6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
An object moves counterclockwise around the circular path shown. At point P its velocity is v⃗ and its total acceleration is a⃗, which makes the angle shown with the line from P to the center of the path. What is the radius of the path?
Answer and reasoning
A3.6 m A student who takes the whole acceleration as centripetal picks this: r = (3.0 m/s)²/(2.5 m/s²). The acceleration is tilted forward from the radius, so part of it changes the speed; only the component toward the center, 2.0 m/s², equals v²/r.
B4.5 mCorrect Only the component of a⃗ toward the center is centripetal: ac = a cos 37° = (2.5 m/s²)(0.80) = 2.0 m/s². Then r = v²/ac = (3.0 m/s)²/(2.0 m/s²) = 4.5 m. The other component, along v⃗, means the object is speeding up; it does not enter v²/r.
C6.0 m A student who finds the component toward the center with sin 37° picks this: ac = (2.5 m/s²)(0.60) = 1.5 m/s², so r = (3.0 m/s)²/(1.5 m/s²). The angle is measured from the radius, so the radial component is the adjacent side, a cos 37°.
D1.5 m A student who uses ac = v/r, without the square, picks this: r = v/ac = (3.0 m/s)/(2.0 m/s²). The units, (m/s)/(m/s²) = s, show the error; ac = v²/r.
Working Centripetal component: ac = a cos 37° = (2.5 m/s²)(0.80) = 2.0 m/s² (the angle is between a⃗ and the radius). ac = v²/r, so r = v²/ac = (3.0 m/s)²/(2.0 m/s²) = 4.5 m. Distractors: whole a as ac → 9.0/2.5 = 3.6 m; sin 37° → 9.0/1.5 = 6.0 m; ac = v/r → 3.0/2.0 = 1.5 m.
A ball on a string moves in a vertical circle. Which statement correctly describes the forces exerted on the ball as it passes the lowest point of the circle?
Answer and reasoning
ATension up and gravity down, equal in magnitude, since the ball moves horizontally there. A student who thinks a supporting force equals the weight picks this. The ball’s velocity is horizontal at that instant, but its acceleration, toward the center, is vertical, so the vertical forces cannot balance.
BTension up, balancing gravity and an outward centrifugal force that both point down. A student who believes in a real centrifugal force picks this. No object exerts an outward force on the ball; the only forces are tension and gravity, and they do not balance, because the ball accelerates toward the center.
CTension up, gravity down, and a third force, the centripetal force, which points up too. A student who treats ‘centripetal force’ as a separate force picks this. The centripetal force is the net of the real forces, here T − mg; no third object pushes on the ball.
DTension up and gravity down, with the tension larger, so that the net force points up.Correct At the lowest point the center of the circle is directly above the ball, so the ball’s acceleration, v²/r, points up. The net force must point up too: the upward tension is larger than the downward gravitational force, T − mg = mv²/r.
A runner moves along a circular track of radius 10 m. The distance she has run along the track is s(t) = ct³, where c = 0.10 m/s³. What is the magnitude of her tangential acceleration at t = 5.0 s?
Answer and reasoning
A3.0 m/s²Correct Her speed is v = ds/dt = 3ct², and the tangential acceleration is the rate of change of speed: at = dv/dt = 6ct = 6(0.10 m/s³)(5.0 s) = 3.0 m/s².
B7.5 m/s² A student who takes the size of the acceleration from the speed picks this: v = 3ct² = 3(0.10)(5.0)² = 7.5 m/s. The tangential acceleration is how fast the speed changes, dv/dt, not the speed itself.
C1.5 m/s² A student who uses a = v/t, as for constant acceleration from rest, picks this: (7.5 m/s)/(5.0 s). Here the acceleration grows with time, so v/t is only its average over the first 5.0 s; the instantaneous value is dv/dt = 6ct.
D5.6 m/s² A student who thinks the tangential acceleration is the part that turns the runner picks this, computing v²/r = (7.5 m/s)²/(10 m). That is the centripetal acceleration; the tangential acceleration, which changes her speed, is dv/dt.
Working v = ds/dt = 3ct² = 3(0.10)(25) = 7.5 m/s. at = dv/dt = 6ct = 6(0.10)(5.0) = 3.0 m/s². Distractors: v = 7.5; v/t = 1.5; v²/r = 56.25/10 = 5.6 m/s².
A car moves counterclockwise around a circular track and is slowing down. The figure shows the car at the bottom of the track at one instant, its velocity v⃗ (dashed) and four arrows, P, Q, R and S. Which arrow best shows the direction of the car’s acceleration at that instant?
Answer and reasoning
AArrow P A student who thinks the acceleration of any object moving in a circle points at the center picks this. That holds only at constant speed; a slowing car also has a tangential component opposite to its velocity.
BArrow SCorrect The acceleration has two perpendicular components: centripetal, toward the center (up), because the direction of motion is changing, and tangential, opposite to v⃗ (left), because the car is slowing down. Their vector sum points up and to the left, inside the circle but behind the line to the center.
CArrow Q A student who thinks the acceleration has a component along the direction of motion, because the car is moving forward, picks this. Q is the direction for a car that is speeding up; slowing down puts the tangential component opposite to v⃗.
DArrow R A student who counts only the change of speed as acceleration picks this. The car’s direction of motion is also changing, which requires a centripetal component toward the center.
A rider on a large amusement ride moves at constant speed in a horizontal circle of radius 16 m. The ride completes 3 revolutions every 30 s. What is the magnitude of the rider’s acceleration?
Answer and reasoning
A0.63 m/s² A student who uses a = v/r picks this: (10 m/s)/(16 m). The centripetal acceleration is v²/r; v/r has units of s⁻¹, not m/s².
B10 m/s² A student who takes the size of the acceleration from the speed picks this: the rider moves at 10 m/s. Acceleration measures how fast the velocity changes; here a = v²/r.
C6.3 m/s²Correct The frequency is f = 3/(30 s) = 0.10 Hz, so the period is T = 1/f = 10 s. The rider’s speed is v = 2πr/T = 2π(16 m)/(10 s) = 10 m/s, and the acceleration is centripetal: a = v²/r = (10 m/s)²/(16 m) = 6.3 m/s².
D0 m/s² A student who thinks an object moving at constant speed has no acceleration picks this. The rider’s direction of motion changes continuously, so there is a centripetal acceleration v²/r.
