10 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 10
Earth's mass is about 81 times the mass of the Moon. How does the magnitude of the gravitational force exerted on the Moon by Earth compare with the magnitude of the gravitational force exerted on Earth by the Moon?
Answer and reasoning
AThe force on the Moon is 81 times as large. A student who believes the more massive body exerts the larger force picks this. In G m1m2/r² the two masses appear as a product, which is the same number for the force on the Moon and the force on Earth, so the forces have equal magnitudes.
BThe force on each has exactly the same magnitude.Correct Newton's law of universal gravitation gives the magnitude of both forces as G mEarth mMoon/r², with the same product of masses and the same distance r, so they are equal; they are the two forces of one interaction, a Newton's third-law pair. The Moon's motion changes far more only because the same size of force acts on a much smaller mass.
CThe force on Earth is 81 times the force on the Moon. A student who thinks the force on a body depends only on that body's own mass, as Fg = mg seems to say, picks this. Both masses appear in G m1m2/r²: the Moon's small mass reduces its pull on Earth by exactly as much as Earth's large mass increases it, so the two forces are equal in magnitude.
DThe force on Earth is zero; the Moon is too small. A student who thinks a small body cannot pull on a much larger one picks this. Every object with mass attracts every other object with mass: the Moon pulls on Earth with a force equal in magnitude to Earth's pull on the Moon. Earth's motion changes only slightly because its mass is so large.
Point P is in empty space near Mars, and there is no object at P. Which statement about the gravitational field of Mars at point P is correct?
Answer and reasoning
AThere is no field at P until an object is placed there to feel it. A student who thinks a field exists only where an object detects it picks this. The field is created by Mars, and GM/r² contains only Mars's mass and the distance from its center; an object placed at P measures the field but does not create it.
BThe field at P would be stronger if a more massive object were placed there. A student who confuses field strength with force picks this. A more massive object at P would have a larger force exerted on it, but in proportion to its mass, so the force per kilogram, the field strength, is the same for any object.
CIt has a value at P giving the force per kilogram on any object put there.Correct A field models the effect of a noncontact force at every point in space. Mars's gravitational field exists at P whether or not anything is there; its value tells us the gravitational force per kilogram that any object placed at P would experience, and it is set by Mars's mass and the distance from Mars's center.
DThere is no field at P if P lies above the top of Mars's atmosphere. A student who thinks gravity needs air, or stops where space begins, picks this. Mars's field extends through empty space, getting weaker with distance from its center but never stopping at the top of the atmosphere.
A book rests on a table on Earth. Which of the following is the book's weight?
Answer and reasoning
AThe downward push that the book exerts on the table A student who takes weight to be the force an object exerts on its support picks this. That push is a contact force exerted on the table, not on the book; it equals mg in magnitude here only because the book is at rest on a level table.
BThe gravitational pull the book exerts on Earth A student who mixes up the force exerted on an object with the force exerted by it picks this. This force is the third-law partner of the book's weight: it has the same magnitude but is exerted on Earth, not on the book.
CThe amount of matter in the book, measured in kilograms A student who treats mass and weight as the same quantity picks this. The amount of matter, in kilograms, is the book's mass; its weight is the gravitational force on it, mg, in newtons, which would be smaller on the Moon.
DThe gravitational force that Earth exerts on the bookCorrect Weight is the gravitational force exerted by an astronomical body, here Earth, on a relatively small nearby object: Fg = mg, directed toward Earth's center. It is a force exerted on the book, measured in newtons.
A ball is thrown straight up and rises 20 m above the ground before falling back. Earth's radius is 6.4 × 10⁶ m. Which statement about the gravitational force exerted on the ball by Earth during its flight is correct?
Answer and reasoning
AIt is noticeably weaker at the top, where the ball is farthest from Earth. A student who judges the change in distance from Earth's surface picks this. What matters is the distance from Earth's center, 6.4 × 10⁶ m, and 20 m more makes a negligible difference to it.
BIt is zero at the top, where the ball is momentarily at rest. A student who thinks no force acts on an object at rest picks this. The gravitational force depends on the masses and the distance between the centers, not on the velocity; at the top it is still mg downward, which is why the ball falls back.
CIt has practically the same magnitude and direction throughout.Correct The ball's distance from Earth's center changes from 6.4 × 10⁶ m by at most 20 m, about 3 parts in a million, so the force, proportional to 1/r², changes by only about 6 parts in a million. The change is negligible, so the force can be treated as constant: mg, directed toward Earth's center, at every point of the flight.
DIt points upward while the ball rises and downward while it falls. A student who thinks the force on a moving object points in its direction of motion picks this. Earth's gravitational force always points toward Earth's center; while the ball rises it slows down precisely because this force points opposite to its velocity.
Near the surface of Earth, the gravitational field strength is g ≈ 10 N/kg. Which statement correctly describes what this value means?
Answer and reasoning
AA freely falling object near Earth's surface falls 10 m in each second. A student who reads 10 m/s² as a distance per second picks this. The free-fall acceleration of about 10 m/s² means the velocity increases by 10 m/s each second; from rest an object falls about 5 m in the first second and 15 m in the next.
BEvery planet and every moon also has a field of about 10 N/kg at its surface. A student who treats g as a universal constant picks this. The surface field is GM/R², which depends on the body's mass and radius: about 1.6 N/kg on the Moon and 3.7 N/kg on Mars.
CEarth exerts about 10 N per kilogram on an object only while it falls. A student who thinks gravity acts only on falling objects picks this. An object at rest on a table has the same gravitational force exerted on it; the table's normal force balances it, which is why the object does not fall.
DEarth pulls on each kilogram of any nearby object with about 10 N.Correct Field strength is gravitational force per unit mass, g = Fg/m. A value of about 10 N/kg means Earth exerts about 10 N on each kilogram of an object near its surface, whatever the object is doing: about 10 N on 1 kg, 50 N on 5 kg.
Working g = |F⃗g|/m, so 10 N/kg means a gravitational force of about 10 N on each 1 kg: Fg ≈ 10 N on a 1 kg object and 50 N on a 5 kg object, whether the object is falling or at rest.
A 60 kg student stands on a scale in an elevator. The elevator is moving upward and speeding up, with an acceleration of magnitude 2.0 m/s². Use g = 10 m/s². What is the reading on the scale?
