3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A large truck traveling at high speed collides with a small parked car. Which statement correctly compares the force exerted on the car by the truck with the force exerted on the truck by the car during the collision?
Answer and reasoning
AThe truck exerts the larger force, since it has more mass. A student who believes the more massive object exerts the larger force picks this. The two forces are one interaction, so their magnitudes are equal whatever the masses. The truck's larger mass means the equal force changes the truck's motion less than the car's.
BThe car exerts no force, since a parked car cannot push back. A student who thinks a passive object cannot exert a force picks this. Being parked does not matter: while the truck pushes on the car, the car pushes back on the truck with a force of equal magnitude. That force is why the front of the truck is dented and the truck slows down.
CThe two forces have equal magnitudes throughout the collision.Correct The truck and the car interact, so by Newton's third law the force exerted on the car by the truck and the force exerted on the truck by the car have equal magnitudes and opposite directions at every instant of the collision. The car is damaged more because the same size of force acts on a smaller, weaker object.
DThe car pushes back later, in reaction to the truck's push. A student who thinks the 'reaction' comes after the 'action' picks this. The two forces of an interaction begin and end at the same instants: the car pushes on the truck for exactly as long as the truck pushes on the car.
A projectile is launched from level ground. If it did not explode, it would land a horizontal distance R from the launch point. At the highest point of its path, it explodes into two fragments. Fragment 1, which has one-third of the projectile's mass, falls straight down and lands directly below the highest point. Both fragments land at the same time, and air resistance is negligible. How far from the launch point does fragment 2 land?
Answer and reasoning
A2.00R A student who weights each position with the other fragment's mass writes R = (2/3)(R/2) + (1/3)x2 and gets 2R. Each position must be multiplied by the mass of the fragment that is there: the heavier fragment 2 lands nearer to R than the lighter fragment 1 does.
B1.00R A student who thinks a fragment keeps to the original path picks R, where the whole projectile would have landed. It is the center of mass of the two fragments, not either fragment, that lands at R; with fragment 1 at R/2, fragment 2 must land beyond R.
C1.25RCorrect The explosion's forces are internal, so the center of mass keeps to the original path and is at R when both fragments land. Fragment 1 lands at R/2, so R = (1/3)(R/2) + (2/3)x2, which gives x2 = 5R/4 = 1.25R. Fragment 2 lands beyond R because it carries more of the mass than fragment 1 and must balance it about R.
D0.50R A student who thinks the explosion stops the forward motion of the system puts the center of mass at R/2, directly below the explosion, and finds that fragment 2 must land there too. The explosion's forces are internal, so the center of mass keeps its horizontal motion and lands at R.
Working The forces in the explosion are internal to the two-fragment system, so they do not change the motion of its center of mass. The center of mass follows the original path and, because both fragments land at the same time, it is at xcm = R when they land. The path is symmetric, so the highest point is at R/2 and fragment 1 lands at x1 = R/2. xcm = (m1x1 + m2x2)/(m1 + m2) with m1 = M/3 and m2 = 2M/3: R = (1/3)(R/2) + (2/3)x2, so (2/3)x2 = R − R/6 = 5R/6 and x2 = 5R/4 = 1.25R.
The diagram shows two setups, each with a spring scale between two light strings. In setup 1, one string is tied to a wall and a student pulls the other string with a horizontal force of 50 N. In setup 2, two students pull the strings in opposite directions, each with a horizontal force of 50 N. Everything is at rest. How do the readings of the two scales compare?
Answer and reasoning
AScale 2 reads 100 N, double scale 1. A student who adds the pulls at the two ends picks this. Each end of scale 2 is pulled with 50 N, so the tension, and the reading, is 50 N. Scale 1 is also pulled with 50 N at each end, one of them by the wall, so the two setups are identical for the scales.
BScale 2 reads 0 N, as its pulls cancel. A student who thinks balanced end forces leave no tension picks this. The two pulls do give zero net force on the scale, but each end of the scale is still pulled with 50 N, and the spring inside is stretched by exactly that tension, so it reads 50 N.
CScale 1 reads 0 N, as a wall cannot pull. A student who thinks a passive object cannot exert a force picks this. The string pulls on the wall, and the wall pulls back on the string with a force of equal magnitude, 50 N. The wall does the same job as the second student in setup 2.
DScale 1 and scale 2 each read 50 N.Correct A scale reads the tension in it, the force it exerts on whatever pulls on each of its ends. In setup 1 the wall pulls on its string with 50 N, just as the second student does in setup 2, so the two scales are pulled in exactly the same way and both read 50 N.
