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Physics · Laws of Motion

Newton's Third Law & Action–Reaction Pairs

Newton's second law tells you what a force does. The third law tells you where every force comes from: another body. Forces are always mutual, always paired, always equal and opposite — and always on two different bodies. This page unpacks the NCERT statement \(\vec F_{AB} = -\vec F_{BA}\), works the billiard-ball example NEET likes to reuse, and clears the four myths that swallow easy marks every year.

Where do forces come from?

Newton's second law, \(\vec F = m\vec a\), tells you the effect of a force, but never where the force itself originates. That is the job of the third law. NCERT's framing is precise: the external force on a body always arises due to some other body. Every push is somebody pushing something, and that "somebody" is always a second material body.

The third law: forces always come in pairs

Press your palm against a spring; the spring pushes back. The interaction is not one-way — it is a mutual exchange between two bodies. NCERT formalises this as the third law:

To every action, there is always an equal and opposite reaction.

The famous phrasing invites trouble in exam halls, so NCERT recommends the cleaner restatement: the force on body A by body B is equal in magnitude and opposite in direction to the force on body B by body A. Symbolically,

\[ \vec F_{AB} = -\vec F_{BA} \]

Two consequences sit inside this equation. Forces never come alone — every interaction has two arrows. And the Earth's gravitational pull on a falling stone has a partner: the stone tugs the Earth upward with exactly the same force. We do not see the Earth move because its mass is so vast that the acceleration is unmeasurably small, but the force itself is symmetric.

Three things NCERT warns about

NCERT flags three subtle errors that every NEET aspirant must internalise before touching a free-body diagram.

1. "Action" and "reaction" are just two names for "force"

The traditional words suggest two categories of thing. They are not. Both are forces; both obey the second law. NCERT's preferred phrasing — "force on A by B" and "force on B by A" — drops the loaded vocabulary entirely. Use it whenever you can.

2. Action does not precede reaction — they are simultaneous

The third law is not a cause-and-effect chain. The two forces come into existence at the same instant and vanish at the same instant. Neither is the "trigger" for the other; either may be called the action. There is no built-in time ordering.

3. The pair acts on different bodies — they do not cancel on either body

The two forces in a third-law pair act on two different bodies. On body A's FBD only \(\vec F_{AB}\) appears; \(\vec F_{BA}\) belongs on B's FBD. Adding them and concluding the "net force on A is zero" is a serious error — they never appear on the same diagram.

They cancel only when you treat A and B as a single system: then they are internal forces and cancel pairwise. This is exactly why the second law works for systems and rigid bodies — internal forces always cancel, leaving only external forces to determine the motion of the centre of mass.

The billiard-ball worked example

NCERT Example 4.5 is the canonical NEET-style demonstration. It uses the second law to find the change in momentum of the ball and the third law to convert that into the force on the wall.

NCERT Example 4.5

Two identical billiard balls strike a rigid wall with the same speed \(u\) and rebound elastically (no loss of speed). Ball (a) strikes head-on along the normal; ball (b) strikes at an angle of \(30^\circ\) to the normal.
(i) What is the direction of the force on the wall in each case?
(ii) What is the ratio of the magnitudes of the impulses imparted to the two balls by the wall?

(i) Set up: let the outward normal be \(+x\). Find \(\Delta\vec p\) of the ball (force on ball, second law), then flip the sign for the wall (third law).

Case (a) head-on: initial \(p_x = -mu\), final \(p_x = +mu\). So \(\Delta p_x = +2mu\), \(\Delta p_y = 0\). Force on ball is along \(+x\); force on wall is along \(-x\) — normal to the wall.

Case (b) at 30° to normal: only the normal component flips. \(\Delta p_x = +2mu\cos 30^\circ\), \(\Delta p_y = 0\). Force on the wall is again along the normal. The naive "force comes in at 30°" is wrong.

(ii) Impulse ratio: \(|J_a| = 2mu\), \(|J_b| = 2mu\cos 30^\circ\). Hence

\[ \frac{|J_a|}{|J_b|} = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}} \approx 1.15. \]

The head-on collision delivers about 15% more impulse to the wall than the oblique one.

Action-reaction pairs in NEET-favourite scenarios

NCERT and NIOS list a recurring set of everyday situations where the third law is the whole story. NEET likes these because they sound like common sense until you write the FBD.

Walking and running

You do not "push yourself forward." Your foot pushes the ground backward; by the third law the ground pushes your foot forward. That forward force is friction. On ice the friction is too small to supply the reaction, and the foot slips.

Swimming

Your hands and legs push water backward; the water pushes you forward. The viscosity of water provides the mechanical handhold needed for the pair to develop a sizeable magnitude.

Rocket propulsion

Hot exhaust gases are expelled downward; by the third law, the exhaust pushes the rocket upward. The pair is between the rocket and its own exhaust — not between the rocket and the surrounding air. That is why rockets accelerate just fine, indeed better, in vacuum.

Recoil of a gun

The expanding gases push the bullet forward; the bullet pushes the gun backward through the same gases. This is recoil. Momentum conservation (a consequence of the second + third laws) gives the recoil speed, but the direction comes from the third law alone.

