Goal of the work: studying the impact of balls, determining the coefficient of speed recovery upon impact.
Devices and accessories: experimental setup, set of balls.
Brief theory
An impact is a short-term interaction of bodies in which, over a short period of time (), a significant change in the velocities of the bodies occurs. In many cases, a system of bodies interacting upon impact can be considered closed, since interaction forces ( strike forces) exceed all external forces acting on the bodies.
A straight line passing through the point of contact of bodies and normal to the surface of their contact is called strike line. If the line of impact passes through the centers of mass of the colliding bodies, then the impact is called central.
There are two limiting cases of impact: absolutely inelastic and absolutely elastic.
Absolutely inelastic impact- this is a collision of bodies, after which the interacting bodies move as a single unit or stop. With such an impact, the mechanical energy of the colliding bodies partially or completely turns into internal energy. Bodies undergo deformations that are inelastic and heat up. In a completely inelastic collision, the law of conservation of momentum is satisfied.
Absolutely elastic impact- a collision in which the mechanical energy of colliding bodies is not converted into other types of energy. During such an impact, the bodies are also deformed, but the deformations are elastic. After the collision, the bodies move at different speeds. With an absolutely elastic impact, the laws of conservation of momentum and mechanical energy are satisfied.
Absolutely elastic impact – idealization. When real bodies collide, mechanical energy is only partially restored by the end of the interaction, due to losses due to the formation of residual deformations and heating.
The degree of impact elasticity is characterized by the value 
, called speed recovery factor.
With a central impact 
is determined by the expression
, (1)
Where 
relative speed of bodies before collision, 
relative speed of bodies after collision.
The velocity recovery coefficient depends on the elastic properties of the material of the colliding bodies. For a completely elastic impact 
= 1, for absolutely inelastic 
= 0, for real hits 
0 <
< 1
(например, при соударении тел из дерева
0.5, steel 
0.55, ivory 
0,9).
In this laboratory work, the central impact of two metal balls is studied and the velocity restitution coefficient is determined.
The setup for studying the collision of balls is shown schematically in Figure 1. It consists of a base 1 with adjustable supports, on which the stand is fixed 2 with two brackets. On the top bracket 3 there is a mechanism for fastening bifilar threads-suspensions 4 for balls 5 . Measuring scales are attached to the bottom bracket 6 , graduated in degrees . On the right scale there is an electromagnet 7 , which can move along the scale and be fixed in a certain position.
Let two balls have the same mass 
hang on threads of the same length, touching each other (Fig. 2). When the right ball is deflected (ball 1
) from the equilibrium position to the angle 
it will gain potential energy 
(
the height of the ball's center of mass, 
acceleration of gravity). If the ball is released, then when the ball returns to the equilibrium position, its potential energy will completely transform into kinetic energy.
According to the law of conservation of mechanical energy
,
(2)
Where 
ball speed 1
when it reaches the equilibrium position (before colliding with the ball 2
).
From formula (2) it follows
. (3)
Height 
can be expressed through 
(deflection angle) and 
(distance from the suspension point to the center of mass of the ball). From Figure 2 it is clear that 
, i.e. 
. Because 
, That
. (4)
Substituting formula (4) into (3), we get 
. If the angle 
small, then 
and therefore
=
.
(5)
Similar formulas can be obtained for 
And 
─ speeds of the balls after impact:
,
, (6)
Where 
And 
Substituting into expression (1) the values 
,
,
(formulas (5), (6)) and, taking into account that the ball 2
was at rest before the collision, i.e. 
= 0, we get
.
(7)
Thus, to determine the speed recovery coefficient it is necessary at a given angle 
measure 
And 
angles of deviation from the vertical of the threads-suspenders of the balls after the impact.
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 of the device
General view of the device for studying the collision of balls FPM -08 is shown in Fig. 1. Base 1 is equipped with adjustable legs (2), which allow you to set the base of the device horizontally. A column 3 is fixed at the base, to which the lower 4 and upper 5 brackets are attached. A rod 6 and a screw 7 are attached to the upper bracket, which are used to set the distance between the balls. On the rods 6 there are movable holders 8 with bushings 9, fixed with bolts 10 and adapted for attaching hangers 11.Wires 12 pass through the hangers 11, supplying voltage to the hangers 13, and through them to the balls 14. After loosening the screws 10 and 11, a central collision of the balls can be achieved.
