Laws of conservation of energy and momentum. Application of the laws of conservation of energy and momentum in mechanical phenomena 3 laws of conservation of momentum and energy

The solution to many practical problems is greatly simplified if we use the laws of conservation - the law of conservation of momentum and the law of conservation and transformation of energy, because these laws can be used even when the forces acting in the system are unknown. So, let's remember the types mechanical energy and solve several problems on the application of conservation laws.

Remembering mechanical energy

Energy (from the Greek “activity”) is a physical quantity that is a general measure of the movement and interaction of all types of matter.

Energy is represented by the symbol E (or W). The SI unit of energy is the joule:

In mechanics we deal with mechanical energy.

mechanical energy is a physical quantity that is a measure of the movement and interaction of bodies and characterizes the ability of bodies to perform mechanical work.

Types of mechanical energy

The sum of the kinetic and potential energies of a body (system of bodies) is the total mechanical energy of the body (system of bodies): E = E k + E p

While studying mechanical energy in the 7th grade physics course, you learned that when a system of bodies is closed, and the bodies of the system interact with each other only by elastic forces and gravitational forces, the total mechanical energy of the system does not change.

This is the law of conservation and transformation of mechanical energy, which can be written mathematically as follows:

where E k0 + E p0 is the total mechanical energy of the system of bodies at the beginning of observation; E k + E p is the total mechanical energy of the system of bodies at the end of the observation.

We recall the algorithm for solving problems on the law of conservation of mechanical energy

Algorithm for solving problems using the law of conservation of mechanical energy

1. Read the problem statement. Determine whether the system is closed and whether the action of resistance forces can be neglected. Write down a brief statement of the problem.

2. Make an explanatory drawing in which you indicate the zero level, the initial and final states of the body (system of bodies).

3. Write down the law of conservation and transformation of mechanical energy. Make this entry more specific using the problem data and the appropriate energy calculation formulas.

4. Solve the resulting equation for the unknown quantity. Check its unit and find the numerical value.

5. Analyze the result, write down the answer.

The law of conservation of mechanical energy greatly simplifies the solution of many practical problems. Let's consider an algorithm for solving such problems using a specific example.

Task 1. A participant in a bungee jumping attraction jumps from a bridge (see picture).

What is the stiffness of the rubber rope to which the athlete is tied if during the fall the cord stretches from 40 to 100 m? The athlete's mass is 72 kg, the initial speed of his movement is zero. Don't take air resistance into account.

Analysis of a physical problem. We do not take air resistance into account, therefore the system of bodies “Earth - man - cord” can be considered closed and to solve the problem we can use the law of conservation of mechanical energy: at the beginning of the jump the athlete has the potential energy of the raised body, at the lowest point this energy is converted into the potential energy of the deformed cord .

Search for a mathematical model, solution Let's make a drawing in which we indicate the initial and final positions of the athlete. For the zero level, we will choose the lowest position of the athlete (the cord is stretched to the maximum, the speed of the athlete’s movement is zero). Let's write down the law of conservation of mechanical energy.

We apply the law of conservation of mechanical energy and the law of conservation of momentum simultaneously

Have you played billiards? One of the types of collision of billiard balls is an elastic central impact - a collision in which there is no loss of mechanical energy, and the speeds of movement of the balls before and after the impact are directed along a straight line passing through the centers of the balls.

Problem 2. A ball moving on a billiard table at a speed of 5 m/s collides with a stationary ball of the same mass (see figure). Determine the speed of the balls after the collision. Consider the impact to be elastic and central.

Analysis of a physical problem. The system of two balls can be considered closed, the impact is elastic and central, which means there is no loss of mechanical energy. Therefore, to solve the problem, you can use both the law of conservation of mechanical energy and the law of conservation of momentum. Let's select the table surface as the zero level. Since the potential energies of the balls before and after the impact are equal to zero, the total mechanical energy of the system is equal to the sum of the kinetic energies of the balls.

Let us write down the law of conservation of momentum and the law of conservation of mechanical energy for a system of two balls, taking into account that v 02 = 0:

Search for a mathematical model, solution. Let's make a drawing in which we indicate the position of the balls before and after the impact.

Analysis of results. We see that the balls “exchanged” speeds: ball 1 stopped, and ball 2 acquired the speed of ball 1 before the collision. Note: during an elastic central impact of two bodies of the same mass, these bodies “exchange” velocities, regardless of what the initial velocities of the bodies were.

We apply the law of conservation of mechanical energy and the law of conservation of momentum alternately

If you're wondering how fast an arrow shoots from a bow or how fast an air rifle bullet travels, a ballistic pendulum—a heavy body suspended from metal rods—can help. Let's find out how to use this device to determine the speed of a bullet.

Problem 3. A bullet weighing 0.5 g hits a wooden block weighing 300 g suspended on rods and gets stuck in it. Determine how fast the bullet was moving if, after the bullet hit, the block rose to a height of 1.25 cm (see figure).

Analysis of a physical problem. When a bullet hits a block, it gains speed. The time it takes for a bullet to penetrate the block is short, so at this time the “bullet - block” system can be considered closed and the law of conservation of momentum can be used. But the law of conservation of mechanical energy cannot be used, since friction is present.

When the bullet stopped its movement inside the block and it began to deflect, then the effect of the air resistance force can be neglected and the law of conservation of mechanical energy for the “Earth - block” system can be used. But the momentum of the block will decrease, since part of the momentum is transferred to the Earth.

Search for a mathematical model, solution Let us write the law of conservation of momentum for positions 1 and 2 (see figure), taking into account that in position 1 the block is at rest, and in position 2 the block and the bullet move together:

Let us write down the law of conservation of mechanical energy for positions 2 and 3 and specify it:

Substituting the expression for speed (2) into formula (1), we obtain a formula for determining the speed of a body using a ballistic pendulum:

Let's check the unit and find the value of the desired quantity:

Instead of results

We have looked at only a few examples of problem solving. At first glance, it seems that both momentum and mechanical energy are not always conserved. As for momentum, this is not true. The law of conservation of momentum is a universal law of the Universe. And the supposed “appearance” of an impulse

(see problem 1 in § 38) and its “disappearance” (see problem 3 in § 38, positions of bodies 2 and 3) are explained by the fact that the Earth also receives an impulse. That is why, when solving problems, we “look for” a closed system.

