Assoc.

LABORATORY WORK No. 1-5: COLLISION OF BALLS.

Student__________________________________________________________________________ group:_________________

Tolerance_________________________________ Execution ________________________________Protection _________________

Goal of the work:Checking the law of conservation of momentum. Verification of the law of conservation of mechanical energy for elastic collisions. Experimental determination of the momentum of the balls before and after the collision, calculation of the coefficient of recovery of kinetic energy, determination of the average force of the collision of two balls, the speed of the balls upon collision.

Devices and accessories: Ball Collision Instrument FPM -08, scales, balls made of different materials.

Description of the experimental setup. Mechanical design device

Squares with scales 15,16 are attached to the lower bracket, and an electromagnet 17 is attached to special guides. After unscrewing the bolts 18,19, the electromagnet can be moved along the right scale and the height of its installation can be fixed, which allows you to change the initial ball. A stopwatch is attached to the base of the device. FRM -16 21, transmitting voltage through connector 22 to the balls and electromagnet.

On front panel stopwatch FRM -16 contains the following manipulation elements:

1.W 1 (Network) - network switch. Pressing this key turns on the supply voltage;

2.W 2 (Reset) – reset the meter. Pressing this key resets the stopwatch circuits FRM -16.

3.W 3 (Start) – electromagnet control. Pressing this key causes the electromagnet to be released and a pulse to be generated in the stopwatch circuit as permission to measure.

COMPLETING OF THE WORK

Exercise No. 1.Verification of the law of conservation of momentum under inelastic central impact. Determination of the coefficient

Restoration of kinetic energy.

To study an inelastic impact, two steel balls are taken, but a piece of plasticine is attached to one ball in the place where the impact occurs.

Table No. 1.

1. Get from your teacher the initial value of the deflection angle of the first ball font-size:10.0pt">2.

3. <ПУСК>and measure the angle of deflection of the second ball . Repeat the experiment five times. Write down the obtained deviation angle values ​​in table No. 1.

4. The masses of the balls are written on the installation.

5. According to the formula find the momentum of the first ball before the collision and write it down in table No. 1.

6. According to the formula find five values ​​of the momentum of the ball system after the collision and write it down in table No. 1.

7. According to the formula

8. According to the formula find the dispersion of the average value of the momentum of the system of balls after the collision..gif" width="40" height="25"> enter it in table No. 1.

9. According to the formula font-size:10.0pt">10. According to the formula font-size:10.0pt">11. font-size:10.0pt">12.Write down the interval for the momentum of the system after the collision in the form font-size:10.0pt">Find the ratio of the projection of the system's momentum after the inelastic impact to the initial value of the projection of the momentum before the impact font-size:10.0pt">Exercise No. 2. Verification of the law of conservation of momentum and mechanical energy during an elastic central impact.

Determination of the force of interaction between balls during a collision.

To study elastic impact, two steel balls are taken. The ball that is deflected towards the electromagnet is considered the first.

Table No. 2.

1. Get from your teacher the initial value of the deflection angle of the first ball DIV_ADBLOCK3">

2. Install the electromagnet so that the deflection angle of the first ball (smaller mass) corresponds to the specified value.

3. Deflect the first ball at a given angle, press the key<ПУСК>and count the deflection angles of the first ball and the second ball and the collision time of the balls font-size:10.0pt">4. According to the formula find the momentum of the first ball before the collision and write it down in table No. 2.

5. According to the formula find five values ​​of the momentum of the ball system after the collision and write it down in table No. 2.

6. According to the formula find the average value of the system's momentum after the collision.

7. According to the formula find the dispersion of the average value of the momentum of the system of balls after the collision..gif" width="40" height="25"> enter it in table No. 2.

8. According to the formula find the initial value of the kinetic energy of the first ball before the collision font-size:10.0pt">9. According to the formula find five values ​​of the kinetic energy of the system of balls after the collision font-size:10.0pt">10.Using the formula, find the average kinetic energy of the system after the collision.

11. According to the formula find the dispersion of the average value of the kinetic energy of the system of balls after the collision..gif" width="36" height="25 src="> enter it in table No. 2.

12. Using the formula, find the kinetic energy recovery coefficient font-size:10.0pt">13. According to the formula find the average value of the interaction force and enter it in table No. 2.

14. Write the interval for the momentum of the system after the collision in the form .

15. Write down the interval for the kinetic energy of the system after the collision in the form font-size: 10.0pt;font-weight:normal">Find the ratio of the projection of the system's momentum after the elastic impact to the initial value of the projection of the momentum before the impact font-size:10.0pt">Find the ratio of the kinetic energy of the system after an elastic impact to the value of the kinetic energy of the system before the impact font-size: 10.0pt">Compare the resulting value of the interaction force with the force of gravity of a ball of greater mass. Draw a conclusion about the intensity of the mutual repulsion forces acting during the impact.

CONTROL QUESTIONS

1. Impulse and energy, types of mechanical energy.

2. The law of change in momentum, the law of conservation of momentum. The concept of a closed mechanical system.

3. The law of change in total mechanical energy, the law of conservation of total mechanical energy.

4. Conservative and non-conservative forces.

5. Impact, types of impacts. Writing conservation laws for absolutely elastic and absolutely inelastic blows.

6. Interconversion of mechanical energy during free fall of a body and elastic vibrations.

Work, power, efficiency. Types of energy.

- Mechanical work constant in magnitude and direction of force

A=FScosα ,

Where A– work of force, J

F- force,

S– displacement, m

α - angle between vectors and

Types of mechanical energy

Work is a measure of the change in energy of a body or system of bodies.

In mechanics there is a distinction the following types energy:

- Kinetic energy

font-size:10.0pt">font-size:10.0pt"> where T is kinetic energy, J

M – point mass, kg

ν – point speed, m/s

peculiarity:

Kinds potential energy

- Potential energy of a material point raised above the Earth

peculiarity:

(see picture)

- Potential energy of a system of material points or an extended body raised above the Earth

P=mghts.T.

