When the impulse is 0. The momentum of the body. Law of conservation of momentum

Having studied Newton's laws, we see that with their help it is possible to solve the basic problems of mechanics if we know all the forces acting on the body. There are situations in which it is difficult or even impossible to determine these values. Let's consider several such situations.When two billiard balls or cars collide, we can state that current forces, that this is their nature, elastic forces act here. However, we will not be able to accurately determine either their modules or their directions, especially since these forces have an extremely short duration of action.With the movement of rockets and jet planes, we also can say little about the forces that set these bodies in motion.In such cases, methods are used that allow one to avoid solving the equations of motion and immediately use the consequences of these equations. In this case, new physical quantities are introduced. Let's consider one of these quantities, called the momentum of the body

An arrow fired from a bow. The longer the contact of the string with the arrow continues (∆t), the greater the change in the arrow's momentum (∆), and therefore, the higher its final speed.

Two colliding balls. While the balls are in contact, they act on each other with forces equal in magnitude, as Newton’s third law teaches us. This means that the changes in their momenta must also be equal in magnitude, even if the masses of the balls are not equal.

After analyzing the formulas, two important conclusions can be drawn:

1. Identical forces acting for the same period of time cause the same changes in momentum different bodies, regardless of the mass of the latter.

2. The same change in the momentum of a body can be achieved either by acting with a small force over a long period of time, or by acting briefly with a large force on the same body.

According to Newton's second law, we can write:

∆t = ∆ = ∆ / ∆t

The ratio of the change in the momentum of a body to the period of time during which this change occurred is equal to the sum of the forces acting on the body.

Having analyzed this equation, we see that Newton's second law allows us to expand the class of problems to be solved and include problems in which the mass of bodies changes over time.

If we try to solve problems with variable mass of bodies using the usual formulation of Newton’s second law:

then attempting such a solution would lead to an error.

An example of this is the already mentioned jet plane or space rocket, which burn fuel while moving, and the products of this combustion are released into the surrounding space. Naturally, the mass of an aircraft or rocket decreases as fuel is consumed.

Despite the fact that Newton’s second law in the form “the resultant force is equal to the product of the mass of a body and its acceleration” allows us to solve a fairly wide class of problems, there are cases of motion of bodies that cannot be fully described by this equation. In such cases, it is necessary to apply another formulation of the second law, connecting the change in the momentum of the body with the impulse of the resultant force. In addition, there are a number of problems in which solving the equations of motion is mathematically extremely difficult or even impossible. In such cases, it is useful for us to use the concept of momentum.

Using the law of conservation of momentum and the relationship between the momentum of a force and the momentum of a body, we can derive Newton's second and third laws.

Newton's second law is derived from the relationship between the impulse of a force and the momentum of a body.

The impulse of force is equal to the change in the momentum of the body:

Having made the appropriate transfers, we obtain the dependence of force on acceleration, because acceleration is defined as the ratio of the change in speed to the time during which this change occurred:

Substituting the values ​​into our formula, we obtain the formula for Newton’s second law:

To derive Newton's third law, we need the law of conservation of momentum.

Vectors emphasize the vector nature of speed, that is, the fact that speed can change in direction. After transformations we get:

Since the period of time in a closed system was a constant value for both bodies, we can write:

We have obtained Newton's third law: two bodies interact with each other with forces equal in magnitude and opposite in direction. The vectors of these forces are directed towards each other, respectively, the modules of these forces are equal in value.

Bibliography

1. Tikhomirova S.A., Yavorsky B.M. Physics ( a basic level of) - M.: Mnemosyne, 2012.
2. Gendenshtein L.E., Dick Yu.I. Physics 10th grade. - M.: Mnemosyne, 2014.
3. Kikoin I.K., Kikoin A.K. Physics - 9, Moscow, Education, 1990.

Homework

1. Define the impulse of a body, the impulse of force.
2. How are the impulse of a body related to the impulse of force?
3. What conclusions can be drawn from the formulas for body impulse and force impulse?

1. Internet portal Questions-physics.ru ().
2. Internet portal Frutmrut.ru ().
3. Internet portal Fizmat.by ().

Newton's second law \(~m \vec a = \vec F\) can be written in a different form, which is given by Newton himself in his main work “Mathematical Principles of Natural Philosophy”.

If a constant force acts on a body (material point), then the acceleration is also constant

\(~\vec a = \frac(\vec \upsilon_2 - \vec \upsilon_1)(\Delta t)\) ,

where \(~\vec \upsilon_1\) and \(~\vec \upsilon_2\) are the initial and final values ​​of the body velocity.

