5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Two carts collide on a level track, and the net external force on the two-cart system is negligible. Which feature distinguishes an elastic collision from an inelastic collision?
Answer and reasoning
AThe system's total momentum is unchanged by the collision. A student who thinks momentum is conserved only in elastic collisions picks this. With negligible net external force, the total momentum is unchanged in every collision, elastic or inelastic, so it cannot tell them apart.
BThe system's total kinetic energy is unchanged by the collision.Correct An elastic collision is one in which the system's total kinetic energy after the collision equals the total before it. In an inelastic collision the total decreases. That comparison is the whole test.
CThe carts move apart afterward instead of sticking to each other. A student who sorts collisions by whether the objects bounce or stick picks this. Carts that bounce apart usually still lose some kinetic energy, so they can collide inelastically. Only the kinetic energy totals decide.
DThe carts' bumpers are made of a springy material, such as rubber. A student who takes 'elastic' to mean stretchy picks this. The physics term describes the system's kinetic energy, not the material; rubber bumpers can still collide inelastically.
Cart 1 rolls along a level track toward an identical cart 2, which is at rest. After they collide, cart 1 is at rest and cart 2 moves off with cart 1's original velocity. A student says that the collision cannot be elastic, because cart 1's kinetic energy changed. Which response to the student is correct?
Answer and reasoning
AOnly the total kinetic energy of the two carts must be unchanged, and it is.Correct An elastic collision requires only that the system's total kinetic energy is unchanged. Cart 2 ends with exactly the kinetic energy cart 1 started with, (1/2)mv², so the total is the same before and after even though each cart's kinetic energy changed.
BThe student is right: in an elastic collision each cart keeps its kinetic energy. A student who applies 'kinetic energy is conserved' to each object picks this. The condition applies to the system's total. Kinetic energy can pass from one object to another; here all of cart 1's passes to cart 2.
CEvery collision conserves kinetic energy, since total energy is conserved. A student who equates conservation of energy with conservation of kinetic energy picks this. In many collisions some kinetic energy becomes thermal energy and sound. This collision is elastic because the total kinetic energy is unchanged, not because every collision is.
DThe total momentum is unchanged, and that shows kinetic energy is too. A student who thinks conserved momentum guarantees conserved kinetic energy picks this. Momentum is conserved in inelastic collisions as well. The collision is elastic because the kinetic energy totals, (1/2)mv² before and after, are equal.
A rubber ball is released from rest at height h above a hard floor. It bounces and rises to a maximum height of 0.60h. Air resistance is negligible. Which statement about the ball's collision with the floor, with its reasoning, is correct?
Answer and reasoning
AIt is inelastic, since the ball has less kinetic energy after than before.Correct With negligible air resistance, the ball's kinetic energy just before the bounce is mgh and just after is mg(0.60h) = 0.60mgh, because it rises only to 0.60h. Earth's kinetic energy is negligible, so the system's total kinetic energy decreased: the collision is inelastic.
BIt is elastic, since the ball bounces back up rather than sticking to the floor. A student who thinks any collision in which the objects separate is elastic picks this. The ball does bounce, but it rises only to 0.60h, so it left the floor with less kinetic energy than it arrived with.
CIt is elastic, since rubber is an elastic material that springs back to its shape. A student who takes 'elastic' to mean springy material picks this. The physics term refers to kinetic energy: the ball rises only to 0.60h, so 40% of its kinetic energy was transformed in the bounce.
DIt is inelastic, since the ball's momentum reverses direction and is not conserved. A student who links conserved momentum with elastic collisions picks this. The ball's momentum reverses in an elastic bounce too, because the floor exerts an external force on the ball. The collision is inelastic because the kinetic energy decreased.
Working Falling from h: K just before the bounce = mgh (ball–Earth system, energy conserved in flight). Rising to 0.60h: K just after = mg(0.60h) = 0.60mgh. K after < K before, so the collision is inelastic (40% of the kinetic energy is transformed). Earth's kinetic energy is negligible.
A fast-moving lump of clay catches up with a slower lump moving in the same direction along a level, low-friction surface, and the two lumps stick together. Which statement correctly describes the kinetic energy of the two-lump system in this collision?
