Relativistic energy of a mechanical system. Basic law of relativistic dynamics. Relativistic energy Total energy of the relativistic elementary
Newton's second law states that the time derivative of the momentum of a particle (material point) is equal to the resulting force acting on the particle (see formula (9.1)). The equation of the second law turns out to be invariant with respect to the Lorentz transformations, if by momentum we mean the quantity (67.5). Therefore, the relativistic expression of Newton's Second Law is
It should be borne in mind that the relation is not applicable in the relativistic case, and the acceleration w and the force F, generally speaking, turn out to be noncollinear.
Note that momentum and force are not invariant quantities. The formulas for the transformation of the momentum components in the transition from one inertial frame of reference to another will be obtained in the next section. We will give the formulas for the transformation of the force components without. output:
(velocity of the particle in the system K). If in frame K the force F acting on the particle is perpendicular to the particle velocity V, the scalar product FV is equal to zero and the first of formulas (68.2) is simplified as follows:
To find a relativistic expression for energy, we will do the same as we did in § 19. We multiply equation (68.1) by the particle displacement . As a result, we get
The right side of this relation gives the work done on the particle in time . In § 19 it was shown that the work of the resultant of all forces goes to the increment of the kinetic energy of the particle (see formula). Therefore, the left side of the relation must be interpreted as an increment of the kinetic energies T of the particle over time. In this way,
We transform the resulting expression, taking into account that (see (2.54)):
Integrating the resulting relation gives
(68.4)
According to the meaning of the kinetic energy, it must vanish at From here, the value for the constant is equal to Therefore, the relativistic expression for the kinetic energy of a particle has the form
In the case of low speeds, formula (68.5) can be transformed as follows:
We have arrived at the Newtonian expression for the kinetic energy of a particle. This was to be expected, since at velocities much less than the speed of light, all formulas of relativistic mechanics must pass into the corresponding formulas of Newtonian mechanics.
Consider a free particle (that is, a particle not subject to external forces) moving with a speed v. We found out that this particle has a kinetic energy defined by formula (68.5). However, there are reasons (see below) to attribute to a free particle, in addition to the kinetic energy (68.5), an additional energy equal to
Thus, the total energy of a free particle is determined by the expression . Taking into account (68.5), we get that
When expression (68.7) goes into (68.6). That is why it is called rest energy. This energy is the internal energy of the particle, not related to the motion of the particle as a whole.
Formulas (68.6) and (68.7) are valid not only for an elementary particle, but also for a complex body consisting of many particles. The energy of such a body contains, in addition to the rest energies of the particles included in its composition, also the kinetic energy of the particles (due to their movement relative to the center of mass of the body) and the energy of their interaction with each other. The rest energy, as well as the total energy (68.7), does not include potential energy body in an external force field.
Eliminating the velocity v from equations (67.5) and (68.7) (equation (67.5) must be taken in scalar form), we obtain an expression for the total energy of the particle in terms of momentum p:
If this formula can be represented as
The resulting expression differs from the Newtonian expression for the kinetic energy by the term
Note that a comparison of expressions (67.5): and (68.7) implies the formula
Let us explain why energy (68.7) and not only kinetic energy (68.5) should be attributed to a free particle. Energy in its sense should be a conserved quantity. An appropriate consideration shows that the sum (over particles) of expressions of the form (68.7) is preserved in collisions of particles, while the sum of expressions (68.5) turns out to be non-conserved. It is impossible to satisfy the requirement of conservation of energy in all inertial frames of reference if the rest energy (68.6) is not taken into account as part of the total energy.
Relativistic momentum: .
Kinetic energy of a relativistic particle: 
.
Relativistic relationship between total energy and momentum: .
Velocity addition theorem in relativistic mechanics: ,
where u and are the velocities in two inertial frames of reference moving relative to each other with the speed , coinciding in the direction with u(sign "-") or opposite to it directed (sign "+").
MOLECULAR PHYSICS AND THERMODYNAMICS
Amount of substance:,
where N is the number of molecules, N A is the Avogadro constant, m is the mass of the substance, m- molar mass.
Klaiperon-Mendeleev equation: ,
where P- gas pressure, V- its volume, R is the paint gas constant, T is the absolute temperature.
Equation of the molecular kinetic theory of gas: 
,
where n is the concentration of molecules, is the average kinetic energy of the translational motion of a molecule, m0 is the mass of the molecule, is the mean square velocity.
Average energy of a molecule: ,
where i is the number of degrees of freedom, k is the Boltzmann constant.
Internal energy of an ideal gas: .
Molecule speeds:
root mean square: 
,
arithmetic mean: 
,
most likely: 
.
Mean free path of a molecule: ,
where d is the effective molecular diameter.
Average number of collisions of a molecule per unit time:
The distribution of molecules in the potential field of forces: ,
where P is the potential energy of the molecule.
Barometric formula: .
Diffusion equation: ,
where D is the diffusion coefficient, r- density, dS is an elementary area perpendicular to the direction along which diffusion occurs.
Heat equation: , æ ,
where æ is thermal conductivity.
Force of internal friction: ,
where h– dynamic viscosity.
Diffusion coefficient: .
Viscosity (dynamic): 
.
Thermal conductivity: æ ,
where C V is the specific isochoric heat capacity.
Molar heat capacity of an ideal gas:
isochoric: ,
isobaric: 
.
First law of thermodynamics:
The work of gas expansion in the process:
isobaric : ,
isothermal: 
,
isochoric:
adiabatic: ,
Poisson's equations:
Efficiency of the Carnot cycle: 
,
where Q and T- the amount of heat received from the heater and its temperature; Q0 and T 0- the amount of heat transferred to the refrigerator, and its temperature.
Entropy change during the transition from state 1 to state 2: .
EXAMPLES OF SOLVING PROBLEMS
1. The movement of a body with a mass of 1 kg is given by the equation s = 6t3 + 3t + 2. Find the dependence of speed and acceleration on time. Calculate the force acting on the body at the end of the second second.
Solution. We find the instantaneous speed as the derivative of the path with respect to time: , . Instantaneous acceleration is determined by the first derivative of the speed with respect to time or the second derivative of the path with respect to time: , . The force acting on the body is determined by Newton's second law: , where , according to the condition of the problem, is the acceleration at the end of the second second. Then, N.
Answer: , , N.
2. A rod 1 m long moves past the observer at a speed 20% less than the speed of light. What will its length seem to the observer?
Solution. The dependence of body length on velocity in relativistic mechanics is expressed by the formula: , where l 0 is the length of the resting rod; - the speed of its movement; With is the speed of light in vacuum. Substituting into the formula for l 0 numerical values, we have: l= 0.6 m.
