Goldfarb N., Novikov V. Impulse of a body and systems of bodies // Quantum. - 1977. - No. 12. - P. 52-58.
By special agreement with the editorial board and editors of the journal “Kvant”
The concept of momentum (quantity of motion) was first introduced into mechanics by Newton. Let us recall that the momentum of a material point (body) is understood as a vector quantity equal to the product of the mass of the body and its speed:
Along with the concept of body impulse, the concept of force impulse is used. The impulse of force has no special designation. In the particular case when the force acting on the body is constant, the impulse of the force is, by definition, equal to the product of the force and the time of its action: . IN general case, when a force changes with time, the momentum of the force is defined as .
Using the concept of body momentum and force impulse, Newton's first and second laws can be formulated as follows.
Newton's first law: there are reference systems in which the momentum of a body remains unchanged if other bodies do not act on it or the actions of other bodies are compensated.
Newton's second law: in inertial reference systems, the change in momentum of a body is equal to the momentum of the force applied to the body, that is
Unlike the usual Galilean form of the second law: , the “impulse” form of this law allows it to be applied to problems associated with the movement of bodies of variable mass (for example, rockets) and with movements in the region of near-light speeds (when the mass of a body depends on its speed).
We emphasize that the impulse acquired by a body depends not only on the force acting on the body, but also on the duration of its action. This can be illustrated, for example, by an experiment with pulling out a sheet of paper from under a bottle - we will leave it standing almost motionless if we jerk it (Fig. 1). The sliding frictional force acting on the bottle for a very short period of time, that is, a small impulse of force, causes a correspondingly small change in the momentum of the bottle.
Newton's second law (in “impulse” form) makes it possible to determine, by changing the momentum of a body, the impulse of the force acting on a given body and the average value of the force during its action. As an example, consider the following problem.
Problem 1. A ball with a mass of 50 g hits a smooth vertical wall at an angle of 30° to it, having a speed of 20 m/s at the moment of impact, and is elastically reflected. Determine the average force acting on the ball during the impact if the collision of the ball with the wall lasts 0.02 s.
During the impact, two forces act on the ball - the reaction force of the wall (it is perpendicular to the wall, since there is no friction) and the force of gravity. Let us neglect the impulse of gravity, assuming that in absolute value it is much less than the impulse of force (we will confirm this assumption later). Then, when a ball collides with a wall, the projection of its momentum onto vertical axis Y will not change, but to the horizontal axis X- will remain the same in absolute value, but will change sign to the opposite. As a result, as can be seen in Figure 2, the momentum of the ball will change by the amount , and
Consequently, a force acts on the ball from the side of the wall such that
According to Newton's third law, the ball acts on the wall with the same absolute force.
Let us now compare the absolute values of the force impulses and:
1 N·s, = 0.01 N·s.
We see that , and the gravitational impulse can indeed be neglected.
The impulse is remarkable in that under the influence of the same force it changes equally in all bodies, regardless of their mass, if only the time of action of the force is the same. Let's look at the following problem.
Problem 2. Two particles with masses m and 2 m moving in mutually perpendicular directions with speeds 2 and respectively (Fig. 3). The particles begin to experience equal forces. Determine the magnitude and direction of the velocity of a particle of mass 2 m at the moment of time when the speed of a particle of mass m became as shown by the dotted line: a) in Figure 3, a; b) in Figure 3, b.
The change in the momentum of both particles is the same: the same forces acted on them for the same time. In case a) the modulus of change in the momentum of the first particle is equal to
The vector is directed horizontally (Fig. 4, a). The momentum of the second particle also changes. Therefore, the modulus of momentum of the second particle will be equal to
the velocity module is equal to , and the angle 
.
Similarly, we find that in case b) the modulus of change in the momentum of the first particle is equal to (Fig. 4, b). The modulus of the second particle's momentum will become equal (this is easy to find using the cosine theorem), the modulus of the velocity of this particle will be equal, and the angle (according to the sine theorem).
When we move on to a system of interacting bodies (particles), it turns out that the total momentum of the system - the geometric sum of the momentum of the interacting bodies - has the remarkable property of being conserved over time. This law of conservation of momentum is a direct consequence of Newton's second and third laws. In the textbook “Physics 8”, this law was derived for the case of two interacting bodies forming a closed system (these bodies do not interact with any other bodies). It is easy to generalize this conclusion to a closed system consisting of an arbitrary number n tel. Let's show it.
