5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Ball 1, of mass m, moving with velocity +v₀, collides head-on with ball 2, of mass 4m, which is at rest on a frictionless surface. The collision is elastic. What is the velocity v₁ of ball 1 immediately after the collision?
Answer and reasoning
Av₁ = −v₀ A student who treats the heavier ball as a fixed wall picks this. Ball 2 would then need momentum 2mv₀, a speed of v₀/2, and the total kinetic energy would double; the free ball 2 takes some of the kinetic energy, so ball 1 rebounds more slowly than it arrived.
Bv₁ = 0 A student who expects ball 1 to stop, as in Newton's cradle, picks this. Ball 2 would then move at v₀/4 with only one quarter of the original kinetic energy; stopping is the elastic result only for equal masses.
Cv₁ = +v₀/5 A student who assumes the balls move off together picks this: mv₀/(5m) = v₀/5. That is the result for balls that stick, which keep only one fifth of the kinetic energy; in an elastic collision the balls separate with different velocities.
A ball moving at speed v collides head-on with a ball of greater mass that is at rest on a frictionless surface. The collision is elastic. Which statement about the collision is correct?
Answer and reasoning
AThe moving ball loses kinetic energy; the struck ball gains the same amount.Correct In an elastic collision the system's total kinetic energy is unchanged, but the balls' individual kinetic energies change: the struck ball starts moving, so it gains kinetic energy, and the moving ball loses exactly that amount.
BEach ball has the same kinetic energy after the collision as it had before. A student who applies 'kinetic energy is conserved' to each ball separately picks this. Only the total is conserved; the struck ball gains kinetic energy, so the moving ball must lose the same amount.
CThe balls exchange velocities, so the struck ball moves off with speed v. A student who generalizes Newton's cradle picks this. Velocities are exchanged only for equal masses; with a heavier struck ball, the moving ball rebounds and the struck ball moves off more slowly than v.
DThe moving ball bounces back at speed v, while the struck ball stays at rest. A student who treats the heavier ball as a fixed wall picks this. The struck ball is free to move, so it gains momentum and kinetic energy, and the moving ball rebounds more slowly than v.
Cart A collides with an identical cart B, which is at rest on a level track with negligible friction. Three trials, I, II and III, are run with the same initial velocity of A. The diagram shows the carts' velocities before the collision and just after it in each trial. Which ranks the kinetic energy transformed into other forms in the three trials?
Answer and reasoning
AI > III > IICorrect With each cart of mass m, the kinetic energy before is ½m(4.0)² = 8.0m (in J, with m in kg). After: I, ½(2m)(2.0)² = 4.0m; II, ½m(4.0)² = 8.0m; III, ½m(1.0)² + ½m(3.0)² = 5.0m. Energy transformed: I, 4.0m; II, 0; III, 3.0m.
BI > II = III A student who thinks only collisions in which the carts stick are inelastic picks this. In III the carts separate, but their total kinetic energy falls from 8.0m to 5.0m, so 3.0m is transformed.
CI = II = III A student who takes kinetic energy to be proportional to speed finds the same total, proportional to 4.0 m/s, in every trial, and picks this. Kinetic energy depends on v², so 1.0² + 3.0² = 10 is less than 4.0² = 16, and 2 × 2.0² = 8 is less again.
DII > III > I A student who counts the kinetic energy lost by A as transformed picks this: A loses all its kinetic energy in II. In II that energy all goes to B as kinetic energy, so none is transformed.
Working K₀ = ½m(4.0)² = 8.0m. After: I = ½(2m)(2.0)² = 4.0m; II = ½m(4.0)² = 8.0m; III = ½m(1.0)² + ½m(3.0)² = 5.0m. Transformed: I 4.0m, II 0, III 3.0m → I > III > II. (Momentum is 4.0m in every trial.)
Two carts with clay bumpers collide and stick together on a level track with negligible friction. Just after the collision, the carts' total kinetic energy is less than it was before. Which statement best accounts for the decrease?
Answer and reasoning
ANonconservative forces in the clay destroyed some of it, so the total energy decreased. A student who thinks energy can be destroyed picks this. The missing kinetic energy has been transformed into other forms, mainly internal energy; the total energy is unchanged.
BNonconservative forces in the clay turned some of it into internal energy and sound.Correct As the clay deforms permanently, nonconservative forces do work that transforms part of the kinetic energy into internal (thermal) energy of the clay and carts, and into sound. The total energy is conserved; that kinetic energy is not restored.
CNonconservative forces in the clay removed some of it along with part of the momentum. A student who thinks momentum is lost along with kinetic energy picks this. With negligible external impulse, the carts' total momentum is the same before and after; only kinetic energy is transformed.
