René Descartes (), French philosopher, mathematician, physicist and physiologist. He expressed the law of conservation of momentum and defined the concept of impulse of force.

Law of conservation of momentum The momentum of a body (amount of motion) is a measure of mechanical motion, equal in classical theory to the product of a body’s mass and its speed. The momentum of a body is a vector quantity, directed in the same way as its speed. The law of conservation of momentum serves as the basis for explaining a wide range of natural phenomena and is used in various sciences.

Elastic impact An absolutely elastic impact is a collision of bodies, as a result of which their internal energies remain unchanged. With an absolutely elastic impact, not only the momentum is conserved, but also the mechanical energy of the system of bodies. Examples: collision of billiard balls, atomic nuclei and elementary particles. The figure shows an absolutely elastic central impact: As a result of the central elastic impact of two balls of equal mass, they exchange velocities: the first ball stops, the second begins to move at a speed equal to the speed of the first ball.

Inelastic impact Absolutely inelastic impact: this is the name of the collision of two bodies, as a result of which they join together and move further as one whole. During an inelastic impact, part of the mechanical energy of the interacting bodies transforms into internal energy, and the momentum of the system of bodies is conserved. Examples of inelastic interaction: collision of sticking plasticine balls, automatic coupling of cars, etc. The figure shows a completely inelastic collision: After an inelastic collision, two balls move as one with a speed less than the speed of the first ball before the collision.

Calculations: A B C As a result of the experiment, we received: m gun = 0.154 kg m projectile = 0.04 kg AC = L gun = 0.1 m L projectile = 1.2 m Using a meter, we determined the time of movement of the projectile and gun , it was: t pistol = 0.6 s t projectile = 1.4 s Now we determine the speed of the projectile and pistol during the shot using the formula: V = L/t We found that V pistol = 0.1: 0.6 = 0 .16 m/s V projectile = 1.2:1.4 = 0.86 m/s And finally we can calculate the momentum of these two bodies using the formula: P=mV We got: P gun = 0.154 * 0.16 = 0.025 kg *m/s P projectile = 0.04 *0.86 = 0.034 kg *m/s m p *V p = m s *V s 0.025 = 0.034 The discrepancy was due to the effect of friction on the projectile and instrument error. 0.1 m 1.2 m projectile pistol

Examples of application of the law of conservation of momentum The law is strictly observed in the phenomena of recoil during a shot, the phenomenon of jet propulsion, explosive phenomena and the phenomena of collision of bodies. The law of conservation of momentum is used: when calculating the velocities of bodies during explosions and collisions; when calculating jet vehicles; in the military industry when designing weapons; in technology - when driving piles, forging metals, etc.

The law of conservation of momentum underlies jet propulsion. Much credit for the development of the theory of jet propulsion belongs to Konstantin Eduardovich Tsiolkovsky. The founder of the theory of space flight is the outstanding Russian scientist Tsiolkovsky (). He gave the general principles of the theory of jet propulsion, developed the basic principles and designs of jet aircraft, and proved the need for using a multi-stage rocket for interplanetary flights. Tsiolkovsky's ideas were successfully implemented in the USSR during the construction of artificial Earth satellites and spacecraft.

Reactive motion The movement of a body resulting from the separation of part of its mass from it at a certain speed is called reactive. All types of motion, except reactive motion, are impossible without the presence of forces external to a given system, that is, without the interaction of the bodies of a given system with the environment, and for reactive motion to occur, interaction of the body with the environment is not required. Initially the system is at rest, i.e. its total momentum is zero. When part of its mass begins to be ejected from the system at a certain speed, then (since the total momentum of a closed system, according to the law of conservation of momentum, must remain unchanged) the system receives a speed directed in the opposite direction.

Conclusions: During interaction, the change in the momentum of a body is equal to the impulse of the force acting on this body. When bodies interact with each other, the change in the sum of their impulses is zero. And if the change in a certain quantity is zero, then this means that this quantity is conserved. The practical and experimental verification of the law was successful and once again it was established that the vector sum of the momenta of the bodies that make up the closed system does not change.

