Newton's Third Law of Motion

Newton’s Third Law of Motion 

SOMNATH DATTA 

656 “Snehalata, 13 th Main, 4 th Stage, T.K. Layout, Mysore 5700009, India 

E.mail: datta.som@gmail.com 

http://sites.google.com/site/physicsforpleasure 

1 A Slide Show on Newton’s Third Law 

of Motion 

1. Newton’s first and second laws tell us that a 

body accelerates due to external forces. The net 

external force equals the mass of the body 

times its acceleration. These laws do not tell us 

where this force comes from. Does the force 

that drives this car come from the engine? 

2. If so, why is this car stuck on wet clay road 

when the engine is running in full throttle? The 

force that pushes the car forward does not 

come from the engine. This force comes from 

the road due to friction between the tyre and 

the road, and is called traction. 

3. Does the force come from his muscles? 

Physics Education • July − September 2009 213

4. If so, why is he slipping? No runner is his 

able to run on a slippery track in spite of his 

muscle power. The force that drives the runner 

forward comes from the ground when there is 

sufficient friction between his feet and the 

ground. 

5. This car requires 40,000 N to speed up at the 

rate of 10m/s 2 . This sprinter requires 300 N to 

speed up at the rate of 5m/s 2 . How is the road 

able to judge the individual force requirements 

for the car and the sprinter? 

6. Answer to this comes from Newton’s Third 

Law of Motion: 

• Nothing in the universe can act without 

being acted upon. 

• To every action there is an equal and 

opposite reaction. 

The first statement is somewhat 

philosophical. The second one is vague. In 

exact terms what the first statement means is 

that whenever any object A offers a force F, it 

itself experiences the same force back on itself 

in the reverse direction. 

214 Physics Education • July − September 2009

7. The second statement can be interpreted as 

follows. Assume two objects A and B. A offers 

a force on B directed from A to B. We denote 

this force by F AB . The 3rd Law says B will 

simultaneously offer a force F BA on A which is 

directed from B to A. These two forces are 

equal and opposite, even if the objects A and B 

are grossly unequal in size and mass. Written 

as vectors, 

F BA = −F AB . (1) 

We can refer to any one of the pair of forces, 

say the force F AB , as the action force and the 

second one, i.e. F BA , as the reaction force. 

Then Eq. (1) is a mathematical expression of 

the second statement. 

In the particular example shown these two 

forces are repulsive forces. 

8. Here they are attractive forces. A pulls B 

with a force F AB . As a result A is itself being 

pulled towards B with a force F BA , and F BA = 

−F AB . 

9. A runner can run only if the ground kicks 

him forward. To get this kick the runner must 

kick the ground backward. It is possible to kick 

a solid and firm object like football or grass 

with sufficient force, but it is difficult to kick a 

loose object like air or slippery road with 

sufficient force. Therefore the runner is unable 

to run on a slippery road. 


10. It is possible to get the necessary force of 

40,000 N from a dry ground because the tyre, 

powered by the engine, can press against the 

road with this force due to friction between the 

two surfaces (of the tyre and the ground.) A 

frictionless surface on the other hand will not 

allow the tyre to rub with a large force and 

because of this the car will not be able to 

accelerate. 

11. There is a common misconception about 

the Third Law especially among the beginners. 

They think that this law holds only when the 

interacting bodies are in equilibrium since the 

forces of action and reaction will then cancel 

each other. This is entirely wrong. Here we 

show two bodies A and B. 

12. But the force F AB is acting on B only. 

Therefore B is not in equilibrium. 

13. The reaction force F BA is acting on A only. 

Therefore A is not in equilibrium. 


14. Now see the complete picture. The forces 

are equal and opposite. But they do not cancel 

each other. Because they are acting on different 

bodies. 

15. We see that the third law holds between 

any two bodies, whether at rest or in motion. 

16. Let us illustrate the Third Law with a few 

more examples. This stone of mass 1 kg is 

falling to the ground in a parabolic trajectory 

with acceleration g equal to 9.81 m/s 2 , like any 

other stone. The gravitational pull on the stone 

is m times g, that is, 9.81 N. Since the earth is 

pulling the stone with force 9.81 N downward, 

the stone is at the same time pulling the earth 

with a force of the same magnitude of 9.81 N, 

but directed upward. 