Working f = 3/30 s = 0.10 Hz ⇒ T = 1/f = 10 s. v = 2πr/T = 2π(16)/10 = 10.05 m/s. a = v²/r = 4π²r/T² = 6.32 ≈ 6.3 m/s². Distractors: v/r = 0.63; speed as acceleration 10; zero.
The International Space Station moves at constant speed in a circular orbit around Earth. Which statement correctly describes the force or forces exerted on the station?
Answer and reasoning
ANo net force, because gravity is negligible at the station’s height. A student who thinks there is no gravity in orbit picks this. At the station’s height the gravitational field is only about 10% weaker than at the surface; with no net force the station would move in a straight line.
BEarth’s gravity inward, balanced by an outward centrifugal force. A student who believes in a real centrifugal force picks this. No object pushes the station outward; if the forces balanced, it would move in a straight line rather than a circle.
CEarth’s gravity inward and a forward force keeping the station moving. A student who thinks motion needs a force along the velocity picks this. The station keeps its speed with no forward force; nothing exerts one on it.
DEarth’s gravitational force alone, toward Earth’s center.Correct Gravity is the only force on the station (air resistance is negligible at that height, and nothing pushes it along). It points toward Earth’s center, perpendicular to the station’s velocity, and provides its centripetal acceleration.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.10.A.1 Centripetal acceleration, acFix
Centripetal acceleration, ac
The component of an object’s acceleration directed toward the center of its circular path, in m/s². It changes the direction of the velocity, not the speed.
Magnitude of centripetal acceleration
ac = v²/r, where v is the object’s speed (tangential speed) and r the radius of its circular path. At a fixed radius it grows with the square of the speed.
Direction of centripetal acceleration
Toward the center of the circular path, perpendicular to the velocity. For uniform circular motion it is the direction of the change in velocity, Δv⃗.
Students often think Acceleration is a change in speed only, so an object whose speed is constant, such as one in uniform circular motion, has zero acceleration, and only the speeding up or slowing down of an object counts as acceleration. In fact Yes. Acceleration is the rate of change of velocity, and velocity is a vector. At constant speed the direction of the velocity still changes continuously, so the acceleration is not zero; it has magnitude v²/r and points toward the center.
Students often think Centripetal acceleration is proportional to the speed itself (ac = v/r), so doubling the speed doubles ac, and the speed for a given acceleration is ac r. In fact No. ac = v²/r: at a fixed radius, doubling the speed makes the centripetal acceleration four times as large. Equivalently, the speed for a given ac and r is √(ac r), not ac r.
2.10.A.2 Net force in circular motion Fix
Net force in circular motion
The sum of the components of the real forces toward the center, which equals m ac = mv²/r. It can come from one force, several forces, or components of forces; ‘centripetal force’ names this sum, not an extra force. SI unit: N.
Minimum speed at the top of a vertical loop
The least speed for which an object on the inside of a vertical loop (or on a string) stays on its circular path at the top: v = √(gr), when the gravitational force alone provides the centripetal acceleration and the normal force (or tension) is zero.
Banked curve
A curved road tilted at angle θ to the horizontal. The horizontal components of the normal force and of any static friction force provide the net force toward the center; with no friction needed, the speed is v = √(gr tan θ).
Conical pendulum
A bob on a string of length L moving in a horizontal circle, with the string at a constant angle θ to the vertical. The vertical component of tension balances Fg; the horizontal component provides the centripetal acceleration; the radius is L sin θ.
Students often think An object moving in a circle has an outward centrifugal force exerted on it, which balances the inward force so that the object stays on its circle. In fact No. In an inertial frame every force on the object is exerted by another object: tension, gravity, a normal force, friction. None is directed outward away from the center unless an object pushes that way. The net force points toward the center, and the object is not in equilibrium.
Students often think Centripetal force is a separate force exerted on an object in circular motion, drawn on the free-body diagram alongside tension, gravity and the normal force and combined with them. In fact No. ‘Centripetal force’ names the net force toward the center, which is the sum of real forces such as tension, gravity, normal force and friction (or their components). It is not an additional force to be added to them.
2.10.A.3 Tangential acceleration, atFix
Tangential acceleration, at
The component of acceleration tangent to the circular path, equal to the rate of change of speed, dv/dt, in m/s². It points along the velocity when the object speeds up and opposite to it when it slows down.
Students often think The centripetal acceleration is the part of the acceleration that changes the speed of an object moving in a circle, and the tangential acceleration is the part that turns it. In fact No, the other way round. The tangential component, along the velocity, changes the speed. The centripetal component, perpendicular to the velocity, changes only its direction.
Students often think Relations for constant acceleration, such as a = v/t for motion from rest, give the acceleration at any instant even when the acceleration changes with time. In fact No. a = Δv/Δt gives only the average acceleration over the interval. When the acceleration changes, its instantaneous value is the derivative dv/dt at that time, which differs from the average.
2.10.A.4 Net (total) acceleration in circular motion Fix
Net (total) acceleration in circular motion
The vector sum of the centripetal and tangential accelerations, which are perpendicular, so its magnitude is √(ac² + at²). It points toward the center only when at = 0.
Students often think The magnitude of an object’s total acceleration is the tangential acceleration plus the centripetal acceleration, added as numbers. In fact No. The two components are perpendicular, so the magnitude of the total acceleration is √(at² + ac²), which is less than at + ac.
Students often think The acceleration of any object moving in a circle, even one whose speed is changing, points exactly toward the center, so the whole acceleration is the centripetal acceleration. In fact Only in uniform circular motion. If the speed is changing, the acceleration also has a tangential component, so the total acceleration does not point at the center.
2.10.A.5 Uniform circular motion Fix
Uniform circular motion
Motion in a circle at constant speed. The velocity changes direction continuously, so the object accelerates (ac = v²/r) although its speed, period and frequency are constant.
Period, T
The time to complete one full circular path, one full rotation, or one full cycle of an oscillatory motion. SI unit: s.
Frequency, f
The number of revolutions (or cycles) completed per unit time, f = 1/T. SI unit: hertz (Hz = s⁻¹).
Speed in uniform circular motion
v = 2πr/T: the circumference 2πr divided by the period. On one rotating rigid object all points share T, so speed is proportional to distance from the axis.
Students often think Because ac = v²/r has r in the denominator, a point at a larger radius always has a smaller centripetal acceleration, even on the same rotating object. In fact No. All points of a rigid rotating object have the same period, so a point farther out moves faster: v = 2πr/T grows in proportion to r. Then ac = v²/r = 4π²r/T² also grows in proportion to r.