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 the front panel of the 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.
Experience no. | ||||||||||
1 | ||||||||||
2 | ||||||||||
3 | ||||||||||
4 | ||||||||||
5 |
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.
Experience no. | |||||||||||||
1 | |||||||||||||
2 | |||||||||||||
3 | |||||||||||||
4 | |||||||||||||
5 |
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, the following types of energy are distinguished:
- 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:
Types of 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 equal to zero depending on the choice of the initial level of potential energy reading
- 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"> - expended (or total power) font-size:10.0pt">whereApoleznaya And Azatris the useful and expended work of force, respectively
The power of a 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. body speed at a given time), m/s
Efficiency factor (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:
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.
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 the internal forces are much greater 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 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. ,
font-size:10.0pt">law of change in angular momentumThe 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,
Goals of work:
1) study of the laws of elastic and inelastic collision of balls,
2) determination of the ratio of velocities and masses of the balls.
Basic concepts and patterns
An example of the application of the laws of conservation of momentum and energy when solving a real physical problem is impact of absolutely elastic and inelastic bodies.
Hit(or collision) is a collision of two or more bodies in which the interaction lasts a very short time. When impacted, bodies experience deformation. The impact phenomenon usually occurs in hundredths, thousandths and millionths of a second. The smaller the deformation of the bodies, the shorter the collision time. Since in this case the momentum of the bodies changes by a finite amount, enormous forces develop during a collision.
The impact process is divided into two phases.
First phase– from the moment the bodies come into contact until the moment when their relative speed becomes zero.
Second phase- from this last moment until the moment when the contact of bodies stops.
From the moment deformations occur, forces directed opposite to the relative velocities of the bodies begin to act at the points of contact of the bodies. In this case, the energy of mechanical motion of bodies transforms into the energy of elastic deformation (the first phase of impact).
In the second phase of the impact, when the relative speed has become zero, partial or complete restoration of the shape of the bodies begins, then the bodies diverge and the impact ends. In this phase, the kinetic energy of the system increases due to the positive work of elastic forces.
For real bodies, the relative speed after an impact does not reach the value that it had before the impact, since part of the mechanical energy irreversibly transforms into internal and other forms of energy.
There are two extreme types of impact:
a) blow absolutely inelastic;
b) blow absolutely elastic.
An absolutely inelastic impact (close to it) occurs when bodies made of plastic materials (clay, plasticine, lead, etc.) collide, the shape of which is not restored after the cessation of the external force.
An absolutely inelastic impact is an impact after which the deformations that occur in the bodies are completely preserved. After a completely inelastic impact, the bodies move with a common speed.
An absolutely elastic impact (close to it) occurs when bodies made of elastic materials (steel, ivory, etc.) collide, the shape of which is completely (or almost completely) restored after the cessation of the external force. With an elastic impact, the shape of the bodies and the value of their kinetic force are restored energy. After an impact, the bodies move at different speeds, but the sum of the kinetic energies of the bodies before the impact is equal to the sum of the kinetic energies after the impact. The straight line coinciding with the normal to the surface of the bodies at the point of their contact is called the line of impact. The impact is called central if the line of impact passes through centers of gravity of bodies. If the velocity vectors of the bodies before the impact lay on the line of impact, then the impact is called direct.
When bodies collide, two conservation laws.
1. Law of conservation of momentum.
In a closed system (a system for which the resultant of all external forces is zero), the vector sum of the momenta of the bodies does not change, i.e. constant value:
= = = const, (4.1)
where is the total momentum of the system,
– impulse i-th body of the system.
2. Law of conservation of energy
In a closed system of bodies, the sum of kinetic, potential and internal energy remains constant:
W k + W n + Q = const, (4.2)
Where W to– kinetic energy of the system,
W n– potential energy of the system,
Q– energy of thermal motion of molecules (thermal energy).
The simplest case of collision of bodies is the central impact of two balls. Consider the impact of balls with masses m i And m 2 .
