3.2. Pulse
3.2.2. Body impulse change
To apply the laws of change and conservation of momentum, it is necessary to be able to calculate the change in momentum.
Change in momentumΔ P → body is determined by the formula
Δ P → = P → 2 - P → 1,
where P → 1 = m v → 1 is the initial impulse of the body; P → 2 = m v → 2 is its final impulse; m is body weight; v → 1 - initial speed of the body; v → 2 is its final speed.
To calculate the change in body momentum, it is advisable to apply the following algorithm:
1) choose a coordinate system and find the projections of the initial P → 1 and final P → 2 impulses of the body onto the coordinate axes:
P 1 x, P 2 x;
P 1 y, P 2 y;
∆P x = P 2 x - P 1 x;
∆P y = P 2 y - P 1 y;
3) calculate the modulus of the vector of change in momentum Δ P → as
Δ P = Δ P x 2 + Δ P y 2.
Example 4. The body falls at an angle of 30 ° to the vertical on the horizontal plane. Determine the modulus of the body impulse change during the impact, if by the moment of contact with the plane the modulus of the impulse of the body is equal to 15 kg m / s. The impact of the body on the plane is considered absolutely elastic.
Solution. A body falling on a horizontal surface at a certain angle α to the vertical and colliding with this surface is absolutely elastic,
- firstly, the modulus of its velocity, and hence the magnitude of the impulse, remains unchanged:
P 1 = P 2 = P;
- secondly, it is reflected from the surface at the same angle at which it falls on it:
α 1 = α 2 = α,
where P 1 = mv 1 - body impulse module before impact; P 2 = mv 2 - body impulse module after impact; m is body weight; v 1 - the value of the speed of the body before the impact; v 2 - the value of the speed of the body after impact; α 1 - angle of incidence; α 2 - angle of reflection.
Specified body impulses, angles and coordinate system are shown in the figure.
To calculate the modulus of the body impulse change, we will use the following algorithm:
1) we write the projections of the impulses before and after the impact of the body against the surface on the coordinate axes:
P 1 x = mv sin α, P 2 x = mv sin α;
P 1 y = −mv cos α, P 2 y = mv cos α;
2) we find the projection of the change in momentum onto the coordinate axes using the formulas
Δ P x = P 2 x - P 1 x = m v sin α - m v sin α = 0;
Δ P y = P 2 y - P 1 y = m v cos α - (- m v cos α) = 2 m v cos α;
Δ P = (Δ P x) 2 + (Δ P y) 2 = (Δ P y) 2 = | Δ P y | = 2 m v cos α.
The value P = mv is given in the problem statement; therefore, we calculate the modulus of change in the impulse by the formula
Δ P = 2 P cos 30 ° = 2 ⋅ 15 ⋅ 0.5 3 ≈ 26 kg ⋅ m / s.
Example 5. A stone weighing 50 g is thrown at an angle of 45 ° to the horizon at a speed of 20 m / s. Find the modulus of the stone's impulse change during the flight. Neglect air resistance.
Solution. If there is no air resistance, then the body moves along a symmetrical parabola; wherein
- firstly, the velocity vector at the point of falling of the body makes an angle β with the horizon, equal to the angle α (α is the angle between the velocity vector of the body at the point of throwing and the horizon):
- secondly, the moduli of velocities at the point of throwing v 0 and at the point of falling of the body v are also the same:
v 0 = v,
where v 0 - the value of the speed of the body at the point of throwing; v is the value of the body's velocity at the point of fall; α is the angle that makes the velocity vector with the horizon at the point of throwing the body; β is the angle that the velocity vector makes with the horizon at the point of falling of the body.
The body's velocity vectors (momentum vectors) and angles are shown in the figure.
To calculate the modulus of the body impulse change during flight, we will use the following algorithm:
1) we write down the projections of the impulses for the throwing point and for the point of incidence on the coordinate axes:
P 1 x = mv 0 cos α, P 2 x = mv 0 cos α;
P 1 y = mv 0 sin α, P 2 y = −mv 0 sin α;
2) we find the projection of the change in momentum onto the coordinate axes using the formulas
Δ P x = P 2 x - P 1 x = m v 0 cos α - m v 0 cos α = 0;
Δ P y = P 2 y - P 1 y = - m v 0 sin α - m v 0 sin α = - 2 m v 0 sin α;
3) calculate the modulus of change in the momentum as
Δ P = (Δ P x) 2 + (Δ P y) 2 = (Δ P y) 2 = | Δ P y | = 2 m v 0 sin α,
where m is body weight; v 0 - the module of the initial velocity of the body.
