Action-reaction pairs · Equal and opposite forces · Applications in swimming, rockets and collisions
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State Newton's Third Law — describe the relationship between action and reaction forces.
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Identify force pairs — recognise that action-reaction forces act on different objects.
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Explain swimming — apply Newton's Third Law to a swimmer pushing against water.
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Explain rocket propulsion — describe how exhaust gases produce thrust via Newton's Third Law.
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Analyse collisions — use Newton's Third Law alongside momentum conservation in collisions.
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Avoid common errors — distinguish Newton's Third Law pairs from balanced forces on a single object.
What is Newton's Third Law?
Newton's Third Law: When object A exerts a force on object B, object B exerts an equal and opposite force on object A. These forces are the same type, equal in magnitude, opposite in direction, and act on different objects.
Newton's Third Law is often stated as: "Every action has an equal and opposite reaction." However, this shorthand can be misleading — the word "action" and "reaction" suggest one force causes the other sequentially, but in reality the two forces exist simultaneously. Neither force is more fundamental than the other; they always come in pairs.
The critical word in Newton's Third Law is different. The two forces in a Newton's Third Law pair never act on the same object. This distinguishes them from balanced forces (Newton's First Law), where two equal and opposite forces act on the same object and result in zero net force.
FA on B = −FB on A
The minus sign indicates the forces are in opposite directions. Their magnitudes are always equal: |FA on B| = |FB on A|.
A Newton's Third Law pair always involves forces of the same type (e.g. both gravitational, both normal contact, both friction). If the types differ, they are not a Newton's Third Law pair.
Identifying Newton's Third Law Pairs
To identify a force pair correctly, you can use this checklist:
The forces act on two different objects.
The forces are equal in magnitude.
The forces are opposite in direction.
The forces are of the same type (gravitational, contact, magnetic, etc.).
A useful language template is: "Object A exerts a [type] force on Object B; Object B exerts an equal and opposite [type] force on Object A."
Common example — a book on a table:
• The book exerts a contact (normal) force downward on the table.
• The table exerts an equal contact (normal) force upward on the book.
These are a Newton's Third Law pair — same type (contact), equal magnitude, opposite direction, different objects.
What is NOT a Newton's Third Law pair? The weight of the book (gravitational pull of Earth on book) and the normal reaction of the table on the book are not a Newton's Third Law pair — they are balanced forces on the same object (the book). They are of different types (gravitational vs contact) and involve different agents, so they fail the checklist.
A Newton's Third Law pair: same type · equal magnitude · opposite direction · different objects — always both present, always simultaneously.
Applications: Swimming and Rockets
🏊 Swimming
When a swimmer pushes backwards on the water with their hands and feet, the water pushes forward on the swimmer with an equal force. This forward reaction force is what propels the swimmer through the pool. The harder and faster the swimmer pushes, the greater the reaction force and the greater their acceleration (by Newton's Second Law).
Force pair in swimming:
• Swimmer exerts a backward force on the water (contact/friction force).
• Water exerts an equal forward force on the swimmer (contact/friction force).
🚀 Rocket Propulsion
A rocket engine burns fuel and expels hot exhaust gases at very high speed downward/backward. By Newton's Third Law, the exhaust gases exert an equal and opposite force on the rocket upward/forward — this is the thrust. Rockets do not need air to push against; the reaction force is between the rocket and its own exhaust gases. This is why rockets can operate in the vacuum of space.
Force pair in rocket propulsion:
• Rocket exerts a force on exhaust gases (pushes them backward/downward).
• Exhaust gases exert an equal and opposite force on the rocket (thrust, forward/upward).
Thrust = Rate of change of momentum of exhaust gases = Δ(mv) ÷ Δt
The greater the mass of gas expelled per second and/or the greater the speed of ejection, the larger the thrust on the rocket. This links Newton's Third Law directly to momentum and Newton's Second Law.
