Newton's Third Law - AP Physics C Mechanics

1. A 0.50 kg ball moving at 8.0 m/s collides with a wall and rebounds at 6.0 m/s in the opposite direction. The collision lasts 0.020 s. If we consider the ball and wall as separate objects, which statement about the forces during the collision is correct according to Newton's third law?
   Answer: The forces are equal in magnitude at 350 N, but opposite in direction throughout the collision The correct answer is C. Newton's third law requires equal and opposite forces regardless of mass or motion. The ball's momentum change is $\Delta p = 0.50(-6.0-8.0) = -7.0 \text{ kg⋅m/s}$. The average force magnitude is $|F| = |\Delta p|/t = 7.0/0.020 = 350 \text{ N}$. By Newton's third law, both forces have this magnitude. A and B are wrong because they violate Newton's third law by suggesting unequal forces. D is wrong because it incorrectly calculates the force as $\Delta p/t = (0.50 \times 4.0)/0.020 = 100 \text{ N}$, using only the speed change rather than the full momentum change.
2. In a tug-of-war on level ground, two students pull on a rope with inline force sensors: left student pulls rightward with $F_{L\to R}=3.5\times10^{2}\,\text{N}$, and the right student pulls leftward; the rope briefly accelerates $0.40\,\text{m/s}^2$ toward the left. According to Newton's Third Law, if the left student exerts $3.5\times10^{2}\,\text{N}$ on the rope, what force does the rope exert on the left student?
   Answer: $3.5\times10^{2}\,\text{N}$ leftward on the left student. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, with these forces acting on different objects. In the provided scenario, the left student pulls on the rope with 350 N rightward, so by Newton's Third Law, the rope must pull on the left student with 350 N in the opposite direction (leftward). Choice A is correct because it correctly identifies that the rope exerts 350 N leftward on the left student, forming an action-reaction pair with the student's rightward pull on the rope. Choice B is incorrect because it has the wrong direction - the rope cannot pull the student in the same direction the student pulls the rope. To help students, emphasize that action-reaction pairs always point in opposite directions and act on different objects. Encourage students to identify the two objects involved and draw force diagrams showing the equal and opposite forces.
3. A model rocket on a launch pad produces a measured thrust of $8.0\times10^{0}\,\text{N}$ upward for $0.60\,\text{s}$. During this time, the exhaust gases are pushed downward by the engine, and the rocket is pushed upward by the gases. Assume the rocket and exhaust interact only through this thrust force during the burn. Explain how Newton's Third Law applies to the interaction between the rocket and the exhaust gases.
   Answer: The rocket pushes the gases down; the gases push the rocket up with equal magnitude This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law applies to all interactions between objects, including between a rocket and its exhaust gases. In the provided scenario, the rocket engine pushes exhaust gases downward, and by Newton's Third Law, the gases must push the rocket upward with equal force. Choice A is correct because it correctly identifies that the rocket and gases exert equal and opposite forces on each other - the rocket pushes gases down, gases push rocket up. Choice C is incorrect because it assumes unequal forces, violating Newton's Third Law; the acceleration occurs because the net force on the rocket (thrust minus weight) is non-zero, not because the action-reaction forces are unequal. To help students, emphasize that Newton's Third Law forces are always equal regardless of acceleration. Use the analogy of pushing against a wall - you and the wall push equally on each other even though only you accelerate.
4. A rocket in space ejects exhaust gases at a rate of 100 kg/s with a velocity of 3000 m/s relative to the rocket. At a particular instant, the rocket has a mass of 5000 kg and is accelerating at $60 \text{ m/s}^2$. What is the magnitude of the force that the exhaust gases exert on the rocket according to Newton's third law?