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Algorithm for solving problems on the topic “Conservation Laws” 1) Carefully study the conditions of the problem, understand the physical essence of the phenomena and processes considered in the problem, understand the main question of the problem. 2) Mentally imagine the situation described in the problem, find out the purpose of the solution, clearly highlight the data and unknown quantities. 3) Write down a brief statement of the problem. Simultaneously express all quantities in SI units. 4) Make a drawing showing the situation before and after the event. 5) Write down the law of conservation of momentum (in projection onto the selected axis) by checking the system for closure and/or the law of conservation of energy in accordance with what is shown in the drawing (on the one hand, what was “before”, on the other, what was “after” ). Select the zero potential energy level. 6) Solve an equation or system of equations for an unknown quantity, i.e. solve the problem in general form. 7) If not all quantities are known, then to find them you can use an algorithm for solving problems on the topic “Dynamics”. 8) Find the required value. 9) Determine the unit of quantity. Check if it makes sense.10) Calculate the number.11) Check the answer for “stupidity” and write it down.
before
m1
m2
Law of conservation of momentum
m1V1+m2V2=(m1+m2)V
X
In projection onto the X axis:
m1V1+m2V2=(m1+m2)V
V=
m1V1+m2V2
m1+m2
m1V1-m2V2=(m1+m2)V
V=
m1V1-m2V2
m1+m2
m1
m2
V1
V2
V
V1
V2
V
1) The boy catches up with the cart (runs towards the cart) and jumps on it. Then they move together. The mass of the boy is m1, the mass of the cart is m2. Boy speed V1, cart speed V2. Algorithm
after
before
after
2) 200 kg of crushed stone was poured on top of a trolley weighing 800 kg, rolling along a horizontal track at a speed of 0.2 m/s. By how much did the speed of the trolley decrease? Algorithm
before:
after:
Solution:
Given: m1=800 kg =0.2 m/s m2=200 kg
X
m1
m2
m1+m2
ox:
Answer: speed decreased by 0.04 m/s
dimension
3) A fisherman weighing 60 kg moves from the bow to the stern of the boat. How much will a boat with a length of 3 m and a mass of 120 kg move relative to the water? algorithm
Solution:
before:
after:
Given: m1= 60 kgl= 3 mm2= 120 kgS - ?
ox: 0=-m1V1+(m1+m2)V2
We assume that the movement of the fisherman and the boat is uniform
substitute into the equation
0= -m1 +(m1+m2)
l
t
s
t
S= =1 m
60*3
180
m1l=(m1+m2)S
S=
m1l
m1+m2
Answer: The boat moved 1 m.
V1=
l
t
V2=
s
t
m2
m1
m1
m2
V1
V2
1
1
=>
=>
x
4) The hunter shoots from a light inflatable boat. What speed does the boat acquire at the moment of the shot if the mass of the hunter with the boat is 70 kg, the mass of the shot is 35 g and the average initial velocity of the shot is 320 m/s? When fired, the barrel of the gun forms an angle of 60° to the horizontal. algorithm
Solution:
before:
after:
Given: m1= 70 kgm2= 0.035 kgV2= 320 m/sα = 60°V1 - ?
ox: 0= -m1V1+m2V2cosα
m1V1=m2V2cosα
V1= =0.08 m/s
0.035*320*S
70
V1=
m2V2 cos α
m1
dimension
Answer: the boat acquires a speed of 0.08 m/s
α
m1
V1
m2
V2
X
5) A grenade flying horizontally at a speed of 10 m/s exploded into two fragments with masses of 1 kg and 1.5 kg. The speed of the large fragment after the rupture horizontally increased to 25 m/s. Determine the speed and direction of movement of the smaller fragment. algorithm
Solution:
after:
before:
(m1+m2)V = - m1V1+m2V2
m1V1 = m2V2 – (m1+m2)V
V1=
m2V2 – (m1+m2)V
m1
V1= = 12.5 m/s
1,5 * 25 – (1+1,5) * 10
1
Answer: the speed of the smaller fragment is 12.5 m/s.
Given:V= 10 m/sm1= 1 kgm2= 1.5 kgV2= 25 m/sV1 - ?
V
m2
m1
V1
V2
X
Law of conservation of momentum in projection onto the X axis:
6) A bullet flies horizontally at a speed of 400 m/s, pierces a box standing on a horizontal rough surface and continues to move in the same direction with a speed * V0. The mass of the box is 40 times the mass of the bullet. The coefficient of sliding friction between the box and the surface is M = 0.15. How far has the box moved before its speed decreases by 20%. Algorithm Dynamics (USE)
Given: = 400m/s = ѕm2= 40m1μ=0.15 =0.8S - ?
=>
I.
Solution:
before:
after:
Let's write down the law of conservation of momentum
=>
=>
Let's write down Newton's II law
II.
ox:
oy:
=>
=>
Answer: 0.75m.
7) A body weighing 3 kg falls freely from a height of 5 m. Find the potential and kinetic energy of the body at a distance of 2 m from the earth’s surface. algorithm
Solution:
before:
after:
Given:m= 3 kgh= 5 kgh’= 2 kgEp’, Ek’- ?
h'
h
Ep=mgh, Ek=0
Ep=3 * 9.8 * 5=150 J
Ep'=mgh'
Ep'= 3 * 9.8 * 2=60
According to the law of conservation of energy:
Ep= Ep’ + Ek’
Ek= 90 J
Answer: Ep’=60 J; Ek’=90 J
Ep=0
8) A stone is thrown vertically upward with an initial speed of 10 m/s. At what height h is the kinetic energy of the stone equal to its potential energy? algorithm
Solution:
before:
after:
Given: = 10 m/s - ?
h
Law of energy conservation
h= = 2.5 m.
100
4 * 9.8
Answer: h= 2.5 m.
=>
Ep=0
9) A load weighing 25 kg hangs on a cord 2.5 m long. To what maximum height can the load be moved to the side so that the cord does not break during further free swings? The maximum tension force that the cord can withstand without breaking is 550 N. algorithm
Solution:
Given: m= 25 kg = 2.5 mTmax= 550 Nh - ?
y
Ep=mgh, Ek=0
mg
1
h
a
T
2
T
According to the law of conservation of energy when moving from point 1 to point 2
Ep=Ek
=>
This means that it is necessary to find the speed at point 2
According to Newton's II law in volume 2
Projection on oy:
, Where
Means
Answer: 1.5 m.
Dynamics
Ep=0
10) A circus performer weighing 60 kg falls into a stretched net from a height of 4 m. With what force does the net act on the performer if it bends by 1 m? algorithm
According to the law of conservation of energy
potential energy of a deformed mesh
before:
after:
Answer: 6000 N
Let's calculate:
Ek=0
Ep=mg(h+x)
Given:m= 60 kgh= 4 m x = 1 mF - ?
X
Ep=0
h
before:
after:
The artist is subject to an elastic force from the mesh
=>
F=
2 * 60 * 9,8(4+1)
1
=6000 N
11) A pendulum of mass m is tilted at an angle α from the vertical. What is the tension in the string as the pendulum passes through its equilibrium position? algorithm
Solution:
Given:mαT - ?
1
h
a
T
mg
y
2
T
According to Newton's second law in volume 2:
Projection on oy:
=>
=>
1
According to the law of conservation of energy during the transition from point 1 to point 2 Ep=Ek
=>
Let's substitute in
=>
We substitute in and we get
1
T=m(2g(1-cosα)+g)=mg(2-2cosα+1)=mg(3-2cosα)
Dynamics
Ep=0
12) A body is thrown from the surface of the earth at a speed of 8 m/s at an angle of 60° to the horizontal. Find the value of its speed at a height of 1.95 m. algorithm
Given: = 8 m/сh= 1.95 mα= 60° - ?
Solution:
=>
h
after:
before:
Let's check whether the body can reach height h1
the problem question makes sense
=>
=>
=>
Answer: 5 m/s.
y
x
Ep=0
13) A body slides without friction along a smooth horizontal surface at a speed of 5 m/s and enters a movable slide with a height of H = 1.2 m. To what height does the body rise on the slide if the mass of the slide is 5 times the body mass? algorithm
Given: = 5m/sH= 1.2 mm2= 5m1h - ?
Solution:
h
after:
To find V1 we write the law of conservation of momentum
=>
=>
Speed ​​on m1
Answer: 1.04 m.
before:
Let's write down the law of conservation of energy
Ep=0