Answer: l= 0.6 m.
3. Two particles move towards each other with speeds: 1) = 0.5 With and u = 0,75With; 2) = With and u = 0,75With. Find their relative speed in the first and second cases.
Solution. According to the theorem on the addition of velocities of bodies moving towards each other, in the theory of relativity: , where , u are the speeds of the first and second bodies, respectively; is their relative speed; With is the speed of light in vacuum. For the first and second cases we find:
This confirms that, firstly, in no inertial reference frame, the speed of the process can exceed the speed of light, and, secondly, the speed of light propagation in vacuum is absolute.
Answer: = 0.91 With; = With.
4. On two cords of the same length, equal to 0.8 m, two lead balls with masses of 0.5 and 1 kg are suspended. The balls are in contact with each other. A ball of smaller mass was taken aside so that the cord deviated through an angle a=60°, and released. How high will both balls rise after the collision? The impact is assumed to be central and inelastic. Determine the energy expended on the deformation of the balls upon impact.
Solution. Since the impact of the balls is inelastic, after the impact the balls will move with a common speed u. The law of conservation of momentum during this impact has the form:
Here and are the speeds of the balls before the impact. The speed of the big ball before the impact is zero (= 0). We find the speed of the smaller ball using the law of conservation of energy. When the smaller ball deviates by an angle, it is given potential energy, which then turns into kinetic energy: . Consequently: . From geometric constructions it follows: , therefore:
. (2)
From equations (1) and (2) we find the speed of the balls after the impact:
. (3)
The kinetic energy possessed by the balls after the impact is converted into potential:
where h is the height of the balls after the collision. From formula (4) we find , or taking into account (3) 
and substituting the numerical data we get h= 0.044 m. During an inelastic impact of balls, part of the energy is spent on their deformation. The strain energy is determined by the difference between the kinetic energies before and after the impact:
. Using equations (2) and (3), we obtain: , J.
Answer: h= 0.044 m, DE D= 1.3 J.
5. A hammer weighing 70 kg falls from a height of 5 m and hits an iron product lying on an anvil. The mass of the anvil together with the product is 1330 kg. Assuming that the impact is absolutely inelastic, determine the energy spent on the deformation of the product. The hammer–workpiece–anvil system is considered closed.
Solution. By the condition of the problem, the hammer–workpiece–anvil system is considered closed, and the impact is inelastic. Based on the law of conservation of energy, we can assume that the energy spent on the deformation of the product is equal to the difference between the values of the mechanical energy of the system before and after the impact. We assume that only the kinetic energy of the bodies changes during the impact, i.e., we neglect the insignificant movement of the bodies along the vertical during the impact. Then for the deformation energy of the product we have:
, (1)
where is the speed of the hammer at the end of the fall from a height h; is the total velocity of all bodies in the system after an inelastic impact. Hammer speed at the end of the fall h is determined without taking into account air resistance and friction by the formula:
We will find the total speed of all bodies of the system after an inelastic impact by applying the law of conservation of momentum: . For the system under consideration, the law of conservation of momentum has the form 
, where:
Substituting expressions (2) and (3) into formula (1), we obtain: 
, J.
Answer: J.
6. A body of mass 1 kg under the action of a constant force moves in a straight line. The dependence of the path traveled by the body on time is given by the equation s = 2t2+4t+1. Determine the work of the force in 10 seconds from the beginning of its action and the dependence of kinetic energy on time.
Solution. The work done by the force is expressed through a curvilinear integral:
The force acting on the body, from Newton's II law is equal to: or (the instantaneous value of acceleration is determined by the first derivative of the speed with respect to time or the second derivative of the path with respect to time). Accordingly, we find:
From expression (2) we determine ds:
Substituting (4) and (5) into equation (1), we get: 
Using this formula, we determine the work done by the force in 10 seconds from the beginning of its action: 
, BUT= 960 J. Kinetic energy is determined by the formula:
Substituting (2) into (6), we have: 
.
Answer: BUT= 960 J, T \u003d m (8t 2 + 16t + 8).
7. Proton moves at a speed of 0.7 With (With is the speed of light). Find the momentum and kinetic energy of the proton.
Solution. The momentum of a proton is determined by the formula:
Since the speed of a proton is comparable to the speed of light, it is necessary to take into account the dependence of mass on speed, using the relativistic expression for mass:
where m is the mass of the moving proton; m0\u003d 1.67 × 10 -27 kg is the rest mass of the proton; v is the speed of the proton; c= 3×10 8 m/s is the speed of light in vacuum; v/c = b is the speed of the proton, expressed in fractions of the speed of light. Substituting equation (2) into (1) we obtain: , kg×m/s. In relativistic mechanics, the kinetic energy of a particle is defined as the difference between the total energy E and rest energy E 0 this particle:
. (3)
Answer: p\u003d 4.91 × 10 -19 kg × m / s, T\u003d 0.6 × 10 -10 J.
8. A thin rod rotates with an angular velocity of 10 s -1 in a horizontal plane around a vertical axis passing through the middle of the rod. In the process of rotation in the same plane, the rod moves so that the axis of rotation passes through its end. Find the angular velocity after the displacement.
Solution. We use the law of conservation of angular momentum: , where J i, is the moment of inertia of the rod relative to the axis of rotation. For an isolated system of bodies, the vector sum of angular momenta remains constant. In this problem, due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod will also change. In accordance with the law of conservation of angular momentum, we write:
It is known that the moment of inertia of the rod about the axis passing through the center of mass and perpendicular to the rod is equal to:
According to the Steiner theorem: where J is the moment of inertia of the body about an arbitrary axis of rotation; J0 is the moment of inertia about a parallel axis passing through the center of mass; d is the distance from the center of mass to the selected axis of rotation. Find the moment of inertia about the axis passing through its end and perpendicular to the rod:
. (3)
Substituting formulas (2) and (3) into (1), we have: , whence .
Answer: w 2= 2.5 s -1 .
9. A flywheel with a mass of 4 kg rotates at a frequency of 720 min -1 around a horizontal axis passing through its center. The mass of the flywheel can be considered uniformly distributed along its rim with a radius of 40 cm. After 30 s, the flywheel stopped under the action of a braking moment. Find the braking torque and the number of revolutions that the flywheel will make before it comes to a complete stop.
Solution. To determine the braking torque M forces acting on the body, you need to apply the basic equation of the dynamics of rotational motion:
where J- the moment of inertia of the flywheel about the axis passing through the center of mass; is the change in the angular velocity over a period of time . By condition, , where is the initial angular velocity, since the final angular velocity = 0. Let us express the initial angular velocity in terms of the flywheel rotation frequency; then and moment of inertia of the flywheel , where m- mass of the flywheel; R is its radius. Formula (1) takes the form: 
where M= -1.61 N×m. The “-” sign indicates that the moment is languid.