Answer and reasoning
A720 NCorrect The scale reads the normal force. With up as positive, FN − mg = ma, so FN = m(g + a) = 60 kg × (10 + 2.0) m/s² = 720 N. The student is accelerating upward, so the scale must push up harder than gravity pulls down, and her apparent weight is greater than her 600 N weight.
B600 N A student who thinks a scale always reads the gravitational force picks mg = 600 N. That is the reading only when the elevator is not accelerating; here the net force must be upward, so the normal force exceeds 600 N.
C480 N A student who takes the acceleration with the wrong sign calculates m(g − a) = 480 N. That would give a net force downward, but the acceleration is upward, so FN − mg = +ma and FN = 720 N.
D120 N A student who takes the scale reading to be the net force calculates ma = 60 kg × 2.0 m/s² = 120 N. The net force is FN − mg; the scale reads FN itself, 600 N + 120 N = 720 N.
Working Up positive: FN − mg = ma → FN = m(g + a) = (60 kg)(10 m/s² + 2.0 m/s²) = 720 N.
Astronauts aboard the International Space Station, about 400 km above Earth's surface, float freely, and a bathroom scale held under an astronaut's feet reads zero. Which statement correctly explains why the astronauts appear weightless?
Answer and reasoning
AThere is no gravity at the station's height to pull them toward Earth. A student who thinks there is no gravity in space picks this. At the station's height the gravitational force on an astronaut is still almost 90% of that at Earth's surface; without it the station would not stay near Earth.
BAn outward force balances Earth's gravity, so the net force on them is zero. A student who believes in a balancing outward force picks this. In an inertial frame, the only force exerted on an orbiting astronaut is gravity, and the net force is not zero. The astronaut appears weightless because nothing else, such as a floor, pushes on her.
CGravity is the only force on them and the station, so no floor pushes on them.Correct At 400 km Earth's field is still almost 90% of its surface value. The astronauts and the station have only the gravitational force exerted on them, so they fall together; the floor and the scale exert no normal force on the astronauts, and an apparent weight of zero is what 'weightless' means.
DTheir mass is zero in orbit, so the scale has nothing to measure. A student who treats mass and weight as the same picks this. An astronaut's mass is the same in orbit as on Earth; the scale reads zero because it exerts no normal force on the astronaut, not because the mass has gone.
The diagram shows two identical closed rooms with no windows. Room 1 is at rest on Earth's surface. Room 2 is in a rocket far from any planet or star, speeding up with the acceleration shown. In each room a student stands on a scale and releases a ball from rest; the diagram gives each scale reading, the ball's height and the time it takes to reach the floor. Use g = 10 m/s². Which claim do these measurements support?
Answer and reasoning
ANeither student can tell from these readings which room she is in.Correct The measurements are identical: each scale reads 600 N and each ball falls 1.8 m in 0.60 s, an acceleration of 10 m/s² relative to the floor in both rooms. This is the equivalence principle: an observer in the accelerating room cannot distinguish the apparent weight caused by the acceleration from the gravitational force of a field.
BThe second student's reading shows a gravitational field acts. A student who thinks only gravity can produce an apparent weight picks this. Room 2 is far from any planet, so the gravitational field there is negligible; the floor pushes up with 600 N because it accelerates the student.
CA heavier ball would fall faster in room 1 only, revealing the difference. A student who believes heavier objects fall faster under gravity picks this. With gravity the only force, every ball in room 1 accelerates at 10 m/s²; in room 2 every released ball also reaches the floor in the same time, so a heavier ball would not reveal anything.
DEach student's weight is 600 N, as weight does not depend on location. A student who thinks weight is a fixed property picks this. Weight is the gravitational force exerted by a nearby astronomical body; in room 2, far from any planet, the student's weight is almost zero, even though her apparent weight is 600 N.
Working In each room the ball falls Δy = 1.8 m in t = 0.60 s from rest: a = 2Δy/t² = 2(1.8 m)/(0.60 s)² = 10 m/s² relative to the floor. Each scale reads 600 N. The measurements in the two rooms are identical.
On the International Space Station, an astronaut has two closed toolboxes that look identical; one is full of tools and the other is empty. Both float freely. Which method would let her tell which toolbox is full?
Answer and reasoning
ARest each one on her palm and feel which one presses down harder. A student who thinks an object always presses on its support with its weight picks this. In the station the toolboxes and her hand fall together, so neither toolbox presses on her palm: both have zero apparent weight.
BNone will work: with no weight, every object is as easy to start and stop. A student who thinks inertia comes from weight picks this. Inertial mass is a property of the object, not of gravity: the full toolbox is as hard to speed up or stop in orbit as it is on a frictionless surface on Earth.
CPush each with the same small force; see which speeds up less.Correct Inertial mass determines how much an object's motion resists change, and it does not depend on gravity. With the same force, the full toolbox, with the greater inertial mass, speeds up less (a = Fnet/m), in orbit just as on Earth.
DRelease both at the same height and see which one drifts down faster. A student who believes heavier objects fall faster picks this. Both toolboxes, the astronaut and the station fall with the same acceleration, so neither toolbox drifts toward the floor; even on Earth, with negligible air resistance, they would fall together.
On Earth, a rock on one pan of an equal-arm balance is balanced by 4.0 kg of standard masses on the other pan. The balance, the rock and the standard masses are taken to the Moon, where the gravitational field strength is 1.6 N/kg. Use g = 10 m/s² on Earth. What mass of standard masses balances the rock on the Moon?
Answer and reasoning
AOnly 0.64 kg of standard masses is needed to balance it. A student who expects the balance to behave like a spring scale scales the result by 0.16 and gets 0.64 kg. The standard masses on the other pan also weigh 0.16 times as much on the Moon, so the same 4.0 kg still balances.
BA much larger total, 25 kg of standard masses, is now needed. A student who expects the balance result to change with g, as a spring-scale reading would, and then uses the ratio of field strengths upside down calculates 4.0 kg × 10/1.6 = 25 kg. The balance compares the gravitational forces on the two pans, which both change by the same factor, 1.6/10, on the Moon, so the result does not change at all: 4.0 kg still balances.
CA total of 6.4 kg of standard masses balances the rock. A student who calculates the rock's weight on the Moon, 4.0 kg × 1.6 N/kg = 6.4 N, and reads it as a mass picks this. A weight in newtons is not a mass in kilograms, and the balance compares masses: 4.0 kg still balances.
