3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
The speed of a 2.0 kg cart moving along a straight track increases from 1.0 m/s to 3.0 m/s. By how much does the cart's translational kinetic energy increase?
Answer and reasoning
A4.0 J A student who substitutes the change in speed into (1/2)mv² calculates (1/2)(2.0 kg)(2.0 m/s)² = 4.0 J. Squaring the change in speed is not the same as finding the change in the squares: 3.0² − 1.0² = 8.0, not (3.0 − 1.0)² = 4.0.
B8.0 JCorrect K depends on the square of the speed, so find K at each speed and subtract: Ki = (1/2)(2.0 kg)(1.0 m/s)² = 1.0 J and Kf = (1/2)(2.0 kg)(3.0 m/s)² = 9.0 J, so ΔK = 9.0 J − 1.0 J = 8.0 J.
C2.0 J A student who treats K as proportional to speed, as if K = (1/2)mv, calculates (1/2)(2.0)(3.0) − (1/2)(2.0)(1.0) = 2.0. K depends on v², so tripling the speed makes K nine times as large: from 1.0 J to 9.0 J.
D9.0 J A student who gives the final kinetic energy as the change picks this. The cart already had 1.0 J of kinetic energy at 1.0 m/s, so the increase is 9.0 J − 1.0 J = 8.0 J.
Working Ki = (1/2)(2.0 kg)(1.0 m/s)² = 1.0 J. Kf = (1/2)(2.0 kg)(3.0 m/s)² = 9.0 J. ΔK = Kf − Ki = 9.0 J − 1.0 J = 8.0 J.
Two identical carts, each of mass m, move toward each other along a level track, each with speed v relative to the track. Which statement about the total translational kinetic energy of the two-cart system, measured in the track's frame, is correct?
Answer and reasoning
AIt is mv², the sum of the two carts' kinetic energies.Correct Kinetic energy is a scalar with no direction, so the kinetic energies of the two carts add: (1/2)mv² + (1/2)mv² = mv². The directions of motion do not matter.
BIt is zero, since the carts move in opposite directions. A student who thinks kinetic energies in opposite directions cancel picks this. Velocities in opposite directions have opposite signs, but kinetic energy is a scalar that is never negative, so each cart's (1/2)mv² adds to the total.
CIts sign depends on which way is chosen to be positive. A student who thinks kinetic energy takes the sign of the velocity picks this. K depends on v², which is the same for +v and −v, so the total kinetic energy is mv² whichever direction is chosen as positive.
DIt is 2mv², since each of the carts has kinetic energy mv². A student who leaves out the factor of 1/2 in K = (1/2)mv² picks this. Each cart's kinetic energy is (1/2)mv², so the total is mv², not 2mv².
Working Each cart has K = (1/2)mv², a positive scalar that does not depend on direction. Total K = (1/2)mv² + (1/2)mv² = mv².
The diagram shows a passenger of mass 60 kg walking toward the front of a train that moves along a straight, level track. The arrows give the passenger's velocity relative to the train and the train's velocity relative to the ground. What is the passenger's translational kinetic energy measured in the ground's frame?
Answer and reasoning
A1.2 × 10⁴ JCorrect In the ground's frame, the passenger's velocity is the train's velocity plus the passenger's velocity relative to the train: 18 m/s + 2.0 m/s = 20 m/s, both toward the front. K = (1/2)(60 kg)(20 m/s)² = 1.2 × 10⁴ J.
B1.2 × 10² J A student who thinks kinetic energy is the same for every observer uses the passenger's walking speed, (1/2)(60 kg)(2.0 m/s)² = 1.2 × 10² J. That is the kinetic energy in the train's frame; in the ground's frame the passenger moves at 20 m/s.
C7.7 × 10³ J A student who subtracts the two speeds uses 18 m/s − 2.0 m/s = 16 m/s and gets (1/2)(60 kg)(16 m/s)² ≈ 7.7 × 10³ J. The passenger walks toward the front, in the direction the train moves, so the velocities add to 20 m/s.
D2.4 × 10⁴ J A student who leaves out the factor of 1/2 calculates (60 kg)(20 m/s)² = 2.4 × 10⁴ J. With K = (1/2)mv², the kinetic energy is half of this, 1.2 × 10⁴ J.