Where P– potential energy, J

m– weight, kg

g– free fall acceleration, m/s2

h– height of the point above the zero level of potential energy reference, m

hc. T. - the height of the center of mass of a system of material points or an extended body above

Zero potential energy reference level, m

peculiarity: can be positive, negative and zero depending on choice entry level potential energy count

- Potential energy of a deformed spring

font-size:10.0pt">where To– spring stiffness coefficient, N/m

Δ X– value of spring deformation, m

Peculiarity: is always a positive quantity.

- Potential energy of gravitational interaction of two material points

https://pandia.ru/text/79/299/images/image057_1.gif" width="47" height="41 src="> , whereG– gravitational constant,

M And m– point masses, kg

r– distance between them, m

peculiarity: is always a negative quantity (at infinity it is assumed to be zero)

Total mechanical energy

(this is the sum of kinetic and potential energy, J)

E = T + P

Mechanical power force N

(characterizes the speed of work)

Where A– work done by force during time t

Watt

distinguish: - useful power font-size:10.0pt"> - spent (or full power) font-size:10.0pt">whereApoleznaya And Azatris the useful and expended work of force, respectively

Power constant force can be expressed through the speed of a uniformly moving

under the influence of this body force:

N = Fv. cosα, where α is the angle between the force and velocity vectors

If the speed of the body changes, then instantaneous power is also distinguished:

N=Fv instantcosα, Where v instantis the instantaneous speed of the body

(i.e. the speed of the body in this moment time), m/s

Coefficient useful action(efficiency)

(characterizes the efficiency of an engine, mechanism or process)

η = font-size:10.0pt">Link A, N and η

LAWS OF CHANGE AND CONSERVATION IN MECHANICS

Momentum of a material point is a vector quantity equal to the product of the mass of this point and its speed:

,

Impulse of the system material points is called a vector quantity equal to:

An impulse of poweris called a vector quantity equal to the product of a force and the time of its action:

,

Law of momentum change:

Impulse change vector mechanical system bodies is equal to the product of the vector sum of all external forces acting on the system and the duration of action of these forces.

font-size:10.0pt">Law of conservation of momentum:

The vector sum of the impulses of the bodies of a closed mechanical system remains constant both in magnitude and direction for any movements and interactions of the bodies of the system.

font-size:10.0pt">Closed is a system of bodies that is not acted upon by external forces or the resultant of all external forces is zero.

Externalare called forces acting on a system from bodies not included in the system under consideration.

Internalare the forces acting between the bodies of the system itself.

For open mechanical systems, the law of conservation of momentum can be applied in the following cases:

1. If the projections of all external forces acting on the system onto any direction in space are equal to zero, then the law of conservation of momentum projection is satisfied in this direction,

(that is, if font-size:10.0pt">2.If internal forces much larger in magnitude than external forces (for example, a rupture

projectile), or the period of time during which they act is very short

External forces (for example, an impact), then the law of conservation of momentum can be applied

In vector form,

(that is, font-size:10.0pt">The law of conservation and transformation of energy:

Energy does not appear from anywhere and does not disappear anywhere, but only passes from one type of energy to another, and in such a way that the total energy of an isolated system remains constant.

(for example, mechanical energy when bodies collide is partially converted into thermal energy, the energy of sound waves, and is spent on work to deform the bodies. However, the total energy before and after the collision does not change)

Law of change in total mechanical energy:

To non-conservative - all other forces.

Features of conservative forces : the work of a conservative force acting on a body does not depend on the shape of the trajectory along which the body moves, but is determined only by the initial and final position of the body.

A moment of powerrelative to a fixed point O is a vector quantity equal to

,

Vector direction M can be determined by gimlet rule:

If the handle of the gimlet is rotated from the first factor to vector product to the second shortest turn, then the translational movement of the gimlet will indicate the direction of vector M. ,

The product of the vector sum of the moments of all external forces relative to a fixed point O acting on a mechanical system by the time of action of these forces is equal to the change in the angular momentum of this system relative to the same point O.

law of conservation of angular momentum of a closed system

The angular momentum of a closed mechanical system relative to a fixed point O does not change either in magnitude or direction during any movements and interactions of the bodies of the system.

If the problem requires finding the work done by a conservative force, then it is convenient to apply the potential energy theorem:

Potential Energy Theorem:

The work of a conservative force is equal to the change in the potential energy of a body or system of bodies, taken with the opposite sign.

(i.e. font-size:10.0pt">Kinetic energy theorem:

The change in the kinetic energy of a body is equal to the sum of the work done by all forces acting on this body.

(that is, font-size:10.0pt">Law of motion of the center of mass of a mechanical system:

The center of mass of a mechanical system of bodies moves as a material point to which all the forces acting on this system are applied.

(that is, font-size:10.0pt"> where m is the mass of the entire system, font-size:10.0pt">The law of motion of the center of mass of a closed mechanical system:

The center of mass of a closed mechanical system is at rest or moves uniformly and rectilinearly for any movements and interactions of the bodies of the system.

(that is, if font-size:10.0pt"> It should be remembered that all laws of conservation and change must be written relative to the same inertial reference frame (usually relative to the earth).

Types of blows

With a blowcalled the short-term interaction of two or more bodies.

Central(or direct) is an impact in which the velocities of the bodies before the impact are directed along a straight line passing through their centers of mass. (otherwise the blow is called non-central or oblique)

Elasticcalled an impact in which bodies, after interaction, move separately from each other.

Inelasticis called an impact in which the bodies, after interaction, move as a single whole, that is, at the same speed.

The limiting cases of impacts are absolutely elastic And absolutely inelastic blows.