DThe same 4.0 kg of standard masses still balances the rock.Correct An equal-arm balance compares the gravitational forces on the two pans, which depend on the gravitational masses. On the Moon both forces are multiplied by the same factor, 1.6/10 = 0.16, so they still balance: mrock g = mstd g for any g. The rock's gravitational mass is 4.0 kg wherever it is.
Working Balance condition: mrock g = mstd g, so mstd = mrock for any g. On the Moon both pans' gravitational forces are 0.16 times their Earth values (4.0 kg × 1.6 N/kg = 6.4 N each), so 4.0 kg still balances.
In preparation: 0 of 12 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.6.A.1 Newton's law of universal gravitation Fix
Newton's law of universal gravitation
|F⃗g| = G m1m2/r²: any two objects or systems with mass attract each other with a force whose magnitude is proportional to each of their masses and inversely proportional to the square of the distance r between their centers of mass. The two forces of the interaction have equal magnitudes whatever the masses. Unit of force: N.
Universal gravitational constant, G
The constant of proportionality in Newton's law of universal gravitation: G = 6.67 × 10⁻¹¹ N·m²/kg², the same for every pair of objects everywhere. Its small size is why the gravitational forces between everyday objects are tiny. It is not the same quantity as g, the gravitational field strength at a particular place.
Inverse-square dependence
A quantity that is inversely proportional to r² changes by the factor (r1/r2)² when the distance changes from r1 to r2: doubling r divides the gravitational force by 4, tripling r divides it by 9. The force becomes smaller and smaller with distance but never exactly zero.
Attractive gravitational force
The gravitational force exerted on an object by another object always points toward the other object: gravity pulls objects with mass together and never pushes them apart. It acts between any two objects with mass, in space as on Earth, and needs no air or contact.
Line of centers
The gravitational force between two objects is exerted along the line joining their centers of mass. The force on an object exerted by Earth points toward Earth's center, the force exerted by the Moon points toward the Moon's center, whatever the direction in which the object is moving.
Center of mass as the point of application of the gravitational force
The gravitational force on a system can be treated as exerted at the system's center of mass. For uniform spheres, such as planets or balls, the distance r in Newton's law is measured between the centers, not between the surfaces; for an object at height h above a planet of radius R, r = R + h.
Students often think In a gravitational interaction, the more massive object exerts the larger force on the other object. In fact No. The forces that the two bodies exert on each other have equal magnitudes. Newton's law gives both as G m1m2/r², with the same product of masses and the same distance, and they form a Newton's third-law pair.
Students often think The gravitational force on an object depends only on that object's own mass (as Fg = mg seems to say), not on the mass of the other object. In fact No. The force between two objects is proportional to the product of both masses. Doubling either mass doubles the force on each object; doubling both multiplies it by 4.
2.6.A.2 Field Fix
Field
A model that assigns to every point in space a quantity describing the noncontact force an object would experience if it were placed there. The gravitational field of a planet exists at a point whether or not an object is there, and does not depend on the object that is later placed there.
Gravitational field strength, g
The gravitational force per unit mass on a test object at a point: |g⃗| = |F⃗g|/m = GM/r², where M is the mass of the system creating the field and r is the distance from its center of mass. Unit: N/kg. It is a vector directed toward the center of mass of the system creating the field.
Test object
An object placed at a point to measure a field, with mass small enough that it does not noticeably change the field it measures. Dividing the gravitational force on it by its mass gives the field strength, a value that is the same whichever test object is used.
Free-fall acceleration
When the gravitational force is the only force exerted on an object, the object's acceleration in m/s² is numerically equal to the gravitational field strength in N/kg at its location (1 N/kg = 1 m/s²), whatever the object's mass. Near Earth's surface this is about 10 m/s²; on the Moon about 1.6 m/s².
Students often think A gravitational field exists at a point only when an object is placed there; empty space has no field. In fact Yes. The field is created by the planet (or other mass) and exists at every point around it; it describes the force an object would experience if it were placed there.
Students often think The gravitational field strength at a point depends on the mass of the object placed there: a heavier object is in a stronger field. In fact No. The gravitational force on a heavier object is larger, but in proportion to its mass, so the force per unit mass, the field strength, is the same for any object placed at that point.
2.6.A.3 Weight Fix
Weight
The gravitational force exerted by an astronomical body, such as Earth, the Moon or Mars, on a relatively small nearby object: Fg = mg, measured in newtons (N). An object's weight depends on where it is (on g), whereas its mass, measured in kilograms (kg), does not.
Students often think An object's weight is the downward force it exerts on whatever supports it (the table, floor or scale). In fact No. Weight is the gravitational force exerted on the object by Earth (or another astronomical body). The push of the object on its support is a contact force exerted on the support; it equals the weight in magnitude only in special cases, such as an object at rest on a level surface.
Students often think The force exerted on an object and the force exerted by that object are treated as the same force; the object on which a force is exerted is not distinguished from the object exerting it. In fact No. They are the two forces of one interaction, a third-law pair: equal in magnitude, opposite in direction, exerted on different objects. The book's weight is the one exerted on the book.
2.6.B.1 Constant gravitational force model Fix
Constant gravitational force model
If the distance between the centers of mass of two systems changes by a negligible fraction as they move, the gravitational force can be treated as constant between the initial and final positions. A change in height of tens of meters near Earth (radius 6.4 × 10⁶ m) is negligible; a change of hundreds of kilometers is not.
Students often think Because gravity is constant near Earth's surface, the gravitational force (or field) stays the same at any distance from the planet. In fact No. It can be treated as constant only when the change in distance from Earth's center is a negligible fraction of that distance. At 400 km above the surface the force is about 11% less than at the surface.
Students often think When an object is momentarily at rest, such as a ball at the top of its flight, no force (or no gravitational force) is exerted on it. In fact No. The gravitational force depends on the masses and the distance between the centers, not on the velocity. At the top the force is the same, mg downward, as everywhere else in the flight.
2.6.B.2 Gravitational field strength near Earth's surface Fix
Gravitational field strength near Earth's surface
g ≈ 10 N/kg near Earth's surface: Earth exerts about 10 N on each kilogram of a nearby object. The Table of Information gives 9.8 N/kg; this bank uses 10 N/kg (and g = 10 m/s²) whenever a numerical value is needed.
