What Is Conserved in a Collision? Momentum vs Energy

Momentum is conserved in every collision, elastic or inelastic, as long as the system is isolated. Kinetic energy is conserved only in elastic collisions. Total energy is always conserved, but in an inelastic collision part of it leaves the kinetic account as heat, sound and permanent deformation.

AP Physics: Unit 4 (topics 4.3 Conservation of Linear Momentum, 4.4 Elastic and Inelastic Collisions). AP Physics 1 Unit 4, Linear Momentum, weighted at 10 to 15 percent of the exam and estimated at about 10 to 15 class periods. Learning objective 4.3.B asks students to describe how the selection of a system determines whether the momentum of that system changes, and learning objective 4.4.A asks students to describe whether an interaction between objects is elastic or inelastic. Total energy conservation comes from Unit 3, essential knowledge 3.4.C.1.

The three answers, one quantity at a time

Three separate questions hide inside the phrase "what is conserved in a collision", and they have three different answers.

Momentum is conserved in every type of collision, elastic, inelastic and perfectly inelastic alike, as long as the system is isolated. Add the momenta of every object in the system as vectors before the impact, add them again after it, and you get the same total. The AP Physics 1 CED states this at essential knowledge 4.3.B.1 in five words: momentum is conserved in all interactions. Isolated means the net external force on the system you chose is zero, or small enough during the brief contact that the external impulse can be neglected.

Kinetic energy is conserved only in elastic collisions. That is not a result you derive about elastic collisions; it is what the words elastic collision mean. EK 4.4.A.1 defines an elastic collision as one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system, and EK 4.4.A.3 defines an inelastic collision as one in which the total kinetic energy of the system decreases. Collisions between everyday objects shed some kinetic energy, so elastic is a model you check for rather than a default you assume.

Total energy is conserved in every collision, with no conditions attached to the collision type. Energy does not disappear in a crash; it changes form. EK 3.4.C.1 puts it as plainly as 4.3.B.1 does for momentum: energy is conserved in all interactions. The lost kinetic energy becomes thermal energy in the objects, sound in the air, and permanent deformation of the materials.

One more quantity is worth naming so you do not go looking for it: the kinetic energy of each individual object is not conserved in any collision type, including elastic ones. EK 4.4.A.2 says so directly. Elastic means the total stays put while the shares move around.

Everything below explains why the first two statements come apart, which is the part a one-line answer cannot carry.

Elastic, inelastic and perfectly inelastic side by side

Read the table by column if you are classifying a collision, and by row if you are asking what happened to one quantity.

Quantity or feature	Elastic	Inelastic	Perfectly inelastic
Total momentum	Conserved	Conserved	Conserved
Total kinetic energy	Conserved, $K_i = K_f$	Decreases	Decreases by the largest amount the momentum allows
Total energy	Conserved	Conserved	Conserved
Kinetic energy of each object	Can change	Can change	Can change
Objects after impact	Separate, generally at new speeds	Separate	Stuck together at one common velocity
Conservation equations available	Two, momentum and kinetic energy	One, momentum	One, momentum, plus $v_1' = v_2'$
CED essential knowledge	4.4.A.1, 4.4.A.2	4.4.A.3, 4.4.A.4	4.4.A.5

Three things in that table are where the classification goes wrong.

Perfectly inelastic is a special case of inelastic, not a fourth category. EK 4.4.A.5 defines it by what the objects do rather than by how much energy is lost: the objects stick together and move with the same velocity after the collision. Any collision that is not elastic is inelastic, and the perfectly inelastic ones are the subset that end up sharing one velocity.
The momentum row never changes. Only the isolation of the system can put a mark against momentum, and that has nothing to do with whether the objects bounce or stick.
The count of conservation equations is why elastic problems feel harder. Elastic hands you two equations, so you can solve for two unknown final velocities. Inelastic hands you one, so you need a second fact from the problem. Perfectly inelastic supplies that fact for free, because the two final velocities are one number.