Absolutely elastic impact Absolutely inelastic impact

1. the conservation law is fulfilled 1. the conservation law is satisfied

Pulse: pulse:

2. law of conservation of complete 2. law of conservation and transformation

Kinetic energy of a rigid body rotating about an axis moving translationally

, font-size:10.0pt">The basic equation for the dynamics of the rotational motion of a mechanical system:

The vector sum of the moments of all external forces acting on a mechanical system relative to a fixed point O is equal to the rate of change of the angular momentum of this system.

font-size:10.0pt">Basic equation for the dynamics of rotational motion of a rigid body:

Vector sum of the moments of all external forces acting on a body relative to a fixed axis Z , is equal to the product of the moment of inertia of this body relative to the axis Z , on its angular acceleration.

font-size:10.0pt">Steiner's theorem :

The moment of inertia of a body relative to an arbitrary axis is equal to the sum of the moment of inertia of the body relative to an axis parallel to the given one and passing through the center of mass of the body, plus the product of the body mass by the square of the distance between these axes

font-size:10.0pt">,

Moment of inertia of a material point https://pandia.ru/text/79/299/images/image108_0.gif" width="60" height="29 src=">

Elementary work of moment of forces during rotation of a body around a fixed axis,

The work of the moment of force when a body rotates around a fixed axis,

Goal of the work:

Experimental and theoretical determination of the value of the momentum of the balls before and after the collision, the coefficient of kinetic energy recovery, and the average force of the collision of two balls. Checking the law of conservation of momentum. Verification of the law of conservation of mechanical energy for elastic collisions.

Equipment: installation “Collision of balls” FM 17, consisting of: base 1, rack 2, in the upper part of which an upper bracket 3 is installed, intended for hanging balls; a housing designed to mount a scale of 4 angular movements; electromagnet 5, designed to fix starting position one of the balls 6; adjustment units ensuring direct central impact of the balls; threads 7 for hanging metal balls; wires to ensure electrical contact of the balls with terminals 8. The control unit 9 is used to launch the ball and calculate the time before impact. Metal balls 6 are made of aluminum, brass and steel. Mass of balls: brass 110.00±0.03 g; steel 117.90±0.03 g; aluminum 40.70±0.03 g.

Brief theory.

When the balls collide, the interaction forces change quite sharply with the distance between the centers of mass; the entire interaction process takes place in a very small space and in a very short period of time. This interaction is called a blow.

There are two types of impacts: if the bodies are absolutely elastic, then the impact is called absolutely elastic. If the bodies are absolutely inelastic, then the impact is absolutely inelastic. In this lab, we will only consider the center shot, that is, a shot that occurs along a line connecting the centers of the balls.

Let's consider absolutely inelastic impact. This blow can be observed on two lead or wax balls suspended on a thread of equal length. The collision process proceeds as follows. As soon as balls A and B come into contact, their deformation will begin, as a result of which resistance forces (viscous friction) will arise, braking ball A and accelerating ball B. Since these forces are proportional to the rate of change in deformation (i.e., the relative speed of the balls ), then as the relative speed decreases, they decrease and become zero as soon as the speeds of the balls level out. From this moment on, the balls, having “merged”, move together.

Let us consider the problem of the impact of inelastic balls quantitatively. We will assume that no third bodies act on them. Then the balls form closed system, in which the laws of conservation of energy and momentum can be applied. However, the forces acting on them are not conservative. Therefore, the law of conservation of energy is applied to the system:

where A is the work of non-elastic (conservative) forces;

E and E′ – total energy two balls, respectively, before and after the impact, consisting of the kinetic energy of both balls and the potential energy of their interaction with each other:

U, (2)

Since the balls do not interact before and after the impact, relation (1) takes the form:

Where are the masses of the balls; - their speed before impact; v′ is the speed of the balls after impact. Since A<0, то равенство (3) показывает, что кинетическая энергия системы уменьшилась. Деформация и нагрев шаров произошли за счет убыли кинетической энергии.

To determine the final speed of the balls, you should use the law of conservation of momentum

Since the impact is central, all velocity vectors lie on the same straight line. Taking this line as the X axis and projecting equation (5) onto this axis, we obtain the scalar equation:

(6)

From this it is clear that if the balls moved in one direction before the impact, then after the impact they will move in the same direction. If the balls were moving towards each other before the impact, then after the impact they will move in the direction where the ball with greater momentum was moving.

Let us put v′ from (6) into equality (4):

(7)

Thus, the work of internal non-conservative forces during deformation of the balls is proportional to the square of the relative speed of the balls.

Absolutely elastic impact proceeds in two stages. The first stage - From the beginning of the contact of the balls to the equalization of velocities - proceeds in the same way as with an absolutely inelastic impact, with the only difference that the interaction forces (as elastic forces) depend only on the magnitude of the deformation and do not depend on the rate of its change. Until the speeds of the balls are equal, the deformation will increase and the interaction forces will slow down one ball and accelerate the other. At the moment when the velocities of the balls become equal, the interaction forces will be greatest, from this moment the second stage of the elastic impact begins: the deformed bodies act on each other in the same direction in which they acted before the velocities equalized. Therefore, the body that was slowing down will continue to slow down, and the one that was accelerating will continue to accelerate, until the deformation disappears. When the shape of bodies is restored, all potential energy again turns into kinetic energy balls, etc. with an absolutely elastic impact, the bodies do not change their internal energy.

We will assume that two colliding balls form a closed system in which the forces are conservative. In such cases, the work of these forces leads to an increase in the potential energy of interacting bodies. The law of conservation of energy will be written as follows:

where are the kinetic energies of the balls at an arbitrary moment of time t (during the impact), and U is the potential energy of the system at the same moment. − the value of the same quantities at another time t′. If time t corresponds to the beginning of the collision, then ; if t′ corresponds to the end of the collision, then Let us write down the laws of conservation of energy and momentum for these two moments of time:

(8)

Let us solve the system of equations (9) and (10) for 1 v′ and 2 v′. To do this, we rewrite it in the following form:

Let's divide the first equation by the second:

(11)

Solving the system from equation (11) and the second equation (10), we obtain:

, (12)

Here the velocities have a positive sign if they coincide with the positive direction of the axis, and a negative sign otherwise.

Installation “Collision of balls” FM 17: design and principle of operation:

1 The installation “Collision of balls” is shown in the figure and consists of: base 1, stand 2, in the upper part of which an upper bracket 3 is installed, intended for hanging balls; a housing designed to mount a scale of 4 angular movements; an electromagnet 5 designed to fix the initial position of one of the balls 6; adjustment units ensuring direct central impact of the balls; threads 7 for hanging metal balls; wires to ensure electrical contact of the balls with terminals 8. The control unit 9 is used to launch the ball and calculate the time before impact. Metal balls 6 are made of aluminum, brass and steel.