The angle of rotation (i.e. angular path) during the time the flywheel rotates to stop can be determined by the formula for uniformly retarded rotation:
where is the angular acceleration. By condition, , , . Then expression (2) can be written as follows: 
. Because j = 2pN, w 0 = 2pn, then the number of complete revolutions of the flywheel: .
Answer: M= 1.61 N×m, N = 180.
10. In a vessel with a volume of 2 m 3 there is a mixture of 4 kg of helium and 2 kg of hydrogen at a temperature of 27 ° C. Determine the pressure and molar mass of the mixture of gases.
Solution. Let's use the Klaiperon-Mendeleev equation, applying it to helium and hydrogen:
where P1 is the partial pressure of helium; m 1 is the mass of helium; is its molar mass; V is the volume of the vessel; T is the gas temperature; R= 8.31 J/(mol×K) – molar gas constant; P2- partial pressure of hydrogen; m2 is the mass of hydrogen; is its molar mass. under partial pressure P1 and P2 is understood to be the pressure that a gas would produce if it were alone in a vessel. According to Dalton's law, the pressure of the mixture is equal to the sum of the partial pressures of the gases that make up the mixture:
From equations (1) and (2) we express P1 and P2 and substitute into equation (3). We have:
. (4)
We find the molar mass of a mixture of gases by the formula: , where v1 and v2 are the number of moles of helium and hydrogen, respectively. The number of moles of gases is determined by the formulas: and . Then: . Substituting numerical values we get: P= 2493 kPa and = 3×10 -3 kg/mol.
Answer: P\u003d 2493 KPa, \u003d 3 × 10 -3 kg / mol.
11. What is the average kinetic energy of the translational and rotational motion of molecules contained in 2 kg of hydrogen at a temperature of 400 K?
Solution. We consider hydrogen to be an ideal gas. The hydrogen molecule is diatomic, the bond between the atoms is considered rigid. Then the number of degrees of freedom of the hydrogen molecule is 5, three of which are translational and two rotational. On average, there is energy per degree of freedom, where k is the Boltzmann constant; T is the thermodynamic temperature. For one molecule: and . The number of molecules contained in the mass of gas: . Then the average kinetic energy of the translational motion of the molecules of two kilograms of hydrogen is: 
. The average kinetic energy of the rotational motion of the same molecules: . Substituting numerical values we have: =4986 KJ and =2324 KJ.
Answer: \u003d 4986 KJ, \u003d 2324 KJ.
12. Determine average length free path of molecules and the number of collisions in 1 s occurring between all oxygen molecules in a vessel with a capacity of 2 liters at a temperature of 27 0 C and a pressure of 100 kPa.
Solution. The mean free path of oxygen molecules is calculated by the formula: , where d is the effective diameter of an oxygen molecule; n is the number of molecules per unit volume, which can be determined from the equation: , where k is the Boltzmann constant. Thus, we have: . Number of collisions Z, occurring between all molecules in 1 s, is equal to: , where N- the number of oxygen molecules in a vessel with a volume of 2×10 -3 m3; 
. Substituting numerical values, we get: Z
Answer: Z\u003d 9 × 10 28 s -1, \u003d 3.56 × 10 8 m.
13. Determine the coefficients of diffusion and internal friction of nitrogen at a temperature T\u003d 300 K and a pressure of 10 5 Pa.
Solution. The diffusion coefficient is determined by the formula: , where<V> is the arithmetic mean velocity of molecules, is the mean free path of molecules. To find, we use the formula from the solution of example 12: 
. The expression for the diffusion coefficient takes the form: 
. Coefficient of internal friction: , where r is the density of the gas at a temperature of 300 K and a pressure of 10 5 Pa. For finding r Let us use the equation of state for an ideal gas. We write it for two states of nitrogen: under normal conditions T 0\u003d 273 K, P\u003d 1.01 × 10 5 Pa and in the conditions of the task: and . Considering that and , we have: . The coefficient of internal friction of the gas can be expressed in terms of the diffusion coefficient: . Substituting numerical values, we get: D\u003d 4.7 × 10 5 m 2 / s and h= 5.23×10 -5 kg/(m×s).
Answer: D\u003d 4.7 × 10 5 m 2 / s and h= 5.23×10 -5 kg/(m×s).
14. Oxygen weighing 160 g is heated at a constant pressure from 320 to 340 K. Determine the amount of heat absorbed by the gas, the change in internal energy and the work of expansion of the gas.
Solution. The amount of heat required to heat a gas at constant pressure: 
. Here with p and C p are the specific and molar heat capacities of the gas at constant pressure; m\u003d 32 × 10 -3 kg / mol - the molar mass of oxygen. For all diatomic gases: , J/(mol×K). The change in the internal energy of the gas is found by the formula: , where C V is the molar heat capacity of the gas at constant volume. For all diatomic gases: With V = = 5/ 2×R; C V= 20.8 J/(mol×K). The work of gas expansion in an isobaric process: , where is the change in gas volume, which can be found from the Claiperon–Mendeleev equation. In the isobaric process: and . By term subtraction of expressions we find: , therefore: . Substituting numerical values, we get: J, J, J.
Answer: J, J, J.
15. The volume of argon at a pressure of 80 kPa increased from 1 to 2 liters. How much will the internal energy of the gas change if the expansion was carried out: a) isobarically; b) adiabatically.
Solution. Let's apply the first law of thermodynamics. According to this law, the amount of heat Q transferred to the system is spent on increasing internal energy and on external mechanical work BUT: . The size of the system can be determined by knowing the mass of the gas, the specific heat capacity at constant volume with V and temperature change : . However, it is more convenient to determine the change in internal energy through the molar heat capacity C V, which can be expressed in terms of the number of degrees of freedom: . Substituting the value C V we get: . The change in internal energy depends on the nature of the process in which the gas expands. With the isobaric expansion of a gas, according to the first law of thermodynamics, part of the amount of heat goes to change the internal energy. It is impossible to find for argon using the obtained formula, since the gas mass and temperature are not given in the condition of the problem. Therefore, it is necessary to transform this formula. Let's write the Klaiperon-Mendeleev equation for the initial and final states of the gas: and , or . Then: . This equation is a calculation equation to be determined under isobaric expansion. During the adiabatic expansion of the gas, heat exchange with the environment does not occur, therefore Q= 0. The first law of thermodynamics can be written as: . This relationship establishes that the work of expanding the gas can only be done by reducing the internal energy of the gas (minus sign before ): . The work formula for an adiabatic process is: 
, where g– adiabatic exponent equal to: . For argon, a monatomic gas ( i= 3) – we have g=1.67. We find the change in internal energy during the adiabatic process for argon: 
. To determine the expansion work of argon, the formula for should be transformed, taking into account the parameters given in the problem statement. Applying the Klaiperon-Mendeleev equation for this case, we obtain an expression for calculating the change in internal energy: 
. Substituting numerical values, we have: a) with isobaric expansion J; b) with adiabatic expansion J.