Substituting this acceleration value into Newton's second law, we get:

\(~\frac(m \cdot (\vec \upsilon_2 - \vec \upsilon_1))(\Delta t) = \vec F\) or \(~m \vec \upsilon_2 - m \vec \upsilon_1 = \vec F \Delta t\) . (1)

This equation introduces a new physical quantity- momentum of a material point.

The impulse of the material points name a quantity equal to the product of the mass of a point and its speed.

Let us denote momentum (it is also sometimes called momentum) by the letter \(~\vec p\) . Then

\(~\vec p = m \vec \upsilon\) . (2)

From formula (2) it is clear that momentum is a vector quantity. Because m> 0, then the momentum has the same direction as the speed.

The unit of impulse does not have a special name. Its name is obtained from the definition of this quantity:

Another form of writing Newton's second law

Let us denote by \(~\vec p_1 = m \vec \upsilon_1\) the momentum of the material point at the initial moment of the interval Δ t, and through \(~\vec p_2 = m \vec \upsilon_2\) - the impulse at the final moment of this interval. Then \(~\vec p_2 - \vec p_1 = \Delta \vec p\) is change in momentum in time Δ t. Now equation (1) can be written as follows:

\(~\Delta \vec p = \vec F \Delta t\) . (3)

Since Δ t> 0, then the directions of the vectors \(~\Delta \vec p\) and \(~\vec F\) coincide.

According to formula (3)

the change in the momentum of a material point is proportional to the force applied to it and has the same direction as the force.

This is exactly how it was first formulated Newton's second law.

The product of a force and the duration of its action is called impulse of force. Do not confuse the impulse \(~m \vec \upsilon\) of a material point and the impulse of force \(\vec F \Delta t\) . These are completely different concepts.

Equation (3) shows that identical changes in the momentum of a material point can be obtained as a result of the action of a large force over a small time interval or a small force over a large time interval. When you jump from a certain height, your body stops due to the action of force from the ground or floor. The shorter the duration of the collision, the greater the braking force. To reduce this force, braking must occur gradually. This is why athletes land on soft mats when high jumping. By bending, they gradually slow down the athlete. Formula (3) can be generalized to the case when the force changes over time. To do this, the entire time period Δ t the actions of the force must be divided into such small intervals Δ t i so that on each of them the value of the force can be considered constant without a large error. For each small time interval, formula (3) is valid. Summing up the changes in pulses over short time intervals, we get:

\(~\Delta \vec p = \sum^(N)_(i=1)(\vec F_i \Delta t_i)\) . (4)

The symbol Σ (Greek letter "sigma") means "sum". Indexes i= 1 (bottom) and N(at the top) mean that it is summed N terms.

To find the momentum of a body, they do this: mentally break the body into individual elements (material points), find the impulses of the resulting elements, and then sum them up as vectors.

The momentum of a body is equal to the sum of the impulses of its individual elements.

Change in the momentum of a system of bodies. Law of conservation of momentum

When considering any mechanical problem, we are interested in the motion of a certain number of bodies. The set of bodies whose motion we study is called mechanical system or just a system.

Changing the momentum of a system of bodies

Let us consider a system consisting of three bodies. These could be three stars experiencing influence from neighboring cosmic bodies. External forces act on the bodies of the system \(~\vec F_i\) ( i- body number; for example, \(~\vec F_2\) is the sum of external forces acting on body number two). Between the bodies there are forces \(~\vec F_(ik)\) called internal forces (Fig. 1). Here is the first letter i in the index means the number of the body on which the force \(~\vec F_(ik)\) acts, and the second letter k means the number of the body from which it acts given power. Based on Newton's third law

\(~\vec F_(ik) = - \vec F_(ki)\) . (5)

Due to the action of forces on the bodies of the system, their impulses change. If the force does not change noticeably over a short period of time, then for each body of the system we can write down the change in momentum in the form of equation (3):

\(~\Delta (m_1 \vec \upsilon_1) = (\vec F_(12) + \vec F_(13) + \vec F_1) \Delta t\) , \(~\Delta (m_2 \vec \upsilon_2) = (\vec F_(21) + \vec F_(23) + \vec F_2) \Delta t\) , (6) \(~\Delta (m_3 \vec \upsilon_3) = (\vec F_(31) + \vec F_(32) + \vec F_3) \Delta t\) .