Practical part

Preparing the device for operation

Before starting work, you need to check whether the impact of the balls is central; to do this, you need to deflect the first ball (of less mass) at a certain angle and press the key Start. The planes of motion of the balls after the collision must coincide with the plane of motion of the first ball before the collision. The center of mass of the balls at the moment of impact must be on the same horizontal line. If this is not observed, then you need to perform the following steps:

1. Using screws 2 achieve vertical position columns 3 (Fig. 1).

2. By changing the length of the suspension thread of one of the balls, it is necessary to ensure that the centers of mass of the balls are on the same horizontal line. When the balls touch, the threads must be vertical. This is achieved by moving screws 7 (see Fig. 1).

3. It is necessary to ensure that the planes of the balls’ trajectories after the collision coincide with the plane of the first ball’s trajectory before the collision. This is achieved using screws 8 and 10.

4. Loosen nuts 20, set angular scales 15,16 so that the angle indicators at the moment when the balls occupy a resting position show zero on the scales. Tighten nuts 20.

Exercise 1.Determine the time of collision of the balls.

1. Insert aluminum balls into the suspension brackets.

2. Enable installation

3. Move the first ball to a corner and fix it with an electromagnet.

4. Press the “START” button. This will cause the balls to hit.

5. Use the timer to determine the time of collision of the balls.

6. Enter the results into the table.

7. Take 10 measurements, enter the results in a table

9. Draw a conclusion about the dependence of the impact time on the mechanical properties of the materials of the colliding bodies.

Task 2. Determine the recovery coefficients of velocity and energy for the case of an elastic impact of balls.

1. Insert aluminum, steel or brass balls into the brackets (as directed by the teacher). Balls material:

2. Take the first ball to the electromagnet and record the throwing angle

3. Press the “START” button. This will cause the balls to hit.

4. Using scales, visually determine the rebound angles of the balls

5. Enter the results into the table.

6. Take 10 measurements and enter the results into the table.

7. Based on the results obtained, calculate the remaining values ​​using the formulas.

The speeds of the balls before and after impact can be calculated as follows:

Where l- distance from the suspension point to the center of gravity of the balls;

Throwing angle, degrees;

Angle of rebound of the right ball, degrees;

Bounce angle of the left ball, degrees.

The speed recovery coefficient can be determined by the formula:

The energy recovery coefficient can be determined by the formula:

The energy loss during a partially elastic collision can be calculated using the formula:

8. Calculate the average values ​​of all quantities.

9. Calculate errors using the formulas:

=

=

=

=

=

=

10. Write down the results, taking into account the error, in standard form.

Task 3. Verification of the law of conservation of momentum under inelastic central impact. Determination of the kinetic energy recovery coefficient.

To study an inelastic impact, two steel balls are taken, but a piece of plasticine is attached to one of them in the place where the impact occurs. The ball that is deflected towards the electromagnet is considered the first.

Table No. 1

1. Obtain from the teacher the initial value of the angle of deflection of the first ball and write it down in table No. 1.

2. Install the electromagnet so that the deflection angle of the first ball corresponds to the specified value

3. Deflect the first ball to the specified angle, press the key<ПУСК>and measure the angle of deflection of the second ball. Repeat the experiment 5 times. Write down the obtained deviation angle values ​​in table No. 1.

4. The mass of the balls is indicated on the installation.

5. Using the formula, find the momentum of the first ball before the collision and write the result in the table. No. 1.

6. Using the formula, find 5 values ​​of the momentum of the ball system after the collision and write the result in the table. No. 1.

7. According to the formula

8. According to the formula find the dispersion of the average value of the momentum of the system of balls after the collision. Find the standard deviation of the average momentum of the system after the collision. Enter the resulting value into table No. 1.

9. According to the formula find the initial value of the kinetic energy of the first ball before the collision, and enter it in table No. 1.

10. Using the formula, find five values ​​of the kinetic energy of the system of balls after a collision, and enter them in the table. No. 1.

11. According to the formula 5 find the average value of the kinetic energy of the system after the collision.

12. According to the formula

13. Using the formula, find the kinetic energy recovery coefficient. Based on the obtained value of the kinetic energy recovery coefficient, draw a conclusion about the conservation of the system’s energy during a collision.

14. Write down the answer for the momentum of the system after the collision in the form

15. Find the ratio of the projection of the system's momentum after the inelastic impact to the initial value of the projection of the system's momentum before the impact. Based on the obtained value of the ratio of the projection of impulses before and after the collision, draw a conclusion about the conservation of the momentum of the system during the collision.

Task 4. Verification of the law of conservation of momentum and mechanical energy during an elastic central impact. Determination of the force of interaction between balls during a collision.

To study elastic impact, two steel balls are taken. The ball that is deflected towards the electromagnet is considered the first.

Table No. 2.

1. Obtain from the teacher the initial value of the angle of deflection of the first ball and write it down in the table. No. 2

2. Install the electromagnet so that the deflection angle of the first ball corresponds to the specified value.

3. Deflect the first ball to the specified angle, press the key<ПУСК>and count the angles of deflection of the first ball and the second ball and the time of collision of the balls. Repeat the experiment 5 times. Write down the obtained values ​​of deflection angles and impact times in the table. No. 2.

4. The masses of the balls are indicated on the installation.

5. Using the formula, find the momentum of the first ball before the collision and write the result in table No. 2.

6. Using the formula, find 3 values ​​of the momentum of the ball system after the collision and write the result in the table. No. 2.

7. According to the formula find the average value of the system's momentum after the collision.

8. according to Formula find the dispersion of the average value of the momentum of the system of balls after the collision. Find the standard deviation of the average momentum of the system after the collision. Enter the resulting value into table No. 2.

9. According to the formula find the initial value of the kinetic energy of the first ball before the collision and enter the result in the table. No. 2.

10. Using the formula, find five values ​​of the kinetic energy of the system of balls after a collision, and enter the results in the table. No. 2.

11. According to the formula find the average kinetic energy of the system after the collision

12. According to the formula find the dispersion of the average kinetic energy of the system of balls after the collision. Find the standard deviation of the mean kinetic energy of the system after the collision. Enter the resulting value in the table. No. 2.