Therefore, the calculation of the modulus of change in the impulse will be performed by the formula
Δ P = 2 m v 0 sin 45 ° = 2 ⋅ 50 ⋅ 10 - 3 ⋅ 20 ⋅ 0.5 2 ≈ 1.4 kg ⋅ m / s.
Themes of the USE codifier: momentum of a body, momentum of a system of bodies, law of conservation of momentum.
Pulse body is a vector quantity equal to the product of the body's mass by its velocity:
There are no special units of measure for impulse. The dimension of momentum is simply the product of the dimension of mass and the dimension of velocity:
Why is the concept of momentum interesting? It turns out that it can be used to give Newton's second law a slightly different, also extremely useful form.
Newton's second law in impulse form
Let be the resultant of the forces applied to the body of mass. We start with the usual writing of Newton's second law:
Taking into account that the acceleration of the body is equal to the derivative of the velocity vector, Newton's second law is rewritten as follows:
We introduce a constant under the derivative sign:
As you can see, the derivative of the impulse is obtained on the left side:
. ( 1 )
Relation (1) is a new form of writing Newton's second law.
Newton's second law in impulse form. The derivative of the momentum of the body is the resultant of the forces applied to the body.
You can also say this: the resulting force acting on the body is equal to the rate of change in the body's momentum.
The derivative in formula (1) can be replaced by the ratio of final increments:
. ( 2 )
In this case, there is an average force acting on the body during the time interval. The smaller the value, the closer the ratio is to the derivative, and the closer the average force is to its instantaneous value at a given moment in time.
In tasks, as a rule, the time interval is rather short. For example, it can be the time the ball hits the wall, and then the average force acting on the ball from the side of the wall during the strike.
The vector on the left-hand side of relation (2) is called change of momentum during . The change in momentum is the difference between the final and initial vectors of the momentum. Namely, if is the momentum of the body at some initial moment of time, is the momentum of the body after a period of time, then the change in momentum is the difference:
We emphasize again that the change in momentum is the difference of vectors (Fig. 1):
For example, let the ball fly perpendicular to the wall (the impulse before the impact is equal) and bounces back without losing speed (the impulse after the impact is equal). Despite the fact that the modulus of the impulse has not changed (), there is a change in the impulse:
Geometrically, this situation is shown in Fig. 2:
The modulus of the impulse change, as we can see, is equal to the doubled modulus of the initial impulse of the ball:.
Let's rewrite formula (2) as follows:
, ( 3 )
or, describing the change in momentum, as above:
The quantity is called impulse of power. There is no special unit of measure for the impulse of force; the dimension of the impulse of force is simply the product of the dimensions of force and time:
(Note that turns out to be another possible unit of measure for body momentum.)
The verbal formulation of equality (3) is as follows: the change in the momentum of the body is equal to the momentum of the force acting on the body for a given period of time. This, of course, is again Newton's second law in impulse form.
Force calculation example
As an example of applying Newton's second law in impulse form, let's consider the following problem.
Task.
A ball of mass g, flying horizontally at a speed of m / s, hits a smooth vertical wall and bounces off it without losing speed. The angle of incidence of the ball (that is, the angle between the direction of movement of the ball and the perpendicular to the wall) is equal to. Strike lasts for. Find the average strength,
acting on the ball during impact.
Solution. Let us show first of all that the angle of reflection is equal to the angle of incidence, that is, the ball will bounce off the wall at the same angle (Fig. 3).
According to (3) we have:. Hence it follows that the vector of change in momentum co-directional with a vector, that is, directed perpendicular to the wall in the direction of the ball's rebound (Fig. 5).