DNonconservative forces in the clay remove exactly half of it whenever two objects stick. A student who generalizes the identical-cart example picks this. The fraction transformed depends on the masses: for a cart striking another at rest, it is m₂/(m₁ + m₂), one half only when the masses are equal.
The diagram is a top view of two pucks sliding on a frictionless horizontal surface just before they collide. The pucks stick together. What is the speed of the combined pucks just after the collision?
Answer and reasoning
A(4/3)v A student who adds the momentum magnitudes picks this: (2mv + 2mv)/(3m) = (4/3)v. The pucks' momenta are perpendicular, so they combine by components into a total of magnitude 2√2 mv.
B√2 v A student who conserves kinetic energy picks this: ½m(2v)² + ½(2m)v² = ½(3m)vf² gives vf = √2 v. The pucks stick together, so kinetic energy is not conserved; only momentum is.
C(√5/2)v A student who averages the two velocity vectors without weighting them by mass picks this: the average of (2v, 0) and (0, v) has magnitude (√5/2)v. The velocities must be weighted by the masses m and 2m.
D(2√2/3)vCorrect Momentum is conserved in each direction. x: m(2v) = 2mv; y: (2m)v = 2mv. The total has magnitude √((2mv)² + (2mv)²) = 2√2 mv, and the combined mass is 3m, so the speed is (2√2/3)v ≈ 0.94v.
Working px = m(2v) = 2mv; py = (2m)(v) = 2mv. |p| = 2√2 mv. vf = 2√2 mv/(3m) = (2√2/3)v ≈ 0.94v. (Checked with sympy; KE-conserving value √2 v, unweighted average (√5/2)v.)
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.4.A.1 Kinetic energy of a system Fix
Kinetic energy of a system
The sum of the translational kinetic energies of the system's objects, Ksys = Σ ½mi vi². It is a scalar and never negative, so the objects' kinetic energies add whatever their directions of motion. Unit: joule (J).
Elastic collision
A collision in which the system's total kinetic energy just after the collision equals its total kinetic energy just before. Kinetic energy may be stored briefly during contact, for example in compressed bumpers, but all of it is returned as kinetic energy by the end of the collision.
Students often think Kinetic energy has a direction, so the kinetic energy of an object moving in the negative direction is subtracted from the total. In fact No. Kinetic energy, ½mv², is a scalar and is never negative, so the kinetic energies of all the objects add, whatever their directions of motion.
Students often think Kinetic energy is proportional to speed, as momentum is, so halving an object's speed halves its kinetic energy. In fact No. K = ½mv², so kinetic energy is proportional to the square of the speed: halving an object's speed reduces its kinetic energy to one quarter.
4.4.A.2 Transfer of kinetic energy in an elastic collision Fix
Transfer of kinetic energy in an elastic collision
In an elastic collision only the system's total kinetic energy is unchanged. The individual objects' kinetic energies generally change: the kinetic energy one object loses, the other gains.
Students often think In an elastic collision, each object has the same kinetic energy after the collision as it had before. In fact No. Only the total is unchanged. Kinetic energy is transferred between the objects, so the individual kinetic energies usually change; a moving ball that strikes a stationary one, for example, gives some of its kinetic energy to it.
Students often think In an elastic collision, the moving object stops and the struck object moves off with its velocity, whatever their masses. In fact Only if the two objects have equal masses. A lighter object rebounds from a heavier one at rest, and a heavier object continues forward after striking a lighter one.
4.4.A.3 Inelastic collision Fix
Inelastic collision
A collision in which the system's total kinetic energy decreases. When the net external impulse is negligible, the system's momentum is still conserved.
Students often think Whenever momentum is conserved in a collision, kinetic energy is conserved as well, so conservation of kinetic energy can be applied to any collision. In fact Not necessarily. Momentum is conserved in a collision whenever the net external impulse is negligible, but the total kinetic energy is unchanged only in an elastic collision; in an inelastic collision it decreases.
Students often think In an inelastic collision, some of the system's momentum is lost along with the kinetic energy. In fact No. Momentum is conserved in all interactions, so in an inelastic collision with negligible external impulse the system's total momentum is unchanged; only kinetic energy is transformed into other forms.
4.4.A.4 Transformation by nonconservative forces Fix
Transformation by nonconservative forces
In an inelastic collision, nonconservative forces during contact, such as those that permanently deform the objects, transform part of the kinetic energy into other forms, such as internal (thermal) energy and sound, and that energy is not restored as kinetic energy. Total energy is conserved.
Students often think The kinetic energy that disappears in an inelastic collision is destroyed, so the total energy of the system decreases. In fact No. Energy is conserved. Nonconservative forces during the collision transform the missing kinetic energy into other forms, such as internal (thermal) energy of the deformed objects and sound.