17. Therefore the earth is also accelerating 

toward the stone! However, since the mass of 

the earth is very LARGE, its acceleration 

(toward the stone) is very small. 


18. Another example: A football player kicks a 

football. Therefore according to the 3rd Law, 

the football kicks the player back. 

19. The kick is an impulsive force, lasting for a 

very short time, say, 1 of a second. Taking 

10 

the mass of the ball to be ¼ kg, we have 

calculated the kick force to be 250 N. 

20. The 3rd Law tells us that the ball also kicks 

the player back in the opposite direction. 

However, since the mass of the player is 100 

kg, i.e., 400 times the mass of the ball, his 

acceleration is 400 times less and lasts exactly 

for 1 

10 second. 

21. Here we see the complete picture. The 

mutual kick forces are equal and opposite. But 

they do not cancel each other, because they act 

on different bodies. 


22. Let us try to understand how this horse is 

able to pull the cart. 

23. The cart requires a force of 2500 N to 

accelerate. But there is a backward frictional 

drag on the wheels. Therefore the horse must 

pull the cart with a force of 3000 N. 

24. The horse needs 1500 N to accelerate. 

According to the 3rd Law, the cart is pulling 

horse back with force 3000 N. Therefore the 

ground must provide the necessary traction 

force of 4500 N in the forward direction. 

25. How does the horse get this 4500 N 

traction force? By pressing against the ground 

with this force, according to the 3rd Law. Also 

since the wheels are experiencing frictional 

drag of 500 N backward, the ground is at the 

same time dragged forward with 500 N, again 

due to the 3rd Law. The resulting acceleration 

of the ground is very small, since the mass of 

the earth is very LARGE. 


26. Now see the complete picture. For every 

force there is an equal and opposite reaction 

force. But the forces on any one of the three 

bodies are not in equilibrium. Therefore, the 

cart, the horse and the ground are all 

accelerating (though in different degrees). 

27. As final example we consider the motion of 

the earth and the moon due to the gravitational 

attraction between the two. Note the distance 

between the centres of the earth and the moon, 

and the masses of the earth and the moon. The 

gravitational force with which the earth is 

pulling the moon is 20.6 × 10 19 N. 

28. As a result the moon is accelerating toward 

the earth and is moving in a circle. 

29. The earth is experiencing a force of exactly 

the same magnitude of 20.6 × 10 19 N but 

directed toward the moon. However, the mass 

of the earth is about 83 times the mass of the 

moon. Therefore the earth is undergoing a 

much smaller acceleration of 3.44 × 10 −5 m/s 2 . 

As a result the centre of the earth is also 

moving in a circle, a much smaller circle. 


30. Both the moon and the earth are 

accelerating toward each other according to the 

Third Law of Motion. The centre of each is 

moving in a circle, one circle being much 

larger than the other because one mass is much 

smaller than the other. 

2 Free Body Diagrams 

Newton’s first law of motion is difficult to 

believe. Why should anyone believe that an 

object keeps moving forever along a straight 

line in the absence of external interferences, 

when our everyday encounters seem to 

contradict the statement? 

Similarly Newton’s third law of motion is 

rarely believed. It is easy to “imagine” equal 

and opposite action-reaction forces when, for 

example, a stone is resting on a table. But how 

to trust the same statement when the table 

breaks and the stone crashes down? 

The confusion is understandable, and 

persists. The reason − our apathy to explain the 

theory properly with effective drawings. For 

understanding concepts and principles of 

mechanics drawing freebody diagrams is an 

essential first step. Without this Newton’s laws 

of motion can never be appreciated. 

2.1 What is a Freebody Diagram? 

A freebody diagram (FBD) is a pictorial 

formulation of the mechanics problem through 

figures in which each object in a system of 

interacting bodies is shown separately and the 

forces acting on this body (and on this body 

alone) are clearly delineated and also shown 

separately. The slides presented in the previous 

section have shown many examples of FBDs. 

Some of the important ones are the following 

• A runner and the ground in Slide 9 (The 

gravity force has been suppressed); 

• A car and the ground in Slide 10; 

• A falling stone and the earth in Slides 16- 

17; 

• A football player, football and the ground 

in Slides 19- 21; 

• A cart, a horse pulling it, and the ground in 

Slides 23-26; 

• The earth and the moon in Slides 27-29. 

