AP Physics C Mechanics Quiz: Newtons Third Law

20 questions · exam conditions

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Newtons Third LawQuestion 1 of 20

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{boat}\to\text{rower}}

F_{\text{water}\to\text{oar}}

F_{\text{rower}\to\text{oar}}

F_{\text{Earth}\to\text{boat}}

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Newtons Third Law

Practice Newtons Third Law in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Newtons Third Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$F_{\text{boat}\to\text{rower}}$ .
$F_{\text{water}\to\text{oar}}$ . (correct answer)
$F_{\text{rower}\to\text{oar}}$ .
$F_{\text{Earth}\to\text{boat}}$ .

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.

Question 2

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 $5.0\times10^{-1}\,\text{N}$ forward on the balloon. (correct answer)
The air exerts $5.0\times10^{-1}\,\text{N}$ backward on the balloon.
The air exerts a smaller force because the air has less mass.
The reaction force is the balloon's weight, not an air 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 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.

Question 3

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?

$9.0\times10^{0}\,\text{N}$ upward. (correct answer)
$9.0\times10^{0}\,\text{N}$ downward.
$1.8\times10^{1}\,\text{N}$ upward.
$0\,\text{N}$ until the rocket leaves the pad.

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.

Question 4

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?

$3.5\times10^{2}\,\text{N}$ leftward on the left student. (correct answer)
$3.5\times10^{2}\,\text{N}$ rightward on the left student.
$1.4\times10^{2}\,\text{N}$ 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.

Question 5

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 $\vec F_{B\to A}$ ?

$\vec F_{A\to B}=+9.0\times10^{1}\,\text{N}$ (right) (correct answer)
The reaction is the normal force from the track on cart A
$\vec F_{A\to B}=-9.0\times10^{1}\,\text{N}$ (left)
$\vec F_{A\to B}=+4.5\times10^{1}\,\text{N}$ (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.

Question 6

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.

The rocket pushes the gases down; the gases push the rocket up with equal magnitude (correct answer)
The rocket pushes the gases down; Earth pushes the rocket up as the reaction force
The gases push up harder than the rocket pushes down because the rocket accelerates
Both forces act upward on the rocket and gases because thrust points 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 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.

Question 7

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?

$2.4\times10^{2}\,\text{N}$ to the right.
$2.4\times10^{2}\,\text{N}$ to the left. (correct answer)
$6.0\times10^{1}\,\text{N}$ to the left.
$0\,\text{N}$ because the wall does not move.

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.

Question 8

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?

They are equal in magnitude and opposite in direction. (correct answer)
The larger force is on the water because it has greater mass.
They act on the same object, so they cancel directly.
They are equal in direction because both point backward.

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.

Question 9

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?

$4.5\times10^{2}\,\text{N}$ rightward. (correct answer)
$4.0\times10^{2}\,\text{N}$ leftward.
$4.5\times10^{2}\,\text{N}$ 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.

Question 10

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 $F_{A\to B}$ ?

$F_{\text{track}\to A}$ .
$F_{B\to A}=-6.0\times10^{1}\,\text{N}$ (left). (correct answer)
$F_{A\to B}=+6.0\times10^{1}\,\text{N}$ (right).
$F_{A\to\text{track}}$ .

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.

Question 11

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.

The rocket's thrust is the same force as its weight, so no reaction pair exists.
The gases push up on the rocket with $8.0\times10^{0}\,\text{N}$ while the rocket pushes down on the gases with $8.0\times10^{0}\,\text{N}$ . (correct answer)
The rocket pushes up on the gases and the gases push up on the rocket.
Earth pushes up on the rocket with $8.0\times10^{0}\,\text{N}$ , which is the reaction to thrust.

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 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.

Question 12

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 $F_{B\to A}$ ?

$F_{A\to B}$ is larger because cart A moves faster.
$F_{A\to B}=-F_{B\to A}$ at each instant of contact. (correct answer)
$F_{A\to B}$ and $F_{B\to A}$ point in the same direction.
$F_{B\to A}=0$ because the track is frictionless.

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.

Question 13

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?

$F_{B\to A}=-1.2\times10^{2}\,\text{N}$ (left) (correct answer)
$F_{B\to A}=+1.2\times10^{2}\,\text{N}$ (right)
$F_{B\to A}=-6.0\times10^{1}\,\text{N}$ (left)
$F_{B\to A}=0\,\text{N}$ because the track is frictionless

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 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.

Question 14

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 $\vec F_{\text{wall}\to \text{skater}}$ ?

$\vec F_{\text{skater}\to \text{wall}}$ is larger because the skater accelerates
$\vec F_{\text{skater}\to \text{wall}}=-\vec F_{\text{wall}\to \text{skater}}$ (correct answer)
$\vec F_{\text{skater}\to \text{wall}}=\vec F_{\text{wall}\to \text{skater}}$
They are not a third-law pair because one object is stationary

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 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.

Question 15

A rower pulls on an oar so the blade pushes water backward; a spring scale on the oar reads $F_{\text{oar}\to \text{water}}=2.5\times10^{2}\,\text{N}$ (backward) during a stroke. The boat speeds up forward while the stroke lasts about $0.40\,\text{s}$ . Neglect air drag for the short interval and focus only on the oar–water interaction. In the scenario, which force is the reaction to $\vec F_{\text{oar}\to \text{water}}$ ?