Newton's Third Law in Collisions
During a collision between two objects, Newton's Third Law tells us that the force each object exerts on the other is equal in magnitude and opposite in direction at every instant during the collision. Because the contact time Δt is the same for both objects, by the impulse–momentum theorem:
Impulse = F × Δt = Δp
If the forces are equal and opposite (Newton's Third Law) and the time of contact is the same, then the impulses are equal and opposite, meaning the changes in momentum are equal and opposite. This is exactly why total momentum is conserved in a collision — Newton's Third Law is the underlying reason for conservation of momentum.
Collision example: Car A (mass 1000 kg) rear-ends Car B (mass 1200 kg).
• A exerts force F on B for time Δt → B gains momentum +FΔt.
• B exerts force −F on A for same time Δt → A loses momentum −FΔt.
Total change in momentum = 0 ✓ (momentum conserved)
This is important: although the forces are equal, the accelerations are not (unless the masses are equal). A lighter object experiences a greater acceleration for the same force (a = F/m). In a car collision, a lighter car changes velocity more than a heavier truck — same force, same time, same impulse in magnitude, but different mass means different change in velocity.
Equal forces → equal and opposite impulses → equal and opposite changes in momentum. This is the microscopic basis for conservation of momentum.
Key Quantities and Units
Quantity
Symbol
Unit
Notes
Force
F
Newton (N)
Vector quantity; direction matters
Mass
m
kilogram (kg)
Scalar; measure of inertia
Acceleration
a
m s⁻²
Produced by net (resultant) force
Momentum
p
kg m s⁻¹
p = mv; vector quantity
Impulse
J
N s = kg m s⁻¹
J = FΔt = Δp
Time of contact
Δt
seconds (s)
Same for both objects in a collision
Newton's Second Law: F = ma (net force on a single object). Newton's Third Law: force pairs on different objects. Never confuse the two!
Example 1 — Identifying a Newton's Third Law Pair
A horse pulls a cart forward with a force of 800 N. Identify the Newton's Third Law force pair and explain why the horse-cart system can still accelerate.
1Identify the action force: The horse exerts a forward contact (tension/friction) force of 800 N on the cart.
2State the reaction force: By Newton's Third Law, the cart exerts an equal and opposite force of 800 N backward on the horse.
3Check the pair: Same type (contact/tension) ✓ · Equal magnitude (800 N each) ✓ · Opposite directions ✓ · Different objects (horse and cart) ✓
4Explain acceleration: The Newton's Third Law pair acts on different objects, so we cannot say they "cancel out." To find the acceleration of the cart, we look only at the forces on the cart. The horse pulls the cart forward with 800 N; if friction on the cart from the ground is less than 800 N, there is a net forward force on the cart → the cart accelerates. Similarly, the horse is pushed forward by friction from the ground; if that exceeds 800 N (the cart pulling back), the horse accelerates.
Newton's Third Law pair: Horse exerts 800 N forward on cart; Cart exerts 800 N backward on horse. The system accelerates because the net force on each object (considering all forces on that object separately) is non-zero.
Example 2 — Rocket Thrust Calculation
A rocket expels 50 kg of exhaust gas every second at a speed of 2000 m s⁻¹ relative to the rocket. Calculate the thrust force on the rocket.
1Recall the relationship: Thrust = rate of change of momentum of exhaust gases = Δp ÷ Δt
2Calculate momentum change per second:
Mass expelled per second: Δm/Δt = 50 kg s⁻¹
Speed of exhaust: v = 2000 m s⁻¹
Δp per second = (Δm/Δt) × v = 50 × 2000 = 100 000 kg m s⁻¹ per second
3Apply Newton's Third Law: The rocket pushes the gas backward (force on gas = 100 000 N backward). The gas pushes the rocket forward with an equal and opposite force.
4State the thrust: Thrust on rocket = 100 000 N = 1.0 × 10⁵ N (forward)
Thrust = (Δm/Δt) × v = 50 × 2000 = 100 000 N (1.0 × 10⁵ N) directed forward on the rocket.
Example 3 — Collision: Forces and Accelerations
A car of mass 800 kg collides with a stationary van of mass 2400 kg. During the collision, the contact force between them is 12 000 N and the collision lasts 0.2 s. Calculate: (a) the impulse on each vehicle, (b) the change in velocity of each vehicle.