Working Both velocities point toward the front of the train, so the passenger's velocity relative to the ground is 18 m/s + 2.0 m/s = 20 m/s. K = (1/2)(60 kg)(20 m/s)² = 12 000 J = 1.2 × 10⁴ J.
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, unit: J) Fix
Translational kinetic energy (K, unit: J)
The energy an object has because its center of mass is moving: K = (1/2)mv², where m is the object's mass and v is its speed. K is zero for an object at rest and positive for a moving object.
Joule (J)
The SI unit of energy. From K = (1/2)mv², 1 J = 1 kg·m²/s²: a 2 kg object moving at 1 m/s has a kinetic energy of 1 J.
Dependence of K on mass and speed
K is proportional to the mass and to the square of the speed. Doubling the mass doubles K; doubling the speed makes K four times as large; halving the speed makes K one quarter as large.
Change in kinetic energy (ΔK, unit: J)
ΔK = Kf − Ki = (1/2)mvf² − (1/2)mvi². Because K depends on v², ΔK must be found from the kinetic energies at the two instants, not by substituting the change in speed into (1/2)mv².
Students often think Translational kinetic energy is proportional to speed, as if K = (1/2)mv, so doubling the speed doubles K and halving it halves K. In fact No. K = (1/2)mv² depends on the square of the speed, so doubling the speed makes K four times as large, and halving the speed makes K one quarter as large.
Students often think The object with more mass has more kinetic energy, whatever the two speeds. In fact No. K depends on both mass and speed, and on the speed squared. A cart of mass m at speed 2v has twice the kinetic energy of a cart of mass 2m at speed v.
3.1.A.2 Scalar quantity Fix
Scalar quantity
A quantity that has a magnitude but no direction. Translational kinetic energy is a scalar: it depends on the speed, not the direction of motion, and it is never negative.
Kinetic energy of a system
The total translational kinetic energy of a system of objects is the sum of the kinetic energies of the objects. Because each is a positive scalar, they add whatever directions the objects move in.
Students often think Kinetic energy has a direction, like velocity: an object moving in the negative direction has negative kinetic energy, and the sign of K depends on which direction is chosen as positive. In fact No. K is a scalar: it has a size but no direction. Because m is positive and v² cannot be negative, K is never negative, and an object moving in the negative direction has the same K as one moving at the same speed in the positive direction.
Students often think A negative acceleration (a downward-sloping velocity–time graph) means the object is slowing down, so its speed and kinetic energy decrease the whole time. In fact Not necessarily. A downward slope means a negative acceleration. The object slows down while its velocity is positive, but once its velocity is negative, the same acceleration makes it speed up in the negative direction.
3.1.A.3 Reference frame Fix
Reference frame
The point of view (a coordinate system attached to an observer) from which positions and velocities are measured. Velocities measured in different frames differ by the velocity of one frame relative to the other.
Frame dependence of kinetic energy
An object's K is calculated from its speed in the observer's frame, so observers in different frames can measure different values of K for the same object. A ball at rest on a moving train has K = 0 in the train's frame and K > 0 in the ground's frame; both values are correct.
Students often think An object's kinetic energy is a fixed property of the object, so every observer must measure the same value of K. In fact No. K is calculated from the object's speed in the observer's frame of reference, and observers in different frames measure different speeds, so they can measure different values of K. A ball at rest on a moving train has K = 0 for a passenger and K > 0 for someone on the ground.
Students often think Kinetic energy must be measured relative to the ground, the true frame of reference; values measured in other frames are wrong. In fact No. Any inertial frame can be used. The ground's frame is a common choice, but a passenger on a train moving at constant velocity measures K correctly in the train's frame; values from different frames are each correct in their own frame.
6 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 6
Cart A has mass m and moves at speed 2v. Cart B has mass 2m and moves at speed v. Which gives the ratio of the translational kinetic energy of cart A to that of cart B?
Answer and reasoning
AKA/KB = 1 A student who treats K as proportional to speed finds that the factor of 2 in A's speed and the factor of 2 in B's mass cancel. K depends on the square of the speed, so A's doubled speed counts four times, not twice.