2.2 The Golden Rules for Understanding 

Newton’s Laws of Motion 

For understanding Newton’s laws of motion 

properly, especially their applications to 

extended objects, it is essential that the 

following golden rules be kept in mind. 

1. Clearly identify the “system” whose 

motion you are going to analyze. 

Sometimes it may be useful to draw a 

boundary around this system in order to 

identify it and isolate it from all 

neighbouring systems. 

2. All the forces that are coming from outside 

this system you have just identified − by 

this I mean the forces that are coming from 

outside the boundary you have drawn - are 

to be called external forces. Forces of 

“interaction” among objects inside the 

system that is, forces coming from objects 

that are confined within the boundary you 

have drawn − are all internal forces. 

3. Therefore draw a freebody diagram (FBD) 

of all the external forces. You must 

remember that FBDs are a must for 

understanding mechanics. It is impossible 


to appreciate the principles of mechanics, 

the forces and how they guide the motion 

of objects, without drawing FBDs. 

4. Add up vectorially all the external forces 

acting on the system. In some cases this 

vector sum may be zero. In that case a 

central point of the object, called the 

Centre of Mass, to be abbreviated as CM, 

will be either stationary or moving with 

constant velocity (i.e., constant speed 

along a straight line.) 

We shall refer to the motion of the CM as 

the bulk translatory motion of the system. 

If the vector sum of the external forces is 

non-zero, then that vector sum F total will be 

called the Resultant Force on the system. 

The bulk translatory motion of the system 

is determined by Newton’s second law of 

motion, which can now be written as 

M r = F total (2) 

where M is the total mass of all the component 

parts of the system, and r is the radius vector of 

the CM of the system, so that r is the 

acceleration of the CM. 

We have illustrated Rules 1,2,3 in Figures 

1. In the Top Box we have shown three objects 

A, B and C interacting with one another, and 

forming a system. We assume that this system 

of three objects is not subjected to any 

“external force”. Individually the objects A, B 

and C are moving and accelerating only due to 

the forces of interaction among themselves. 

In frame (a) we have depicted each one of 

them as a single system, by drawing 

boundaries (with dotted lines) around each. 

The forces of interaction are (F 1 ,F′ 1 ), (F 2 ,F′ 2 ), 

(F 3 ,F′ 3 ). The pair of forces within parentheses 

are the “equal and opposite” action-reaction 

pairs. For example if F′ 1 is the action force, 

then its counterpart F 1 is the corresponding 

(equal and opposite) reaction force. Similarly if 

F 3 is the action force, then its counterpart F′ 3 is 

the corresponding (equal and opposite) reaction 

force. Note also that we have drawn “shooting” 

lines on each one of the three objects indicating 

that the objects are all moving objects having 

velocities v a , v b , v c respectively. 

Each one of the three systems A,B,C is 

subjected to the following external forces: 

(1) external forces F 1 ,F 2 on A; 

(2) external forces F′ 1 ,F′ 3 on B; 

(3) external forces F′ 2 ,F 3 on C. 

In frame (b) we have clubbed A and B 

forming a single system AB. Now it is an 

interaction between two systems namely, AB 

and C. The only external force on AB is F′, and 

the only external force on C is F = −F′. 

In frame (c) we have taken A, B and C 

together as a single system ABC. There are no 

external forces on this system ABC. 

We shall now illustrate Rule 4 in the form 

of the following equations of motion. 

System Net external force on the system Eqn of Motion 

A F total = F 1 + F 2 m a ra 

= F1 + F 2 

B F total = F′ 1 + F′ 3 m b 

rb 

= F′ 1 + F′ 3 

C F total = F′ 2 + F 3 m c rc 

= F′ 2 + F 3 

AB F total = −F = −(F′ 2 +F 3 ) = F 2 + F′ 3 (m a +m b ) r ab 

= F′=−F 

ABC F total = 0 (m a +m b +m c ) r abc 

=0 

(3) 


Figure 1. Forces of interaction between systems of material objects. 

In the above table m a ,m b ,m c represent the 

masses of the systems A, B, C respectively, 

and r a , r b , r c , r ab , r abc represent the locations of 

CMs of the systems A, B, C, AB, ABC 

respectively. 

In the Bottom Box of Figure 1 we have 

shown a stone falling under the force of 

gravity. The bulk motion of the stone is 

determined by the gravity force mg. We had 

shown an FBD of this object in slide 16 of the 

previous section. However, we have repeated 

the FBD in frame (a). 