Working Each scale reads the tension in it, which for light strings and a light scale equals the pull at each end. Setup 1: the student's string pulls the scale with 50 N; the scale pulls on the other string, which pulls on the wall, and the wall pulls back on that string with 50 N (Newton's third law), so the scale is pulled with 50 N at each end and reads 50 N. Setup 2: each end is pulled with 50 N, so the scale reads 50 N. The pulls are not added (that would give 100 N), and balanced end forces give zero net force on the scale, not zero tension.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.3.A.1 Newton's third law Fix
Newton's third law
When object A exerts a force on object B, object B exerts a force on object A of equal magnitude and opposite direction: F⃗A on B = −F⃗B on A. The two forces exist at the same instants, and the law holds whether the objects are at rest, moving steadily or speeding up, and whatever their masses.
Interaction
A mutual action between two objects or systems in which each exerts a force on the other. In AP Physics 1 an interaction is either by contact (normal, friction, tension and spring forces) or at a distance, and the only interaction at a distance is gravitational.
Third-law force pair
The two forces that make up one interaction. They act on different objects, so they never appear on the same free-body diagram and never cancel each other. They are of the same kind (both contact forces, or both gravitational) and last exactly as long as the interaction.
Force (unit: N)
A vector quantity that describes an interaction between two objects. F⃗A on B denotes the force exerted on object B by object A. The SI unit is the newton: 1 N = 1 kg·m/s².
Students often think In an interaction, the object with more mass exerts the larger force on the other object. In fact No. The two forces of an interaction always have equal magnitudes. The lighter object's motion changes more because a force of the same size acts on a smaller mass, not because it receives a larger force.
Students often think Passive objects, such as walls, floors, parked cars and loads being pulled, have forces exerted on them but do not exert forces themselves. In fact Yes. Every force is one side of an interaction, so an object that has a force exerted on it always exerts a force of equal magnitude and opposite direction on the other object, whether or not it is 'doing' anything.
2.3.A.2 Internal force Fix
Internal force
A force exerted by one object in a chosen system on another object in the same system. Internal forces occur in third-law pairs within the system, so they add to zero and cannot change the motion of the system's center of mass.
External force
A force exerted on an object in the system by an object outside the system. Only external forces can change the motion of a system's center of mass. Whether a force is internal or external depends on the choice of system: the rope's pull on a skater is internal to the system of both skaters and the rope, but external to that skater alone.
Center of mass of a system (xcm, unit: m)
The mass-weighted average position of the objects in a system: xcm = (m1x1 + m2x2 + …)/(m1 + m2 + …). Internal forces cannot change its motion, so if the external forces on a system are zero or balance, its center of mass keeps moving as before (or stays at rest), however the parts move relative to each other.
Students often think Internal forces, such as the forces in an explosion or between two people pushing off each other, change the motion of the system's center of mass. In fact No. Internal forces occur in third-law pairs within the system and add to zero, so they cannot change the motion of the center of mass. Only external forces can do that.
Students often think Because the two objects exert equal forces on each other, they move equal distances. In fact No. The forces on the two objects have equal magnitudes, but the center of mass stays fixed, so the heavier object moves a shorter distance: the displacements are inversely proportional to the masses.
2.3.A.3 Tension (unit: N) Fix
Tension (unit: N)
The magnitude of the pull exerted by a string, cable, chain or rope. It is the macroscopic result of the pulls that neighboring segments of the string exert on each other when the string is pulled at its ends by external forces. A string pulls along its own length; it cannot push.
Ideal string
A model of a string that has negligible mass and does not stretch when under tension. Because it does not stretch, the objects at its two ends move together along the string, with equal speeds.
Uniform tension in an ideal string
The tension in an ideal string has the same value at every point along it, whether the string is at rest or accelerating, so the string pulls on the objects at its two ends with forces of equal magnitude.
String with nonnegligible mass
When the mass of a string, rope or chain is not negligible, the tension can differ from point to point. In a chain hanging at rest, each point supports the part of the chain below it, so the tension is greater toward the top. AP Physics 1 treats this only qualitatively.
Ideal pulley
A pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction. An ideal string passing over an ideal pulley has the same tension on both sides: the pulley changes the direction of the string's pull but not its magnitude.
Students often think The tension in a string is the sum of the forces pulling on its two ends. In fact No. The tension is 50 N. Each end of the string pulls with the tension on whatever holds it, so a string pulled with 50 N at each end has a tension of 50 N; the two pulls are not added.
Students often think When the forces at the two ends of a string or chain balance, they cancel, so the tension in it is zero. In fact No. The equal pulls at the ends make the net force on the string zero, but each segment of the string still pulls on its neighbors, and the tension equals the size of each end pull.
