Details Category: Mechanics Published 04/21/2014 14:29 Views: 53268

In classical mechanics, there are two conservation laws: the law of conservation of momentum and the law of conservation of energy.

Body impulse

The concept of momentum was first introduced by a French mathematician, physicist, and mechanic. and the philosopher Descartes, who called impulse amount of movement .

From Latin, “impulse” is translated as “push, move.”

Any body that moves has momentum.

Let's imagine a cart standing still. Its momentum is zero. But as soon as the cart starts moving, its momentum will no longer be zero. It will begin to change as the speed changes.

Momentum of a material point, or amount of movement – a vector quantity equal to the product of the mass of a point and its speed. The direction of the point's momentum vector coincides with the direction of the velocity vector.

If we are talking about a solid physical body, then the momentum of such a body is called the product of the mass of this body and the speed of the center of mass.

How to calculate the momentum of a body? One can imagine that a body consists of many material points, or a system of material points.

If - the impulse of one material point, then the impulse of a system of material points

That is, momentum of a system of material points is the vector sum of the momenta of all material points included in the system. It is equal to the product of the masses of these points and their speed.

The unit of impulse in the international system of units SI is kilogram-meter per second (kg m/sec).

Impulse force

In mechanics, there is a close connection between the momentum of a body and force. These two quantities are connected by a quantity called impulse of force .

If a constant force acts on a bodyF over a period of time t , then according to Newton's second law

This formula shows the relationship between the force that acts on a body, the time of action of this force and the change in the speed of the body.

The quantity equal to the product of the force acting on a body and the time during which it acts is called impulse of force .

As we see from the equation, the impulse of force is equal to the difference in the impulses of the body at the initial and final moments of time, or the change in impulse over some time.

Newton's second law in momentum form is formulated as follows: the change in the momentum of a body is equal to the momentum of the force acting on it. It must be said that Newton himself originally formulated his law in exactly this way.

Force impulse is also a vector quantity.

The law of conservation of momentum follows from Newton's third law.

It must be remembered that this law operates only in a closed, or isolated, physical system. A closed system is a system in which bodies interact only with each other and do not interact with external bodies.

Let us imagine a closed system of two physical bodies. The forces of interaction of bodies with each other are called internal forces.

The force impulse for the first body is equal to

According to Newton's third law, the forces that act on bodies during their interaction are equal in magnitude and opposite in direction.

Therefore, for the second body the momentum of the force is equal to

By simple calculations we obtain a mathematical expression for the law of conservation of momentum:

Where m 1 And m 2 – body masses,

v 1 And v 2 – velocities of the first and second bodies before interaction,

v 1" And v 2" – velocities of the first and second bodies after interaction .

p 1 = m 1 · v 1 - momentum of the first body before interaction;

p 2 = m 2 · v 2 - momentum of the second body before interaction;

p 1 "= m 1 · v 1" - momentum of the first body after interaction;

p 2 "= m 2 · v 2" - momentum of the second body after interaction;

That is

p 1 + p 2 = p 1" + p 2"

In a closed system, bodies only exchange impulses. And the vector sum of the momenta of these bodies before their interaction is equal to the vector sum of their momenta after the interaction.

So, as a result of firing a gun, the momentum of the gun itself and the momentum of the bullet will change. But the sum of the impulses of the gun and the bullet in it before the shot will remain equal to the sum of the impulses of the gun and the flying bullet after the shot.

When firing a cannon, there is recoil. The projectile flies forward, and the gun itself rolls back. The projectile and the gun are a closed system in which the law of conservation of momentum operates.

The momentum of each body in a closed system can change as a result of their interaction with each other. But the vector sum of the impulses of bodies included in a closed system does not change when these bodies interact over time, that is, it remains constant. This is it law of conservation of momentum.

More precisely, the law of conservation of momentum is formulated as follows: the vector sum of the impulses of all bodies of a closed system is a constant value if there are no external forces acting on it, or their vector sum is equal to zero.

The momentum of a system of bodies can change only as a result of the action of external forces on the system. And then the law of conservation of momentum will not apply.

It must be said that closed systems do not exist in nature. But, if the time of action of external forces is very short, for example, during an explosion, shot, etc., then in this case the influence of external forces on the system is neglected, and the system itself is considered as closed.

In addition, if external forces act on the system, but the sum of their projections onto one of the coordinate axes is zero (that is, the forces are balanced in the direction of this axis), then the law of conservation of momentum is satisfied in this direction.

The law of conservation of momentum is also called law of conservation of momentum .

The most striking example of the application of the law of conservation of momentum is jet motion.

Jet propulsion

Reactive motion is the movement of a body that occurs when some part of it is separated from it at a certain speed. The body itself receives an oppositely directed impulse.

The simplest example of jet propulsion is the flight of a balloon from which air escapes. If we inflate a balloon and release it, it will begin to fly in the direction opposite to the movement of the air coming out of it.

An example of jet propulsion in nature is the release of liquid from the fruit of a crazy cucumber when it bursts. At the same time, the cucumber itself flies in the opposite direction.

Jellyfish, cuttlefish and other inhabitants of the deep sea move by taking in water and then throwing it out.

Jet thrust is based on the law of conservation of momentum. We know that when a rocket with a jet engine moves, as a result of fuel combustion, a jet of liquid or gas is ejected from the nozzle ( jet stream ). As a result of the interaction of the engine with the escaping substance, reaction force . Since the rocket, together with the emitted substance, is a closed system, the momentum of such a system does not change with time.

Reactive force arises from the interaction of only parts of the system. External forces have no influence on its appearance.

Before the rocket began to move, the sum of the impulses of the rocket and the fuel was zero. Consequently, according to the law of conservation of momentum, after turning on the engines, the sum of these impulses is also zero.

where is the mass of the rocket

Gas flow rate

Changing rocket speed

∆mf - fuel consumption

Suppose the rocket operated for a period of time t .

Dividing both sides of the equation by ∆ t, we get the expression

According to Newton's second law, the reactive force is equal to

Reaction force, or jet thrust, ensures the movement of the jet engine and the object associated with it in the direction opposite to the direction of the jet stream.

Jet engines are used in modern aircraft and various missiles, military, space, etc.