Here on the left side of each equation is the change in the momentum of the body \(~\vec p_i = m_i \vec \upsilon_i\) for a short time Δ t. In more detail\[~\Delta (m_i \vec \upsilon_i) = m_i \vec \upsilon_(ik) - m_i \vec \upsilon_(in)\] where \(~\vec \upsilon_(in)\) is the speed in beginning, and \(~\vec \upsilon_(ik)\) - at the end of the time interval Δ t.

Let us add the left and right sides of equations (6) and show that the sum of changes in the impulses of individual bodies is equal to the change in the total impulse of all bodies of the system, equal to

\(~\vec p_c = m_1 \vec \upsilon_1 + m_2 \vec \upsilon_2 + m_3 \vec \upsilon_3\) . (7)

Really,

\(~\Delta (m_1 \vec \upsilon_1) + \Delta (m_2 \vec \upsilon_2) + \Delta (m_3 \vec \upsilon_3) = m_1 \vec \upsilon_(1k) - m_1 \vec \upsilon_(1n) + m_2 \vec \upsilon_(2k) - m_2 \vec \upsilon_(2n) + m_3 \vec \upsilon_(3k) - m_3 \vec \upsilon_(3n) =\) \(~=(m_1 \vec \upsilon_( 1k) + m_2 \vec \upsilon_(2k) + m_3 \vec \upsilon_(3k)) -(m_1 \vec \upsilon_(1n) + m_2 \vec \upsilon_(2n) + m_3 \vec \upsilon_(3n)) = \vec p_(ck) - \vec p_(cn) = \Delta \vec p_c\) .

Thus,

\(~\Delta \vec p_c = (\vec F_(12) + \vec F_(13) + \vec F_(21) + \vec F_(23) + \vec F_(31) + \vec F_(32 ) + \vec F_1 + \vec F_2 + \vec F_3) \Delta t\) . (8)

But the interaction forces of any pair of bodies add up to zero, since according to formula (5)

\(~\vec F_(12) = - \vec F_(21) ; \vec F_(13) = - \vec F_(31) ; \vec F_(23) = - \vec F_(32)\) .

Therefore, the change in the momentum of the system of bodies is equal to the momentum of external forces:

\(~\Delta \vec p_c = (\vec F_1 + \vec F_2 + \vec F_3) \Delta t\) . (9)

We came to an important conclusion:

The momentum of a system of bodies can only be changed by external forces, and the change in the momentum of the system is proportional to the sum of the external forces and coincides with it in direction. Internal forces, changing the impulses of individual bodies of the system, do not change the total impulse of the system.

Equation (9) is valid for any time interval if the sum of external forces remains constant.

Law of conservation of momentum

An extremely important consequence follows from equation (9). If the sum of external forces acting on the system is zero, then the change in the momentum of the system is also equal to zero\[~\Delta \vec p_c = 0\] . This means that, no matter what time interval we take, the total impulse at the beginning of this interval \(~\vec p_(cn)\) and at its end \(~\vec p_(ck)\) is the same\ [~\vec p_(cn) = \vec p_(ck)\] . The momentum of the system remains unchanged, or, as they say, conserved:

\(~\vec p_c = m_1 \vec \upsilon_1 + m_2 \vec \upsilon_2 + m_3 \vec \upsilon_3 = \operatorname(const)\) . (10)

Law of conservation of momentum is formulated as follows:

if the sum of external forces acting on the bodies of the system is equal to zero, then the momentum of the system is conserved.

Bodies can only exchange impulses, but the total value of the impulse does not change. You just need to remember that the vector sum of the pulses is preserved, and not the sum of their modules.

As can be seen from our conclusion, the law of conservation of momentum is a consequence of Newton’s second and third laws. A system of bodies that is not acted upon by external forces is called closed or isolated. In a closed system of bodies, momentum is conserved. But the scope of application of the law of conservation of momentum is wider: even if external forces act on the bodies of the system, but their sum is zero, the momentum of the system is still conserved.

The obtained result is easily generalized to the case of a system containing an arbitrary number N of bodies:

\(~m_1 \vec \upsilon_(1n) + m_2 \vec \upsilon_(2n) + m_3 \vec \upsilon_(3n) + \ldots + m_N \vec \upsilon_(Nn) = m_1 \vec \upsilon_(1k) + m_2 \vec \upsilon_(2k) + m_3 \vec \upsilon_(3k) + \ldots + m_N \vec \upsilon_(Nk)\) . (eleven)

Here \(~\vec \upsilon_(in)\) is the speed of bodies at the initial moment of time, and \(~\vec \upsilon_(ik)\) - at the final moment. Since momentum is a vector quantity, equation (11) is a compact representation of three equations for the projections of the system’s momentum onto the coordinate axes.

