6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
Car X rounds a curve of radius 50 m and car Y rounds a curve of radius 100 m. Both cars move at the same constant speed. How does the magnitude of car X's centripetal acceleration compare with that of car Y?
Answer and reasoning
AIt is half as large, since ac increases as the radius increases. A student who puts the radius in the numerator, as if ac = v²r, picks this. At a fixed speed a larger radius gives a SMALLER centripetal acceleration, so car X, on the tighter curve, has the larger one.
BIt is four times as large, since ac varies as 1/r², like gravity. A student who carries the inverse-square law of gravitation over to centripetal acceleration picks this. ac = v²/r depends on 1/r, so halving the radius at the same speed doubles the acceleration; it does not quadruple it.
CThey are equal, since both cars travel at a constant speed. A student who thinks constant speed means zero acceleration picks this, reasoning that both accelerations are zero. Both cars change direction continuously, so both accelerate toward the centers of their curves, and the magnitudes differ.
DIt is twice as large, as ac = v²/r and car X's radius is half.Correct At the same speed, ac = v²/r is inversely proportional to the radius. Car X's radius, 50 m, is half of car Y's, so car X's centripetal acceleration is twice as large: the tighter curve needs the larger acceleration.
A cart moves around the inside of a vertical circular loop of radius r. Friction is negligible. As the cart passes the top of the loop, the normal force exerted on it by the track has a magnitude equal to the cart's weight, mg. Which expression gives the cart's speed at the top of the loop?
Answer and reasoning
Av = √(gr) A student who takes the normal force alone to be the centripetal force sets FN = mv²/r, so mg = mv²/r and v = √(gr). The gravitational force also points toward the center at the top, so the net force is FN + mg = 2mg.
Bv = √(2g/r) A student who writes the centripetal acceleration as v²r, with the radius in the numerator, gets v²r = 2g and v = √(2g/r). The centripetal acceleration is v²/r, so 2g = v²/r and v = √(2gr); √(2g/r) does not even have the unit of speed.
Cv = √(2gr)Correct At the top, the normal force and the gravitational force both point down, toward the center of the loop. Newton's second law toward the center: FN + mg = mv²/r. With FN = mg, 2mg = mv²/r, so v = √(2gr).
Dv = 2gr A student who forgets the square root at the end, writing v = 2gr instead of v² = 2gr, picks this. The expression 2gr has the unit m²/s², not m/s; the speed is its square root, √(2gr).
Working Toward the center (down) positive at the top: FN + mg = mv²/r. With FN = mg: 2mg = mv²/r → v² = 2gr → v = √(2gr). Units: √((m/s²)(m)) = m/s.
A cart moves along a circular track of radius 9.0 m. The graph shows the cart's speed v as a function of time t. What is the magnitude of the cart's tangential acceleration during the 1.0 s interval shown?
Answer and reasoning
A2.0 m/s²Correct Tangential acceleration is the rate at which the speed changes, the slope of the speed–time graph: (6.0 m/s − 4.0 m/s) ÷ 1.0 s = 2.0 m/s². It points along the cart's velocity, since the cart is speeding up.
B6.0 m/s² A student who divides the speed at the end of the interval by the time, (6.0 m/s) ÷ (1.0 s), picks this. That reads the height of the graph, not its slope; the cart did not start from rest, so its speed changed by only 2.0 m/s.
C5.0 m/s² A student who finds the area under the graph, (4.0 + 6.0)/2 × 1.0 = 5.0, picks this. That area is the distance traveled, 5.0 m; the acceleration is the slope.
D4.0 m/s² A student who takes every acceleration in circular motion to be v²/r calculates (6.0 m/s)² ÷ 9.0 m = 4.0 m/s². That is the centripetal acceleration at t = 1.0 s; the tangential acceleration, the rate of change of speed, is a separate component.
Working at = Δv/Δt = slope = (6.0 − 4.0) m/s ÷ (1.0 − 0) s = 2.0 m/s². (Centripetal acceleration at t = 1.0 s is v²/r = 36/9.0 = 4.0 m/s², a different component.)
A car moves around a circular track of radius 12 m. At one instant its speed is 6.0 m/s, and its speed is increasing at a rate of 4.0 m/s². What is the magnitude of the car's acceleration at that instant?
Answer and reasoning
A7.0 m/s² A student who adds the two components as numbers, 3.0 + 4.0, picks this. The centripetal and tangential accelerations are perpendicular vectors, so their sum has magnitude √(3.0² + 4.0²) = 5.0 m/s².
B5.0 m/s²Correct The centripetal acceleration is ac = v²/r = (6.0 m/s)² ÷ 12 m = 3.0 m/s², toward the center; the tangential acceleration is 4.0 m/s², along the velocity. They are perpendicular, so the net acceleration has magnitude √(3.0² + 4.0²) = 5.0 m/s².