Jumping off a boat

When you jump forward off a boat, you push the boat backward; the boat pushes you forward. The boat slides backward in the water. Again, the pair lives on two different bodies.

Horse and cart

The classic "paradox" — if the horse pulls the cart, doesn't the cart pull the horse back equally, so how does anything move? — dissolves once you note that the two members act on different bodies. The horse moves forward because of a different pair: the horse pushes the ground backward through its hooves, and the ground pushes the horse forward. That second pair is the unbalanced external force on the horse–cart system. The harness pair is internal and cancels.

Mass asymmetry and the "we don't notice" effect

Equal forces do not mean equal accelerations. When the Earth pulls a 1 kg stone with \(\approx 10\) N, the stone pulls the Earth with the same 10 N. But the stone's acceleration is \(10\;\mathrm{m\,s^{-2}}\), while the Earth's is \(10 / (6 \times 10^{24}) \approx 1.7 \times 10^{-24}\;\mathrm{m\,s^{-2}}\) — undetectable.

The same logic explains why a fly hitting the windscreen of a fast bus dies while the bus is undisturbed. The forces are equal. The bus is so much more massive that its velocity barely changes. The fly, with negligible inertia, decelerates to zero in milliseconds — requiring a brutal force, which by the third law is exactly what it delivers to the windscreen.

Worked examples

Example 1 — NCERT Exercise 4.4

A particle of mass \(m\) attached to a string of length \(l\) is whirled in a horizontal circle at constant speed \(v\). The net force on the particle directed toward the centre of the circle is:
(i) \(T\) (ii) \(T - \dfrac{mv^2}{l}\) (iii) \(T + \dfrac{mv^2}{l}\) (iv) \(0\)

Answer: (i) \(T\). The only horizontal force on the particle is the string tension, pulling inward. \(mv^2/l\) is not a separate force — it is the value the net inward force must take to maintain circular motion. The second law gives \(T = mv^2/l\). The third-law partner is the particle's outward pull on the central peg.

Example 2 — Bullet and block

A bullet of mass \(0.05\) kg moving at \(200\;\mathrm{m\,s^{-1}}\) strikes a heavy block and is brought to rest in \(0.01\) s. Find the magnitude of the average force exerted by the bullet on the block during contact.

Step 1 — change in momentum of the bullet:

\[ \Delta p = m(v_f - v_i) = 0.05 \times (0 - 200) = -10\;\mathrm{N\,s}. \]

Step 2 — force on the bullet (second law):

\[ F_{\text{bullet}} = \frac{\Delta p}{\Delta t} = \frac{-10}{0.01} = -1000\;\mathrm{N}. \]

The negative sign says the force on the bullet is opposite to its motion — the block decelerates the bullet.

Step 3 — force on the block (third law): by \(\vec F_{AB} = -\vec F_{BA}\), the bullet exerts \(+1000\;\mathrm{N}\) on the block in the direction of motion. So the average force on the block is \(\boxed{1000\;\mathrm{N}}\) along the bullet's original direction. Single unknown \((F)\), single equation, no shortcuts needed.

Example 3 — Boy on a boat

A 50 kg boy jumps horizontally off a stationary 100 kg boat with a speed of 2 m/s relative to the ground. What is the boat's recoil speed?

No horizontal external force acts on the boy + boat system (water resistance neglected), so total momentum is conserved. Initial momentum is zero; final must be zero too:

\[ m_{\text{boy}} v_{\text{boy}} + m_{\text{boat}} v_{\text{boat}} = 0. \]

\[ 50(2) + 100 \, v_{\text{boat}} = 0 \quad\Rightarrow\quad v_{\text{boat}} = -1\;\mathrm{m\,s^{-1}}. \]

The boat moves at \(1\;\mathrm{m\,s^{-1}}\) in the direction opposite to the boy. The underlying mechanism is third-law: the boy pushes the boat backward, the boat pushes the boy forward, and the impulses are equal and opposite.

Quick Recap

Six facts to lock in

  • \(\vec F_{AB} = -\vec F_{BA}\) — equal magnitude, opposite direction, two different bodies.
  • The pair members never appear on the same FBD; they do not cancel on a single body.
  • No time ordering — action and reaction are simultaneous; either label is valid.
  • "Action" and "reaction" are just two names for force; prefer "force on A by B."
  • Equal force, unequal acceleration — the small body accelerates visibly, the large body almost not at all.
  • Internal forces of a system cancel in pairs; only external forces accelerate the centre of mass — that is the third law extended to bodies.

NEET PYQ Snapshot — Newton's Third Law

The third law surfaces directly in only one NEET PYQ in the recent bank — NEET 2018 Q.28 — and indirectly in any FBD-pair problem. We pair it with NCERT Example 4.5 (billiard balls) and NCERT Exercise 4.4 (single-unknown identification).

NEET 2018 Q.28

A block of mass \(m\) rests on a smooth inclined wedge of angle \(\theta\). The wedge is accelerated horizontally with acceleration \(a\). What relation between \(a\) and \(\theta\) keeps the block stationary on the wedge?

