Identify Action-Reaction Force Pairs - Grade 11 Physics

Which force is the Newton's-third-law partner of the student's $70\ \text{N}$ force on the rope?
Answer: The rope pulls the student westward with a force of $70\ \text{N}$ . Whenever you see a question about Newton's third law, remember the core rule: the third-law partner of a force must be equal in magnitude, opposite in direction, act on a different object, and involve the same two objects — just swapped. The pair is always structured as: "Object A pushes Object B" ↔ "Object B pushes Object A." Here, the force in question is: *the student pushes the rope eastward with $70\ \text{N}$ *. To find its Newton's-third-law partner, simply swap the two objects and reverse direction: *the rope pulls the student westward with $70\ \text{N}$ *. That's exactly what D states, making it correct. The magnitude must stay $70\ \text{N}$ — Newton's third law pairs are always equal in magnitude, no exceptions. A is wrong because it involves the sled and the rope — a completely different interaction from the student-rope interaction. The sled pulling the rope westward with $50\ \text{N}$ is the third-law partner of the rope pulling the sled eastward, not of anything the student does. B describes the rope acting on the sled, which again belongs to the sled-rope interaction — unrelated to the student. C is a tempting trap: it correctly identifies the student and reverses the direction, but the force acts on the student from the sled, which isn't even a direct interaction — the student and sled aren't touching. Third-law pairs only exist between objects in direct contact (or direct interaction). A reliable strategy: always identify the two objects in the original force, then write the partner as those same two objects with roles and direction reversed. If the objects change, it's not a third-law pair.
The student exerts a downward force on the scale. Which force is the Newton's-third-law partner of this force?
Answer: The scale pushes the student upward with a force of $708\ \text{N}$ . When a question asks for a Newton's third law "partner" force, you need to identify an action-reaction pair — two forces that are equal in magnitude, opposite in direction, and act on different objects due to the same interaction. The key rule: if Object A exerts a force on Object B, then Object B exerts an equal and opposite force back on Object A. Here, the student exerts a downward force on the scale. The interaction is between the student and the scale. By Newton's third law, the reaction force must be the scale pushing back on the student — upward. The magnitude of this normal force equals $mg + ma = 60.0(9.8 + 2.0) = 60.0 \times 11.8 = 708\ \text{N}$ . That makes B correct: the scale pushes the student upward with $708\ \text{N}$ . A is wrong because Earth's gravitational pull on the student ( $588\ \text{N}$ ) is the partner of the student pulling Earth upward, not the partner of a student-scale contact force — it involves a completely different interaction. C is wrong because the elevator floor pushing the scale is a separate contact force between the floor and the scale, not a reaction to the student-scale interaction, even though it has the same magnitude. D describes the correct Newton's third law partner of Earth's gravity on the student, but that's irrelevant here — the question asks about the student-scale interaction, not the student-Earth gravitational interaction. Study tip: Always identify the two objects involved in the original force. The third-law partner must involve the same two objects, just reversed — different object, same interaction.
The exhaust gas pushes the rocket forward. Which force is the Newton's-third-law partner of this thrust force?
Answer: The rocket pushes the exhaust backward with a force of $36.0\ \text{kN}$ . Whenever you see a question about Newton's third law, remember the golden rule: action-reaction pairs always involve the same two objects, act in opposite directions, and are equal in magnitude. The force pair must be between the same two objects — just swapped. Here, the interaction is between the rocket and the exhaust gas. The exhaust pushes the rocket forward (thrust), so by Newton's third law, the rocket must push the exhaust backward with an equal and opposite force. That's exactly what answer C states. The magnitude confirms it: $F = \dot{m} \cdot v_{ex} = 120\ \text{kg/s} \times 300\ \text{m/s} = 36{,}000\ \text{N} = 36.0\ \text{kN}$ . ✓ A is wrong because it introduces a third object — "space" — which has nothing to do with this interaction. Newton's third law pairs only involve the two objects directly interacting, not the surrounding environment. B is wrong for a similar reason: the rocket pulling "nearby matter" describes gravitational attraction, a completely different interaction. The third-law partner must involve the same force type (here, the contact/pressure force between rocket and exhaust). D is wrong because an object cannot exert a Newton's-third-law force on itself. Every action-reaction pair requires two distinct objects. A helpful memory trick: to find the third-law partner of any force, just swap the subject and object. "Exhaust pushes rocket forward" → partner is "rocket pushes exhaust backward." Same two objects, opposite direction, same magnitude — every time.
Which force is the Newton's-third-law partner of the road's friction force on the tires?