14) Two bodies with a mass of 1/18 kg are moving towards each other. The speed of the first body is 4 m/s, the second is 8 m/s. How much heat will be released as a result of a completely inelastic collision of bodies? algorithm
Given: = 4 m/s = 8 m/sm1=m2= 1/18 kg Q - ?
Solution:
x
before:
after:
According to the law of conservation of energy, the amount of heat released is equal to the loss of mechanical energy (in our case, kinetic energy)
Let's find the final speed from the law of conservation of momentum
In projection onto OX:
=>
because m1=m2
From here:
Replace m1=m2=m
Answer: 2 J
15) At a certain height the glider had a speed of 10 m/s. Determine the speed of the glider when it descends by 40 meters. Neglect air resistance Algorithm
Given: = 10 m/сh1-h2= 40 m - ?
Solution:
before:
after:
Law of energy conservation
Answer: 30 m/s.
Ep=0
16) Two bodies of mass m1 and m2 are attached to threads of the same length with a common suspension point and are deflected - one to the left, the other to the right - at the same angle. The bodies are released at the same time. When they hit each other they stick together. Determine the ratio of the height to which the bodies rise after sticking together to the height from which they began their downward movement. Algorithm
Given:m1m2
before:
after:
Balls raised => reported Ep
because Angles are equal, then heights are equal
Before the impact, Ep of the balls turned into Ek
Means
The impact is inelastic, meaning at the moment of impact
Further
x
Ep=0
Law of conservation of momentum in projection onto OX:
=>
At the moment of lifting the balls, the ZSE is performed
From here:
Answer:
back
17) Elastic impact algorithm
before:
after:
I
The body m2 was raised to a height h and was told Ep. Before the impact, Ep turned into Ek.
=>
II
At the moment of impact, ZSI and ZSE are performed
1
=>
Solving the system
Means,
=>
let's substitute in 1
Further
Ep=0
=>
III
After the impact, the balls rise to heights h1 and h2. ZSE in progress
=>
=>
back
18) A heavy ball slides without friction along an inclined chute, forming a “dead loop” of radius R. From what height should the ball begin to move so as not to break away from the chute at the top point of the trajectory? Algorithm
Given:Rh - ?
Solution:
According to the law of conservation of energy
=>
=>
=>
To find V at point 2, we write Newton’s II law
Let's project onto OY:
- centripetal acceleration
in the limiting case
=>
let's substitute in 1
Answer: 2.5R
Dynamics
Ep=0
19) Two elastic balls are suspended on thin threads next to each other so that they are at the same height and touch. The masses of the balls are m1 = 10g and m2 = 15g. A ball of mass m1 is deflected by an angle α= 60°. Determine what the ratio of the lengths of the threads should be so that the second pendulum deviates by a larger angle. The collision is considered to be absolutely elastic. Algorithm (Olympiad)
Given:m1 = 0.01kgm2 = 0.015kgα = 60°
Solution:
Let's divide the task into 3 stages:
I Deflect a ball of mass m1
Law S.E.
=>
We'll find
=>
Means
II At the moment of impact, ZSE and ZSI are performed
=>
Further
Ep=0
Let's solve the system
=>
2
1
Let's substitute in
1
2
=>
=>
Means
=>
III Balls rise after impact
Z.S.E.
Means
from here
or
Answer:
back
20) A weightless spring with stiffness k = 100 N/m is attached to the dynamometer, on which a stationary weightless bowl hangs. A piece of plasticine falls onto a bowl from a height h = 20 cm with zero initial velocity. The plasticine sticks to the bowl, and the maximum reading of the dynamometer is F = 5 N. What is the mass of the plasticine? Algorithm (USE)
Given: k= 100N/mh= 0.2 m = 0F= 5 nm- ?
Solution:
Select zero potential energy level
Ep=0
before:
after:
Let's write down the law of conservation of energy
from here
Where
Answer: 0.05 kg