Mechanical energy is indeed not always conserved: the system can gain additional mechanical energy if external forces perform positive work (for example, you threw a ball); the system may lose some mechanical energy if external forces perform negative work (for example, a bicycle stops due to friction). However total energy(the sum of the energies of the bodies of the system and the particles of which these bodies are composed) always remains unchanged. The law of conservation of energy is a universal law of the Universe.

Exercise No. 38

When performing tasks 2-4, air resistance should be neglected.

1. A load weighing 40 kg was dropped from an airplane. After the speed of the load reached 20 m/s at an altitude of 400 m, it began to move uniformly. Determine: 1) the total mechanical energy of the load at an altitude of 400 m; 2) the total mechanical energy of the load at the moment of landing; 3) energy into which part of the mechanical energy of the load was converted.

2. A ball was thrown horizontally from a height of 4 m at a speed of 8 m/s. Determine the speed of the ball at the moment of its fall. Solve the problem in two ways: 1) considering the movement of the ball as the movement of a body thrown horizontally; 2) using the law of conservation of mechanical energy. Which method is more convenient in this case?

3. Plasticine ball 1 weighing 20 g and ball 2 three times larger in mass are suspended on threads. Ball 1 was deviated from its equilibrium position to a height of 20 cm and released.

Ball 1 collided with ball 2 and stuck to it (Fig. 1). Determine: 1) the speed of movement of ball 1 before the collision; 2) the speed of movement of the balls after the collision; 3) the maximum height to which the balls will rise after a collision.

4. A ball weighing 10 g flies out of a spring gun, hits the center of a plasticine bar suspended on threads, and sticks to it. To what height will the block rise if before the shot the spring was compressed by 4 cm, the spring stiffness was 256 N/m, and the mass of the block was 30 g?

Experimental task

"Ballistic Pendulum". Make a ballistic pendulum (Fig. 2).

Take a paper box and mold another box out of plasticine, a little smaller in size. Insert the plasticine box into the paper one and hang the device on threads.

Test the device by measuring, for example, the speed of the ball of a children's spring gun. For calculations, use the formula obtained when solving problem 3 in § 38.

LABORATORY WORK No. 7

Subject. Study of the law of conservation of mechanical energy.

Goal: to verify experimentally that the total mechanical energy of a closed system of bodies remains unchanged if only gravity and elastic forces act in the system.

Equipment: tripod with coupling and foot,

dynamometer, set of weights, ruler 4050 cm long, rubber cord 15 cm long with a pointer and loops at the ends, pencil, strong thread.

theoretical information

To perform the work, you can use the experimental setup shown in Fig. 1. Having marked on the ruler the position of the pointer when the cord is unloaded (mark 0), a load is suspended from the loop of the cord. The load is pulled down (position 1), giving the cord some elongation (Fig. 2). In position 1, the total mechanical energy of the “cord-load-Earth” system is equal to the potential energy of the stretched cord:

where F 1 = kx 1 is the modulus of the elastic force of the cord when it is stretched by x 1.

Then the load is released and the position of the pointer is noted at the moment when the load reaches its maximum height (position 2). In this position, the total mechanical energy of the system is equal to the sum of the potential energy of the load raised to a height h and the potential energy of the stretched cord:

instructions for work

preparation for the experiment

1. Before starting work, remember:

1) safety requirements when performing laboratory work;

2) the law of conservation of total mechanical energy.

2. Analyze formulas (1) and (2). What measurements should be made to determine the total mechanical energy of the system in position 1; in position 2? Make a plan for the experiment.

3. Assemble the installation as shown in Fig. 1.

4. Pull the bottom loop of the cord vertically downward to straighten the cord without pulling it. Mark on the ruler with a pencil the position of the pointer when the cord is not loaded and mark 0.

Experiment

Strictly adhere to the safety instructions (see flyleaf).

Immediately enter the measurement results into the table.

1. Using a dynamometer, determine the weight P of the load.

2. Hang the weight from the loop. Having pulled the load down, mark the position of the 1 pointer on the ruler, and put the number 1 next to the mark.

3. Release the load. Noticing the position of the pointer at the moment when the load reached greatest height(position 2), place mark 2 in the appropriate place. Please note: if mark 2 is higher than mark 0, the experiment must be repeated, reducing the stretch of the cord and accordingly changing the location of mark 1.

4. Measure the elastic forces F 1 and F 2 that arise in the cord when it is stretched by x 1 and x 2, respectively. To do this, remove the weight and, hooking the loop of the cord with the hook of the dynamometer, stretch the cord first to mark 1, and then to mark 2.

5. Having measured the distances between the corresponding marks, determine the extensions x 1 and x 2 of the cord, as well as the maximum height h of lifting the load (see Fig. 2).

6. Repeat the steps described in steps 1-5, hanging two weights together on the cord.

Processing the experiment results

1. For each experiment, determine:

1) total mechanical energy of the system in position 1;

2) total mechanical energy of the system in position 2.

2. Finish filling out the table.

Analysis of experimental results

Analyze the experiment and its results. Formulate a conclusion in which: 1) compare the values ​​you obtained for the total mechanical energy of the system in position 1; in position 2; 2) indicate the reasons for possible discrepancies in results; 3) indicate the physical quantities whose measurement, in your opinion, gave the greatest error.

Assignment with an asterisk

According to the formula

experiment.

Creative task

Take a small ball on a long strong thread. Tie a rubber cord to the thread and secure it so that the ball hangs at a distance of 20-30 cm from the floor. Pull the ball down and measure the extension of the cord. After releasing the ball, measure the height to which it rose. Determine the stiffness of the cord and calculate this height theoretically. Compare the result of the calculation with the result of the experiment. What are the possible reasons for the discrepancies?

This is textbook material

Movement in nature does not arise from nothing and does not disappear - it is transmitted from one object to another. Under certain conditions, movement is able to accumulate, but when released, it reveals its ability to be preserved.

Have you ever wondered why:

• A football player can stop a ball flying at high speed with his foot or head, but a person cannot stop a carriage moving on the rails even very slowly (the mass of the carriage is much greater than the mass of the ball).
• A glass of water is placed on a long strip of strong paper. If you pull the strip slowly, the glass moves along with the paper. and if you sharply pull the strip of paper, the glass remains motionless. (the glass will remain motionless due to inertia - the phenomenon of maintaining the speed of a body constant in the absence of the action of other bodies on it)
• A tennis ball, hitting a person, does not cause any harm, but a bullet, which is less in mass, moves at high speed (600-800 m/s), turns out to be deadly (the speed of the bullet is much higher than that of the ball).