C3.0 m/s² A student who takes the acceleration in circular motion to be v²/r alone picks this. The car is also speeding up, so its acceleration has a tangential component of 4.0 m/s² as well.
D4.0 m/s² A student who thinks acceleration means only a change of speed picks the tangential value. The car's direction is also changing, which adds a perpendicular centripetal component of v²/r = 3.0 m/s².
Working ac = v²/r = (6.0 m/s)²/(12 m) = 3.0 m/s² (toward center). at = 4.0 m/s² (along velocity). Perpendicular: a = √(ac² + at²) = √(9.0 + 16) m/s² = 5.0 m/s².
A pendulum bob is released from rest at point X. It swings through its lowest point to point Y on the other side, then swings back toward X, and the motion repeats. Which of the following is the period of the pendulum?
Answer and reasoning
AThe time for the bob to swing out from X to Y and back to XCorrect The period is the time for one full cycle of the motion. The cycle starts at X and is complete when the bob is back at X, ready to repeat the same motion: a swing out to Y and back.
BThe time the bob takes to travel from X to Y for the first time A student who counts one swing as a cycle picks this. The swing from X to Y is only half the cycle; the motion does not start to repeat until the bob has swung back to X.
CThe number of full swings the bob makes each second A student who confuses period with frequency picks this. The number of full swings (cycles) each second is the frequency, f; the period is the time for one cycle, T = 1/f.
DThe total time the bob keeps swinging until it comes to rest A student who reads 'period' as a whole stretch of time picks this. The period is the time for ONE cycle; a pendulum that swings for a long time completes many periods.
An astronaut floats inside a space station that moves in a circular orbit about 400 km above Earth's surface. Which statement about the forces exerted on the astronaut is correct?
Answer and reasoning
AEarth's gravitational force acts on the astronaut and causes the centripetal acceleration.Correct At 400 km above the surface the astronaut is only about 6% farther from Earth's center than at the surface, so Earth's gravitational force is still large (about 90% of its surface value). It is the force that makes the astronaut, like the station, move in a circle; both fall around Earth together, which is why the astronaut floats.
BNo gravitational force acts on the astronaut, as the station is so far from Earth. A student who takes 'weightless' literally picks this. Earth's gravitational force on the astronaut is almost as large as at the surface. Without it the astronaut would move in a straight line, not in a circle around Earth.
CEarth's gravitational force is balanced by an outward centrifugal force on it. A student who believes in a real outward force picks this. No object exerts an outward force on the astronaut. If the forces were balanced, the astronaut would move in a straight line at constant speed; the net force points toward Earth's center.
DEarth's gravitational force acts along with a forward force that keeps it moving. A student who thinks a moving object needs a forward force picks this. No force along the path is needed to keep the astronaut moving at constant speed; the only force is gravity, perpendicular to the motion, which changes the direction of motion.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.9.A.1 Centripetal acceleration, acFix
Centripetal acceleration, ac
The component of an object's acceleration directed toward the center of its circular path. Its magnitude is ac = v²/r, where v is the tangential speed and r the radius of the path. SI unit: m/s².
Tangential speed, v
The speed of an object moving along a circular path, measured along the path. The velocity at each instant is tangent to the circle. SI unit: m/s.
Radius of the circular path, r
The distance from the center of the circular path to the object; half the diameter of the circle. For an orbit, it is measured from the center of the central body, not from its surface. SI unit: m.
Direction of centripetal acceleration
Toward the center of the circular path at every point, and so perpendicular to the velocity. The direction changes continuously as the object moves around the circle.
Students often think A pendulum bob has zero acceleration at the lowest point of its swing, because it is moving fastest there (and that is its resting position). In fact No. At the lowest point the bob's speed is at its greatest, so the rate of change of speed (the tangential acceleration) is zero there, but the bob is moving along a circular arc, so it has a centripetal acceleration v²/L straight up toward the pivot.
Students often think A swinging or falling object always accelerates straight down, in the direction of the gravitational force. In fact No. The gravitational force is only one of the forces on a pendulum bob; the tension also acts. At the lowest point the tension exceeds the weight and the acceleration points straight up, toward the pivot.
2.9.A.2 Net force in circular motion Fix
Net force in circular motion
The vector sum of all the forces exerted on the object. In uniform circular motion it points toward the center and has magnitude mac = mv²/r. It can come from a single force (gravity on a satellite), from several forces (tension and the gravitational force at the bottom of a swing) or from components of forces (the normal force on a banked road). "Centripetal force" is a name for this net inward force, not an additional force to be drawn on a free-body diagram. SI unit: N.
