Tasks: verification of the laws of conservation of momentum and energy in absolutely elastic and inelastic collisions of balls.
Equipment: device for investigation of collisions of balls FPM-08.
Brief theory:
Rectilinear motion:
A vector quantity numerically equal to the product of the mass of a material point and its velocity and having the direction of velocity is called momentum (momentum)) material point.
Law of conservation of momentum: = const- the momentum of a closed system does not change over time.
Law of energy conservation: in a system of bodies between which only conservative forces act, the total mechanical energy remains constant over time. E = T + P = const ,
where E - total mechanical energy, T - kinetic energy, R - potential energy.
Kinetic energy mechanical system is the energy of the mechanical movement of the system. Kinetic energy for
forward movement: 
, rotational motion 
where J - moment of inertia, ω - cyclic frequency).
Potential energy systems of bodies is the energy of interaction between the bodies of the system (it depends on the relative position of the bodies and the type of interaction between the bodies) Potential energy of an elastically deformed body: 
; in torsion deformation 
where k is the stiffness coefficient (torsional modulus), X - deformation, α - twist angle).
Absolutely elastic impact- a collision of two or more bodies, as a result of which no deformations remain in the interacting bodies and all the kinetic energy that the bodies possessed before the impact is again converted into kinetic energy after the impact.
Absolutely inelastic impact - a collision of two or more bodies, as a result of which the bodies are combined, moving further as a whole, part of the kinetic energy is converted into internal energy.
Derivation of the working formula:
In this setup, two balls with masses m 1 and m 2 are suspended on thin threads of the same length L. ball with mass m 1 deflected at an angle α 1 and let go. On installation angle α 1 you set it yourself, measuring it on a scale and fixing the ball with an electromagnet, the angles of deviation α 1 ’ and α 2 ’ balls after the collision is also measured on a scale.
1 . Let us write down the laws of conservation of momentum and energy for an absolutely elastic collision
before the collision first ball speed V 1, second ball speed V 2 =0;
momentum of the first ball p 1 = m 1 V 1 , momentum of the second R 2 = 0 ,
after impact-velocities of the first and second balls V 1 ’ and V 2 ’
balls momentum p 1
’
=
m 1
V 1
’
and p 2
=
m 2
V 2
’
m
1
V 1
=
m 1
V 1
’+
m 2
V 2
’
law of conservation of momentum;
the law of conservation of energy of the system before and after the collision of balls
h, it acquires potential energy
R= m 1
gh,
- this energy is completely converted into the kinetic energy of the same ball 
, hence the speed of the first ball before the collision 
Express h through the length of the thread L and angle of impact α , from fig. 2 shows that
h + L cos α 1 = L
h = L( 1-cosα 1 ) = 2 L sin 2 (α 1 /2),
then 
If the corners α one ! and α 2! the angles of deflection of the balls after the collision, then, arguing similarly, we can write down the velocities after the collision for the first and second balls:
We substitute the last three formulas into the law of conservation of momentum
( working formula 1)
This equation includes quantities that can be obtained by direct measurements. If, when substituting the measured values, equality is satisfied, then the law of conservation of momentum in the system under consideration is also satisfied, as well as the law of conservation of energy, since these laws were used in the derivation of the formula.
2 . Let us write down the laws of conservation of momentum and energy for a perfectly inelastic collision
m 1
V 1
=
(m 1
+
m 2
)
V 2
law of conservation of momentum; where V 1
- the speed of the first ball before the collision; V 2
- the total speed of the first and second balls after the collision. 
the law of conservation of energy of the system before and after the collision of the balls, where W -
part of the energy that is converted into internal energy (heat).
The law of conservation of energy of the system until the moment of impact, when the first ball is raised to a height h corresponding to the angle α
1.
(see fig.3) 
- the law of conservation of energy of the system after the moment of impact, corresponding to the angle 
.
Let's express the speed V and V’ from the laws of conservation of energy:
, 
, 
We substitute these formulas into the law of conservation of momentum and get:
working formula 2
Using this formula, you can check the law of conservation of momentum and the law of conservation of energy for a perfectly inelastic impact.
Average interaction strength between two balls at the moment of elastic impact can be determined by the change in the momentum of one (first) ball
Substituting into this formula the values of the velocities of the first ball before and after the impact
And 
we get:
working formula 3
where ∆ t = t- the time of collision of the balls, which can be measured using a microstopwatch.
Description of the experimental
settings:
A general view of the FPM-08 instrument for studying ball collisions is shown in fig. 4.