Simple observations and experiments prove that rest and motion are relative, the speed of a body depends on the choice of the reference system; according to Newton's second law, regardless of whether the body was at rest or moving, a change in the speed of its movement can only occur under the influence of force, i.e. as a result of interaction with other bodies. However, there are quantities that can be conserved during the interaction of bodies. These quantities are energy And pulse.

Body impulse is called a vector physical quantity, which is a quantitative characteristic of the translational motion of bodies. The impulse is indicated by . The momentum of a body is equal to the product of the mass of the body and its speed: . The direction of the momentum vector p coincides with the direction of the body's velocity vector. The unit of impulse is .

For the momentum of a system of bodies, the conservation law is satisfied, which is valid only for closed physical systems. In general, a closed system is a system that does not exchange energy and mass with bodies and fields that are not part of it. In mechanics closed called a system that is not affected by external forces or the action of these forces is compensated. In this case, where is the initial impulse of the system, and is the final one. In the case of two bodies included in the system, this expression has the form , where are the masses of the bodies, and are the velocities before interaction, and are the velocities after interaction (Fig. 4). This formula is the mathematical expression of the law of conservation of momentum: the momentum of a closed physical system is conserved during any interactions occurring within this system. In other words: in a closed physical system, the geometric sum of the momenta of bodies before interaction is equal to the geometric sum of the momenta of these bodies after interaction. In the case of an open system, the momentum of the bodies of the system is not conserved. However, if the system also has a direction in which external forces do not act or their action is compensated, then the projection of the impulse in this direction is preserved. In addition, if the interaction time is short (shot, explosion, impact), then during this time, even in the case of an open system, external forces slightly change the impulses of the interacting bodies. Therefore, for practical calculations in this case, the law of conservation of momentum can also be applied.

Experimental studies of the interactions of various bodies - from planets and stars to atoms and elementary particles - have shown that in any system of interacting bodies, in the absence of action on the part of other bodies not included in the system, or the sum of the acting forces is equal to zero, the geometric sum of the momenta of the bodies really remains unchanged .

In mechanics, the law of conservation of momentum and Newton's laws are interconnected. If a force acts on a body of mass over time and the speed of its movement changes from to , then the acceleration of motion a of the body is equal to . Based on Newton's second law for force, we can write , it follows

. - a vector physical quantity that characterizes the action of a force on a body over a certain period of time and is equal to the product of the force and the time of its action is called impulse of force. The unit of force impulse is .

The law of conservation of momentum underlies jet propulsion. Jet propulsion- this is the movement of the body that occurs after the separation of its part from the body.

Let the body mass be at rest. Some part of it has been separated from the body by mass with speed. Then the remaining part will move in the opposite direction with speed, the mass of the remaining part. Indeed, the sum of the impulses of both parts of the body before separation was equal to zero and after separation will be equal to zero:

From here.

Much credit for the development of the theory of jet propulsion belongs to K. E. Tsiolkovsky.

He developed the theory of flight of a body of variable mass (a rocket) in a uniform gravitational field and calculated the fuel reserves necessary to overcome the force of gravity; fundamentals of the theory of a liquid jet engine, as well as elements of its design; the theory of multistage rockets, and proposed two options: parallel (several jet engines operate simultaneously) and sequential (jet engines operate one after another). K. E. Tsiolkovsky strictly scientifically proved the possibility of flying into space using rockets with a liquid jet engine, proposed special trajectories for landing spacecraft on Earth, put forward the idea of ​​​​creating interplanetary orbital stations and examined in detail the living conditions and life support on them. Tsiolkovsky's technical ideas are used in the creation of modern rocket and space technology. Movement using a jet stream according to the law of conservation of momentum is the basis of a hydrojet engine. The movement of many marine mollusks (octopus, jellyfish, squid, cuttlefish) is also based on the reactive principle.

Common Mistakes

1. There were applicants who made a serious mistake when explaining the principle of operation of a jet engine. They argued that the movement of a jet plane is due to the interaction of emitted gases and air: the plane acts on the air, and the air, according to Newton's third law, acts on the plane, as a result of which it moves. This is, of course, not true. The real reason for the movement of a jet aircraft is the interaction of gases escaping from the nozzle, which are formed during the combustion of fuel. Due to the high pressure in the combustion chamber, these gases acquire some momentum, therefore, according to the law of conservation of momentum, the aircraft receives an impulse of the same magnitude, but opposite in direction. So the plane doesn't push away from the air. On the contrary, atmospheric air is only an obstacle to the movement of the aircraft.

2. Some students cannot give a complete and correct answer to the question: in what cases can the law of conservation of momentum be applied? It is useful to remember the following criteria for its applicability:

1. the system of bodies is closed, i.e. the bodies of this system are not acted upon by external forces;
2. external forces act on the bodies of the system, but their vector sum is zero
3. the system is not closed, but the sum of the projections of all external forces onto any coordinate axis is equal to zero; then the sum of the projections of the impulses of all bodies of the system onto this axis remains constant.
4. the time of interaction between bodies is short (for example, the time of impact, shot, explosion); in this case, the impulse of external forces can be neglected and the system can be considered as closed.

Lesson objectives:

1. educational: formation of the concepts “body impulse”, “force impulse”; the ability to apply them to the analysis of the phenomenon of interaction of bodies in the simplest cases; to ensure that students understand the formulation and derivation of the law of conservation of momentum;
2. developing: to develop the ability to analyze, establish connections between elements of the content of previously studied material on the basics of mechanics, skills of search cognitive activity, and the ability for self-analysis;
3. educational: development of students’ aesthetic taste, arousing a desire to constantly expand their knowledge; maintain interest in the subject.

Equipment: metal balls on strings, demonstration carts, weights.

Learning tools: test cards.

Lesson progress

1. Organizational stage (1 min)

2. Repetition of the studied material. (10 min)

Teacher: You will learn the topic of the lesson by solving a small crossword puzzle, the key word of which will be the topic of our lesson. (We solve from left to right, we write the words vertically one by one).