Answer: The tires push the road eastward with a friction force of $3600\ \text{N}$ . Whenever you see a question about Newton's Third Law, remember the golden rule: the action-reaction pair must involve the same two objects, with forces equal in magnitude, opposite in direction, and swapped in terms of who is pushing whom. Here, the original force is: the road exerts a westward friction force on the tires. To find its Newton's Third Law partner, simply reverse the two objects and reverse the direction: the tires exert an eastward friction force on the road. That makes D correct — the tires push the road eastward with $3600\ \text{N}$ . Same magnitude, opposite direction, same two objects (road and tires), just swapped roles. Choice A describes gravity from Earth on the car. Its Newton's Third Law partner would be the car pulling Earth upward — a completely different force pair that has nothing to do with friction or the road. Choice B says the tires push the car eastward. This is a tempting distractor because it sounds like an internal reaction, but "the road pushes the tires" and "the tires push the car" involve different object pairs — that's not a Third Law pair, that's a force being transmitted through the car's system. Choice C says the road pushes Earth eastward. This incorrectly replaces "tires" with "Earth," mixing up the interacting objects. The tires are what directly contact the road, not the entire Earth. Study tip: Always identify the two objects in the original force, then write the partner as: Object 2 pushes Object 1 in the opposite direction. If the objects change, it's not a Third Law pair.
Which statement identifies the Newton's-third-law reaction to the spring's force on the block at that instant?
Answer: The block pushes the spring westward with a force of $60\ \text{N}$ . When tackling Newton's third law questions, always ask: which two objects are interacting, and what is each doing to the other? A third-law pair must involve the same two objects, the same type of force, and opposite directions. This is where most students go wrong — they confuse "forces that balance" with "action-reaction pairs." Here, the interaction is between the spring and the block. The spring pushes the block eastward, so the reaction force must be the block pushing the spring in the opposite direction — westward. The magnitude comes from Hooke's Law: $F = kx = 400 \times 0.15 = 60\ \text{N}$ . So the block pushes the spring westward with $60\ \text{N}$ , making B correct. Now let's disqualify the distractors. A describes the wall pushing the spring — that's a different pair of objects (wall and spring), not the reaction to the spring-on-block force. It's also only $30\ \text{N}$ , which is a calculation error. C describes the block pushing the surface downward — that's a gravity/normal-force interaction, a completely different force pair unrelated to the spring. D comes closest in spirit (spring pushes wall eastward is real), but again it's the wrong pair of objects — the wall is not the block — and the magnitude $30\ \text{N}$ is wrong. A reliable memory trick: to find the Newton's third law reaction, swap the subject and object of the original force and reverse the direction. "Spring pushes block east" → "Block pushes spring west." Same magnitude, same force type, opposite direction, swapped objects — every time.
Which statement correctly describes the electric action-reaction pair immediately after both spheres are released?
Answer: Each sphere attracts the other with $1.8\ \text{N}$ , although their acceleration magnitudes may differ. Whenever you see a question involving two charged objects exerting forces on each other, your first instinct should be Newton's Third Law: every action force has an equal and opposite reaction force — no exceptions, regardless of mass, charge size, or sign. Start with Coulomb's Law to find the force magnitude: $F = k\frac{|q_X||q_Y|}{r^2} = (9.0\times10^9)\frac{(2.0\times10^{-6})(4.0\times10^{-6})}{(0.20)^2} = 1.8\ \text{N}$ Since the charges are opposite in sign, both forces are attractive. By Newton's Third Law, sphere X pulls on Y with 1.8 N, and sphere Y pulls on X with 1.8 N — same magnitude, opposite direction. This makes A correct: both forces are 1.8 N attractive, and because the spheres have different masses, their accelerations ( $a = F/m$ ) will differ. B is wrong because it assigns different force magnitudes (1.8 N and 0.90 N) to the two spheres — a direct violation of Newton's Third Law. The force pair is always equal in magnitude. C gets the force wrong (3.6 N instead of 1.8 N — likely from forgetting to square the distance or doubling the force) and falsely claims equal accelerations must result. Equal forces do not require equal accelerations when masses differ. D is wrong because the reaction to sphere X's force on Y is Y's force on X — not Earth's gravity. Earth is irrelevant to this electric interaction. Study tip: On any forces question involving two objects, immediately ask: "What does Newton's Third Law say here?" The action-reaction pair always involves the same two objects, same force type, same magnitude.
Which force is the Newton's-third-law partner of Earth's gravitational force on the book?
Answer: The book pulls Earth upward with a force of $20\ \text{N}$ . Whenever you see a question about Newton's Third Law, use this mental checklist: the action-reaction pair must involve the same two objects, the same type of force, and opposite directions. A common trap is confusing Newton's Third Law pairs with forces that merely balance each other (Newton's First Law equilibrium). Here, Earth exerts a gravitational pull on the book ( $20\ \text{N}$ downward). To find the Third Law partner, simply reverse the relationship: the book must pull Earth with the same gravitational force ( $20\ \text{N}$ upward). That's exactly what D describes — the book pulls Earth upward with $20\ \text{N}$ . The pair is: Earth pulls book down ↔ book pulls Earth up. Same two objects (Earth and book), same force type (gravity), opposite directions. D is correct. A is wrong because it introduces the table pushing Earth — this pairs the table with Earth, not Earth with the book, so it can't be the partner of Earth's force on the book. B describes the book pushing the table downward. This is actually the reaction to the table's normal force on the book — a completely separate Newton's Third Law pair involving the table and book, not Earth and book. C — the table pushing the book upward — is the normal force. This balances gravity in a Newton's First Law sense (net force = 0), but it's a different force type acting on a different object pair, so it is not the Third Law partner. Your go-to strategy: always identify the two objects and the force type first, then swap them to find the Third Law pair. If the object pair changes, you've left the correct pairing.