  1. \(a = g\sin\theta\)
  2. \(a = g\cos\theta\)
  3. \(a = g/\tan\theta\)
  4. \(a = g\tan\theta\)
Answer: (4) \(a = g\tan\theta\)

Why: In the wedge's frame, gravity \(mg\) acts down and a pseudo-force \(ma\) acts horizontally. For the block to stay put on the smooth incline, the resultant must be perpendicular to the incline, so \(\tan\theta = ma/mg\), giving \(a = g\tan\theta\). The third law connects this to the normal pair — the block's push on the wedge equals the wedge's normal on the block, on different bodies.

NCERT Example 4.5

Two identical billiard balls strike a rigid wall with speed \(u\); ball (a) head-on, ball (b) at \(30^\circ\) to the normal. Both rebound elastically. What is the ratio of impulses delivered to the two balls by the wall?

  1. \(1 : 1\)
  2. \(\cos 30^\circ : 1\)
  3. \(1 : \cos 30^\circ\)
  4. \(\sqrt{3} : 1\)
Answer: (3) \(1 : \cos 30^\circ\) (\(\approx 1.15 : 1\))

Why: Only the normal component of momentum is reversed in an elastic collision with a rigid wall. Case (a): \(|\Delta p| = 2mu\). Case (b): \(|\Delta p| = 2mu\cos 30^\circ\). Impulse equals \(|\Delta p|\), so the ratio is \(2mu : 2mu\cos 30^\circ = 1 : \cos 30^\circ\). The third law then converts the impulse on the ball into the equal-and-opposite impulse on the wall — in both cases directed along the normal to the wall.

NCERT Exercise 4.4

A stone of mass \(m\) tied to a string of length \(l\) is whirled in a horizontal circle at constant speed \(v\). The net force on the particle directed toward the centre of the circle is:

  1. \(T\)
  2. \(T - \dfrac{mv^2}{l}\)
  3. \(T + \dfrac{mv^2}{l}\)
  4. \(0\)
Answer: (1) \(T\)

Why: The only physical inward force on the particle is the string tension. The quantity \(mv^2/l\) is not a separate force — it is the value the net inward force must take to maintain circular motion. Setting \(T = mv^2/l\) is the application of the second law, not the addition of a new force. The third-law partner of \(T\) is the outward pull of the particle on the central peg or hand — on a different body.

FAQs — Newton's Third Law

The six questions NEET aspirants ask most about action-reaction pairs.

Why don't action and reaction cancel each other out?
Because they act on two different bodies. The force on A by B acts on A; the force on B by A acts on B. Cancellation in Newton's first law requires multiple forces on the same body. Action-reaction pairs only cancel when you treat A and B together as one system, where they become internal forces.
Does the heavier body in a pair always feel the larger force?
No. The force magnitudes are equal by the third law. What differs is the acceleration. When the Earth pulls a stone with force F, the stone pulls the Earth with the same force F; but the Earth's mass is so large that its acceleration F/M is unmeasurably small while the stone's F/m is g.
What is the third-law pair of the normal reaction on a book on a table?
The pair of the table pushing the book up (normal reaction R) is the book pushing the table down with the same force R. It is NOT the weight of the book. The weight (Earth pulling book) has its own pair: the book pulling the Earth upward.
Is gravity a third-law pair force?
Yes. Newton's law of gravitation gives equal-and-opposite mutual attractions. If Earth pulls an apple down with force mg, the apple pulls the Earth up with exactly mg. Same magnitude, opposite direction, different bodies — a textbook third-law pair.
How does a rocket work in vacuum if there's nothing to push against?
The rocket pushes the exhaust gases out backward; the exhaust pushes the rocket forward. The third-law pair is between rocket and its own exhaust, not between rocket and air. Rockets work better in vacuum precisely because there is no drag.
Why does pushing a wall not move it?
The wall pushes you back with the same force (third law). You don't move because friction and other forces on you balance the wall's push. The wall doesn't move because it's anchored to a much larger system (the building, the Earth) whose effective mass makes its acceleration negligible.
Related Topics
Laws of Motion — Parent chapter Full chapter overview: inertia, F=ma, third law, friction, circular motion, PYQs. Inertia & Newton's First Law Inertial frames, the meaning of "no net force," everyday inertia examples. Newton's Second Law Momentum, F = dp/dt, impulse, and the route from second law to F = ma. Conservation of Linear Momentum Derivation from second + third laws; isolated systems, collisions, recoil. Equilibrium of a Particle ΣF = 0 statics, concurrent forces, Lami's theorem, NEET-style setups. Free-Body Diagrams The drawing skill: isolating bodies, marking forces, spotting third-law partners. Friction: Static, Kinetic, Rolling Friction laws; the third-law pair of the walker's push; rolling without slipping. Pulley & Connected Body Problems Tension, constraints, Atwood machines, system-equation discipline. Circular Motion & Centripetal Force v²/r requirements; tension, friction, normal as centripetal providers. Banking of Roads Tilted track geometry; maximum safe speed with and without friction. Apparent Weight in Lifts Accelerating frames; normal reaction as the scale reading; free-fall feels weightless.