On the basis of the installation there is an electric microstopwatch RM-16, designed to measure short time intervals.
On the front panel of the microstopwatch there is a “time” display (time is counted in microseconds), as well as “NETWORK”, “RESET”, “START” buttons.
A column with a scale is also attached to the base, on which the upper and lower brackets are installed. Two rods and a knob are installed on the upper bracket, which serves to adjust the distance between the balls. Through the suspensions, wires are drawn, through which voltage is supplied to the balls from the microstopwatch.
On the lower bracket there are scales for reading the angles that the balls have relative to the vertical. These scales can be moved along the bracket. Also on the bracket on a special stand is an electromagnet, which serves to fix one of the balls in a certain position. The electromagnet can be moved along the right scale by unscrewing the nuts securing it to the scale. At the end of the body of the electromagnet there is a screw for adjusting the strength of the electromagnet.
Work Instructions
1 task: verification of the law of conservation of momentum and the law of conservation of energy for a perfectly elastic impact.
To complete this task, it is necessary to measure the masses of the balls and the angles of deviation relative to the vertical. 
2 task:
verification of the law of conservation of momentum and the law of conservation of energy for a perfectly inelastic impact
| m 1 | m2 | № | α 1 |  |  |  | Before the impact
| After the impact  |
| 1 | ||||||||
| 2 | ||||||||
| 3 | ||||||||
| 4 | ||||||||
| 5 | ||||||||
| Wed |
Repeat steps 1-9 for plasticine balls and substitute the results into the working formula 2.
3 task: exploreforce of interaction of balls in elastic collision
We need to graph the function F Wed = f(α 1 ). For this task, the working formula 3 is used, To plot the function F Wed = f(α 1 ), measurements need to be taken - kickback angle of the first ball after impact and t- impact time at various values α 1 .
Press the "RESET" button on the microstopwatch;
Set the right ball at an angle α 1 = 14º, make collisions of the balls, measure on the angular scale and read the microstopwatch. Calculate F cp for each measurement according to the working formula 3;
Enter the result of the measurement in the table;
m 1
L
№
α 1
Δ t
Fcp
1
14º
2
14º
3
14º
4
10º
5
10º
6
10º
7
7º
8
7º
Plot the Function F Wed = f(α 1 ),
Draw conclusions about the obtained dependence:
How strength depends F cp α 1) ?
How does the time Δ t collisions from the initial speed ( α 1) ?
test questions:
What is called collision?
Absolutely elastic and absolutely inelastic collisions.
What forces arise when two balls come into contact.
What is called the coefficient of recovery of speed and energy. And how do they change in the case of absolutely elastic and absolutely inelastic collisions?
What conservation laws are used in doing this work? Formulate them.
How does the magnitude of the final momentum depend on the ratio of the masses of the colliding balls?
How does the value of the kinetic energy transferred from the first ball to the second depends on the mass ratio?
What is the impact time for?
What is the center of inertia (or center of mass)?
Literature:
Trofimova T.I. Physics course. Moscow: Higher school, 2000
Matveev A.N.: Mechanics and the theory of relativity. - M., Higher School, 1986, pp. 219-228.
4. Gabyshev H.H. Methodical manual on mechanics - Yakutsk., YSU, 1989
The purpose of the work: to get acquainted with the phenomenon of impact on the example of the collision of balls, calculate the energy recovery coefficient, check the law of conservation of momentum.
Theoretical information
Let us deflect ball A with mass at an angle
where and the readings on the measurement scale. In this case, the ball will rise to a height (see Fig. 1). As can be seen from the figure, the lifting height can be expressed in terms of the length of the suspension and the deflection angle:
After the ball is released without an initial speed, it will accelerate and at the bottom of its trajectory it will acquire a horizontal speed, which can be found from the law of conservation of energy:
At the lowest point of its trajectory, ball A collides with ball B, and after a very short impact, they fly apart in opposite directions with horizontal velocities and (see Fig. 2). Since during the impact the tension forces of the threads and the gravity forces acting on the balls are directed vertically, the law of conservation of the horizontal projection of the system momentum must be satisfied:
In most cases, real impacts of bodies are not elastic due to the occurrence of dissipative forces inside these bodies (internal friction), so the kinetic energy of the system as a whole decreases upon impact. The kinetic energy recovery coefficient is a value equal to:
The speed recovery factor is always less than one:. Equality to unity means complete conservation of energy, which can only be in the ideal case of the absence of dissipative forces in the system.
