2nd law in impulse form. Force, Newton's second law

Newton's second law in impulse form. Basic equation of dynamics. Body Impulse: Increment body impulse equal to the impulse of the force acting on it.

Momentum of the particle system and - internal forces Particle system The momentum of a particle system can only change under the influence of external forces

Center of mass of a particle system. Law of motion of the center of mass. 1). Radius vector of the center of mass: 2). Center of mass speed: 3). Law of motion of the center of mass of a particle system:

Law of conservation of momentum The momentum of a closed system of particles does not change over time 1). In classical mechanics, the law of conservation of momentum is a consequence of Newton's laws: In a closed system of particles 2). The law of conservation of momentum is a fundamental law of nature.

The law of conservation of momentum can be applied 1). If the particle system is closed 2). If 3). If, then 4). If short-term interaction forces in the system are many times greater in magnitude external forces

Jet propulsion The speed of the reference frame is equal to the speed of the rocket at time t=0: - the mass of the rocket - the speed of the gas relative to the rocket

Force is a measure of interaction (mutual action). If the action is large (small), then they speak of a large (small) force. Strength is represented by the letter $$ F$$ (the first letter of the word force).

Etc and interaction, the greater the force, the greater the acceleration of the body on which this force acts. Consequently, the acceleration is directly proportional to the acting force: a ∼ F a\sim F .

But it has already been said that acceleration depends on the mass of the body: a ∼ 1 m a \sim \frac 1m

Generalizing these dependencies we get:

Now let's consider the properties of force established experimentally:

1) The result of the action (manifestation) of force depends on the direction acting force, therefore, the force isvector quantity.

2) The result of the action (manifestation) of force depends on the magnitude of the applied force.

3) Result of action(manifestation) of force depends on the point of application of the force.

4) The unit of force is taken to be the value of the force that causes an acceleration of 1 m / s 2 1\ \mathrm(m)/\mathrm(s)^2for a body weighing 1 kg 1\ \mathrm(kg) . The unit of force was named after Is aka Newton 1 new" tone. (Pronounce the surname schiseems right like thisthe way the surname is pronounced in the state where it is o the scientist lived or lives. )

[ F → ] = 1 N = 1 kg m s 2 (newton). [\overset(\rightarrow)(F)] = 1\ \mathrm(N) = 1\ \mathrm(kg)\cdot\frac(\mathrm(m))(\mathrm(s)^2)\quad \ mathrm((newton)).

5) If several forces act on a body at the same time, then each force acts independently of the others. (Principle of superposition of forces). Then all the forces need to be added vectorially and get the resulting force(Fig. 4) .

Rice. 4

From the above properties of force follows, as a generalization of experimental facts, Newton’s second law:

Second Law Newton: The sum of all forces acting on a body is equal to the product of the mass of the body and the acceleration imparted by this sum of forces:

∑ F → = m a → . \boxed(\sum \vec(F) = m\vec(a)).

This expression can be presented in another form: since a → = v → k - v → 0 t \vec a = \frac(\vec v_\mathrm(k) - \vec v_0)(t) , then Newton’s second law takes the form:∑ F → = m v → k - v → 0 t \sum \vec F = m\frac(\vec v_\mathrm(k) - \vec v_0)(t) .

The product of a body's mass and its speed is called the body's momentum:

p → = m v → \vec p = m\vec v ,

then we get a new expression for Newton’s second law:

∑ F → = m v → k - m v → 0 t = p → k - p → 0 t = Δ p → t \boxed(\sum \vec F = \frac(m\vec v_\mathrm(k) - m\ vec v_0)(t)) = \frac(\vec p_\mathrm(k) - \vec p_0)(t) = \frac(\Delta \vec p)(t) .

∑ F → = p → k - p → 0 t \boxed(\sum \vec F = \frac(\vec p_\mathrm(k) - \vec p_0)(t)) - - Newton's second law in impulse form for the average value of force. Here p → k - p → 0 = Δ p → \vec p_\mathrm(k) - \vec p_0 = \Delta \vec p - - change in body momentum, t - t\ - time of change in the body's momentum.