Students often think An acceleration (or field strength) of 10 m/s² means that a falling object moves 10 m in each second, or moves at 10 m/s. In fact No. It means the velocity changes by 10 m/s in each second. The distance fallen in each second increases: from rest, 5 m in the first second, 15 m in the next.
Students often think Gravity acts on an object only while it is falling; an object at rest on a support has no gravitational force exerted on it. In fact On both. Earth exerts a gravitational force of about 10 N per kilogram on any object near its surface, whether it is falling, at rest on a support, or moving upward.
2.6.C.1 Apparent weight Fix
Apparent weight
The magnitude of the normal force exerted on a system by the surface supporting it, which is what a bathroom scale reads. Unit: N. It equals the magnitude of the gravitational force when the surface is level, no other force on the system has a vertical component, and the system has no vertical acceleration; on a ramp or in a vertically accelerating elevator it differs from mg.
Normal force, FN
The contact force exerted on an object by a surface, perpendicular to the surface and directed away from it. A scale under an object reads the magnitude of the normal force it exerts on the object (by Newton's third law, also the magnitude of the force the object exerts on the scale). Unit: N.
Students often think A scale always reads an object's weight, the gravitational force mg; its reading cannot change unless the gravitational force changes. In fact No. A scale reads the normal force it exerts on the object, the apparent weight. This equals mg when the scale is level, no other force on the object has a vertical component, and the object has no vertical acceleration.
Students often think On an incline, the component of the gravitational force perpendicular to the surface is mg sin θ (sine and cosine are swapped). In fact No. The perpendicular component is mg cos θ and the component along the ramp is mg sin θ. At θ = 0 (level ground) the perpendicular component must be the whole weight, and cos 0 = 1.
2.6.C.2 Apparent weight of an accelerating system Fix
Apparent weight of an accelerating system
For a system on a scale that accelerates vertically, FN − mg = ma with up positive, so FN = m(g + a). An upward acceleration (speeding up upward or slowing down while moving downward) makes the apparent weight larger than mg; a downward acceleration makes it smaller. What matters is the direction of the acceleration, not of the velocity.
Students often think When applying Fnet = ma to a vertically accelerating object, the sign of the acceleration is chosen inconsistently with the chosen positive direction, giving FN = m(g − a) for an upward acceleration. In fact No. With up as positive, FN − mg = ma, so FN = m(g + a): an upward acceleration requires the scale to push up harder than gravity pulls down.
Students often think The reading of a scale under an accelerating object is the net force on the object, ma. In fact No. It reads the normal force. The net force is the difference between the normal force and the gravitational force, FN − mg = ma, which is much smaller than the reading.
2.6.C.3 Apparent weightlessness Fix
Apparent weightlessness
A system appears weightless (apparent weight zero) when no forces are exerted on it, or when the gravitational force is the only force exerted on it, as for astronauts in an orbiting space station or a person in a freely falling elevator. The gravitational force on the system is not zero in the second case.
Students often think An object in orbit is weightless because an outward (centrifugal) force balances the gravitational force, so the net force on it is zero. In fact No. In an inertial frame no outward force acts on them. Gravity is the only force on the astronauts and the station, so both fall together, the floor exerts no normal force, and the astronauts' apparent weight is zero.
2.6.C.4 Noninertial reference frame Fix
Noninertial reference frame
A reference frame that is accelerating, such as a room in a rocket that is speeding up; in it, an observer would not verify Newton's first law. Objects released from rest in the frame appear to accelerate relative to it even though no additional force is exerted on them.
Equivalence principle
An observer in a noninertial reference frame cannot distinguish an object's apparent weight from the gravitational force exerted on it by a gravitational field: inside a closed room, being at rest in a field of 10 N/kg and accelerating at 10 m/s² far from any planet give the same scale readings and the same falls.
Students often think An apparent weight can only be produced by gravity, so a nonzero scale reading shows that a gravitational field acts on the object. In fact No. By the equivalence principle, an observer in an accelerating room cannot distinguish the apparent weight produced by the acceleration from the gravitational force produced by a field; the same reading can arise either way.
2.6.D.1 Inertial mass Fix
Inertial mass
The property of an object that determines how much its motion resists change when it interacts with another object: for a given net force, the larger the inertial mass, the smaller the acceleration (m = Fnet/a). Unit: kg. It does not depend on gravity, so it is the same in orbit as on Earth.
Students often think An object's inertia comes from its weight (or from friction), so in weightless conditions every object is equally easy to start moving and to stop. In fact No. An object's inertial mass does not depend on gravity: a full toolbox in orbit takes a larger force to give it the same acceleration as an empty one, just as on Earth.
Students often think The slope of a graph is taken to be the quantity sought, even when the graph's axes make the slope its reciprocal (for example, the slope of an a–Fnet graph read as the mass). In fact No. From a = Fnet/m, the slope Δa/ΔFnet is 1/m, so the mass is the reciprocal of the slope: m = ΔFnet/Δa.
2.6.D.2 Gravitational mass Fix
Gravitational mass
The property of an object that determines the size of the gravitational force between it and another system with mass (Fg = mg in a field g). Unit: kg. An equal-arm balance compares gravitational masses, so it gives the same result wherever it is used.
Students often think An equal-arm balance measures weight like a spring scale, so its result falls in proportion to g in a weaker field. In fact No. An equal-arm balance compares the gravitational forces on the two pans. Both forces change by the same factor on the Moon, so the same standard masses balance the object: it compares gravitational masses.
2.6.D.3 Equivalence of inertial and gravitational mass Fix
Equivalence of inertial and gravitational mass
Experiments have verified that an object's inertial mass and gravitational mass are equivalent. As a result the acceleration produced by gravity alone, a = (mgrav g)/minertial = g, is the same for every object at a given place, whatever its mass or material.
Students often think Objects that fall with the same acceleration must have equal gravitational forces exerted on them. In fact No. The gravitational force on the hammer is much larger, in proportion to its larger gravitational mass. They have the same acceleration because each force is in the same proportion to each object's inertial mass.
18 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 18
Two objects exert gravitational forces of magnitude F on each other. The mass of each object is doubled, and the distance between their centers is tripled. What is the new magnitude of the gravitational force that each object exerts on the other?
Answer and reasoning
A(4/9)FCorrect The force is proportional to the product of the masses and inversely proportional to the square of the distance. Doubling both masses multiplies the product by 2 × 2 = 4; tripling the distance divides by 3² = 9. The new force is (4/9)F.
