Newton's Third Law

Push against a wall as hard as you like — the wall pushes back, exactly as hard, every single time. Slap the water while swimming and the water slaps your hand back. Fire a rifle and the stock kicks into your shoulder with the same force that hurled the bullet forward. This stubborn, universal symmetry is Newton's Third Law, and it is arguably the most misunderstood of the three laws precisely because it sounds so simple: "to every action there is an equal and opposite reaction." The danger is that the word "action" tempts you to picture one force causing another a moment later, like a domino toppling the next. Newton meant something stranger and cleaner: the two forces are not cause and effect at all. They are two faces of a single interaction, born together, dying together, and — this is the part exam papers hunt for — always acting on two different bodies. Because they act on different bodies, they never cancel each other, and that single fact is the seed from which the entire idea of momentum conservation grows. This article pins the Third Law down precisely: what counts as a genuine action-reaction pair, why a book resting on a table involves forces that look similar but are not partners at all, how the law explains walking, swimming, rocket propulsion and recoil, and how it connects forward to the conservation of linear momentum. Expect crisp definitions, a labelled force diagram, one full derivation connecting the Third Law to momentum conservation, and the traps that cost CBSE students marks year after year.

Key takeaways

Newton's Third Law: for every force one body exerts on another, the second body exerts an equal and opposite force on the first.
The two forces of a pair act on different bodies, at the same instant, and are of the same type (both contact, or both gravitational).
Because the two forces act on different bodies, they never cancel each other — cancellation only happens between forces acting on the same body (equilibrium).
A book's weight and the normal force on it are NOT a Third-Law pair — both act on the book; they merely balance.
The Third Law is the physical reason walking, swimming, and rocket propulsion work: you push something, it pushes you back, and that reaction accelerates you.
The Third Law is the seed of the law of conservation of linear momentum: internal forces between two bodies always come in equal-and-opposite pairs, so they cannot change the total momentum of the system.
Key idea: every force comes paired with an equal, opposite, simultaneous force exerted by the second body back on the first, the two never cancel because they act on different bodies, and this pairing is exactly what makes total momentum conserved when no external force acts.

The core insight

If body A exerts a force on body B, body B simultaneously exerts an equal and opposite force of the same kind back on A, and because the two forces act on different bodies they never cancel — a fact that directly forces the total momentum of an isolated system to stay constant.

Where it fits in the syllabus

Newton's Third Law sits inside CBSE Class 11 Physics Unit III, "Laws of Motion," directly after the First and Second Laws and just before the law of conservation of linear momentum, which it exists to justify. The board expects you to state the law precisely, identify genuine action-reaction pairs in everyday situations (walking, swimming, recoil, a block resting on a surface), and clearly distinguish a Third-Law pair from a pair of balanced forces acting on one body. This is qualitative-plus-simple-numerical material — no calculus is required, and the arithmetic never goes beyond $F = ma$ applied to isolated bodies.

The physical picture

Newton's Third Law is written most compactly as a statement about two forces between two bodies. If body A exerts a force on body B, call it $\vec{F}_{BA}$ (read as "force ON B, DUE TO A"), then body B simultaneously exerts a force back on A, $\vec{F}_{AB}$ , and the law says

\vec{F}_{AB} = -\,\vec{F}_{BA}.

where: $\vec{F}_{BA}$ is the force exerted on B by A, and $\vec{F}_{AB}$ is the force exerted on A by B; the minus sign means the two vectors have equal magnitude but point in opposite directions. The physical reading is that these two forces are never separated in time — there is no delay between "action" and "reaction," they switch on and off together, as if they were two views of a single push-pull event rather than two separate events. The subtlety that decides most exam marks is this: the two forces of a pair act on two different bodies. $\vec{F}_{BA}$ acts on B; $\vec{F}_{AB}$ acts on A. Because they act on different bodies, you can never add them together to find a net force on a single object — they belong in two separate free-body diagrams, one for A and one for B. This is precisely why they can never cancel: cancellation (balancing to zero) only makes sense for forces that act on the same body.

Figure

A book resting on a table showing the true Newton third law pair between book and Earth alongside the separate balanced pair of weight and normal force acting on the book alone

Consider the book resting on the table. Two forces act on the book: its weight $W$ (pulled down by the Earth) and the normal reaction $N$ (pushed up by the table). Because both act on the same body and the book is not accelerating, $N = W$ — they balance, an application of the First Law, not the Third. The true Third-Law partner of $W$ is the gravitational pull the book exerts on the Earth, upward, invisible only because the Earth's mass is so huge that its resulting acceleration is unmeasurably tiny. The true Third-Law partner of $N$ is the force the book exerts downward on the table. Four forces exist in this picture, in two genuine pairs, but only two of them (both acting on the book) ever get compared for balance.