   Answer: 300,000 N When you encounter rocket propulsion problems, you're dealing with Newton's third law and the momentum principle. The key insight is that the force exerted by exhaust gases on the rocket equals the rate of momentum change of the expelled gases. To find this force, use the relationship $F = \frac{dm}{dt} \times v_{exhaust}$, where $\frac{dm}{dt}$ is the mass flow rate and $v_{exhaust}$ is the exhaust velocity relative to the rocket. Given: mass flow rate = 100 kg/s and exhaust velocity = 3000 m/s $F = 100 \text{ kg/s} \times 3000 \text{ m/s} = 300,000 \text{ N}$ This is the thrust force the exhaust gases exert on the rocket, making B correct. Now for the distractors: Choice A (600,000 N) likely comes from incorrectly doubling the thrust force, perhaps by misapplying Newton's third law. Choice C (900,000 N) might result from adding the rocket's weight to the thrust, but remember we're in space where gravitational effects are negligible. Choice D (200,000 N) could come from calculation errors or using incorrect formulas. Notice that the rocket's mass (5000 kg) and acceleration (60 m/s²) are red herrings in this problem. While these could help you find the net force on the rocket, the question specifically asks for the exhaust force according to Newton's third law. Remember: In rocket problems, focus on the momentum transfer rate of the exhaust. The thrust force depends only on how fast mass is ejected and at what velocity, not on the rocket's current state of motion.
5. On a frictionless track, cart A ($m_A=0.50\,\text{kg}$) moving right at $2.0\,\text{m/s}$ collides with cart B ($m_B=1.00\,\text{kg}$) moving left at $1.0\,\text{m/s}$; during contact, sensors read $F_{A\to B}=+1.2\times10^{2}\,\text{N}$ (right) for $0.020\,\text{s}$ and $F_{B\to A}=-1.2\times10^{2}\,\text{N}$ (left) for the same interval, showing equal-magnitude opposite forces on different carts. What is the relationship between $F_{A\to B}$ and $F_{B\to A}$?
   Answer: $F_{A\to B}=-F_{B\to A}$ at each instant of contact. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, cart A exerts a force on cart B while cart B simultaneously exerts an equal and opposite force on cart A during their collision. Choice B is correct because it correctly identifies that $F_{A\to B}=-F_{B\to A}$, meaning the forces are equal in magnitude (both 120 N) but opposite in direction (one is positive/right, the other is negative/left). Choice A is incorrect because it assumes the faster-moving object exerts a larger force, which violates Newton's Third Law. To help students, emphasize that action-reaction pairs always have equal magnitudes regardless of the masses or velocities of the objects. Practice identifying which forces form action-reaction pairs by checking if they act on different objects.
6. A rower pulls an oar handle with $F_{\text{rower}\to\text{oar}}=1.8\times10^{2}\,\text{N}$ toward the stern; the blade pushes water backward with $F_{\text{oar}\to\text{water}}=1.8\times10^{2}\,\text{N}$, and the water pushes the blade forward with equal magnitude. In the scenario, which force is the reaction to $F_{\text{oar}\to\text{water}}$?
   Answer: $F_{\text{water}\to\text{oar}}$. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, acting on different objects. In the provided scenario, the oar pushes water backward with force $F_{\text{oar}\to\text{water}}$, so by Newton's Third Law, the water must push the oar forward with an equal and opposite force. Choice B is correct because $F_{\text{water}\to\text{oar}}$ is the reaction force to $F_{\text{oar}\to\text{water}}$ - these forces form an action-reaction pair with equal magnitudes but opposite directions. Choice C is incorrect because $F_{\text{rower}\to\text{oar}}$ forms a different action-reaction pair with the oar pushing back on the rower. To help students, emphasize identifying which two objects are interacting directly - here it's the oar and water. Practice recognizing that action-reaction pairs always involve the same two objects exerting forces on each other.