13. Using the formula, find the kinetic energy recovery coefficient.

14. According to the formula find the average value of the interaction force and enter the result in table No. 2.

15. Write down the answer for the momentum of the system after the collision in the form: .

16. Write down the interval for the kinetic energy of the system after the collision as: .

17. Find the ratio of the projection of the impulse of the system after the elastic impact to the initial value of the projection of the impulse before the impact. Based on the obtained value of the ratio of the projection of impulses before and after the collision, draw a conclusion about the conservation of the momentum of the system during the collision.

18. Find the ratio of the kinetic energy of the system after an elastic impact to the value of the kinetic energy of the system before the impact. Based on the obtained value of the ratio of kinetic energies before and after the collision, draw a conclusion about the conservation of the mechanical energy of the system during the collision.

19. Compare the resulting value of the interaction force with the force of gravity of a ball of greater mass. Draw a conclusion about the intensity of the mutual repulsion forces acting during the impact.

Control questions:

1. Describe the types of impacts, indicate which laws are followed during an impact?

2. Mechanical system. The law of change in momentum, the law of conservation of momentum. The concept of a closed mechanical system. When can the law of conservation of momentum be applied to an open mechanical system?

3. Determine the velocities of bodies of the same mass after impact in the following cases:

1) The first body is moving, the second is at rest.

2) both bodies move in the same direction.

3) both bodies are moving in the opposite direction.

4. Determine the magnitude of the change in momentum of a point of mass m uniformly rotating in a circle. In one and a half, in a quarter period.

5. Form the law of conservation of mechanical energy, in which cases it is not satisfied.

6. Write down formulas for determining the recovery coefficients of speed and energy, explain the physical meaning.

7. What determines the amount of energy loss during a partially elastic impact?

8. Body impulse and force impulse, types of mechanical energy. Mechanical work of force.

Laboratory work No. 1-5: collision of balls. Student group - page No. 1/1

Assoc. Mindolin S.F.
LABORATORY WORK No. 1-5: COLLISION OF BALLS.
Student__________________________________________________________________________ group:_________________

Tolerance_________________________________ Execution ________________________________Protection _________________
Goal of the work: Checking the law of conservation of momentum. Verification of the law of conservation of mechanical energy for elastic collisions. Experimental determination of the momentum of the balls before and after the collision, calculation of the coefficient of recovery of kinetic energy, determination of the average force of the collision of two balls, the speed of the balls upon collision.

Devices and accessories: device for studying the collision of balls FPM-08, scales, balls made of different materials.

Description of the experimental setup. Mechanical design of the device

Squares with scales 15,16 are attached to the lower bracket, and an electromagnet 17 is attached to special guides. After unscrewing the bolts 18,19, the electromagnet can be moved along the right scale and the height of its installation can be fixed, which allows you to change the initial ball. A stopwatch FRM-16 21 is attached to the base of the device, transmitting voltage through connector 22 to the balls and electromagnet.

The front panel of the FRM-16 stopwatch contains the following manipulation elements:

1. 
   W1 (Network) - network switch. Pressing this key turns on the supply voltage;
2. 
   W2 (Reset) – reset the meter. Pressing this key resets the FRM-16 stopwatch circuits.
3. 
   W3 (Start) – electromagnet control. Pressing this key causes the electromagnet to be released and a pulse to be generated in the stopwatch circuit as permission to measure.

COMPLETING OF THE WORK
Exercise No. 1. Verification of the law of conservation of momentum under inelastic central impact. Determination of the coefficient

recovery of kinetic energy.

Table No. 1.

№ experience
1
2
3
4
5

Find the ratio of the projection of the momentum of the system after an inelastic impact

Exercise No. 2. Verification of the law of conservation of momentum and mechanical energy during an elastic central impact.

Determination of the force of interaction between balls during a collision.

Table No. 2.

№ experience
1
2
3
4
5

Find the ratio of the projection of the momentum of the system after an elastic impact to the initial value of the projection of the impulse before the impact
. Based on the obtained value of the ratio of the projection of impulses before and after the collision, draw a conclusion about the conservation of the momentum of the system during the collision.

Find the ratio of the kinetic energy of the system after an elastic impact to the value of the kinetic energy of the system before impact . Based on the obtained value of the ratio of kinetic energies before and after the collision, draw a conclusion about the conservation of the mechanical energy of the system during the collision.

Compare the resulting value of the interaction force
with the gravity of a ball of greater mass. Draw a conclusion about the intensity of the mutual repulsion forces acting during the impact.

CONTROL QUESTIONS

1. 
   Impulse and energy, types of mechanical energy.
2. 
   The law of change in momentum, the law of conservation of momentum. The concept of a closed mechanical system.
3. 
   The law of change in total mechanical energy, the law of conservation of total mechanical energy.
4. 
   Conservative and non-conservative forces.
5. 
   Impact, types of impacts. Writing conservation laws for absolutely elastic and absolutely inelastic impacts.
6. 
   Interconversion of mechanical energy during free fall of a body and elastic vibrations.

Work, power, efficiency. Types of energy.

- Mechanical work constant in magnitude and direction of force

A= FScosα ,
Where A– work of force, J

F- force,

S– displacement, m

α - angle between vectors And

Work is a measure of the change in energy of a body or system of bodies.