The AP Physics 1 formula sheet prints no elastic-collision formula. For this whole topic it gives you $\vec{p} = m\vec{v}$ , $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ , $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ , $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i}$ and $K = \frac{1}{2}mv^2$ . The reasoning that decides what is conserved has to come from you.

Why momentum is conserved but kinetic energy is not

This is the question the table cannot answer, and it comes down to one asymmetry: the internal forces cancel in a vector sum, but the work those same forces do does not cancel.

Start with momentum. During the contact, the only forces that matter are the two objects pushing on each other. Newton's third law makes those forces equal in magnitude and opposite in direction at every instant, and they act for the same time interval, because contact starts and ends for both objects at the same moment. Impulse is force times the time over which it acts, so the two impulses are equal and opposite:

\vec{J}_{1 \to 2} = -\vec{J}_{2 \to 1}

The CED states that at EK 4.3.A.3.i: the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and this is a direct result of Newton's third law. Add them to get the change in the system's total momentum:

\Delta \vec{p}_{\text{system}} = \vec{J}_{1 \to 2} + \vec{J}_{2 \to 1} = 0

That cancellation is exact and unconditional. It does not care how squishy the objects are, how long the contact lasts, or how much the surfaces heat up. A vector plus its own negative is zero, always. That is why momentum survives.

Now run the same argument on energy and watch it fail. Work is not force times time; it is the component of force parallel to the displacement of its point of application, times that displacement. EK 3.2.A.3.i: only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy.

The two forces are equal and opposite. The two displacements are not. During the collision the objects move through different distances, in different directions, and their contact surfaces deform by different amounts, so the two works are computed against different displacements and no identity forces them to cancel:

W_{\text{on }1} + W_{\text{on }2} \ne 0 \ \text{in general}

A fast cart running into a slow one makes this concrete. The slow cart is pushed forward while it moves forward, so positive work is done on it; the fast cart is pushed backward while still moving forward, so negative work is done on it. Nothing makes those two amounts equal.

The physical story behind the algebra is deformation. Every collision has a compression phase, when the objects squash into each other, and a restitution phase, when they push apart again. During compression the internal forces store energy in the deformed material, and what happens next is the entire difference between the two collision types.

If the material springs all the way back, every joule stored during compression is returned as kinetic energy during restitution. Total kinetic energy comes out as it went in, and the collision is elastic.
If it does not spring all the way back, some of that stored energy stays behind as thermal energy, sound and a permanent dent. Total kinetic energy comes out lower, and the collision is inelastic. EK 4.4.A.4 is the CED's version: in an inelastic collision, some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.

So the short answer to "why is momentum conserved but not kinetic energy" is that momentum conservation rests on a cancellation of vectors with no escape clause, while kinetic energy conservation rests on a physical question, whether the deformation reverses, that the objects are free to answer either way.

What an isolated system actually requires

The isolation condition on momentum is the one caveat on this page, so it is worth getting exact rather than waving at.

EK 4.3.B.2 sets the requirement: if the net external force on the selected system is zero, the total momentum of the system is constant. EK 4.3.B.3 gives the alternative: if the net external force on the selected system is nonzero, momentum is transferred between the system and the environment. Both statements are about the system you selected, not about the universe. Momentum bookkeeping is a choice you make before you write an equation.

Two moves keep the condition satisfied in the collision problems you will meet.

Put every colliding object inside the system. If the system is both carts, the forces they exert on each other are internal and cancel. If the system is one cart, the other cart's push is external and that cart's momentum changes. Same physics, different bookkeeping, and only one of them lets you write $p_i = p_f$ .
Use the shortness of the collision. Gravity and friction do not switch off during an impact, so strictly the net external force is not zero. The CED licences the usual approximation at EK 4.1.A.3.i, which defines a collision as a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.

That second point is why a car crash on a road with friction, in a gravitational field, still conserves momentum across the impact.