Students often think Period and frequency can be used interchangeably, so a rate such as ‘revolutions per second’ can be substituted where the period is needed, or the period where the frequency is needed. In fact No. The period T is the time for one revolution, in seconds; the frequency f is the number of revolutions per unit time, in hertz (s⁻¹). They are reciprocals: T = 1/f.
2.10.B.1 Circular orbit Fix
Circular orbit
The path of a satellite moving in a circle around a central body, with the gravitational force of the central body the only force providing its centripetal acceleration: GMm/R² = mv²/R. R is measured from the central body’s center.
Kepler’s third law (circular orbits)
T² = (4π²/(GM))R³: the square of the orbital period is proportional to the cube of the orbital radius, with a constant set by the central body’s mass M (not the satellite’s).
Students often think A moving object needs a force in its direction of motion to keep it moving, so a satellite in orbit must have a forward force on it in addition to the force pulling it inward. In fact No. With no resistive forces, an object keeps its speed without any force along its velocity. A satellite in a circular orbit has only the gravitational force exerted on it, directed toward the center of the planet, perpendicular to its velocity.
Students often think The period of a satellite in a circular orbit depends on the satellite’s own mass: a more massive satellite, pulled harder, orbits faster (or, having more inertia, more slowly). In fact No. The gravitational force on the satellite and the net force it needs, mv²/R, are both proportional to its mass, so the mass cancels: the speed and the period at a given radius depend only on the central body’s mass M and on R.
23 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 23
A car moves at constant speed around a circular track, and its centripetal acceleration is 5.0 m/s². The car then moves at constant speed around a second circular track whose radius is half as large, at twice the speed. What is the car’s centripetal acceleration on the second track?
Answer and reasoning
A20 m/s² A student who takes ac proportional to v rather than v² picks this: (2)/(1/2) = 4, and 4 × 5.0 m/s² = 20 m/s². The speed is squared in ac = v²/r, so doubling it alone makes ac four times as large.
B10 m/s² A student who applies the radius change the wrong way round, as if ac were proportional to r, picks this: 2² × (1/2) = 2. The radius is in the denominator: halving r doubles ac.
C80 m/s² A student who assumes ac falls off as 1/r², like the gravitational force, picks this: 2²/(1/2)² = 16. In ac = v²/r the radius appears to the first power, so halving it only doubles ac.
D40 m/s²Correct ac = v²/r. Doubling v multiplies v² by 4, and halving r multiplies 1/r by 2, so ac becomes 4 × 2 = 8 times as large: 8 × 5.0 m/s² = 40 m/s².
Three carts, X, Y and Z, each move at constant speed around a circular track. The figure, not drawn to scale, gives each cart’s speed and the radius of its track. Which ranking of the magnitudes of the carts’ accelerations is correct?
Answer and reasoning
AaX > aY > aZ A student who uses v/r instead of v²/r picks this: 2.5, 1.0 and 0.40 for X, Y and Z. Squaring the speed matters here: Y’s speed is three times X’s, which outweighs its larger radius.
BaY > aZ > aX A student who ranks the accelerations by speed picks this. Acceleration measures how fast the velocity changes, not how fast the cart moves: X is the slowest, but its very small radius gives it a larger acceleration than Z.
CaY > aX > aZCorrect Each cart’s acceleration is centripetal, a = v²/r: aX = (10 m/s)²/(4.0 m) = 25 m/s², aY = (30 m/s)²/(30 m) = 30 m/s², aZ = (20 m/s)²/(50 m) = 8.0 m/s². So aY > aX > aZ.
DaZ > aY > aX A student who thinks a larger circle means a larger centripetal acceleration picks this, ranking by radius. At given speeds a larger radius gives a smaller acceleration, since a = v²/r.
Working a = v²/r: X: 100/4.0 = 25 m/s²; Y: 900/30 = 30 m/s²; Z: 400/50 = 8.0 m/s². Y > X > Z. (v/r: 2.5, 1.0, 0.40 → X > Y > Z; by speed: Y > Z > X; by radius: Z > Y > X.)
An object moves at constant speed counterclockwise around a circle centered at O. The figure shows its velocity v⃗₁ at point P₁ and its velocity v⃗₂ at point P₂, a short time later. Which conclusion about the object’s average acceleration between P₁ and P₂ is supported by the figure?
Answer and reasoning
AIt is not zero and points along v⃗₂ − v⃗₁, into the circle.Correct Average acceleration is Δv⃗/Δt. The arrows have equal lengths but different directions, so Δv⃗ = v⃗₂ − v⃗₁ is not zero. Drawing v⃗₂ and −v⃗₁ tip to tail gives a vector pointing up and to the left, toward O from the part of the circle between P₁ and P₂.
BIt is zero, because the two velocity arrows are equally long. A student who thinks acceleration means only a change in speed picks this. Equal lengths show that the speed is unchanged, but the directions differ, so the velocity, a vector, has changed and the object has accelerated.
CIt is not zero and points away from O, as the object tends outward. A student who believes in an outward centrifugal effect picks this. In an inertial frame Δv⃗ = v⃗₂ − v⃗₁ points inward, toward O; the object’s tendency to continue in a straight line is not an outward acceleration.
DIt is not zero and points along v⃗₂, the way the object now moves. A student who thinks acceleration points along the velocity picks this. Acceleration points along the change in velocity, v⃗₂ − v⃗₁, which here is perpendicular to the direction of motion at the midpoint of the arc, not along v⃗₂.
Working v⃗₁ (at 60° below the horizontal radius) points up-right; v⃗₂ (on the horizontal radius) points straight up; |v⃗₁| = |v⃗₂|. v⃗₂ − v⃗₁ points up and to the left, along the line from the midpoint of the arc to O. Average acceleration Δv⃗/Δt is in that direction.
A car of mass m drives at speed v over the top of a hill whose vertical cross section is a circular arc of radius R. The car stays in contact with the road. What is the magnitude of the normal force exerted by the road on the car at the top of the hill?
Answer and reasoning
Amg A student who thinks the normal force always equals the weight picks this. The car accelerates downward, toward the center of the arc, so the downward gravitational force must exceed the upward normal force.
