Newton's third law and action-reaction pairs

Push against a wall and the wall pushes back — hard enough that your hand feels it, even though the wall never moved. That silent shove back is Newton's third law in its purest form: forces never occur alone, they occur in matched pairs acting on two different bodies at once. Every walk you take, every rocket launch, every recoiling gun relies on this single idea, yet it is one of the most misquoted laws in physics — "equal and opposite" gets repeated without ever asking equal and opposite what, acting on whom. This article pins down the precise definition of an action-reaction pair, shows how the law falls directly out of momentum conservation for an isolated two-body system, and works through a recoil problem where the law becomes a genuine calculation, not just a slogan.

The core insight

Whenever body A exerts a force on body B, body B simultaneously exerts an equal and opposite force back on A, and because these two forces act on different bodies they never cancel within either body's own equation of motion.

The core idea

Newton's third law states that to every action there is an equal and opposite reaction, with the action and reaction forces acting on two different bodies, simultaneously, along the same line of action. If body A exerts F⃗AB\vec F_{AB} on body B, then B exerts F⃗BA\vec F_{BA} back on A such that

F⃗AB=−F⃗BA\vec F_{AB} = -\vec F_{BA}

where: F⃗AB\vec F_{AB} is the force exerted by A on B, and F⃗BA\vec F_{BA} is the force exerted by B on A. A genuine action-reaction pair must satisfy three conditions together: the two forces act on different bodies, they are equal in magnitude, and they are opposite in direction along the same line. Miss any one of these — most often the "different bodies" condition — and the forces in question are not a third-law pair at all, merely two forces that happen to balance on a single object. The law is not an independent postulate tacked onto mechanics; it is a direct requirement of momentum conservation. For an isolated two-body system with no external force, the total momentum p⃗A+p⃗B\vec p_A + \vec p_B stays constant, so its time derivative vanishes:

dp⃗Adt+dp⃗Bdt=0  ⟹  F⃗BA=−F⃗AB\frac{d\vec p_A}{dt} + \frac{d\vec p_B}{dt} = 0 \implies \vec F_{BA} = -\vec F_{AB}

where: dp⃗A/dtd\vec p_A/dt is the force on A due to B, and dp⃗B/dtd\vec p_B/dt is the force on B due to A. Momentum conservation for the isolated pair forces the mutual interaction to be equal and opposite — the third law is momentum conservation viewed from the two individual bodies rather than the system as a whole. The recurring puzzle students raise is: if every force is cancelled by an equal opposite force, how does anything ever accelerate? The resolution is that the two forces of a pair act on different objects, so each body's own equation of motion, F⃗net=ma⃗\vec F_{net} = m\vec a, includes only the force acting on that body, never both members of the pair together.

Figure
Two skaters push apart on ice with equal and opposite force arrows shown on each skater and equal length recoil velocity arrows pointing away from each other
Two skaters push apart — equal and opposite forces, different bodies A B F on A from B F on B from A recoil velocity of A recoil velocity of B

Derivation

Derivation ANewton's third law from conservation of momentum

The goal is to show, starting only from the definition of an isolated system's total momentum and the fact that it stays constant, that the mutual force one body exerts on another must equal the mutual force the second exerts back on the first, in magnitude and in the opposite direction — this is the third law derived, not merely asserted, and it is the reasoning ISC Class 11 expects for full marks on this topic.

Step 1. Set up an isolated system of two bodies A and B, with no external force acting on the pair. Their total momentum is p⃗=p⃗A+p⃗B\vec p = \vec p_A + \vec p_B, where p⃗A=mAv⃗A\vec p_A = m_A \vec v_A and p⃗B=mBv⃗B\vec p_B = m_B \vec v_B are the individual momenta.

Step 2. Since the system is isolated, its total momentum is conserved, meaning p⃗\vec p does not change with time. Differentiating, dp⃗dt=dp⃗Adt+dp⃗Bdt=0\dfrac{d\vec p}{dt} = \dfrac{d\vec p_A}{dt} + \dfrac{d\vec p_B}{dt} = 0.

Step 3. Identify each derivative as a force. By Newton's second law, dp⃗A/dtd\vec p_A/dt is the net force acting on A, which here is entirely due to B, so dp⃗A/dt=F⃗BAd\vec p_A/dt = \vec F_{BA}; similarly dp⃗B/dt=F⃗ABd\vec p_B/dt = \vec F_{AB}, the force on B due to A.

Step 4. Substitute into the Step 2 result: F⃗BA+F⃗AB=0\vec F_{BA} + \vec F_{AB} = 0, so F⃗BA=−F⃗AB\vec F_{BA} = -\vec F_{AB}. This is exactly Newton's third law — the two mutual forces are equal in magnitude, opposite in direction, and each acts on a different body, all falling out of momentum conservation alone. Verified: for the astronaut-and-box worked example below, 70×(−0.5)+10×3.5=−35+35=070 \times (-0.5) + 10 \times 3.5 = -35 + 35 = 0, confirming total momentum stayed zero as required by Step 2.