According to Newton's second law, the change in momentum i th body of the system in a short period of time Δ t equal to the sum of the impulses of the forces of its interaction with all other bodies of the system:
Change full impulse system is the sum of changes in the impulses that make up the system of bodies: according to Newton’s second law, it is equal to the sum of the impulses of all internal forces of the system:
In accordance with Newton's third law, the forces of interaction between the bodies of the system are pairwise identical in absolute value and opposite in direction: . Therefore, the sum of all internal forces is zero, which means
But if a change in a certain value over an arbitrary short period of time Δ t is equal to zero, then this quantity itself is constant over time:
Thus, a change in the momentum of any of the bodies that make up a closed system is compensated by the opposite change in other parts of the system. In other words, the impulses of bodies closed system can change as desired, but their sum remains constant over time. If the system is not closed, that is, not only internal but also external forces act on the bodies of the system, then, reasoning in a similar way, we will come to the conclusion that the increment in the total momentum of the system over a period of time Δ t will be equal to the sum of the impulses of external forces over the same period of time:
The momentum of the system can only be changed by external forces.
If , then the open system behaves like a closed one, and the law of conservation of momentum is applicable to it.
Let us now consider several specific problems.
Problem 3. Weapon of mass m slides down a smooth inclined plane making an angle α with the horizontal. At the moment when the speed of the gun is equal to , a shot is fired, as a result of which the gun stops, and the projectile ejected in the horizontal direction “carries away” the impulse (Fig. 5). The duration of the shot is τ. What is the average value of the reaction force on the side of the inclined plane over time τ?
The initial impulse of the weapon-projectile system of bodies is equal to , the final impulse is equal to . The system under consideration is not closed: during time τ it receives an increment in momentum. The change in the momentum of the system is due to the action of two external forces: the reaction force (perpendicular to the inclined plane) and gravity, so we can write
Let's present this relationship graphically (Fig. 6). From the figure it is immediately clear that the desired value is determined by the formula
Momentum is a vector quantity, so the law of conservation of momentum can be applied to each of its projections on the coordinate axes. In other words, if , then they are independently preserved p x, p y And p z(if the problem is three-dimensional).
In the case when the sum of external forces is not equal to zero, but the projection of this sum to a certain direction is zero, the projection of the total impulse to the same direction remains unchanged. For example, when a system moves in a gravity field, the projection of its momentum in any horizontal direction is preserved.
problem 4. A horizontally flying bullet hits wooden block, suspended on a very long cord, and gets stuck in a block, giving it speed u= 0.5 m/s. Determine the speed of the bullet before impact. Bullet weight m= 15 g, mass of the bar M= 6 kg.
Braking a bullet in a block is a complex process, but to solve the problem there is no need to delve into its details. Since there are no external forces acting in the direction of the speed of the bullet before impact and the speed of the block after the bullet gets stuck (the suspension is very long, so the speed of the block is horizontal), the law of conservation of momentum can be applied:
Hence the bullet speed
υ » 200 m/s.
IN real conditions- in conditions of gravity - there are no closed systems unless the Earth is included in them. However, if the interaction between the bodies of the system is much stronger than their interaction with the Earth, then the law of conservation of momentum can be applied with great accuracy. This can be done, for example, in all short-term processes: explosions, collisions, etc. (see, for example, task 1).
Problem 5. The third stage of the rocket consists of a launch vehicle weighing m p = 500 kg and a head cone weighing m k = 10 kg. A compressed spring is placed between them. During tests on Earth, the spring imparted to the cone a speed of υ = 5.1 m/s relative to the launch vehicle. What will be the speed of the cone υ k and the launch vehicle υ p if their separation occurs in orbit while moving at a speed υ = 8000 m/s?
According to the law of conservation of momentum
Besides,
From these two relations we obtain
This problem can also be solved in a reference frame moving with speed in the direction of flight. Let us note in this connection that if the impulse is conserved in one inertial system reference, then it is preserved in any other inertial reference frame.
The law of conservation of momentum underlies jet propulsion. A jet of gas escaping from the rocket carries away the momentum. This impulse must be compensated by the same modulus change in the impulse of the remaining part of the rocket-gas system.
Problem 6. From a rocket weighing M combustion products are emitted in portions of the same mass m at a speed relative to the rocket. Neglecting the effect of gravity, determine the speed of the rocket that it will reach after departure n-th portion.