∑ F → = d p → d t - \boxed(\sum \vec F = \frac(d\vec p)(dt))\ - Newton's second law in impulse form for the instantaneous value of force.

From the second law, in particular, it follows that the acceleration of a body subjected to the action of several forces is equal to the sum of the accelerations imparted by each force:

A → = ∑ a → i = a → 1 + a → 2 + … + a → i = ∑ F → m = F → 1 + F → 2 + … + F → i m = F → 1 m + F → 2 m + … + F → i m \boxed(\vec a = \sum \vec a_i = \vec a_1 + \vec a_2 + \dots + \vec a_i = \frac(\sum \vec F)(m) = \frac( \vec F_1 + \vec F_2 + \dots + \vec F_i)(m) = \frac(\vec F_1)(m) + \frac(\vec F_2)(m) + \dots + \frac(\vec F_i )(m)) .

The first form of writing the second law (∑ F → = m a →) (\sum \vec F = m\vec a) fair only at low speeds compared to speed Sveta. And, of course, Newton's second law is satisfied onlyV inertial reference systems . It should also be noted that Newton’s second law is valid for bodies of constant mass, finite sizes and moving progressively.

IN the second (impulse) expression has more general character and is valid at any speed.

As a rule, in school course physics, force does not change over time. However, the last pulse form of recording makes it possible to take into account the dependence of force on time, andthen the change in the momentum of the body will be found using definite integral over the time interval under study. In more simple cases(force changes linearly with time) you can take the average value of the force.


Rice. 5

Sometimes it is very useful to know that the product F → t \vec F \cdot tis called a force impulse, and its value F → · t = Δ p → \vec F \cdot t = \Delta \vec pequal to the change in momentum of the body.

For a constant force on a graph of force versus time, we can find that the area of ​​the figure under the graph is equal to the change in momentum(Fig. 5) .

But even if the force changes with time, then in this case, dividing time into small intervals Δ t \Delta tsuch that the magnitude of the force remains unchanged over this interval(Fig. 6), and then, summing up the resulting “columns”, we get:

The area of ​​the figure under the graph F (t) F (t) is numerically equal to the change in momentum.

IN observed natural phenomena strength tends to change over time. We often Using simple process models, we consider the forces to be constant. The very possibility of using simple models arises from the possibility of countingmedium strength, i.e. that is, such a constant force for which the area under the graph versus time will be equal to the area under the graph of the real force.

Rice. 6

One more very important consequence of Newton’s second law should be added, related to the equality of inertial and gravitational masses.









The indistinguishability of gravitational and inertial masses means that the accelerations caused by gravitational interaction (the law universal gravity) and any others are also indistinguishable.

Example 2. A ball weighing 0.5 kg 0.5\ \mathrm(kg) after an impact lasting 0.02 s 0.02\ \mathrm(s) acquires a speed of 10 m/s 10\ \mathrm(m)/\mathrm( With) . Find average strength blow.

Solution. IN in this case It’s more rational to choose Newton’s second law in impulse form, i.e.because the initial and final velocities are known, not the acceleration, and the time of action of the force is known. It should also be noted that the force acting on the ball does not remainconstant. According to what law does force change over time?, Not known. For simplicity, we will use the assumption that the force is constant, and itswe will call it average.

Then ∑ F → = Δ p → t \sum \vec F = \frac(\Delta \vec p)(t), i.e. F → avg t = Δ p → \vec F_\mathrm(average)\ cdot t = \Delta \vec p . In the projection onto the axis directed along the line of action of the force, we obtain: F cf t = p k - p 0 = m v k F_\mathrm(cf)\cdot t = p_\mathrm(k)-p_0 = mv_\mathrm(k ) . Finally, for the required force we obtain:

Quantitatively, the answer will be: F avg = 0.5 kg 10 m s 0.02 s = 250 N F_\mathrm(avg) = \frac(0.5\ \mathrm(kg)\cdot 10\ \frac(\ mathrm(m))(\mathrm(s)))(0.02\ \mathrm(s)) = 250\ \mathrm(N) .