7. On a frictionless track, cart A ($m_A=0.50\,\text{kg}$) moving $+2.0\,\text{m/s}$ collides with cart B ($m_B=1.0\,\text{kg}$) moving $-1.0\,\text{m/s}$; force sensors read $F_{A\to B}=+1.2\times10^{2}\,\text{N}$ (right) for $0.050\,\text{s}$. At the same time, cart B pushes on cart A during contact. Both carts remain in contact only during the sensor spike, and friction is negligible. If cart A exerts a force of $+1.2\times10^{2}\,\text{N}$ on cart B, what force does cart B exert on cart A?
   Answer: $F_{B\to A}=-1.2\times10^{2}\,\text{N}$ (left) This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, cart A pushes on cart B with +120 N (to the right), which illustrates the action force in this interaction. Choice A is correct because it correctly identifies that cart B must push back on cart A with -120 N (to the left), forming an action-reaction pair with equal magnitude but opposite direction. Choice B is incorrect because it suggests the forces have the same direction, violating Newton's Third Law. To help students, emphasize that action-reaction pairs always act on different objects and are always equal and opposite. Practice identifying force pairs by labeling which object exerts the force and which object experiences it.
8. A model rocket on a launch pad produces a measured thrust of $T=8.0\times10^{0}\,\text{N}$ upward for $0.50\,\text{s}$ as exhaust gases are expelled downward; simultaneously, the rocket pushes down on the gases with $8.0\times10^{0}\,\text{N}$. Explain how Newton's Third Law applies to the interaction between the rocket and the exhaust gases.
   Answer: The gases push up on the rocket with $8.0\times10^{0}\,\text{N}$ while the rocket pushes down on the gases with $8.0\times10^{0}\,\text{N}$. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, the rocket pushes downward on the exhaust gases with 8.0 N, and simultaneously the gases push upward on the rocket with 8.0 N, forming an action-reaction pair. Choice B is correct because it correctly identifies both forces in the action-reaction pair: the gases push up on the rocket while the rocket pushes down on the gases, with equal magnitudes. Choice C is incorrect because it states both forces point in the same direction, which violates Newton's Third Law. To help students, emphasize that rocket propulsion works through action-reaction pairs between the rocket and expelled gases, not between the rocket and Earth. Practice identifying the two interacting objects and ensuring the forces point in opposite directions.
9. Two identical springs with spring constant $k$ are connected in series between a wall and a 3.0 kg mass. When the system is in equilibrium, the springs are compressed by a total of 0.20 m. If the mass is displaced an additional 0.10 m and released, what is the magnitude of the force that the mass exerts on the springs at the moment of release?
   Answer: The same as the force the springs exert on the mass The correct answer is A. By Newton's third law, the force that the mass exerts on the springs is always equal in magnitude and opposite in direction to the force the springs exert on the mass, regardless of the system's configuration or motion state. This is a fundamental property of action-reaction pairs. B and C are wrong because they suggest the third law forces can have different magnitudes. D is wrong because it confuses the instantaneous velocity (zero) with the force, and because Newton's third law applies at all times, not just when objects are moving.
10. A skater of mass $60\,\text{kg}$ pushes on a rigid wall and accelerates away at $0.80\,\text{m/s}^2$ for $0.50\,\text{s}$. During the push, the wall exerts a horizontal force on the skater of $F_{\text{wall}\to \text{skater}}=4.8\times10^{1}\,\text{N}$ away from the wall. The skater's hands remain in contact with the wall only during this interval. What is the relationship between $\vec F_{\text{skater}\to \text{wall}}$ and $\vec F_{\text{wall}\to \text{skater}}$?
    Answer: $\vec F_{\text{skater}\to \text{wall}}=-\vec F_{\text{wall}\to \text{skater}}$ This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law establishes that forces between interacting objects are always equal in magnitude but opposite in direction, regardless of the motion of either object. In the provided scenario, the wall pushes the skater away with 48 N, and by Newton's Third Law, the skater must push the wall with equal force in the opposite direction. Choice B is correct because it correctly states that the skater's force on the wall is the negative (opposite) of the wall's force on the skater, satisfying Newton's Third Law. Choice A is incorrect because it assumes the accelerating object exerts a larger force, which is a common misconception - Newton's Third Law forces are always equal regardless of acceleration. To help students, emphasize that the wall doesn't accelerate not because forces are unequal, but because Earth (to which the wall is attached) has enormous mass. Use examples like a person jumping off Earth to show equal forces produce different accelerations due to different masses.