The stone is composed of a large number of 

atoms, of the order of, say, 10 25 . These atoms 

are pulling or pushing each other. Yet, they are 

all internal forces and, therefore, have no effect 

on the bulk motion of the stone. 

Now, “imagine” a boundary line Σ dividing 

the stone into parts A and B, which we shall 

refer to as A-stone and B-stone respectively. 

Then there are action-reaction forces (F,F′) 

between the two parts, which are always equal 

and opposite. We have shown the FBD of both 

the parts in frame (b). The CM of A-stone is 

moving under the effect of (1) the gravity force 

m a g, and (2) the force F on it coming from B. 

These two forces cause the falling of the A- 

stone under gravity, along with rotational 

motion of the CM. 1 

In the same manner the CM of the B-stone 

is moving under the effect of (1) the gravity 

force m b g, and (2) the reaction force F′ = −F on 

it coming from A. These two forces together 

cause the falling under gravity along with 

rotation of the CM of B. 

We shall now apply Rule 4 to the falling 

stone and its two parts and obtain the following 

equations of motion. 

1. There can also be oscillatory motion of the CMs 

of A and B, due to vibration induced by the forces F 

and F′ respectively. 


System Net external force on the system Eqn of Motion 

Whole stone F total = Mg M r = Mg 

A-stone F total =m a g+ F m a r a = m a g+F 

B-stone F total = m b g − F m b r b = m b g+F′ = m b g−F 

(4) 

In the above table (M = m a + m b , r), (m a , 

r a ), (m b , r b ) represent (mass, location of the 

CM) of the whole stone, A-stone, B-stone 

respectively. 

The reader should note here an important 

point. The only external force on the whole 

stone is the gravitational force. Hence the point 

C (representing the CM of the stone) falls in a 

parabolic path like an ordinary projectile. On 

the other hand A-stone and B-stone are 

subjected to extra forces F and −F 

respectively, besides the force of gravity. Their 

CMs C a and C b follow “nonparabolic” paths. 

3 Further Examples of FBDs 

In Figures 2-5 we have cited a number of 

instances in which the third law presents us 

with a paradox. We have illustrated graphically 

each situation by identifying the forces with 

FBDs, and in two cases listing them in a 

tabular form. This, we hope, will facilitate 

clarification of the Second and the Third Laws 

of Motion. 

We have presented in Tables 1 and 2 the 

“Forces at Work” corresponding to two of the 

examples cited, namely, (a) Diving, and (b) 

Tug of War. In the right column of each table 

you will see the forces on the subsystems. The 

action-reaction forces on these subsystems 

taken together add up to zero, so that the net 

force on the “Total” system is nil. 

In writing the vectors in Table 2 we have 

taken the forces T 1 ,T 2 ,F 1 ,F 2 to be in the 

direction of the positive X axis (i.e., 

rightward), and the forces N 1 ,N 2 ,M 1 g,M 2 g to be 

in the direction of the positive Z axis (i.e., 

upward). 

4 Every Real Force has a Parent 

Newton’s first law of motion gives a 

qualitative meaning of force. A force is 

something which makes a particle, a system, 

deviate from its “most natural” motion, 

namely, moving in a straight line with constant 

velocity. The second law pinpoints the concept 

further by painting force in the colour of a 

vector, a directed straight line with a specified 

length − equal to the rate of change of 

momentum of the object. The third law refines 

the concept of force further by making the 

grand statement, “force is an interaction 

between two objects in which each one is 

equal.” 

The earth pulls a mosquito with the same 

force with which the mosquito pulls the earth. 

A proton pulls an electron with the same force 

with which the electron (about two thousand 

times lighter) pulls the proton. These 

statements are as profound as the shock and 

disbelief they generate in our minds. Yet the 

statements are true. 

However, the reader must not miss a crucial 

assumption in the statements of the laws of 

motion. Velocity, momentum, acceleration are 

frame dependent quantities. They have no 

meaning unless you have specified a frame of 

reference. What frame? The frame in which the 

statements are valid. 

That reads like the old joke: which came 

first, the egg or the chicken? 

However, the assumption of an inertial 

frame is not an absurdity. The concept of a 

family of inertial frames (the name of the 

frames in which the laws of motion hold) lies 

at the foundation of classical mechanics. 