1. The phenomenon of maintaining a constant speed in the absence of external influences or when they are compensated.
2. The phenomenon of changing the volume or shape of the body.
3. The force generated during deformation tends to return the body to its original position.
4. An English scientist, a contemporary of Newton, established the dependence of the elastic force on deformation.
5. Unit of mass.
6. English scientist who discovered the basic laws of mechanics.
7. Vector physical quantity, numerically equal to the change in speed per unit time.
8. The force with which the Earth attracts all bodies to itself.
9. A force that arises due to the existence of interaction forces between molecules and atoms of contacting bodies.
10. Measure of interaction between bodies.
11. A branch of mechanics that studies the laws of mechanical motion of material bodies under the influence of forces applied to them.

3. Studying new material. (18 min)

Guys, the topic of our lesson “Body impulse. Law of conservation of momentum"

Lesson Objectives: learn the concept of momentum of a body, the concept of a closed system, study the law of conservation of momentum, learn to solve problems on the law of conservation.

Today in the lesson we will not only perform experiments, but also prove them mathematically.

Knowing the basic laws of mechanics, primarily Newton’s three laws, it would seem that one can solve any problem about the motion of bodies. Guys, I’ll show you some experiments, and you think, is it possible in these cases to solve problems using only Newton’s laws?

Problem experiment.

Experiment No. 1. Rolling a light-moving cart down an inclined plane. She moves a body that is in her path.

Is it possible to find the force between the cart and the body? (no, since the collision between the cart and the body is short-lived and the force of their interaction is difficult to determine).

Experience No. 2. Rolling a loaded cart. Moves the body further.

Is it possible to find the force of interaction between the cart and the body in this case?

Draw a conclusion: what physical quantities can be used to characterize the movement of a body?

Conclusion: Newton's laws make it possible to solve problems related to finding the acceleration of a moving body if all the forces acting on the body are known, i.e. resultant of all forces. But it is often very difficult to determine the resultant force, as was the case in our cases.

If a toy cart is rolling towards you, you can stop it with your toe, but what if a truck is rolling towards you?

Conclusion: to characterize movement, you need to know the mass of the body and its speed.

Therefore, to solve problems, they use another important physical quantity - body impulse.

The concept of momentum was introduced into physics by the French scientist René Descartes (1596-1650), who called this quantity “quantity of motion”: “I accept that in the universe ... there is a certain quantity of motion, which never increases, does not decrease, and, thus, if one body sets another in motion, it loses as much of its motion as it imparts.”

Let's find the relationship between the force acting on the body, the time of its action, and the change in the speed of the body.

Let the body mass m force begins to act F. Then, from Newton’s second law, the acceleration of this body will be A.

Remember how Newton's 2nd law is read?

Let us write the law in the form

On the other side:

Or We obtained the formula for Newton's second law in impulse form.

Let us denote the product through r:

The product of a body's mass and its speed is called the body's momentum.

Pulse r– vector quantity. It always coincides in direction with the body's velocity vector. Any body that moves has momentum.

Definition: The momentum of a body is a vector physical quantity equal to the product of the mass of the body and its speed and having the direction of speed.

Like any physical quantity, momentum is measured in certain units.

Who wants to derive the unit of measurement for impulse? (The student takes notes at the blackboard.)

(p) = (kg m/s)

Let's return to our equality . In physics, the product of force and time of action is called impulse of power.

Impulse force shows how the momentum of a body changes over a given time.

Descartes established the law of conservation of momentum, but he did not clearly understand that momentum is a vector quantity. The concept of momentum was clarified by the Dutch physicist and mathematician Huygens, who, by studying the impact of balls, proved that when they collide, it is not the arithmetic sum that is preserved, but the vector sum of momentum.

Experiment (two balls are suspended on threads)

The right one is rejected and released. Returning to its previous position and hitting a stationary ball, it stops. In this case, the left ball begins to move and deflects at almost the same angle as the right ball was deflected.

Momentum has an interesting property that only a few physical quantities have. This is a conservation property. But the law of conservation of momentum is satisfied only in a closed system.

A system of bodies is called closed if the bodies interacting with each other do not interact with other bodies.

The momentum of each of the bodies that make up a closed system can change as a result of their interaction with each other.

The vector sum of the impulses of the bodies that make up a closed system does not change over time for any movements and interactions of these bodies.

This is the law of conservation of momentum.

Examples: a gun and a bullet in its barrel, a cannon and a shell, a rocket shell and the fuel in it.

Law of conservation of momentum.

The law of conservation of momentum is derived from Newton's second and third laws.

Let's consider a closed system consisting of two bodies - balls with masses m 1 and m 2, which move along a straight line in the same direction with a speed? 1 and? 2. With a slight approximation, we can assume that the balls represent a closed system.

From experience it is clear that the second ball moves at a higher speed (the vector is depicted by a longer arrow). Therefore, he will catch up with the first ball and they will collide. ( View the experiment with teacher's comments).

Mathematical derivation of the conservation law

And now we will motivate the “commanders”, using the laws of mathematics and physics, to make a mathematical derivation of the law of conservation of momentum.

5) Under what conditions is this law fulfilled?

6) What system is called closed?

7) Why does recoil occur when firing a gun?

5. Problem solving (10 min.)

No. 323 (Rymkevich).

Two inelastic bodies, the masses of which are 2 and 6 kg, move towards each other at speeds of 2 m/s each. At what speed and in what direction will these bodies move after the impact?

The teacher comments on the drawing for the problem.

7. Summing up the lesson; homework (2 min)

Homework: § 41, 42 ex. 8 (1, 2).

Literature:

1. V. Ya. Lykov. Aesthetic education in teaching physics. Book for teachers. -Moscow “ENLIGHTENMENT” 1986.
2. V. A. Volkov. Lesson developments in physics, grade 10. - Moscow “VAKO” 2006.
3. Edited by Professor B.I. Spassky. Reader on physics. -MOSCOW “ENLIGHTENMENT” 1987.
4. I. I. Mokrova. Lesson plans based on the textbook by A.V. Peryshkin “Physics. 9th grade.” - Volgograd 2003.

Body impulse

The momentum of a body is a quantity equal to the product of the mass of the body and its speed.

It should be remembered that we are talking about a body that can be represented as a material point. The momentum of the body ($p$) is also called the momentum. The concept of momentum was introduced into physics by René Descartes (1596–1650). The term “impulse” appeared later (impulsus in Latin means “push”). Momentum is a vector quantity (like speed) and is expressed by the formula:

$p↖(→)=mυ↖(→)$

The direction of the momentum vector always coincides with the direction of the velocity.