B(4/3)F A student who treats the force as inversely proportional to the distance, not to its square, divides by 3 instead of 9 and gets (4/3)F. Tripling the distance divides the force by 3² = 9.
C(2/9)F A student who thinks the force on an object depends only on that object's own mass counts just one doubling and gets (2/9)F. Both masses appear in G m1m2/r², so doubling each of them multiplies the force by 4.
D(2/3)F A student who 'squares' 3 by doubling it divides by 6 instead of 9: (4/6)F = (2/3)F. Squaring means multiplying a number by itself, so tripling the distance divides the force by 9.
Working |F⃗g| = G m1m2/r². New force = G(2m1)(2m2)/(3r)² = (2 × 2)/(3²) × G m1m2/r² = (4/9)F.
A space probe is placed at different distances r from the center of a planet. The table shown gives the magnitude F of the gravitational force exerted on the probe by the planet at each distance. Which claim do the data support?
Answer and reasoning
AF falls to one-half of its value each time r doubles. A student who expects the force to be inversely proportional to the distance picks this. The table contradicts it: from r = 1.0 to 2.0 (× 10⁷ m), F falls from 3600 N to 900 N, one-fourth, not one-half (which would be 1800 N).
BF will reach zero a little beyond r = 4.0 × 10⁷ m. A student who thinks gravity has a limited range picks this, extending the trend in a straight line. The decreases get smaller and smaller (2700 N, 500 N, 175 N), and under the inverse-square pattern the data follow, F never reaches zero: at 8.0 × 10⁷ m it is still 3600 N/64 ≈ 56 N.
CThe probe exerts no gravitational force on the planet. A student who thinks a small object cannot pull on a large one picks this. The data give only the force on the probe, so they cannot support this claim, and it is false: the probe pulls on the planet with a force of the same magnitude as the planet's pull on the probe.
DF falls to one-fourth of its value each time r doubles.Correct When r doubles from 1.0 to 2.0 (× 10⁷ m), F falls from 3600 N to 900 N, one-fourth; when r doubles from 2.0 to 4.0, F falls from 900 N to 225 N, again one-fourth. The row at 3.0 fits too: 3600 N/3² = 400 N. F × r² is the same for every row, so F is inversely proportional to r², as Newton's law of universal gravitation predicts.
An astronaut floats at rest near her spacecraft, far from any planet or star. Which statement about gravitational forces between the astronaut and the spacecraft is correct?
Answer and reasoning
AEach pulls the other, but the craft pulls harder. A student who thinks the more massive object exerts the larger force picks this. The two pulls form a Newton's third-law pair: each has magnitude G mastronaut mcraft/r², so they are equal in magnitude. The same force changes the astronaut's motion more only because her mass is smaller.
BNeither pulls; gravity needs a planet or a star. A student who thinks only astronomical bodies exert gravitational forces picks this. Every object with mass attracts every other object with mass; the force between an astronaut and a spacecraft is small because G is small, not because it is absent.
CEach one pulls the other toward itself.Correct Gravitational forces are attractive and act between any two objects with mass, in space as anywhere else. The astronaut pulls the spacecraft toward her and the spacecraft pulls her toward it, with forces of equal magnitude. With everyday masses and distances the forces are tiny, because G is so small, but they are not zero.
DThe craft pulls her, but she does not pull it. A student who thinks a small object cannot pull on a larger one picks this. Gravitational attraction is mutual: the astronaut pulls on the spacecraft with a force of the same magnitude as the spacecraft's pull on her.
The diagram shows a space probe moving up and to the left along the dashed path, with Earth and the Moon also shown. Four arrows are drawn from the probe. Which arrow shows the direction of the gravitational force exerted on the probe by the Moon?
Answer and reasoning
AArrow 1 A student who thinks the force on a moving object points along its motion picks this. The gravitational force does not depend on the direction of the velocity; the force exerted by the Moon points along the line from the probe to the Moon's center.
BArrow 3 A student who thinks every gravitational force points toward Earth picks this. Arrow 3 shows the direction of the force exerted on the probe by Earth. The force exerted by the Moon points toward the Moon's center.
CArrow 4 A student who thinks gravity always points 'down' picks this. Down the page is not a special direction here: the Moon's gravitational force on the probe points toward the Moon's center, up and to the right.
DArrow 2Correct The gravitational force is attractive and is exerted along the line joining the centers of mass of the two objects. The force exerted on the probe by the Moon therefore points from the probe directly toward the Moon's center, which is arrow 2.
Working The gravitational force is attractive and is exerted along the line joining the centers of mass, so the Moon's force on the probe points from the probe toward the Moon's center: arrow 2. Arrow 3 is the direction of Earth's force on the probe; the probe's velocity (arrow 1) and 'down the page' (arrow 4) do not set the direction of either gravitational force.
Two uniform bowling balls, each of mass 7.0 kg, rest on a level floor with their centers on a horizontal line. The diagram shows the radius of each ball and the distance between their facing surfaces. What is the magnitude of the gravitational force that each ball exerts on the other? Use G = 6.67 × 10⁻¹¹ N·m²/kg².
Answer and reasoning
A4.2 × 10⁻⁸ N A student who uses the gap between the surfaces, 0.28 m, as r gets (6.67 × 10⁻¹¹)(7.0)(7.0)/(0.28)² = 4.2 × 10⁻⁸ N. The distance in Newton's law is measured between the centers of mass, which here is 0.50 m.
B1.3 × 10⁻⁸ NCorrect The gravitational force on each uniform ball can be treated as exerted at its center, so r is the center-to-center distance: 0.11 m + 0.28 m + 0.11 m = 0.50 m. F = (6.67 × 10⁻¹¹)(7.0)(7.0)/(0.50)² = 1.3 × 10⁻⁸ N.
C6.5 × 10⁻⁹ N A student who divides by r instead of r² gets (6.67 × 10⁻¹¹)(7.0)(7.0)/0.50 = 6.5 × 10⁻⁹ N, and the unit would not be newtons. The force is inversely proportional to the square of the distance.
D3.7 × 10⁻⁹ N A student who adds the masses gets (6.67 × 10⁻¹¹)(7.0 + 7.0)/(0.50)² = 3.7 × 10⁻⁹, which has the unit of a field strength, N/kg, not of a force. The masses are multiplied: G m1m2/r².
