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 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. On active site During 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 weighing 100 kg, flying horizontally along a 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 wagon + projectile system is closed, therefore in 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 into the resulting formula numerical values physical quantities, we 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: |
Momentum in physics
Translated from Latin, “impulse” means “push.” This physical quantity is also called “quantity of motion”. It was introduced into science around the same time when Newton's laws were discovered (at the end of the 17th century).
The branch of physics that studies the movement and interaction of material bodies is mechanics. Impulse in mechanics is vector quantity, equal to the product of a body’s mass and its speed: p=mv. The directions of the momentum and velocity vectors always coincide.
In the SI system, the unit of impulse is the impulse of a body weighing 1 kg, which moves at a speed of 1 m/s. Therefore, the SI unit of impulse is 1 kg∙m/s.
In calculation problems, projections of velocity and momentum vectors onto any axis are considered and equations for these projections are used: for example, if the x axis is selected, then the projections v(x) and p(x) are considered. By definition of momentum, these quantities are related by the relation: p(x)=mv(x).
Depending on which axis is selected and where it is directed, the projection of the momentum vector onto it can be either positive or negative.
Law of conservation of momentum
The impulses of material bodies during their physical interaction can change. For example, when two balls suspended on threads collide, their impulses change mutually: one ball can start moving from a stationary state or increase its speed, and the other, on the contrary, reduce its speed or stop. However, in a closed system, i.e. when bodies interact only with each other and are not exposed to external forces, the vector sum of the impulses of these bodies remains constant during any of their interactions and movements. This is the law of conservation of momentum. Mathematically it can be derived from Newton's laws.
The law of conservation of momentum is also applicable to systems where some external forces act on bodies, but their vector sum is zero (for example, the force of gravity is balanced by the elastic force of the surface). Conventionally, such a system can also be considered closed.
IN mathematical form the law of conservation of momentum is written as follows: p1+p2+…+p(n)=p1’+p2’+…+p(n)’ (pulses p are vectors). For a two-body system, this equation looks like p1+p2=p1’+p2’, or m1v1+m2v2=m1v1’+m2v2’. For example, in the considered case with balls, the total impulse of both balls before the interaction will be equal to the total impulse after the interaction.
Let the body mass m for some short period of time Δ t force acted Under the influence of this force, the speed of the body changed by 
Therefore, during the time Δ t the body moved with acceleration
From the basic law of dynamics ( Newton's second law) follows:
A physical quantity equal to the product of the mass of a body and the speed of its movement is called body impulse(or amount of movement). The momentum of a body is a vector quantity. The SI unit of impulse is kilogram meter per second (kg m/s).
A physical quantity equal to the product of a force and the time of its action is called impulse of force . Force impulse is also a vector quantity.
In new terms Newton's second law can be formulated as follows:
ANDThe change in the momentum of the body (amount of motion) is equal to the impulse of force.
Denoting the momentum of a body with a letter, Newton’s second law can be written in the form
Exactly in this general view Newton himself formulated the second law. The force in this expression represents the resultant of all forces applied to the body. This vector equality can be written in projections onto the coordinate axes:
Thus, the change in the projection of the body's momentum onto any of the three mutually perpendicular axes is equal to the projection of the force impulse onto the same axis. Let's take as an example one-dimensional movement, i.e. the movement of a body along one of the coordinate axes (for example, the axis OY). Let the body fall freely with an initial speed v 0 under the influence of gravity; the falling time is t. Let's direct the axis OY vertically down. Gravity impulse F t = mg during t equals mgt. This impulse is equal to the change in the momentum of the body
This simple result coincides with the kinematicformulafor speed uniformly accelerated motion . In this example, the force remained unchanged in magnitude throughout the entire time interval t. If the force changes in magnitude, then the average value of the force must be substituted into the expression for the impulse of force F cf over the period of time of its action. Rice. 1.16.1 illustrates the method for determining the time-dependent force impulse.
Let us choose a small interval Δ on the time axis t, during which the force F (t) remains virtually unchanged. Impulse force F (t) Δ t in time Δ t will be equal to the area of the shaded column. If the entire time axis is in the interval from 0 to t split into small intervals Δ ti, and then sum the force impulses at all intervals Δ ti, then the total impulse of force will be equal to the area formed by the stepped curve with the time axis. In the limit (Δ ti→ 0) this area is equal to the area limited by the graph F (t) and axis t. This method of determining the force impulse from a graph F (t) is general and applicable to any laws of force change over time. Mathematically, the problem reduces to integration functions F (t) on the interval .