Figure 2: Free body diagrams of diving. The action-reaction pair of forces are shown with the same 

roman bold letter appearing twice, once with prime and once again without it. The forces at play are 

the tension forces (T,T′) between the board and the left pillar; the compressive (pressure) forces 

(P,P′) between the board and the left pillar; the force R′ with which the diver kicks the board and the 

kickback reaction force R he receives, the gravitational forces (G 1 ,G′ 1 ), (G 2 ,G′ 2 ), (G 3 ,G′ 3 ), (G 4 ,G′ 4 ). 

Drawings on Left show Normal Diving. Drawings on Right show accident in diving. The board gets 

uprooted from one of the supporting pillars, and the diver tumbles into the water. In this case the 

pair of forces T,T′ disappear, and they disappear simultaneously. 


Figure 3. We have now written each pair of action reaction forces with the same symbol (even 

though they have opposite signs.) For example (T 1 ,T 1 ) are action-reaction forces of tension. Top : 

Tug of War. Note that the tension forces T 1 ,T 2 are not equal in magnitude, because the rope has 

considerable mass m and it is accelerating. The contestant on the right is pushing the ground with 

friction force F 1 , and a downward force N 1 , and at the same time pulling the earth upward with a 

gravitational force M 1 g. The earth is repaying him back with the same force in opposite direction. 

Bottom : Boy jumping off a boat. In Figure (b) we have shown the forces acting on the “boat + boy” 

system. Figures (c) and (d) show the FBDs of the boat and the boy separately. We have removed 

the earth from our “system”. 


Figure 4. Top: FBD of an unfortunate man who is sinking into a quicksand. He sinks because the 

gravitational force downward is larger in magnitude than the upward reaction force provided by the 

ground. We have not shown the action-reaction counterpart of this gravitational force (pulling the 

earth upward.) Bottom : Action-reaction forces between a balloon and the escaping gas. We have 

marked out a rectangular column of the gas which is pushed downward with a gas-pressure force 

of magnitude F 1 . This causes an upward reaction force of the same magnitude acting on the 

balloon. The balloon accelerates upward because this reaction force F 1 is larger than the weight of 

the balloon (same as the downward force of gravity on it. The lateral pressure forces F 2 , F 3 , F 4 , F 5 

are balanced causing no sidewise acceleration of the gas. Their reaction counterparts are similarly 

balanced causing no sidewise acceleration of the balloon. 


Figure 5. A locomotive steam engine hauling a train consisting of only two coaches. We assume 

that the train is moving with constant speed, so that the net force on the train is zero. (a) General 

picture. The train is moving from right to left as indicated by the arrow at the front and by the 

direction in which the Driving Wheels are turning. (b) FBD of the engine and the coaches. Note that 

the driving wheels, connected to the piston (reciprocating inside the steam chamber), shown as a 

box) by the connecting rod, are pressing the rails backward thereby getting the traction force F t , 

directed forward. It is this traction force that moves the train. The other wheels are subjected to 

rolling frictions F r1 ,F r2 ,F r3 ,F r4 , directed backward, and trying to slow down the train. T 1 ,T 2 are 

tensions in the chains that connect the engine to coach 1, and coach 1 to coach 2 respectively. (c) 

FBD of the whole train. (d) FBD of the rails. We have suppressed the gravity forces. In the boxes 

we have indicated the force equalities. When the train accelerates, the equality symbol “=” 

becomes “greater than” symbol “>”. 


Table 1. Forces at work in diving 

System External force on the system 

Diver R+G 1 

Diving board 

T+G 2 +P+R′ 

Left pillar T′+G 4 

Right pillar P′+G 3 

Earth G′ 1 +G′ 2 +G′ 3 +G′ 4 

Total 0 

(5) 

Table 2. Forces at work in tug of war 

System Net external force on the system 

Player-right 

−T 1 +F 1 +N 1 −M 1 g 

Player-left 

T 2 −F 2 +N 2 −M 2 g 

Rope T 1 −T 2 

Earth 

−F 1 +F 2 −N 1 −N 2 +M 1 g+M 2 g 

Total 0 

(6) 

Anyone who applies Newton’s laws of 

motion must be standing in an inertial frame. 