Working The gravitational force on a uniform sphere can be treated as exerted at its center, so r = 0.11 m + 0.28 m + 0.11 m = 0.50 m. F = Gm1m2/r² = (6.67 × 10⁻¹¹ N·m²/kg²)(7.0 kg)(7.0 kg)/(0.50 m)² = 1.3 × 10⁻⁸ N.
A satellite of mass ms is at height h above the surface of a planet of mass M and radius R. Which expression gives the magnitude of the planet's gravitational field at the satellite's location?
Answer and reasoning
AGM/(R + h)²Correct The field strength is the gravitational force per unit mass on a test object: (GMms/r²)/ms = GM/r², where M is the mass creating the field and r is measured from the planet's center. The satellite is R + h from the center, so |g⃗| = GM/(R + h)².
BGMms/(R + h)² A student who confuses the field with the force picks this: GMms/(R + h)² is the gravitational force on the satellite, in newtons. The field is the force per unit mass, so the satellite's mass cancels out.
CGM/(R² + h²) A student who squares R and h separately picks this. (R + h)² = R² + 2Rh + h², not R² + h²; the whole distance from the center, R + h, must be squared.
DGms/(R + h)² A student who puts the satellite's own mass into the field expression picks this. The field at the satellite's location is created by the planet, so the planet's mass M is used; the satellite's mass cancels when the force is divided by ms.
Working |g⃗| = |F⃗g|/ms = (GMms/r²)/ms = GM/r², with r measured from the planet's center: r = R + h. So |g⃗| = GM/(R + h)².
Planet X has mass aME and radius bRE, where ME and RE are Earth's mass and radius and a and b are positive numbers. The gravitational field strength at Earth's surface is g. An object of mass m rests on the surface of Planet X. Which expression gives the gravitational field strength at the surface of Planet X?
Answer and reasoning
A(a/b)g A student who takes the field to be inversely proportional to the radius, not its square, picks this. The field at the surface is GM/R², so the radius factor b must be squared.
B(a/b²)mg A student who confuses the field with the force picks this: (a/b²)mg is the gravitational force on the object, its weight on Planet X, in newtons. The field strength is the force per unit mass, so m does not appear.
C(a/(2b))g A student who 'squares' b by doubling it picks this. (bRE)² = b²RE², so the field is divided by b², not by 2b.
D(a/b²)gCorrect At a planet's surface the field strength is GM/R². For Planet X: G(aME)/(bRE)² = (a/b²)(GME/RE²) = (a/b²)g. The larger mass increases the field, but a larger radius puts the surface farther from the center. The object's mass does not affect the field.
Working gX = GMX/RX² = G(aME)/(bRE)² = (a/b²)(GME/RE²) = (a/b²)g. The object's mass m cancels from Fg/m, so it does not appear in the field strength.
The diagram shows a planet of radius R and two points, P and Q, on a line through the planet's center. The arrow at P represents the planet's gravitational field at P, of magnitude 4.5 N/kg. What is the magnitude of the planet's gravitational field at Q?
Answer and reasoning
A2.0 N/kgCorrect The field strength is GM/r², with r measured from the planet's center. From the diagram, rP = 2R and rQ = 3R, so gQ = 4.5 N/kg × (2R/3R)² = 4.5 × 4/9 = 2.0 N/kg.
B3.0 N/kg A student who scales the field by 1/r instead of 1/r² gets 4.5 N/kg × 2/3 = 3.0 N/kg. The field is inversely proportional to the square of the distance from the center, so the factor is (2/3)² = 4/9.
C1.1 N/kg A student who measures distances from the planet's surface (R for P, 2R for Q) gets 4.5 N/kg × (1/2)² = 1.1 N/kg. The distance in GM/r² is measured from the planet's center: 2R and 3R.
D4.5 N/kg A student who treats the field as the same at every distance picks this. Q is 1.5 times as far from the center as P, a large change, so the field there is smaller by the factor (2/3)².
Working |g⃗| = GM/r², so gQ/gP = (rP/rQ)². From the diagram rP = 2R and rQ = 3R: gQ = 4.5 N/kg × (2/3)² = 2.0 N/kg.
An astronaut on the Moon, where there is no air, releases a hammer from rest. The graph shows the hammer's velocity v as a function of time t until it reaches the ground, with downward as positive. What is the magnitude of the gravitational field at the Moon's surface?
Answer and reasoning
A2.40 N/kg A student who reads the height of the graph instead of its slope picks this. 2.40 m/s is the hammer's velocity when it reaches the ground; the acceleration is the slope, 2.40 m/s ÷ 1.50 s = 1.60 m/s².
B1.60 N/kgCorrect The gravitational force is the only force on the hammer, so its acceleration in m/s² equals the field strength in N/kg. The acceleration is the slope of the v–t graph: (2.40 m/s − 0)/(1.50 s − 0) = 1.60 m/s², so the field is 1.60 N/kg.
C1.80 N/kg A student who finds the area under the graph, (1/2)(1.50 s)(2.40 m/s) = 1.80 m, picks this. That area is the distance the hammer falls, not its acceleration; the acceleration is the slope.
D10.0 N/kg A student who treats 10 N/kg as the field strength everywhere picks this. That value applies near Earth's surface; the graph gives the Moon's value as the slope, 1.60 m/s², so the field there is 1.60 N/kg.
Working Only the gravitational force acts, so a (m/s²) = g (N/kg). a = slope = (2.40 m/s − 0)/(1.50 s − 0) = 1.60 m/s², so g = 1.60 N/kg. (Area = (1/2)(1.50 s)(2.40 m/s) = 1.80 m is the distance fallen.)
On Mars the gravitational field strength is 3.7 N/kg. An astronaut releases a 2.0 kg rock and a 6.0 kg rock from rest at the same height at the same instant. Air resistance is negligible. How do the accelerations of the two rocks compare?
Answer and reasoning
AThe 6.0 kg rock's acceleration is three times the other's. A student who believes heavier objects fall faster picks this. The gravitational force on the 6.0 kg rock is three times as large, but so is its mass, so its acceleration, Fg/m, is the same as the lighter rock's.
BThe 2.0 kg rock accelerates more, since it has less inertia. A student who considers only inertia picks this. The lighter rock is easier to accelerate, but the gravitational force on it is smaller in exactly the same proportion, so both rocks accelerate at 3.7 m/s².