This means that the result of the interaction of bodies depends on both the mass of the bodies and their speed at the same time.

Another great French philosopher, mathematician, physicist and physiologist, founder of modern European rationalism and one of the most influential metaphysicians of modern times, introduced the concept of “quantity of motion”. He also expressed the law of conservation of momentum and gave the concept of impulse of force.

“I accept that in the Universe... there is a certain amount of motion that never increases or decreases, and thus, if one body sets another in motion, it loses as much of its motion as it imparts.” R. Descartes

Descartes, judging by his statements, understood the fundamental significance of the concept of momentum - or momentum of a body - introduced by him in the 17th century - as the product of the mass of a body by the value of its speed. And although he made the mistake of not considering momentum as a vector quantity, the law of conservation of momentum that he formulated has stood the test of time with honor. At the beginning of the 18th century, the error was corrected, and the triumphant march of this law in science and technology continues to this day.

As one of the fundamental laws of physics, it has given scientists an invaluable research tool, prohibiting some processes and opening the way for others. Explosion, jet motion, atomic and nuclear transformations - this law works perfectly everywhere. And in how many everyday situations the concept of impulse helps to understand, today, we hope, you will see for yourself.

Quantity of motion is a measure of mechanical motion equal to material point product of its mass m for speed v. Quantity of movement mv- a vector quantity, directed in the same way as the speed of a point. Sometimes the quantity of motion is also called impulse. The amount of movement at any moment in time is characterized by speed object of a certain masses when moving it from one point in space to another.

Body impulse (or amount of movement) called a vector quantity equal to the product of a body’s mass and its speed:

Body impulse directed in the same direction as the speed of the body.

Unit of measurement momentum in SI is 1 kg m/s.

A change in the momentum of a body occurs when bodies interact, for example, during impacts. (Video "Billiard balls") When bodies interact pulse one body can be partially or completely transferred to another body.

Types of collisions:

Absolutely inelastic impact- this is an impact interaction in which bodies connect (stick together) with each other and move further as one body.

The bullet gets stuck in the block and then they move as one piece. A piece of plasticine sticks to the wall.

Absolutely elastic impact- this is a collision in which the mechanical energy of a system of bodies is conserved.

After a collision, the balls bounce off each other in different directions. The ball bounces off the wall.

Let a body of mass m be acted upon by a force F for some short period of time Δt.

Under the influence of this force, the speed of the body changed by

Consequently, during the time Δt the body moved with acceleration

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

Physical quantity equal to the product of force and the time of its action, called impulse of force:

The impulse of force is also vector quantity.

The impulse of force is equal to the change in the momentum of the body (Newton's II law in impulse form):

Denoting the momentum of the body with the letter p, Newton’s second law can be written as:

Exactly in this general view Newton himself formulated the second law. The force in this expression represents the resultant of all forces applied to the body.

To determine the change in momentum, it is convenient to use a pulse diagram, which depicts the pulse vectors, as well as the vector of the sum of pulses, constructed according to the parallelogram rule.

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.

In mechanics, there are often problems when it is necessary to simultaneously consider several bodies moving in different ways. These are, for example, problems about the movement of celestial bodies, about the collision of bodies, about the recoil of a firearm, where both the projectile and the gun begin to move after the shot, etc. In these cases, we speak about the movement of a system of bodies: the solar system, a system of two colliding bodies , “gun - projectile” systems, etc. Some forces act between the bodies of the system. IN solar system these are the forces of universal gravity, in a system of colliding bodies - elastic forces, in the "gun - projectile" system - forces created by powder gases.

The impulse of the system of bodies will be equal to the sum of the impulses of each of the bodies. included in the system.

In addition to the forces acting from some bodies of the system on others (“internal forces”), bodies can also be acted upon by forces from bodies that do not belong to the system (“external” forces); for example, the force of gravity and the elasticity of the table also act on colliding billiard balls, the force of gravity also acts on the cannon and the projectile, etc. However, in a number of cases, all external forces can be neglected. Thus, when studying the collision of rolling balls, the forces of gravity are balanced for each ball separately and therefore do not affect their movement; When fired from a cannon, gravity will have an effect on the flight of the projectile only after it leaves the barrel, which will not affect the magnitude of the recoil. Therefore, one can often consider the movements of a system of bodies, assuming that there are no external forces.

If a system of bodies is not affected by external forces from other bodies, such a system is called closed.

CLOSED SYSTEM – THIS IS A SYSTEM OF BODIES THAT INTERACT ONLY WITH EACH OTHER.

Law of conservation of momentum.

In a closed system, the vector sum of the impulses of all bodies included in the system remains constant for any interactions of the bodies of this system with each other.

The law of conservation of momentum serves as the basis for explaining a wide range of natural phenomena and is used in various sciences:

1. The law is strictly observed in the phenomena of recoil when fired, the phenomenon jet propulsion, explosive phenomena and phenomena of collision of bodies.
2. The law of conservation of momentum is used: when calculating the velocities of bodies during explosions and collisions; when calculating jet vehicles; in the military industry when designing weapons; in technology - when driving piles, forging metals, etc.

The energy characteristics of motion are introduced on the basis of the concept of mechanical work or work of force.

If a force acts on a body and the body moves under the influence of this force, then the force is said to do work.

Mechanical work – this is a scalar quantity equal to the product of the modulus of force acting on the body, the modulus of displacement and the cosine of the angle between the force vector and the displacement (or velocity) vector.

Work is a scalar quantity. It can be either positive (0° ≤ α< 90°), так и отрицательна (90° < α ≤ 180°). При α = 90° работа, совершаемая силой, равна нулю.

In the SI system, work is measured in joules (J). A joule is equal to the work done by a force of 1 N to move 1 m in the direction of the force.

The work done by a force per unit time is called power.

Power N – physical quantity equal to the ratio of work A to the time period t during which this work was performed:

In the International System (SI), the unit of power is called watt (W). A watt is equal to the power of a force that does 1 J of work in 1 s.