Answer and reasoning
AIt decreases, and the kinetic energy that is lost then no longer exists in any form. A student who thinks energy is used up picks this. Energy is never destroyed: the lost kinetic energy becomes other forms, such as thermal energy in the warmed, deformed clay and sound.
BIt decreases, and the lost kinetic energy becomes extra momentum of the lumps. A student who thinks kinetic energy can turn into momentum picks this. Momentum and energy are different quantities with different units; the total momentum is unchanged, and the lost kinetic energy becomes thermal energy and sound.
CIt decreases; nonconservative forces turn the loss into thermal energy and sound.Correct The collision is inelastic, so the total kinetic energy decreases. The kinetic energy that is not restored is transformed by nonconservative forces between the deforming lumps into other forms, mainly thermal energy, with some sound. None of it is destroyed.
DAll of it is transformed into forms such as thermal energy, since the lumps stick. A student who thinks sticking removes all the kinetic energy picks this. Both lumps were moving the same way, so the system has momentum, and the combined lump keeps moving with it. Only part of the kinetic energy is transformed.
A cart of mass m moving at speed v collides with a cart of mass 2m that is at rest, and the two carts stick together. Friction is negligible. Which expression gives the kinetic energy of the system that is transformed into other forms of energy in the collision?
Answer and reasoning
Amv²/2 A student who thinks sticking removes all the kinetic energy picks this, the whole initial kinetic energy. The system has momentum mv, so the stuck carts keep moving at v/3 with kinetic energy mv²/6.
Bmv²/4 A student who applies the equal-mass result, half the kinetic energy transformed, picks this. That holds only for equal masses. Here the struck cart has mass 2m, and mv²/3, two-thirds of the initial kinetic energy, is transformed.
Cmv²/8 A student who takes the common speed to be the average of v and 0, v/2, picks this: K after = (1/2)(3m)(v/2)² = 3mv²/8, so mv²/2 − 3mv²/8 = mv²/8. Momentum gives v/3, not v/2, because the masses differ.
Dmv²/3Correct The carts stick, so they share one velocity: mv = (3m)vf gives vf = v/3. The kinetic energy falls from (1/2)mv² to (1/2)(3m)(v/3)² = mv²/6, so mv²/2 − mv²/6 = mv²/3 is transformed.
Working Momentum: mv = (m + 2m)vf, so vf = v/3. K before = (1/2)mv². K after = (1/2)(3m)(v/3)² = mv²/6. Transformed = mv²/2 − mv²/6 = mv²/3.
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 Elastic collision Fix
Elastic collision
A collision in which the total kinetic energy of the system just after the collision equals the total kinetic energy just before it. Whether a collision is elastic is decided by comparing these two totals, not by whether the objects bounce, stick or are made of a springy material.
Kinetic energy of a system (unit: J)
The sum of the translational kinetic energies, K = (1/2)mv², of all the objects in the system. Kinetic energy is a scalar, so every object's contribution is positive whatever its direction of motion.
Students often think Collisions are sorted by what the objects do afterward: if the objects bounce or move apart the collision is elastic, and only if they stick together is it inelastic. In fact No. Whether a collision is elastic depends only on whether the system's total kinetic energy is unchanged. Objects that bounce apart usually still lose some kinetic energy, so most collisions in which objects separate are inelastic.
Students often think Energy is conserved, so the kinetic energy of the colliding objects is the same before and after every collision. In fact No. The total energy of an isolated system is conserved, but the kinetic energy can be transformed into other forms, such as thermal energy and sound. Kinetic energy is unchanged only in an elastic collision.
4.4.A.2 Kinetic energy of each object in an elastic collision Fix
Kinetic energy of each object in an elastic collision
In an elastic collision only the system's total kinetic energy is unchanged. Kinetic energy can be transferred from one object to another, so each object's final kinetic energy may differ from its initial kinetic energy.
Students often think In an elastic collision each object's kinetic energy is the same after the collision as before; if any object's kinetic energy changes, the collision is inelastic. In fact No. Only the total kinetic energy of the system is unchanged. Kinetic energy is usually transferred between the objects, so each object's kinetic energy can change: one object's loss equals the other's gain.