The SI unit of impulse is the impulse of a body with a mass of $1$ kg moving at a speed of $1$ m/s; therefore, the unit of impulse is $1$ kg $·$ m/s.

If a constant force acts on a body (material point) during a period of time $∆t$, then the acceleration will also be constant:

$a↖(→)=((υ_2)↖(→)-(υ_1)↖(→))/(∆t)$

where $(υ_1)↖(→)$ and $(υ_2)↖(→)$ are the initial and final velocities of the body. Substituting this value into the expression of Newton's second law, we get:

$(m((υ_2)↖(→)-(υ_1)↖(→)))/(∆t)=F↖(→)$

Opening the brackets and using the expression for the momentum of the body, we have:

$(p_2)↖(→)-(p_1)↖(→)=F↖(→)∆t$

Here $(p_2)↖(→)-(p_1)↖(→)=∆p↖(→)$ is the change in momentum over time $∆t$. Then the previous equation will take the form:

$∆p↖(→)=F↖(→)∆t$

The expression $∆p↖(→)=F↖(→)∆t$ is a mathematical representation of Newton's second law.

The product of a force and the duration of its action is called impulse of force. That's why the change in the momentum of a point is equal to the change in the momentum of the force acting on it.

The expression $∆p↖(→)=F↖(→)∆t$ is called equation of body motion. It should be noted that the same action - a change in the momentum of a point - can be achieved by a small force over a long period of time and by a large force over a short period of time.

Impulse of the system tel. Law of Momentum Change

The impulse (amount of motion) of a mechanical system is a vector equal to the sum of the impulses of all material points of this system:

$(p_(syst))↖(→)=(p_1)↖(→)+(p_2)↖(→)+...$

The laws of change and conservation of momentum are a consequence of Newton's second and third laws.

Let us consider a system consisting of two bodies. The forces ($F_(12)$ and $F_(21)$ in the figure with which the bodies of the system interact with each other are called internal.

Let, in addition to internal forces, external forces $(F_1)↖(→)$ and $(F_2)↖(→)$ act on the system. For each body we can write the equation $∆p↖(→)=F↖(→)∆t$. Adding the left and right sides of these equations, we get:

$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_(12))↖(→)+(F_(21))↖(→)+(F_1)↖(→)+ (F_2)↖(→))∆t$

According to Newton's third law, $(F_(12))↖(→)=-(F_(21))↖(→)$.

Hence,

$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$

On the left side there is a geometric sum of changes in the impulses of all bodies of the system, equal to the change in the impulse of the system itself - $(∆p_(syst))↖(→)$. Taking this into account, the equality $(∆p_1)↖(→)+(∆p_2) ↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$ can be written:

$(∆p_(syst))↖(→)=F↖(→)∆t$

where $F↖(→)$ is the sum of all external forces acting on the body. The result obtained means that the momentum of the system can only be changed by external forces, and the change in the momentum of the system is directed in the same way as the total external force. This is the essence of the law of change in momentum of a mechanical system.

Internal forces cannot change the total momentum of the system. They only change the impulses of individual bodies of the system.

Law of conservation of momentum

From the equation $(∆p_(syst))↖(→)=F↖(→)∆t$ the law of conservation of momentum follows. If no external forces act on the system, then the right side of the equation $(∆p_(syst))↖(→)=F↖(→)∆t$ becomes zero, which means the total momentum of the system remains unchanged:

$(∆p_(syst))↖(→)=m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=const$

A system on which no external forces act or the resultant of external forces is zero is called closed.

The law of conservation of momentum states:

The total momentum of a closed system of bodies remains constant for any interaction of the bodies of the system with each other.

The result obtained is valid for a system containing an arbitrary number of bodies. If the sum of external forces is not equal to zero, but the sum of their projections to some direction is equal to zero, then the projection of the system’s momentum to this direction does not change. So, for example, a system of bodies on the surface of the Earth cannot be considered closed due to the force of gravity acting on all bodies, however, the sum of the projections of impulses on the horizontal direction can remain unchanged (in the absence of friction), since in this direction the force of gravity does not works.

Jet propulsion

Let us consider examples that confirm the validity of the law of conservation of momentum.

Let's take a children's rubber ball, inflate it and release it. We will see that when the air begins to leave it in one direction, the ball itself will fly in the other. The motion of a ball is an example of jet motion. It is explained by the law of conservation of momentum: the total momentum of the “ball plus air in it” system before the air flows out is zero; it must remain equal to zero during movement; therefore, the ball moves in the direction opposite to the direction of flow of the jet, and at such a speed that its momentum is equal in magnitude to the momentum of the air jet.

Jet motion call the movement of a body that occurs when some part of it is separated from it at any speed. Due to the law of conservation of momentum, the direction of movement of the body is opposite to the direction of movement of the separated part.

Rocket flights are based on the principle of jet propulsion. A modern space rocket is a very complex aircraft. The mass of the rocket consists of the mass of the working fluid (i.e., hot gases formed as a result of fuel combustion and emitted in the form of a jet stream) and the final, or, as they say, “dry” mass of the rocket remaining after the working fluid is ejected from the rocket.

When a jet of gas is ejected from a rocket at high speed, the rocket itself rushes in the opposite direction. According to the law of conservation of momentum, the momentum $m_(p)υ_p$ acquired by the rocket must be equal to the momentum $m_(gas)·υ_(gas)$ of the ejected gases:

$m_(p)υ_p=m_(gas)·υ_(gas)$

It follows that the speed of the rocket

$υ_p=((m_(gas))/(m_p))·υ_(gas)$

From this formula it is clear that the greater the speed of the rocket, the greater the speed of the emitted gases and the ratio of the mass of the working fluid (i.e., the mass of the fuel) to the final (“dry”) mass of the rocket.

The formula $υ_p=((m_(gas))/(m_p))·υ_(gas)$ is approximate. It does not take into account that as the fuel burns, the mass of the flying rocket becomes less and less. The exact formula for rocket speed was obtained in 1897 by K. E. Tsiolkovsky and bears his name.