CBoth rocks accelerate downward at the same rate, 3.7 m/s² each.Correct Gravity is the only force on each rock. The 6.0 kg rock has three times the gravitational force exerted on it (22 N against 7.4 N) but also three times the inertial mass, so a = Fg/m = 3.7 m/s² for both, numerically equal to the field strength.
DBoth accelerate at 10 m/s², as all falling objects do. A student who treats 10 m/s² as the free-fall acceleration everywhere picks this. That value applies near Earth's surface; on Mars an object with only gravity acting on it accelerates at 3.7 m/s², the local field strength.
Working Fg = mgMars: 2.0 kg × 3.7 N/kg = 7.4 N; 6.0 kg × 3.7 N/kg = 22 N. With gravity the only force, a = Fg/m = 7.4 N/2.0 kg = 22 N/6.0 kg = 3.7 m/s² for both.
An object of mass m is released from rest at a small height h above the surface of a planet of mass M and radius R. The planet has no atmosphere. Which expression gives the time the object takes to reach the surface?
Answer and reasoning
AR√(h/(2GM)) A student who moves the 1/2 in Δy = (1/2)at² to the wrong side writes t² = h/(2a) and picks this. Multiplying both sides by 2 gives t² = 2h/a.
BR√(2h/(GM))Correct Gravity is the only force, so the object's acceleration equals the field strength, GM/R² (constant, as h is small compared with R). From Δy = (1/2)at² starting from rest, t = √(2h/a) = √(2hR²/(GM)) = R√(2h/(GM)).
C√(2hR/(GM)) A student who takes the field strength as GM/R instead of GM/R² picks this. The field, and so the acceleration, is inversely proportional to the square of the distance from the center, which gives a factor R², not R, in t².
DR√(2h/(Gm)) A student who puts the falling object's mass into GM/R² picks this, and would then predict that heavier objects fall faster. The field is created by the planet, so M is used; the object's mass cancels out of its acceleration.
Working Only gravity acts, so a = g = GM/R² (constant because h ≪ R). From rest, h = (1/2)at², so t = √(2h/a) = √(2hR²/(GM)) = R√(2h/(GM)).
A rover weighs 1600 N at Earth's surface. Use g = 10 m/s². The gravitational field strength at the surface of Mars is 3.7 N/kg. What is the rover's weight on Mars?
Answer and reasoning
A5.9 × 10² NCorrect The rover's mass is the same everywhere: m = W/g = 1600 N ÷ 10 N/kg = 160 kg. On Mars its weight is mgMars = 160 kg × 3.7 N/kg = 592 N ≈ 5.9 × 10² N.
B1.6 × 10³ N A student who thinks weight is a fixed property of the rover picks this. Weight is the gravitational force exerted by the planet the rover is on, mg; the mass stays at 160 kg, but g on Mars is 3.7 N/kg, so the weight is smaller.
C5.9 × 10³ N A student who treats the weight in newtons as a mass in kilograms calculates 1600 × 3.7 = 5.9 × 10³ N. The rover's mass is 1600 N ÷ 10 N/kg = 160 kg, so its weight on Mars is 160 kg × 3.7 N/kg.
D4.3 × 10³ N A student who uses the ratio of field strengths upside down calculates 1600 N × 10/3.7 = 4.3 × 10³ N, which would make the rover heavier where gravity is weaker. WMars = 1600 N × 3.7/10.
Working m = WEarth/g = 1600 N ÷ 10 N/kg = 160 kg (unchanged on Mars). WMars = mgMars = 160 kg × 3.7 N/kg = 592 N ≈ 5.9 × 10² N.
The International Space Station orbits 4.0 × 10⁵ m above Earth's surface. Earth's radius is 6.4 × 10⁶ m. The gravitational force exerted by Earth on an astronaut at the station is what multiple of the gravitational force exerted by Earth on the same astronaut at Earth's surface?
Answer and reasoning
A0.94 A student who uses 1/r instead of 1/r² gets 6.4/6.8 = 0.94. The force is inversely proportional to the square of the distance from Earth's center, so the ratio is (6.4/6.8)².
B0.89Correct Distances are measured from Earth's center: 6.4 × 10⁶ m at the surface and 6.4 × 10⁶ m + 0.40 × 10⁶ m = 6.8 × 10⁶ m at the station. F ∝ 1/r², so the ratio is (6.4/6.8)² = 0.89. A change of about 6% in r is not negligible, so the force cannot be treated as constant between these positions.
C1.00 A student who treats the gravitational force near Earth as constant at any height picks this. The station is 400 km up, which changes the distance from Earth's center by about 6%, and the force by about 11%.
D0.00 A student who thinks there is no gravity in orbit picks this. The astronaut floats because gravity is the only force on her and the station, not because the force is zero: at this height it is still almost 90% of its value at the surface.
Working rsurface = 6.4 × 10⁶ m; rstation = 6.4 × 10⁶ m + 0.40 × 10⁶ m = 6.8 × 10⁶ m. F ∝ 1/r², so Fstation/Fsurface = (6.4/6.8)² = 0.89.
A student of mass 50 kg stands at rest on a bathroom scale that lies on a ramp inclined at 37° to the horizontal, as shown in the diagram. The scale does not slide. Use g = 10 m/s², sin 37° = 0.60 and cos 37° = 0.80. What is the reading on the scale?
Answer and reasoning
A500 N A student who thinks a scale always reads the gravitational force picks mg = 500 N. The scale reads the normal force, which on the ramp balances only the component of gravity perpendicular to the surface, mg cos 37° = 400 N.
B300 N A student who uses sin 37° for the perpendicular component gets 500 N × 0.60 = 300 N. mg sin 37° is the component along the ramp; the perpendicular component, which the normal force balances, is mg cos 37°.
C400 NCorrect The scale reads the magnitude of the normal force it exerts on the student, her apparent weight. Perpendicular to the ramp, only the normal force and the component of gravity mg cos 37° act, and the student does not accelerate, so FN = mg cos 37° = 50 kg × 10 m/s² × 0.80 = 400 N, less than her weight.
D625 N A student who makes the vertical component of the normal force balance the whole weight gets 500 N ÷ 0.80 = 625 N. That ignores the force along the ramp that keeps the student from sliding. Using axes perpendicular to the ramp, FN = mg cos 37° = 400 N.