Bmg+mv²/R A student who adds a downward ‘centripetal force’ to the diagram and then balances the forces picks this: FN = mg + mv²/R. The centripetal force is the net force, mg − FN, not an extra force to be balanced.
Cmg − mv²/RCorrect At the top, the center of the arc is below the car, so the car’s acceleration, v²/R, points down. The gravitational force and the normal force together give the net force toward the center. Taking down as positive: mg − FN = mv²/R, so FN = mg − mv²/R.
Dmv²/R A student who takes the normal force alone as the centripetal force picks this. Two forces act along the line to the center, gravity (down) and the normal force (up); their difference equals mv²/R.
Working Down (toward center) positive: ΣF = mg − FN = mv²/R ⇒ FN = mg − mv²/R (requires v ≤ √(gR)). Distractors: FN = mg; extra Fc balanced: FN = mg + mv²/R; normal force alone: FN = mv²/R.
A car drives at constant speed along the road shown in the figure, passing points X and Y, where the road is a circular arc. FX and FY are the magnitudes of the normal force exerted by the road on the car at X and Y, and Fg is the magnitude of the gravitational force on the car. Use g = 10 m/s². Which is correct?
Answer and reasoning
AFX = FY = Fg A student who thinks the normal force always equals the weight picks this. At X the car accelerates downward and at Y upward, so the normal force is less than Fg at X and greater at Y.
BFY > Fg > FXCorrect ac = v²/r = (10 m/s)²/(20 m) = 5.0 m/s². At X the center is below the car: Fg − FX = m(5.0 m/s²), so FX = 0.5Fg. At Y the center is above: FY − Fg = m(5.0 m/s²), so FY = 1.5Fg.
CFX > Fg > FY A student who adds a separate ‘centripetal force’ and balances it picks this: at X a downward centripetal force makes FX = Fg + mv²/r; at Y an upward one makes FY = Fg − mv²/r. The net force itself points to the center, which reverses both results.
DFg > FX = FY A student who takes the normal force alone as the centripetal force picks this: FX = FY = mv²/r = 0.5Fg. The gravitational force also lies along the line to the center, so it must be included in the net force at both points.
Working ac = v²/r = 100/20 = 5.0 m/s² = 0.5g. X (center below): Fg − FX = 0.5mg ⇒ FX = 0.5Fg. Y (center above): FY − Fg = 0.5mg ⇒ FY = 1.5Fg. FY > Fg > FX.
A cart moves on the inside of a vertical circular loop of radius R. As it passes the top of the loop, the track exerts on the cart a force equal in magnitude to three times the cart’s weight. Friction is negligible. What is the cart’s speed at the top of the loop?
Answer and reasoning
A2√(gR)Correct At the top, the track is above the cart, so the normal force points down, toward the center, like gravity. FN + mg = mv²/R with FN = 3mg gives 4mg = mv²/R, so v² = 4gR and v = 2√(gR).
B√(3gR) A student who takes the normal force alone as the centripetal force picks this: 3mg = mv²/R. Gravity also points toward the center at the top, so the two forces add to give the net force, 4mg.
C√(2gR) A student who draws the normal force upward, as on a floor, picks this: the two forces then oppose, and the net force is taken as 3mg − mg = 2mg = mv²/R. On the inside of the loop the track is above the cart at the top, so the normal force points down, the same way as gravity.
D4√(gR) A student who scales the speed by the factor of change in the force picks this: at the least speed, √(gR), the net force is mg; here it is 4mg, so the speed is taken to be 4 times as large. The net force is proportional to v², so 4 times the net force gives 2 times the speed.
Working Down positive at the top: FN + mg = mv²/R with FN = 3mg ⇒ v² = 4gR ⇒ v = 2√(gR). Distractors: normal force alone, 3mg = mv²/R ⇒ √(3gR); normal force drawn upward, net force 3mg − mg = 2mg = mv²/R ⇒ √(2gR); speed scaled by the force factor 4 from the least speed √(gR) (net force mg) ⇒ 4√(gR) (sympy-checked).
A puck of mass 0.30 kg slides on a frictionless horizontal table. It is tied to a string that passes through a small hole in the table and is attached to a block, which hangs at rest below the table. The puck moves in a circle of radius 0.40 m, centered on the hole, at a constant speed of 1.6 m/s. Use g = 10 m/s². What is the mass of the hanging block?
Answer and reasoning
A0.49 kg A student who takes the tension to be mg + mv²/r, as at the bottom of a vertical circle, picks this: Mg = (0.30 kg)(10 m/s²) + (0.30 kg)(1.6 m/s)²/(0.40 m) = 4.92 N. The table supports the puck’s weight, so the tension alone gives the puck’s centripetal acceleration: T = mv²/r.
B0.53 kg A student who lets the block’s weight accelerate both objects picks this: Mg = (m + M)v²/r, so M = mv²/(gr − v²) = 0.53 kg. The block hangs at rest, so it does not accelerate; only the puck has the acceleration v²/r.
C0.12 kg A student who uses ac = v/r picks this: Mg = mv/r = (0.30 kg)(1.6 m/s)/(0.40 m) = 1.2 N. The centripetal acceleration is v²/r = 6.4 m/s², not v/r, whose units are 1/s.
D0.19 kgCorrect The block is at rest, so the tension is T = Mg. The tension is the only horizontal force on the puck and points toward the hole, so T = mv²/r = (0.30 kg)(1.6 m/s)²/(0.40 m) = 1.92 N. Then M = T/g = 0.19 kg.
Working Block at rest: T = Mg. Puck, horizontal: T = mv²/r. So M = mv²/(gr) = (0.30 kg)(1.6 m/s)²/((10 m/s²)(0.40 m)) = 0.192 kg ≈ 0.19 kg. Distractors: T = mg + mv²/r ⇒ M = m + mv²/(gr) = 0.49 kg; Mg = (m + M)v²/r ⇒ M = mv²/(gr − v²) = 0.768/1.44 = 0.53 kg; ac = v/r ⇒ M = mv/(gr) = 0.12 kg.
A cart on the inside of a vertical loop of radius 3.60 m passes the top at 6.00 m/s, the least speed at which it stays in contact with the track there. What is the least speed at which a cart stays in contact with the track at the top of a vertical loop of radius 14.4 m?
Answer and reasoning
A6.00 m/s A student who thinks equal accelerations mean equal speeds picks this: at the least speed the centripetal acceleration is g at the top of either loop. But ac = v²/r, so the same acceleration on a larger circle needs a larger speed.
