of bodies is preserved, we introduce the concept of a closed system of bodies. This is the name of a system of bodies that interact only with each other (that is, they do not interact with bodies that are not included in this system).

Exactly closed systems of bodies do not exist in nature, if only because it is impossible to “turn off” the forces of universal gravitation.

But in many cases the system of bodies can be considered closed with good accuracy. For example, when external forces (forces acting on the bodies of the system from other bodies) balance each other or they can be neglected.

This is exactly what happened in our experiments with carts: the external forces acting on them (gravity and normal reaction force) balanced each other, and the friction force could be neglected. Therefore, the speeds of the carts changed only due to their interaction with each other.

The experiments described, like many others like them, indicate that
momentum conservation law: the vector sum of the momenta of the bodies that make up a closed system does not change with any interactions between the bodies of the system:
The law of conservation of momentum is satisfied only in inertial frames of reference.

Law of conservation of momentum as a consequence of Newton's laws

Let us show by the example of a closed system of two interacting bodies that the law of conservation of momentum is a consequence of Newton's second and third laws.

Let's designate masses of bodies m 1 and m 2 , and their initial speeds 1 and 2 . Then the vector sum of the momenta of the bodies

Let the interacting bodies move with accelerations 1 and 2 during the time interval ∆t.

5. Explain why the change in the total momentum of bodies can be written as

Clue. Use the fact that for each body ∆ = m∆, and also the fact that ∆ = ∆t.

6. Denote by 1 and 2 the forces acting respectively on the first and second bodies. Prove that

Clue. Take advantage of Newton's second law and the fact that the system is closed, as a result of which the accelerations of bodies are due only to the forces with which these bodies act on each other.

7. Prove that

Clue. Use Newton's third law.

So, the change in the total momentum of the interacting bodies is zero. And if the change in some value is zero, then this means that this value is conserved.

8. Why does it follow from the above reasoning that the law of conservation of momentum is satisfied only in inertial frames of reference?

3. Impulse of force

There is a saying: “If you knew where you would fall, you would lay straws.” Why do you need a "straw"? Why do athletes in training and competitions fall or jump on soft mats and not on hard floors? Why, after a jump, you need to land on bent legs, and not on straightened ones? Why do cars need seat belts and airbags?
We will be able to answer all these questions by getting acquainted with the concept of “impulse of force”.

The impulse of a force is the product of a force and the time interval ∆t during which this force acts.

The name "impulse of force" is not accidentally "echoes" with the concept of "impulse". Let us consider the case when a force acts on a body of mass m during a time interval ∆t.

9. Prove that the change in the body's momentum ∆ is equal to the momentum of the force acting on this body:

Clue. Use the fact that ∆ = m∆ and Newton's second law.

Let us rewrite formula (6) in the form

This formula is another form of Newton's second law. (It is in this form that Newton himself formulated this law.) It follows from it that a large force acts on a body if its momentum changes significantly over a very short period of time ∆t.

That is why large forces arise during impacts and collisions: impacts and collisions are characterized by precisely a small interaction time interval.

In order to weaken the force of impact or reduce the forces arising from the collision of bodies, it is necessary to lengthen the period of time during which the impact or collision occurs.

10. Explain the meaning of the saying given at the beginning of this section, and also answer other questions placed in the same paragraph.

11. A ball of mass 400 g hit the wall and bounced off it with the same modulo speed equal to 5 m/s. Before the impact, the speed of the ball was directed horizontally. What is the average pressure force of the ball on the wall if it was in contact with the wall for 0.02 s?

12. A cast-iron blank weighing 200 kg falls from a height of 1.25 m into the sand and plunges into it by 5 cm.
a) What is the momentum of the blank just before the impact?
b) What is the change in the momentum of the blank during the impact?
c) How long did the blow last?
d) What is the average impact force?

Additional questions and tasks

13. A ball of mass 200 g moves with a speed of 2 m/s to the left. How should another ball of mass 100 g move in order for the total momentum of the balls to be zero?

14. A ball of mass 300 g moves uniformly along a circle with a radius of 50 cm at a speed of 2 m/s. What is the modulus of change in the momentum of the ball:
a) for one full circulation period?
b) for half of the circulation period?
c) in 0.39 s?

15. The first board lies on the asphalt, and the second is the same - on loose sand. Explain why it is easier to drive a nail into the first board than into the second?