Off-system power unit 1 hp = 735 W

Relationship between power and speed in uniform motion:

N=A/t since A=FScosα then N=(FScosα)/t, but S/t = v therefore

N=Fvcosα

The units of work and power used in technology are:

1 W s = 1 J; 1Wh = 3.6·10 3 J; 1 kWh = 3.6 10 6 J

If a body is capable of doing work, then it is said to have energy.

Mechanical energy of the body -it is a scalar quantity equal to the maximum work that can be done under given conditions.

Designated E SI unit of energy

Mechanical work is a measure of the change in energy in various processesA =ΔE.

There are two types of mechanical energy - kinetic Ek And potential Ep energy.

The total mechanical energy of a body is equal to the sum of its kinetic and potential energies

E = Ek + Ep

Kinetic energy - This is the energy of a body due to its movement.

A physical quantity equal to half the product of a body’s mass and the square of its speed is called kinetic energybody:

Kinetic energy is the energy of motion. Kinetic energy of a body of mass m, moving with a speed equal to the work that must be done by a force applied to a body at rest in order to impart this speed to it:

If a body moves with a speed, then to stop it completely it is necessary to do work

Along with kinetic energy or energy of motion, the concept plays an important role in physics potential energy or energy of interaction between bodies.

Potential energy– body energy due to relative position interacting bodies or parts of one body.

The concept of potential energy can be introduced only for forces whose work does not depend on the trajectory of the body and is determined only by the initial and final positions. Such forces are called conservative. The work done by conservative forces on a closed trajectory is zero.

They have the property of conservatism gravity And elastic force. For these forces we can introduce the concept of potential energy.

Ppotential energy bodies in a gravity field(potential energy of a body raised above the ground):

Ep = mgh

It is equal to the work done by gravity when lowering the body to zero level.

The concept of potential energy can also be introduced for elastic force. This force also has the property of being conservative. When stretching (or compressing) a spring, we can do this in various ways.

You can simply extend the spring by an amount x, or first extend it by 2x, and then reduce the extension to a value of x, etc. In all these cases, the elastic force does the same work, which depends only on the extension x of the spring in the final state , if the spring was initially undeformed. This work is equal to the work of the external force A, taken with the opposite sign:

where k is the spring stiffness.

An extended (or compressed) spring can set a body attached to it in motion, that is, impart kinetic energy to this body. Consequently, such a spring has a reserve of energy. The potential energy of a spring (or any elastically deformed body) is the quantity

Potential energy of an elastically deformed body is equal to the work done by the elastic force during the transition from a given state to a state with zero deformation.

If in the initial state the spring was already deformed, and its elongation was equal to x1, then when transitioning to a new state with elongation x2, the elastic force will do work equal to the change in potential energy, taken with the opposite sign:

Potential energy during elastic deformation is the energy of interaction of individual parts of the body with each other by elastic forces.

If the bodies that make up closed mechanical system, interact with each other only by the forces of gravity and elasticity, then the work of these forces is equal to the change in the potential energy of the bodies, taken with the opposite sign:

A = –(Ep2 – Ep1).

According to the kinetic energy theorem, this work is equal to the change in the kinetic energy of bodies:

Therefore Ek2 – Ek1 = –(Ep2 – Ep1) or Ek1 + Ep1 = Ek2 + Ep2.

The sum of the kinetic and potential energy of the bodies that make up a closed system and interact with each other by gravitational and elastic forces remains unchanged.

This statement expresses law of energy conservation in mechanical processes. It is a consequence of Newton's laws.

The sum E = Ek + Ep is called total mechanical energy.

The total mechanical energy of a closed system of bodies interacting with each other only by conservative forces does not change with any movements of these bodies. There are only mutual transformations of the potential energy of bodies into their kinetic energy, and vice versa, or the transfer of energy from one body to another.

E = Ek + Ep = const

The law of conservation of mechanical energy is satisfied only when bodies in a closed system interact with each other by conservative forces, that is, forces for which the concept of potential energy can be introduced.

IN real conditions Almost always, moving bodies, along with gravitational forces, elastic forces and other conservative forces, are acted upon by frictional forces or environmental resistance forces.

The friction force is not conservative. The work done by the friction force depends on the length of the path.

If friction forces act between the bodies that make up a closed system, then mechanical energy is not conserved. Part of the mechanical energy is converted into internal energy of bodies (heating).

The figure shows graphs of the dependence of momentum on the speed of movement of two bodies. Which body has more mass and by how many times?

1) The masses of the bodies are the same

2) Body weight 1 is 3.5 times greater

3) Body weight 2 is greater

4) According to schedules it is impossible

compare body masses

Plasticine ball weighing T, moving at speed V , collides with a resting plasticine ball of mass 2t. After the impact, the balls stick together and move together. What is their speed?

1) v /3

3) v /2

4) There is not enough data to answer

Cars weighing m = 30 t and m= 20 tons move along a straight railway track at speeds, the time dependence of whose projections on an axis parallel to the tracks is shown in the figure. After 20 seconds, automatic coupling occurred between the cars. At what speed and in which direction will the coupled cars travel?

1) 1.4 m/s, in the direction of initial movement 1.

2) 0.2 m/s, in the direction of initial movement 1.

3) 1.4 m/s, towards the initial movement 2 .

4) 0.2 m/s, towards the initial movement 2 .

Energy (E) is a physical quantity that shows how much work a body can do

Work done is equal to the change in body energy

The body coordinate changes according to the equation x : = 2 + 30 t - 2 t 2 , written in SI. Body weight 5 kg. What is the kinetic energy of the body 3 s after the start of movement?

1) 810 J

2) 1440 J

3) 3240 J

4) 4410 J

The spring is stretched by 2cm . At the same time, work is done 2 J. How much work must be done to stretch the spring another 4 cm.

1) 16 J

2) 4 J

3) 8 J

4) 2 J

Which formula can be used to determine the kinetic energy E k that the body has at the top point of the trajectory (see figure)?

2) E K =m(V 0) 2 /2 + mgh-mgH

4) E K =m(V 0) 2 /2 + mgH

A ball was thrown from a balcony 3 times with the same initial speed. The first time the ball's velocity vector was directed vertically downward, the second time - vertically upward, and the third time - horizontally. Neglect air resistance. The modulus of the ball's speed when approaching the ground will be:

1) more in the first case

2) more in the second case

3) more in the third case

4) the same in all cases

The skydiver descends uniformly from point 1 to point 3 (Fig.). At which point in the trajectory does its kinetic energy have the greatest value?