Themes Unified State Exam 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 body mass to its speed:

There are no special units for measuring 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 with its help you can give Newton's second law a slightly different, also extremely useful form.

Newton's second law in impulse form

Let be the resultant of forces applied to a body of mass . We start with the usual notation 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 we can see, the derivative of the impulse is obtained on the left side:

. ( 1 )

The ratio (1) is new form records of Newton's second law.

Newton's second law in impulse form. The derivative of the momentum of a body is the resultant of the forces applied to the body.

We can say this: the resulting force acting on a body is equal to the rate of change of 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 time.

In tasks, as a rule, the time interval is quite small. For example, this could be the time of impact of the ball with the wall, and then the average force acting on the ball from the wall during the impact.

The vector on the left side of relation (2) is called change in impulse during . The change in momentum is the difference between the final and initial momentum vectors. Namely, if is the momentum of the body at some starting moment time, is the momentum of the body after a period of time, then the change in momentum is the difference:

Let us emphasize once again that the change in momentum is the difference between vectors (Fig. 1):

Let, for example, the ball fly perpendicular to the wall (the momentum before the impact is equal to ) and bounce back without losing speed (the momentum after the impact is equal to ). Despite the fact that the impulse has not changed in absolute value (), there is a change in the impulse:

Geometrically, this situation is shown in Fig.

2: The modulus of change in momentum, as we see, is equal to twice the modulus initial impulse

ball: .

, ( 3 )

Let us rewrite formula (2) as follows:

or, describing the change in momentum, as above: The quantity is called impulse of power.

There is no special unit of measurement for force impulse; the dimension of the force impulse is simply the product of the dimensions of force and time:

(Note that this turns out to be another possible unit of measurement for a body's momentum.) The verbal formulation of equality (3) is as follows: the change in the momentum of a body is equal to the momentum of the force acting on the body over a given period of time.

This, of course, is again Newton's second law in momentum form.

Example of force calculation

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 . The blow lasts for s. Find the average force,

Solution. acting on the ball during impact. Let us first show that the angle of reflection equal to angle

fall, that is, the ball will bounce off the wall at the same angle (Fig. 3). According to (3) we have: . It follows that the vector of momentum change co-directed

with vector, that is, directed perpendicular to the wall in the direction of the ball’s rebound (Fig. 5).

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 to , that is, the angle of reflection is really equal to the angle of incidence. Now let us note in addition that in our isosceles triangle

there is an angle (this is the angle of incidence); therefore, this triangle is equilateral. From here:

And then the desired average force acting on the ball is:

Impulse of a system of bodies

Let's start with a simple situation of a two-body system. Namely, let there be body 1 and body 2 with impulses and, respectively. The impulse of the system of these bodies is the vector sum of the impulses of each body: similar to the second Newton's law in the form (1). Let's derive this formula.

We will call all other objects with which the bodies 1 and 2 we are considering interact external bodies. The forces with which external bodies act on bodies 1 and 2 are called by external forces. Let be the resultant external force acting on body 1. Similarly, let be the resultant 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 a force. Then body 1 acts on body 2 with a force. According to Newton's third law, the forces are equal in magnitude and opposite in direction: . Forces and are internal forces, operating in the system.

Let us write for each body 1 and 2 Newton’s second law in the form (1):

, ( 4 )

. ( 5 )

Let's add equalities (4) and (5):

On the left side of the resulting equality there is a sum of derivatives equal to the derivative of the sum of the 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 us also denote - this is the resultant of external forces acting on the system. We get:

. ( 6 )

Thus, the rate of change of momentum of a system of bodies is the resultant of external forces applied to the system. We wanted to obtain equality (6), which plays the role of Newton’s second law for a system of bodies.

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.

By the impulse of the system of bodies bodies is the vector sum of the momenta of all bodies included in the system. If a system consists of bodies, then the momentum of this system is equal to:

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 obtain the derivative of the momentum of the system, and on the right side there remains only the sum of external forces (internal forces, adding in pairs, will give zero due to Newton's third law). Therefore, equality (6) will remain valid in the general case.