Minimum speed at the top of a vertical loop
The smallest speed with which an object can pass the top of a vertical circular loop and still follow the circle. At this speed the track (or string) exerts no force, the gravitational force alone causes the centripetal acceleration, g = v²/r, and v = √(gr). SI unit: m/s.
Banked curve
A curved road whose surface is tilted at an angle θ to the horizontal, lower on the inside of the curve. The horizontal component of the normal force points toward the center of the curve. At one speed, found from tanθ = v²/(rg), no friction is needed; at other speeds static friction exerted along the road surface also contributes.
Conical pendulum
An object on a string that moves in a horizontal circle while the string sweeps out a cone at a constant angle θ to the vertical. The vertical component of the tension balances the gravitational force (T cosθ = mg), and the horizontal component, T sinθ, is the net force toward the center of the circle.
Students often think An object moving in a circle has a centripetal force exerted on it in addition to its other forces, and this force belongs on its free-body diagram. In fact No. 'Centripetal force' is the name for the net force toward the center. It is provided by forces already on the diagram, such as friction, tension, the normal force or gravity, or by their components.
Students often think An object moving in a circle has a real outward (centrifugal) force exerted on it that balances the inward force. In fact No. In an inertial frame, the forces on an object moving in a circle are the real forces exerted by other objects (gravity, tension, normal, friction), and their sum points toward the center. The feeling of being pushed outward is the body's tendency to keep moving in a straight line.
2.9.A.3 Tangential acceleration, atFix
Tangential acceleration, at
The component of an object's acceleration tangent to its circular path. Its magnitude is the rate at which the object's speed changes; it points along the velocity when the object speeds up and opposite to it when the object slows down, and it is zero when the speed is constant. On a speed–time graph it is the slope. SI unit: m/s².
Students often think The acceleration can be found by dividing the speed at an instant by the time, reading the height of a speed–time graph instead of its slope. In fact No. Tangential acceleration is the rate of change of speed, Δv/Δt, which is the slope of a speed–time graph, not the speed at one instant divided by the time.
Students often think The area under a speed–time graph gives the acceleration. In fact No. The area under a speed–time graph is the distance traveled; the acceleration is the slope.
2.9.A.4 Net acceleration in circular motion Fix
Net acceleration in circular motion
The vector sum of the centripetal and tangential accelerations. The two are perpendicular, so the magnitude is √(ac² + at²). The net acceleration points exactly toward the center only at instants when the speed is not changing (at = 0): throughout uniform circular motion, or at the lowest point of a pendulum's swing.
Students often think The acceleration of an object moving in a circle is always the centripetal acceleration, v²/r toward the center, even when its speed is changing. In fact Only at instants when its speed is not changing. If the speed is increasing or decreasing, the acceleration also has a tangential component, and the net acceleration points between the center and the direction of the tangential component.
Students often think Acceleration means a change of speed only, so an object in circular motion accelerates only along its path, when and as fast as its speed changes. In fact No. Acceleration is the rate of change of velocity, which includes changes in direction. An object that speeds up while moving in a circle has a tangential component (change of speed) and a centripetal component (change of direction).
2.9.A.5 Uniform circular motion Fix
Uniform circular motion
Motion of an object around a circular path at constant speed. The direction of the velocity changes continuously, so the object accelerates even though its speed does not change. Its revolutions are described by the period T and the frequency f.
Period, T
The time to complete one full revolution around a circular path, one full rotation, or one full cycle of an oscillation (for a pendulum, a swing out and back to the starting point). SI unit: s.
Frequency, f
The number of revolutions (or cycles) completed per unit time: f = 1/T. SI unit: hertz, Hz, where 1 Hz = 1 s⁻¹, one revolution per second.
Period of motion at constant speed
At constant speed, one revolution covers the circumference 2πr, so the period is T = 2πr/v and, equivalently, v = 2πr/T.
Students often think The period is the time for one swing from one side to the other, half of the full back-and-forth cycle. In fact No. The period is the time for a full cycle, which for a pendulum is a swing out to the far side and back to the starting point. A swing from one side to the other is half a period.
Students often think Period and frequency are interchangeable: the period is the number of cycles per second, or the frequency is the time for one cycle, and the two increase together. In fact No. The period T is a time (seconds per revolution); the frequency f is a rate (revolutions per second, Hz). They are reciprocals, T = 1/f: a larger frequency means a shorter period.
2.9.B.1 Satellite in circular orbit Fix
Satellite in circular orbit
An object moving in a circle around a central body, with the gravitational force exerted by the central body as the only force, so ac = GM/R², where M is the central body's mass and R the orbital radius measured from its center. The satellite's own mass cancels, so its speed and period do not depend on it.
