Newton's Third Law of Motion

Push against a wall and something strange happens: the wall pushes back, exactly as hard, in the opposite direction. Nobody arranged this — it is a rule that every force in the universe obeys. Walk, swim, fire a gun, launch a rocket, and the same silent partnership is at work: one push always comes paired with an equal, opposite push in return, on a different body. Newton's Third Law of Motion names this pairing precisely, and it is the last of the three laws in the Class 9 chapter Force and Laws of Motion, following inertia (First Law) and F=maF = ma (Second Law). This article restates the law carefully, shows why action and reaction forces never cancel despite being equal and opposite, works through the classic wall-push and boy-throwing-a-ball scenarios with full numbers, and connects the law — qualitatively and then numerically — to the conservation of momentum that closes out the chapter.

The core insight

Every force comes in a pair — the reaction is always equal in magnitude and opposite in direction to the action, but the two forces act on two different bodies, so they never cancel and, over an interaction, they conserve the total momentum of the system.

The core idea

Newton's Third Law of Motion: to every action there is an equal and opposite reaction, and the action and reaction act on two different bodies. If body A exerts a force on body B, then B simultaneously exerts an equal force on A, directed the opposite way. Written as a formula, with A pushing B:

F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}

where: F⃗AB\vec{F}_{AB} is the force exerted by A on B (the "action") and F⃗BA\vec{F}_{BA} is the force exerted by B on A (the "reaction"); the minus sign carries the "opposite direction" and the equal magnitudes are implied by the equals sign. Both forces exist at the same instant — there is no first mover and no delay between action and reaction. Two features decide almost every exam question on this law:

  • Equal magnitude, opposite direction. The two forces are always identical in size. There is no such thing as a "weaker" reaction — a fly striking a windshield exerts a force on the glass exactly equal to the force the glass exerts back on the fly, even though the visible effects are wildly different because the fly's mass is tiny.
  • Acting on two different bodies. This is the detail examiners target hardest. The action acts on B; the reaction acts on A. Because the two forces sit on two different objects, they can never add up to zero on either object alone — cancellation (balanced forces) only happens when equal and opposite forces act on the same body.

The figure below makes the "two different bodies" point concrete: a person pushing a wall feels a return push on themselves, not on the wall.

Figure
A person pushing a wall with an action force arrow toward the wall and a reaction force arrow of equal length pointing back toward the person, labelled as acting on two different bodies
Action on the wall, reaction on the person — never on the same body wall person action: 100 N on wall reaction: 100 N on person
Derivation AWhy forces come as pairs — from the Third Law to conservation of momentum

The Third Law is not just a rule about magnitudes — it is the reason momentum can be tracked from one interaction to the next; the goal here is to show, using only the Third Law and the Second Law, that whenever two isolated bodies interact, the total momentum of the pair before the interaction equals the total momentum after, which is the qualitative bridge to the numerical conservation-of-momentum problems that close this chapter.

Step 1. Let body A exert a force F⃗AB\vec{F}_{AB} on body B for the same duration Δt\Delta t that B exerts F⃗BA\vec{F}_{BA} on A — the two forces act simultaneously, for the same interval, by the Third Law.

Step 2. By the Third Law, F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}, so multiplying both sides by the common time Δt\Delta t gives F⃗AB Δt=−F⃗BA Δt\vec{F}_{AB}\,\Delta t = -\vec{F}_{BA}\,\Delta t.

Step 3. A force acting for a time Δt\Delta t produces a change in momentum, F⃗ Δt=Δp⃗\vec{F}\,\Delta t = \Delta \vec{p} (from the Second Law). So the change in B's momentum, Δp⃗B\Delta \vec{p}_B, equals −Δp⃗A-\Delta \vec{p}_A, the negative of the change in A's momentum.

Step 4. Rearranging, Δp⃗A+Δp⃗B=0\Delta \vec{p}_A + \Delta \vec{p}_B = 0, meaning the total momentum of the two-body system, p⃗A+p⃗B\vec{p}_A + \vec{p}_B, does not change during the interaction — provided no outside force acts on the pair. This is the law of conservation of momentum: in a collision or an explosion between two isolated bodies, total momentum before equals total momentum after, and it follows directly from the Third Law.

Worked example

A boy of mass 40 kg40\ \text{kg} stands at rest on frictionless ice and throws a ball of mass 2 kg2\ \text{kg} horizontally at 10 m/s10\ \text{m/s}. Find the boy's recoil speed, and identify the action–reaction pair responsible.

