"the force of friction can be neglected." Handbook for practical exercises Ignore friction

This is how I see the expression of the main principle, which has always provided humanity with tremendous speed, at which it serenely and freely rushes towards the sign "Stop!". This principle can be expressed, of course, in another way: " take object "A" as a point" or " thirty percent of the votes for other ideas can be neglected". You can neglect anything, if only this something makes it difficult to give a clear answer to our questions. And there are many such questions that have been of interest to us for many, many years, and if we are not able to get an unambiguous answer to them, then we simply manipulate the data so that we still get an unambiguous answer. Data manipulation most often occurs through system simplification; we neglect some data and get a certain result. So the next question is born. And why and based on what considerations did we neglect these particular data? Maybe we are just trying to adjust the conditions of the problem to the pre-expected result? Why do we pretend that the data we are neglecting does not exist in nature and does not affect anything?

"Of course they influence- any mathematician or physicist will tell you - but their influence is absolutely insignificant. And we take into account even this insignificant influence, if we suddenly want to take it into account at all, using such a concept as an error".(very interesting word by the way). But after all, even from mathematics it is known that the error grows with an increase in the number of actions performed, which implied the neglect of some data. ( If, for example, we multiply two numbers that have been rounded to within seven correct decimal places, we will get a number that will no longer contain seven correct decimal places. Those. the error is growing.) There is an unaccounted for, tiny little such factor, here we will also neglect it, and so many, many times. And in the end, we get not just an inaccurate result; it will already contain an unacceptable inaccuracy from the point of view of using this result to solve many other problems. But such results are often universally adopted, and rarely anyone notices that the inaccuracy is unacceptably large. He is immediately given an example for which the use of an inexact result does not create any problems. At least that's how it seems at first glance. When an error comes out, and this can happen after solving a couple of equations arising from the previous logic, or it can happen after several hundred years, we will have to go back down the calculations and paradigms until we see an unacceptable simplification at some stage .

So it's not for nothing that they say the devil is in the details"and, it is quite possible that not in vain the word "error"has a clear indication of something ungrateful. We are doing something bad, it turns out?

Now look at the amount of text that just describes the skillful twists and tricks that humanity uses in an attempt to answer questions of interest to it. After all, there was an opportunity to act differently and not try to distort the surrounding reality by juggling the data, and not try to look for answers to all questions. One could simply understand and come to terms with the fact that we are not allowed to know the answer to some questions, if only because a person has not even learned how to formulate them correctly. One could finally come to terms with the fact that the world is much more complicated than our schematic ideas about it. It was already possible to come to terms with the fact that the technogenic world created by us is based on simplifications, and therefore it is obviously simpler, and therefore not ideal, and this can be forgiven. And we can forgive that too. One could finally come to terms with the fact that the smaller is not capable of cognizing the larger, that a less complex system is not capable of cognizing a more complex one. And you could just live, loving this world as it is. And love yourself in this world and in general everything that is in this world. And there are such people, believe me =). But there are also those who do not want to love - they want to explore, and the subject of research, meanwhile, is in no hurry to leave the category " underexplored" and " don't go to grandma"on a very, very long time will remain unknown to people. There are different people in general.

The process described in the problem can be divided into two stages:

1) the bullet hits the body, giving it some speed;

2) having this initial speed, the body deviates on the threads at a certain angle .

At the first stage there is an inelastic interaction of bodies. In this case, non-conservative forces act in the system (body + bullet) (friction force, or resistance to the movement of a bullet in the body), and part of the bullet's energy is converted into heat. Under these conditions, the law of conservation of mechanical energy is not satisfied, so we use only the law of conservation of momentum. At this stage, the horizontal projection of the momentum must be preserved, from which it is possible to find the initial momentum of the body after the bullet hits it.

At the second stage, there are no non-conservative forces. Therefore, we apply the law of conservation of energy, relating the angle of deflection of the threads with the change in the energy of the body in the field of gravity. From here we find the desired speed.