Kepler's third law (circular orbits)
For circular orbits around a central body of mass M, T² = (4π²/(GM))R³. The ratio T²/R³ is the same for every satellite of that body, and a graph of T² against R³ is a straight line through the origin with slope 4π²/(GM). G = 6.67 × 10⁻¹¹ N·m²/kg².
Students often think Astronauts in orbit float because Earth's gravity does not reach them there, or has become negligible. In fact No. At the height of a typical space station Earth's gravitational force on an astronaut is still about 90% of its value at the surface. It is the force that makes the astronaut and the station move in a circle; they appear weightless because they are in free fall together.
Students often think Satellites in larger orbits move faster, as points farther from the axis of a turning wheel do. In fact No. Points on a rigid wheel share one period, so outer points move faster. Satellites are not joined together: gravity weakens with distance, and v = √(GM/R) decreases as R increases, so outer satellites move more slowly and have much longer periods.
16 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 16
A pendulum bob swings back and forth on a light string in a vertical plane. Which statement correctly describes the bob's acceleration at the instant it passes through the lowest point of its swing?
Answer and reasoning
AIt is zero, since the bob is moving at its greatest speed there. A student who treats the lowest point as a point of equilibrium picks this. Only the tangential component, the rate of change of speed, is zero there. The bob is moving on a circle, so its velocity is changing direction and its centripetal acceleration v²/L is at its greatest.
BIt points straight up, toward the string's point of support.Correct At the lowest point the bob's speed has stopped increasing and not yet begun to decrease, so the tangential acceleration is zero. The bob is still moving along a circular arc, so it has a centripetal acceleration, v²/L, toward the center of that arc: straight up, toward the point of support.
CIt points straight down, the direction of the gravitational force. A student who takes every swinging object to accelerate in the direction of gravity picks this. The tension also acts on the bob, and at the lowest point it is greater than the gravitational force, so the net force and the acceleration point up.
DIt points horizontally, in the direction the bob is moving. A student who thinks acceleration points along the motion picks this. At the lowest point the speed is momentarily not changing, so there is no acceleration along the motion; the acceleration there is entirely centripetal, perpendicular to the velocity.
A carousel turns at a constant rate. Horse P is a distance r from the axis of rotation, and horse Q is a distance 2r from the axis. What is the ratio aQ/aP of the magnitudes of the horses' centripetal accelerations?
Answer and reasoning
A2Correct Both horses complete each revolution in the same time T, so v = 2πr/T gives vQ = 2vP. Then ac = v²/r gives aQ/aP = (2²)/2 = 2. (Equivalently, ac = 4π²r/T² is proportional to r when T is fixed.)
B½ A student who thinks all points on the carousel move at the same speed uses ac = v²/r with equal speeds and gets aQ/aP = r/(2r) = ½. Horse Q travels twice as far in each revolution in the same time, so it moves twice as fast as P.
C1 A student who uses ac ∝ v²/r², squaring the radius as in the gravitational force law, gets (2²)/(2²) = 1. Centripetal acceleration is v²/r: the speed ratio is squared but the radius ratio is not, giving 4/2 = 2.
D8 A student who uses ac = v²r, with the radius in the numerator, gets (2²)(2) = 8. The radius belongs in the denominator: ac = v²/r gives 4/2 = 2.
Working Same period T for both horses (they turn together). v = 2πr/T, so vP = 2πr/T and vQ = 2π(2r)/T = 2vP. ac = v²/r: aQ/aP = [(2vP)²/(2r)] ÷ [vP²/r] = 4/2 = 2. Check: ac = (2πr/T)²/r = 4π²r/T², proportional to r at fixed T.
A car moves counterclockwise at constant speed around a circle with center C. The diagram shows its velocity v⃗P at point P and its velocity v⃗Q at a later instant, at point Q; P and Q are the same distance from the top of the circle. Which claim about the change in velocity, Δv⃗ = v⃗Q − v⃗P, do the vectors in the diagram support?
Answer and reasoning
AIt is zero, since the two arrows have equal lengths. A student who equates equal speeds with equal velocities picks this. The arrows have equal lengths but different directions, so v⃗Q ≠ v⃗P and Δv⃗ is not zero.
BIt points away from C, the center of the circle. A student who subtracts in the wrong order, v⃗P − v⃗Q, picks this. That arrow points up, away from C. The change in velocity is final minus initial: v⃗P + Δv⃗ = v⃗Q, which requires Δv⃗ to point down, toward C.
CIt points toward C, the center of the circle.Correct v⃗P and v⃗Q have the same horizontal component (to the left) and opposite vertical components (up at P, down at Q). Their difference, v⃗Q − v⃗P, is therefore straight down, from the top of the circle toward C. This is why the car's acceleration points toward the center even though its speed is constant.