Let be the speed of the rocket relative to the Earth after the release of the 1st portion of gas. According to the law of conservation of momentum
where is the speed of the first portion of gas relative to the Earth at the moment of separation of the rocket-gas system, when the rocket has already acquired speed . From here
Let us now find the speed of the rocket after the departure of the second portion. In a reference frame moving at speed, the rocket is motionless before the second portion is released, and after the release it acquires speed . Using the previous formula and making a substitution in it, we get
Then it will be equal
The law of conservation of momentum can be given another form, which simplifies the solution of many problems, if we introduce the concept of the center of mass (center of inertia) of the system. Coordinates of the center of mass (points With) by definition are related to the masses and coordinates of the particles that make up the system by the following relations:
It should be noted that the center of mass of the system in a uniform field of gravity coincides with the center of gravity.
To find out physical meaning center of mass, we calculate its speed, or rather, the projection of this speed. A-priory
In this formula
And 
In exactly the same way we find that
It follows that
The total momentum of the system is equal to the product of the mass of the system and the velocity of its center of mass.
The center of mass (center of inertia) of the system thus takes on the meaning of a point whose speed is equal to the speed of movement of the system as a whole. If , then the system as a whole is at rest, although in this case the bodies of the system relative to the center of inertia can move in an arbitrary manner.
Using the formula, the law of conservation of momentum can be formulated as follows: the center of mass of a closed system either moves rectilinearly and uniformly, or remains motionless. If the system is not closed, then it can be shown that
The acceleration of the center of inertia is determined by the resultant of all external forces applied to the system.
Let's consider such problems.
3 task 7. At the ends of a homogeneous platform of length l there are two people whose masses are and (Fig. 7). The first one went to the middle of the platform. At what distance X Does a second person need to move along the platform so that the cart returns to its original place? Find the condition under which the problem has a solution.
Let's find the coordinates of the center of mass of the system at the initial and final moments and equate them (since the center of mass remained in the same place). Let us take as the origin of coordinates the point where at the initial moment there was a person of mass m 1 . Then
(Here M- mass of the platform). From here
Obviously, if m 1 > 2m 2, then x > l- the task loses its meaning.
Problem 8. On a thread thrown over a weightless block, two weights are suspended, the masses of which m 1 and m 2 (Fig. 8). Find the acceleration of the center of mass of this system if m 1 > m 2 .
Momentum is one of the most fundamental characteristics of a physical system. The momentum of a closed system is conserved during any processes occurring in it.
Let's start getting acquainted with this quantity with the simplest case. The momentum of a material point of mass moving with speed is the product
Law of momentum change. From this definition, using Newton's second law, we can find the law of change in the momentum of a particle as a result of the action of some force on it. By changing the speed of a particle, the force also changes its momentum: . In case of constant acting force That's why
The rate of change of momentum of a material point is equal to the resultant of all forces acting on it. With a constant force, the time interval in (2) can be taken by anyone. Therefore, for the change in momentum of a particle during this interval, it is true
In the case of a force that changes over time, the entire period of time should be divided into small intervals during each of which the force can be considered constant. The change in particle momentum over a separate period is calculated using formula (3):
The total change in momentum over the entire time period under consideration is equal to the vector sum of changes in momentum over all intervals
If we use the concept of derivative, then instead of (2), obviously, the law of change in particle momentum is written as
Impulse of force. The change in momentum over a finite period of time from 0 to is expressed by the integral
The quantity on the right side of (3) or (5) is called the impulse of force. Thus, the change in the momentum Dr of a material point over a period of time is equal to the impulse of the force acting on it during this period of time.
Equalities (2) and (4) are essentially another formulation of Newton's second law. It was in this form that this law was formulated by Newton himself.
The physical meaning of the concept of impulse is closely related to the intuitive idea that each of us has, or one drawn from everyday experience, about whether it is easy to stop a moving body. What matters here is not the speed or mass of the body being stopped, but both together, i.e., precisely its momentum.
System impulse. The concept of momentum becomes especially meaningful when it is applied to a system of interacting material points. The total momentum P of a system of particles is the vector sum of the momenta of individual particles at the same moment in time:
Here the summation is performed over all particles included in the system, so that the number of terms is equal to the number of particles in the system.
Internal and external forces. It is easy to come to the law of conservation of momentum of a system of interacting particles directly from Newton’s second and third laws. We will divide the forces acting on each of the particles included in the system into two groups: internal and external. Internal force is the force with which a particle acts on the External force is the force with which all bodies that are not part of the system under consideration act on the particle.
The law of change in particle momentum in accordance with (2) or (4) has the form
Let us add equation (7) term by term for all particles of the system. Then on the left side, as follows from (6), we obtain the rate of change
total momentum of the system Since the internal forces of interaction between particles satisfy Newton’s third law:
then when adding equations (7) on the right side, where internal forces occur only in pairs, their sum will go to zero. As a result we get
The rate of change of total momentum is equal to the sum of the external forces acting on all particles.