Newton's third law and momentum

Derivation AThird Law forces cannot change the total momentum of a two-body system

The goal is to show, using only the Second and Third Laws, why the internal forces two bodies exert on each other during a collision or any interaction can never alter the total momentum of the pair — a result that is the direct ancestor of the conservation of linear momentum. We use only vector addition and the definition $F = dp/dt$ , with no assumption about what kind of force is involved (contact, spring, gravity) beyond the fact that it obeys Newton's Third Law.

Step 1. Consider two bodies, A and B, isolated from every other force (no external push, no friction, nothing outside the pair). Let A exert a force $\vec{F}_{BA}$ on B, and by the Third Law B exerts $\vec{F}_{AB} = -\vec{F}_{BA}$ on A.

Step 2. Apply Newton's Second Law to each body separately. For A, the rate of change of its momentum equals the only force acting on it: $\dfrac{d\vec{p}_A}{dt} = \vec{F}_{AB}$ . For B, similarly, $\dfrac{d\vec{p}_B}{dt} = \vec{F}_{BA}$ .

Step 3. Add the two rates of change together to get the rate of change of the total momentum of the system, $\vec{p} = \vec{p}_A + \vec{p}_B$ . This gives $\dfrac{d\vec{p}}{dt} = \dfrac{d\vec{p}_A}{dt} + \dfrac{d\vec{p}_B}{dt} = \vec{F}_{AB} + \vec{F}_{BA}$ .

Step 4. Substitute the Third Law relation from Step 1: $\vec{F}_{AB} + \vec{F}_{BA} = -\vec{F}_{BA} + \vec{F}_{BA} = 0$ . So $\dfrac{d\vec{p}}{dt} = 0$ , meaning the total momentum $\vec{p}_A + \vec{p}_B$ stays exactly constant however the two bodies push, pull, collide with, or recoil from each other, as long as no external force intrudes. Verified: for any pair with $\vec{F}_{AB} = -\vec{F}_{BA}$ , the sum $d\vec{p}_A/dt + d\vec{p}_B/dt$ collapses to zero by direct substitution — the algebra is exact, not approximate, and holds for every instant of the interaction, not just before and after.

This is the deep reason the Third Law is taught right before momentum conservation: internal forces, being equal-and-opposite pairs by the Third Law, always cancel out of the total momentum bookkeeping, even though they very much do not cancel out of each individual body's own momentum. It is also why the recoil of a gun and the qualitative behaviour of a rocket both trace back to this law — the gun pushes the bullet one way, the bullet's reaction pushes the gun (and shooter) the other way, and the two momentum changes are equal and opposite.

Worked examples

Board level - identifying the reaction force in a swimmer's push

A swimmer of mass $60\ \text{kg}$ pushes horizontally against the pool wall with a force of $120\ \text{N}$ . Find the swimmer's acceleration immediately after the push, and state clearly which force in the Third-Law pair is responsible.

Step 1. Identify the pair. The swimmer pushes the wall with $120\ \text{N}$ ; by the Third Law the wall pushes back on the swimmer with an equal and opposite force of $120\ \text{N}$ , directed away from the wall.

Step 2. Decide which force accelerates the swimmer. Only forces acting on the swimmer enter the swimmer's own $F = ma$ equation. The swimmer's push acts on the wall, not on the swimmer, so it is irrelevant to the swimmer's motion; only the wall's reaction, acting on the swimmer, counts.

Step 3. Apply the Second Law to the swimmer using the reaction force: $a = \dfrac{F}{m} = \dfrac{120}{60} = 2.0\ \text{m/s}^2$ , directed away from the wall. The swimmer accelerates at $2.0\ \text{m/s}^2$ , produced entirely by the wall's reaction force, not by the swimmer's own push.

Competitive level - two blocks in contact, finding the internal Third Law pair force

Two blocks, A ( $4.0\ \text{kg}$ ) and B ( $2.0\ \text{kg}$ ), sit touching on a frictionless floor. A horizontal push of $12\ \text{N}$ is applied to A, driving both blocks together. Find the common acceleration, the contact force between the blocks, and identify the Third-Law pair among the internal forces.

Step 1. Treat A and B as a single combined system to find the common acceleration, since the applied $12\ \text{N}$ is the only external horizontal force on the pair: $a = \dfrac{F}{m_A + m_B} = \dfrac{12}{4.0 + 2.0} = 2.0\ \text{m/s}^2$ .

Step 2. Isolate block B alone. The only horizontal force on B is the contact push $N$ from A, so by the Second Law $N = m_B a = 2.0 \times 2.0 = 4.0\ \text{N}$ .