In mechanics, the following types of energy are distinguished:

- Kinetic energy

- kinetic energy of a material point

- kinetic energy of a system of material points.

where T is kinetic energy, J

m – point mass, kg

ν – point speed, m/s

peculiarity:
Types of potential energy

- Potential energy of a material point raised above the Earth
P=mgh
peculiarity:

(see picture)

m– weight, kg

g– free fall acceleration, m/s 2

h– height of the point above the zero level of potential energy reference, m

h c.t.. - the height of the center of mass of a system of material points or an extended body above

zero potential energy reference level, m

- Potential energy of a deformed spring

, Where To– spring stiffness coefficient, N/m

Δ X– value of spring deformation, m

- Potential energy of gravitational interaction of two material points

M And m– point masses, kg

r– distance between them, m

peculiarity: is always a negative quantity (at infinity it is assumed to be zero)

Total mechanical energy

E = T + P

Mechanical power force N

Where A– work done by force during time t

Watt

distinguish: - useful power

Expended (or total power)

Where A useful And A cost is the useful and expended work of force, respectively

under the influence of this body force:

N = Fv instant . cosα, Where v instant is the instantaneous speed of the body

(i.e. body speed at a given time), m/s

(characterizes the efficiency of an engine, mechanism or process)

LAWS OF CHANGE AND CONSERVATION IN MECHANICS

Momentum of a material point is a vector quantity equal to the product of the mass of this point and its speed:

,

Impulse of the system material points is called a vector quantity equal to:

An impulse of power is called a vector quantity equal to the product of a force and the time of its action:

,

Law of momentum change:

The vector of change in the momentum of a mechanical system of bodies is equal to the product of the vector sum of all external forces acting on the system and the duration of action of these forces.

Law of conservation of momentum:

The vector sum of the impulses of the bodies of a closed mechanical system remains constant both in magnitude and direction for any movements and interactions of the bodies of the system.

Closed is a system of bodies that is not acted upon by external forces or the resultant of all external forces is zero.

External are called forces acting on a system from bodies not included in the system under consideration.

Internal are the forces acting between the bodies of the system itself.
For open mechanical systems, the law of conservation of momentum can be applied in the following cases:

1. 
   If the projections of all external forces acting on the system onto any direction in space are equal to zero, then the law of conservation of momentum projection is satisfied in this direction,

1. 
   If the internal forces are much greater in magnitude than external forces (for example, a rupture

external forces (for example, an impact), then the law of conservation of momentum can be applied

in vector form,

(that is )

Law of conservation and transformation of energy:

Energy does not appear from anywhere and does not disappear anywhere, but only passes from one type of energy to another, and in such a way that the total energy of an isolated system remains constant.

The change in the total mechanical energy of a system of bodies is equal to the sum of the work done by all non-conservative forces acting on the bodies of this system.

(that is )

Law of conservation of total mechanical energy:

The total mechanical energy of a system of bodies, the bodies of which are acted upon only by conservative forces or all non-conservative forces acting on the system do no work, does not change over time.

(that is
)

Towards conservative forces include:
,
,
,
,
.

To non-conservative- all other forces.

Features of conservative forces : the work of a conservative force acting on a body does not depend on the shape of the trajectory along which the body moves, but is determined only by the initial and final position of the body.

A moment of power relative to a fixed point O is a vector quantity equal to

,

Vector direction M can be determined by gimlet rule:

If the handle of the gimlet is rotated from the first factor in the vector product to the second by the shortest rotation, then the translational movement of the gimlet will indicate the direction of vector M.

Modulus of the moment of force relative to a fixed point
,

M moment of impulse body relative to a fixed point

,

The direction of the vector L can be determined using the gimlet rule.

If the handle of the gimlet is rotated from the first factor in the vector product to the second by the shortest rotation, then the translational movement of the gimlet will indicate the direction of vector L.
Module of angular momentum of a body relative to a fixed point
,

law of change of angular momentum

The product of the vector sum of the moments of all external forces relative to a fixed point O acting on a mechanical system by the time of action of these forces is equal to the change in the angular momentum of this system relative to the same point O.

law of conservation of angular momentum of a closed system

The angular momentum of a closed mechanical system relative to a fixed point O does not change either in magnitude or direction during any movements and interactions of the bodies of the system.

If the problem requires finding the work done by a conservative force, then it is convenient to apply the potential energy theorem:

Potential Energy Theorem:

The work of a conservative force is equal to the change in the potential energy of a body or system of bodies, taken with the opposite sign.

(that is )

Kinetic energy theorem:

The change in the kinetic energy of a body is equal to the sum of the work done by all forces acting on this body.

(that is
)

Law of motion of the center of mass of a mechanical system:

The center of mass of a mechanical system of bodies moves as a material point to which all the forces acting on this system are applied.

(that is
),

where m is the mass of the entire system,
- acceleration of the center of mass.

Law of motion of the center of mass of a closed mechanical system:

The center of mass of a closed mechanical system is at rest or moves uniformly and rectilinearly for any movements and interactions of the bodies of the system.

(that is, if)

It should be remembered that all laws of conservation and change must be written relative to the same inertial frame of reference (usually relative to the earth).

Types of blows

With a blow called the short-term interaction of two or more bodies.

Central(or direct) is an impact in which the velocities of the bodies before the impact are directed along a straight line passing through their centers of mass. (otherwise the blow is called non-central or oblique)

Elastic called an impact in which bodies, after interaction, move separately from each other.

Inelastic is called an impact in which the bodies, after interaction, move as a single whole, that is, at the same speed.

The limiting cases of impacts are absolutely elastic And absolutely inelastic blows.

1. the conservation law is fulfilled 1. the conservation law is satisfied

pulse: pulse:

2. law of conservation of complete 2. law of conservation and transformation

mechanical energy: energy:

Where Q- quantity of heat,

released as a result of the impact.

Δ U– change in internal energy of bodies in

as a result of the impact
DYNAMICS OF A RIGID BODY

Momentum of a rigid body rotating about a fixed axis
,

Kinetic energy of a rigid body rotating about a fixed axis
,

Kinetic energy of a rigid body rotating about an axis moving translationally

The basic equation for the dynamics of rotational motion of a mechanical system:

The vector sum of the moments of all external forces acting on a mechanical system relative to a fixed point O is equal to the rate of change of the angular momentum of this system.

The basic equation for the dynamics of rotational motion of a rigid body:

The vector sum of the moments of all external forces acting on a body relative to the stationary Z axis is equal to the product of the moment of inertia of this body relative to the Z axis and its angular acceleration.

Steiner's theorem:

The moment of inertia of a body relative to an arbitrary axis is equal to the sum of the moment of inertia of the body relative to an axis parallel to the given one and passing through the center of mass of the body, plus the product of the body mass by the square of the distance between these axes

,

Moment of inertia of a material point
,

Elementary work of moment of forces during rotation of a body around a fixed axis
,

The work of the moment of force when a body rotates around a fixed axis
,

Laboratory work

Measuring the collision time of elastic balls

Goal of the work: Measurement of the collision time of elastic balls, determination of the law of elastic force arising when the balls collide.

