3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A cart has a translational kinetic energy of 12 J. Cargo is then removed so that the cart’s total mass is halved, and the cart moves at twice its original speed. What is the cart’s new translational kinetic energy?
Answer and reasoning
A12 J A student who takes K as proportional to v rather than v² picks this: (1/2)(2) = 1, so K is unchanged. The speed is squared, so doubling it alone makes K four times as large.
B48 J A student who thinks mass does not affect kinetic energy picks this, applying only the speed factor of 4. K is proportional to m, so halving the mass halves K.
C96 J A student who applies the mass factor the wrong way round picks this: 4 × 2 = 8, as if halving the mass doubled K. K ∝ m, so halving m multiplies K by 1/2.
D24 JCorrect K = (1/2)mv². Halving m multiplies K by 1/2; doubling v multiplies it by 2² = 4. The new K is (1/2)(4)(12 J) = 24 J.
Working K₂/K₁ = (m₂/m₁)(v₂/v₁)² = (1/2)(2)² = 2 ⇒ 24 J. Distractors: K ∝ mv ⇒ (1/2)(2) = 1 ⇒ 12 J; mass ignored ⇒ 4 ⇒ 48 J; mass factor inverted ⇒ 2 × 4 = 8 ⇒ 96 J.
A ball moves at constant speed in a horizontal circle. Which statement about the ball is correct?
Answer and reasoning
AIts K keeps changing direction, just as its velocity does. A student who treats kinetic energy as a vector picks this. K is a scalar with no direction; only the velocity turns.
BIts K and its velocity both stay constant, as its speed does too. A student who equates constant speed with constant velocity picks this. K is constant, but the velocity changes because its direction changes.
CIts K is constant, but its velocity changes direction.Correct Kinetic energy depends only on speed, K = (1/2)mv², and the speed is constant, so K is constant. Velocity is a vector, and its direction changes continuously as the ball goes around.
DIts K changes sign each time the ball reverses its direction. A student who thinks motion in a negative direction gives negative kinetic energy picks this. The velocity is squared in K = (1/2)mv², so K is positive whichever way the ball moves.
A train moves along a straight, level track. The figure shows a ball rolling along the train’s floor, opposite to the train’s motion, with the speeds and the ball’s mass labeled. Treat the ball as a particle. What is the ball’s translational kinetic energy measured by an observer at rest on the ground?
Answer and reasoning
A1.2 × 10² J A student who adds the two speeds whatever their directions picks this: (1/2)(0.40)(24 m/s)². The ball moves backward relative to the train, so its speed relative to the ground is 20 − 4.0 = 16 m/s.
B3.2 × 10⁰ J A student who thinks every observer measures the same kinetic energy picks this, using the speed relative to the train: (1/2)(0.40)(4.0)² = 3.2 J. That is the value for an observer on the train; relative to the ground the ball moves at 16 m/s.
C5.1 × 10¹ JCorrect Relative to the ground, the velocities add as vectors: taking the train’s direction as positive, v = +20 m/s − 4.0 m/s = +16 m/s. K = (1/2)mv² = (1/2)(0.40 kg)(16 m/s)² = 51 J.
D8.3 × 10¹ J A student who adds kinetic energies from the two frames picks this: (1/2)(0.40)(20)² + (1/2)(0.40)(4.0)² = 80 J + 3.2 J. Velocities add between frames; kinetic energies do not, because of the square.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.1.A.1 Translational kinetic energy, K Fix
Translational kinetic energy, K
The energy an object has because of the motion of its center of mass: K = (1/2)mv², where m is the mass and v the speed. SI unit: joule (J).
Joule, J
The SI unit of energy: 1 J = 1 kg·m²/s², the kinetic energy of a 2 kg object moving at 1 m/s.
Students often think Translational kinetic energy is proportional to speed, so K changes by the same factor as v, as momentum-like quantities such as mv do. In fact No. K = (1/2)mv²: kinetic energy is proportional to the square of the speed, so doubling the speed makes K four times as large.
Students often think Translational kinetic energy can be written as K = mv², the factor 1/2 being unimportant. In fact No. K = (1/2)mv² exactly; leaving out the 1/2 doubles every kinetic energy, and the error does not cancel when kinetic energies are compared with other energies.
3.1.A.2 Scalar quantity Fix
Scalar quantity
A quantity with magnitude only and no direction. Translational kinetic energy is a scalar: it is never negative and is the same for any direction of motion at a given speed.
Speed from velocity components
For perpendicular components, v² = vx² + vy², so K = (1/2)m(vx² + vy²). The components’ directions do not affect K.
Students often think Translational kinetic energy is a vector with the direction of the velocity, so it has components (Kx, Ky) that are combined like vectors, and it changes when the direction of motion changes. In fact No. K is a scalar: it has a magnitude only. It depends on the speed, v² = vx² + vy², not on the direction of motion, and it has no components to be combined as vectors.
Students often think An object moving in the negative direction has negative kinetic energy, because its velocity is negative. In fact No. K = (1/2)mv² with m > 0 and v² ≥ 0, so K is never negative. An object moving in the −x direction has the same kinetic energy as one moving in the +x direction at the same speed.
3.1.A.3 Frame of reference Fix
Frame of reference
The viewpoint (an observer with coordinate axes) relative to which positions and velocities are measured. Since K depends on speed, its value depends on the frame.
Relative velocity
The velocity of an object measured in a given frame. For a frame moving at velocity u⃗ relative to the ground, the object’s velocity relative to the ground is v⃗ground = v⃗rel + u⃗ (vector addition).