Work of force

The term “work” was introduced into physics in 1826 by the French scientist J. Poncelet. If in everyday life only human labor is called work, then in physics and, in particular, in mechanics it is generally accepted that work is performed by force. The physical quantity of work is usually denoted by the letter $A$.

Work of force is a measure of the action of a force, depending on its magnitude and direction, as well as on the displacement of the point of application of the force. For a constant force and linear displacement, the work is determined by the equality:

$A=F|∆r↖(→)|cosα$

where $F$ is the force acting on the body, $∆r↖(→)$ is the displacement, $α$ is the angle between the force and the displacement.

The work of force is equal to the product of the moduli of force and displacement and the cosine of the angle between them, i.e., the scalar product of the vectors $F↖(→)$ and $∆r↖(→)$.

Work is a scalar quantity. If $α 0$, and if $90°

When several forces act on a body, the total work (the sum of the work of all forces) is equal to the work of the resulting force.

The unit of work in SI is joule($1$ J). $1$ J is the work done by a force of $1$ N along a path of $1$ m in the direction of action of this force. This unit is named after the English scientist J. Joule (1818-1889): $1$ J = $1$ N $·$ m. Kilojoules and millijoules are also often used: $1$ kJ $= 1,000$ J, $1$ mJ $= $0.001 J.

Work of gravity

Let us consider a body sliding along an inclined plane with an angle of inclination $α$ and a height $H$.

Let us express $∆x$ in terms of $H$ and $α$:

$∆x=(H)/(sinα)$

Considering that the force of gravity $F_т=mg$ makes an angle ($90° - α$) with the direction of movement, using the formula $∆x=(H)/(sin)α$, we obtain an expression for the work of gravity $A_g$:

$A_g=mg cos(90°-α) (H)/(sinα)=mgH$

From this formula it is clear that the work done by gravity depends on the height and does not depend on the angle of inclination of the plane.

It follows that:

1. the work of gravity does not depend on the shape of the trajectory along which the body moves, but only on the initial and final position of the body;
2. when a body moves along a closed trajectory, the work done by gravity is zero, i.e., gravity is a conservative force (forces that have this property are called conservative).

Work of reaction forces, is equal to zero, since the reaction force ($N$) is directed perpendicular to the displacement $∆x$.

Work of friction force

The friction force is directed opposite to the displacement $∆x$ and makes an angle of $180°$ with it, therefore the work of the friction force is negative:

$A_(tr)=F_(tr)∆x·cos180°=-F_(tr)·∆x$

Since $F_(tr)=μN, N=mg cosα, ∆x=l=(H)/(sinα),$ then

$A_(tr)=μmgHctgα$

Work of elastic force

Let an external force $F↖(→)$ act on an unstretched spring of length $l_0$, stretching it by $∆l_0=x_0$. In position $x=x_0F_(control)=kx_0$. After the force $F↖(→)$ ceases to act at point $x_0$, the spring is compressed under the action of force $F_(control)$.

Let us determine the work of the elastic force when the coordinate of the right end of the spring changes from $x_0$ to $x$. Since the elastic force in this area changes linearly, Hooke’s law can use its average value in this area:

$F_(control av.)=(kx_0+kx)/(2)=(k)/(2)(x_0+x)$

Then the work (taking into account the fact that the directions $(F_(control av.))↖(→)$ and $(∆x)↖(→)$ coincide) is equal to:

$A_(control)=(k)/(2)(x_0+x)(x_0-x)=(kx_0^2)/(2)-(kx^2)/(2)$

It can be shown that the form of the last formula does not depend on the angle between $(F_(control av.))↖(→)$ and $(∆x)↖(→)$. The work of elastic forces depends only on the deformations of the spring in the initial and final states.

Thus, the elastic force, like gravity, is a conservative force.

Power power

Power is a physical quantity measured by the ratio of work to the period of time during which it is produced.

In other words, power shows how much work is done per unit of time (in SI - per $1$ s).

Power is determined by the formula:

where $N$ is power, $A$ is work done during time $∆t$.

Substituting into the formula $N=(A)/(∆t)$ instead of the work $A$ its expression $A=F|(∆r)↖(→)|cosα$, we obtain:

$N=(F|(∆r)↖(→)|cosα)/(∆t)=Fυcosα$

Power is equal to the product of the magnitudes of the force and velocity vectors and the cosine of the angle between these vectors.

Power in the SI system is measured in watts (W). One watt ($1$ W) is the power at which $1$ J of work is done for $1$ s: $1$ W $= 1$ J/s.

This unit is named after the English inventor J. Watt (Watt), who built the first steam engine. J. Watt himself (1736-1819) used another unit of power - horsepower (hp), which he introduced so that he could compare the performance of a steam engine and a horse: $1$ hp. $= 735.5$ W.

In technology, larger power units are often used - kilowatt and megawatt: $1$ kW $= 1000$ W, $1$ MW $= 1000000$ W.

Kinetic energy. Law of change of kinetic energy

If a body or several interacting bodies (a system of bodies) can do work, then they are said to have energy.

The word “energy” (from the Greek energia - action, activity) is often used in everyday life. For example, people who can do work quickly are called energetic, having great energy.

The energy possessed by a body due to motion is called kinetic energy.

As in the case of the definition of energy in general, we can say about kinetic energy that kinetic energy is the ability of a moving body to do work.

Let us find the kinetic energy of a body of mass $m$ moving with a speed $υ$. Since kinetic energy is energy due to motion, its zero state is the state in which the body is at rest. Having found the work necessary to impart a given speed to a body, we will find its kinetic energy.

To do this, let us calculate the work in the area of ​​displacement $∆r↖(→)$ when the directions of the force vectors $F↖(→)$ and displacement $∆r↖(→)$ coincide. In this case the work is equal

where $∆x=∆r$

For the motion of a point with acceleration $α=const$, the expression for displacement has the form:

$∆x=υ_1t+(at^2)/(2),$

where $υ_1$ is the initial speed.