11. A rower's oar blade pushes water backward with $2.2\times10^{2}\,\text{N}$ for $0.15\,\text{s}$; the water simultaneously pushes the blade forward with $2.2\times10^{2}\,\text{N}$, and the boat speeds up. What is the relationship between the force water exerts on the oar and the force the oar exerts on the water?
    Answer: They are equal in magnitude and opposite in direction. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, the oar pushes water backward with 220 N while the water pushes the oar forward with 220 N, demonstrating a perfect action-reaction pair. Choice A is correct because it correctly identifies that the forces are equal in magnitude (both 220 N) and opposite in direction (one backward, one forward). Choice C is incorrect because it claims the forces act on the same object - action-reaction pairs always act on different objects (one on the oar, one on the water). To help students, emphasize that action-reaction pairs never cancel because they act on different objects. The boat accelerates because the forward force on the oar is transmitted to the boat, while the backward force on the water doesn't affect the boat directly.
12. A model rocket engine test stand shows the engine pushes exhaust gases downward with $1.5\times10^{1}\,\text{N}$ during a steady burn. Simultaneously, a load cell on the stand measures an upward force on the rocket of $1.5\times10^{1}\,\text{N}$. Assume the interaction of interest is only rocket–exhaust, not rocket–stand. What is the relationship between the force the rocket exerts on the exhaust and the force the exhaust exerts back?
    Answer: They are equal magnitude and opposite direction This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law requires that the force the rocket exerts on exhaust gases must be equal in magnitude but opposite in direction to the force the exhaust exerts on the rocket. In the provided scenario, the rocket pushes exhaust downward with 15 N, and the exhaust pushes the rocket upward with 15 N, perfectly demonstrating Newton's Third Law. Choice A is correct because it correctly identifies that these forces are equal in magnitude (15 N) but opposite in direction (one up, one down). Choice B is incorrect because it assumes force depends on mass, but Newton's Third Law forces are always equal regardless of the masses involved. To help students, emphasize that Newton's Third Law is about the interaction between two objects, not about their individual properties like mass. Use rocket demonstrations to show how exhaust and rocket push equally on each other, with different accelerations due to their different masses.
13. A model rocket produces thrust $9.0\times10^{0}\,\text{N}$ upward as exhaust is expelled downward; a high-speed sensor indicates the rocket exerts $9.0\times10^{0}\,\text{N}$ downward on the exhaust at that instant. If the rocket exerts a force of $9.0\times10^{0}\,\text{N}$ on the exhaust, what force does the exhaust exert on the rocket?
    Answer: $9.0\times10^{0}\,\text{N}$ upward. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, the rocket exerts 9.0 N downward on the exhaust gases, so by Newton's Third Law, the exhaust must exert 9.0 N upward on the rocket. Choice A is correct because it correctly identifies that the exhaust exerts 9.0 N upward on the rocket, forming an action-reaction pair with the rocket's downward push on the exhaust. Choice D is incorrect because it assumes no force exists until motion begins - Newton's Third Law applies at every instant of contact, regardless of whether the rocket has started moving. To help students, emphasize that thrust is generated through action-reaction pairs between rocket and exhaust, not between rocket and launch pad. Practice analyzing rocket propulsion as a continuous exchange of forces between rocket and expelled gases.
14. On a frictionless track, cart A ($0.80\,\text{kg}$) and cart B ($0.80\,\text{kg}$) collide head-on; a sensor on A records $F_{B\to A}=-9.0\times10^{1}\,\text{N}$ (left) for $0.040\,\text{s}$. A second sensor on B records the interaction simultaneously. The carts are the only interacting bodies during contact, and the force spike is symmetric in time. In the scenario, which force is the reaction to $\vec F_{B\to A}$?