Students often think The kinetic energy transformed in a collision equals the kinetic energy lost by the moving object, ignoring the kinetic energy the other object gains. In fact No. Much of the kinetic energy the moving object loses is transferred to the other object as kinetic energy. The energy transformed into other forms is the decrease in the system's total kinetic energy.
4.4.A.5 Perfectly inelastic collision Fix
Perfectly inelastic collision
A collision in which the objects stick together and move with the same velocity afterward, v⃗ = (m₁v⃗₁ + m₂v⃗₂)/(m₁ + m₂) when momentum is conserved. For given masses and initial velocities, it gives the largest decrease in kinetic energy that conservation of momentum allows.
Students often think When a moving object hits and sticks to a stationary one, the pair moves off at the moving object's original speed, the moving object simply carrying the other along. In fact No. The pair shares the original momentum, so its speed is v = m₁v₀/(m₁ + m₂), smaller than v₀ by the factor m₁/(m₁ + m₂).
Students often think Objects that collide and stick together move off with the plain average of their initial velocities, each counted equally. In fact Only if the objects have equal masses. The common velocity is the mass-weighted average, (m₁v⃗₁ + m₂v⃗₂)/(m₁ + m₂).
5 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 5
The diagram shows two carts on a level track just before and just after they collide. Which claim about the collision do the data support?
Answer and reasoning
AIt is elastic, as the carts bounced apart instead of sticking together. A student who classifies collisions by whether the objects bounce or stick picks this. A collision is elastic only if the total kinetic energy is unchanged; here it falls from 18 J to 10 J although the carts bounce apart.
BIt is elastic, as the carts' total momentum is 6.0 kg·m/s before and after. A student who thinks kinetic energy is conserved whenever momentum is picks this. Momentum is conserved in this collision, as in any collision with negligible external impulse, but the total kinetic energy falls from 18 J to 10 J.
CIt is inelastic, as the total kinetic energy fell from 18 J to 10 J.Correct Before: K = ½(1.0)(6.0)² = 18 J. After: K = ½(1.0)(2.0)² + ½(4.0)(2.0)² = 2.0 + 8.0 = 10 J. The total kinetic energy decreased, so the collision is inelastic, even though the carts bounced apart and momentum (6.0 kg·m/s) was conserved.
DIt is inelastic, as the total kinetic energy fell from 18 J to 6.0 J. A student who subtracts the kinetic energy of the cart moving left picks this: 8.0 J − 2.0 J = 6.0 J. Kinetic energy is never negative, so the two carts' kinetic energies add to 10 J.
Working Right positive. p before = (1.0)(6.0) = 6.0 kg·m/s; p after = (1.0)(−2.0) + (4.0)(2.0) = 6.0 kg·m/s. K before = ½(1.0)(6.0)² = 18 J; K after = ½(1.0)(2.0)² + ½(4.0)(2.0)² = 2.0 + 8.0 = 10 J. K decreased by 8.0 J: inelastic.
A 2.0 kg cart moving at 3.0 m/s collides with a 1.0 kg cart at rest on a level track with negligible friction. Immediately after the collision, the 2.0 kg cart moves at 1.2 m/s and the 1.0 kg cart moves at 3.6 m/s, both in the original direction of motion. How much of the system's kinetic energy is transformed into other forms in the collision?
Answer and reasoning
A0.0 J A student who thinks kinetic energy is conserved whenever momentum is picks this. Momentum is conserved here (6.0 kg·m/s before and after), but the total kinetic energy falls from 9.0 J to 7.92 J.
B1.1 JCorrect Before: K = ½(2.0)(3.0)² = 9.0 J. After: K = ½(2.0)(1.2)² + ½(1.0)(3.6)² = 1.44 + 6.48 = 7.92 J. The decrease, 9.0 − 7.92 ≈ 1.1 J, is transformed by nonconservative forces into other forms.
C2.2 J A student who writes kinetic energy as mv² picks this: 18 J − (2.88 J + 12.96 J) ≈ 2.2 J. With K = ½mv², the totals are 9.0 J and 7.92 J, a decrease of 1.1 J.
D7.6 J A student who counts all the kinetic energy lost by the 2.0 kg cart as transformed picks this: 9.0 − ½(2.0)(1.2)² ≈ 7.6 J. Most of that energy, 6.48 J, became kinetic energy of the 1.0 kg cart.
Cart X, of mass m, moving at speed v₀ along a level track with negligible friction, collides with and sticks to cart Y, which is at rest. In trial 1, Y has mass m. In trial 2, Y has mass 2m, and X has the same initial speed. What is the ratio of the kinetic energy transformed into other forms in trial 2 to that in trial 1?