The condition genuinely fails when an external force is comparable to the internal ones. A ball bouncing off the floor is the standard case: treat the ball alone and its momentum reverses, because the floor is outside the system and delivers a large external impulse. Treat the ball and the Earth as one system and momentum is conserved again, with the Earth taking an equal and opposite momentum change that its mass makes unmeasurable. See conservation of linear momentum for the system-choice treatment and the conservation of momentum guide for the solving routine.

Where the missing kinetic energy goes

The sentence "kinetic energy is not conserved" sounds like energy went missing. It did not. Only the kinetic account shrank, and the CED tells you which accounts grew.

EK 4.4.A.4 names the mechanism: the initial kinetic energy that is not restored to kinetic energy is transformed by nonconservative forces into other forms of energy. A boundary statement under Topic 3.4 names the forms, saying that AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. A second boundary statement, under Topic 3.2, sets the depth: AP Physics 1 only expects students to analyze the transfer of mechanical energy, as defined in Unit 3, Topic 4: Conservation of Energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound. In AP Physics 2, students will also study how thermal energy can be transferred between systems through heating or cooling.

The destinations, then, are:

Thermal energy. The objects get warmer. Bending a paper clip until the fold is hot to the touch is the same effect at a speed you can feel.
Sound. The bang is energy radiating away as a pressure wave. Usually a small share, but real, and the one destination you can hear.
Permanent deformation. A crumpled bumper stays crumpled. The work that reshaped the metal is not coming back as motion.
Rotation and vibration. A struck object can leave the collision spinning or ringing. Still mechanical energy, but not the translational $\frac{1}{2}mv^2$ a collision problem tracks.

What AP Physics 1 asks here is qualitative: say the missing kinetic energy went to thermal energy and sound, and find how much by subtracting $K_f$ from $K_i$ . You are not asked for the temperature rise, which needs specific heat capacity and belongs to AP Physics 2.

For a free-response answer: never write that energy was lost or destroyed. Write that kinetic energy was transformed into thermal energy, sound and deformation, and that total energy was conserved. The conservation of energy guide and Topic 3.4 carry the full energy framing.

Perfectly inelastic collisions: the largest loss the momentum allows

The question "what is conserved in a perfectly inelastic collision" has the same answer as every other collision type, and the fact that it feels like it should not is worth confronting head on. Momentum is conserved. Kinetic energy is not. Total energy is. Sticking together is the maximum-damage case for kinetic energy, and it changes nothing about momentum.

EK 4.4.A.5 defines the case: in a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision. That single common velocity is determined by momentum conservation alone. For two objects with the second at rest:

m_1 v_1 = (m_1 + m_2)v_f \quad \Rightarrow \quad v_f = \frac{m_1 v_1}{m_1 + m_2}

That common velocity is the velocity of the system's centre of mass, which the sheet writes $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i}$ and which EK 4.3.A.1.ii says is constant in the absence of a net external force. Useful as a cross-check: the centre of mass of an isolated system moves at the same velocity before, during and after any collision, and in a perfectly inelastic one the objects end up moving with it.

Two consequences follow.

A perfectly inelastic collision loses more kinetic energy than any other outcome with the same masses and the same initial velocities. Momentum conservation fixes the total $\sum m_i v_i$ , and for that fixed total the kinetic energy $\sum \frac{1}{2}m_i v_i^2$ is smallest when every object moves at the same velocity. Any other split leaves the objects moving relative to each other, and that relative motion is extra kinetic energy.

It never loses all of the kinetic energy, unless the total momentum was zero to begin with. Students reason that perfectly inelastic must mean all the kinetic energy is gone, and that cannot be right: the combined object is still moving. Final kinetic energy is zero only if the final velocity is zero, which momentum conservation permits only when the momenta cancelled beforehand, as in a head-on collision between equal and opposite momenta.

With the second object initially at rest, the surviving fraction has a clean closed form, derived below:

\frac{K_f}{K_i} = \frac{m_1}{m_1 + m_2}

Which says something you can feel: a heavy object hitting a light one keeps most of its kinetic energy, and a light object burying itself in a heavy one keeps almost none.