16. A bullet with a mass of 10 g, flying at a speed of 700 m / s, pierced the board, after which the speed of the bullet became equal to 300 m / s. Inside the board, the bullet moved for 40 μs.
a) What is the change in momentum of the bullet due to passing through the board?
b) With what average force did the bullet act on the board when passing through it?

Let the body mass m for some small time interval Δ t force acted Under the influence of this force, the speed of the body changed by Therefore, during the time Δ t the body moves with acceleration

From the basic law of dynamics ( Newton's second law) follows:

The physical quantity equal to the product of the mass of the body and the speed of its movement is called body momentum(or amount of movement). The momentum of the body is a vector quantity. The SI unit of momentum is kilogram-meter per second (kg m/s).

The physical quantity equal to the product of the force and the time of its action is called momentum of force . The momentum of a force is also a vector quantity.

In new terms Newton's second law can be formulated as follows:

ANDthe change in the momentum of the body (momentum) is equal to the momentum of the force.

Denoting the momentum of the body by the letter Newton's second law can be written as

It was in this general form that Newton himself formulated the second law. The force in this expression is the resultant of all forces applied to the body. This vector equality can be written in projections onto the coordinate axes:

Thus, the change in the projection of the momentum of the body on any of the three mutually perpendicular axes is equal to the projection of the momentum of the force on the same axis. Consider as an example one-dimensional movement, i.e., the movement of the body along one of the coordinate axes (for example, the axis OY). Let the body fall freely with an initial velocity υ 0 under the action of gravity; the fall time is t. Let's direct the axis OY vertically down. The momentum of gravity F t = mg during t equals mgt. This momentum is equal to the change in momentum of the body

This simple result coincides with the kinematicformulafor the speed of uniformly accelerated motion. In this example, the force remained unchanged in absolute value over the entire time interval t. If the force changes in magnitude, then the average value of the force must be substituted into the expression for the impulse of the force F cf on the time interval of its action. Rice. 1.16.1 illustrates a method for determining the impulse of a time-dependent force.

Let us choose a small interval Δ on the time axis t, during which the force F (t) remains virtually unchanged. Impulse of force F (t) Δ t in time Δ t will be equal to the area of the shaded bar. If the entire time axis on the interval from 0 to t split into small intervals Δ ti, and then sum the force impulses on all intervals Δ ti, then the total impulse of the force will be equal to the area formed by the step curve with the time axis. In the limit (Δ ti→ 0) this area is equal to the area bounded by the graph F (t) and axis t. This method for determining the momentum of a force from a graph F (t) is general and applicable to any laws of force change with time. Mathematically, the problem is reduced to integration functions F (t) on the interval .

The impulse of force, the graph of which is shown in fig. 1.16.1, on the interval from t 1 = 0 s to t 2 = 10 s is equal to:

In this simple example

In some cases, the average force F cp can be determined if the time of its action and the impulse imparted to the body are known. For example, a strong impact of a football player on a ball weighing 0.415 kg can give him a speed υ = 30 m/s. The impact time is approximately equal to 8·10 -3 s.

Pulse p acquired by the ball as a result of a stroke is:

Therefore, the average force F cf, with which the football player's foot acted on the ball during the kick, is:

This is a very big power. It is approximately equal to the weight of a body weighing 160 kg.

If the movement of the body during the action of the force occurred along a certain curvilinear trajectory, then the initial and final momenta of the body may differ not only in absolute value, but also in direction. In this case, to determine the change in momentum, it is convenient to use pulse diagram , which depicts the vectors and , as well as the vector constructed according to the parallelogram rule. As an example, in fig. 1.16.2 shows an impulse diagram for a ball bouncing off a rough wall. ball mass m hit the wall with a speed at an angle α to the normal (axis OX) and rebounded from it with a speed at an angle β. During contact with the wall, a certain force acted on the ball, the direction of which coincides with the direction of the vector

With a normal fall of a ball with a mass m on an elastic wall with a speed , after the rebound the ball will have a speed . Therefore, the change in the momentum of the ball during the rebound is

In projections on the axis OX this result can be written in the scalar form Δ px = –2mυ x. Axis OX directed away from the wall (as in Fig. 1.16.2), so υ x < 0 и Δpx> 0. Therefore, the module Δ p momentum change is related to the modulus υ of the ball speed by the relation Δ p = 2mυ.