Newton's Third Law
Help Questions
AP Physics C: Mechanics › Newton's Third Law
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 $F_{\text{oar}\to\text{water}}$?
$F_{\text{water}\to\text{oar}}$.
$F_{\text{Earth}\to\text{boat}}$.
$F_{\text{boat}\to\text{rower}}$.
$F_{\text{rower}\to\text{oar}}$.
Explanation
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.
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?
The air exerts a smaller force because the air has less mass.
The air exerts $5.0\times10^{-1},\text{N}$ backward on the balloon.
The reaction force is the balloon’s weight, not an air force.
The air exerts $5.0\times10^{-1},\text{N}$ forward on the balloon.
Explanation
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.
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?
$9.0\times10^{0},\text{N}$ downward.
$0,\text{N}$ until the rocket leaves the pad.
$9.0\times10^{0},\text{N}$ upward.
$1.8\times10^{1},\text{N}$ upward.
Explanation
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.
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?
$1.4\times10^{2},\text{N}$ leftward on the left student.
$3.5\times10^{2},\text{N}$ rightward on the left student.
$3.5\times10^{2},\text{N}$ leftward on the left student.
$0,\text{N}$ because the rope is massless.
Explanation
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.
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 $\vec F_{B\to A}$?
$\vec F_{A\to B}=-9.0\times10^{1},\text{N}$ (left)
The reaction is the normal force from the track on cart A
$\vec F_{A\to B}=+4.5\times10^{1},\text{N}$ (right)
$\vec F_{A\to B}=+9.0\times10^{1},\text{N}$ (right)
Explanation
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.
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.
Both forces act upward on the rocket and gases because thrust points upward
The gases push up harder than the rocket pushes down because the rocket accelerates
The rocket pushes the gases down; the gases push the rocket up with equal magnitude
The rocket pushes the gases down; Earth pushes the rocket up as the reaction force
Explanation
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.
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?
$6.0\times10^{1},\text{N}$ to the left.
$2.4\times10^{2},\text{N}$ to the right.
$0,\text{N}$ because the wall does not move.
$2.4\times10^{2},\text{N}$ to the left.
Explanation
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.
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?
The larger force is on the water because it has greater mass.
They are equal in direction because both point backward.
They are equal in magnitude and opposite in direction.
They act on the same object, so they cancel directly.
Explanation
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.
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?
$4.0\times10^{2},\text{N}$ leftward.
$4.5\times10^{2},\text{N}$ rightward.
$4.5\times10^{2},\text{N}$ leftward.
$5.0\times10^{1},\text{N}$ rightward.
Explanation
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.
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 $F_{A\to B}$?
$F_{A\to B}=+6.0\times10^{1},\text{N}$ (right).
$F_{A\to\text{track}}$.
$F_{\text{track}\to A}$.
$F_{B\to A}=-6.0\times10^{1},\text{N}$ (left).
Explanation
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.