Elastic collisions, and the explosion that runs the other way

Elastic collisions are the only ones where you get a second conservation equation, and getting the definition exactly right is what stops you from overusing it.

EK 4.4.A.1: an elastic collision between objects is one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system. Two cautions come with that sentence.

It is about the system total, not about each object. EK 4.4.A.2 spells this out: in an elastic collision, the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies. A cue ball that stops dead after striking an identical ball has given away all of its kinetic energy, and the collision can still be elastic, because the other ball received exactly that amount.
It is a test you apply, not a property you assume. Compute $K_i$ , compute $K_f$ , compare. If a problem does not tell you a collision is elastic and the numbers do not come out equal, it is not elastic. Hard steel ball bearings, gas molecules and magnetically repelling carts come close. Most everyday collisions do not.

Explosions are the same physics with the sign of the energy change flipped. EK 4.1.A.3.iii models an explosion as an interaction in which forces internal to the system move objects within that system apart. A compressed spring released between two carts, a rocket firing, a firework: the objects fly apart and the total kinetic energy afterwards is larger than before. Momentum does not care. A system that started at rest with zero total momentum has zero total momentum afterwards, which is why two carts released from a compressed spring must carry equal and opposite momenta.

That case makes the asymmetry unmistakable. Kinetic energy went up, so it was plainly not conserved, and momentum was still exactly conserved. The extra kinetic energy came from elastic potential energy in the spring or chemical energy in the propellant, so total energy was conserved throughout.

The collision lab on the interactives page lets you drag the two masses, the two velocities and the elasticity, and draws momentum and kinetic energy as signed bars before and after, so you can hunt for settings that leave the momentum bars alone while the energy bars collapse.

What the CED requires, and how the exam asks it

This material sits in AP Physics 1 Unit 4, Linear Momentum, weighted at 10 to 15 percent of the multiple-choice section and estimated at about 10 to 15 class periods. Two topics carry the conserved-quantity question.

Topic 4.3, Conservation of Linear Momentum. Learning objective 4.3.B asks you to describe how the selection of a system determines whether the momentum of that system changes, and the essential knowledge under it is the three-line summary of this page: momentum is conserved in all interactions (4.3.B.1); if the net external force on the selected system is zero, the total momentum of the system is constant (4.3.B.2); if it is nonzero, momentum is transferred between the system and the environment (4.3.B.3).

Topic 4.4, Elastic and Inelastic Collisions. One learning objective, 4.4.A: describe whether an interaction between objects is elastic or inelastic. All five pieces of essential knowledge under it are quoted above. The CED lists no boundary statement under Topic 4.4.

Topic 4.3 does carry a boundary statement, and it sets the difficulty ceiling for momentum questions. In full, exception clause included:

"AP Physics 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the AP Physics 1 Exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 includes a full treatment of conservation of momentum in two dimensions for problems that include one unknown final velocity."

That middle clause is the part students miss when the statement gets summarised as "no simultaneous equations". You will not be asked to grind through two equations in two unknowns, and you will be asked to set them up and to reason about what happens to one quantity when another changes. Topic 4.3 also lists skill 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws: that is what "is momentum conserved here, and how do you know" is testing, and the answer names the system and the absence of a net external force on it, not a number.

For the procedures rather than the principles, the conservation of momentum guide has the before-and-after solving routine, the impulse-momentum theorem guide covers $\vec{J} = \Delta \vec{p}$ and force-time graphs, and the momentum and collision calculator checks final velocities and energy audits step by step. The CED framing lives on Topic 4.3 and Topic 4.4, with the definition of momentum on Topic 4.1.

One collision, three endings, three energy audits

A $2.0$ kg cart moving at $6.0$ m/s to the right strikes a $4.0$ kg cart at rest on a frictionless track. Work out the final velocities and the kinetic-energy audit for three outcomes: perfectly inelastic, elastic, and a partially inelastic case in which the $2.0$ kg cart rebounds at $1.0$ m/s. Take rightward as positive throughout.