1) At point 1.

2) At point 2 .

3) At point 3.

4) At all points the values

energies are the same.

Having slid down the slope of the ravine, the sled rises along its opposite slope to a height of 2 m (to the point 2 in the figure) and stop. Sled weight 5 kg. Their speed at the bottom of the ravine was 10 m/s. How did the total mechanical energy of the sled change when moving from point 1 to point 2?

1) Has not changed.

2) Increased by 100 J.

3) Decreased by 100 J.

4) Decreased by 150 J.

E full = E kin + U

E kin = mv 2 /2 + Jw 2 /2 – kinetic energy of translational and rotational motion,

U = mgh – potential energy of a body of mass m at a height h above the Earth’s surface.

Ftr = kN – sliding friction force, N – normal pressure force, k – friction coefficient.

In the case of an off-center impact, the law of conservation of momentum

S p i= const is written in projections on the coordinate axes.

The law of conservation of angular momentum and the law of dynamics of rotational motion

S L i= const – law of conservation of angular momentum,

L os = Jw - axial angular momentum,

L orb = [ rp] – orbital angular momentum,

dL/dt=SM ext – law of dynamics of rotational motion,

M= [rF] = rFsina – moment of force, F – force, a – angle between radius – vector and force.

A = òМdj - work during rotational motion.

Mechanics section

Kinematics

Task

Task. The dependence of the distance traveled by a body on time is given by the equation s = A–Bt+Ct 2. Find the speed and acceleration of the body at time t.

Example solution

v = ds/dt = -B + 2Ct, a = dv/dt =ds 2 /dt 2 = 2C.

Options

1.1. The dependence of the distance traveled by the body on time is given

equation s = A + Bt + Ct 2, where A = 3 m, B = 2 m/s, C = 1 m/s 2.

Find the speed in the third second.

2.1. The dependence of the distance traveled by the body on time is given

equation s= A+Bt+Ct 2 +Dt 3, where C = 0.14 m/s 2 and D = 0.01 v/s 3.

How long after the start of motion does the body accelerate?

will be equal to 1 m/s 2.

3.1. The wheel, rotating uniformly accelerated, reached angular velocity

20 rad/s after N = 10 revolutions after the start of movement. Find

angular acceleration of the wheel.

4.1. A wheel with a radius of 0.1 m rotates so that the dependence of the angle

j =A +Bt +Ct 3, where B = 2 rad/s and C = 1 rad/s 3. For points lying

on the wheel rim, find 2 s after the start of movement:

1) angular speed, 2) linear speed, 3) angular

acceleration, 4) tangential acceleration.

5.1. A wheel with a radius of 5 cm rotates so that the dependence of the angle

The rotation of the wheel radius versus time is given by the equation

j =A +Bt +Ct 2 +Dt 3, where D = 1 rad/s 3. Find for points lying

on the wheel rim, the change in tangential acceleration for

every second of movement.

6.1. A wheel with a radius of 10 cm rotates so that the dependence

linear speed of points lying on the wheel rim, from

time is given by the equation v = At ​​+ Bt 2, where A = 3 cm/s 2 and

B = 1 cm/s 3. Find the angle made by the vector of the total

acceleration with the wheel radius at time t = 5s after

start of movement.

7.1. The wheel rotates so that the dependence of the angle of rotation of the radius

wheel versus time is given by the equation j =A +Bt +Ct 2 +Dt 3, where

B = 1 rad/s, C = 1 rad/s 2, D = 1 rad/s 3. Find the radius of the wheel,

if it is known that by the end of the second second of movement

the normal acceleration of points lying on the wheel rim is

and n = 346 m/s 2.

8.1.The radius vector of a material point changes over time according to

law R=t 3 I+ t 2 j. Determine for time t = 1 s:

speed module and acceleration module.

9.1.The radius vector of a material point changes over time according to

law R=4t 2 I+ 3t j+2To. Write down the expression for the vector

speed and acceleration. Determine for time t = 2 s

speed module.

10.1. A point moves in the xy plane from a position with coordinates

x 1 = y 1 = 0 with speed v= A i+Bx j. Define Equation

trajectories of the point y(x) and the shape of the trajectory.

Moment of inertia

distance L/3 from the beginning of the rod.

Example solution.

M - mass of the rod J = J st + J gr

L – rod length J st1 = mL 2 /12 – moment of inertia of the rod

2m is the mass of the bob relative to its center. By theorem

Steiner we find the moment of inertia

J = ? the rod relative to the o axis, spaced from the center at a distance a = L/2 – L/3 = L/6.

J st = mL 2 /12 + m(L/6) 2 = mL 2 /9.

According to the principle of superposition

J = mL 2 /9 + 2m(2L/3) 2 = mL 2.

Options

1.2. Determine the moment of inertia of a rod of mass 2m relative to an axis located at a distance L/4 from the beginning of the rod. At the end of the rod there is a concentrated mass m.

2.2. Determine the moment of inertia of a rod of mass m relative to

axis spaced from the beginning of the rod at a distance of L/5. At the end

the concentrated mass of the rod is 2m.

3.2. Determine the moment of inertia of a rod of mass 2m relative to an axis located at a distance L/6 from the beginning of the rod. At the end of the rod there is a concentrated mass m.

4.2. Determine the moment of inertia of a rod of mass 3m relative to an axis located at a distance L/8 from the beginning of the rod. At the end of the rod there is a concentrated mass of 2m.

5.2. Determine the moment of inertia of a rod of mass 2m relative to an axis passing through the beginning of the rod. Concentrated masses m are attached to the end and middle of the rod.

6.2. Determine the moment of inertia of a rod of mass 2m relative to an axis passing through the beginning of the rod. A concentrated mass 2m is attached to the end of the rod, and a concentrated mass 2m is attached to the middle.

7.2. Determine the moment of inertia of a rod of mass m relative to an axis located L/4 from the beginning of the rod. Concentrated masses m are attached to the end and middle of the rod.

8.2. Find the moment of inertia of a thin homogeneous ring of mass m and radius r relative to an axis lying in the plane of the ring and spaced from its center by r/2.

9.2. Find the moment of inertia of a thin homogeneous disk of mass m and radius r relative to an axis lying in the plane of the disk and spaced from its center by r/2.