When is the law of conservation of momentum satisfied?

All real systems, of course, are not closed; the sum of external forces can quite rarely turn out to be equal to zero. Nevertheless, in many cases the law of conservation of momentum can be applied.

If the sum of external forces is not equal to zero, but the sum of the projections of forces on some direction is equal to zero, then the projection of the system’s momentum on this direction is preserved. For example, a system of bodies on Earth or near its surface cannot be closed, since all bodies are affected by the force of gravity, which changes the momentum vertically according to equation (9). However, along the horizontal direction, the force of gravity cannot change the momentum, and the sum of the projections of the impulses of bodies onto the horizontally directed axis will remain unchanged if the action of resistance forces can be neglected.

In addition, during fast interactions (a projectile explosion, a gun shot, collisions of atoms, etc.), the change in the impulses of individual bodies will actually be due only to internal forces. The momentum of the system is preserved with great accuracy, because such external forces as the force of gravity and the friction force, which depends on speed, do not noticeably change the momentum of the system. They are small compared to internal forces. Thus, the speed of projectile fragments during an explosion, depending on the caliber, can vary within the range of 600 - 1000 m/s. The time interval during which gravity could impart such a speed to bodies is equal to

\(~\Delta t = \frac(m \Delta \upsilon)(mg) \approx 100 s\)

Internal gas pressure forces impart such velocities in 0.01 s, i.e. 10,000 times faster.

Jet propulsion. Meshchersky equation. Reactive force

Under jet propulsion understand the movement of a body that occurs when some part of it is separated at a certain speed relative to the body,

for example, when combustion products flow out of a jet nozzle aircraft. In this case, a so-called reactive force appears, imparting acceleration to the body.

Observing jet motion is very simple. Inflate a child's rubber ball and release it. The ball will quickly rise up (Fig. 2). The movement, however, will be short-lived. The reactive force acts only as long as the outflow of air continues.

The main feature of the reactive force is that it occurs without any interaction with external bodies. There is only interaction between the rocket and the stream of matter flowing out of it.

The force that imparts acceleration to a car or pedestrian on the ground, a steamship on the water or a propeller-driven airplane in the air arises only due to the interaction of these bodies with the ground, water or air.

When the fuel combustion products flow out, due to the pressure in the combustion chamber, they acquire a certain speed relative to the rocket and, therefore, a certain momentum. Therefore, in accordance with the law of conservation of momentum, the rocket itself receives an impulse of the same magnitude, but directed in the opposite direction.

The mass of the rocket decreases over time. A rocket in flight is a body of variable mass. To calculate its motion, it is convenient to apply the law of conservation of momentum.

Meshchersky equation

Let us derive the equation of motion of the rocket and find an expression for the reactive force. We will assume that the speed of the gases flowing out of the rocket relative to the rocket is constant and equal to \(~\vec u\) . External forces do not act on the rocket: it is in outer space far from stars and planets.

Let at some moment of time the speed of the rocket relative to the inertial system associated with the stars be equal to \(~\vec \upsilon\) (Fig. 3), and the mass of the rocket be equal M. After a short time interval Δ t the mass of the rocket will become equal

\(~M_1 = M - \mu \Delta t\) ,

Where μ - fuel consumption ( fuel consumption is called the ratio of the mass of burned fuel to the time of its combustion).

During the same period of time, the speed of the rocket will change by \(~\Delta \vec \upsilon\) and become equal to \(~\vec \upsilon_1 = \vec \upsilon + \Delta \vec \upsilon\) . The speed of gas outflow relative to the chosen inertial reference frame is equal to \(~\vec \upsilon + \vec u\) (Fig. 4), since before the start of combustion the fuel had the same speed as the rocket.

Let us write down the law of conservation of momentum for the rocket-gas system:

\(~M \vec \upsilon = (M - \mu \Delta t)(\vec \upsilon + \Delta \vec \upsilon) + \mu \Delta t(\vec \upsilon + \vec u)\) .

Opening the brackets, we get:

\(~M \vec \upsilon = M \vec \upsilon - \mu \Delta t \vec \upsilon + M \Delta \vec \upsilon - \mu \Delta t \Delta \vec \upsilon + \mu \Delta t \vec \upsilon + \mu \Delta t \vec u\) .

The term \(~\mu \Delta t \vec \upsilon\) can be neglected in comparison with the others, since it contains the product of two small quantities (this quantity is said to be of the second order of smallness). After bringing similar terms we will have:

\(~M \Delta \vec \upsilon = - \mu \Delta t \vec u\) or \(~M \frac(\Delta \vec \upsilon)(\Delta t) = - \mu \vec u\ ) . (12)

This is one of Meshchersky’s equations for the motion of a body of variable mass, obtained by him in 1897.