Students often think The equal-mass results hold for all collisions: in an elastic collision the moving object always stops and hands on all its kinetic energy, and in a sticking collision half the kinetic energy is always lost. In fact No. For equal masses, an elastic collision with an object at rest leaves the moving object at rest, and a sticking collision halves the speed and transforms half the kinetic energy. With different masses, the moving object keeps moving or rebounds, and the fraction of kinetic energy transformed is different.
4.4.A.3 Inelastic collision Fix
Inelastic collision
A collision in which the total kinetic energy of the system decreases. The objects may stick together or move apart afterward; the decrease in total kinetic energy is what makes the collision inelastic.
Momentum and kinetic energy in collisions
Two separate accounts. The total momentum of a system with negligible net external force is the same before and after every collision, elastic or inelastic (Topic 4.3). The total kinetic energy is unchanged only in an elastic collision. Finding that momentum is conserved says nothing about whether kinetic energy is.
Students often think Momentum is conserved only in elastic collisions, so a collision in which momentum is conserved is elastic and one in which it is not conserved is inelastic. In fact No. The total momentum of a system with negligible net external force is the same before and after every collision, elastic or inelastic. What distinguishes an elastic collision is that the total kinetic energy is also unchanged.
Students often think Momentum and kinetic energy go together: whenever the total momentum is conserved, the total kinetic energy is conserved as well. In fact No. Momentum is conserved in every collision with negligible net external force, elastic or not, while kinetic energy is conserved only in elastic collisions. The same total momentum is consistent with any decrease in kinetic energy.
4.4.A.4 Energy transformed in an inelastic collision Fix
Energy transformed in an inelastic collision
In an inelastic collision, part of the initial kinetic energy is not restored to kinetic energy: nonconservative forces between the objects transform it into other forms, such as thermal energy and sound. The amount transformed is the system's initial kinetic energy minus its final kinetic energy; no energy is destroyed.
Students often think The kinetic energy lost in an inelastic collision is used up and no longer exists in any form. In fact No. Energy is never destroyed. The kinetic energy that is not restored is transformed by nonconservative forces between the objects into other forms, such as thermal energy and sound.
Students often think The kinetic energy transformed into other forms in a collision is whatever kinetic energy the moving object loses; what the other object gains does not need to be counted. In fact No. The energy transformed is the decrease in the total kinetic energy of the system. Part of the incoming object's loss is usually gained by the other object as kinetic energy, and that part is not transformed.
4.4.A.5 Perfectly inelastic collision Fix
Perfectly inelastic collision
An inelastic collision in which the objects stick together and move with the same velocity afterward. Combined with conservation of momentum, the common final velocity is the total momentum divided by the total mass. It does not mean that all the kinetic energy is lost: the stuck objects keep moving whenever the total momentum is not zero.
Students often think In a perfectly inelastic collision all of the kinetic energy is transformed, because 'perfectly' inelastic means the objects end up with no kinetic energy. In fact Not in general. The objects stick together, but they keep moving with a common velocity unless the total momentum is zero. The kinetic energy decreases, but only by the amount the sticking removes.
Students often think After two objects stick together, their common speed is the average of their speeds before the collision, whatever their masses. In fact Only if the masses are equal. The common velocity is the total momentum divided by the total mass, so it depends on the masses: a cart of mass m moving at v that sticks to a cart of mass 2m at rest moves off at v/3, not v/2.
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
Carts A and B collide on a level track with negligible friction. The table shows the mass m of each cart and its velocity v just before and just after the collision, with positive to the right. Which claim about the collision, with its reasoning, is correct?
Answer and reasoning
AElastic, since the carts' total momentum is the same before and after. A student who thinks conserved momentum means conserved kinetic energy picks this. The total momentum is 6.0 kg·m/s before and after, but it would be for an inelastic collision too. The conclusion is right only because the kinetic energy totals, 9.0 J and 9.0 J, are equal.
BElastic, since the carts separate instead of sticking together. A student who classifies collisions by whether the objects bounce apart picks this. Carts that separate can still lose kinetic energy. This collision is elastic because the total kinetic energy is 9.0 J before and after, not because the carts separate.