Substituting the expression for $∆x$ into the equation $A=F·∆x$ from $∆x=υ_1t+(at^2)/(2)$ and using Newton’s second law $F=ma$, we obtain:

$A=ma(υ_1t+(at^2)/(2))=(mat)/(2)(2υ_1+at)$

Expressing the acceleration through the initial $υ_1$ and final $υ_2$ velocities $a=(υ_2-υ_1)/(t)$ and substituting in $A=ma(υ_1t+(at^2)/(2))=(mat)/ (2)(2υ_1+at)$ we have:

$A=(m(υ_2-υ_1))/(2)·(2υ_1+υ_2-υ_1)$

$A=(mυ_2^2)/(2)-(mυ_1^2)/(2)$

Now equating the initial speed to zero: $υ_1=0$, we obtain an expression for kinetic energy:

$E_K=(mυ)/(2)=(p^2)/(2m)$

Thus, a moving body has kinetic energy. This energy is equal to the work that must be done to increase the speed of the body from zero to the value $υ$.

From $E_K=(mυ)/(2)=(p^2)/(2m)$ it follows that the work done by a force to move a body from one position to another is equal to the change in kinetic energy:

$A=E_(K_2)-E_(K_1)=∆E_K$

The equality $A=E_(K_2)-E_(K_1)=∆E_K$ expresses theorem on the change in kinetic energy.

Change in body kinetic energy(material point) for a certain period of time is equal to the work done during this time by the force acting on the body.

Potential energy

Potential energy is the energy determined by the relative position of interacting bodies or parts of the same body.

Since energy is defined as the ability of a body to do work, potential energy is naturally defined as the work done by a force, depending only on the relative position of the bodies. This is the work of gravity $A=mgh_1-mgh_2=mgH$ and the work of elasticity:

$A=(kx_0^2)/(2)-(kx^2)/(2)$

Potential energy of the body interacting with the Earth, they call a quantity equal to the product of the mass $m$ of this body by the acceleration of free fall $g$ and the height $h$ of the body above the Earth’s surface:

The potential energy of an elastically deformed body is a value equal to half the product of the elasticity (stiffness) coefficient $k$ of the body and the square of the deformation $∆l$:

$E_p=(1)/(2)k∆l^2$

The work of conservative forces (gravity and elasticity), taking into account $E_p=mgh$ and $E_p=(1)/(2)k∆l^2$, is expressed as follows:

$A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$

This formula allows us to give a general definition of potential energy.

The potential energy of a system is a quantity that depends on the position of the bodies, the change in which during the transition of the system from the initial state to the final state is equal to the work of the internal conservative forces of the system, taken with the opposite sign.

The minus sign on the right side of the equation $A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$ means that when work is performed by internal forces (for example, a fall bodies on the ground under the influence of gravity in the “rock-Earth” system), the energy of the system decreases. Work and changes in potential energy in a system always have opposite signs.

Since work determines only the change in potential energy, then only the change in energy has physical meaning in mechanics. Therefore, the choice of the zero energy level is arbitrary and determined solely by considerations of convenience, for example, the ease of writing the corresponding equations.

The law of change and conservation of mechanical energy

Total mechanical energy of the system the sum of its kinetic and potential energies is called:

It is determined by the position of bodies (potential energy) and their speed (kinetic energy).

According to the kinetic energy theorem,

$E_k-E_(k_1)=A_p+A_(pr),$

where $A_p$ is the work of potential forces, $A_(pr)$ is the work of non-potential forces.

In turn, the work of potential forces is equal to the difference in the potential energy of the body in the initial $E_(p_1)$ and final $E_p$ states. Taking this into account, we obtain an expression for law of change of mechanical energy:

$(E_k+E_p)-(E_(k_1)+E_(p_1))=A_(pr)$

where the left side of the equality is the change in total mechanical energy, and the right side is the work of non-potential forces.

So, law of change of mechanical energy reads:

The change in the mechanical energy of the system is equal to the work of all non-potential forces.

A mechanical system in which only potential forces act is called conservative.

In a conservative system $A_(pr) = 0$. It follows law of conservation of mechanical energy:

In a closed conservative system, the total mechanical energy is conserved (does not change with time):

$E_k+E_p=E_(k_1)+E_(p_1)$

The law of conservation of mechanical energy is derived from Newton's laws of mechanics, which are applicable to a system of material points (or macroparticles).

However, the law of conservation of mechanical energy is also valid for a system of microparticles, where Newton’s laws themselves no longer apply.

The law of conservation of mechanical energy is a consequence of the uniformity of time.

Uniformity of time is that, under the same initial conditions, the occurrence of physical processes does not depend on at what point in time these conditions are created.

The law of conservation of total mechanical energy means that when the kinetic energy in a conservative system changes, its potential energy must also change, so that their sum remains constant. This means the possibility of converting one type of energy into another.

In accordance with the various forms of motion of matter, various types of energy are considered: mechanical, internal (equal to the sum of the kinetic energy of the chaotic movement of molecules relative to the center of mass of the body and the potential energy of interaction of molecules with each other), electromagnetic, chemical (which consists of the kinetic energy of the movement of electrons and electrical the energy of their interaction with each other and with atomic nuclei), nuclear, etc. From the above it is clear that the division of energy into different types is quite arbitrary.

Natural phenomena are usually accompanied by the transformation of one type of energy into another. For example, friction of parts of various mechanisms leads to the conversion of mechanical energy into heat, i.e. internal energy. In heat engines, on the contrary, internal energy is converted into mechanical energy; in galvanic cells, chemical energy is converted into electrical energy, etc.

Currently, the concept of energy is one of the basic concepts of physics. This concept is inextricably linked with the idea of ​​​​the transformation of one form of movement into another.

This is how the concept of energy is formulated in modern physics:

Energy is a general quantitative measure of movement and interaction of all types of matter. Energy does not appear from nothing and does not disappear, it can only move from one form to another. The concept of energy links together all natural phenomena.

Simple mechanisms. Mechanism efficiency

Simple mechanisms are devices that change the magnitude or direction of forces applied to a body.

They are used to move or lift large loads with little effort. These include the lever and its varieties - blocks (movable and fixed), gates, inclined plane and its varieties - wedge, screw, etc.

Lever. Leverage rule

A lever is a rigid body capable of rotating around a fixed support.