    Answer: $\vec F_{A\to B}=+9.0\times10^{1}\,\text{N}$ (right) This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law requires that when cart B pushes cart A with a certain force, cart A must push cart B with equal magnitude in the opposite direction. In the provided scenario, cart B pushes cart A with -90 N (to the left), so cart A must push cart B with the reaction force. Choice A is correct because it correctly identifies that cart A pushes cart B with +90 N (to the right), equal in magnitude but opposite in direction to the force B exerts on A. Choice B is incorrect because the normal force from the track is not the reaction to the horizontal collision force - it's the reaction to the cart's weight. To help students, emphasize that action-reaction pairs must be between the same two objects and along the same line of action. Draw separate free body diagrams for each cart to show how the collision forces form a Newton's Third Law pair.
15. A skater of mass $60\,\text{kg}$ pushes on a rigid wall; a force probe shows the skater exerts $F_{\text{skater}\to\text{wall}}=2.4\times10^{2}\,\text{N}$ to the right for $0.30\,\text{s}$, and the skater accelerates left at $4.0\,\text{m/s}^2$ during the push. If the skater exerts $2.4\times10^{2}\,\text{N}$ on the wall, what force does the wall exert on the skater?
    Answer: $2.4\times10^{2}\,\text{N}$ to the left. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, the skater pushes on the wall with 240 N to the right, so by Newton's Third Law, the wall must push back on the skater with 240 N to the left. Choice B is correct because it correctly identifies that the wall exerts 240 N to the left on the skater, forming an action-reaction pair with the skater's rightward push. Choice D is incorrect because it assumes a stationary object cannot exert force, which is a common misconception - Newton's Third Law applies regardless of whether objects move. To help students, emphasize that action-reaction pairs exist even when one object (like a wall) doesn't accelerate. The wall's inability to move is due to other forces, not the absence of the reaction force.
16. In a tug-of-war, the left student pulls on the rope with $4.0\times10^{2}\,\text{N}$ rightward while the right student pulls with $4.5\times10^{2}\,\text{N}$ leftward, and the rope accelerates left. According to Newton's Third Law, what force does the rope exert on the right student?
    Answer: $4.5\times10^{2}\,\text{N}$ rightward. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, acting on different objects. In the provided scenario, the right student pulls on the rope with 450 N leftward, so by Newton's Third Law, the rope must pull on the right student with 450 N in the opposite direction (rightward). Choice A is correct because it correctly identifies that the rope exerts 450 N rightward on the right student, forming an action-reaction pair with the student's leftward pull on the rope. Choice B is incorrect because it uses the wrong magnitude (400 N from the left student) - each student forms their own action-reaction pair with the rope. To help students, emphasize that each person-rope interaction forms its own action-reaction pair, and the forces in each pair must be equal. The rope experiences different forces from each student, but each student-rope pair follows Newton's Third Law independently.
17. A book rests on a table, which stands on the floor. Consider the force that the book exerts on the table and the force that the table exerts on the book. According to Newton's third law, these forces:
    Answer: Are equal in magnitude, act on different objects, and form a true action-reaction pair The correct answer is D. Newton's third law requires action-reaction pairs to be equal in magnitude, opposite in direction, and act on different objects. The book pushes down on the table, and the table pushes up on the book - these forces act on different objects and form a true action-reaction pair. A is wrong because Newton's third law applies regardless of motion state. B is wrong because action-reaction pairs always act on different objects, so they cannot directly cancel to prevent acceleration of a single system. C is wrong because it incorrectly states the forces act on the same object.
18. A rope of negligible mass passes over an ideal pulley. A 4.0 kg mass hangs from one end, and a 6.0 kg mass hangs from the other end. When the system is released, what is the magnitude of the force that the 4.0 kg mass exerts on the rope?