Set the before state once, because it is shared by all three endings. Total momentum $p_i = (2.0)(6.0) + (4.0)(0) = 12\ \text{kg}\cdot\text{m/s}$ to the right. Total kinetic energy $K_i = \frac{1}{2}(2.0)(6.0)^2 + 0 = 36\ \text{J}$ .
Perfectly inelastic. The carts stick, so one common velocity: $v_f = \frac{12}{2.0 + 4.0} = 2.0\ \text{m/s}$ . Then $K_f = \frac{1}{2}(6.0)(2.0)^2 = 12\ \text{J}$ . Kinetic energy lost: $36 - 12 = 24\ \text{J}$ , which is $\frac{24}{36} = 66.7$ percent of the original.
Elastic. Both momentum and kinetic energy hold, giving $v_1' = -2.0\ \text{m/s}$ and $v_2' = +4.0\ \text{m/s}$ . Check the momentum: $(2.0)(-2.0) + (4.0)(4.0) = -4.0 + 16 = 12\ \text{kg}\cdot\text{m/s}$ , unchanged. Check the kinetic energy: $\frac{1}{2}(2.0)(2.0)^2 + \frac{1}{2}(4.0)(4.0)^2 = 4.0 + 32 = 36\ \text{J}$ , unchanged to every digit. Nothing was lost.
Partially inelastic. Now only momentum is available. With $v_1' = -1.0$ m/s given, $(2.0)(-1.0) + (4.0)v_2' = 12$ , so $v_2' = \frac{14}{4.0} = 3.5\ \text{m/s}$ . Then $K_f = \frac{1}{2}(2.0)(1.0)^2 + \frac{1}{2}(4.0)(3.5)^2 = 1.0 + 24.5 = 25.5\ \text{J}$ , a loss of $10.5\ \text{J}$ , or $29.2$ percent.
Line the three up. Momentum after impact: $12$ , $12$ , $12\ \text{kg}\cdot\text{m/s}$ . Kinetic energy after impact: $12$ , $36$ , $25.5\ \text{J}$ . The momentum row is identical across all three endings and the kinetic-energy row is not, which is the whole answer to what is conserved in a collision. The perfectly inelastic ending sits at the bottom of the energy range and the elastic one at the top, with every physically possible outcome for these two carts in between.

All three endings conserve momentum at $12\ \text{kg}\cdot\text{m/s}$ . Kinetic energy finishes at $12$ J (perfectly inelastic, $24$ J lost), $36$ J (elastic, nothing lost) and $25.5$ J (partially inelastic, $10.5$ J lost).

The impulses cancel, the works do not

Using the same two carts, audit the perfectly inelastic ending and the elastic ending object by object. Compute the impulse delivered to each cart and add them, then compute the work done on each cart, using the work-energy theorem $\Delta K = \sum W$ , and add those. Show which sum is forced to zero and which is not.