CElastic, since the carts' total kinetic energy is the same before and after.Correct Before: (1/2)(2.0 kg)(3.0 m/s)² = 9.0 J, with cart B at rest. After: (1/2)(2.0 kg)(1.0 m/s)² + (1/2)(1.0 kg)(4.0 m/s)² = 1.0 J + 8.0 J = 9.0 J. The system's total kinetic energy is unchanged, which is the definition of an elastic collision.
DInelastic, since cart A does not stop dead when it strikes cart B. A student who expects the moving cart always to stop in an elastic collision picks this. That happens only when the carts have equal masses. Here cart A has twice cart B's mass, so it keeps moving, and the total kinetic energy is still 9.0 J before and after.
Working K before = (1/2)(2.0)(3.0)² + 0 = 9.0 J. K after = (1/2)(2.0)(1.0)² + (1/2)(1.0)(4.0)² = 1.0 + 8.0 = 9.0 J. Equal, so elastic. Momentum before (2.0)(3.0) = 6.0 kg·m/s; after (2.0)(1.0) + (1.0)(4.0) = 6.0 kg·m/s (conserved, as in any collision with negligible net external force).
Two carts of equal mass collide on a level track with negligible friction. The bar chart shows the kinetic energy of each cart just before and just after the collision. Which claim about the collision, with its reasoning, do the data support?
Answer and reasoning
AIt is inelastic, since each cart's own kinetic energy has changed. A student who thinks each object must keep its kinetic energy in an elastic collision picks this. Each cart's kinetic energy would change in an elastic collision too. The claim is right for a different reason: the total fell from 8 J to 5 J.
BIt is elastic, since energy is conserved in every collision. A student who equates conservation of energy with conservation of kinetic energy picks this. Total energy is conserved, but the chart shows the kinetic energy total falling from 8 J to 5 J: 3 J became other forms, such as thermal energy and sound.
CIt is elastic, since both carts are still moving afterward. A student who thinks any collision in which the objects do not stick is elastic picks this. Carts that move apart can still lose kinetic energy, and these did: the total fell from 8 J to 5 J.
DIt is inelastic, since the carts' total kinetic energy decreased.Correct Add the bars. Before: 8 J + 0 = 8 J. After: 0.5 J + 4.5 J = 5 J. The system's total kinetic energy decreased by 3 J, which is what makes a collision inelastic.
Working Total K before = 8 J + 0 = 8 J. Total K after = 0.5 J + 4.5 J = 5 J. The total decreased by 3 J, so the collision is inelastic. (Consistent with carts of 1.0 kg each: 4.0 m/s and 0 before; 1.0 m/s and 3.0 m/s after; momentum 4.0 kg·m/s before and after.)
Cart A rolls along a level track toward cart B, which is at rest. Cart B has twice the mass of cart A. The graph shows the velocity v of each cart as a function of time t; the carts collide at t = 0.50 s. Which claim about the collision is supported by the graph?
Answer and reasoning
AIt is not perfectly inelastic, as the carts are still moving after the collision. A student who thinks a perfectly inelastic collision removes all the kinetic energy picks this. Stuck objects keep moving whenever the system has momentum. Here the carts move off together at 1.0 m/s.
BIt is perfectly inelastic, as after the collision the two carts move with the same velocity.Correct After the collision the two lines coincide at 1.0 m/s: the carts move together with one velocity, which is what defines a perfectly inelastic collision. The system keeps kinetic energy, (1/2)(3m)(1.0 m/s)², but less than the (1/2)m(3.0 m/s)² it had before.
CIt is elastic, as the carts' total momentum is the same before and after the collision. A student who thinks conserved momentum means conserved kinetic energy picks this. The momentum is 3.0m before and after, but the kinetic energy falls from 4.5m to 1.5m (in joules, with m in kilograms), so the collision is not elastic.
DIt is not perfectly inelastic, as the carts' speed afterward is not half of A's speed. A student who takes the common speed of stuck objects to be the average of their speeds before, (3.0 m/s + 0)/2 = 1.5 m/s, picks this and concludes the carts did not stick. The average holds only for equal masses: momentum gives m(3.0 m/s) = (3m)vf, so vf = 1.0 m/s, and the two lines coincide there, as the graph shows.
Working After the collision both carts move at 1.0 m/s: same velocity, so they move together and the collision is perfectly inelastic. Momentum: m(3.0) = (m + 2m)(1.0). K before = (1/2)m(3.0)² = 4.5m; K after = (1/2)(3m)(1.0)² = 1.5m, so two-thirds is transformed.