Which force is the direct Newton's-third-law partner of the force exerted by the swimmer during the stroke?
Answer: The pushed water drives the swimmer eastward with an average force of $90\ \text{N}$ . When you see a question asking for a Newton's Third Law "partner" force, your job is to identify the reaction to a specific action — and the pair must satisfy three strict conditions: equal magnitude, opposite direction, and acting on different objects of the same interaction. Here, the action force is: the swimmer pushes water westward with 90 N. To find its partner, simply flip the objects and the direction: the pushed water pushes the swimmer eastward with 90 N. That's exactly what B states, making it the correct answer. This is the direct reaction force — same interaction, same magnitude, reversed direction, acting on the swimmer instead of the water. A is a tempting trap because it mentions 90 N and involves water acting on the swimmer, but water resistance is a separate interaction (drag from surrounding water), not the reaction to the swimmer's stroke. Two forces can be equal and opposite without being a Newton's Third Law pair. C pairs the swimmer's weight with a force on Earth — that's actually a valid Newton's Third Law pair for gravity, but it has nothing to do with the swimming stroke interaction. It's a completely different force altogether. D describes surrounding water pushing displaced water eastward. This force acts between two water masses — neither object is the swimmer — so it cannot be the partner to the swimmer's push. A useful memory trick: Newton's Third Law pairs always follow the pattern "A pushes B, so B pushes A." Swap the two objects, flip the direction, keep the magnitude — if your answer doesn't fit that template exactly, it's not the true action-reaction pair.
Which statement correctly identifies the reaction force and compares the vehicles' resulting accelerations during the interaction?
Answer: The car pushes the truck westward with $12.0\ \text{kN}$ , but the car has the greater acceleration magnitude. When you see a collision problem involving forces between two objects, your first instinct should be Newton's Third Law: action-reaction pairs are always equal in magnitude, opposite in direction, and act on different objects. Then use Newton's Second Law ( $a = F/m$ ) to compare accelerations. Here, the truck exerts $12.0\ \text{kN}$ eastward on the car — so by Newton's Third Law, the car simultaneously pushes the truck with exactly $12.0\ \text{kN}$ westward. Same force magnitude, opposite direction. Now compare accelerations using $a = F/m$ : - Car: $a = \frac{12{,}000\ \text{N}}{1200\ \text{kg}} = 10\ \text{m/s}^2$ - Truck: $a = \frac{12{,}000\ \text{N}}{6000\ \text{kg}} = 2\ \text{m/s}^2$ The car experiences the greater acceleration because it has the smaller mass. This confirms A is correct. B is wrong on two counts: the reaction force comes from the car pushing the truck, not the road, and the accelerations are definitely not equal. C misapplies Newton's Third Law by changing the force magnitude to $2.4\ \text{kN}$ — reaction forces never differ in magnitude from their action forces. D incorrectly identifies the reaction force as the truck pushing the road, which is a completely different interaction pair, and also wrongly claims the truck has the greater acceleration (the heavier object accelerates less). Study tip: Newton's Third Law pairs always involve the same two objects swapping roles — if Object A pushes Object B, then Object B pushes Object A. Never mix in a third object, and never change the force magnitude.
Which force is the Newton's-third-law partner of the cone's force on that layer of air?
Answer: The nearby layer of air pushes the cone westward with a force of $0.40\ \text{N}$ . Whenever you see a question about Newton's third law, use the partner-force checklist: the two forces must involve the same two objects, act in opposite directions, have equal magnitudes, and be of the same type. A common trap is mixing up related-but-not-partner forces from nearby events in the same scenario. Here, the action force is: *the cone pushes the nearby air layer eastward with $0.40\ \text{N}$ *. To find its Newton's-third-law partner, simply swap the subject and object and reverse the direction: *the nearby air layer pushes the cone westward with $0.40\ \text{N}$ *. That's exactly what C says — same two objects (cone and nearby air), opposite direction (westward), same magnitude. C is correct. A is wrong because it introduces a different object — a distant air layer — which has nothing to do with the original interaction. Forces between different pairs of objects are never Newton's-third-law partners, even if they're part of the same wave. B describes the coil pushing the cone, which is a real force in the speaker, but it's part of a different interaction (coil ↔ cone), not the cone ↔ air interaction. This is a classic "chain of forces" trap — just because force B causes force A doesn't make them third-law partners. D is nonsensical because a sound wave is not an object that can exert a force; Newton's third law requires two distinct physical objects acting on each other. Your go-to strategy: identify the two objects in the original force, then write the partner force by swapping them and flipping the direction. If either object changes, it's not the third-law partner.

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