If we introduce the notation \(~\vec F_r = - \mu \vec u\) , then equation (12) will coincide in form with Newton’s second law. However, body weight M here it is not constant, but decreases with time due to the loss of matter.

The quantity \(~\vec F_r = - \mu \vec u\) is called reactive force. It appears as a result of the outflow of gases from the rocket, is applied to the rocket and is directed opposite to the speed of the gases relative to the rocket. Reactive force is determined only by the speed of gas flow relative to the rocket and fuel consumption. It is important that it does not depend on the details of the engine design. It is only important that the engine ensures the outflow of gases from the rocket at a speed of \(~\vec u\) with fuel consumption μ . Reactive force space rockets reaches 1000 kN.

If external forces act on a rocket, then its movement is determined by the reactive force and the sum of external forces. In this case, equation (12) will be written as follows:

\(~M \frac(\Delta \vec \upsilon)(\Delta t) = \vec F_r + \vec F\) . (13)

Jet engines

Jet engines are currently widely used in connection with the development of outer space. They are also used for meteorological and military missiles of various ranges. In addition, all modern high-speed aircraft are equipped with air-breathing engines.

In outer space, it is impossible to use any engines other than jet engines: there is no support (solid, liquid or gaseous) from which the spacecraft could receive acceleration. The use of jet engines for aircraft and rockets that do not go beyond the atmosphere is due to the fact that it is jet engines that are capable of providing maximum flight speed.

Jet engines are divided into two classes: rocket And air-jet.

In rocket engines, the fuel and the oxidizer necessary for its combustion are located directly inside the engine or in its fuel tanks.

Figure 5 shows a diagram of a solid fuel rocket engine. Gunpowder or some other solid fuel, capable of burning in the absence of air, is placed inside the combustion chamber of the engine.

When fuel burns, gases are formed that have a very high temperature and exert pressure on the walls of the chamber. The pressure on the front wall of the chamber is greater than on the back wall, where the nozzle is located. The gases flowing through the nozzle do not encounter a wall on their way on which they could exert pressure. The result is a force that pushes the rocket forward.

The narrowed part of the chamber - the nozzle - serves to increase the rate of flow of combustion products, which in turn increases the reactive force. The narrowing of the gas stream causes an increase in its speed, since in this case the same mass of gas must pass through a smaller cross section per unit time as with a larger cross section.

Also applicable rocket engines, operating on liquid fuel.

In liquid-propellant jet engines (LPRE), kerosene, gasoline, alcohol, aniline, liquid hydrogen, etc. can be used as fuel, and liquid oxygen, nitric acid, liquid fluorine, hydrogen peroxide, etc. can be used as an oxidizing agent necessary for combustion. The fuel and oxidizer are stored separately in special tanks and, using pumps, are supplied to the chamber, where, during fuel combustion, a temperature of up to 3000 ° C and a pressure of up to 50 atm develops (Fig. 6). Otherwise the engine operates in the same way as a solid fuel engine.

Hot gases (combustion products), exiting through the nozzle, rotate the gas turbine, which drives the compressor. Turbocompressor engines are installed in our airliners Tu-134, Il-62, Il-86, etc.

Not only rockets are equipped with jet engines, but also most of modern aircraft.

Successes in space exploration

Basic theory jet engine And scientific proof the possibilities of flights in interplanetary space were first expressed and developed by the Russian scientist K.E. Tsiolkovsky in his work “Exploration of world spaces using reactive instruments.”

K.E. Tsiolkovsky also came up with the idea of ​​​​using multi-stage rockets. The individual stages that make up the rocket are supplied with their own engines and fuel supply. As the fuel burns out, each successive stage is separated from the rocket. Therefore, in the future, fuel is not consumed to accelerate its body and engine.

Tsiolkovsky's idea of ​​​​building a large satellite station in orbit around the Earth, from which rockets will be launched to other planets solar system, has not yet been implemented, but there is no doubt that sooner or later such a station will be created.

Currently, Tsiolkovsky’s prophecy is becoming a reality: “Humanity will not remain forever on Earth, but in pursuit of light and space, it will first timidly penetrate beyond the atmosphere, and then conquer the entire circumsolar space.”

Our country has the great honor of launching the first artificial satellite Earth. Also, for the first time in our country, on April 12, 1961, a flight was carried out spaceship with cosmonaut Yu.A. Gagarin on board.