Let us choose an inertial frame of reference associated with the Earth's surface. We direct the axis along the direction of the bullet movement, and the axis vertically upwards. Let us write the law of conservation of momentum in the projection onto the axis for the collision of bodies:

where is the velocity of the body immediately after the collision. We neglect the mass of the bullet, because by condition . For the second stage, we write the law of conservation of energy in the form

,(2)

where is the height of the body. The kinetic energy of the body at the beginning of the stage is equal to the potential energy in the gravity field at the end. From the figure we find , where

It is advisable to apply the law of conservation of momentum to solve those problems in which it is required to determine speeds, and not forces or accelerations. Of course, such problems can be solved using Newton's laws. But applying the law of conservation of momentum simplifies the solution.

Before solving the problem using the law of conservation of momentum, it is necessary to find out whether it can be applied in this case. The law can be applied to a closed system or in the case when the sum of the projections of forces on any direction is equal to zero, and also when the impulse external forces can be neglected.

To solve the problem, it is necessary to write the law in vector form (5.3.7).

After that, the vector equation is written in projections on the axes of the selected coordinate system (1).

The choice of the direction of the axes is dictated by the convenience of solving the problem. If, for example, all bodies move along one straight line, then it is advisable to direct the coordinate axis along this straight line.

When solving some problems, it is necessary to use additional equations of kinematics.

Some problems are solved using the momentum change equation in the form (5.3.5).

Task 1

A steel ball of mass 0.05 kg is dropped from a height of 5 m onto a steel plate. After the collision, the ball bounces off the plate with the same speed modulo. Find the force acting on the plate upon impact, assuming it to be constant. The impact time is 0.01 s.

Decision. Upon impact, the ball and the plate act on each other with forces equal in magnitude but opposite in direction. Having determined the force acting on the ball from the side of the plate, we will thereby find the force with which the ball acted on the plate during the time Δt during which the collision lasts.

During the collision, two forces act on the ball: the force of gravity m and the force from the side of the plate (Fig. 5.13).

Rice. 5.13

According to equation (5.2.3)

Denote by 1 the speed of the ball immediately before hitting the plate, and by 2 - the speed after the hit, then the change in the momentum of the ball Δ \u003d m 2 - m 1, therefore

In projections on the Y-axis, this equation will be written as follows:

Considering that v 2 = v 1 = v, we get

The modulus of the ball's velocity when it falls from a height h is determined by the formula v = 10 m/s. Now, using the expression (5.7.1), we find the force modulus:

According to Newton's third law

Therefore, F 1 \u003d 100.5 N; this force is applied to the plate and directed downward.

Note that the shorter the interaction time Δt, the greater will be the value of the quantity in formula (5.7.1) compared to mg. Therefore, the impact force can be ignored. If the ball were made of plasticine, then it would stick to the plate and the modulus of change in its momentum would be half as much. Accordingly, the force acting on the plate would also be half as much.

Task 2

During maneuvers at the railway station, two platforms with masses m 1 \u003d 2.4 10 4 kg and m 2 \u003d 1.6 10 4 kg moved towards each other with speeds whose modules are equal to v 1 \u003d 0.5 m / s and v 2 = 1 m/s. Find the speed of their joint movement after the automatic coupler has worked.

Decision. Let's schematically depict the moving platforms before the collision (Fig. 5.14). External forces 1 and m 1 , 2 and m 2 acting on the bodies of the system are mutually balanced. Friction forces also act on the platforms, which are external to the system.

Rice. 5.14

When the platforms roll along the rails, the friction forces are small, so they will not noticeably change the momentum of the system over a short collision time interval. Therefore, we can apply the law of conservation of momentum:

where is the speed of the platforms after coupling.

In projections onto the X axis, we have:

Since v 1x \u003d v 1 a v 2x \u003d -v 2, then

The negative sign of the velocity projection shows that the velocity is directed opposite to the X axis (from right to left).