10.2. Find the moment of inertia of a homogeneous ball of mass m and radius

r relative to an axis spaced from its center by r/2.

Presnyakova I.A. 1Bondarenko M.A. 1

Atayan L.A. 1

1 Municipal educational institution“Secondary school No. 51 named after Hero Soviet Union A. M. Chislov Traktorozavodsky district of Volgograd"

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Introduction

In the world in which we live, everything flows and changes, but a person always hopes to find something unchanged. This unchangeable must be the primary source of any movement - this is energy.

Relevance of the problem stems from an increased interest in the exact sciences. Objective possibilities for the formation of cognitive interest - experimental justification as the main condition for scientific knowledge.

Object of study- energy and impulse.

Item: laws of conservation of energy and momentum.

Goal of the work:

Investigate the implementation of the laws of conservation of energy and momentum in various mechanical processes;

Develop skills research work, learn to analyze the results obtained.

To achieve this goal, the following were completed: tasks:

- conducted an analysis of theoretical material on the research topic;

We studied the specifics of the action of conservation laws;

Considered practical significance these laws.

Hypothesis research is that the laws of conservation and transformation of energy and momentum are universal laws of nature.

Significance of the work consists in using research results in physics lessons, which determines the possibility of increasing new skills and abilities; The development of the project is expected through the creation of a website where further experimental studies will be revealed.

Chapter I.

1. 1 Types of mechanical energy

Energy is a general measure of various processes and types of interaction. Mechanical energy is a physical quantity that characterizes the ability of a body or system of bodies to do work. The energy of a body or system of bodies is determined by the maximum work that they are capable of performing under given conditions. Mechanical energy includes two types of energy - kinetic and potential. Kinetic energy is the energy of a moving body. To calculate kinetic energy, assume that per body of mass m for a time t constant force acts F, which causes a change in speed by the amount v-v 0 , and at the same time work is done A = Fs(1), where s is the path traveled by the body in time t in the direction of the force. According to Newton's second law, we write Ft = m(v - v 0), from where F = m.The path traveled by the body during time will be determined through the average speed: s =v Wed t.Since the motion is uniformly variable, then s = t.We can conclude that the kinetic energy of a body of mass m, moving forward at a speed v, provided that v 0 = 0, equal to: E k = (3). Under appropriate conditions, it is possible to change the potential energy, due to which work is done.

Let's conduct an experiment: Let's compare the potential energy of the spring with the potential energy of the raised body. Equipment: tripod, training dynamometer, ball weighing 50 g, threads, measuring ruler, training scales, weights. Let's determine the height of the ball lifting due to the potential energy of the stretched spring, using the law of conservation of mechanical energy. Let's conduct an experiment and compare the results of calculation and experiment.

Work order .

1. Let's measure the mass using scales m ball.

2. Mount the dynamometer on a tripod and tie a ball to the hook. Let's notice the initial deformation x 0 springs corresponding to the dynamometer reading F 0 =mg.

3. Hold the ball on the table surface, raise the leg of the tripod with the dynamometer so that the dynamometer shows the force F 0 +F 1 , Where F 1 = 1 N, with an extension of the dynamometer spring equal to x 0 + x 1 .

4. Calculate the height H T, to which the ball should rise under the action of the elastic force of a stretched spring in the field of gravity: H T =

5. Let's release the ball and use a ruler to note the height H E, to which the ball rises.

6. Let's repeat the experiment, raising the dynamometer so that its elongation is equal to x 0 + x 2 , x 0 + x 3 , which corresponds to the dynamometer readings F 0 +F 2 And F 0 +F 3 , Where F 2 = 2 N, F 3 = 3 N.

7. Calculate the height of the ball in these cases and make the corresponding height measurements using a ruler.

8. The results of measurements and calculations are entered into the reporting table.

				H T, m	H E, m

kx 2 /2= mgH (0.0125 J= 0,0125J)

9. For one of the experiments, we will evaluate the reliability of testing the law of conservation of energy = mgH .

1.2. Law of energy conservation

Let us consider the process of changing the state of a body raised to a height h. Moreover, its potential energy E p = mh. The body began to fall freely ( v 0 = 0). At the beginning of the fall E p = max, and E k = 0. However, the sum of kinetic and potential energy at all intermediate points along the path remains unchanged if energy is not dissipated by friction, etc. therefore, if there is no conversion of mechanical energy into other types of energy, then Ep+E k = const. Such a system is conservative. The energy of a closed conservative system remains constant during all processes and transformations occurring in it. Energy can move from one type to another (mechanical, thermal, electrical, etc.), but its total amount remains constant. This position is called the law of conservation and transformation of energy .

Let's conduct an experiment: Let us compare the changes in the potential energy of a stretched spring with the change in the kinetic energy of the body.

	F at							E k	Δ E k

Equipment : two tripods for frontal work, a training dynamometer, a ball, threads, sheets of white and carbon paper, a measuring ruler, training scales with a tripod, weights. Based on the law of conservation and transformation of energy when bodies interact with elastic forces, the change in the potential energy of a stretched spring should be equal to the change kinetic energy of the body associated with it, taken with the opposite sign: Δ E p= - Δ E k To experimentally verify this statement, you can use the setup. We fix a dynamometer in the leg of the tripod. We tie a ball to its hook on a thread 60-80 cm long. On another tripod, at the same height as the dynamometer, we strengthen the groove in the foot. Having placed the ball on the edge of the gutter and holding it, we move the second tripod away from the first by the length of the thread. If you move the ball away from the edge of the groove by x, then as a result of deformation the spring will acquire a reserve of potential energy Δ E p = , where k- spring stiffness. Then release the ball. Under the influence of elastic force, the ball acquires speed υ . Neglecting the losses caused by the action of friction, we can assume that the potential energy of the stretched spring will be completely converted into the kinetic energy of the ball: The speed of the ball can be determined by measuring its flight range s when free falling from a height h. From expressions v= and t= it follows that v= s. Then Δ E k= = . Subject to equality F at = kx we get: =.

kx2/2 = (mv) 2 /2

0.04 = 0.04. Let us estimate the limits of error in measuring the potential energy of a stretched spring. Since E p =, then the relative error limit is equal to: = + = +. The absolute error limit is equal to: Δ Ep = E p. Let us estimate the error limits for measuring the kinetic energy of the ball. Because E k = , then the relative error limit is equal to: = + ? + ? g + ? h.Errors? g And? h compared to the error?s can be neglected. In this case ≈ 2? = 2. The experimental conditions for measuring the flight range are such that the deviations of the results of individual measurements from the average are significantly higher than the systematic error limit (Δs case Δ s syst), therefore we can assume that Δs av ≈ Δs random. The limit of the random error of the arithmetic mean with a small number of measurements N is found by the formula: Δs av = ,

where is calculated by the formula:

Thus, = 6. The absolute error limit for measuring the kinetic energy of the ball is equal to: Δ E k = E k .