Step 1. Identify the action–reaction pair. Action: the boy's arms push the ball forward. Reaction: the ball pushes back on the boy's arms, equal in magnitude, opposite in direction, for the same instant of time.

Step 2. Apply conservation of momentum. Before the throw, the boy-and-ball system is at rest, so total momentum is zero: pinitial=0p_{\text{initial}} = 0. Taking the ball's direction as positive, after the throw the total momentum must still be zero, so mboy vboy+mball vball=0m_{\text{boy}}\,v_{\text{boy}} + m_{\text{ball}}\,v_{\text{ball}} = 0.

Step 3. Solve for the boy's recoil velocity. vboy=−mball vballmboy=−2×1040=−0.5 m/sv_{\text{boy}} = -\dfrac{m_{\text{ball}}\,v_{\text{ball}}}{m_{\text{boy}}} = -\dfrac{2 \times 10}{40} = -0.5\ \text{m/s}. The negative sign shows the boy moves opposite to the ball.

Step 4. Check the direction against the Third Law. The reaction force on the boy points opposite to the action force he applied on the ball, so his recoil in the direction opposite to the throw is exactly what the law predicts, not a separate coincidence. The boy recoils backward at 0.5 m/s0.5\ \text{m/s} — much slower than the ball's 10 m/s10\ \text{m/s} because his mass (40 kg40\ \text{kg}) is twenty times the ball's (2 kg2\ \text{kg}), the same pattern seen in a gun's recoil against a much lighter bullet.

Competitive stretch

If the same boy instead threw two identical 2 kg2\ \text{kg} balls one after another, each at 10 m/s10\ \text{m/s} relative to the ice, his final recoil speed would not simply double, because his own mass decreases slightly after the first throw (he is 2 kg2\ \text{kg} lighter while throwing the second ball) — a reminder that conservation of momentum must be applied in stages when mass is exchanged between bodies mid-problem, a subtlety Class 9 numericals avoid by using single-throw or single-collision setups only.

Common traps

  • Thinking action and reaction cancel out. They never do, because they act on two different bodies; cancellation (balanced forces) only happens for forces on the same body.
  • Assuming the "stronger" object only gives the action. Both forces are always exactly equal in magnitude, regardless of mass or apparent strength — only the resulting acceleration differs, because acceleration also depends on mass.
  • Treating action and reaction as sequential. They occur at the same instant, not one after the other.
  • Forgetting momentum conservation needs zero external force. The boy-and-ball total momentum stays zero only because the ice is frictionless; a rough floor would introduce an external friction force and the simple equality would not hold.

Ask the tutor: Want to see why a rocket keeps accelerating in empty space with nothing to push off, or try another recoil numerical with different masses? Ask the AI tutor.

If action and reaction forces are always equal, why does a bullet hurt more than the recoil hurts the shooter's shoulder?

The forces on the bullet and on the gun really are equal in magnitude at every instant, by the Third Law. What differs is the resulting acceleration, because acceleration depends on mass through F=maF = ma. The bullet has a tiny mass, so the same force gives it an enormous acceleration and a very high speed; the gun (and shooter) has a much larger mass, so the identical force gives a much smaller acceleration and a gentler recoil. Equal forces, very unequal outcomes — that is the mass difference showing up, not a violation of the law.

Why doesn't a person pushing a wall move, if the wall pushes back with an equal force?

If the person is standing firmly on ordinary ground with friction, the reaction from the wall is balanced by friction from the floor, so the net force on the person is zero and they stay still. But change the setup — put the same person on a frictionless skateboard, as in the classic scenario — and there is no floor friction to balance the wall's push, so the unbalanced reaction accelerates the person backward. The Third Law always supplies an equal and opposite reaction; whether the person moves depends on the net force acting on the person alone, which can include other forces besides the reaction.

Is conservation of momentum a separate law from Newton's Third Law, or the same thing?

They are connected but not identical. The Third Law is a statement about the forces two interacting bodies exert on each other. Conservation of momentum is a consequence that follows once you combine the Third Law with the Second Law (FΔt=ΔpF\Delta t = \Delta p): because the interaction forces are equal, opposite, and act for the same time, the momentum changes of the two bodies cancel, so the system's total momentum stays constant. In Class 9, momentum conservation is usually treated as its own named law, but the derivation above shows it is really the Third Law expressed in the language of momentum.

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