B12.0 m/sCorrect At the least speed the normal force is zero and gravity alone gives the centripetal acceleration: g = v²/r, so v = √(gr). The least speed is proportional to √r; the radius is 4 times as large, so the speed is √4 = 2 times as large: 2 × 6.00 m/s = 12.0 m/s.
C24.0 m/s A student who takes ac proportional to v picks this: g = v/r makes the least speed proportional to r, 4 × 6.00 m/s. With ac = v²/r, the speed goes as √r.
D96.0 m/s A student who scales a square-root dependence by the square picks this: (4)² × 6.00 m/s. Since v = √(gr), multiplying r by 4 multiplies v by √4 = 2.
Working Least speed at top: FN = 0, mg = mv²/r ⇒ v = √(gr) ∝ √r. Ratio √(14.4/3.60) = √4 = 2 ⇒ 12.0 m/s. (Check: √(10 × 3.60) = 6.00 m/s.) Distractors: same a ⇒ same v, 6.00; v ∝ r ⇒ 24.0; v ∝ r² ⇒ 96.0.
A cart on the inside of a vertical loop passes the top of the loop at exactly the least speed for which it stays in contact with the track. Which statement describes the forces exerted on the cart at that instant?
Answer and reasoning
AGravity and a downward normal force equal in size to the cart’s weight act. A student who thinks a normal force equals the weight picks this. With FN = mg the cart would need v = √(2gr), more than the least speed; at the least speed the normal force has fallen to zero.
BGravity acts, balanced by an upward centrifugal force; the net force is zero. A student who believes in a real outward force picks this. No object exerts such a force; the net force on the cart is mg, downward, toward the center, as its circular motion requires.
COnly the gravitational force acts; the track exerts no force on it.Correct At the least speed the cart is just on the point of losing contact, so the normal force is zero. The gravitational force alone points toward the center and provides the centripetal acceleration: mg = mv²/r, which gives v = √(gr).
DGravity, the normal force and a separate centripetal force all act downward. A student who treats the centripetal force as an extra force picks this. The centripetal force is the net force itself; at the least speed it is gravity alone, with no normal force.
A car rounds a banked curve at a speed greater than the speed at which no friction would be needed. The car does not slide. Which describes the static friction force exerted on the car by the road?
Answer and reasoning
ADown the slope, adding to the normal force’s inward horizontal component.Correct Above the no-friction speed, the normal force’s horizontal component is too small to give mv²/R; without friction the car would slide up and outward. Static friction opposes that tendency, pointing down the slope, and its horizontal component adds to the net force toward the center.
BBackward along the road surface, opposite to the direction of the car’s velocity. A student who thinks friction always opposes the direction of motion picks this. Static friction opposes the slipping that would occur without it; here the car would slip up the bank, not backward, and a free-rolling car needs no backward force.
CUp the slope, keeping the car from sliding down the banked road surface. A student who thinks friction on a slope always points up it picks this. That is true at speeds below the no-friction speed; above it the car tends to slide up the bank, so friction points down the slope.
DHorizontally toward the curve’s center, because friction is the centripetal force. A student who takes one force as ‘the centripetal force’ picks this. Friction is parallel to the road surface, which is tilted; the net force toward the center is the sum of the horizontal components of the normal force and friction.
A curve of radius 40 m is banked at 37° to the horizontal. The coefficient of static friction between a car’s tires and the road is 0.25. Use g = 10 m/s². What is the greatest speed at which the car can round the curve without sliding?
Answer and reasoning
A17 m/s A student who takes the speed at which no friction is needed as the greatest speed picks this: v = √(gR tan θ) = √(300) m/s. At higher speeds friction down the slope adds to the inward force, so the car can go faster before sliding.
B22 m/sCorrect At the greatest speed the car is about to slide up the bank, so static friction points down the slope with magnitude μs FN. Vertically: FN cos θ − μs FN sin θ = mg. Horizontally: FN sin θ + μs FN cos θ = mv²/R. Dividing: v² = gR(sin θ + μs cos θ)/(cos θ − μs sin θ) = (400 m²/s²)(0.80)/(0.65), so v = 22 m/s.
C16 m/s A student who uses FN = mg cos θ, as on an incline, picks this: horizontally (mg cos θ)(sin θ + μs cos θ) = mv²/R gives v² = (400)(0.80)(0.80). The car’s acceleration is horizontal, not along the slope, so the normal force is found from the vertical forces.
D13 m/s A student who takes friction up the slope picks this: v² = gR(sin θ − μs cos θ)/(cos θ + μs sin θ) = (400)(0.40)/(0.95). That is the least speed; at the greatest speed the car tends to slide up the bank, so friction points down the slope.
Working Friction down slope, f = μs FN. Vertical: FN(cos θ − μs sin θ) = mg. Horizontal: FN(sin θ + μs cos θ) = mv²/R. v² = gR(sin θ + μs cos θ)/(cos θ − μs sin θ) = 10 × 40 × (0.60 + 0.20)/(0.80 − 0.15) = 492 m²/s² ⇒ v = 22 m/s. Distractors: √(400 × 0.75) = 17; √(400 × 0.80 × 0.80) = 16; √(400 × 0.40/0.95) = 13.
The figure shows a conical pendulum: a small ball on a string moves at constant speed in the horizontal circle drawn dashed, with the string length and angle labeled. Use g = 10 m/s². How long does the ball take to complete one revolution?
Answer and reasoning
A3.2 s A student who takes the string length as the circle’s radius picks this: v² = gL tan θ = 15 m²/s² and the time is 2π(2.0 m)/v. The ball circles a point directly below the support, so the radius is L sin θ = 1.2 m.
B2.8 s A student who sets the tension equal to the ball’s weight picks this: mg sin θ = mv²/r gives v² = gr sin θ = 7.2 m²/s², and then 2πr/v. Only the tension’s vertical component balances mg, so the tension is mg/cos θ, larger than mg.
C2.5 sCorrect The circle’s radius is r = L sin θ = 1.2 m. Vertically, FT cos θ = mg; horizontally, FT sin θ = mv²/r (FT is the tension). Dividing: v² = gr tan θ = (10)(1.2)(0.75) = 9.0 m²/s², so v = 3.0 m/s. The time for one revolution is 2πr/v = 2π(1.2 m)/(3.0 m/s) = 2.5 s.