Perfectly inelastic, impulses. Cart 1 goes from $6.0$ m/s to $2.0$ m/s: $\Delta p_1 = (2.0)(2.0 - 6.0) = -8.0\ \text{kg}\cdot\text{m/s}$ . Cart 2 goes from $0$ to $2.0$ m/s: $\Delta p_2 = (4.0)(2.0 - 0) = +8.0\ \text{kg}\cdot\text{m/s}$ . Sum: $-8.0 + 8.0 = 0$ . Each cart's momentum changed a lot; the system's did not change at all.
Perfectly inelastic, works. By the work-energy theorem the net work on each cart equals its change in kinetic energy. Cart 1: $\Delta K_1 = \frac{1}{2}(2.0)(2.0)^2 - \frac{1}{2}(2.0)(6.0)^2 = 4.0 - 36 = -32\ \text{J}$ . Cart 2: $\Delta K_2 = \frac{1}{2}(4.0)(2.0)^2 - 0 = +8.0\ \text{J}$ . Sum: $-32 + 8.0 = -24\ \text{J}$ . Not zero, and exactly the $24$ J the first example found missing.
Elastic, impulses. Cart 1 goes from $6.0$ to $-2.0$ m/s: $\Delta p_1 = (2.0)(-2.0 - 6.0) = -16\ \text{kg}\cdot\text{m/s}$ . Cart 2 goes from $0$ to $4.0$ m/s: $\Delta p_2 = (4.0)(4.0) = +16\ \text{kg}\cdot\text{m/s}$ . Sum: $0$ again. The individual impulses are twice as large as in the sticking case, and they still cancel exactly.
Elastic, works. Cart 1: $\Delta K_1 = \frac{1}{2}(2.0)(2.0)^2 - \frac{1}{2}(2.0)(6.0)^2 = 4.0 - 36 = -32\ \text{J}$ . Cart 2: $\Delta K_2 = \frac{1}{2}(4.0)(4.0)^2 - 0 = +32\ \text{J}$ . Sum: $0$ . Here the works do cancel, and that is what makes the collision elastic, not a general rule. ( $\Delta K_1$ coming out as $-32$ J in both endings is a coincidence of these masses and speeds: $v_1'$ is $+2.0$ m/s in one case and $-2.0$ m/s in the other, and kinetic energy uses $v^2$ .)
Read the four sums together. The impulse sums are zero in both endings, forced there by Newton's third law acting over a shared contact time. The work sums are $0$ in one ending and $-24$ J in the other, because work depends on each cart's own displacement and nothing forces those numbers to be opposites. That asymmetry is why momentum is conserved in every collision and kinetic energy is not.

Internal impulses sum to zero in both endings ( $-8.0 + 8.0$ and $-16 + 16$ ), so momentum is conserved either way. Internal works sum to $-24$ J for the perfectly inelastic ending and to $0$ for the elastic one, which is exactly the difference between the two collision types.

How much kinetic energy a perfectly inelastic collision must lose

Derive the fraction of kinetic energy that survives a perfectly inelastic collision when a mass $m_1$ moving at $v_1$ strikes a stationary mass $m_2$ , then evaluate it for equal masses, for the $2.0$ kg into $4.0$ kg case above, and for a $0.010$ kg bullet embedding in a $2.00$ kg block.

Momentum conservation gives the common final velocity: $m_1 v_1 = (m_1 + m_2)v_f$ , so $v_f = \frac{m_1 v_1}{m_1 + m_2}$ .
Write the two kinetic energies. Before: $K_i = \frac{1}{2}m_1 v_1^2$ . After: $K_f = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2}(m_1+m_2)\left(\frac{m_1 v_1}{m_1+m_2}\right)^2 = \frac{m_1^2 v_1^2}{2(m_1+m_2)}$ .
Divide, and $v_1^2$ and the factor of $\frac{1}{2}$ both cancel: $\frac{K_f}{K_i} = \frac{m_1^2 v_1^2 / [2(m_1+m_2)]}{m_1 v_1^2 / 2} = \frac{m_1}{m_1 + m_2}$ . The fraction lost is therefore $\frac{m_2}{m_1 + m_2}$ . The initial speed has dropped out entirely: the percentage lost depends only on the mass ratio.
Equal masses. $\frac{m_2}{m_1+m_2} = \frac{1}{2}$ , so exactly $50$ percent of the kinetic energy is lost, whatever the speed. The $2.0$ kg into $4.0$ kg case. $\frac{4.0}{2.0+4.0} = \frac{2}{3} = 66.7$ percent lost, matching the $24$ J out of $36$ J computed directly in the first example.
Bullet into block. $\frac{2.00}{0.010 + 2.00} = \frac{2.00}{2.010} = 0.995$ , so $99.5$ percent of the bullet's kinetic energy is lost and only $0.5$ percent survives as motion of the block. That is why a ballistic pendulum must be solved with momentum across the impact and only then with energy for the swing: conserving kinetic energy through the impact would overstate the swing by a factor of about $200$ .
Check the limits. As $m_2 \to 0$ the fraction lost goes to zero, since barely anything was hit; as $m_2 \to \infty$ it goes to $1$ , since the combined object barely moves. It reaches a full $100$ percent only in that limit or when the total momentum was zero to start with.