DIt points in the car's direction of motion at the top. A student who expects any change in velocity to point along the motion picks this. At the top the car moves to the left, but the horizontal components of v⃗P and v⃗Q are equal, so there is no change to the left; the whole change is vertical, toward C.
A car travels at constant speed around a curve on a flat, horizontal road. Air resistance is negligible. The diagram shows a student's free-body diagram for the car, viewed from behind, with the center of the curve to the left. The student has labeled Fc as the centripetal force. Which statement about the student's diagram is correct?
Answer and reasoning
AIt is correct, since a car on a curve needs a centripetal force. A student who treats the centripetal force as a separate force picks this. The car does need a net force toward the center, but that net force is the friction force already drawn. Adding Fc counts it twice, and no object exerts Fc.
BAn outward force equal to Fc should be added to the diagram. A student who believes in a real outward force picks this. No object exerts an outward force on the car. If the horizontal forces balanced, the car would travel in a straight line; the net horizontal force must point toward the center.
CThe arrow Ff should be removed, as the car is not skidding. A student who thinks friction acts only during sliding picks this. Static friction acts precisely because the tires do not slide; without Ff nothing would push the car toward the center, and it would travel off in a straight line.
DThe arrow labeled Fc should be removed from the diagram.Correct Every force on a free-body diagram must be exerted by an identifiable object. The centripetal force is not an extra force; it is the net force toward the center, which here is provided entirely by the static friction force Ff. FN and Fg balance, and Ff alone causes the centripetal acceleration.
A toy car moves around the inside of a vertical circular loop of track with a diameter of 0.80 m. Friction is negligible. What is the minimum speed the car must have at the top of the loop to stay in contact with the track there? Use g = 10 m/s².
Answer and reasoning
A2.8 m/s A student who substitutes the diameter, 0.80 m, as r gets √((10)(0.80)) = 2.8 m/s. The radius of the loop is half the diameter, 0.40 m.
B2.0 m/sCorrect r = 0.80 m ÷ 2 = 0.40 m. At the minimum speed the track exerts no force at the top, so the gravitational force alone causes the centripetal acceleration: g = v²/r, and v = √(gr) = √((10 m/s²)(0.40 m)) = 2.0 m/s.
C4.0 m/s A student who calculates gr = (10)(0.40) = 4.0 and stops there picks this. That is v², in m²/s²; the speed is √4.0 = 2.0 m/s.
D0.0 m/s A student who pictures the car pausing at the top, like a ball thrown straight up, picks this. At the top the car is still moving in a circle and needs a centripetal acceleration of at least g, which requires v = √(gr).
Working r = 0.80 m/2 = 0.40 m. At minimum speed FN = 0 at the top, so mg = mv²/r → v = √(gr) = √((10 m/s²)(0.40 m)) = √(4.0 m²/s²) = 2.0 m/s.
For each of several vertical loops of different radius r, students release a cart from a height they choose and measure its speed v at the top of the loop. The graph shows v² plotted against r, with a best-fit line. Which claim about the cart at the top of each loop do the data support?
Answer and reasoning
AAt the top, the cart's centripetal acceleration equals g, so the track exerts no force.Correct The best-fit line passes through the origin with a slope of about 10 m²/s² per meter, so v²/r ≈ 10 m/s² = g for every loop. A centripetal acceleration of g is exactly what the gravitational force alone provides, so the track exerts no force on the cart at the top: each cart was moving at the minimum speed for its loop.
BThe cart's speed at the top is directly proportional to the loop's radius. A student who reads the straight line without noticing the square on v² picks this. The graph shows v² proportional to r, so v is proportional to √r: the cart in the 0.80 m loop moves only about twice as fast as the one in the 0.20 m loop, not four times as fast.
CThe cart's centripetal acceleration at the top is greater in the larger loops. A student who puts r in the numerator of the centripetal acceleration, ac = v²r, finds that v²r grows with the radius, from about 0.42 for the 0.20 m loop to about 6.3 for the 0.80 m loop. The centripetal acceleration is v²/r, the slope of the graph, which is about 10 m/s² for every loop.
DAt the top, the track exerts a force on the cart equal to the cart's weight. A student who takes the normal force always to equal the weight picks this. Then the net force would be 2mg and v²/r would be 20 m/s², twice the slope of the graph; the data give v²/r ≈ 10 m/s², so the track exerts no force on the cart at the top.
Working Slope of best-fit line ≈ (9.5 − 0) m²/s² ÷ (0.95 − 0) m = 10 m/s². v²/r = ac ≈ 10 m/s² = g, so Fnet = mg = Fg and FN = 0.