D2.2 s A student who always takes horizontal components with cos θ and vertical ones with sin θ picks this. That student takes the radius as L cos θ = 1.6 m and writes FT sin θ = mg and FT cos θ = mv²/r, so v² = gr cos θ/sin θ = 21 m²/s² and the time is 2π(1.6 m)/v. The angle is measured from the vertical, so the radius is L sin θ and the vertical component of the tension is FT cos θ.
Working r = L sin θ = 2.0 × 0.60 = 1.2 m. FT cos θ = mg; FT sin θ = mv²/r ⇒ v² = g r tan θ = 10 × 1.2 × 0.75 = 9.0 ⇒ v = 3.0 m/s. Time = 2πr/v = 2π(1.2)/3.0 = 2.51 s ≈ 2.5 s (equivalently 2π√(L cos θ/g)). Distractors: r = L: v² = 10 × 2.0 × 0.75 = 15, 2π(2.0)/√15 = 3.2 s; FT = mg: v² = g r sin θ = 7.2, 2π(1.2)/√7.2 = 2.8 s; horizontal always with cos θ, vertical with sin θ: r = L cos θ = 1.6 m, FT sin θ = mg and FT cos θ = mv²/r, v² = g r cos θ/sin θ = 10 × 1.6 × 0.8/0.6 = 21.3, 2π(1.6)/√21.3 = 2.2 s.
A ball on a string moves as a conical pendulum: it travels in a horizontal circle at constant speed, with the string at a constant angle to the vertical. What provides the ball’s centripetal acceleration?
Answer and reasoning
AThe tension, which an outward centrifugal force on the ball balances. A student who believes in a real centrifugal force picks this. No object pushes the ball outward; the ball is not in equilibrium, because it accelerates toward the center.
BA centripetal force on the ball in addition to the tension and Fg. A student who treats ‘centripetal force’ as an extra force picks this. The only objects acting on the ball are the string and Earth; the net of their forces is what is called the centripetal force.
CThe whole tension, the single force pointing to the circle’s center. A student who takes one force as ‘the centripetal force’ picks this. The string slopes upward to the support, so the tension does not point at the circle’s center; only its horizontal component does.
DThe tension’s horizontal part; its vertical part balances Fg.Correct Only two forces act: the gravitational force (vertical) and the tension (along the string). The ball has no vertical acceleration, so the tension’s vertical component balances Fg; its horizontal component, toward the center of the circle, is the net force, mv²/r.
An object starts from rest and moves along a circle of radius R. Its tangential acceleration has constant magnitude at. What is the magnitude of the object’s total acceleration when it has traveled one quarter of the way around the circle?
Answer and reasoning
Aat(1 + π) A student who adds the two components as numbers picks this: at + πat. The tangential component is along the velocity and the centripetal component perpendicular to it, so they add as vectors, by the Pythagorean theorem.
Bπat A student who takes the whole acceleration of an object on a circle as centripetal picks this. The object is still speeding up, so its acceleration also has the tangential component at.
Cat A student who counts only the change of speed as acceleration picks this. The direction of the velocity is also changing, which adds the centripetal component v²/R = πat.
Dat√(1 + π²)Correct After a quarter circle, a distance πR/2, the speed satisfies v² = 2at(πR/2) = πRat, so the centripetal acceleration is v²/R = πat. The tangential and centripetal accelerations are perpendicular, so |a⃗| = √(at² + π²at²) = at√(1 + π²).
Working s = πR/2; v² = 2at s = πRat (constant tangential acceleration from rest). ac = v²/R = πat. |a| = √(at² + ac²) = at√(1 + π²) (sympy-checked). Distractors: scalar sum at(1 + π); ac only πat; at only.
A point on a rotating turntable moves at constant speed in a circle centered on the origin. The graph shows the point’s x-coordinate as a function of time. What is the point’s speed?
Answer and reasoning
A1.0 m/sCorrect The radius is the largest value of x, 0.40 m. One full cycle, from one maximum to the next, takes 2.4 s, so T = 2.4 s. Then v = 2πr/T = 2π(0.40 m)/(2.4 s) = 1.0 m/s.
B2.1 m/s A student who takes the period as the time between successive crossings of x = 0, 1.2 s, picks this. Between those crossings the point moves in opposite directions; only half a revolution has been completed.
C4.2 m/s A student who reads the period from t = 0 to the first maximum, 0.6 s, picks this. The graph starts at x = 0, so the first maximum comes after a quarter of a revolution.
D6.0 m/s A student who mixes up period and frequency picks this, using the period, 2.4 s, in place of f in v = 2πrf: 2π(0.40)(2.4) = 6.0. The frequency is f = 1/T = 0.42 Hz, which gives v = 2πrf = 1.0 m/s.
Working Read r = 0.40 m (amplitude of x) and T = 2.4 s (maximum to maximum). v = 2πr/T = 2π(0.40)/2.4 = 1.05 ≈ 1.0 m/s. Distractors: T = 1.2 s ⇒ 2.1; T = 0.6 s ⇒ 4.2; 2πr·(2.4) = 6.0.
The figure shows a turntable, seen from above, rotating at a constant rate about its axis O. Two coins, X and Y, rest on it without slipping at the distances from O shown. Which statement comparing the coins’ motions is correct?
Answer and reasoning
AEqual speeds; X has the larger acceleration. A student who thinks every point of a rotating object moves at the same speed picks this. Y travels twice as far as X in each revolution in the same time, so Y moves twice as fast.
BEqual periods; Y has the greater acceleration.Correct Both coins turn with the turntable, so they share one period T. Y’s circle is twice as long, so v = 2πr/T is twice as large for Y. Then a = v²/r = 4π²r/T² is proportional to r: Y’s acceleration is twice X’s.
CEqual periods; X has the larger acceleration. A student who reads a = v²/r as ‘larger r, smaller a’ without letting v change picks this. With equal periods, v ∝ r, so a = 4π²r/T² is larger for Y.
DY has the shorter period and the greater speed. A student who thinks the faster coin goes round more often picks this. Both coins complete each revolution together with the turntable; Y is faster only because its circle is longer.
Working Same T (rigid rotation). v = 2πr/T ⇒ vY = 2vX. a = v²/r = 4π²r/T² ⇒ aY = 2aX.
A satellite moves in a circular orbit 1.0 × 10⁶ m above the surface of a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N·m²/kg². What is the period of the orbit?