Answer: a) \u003d 121 J; b) = -44.6 J.
16. The temperature of the heater of the heat engine is 500 K. The temperature of the refrigerator is 400 K. Determine the efficiency. a heat engine operating according to the Carnot cycle, and the full power of the machine if the heater transfers 1675 J of heat to it every second.
Solution. The efficiency of the machine is determined by the formula: or . From these expressions we find: 
. Let's do the calculations: A\u003d 335 J. This work is done in 1 s, therefore, the total power of the machine is 335 watts.
Answer: = 0.2, N\u003d 335 W.
17. Hot water of a certain mass gives off heat to cold water of the same mass and their temperatures become the same. Show that the entropy increases.
Solution. Let the hot water temperature T 1, cold T 2, and the temperature of the mixture. Let us determine the temperature of the mixture based on the heat balance equation: or 
, where: . The change in entropy that occurs when hot water is cooled: 
. The change in entropy that occurs when cold water is heated: 
. The change in the entropy of the system is: 
or 
; as well as 4T1T2>0, then .
CONTROL WORK No. 1
101. Under the influence of what force in a rectilinear motion of a body, a change in its coordinates over time occurs according to the law x \u003d 10 + 5t - - 10t 2? Body weight 2 kg.
102. Find the law of motion of a body with a mass of 1 kg under the action of a constant force of 10 N, if at the moment t = 0 the body was at rest at the origin ( x = 0).
103. Find the law of motion of a body with a mass of 1 kg under the action of a constant force of 1 N, if at the moment t = 0 start coordinate x = 0 and v 0 = 5m/s.
104. Find the law of motion of a body with a mass of 1 kg under the action of a constant force of 2 N, if at the moment t = 0 we have x 0 = 1 m and v 0 =2 m/s.
105. A body of mass 2 kg moves with an acceleration that changes according to the law a = 5t-10. Determine the force acting on the body 5 s after the start of the action, and the speed at the end of the fifth second.
106. A solid ball with a mass of 1 kg and a radius of 5 cm rotates around an axis passing through its center. The law of rotation of the ball is expressed by the equation. At the point farthest from the axis of rotation, the force acting on the ball is tangent to the surface. Determine this force and the braking moment.
107. A car moves along a curvature of a highway with a radius of curvature of 100 m. The law of car movement is expressed by the equation. Find the speed of the car, its tangential, normal and total acceleration at the end of the fifth second.
108. A material point moves along a circle with a radius of 20 m. The dependence of the path traveled by the point on time is expressed by the equation. Determine the distance traveled, the angular velocity and angular acceleration of the point after 3 s from the beginning of its movement.
109. A material point moves along a circle with a radius of 1 m according to the equation. Find the speed, tangential, normal and total acceleration at time 3 s.
110. The body rotates uniformly accelerated with an initial angular velocity of 5 s -1 and an angular acceleration of 1 rad / s 2. How many revolutions does the body make in 10 s?
111. A 2x2x4 cm 3 parallelepiped moves parallel to the larger edge. At what speed will it appear as a cube.
112. What speed must a moving body have in order for its longitudinal dimensions to be halved?
113. π-meson is an unstable particle. Its own lifetime is 2.6×10 -8 s. How far will the π-meson fly before decay, if it moves at a speed of 0.9 With?
114. Find the proper lifetime of an unstable particle - a meson moving at a speed of 0.99 With, if the distance it flew before disintegration is 0.1 km.
115. Own lifetime of the π-meson 2.6×10 -8 s. What is the lifetime of a π-meson for an observer, relative to which this particle moves at a speed of 0.8 With?
116. An electron whose speed is 0.9 With, moves towards the proton, which has a speed of 0.8 With
117. A radioactive nucleus that flew out of the accelerator at a speed of 0.8 With, ejected in the direction of its motion -particle with a speed of 0.7 With regarding the accelerator. Find the speed of the particle relative to the nucleus.
118. Two particles move towards each other at a speed of 0.8 With. Determine the speed of their relative movement.
119. At what speed of motion will the relativistic contraction of the length of a moving body be 25%.
120. What speed should a moving body have in order for its longitudinal dimensions to decrease by 75%.
121. A solid cylinder weighing 0.1 kg rolls without slipping at a constant speed of 4 m/s. Determine the kinetic energy of the cylinder, the time until it stops, if a friction force of 0.1 N acts on it.
122. A solid ball rolls down an inclined plane, the length of which is 1 m and the angle of inclination is 30°. Determine the speed of the ball at the end of the inclined plane. The friction of the ball on the plane is ignored.
123. A hollow cylinder of mass 1 kg rolls on a horizontal surface at a speed of 10 m/s. Determine the force that must be applied to the cylinder to stop it on a path of 2 m.
124. The flywheel, having the shape of a disk with a mass of 10 kg and a radius of 0.1 m, was spun up to a frequency of 120 min -1. Under the action of the friction force, the disk stopped after 10 With. Find the moment of the forces of friction, assuming it to be constant.
125. A hoop and a disk roll down an inclined plane making an angle of 30° with the horizon. What are their accelerations at the end of the descent? Ignore the force of friction.
126. A ball at rest with a mass of 2 kg collides with the same ball moving at a speed of 1 m / s. Calculate the work done due to deformation during a direct central inelastic impact.
127. The mass of the projectile is 10 kg, the mass of the gun barrel is 500 kg. When fired, the projectile receives a kinetic energy of 1.5 × 10 6 J. What kinetic energy does the gun barrel receive due to recoil?
128. A skater with a mass of 60 kg, standing on skates on ice, throws a stone with a mass of 2 kg in a horizontal direction at a speed of 10 m / s. At what distance will the skater roll back if the friction coefficient of the skates on the ice is 0.02.
129. A hydrogen molecule moving at a speed of 400 m / s flies up to the vessel wall at an angle of 60 ° and elastically hits it. Determine the momentum received by the wall. Take the mass of molecules equal to 3×10 -27 kg.
130. A steel ball weighing 50 g fell from a height of 1 m onto a large plate, transferring to it an impulse of force equal to 0.27 N × s. Determine the amount of heat released upon impact and the height to which the ball rises.