Answer and reasoning
A1.19 A student who counts the kinetic energy lost by X as transformed picks this: X loses K₀(1 − 1/4) in trial 1 and K₀(1 − 1/9) in trial 2, a ratio of 32/27 ≈ 1.19. Part of what X loses becomes Y's kinetic energy, which is not transformed.
B1.00 A student who thinks exactly half the kinetic energy is transformed whenever objects stick picks this. That is true only for equal masses; with Y of mass 2m, two thirds is transformed.
C1.33Correct Initial kinetic energy K₀ = ½mv₀² in both trials. Trial 1: v = v₀/2, K after = ½(2m)(v₀/2)² = K₀/2, so K₀/2 is transformed. Trial 2: v = v₀/3, K after = ½(3m)(v₀/3)² = K₀/3, so 2K₀/3 is transformed. The ratio is (2/3)/(1/2) = 1.33.
D0.50 A student who takes the common velocity to be the plain average, v₀/2, in both trials picks this: K₀/2 is transformed in trial 1 but only K₀/4 in trial 2. The common velocity is the mass-weighted average, v₀/3 in trial 2.
Working K₀ = ½mv₀². Trial 1: v = mv₀/(2m) = v₀/2; Kf = ½(2m)(v₀²/4) = K₀/2; transformed K₀/2. Trial 2: v = v₀/3; Kf = ½(3m)(v₀²/9) = K₀/3; transformed 2K₀/3. Ratio = (2/3)/(1/2) = 4/3 ≈ 1.33. (Checked with sympy.)
A 3.0 kg cart moving at +4.0 m/s and a 1.0 kg cart moving at −6.0 m/s collide on a level track with negligible friction and stick together. What is the velocity of the carts immediately after the collision?
Answer and reasoning
A+4.5 m/s A student who adds the momentum magnitudes picks this: (12 + 6.0)/4.0 = 4.5 m/s. The 1.0 kg cart moves in the negative direction, so its momentum, −6.0 kg·m/s, is subtracted.
B+1.5 m/sCorrect Total momentum before: (3.0)(+4.0) + (1.0)(−6.0) = +6.0 kg·m/s. The stuck carts have mass 4.0 kg, so v = (+6.0)/(4.0) = +1.5 m/s.
C−1.0 m/s A student who averages the two velocities without weighting them by mass picks this: (4.0 − 6.0)/2 = −1.0 m/s. The 3.0 kg cart has three times the mass, so its velocity counts three times as much.
D+4.0 m/s A student who expects the stuck carts to keep the more massive cart's velocity picks this. The 1.0 kg cart's momentum, −6.0 kg·m/s, reduces the total, so the pair moves at only +1.5 m/s.
Working p = (3.0 kg)(+4.0 m/s) + (1.0 kg)(−6.0 m/s) = +6.0 kg·m/s. v = 6.0/(3.0 + 1.0) = +1.5 m/s.
A lump of clay of mass m moving horizontally strikes and sticks to a block of mass 3m that hangs at rest from light strings. The block and clay then swing up to a maximum height h above the block's starting position. Air resistance is negligible. What was the speed of the clay just before it struck the block?
Answer and reasoning
A5.7√(gh)Correct After the collision, the 4m clay–block system swings up with its mechanical energy conserved: ½(4m)V² = (4m)gh, so V = √(2gh). During the brief collision, momentum is conserved: mv₀ = (4m)V, so v₀ = 4√(2gh) = √32·√(gh) ≈ 5.7√(gh).
B2.8√(gh) A student who conserves kinetic energy in the collision picks this: ½mv₀² = (4m)gh gives v₀ = √(8gh) ≈ 2.8√(gh). The clay sticks, so most of its kinetic energy is transformed in the collision; only momentum is conserved through it.
C4.0√(gh) A student who thinks half the clay's kinetic energy is always transformed when objects stick picks this: ½(½mv₀²) = (4m)gh gives v₀ = 4√(gh). With a block of mass 3m, three quarters of the kinetic energy is transformed, not half.
D1.4√(gh) A student who thinks the block and clay move off at the clay's original speed picks this: v₀ = V = √(2gh) ≈ 1.4√(gh). The clay shares its momentum with a block three times as massive, so the pair moves at only a quarter of the clay's speed.
Working Swing: ½(4m)V² = 4mgh → V = √(2gh). Collision: mv₀ = 4mV → v₀ = 4√(2gh) = √32 √(gh) ≈ 5.66√(gh) → 5.7√(gh). Fraction of K transformed in the collision = 3m/(4m) = 3/4. (Checked with sympy.)
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