A road is banked at 37° to the horizontal on a curve, lower on the inside of the curve. A car travels around the curve in a horizontal circle of radius 120 m. At what speed can the car round the curve with no friction force exerted on it by the road? Use g = 10 m/s², sin 37° = 0.60, and cos 37° = 0.80.
Answer and reasoning
A24 m/s A student who takes FN = mg cos37°, as for a block on an incline, gets mv²/r = mg cos37° sin37°, so v² = (120)(10)(0.80)(0.60) = 576 and v = 24 m/s. The car accelerates horizontally, so it is the vertical components that balance: FN cos37° = mg, which makes FN greater than mg.
B27 m/s A student who takes the net force to act along the slope, mg sin37° = mv²/r, gets v² = (120)(10)(0.60) = 720 and v = 27 m/s. The car's circle is horizontal, so its acceleration is horizontal, and it is the horizontal component of FN that provides mv²/r.
C30 m/sCorrect With no friction, only FN and Fg act. The circle is horizontal, so resolve horizontally and vertically: FN cos37° = mg and FN sin37° = mv²/r. Dividing, tan37° = v²/(rg), so v² = (120 m)(10 m/s²)(0.75) = 900 m²/s² and v = 30 m/s.
D40 m/s A student who swaps the components of the normal force, FN sin37° = mg and FN cos37° = mv²/r, gets v² = rg/tan37° = 1600 and v = 40 m/s. The angle between FN and the vertical is 37°, so the vertical component is FN cos37°.
A highway curve is banked so that a car moving at 25 m/s can round it with no friction force from the road. Another car rounds the same curve at a constant 15 m/s without sliding. Air resistance and rolling friction are negligible. Which statement describes the static friction force exerted by the road on this slower car?
Answer and reasoning
AIt is directed down the slope of the road, toward the center of the curve. A student who thinks friction always helps a car turn picks this. Friction points down the slope only when the car goes FASTER than the speed for which the curve is banked; this car is slower, so it would slide down the bank, and friction points up the slope.
BIt is directed along the road, opposite to the direction of the car's velocity. A student who thinks friction always opposes the motion picks this. With air resistance and rolling friction negligible, no force along the road is needed at constant speed; the static friction here acts across the road, to stop the car sliding down the bank.
CIt is zero, since the car's tires do not slide on the road surface. A student who thinks friction acts only while surfaces slide picks this. Static friction acts precisely when the tires do not slide; here it is what keeps the slower car from sliding down the bank.
DIt is directed up the slope of the road, away from the center of the curve.Correct At 25 m/s the horizontal component of the normal force alone provides mv²/r. At 15 m/s much less net force toward the center is needed, so without friction the car would slide down the bank toward the center. Static friction opposes that sliding, so it points up the slope.
A small ball of mass m hangs from a light string and moves at constant speed in a horizontal circle, as shown in the diagram, where the angle θ is marked. Which expression gives the tension in the string?
Answer and reasoning
Amg cosθ A student who resolves the gravitational force along the string, as for a block on an incline, picks this. The ball accelerates horizontally, not along the string, so it is the vertical components that balance: T cosθ = mg. The tension is larger than mg, not smaller.
Bmg/cosθCorrect The ball moves in a horizontal circle, so it has no vertical acceleration: the vertical component of the tension balances the gravitational force, T cosθ = mg, and T = mg/cosθ. The horizontal component, T sinθ, is the net force toward the center of the circle.
Cmg/sinθ A student who takes the vertical component of the tension to be T sinθ picks this. The angle θ is measured from the vertical, so the vertical component, adjacent to θ, is T cosθ.
Dmg tanθ A student who takes the tension to be the centripetal force picks this. mg tanθ is the NET force on the ball, which is the horizontal component of the tension, T sinθ; the tension itself also has a vertical component that supports the ball.
Working Vertical: T cosθ − mg = 0 → T = mg/cosθ. Horizontal (toward center): T sinθ = mv²/r, so the net force is T sinθ = mg tanθ.
The diagram shows a top view of a car moving counterclockwise around a circular track with center C. At point P the car is speeding up. Which numbered arrow best shows the direction of the car's acceleration at P?
Answer and reasoning
AArrow 1 A student who takes the acceleration in circular motion always to point toward the center picks the arrow pointing straight at C. That is correct only at instants when the speed is not changing; here the car is speeding up, so there is also a tangential component along the velocity.
BArrow 2 A student who thinks acceleration means only a change of speed picks the arrow pointing straight along the motion. The car's direction of motion is also changing, so its acceleration has a component toward C as well.
CArrow 3Correct The car's acceleration has two perpendicular components: a centripetal component toward C, because the car moves in a circle, and a tangential component along the velocity (up the page), because the car is speeding up. Their vector sum points between the direction toward C and the forward direction.