The force impulse, the graph of which is presented in Fig. 1.16.1, in the interval from t 1 = 0 s to t 2 = 10 s is equal to:
In this simple example
In some cases, medium strength F cp can be determined if the time of its action and the impulse imparted to the body are known. For example, a strong hit by a football player on a ball with a mass of 0.415 kg can give him a speed of υ = 30 m/s. The impact time is approximately 8·10 -3 s.
Pulse p, acquired by the ball as a result of a strike is:
Therefore, the average force F the average with which the football player’s foot acted on the ball during the kick is:
This is a very big power. It is approximately equal to the weight of a body weighing 160 kg.
If the movement of a body during the action of a force occurred along a certain curvilinear trajectory, then the initial and final impulses of the body may differ not only in magnitude, but also in direction. In this case, to determine the change in momentum it is convenient to use pulse diagram
, which depicts the vectors and , as well as the vector 
built according to the parallelogram rule. As an example in Fig. Figure 1.16.2 shows a diagram of impulses for a ball bouncing off a rough wall. Ball mass m hit the wall with a speed at an angle α to the normal (axis OX) and bounced off it with a speed at an angle β. During contact with the wall, a certain force acted on the ball, the direction of which coincides with the direction of the vector
During a normal fall of a ball with a mass m on an elastic wall with speed, after the rebound the ball will have speed. Therefore, the change in momentum of the ball during the rebound is equal to 
In projections onto the axis OX this result can be written in scalar form Δ px = -2mυ x. Axis OX directed away from the wall (as in Fig. 1.16.2), therefore υ x < 0 и Δpx> 0. Therefore, the module Δ p the change in momentum is related to the modulus υ of the ball speed by the relation Δ p = 2mυ.
Vector physical quantity, equal to the product of the mass of the body and its speed, is called the momentum of the body: p - mv. The impulse of a system of bodies is understood as the sum of the impulses of all bodies of this system: ?p=p 1 +p 2 +... .
Law of conservation of momentum: in a closed system of bodies, during any processes, its momentum remains unchanged, i.e.
?p = const.
The validity of this law is easy to prove, for simplicity, considering a system of two bodies. When two bodies interact, the momentum of each of them changes, and these changes are equal to ?p = F 1 ?t and?p 2 = F 2 ?t, respectively. At the same time, the change full impulse system is equal to: ?р = ?р 1 + ?р 2 =F 1 ?t + F 2 ?t = (F 1 + F 2) ?t.
However, according to Newton's third law, F 1 = -F 2. Thus, ?р = 0.
One of the most important consequences of the law of conservation of momentum is the existence of reactive motion. Jet motion occurs when any part of it is separated from a body at a certain speed.
For example, jet propulsion is made by a rocket. Before launch, the momentum of the rocket is zero, and it should remain so after launch. Applying the law of conservation of momentum (we do not take into account the effect of gravity), we can calculate what speed the rocket will develop after burning all the fuel in it: m r v r + mv = 0, where V r is the speed of gases emitted in the form of a jet stream, tg is the mass of burned fuel , v is the speed of the rocket, and m is its mass. From here we calculate the speed of the rocket:
Schemes for various rockets were developed by K. E. Tsiolkovsky, who is considered the founder of the theory space flights. In practice, the ideas of K. E. Tsiolkovsky began to be implemented by scientists, engineers and cosmonauts under the leadership of S. P. Korolev.
The problem is to apply the law of conservation of momentum. A boy with a mass tg = 50 kg runs at a speed vx = 5 m/s, catches up with a cart with a mass t2 = 100 kg, moving at a speed i>2 = 2 m/s, and jumps on it. At what speed v will the cart move with the boy? Ignore friction.
Solution. The boy-cart system of bodies can be considered closed, since the gravity forces of the boy and the cart are balanced by the reaction forces of the supports, and friction is not taken into account.
Let's connect the reference frame to the Earth and direct the OX axis in the direction of movement of the boy and the cart. In this case, the projections of impulses and velocities onto the axis will be equal to their modules. Therefore, we can write the relations in scalar form.