10 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 10
A student stands at rest on a level floor. Earth exerts a downward gravitational force on the student. Which force forms a Newton's third-law pair with this force?
Answer and reasoning
AThe upward normal force exerted on the student by the floor A student who pairs any two equal and opposite forces picks this. The normal force and Earth's gravitational force both act on the student and come from different interactions (contact and gravitational). They balance because the student is at rest, not because of the third law.
BThe downward contact force the student exerts on the floor A student who treats the push on the floor as the student's weight picks this. That push is a contact force; its third-law partner is the floor's normal force on the student. The partner of a gravitational force exerted by Earth must be a gravitational force exerted on Earth.
CNo force, as the student is far too small to exert a force on Earth A student who thinks a small object cannot pull on a large one picks this. The student pulls on Earth with a gravitational force equal in magnitude to Earth's pull on the student. Earth's motion changes only negligibly because its mass is enormous, not because the force is absent.
DThe upward gravitational force exerted on Earth by the studentCorrect The downward force is one side of the gravitational interaction between Earth and the student. The other side is the student's gravitational force on Earth: equal in magnitude, upward, of the same kind, and exerted on the other object in the interaction.
A hand pushes block L to the right, and L pushes a smaller block S ahead of it. Both blocks speed up to the right along a level surface with negligible friction. The diagram shows a student's representation of the forces that L and S exert on each other: FL on S is the force exerted on S by L, and FS on L is the force exerted on L by S. Which statement about the student's diagram is correct?
Answer and reasoning
AThe two arrows should be drawn with the same length.Correct FL on S and FS on L are the two forces of one interaction, so by Newton's third law they have equal magnitudes and opposite directions at every instant, even while the blocks speed up. The directions in the diagram are right, but the arrows must be the same length.
BIt is correct, since block L has more mass than block S. A student who believes the more massive object exerts the larger force picks this. L and S exert forces of equal magnitude on each other whatever their masses. The blocks speed up because of the hand's push on L, which is not part of this interaction.
CBoth arrows should point to the right, the direction of motion. A student who believes the forces on a moving object point in its direction of motion picks this. S is pushed forward by L, but S in turn pushes backward on L; the force on L by S points to the left even though L moves to the right.
DThe arrow on L should be removed, since S is pushed, not pushing. A student who thinks a passive object exerts no force picks this. A block that is pushed pushes back: S exerts a force on L, to the left, with the same magnitude as the force L exerts on S.
Cart 1 rolls along a level track and collides with cart 2. Cart 1 has three times the mass of cart 2. Force sensors record the force exerted on cart 1 by cart 2 and the force exerted on cart 2 by cart 1; the graph shows the data, with the direction of cart 1's initial motion as positive. Which claim do the data support?
Answer and reasoning
ACart 1 exerted the larger force, since cart 1 has more mass. A student who believes the more massive object exerts the larger force picks this. The data contradict it: the two curves have the same size at every instant, with peaks of 12 N in magnitude, even though cart 1 has three times the mass.
BAt each instant, the forces had equal sizes and opposite directions.Correct At every instant the dashed curve (force on cart 1) is the mirror image of the solid curve (force on cart 2): the values have the same size and opposite signs, from start to finish, with peaks of +12 N and −12 N. This is the evidence for Newton's third law, F⃗1 on 2 = −F⃗2 on 1.
CThe forces cancel, so the net force on each cart was zero. A student who thinks third-law forces cancel picks this. The two forces act on different carts, so they cannot be added to give a net force on either cart. During the collision each cart has a nonzero force exerted on it, and each cart's velocity changes.
DThe force on cart 1 was the smaller, since its values are negative. A student who reads a negative value as a smaller force picks this. The negative sign shows only that the force on cart 1 points opposite to its initial motion. Ignoring the signs, the two curves reach the same size, 12 N, at the same instant.
A horse pulls a cart along a level road, and both speed up from rest. A student argues: “The cart pulls back on the horse with a force equal in magnitude to the horse’s pull on the cart, so these forces cancel and the cart cannot speed up.” Which statement correctly identifies the flaw in the argument?
Answer and reasoning
AThe two forces act on different objects, so they cannot cancel.Correct The horse's pull acts on the cart and the cart's pull acts on the horse. Forces cancel only when they act on the same object, so whether the cart speeds up depends on the horizontal forces exerted on the cart itself: the horse's forward pull and the road's backward force on the cart.