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| Rice. 5. To the task |
Vectors and
equal in modulus
(since the speed of the ball has not changed). Therefore, a triangle composed of vectors and is isosceles. This means that the angle between the vectors and is equal, that is, the angle of reflection is really equal to the angle of incidence.
Now note, in addition, that our isosceles triangle has an angle (this is the angle of incidence); therefore, this triangle is equilateral. Hence:
And then the required average force acting on the ball:
The impulse of the system of bodies
Let's start with a simple situation for a two-body system. Namely, let there be body 1 and body 2 with impulses and, respectively. The momentum of the system of these bodies is the vector sum of the impulses of each body:
It turns out that for the momentum of a system of bodies there is a formula similar to Newton's second law in the form (1). Let's deduce this formula.
All other objects with which the bodies 1 and 2 we are considering interact, we will call external bodies. The forces with which external bodies act on bodies 1 and 2 are called external forces. Let be the resulting external force acting on body 1. Similarly, the resulting external force acting on body 2 (Fig. 6).
In addition, bodies 1 and 2 can interact with each other. Let body 2 act on body 1 with force. Then body 1 acts on body 2 with force. According to Newton's third law, the forces and are equal in magnitude and opposite in direction:. Forces and is internal forces, operating in the system.
Let's write for each body 1 and 2 Newton's second law in the form (1):
, ( 4 )
. ( 5 )
Let us add equalities (4) and (5):
On the left side of the obtained equality is the sum of derivatives, equal to the derivative of the sum of vectors and. On the right side, we have by virtue of Newton's third law:
But - this is the impulse of the system of bodies 1 and 2. Let's also designate - this is the resultant of external forces acting on the system. We get:
. ( 6 )
In this way, the rate of change of the momentum of a system of bodies is the resultant of external forces applied to the system. Equality (6), which plays the role of Newton's second law for a system of bodies, is what we wanted to obtain.
Formula (6) was derived for the case of two bodies. Now let us generalize our reasoning to the case of an arbitrary number of bodies in the system.
The impulse of the system of bodies bodies is called the vector sum of the impulses of all bodies included in the system. If the system consists of bodies, then the momentum of this system is:
Then everything is done in exactly the same way as above (only technically it looks a little more complicated). If for each body we write down equalities similar to (4) and (5), and then add all these equalities, then on the left side we again get the derivative of the impulse of the system, and on the right side there will be only the sum of external forces (internal forces, adding in pairs, will give zero in view of Newton's third law). Therefore, equality (6) remains valid in the general case.
Momentum conservation law
The system of bodies is called closed, if the actions of external bodies on the bodies of a given system are either negligibly small or cancel each other out. Thus, in the case of a closed system of bodies, only the interaction of these bodies with each other, but not with any other bodies, is essential.
The resultant of external forces applied to the closed system is zero:. In this case, from (6) we obtain:
But if the derivative of the vector vanishes (the rate of change of the vector is zero), then the vector itself does not change with time:
Impulse conservation law. The momentum of a closed system of bodies remains constant over time for any interactions of bodies within this system.
The simplest problems on the law of conservation of momentum are solved according to the standard scheme, which we will now show.
Task. A body of mass g moves at a speed of m / s on a smooth horizontal surface. A body of mass r is moving towards it with a speed of m / s. An absolutely inelastic shock occurs (the bodies stick together). Find the speed of bodies after impact.
Solution. The situation is shown in Fig. 7. The axis is directed towards the movement of the first body.
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| Rice. 7. To the task |
Since the surface is smooth, there is no friction. Since the surface is horizontal and movement occurs along it, the force of gravity and the reaction of the support balance each other:
Thus, the vector sum of the forces applied to the system of these bodies is equal to zero. This means that the system of bodies is closed. Therefore, the law of conservation of momentum is fulfilled for it:
. ( 7 )
The impulse of the system before the impact is the sum of the impulses of the bodies:
After an inelastic impact, one body of mass was obtained, which moves with the required speed:
From the law of conservation of momentum (7) we have:
From here we find the speed of the body formed after the impact:
Let's move on to the projections on the axis:
By condition, we have: m / s, m / s, so that
The minus sign indicates that the stuck together bodies move in the direction opposite to the axis. Seeking speed: m / s.