1State Newton's Third Law for the collision: The car exerts 12 000 N on the van (forward). The van exerts 12 000 N on the car (backward) — equal magnitude, opposite direction.
2Calculate impulse on each vehicle:
Impulse on van = F × Δt = 12 000 × 0.2 = 2400 N s (forward)
Impulse on car = 12 000 × 0.2 = 2400 N s (backward)
3Change in velocity of the van:
Δp = mΔv → Δv = Δp ÷ m = 2400 ÷ 2400 = 1.0 m s⁻¹ (forward)
4Change in velocity of the car:
Δv = Δp ÷ m = 2400 ÷ 800 = 3.0 m s⁻¹ (backward — car slows down)
5Key observation: Equal forces, equal impulses, but the lighter car (800 kg) experiences 3× the change in velocity compared to the heavier van (2400 kg) — consistent with a = F/m.
(a) Impulse on each = 2400 N s (equal magnitude, opposite directions).
(b) Δv of van = 1.0 m s⁻¹ forward; Δv of car = 3.0 m s⁻¹ backward.
Example 4 — Swimming Force Analysis
A swimmer of mass 65 kg pushes backward on the water with a force of 195 N. Ignoring water resistance, calculate the swimmer's initial acceleration and state the Newton's Third Law pair involved.
1Identify the Newton's Third Law pair:
• Swimmer's hands/feet exert 195 N backward on the water.
• Water exerts 195 N forward on the swimmer (the reaction force that propels them).
2Identify the net force on the swimmer: The forward reaction force from the water = 195 N. (Water resistance ignored as stated.)
3Apply Newton's Second Law to the swimmer:
F = ma → a = F ÷ m = 195 ÷ 65
4Calculate:
a = 195 ÷ 65 = 3.0 m s⁻² forward
Newton's Third Law pair: 195 N backward on water (by swimmer) and 195 N forward on swimmer (by water).
Swimmer's acceleration = F ÷ m = 195 ÷ 65 = 3.0 m s⁻² forward.
Question 1 — Which of the following is a correct Newton's Third Law pair when a book rests on a table?
Question 2 — A rocket in space fires its engines. Which statement best explains how the rocket accelerates using Newton's Third Law?
Question 3 — A rocket expels 80 kg of gas per second at 1500 m s⁻¹. Calculate the thrust force on the rocket in Newtons.
Question 4 — Two ice skaters, A (mass 60 kg) and B (mass 90 kg), push off each other from rest. Skater A moves away at 3 m s⁻¹. What is the speed of skater B? (Momentum is conserved.)
Question 5 — During a collision lasting 0.05 s, a force of 600 N acts between two trolleys. Calculate the magnitude of the impulse experienced by each trolley (in N s).
Challenge 1 — Extended Analysis
A student claims: "A horse cannot pull a cart because by Newton's Third Law, the cart pulls the horse backward with an equal force. These forces cancel, so nothing can ever accelerate." Explain in detail why this reasoning is incorrect. Your answer should reference Newton's laws correctly and identify the forces on each object separately. [6 marks]
Mark scheme:
✓ The student has confused Newton's Third Law pairs with balanced forces on a single object.
✓ Newton's Third Law pairs act on different objects and cannot be added together to find the net force on one object.
✓ To find whether the cart accelerates, consider only the forces acting on the cart: (1) the horse pulls the cart forward, (2) friction from the ground acts backward on the cart. If the forward pull exceeds friction, the net force on the cart is forward → cart accelerates.
✓ To find whether the horse accelerates, consider forces on the horse: (1) friction from the ground pushes the horse forward, (2) the cart pulls the horse backward. If ground friction > cart's pull, the horse accelerates forward.
✓ The two forces in a Newton's Third Law pair are equal and opposite but they never act on the same object, so they never "cancel."
✓ Balanced forces (Newton's First Law) are two forces equal and opposite acting on the same object — not the same situation.