Chapter II.

2.1. Law of conservation of momentum

The momentum of a body (amount of motion) is the product of the body's mass and its speed. Impulse is a vector quantity. SI unit of impulse: = kg*m/s = N*s. If p is the momentum of the body, m- body mass, v- body speed, then = m(1). A change in the momentum of a body of constant mass can only occur as a result of a change in speed and is always due to the action of a force. If Δp is a change in momentum, m- body weight, Δ v = v 2 -v 1 - speed change, F- constant force accelerating the body, Δ t is the duration of the force, then according to the formulas = m And = . We have = m= m,

Taking into account expression (1) we obtain: Δ = mΔ = Δ t (2).

Based on (6), we can conclude that the changes in the impulses of two interacting bodies are identical in magnitude, but opposite in direction (if the impulse of one of the interacting bodies increases, then the impulse of the other body decreases by the same amount), and based on (7) - that the sums the momenta of the bodies before and after the interaction are equal, i.e. the total momentum of bodies does not change as a result of interaction. The law of conservation of momentum is valid for a closed system with any number of bodies: = = constant. The geometric sum of the impulses of a closed system of bodies remains constant for any interactions of the bodies of this system with each other, i.e. the momentum of a closed system of bodies is conserved.,

Let's conduct an experiment: Let's check the fulfillment of the law of conservation of momentum.

Equipment: tripod for frontal work; arched tray; balls with a diameter of 25mm - 3 pcs.; measuring ruler 30 cm long with millimeter divisions; sheets of white and carbon paper; training scales; weights. Let us check the fulfillment of the law of conservation of momentum during a direct central collision of balls. According to the law of conservation of momentum for any interaction of bodies, the vector sum

m 1 kg

m 2 kg

l 1. m

v 1 .m/s

p 1. kg*m/s

l ′ 1

l ′ 2

v ′ 1

v ′ 2

p ′ 1

p ′ 2

central

impulses before interaction is equal to the vector sum of impulses of bodies after interaction. The validity of this law can be verified experimentally by studying collisions of balls in an installation. To impart a certain impulse to the ball in the horizontal direction, we use an inclined tray with a horizontal section. The ball, having rolled off the tray, moves along a parabola until it hits the table surface. Velocity projections

the ball and its momentum on the horizontal axis do not change during free fall, since there are no forces acting on the ball in the horizontal direction. Having determined the momentum of one ball, we conduct an experiment with two balls, placing the second ball on the edge of the tray, and launch the first ball in the same way as in the first experiment. After the collision, both balls fly off the tray. According to the law of conservation of momentum, the sum of the impulses of the first and second balls before the collision must be equal to the sum of the impulses of these balls after the collision: + = + (1). If a central impact occurs during the collision of the balls (in which the velocity vectors of the balls at the moment of collision are parallel to the line connecting the centers balls), and both balls after the collision move along the same straight line and in the same direction in which the first ball moved before the collision, then from vector form writing the law of conservation of momentum, you can go to the algebraic form:p 1 +p 2 = p ′ 1 +p ′ 2 , or m 1 v 1 + m 2 v 2 = m 1 v ′ 1 + m 2 v ′ 2 (2). Since the speed v 2 of the second ball before the collision was equal to zero, then expression (2) is simplified: m 1 v 1 = m 1 v ′ 1 + m 2 v ′ 2 (3)

To check the fulfillment of equality (3), we measure the masses m 1 And m 2 balls and calculate the speed v 1 , v ′ 1 And v ′ 2 . While the ball is moving along a parabola, the projection of speed on the horizontal axis will not change; it can be found by range l flight of the ball in the horizontal direction and time t its free fall ( t=):v= = l(4). p1 = p′1 + p′2

0.06 kg*m/s = (0.05+0.01) kg*m/s

0.06 kg*m/s=0.06 kg*m/s

We are convinced of the fulfillment of the law of conservation of momentum during a direct central collision of balls.

Let's conduct an experiment: let's compare the impulse of the elastic force of the spring with the change in the impulse of the projectile. Equipment: double-sided ballistic pistol; technical scales with weights; calipers; level; measuring tape; plumb line; spring dynamometer for a load of 4 N; laboratory tripod with coupling; plate with wire loop; two sheets of writing and copy paper each. It is known that the impulse of a force is equal to the change in the impulse of a body on which a constant force acts, i.e. Δ t = m- m. In this work, the elastic force of the spring acts on a projectile, which at the beginning of the experiment is at rest ( v 0 = 0): the shot is fired by projectile 2, and projectile 1 at this time is firmly held by hand on the platform. Therefore, this relationship in scalar form can be rewritten as follows: Ft = mv, Where F- the average elastic force of the spring, equal to t- time of action of the elastic force of the spring, m- projectile mass 2, v-horizontal component of projectile speed. We measure the maximum elastic force of the spring and the mass of projectile 2. Speed v calculated from the relation v=, Where - constant, A h- height and s - projectile flight range are taken from experience. The time of force action is calculated from two equations: v = at And v 2 = 2ax, i.e. t=, Where x- the amount of spring deformation. To find the value x measure the length of the protruding part of the spring at the first projectile l, and for the second - the length of the protruding rod and add them up: x = l 1 +l 2 . We measure the flight range s (the distance from the plumb line to the average point) and the fall height h. Then we determine the mass of the projectile on the scales m 2 and, measuring with a caliper l 1 And l 2 , calculate the amount of spring deformation x. After this, we unscrew the ball from projectile 1 and clamp it onto a plate with a wire loop. We connect the shells and hook the hook of the dynamometer to the loop. Holding the projectile with hand 2, we compress the spring using a dynamometer (in this case the projectiles should connect) and determine the elastic force of the spring. Knowing the flight range and the fall height, we calculate the speed of the projectile

						mv, 10 -2 kg*m/sec	Ft, 10 -2 kg*m/sec

v=, and then the time of action of the force t = . Finally, we calculate the change in projectile momentum mv and impulse of force Ft. We repeat the experiment three times, changing the elastic force of the spring, and enter all the results of measurements and calculations into a table. The results of the experiment with h= 0.2 m and m= 0.28 kg will be: mv=Ft (3.47*10-2 kg*m/s =3.5*10-2 kg*m/s)