BKA/KB = ½ A student who thinks the heavier object has more kinetic energy, whatever the speeds, compares only the masses and picks this. Cart B has twice the mass, but cart A's doubled speed makes its kinetic energy four times as large, which outweighs the mass.
CKA/KB = 4 A student who thinks mass does not affect kinetic energy compares only the speeds: (2v)²/v² = 4. K is proportional to mass as well, so B's doubled mass doubles B's kinetic energy, and the ratio is 4/2 = 2.
DKA/KB = 2Correct KA = (1/2)m(2v)² = 2mv² and KB = (1/2)(2m)v² = mv², so KA/KB = 2. Doubling the speed multiplies K by 4, while doubling the mass multiplies it by only 2, so the faster, lighter cart has more kinetic energy.
Working KA = (1/2)m(2v)² = 2mv². KB = (1/2)(2m)v² = mv². KA/KB = 2mv²/mv² = 2.
A cart of mass m moves along a straight track with translational kinetic energy K. Which expression gives the cart's speed?
Answer and reasoning
Av = √(K/m) A student who uses K = mv², leaving out the factor of 1/2, gets v = √(K/m). With the 1/2 kept, v² = 2K/m, so this expression is too small by a factor of √2.
Bv = 2K/m A student who treats K as proportional to speed, as if K = (1/2)mv, solves to get v = 2K/m. K depends on v², so solving for v needs a square root; 2K/m has the units of speed squared, not speed.
Cv = √(2K/m)Correct Rearranging K = (1/2)mv²: multiplying both sides by 2 gives 2K = mv², so v² = 2K/m and v = √(2K/m). Check: (1/2)m(2K/m) = K.
Dv = √(2K) A student who thinks mass does not affect kinetic energy leaves m out, as if K = (1/2)v², and gets v = √(2K). K is proportional to m, so for the same K a cart with more mass moves more slowly: the speed must depend on m, as in v = √(2K/m).
Working K = (1/2)mv². Multiply both sides by 2: 2K = mv². Divide by m: v² = 2K/m. Take the square root: v = √(2K/m).
The graph shows the velocity v of a cart moving along a straight track as a function of time t. Which statement about the cart's translational kinetic energy K is correct?
Answer and reasoning
AK is negative from t = 2.0 s to t = 4.0 s. A student who thinks kinetic energy takes the sign of the velocity picks this. The velocity is negative after t = 2.0 s, but K = (1/2)mv² depends on v², which is positive; the cart is speeding up in the negative direction, so K increases from zero.
BK is the same at t = 0 and at t = 4.0 s.Correct The cart's velocity is +4.0 m/s at t = 0 and −4.0 m/s at t = 4.0 s. K = (1/2)mv² depends on v², and (+4.0 m/s)² = (−4.0 m/s)², so K has the same value at the two instants: kinetic energy is a scalar and does not depend on the direction of motion.
CK decreases throughout, from t = 0 to t = 4.0 s. A student who thinks a downward-sloping velocity–time graph means slowing down the whole time picks this. The cart slows down only until t = 2.0 s; after that its velocity is negative and growing in size, so its speed and K increase.
DK at t = 1.0 s is half of K at t = 0, as v halves. A student who treats kinetic energy as proportional to speed picks this. The velocity halves from 4.0 m/s to 2.0 m/s, but K depends on v², so K at t = 1.0 s is (1/2)² = 1/4 of K at t = 0.
Working From the graph: v = +4.0 m/s at t = 0, +2.0 m/s at t = 1.0 s, 0 at t = 2.0 s and −4.0 m/s at t = 4.0 s. K = (1/2)mv² depends on v², so K(0) = K(4.0 s) = (1/2)m(16 m²/s²); K(1.0 s) = (1/2)m(4.0 m²/s²) = K(0)/4. K decreases to zero at t = 2.0 s and then increases again; it is never negative.
A 0.50 kg ball rests on the floor of a train car that moves along a straight, level track at a constant 20 m/s relative to the ground. Student X, riding in the train, claims that the ball's translational kinetic energy is 0 J. Student Y, standing on the ground beside the track, claims that it is 100 J. Which statement about the two claims is correct?