A 0.50 kg cart moving at 5.0 m/s collides with a 3.0 kg cart at rest on a level track with negligible friction. After the collision, the 0.50 kg cart moves backward at 1.0 m/s and the 3.0 kg cart moves forward at 1.0 m/s. How much of the system's kinetic energy is transformed into other forms of energy?
Answer and reasoning
A6.0 J A student who counts only the kinetic energy the moving cart loses picks this: 6.25 J − 0.25 J. The 3.0 kg cart gains 1.5 J of kinetic energy, and that part is not transformed; the system loses 6.25 J − 1.75 J = 4.5 J.
B9.0 J A student who calculates kinetic energy as mv² picks this: 12.5 J − (0.50 J + 3.0 J). Leaving out the factor 1/2 doubles the answer; with K = (1/2)mv² the loss is 4.5 J.
C4.5 JCorrect Before: (1/2)(0.50 kg)(5.0 m/s)² = 6.25 J. After: (1/2)(0.50 kg)(1.0 m/s)² + (1/2)(3.0 kg)(1.0 m/s)² = 0.25 J + 1.5 J = 1.75 J. The kinetic energy transformed is 6.25 J − 1.75 J = 4.5 J.
D5.0 J A student who gives the backward-moving cart negative kinetic energy picks this: after = 1.5 J − 0.25 J = 1.25 J, so 6.25 J − 1.25 J = 5.0 J. Kinetic energy is a scalar and is never negative, so the total after is 1.75 J and 4.5 J is transformed.
Working K before = (1/2)(0.50)(5.0)² = 6.25 J. K after = (1/2)(0.50)(1.0)² + (1/2)(3.0)(1.0)² = 0.25 J + 1.5 J = 1.75 J. Transformed = 6.25 J − 1.75 J = 4.5 J. (Momentum check: (0.50)(5.0) = 2.5 kg·m/s before; (0.50)(−1.0) + (3.0)(1.0) = 2.5 kg·m/s after.)
In trial 1, cart A, of mass m, moves at speed v along a level track with negligible friction and collides with cart B, of mass m, which is at rest; the carts stick together. In trial 2, cart A again moves at speed v, but cart B is replaced by a cart of mass 3m at rest, and again the carts stick together. The total kinetic energy of the two-cart system just after the collision in trial 2 is how many times that in trial 1?
Answer and reasoning
A1.00 A student who thinks a sticking collision always loses half the kinetic energy picks this, giving (1/2)(1/2)mv² in both trials. That is true only when the masses are equal; the fraction kept is mA/(mA + mB), which is 1/4 in trial 2, so K2 is 0.50 times K1.
B0.50Correct The carts stick, so they share one velocity, found from conservation of momentum: v/2 in trial 1 and v/4 in trial 2. Then K1 = (1/2)(2m)(v/2)² = mv²/4 and K2 = (1/2)(4m)(v/4)² = mv²/8, so K2 is 0.50 times K1. The heavier target leaves only 1/4 of A's kinetic energy instead of 1/2.
C2.00 A student who takes the common speed to be the average of the carts' speeds, v/2 in both trials, picks this: (1/2)(4m)(v/2)² is twice (1/2)(2m)(v/2)². Momentum conservation gives v/4 in trial 2, and then K2 = mv²/8, which is 0.50 times K1 = mv²/4.
D0.25 A student who counts only cart A's kinetic energy after the collision, leaving out what cart B gains, picks this: (1/2)m(v/4)² ÷ (1/2)m(v/2)² = 0.25. The system's kinetic energy includes cart B's; with both carts counted, K2 is 0.50 times K1.
Working Momentum is conserved and the carts move with one velocity afterward. Trial 1: mv = (2m)v1, v1 = v/2; K1 = (1/2)(2m)(v/2)² = mv²/4. Trial 2: mv = (4m)v2, v2 = v/4; K2 = (1/2)(4m)(v/4)² = mv²/8. Ratio K2/K1 = (1/8)/(1/4) = 0.50. In general Kafter = Kbefore × mA/(mA + mB): 1/2 in trial 1, 1/4 in trial 2.
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