21) A block of mass m1=500g slides down an inclined plane from a height of 0.8 m and, moving along a horizontal plane, collides with a stationary block of mass m2=300g. Assuming the collision to be completely inelastic, determine the change in the total kinetic energy of the block as a result of the collision. Friction during movement can be neglected. Assume that the inclined plane smoothly turns into a horizontal one. Algorithm (USE)
Given: m1= 0.5 kgm2= 0.3 kgh= 0.8 m = 0Ftr= 0ΔEk= ?
before:
after:
I. When a block slides, the law of conservation of energy
=>
II. Collision. Inelastic impact
ZSI is executed, ZSE is not executed
=>
Answer: Ek decreased by 1.5 J
Ep=0
22) A ball suspended on a thread 90 cm long is moved from its equilibrium position to an angle of 60° and released. At the moment the ball passes the equilibrium position, it is hit by a bullet flying towards the ball at a speed of 300 m/s, which pierces the ball and flies out horizontally at a speed of 200 m/s, after which the ball continues to move in the same direction and deflects at an angle of 39°. Determine the ratio of the masses of the ball and the bullet (The mass of the ball is considered unchanged, the diameter of the ball is negligibly small compared to the length of the thread, cos 39° = 7/9) Algorithm (USE)
Given: = 0.9α = 60° =300m/s =200m/sβ = 39°cos 39° = 7/9
Solution:
Ep=0
Set the potential energy level to zero
Let's divide the task into 3 stages:
I. Ball from state I to state II.
Law of energy conservation:
=>
=>
=>
Let's define h1
=>
=>
=>
=>
Further
II. Impact moment:
ZSI is carried out, ZSE is not carried out
FSI in projection onto the X axis
III. The ball rises and deflects at an angle of 39°
ZSE:
Let's calculate
Answer: 100
back
23) A ball with mass M = 240 g lies on a smooth horizontal table, attached to a spring with stiffness k = 40 kN/m. The other end of the spring is fixed. A bullet of mass m = 10 g hits the ball, which at the moment of impact has an initial speed of 400 m/s, directed along the axis of the spring. The bullet gets stuck in the ball. Determine the amplitude of the ball's oscillations after impact. Algorithm
Given:M= 0.24 kgk= N/mmm= 0.01 kg =400 m/s xm - ?
Solution:
Ep=0
Let's determine the zero level of potential energy
Let's divide the task into 2 stages:
I. Moment of impact
ZSI is carried out, ZSE is not carried out
=>
Ball and bullet speed
=>
II. When the ball moves, its Ek turns into the moment of complete compression of the spring in Ep
=>
=>
Answer: 0.04 m.
after:
before:
24) The initial speed of a projectile fired vertically upward is 160 m/s. At the point of maximum lift, the projectile exploded into 2 fragments, the masses of which are in the ratio 1:4. The fragments scattered in vertical directions, with the smaller fragment flying down and falling to the ground at a speed of 200 m/s. Determine the speed that the larger fragment had at the moment it hit the ground. Neglect air resistance. Algorithm (USE)
Given: =160 m/s = = 200 m/s
Solution:
We divide the solution of the problem into 3 stages
I. The projectile flies upward
after:
before:
Law of energy conservation
=>
=>
II. Moment of shell burst
Law of conservation of momentum:
=>
=>
=>
after:
before:
For the first fragment, the law of conservation of energy
=>
=>
=>
Further
Ep=0
=>
=>
III. For the second fragment (without wind resistance)
after:
before:
Answer: 162.8 m/s
back
Algorithm for solving “Dynamics” problems Make a drawing. Draw the body, all the forces acting on it, show the direction of acceleration, select the axes. Write down Newton’s second law in vector form. Project the vectors of the resulting equation onto the axes and obtain scalar equations. Solve the equation (system of equations) for the desired quantity.
6, 9, 11, 18