A perfectly inelastic collision with the target at rest keeps $\frac{K_f}{K_i} = \frac{m_1}{m_1+m_2}$ and loses $\frac{m_2}{m_1+m_2}$ , independent of speed: $50$ percent for equal masses, $66.7$ percent for $2.0$ kg into $4.0$ kg, and $99.5$ percent for a $0.010$ kg bullet in a $2.00$ kg block.

Frequently asked questions

What is conserved in a perfectly inelastic collision?

Momentum and total energy are conserved; kinetic energy is not. The objects stick together and move off with one common velocity that momentum conservation alone determines, and the total kinetic energy afterwards is lower than before. A perfectly inelastic collision loses more kinetic energy than any other outcome with the same masses and initial velocities, but it never loses all of it unless the total momentum was zero to begin with, because the combined object is still moving. The missing kinetic energy becomes thermal energy, sound and permanent deformation, so total energy is conserved.

Is momentum conserved in an elastic collision?

Yes. Momentum is conserved in an elastic collision, in an inelastic one and in a perfectly inelastic one, provided the system is isolated so the net external force on it is zero or negligible during the impact. What makes a collision elastic is an extra property on top of that: the total kinetic energy of the system is the same before and after. So an elastic collision gives you two conservation equations, momentum and kinetic energy, while an inelastic collision gives you only momentum.

Is momentum conserved in an inelastic collision?

Yes. Momentum conservation does not depend on the collision type. An inelastic collision is defined by its total kinetic energy decreasing, and that has no bearing on momentum: the forces the two objects exert on each other are equal and opposite over the same contact time, so the impulses they deliver cancel exactly in the vector sum. The only condition is that the system be isolated, which means putting every colliding object inside the system and treating the brief external impulse from gravity or friction as negligible.

Why is momentum conserved but not kinetic energy?

Because the internal forces cancel when you add impulses, but not when you add work. The two objects push on each other with equal and opposite forces for the same length of time, so their impulses are exactly opposite and the momentum changes cancel to zero whatever the objects are made of. Work is force times the displacement of its point of application, and the two objects move through different displacements during the impact, so the two amounts of work have no reason to cancel. If the deformation springs back completely they do cancel, and the collision is elastic; if not, some kinetic energy stays behind as thermal energy, sound and permanent deformation.

Is kinetic energy conserved in an inelastic collision?

No. That is the definition of an inelastic collision: the total kinetic energy of the system decreases. The AP Physics 1 CED states it at essential knowledge 4.4.A.3, and 4.4.A.4 adds that the missing kinetic energy is transformed by nonconservative forces into other forms of energy. To find how much was lost, compute the total kinetic energy before and after and subtract. Never use kinetic energy conservation as an equation in an inelastic problem; use momentum conservation instead.

Is energy conserved in an inelastic collision?

Total energy is conserved in an inelastic collision, even though kinetic energy is not. Energy is never destroyed; it changes form. The kinetic energy that disappears from the motion reappears as thermal energy in the objects, sound radiating into the air, permanent deformation of the materials, and sometimes rotation or vibration. The AP Physics 1 CED states the principle at essential knowledge 3.4.C.1: energy is conserved in all interactions. In a free-response answer, write that kinetic energy was transformed, not that energy was lost.

When is momentum not conserved in a collision?

When the system you chose is not isolated, meaning a significant external force acts on it during the impact. Two cases account for most of them: choosing a system that leaves out one of the colliding objects, so that object's push counts as external, and a collision with something anchored to the Earth, such as a ball bouncing off the floor or a car hitting a wall. Both are fixed by enlarging the system. Include the floor and the Earth and momentum is conserved again, with the Earth's share of the change unmeasurable because of its mass.