BThe horse pulls harder than the cart pulls back while they speed up. A student who thinks equal and opposite forces hold only at rest or constant velocity picks this. The horse and cart pull on each other with equal magnitudes at every instant, including while they speed up. The student's error is in what the forces act on, not in their sizes.
CThe cart pulls back after the horse's pull has set it moving. A student who thinks the 'reaction' comes after the 'action' picks this. The two forces begin and end together: from the first instant the horse pulls on the cart, the cart pulls back on the horse with a force of equal magnitude.
DA cart cannot pull on anything, so no force acts back on the horse. A student who thinks a passive object exerts no force picks this. The cart does pull back on the horse through the harness, with a force equal in magnitude to the horse's pull; the horse feels it as the load it is pulling.
Carts X and Y are held at rest on a level track with a compressed spring between them. Friction is negligible. When the carts are released, the spring pushes them apart. Cart X has twice the mass of cart Y. When cart Y has moved 0.40 m from its starting position, how far has cart X moved from its starting position?
Answer and reasoning
A0.40 m A student who reasons that equal forces produce equal distances picks this. The spring does push on the two carts with forces of equal magnitude, but cart X has twice the mass, and because the center of mass stays fixed it moves half as far as cart Y.
B0.20 mCorrect The spring's forces are internal to the system of both carts and the spring, and friction is negligible, so the center of mass stays at rest. mXΔxX + mYΔxY = 0 gives |ΔxX| = (mY/mX)(0.40 m) = (1/2)(0.40 m) = 0.20 m: doubling the mass halves the distance moved.
C0.80 m A student who multiplies by the wrong mass ratio gets (mX/mY)(0.40 m) = 0.80 m, which would have the heavier cart moving farther. With the center of mass fixed, the heavier cart moves the shorter distance: mXΔxX = −mYΔxY.
D0.13 m A student who carries the total mass into the ratio calculates 0.40 m × mY/(mX + mY) = 0.40 m × 1/3 = 0.13 m. The total mass is used only to locate the center of mass; the ratio of the two distances is mY/mX = 1/2.
Working System: carts X and Y and the spring. The spring's forces on the carts are internal, and friction is negligible, so the center of mass of the system stays at rest where it started. Then mXΔxX + mYΔxY = 0, so |ΔxX| = (mY/mX)|ΔxY| = (1/2)(0.40 m) = 0.20 m. The displacement of each cart is inversely proportional to its mass.
In an experiment, a cart on a level track is connected by a string that passes over a pulley to an object hanging below the pulley. The students want to model the string as an ideal string. Which observation would be evidence that this model is not appropriate?
Answer and reasoning
AThe cart and the hanging object speed up instead of moving steadily. A student who thinks the pulls at the two ends of a string are equal only at constant velocity picks this. An ideal string has negligible mass, so it pulls equally on the cart and the hanging object even while they speed up. Speeding up is fully consistent with the model.
BThe string becomes noticeably longer once the object is released.Correct An ideal string does not stretch when under tension. A string that visibly lengthens as the tension in it builds up breaks that assumption, so the cart and the hanging object no longer move together through equal distances, and the ideal-string model does not apply.
CThe tensions measured on the two sides of the pulley are equal. A student who thinks a pulley changes the tension in a string picks this, expecting different tensions on the two sides. With an ideal string passing over an ideal pulley the tension is the same on both sides, so equal readings are exactly what the ideal-string model predicts, not evidence against it.
DThe tension in the string is less than the hanging object's weight. A student who thinks the tension must always equal the weight of the hanging object picks this. The ideal-string model is about the string's mass and stretch. How the tension compares with the hanging object's weight depends on how the object moves, not on whether the string is ideal, so this observation says nothing about the model.
A student pulls a sled across level snow using a light rope that can be modeled as an ideal string, and the sled speeds up. Light force sensors measure the tension at point X, next to the student's hands, and at point Y, next to the sled. How do the two readings compare?
Answer and reasoning
AThey are equal at every instant while the sled speeds up.Correct The rope is modeled as an ideal string, and the tension in an ideal string is the same at every point, whether the rope is at rest or speeding up. With negligible mass, no difference in tension between X and Y is needed to accelerate the rope itself.
BX reads more, since the sled speeds up the whole time. A student who thinks the leading end must pull harder while things speed up picks this. That would be true of a rope with significant mass, which needs a net force of its own to speed up; an ideal rope does not, so the tension is the same at X and Y.
CY reads zero, since the sled does not pull back on the rope. A student who thinks a passive object exerts no force picks this. The rope pulls forward on the sled, and the sled pulls back on the rope with a force of equal magnitude; that pull is the tension the sensor at Y measures.