Step 3. Check using block A. The forces on A are the applied $12\ \text{N}$ forward and B's reaction push backward on A, of magnitude $N$ (by the Third Law, since A pushes B with $N$ forward, B pushes A with $N$ backward). Net force on A is $12 - N = m_A a$ , so $12 - 4.0 = 8.0 = 4.0 \times 2.0$ , which checks out exactly.

Step 4. Name the pair explicitly. "A pushes B forward with $4.0\ \text{N}$ " and "B pushes A backward with $4.0\ \text{N}$ " are the two members of the genuine Third-Law pair — equal magnitude, opposite direction, acting on two different blocks, and neither one is the $12\ \text{N}$ applied force (which acts only on A, from an outside agent, and has no Third-Law partner within this two-block system). The common acceleration is $2.0\ \text{m/s}^2$ and the contact force is $4.0\ \text{N}$ . Timing note: this pattern (system-then-isolate-one-body) is standard in competitive papers and should take under two minutes. The fast check is always: does my "reaction force" act on a different body than the original force, and is it the same type of force? If either answer is no, you have picked the wrong partner.

Common traps

Treating an action-reaction pair as a pair of balanced forces. The two forces of a Third-Law pair act on different bodies and therefore never cancel. Only forces acting on the same body can balance each other to give zero net force.
Pairing the weight and normal force on a book as a Third-Law pair. Both act on the same body (the book), so they can balance but are not partners. The weight's true partner is the book's pull on the Earth; the normal force's true partner is the book's push down on the table.
Believing the heavier or stronger body exerts a bigger force. In any collision, however unequal the masses, the two bodies exert exactly equal and opposite forces on each other. The lighter body simply accelerates more, because $a = F/m$ with the same $F$ but smaller $m$ .
Adding a "centrifugal force" to balance things in an inertial frame. That belongs to circular motion, not the Third Law, and should never appear when you are working entirely from the ground (an inertial frame); real forces here are only weight, normal reaction, tension, spring force and friction as needed.
Forgetting that Third-Law pairs must be the same type of force. A contact push pairs with a contact push; a gravitational pull pairs with a gravitational pull. You cannot pair a normal reaction with a weight, because one is a contact force and the other is gravitational.

You should be able to…

State Newton's Third Law precisely and write it as $\vec{F}_{AB} = -\vec{F}_{BA}$ .
Identify both members of a genuine action-reaction pair and name the two different bodies each force acts on.
Explain, using a labelled example, why weight and normal force on the same body are balanced but are not a Third-Law pair.
Use the reaction force (not the applied force) to find the acceleration of the body that is pushing against something else.
Derive, using $F = dp/dt$ , why Third-Law force pairs cannot change the total momentum of an isolated two-body system.
Recognize when a "centrifugal force" argument is being wrongly smuggled into an inertial-frame Third-Law question.

Ask the tutor: if you are still unsure whether a pair of forces you have spotted is a genuine Newton third law pair or just two balanced forces on the same body, open the AI tutor from this page and walk through your specific example together.

If action and reaction forces are always equal and opposite, why does anything ever move at all?

Because the two forces act on two different bodies, so each body only feels the force acting on it, not the one it exerts on something else. When you push off a wall, the wall's push back on you is a real, unbalanced force on your body (nothing else is pushing you the opposite way at that instant), so you accelerate. Meanwhile your push acts entirely on the wall, which is anchored to the Earth and does not move. Motion happens because we always analyze one body's own set of forces, never lump a body's push on something else together with the forces acting back on that body.

Is the normal force always equal to the weight, since they look like a pair?

No, and this trips up many students. Weight and normal force are equal only when the surface is horizontal and the body is not accelerating vertically — that is a case of balanced forces on one body, not a Third-Law statement. On an inclined surface, or when the body accelerates vertically (like in a lift), the normal force differs from the weight. The genuine Third-Law partner of the weight is always the pull the body exerts back on the Earth, which is never equal to the normal force from a table.

How does the Third Law lead to conservation of momentum, exactly?

Whenever two bodies interact only with each other, Newton's Second Law says each one's momentum changes at a rate equal to the force acting on it. The Third Law guarantees those two forces are equal and opposite at every instant, so the two rates of change of momentum are also equal and opposite, and they cancel when you add them up for the whole system. That means the total momentum of the two bodies together never changes during the interaction, however violent the push, pull, or collision — this is exactly the law of conservation of linear momentum for an isolated system.

Does the Third Law apply to non-contact forces like gravity too, or only to pushes and pulls?

It applies to every kind of force, contact or non-contact. Gravity is a perfect example: the Earth pulls you down with your weight, and by the Third Law you pull the Earth up with an equal and opposite force. You never notice the Earth accelerating toward you because its mass is enormous ( $a = F/m$ gives a minuscule acceleration for such a huge $m$ ), but the force pair is exactly as real and exactly as equal as any contact push. The same logic applies to magnetic and electric forces between two charged or magnetic bodies.