131. With what speed does an electron move if its kinetic energy is 1.02 MeV? Determine the momentum of an electron.
132. The kinetic energy of a particle turned out to be equal to its rest energy. What is the speed of this particle?
133. The mass of a moving proton is 2.5 × 10 -27 kg. Find the speed and kinetic energy of the proton.
134. A proton passed through an accelerating potential difference of 200 MV. How many times greater is its relativistic mass than its rest mass? What is the speed of a proton?
135. Determine the speed of an electron if its relativistic mass is three times the rest mass. Calculate the kinetic and total energy of an electron.
136. Calculate the speed, kinetic and total energy of the proton at the moment when its mass is equal to the rest mass of the -particle.
137. Find the momentum, total and kinetic energy of an electron moving at a speed equal to 0.7 With.
138. A proton and a -particle pass through the same accelerating potential difference, after which the mass of the proton is half the rest mass of the -particle. Determine the potential difference.
139. Find the momentum, total and kinetic energy of a neutron moving at a speed of 0.6 With.
140. How many times the mass of a moving deuteron is greater than the mass of a moving electron if their speeds are respectively equal to 0.6 With and 0.9 With. What are their kinetic energies.
141. Find the average kinetic energy of the rotational motion of all molecules contained in 0.20 g of hydrogen at a temperature of 27 °C.
142. The pressure of an ideal gas is 10 MPa, the concentration of molecules is 8 × 10 10
cm -3 . Determine the average kinetic energy of the translational motion of one molecule and the temperature of the gas.
143. Determine the average value of the total kinetic energy of one molecule of argon and water vapor at a temperature of 500 K.
144. The average kinetic energy of the translational motion of gas molecules is 15 × 10 -21 J. The concentration of molecules is 9 × 10 19 cm -3. Determine the gas pressure.
145. In a cylinder with a capacity of 50 liters is compressed hydrogen at 27 ° C. After part of the air was released, the pressure dropped by 10 5 Pa. Determine the mass of released hydrogen. The process is considered isothermal.
146. In a spherical vessel with a radius of 0.1 m, there is 56 g of nitrogen. To what temperature can the gas be heated if the walls of the vessel can withstand a pressure of 5 10 5 Pa?
147. At a temperature of 300 K and a pressure of 1.2 × 10 5 Pa, the density of a mixture of hydrogen and nitrogen is 1 kg / m 3. Determine the molar mass of the mixture.
148. In a cylinder with a capacity of 0.8 m 3 there are 2 kg of hydrogen and 2.9 kg of nitrogen. Determine the pressure of the mixture if the ambient temperature is 27 °C.
149. To what temperature can a sealed vessel containing 36 g of water be heated so that it does not burst, if it is known that the walls of the vessel can withstand a pressure of 5 × 10 6 Pa. The volume of the vessel is 0.5 l.
150. At a temperature of 27 ° C and a pressure of 10 6 Pa, the density of a mixture of oxygen and nitrogen is 15 g / dm 3. Determine the molar mass of the mixture.
151. A vessel with a capacity of 1 liter contains oxygen weighing 32 g. Determine the average number of collisions of molecules per second at a temperature of 100 K.
152. Determine the average length and average free path of carbon dioxide molecules at a temperature of 400 K and a pressure of 1.38 Pa.
153. In a vessel with a capacity of 1 liter is 4.4 g of carbon dioxide. Determine the mean free path of the molecules.
154. Determine the diffusion coefficient of helium at a pressure of 1 10 6 Pa and a temperature of 27 ° C.
155. Determine the coefficient of internal friction of oxygen at a temperature of 400 K.
156. A vessel with a capacity of 5 liters contains 40 g of argon. Determine the average number of molecular collisions per second at a temperature of 400 K.
157. Determine the coefficient of internal friction of air at a temperature of 100 K.
158. Determine the diffusion coefficient of nitrogen at a pressure of 0.5 × 10 5 Pa and a temperature of 127 ° C.
159. The coefficient of internal friction of oxygen under normal conditions is 1.9 × 10 -4 kg / m × s. Determine the thermal conductivity of oxygen.
160. Diffusion coefficient of hydrogen under normal conditions
9.1×10 -5 m 2 /s. Determine the thermal conductivity of hydrogen.
161. Determine how much heat must be imparted to argon with a mass of 400 g in order to heat it by 100 K: a) at a constant volume; b) at constant pressure.
162. How many times will the volume of 2 moles of oxygen increase during isothermal expansion at a temperature of 300 K, if 4 kJ of heat was imparted to the gas.
163. How much heat must be imparted to 2 moles of air in order for it to do work of 1000 J: a) in an isothermal process; b) in an isobaric process.
164. Find the work and change in internal energy during the adiabatic expansion of 28 g of nitrogen, if its volume has doubled. The initial nitrogen temperature is 27 °C.
165. Oxygen, occupying a volume of 10 liters and under a pressure of 2 10 5 Pa, is adiabatically compressed to a volume of 2 liters. Find the work of compression and the change in the internal energy of oxygen.
166. Determine the amount of heat reported by 88 g of carbon dioxide if it was heated isobarically from 300 K to 350 K. What work can the gas do in this case and how will its internal energy change?
167. In which process is it more profitable to expand the air: isobaric or isothermal, if the volume increases five times. The initial gas temperature is the same in both cases.
168. In what process is it more profitable to heat 2 moles of argon per 100 K: a) isobaric; b) isochoric.
169. Nitrogen weighing 20 g during isobaric heating reported 3116 J of heat. How did the temperature and internal energy of the gas change.
170. During the isothermal expansion of one mole of hydrogen, heat of 4 kJ was expended, while the volume of hydrogen increased five times. At what temperature does the process take place? What is the change in the internal energy of the gas, what is the work done by the gas?
171. Determine the change in the entropy of 14 g of nitrogen when it is heated isobarically from 27 ° C to 127 ° C.
172. How will the entropy of 2 moles of carbon dioxide change during isothermal expansion if the volume of the gas increases four times.
173. Performing the Carnot cycle, the gas gave the refrigerator 25% of the heat received from the heater. Determine the temperature of the refrigerator if the temperature of the heater is 400 K.
174. A heat engine operates according to the Carnot cycle, efficiency which is 0.4. What will be the efficiency this machine if it goes through the same cycle in the opposite direction?
175. The refrigerating machine operates on the reverse Carnot cycle, efficiency. which 40%. What will be the efficiency this machine, if it works according to the direct Carnot cycle.