$\vec F_{\text{water}\to \text{oar}}$ forward, $2.5\times10^{2}\,\text{N}$ (correct answer)
$\vec F_{\text{boat}\to \text{water}}$ forward, $2.5\times10^{2}\,\text{N}$
$\vec F_{\text{water}\to \text{oar}}$ backward, $2.5\times10^{2}\,\text{N}$
The reaction is the boat's weight on Earth, $mg$

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 identifying the correct force pairs - if the oar pushes water backward, then water must push the oar forward with equal magnitude. In the provided scenario, the oar blade pushes water backward with 250 N, creating the action force in this interaction. Choice A is correct because it correctly identifies that the water pushes the oar forward with 250 N, forming the reaction force that propels the boat forward. Choice C is incorrect because it suggests the water pushes the oar backward, which would oppose the boat's forward motion and violate Newton's Third Law. To help students, emphasize that the direction of reaction forces must make physical sense - the water's push on the oar is what drives the boat forward. Practice identifying force pairs in propulsion scenarios like swimming, where hands push water backward and water pushes hands forward.

Question 16

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?

They are equal magnitude and opposite direction (correct answer)
The exhaust force is smaller because gases have little mass
They are equal magnitude and same direction
The reaction to thrust is the rocket's weight on Earth

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 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.

Question 17

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?

The same as the force the springs exert on the mass (correct answer)
Half the force the springs exert on the mass
Twice the force the springs exert on the mass
Zero, since the mass is momentarily at rest

Explanation: 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.

Question 18

A 1500 kg car traveling at 20 m/s collides head-on with a 3000 kg truck traveling at 10 m/s in the opposite direction. During the 0.15 s collision, the average force exerted by the car on the truck is 80,000 N. What is the change in momentum of the car during this collision?

-12,000 kg⋅m/s (correct answer)
+12,000 kg⋅m/s
-24,000 kg⋅m/s
+6,000 kg⋅m/s

Explanation: The correct answer is A. By Newton's third law, if the car exerts 80,000 N on the truck, the truck exerts 80,000 N on the car in the opposite direction. Using the impulse-momentum theorem:

\Delta p_{car} = F \cdot t = (-80,000)(0.15) = -12,000 \text{ kg⋅m/s}

. The negative sign indicates the force opposes the car's initial motion. B is wrong because it has the wrong sign. C is wrong because it incorrectly doubles the impulse. D is wrong because it uses half the correct impulse magnitude.

Question 19

A person stands in an elevator that is accelerating upward at $2.0 \text{ m/s}^2$ . The person has a mass of 70 kg. A scale beneath the person reads 840 N. According to Newton's third law, which statement correctly describes the forces between the person and the scale?

The person exerts 140 N downward on the scale, and the scale exerts 840 N upward on the person
The person exerts 686 N downward on the scale, and the scale exerts 840 N upward on the person
The person exerts 840 N downward on the scale, and the scale exerts 686 N upward on the person
The person exerts 840 N downward on the scale, and the scale exerts 840 N upward on the person (correct answer)

Explanation: When analyzing forces in accelerating reference frames, you need to carefully apply Newton's laws while remembering that Newton's third law always creates equal and opposite force pairs between interacting objects. Let's examine what's happening here. The scale reads 840 N, which represents the normal force the scale exerts upward on the person. This reading makes physical sense because the person is accelerating upward, so the scale must push harder than just supporting their weight (

mg = 70 \times 9.8 = 686 \text{ N}

). According to Newton's third law, forces always come in equal and opposite pairs. If the scale pushes up on the person with 840 N, then by Newton's third law, the person must push down on the scale with exactly 840 N. These are action-reaction pairs that must be equal in magnitude. Choice A incorrectly uses 140 N downward, which has no physical basis in this problem. Choice B suggests the person exerts only 686 N downward (their weight) while receiving 840 N upward, violating Newton's third law by having unequal action-reaction forces. Choice C reverses the logic, incorrectly claiming the scale only exerts 686 N upward when we know it reads 840 N. Choice D correctly identifies that both forces in the action-reaction pair are 840 N: the person exerts 840 N downward on the scale, and the scale exerts 840 N upward on the person. Study tip: Newton's third law force pairs are always equal in magnitude, regardless of acceleration. When you see a scale reading, that's the normal force—and its reaction partner has the same magnitude.

Question 20

Two ice skaters, initially at rest, push off from each other. Skater A has a mass of 60 kg and moves away at 4.0 m/s. Skater B has a mass of 80 kg. During the 0.50 s push-off period, what was the average magnitude of the force that skater A exerted on skater B?

640 N
480 N (correct answer)
360 N
240 N

Explanation: This problem tests conservation of momentum combined with Newton's second and third laws. When you see two objects pushing apart from rest, immediately think about momentum conservation and the force-time relationship. Since both skaters start at rest, the total initial momentum is zero. By conservation of momentum, the final momenta must be equal and opposite. For skater A:

p_A = (60 \text{ kg})(4.0 \text{ m/s}) = 240 \text{ kg⋅m/s}

. Therefore, skater B must have momentum

p_B = -240 \text{ kg⋅m/s}

, giving skater B a velocity of

v_B = -240/80 = -3.0 \text{ m/s}

. Now apply the impulse-momentum theorem to skater B. The change in momentum is

\Delta p_B = 80(-3.0) - 0 = -240 \text{ kg⋅m/s}

. Using

F\Delta t = \Delta p

, the force from A on B is:

F = \frac{-240 \text{ kg⋅m/s}}{0.50 \text{ s}} = -480 \text{ N}

. The magnitude is 480 N, which is choice B. Choice A (640 N) likely comes from incorrectly using skater A's mass in the force calculation. Choice C (360 N) might result from arithmetic errors or confusion about which skater to analyze. Choice D (240 N) represents the momentum value rather than force, showing confusion between impulse and momentum concepts. Remember: In push-off problems, always use conservation of momentum first to find both velocities, then apply impulse-momentum theorem to either skater. Newton's third law ensures the force magnitudes are equal regardless of which skater you analyze.