Working Axes perpendicular and parallel to the ramp. Perpendicular: FN − mg cos 37° = 0, so FN = (50 kg)(10 m/s²)(0.80) = 400 N. The scale reads FN, the apparent weight (the component along the ramp, mg sin 37° = 300 N, is balanced by the scale's friction on the student's feet).
A student stands on a scale in an elevator in three situations. 1: the elevator moves upward at constant speed. 2: the elevator moves upward and is slowing down. 3: the elevator moves downward and is slowing down. The acceleration has the same magnitude in situations 2 and 3. The scale readings are R1, R2 and R3. Which ranking is correct?
Answer and reasoning
AR1 = R2 > R3 A student who thinks the reading depends on the direction of motion picks this: 'moving up, heavier; moving down, lighter'. The reading depends on the direction of the acceleration: in 2 it is downward (R2 < mg), in 3 upward (R3 > mg).
BR3 > R1 > R2Correct The reading is the normal force: FN = m(g + a) with up positive. In 1, a = 0, so R1 = mg. In 2, the elevator moves up and slows, so a is downward and R2 < mg. In 3, it moves down and slows, so a is upward and R3 > mg. Only the direction of the acceleration matters.
CR1 = R2 = R3 A student who thinks a scale always reads the gravitational force picks this. The gravitational force is the same in all three, but in 2 and 3 the elevator accelerates, so the normal force, which the scale reads, differs from mg.
DR1 > R2 = R3 A student who thinks slowing down always means a downward acceleration picks this. Slowing down means the acceleration is opposite to the velocity: downward in 2 but upward in 3, so R3 is the largest reading.
Working Up positive: FN − mg = ma → FN = m(g + a). 1: a = 0 → R1 = mg. 2: moving up, slowing → a downward → R2 = m(g − |a|) < mg. 3: moving down, slowing → a upward → R3 = m(g + |a|) > mg. So R3 > R1 > R2.
A student whose weight is 500 N stands on a scale in an elevator. The elevator starts from rest at t = 0 and moves upward until it stops at t = 10 s. The graph shows the scale reading F as a function of time t. During which time interval is the elevator's acceleration directed downward?
Answer and reasoning
AOnly between 0 and 2 s A student who thinks a larger reading means she is pushed downward picks this. From 0 to 2 s the scale pushes up on her with 600 N, more than her 500 N weight, so the net force and the acceleration are upward: the elevator is speeding up.
BOnly between 2 and 10 s A student who thinks the elevator slows once the extra push stops picks this: at 2 s the reading falls to her weight, so she concludes the elevator slows from then until it stops. From 2 s to 8 s the reading equals her weight, so the net force is zero and the elevator moves upward at constant velocity (Newton's first law); only from 8 s to 10 s, when the reading is 400 N, is the net force, and so the acceleration, downward.
CAt no time during the ride A student who takes the acceleration to point the way the elevator moves picks this. The elevator moves upward throughout, but from 8 s to 10 s it slows down, so its acceleration is downward, as the 400 N reading shows.
DOnly between 8 s and 10 sCorrect The reading is the normal force on the student. From 8 s to 10 s it is 400 N, less than her 500 N weight, so the net force on her, and her acceleration, is downward: the elevator is slowing down while still moving upward, and it stops at 10 s.
Working Up positive: FN − Fg = ma, with m = 500 N ÷ 10 N/kg = 50 kg. 0–2 s: (600 − 500) N = 100 N → a = +2.0 m/s² (upward). 2–8 s: 500 − 500 = 0 → a = 0. 8–10 s: (400 − 500) N = −100 N → a = −2.0 m/s² (downward).
A student exerts different horizontal net forces on a cart and measures the cart's acceleration each time. The graph shows the cart's acceleration a as a function of the net force. What is the cart's inertial mass?
Answer and reasoning
A0.067 kg A student who takes the slope of the graph as the mass gets 0.40 ÷ 6.0 = 0.067. With a on the vertical axis, a = (1/m)Fnet, so the slope is 1/m, in kg⁻¹; the mass is its reciprocal, 15 kg.
B15 kgCorrect Inertial mass is the ratio of the net force to the acceleration it produces: m = Fnet/a. From the line, a net force of 6.0 N gives 0.40 m/s², so m = 6.0 N ÷ 0.40 m/s² = 15 kg. (The slope of this graph is 1/m.)
C0.60 kg A student who divides the force by g, as when finding a mass from a weight, gets 6.0 N ÷ 10 N/kg = 0.60 kg. The force here is a horizontal net force producing 0.40 m/s², not the gravitational force, so m = Fnet/a.
D1.2 kg A student who takes the area under the graph, (1/2)(6.0)(0.40) = 1.2, picks this. The relationship between net force and acceleration is a ratio, m = Fnet/a, which the slope of the graph (its reciprocal) gives, not the area.
Working Straight line through the origin: a = Fnet/m, slope = 1/m. At Fnet = 6.0 N, a = 0.40 m/s²: m = 6.0 N ÷ 0.40 m/s² = 15 kg.
In 1971, an Apollo 15 astronaut on the Moon, where there is no air, released a hammer and a feather from rest at the same height at the same instant. They reached the ground at the same time. Which claim does this observation support?
Answer and reasoning
AEach object's gravitational mass is in the same ratio to its inertial mass.Correct Each object's acceleration is a = Fg/minertial = (mgrav g)/minertial. The hammer and the feather differ greatly in mass and material yet had equal accelerations, so mgrav/minertial is the same for both. This is the equivalence of gravitational and inertial mass, which experiments have verified.
BThe Moon exerts gravitational forces of equal size on both objects. A student who thinks equal accelerations mean equal forces picks this. The force on the hammer is much larger, in proportion to its larger gravitational mass; the hammer also has a proportionally larger inertial mass, so the two accelerations are equal.
CHeavier objects fall faster only where gravity is strong, as on Earth. A student who believes heavier objects fall faster under gravity picks this. On Earth, too, objects with only gravity acting on them fall with the same acceleration; a feather falls slowly in air because of air resistance.
DEvery object falls at 10 m/s² on the Moon, just as it does on Earth. A student who treats 10 m/s² as the free-fall acceleration everywhere picks this. The observation shows only that the two accelerations were equal; the Moon's field strength, and so the free-fall acceleration there, is about 1.6 m/s².
Working a = Fg/minertial = (mgrav/minertial)gMoon. Equal accelerations for the hammer and the feather ⇒ mgrav/minertial is the same for both objects.
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