176. In a direct Carnot cycle, a heat engine does work of 1000 J. The temperature of the heater is 500 K, the temperature of the refrigerator is 300 K. Determine the amount of heat received by the machine from the heater.
177. Find the change in entropy when 2 kg of water is heated from 0 to 100 ° C and then turns into steam at the same temperature.
178. Find the change in entropy during the melting of 2 kg of lead and its further cooling from 327 to 0 ° C.
179. Determine the change in entropy that occurs when mixing 2 kg of water at a temperature of 300 K and 4 kg of water at a temperature of 370 K.
180. Ice weighing 1 kg, located at a temperature of 0 ° C, is heated to a temperature of 57 ° C. Determine the change in entropy.
> Relativistic kinetic energy
Learn the formula for kinetic energy of a relativistic particle. Learn how to determine relativistic kinetic energy, momentum relation, total energy.
In the form of a formula, the relativistic kinetic energy is given as: (m is the rest mass, v is the speed, c is the speed of light).
Learning task
- Compare the classical and kinetic relativistic energies for objects whose speed is less than or close to the speed of light.
Key Points
- The formula shows that the energy of an object approaches infinity if the speed approaches the speed of light. Therefore, you can not accelerate the object on the border.
- Calculations of kinetic energy are carried out according to the formula: E rest \u003d E 0 \u003d mc 2.
- At a low velocity index, the relativistic kinetic energy can be approximated by the classical one. Therefore, the total energy is divided by the energy of the mass at rest, with the addition of the traditional kinetic energy.
Terms
- The Lorentz coefficient is a factor for determining the degree of temporal slowdown, length contraction and relativistic mass of a moving object.
- Classical mechanics - all the physical laws of nature that characterize the behavior of the ordinary world.
- Special Relativity: The speed of light remains stable in any frame of reference.
Kinetic energy is based on body mass and speed. Given by the formula: 
(m is the mass, v is the speed of the body).
Classical kinetic energy is related to momentum by the equation:
(p is the momentum).
If the speed of an object is a remarkable fraction of the speed of light, then special relativity must be used to determine the kinetic energy. Here it is necessary to change the expression for the linear momentum. Formula:
p = mγv, where γ is the Lorentz coefficient:
Kinetic energy has a relationship with momentum, so the relativistic expression differs from the classical one:
It can be seen from the formula that the energy of an object approaches infinity when the speed approaches the speed of light. Therefore, it is impossible to accelerate an object on this line.
A mathematical by-product is the mass-energy equivalence equation. A body in a resting position must have energy:
The popular connection between Einstein,E = mc 2 and the atomic bomb displayed on the cover of a magazine
E rest \u003d E 0 \u003d mc 2.
The general formula for the energy of an object that is not at rest is:
KE = mc 2 - m 0 c 2 (m is the relativistic mass of the object, and m 0 is the mass of the object at rest).
At low speeds, the relativistic kinetic energy can be approximated by the classical one. This is shown in the Taylor expansion:
E to ≈ mc 2 (1 + 0.5 v 2 / s 2) - mc 2 = 0.5 mv 2.
It turns out that the total energy can be divided by the rest mass energy with the addition of classical kinetic energy at low speed indicators.
It can only partially satisfy researchers in the implementation of mathematical calculations and the compilation of certain mathematical models. Newtonian laws are only valid for Galilean transformations, but for all other cases new transformations are required, which are reflected in the presented Lorentz transformations. He introduced such principles and concepts in order to make accurate calculations for interacting objects that carry out similar processes at ultra-high speeds close to the speed of light.
Figure 1. Momentum and energy in relativistic mechanics. Author24 - online exchange of student papers
The very theory of relativity, which was formulated by Albert Einstein, requires a serious revision of the dogmas of classical mechanics. Lorentz introduced additional equations of dynamics, the purpose of which were the same transformations of classical ideas about ongoing physical processes. It was necessary to change the formulas in such a way that they remained true when moving from one inertial frame of reference to another.
Relativistic momentum
Figure 2. Relativistic momentum. Author24 - online exchange of student papers
In order to introduce the concept of energy in relativistic mechanics, it is necessary to consider:
- relativistic momentum;
- matching principle.
When obtaining a relativistic expression for momentum, it is necessary to apply the correspondence principle. In relativistic mechanics, the momentum of a particle can be determined by the speed of that particle. However, the dependence of momentum on velocity seems to be a more complex mechanism than similar processes in classical mechanics. This can no longer be reduced to a simple proportionality, and the effectiveness of the calculations is made up of additional parameters and quantities. The momentum is represented as a vector, where its direction must completely coincide with the direction of the velocity of a certain particle. This is provided for in the symmetry version, since the equivalence comes into effect when the free space is isotropic.
Remark 1
In this case, the momentum of the free particle is directed to the only preferred direction of its velocity. If the particle speed is zero, then the momentum of the particle is also zero.
The speed of a particle in any frame of reference has a finite value. It should always be less than the speed of light, which is displayed as the letter C, but this fact is not able to impose some restrictions on the entire magnitude of the momentum of this particle and the momentum can increase indefinitely.
Relativistic energy
Comparing various calculation methods and techniques, one can find the relativistic energy of particles. It is known that a very important property of energy is its ability to transform from one form to another and vice versa. This occurs in equivalent quantities and under different external conditions. In these metamorphoses is one of the basic laws of conservation and transformation of energy. With such phenomena, researchers have established an increase in the relativistic mass. Similar processes occur with any increase in the energy of bodies, and this does not depend on a certain type of energy, including kinetic energy. It is established that the total energy of a body is proportional to its relativistic mass. This happens regardless of what specific types of energy it consists of.
Visually, such processes can be represented in the form of simple examples:
- a heated body will have a greater rest mass than a cold object;
- a mechanically deformed part also has a greater mass than an unmachined part.
Einstein grasped this relationship between the mass and energy of a body. Accordingly, during an inelastic collision of various particles, certain processes occur for the conversion of kinetic energy into internal energy. It is also called the energy of the thermal motion of particles. With this type of interaction, it is clear that the rest mass of the body will become greater than the total rest mass of the bodies at the beginning of the experiment. The internal energy of a certain body may be accompanied by an increase in mass proportionally. The same process is natural for increasing the value of kinetic energy. According to classical mechanics, such collisions did not imply the formation of internal energy, since they were not included in the concept of mechanical energy.
Proportionality of mass and energy
For the logical operation of the law of relativistic energy, it is necessary to introduce the concept of the laws of conservation of momentum and its relationship with the principle of relativity. This requires that the law of conservation of energy be satisfied in various inertial frames of reference.