Law of conservation of momentum

The system of bodies is called closed, if the actions of external bodies on the bodies of a given system are either negligible or compensate each other. 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 a closed system is equal to zero: . In this case, from (6) we obtain:

But if the derivative of a vector goes to zero (the rate of change of the vector is zero), then the vector itself does not change over time:

Law of conservation of momentum. 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.

As an example of applying Newton's second law in impulse form, let's consider the following problem. A body of mass g moves with a speed m/s on a smooth horizontal surface. A body of mass g moves towards it with a speed of m/s. An absolutely inelastic impact occurs (the bodies stick together). Find the speed of the bodies after the impact.

Solution. The situation is shown in Fig.

7. Let's direct the axis in the direction of movement of the first body.

Rice. 7. To the task

Because 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:

. ( 7 )

Thus, the vector sum of 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 satisfied for it:

The impulse of the system before the impact is the sum of the impulses of the bodies:

After the inelastic impact, one body of mass is obtained, which moves with the desired 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 projections onto the axis:

By condition we have: m/s, m/s, so

The minus sign indicates that the stuck together bodies move in the direction opposite to the axis. Required speed: m/s.

Law of conservation of momentum projection The following situation often occurs in problems. The system of bodies is not closed (the vector sum of external forces acting on the system is not equal to zero), but there is such an axis, the sum of the projections of external forces onto the axis is zero

at any given time. Then we can say that along this axis our system of bodies behaves as closed, and the projection of the system’s momentum onto the axis is preserved.

Let's show this more strictly. Let's project equality (6) onto the axis:

If the projection of the resultant external forces vanishes, then

Therefore, the projection is a constant: Law of conservation of momentum projection.

If the projection onto the axis of the sum of external forces acting on the system is equal to zero, then the projection of the system’s momentum does not change over time.

As an example of applying Newton's second law in impulse form, let's consider the following problem. Let's look at an example of a specific problem to see how the law of conservation of momentum projection works. Mass boy standing on skates on, throws a stone of mass at an angle to the horizontal. Find the speed with which the boy rolls back after the throw.

Solution. The situation is shown schematically in Fig.

8 . The boy is depicted as straight-laced.

Rice. 8. To the task

The momentum of the “boy + stone” system is not conserved. This can be seen from the fact that after the throw, a vertical component of the system’s momentum appears (namely, the vertical component of the stone’s momentum), which was not there before the throw.

Therefore, the system that the boy and the stone form is not closed. Why? The fact is that the vector sum of 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 system’s momentum appears.

However, external forces act only vertically (there is no friction). Therefore, the projection of the impulse onto the horizontal axis is preserved. Before the throw, this projection was zero. Directing the axis in the direction of the throw (so that the boy went in the direction of the negative semi-axis), we get.

Force is a measure of interaction (mutual action). If the action is large (small), then they speak of a large (small) force. Force is denoted by the letter `F` (the first letter of the word force).

When interacting, the greater the force, the greater the acceleration of the body on which this force acts. Therefore, acceleration is directly proportional to the acting force: `a~F`.

But it has already been said that acceleration depends on the mass of the body: `a~1/m`.

Summarizing these dependencies we get:

`a=F/m`, or `F=ma`.

Now let's consider the properties of force established experimentally:

properties of force

1) The result of the action (manifestation) of force depends on the direction of the acting force, therefore, force is a vector quantity.

2) The result of the action (manifestation) of force depends on the magnitude of the applied force.

3) The result of the action (manifestation) of force depends on the point of application of the force.

4) The unit of force is taken to be the value of the force that causes an acceleration of `1 "m"//"c"^2` in a body weighing `1` kg. The unit of force was named after Isaac Newton `1` newton. (It is considered correct to pronounce the surname in the same way as the surname is pronounced in the state where the scientist lived or lives.)

`=1"H"=1 "kg"*"m"/("s"^2)` (newton).

5) If several forces act on a body at the same time, then each force acts independently of the others. (Principle of superposition of forces). Then all forces must be added vectorially and the resulting force is obtained (Fig. 4).