The rule of leverage says:

A lever is in equilibrium if the forces applied to it are inversely proportional to their arms:

$(F_2)/(F_1)=(l_1)/(l_2)$

From the formula $(F_2)/(F_1)=(l_1)/(l_2)$, applying the property of proportion to it (the product of the extreme terms of a proportion is equal to the product of its middle terms), we can obtain the following formula:

But $F_1l_1=M_1$ is the moment of force tending to turn the lever clockwise, and $F_2l_2=M_2$ is the moment of force trying to turn the lever counterclockwise. Thus, $M_1=M_2$, which is what needed to be proven.

The lever began to be used by people in ancient times. With its help, it was possible to lift heavy stone slabs during the construction of pyramids in Ancient Egypt. Without leverage this would not be possible. After all, for example, for the construction of the Cheops pyramid, which has a height of $147$ m, more than two million stone blocks were used, the smallest of which weighed $2.5$ tons!

Nowadays, levers are widely used both in production (for example, cranes) and in everyday life (scissors, wire cutters, scales).

Fixed block

The action of a fixed block is similar to the action of a lever with equal arms: $l_1=l_2=r$. The applied force $F_1$ is equal to the load $F_2$, and the equilibrium condition is:

Fixed block used when you need to change the direction of a force without changing its magnitude.

Movable block

The moving block acts similarly to a lever whose arms are: $l_2=(l_1)/(2)=r$. In this case, the equilibrium condition has the form:

where $F_1$ is the applied force, $F_2$ is the load. The use of a moving block gives a double gain in strength.

Pulley hoist (block system)

An ordinary chain hoist consists of $n$ moving and $n$ fixed blocks. Using it gives a gain in strength of $2n$ times:

$F_1=(F_2)/(2n)$

Power chain hoist consists of n movable and one fixed block. The use of a power pulley gives a gain in strength of $2^n$ times:

$F_1=(F_2)/(2^n)$

Screw

A screw is an inclined plane wound around an axis.

The equilibrium condition for the forces acting on the propeller has the form:

$F_1=(F_2h)/(2πr)=F_2tgα, F_1=(F_2h)/(2πR)$

where $F_1$ is the external force applied to the propeller and acting at a distance $R$ from its axis; $F_2$ is the force acting in the direction of the propeller axis; $h$ — propeller pitch; $r$ is the average thread radius; $α$ is the angle of inclination of the thread. $R$ is the length of the lever (wrench) rotating the screw with a force of $F_1$.

Efficiency

Coefficient of efficiency (efficiency) is the ratio of useful work to all work expended.

Efficiency is often expressed as a percentage and is denoted by the Greek letter $η$ (“this”):

$η=(A_p)/(A_3)·100%$

where $A_n$ is useful work, $A_3$ is all expended work.

Useful work always constitutes only a part of the total work that a person expends using one or another mechanism.

Part of the work done is spent on overcoming frictional forces. Since $A_3 > A_n$, the efficiency is always less than $1$ (or $< 100%$).

Since each of the works in this equality can be expressed as a product of the corresponding force and the distance traveled, it can be rewritten as follows: $F_1s_1≈F_2s_2$.

It follows that, winning with the help of a mechanism in force, we lose the same number of times along the way, and vice versa. This law is called the golden rule of mechanics.

The golden rule of mechanics is an approximate law, since it does not take into account the work of overcoming friction and gravity of the parts of the devices used. Nevertheless, it can be very useful in analyzing the operation of any simple mechanism.

So, for example, thanks to this rule, we can immediately say that the worker shown in the figure, with a double gain in the force of lifting the load by $10$ cm, will have to lower the opposite end of the lever by $20$ cm.

Collision of bodies. Elastic and inelastic impacts

The laws of conservation of momentum and mechanical energy are used to solve the problem of the motion of bodies after a collision: from the known impulses and energies before the collision, the values ​​of these quantities after the collision are determined. Let us consider the cases of elastic and inelastic impacts.

An impact is called absolutely inelastic, after which the bodies form a single body moving at a certain speed. The problem of the speed of the latter is solved using the law of conservation of momentum of a system of bodies with masses $m_1$ and $m_2$ (if we are talking about two bodies) before and after the impact:

$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=(m_1+m_2)υ↖(→)$

It is obvious that the kinetic energy of bodies during an inelastic impact is not conserved (for example, for $(υ_1)↖(→)=-(υ_2)↖(→)$ and $m_1=m_2$ it becomes equal to zero after the impact).

An impact in which not only the sum of impulses is conserved, but also the sum of the kinetic energies of the impacting bodies is called absolutely elastic.

For an absolutely elastic impact, the following equations are valid:

$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=m_1(υ"_1)↖(→)+m_2(υ"_2)↖(→);$

$(m_(1)υ_1^2)/(2)+(m_(2)υ_2^2)/(2)=(m_1(υ"_1)^2)/(2)+(m_2(υ"_2 )^2)/(2)$

where $m_1, m_2$ are the masses of the balls, $υ_1, υ_2$ are the velocities of the balls before the impact, $υ"_1, υ"_2$ are the velocities of the balls after the impact.

As we have already said, there are no exactly closed systems of bodies. Therefore, the question arises: in what cases can the law of conservation of momentum be applied to open systems of bodies? Let's consider these cases.

1. External forces balance each other or can be neglected

We have already met this case in the previous paragraph using the example of two interacting carts.

As a second example, consider a first-grader and a tenth-grader competing in a tug of war while standing on skateboards (Figure 26.1). In this case, external forces also balance each other, and the friction force can be neglected. Therefore, the sum of the opponents' impulses is preserved.

Let the schoolchildren be at rest at the initial moment. Then their total momentum at the initial moment is zero. According to the law of conservation of momentum, it will remain equal to zero even when they move. Hence,

where 1 and 2 are the speeds of schoolchildren at an arbitrary moment (while the actions of all other bodies are compensated).

1. Prove that the ratio of the boys’ velocity modules is inverse to the ratio of their masses:

v 1 /v 2 = m 2 /m 1. (2)

Please note that this relationship will hold regardless of how the opponents interact. For example, it doesn’t matter whether they pull the rope jerkily or smoothly; only one of them or both of them moves the rope with their hands.

2. There is a platform weighing 120 kg on the rails, and on it is a person weighing 60 kg (Fig. 26.2, a). The friction between the platform wheels and the rails can be neglected. The person begins to walk along the platform to the right at a speed of 1.2 m/s relative to the platform (Fig. 26.2, b).