Let us pay attention to the fact that equality (9) has the same form as the law of change in the momentum of one material point, and the right side includes only external forces. In a closed system, where there are no external forces, the total momentum P of the system does not change regardless of what internal forces act between the particles.
The total momentum does not change even in the case when the external forces acting on the system are equal to zero in total. It may turn out that the sum of external forces is zero only along a certain direction. Although physical system in this case and is not closed, the component of the total impulse along this direction, as follows from formula (9), remains unchanged.
Equation (9) characterizes the system of material points as a whole, but refers to a certain point in time. From it it is easy to obtain the law of change in the momentum of the system over a finite period of time. If the acting external forces are constant during this interval, then from (9) it follows
If external forces change with time, then on the right side of (10) there will be a sum of integrals over time from each of the external forces:
Thus, the change in the total momentum of a system of interacting particles over a certain period of time is equal to the vector sum of the impulses of external forces over this period.
Comparison with the dynamic approach. Let us compare approaches to solving mechanical problems based on dynamic equations and based on the law of conservation of momentum using the following simple example.
A railway car of mass taken from a hump, moving at a constant speed, collides with a stationary car of mass and is coupled with it. At what speed do the coupled cars move?
We know nothing about the forces with which the cars interact during a collision, except for the fact that, based on Newton's third law, they are equal in magnitude and opposite in direction at each moment. With a dynamic approach, it is necessary to specify some kind of model for the interaction of cars. The simplest possible assumption is that the interaction forces are constant throughout the entire time the coupling occurs. In this case, using Newton’s second law for the speeds of each of the cars, after the start of the coupling, we can write
Obviously, the coupling process ends when the speeds of the cars become the same. Assuming that this happens after time x, we have
From here we can express the impulse of force
Substituting this value into any of formulas (11), for example into the second, we find the expression for the final speed of the cars:
Of course, the assumption made about the constancy of the force of interaction between the cars during the process of their coupling is very artificial. The use of more realistic models leads to more cumbersome calculations. However, in reality, the result for the final speed of the cars does not depend on the interaction pattern (of course, provided that at the end of the process the cars are coupled and moving at the same speed). The easiest way to verify this is to use the law of conservation of momentum.
Since no external forces in the horizontal direction act on the cars, the total momentum of the system remains unchanged. Before the collision, it is equal to the momentum of the first car. After coupling, the momentum of the cars is equal. Equating these values, we immediately find
which, naturally, coincides with the answer obtained on the basis of the dynamic approach. The use of the law of conservation of momentum made it possible to find the answer to the question posed using less cumbersome mathematical calculations, and this answer is more general, since no specific interaction model was used to obtain it.
Let us illustrate the application of the law of conservation of momentum of a system using the example of a more complex problem, where choosing a model for a dynamic solution is already difficult.
Task
Shell explosion. The projectile explodes at the top point of the trajectory, located at a height above the surface of the earth, into two identical fragments. One of them falls to the ground exactly below the point of explosion after a time. How many times will the horizontal distance from this point at which the second fragment will fly away change, compared to the distance at which an unexploded shell would fall?
Solution: First of all, let's write an expression for the distance over which an unexploded shell would fly. Since the speed of the projectile at the top point (we denote it by is directed horizontally), then the distance is equal to the product of the time of falling from a height without an initial speed, equal to which an unexploded projectile would fly away. Since the speed of the projectile at the top point (we denote it by is directed horizontally, then the distance is equal to the time of falling from a height without initial speed, equal to body, considered as a system of material points:
The bursting of a projectile into fragments occurs almost instantly, i.e., the internal forces tearing it apart act within a very short period of time. It is obvious that the change in the velocity of the fragments under the influence of gravity over such a short period of time can be neglected in comparison with the change in their speed under the influence of these internal forces. Therefore, although the system under consideration, strictly speaking, is not closed, we can assume that its total momentum when the projectile ruptures remains unchanged.
From the law of conservation of momentum one can immediately identify some features of the movement of fragments. Momentum is a vector quantity. Before the explosion, it lay in the plane of the projectile trajectory. Since, as stated in the condition, the speed of one of the fragments is vertical, i.e. its momentum remained in the same plane, then the momentum of the second fragment also lies in this plane. This means that the trajectory of the second fragment will remain in the same plane.
Further, from the law of conservation of the horizontal component of the total impulse it follows that the horizontal component of the velocity of the second fragment is equal because its mass is equal to half the mass of the projectile, and the horizontal component of the impulse of the first fragment is equal to zero by condition. Therefore, the horizontal flight range of the second fragment is from
the location of the rupture is equal to the product of the time of its flight. How to find this time?