These flights were carried out on rockets designed by domestic scientists and engineers under the leadership of S.P. Queen. American scientists, engineers and astronauts have made great contributions to space exploration. Two American astronaut from the crew of the Apollo 11 spacecraft - Neil Armstrong and Edwin Aldrin - landed on the Moon for the first time on July 20, 1969. Man took his first steps on the cosmic body of the solar system.

With the entry of man into space, not only the possibilities of exploring other planets opened up, but also truly fantastic opportunities for studying natural phenomena and Earth's resources that one could only dream of. Cosmic natural history emerged. Previously, a general map of the Earth was compiled bit by bit, like a mosaic panel. Now images from orbit, covering millions of square kilometers, make it possible to select the most interesting areas of the earth's surface for study, thereby saving effort and money. From space, large geological structures are better distinguished: plates, deep faults earth's crust- places of the most probable occurrence of minerals. Found from space new type geological formations ring structures similar to craters of the Moon and Mars,

Nowadays, orbital complexes have developed technologies for producing materials that cannot be produced on Earth, but only in a state of prolonged weightlessness in space. The cost of these materials (ultra-pure single crystals, etc.) is close to the cost of launching spacecraft.

Literature

1. Physics: Mechanics. 10th grade: Textbook. for in-depth study of physics / M.M. Balashov, A.I. Gomonova, A.B. Dolitsky and others; Ed. G.Ya. Myakisheva. - M.: Bustard, 2002. - 496 p.

A 22-caliber bullet has a mass of only 2 g. If you throw such a bullet to someone, he can easily catch it even without gloves. If you try to catch such a bullet flying out of the muzzle at a speed of 300 m/s, then even gloves will not help.

If a toy cart is rolling towards you, you can stop it with your toe. If a truck is rolling towards you, you should move your feet out of its path.

Let's consider a problem that demonstrates the connection between a force impulse and a change in the momentum of a body.

Example. The mass of the ball is 400 g, the speed that the ball acquired after impact is 30 m/s. The force with which the foot acted on the ball was 1500 N, and the impact time was 8 ms. Find the impulse of force and the change in momentum of the body for the ball.

Change in body momentum

Example. Estimate the average force from the floor acting on the ball during impact.

1) During a strike, two forces act on the ball: ground reaction force, gravity.

The reaction force changes during the impact time, so it is possible to find the average reaction force of the floor.

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

Quantity of movement

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

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

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

Please note that we're talking about about the vector value (p¯), directed in the direction of the body’s movement, which is proportional to the velocity modulus, and the role of the proportionality coefficient is played by the mass of the body.

Relationship between force impulse and change in p¯ value

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

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

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 force, that is, a value equal to the product of force by the period of time 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 momentum changes over a period of time dt. Note that this change (dp¯) is completely independent of the total value of the momentum of the body. An impulse of force is the cause of a change in 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 to its decrease (the angle between F¯ and v¯ is more than 90 o).

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

Time-dependent and independent forces

Above, the law of force impulse was presented in differential form. To calculate the value of this quantity, it is necessary to integrate 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 get simple formula:

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

When deciding real problems to change the amount of motion, despite the fact that the force in the general case depends on the time of action, it is assumed to be constant and some effective force is calculated average value F¯.

Examples of the manifestation of a force impulse in practice

The role this quantity plays is easiest to understand using specific examples from practice. Before presenting them, let us write out the corresponding formula again:

Note that if Δp¯ is a constant value, then the magnitude of the force impulse is also a constant, so the greater Δt, the smaller F¯, and vice versa.

Now let's give specific examples impulse 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 exposure to the ground surface (ground reaction force F¯), thereby reducing its force.
• The boxer, deflecting his head from the blow, prolongs the contact time Δt of the opponent’s glove with his face, reducing impact force.
• Modern cars are trying to be designed 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 reduction in the force of the collision and, as a result, a reduction in the risk of injury to passengers).

The concept of moment of force and its impulse

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

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

By analogy with the linear case, the impulse M¯ can be obtained by multiplying it by the period of time during which it acts on the 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 creating 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-plane collision problem

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 the bird if an airplane flying at a speed of v = 800 km/h meets on its path of movement.

Before proceeding with 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 (this could be, for example, a drake or a goose).

Let us neglect the speed of the bird (it is small compared to that of the airplane), and we will also assume that the mass of the airplane is much greater than that of the bird. These approximations allow us to say that the change in 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 obtain the desired expression:

F = Δp/Δt = m*v 2 /l.

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

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

Topics of the Unified State Examination codifier: momentum of a body, momentum of a system of bodies, law of conservation of momentum.