Answer and reasoning
A3.1 × 10² s A student who uses the satellite’s altitude, 1.0 × 10⁶ m, as the orbital radius picks this. In Kepler’s third law R is the distance from the planet’s center: altitude plus the planet’s radius.
B5.5 × 10³ s A student who uses the surface value of the gravitational field, GM/(6.4 × 10⁶ m)² = 9.8 m/s², as the centripetal acceleration in orbit picks this: T = 2π√(R/g). At R = 7.4 × 10⁶ m the field is weaker, GM/R² = 7.3 m/s², so the period is longer.
C4.0 × 10⁷ s A student who drops the square on T picks this: T = 4π²R³/(GM). Kepler’s third law relates T², not T, to R³; T is the square root of 4π²R³/(GM).
D6.3 × 10³ sCorrect The orbital radius is measured from the planet’s center: R = 6.4 × 10⁶ m + 1.0 × 10⁶ m = 7.4 × 10⁶ m. From T² = (4π²/(GM))R³, T = 2π√(R³/(GM)) = 2π√((7.4 × 10⁶ m)³/(4.0 × 10¹⁴ m³/s²)) = 6.3 × 10³ s.
Working R = 6.4e6 + 1.0e6 = 7.4e6 m. GM = 6.67e-11 × 6.0e24 = 4.0e14 m³/s². T = 2π√(R³/GM) = 2π√(4.05e20/4.0e14) = 2π(1006 s) = 6.3 × 10³ s. Distractors: R = 1.0e6 ⇒ 3.1 × 10² s; gsurface = GM/(6.4e6)² = 9.77 m/s², T = 2π√(R/gsurface) = 5.5 × 10³ s; T = 4π²R³/(GM) = 4.0 × 10⁷ s.
In each part of the figure, a moon moves in a circular orbit around a planet. The planets’ masses and the orbits’ radii are labeled. TA and TB are the orbital periods of the moons in A and B. Which is correct?
Answer and reasoning
ATB = 2.8TA A student who treats T²/R³ as the same constant for every orbit, leaving out the planet’s mass, picks this: TB = √8 TA. The constant 4π²/(GM) depends on the central body, so B’s larger planet shortens the period.
BTB = 5.7TA A student who applies the mass dependence the wrong way round, as if T² were proportional to M, picks this: √(8 × 4) = 5.7. A more massive planet pulls harder and gives a shorter period at a given radius: T² ∝ R³/M.
CTB = 1.4TACorrect From T² = (4π²/(GM))R³, T ∝ √(R³/M). For B, R³ is 2³ = 8 times as large and M is 4 times as large, so T² is 8/4 = 2 times as large and TB = √2 TA ≈ 1.4TA.
DTB = 2.0TA A student who drops the square on T picks this: T ∝ R³/M = 8/4 = 2. Kepler’s third law gives T² ∝ R³/M, so T changes by the square root of that factor, √2.
Satellite S₁ moves in a circular orbit around Earth. Satellite S₂, which has twice the mass of S₁, moves in a circular orbit of the same radius. Which claim about the period of S₂, with its reasoning, is correct?
Answer and reasoning
ASame as S₁’s: both the gravitational force on S₂ and the net force it needs, mv²/R, double.Correct Doubling the satellite’s mass doubles the gravitational force GMm/R², and it doubles the net force needed for a given speed, mv²/R. The mass cancels from GMm/R² = mv²/R, so the speed, and with it the period, at a given radius are the same.
BShorter than S₁’s: twice the gravitational force gives S₂ twice the centripetal acceleration. A student who thinks a heavier satellite orbits faster picks this, reasoning from the force alone. The acceleration is the force divided by the mass: (2F)/(2m) is the same as F/m, so S₂ has the same acceleration and period.
CSame as S₁’s: there is no gravity at orbital heights, so a satellite’s mass has no effect. A student who thinks there is no gravity in orbit picks this; the claim is right but the reasoning is wrong. Gravity is the only force on each satellite, and it is what provides the centripetal acceleration.
DLonger than S₁’s: S₂ needs twice the forward force to keep it moving along its orbit. A student who thinks a moving object needs a force along its direction of motion picks this. No forward force acts on either satellite; the only force, gravity, points toward Earth’s center and changes the direction of the velocity.
A planet has radius R, and the magnitude of the gravitational field at its surface is g. A satellite moves in a circular orbit at a height R above the planet’s surface. In terms of g and R, what is the satellite’s speed?
Answer and reasoning
A√(gR/2)Correct At the surface, g = GM/R², so GM = gR². The orbital radius is R + R = 2R, and gravity alone provides the centripetal acceleration: GM/(2R)² = v²/(2R). So v² = GM/(2R) = gR²/(2R) = gR/2.
B√(gR) A student who uses the height above the surface, R, as the orbital radius picks this: v² = GM/R = gR. The orbit’s radius is measured from the planet’s center: 2R.
C√(2gR) A student who takes the gravitational field at the orbit to be g, its surface value, picks this: g = v²/(2R). At twice the distance from the center the field is g/4, which gives v² = gR/2.
DR√g A student who takes centripetal acceleration to fall off as 1/r², like gravity, picks this: with the correct field g/4 at radius 2R, g/4 = v²/(2R)² gives v² = gR², so v = R√g. Centripetal acceleration is v²/r; R√g does not have the units of speed.
Working GM = gR². r = 2R. GM/r² = v²/r ⇒ v² = GM/r = gR²/(2R) = gR/2 ⇒ v = √(gR/2) (sympy-checked). Distractors: r = R ⇒ √(gR); ac = g ⇒ v² = 2gR; ac ∝ 1/r²: g/4 = v²/(2R)² ⇒ v = R√g.
A car of mass m rounds a curve of radius r on a road banked at angle θ to the horizontal. It moves at constant speed v, which is large enough that the static friction force exerted on the car by the road is directed down the slope. The car does not slide, and g is the magnitude of the gravitational field. What is the magnitude of the static friction force?
Answer and reasoning
Am(v²/(r cos θ) − g sin θ) A student who uses FN = mg cos θ, the result for a block whose acceleration is along an incline, picks this: mg cos θ sin θ + Ff cos θ = mv²/r. The car accelerates horizontally, so the normal force is not mg cos θ; it must be found from the vertical and horizontal equations together.