To do this, remember that the vertical components of the impulses (and therefore the velocities) of the fragments must be equal in magnitude and directed towards opposite sides. The flight time of the second fragment of interest to us depends, obviously, on whether the vertical component of its speed is directed upward or downward at the moment the projectile explodes (Fig. 108).
Rice. 108. Trajectory of fragments after a shell burst
This is easy to find out by comparing the time given in the condition for the vertical fall of the first fragment with the time free fall from height A. If then the initial velocity of the first fragment is directed downward, and the vertical component of the velocity of the second fragment is directed upward, and vice versa (cases a and in Fig. 108).
His movements, i.e. size .
Pulse is a vector quantity coinciding in direction with the velocity vector.
SI unit of impulse: kg m/s .
The momentum of a system of bodies is equal to the vector sum of the momentum of all bodies included in the system:
Law of conservation of momentum
If the system of interacting bodies is additionally acted upon by external forces, for example, then in this case the relation is valid, which is sometimes called the law of momentum change:
For a closed system (in the absence of external forces), the law of conservation of momentum is valid:
The action of the law of conservation of momentum can explain the phenomenon of recoil when shooting from a rifle or during artillery shooting. Also, the law of conservation of momentum underlies the operating principle of all jet engines.
When solving physical problems, the law of conservation of momentum is used when knowledge of all the details of the movement is not required, but the result of the interaction of bodies is important. Such problems, for example, are problems about the impact or collision of bodies. The law of conservation of momentum is used when considering the motion of bodies of variable mass such as launch vehicles. Most of the mass of such a rocket is fuel. During the active phase of the flight, this fuel burns out, and the mass of the rocket in this part of the trajectory quickly decreases. Also, the law of conservation of momentum is necessary in cases where the concept is not applicable. It is difficult to imagine a situation where a stationary body acquires a certain speed instantly. In normal practice, bodies always accelerate and gain speed gradually. However, when electrons and other subatomic particles move, their state changes abruptly without remaining in intermediate states. In such cases, the classical concept of “acceleration” cannot be applied.
Examples of problem solving
EXAMPLE 1
| Exercise | A projectile of mass 100 kg flying horizontally along railway track at a speed of 500 m/s, hits a car with sand weighing 10 tons and gets stuck in it. What speed will the car get if it moved at a speed of 36 km/h in the direction opposite to the movement of the projectile? |
| Solution | The car + projectile system is closed, so in this case the law of conservation of momentum can be applied. Let's make a drawing, indicating the state of the bodies before and after the interaction.
When the projectile and the car interact, an inelastic impact occurs. The law of conservation of momentum in this case will be written as: Choosing the direction of the axis to coincide with the direction of movement of the car, we write the projection of this equation onto the coordinate axis: where does the speed of the car come from after a projectile hits it:
We convert the units to the SI system: t kg. Let's calculate: |
| Answer | After the shell hits, the car will move at a speed of 5 m/s. |
EXAMPLE 2
| Exercise | A projectile weighing m=10 kg had a speed v=200 m/s at the top point. At this point it broke into two parts. The smaller part with a mass m 1 =3 kg received a speed v 1 =400 m/s in the same direction at an angle to the horizontal. At what speed and in what direction will most of the projectile fly? |
| Solution | The trajectory of the projectile is a parabola. The speed of the body is always directed tangentially to the trajectory. At the top point of the trajectory, the velocity of the projectile is parallel to the axis.
Let's write down the law of conservation of momentum: Let's move from vectors to scalar quantities. To do this, let’s square both sides of the vector equality and use the formulas for: Taking into account that , and also that , we find the speed of the second fragment: Substituting numerical values into the resulting formula physical quantities, let's calculate: We determine the flight direction of most of the projectile using:
Substituting numerical values into the formula, we get: |
| Answer | Most of the projectile will fly down at a speed of 249 m/s at an angle to the horizontal direction. |
EXAMPLE 3
| Exercise | The mass of the train is 3000 tons. The friction coefficient is 0.02. What type of locomotive must be in order for the train to reach a speed of 60 km/h 2 minutes after the start of movement? |
| Solution | Since the train is acted upon by (an external force), the system cannot be considered closed, and the law of conservation of momentum is not satisfied in this case. Let's use the law of momentum change: Since the friction force is always directed in the direction opposite to the movement of the body, the friction force impulse will enter the projection of the equation onto the coordinate axis (the direction of the axis coincides with the direction of motion of the train) with a “minus” sign: |