Answer and reasoning
AAt most one can be correct, since K is the same for every observer. A student who thinks an object's kinetic energy is a fixed property that all observers agree on picks this. K depends on the speed in the observer's frame, and the ball's speed is 0 in the train's frame and 20 m/s in the ground's frame.
BX is wrong, since K must be measured relative to the ground. A student who treats the ground as the true frame picks this. No inertial frame is privileged: X, in a frame moving at constant velocity with the ball, correctly measures the ball at rest, so K = 0 is the correct value in X's frame.
CY's value would become −100 J if the train moved the other way. A student who thinks kinetic energy takes the sign of the velocity picks this. K = (1/2)mv² depends on v², so the ball's kinetic energy in the ground's frame is +100 J whichever direction the train moves and whichever direction is chosen as positive.
DBoth are correct: each uses the ball's speed in their own frame.Correct K is calculated from the speed measured in the observer's frame. In the train's frame the ball is at rest, so K = 0; in the ground's frame it moves at 20 m/s, so K = (1/2)(0.50 kg)(20 m/s)² = 100 J. Different observers can measure different values of K, and both are correct.
Working In the train's frame the ball is at rest: K = 0. In the ground's frame it moves at 20 m/s: K = (1/2)(0.50 kg)(20 m/s)² = 100 J. Both are correct values of K, each in its own frame; K is never negative, whichever way the train moves.
A ball of mass m is launched from level ground with speed v₀ at an angle of 30° above the horizontal. Air resistance is negligible. Which expression gives the ball's translational kinetic energy at the highest point of its path?
Answer and reasoning
A0.125 mv₀² A student who takes the horizontal component of the velocity as v₀ sin 30° = v₀/2 picks this: (1/2)m(v₀/2)² = 0.125 mv₀². With the angle measured from the horizontal, the horizontal component is v₀ cos 30°; v₀ sin 30° is the initial vertical component, which is zero at the top.
B0.375 mv₀²Correct At the top the ball still moves horizontally at v₀ cos 30°, because the horizontal component of its velocity never changes. K = (1/2)m(v₀ cos 30°)² = (1/2)(3/4)mv₀² = 0.375 mv₀².
C0.433 mv₀² A student who thinks kinetic energy is proportional to speed scales the launch value, (1/2)mv₀², by the speed ratio cos 30° = 0.866 and picks this. K depends on the square of the speed, so the launch value must be multiplied by cos² 30° = 0.75.
D0.750 mv₀² A student who drops the factor 1/2 picks this: m(v₀ cos 30°)² = 0.750 mv₀². Translational kinetic energy is (1/2)mv², so K = 0.375 mv₀².
Working At the highest point the vertical component of the velocity is zero. The horizontal component does not change during projectile motion (zero horizontal acceleration), so it is still v₀ cos 30° = 0.866v₀, and this is the ball's speed at the top. K = (1/2)m(v₀ cos 30°)² = (1/2)(3/4)mv₀² = 0.375 mv₀².
Car P has mass m and moves along a straight, level road. Truck Q has mass 2m and moves along the same road. Measured relative to the road, the two vehicles have equal translational kinetic energies. Q's speed relative to the road is how many times P's speed?
Answer and reasoning
A0.50 A student who treats kinetic energy as proportional to speed picks this: with twice the mass, Q would need half the speed. K depends on v², so halving v² requires multiplying the speed by 1/√2 ≈ 0.71, not by 0.50.
B1.00 A student who thinks kinetic energy depends only on speed picks this: equal kinetic energies would then mean equal speeds. K = (1/2)mv² depends on mass too; Q has twice the mass, so it has the same K at a lower speed, 0.71 times P's.
C0.71Correct Setting the kinetic energies equal, (1/2)m vP² = (1/2)(2m)vQ², gives vQ² = vP²/2. Kinetic energy depends on the square of the speed, so the speed ratio is the square root: vQ/vP = 1/√2 ≈ 0.71.
D0.25 A student who squares the factor instead of taking its square root picks this: v² must be halved, and squaring 1/2 gives 1/4. The factor on the speed is the square root of the factor on v²: √(1/2) ≈ 0.71.
Working KP = KQ: (1/2)m vP² = (1/2)(2m) vQ². So vQ² = vP²/2 and vQ/vP = √(1/2) = 1/√2 ≈ 0.71.
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