DBoth read zero, since the pulls at the two ends cancel. A student who thinks equal end pulls leave a string with no tension picks this. The student's hands pull the rope forward and the sled pulls it back, but each segment of the rope still pulls on its neighbors: the tension is the size of each end pull, not zero.
A uniform heavy chain hangs at rest from a hook in the ceiling, as shown in the diagram. Point P is near the top of the chain, Q is at its middle, and R is near its bottom. Which statement about the tension in the chain is correct?
Answer and reasoning
AIt is the same at P, Q and R, as it is along any string. A student who applies the ideal-string rule to every string picks this. The tension is uniform only in a string of negligible mass. A heavy chain has mass all along it, so points higher up support more chain and have greater tension.
BIt is greatest at R, where the weight of the chain pulls down. A student who pictures the chain's weight acting at its bottom picks this. Weight acts on every link, and R holds up only the short piece of chain below it, so the tension at R is the smallest of the three.
CIt is greatest at P, which supports the most chain below it.Correct The chain's mass is not negligible, so its tension can vary. The part of the chain below each point is held up by the pull of the chain just above that point. P supports almost the whole chain and R very little, so the tension is greatest at P, less at Q and least at R.
DIt is zero at P, Q and R, as the hook's pull and gravity cancel. A student who thinks balanced forces leave no tension picks this. The net force on the chain as a whole is zero, but every link is pulled up by the link above it and down by the link below it, so the tension is not zero anywhere except at the free bottom end.
Block A slides on a horizontal table with negligible friction. An ideal string attached to A passes over an ideal pulley and supports block B, as shown in the diagram, which also shows the weight of B. As the blocks speed up, the tension in the string at point X is 4.0 N. What is the tension in the string at point Y?
Answer and reasoning
A8.0 N A student who thinks a pulley doubles the force in a string picks 2 × 4.0 N = 8.0 N. A single fixed pulley only changes the direction of the pull; with an ideal pulley and string, the tension at Y equals the tension at X.
B2.0 N A student who thinks the two parts of the string share the tension picks 4.0 N ÷ 2 = 2.0 N. The string is one ideal string, so each part carries the full tension of 4.0 N.
C6.0 N A student who takes the tension to equal the weight of the hanging block picks 6.0 N. That is true only if B is at rest or moves at constant velocity. B is speeding up as it moves down, and the tension, 4.0 N all along the string, is less than B's weight.
D4.0 NCorrect An ideal pulley has negligible mass and turns without friction, and the string is ideal, so the tension is the same on both sides of the pulley: 4.0 N at Y, as at X. The pulley changes the direction of the string's pull, not its size.
Working The string is ideal (negligible mass) and the pulley is ideal (negligible mass, frictionless axle), so the tension is the same at every point of the string, on both sides of the pulley. Tension at Y = tension at X = 4.0 N. (It is less than B's 6.0 N weight because B speeds up as it moves down.)
A person of mass m stands at one end of a uniform raft of mass 4m and length L. The raft floats at rest on still water, and the water exerts a negligible horizontal force on it. The person walks to the other end of the raft and stops. How far does the raft move relative to the water?
Answer and reasoning
A0.80L A student who weights each displacement by the other object's mass writes 4m(L − D) = mD and picks this, so that the more massive raft moves farther. Each displacement must be multiplied by its own object's mass: the raft, four times as massive, moves a quarter as far as the person.
B0.50L A student who reasons that the person and the raft push on each other with equal forces, and so move equal distances, splits L into two halves. For the center of mass to stay at rest, the less massive person must move farther than the raft: four times as far.
C0.25L A student who takes the person's displacement relative to the water to be L writes mL = 4mD and picks this. The person walks L relative to the raft; because the raft moves D backward, the person moves only L − D relative to the water.
D0.20LCorrect The forces between the person and the raft are internal, so the center of mass of the person–raft system stays at rest. If the raft moves D backward, the person moves L − D forward relative to the water, and m(L − D) = 4mD gives D = L/5 = 0.20L.
Working The forces between the person's feet and the raft are internal to the person–raft system, and no net external horizontal force is exerted on the system, so its center of mass stays at rest horizontally. Let the raft move a distance D backward (opposite to the person's walk). Relative to the water the person then moves L − D forward. Center of mass fixed: m(L − D) = 4m·D, so L = 5D and D = L/5 = 0.20L. (Each displacement weighted by the other object's mass: 4m(L − D) = mD gives D = 0.80L. Equal distances: 0.50L. Person's displacement relative to the water taken as L: mL = 4mD gives D = 0.25L.)
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