Conservation of momentum is closely related to the proportionality of the energy and mass of the body in all its forms and manifestations. Conservation of momentum is not possible in a closed frame of reference, when there is a transition of energy from the usual form to another. In this case, the mass of the body begins to change, and the law ceases to operate correctly. The law of proportionality of mass and energy is expressed as the most approximate conclusion of the whole theory of relativity.
The inert properties of the body in quantitative terms characterize the mechanics of body mass. Such an inertial mass can represent a measure of the inertia of the entire body. The antipode of the inertial mass is the gravitational mass. It is characterized by the ability of a body to create a certain gravitational field around itself and thus act on other bodies.
At present, the equality of gravitational and inertial mass is confirmed large quantity experimental research. In the theory of relativity, the question also arises of where the concepts of energy and mass of a body appear. This is due to the manifestation of various properties of matter. If they are considered in detail in the indicated plane, then the mass and energy in matter will differ significantly. However, these properties of matter, no doubt, are strongly interconnected. In this context, it is customary to talk about the equivalence of mass and energy, since they are proportional to each other.
The theory of relativity requires a revision and refinement of the laws of mechanics. As we have seen, the equations of classical dynamics (Newton's second law) satisfy the principle of relativity with respect to Galilean transformations. But the transformations of Galileo must be replaced by the transformations of Lorentz! Therefore, the equations of dynamics should be changed so that they remain unchanged during the transition from one inertial frame of reference to another according to the Lorentz transformations. At low velocities, the equations of relativistic dynamics must pass into classical ones, because in this region their validity is confirmed by experience.
Momentum and energy. In the theory of relativity, as in classical mechanics, momentum and energy E are conserved for a closed physical system, but the relativistic expressions for them differ from the corresponding classical ones:
where is the mass of the particle. This is the mass in the reference frame where the particle is at rest. It is often called the rest mass of the particle. It coincides with the particle mass in nonrelativistic mechanics.
It can be shown that the dependence of the momentum and energy of a particle on its velocity in the theory of relativity expressed by formulas (1) inevitably follows from the relativistic effect of time dilation in a moving frame of reference. This will be done below.
Relativistic energy and momentum (1) satisfy equations similar to the corresponding equations of classical mechanics:
relativistic mass. Sometimes the coefficient of proportionality in (1) between the speed of a particle and its momentum
is called the relativistic mass of the particle. With its help, expressions (1) for the momentum and energy of the particle can be written in a compact form
If a relativistic particle, i.e., a particle moving at a speed close to the speed of light, is given additional energy in order to increase its momentum, then its speed will increase very slightly. We can say that the energy of the particle and its momentum now increase due to the growth of its relativistic mass. This effect is observed in the work of high-energy charged particle accelerators and serves as the most convincing experimental confirmation of the theory of relativity.
Peace energy. The most remarkable thing about the formula is that a body at rest has energy: putting in we get
Energy is called rest energy.
Kinetic energy. The kinetic energy of a particle in some frame of reference is defined as the difference between its total energy and the rest energy Using (1), we have
If the speed of the particle is small compared to the speed of light, formula (6) turns into the usual expression for the kinetic energy of the particle in nonrelativistic physics.
The difference between the classical and relativistic expressions for kinetic energy becomes especially significant when the speed of the particle approaches the speed of light. At , the relativistic kinetic energy (6) increases indefinitely: a particle with a nonzero rest mass and
Rice. 10. Dependence of the kinetic energy of the body on the speed
moving at the speed of light would have to have infinite kinetic energy. The dependence of the kinetic energy on the particle velocity is shown in fig. ten.
Proportionality of mass and energy. It follows from formula (6) that when a body accelerates, the increment in kinetic energy is accompanied by a proportional increment in its relativistic mass. Recall that the most important property of energy is its ability to transform from one form to another in equivalent quantities during various physical processes - this is precisely the content of the law of conservation of energy. Therefore, it is natural to expect that an increase in the relativistic mass of a body will occur not only when kinetic energy is imparted to it, but also with any other increase in the body's energy, regardless of the specific type of energy. From here we can make a fundamental conclusion that the total energy of a body is proportional to its relativistic mass, regardless of what specific types of energy it consists of.
Let us explain what has been said in the following simple example. Let us consider an inelastic collision of two identical bodies moving towards each other with the same velocities, so that as a result of the collision one body is formed, which is at rest (Fig. 11a).
Rice. 11. Inelastic collision observed in different systems reference
Let the speed of each of the bodies before the collision be equal to and the rest mass Let us denote the rest mass of the formed body as Now let's consider the same collision from the point of view of an observer in a different frame of reference K, moving relative to the original frame K to the left (Fig. 11b) with a small (nonrelativistic) speed - and.
Since then to convert the speed in the transition from K to K, you can use the classical law of addition of speeds. The law of conservation of momentum requires that the total momentum of the bodies before the collision be equal to the momentum of the formed body. Before the collision, the total momentum of the system is where is the relativistic mass of the colliding bodies; after a collision, it is equal because, as a result, the mass of the formed body and in K can be considered equal to the rest mass. Thus, it follows from the law of conservation of momentum that the rest mass of the body formed as a result of inelastic collision is equal to the sum of the relativistic masses of the colliding particles, i.e., it is greater than the sum of the rest masses of the original particles:
The considered example of an inelastic collision of two bodies, in which kinetic energy is converted into internal energy, shows that an increase in the internal energy of a body is also accompanied by a proportional increase in mass. This conclusion should be extended to all types of energy: a heated body has a greater mass than a cold one, a compressed spring has a greater mass than an uncompressed one, etc.
Equivalence of energy and mass. The law of proportionality of mass and energy is one of the most remarkable conclusions of the theory of relativity. The relationship between mass and energy deserves a detailed discussion.
In classical mechanics, the mass of a body is a physical quantity, which is a quantitative characteristic of its inert properties, i.e., a measure of inertia. This is an inert mass. On the other hand, mass characterizes the body's ability to create a gravitational field and experience force in the gravitational field. It is a gravitating, or gravitational, mass. Inertia and the ability to gravitational interactions are completely different manifestations of the properties of matter. However, the fact that the measures of these different manifestations are denoted by the same word is not accidental, but is due to the fact that both properties always exist together and are always proportional to each other, so that the measures of these properties can be expressed by the same number with an appropriate choice of units. measurements.
The equality of inertial and gravitational masses is an experimental fact, confirmed with a great degree of accuracy in the experiments of Eötvös, Dicke, and others. How should one answer the question: are inertial and gravitational mass the same or not? In their manifestations they are different, but their numerical characteristics are proportional to each other. This state of affairs is characterized by the word "equivalence".