After a collision (see Fig. 3), the action of dissipative forces of internal friction ceases, and if we neglect the energy loss during motion due to air resistance, we can use the energy conservation law for each ball separately. Ball A will deviate by an angle and rise to a height, and ball B will deviate by an angle and rise to a height
Using equations similar to equations (1) and (2), we express the speed of the balls after the impact:
Substituting (2) and (5) into (4), we obtain an expression for calculating the energy recovery coefficient:
Substituting (2) and (5) into (3), we obtain the momentum conservation law in the form:
Equipment: a rack with two weights (balls) hung on a bifilar suspension.
Task: to determine the coefficient of recovery of the body's velocity during inelastic impact of balls.
Work order
Write down the initial positions 0 and 0, corresponding to the points of intersection of the threads of the bifilar suspensions with the scale division line, when the balls are stationary. Here and in what follows, the designation "" refers to ball A with a smaller mass m1, and "" refers to ball B with a smaller mass m2.
Deflect ball A at angle 1 from 10º to 15º and release it without initial velocity. Count the first discard of both balls 2 and 2 (since it is almost impossible to take two counts at once, they do this: first they take a count for one ball, then they make a second hit from the same position of ball A and take a count for the second ball). A blow from this position is made at least 10 times in order to obtain at least five values of thread discards after a blow for each ball (2 and 2). Find the mean<2>and<2>.
Experiment to do for two other values 1. (from 20 to 25, from 30 to 35). Complete table 1.
Check the momentum conservation law (7). To do this, calculate the speeds and using formulas (2) and (5), taking into account that
and the right side of equation (7)
Record the results of measurements and calculations in table. 1 and 2. Calculate the energy recovery factor using formula (6).
Table 1
test questions
Will the system of spheres be closed?
Formulate the law of conservation of momentum of the system.
Is the momentum of the system of balls conserved after the impact? Why?
Type of impact in this work. Analyze the resulting energy recovery factor.
When is the total mechanical energy of the system conserved? Are the kinetic energies of the system of balls equal before and after the impact?
Can mechanical energy not be conserved in some system and the angular momentum remain constant?
Obtain calculated formulas for ball velocities after impact.
List of sources used
Saveliev I.V. Course of general physics. T.1. Mechanics. Molecular physics. - St. Petersburg: Lan, 2007. - 432 p. - ch. II, §23, p.75-77, ch. III, §27-30, p.89-106
LAB #1_5
COLLISIONS OF ELASTIC BALLS
Familiarize yourself with the lecture notes and the textbook (Saveliev, vol. 1, § 27, 28). Run the program "Mechanics. Mol.physics. Select Mechanics and Elastic Ball Collisions. Press the button with the image of the page at the top of the inner window. Read the brief theoretical information. Write down what you need in your notes. (If you have forgotten how to work with the computer simulation system, read the INTRODUCTION again)
GOAL OF THE WORK :
The choice of physical models for the analysis of the interaction of two balls in a collision.
Investigation of , preserved during collisions of elastic balls.
Familiarize yourself with the text in the Handbook and in the computer program (button “Physics”). Review the following material:
blow (collision, collision) - model of interaction of two bodies, the duration of which is equal to zero (instantaneous event). It is used to describe real interactions, the duration of which can be neglected under the conditions of a given problem.
ABSOLUTELY ELASTIC IMPACT - a collision of two bodies, after which the shape and dimensions of the colliding bodies are completely restored to the state that preceded the collision. The total momentum and kinetic energy of a system of two such bodies are conserved (they are the same after the collision as they were before the collision):
Let the second ball rest before the impact. Then, using the definition of momentum and the definition of an absolutely elastic impact, we transform the law of conservation of momentum by projecting it onto the OX axis, along which the body moves, and the OY axis, perpendicular to OX, into the following equation:
aiming distance d is the distance between the line of motion of the first ball and the 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
METHOD AND ORDER OF MEASUREMENTS
Carefully examine the drawing, find all the regulators and other main elements and sketch them in an outline.
Look at the picture on the screen. By setting the aiming 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
Get permission from your instructor to take measurements.
MEASUREMENTS:
Set, by moving the cursors of the regulators, the masses of the balls and the initial speed of the first ball (the first value), indicated in the table. 1 for your team. Target distance d set equal to zero. By clicking on the "START" button on the monitor screen, follow 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 target distance d to (0.2d/R, where R is the radius of the ball) and repeat the measurements.