Impulse projection conservation law
The following situation is often encountered in tasks. The system of bodies is not closed (the vector sum of external forces acting on the system is not zero), but there is such an axis, the sum of the projections of external forces on the axis is zero at any given time. Then we can say that along a given axis, our system of bodies behaves like a closed one, and the projection of the momentum of the system onto the axis is preserved.
Let us show this more rigorously. Let's project equality (6) onto the axis:
If the projection of the resultant external forces vanishes, then
Therefore, the projection is a constant:
Impulse projection conservation law. If the projection onto the axis of the sum of external forces acting on the system is zero, then the projection of the momentum of the system does not change over time.
Let's look at an example of a specific problem, how the law of conservation of momentum projection works.
Task. A mass boy, skating on smooth ice, throws a mass stone at an angle to the horizon. Find the speed with which the boy rolls back after being thrown.
Solution. The situation is shown schematically in Fig. eight . The boy is depicted as a straightforward.
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| Rice. 8. To the task |
The impulse of the "boy + stone" system is not saved. This can be seen at least from the fact that after the throw, the vertical component of the impulse of the system appears (namely, the vertical component of the impulse of the stone), which was not there before the throw.
Therefore, the system formed by the boy and the stone is not closed. Why? The fact is that the vector sum of the external forces is not equal to zero during the throw. The value is greater than the sum, and due to this excess, the vertical component of the momentum of the system appears.
However, external forces act only vertically (no friction). Therefore, the projection of the momentum on the horizontal axis is preserved. Before the throw, this projection was zero. Directing the axis towards the throw (so that the boy went in the direction of the negative semiaxis), we get.
A .22 caliber bullet has a mass of only 2 g. If you throw such a bullet at someone, he can easily catch it even without gloves. If you try to catch such a bullet that flew out of the muzzle at a speed of 300 m / s, then even gloves will not help here.
If a toy cart rolls on you, you can stop it with your toe. If a truck rolls on you, you should get out of the way.
Consider a problem that demonstrates the relationship between the impulse of force and the change in impulse of the body.
Example. The mass of the ball is 400 g, the speed that the ball acquired after impact is 30 m / s. The force with which the leg acted on the ball was 1500 N, and the impact time was 8 ms. Find the momentum of the force and the change in momentum of the body for the ball.
Body impulse change
Example. Estimate the average force from the floor on the ball during the kick.
1) During the impact, two forces act on the ball: the reaction force of the support, the force of gravity.
The reaction force changes over the time of the impact, so it is possible to find the average sex reaction force.
The definition is:
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The history of the appearance of the term
Formal definition of momentum
Impulse is called a conserved physical quantity associated with the homogeneity of space (invariant with respect to translations).
Electromagnetic pulse
An electromagnetic field, like any other material object, has an impulse, which can be easily found by integrating the Poynting vector over the volume:
p = 1 c 2 ∫ S d V = 1 c 2 ∫ [E × H] d V (\ displaystyle \ mathbf (p) = (\ frac (1) (c ^ (2))) \ int \ mathbf (S ) dV = (\ frac (1) (c ^ (2))) \ int [\ mathbf (E) \ times \ mathbf (H)] dV)(in SI system).The existence of an impulse in the electromagnetic field explains, for example, such a phenomenon as the pressure of electromagnetic radiation.
Impulse in quantum mechanics
Formal definition
Pulse modulus is inversely proportional to wavelength λ (\ displaystyle \ lambda):
p = h λ, (\ displaystyle p = (\ frac (h) (\ lambda)),)where h (\ displaystyle h) is Planck's constant.
For particles of not very high energy moving at a speed v ≪ c (\ displaystyle v \ ll c)(speed of light), the modulus of the impulse is p = m v (\ displaystyle p = mv)(where m (\ displaystyle m) is the mass of the particle), and
λ = h p = h m v. (\ displaystyle \ lambda = (\ frac (h) (p)) = (\ frac (h) (mv)).)Consequently, the de Broglie wavelength is shorter, the greater the pulse modulus.