Students often think An object has one true kinetic energy, the value found from its speed relative to the ground, which is truly at rest; values measured by moving observers are wrong. In fact No. Velocity, and with it kinetic energy, is defined relative to a frame of reference. The ground is one frame among many; a value measured in a frame moving relative to the ground (at constant velocity) is equally valid.
Students often think All observers must measure the same kinetic energy for an object, because energy is conserved, so the kinetic energy measured in one frame holds in every frame. In fact No. Observers in different frames measure different speeds and so different kinetic energies for the same object. Energy conservation applies within one frame; it does not require different frames to agree.
4 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 4
A 2.0 kg object moves in a horizontal plane with velocity components vx = 3.0 m/s and vy = 4.0 m/s. What is its translational kinetic energy?
Answer and reasoning
A18 J A student who treats kinetic energy as a vector picks this: ‘components’ of 9.0 J and 16 J combined as √(9.0² + 16²). K is a scalar; the two contributions add as numbers to 25 J.
B25 JCorrect Kinetic energy depends on the speed: v² = vx² + vy² = (3.0 m/s)² + (4.0 m/s)² = 25 m²/s². K = (1/2)mv² = (1/2)(2.0 kg)(25 m²/s²) = 25 J. Equivalently, (1/2)m vx² + (1/2)m vy² = 9.0 J + 16 J, added as numbers, because K is a scalar.
C49 J A student who adds the velocity components to get the speed picks this: v = 3.0 + 4.0 = 7.0 m/s. The components are perpendicular, so v = √(3.0² + 4.0²) = 5.0 m/s.
D50 J A student who leaves out the factor 1/2 picks this: mv² = (2.0 kg)(25 m²/s²). K = (1/2)mv².
Working v² = 3.0² + 4.0² = 25 m²/s². K = ½(2.0)(25) = 25 J. Distractors: √(9.0² + 16²) = 18.4 ≈ 18 J; ½(2.0)(7.0)² = 49 J; (2.0)(25) = 50 J.
A cart moves along a straight track at constant speed u relative to the ground. A small puck of mass m slides on the cart’s frictionless floor at speed v relative to the cart, in the same direction as the cart moves. By how much does the puck’s translational kinetic energy measured by an observer at rest on the ground exceed the value measured by an observer riding on the cart?
Answer and reasoning
Amu²/2 A student who adds kinetic energies from frame to frame picks this: Kground = Kcart + (1/2)mu², so the difference is (1/2)mu². Velocities add between frames, but kinetic energy depends on the square of the total speed, (u + v)², which adds the cross term muv.
Bmu(u + 2v)/2Correct Relative to the ground the velocities add: the puck’s speed is u + v, since both motions are in the same direction. The difference is (1/2)m(u + v)² − (1/2)mv² = (1/2)mu² + muv = mu(u + 2v)/2.
Cmu(u + 2v) A student who writes kinetic energy as mv², dropping the factor 1/2, picks this: m(u + v)² − mv² = mu² + 2muv. With K = (1/2)mv² in each frame, the difference is half of this.
Dmu/2 A student who takes kinetic energy as proportional to speed picks this: (1/2)m(u + v) − (1/2)mv = (1/2)mu. Its units, kg·m/s, are not those of energy; the speeds must be squared before subtracting.
The figure shows three objects, A, B and C, sliding along a straight horizontal line, with their masses and velocities labeled. KA, KB and KC are their translational kinetic energies. Which ranking is correct?
Answer and reasoning
AKB > KC > KACorrect K = (1/2)mv²: KA = (1/2)(1.0)(3.0)² = 4.5 J, KB = (1/2)(1.0)(5.0)² = 12.5 J, KC = (1/2)(4.0)(2.0)² = 8.0 J. B’s direction does not matter, because K depends on speed only.
BKC > KB > KA A student who takes kinetic energy as proportional to speed (mv) picks this: 3.0, 5.0 and 8.0 for A, B and C. The speed is squared in K, which favors the fast, light object B over the slow, heavy C.
CKB > KA > KC A student who thinks mass does not affect kinetic energy picks this, ranking by speed alone. C’s mass is four times A’s, which more than makes up for its lower speed: 8.0 J against 4.5 J.
DKC > KA > KB A student who gives B a negative kinetic energy because it moves to the left picks this. The velocity is squared in K = (1/2)mv², so K is positive whichever way an object moves.
Working KA = 4.5 J, KB = 12.5 J, KC = 8.0 J ⇒ KB > KC > KA. mv: 3.0, 5.0, 8.0 ⇒ C > B > A; speed: B > A > C; with B negative: C > A > B.
A suitcase rests on a rack in a train moving at constant velocity. A passenger on the train says that the suitcase’s translational kinetic energy is zero. A person standing on the platform says that it is (1/2)mv², where m is the suitcase’s mass and v the train’s speed. Which claim, with its reasoning, is correct?
Answer and reasoning
AThe platform observer alone is correct, since speeds must all be measured relative to the ground. A student who thinks the ground is the one true frame picks this. A frame moving at constant velocity relative to the ground is just as valid; the passenger’s value is correct in the train’s frame.
BNeither can be right unless they agree, since energy is conserved and the same for every observer. A student who thinks energy conservation forces all observers to agree picks this. Conservation applies within one frame; different frames measure different speeds and so different kinetic energies.
CThe passenger alone is correct, since the suitcase is at rest on the rack that supports it. A student who takes motion relative to an object’s support as its true motion picks this. Relative to the platform the suitcase moves with the train, and the platform observer’s value is correct in that frame.
DBoth are correct, since K depends on the speed measured in each observer’s frame of reference.Correct Kinetic energy is (1/2)mv² with v the speed in a chosen frame. The suitcase is at rest relative to the train and moves at the train’s speed relative to the platform, so both values are right, each in its own frame.
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