The initial impulse of the system is the sum of the initial impulses of the boy and the cart, respectively equal to m v and m v When the boy rides on the cart, the impulse of the system is equal to (m1 + m2)v. According to the law of conservation of momentum
m 1 v 1 +m 2 v 2 =(m 1 +m 2) v
Let the body mass m for some short period of time Δ t force acted Under the influence of this force, the speed of the body changed by 
Therefore, during the time Δ t the body moved with acceleration
From the basic law of dynamics ( Newton's second law) follows:
A physical quantity equal to the product of the mass of a body and the speed of its movement is called body impulse(or amount of movement). The momentum of a body is a vector quantity. The SI unit of impulse is kilogram meter per second (kg m/s).
A physical quantity equal to the product of a force and the time of its action is called impulse of force . Force impulse is also a vector quantity.
In new terms Newton's second law can be formulated as follows:
ANDThe change in the momentum of the body (amount of motion) is equal to the impulse of force.
Denoting the momentum of a body with a letter, Newton’s second law can be written in the form
It was in this general form that Newton himself formulated the second law. The force in this expression represents the resultant of all forces applied to the body. This vector equality can be written in projections onto the coordinate axes:
Thus, the change in the projection of the body's momentum onto any of the three mutually perpendicular axes is equal to the projection of the force impulse onto the same axis. Let's take as an example one-dimensional movement, i.e. the movement of a body along one of the coordinate axes (for example, the axis OY). Let the body fall freely with an initial speed v 0 under the influence of gravity; the falling time is t. Let's direct the axis OY vertically down. Gravity impulse F t = mg during t equals mgt. This impulse is equal to the change in the momentum of the body
This simple result coincides with the kinematicformulafor speed of uniformly accelerated motion. In this example, the force remained unchanged in magnitude throughout the entire time interval t. If the force changes in magnitude, then the average value of the force must be substituted into the expression for the impulse of force F cf over the period of time of its action. Rice. 1.16.1 illustrates the method for determining the time-dependent force impulse.
Let us choose a small interval Δ on the time axis t, during which the force F (t) remains virtually unchanged. Impulse force F (t) Δ t in time Δ t will be equal to the area of the shaded column. If the entire time axis is in the interval from 0 to t split into small intervals Δ ti, and then sum the force impulses at all intervals Δ ti, then the total impulse of force will be equal to the area formed by the stepped curve with the time axis. In the limit (Δ ti→ 0) this area is equal to the area limited by the graph F (t) and axis t. This method of determining the force impulse from a graph F (t) is general and applicable to any laws of force change over time. Mathematically, the problem reduces to integration functions F (t) on the interval .
The force impulse, the graph of which is presented in Fig. 1.16.1, in the interval from t 1 = 0 s to t 2 = 10 s is equal to:
In this simple example
In some cases, medium strength F cp can be determined if the time of its action and the impulse imparted to the body are known. For example, a strong hit by a football player on a ball with a mass of 0.415 kg can give him a speed of υ = 30 m/s. The impact time is approximately 8·10 –3 s.
Pulse p, acquired by the ball as a result of a strike is:
Therefore, the average force F the average with which the football player’s foot acted on the ball during the kick is:
This is a very big power. It is approximately equal to the weight of a body weighing 160 kg.
If the movement of a body during the action of a force occurred along a certain curvilinear trajectory, then the initial and final impulses of the body may differ not only in magnitude, but also in direction. In this case, to determine the change in momentum it is convenient to use pulse diagram
, which depicts the vectors and , as well as the vector 
built according to the parallelogram rule. As an example in Fig. Figure 1.16.2 shows a diagram of impulses for a ball bouncing off a rough wall. Ball mass m hit the wall with a speed at an angle α to the normal (axis OX) and bounced off it with a speed at an angle β. During contact with the wall, a certain force acted on the ball, the direction of which coincides with the direction of the vector
During a normal fall of a ball with a mass m on an elastic wall with speed, after the rebound the ball will have speed. Therefore, the change in momentum of the ball during the rebound is equal to 
In projections onto the axis OX this result can be written in scalar form Δ px = –2mυ x. Axis OX directed away from the wall (as in Fig. 1.16.2), therefore υ x < 0 и Δpx> 0. Therefore, the module Δ p the change in momentum is related to the modulus υ of the ball speed by the relation Δ p = 2mυ.