Bm((v²/r)cos θ − g sin θ)Correct The acceleration is horizontal, toward the center, so the vertical components balance: FN cos θ − Ff sin θ = mg, and the horizontal components give FN sin θ + Ff cos θ = mv²/r. Eliminating FN gives Ff = m((v²/r)cos θ − g sin θ): the component of the required acceleration along the slope, (v²/r)cos θ, minus the part that gravity’s component down the slope already provides.
Cm(v²/(r cos θ) − g/sin θ)/2 A student who always takes a horizontal component with cos θ and a vertical one with sin θ picks this. For the normal force, which makes angle θ with the vertical, that gives a horizontal component FN cos θ and a vertical component FN sin θ, so (FN + Ff) cos θ = mv²/r and (FN − Ff) sin θ = mg. The normal force’s horizontal component is FN sin θ and its vertical component is FN cos θ.
Dm(v²/r − g sin θ)/cos θ A student who takes the normal force to equal the car’s weight, mg, picks this: mg sin θ + Ff cos θ = mv²/r. The car accelerates toward the center, so the normal force is not mg; it must be found from the vertical and horizontal equations together, and it comes out larger than mg.
Working Forces: FN perpendicular to the road, Ff down the slope, Fg = mg down. Acceleration v²/r horizontal, toward the center. Horizontal: FN sin θ + Ff cos θ = mv²/r. Vertical: FN cos θ − Ff sin θ = mg. Multiply the first by cos θ and the second by sin θ and subtract: Ff = m((v²/r)cos θ − g sin θ) (sympy-checked; positive because v² > gr tan θ). Distractors: FN = mg cos θ put into the horizontal equation, Ff = m(v²/(r cos θ) − g sin θ); FN = mg put into the horizontal equation, Ff = m(v²/r − g sin θ)/cos θ; every horizontal component taken with cos θ and every vertical one with sin θ, so the normal force’s horizontal component becomes FN cos θ and its vertical component FN sin θ, giving (FN + Ff) cos θ = mv²/r and (FN − Ff) sin θ = mg, so Ff = m(v²/(r cos θ) − g/sin θ)/2 (sympy-checked).
A small ball of mass m hangs from a light string of length L and moves as a conical pendulum: it travels at constant speed in a horizontal circle, with the string at a constant angle θ to the vertical. Air resistance is negligible, and g is the magnitude of the gravitational field. What is the ball’s speed?
Answer and reasoning
A√(gL sin θ/cos θ) A student who takes the radius of the circle to be the string length L picks this: FT sin θ = mv²/L with FT cos θ = mg gives v² = gL sin θ/cos θ. The ball circles a point directly below the support, at radius L sin θ.
B√(gL cos²θ/sin θ) A student who always takes horizontal components with cos θ and vertical ones with sin θ picks this. That student takes the radius as L cos θ and writes FT sin θ = mg and FT cos θ = mv²/(L cos θ), which gives v² = gL cos²θ/sin θ. The angle is measured from the vertical, so the radius is L sin θ and the vertical component of the tension is FT cos θ.
C√(gL sin²θ/cos θ)Correct The tension’s vertical component balances gravity, FT cos θ = mg, and its horizontal component gives the centripetal acceleration on a circle of radius L sin θ: FT sin θ = mv²/(L sin θ). Dividing one equation by the other gives v² = gL sin²θ/cos θ.
DL sin θ√(g tan θ) A student who takes centripetal acceleration to fall off as 1/r², like gravity, picks this: FT sin θ = mv²/(L sin θ)² with FT cos θ = mg gives v² = g tan θ (L sin θ)². Centripetal acceleration is v²/r; this expression does not even have the units of speed.
Working Radius of the circle r = L sin θ. Vertical (no vertical acceleration): FT cos θ = mg. Horizontal, toward the center: FT sin θ = mv²/(L sin θ). Dividing: tan θ = v²/(gL sin θ), so v² = gL sin²θ/cos θ and v = √(gL sin²θ/cos θ) (sympy-checked). Distractors: r = L gives v² = gL sin θ/cos θ; horizontal always with cos θ and vertical with sin θ gives r = L cos θ, FT sin θ = mg and FT cos θ = mv²/(L cos θ), so v² = gL cos²θ/sin θ (sympy-checked); ac = v²/r² gives v² = g tan θ (L sin θ)², v = L sin θ√(g tan θ).
Two small balls, A and B, each of mass m, move on a frictionless horizontal table. String 1, of length L, joins a fixed peg to ball A, and string 2, also of length L, joins ball A to ball B. The balls move in circles about the peg with both strings taut and in one straight line, and ball B moves at constant speed v. The magnitude of the force exerted on ball A by string 1 can be written as F₁ = kmv²/L. What is the value of k?
Answer and reasoning
A0.75Correct Both balls go around in the same time, so B, at radius 2L, moves at v and A, at radius L, moves at v/2. String 2 alone gives B its centripetal acceleration: F₂ = mv²/(2L). On A, string 1 pulls inward and string 2 pulls outward, so F₁ − F₂ = m(v/2)²/L = mv²/(4L), and F₁ = 0.75 mv²/L.
B0.25 A student who treats string 1 as the one force that alone gives A its centripetal acceleration picks this: F₁ = m(v/2)²/L. String 2 also pulls on A, outward, so string 1 must supply A’s centripetal force plus that pull.
C1.50 A student who thinks every part of a system turning as one piece moves at the same speed gives A the speed v: F₁ = mv²/(2L) + mv²/L. The balls share a period, so A, at half B’s radius, moves at v/2.
D0.50 A student who thinks every string in the system has the same tension sets F₁ = F₂ = mv²/(2L). Ball A also accelerates toward the peg, so string 1 must pull harder than string 2 by mv²/(4L).
Working The strings stay in a straight line, so both balls complete a revolution in the same period T. v = 2π(2L)/T for B and vA = 2πL/T = v/2 for A. Ball B: string 2 is the only horizontal force, F₂ = mv²/(2L). Ball A: string 1 pulls inward, string 2 pulls outward, F₁ − F₂ = mvA²/L = mv²/(4L). F₁ = mv²/(2L) + mv²/(4L) = 0.75 mv²/L (sympy-checked). Distractors: string 1 alone taken as A’s centripetal force, F₁ = m(v/2)²/L → 0.25; both balls given speed v, F₁ = mv²/(2L) + mv²/L → 1.50; same tension in both strings, F₁ = F₂ = mv²/(2L) → 0.50.
Compiled from the AP Physics C: Mechanics 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