	F max, N					s(from experience)m

The agreement of the final results within the limits of measurement accuracy is confirmed by the law of conservation of momentum. mv=Ft (3.47*10 -2 kg*m/s =3.5*10 -2 kg*m/s). Substituting these expressions into formula (1) and expressing the acceleration through the average elastic force of the spring, i.e. a=, we get the formula for calculating the projectile range: s = . Thus, by measuring F max, projectile mass m, fall height h and spring deformation x = l 1 +l 2 , we calculate the projectile’s flight range and check it experimentally. We perform the experiment at least twice, changing the elasticity of the spring, the mass of the projectile, or the height of the fall.

Chapter III.

3.1. Devices based on the laws of conservation of energy and momentum

Newton's pendulum

Newton's cradle (Newton's pendulum) - mechanical system, named after Isaac Newton to demonstrate the conversion of energy of different types into each other: kinetic to potential and vice versa. In the absence of counteracting forces (friction), the system could operate forever, but in reality this is unattainable. If you deflect the first ball and release it, then its energy and momentum will be transferred without change through the three middle balls to the last one, which will acquire the same speed and rise to the same height. According to Newton's calculations, two balls with a diameter of 30 cm, located at a distance of 0.6 cm, will converge under the influence of the force of mutual attraction a month after the start of motion (the calculation is made in the absence of external resistance). Newton took the density of the balls equal to the average density of the earth: p 5 * 10^3 kg/m^3.

At a distance l = 0.6 cm = 0.006 m between the surfaces of balls of radius R = 15 cm = 0.15 m, a force acts on the balls

F? = GM²/(2R+l)². When the balls touch, a force acts on them

F? = GM²/(2R)². F?/F? = (2R)²/(2R+l)² = (2R/(2R+l))² = (0.3/(0.3 + 0.006))² = 0.996 ≈ 1 so the assumption is valid. The mass of the ball is:

M = ρ(4/3)пR³ = 5000*4*3.14*0.15³/3 = 70.7 kg. The interaction force is

F = GM²/(2R)² = 6.67.10?¹¹.70.7²/0.3² = 3.70.10?? N. The acceleration due to gravity is:a = F/M = 3.70.10??/70.7 = 5.24.10?? m/s². Distance: s = l/2 = 0.6/2 = 0.3 cm = 0.003 m the ball will travel in time t equal to t = √2S/a = √(2*0.003/5.24.10??) = 338 s = 5.6 min. So Newton was wrong: it seems that the balls will come together quickly enough - in 6 minutes.

Maxwell's pendulum

The Maxwell pendulum is a disk (1), tightly mounted on a rod (2), on which threads (3) are wound (Fig. 2.1). The pendulum disk consists of the disk itself and replaceable rings that are fixed on the disk. When the pendulum is released, the disk begins to move: translational downward and rotational around its axis of symmetry. The rotation, continuing by inertia at the lowest point of movement (when the threads are already unwound), again leads to the winding of the threads around the rod, and, consequently, to the rise of the pendulum. The movement of the pendulum then slows down again, the pendulum stops and again begins its downward movement, etc. The acceleration of the translational motion of the center of mass of the pendulum (a) can be obtained from the measured time t and the distance h traveled by the pendulum from the equation. The mass of the pendulum m is the sum of the masses of its parts (axis m0, disk md and ring mk):

The moment of inertia of the pendulum J is also an additive quantity and is determined by the formula

Where, are the moments of inertia of the axis, disk and ring of the pendulum, respectively.

The moment of inertia of the pendulum axis is equal to, where r- axis radius, m 0 = 0.018 kg - axle mass. The moments of inertia of the disk can be found as

Where R d - radius of the disk, m d = 0.018 kg - disk mass. The moment of inertia of the ring is calculated using the formula average radius of the ring, m k is the mass of the ring, b is the width of the ring. Knowing the linear acceleration A and angular acceleration ε(ε · r), you can find the angular velocity of its rotation ( ω ):,The total kinetic energy of the pendulum consists of the energy of translational movement of the center of mass and the energy of rotation of the pendulum around the axis:

Conclusion.

Conservation laws form the foundation on which the continuity of physical theories is based. Indeed, considering the evolution of the most important physical concepts in the field of mechanics, electrodynamics, heat theory, modern physical theories, we were convinced that these theories invariably contain either the same classical conservation laws (energy, momentum, etc.), or together with them new laws appear, forming the core around which the interpretation of experimental facts occurs. “The commonality of conservation laws in old and new theories is another form of internal interconnection of the latter.” It is difficult to overestimate the role of the law of conservation of momentum. It is a general rule obtained by man on the basis of long experience. Skillful use of the law makes it possible to relatively easily solve such practical problems as forging products in a forge shop or driving piles during the construction of buildings.

Application.

Our compatriots I.V. Kurchatov and L.A. Artsimovich investigated one of the first nuclear reactions and proved the validity of the law of conservation of momentum in this type of reaction. Currently, controlled nuclear chain reactions solve humanity's energy problems.

Literature

1. World Encyclopedia

2. Dik Yu.I., Kabardin O.F. "Physics workshop for classes with in-depth study of physics." Moscow: “Enlightenment”, 1993 - p. 93.

3.Kuhling H. Handbook of Physics; translated from German 2nd ed. M, Mir, 1985 - p.120.

4. Pokrovsky A.A. "Workshop on physics in high school" Moscow: “Enlightenment”, 1973, p. 45.

5. Pokrovsky A.A. "Workshop on physics in high school." Moscow: edition 2e, “Enlightenment”, 1982 - p.76.

6. Rogers E. “Physics for the curious. Volume 2.”Moscow: “Mir”, 1969, page 201.

7. Shubin A.S. "General Physics Course". Moscow: " graduate School", 1976 - p. 224.