DArrow 4 A student who thinks the car is flung outward on the curve picks the arrow pointing forward and away from C, combining the speeding up with an outward component. The component of the acceleration perpendicular to the motion is the centripetal acceleration, which points toward C, not away from it.
Working Two perpendicular components at P: ac = v²/r toward C (to the left) and at along the velocity (up the page, since the car speeds up). Their vector sum points up and to the left, between the direction toward C and the forward direction; only this arrow has both components.
A wheel turns at a constant rate with a frequency of 4.0 Hz. Which statement about the wheel's motion is correct?
Answer and reasoning
AIt takes 4.0 s to complete each revolution. A student who confuses frequency with period picks this. The time for one revolution is the period, T = 1/f = 1/(4.0 Hz) = 0.25 s.
BA point on its rim moves at a speed of 4.0 m/s. A student who reads the frequency as a speed picks this. The rim speed is 2πr multiplied by the frequency and depends on the wheel's radius, which is not given.
CIt completes 4.0 revolutions in each minute. A student who takes hertz to mean revolutions per minute picks this. 1 Hz is one revolution per SECOND, so 4.0 Hz is 240 revolutions per minute.
DIt completes 4.0 revolutions every second.Correct Frequency is the number of revolutions completed per unit time, and 1 Hz = 1 s⁻¹, one revolution per second. So 4.0 Hz means 4.0 revolutions each second, and each revolution takes T = 1/f = 0.25 s.
Working f = 4.0 Hz = 4.0 s⁻¹: 4.0 revolutions each second. T = 1/f = 1/(4.0 s⁻¹) = 0.25 s per revolution; 4.0 rev/s × 60 s/min = 240 revolutions per minute.
A runner jogs at a constant speed around a circular track, completing each lap in time T. She then jogs around a second circular track with 4 times the radius of the first, at twice her original speed. By what factor is the time for one lap multiplied?
Answer and reasoning
A½ A student who treats the period as a rate, like a frequency, takes it to be proportional to v/r and gets 2/4 = ½. The period is a time, T = 2πr/v: a longer lap increases it and a greater speed decreases it.
B2Correct One lap is a distance 2πr, so T = 2πr/v. Multiplying r by 4 and v by 2 multiplies T by 4/2 = 2: the lap is four times as long, but she covers it twice as fast.
C1 A student who blends T = 2πr/v with ac = v²/r takes T to be proportional to r/v² and gets 4/2² = 1. The period depends on the first power of the speed: T = 2πr/v.
D4 A student who considers only the size of the track picks this. The lap is four times as long, but she runs it at twice the speed, so T = 2πr/v is multiplied by 4/2 = 2.
Working T = 2πr/v. New: T' = 2π(4r)/(2v) = (4/2)(2πr/v) = 2T.
Satellites X and Y move in circular orbits around the same planet. The orbital radius of satellite Y is twice the orbital radius of satellite X. What is the ratio TY/TX of their orbital periods?
Answer and reasoning
A2.0 A student who takes the period to be proportional to the radius, as if both satellites moved at the same speed, picks this. The outer satellite moves more slowly as well as farther, so T ∝ R3/2 gives 23/2 ≈ 2.8.
B1.0 A student who expects the outer satellite to move faster in proportion to its orbital radius, as points farther out on a turning wheel do, finds that it covers twice the distance at twice the speed and gets a ratio of 1.0. Satellites are not joined like points on a wheel: gravity is weaker farther out, so the outer satellite moves more slowly, v = √(GM/R), and T ∝ R3/2 gives 23/2 ≈ 2.8.
C2.8Correct For orbits around the same central body, T² = (4π²/(GM))R³, so T² ∝ R³ and T ∝ R3/2. Doubling R multiplies T by 23/2 = 2√2 ≈ 2.8: the outer satellite travels twice as far each orbit, and it also moves more slowly.
D4.0 A student who reasons that at twice the distance the gravitational pull is a quarter as strong, so the orbit takes four times as long, picks this. The period follows T² ∝ R³, not T ∝ R², so the factor is 23/2 ≈ 2.8.
Working T² = (4π²/(GM))R³ with the same M: (TY/TX)² = (RY/RX)³ = 2³ = 8, so TY/TX = √8 = 2√2 ≈ 2.8.
Astronomers measure the orbital period T and orbital radius R of several moons that move in circular orbits around a planet. The graph shows T² plotted against R³, with a best-fit line through the origin. What is the mass of the planet? Use G = 6.67 × 10⁻¹¹ N·m²/kg².