When the possible values of d/R are exhausted, increase the initial speed of the first ball and repeat the measurements, starting from zero impact distance d. Record 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 | m2 | V0 (m/s) | V0 (m/s) | Number brigades | m 1 | m2 | V0 (m/s) | V0 (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 \u003d ___ (kg), m 2 \u003d ___ (kg), V 0 \u003d ___ (m / s), (V 0) 2 \u003d _____ (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 THE RESULTS AND PREPARATION OF THE REPORT:
Calculate the required values and fill in tables 2 and 3.
Plot dependency graphs (in three figures)
For each graph, determine the ratio of the masses 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 target 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 perfectly 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 the system of bodies on 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 body system?
What kind of collision releases heat energy?
At what collision is the shape of the bodies restored?
In which collision the shape of the bodies is not restored?
What is the aiming distance (parameter) when balls collide?
1.LITERATURE
Saveliev I.V. Course of general physics. T.1. M.: "Nauka", 1982.
Saveliev I.V. Course of general physics. T.2. M.: "Nauka", 1978.
Saveliev I.V. Course of general physics. T.3. M.: "Nauka", 1979.
2.SOME USEFUL INFO
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 | g0 | 9.8 | m s -2 |
| The speed of light in a vacuum | c | 3 10 8 | m s -1 |
| Avogadro constant | N A | 6.02 10 26 | kmol -1 |
| Universal gas constant | R | 8.31 10 3 | J kmol -1 K -1 |
| Boltzmann constant | k | 1.38 10 -23 | J K -1 |
| elementary charge | e | 1.6 10 -19 | cl |
| Mass of an electron | me | 9.11 10 -31 | kg |
| Faraday constant | F | 9.65 10 4 | Cl mol -1 |
| Electrical constant | about | 8.85 10 -12 | F m -1 |
| Magnetic constant | about | 4 10 -7 | H m -1 |
| Planck's constant | h | 6.62 10 -34 | J s |
SUBSCRIPTIONS AND MULTIPLIERS
for the formation of decimal multiples and submultiples
| Prefix | Symbol | Factor | Prefix | 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 |
Objective: study of the impact of balls, determination of the coefficient of recovery of speed upon impact.
Instruments and accessories: experimental setup, set of balls.
Brief theory
A blow is a short-term interaction of bodies, in which a significant change in the velocities of bodies occurs over a short period of time (). In many cases, the system of bodies interacting upon impact can be considered closed, because the interaction forces ( strike force) exceed all external forces acting on the body.
The straight line passing through the point of contact of the 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 whole or stop. With such an impact, the mechanical energy of the colliding bodies is partially or completely converted into internal energy. Bodies undergo deformations that are inelastic and heat up. In a perfectly inelastic impact, the law of conservation of momentum is satisfied.
Absolutely elastic impact- a collision in which the mechanical energy of the colliding bodies is not converted into other types of energy. In the process of 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, the 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.
On center hit 
is defined by the expression
, (1)
where 
relative speed of bodies before impact, 
relative velocity of bodies after collision.
The velocity recovery coefficient depends on the elastic properties of the material of the colliding bodies. For a perfectly elastic impact 
= 1, for absolutely inelastic 
= 0, for real beats 
0 <
< 1
(например, при соударении тел из дерева
0.5, steel 
0.55, ivory 
0,9).
In this lab, we study the central impact of two metal balls and determine the velocity recovery factor.
The installation for studying the collision of balls is schematically shown in Figure 1. It consists of a base 1 with adjustable legs, on which the rack is fixed 2 with two brackets. On top bracket 3 there is a mechanism for fixing bifilar suspension threads 4 for balls 5 . Measuring scales are fixed on the bottom bracket 6 , graduated in degrees . On the right scale is an electromagnet 7 , which can move along the scale and be fixed in a certain position.
Let two balls of the same mass 
hang on threads of the same length, touching each other (Fig. 2). When deflecting the right ball (ball 1
) from the equilibrium position to the angle 
it will acquire potential energy 
(
the height of the center of mass of the ball, 
acceleration of gravity). If the ball is released, then when the ball returns to the equilibrium position, its potential energy will completely turn into kinetic energy.
According to the law of conservation of mechanical energy
,
(2)
where 
ball speed 1
when it reaches the equilibrium position (before the collision with the ball 2
).
From formula (2) it follows
. (3)
Height 
can be expressed through 
(angle of deflection) and 
(distance from the point of suspension to the center of mass of the ball). Figure 2 shows that 
, i.e. 
. As 
, then
. (4)
Substituting formula (4) into (3), we obtain 
. If the angle 
small, then 
and hence
=
.