    Answer: 47 N When you encounter an Atwood machine problem like this, you're dealing with connected objects where Newton's second law must be applied to the entire system, then to individual components to find internal forces. Since the rope is massless and the pulley is ideal, the tension is uniform throughout the rope. The 6.0 kg mass will accelerate downward while the 4.0 kg mass accelerates upward at the same rate. First, find the system's acceleration by applying Newton's second law to the entire system: the net force is $(6.0 - 4.0)g = 2.0g = 19.6 \text{ N}$ downward, and the total mass is $10.0 \text{ kg}$. Therefore, $a = \frac{19.6}{10.0} = 1.96 \text{ m/s}^2$. Now analyze the 4.0 kg mass individually. Two forces act on it: weight (39.2 N downward) and tension (upward). Since it accelerates upward at 1.96 m/s², Newton's second law gives: $T - mg = ma$, so $T = m(g + a) = 4.0(9.8 + 1.96) = 47.1 \text{ N}$. By Newton's third law, the mass exerts an equal and opposite 47 N force on the rope. Choice A (59 N) incorrectly assumes the entire weight difference creates tension. Choice B (39 N) represents just the weight of the 4.0 kg mass, ignoring acceleration. Choice D (49 N) likely results from calculation errors in determining the acceleration. Strategy tip: In Atwood machine problems, always find the system acceleration first, then analyze individual objects. Remember that tension in an accelerating system differs from the weight of hanging objects due to the additional force needed for acceleration.
19. A balloon is released with internal pressure $1.3\times10^{5}\,\text{Pa}$ while ambient pressure is $1.0\times10^{5}\,\text{Pa}$; escaping air jets backward and a force sensor estimates the balloon exerts $F_{\text{balloon}\to\text{air}}=5.0\times10^{-1}\,\text{N}$ backward on the air as it accelerates out. What is the relationship between the force the balloon exerts on the air and the force the air exerts on the balloon?
    Answer: The air exerts $5.0\times10^{-1}\,\text{N}$ forward on the balloon. This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, acting on different objects. In the provided scenario, the balloon exerts 0.5 N backward on the escaping air, so by Newton's Third Law, the air must exert 0.5 N forward on the balloon. Choice A is correct because it correctly identifies that the air exerts 0.5 N forward on the balloon, forming an action-reaction pair with the balloon's backward push on the air. Choice C is incorrect because it assumes the force depends on mass - Newton's Third Law guarantees equal magnitudes regardless of the masses involved. To help students, emphasize that action-reaction pairs always have equal magnitudes even when the objects have very different masses. The different accelerations result from the different masses, not different forces.
20. Two carts collide on a frictionless track: cart A ($0.80\,\text{kg}$) moves right at $1.5\,\text{m/s}$ and cart B ($0.20\,\text{kg}$) is initially at rest; during the $0.010\,\text{s}$ collision, a sensor reports $F_{A\to B}=+6.0\times10^{1}\,\text{N}$ (right). In the scenario, which force is the reaction to $F_{A\to B}$?
    Answer: $F_{B\to A}=-6.0\times10^{1}\,\text{N}$ (left). This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, and these forces act on different objects. In the provided scenario, cart A exerts a force $F_{A\to B}=+60$ N (rightward) on cart B during the collision, so by Newton's Third Law, cart B must exert an equal and opposite force on cart A. Choice B is correct because $F_{B\to A}=-60$ N (leftward) is the reaction force to $F_{A\to B}$, forming an action-reaction pair with equal magnitude but opposite direction. Choice C is incorrect because it's the same force already given, not its reaction - the reaction must act on a different object and point in the opposite direction. To help students, emphasize using clear notation showing which object exerts force on which, and that action-reaction pairs always involve forces between the same two objects. Practice identifying both forces in the pair and verifying they have opposite signs.

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