From the given properties of force it follows, as a generalization of experimental facts, Newton’s second law:

The sum of all forces acting on a body is equal to the product of the mass of the body and the acceleration imparted by this sum of forces:

`sumvecF=mveca`.

This expression can be presented in another form: since `veca=(vecv_"к"-vecv_0)/t`, then Newton's second law will take the form: `sumvecF=m(vecv_"к"-vcv_0)/t`.

The product of a body's mass and its speed is called the body's momentum: `vecp=mvecv`,

then we get a new expression for Newton’s second law:

`sumvecF=(mvecv_"к"-mvecv_0)/t=(vecp_"к"-vecp_0)/t=(Deltavecp)/t`.

`sum vecF=(vecp_"к"-vecp_0)/t` - Newton's second law in impulse form for the average value of force. Here `vecp_"к"-vecp_0=Deltavecp` is the change in the momentum of the body, `t` is the time of change in the momentum of the body.

`sumvecF=(dvecp)/(dt)` - Newton's second law in impulse form for the instantaneous value of force.

From the second law, in particular, it follows that the acceleration of a body subjected to the action of several forces is equal to the sum of the accelerations imparted by each force:

`veca=sumveca_i=veca_1+veca_2+...+veca_i=(sumvecF)/m=`

`=(vecF_1+vecF_2+...+vecF_i)/m=(vecF_1)/m+(vecF_2)/m+...+(vecF_i)/m`.

The first form of writing the second law `(sumvecF=mveca)` is valid only at low speeds compared to the speed of light. And, of course, Newton's second law is satisfied only in inertial reference systems. It should also be noted that Newton’s second law is valid for bodies of constant mass, finite dimensions and moving translationally.

The second (pulse) expression is more general and valid at any speed.

As a rule, in a school physics course, force does not change over time. However, the last pulse form of recording allows us to take into account the dependence of force on time, and then the change in the momentum of the body will be found using a certain integral over the time interval under study. In simpler cases (the force changes linearly over time), you can take the average value of the force.

Sometimes it is very useful to know that the product `vecF*t` is called the impulse of force, and its value `vecF*t=Deltavecp` is equal to the change in the momentum of the body.

For a constant force on the graph of force versus time, we can obtain that the area of ​​the figure under the graph is equal to the change in momentum (Fig. 5).

But even if the force changes with time, then in this case, dividing the time into small intervals `Deltat` such that the magnitude of the force in this interval remains unchanged (Fig. 6), and then, summing up the resulting “columns”, we get:

The area of ​​the figure under the graph `F(t)` is numerically equal to the change in momentum.

In observed natural phenomena, force tends to change over time. We often, using simple process models, consider forces to be constant. The very possibility of using simple models arises from the possibility of calculating the average force, i.e. such a constant force for which the area under the graph versus time will be equal to the area under the graph of the real force.

One more very important consequence of Newton’s second law should be added, related to the equality of inertial and gravitational masses.

The indistinguishability of gravitational and inertial masses means that accelerations caused by gravitational interaction (the law of universal gravitation) and any other are also indistinguishable.

A ball with a mass of `0.5` kg after an impact lasting `0.02` s acquires a speed of `10` m/s. Find the average impact force.

In this case, it is more rational to choose Newton’s second law in impulse form, since the initial and final velocities, rather than acceleration, are known, and the time of action of the force is known. It should also be noted that the force acting on the ball does not remain constant. According to what law the force changes over time, it is not known. For simplicity, we will use the assumption that the force is constant, and we will call it average.

Then `sumvecF=(Deltavecp)/t`, i.e. `vecF_("avg")*t=Deltavecp`. In the projection onto the axis directed along the line of action of the force, we obtain: `F_"ср"*t=p_"к"-p_0=mv_"к"`. Finally, for the required force we obtain:

`F_"sr"=(mv_"k")/t`.

The quantitative answer will be:

`F_"av"=(0.5"kg"*10"m"/"s")/(0.02"s")=250"H"`.