(5)
Similar formulas can be obtained for 
and 
─ velocities of the balls after the collision:
,
, (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 velocity recovery factor, it is necessary at a given angle 
measure 
and 
angles of deviation from the vertical of the threads-suspensions of the balls after the impact.
Objective:
Experimental and theoretical determination of the value of the momentum of the balls before and after the collision, the coefficient of recovery of kinetic energy, the average force of the collision of two balls. Verification of 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 the upper bracket 3 is installed, intended for suspension of balls; a housing designed to mount a scale of 4 angular displacements; an electromagnet 5 designed to fix the initial position of one of the balls 6; adjustment nodes that provide a direct central impact of the balls; threads 7 for hanging metal balls; wires to ensure the electrical contact of the balls with the terminals 8. To start the ball and count the time to impact, the control unit 9 is used. The 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 impact.
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 consider only the central impact, that is, the impact that occurs along the line connecting the centers of the balls.
Consider absolutely inelastic impact. This impact can be observed on two lead or wax balls suspended from a thread of the same 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, decelerating ball A and accelerating ball B. Since these forces are proportional to the rate of change in deformation (i.e., the relative speed of movement of the balls ), then as the relative velocity decreases, they decrease and vanish as soon as the speeds of the balls equalize. From this point on, the balls, "merged", move together.
Let us consider the problem of the impact of inelastic balls quantitatively. We 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 applies 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 the collision; v′ is the speed of the balls after the 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, one should use the law of conservation of momentum
Since the impact is central, then all the velocity vectors lie on one straight line. Taking this straight line as the axis X and projecting equation (5) onto this axis, we obtain the scalar equation:
(6)
This shows that if the balls before the impact moved in one direction, then after the impact they will move in the same direction. If the balls before the impact moved towards each other, then after the impact they will move in the direction where the ball with the greater momentum was moving.
Let us put v′ from (6) into equality (4):
(7)
Thus, the work of internal non-conservative forces during the 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 alignment of the velocities - proceeds in the same way as in a completely 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 speeds of the balls are equal, the interaction forces will be the 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 equalization of the velocities. Therefore, the body that was slowing down will continue to slow down, and the one that was accelerating will accelerate until the deformation disappears. When the shape of the bodies is restored, all the potential energy again passes into the kinetic energy of the balls, i.e. in a perfectly elastic impact, 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 the 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 (in the process of 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 the moment of 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) with respect to 1 v′ and 2 v′. To do this, we rewrite it in the following form:
Divide the first equation by the second:
(11)
Solving the system from equation (11) and the second equation (10), we get:
, 
(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: device and principle of operation:
1 The "Ball Collision" installation is shown in the figure and consists of: base 1, stand 2, in the upper part of which the upper bracket 3 is installed, designed for hanging the balls; housing designed to mount a scale of 4 angular displacements; electromagnet 5, designed to fix the initial position of one of the balls 6; adjustment nodes that provide a direct central impact of the balls; threads 7 for hanging metal balls; wires to ensure the electrical contact of the balls with the terminals 8. To start the ball and count the time to impact, the control unit 9 is used. The metal balls 6 are made of aluminum, brass and steel.
Practical part
Preparing the device for work
Before starting work, it is necessary to check whether the impact of the balls is central, for this you need to deflect the first ball (of smaller mass) at a certain angle and press the key Start. The planes of the trajectories of the balls after the collision must coincide with the plane 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 the following steps must be performed:
1. Use screws 2 to achieve the 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 the screws 7 (see Fig. 1).
3. It is necessary to ensure that the planes of the trajectories of the balls after the collision coincide with the plane of the trajectory of the first ball before the collision. This is achieved with screws 8 and 10.
4. Loosen nuts 20, set the angle scales 15,16 in such a way that the angle indicators show zero on the scales at the moment when the balls are at rest. Tighten nuts 20.
Exercise 1.Determine the time of collision of the balls.
1. Insert the aluminum balls into the suspension brackets.
2. Enable installation
3. Take the first ball to the 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. Record the results in a table.
7. Make 10 measurements, enter the results in a table
9. Make a conclusion about the dependence of the impact time on the mechanical properties of the materials of the colliding bodies.
Task 2. Determine the coefficients of recovery of speed and energy for the case of 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 throw angle
3. Press the START button. This will cause the balls to hit.
4. Using the scales, visually determine the rebound angles of the balls
5. Record the results in a table.
| No. p / p | W | ||||||||
| ……… | |||||||||
| Mean |
6. Take 10 measurements and enter the results in a 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 point of suspension to the center of gravity of the balls;
Throwing angle, degrees;
Rebound angle of the right ball, degrees;
Rebound angle of the left ball, degrees.