BRIEF THEORY

The collision of elastic balls is not instantaneous. The contact of the balls lasts, although small, but a finite period of time, and the forces arising during the impact, although great, are also finite.

From the moment the balls touch, the process of their deformation begins. The point of contact transforms into a circular area, and the kinetic energy transforms into the energy of elastic deformation. Elastic forces arise, which reach their greatest magnitude at the moment of greatest compression of the balls. Then there is a reverse process of transformation of potential deformation energy into kinetic energy of motion, ending at the moment the balls diverge. All these processes of mutual energy transfer unfold over a very short period of time, called the collision time. In general, the impact time depends on the elastic properties of the material of the balls, their relative speed at the moment the impact begins, and their size.

The collision time is determined by the law of elastic force arising when the balls collide. It is known that during elastic deformation of linear springs and rods, the elastic force F determined by Hooke's law F = -kh, Where h- the amount of spring deformation. When deforming bodies of complex shape, the dependence of the elastic force on the amount of compression can be represented in the following form

This kind of addiction F from h follows from the solution of the so-called contact problem of the theory of elasticity, solved by G. Hertz. It was found that the indicator n=3/2, and the value k upon collision of balls of radius R And R" is determined by the formula

. (2)

Where D depends on the elastic properties of the ball material.

N
It should be noted that upon impact both balls are deformed, therefore, under the compression value h in formula (1) we should understand the difference between the sum R+R" and the distance between the centers of the balls upon contact (see Fig. 1).

The potential energy of contacting deformed balls can be determined using the well-known formula F=-dU/dh.

. (3)

Dependence of the collision time of balls  from parameters k And n in the law of elastic force (1) can be obtained using the law of conservation of energy. In a frame of reference in which the center of inertia of the balls is at rest, the energy before the collision is equal to the kinetic energy of the relative motion  V2/2, Where V is the relative speed of the colliding balls, and  =m1m2 /(m1+m2) their reduced mass.

During the collision the relative speed V=dh/dt will initially decrease to zero. The kinetic energy will also decrease, equal to ( /2)(dh/ dt)2 . At the same time, the compression amount will increase and reach the value h0 at the moment when the relative speed is equal to zero. After reaching maximum compression, the processes will go in the opposite direction. A system of colliding elastic balls can be considered closed, therefore the law of conservation of energy must be satisfied in it, due to which the sum of kinetic energy is  V2/2 and potential energy - (k/ n+1) hn+1 during deformation is constant and equal to the energy of the balls before contact, that is

. (4)

From this equation we can determine the maximum approach of the balls h0, which is achieved when the speed dh/dt=0. We get from (4)

. (5)

Equation (4) is differential equation with separable variables. Solving it relatively dt, we get

Time  , during which the collision lasts (i.e. h varies from 0 before h0$ and back to zero), equals

It is convenient to take this integral if we introduce a new variable

It is also easy to see that x0- the value of the new variable at the point of maximum compression is 1. We have

The last integral is tabular, its value depends only on the number n. Thus, the dependence of impact time on speed takes on the following form.

, (6)

Where I(n)-- the value of the integral depending on n.

EXPERIMENTAL PROCEDURE

The form of formula (6) suggests an experimental technique for determining the parameters in the law of elastic force (1). Let us present formula (6) in the following form

Where (7)

Let's take logarithms of both sides of this expression

This shows that if we experimentally measure the collision time  at different meanings relative speed V and using this data to construct the dependence ln  from ln V, then, according to (8), it is a straight line. Moreover, the tangent of the angle of inclination of this straight line is equal to b, and the cut off part is ln A. By size b, we can determine the exponent n in the law of elastic force. Further on known values n And A, knowing the mass of the balls (i.e. the size  ), you can also calculate the value k.

Dependency Measuring Setup  from V that's how it is . A column is installed on the base, on which two brackets are attached. The upper bracket is equipped with rods that serve to hang the balls. The distance between these rods can be changed using a knob. Mobile holders for hanging balls are placed on the rods. Through these suspensions, voltage is supplied to the lower suspensions, and through them to the balls. The length of the hangers can be adjusted using special bushings with screws. An angular scale is attached to the lower bracket, along which you can move the electromagnet and fix the height of its installation.

An electronic stopwatch is screwed to the base of the device, on the rear panel of which there is a connector that supplies voltage to the balls and the electromagnet. On the front panel of the stopwatch there is a digital display, a button Net", as well as control buttons " Start" And " Reset".

The electronic part of the installation works as follows. When you press the " Start"the voltage supplying the electromagnet is turned off. The right ball, previously held by the electromagnet at a certain angle to the vertical, breaks away from it and comes into contact with the resting left ball. The balls are connected to the contacts of the pulse generating unit. Thus, at the moment the collision begins, a short circuit occurs these contacts, and the formation unit generates an electrical signal. This signal connects a quartz oscillator to the pulse counter, the frequency of which is very stable and equal to 1000000  1Hz, i.e. The duration of one pulse is 1 μs. These pulses, if their number is less than 999, are counted by a counter, that is, time intervals up to 999 μs can be measured. At the end of the collision, when the balls move apart, the formation unit generates a new pulse, which disconnects the quartz oscillator from the pulse counter. The number of pulses counted by the counter during the contact time of the balls, or, what is the same, the duration of the collision in microseconds is displayed on the digital display. If the duration of contact of the balls exceeds 999 microseconds, the light "" lights up on the front panel of the stopwatch overflow". When you press the " button Reset"The stopwatch readings are reset to zero, all electronic circuits are returned to their original state, the device is ready for the next measurements.

Thus, it is clear that the measurement of time in this work is a direct measurement. The systematic measurement error is 1 µs. The measurement of speed in this work, on the contrary, is an indirect measurement. She about
is determined as follows.