Pulse of a body is a vector quantity equal to the product of the body’s mass and its speed:

There are no special units for measuring impulse. The dimension of momentum is simply the product of the dimension of mass and the dimension of velocity:

Why is the concept of momentum interesting? It turns out that with its help you can give Newton's second law a slightly different, also extremely useful form.

Newton's second law in impulse form

Let be the resultant of forces applied to a body of mass . We start with the usual notation of Newton's second law:

Taking into account that the acceleration of the body is equal to the derivative of the velocity vector, Newton’s second law is rewritten as follows:

We introduce a constant under the derivative sign:

As you can see, the derivative of the impulse is obtained on the left side:

. ( 1 )

The ratio (1) is new form records of Newton's second law.

Newton's second law in impulse form. The derivative of the momentum of a body is the resultant of the forces applied to the body.

We can say this: the resulting force acting on a body is equal to the rate of change of the body’s momentum.

The derivative in formula (1) can be replaced by the ratio of final increments:

. ( 2 )

In this case, there is an average force acting on the body during the time interval. The smaller the value, the closer the ratio is to the derivative, and the closer the average force is to its instantaneous value in this moment time.

In tasks, as a rule, the time interval is quite small. For example, this could be the time of impact of the ball with the wall, and then - the average force acting on the ball from the wall during the impact.

The vector on the left side of relation (2) is called change in impulse during . The change in momentum is the difference between the final and initial momentum vectors. Namely, if is the momentum of the body at some initial moment of time, is the momentum of the body after a period of time, then the change in momentum is the difference:

Let us emphasize once again that the change in momentum is the difference between vectors (Fig. 1):

Let, for example, the ball fly perpendicular to the wall (the momentum before the impact is equal to ) and bounce back without losing speed (the momentum after the impact is equal to ). Despite the fact that the impulse has not changed in absolute value (), there is a change in the impulse:

Geometrically, this situation is shown in Fig. 2:

The modulus of change in momentum, as we see, is equal to twice the modulus of the initial impulse of the ball: .

Let us rewrite formula (2) as follows:

, ( 3 )

or, describing the change in momentum, as above:

The quantity is called impulse of power. There is no special unit of measurement for force impulse; the dimension of the force impulse is simply the product of the dimensions of force and time:

(Note that this turns out to be another possible unit of measurement for a body's momentum.)

The verbal formulation of equality (3) is as follows: the change in the momentum of a body is equal to the momentum of the force acting on the body over a given period of time. This, of course, is again Newton's second law in momentum form.

Example of force calculation

As an example of applying Newton's second law in impulse form, let's consider the following problem.

Task. A ball of mass g, flying horizontally at a speed of m/s, hits a smooth vertical wall and bounces off it without losing speed. The angle of incidence of the ball (that is, the angle between the direction of movement of the ball and the perpendicular to the wall) is equal to . The blow lasts for s. Find the average force,
acting on the ball during impact.

Solution. Let us show first of all that the angle of reflection is equal to the angle of incidence, that is, the ball will bounce off the wall at the same angle (Fig. 3).

According to (3) we have: . It follows that the vector of momentum change co-directed with vector, that is, directed perpendicular to the wall in the direction of the ball’s rebound (Fig. 5).

Rice. 5. To the task

Vectors and
equal in modulus
(since the speed of the ball has not changed). Therefore, a triangle composed of vectors , and , is isosceles. This means that the angle between the vectors and is equal to , that is, the angle of reflection is really equal to the angle of incidence.

Now notice in addition that in our isosceles triangle there is an angle (this is the angle of incidence); therefore, this triangle is equilateral. From here:

And then the desired average force acting on the ball is:

Impulse of a system of bodies

Let's start with a simple situation of a two-body system. Namely, let there be body 1 and body 2 with impulses and, respectively. The impulse of the system of these bodies is the vector sum of the impulses of each body:

It turns out that for the momentum of a system of bodies there is a formula similar to Newton’s second law in the form (1). Let's derive this formula.

We will call all other objects with which the bodies 1 and 2 we are considering interact external bodies. The forces with which external bodies act on bodies 1 and 2 are called by external forces. Let be the resultant external force acting on body 1. Similarly, let be the resultant external force acting on body 2 (Fig. 6).

In addition, bodies 1 and 2 can interact with each other. Let body 2 act on body 1 with a force. Then body 1 acts on body 2 with a force. According to Newton's third law, the forces are equal in magnitude and opposite in direction: . Forces and are internal forces, operating in the system.