Answer and reasoning
A2.0 × 10¹² kg A student who reads the slope as 1.20 ÷ 4.0 = 0.30, ignoring the powers of ten in the axis labels, gets 4π²/(6.67 × 10⁻¹¹ × 0.30) = 2.0 × 10¹² kg. The slope is (1.20 × 10⁹ s²) ÷ (4.0 × 10²⁴ m³) = 3.0 × 10⁻¹⁶ s²/m³.
B1.8 × 10⁻⁴ kg A student who calculates the slope as run over rise, (4.0 × 10²⁴ m³) ÷ (1.20 × 10⁹ s²) = 3.3 × 10¹⁵, gets 4π²/(6.67 × 10⁻¹¹ × 3.3 × 10¹⁵) = 1.8 × 10⁻⁴ kg, far less than the mass of a small stone. The slope is rise over run: T² divided by R³.
C3.1 × 10²⁶ kg A student who writes 2π in place of (2π)² = 4π² gets 2π ÷ (6.67 × 10⁻¹¹ × 3.0 × 10⁻¹⁶) = 3.1 × 10²⁶ kg. Kepler's third law has 4π² in the numerator, which comes from squaring 2πR/T.
D2.0 × 10²⁷ kgCorrect Kepler's third law, T² = (4π²/(GM))R³, makes the slope of the T²–R³ graph equal to 4π²/(GM). From the line, slope = (1.20 × 10⁹ s²) ÷ (4.0 × 10²⁴ m³) = 3.0 × 10⁻¹⁶ s²/m³. So M = 4π²/(G × slope) = 39.5 ÷ (6.67 × 10⁻¹¹ × 3.0 × 10⁻¹⁶) = 2.0 × 10²⁷ kg.
Working Slope = ΔT²/ΔR³ = (1.20 × 10⁹ s²)/(4.0 × 10²⁴ m³) = 3.0 × 10⁻¹⁶ s²/m³ = 4π²/(GM). M = 4π²/(G·slope) = 39.48/((6.67 × 10⁻¹¹ N·m²/kg²)(3.0 × 10⁻¹⁶ s²/m³)) = 1.97e+27 kg ≈ 2.0 × 10²⁷ kg.
Two satellites move in circular orbits of the same radius around Earth. Satellite 1 has twice the mass of satellite 2. How does the orbital period of satellite 1 compare with that of satellite 2?
Answer and reasoning
AIt is equal, as the period depends on Earth's mass but not the satellite's.Correct Earth's gravitational force on each satellite, GMm/R², is proportional to the satellite's mass, and so is the force needed for its circular motion, mv²/R. The mass cancels, so T² = (4π²/(GM))R³ depends only on the orbital radius and Earth's mass.
BIt is shorter, as Earth pulls harder on the more massive satellite. A student who reasons that a heavier object is pulled harder and so moves faster picks this. The pull is larger in proportion to the mass, but so is the force needed to keep the larger mass moving in the circle, so the speed and period are unchanged.
CIt is longer, as the more massive satellite is harder to accelerate. A student who applies 'more mass, less acceleration' picks this, forgetting that the gravitational force is also twice as large. The acceleration GM/R² is the same for both satellites, so their periods are equal.
DIt cannot be found without knowing the orbital speed of each satellite. A student who thinks a satellite can orbit at any speed picks this. At a given radius, gravity must provide exactly v²/R, so the speed, √(GM/R), and the period are fixed by the radius and Earth's mass.
A satellite of mass mₛ moves in a circular orbit at a height h above the surface of a planet of mass M and radius R. Which expression gives the satellite's orbital speed?
Answer and reasoning
A√(Gmₛ/(R+h)) A student who puts the satellite's own mass into the gravitational field at its location, writing Gmₛ/(R + h)² instead of GM/(R + h)², picks this. The field there is produced by the planet, so it depends on M; the satellite's mass appears on both sides of Newton's second law and cancels.
B√(GM/(R+h))Correct The planet's gravitational force is the only force on the satellite, so it causes the centripetal acceleration. The orbital radius is the distance from the planet's center, R + h: GMmₛ/(R + h)² = mₛ v²/(R + h). The satellite's mass cancels, v² = GM/(R + h), and v = √(GM/(R + h)).
CGM/(R+h) A student who stops at v² = GM/(R + h) and reports it as the speed picks this. That expression has the unit m²/s²; the speed is its square root.
D√(GM/h) A student who uses the satellite's height above the surface as the orbital radius picks this. Newton's law of gravitation and ac = v²/r both use the distance from the planet's center, which is R + h.
Working The orbital radius is the distance from the planet's center, r = R + h. Fg = GMmₛ/(R + h)² is the only force on the satellite, so GMmₛ/(R + h)² = mₛ v²/(R + h) → v² = GM/(R + h) → v = √(GM/(R + h)). Units: √((N·m²/kg²)(kg)/m) = √(m²/s²) = m/s.
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