In vector form, this is written as:
ρ (\ displaystyle \ \ rho). And instead of the momentum, there is a vector of the momentum density, which coincides in meaning with the vector of the mass flux density p → = ρ v →. (\ displaystyle (\ vec (p)) = \ rho (\ vec (v)).)Having studied Newton's laws, we see that with their help it is possible to solve the basic problems of mechanics if we know all the forces acting on the body. There are situations in which it is difficult or impossible to determine these values. Let's consider a few of these situations.When two billiard balls or cars collide, we can assert about the acting forces that this is their nature, elastic forces act here. However, we will not be able to establish precisely their modules or their directions, especially since these forces have an extremely short duration of action.When rockets and jet planes move, we also have little to say about the forces that set these bodies in motion.In such cases, methods are used that make it possible to get away from solving the equations of motion, and immediately use the consequences of these equations. At the same time, new physical quantities are introduced. Consider one of these quantities, called the momentum of the body
An arrow shot from a bow. The longer the contact of the bowstring with the arrow (∆t) lasts, the greater the change in the momentum of the arrow (∆), and, consequently, the higher its final speed.
Two colliding balls. While the balls are in contact, they act on each other with forces equal in magnitude, as Newton's third law teaches us. This means that the changes in their impulses must also be equal in magnitude, even if the masses of the balls are not equal.
After analyzing the formulas, two important conclusions can be drawn:
1. Identical forces acting during the same period of time cause the same changes in momentum in different bodies, regardless of the mass of the latter.
2. One and the same change in the momentum of a body can be achieved either by acting with a small force for a long period of time, or by acting with a short-term large force on the same body.
According to Newton's second law, we can write:
∆t = ∆ = ∆ / ∆t
The ratio of the change in the momentum of the body to the period of time during which this change occurred is equal to the sum of the forces acting on the body.
After analyzing this equation, we see that Newton's second law allows us to expand the class of problems to be solved and include problems in which the mass of bodies changes over time.
If we try to solve problems with variable mass of bodies using the usual formulation of Newton's second law:
then attempting such a solution would lead to an error.
An example of this is the already mentioned jet plane or space rocket, which, when moving, burn fuel, and the products of this burned are thrown into the surrounding space. Naturally, the mass of an aircraft or rocket decreases as fuel is consumed.
Despite the fact that Newton's second law in the form "the resultant force is equal to the product of the mass of a body and its acceleration" allows solving a fairly wide class of problems, there are cases of motion of bodies that cannot be fully described by this equation. In such cases, it is necessary to apply another formulation of the second law, linking the change in the momentum of the body with the impulse of the resultant force. In addition, there are a number of problems in which the solution of the equations of motion is mathematically extremely difficult or even impossible. In such cases, it is useful for us to use the concept of momentum.
Using the law of conservation of momentum and the relationship between the momentum of the force and the momentum of the body, we can derive the second and third laws of Newton.
Newton's second law is derived from the ratio of the momentum of the force and the momentum of the body.
The impulse of force is equal to the change in the impulse of the body:
Having made the appropriate transfers, we get the dependence of the force on the acceleration, because the acceleration is defined as the ratio of the change in speed to the time during which this change took place:
Substituting the values into our formula, we get the formula for Newton's second law:
To derive Newton's third law, we need the law of conservation of momentum.
Vectors emphasize the vectoriality of speed, that is, the fact that the speed can change in direction. After transformations we get:
Since the time interval in a closed system was a constant value for both bodies, we can write:
We got Newton's third law: two bodies interact with each other with forces equal in magnitude and opposite in direction. The vectors of these forces are directed towards each other, respectively, the modules of these forces are equal in value.
Bibliography
- Tikhomirova S.A., Yavorskiy B.M. Physics (basic level) - M .: Mnemosina, 2012.
- Gendenshtein L.E., Dick Yu.I. Physics grade 10. - M .: Mnemosina, 2014.
- Kikoin I.K., Kikoin A.K. Physics - 9, Moscow, Education, 1990.
Homework
- Give a definition of the impulse of the body, the impulse of power.
- How are the impulse of the body connected with the impulse of force?
- What conclusions can be drawn from the formulas for body impulse and force impulse?
- Internet portal Questions-physics.ru ().
- Internet portal Frutmrut.ru ().
- Internet portal Fizmat.by ().