Task 3

Two plasticine balls, whose mass ratio = 4, stuck together after the collision and began to move along a smooth horizontal surface with a speed of . (Fig. 5.15, top view).

Rice. 5.15

Determine the speed of the light ball before the collision (2) if it moved three times faster than the heavy one (v 1 = Zv 2), and the directions of the balls were mutually perpendicular. Ignore friction.

Decision. Since the speeds of balls 1 and 2 are mutually perpendicular, it is convenient to direct the axes of the rectangular coordinate system parallel to these speeds.

According to the law of conservation of momentum, we have:

We write this equation in projections on the X and Y axes, drawn as shown in Figure 5.15:

Since v 1x = v 1 , v 2x = 0, v 1y = 0 and v 2y = v 2 , then

The modulus of speed is:

So, v 1 = u, therefore, v 1 = Зu.

Task 4

A grasshopper sits on the end of a straw of length l, which lies on a smooth floor. The grasshopper jumps and hits the other end of the straw. With what minimum initial speed relative to the floor min, he must jump if his mass is M, and the mass of the straw is m. Air resistance and friction are ignored.

Decision. Let's direct the Y-axis up, and the X-axis along the straw in the direction of the grasshopper's jump (Fig. 5.16). The projections of the speed v of the grasshopper on the coordinate axes are respectively equal to:

v x = v cos α and v y = vsin α.

Rice. 5.16

Consider the grasshopper-straw system. External forces act on the bodies of the system only in the vertical direction (there is no friction).

Since the sum of the projections of external forces on the X axis is equal to zero, the sum of the projections of the momentum of the grasshopper and the straw on the X axis is preserved:

where v 1x is the projection of the speed of the straw relative to the floor. From here

In the horizontal direction, the grasshopper will fly a distance l relative to the straw.

Therefore, the modulus of the horizontal component of its velocity relative to the moving straw is:

But on the other side,

Thus,

Obviously, the grasshopper's speed modulus is minimal when the denominator of the fraction of the resulting expression is maximum. As you know, the value of the sine cannot be greater than 1. So,

Task 5

At the initial moment of time, the rocket with mass M had a speed v0. At the end of each second, a portion of gas of mass m is ejected from the rocket. The speed of a portion of gas differs from the speed of the rocket before the combustion of a given mass of gas by a constant value equal to u, i.e., the speed of gas outflow is constant. Determine the speed of the rocket after n seconds. The effect of gravity is ignored.

Decision. Denote by v k the speed of the rocket in end of k-th seconds. At the end of the (k + 1)th second, a gas of mass m is ejected from the rocket, which carries with it an impulse equal to m(-u + v k). From the law of conservation of momentum, written for the modules of vectors, it follows that

The change in rocket speed in 1 s is:

Knowing the change in speed for 1 s, you can write an expression for the speed in end of nth seconds:

Exercise 10

1. A lead ball of mass 200 g moves perpendicular to the wall with a speed of 10 m/s and collides with it. Find the force acting on the wall upon impact, assuming it to be constant. The collision time is 0.01 s. The ball does not bounce off the wall.
2. A steel ball of mass 100 g moves along a horizontal surface without friction in a direction perpendicular to the wall. The speed of the ball before impact is 10 m/s. After the collision, the ball bounces off the wall with the same speed, but in the opposite direction. Find the force acting on the wall upon impact, assuming it to be constant. Impact time 0.01 s.
3. A cart filled with sand rolls along the rails in a horizontal direction. Through a hole in the bottom, sand is poured between the rails. Does the speed of the cart change? Friction is ignored.
4. On a platform with a mass of 600 kg, moving horizontally at a speed of 1 m/s, 200 kg of crushed stone was poured on top. What is the speed of the platform?
5. A rocket, the mass of which together with the charge is 250 g, takes off vertically upwards and reaches a height of 150 m. Determine the speed of the outflow of gases from the rocket, assuming that the combustion of the charge occurs instantly. The mass of the charge is 50 g.
6. A prism of mass M with an angle of inclination a is on smooth ice. A dog of mass m stands on a prism at its base. With what speed will the prism move if the dog runs up the prism with a speed v relative to it?
7. A grenade thrown from the surface of the Earth breaks into two identical fragments in highest point trajectories at a distance a from the place of throwing, counting horizontally. One of the fragments flies in the opposite direction with the same modulo speed that the grenade had before the explosion. At what distance l from the point of throwing will the second fragment fall?
8. Two rockets of mass M each fly in the same direction: one with a speed v, and the other with a speed v 1 = 1.1v. When one rocket caught up with another, a short time the engine of the first rocket was turned on. What mass of spent fuel should it throw out at a speed v 2 = Zu relative to the rocket, so that the speeds of the rockets to make a safe docking become equal?
9. Two boats go in parallel courses towards each other with the same modulo speeds. When the boats meet, they exchange goods that have the same mass. The exchange can take place in two ways: 1) first, cargo is transferred from one boat to another, and then cargo is transferred from the second boat back to the first; 2) loads are transferred from boat to boat at the same time. In which method will the speed of the boats after the transfer of goods be greater?
10. Three boats with the same masses M move by inertia one after the other with the same speeds v. Loads of mass m are simultaneously transferred from the middle boat to the outer ones with a speed u relative to the boats. What speed will the boats have after transferring cargo? Ignore water resistance and added mass.
11. The projectile breaks at the top of the trajectory into two equal parts. One half of the projectile receives a velocity directed vertically downwards and falls under the gap, and the second half of the projectile is at a distance l horizontally from this place. Determine the modulus of the velocity of the projectile before the explosion and the modulus of the velocity of the second fragment, if it is known that the explosion occurred at a height H and the first fragment reached the Earth's surface after a time interval equal to t.
12. A person in a boat moves from its bow to the stern. At what distance relative to the water will a boat of length l move if the mass of a person is m 1 and the mass of the boat is m 2? Ignore water resistance and added mass.

(1) Sometimes it is expedient to solve the problem using the law of vector addition.

(2) If, after the collision, the bodies move with the same speed, then such a collision is called absolutely inelastic.

What does this mean and how will it affect the solution of the problem.

First, let's figure out what air resistance is and why it occurs. As you know (you should know, you go to school), all substances are made up of molecules or atoms. Atoms are the smallest particles (let's imagine that these are small, small balls), and molecules are also small, but already consisting of several atoms.

For example, a water molecule H 2 O consists of two hydrogen atoms H and one oxygen atom O (that is, three balls stuck together into one piece).

Since we said before that "all substances" consist of them, then the air also consists of atoms and molecules (we breathe oxygen, which means 100% of it is in the air). When we throw a ball or some object down, it begins to collide with the smallest balls (atoms and molecules) of air. It is these collisions that are called air resistance.

Now let's try to ignore this resistance. To do this, we simply remove all these smallest balls (atoms and molecules) from the air. Agree reminds vacuum (or airless space)? That is, when falling, the bodies will not collide with anyone, but will simply fly down.

Now let's figure out how this will affect the solution of the problem?

Imagine that we are throwing a ball and a feather from the same height. What will fall faster? Ball? Not. Feather? Not. Will they fall the same? Not. Why not? Yes, because we do not know in the air it happens (where there is resistance) or in a vacuum (there is no resistance). In the air, the ball will fall faster, since it is heavier, and it is easier for it to knock the atoms / molecules of the air out of its way. And the feather is lighter, it will slow down a little during these collisions. If we throw them in a vacuum, then they will fall the same way, since they do not have to collide with anyone.

Don't believe? Watch the video (it is not necessary to listen, it is in English).

Here is another video on the same topic