Challenge 2 — Multi-step Calculation
A stationary gun of mass 3.0 kg fires a bullet of mass 0.015 kg. The bullet leaves the barrel at 400 m s⁻¹. (a) Calculate the momentum of the bullet after firing. (b) Using Newton's Third Law and conservation of momentum, calculate the recoil speed of the gun. (c) The bullet travels the 0.60 m barrel length in 0.003 s. Calculate the average force the expanding gas exerts on the bullet, and state the force the bullet exerts on the gun barrel. [6 marks]
(a) pbullet = mv = 0.015 × 400 = 6.0 kg m s⁻¹
(b) Total initial momentum = 0 (system at rest).
By conservation of momentum: pgun + pbullet = 0
pgun = −6.0 kg m s⁻¹
Recoil speed = 6.0 ÷ 3.0 = 2.0 m s⁻¹ (in opposite direction to bullet)
(c) Impulse on bullet = Δp = 6.0 N s (from rest to 400 m s⁻¹)
F = Δp ÷ Δt = 6.0 ÷ 0.003 = 2000 N forward on the bullet.
By Newton's Third Law: The bullet exerts an equal and opposite force of 2000 N backward on the gun barrel.
Challenge 3 — Collision Comparison
Car X (mass 1000 kg, initial velocity 20 m s⁻¹) collides head-on with Car Y (mass 1000 kg, initial velocity −15 m s⁻¹). The cars stick together after the collision. (a) Calculate the velocity of the combined cars after the collision. (b) Calculate the change in momentum of Car X. (c) If the collision lasts 0.1 s, calculate the average force Car Y exerted on Car X during the collision. (d) What force did Car X exert on Car Y? Justify using Newton's Third Law. [8 marks]
(a) Conservation of momentum:
Total p before = (1000 × 20) + (1000 × −15) = 20000 − 15000 = 5000 kg m s⁻¹
Total mass after = 2000 kg
v = 5000 ÷ 2000 = 2.5 m s⁻¹ (in the original direction of Car X)
(b) Δp of Car X = m(v − u) = 1000 × (2.5 − 20) = 1000 × (−17.5) = −17 500 kg m s⁻¹
(Car X lost 17 500 kg m s⁻¹ of momentum — it slowed and changed direction relative to its initial motion.)
(c) F = Δp ÷ Δt = 17 500 ÷ 0.1 = 175 000 N = 1.75 × 10⁵ N (the force on Car X was in the negative/backward direction, decelerating it).
(d) By Newton's Third Law, the force Car X exerted on Car Y = 175 000 N = 1.75 × 10⁵ N in the opposite direction — equal magnitude, opposite direction. This is confirmed by calculating Δp of Car Y: 1000 × (2.5 − (−15)) = 1000 × 17.5 = 17 500 kg m s⁻¹; F = 17 500 ÷ 0.1 = 175 000 N ✓
Challenge 4 — Conceptual and Quantitative
A swimmer of mass 70 kg pushes off a wall. The wall exerts a force of 420 N on the swimmer for 0.25 s. (a) Calculate the impulse the wall exerts on the swimmer. (b) Calculate the swimmer's velocity immediately after pushing off (assume they started from rest). (c) By Newton's Third Law, what force does the swimmer exert on the wall, and why does the wall not accelerate? [5 marks]
(a) Impulse = F × Δt = 420 × 0.25 = 105 N s
(b) Δp = 105 N s; swimmer starts from rest (u = 0)
mv = 105 → v = 105 ÷ 70 = 1.5 m s⁻¹ (away from the wall)
(c) By Newton's Third Law, the swimmer exerts an equal and opposite force on the wall = 420 N toward the wall.
The wall does not accelerate because it is attached to (and part of) a very massive structure (the pool/building/Earth). The resultant force on the Earth-pool-wall system is negligible because of its enormous mass: a = F/m ≈ 420 ÷ 6×10²⁴ ≈ 7×10⁻²³ m s⁻² — essentially zero. The wall also has internal structural forces (from its foundations) that provide a reaction, keeping it stationary.