The space capsule in which the modern day 

heroes − the astronauts − are floating is the best 

example of an inertial frame. In contrast, the 

earth is not. 

What are the forces acting on a missile shot 

out from an earth station? Along with the main 

force, the force of gravity, there are other 

forces, the centrifugal force and the Coriolis 

force. These “other forces” are not forces of 

interaction. And Newton’s third law of motion 

does not apply to them. 

Imagine yourself sitting in a fast 

accelerating train (better still imagine yourself 

sitting inside a jet plane as it accelerates along 

the runway before taking off.) You feel as if 

someone is pushing you backward, against the 

chair in which you are sitting. Let us refer to 

such forces as force type l. 

Think of another sensation. You are on a 

rotating platform of a merry −go-round − your 

first stop on entering a carnival park. If you are 

sitting on a chair on the go-round, you feel that 

someone is trying to throw you outward, away 

from the centre, but the chair is saving you 

from such disaster. Even stranger will be your 

feeling if you try to walk on the platform. 

Besides the outward “throw out” force, called 

the centrifugal force, you will also feel a 

“transverse force” (i.e. perpendicular to the 

direction you are walking in) making you feel 

dizzy. This sideward force is called Coriolis 

force. Let us refer to these two forces as force 

types 2 and 3) 

The force types 1,2 and 3, are not forces of 

action-reaction. They completely disappear 

when the platform you are on (train, aeroplane, 

merry-go-round) comes to a stop. People club 

such forces under fancy names : fictitious 

forces, pseudo-forces. 

The pseudo forces are clearly distinct from 

the real forces. 

All real forces have parents. They originate 

from some material source, and reach out to 

other material objects altering their motions in 

accordance with Newton’s laws. 

In contrast, the pseudo-forces do not have 

parents, i.e., they do not originate from 

material sources. We mistake them to be forces 


ecause we are always thinking our frames to 

be inertial even when they are not. 

Physicists do not feel shy of applying 

Newton’s laws of motion even when sitting in 

noninertial frames. They make up for their 

errors by inventing the fictitious pseudo-forces. 

5 Mach’s Principle 

Pseudo-forces are not as untouchable, bereft of 

respectability, as their category may indicate. 

They have a trail of glorious history. It was 

Ernst Mach’s deep insight that brought 

pseudoforces on par with real forces by 

suggesting that the pseudo-forces originate 

from the mighty galaxies of an accelerating 

universe (when your frame is accelerating with 

respect to an inertial frame, the universe is also 

accelerating backward with respect to your 

frame.) Albert Einstein was so influenced by 

this profound message that he removed all 

distinctions between inertial and non-inertial 

frames and proposed a theory of gravitation in 

which pseudo-forces are no different from 

gravitational forces. 

Acknowledgment 

The author gratefully acknowledges the help 

provided by Michael Murphy and Professor 

A.V. Gopala Rao help in installing Linux for 

preparation of the manuscript. 

References 

There is one full chapter (18 pages long) on FBDs, 

with a wide variety of examples and illustrated 

drawings in the following book which the author 

found in the “References” of the article “Friction” in 

Wikipedia. http://en.wikipedia.org/wiki/Friction. 

1. Andy Ruina, Rudra Pratap, Introduction to 

Statics and Dynamics, Pre-print for Oxford 

Univ Press, January 2002. 

Books on engineering mechanics usually contain 

many examples of FBDs. See for example the 

following book. 

2. F.P. Beer, E.R. Johnston, Vector Mechanics for 

Engineers, McGraw Hill, N.Y. (1962). 

Rolling friction and traction forces are well 

explained in 

3. A.N. Mateev, Mechanics and Theory of 

Relativity, Mir Publishers, Moscow (1989). 

The drawings in this article were all drawn by the 

author. In particular the drawings appearing in 

Section 1 (Slide Show) and the drawings appearing 

in Fig. 2 were published in this journal in 1986. 

4. S. Datta, A Film Strip on Newton’s Third Law 

of Motion, Physics Education, Vol 2, No. 4 

(January-March 1986), pp 30-40. 

The drawings appearing in Figures 3 and 4 were 

part of a Teacher’s Guide handwritten by the author 

in December 1989 at Kavaratti, capital of 

Lakshadweep islands, as part of a teacher training 

programme organized by the Regional College of 

Education (NCERT), Mysore.

Newton's Third Law of Motion

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