Ball velocities before impact and after impact and . For them, the laws of conservation of momentum and energy will be written as follows:
. (4.4)
The impact of the balls is characterized by the coefficient of restitution TO, which is determined by the ratio of the relative speed of the balls after the impact to the relative speed of the balls before the impact. , taken by absolute value i.e.
The velocities of the first ball relative to the second before and after the impact are equal:
, 
. (4.6)
Then the recovery coefficient of the balls is:
. (4.7)
With an absolutely elastic impact, the law of conservation of mechanical energy is satisfied, Q= 0, the relative velocities of the balls before and after the interaction are equal and the recovery coefficient is 1.
During an absolutely inelastic impact, the mechanical energy of the system is not conserved; part of it is converted into internal energy. Bodies are deformed. After interaction, the bodies move at the same speed, i.e. their relative speed is 0, therefore the coefficient of restoration is also zero, K = 0. The law of conservation of momentum will be written as
where is the speed of the bodies after interaction.
The law of conservation of energy will take the form:
. (4.9)
From equation (4.9) we can find Q– mechanical energy converted into internal energy.
In practice, extreme cases of interaction are rarely realized. More often the interaction is intermediate in nature, and the recovery coefficient TO has the meaning.
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; 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. 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 of 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 a 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′ are the total energy of two balls before and after the impact, respectively, 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 the bodies is restored, all the potential energy again turns into the kinetic energy of the balls, i.e. 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 a vertical position of column 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.
| No. | W | ||||||||
| ……… | |||||||||
| Average value |
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
| Experience no. | |||||||||||
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.
| Experience no. | |||||||||||||
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
COLLISIONS OF ELASTIC BALLS
Read the lecture notes and textbook (Savelyev, vol. 1, § 27, 28). Launch the Mechanics program. Mol.physics". Select "Mechanics" and "Collisions of elastic balls". Click the button with the page image at the top of the inner window. Read the brief theoretical information. Write down what is necessary in your notes. (If you have forgotten how to operate the computer simulation system, read the INTRODUCTION again)
GOAL OF THE WORK :
Selection of physical models for analyzing the interaction of two balls in a collision.
Study of conservation of elastic balls during collisions.
Read the text in the Manual and in the computer program (the “Physics” button). Take notes on the following material:
impact (collision, collision)) - a model of interaction of two bodies, the duration of which is zero (instantaneous event). It is used to describe real interactions, the duration of which can be neglected in the conditions of a given problem.
ABSOLUTELY ELASTIC IMPACT - a collision of two bodies, after which the shape and size of the colliding bodies are restored completely to the state that preceded the collision. The total momentum and kinetic energy of a system of two such bodies are conserved (after the collision they are the same as they were before the collision):
Let the second ball be at rest before impact. Then, using the definition of momentum and the definition of an absolutely elastic impact, we transform the law of conservation of momentum, projecting it onto the OX axis, along which the body moves, and the OY axis, perpendicular to OX, into the following equation:
Sighting distance d is the distance between the line of motion of the first ball and a line parallel to it passing through the center of the second ball. We transform the conservation laws for kinetic energy and momentum and obtain:
TASK: Derive formulas 1, 2 and 3
METHODOLOGY and PROCEDURE OF MEASUREMENTS
Carefully examine the drawing, find all the controls and other main elements and sketch them out.
Look at the picture on the screen. Having established the impact distance d 2R (the minimum distance at which no collision is observed), determine the radius of the balls.
By setting the aiming distance to 0
Obtain permission from your teacher to take measurements.
MEASUREMENTS:
Set, by moving the controller sliders with the mouse, the masses of the balls and the initial speed of the first ball (first value), indicated in the table. 1 for your team. Set the aiming distance d equal to zero. By clicking the “START” button on the monitor screen with your mouse, watch the movement of the balls. Record the results of measurements of the required quantities in Table 2, a sample of which is given below.
Change the value of the aiming distance d by the value (0.2d/R, where R is the radius of the ball) and repeat the measurements.
When the possible d/R values have been exhausted, increase the initial speed of the first ball and repeat the measurements starting with zero target distance d. Write the results in a new table 3, similar to table. 2.