The speed recovery factor can be determined by the formula:
The energy recovery factor can be determined by the formula:
The energy loss in a partially elastic collision can be calculated by the formula:
8. Calculate the average values of all quantities.
9. Calculate the errors using the formulas:
=
=
=
=
=
=
10. Record the results, taking into account the error in the standard form.
Task 3. Verification of the Law of Conservation of Momentum for an Inelastic Central Impact. Determination of the coefficient of recovery of kinetic energy.
To study an inelastic impact, two steel balls are taken, but on one of them, in the place where the impact occurs, a piece of plasticine is attached. The ball that is deflected towards the electromagnet is considered first.
Table #1
| experience number | |||||||||||
1. Get the initial value of the deflection angle of the first ball from the teacher and write it down in table No. 1.
2. Set the electromagnet so that the deflection angle of the first ball corresponds to the specified value
3. Deviate the first ball to the specified angle, press the key<ПУСК>and count the deflection angle of the second ball . Repeat the experiment 5 times. Record the obtained values of the deviation angle 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 Table. No. 1.
6. Using the formula, find 5 values of the momentum of the system of balls after the collision and write the result in Table. No. 1.
7. By formula 
8. By formula 
find the variance of the average value of the momentum of the system of balls after the collision. Find the standard deviation of the mean momentum of the system after the collision. Enter the resulting value in table No. 1.
9. By 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 table. No. 1.
11. According to the formula 
5 find the average value of the kinetic energy of the system after the collision.
12. By formula 
13. Using the formula, find the kinetic energy recovery factor. Based on the obtained value of the kinetic energy recovery factor, draw a conclusion about the conservation of the system's energy during a collision.
14. Write the response for the impulse of the system after the collision as
15. Find the ratio of the projection of the momentum of the system after an inelastic impact to the initial value of the projection of the momentum of the system before the impact. Based on the obtained value of the ratio of the projection of impulses before and after the collision, make 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 under Elastic Central Impact. Determination of the force of interaction of balls in a collision.
To study the elastic impact, two steel balls are taken. The ball that is deflected towards the electromagnet is considered first.
Table number 2.
| experience number | |||||||||||||
1. Get the initial value of the deflection angle of the first ball from the teacher and write it down in the table. #2
2. Set the electromagnet so that the deflection angle of the first ball corresponds to the specified value .
3. Reject the first ball to the specified angle, press the key<ПУСК>and count the deflection angles of the first ball and the second ball and the time of collision of the balls . Repeat the experiment 5 times. Record the obtained values of the deflection angles and impact time 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 system of balls after the collision and write the result in Table. No. 2.
7. By formula 
find the average momentum of the system after the collision.
8. Formula 
find the variance of the average value of the momentum of the system of balls after the collision. Find the standard deviation of the mean momentum of the system after the collision. Enter the resulting value in table No. 2.
9. By 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 Table. No. 2.
11. According to the formula 
find the average value of the kinetic energy of the system after the collision
12. By formula 
find the dispersion of the average value of the 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 factor.
14. By formula 
find the average value of the interaction force and enter the result in table No. 2.
15. Write down the response for the impulse 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 momentum of the system after the elastic impact to the initial value of the projection of the momentum before the impact. Based on the obtained value of the ratio of the projection of impulses before and after the collision, make 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 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, make a conclusion about the conservation of the mechanical energy of the system during the collision.
19. Compare the obtained value of the magnitude of the interaction force with the force of gravity of a ball of greater mass. Make a conclusion about the intensity of the forces of mutual repulsion acting during the impact.
Test questions:
1. Describe the types of impacts, indicate what laws are followed during impact?
2. Mechanical system. The law of change of 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 the impact in the following cases:
1) the first body is moving the second is at rest.
2) Both bodies are moving in the same direction.
3) Both bodies are moving in the opposite direction.
4. Determine the magnitude of the change in the momentum of a point of mass m uniformly rotating around the circle. Through one and a half, through a quarter of the period.
5. Form the law of conservation of mechanical energy, in which cases it is not fulfilled.
6. Write down the formulas for determining the coefficients of recovery of speed and energy, explain the physical meaning.
7. What determines the amount of energy loss in a partially elastic impact?
8. Body impulse and force impulse, types of mechanical energy. Mechanical work of force.