Worked example

An astronaut of mass 70 kg70\ \text{kg}, initially at rest in space, pushes off a stationary tool box of mass 10 kg10\ \text{kg} with a force that gives the box a velocity of 3.5 m/s3.5\ \text{m/s} away from her. Use Newton's third law together with momentum conservation to find the astronaut's recoil velocity, and confirm that the mutual forces during the push satisfy the third law.

Step 1. Before the push, the astronaut and the box are both at rest, so the total momentum of the isolated (astronaut + box) system is zero. During the push, the astronaut exerts a force on the box, and by the third law the box exerts an equal and opposite force back on the astronaut — an internal action-reaction pair that cannot change the system's total momentum.

Step 2. Apply conservation of momentum, taking the box's direction of motion as positive: mastvast+mboxvbox=0m_{ast}v_{ast} + m_{box}v_{box} = 0, so 70×vast+10×3.5=070 \times v_{ast} + 10 \times 3.5 = 0.

Step 3. Solve for the astronaut's recoil velocity: vast=−3570=−0.5 m/sv_{ast} = -\dfrac{35}{70} = -0.5\ \text{m/s}, meaning the astronaut recoils at 0.5 m/s0.5\ \text{m/s} in the direction opposite to the box.

Step 4. Check the third law directly. If the push lasted t=0.2 st = 0.2\ \text{s}, the force on the box from the astronaut is Fast→box=mboxabox=10×(3.5/0.2)=175 NF_{ast \to box} = m_{box}a_{box} = 10 \times (3.5/0.2) = 175\ \text{N}, and the force on the astronaut from the box is Fbox→ast=mastaast=70×(0.5/0.2)=175 NF_{box \to ast} = m_{ast}a_{ast} = 70 \times (0.5/0.2) = 175\ \text{N}, in the opposite direction. The two forces come out equal and opposite even though the resulting accelerations (and velocities) differ, because the two bodies have different masses. The astronaut recoils at 0.5 m/s0.5\ \text{m/s}, and the mutual push-off forces are equal in magnitude at 175 N175\ \text{N} each, confirming the third law.

Common traps

  • Pairing an object's weight with the normal reaction acting on the same object and calling them "action-reaction." A true third-law pair always spans two different bodies; weight and normal reaction on one body are merely a balanced pair (first-law equilibrium).
  • Believing that equal-and-opposite forces mean nothing can ever move. Each force in the pair acts on a different body, so the two bodies generally accelerate differently whenever their masses differ.
  • Assuming the third law needs physical contact. It applies equally to non-contact forces such as gravity — the Earth pulls a falling object down, and the object pulls the Earth up with an equal force.
  • Thinking the reaction force arrives after a delay. Action and reaction are simultaneous, never sequential, however small the time gap seems.

Ask the tutor: Not sure whether two forces in a diagram form a genuine third-law pair? Ask the AI tutor to help you check all three conditions on a specific example.

Why does a horse manage to pull a cart forward if the cart pulls back on the horse with an equal force?

The cart's backward pull acts on the horse, and the horse's forward pull acts on the cart — two different bodies, so neither force appears alone in the other's equation of motion. The horse also pushes backward against the ground, and by the third law the ground pushes forward on the horse's hooves with a reaction (friction) force. As long as that forward ground-reaction on the horse exceeds the backward pull of the cart, the horse accelerates forward, dragging the cart with it. The cart itself moves because the only horizontal force acting directly on the cart is the horse's forward pull, which exceeds the cart's own friction with the ground.

If gravity is a third-law pair too, does the Earth actually accelerate upward when I drop a ball?

Yes, in principle, though the effect is utterly unmeasurable. The falling ball pulls the Earth upward with a force equal in magnitude to the Earth's pull on the ball, mgmg. By F=MaF = Ma, the Earth's acceleration is mg/MEarthmg/M_{Earth}, where MEarth≈6×1024 kgM_{Earth} \approx 6 \times 10^{24}\ \text{kg} is enormously larger than any everyday mass mm, so the Earth's resulting acceleration is vanishingly small — far too tiny to detect, even though the third law guarantees it is exactly nonzero.

Does the third law mean the net force on a system of two interacting bodies is always zero?

Only if the two bodies are the entire system and there is no external force. Internal action-reaction pairs between the two bodies always cancel when you add up the forces on the whole system, which is exactly why an isolated system's total momentum is conserved. But if you look at either body alone, only the force acting on that body counts in its own equation of motion, and that single force is generally not zero — which is precisely why individual bodies in a pair can still accelerate.

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