Table 1. Ball masses and initial velocities(do not redraw) .
| Number brigades | m 1 | m 2 | V 0 (m/s) | V 0 (m/s) | Number brigades | m 1 | m 2 | V 0 (m/s) | V 0 (m/s) |
|
| 1 | 1 | 5 | 4 | 7 | 5 | 1 | 4 | 6 | 10 |
|
| 2 | 2 | 5 | 4 | 7 | 6 | 2 | 4 | 6 | 10 |
|
| 3 | 3 | 5 | 4 | 7 | 7 | 3 | 4 | 6 | 10 |
|
| 4 | 4 | 5 | 4 | 7 | 8 | 4 | 4 | 6 | 10 |
Tables 2 and 3. Results of measurements and calculations (number of measurements and rows = 10)
| m 1 =___(kg), m 2 =___(kg), V 0 = ___(m/s), (V 0) 2 = _____(m/s) 2 |
|||||||||||
| № | d/R | V 1 | V 2 | 1 hail | 2 hail | V 1 Cos 1 | V 1 Sin 1 | V 2 Cos 2 | V 2 Sin 2 | (m/s) 2 | (m/s) 2 |
| 1 | 0 | ||||||||||
| 2 | 0.2 | ||||||||||
| ... | |||||||||||
PROCESSING RESULTS AND PREPARING A REPORT:
Calculate the required values and fill out tables 2 and 3.
Build dependency graphs (in three figures)
For each graph, determine the mass ratio m 2 /m 1 of the balls. Calculate the mean of this ratio and the absolute error of the mean.
Analyze and compare measured and specified mass ratio values.
Questions and tasks for self-control
What is an impact (collision)?
For what interaction of two bodies can the collision model be used?
Which collision is called absolutely elastic?
In which collision is the law of conservation of momentum satisfied?
Give a verbal formulation of the law of conservation of momentum.
Under what conditions is the projection of the total momentum of a system of bodies onto a certain axis preserved?
In which collision is the law of conservation of kinetic energy satisfied?
Give a verbal formulation of the law of conservation of kinetic energy.
Define kinetic energy.
Define potential energy.
What is total mechanical energy.
What is a closed system of bodies?
What is an isolated system of bodies?
Which collision releases thermal energy?
In what collision is the shape of the bodies restored?
In what collision is the shape of the bodies not restored?
What is the impact distance (parameter) when balls collide?
1. LITERATURE
Savelyev I.V. General physics course. T.1. M.: “Science”, 1982.
Savelyev I.V. General physics course. T.2. M.: “Science”, 1978.
Savelyev I.V. General physics course. T.3. M.: “Science”, 1979.
2.SOME USEFUL INFORMATION
PHYSICAL CONSTANTS
| Name | Symbol | Meaning | Dimension |
| Gravitational constant | or G | 6.67 10 -11 | N m 2 kg -2 |
| Acceleration of free fall on the surface of the Earth | g 0 | 9.8 | m s -2 |
| Speed of light in vacuum | c | 3 10 8 | m s -1 |
| Avogadro's constant | N A | 6.02 10 26 | kmol -1 |
| Universal gas constant | R | 8.31 10 3 | J kmol -1 K -1 |
| Boltzmann's constant | k | 1.38 10 -23 | JK -1 |
| Elementary charge | e | 1.6 10 -19 | Cl |
| Electron mass | m e | 9.11 10 -31 | kg |
| Faraday's constant | F | 9.65 10 4 | Cl mol -1 |
| Electrical constant | o | 8.85 10 -12 | F m -1 |
| Magnetic constant | o | 4 10 -7 | Hm -1 |
| Planck's constant | h | 6.62 10 -34 | J s |
PRECISIONS AND MULTIPLIERS
to form decimal multiples and submultiples
| Console | Symbol | Factor | Console | Symbol | Factor |
|
| soundboard | Yes | 10 1 | deci | d | 10 -1 |
|
| hecto | G | 10 2 | centi | With | 10 -2 |
|
| kilo | To | 10 3 | Milli | m | 10 -3 |
|
| mega | M | 10 6 | micro | mk | 10 -6 |
|
| giga | G | 10 9 | nano | n | 10 -9 |
|
| tera | T | 10 12 | pico | P | 10 -12 |