Speed V ball at the moment of impact is the same as that of a ball falling vertically from a height H, that is V= 2gH. From Fig. 2 it is clear that H=l-a, Where l- suspension length. But a=l cos  Means H=l(1- cos  ) $. From trigonometry it is known that 1- cos  =2 sin 2( /2), where H=2l sin 2( /2) .Thus, . (9)

The length of the suspension is measured directly with a ruler, the value is read off on a scale with precision  0,5  .

PERFORMANCE OF THE WORK AND EXPERIMENTAL CONDITIONS

1. Adjust the installation of the balls. To do this, use a knob located on the upper bracket to set such a distance between the rods so that the balls are in contact with each other. Adjust the height of the suspension so that the centers of the balls are at the same level.

2. Connect the microstopwatch to the network. Press the button " Net". At the same time, zeros should light up on the digital display. Button " Start" must be released.

3. Install the electromagnet so that the right ball, held by the electromagnet, is deflected to the maximum angle. By pressing the buttons " Reset", and then " Start"carry out a test measurement. In this case, it is necessary to ensure that the collision is central, that is, the trajectory of the left ball after the collision must be in the plane of motion of the right ball before the collision.

4. Using an electromagnet, set the ball at the maximum possible angle to the vertical. Measure the impact time for a given angle at least 5 times. Make sure that the left ball does not move at the moment of impact. Calculate the speed of the right ball before impact using formula (9), calculate the error of determination V. Process the collision time measurement results, that is, calculate the average value, standard deviation, and confidence limits. Analyze the results of measuring time to miss.

5. By changing the suspension angle of the balls in the range to the minimum possible, measure the impact time similarly to point 4. Present the results in the form of a table. Plot ln dependence  from ln V.

PROCESSING OF EXPERIMENTAL RESULTS

Further processing of the experimental dependence ln  from ln V involves the use of formula (8). To emphasize the linear nature of the ln dependence  from ln V, let us introduce new notations x=ln V, y=ln  , a=ln A. Then (8) will take the form usual for a linear function

The task is to find such values a And b, for which the function y=a+bx best matches experimental data. (The meaning of the vague expression “in the best possible way” will become clear later).

For the measure of deviation of function (10) from experimental data for i th experiment, the value is selected (yi-a-bxi)2. Why is this particular value taken, and not just (yi-a-bxi)? It is clear that both signs of evasion a+bxi from yi not good: bad if a And b, are such that yi , but it's also not good if a And b, are such that yi>a+bxi. If the measure of deviation were taken to be yi-a-bxi, and then the sum of the deviations in several experiments would be found, then it would be possible to obtain a very small value due to the mutual destruction of individual terms of large magnitude, but of different signs. This, however, would not mean at all that the parameters a And b well chosen. If the measure of deviation is taken (yi-a-bxi)2, then such mutual destruction will not occur, since all quantities (yi-a-bxi)2>0.

As a measure of overall error S in the description of experimental data by the function y=a+bx the sum of deviation measures for all experiments is taken (we denote their number l), i.e.

. (11)

Method for determining constants a And b included in formula (10), from the requirement of minimal total deviation, is called the least squares method.

Thus, you need to choose a And b, so that the value is minimal. For this purpose, the rules for finding extrema are used, known from mathematical analysis. If a has already been found, then on the right side of (11) it would be possible to change only b, so it should be like this -

Likewise, if it were found b, That -

These two conditions give the following system of equations to determine a And b

. (12)

Values ​​ xi, yi, xi2 and  xiyi can simply be calculated from experimental data. Then system (12) is a system of 2 linear equations relative to 2 unknowns a And b. Solving it in any way, it is not difficult to obtain

. (13)

So the parameters a And b, calculated using formulas (13) provide the best approximation of function (10) to experimental data.

Having determined the quantities a And b, you can calculate the standard deviation S0, characterizing the degree of deviation of the data from the calculated straight line, according to the formula

. (14)

Here a And b- straight line parameters calculated using formulas (13). The root mean square errors of each parameter are determined by the formulas

. (15)

Finally, the confidence limits  a and  b straight line parameters with confidence probability  are calculated as follows

that is, the Student coefficient is selected from tables for some effective probability equal to (1+  )/2 and for a number of points equal to l-2. For example, if you need to find confidence intervals for the parameters of a line obtained by the least squares method of 10 points ( l=10) with confidence probability  =0.9 , then it is necessary to substitute the Student coefficient into formulas (16) t0.95, 8 = 2.36.

Having defined the parameter b, it is possible to restore the indicator in the law by elastic force. To do this, we remember that b=(1-n)/(1+n). Then for n we get

. (17)

Accuracy  n is defined as the error of indirect measurement according to the formula

. (18)

where  b calculated using formula (16). Received value n can now be compared with the theoretical one, equal for balls 3/2 .

Definition of a constant k in law (1) represents a significantly more complex problem. Considering that a=ln A, we have A=ea and, according to formula (7), we obtain.

. (19)

Calculation complexity k According to this formula, the integral is taken quite simply only for n, multiples ½ . This is for experimentally determined n It's hard to expect. For arbitrary n this integral can be expressed through the so-called gamma function, depending on n. Using tables for the gamma function, you can obtain the value of the integral. Another way to calculate the value I(n) is numerical integration on a computer. Having received the value I(n) one way or another, then the value is simply calculated k. Note that, in principle, it is possible to determine the error  k, knowing  n and  a. But this task is very difficult and is not considered here.

Thus, the parameters in the law of elastic force (1) are determined. According to known k And n Next, the value of the maximum approach of the balls is calculated h0 according to formula (5). Such calculations must be carried out for the maximum and minimum speeds in this experiment. After this, the forces acting in these cases at maximum compression of the balls can be calculated using formula (1).

It is of interest to estimate the contact area of ​​the balls at the moment of maximum compression, which can be done if we know the value h, from geometric considerations. Obviously, the contact patch is a circle, the area of ​​which can be considered equal to the area of ​​the base of a spherical segment of radius R and height h.

CONTROL QUESTIONS

... collisions. General view of the collision research device balls... depends on elastic properties of materials balls. In a collision ball from a stationary... to an angle 1. Work order Measurement time interactions balls and angles , β, γ, γ1. 1) ...