Slide 1

Pulse. Law of conservation of momentum.

Physics lesson in 10th grade

Physics teacher of the Nikolaev Secondary School Saushkina T.A.

Slide 2

Newton's laws are satisfied in inertial reference systems. The force of gravity is applied to the Earth. The weight of a body is always directed downward. The acceleration of a body is inversely proportional to the mass of the body. The friction force depends on the area of ​​the contacting surfaces. The force is a vector quantity. The force of gravity is of an electromagnetic nature. The support reaction force is an elastic force.

Task with key Answer: 10010101

Slide 3

Force impulse - force - time

vector physical quantity that is a measure of the action of a force over a certain period of time

Impulse force

Slide 4

Body impulse

Body momentum - mass - body speed

vector physical quantity that is a measure of mechanical motion

Slide 5

Law of conservation of momentum

The vector sum (geometric) of the momenta of bodies in a closed system remains a constant value

The law can be applied: a) if the resultant of external forces is zero; b) for projection onto any axis, if the projection of the resultant onto this axis is zero

Slide 6

Application of the law of conservation of momentum

Slide 7

From the history of jet propulsion

The first gunpowder fireworks and flares were used in China in the 10th century. In the 18th century, combat missiles were used during hostilities between India and England, as well as in the Russian-Turkish wars.

Slide 8

Living rockets

Jet propulsion, now used in airplanes, rockets and spacecraft, is characteristic of octopuses, squids, cuttlefish, jellyfish - all of them, without exception, use the reaction (recoil) of an ejected stream of water for swimming.

Slide 9

In the world of plants

In southern countries (and here on the Black Sea coast too) a plant called “mad cucumber” grows. The mad cucumber (otherwise called the “ladies’ pistol”) shoots at more than 12 m.

Slide 10

There is an old legend about a rich man with a bag of gold, who, finding himself on the absolutely smooth ice of a lake, froze, but did not want to part with his wealth. But he could have been saved if he had not been so greedy! It was enough to push the bag of gold away from himself, and the rich man himself would slide across the ice in the opposite direction according to the law of conservation of momentum.

What would you do in his place?

Slide 11

Getting ready for the Unified State Exam

Slide 12

Part A. On a horizontal surface there is a cart with a mass of 20 kg, on which a person with a mass of 60 kg stands. The person begins to move along the cart at a constant speed, and the cart begins to roll without friction. Module of speed of the cart relative to the surface

greater than the modulus of human velocity relative to the surface less than the modulus of human velocity relative to the surface equal to the modulus of human velocity relative to the surface can be either greater or less than the modulus of human velocity relative to the surface

Slide 13

Part A. A car and a truck are moving at speeds 1= 108 km/h and 2= 54 km/h respectively. Their masses are respectively = 1000 kg and = 3000 kg. How much greater is the momentum of the truck than the momentum of the car?

at 15000 kgm/s at 45000 kgm/s at 30000 kgm/s at 60000 kgm/s

Slide 14

Part A. Two balls of the same mass move with equal velocities along the horizontal XY plane. It is known that for a system of bodies that includes both balls, the projection of momentum onto the OY axis is greater than zero, and the modulus of the projection of momentum onto the OX axis is greater than the modulus of the projection of momentum onto the OY axis. In this case, the direction of the speed of the second ball must coincide with the direction indicated by the number 1 2 3 4