The initial total momentum of the platform and the person is zero in the reference frame associated with the ground. Therefore, we apply the law of conservation of momentum in this reference frame.

a) What is the ratio of the person’s speed to the speed of the platform relative to the ground?
b) How are the modules of the speed of a person relative to the platform, the speed of a person relative to the ground, and the speed of the platform relative to the ground related?
c) At what speed and in what direction will the platform move relative to the ground?
d) What will be the speed of the person and the platform relative to the ground when he reaches its opposite end and stops?

2. The projection of external forces onto a certain coordinate axis is zero

Let, for example, let a cart with sand of mass mt roll along the rails at speed. We will assume that the friction between the wheels of the cart and the rails can be neglected.

A load of mass m g falls into the cart (Fig. 26.3, a), and the cart rolls further with the load (Fig. 26.3, b). Let us denote the final speed of the cart with the load k.

Let's enter the coordinate axes as shown in the figure. The bodies were acted upon only by vertically directed external forces (gravity and the normal reaction force from the rails). These forces cannot change the horizontal projections of the impulses of bodies. Therefore, the projection of the total momentum of the bodies onto the horizontally directed x axis remained unchanged.

3. Prove that the final speed of the loaded cart is

v k = v(m t /(m t + m g)).

We see that the speed of the cart decreased after the load fell.

The decrease in the speed of the cart is explained by the fact that it transferred part of its initial horizontally directed impulse to the load, accelerating it to speed k. When the cart accelerated the load, it, according to Newton’s third law, slowed down the cart.

Please note that in the process under consideration, the total momentum of the cart and the load was not conserved. Only the projection of the total momentum of the bodies onto the horizontally directed x axis remained unchanged.

The projection of the total momentum of the bodies onto the vertically directed axis y in this process changed: before the load fell, it was different from zero (the load was moving down), and after the load fell, it became equal to zero (both bodies were moving horizontally).

4. A load weighing 10 kg flies into a cart with sand weighing 20 kg standing on rails. The speed of the load immediately before hitting the cart is 6 m/s and is directed at an angle of 60º to the horizontal (Fig. 26.4). The friction between the cart wheels and the rails can be neglected.

a) What projection of the total momentum is conserved in this case?
b) What is the horizontal projection of the load’s momentum immediately before it hits the cart?
c) At what speed will the cart with the load move?

3. Impacts, collisions, explosions, shots

In these cases, a significant change in the speed of the bodies (and therefore their momentum) occurs in a very short period of time. As we already know (see the previous paragraph), this means that during this period of time the bodies act on each other with great forces. Typically these forces are much greater than the external forces acting on the bodies of the system.
Therefore, the system of bodies during such interactions can be considered closed with a good degree of accuracy, due to which the law of conservation of momentum can be used.

For example, when a cannonball moves inside a cannon barrel during a cannon shot, the forces exerted on each other by the cannon and the cannonball far exceed the horizontally directed external forces acting on these bodies.

5. A cannon weighing 200 kg fired a cannonball weighing 10 kg in a horizontal direction (Fig. 26.5). The cannonball flew out of the cannon at a speed of 200 m/s. What is the speed of the gun during recoil?

During collisions, bodies also act on each other with fairly large forces for a short period of time.

The easiest to study is the so-called absolutely inelastic collision (or absolutely inelastic impact). This is the name for the collision of bodies, as a result of which they begin to move as a single whole. This is exactly how the carts interacted in the first experiment (see Fig. 25.1), discussed in the previous paragraph. Finding the total speed of the bodies after a completely inelastic collision is quite simple.

6. Two plasticine balls of mass m 1 and m 2 move with speeds 1 and 2. As a result of the collision, they began to move as one. Prove that their total speed can be found using the formula

Typically, cases are considered when bodies move along one straight line before a collision. Let's direct the x axis along this line. Then, in projections onto this axis, formula (3) takes the form

The direction of the total velocity of bodies after an absolutely inelastic collision is determined by the sign of the projection v x .

7. Explain why it follows from formula (4) that the speed of the “united body” will be directed in the same way as the initial speed of a body with a large impulse.

8. Two carts are moving towards each other. When they collide, they interlock and move as one. Let us denote the mass and speed of the cart, which initially moved to the right, m ​​p and p, and the mass and speed of the cart, which initially moved to the left, m l and l. In what direction and at what speed will the coupled carts move if:
a) m p = 1 kg, v p = 2 m/s, m ​​l = 2 kg, v l = 0.5 m/s?
b) m p = 1 kg, v p = 2 m/s, m ​​l = 4 kg, v l = 0.5 m/s?
c) m p = 1 kg, v p = 2 m/s, m ​​l = 0.5 kg, v l = 6 m/s?

Additional questions and tasks

In the tasks for this section it is assumed that friction can be neglected (if the friction coefficient is not specified).

9. A cart weighing 100 kg stands on the rails. A schoolboy weighing 50 kg running along the rails jumped onto this cart with a running start, after which it, together with the schoolboy, began to move at a speed of 2 m/s. What was the student's speed immediately before the jump?

10. Two bogies of mass M each stand on the rails not far from each other. On the first of them stands a man of mass m. A man jumps from the first cart to the second.
a) Which cart will have the greater speed?
b) What will be the ratio of the speeds of the carts?

11. An anti-aircraft gun mounted on a railway platform fires a projectile of mass m at an angle α to the horizontal. The initial velocity of the projectile is v0. What speed will the platform acquire if its mass together with the gun is equal to M? At the initial moment the platform was at rest.

12. A puck with a mass of 160 g sliding on ice hits a lying piece of ice. After the impact, the puck slides in the same direction, but its velocity modulus has been halved. The speed of the ice floe became equal to the initial speed of the puck. What is the mass of the ice cube?

13. A person weighing 60 kg stands at one end of a platform 10 m long and weighing 240 kg. What will be the displacement of the platform relative to the ground when a person moves to its opposite end?
Clue. Assume that the person walks at a constant speed v relative to the platform; express in terms of v the speed of the platform relative to the ground.

14. A wooden block of mass M lying on a long table is hit by a bullet of mass m flying horizontally with speed and gets stuck in it. How long after this will the block slide on the table if the coefficient of friction between the table and the block is μ?