Let us write for each body 1 and 2 Newton’s second law in the form (1):

, ( 4 )

. ( 5 )

Let's add equalities (4) and (5):

On the left side of the resulting equality there is a sum of derivatives equal to the derivative of the sum of the vectors and . On the right side we have, by virtue of Newton’s third law:

But - this is the impulse of the system of bodies 1 and 2. Let us also denote - this is the resultant of external forces acting on the system. We get:

. ( 6 )

Thus, the rate of change of momentum of a system of bodies is the resultant of external forces applied to the system. We wanted to obtain equality (6), which plays the role of Newton’s second law for a system of bodies.

Formula (6) was derived for the case of two bodies. Now let us generalize our reasoning to the case of an arbitrary number of bodies in the system.

Impulse of the system of bodies bodies is the vector sum of the momenta of all bodies included in the system. If a system consists of bodies, then the momentum of this system is equal to:

Then everything is done in exactly the same way as above (only technically it looks a little more complicated). If for each body we write down equalities similar to (4) and (5), and then add all these equalities, then on the left side we again obtain the derivative of the momentum of the system, and on the right side there remains only the sum of external forces (internal forces, adding in pairs, will give zero due to Newton's third law). Therefore, equality (6) will remain valid in the general case.

Law of conservation of momentum

The system of bodies is called closed, if the actions of external bodies on the bodies of a given system are either negligible or compensate each other. Thus, in case closed system bodies, only the interaction of these bodies with each other, but not with any other bodies, is essential.

The resultant of external forces applied to a closed system is equal to zero: . In this case, from (6) we obtain:

But if the derivative of a vector goes to zero (the rate of change of the vector is zero), then the vector itself does not change over time:

Law of conservation of momentum. The momentum of a closed system of bodies remains constant over time for any interactions of bodies within this system.

The simplest problems on the law of conservation of momentum are solved according to the standard scheme, which we will now show.

Task. A body of mass g moves with a speed m/s on a smooth horizontal surface. A body of mass g moves towards it with a speed of m/s. An absolutely inelastic impact occurs (the bodies stick together). Find the speed of the bodies after the impact.

Solution. The situation is shown in Fig. 7. Let's direct the axis in the direction of movement of the first body.

Rice. 7. To the task

Because the surface is smooth, there is no friction. Since the surface is horizontal and movement occurs along it, the force of gravity and the reaction of the support balance each other:

Thus, the vector sum of forces applied to the system of these bodies is equal to zero. This means that the system of bodies is closed. Therefore, the law of conservation of momentum is satisfied for it:

. ( 7 )

The impulse of the system before impact is the sum of the impulses of the bodies:

After the inelastic impact, one body of mass is obtained, which moves with the desired speed:

From the law of conservation of momentum (7) we have:

From here we find the speed of the body formed after the impact:

Let's move on to projections onto the axis:

By condition we have: m/s, m/s, so

The minus sign indicates that the stuck together bodies move in the direction opposite to the axis. Required speed: m/s.

Law of conservation of momentum projection

The following situation often occurs in problems. The system of bodies is not closed (the vector sum of external forces acting on the system is not equal to zero), but there is such an axis, the sum of the projections of external forces onto the axis is zero at any given time. Then we can say that along this axis our system of bodies behaves as closed, and the projection of the system’s momentum onto the axis is preserved.

Let's show this more strictly. Let's project equality (6) onto the axis:

If the projection of the resultant external forces vanishes, then

Therefore, the projection is a constant:

Law of conservation of momentum projection. If the projection onto the axis of the sum of external forces acting on the system is equal to zero, then the projection of the system’s momentum does not change over time.

Let's look at an example of a specific problem to see how the law of conservation of momentum projection works.

Task. Mass boy standing on skates on smooth ice, throws a stone of mass at an angle to the horizontal. Find the speed with which the boy rolls back after the throw.

Solution. The situation is shown schematically in Fig. 8 . The boy is depicted as straight-laced.

Rice. 8. To the task

The momentum of the “boy + stone” system is not conserved. This can be seen from the fact that after the throw, a vertical component of the system’s momentum appears (namely, the vertical component of the stone’s momentum), which was not there before the throw.

Therefore, the system that the boy and the stone form is not closed. Why? The fact is that the vector sum of external forces is not equal to zero during the throw. The value is greater than the sum, and due to this excess, the vertical component of the system’s momentum appears.

However, external forces act only vertically (there is no friction). Therefore, the projection of the impulse onto the horizontal axis is preserved. Before the throw, this projection was zero. Directing the axis in the direction of the throw (so that the boy went in the direction of the negative semi-axis), we get.