A similar question arises in connection with the concepts of rest mass and rest energy in the theory of relativity. Manifestations of the properties of matter corresponding to mass and energy are indisputably different. But the theory of relativity states that these properties are inextricably linked, proportional to each other. Therefore, in this sense, one can speak of the equivalence of rest mass and rest energy. The relation (5) expressing this equivalence is called the Einstein formula. It means that any change in the energy of the system is accompanied by an equivalent change in its mass. It refers to changes various kinds internal energy at which the rest mass changes.
On the law of conservation of mass. Experience shows us that in the vast majority of physical processes in which the internal energy changes, the rest mass remains unchanged. How can this be reconciled with the law of proportionality of mass and energy? The fact is that usually the overwhelming majority of internal energy (and the rest mass corresponding to it) does not participate in transformations, and as a result, it turns out that the mass determined from weighing is practically conserved, despite the fact that the body releases or absorbs energy. This is simply due to insufficient weighing accuracy. To illustrate, consider a few numerical examples.
1. The energy released during the combustion of oil, during the explosion of dynamite and during other chemical transformations, seems to us enormous on the scale of everyday experience. However, if we translate its value into the language of equivalent mass, then it turns out that this mass does not even constitute the full value of the rest mass. For example, when hydrogen combines with oxygen, about an energy is released. The rest mass of the resulting water is less than the mass of the starting materials. This change in mass is too small to be detected with modern instruments.
2. In an inelastic collision of two particles accelerated towards each other to a speed, the additional rest mass of the stuck together pair is
(At this speed, one can use the non-relativistic expression for the kinetic energy.) This value is much less than the error with which the mass can be measured
Rest mass and quantum regularities. It is natural to ask the question: why normal conditions the vast majority of energy is in a completely passive state and does not participate in transformations? The theory of relativity cannot answer this question. The answer must be sought in the field of quantum laws,
one of the characteristic features of which is the existence of stable states with discrete energy levels.
For elementary particles, the energy corresponding to the rest mass either transforms into an active form (radiation) entirely, or does not transform at all. An example is the transformation of an electron-positron pair into gamma radiation.
In atoms, the overwhelming majority of the mass is in the form of the rest mass of elementary particles, which does not change in chemical reactions. Even in nuclear reactions, the energy corresponding to the rest mass of heavy particles (nucleons) that make up nuclei remains passive. But here the active part of the energy, i.e., the energy of interaction of nucleons, already constitutes a noticeable fraction of the rest energy.
Thus, experimental confirmation of the relativistic law of proportionality of rest energy and rest mass should be sought in the world of elementary particle physics and nuclear physics. For example, in nuclear reactions that proceed with the release of energy, the rest mass of the final products is less than the rest mass of the nuclei entering into the reaction. The energy corresponding to this change in mass coincides with good accuracy with the experimentally measured kinetic energy of the formed particles.
How do the momentum and energy of a particle depend on its speed in relativistic mechanics?
What physical quantity is called the mass of a particle? What is rest mass? What is relativistic mass?
Show that the relativistic expression (6) for the kinetic energy transforms into the usual classical one at .
What is rest energy? What is the fundamental difference between the relativistic expression for the energy of a body and the corresponding classical one?
In what physical phenomena does rest energy reveal itself?
How to understand the statement about the equivalence of mass and energy? Give examples of manifestation of this equivalence.
Is the mass of a substance conserved during chemical transformations?
Derivation of the expression for momentum. Let us justify formulas (1) given above without proof by analyzing a simple mental experience. To elucidate the dependence of the particle momentum on the velocity, let us consider the picture of an absolutely elastic "sliding" collision of two identical particles. In the center of mass system, this collision has the form shown in Fig. 12a: before the collision, the particles Y and 2 move towards each other with the same velocities in absolute value, after the collision the particles scatter in opposite directions with the same velocities in absolute value as before the collision. In other words,
in a collision, only the rotation of the velocity vectors of each of the particles occurs by the same small angle
What will the same collision look like in other frames of reference? Let us direct the x-axis along the bisector of the angle and introduce a reference frame K moving along the x-axis relative to the center-of-mass frame with a velocity equal to the x-component of the velocity of particle 1. In this reference frame, the collision pattern will look as shown in Fig. 12b: particle 1 moves parallel to the y axis, changing the direction of velocity and momentum to the opposite during the collision.
The conservation of the x-component of the total momentum of a system of particles in a collision is expressed by the relation
where are the momenta of the particles after the collision. Since (Fig. 126), the requirement of conservation of momentum means the equality of the x-components of the momentum of particles 1 and 2 in the reference frame K:
Now, along with K, we introduce into consideration the reference frame K, which moves relative to the center-of-mass system with a velocity equal to the x-component of the velocity of particle 2.
Rice. 12. To the conclusion of the dependence of body mass on speed
In this system, particle 2 before and after the collision moves parallel to the y axis (Fig. 12c). Applying the law of conservation of momentum, we are convinced that in this frame of reference, as in frame K, there is an equality of -components of the particle momentum
But from the symmetry of the collision patterns in Fig. 12b,c it is easy to conclude that the modulus of momentum of particle 1 in frame K is equal to the modulus of momentum of particle 2 in frame of reference, therefore
Comparing the last two equalities, we find, i.e., the y-component of the momentum of particle 1 is the same in the frames of reference K and K. In the same way, we find In other words, the y-component of the momentum of any particle, perpendicular to the direction of the relative velocity of the frames of reference, is the same in these frames. This is the main conclusion from the considered thought experiment.
But the y-component of the particle velocity has a different value in the frames of reference K and K. According to the velocity transformation formulas
where is the velocity of the system K with respect to K. Thus, in K the y-component of the velocity of particle 1 is less than in K.
This decrease in the y-component of the velocity of particle 1 during the transition from K to K is directly related to the relativistic time transformation: the same distance in K and K between the dashed lines A and B (Fig. 12b, c) particle 1 in the system K takes a longer time, than in K. If in K this time is equal to (proper time, since both events - the intersection of strokes A and B - occur in K at the same coordinate value, then in the system K this time is greater and equal to
Recalling now that the y-component of the momentum of particle 1 is the same in the systems K and K, we see that in the system K, where the y-component of the velocity of the particle is smaller, this particle must be assigned, as it were, a larger mass, if mass is understood, as in non-relativistic physics, the coefficient of proportionality between velocity and momentum. As already noted, this coefficient is sometimes called the relativistic mass. The relativistic mass of a particle depends on the frame of reference, i.e., it is a relative quantity. In that frame of reference, where the speed of the particle is much less than the speed of light, for the relationship between the speed and momentum of the particle, the usual classical expression is true where is the mass of the particle in the sense that it is understood in non-relativistic physics (rest mass).
