Momentum vs Kinetic Energy: The Difference

Momentum is a vector, mass times velocity, and it doubles when the speed doubles. Kinetic energy is a scalar, one half m v squared, and it quadruples. In an isolated collision the total momentum is always conserved, while kinetic energy is conserved only when the collision is elastic.

AP Physics: Unit 4 (topics 3.1 Translational Kinetic Energy, 4.1 Linear Momentum, 4.4 Elastic and Inelastic Collisions). This comparison spans two AP Physics 1 units. Kinetic energy sits in Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods; Topic 3.1 carries learning objective 3.1.A and essential knowledge 3.1.A.1 through 3.1.A.3, with suggested skills 1.C, 2.B, 3.B and 3.C. Momentum sits in Unit 4, Linear Momentum, weighted at 10 to 15 percent over about 10 to 15 class periods; Topic 4.1 carries learning objective 4.1.A and EK 4.1.A.1 through 4.1.A.3, with suggested skills 1.C, 2.B, 2.C and 3.B, and a boundary statement that unless otherwise stated the general term momentum refers specifically to linear momentum. The conservation contrast comes from Topic 4.3 (EK 4.3.B.1 through 4.3.B.3) and Topic 4.4 (EK 4.4.A.1 through 4.4.A.5).

Two definitions, two disagreements

Both quantities are built from the same two ingredients, a mass and a velocity, and they combine them differently in two ways that matter. Everything else on this page follows from those two.

The AP Physics 1 CED gives the definitions in separate units. Essential knowledge 4.1.A.1 says linear momentum is defined by $\vec{p} = m\vec{v}$ , and EK 4.1.A.2 adds that momentum is a vector quantity with the same direction as the velocity. EK 3.1.A.1 says an object's translational kinetic energy is $K = \frac{1}{2}mv^2$ , and EK 3.1.A.2 adds that translational kinetic energy is a scalar quantity.

Disagreement one: direction. Momentum carries the direction of the velocity. Kinetic energy carries none, and because $v$ is squared it can never come out negative. Reverse an object's velocity and its momentum flips sign while its kinetic energy does not move.

Disagreement two: the power of $v$ . Momentum is linear in speed, kinetic energy is quadratic. Triple the speed and the momentum triples while the kinetic energy goes up ninefold.

Those two facts are enough to predict the exam behaviour of both quantities. Direction is why momentum can cancel across a system and kinetic energy cannot, which is why a head-on collision can destroy all of a system's kinetic energy while its momentum bookkeeping stays exact. The power of $v$ is why the two quantities rank objects differently the moment their masses differ.

Momentum vs kinetic energy, side by side

Question you are asking	Momentum	Kinetic energy
Symbol	$\vec{p}$	$K$
Defining equation	$\vec{p} = m\vec{v}$	$K = \frac{1}{2}mv^2$
SI unit	$\text{kg}\cdot\text{m/s}$	$\text{J}$
Vector or scalar	Vector, along the velocity	Scalar, no direction at all
How it scales with speed	Linear: double $v$ , double $\vec{p}$	Quadratic: double $v$ , four times $K$
Can it be negative	Yes, a component can be	No, $v^2$ is never negative
Reverse the velocity	Sign flips	Value unchanged
Total for a system	Vector sum, parts can cancel	Ordinary sum, parts cannot cancel
Can a moving system total zero	Yes	No
What changes it	Impulse, $\vec{J} = \vec{F}_{\text{avg}}\Delta t$	Net work, $\Delta K = \sum W_i$
Which theorem links it to forces	Impulse-momentum theorem	Work-energy theorem
Conserved in an isolated collision	Always	Only if the collision is elastic
Depends on the observer's frame	Yes, through $\vec{v}$	Yes, EK 3.1.A.3 says so directly
CED essential knowledge	4.1.A.1 through 4.1.A.3	3.1.A.1 through 3.1.A.3

The row about totals is the one to memorise. Two carts rolling toward each other with equal masses and equal speeds have a total momentum of zero and a total kinetic energy that is emphatically not zero. Add vectors and they can annihilate; add scalars and they can only pile up. Every difference the exam tests about collisions comes back to that line.

The case that separates them: equal in one, unequal in the other

Give two objects the same momentum and their kinetic energies will differ unless their masses are equal. Give them the same kinetic energy and their momenta will differ. The first worked example below runs both directions with numbers.

Take $p = 24\ \text{kg}\cdot\text{m/s}$ for both an $8.0$ kg object and a $2.0$ kg object. The heavy one is doing $3.0$ m/s and carries $36$ J. The light one is doing $12$ m/s and carries $144$ J, four times as much, from the same momentum.

The two bridging relations make the pattern obvious:

K = \frac{p^2}{2m} \qquad\text{and}\qquad p = \sqrt{2mK}

At fixed momentum, kinetic energy runs inversely with mass, so the lighter object always holds more of it. At fixed kinetic energy, momentum runs with the square root of mass, so the heavier object always carries more of it.

Neither relation is printed on the AP Physics 1 or AP Physics 2 equation sheet. The mechanics panel gives you $\vec{p} = m\vec{v}$ and $K = \frac{1}{2}mv^2$ separately, and you combine them yourself. It is a two-line derivation: solve the first for $v = p/m$ , substitute into the second, and $K = \frac{1}{2}m(p/m)^2 = p^2/2m$ . Doing it once in the margin is faster than remembering it wrong.

Double the speed and the two answers disagree by a factor of two

This is the difference students feel first, usually while driving.

Speed multiplied by	Momentum multiplied by	Kinetic energy multiplied by
$2$	$2$	$4$
$3$	$3$	$9$
$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{4}$
$1.5$	$1.5$	$2.25$

The practical consequence is that the two quantities answer different questions about bringing something to a halt, and the third worked example puts numbers on it.

Ask how long it takes a given braking force to stop the car and you are asking about momentum, because $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ contains a time. Double the speed and the stopping time doubles.
Ask how far the car travels while stopping and you are asking about kinetic energy, because $\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$ contains a distance. Double the speed and the stopping distance quadruples.

Same car, same brakes, same event, two questions, two different scalings. A question that says "how long" and a question that says "how far" are not paraphrases of each other, and the whole difficulty of a two-part free-response question is often that it asks one then the other.

In a collision, one is guaranteed and the other is a question

The CED is unusually blunt here, and the wording is worth having exactly.

EK 4.3.B.1 states that momentum is conserved in all interactions. EK 4.3.B.2 gives the condition under which a chosen system's total momentum stays constant: if the net external force on the selected system is zero, the total momentum of the system is constant. EK 4.3.B.3 covers the other case: if the net external force is nonzero, momentum is transferred between the system and the environment.

Kinetic energy gets no such guarantee. EK 4.4.A.1 defines an elastic collision as one in which the initial kinetic energy of the system equals the final kinetic energy. EK 4.4.A.3 defines an inelastic collision as one in which the total kinetic energy of the system decreases, and EK 4.4.A.4 says the missing kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.

So the collision's label is a statement about kinetic energy and nothing else. Momentum does not care what you call the collision.

The reason sits in the vector row of the table. Momentum is conserved because the two objects exert equal and opposite impulses on each other, and equal and opposite vectors cancel in the system total. Kinetic energies are scalars, all positive, and there is nothing for them to cancel against, so there is no bookkeeping law forcing the sum to hold still. Energy overall is still conserved; the kinetic account is simply not closed.

For the classification procedure, read elastic vs inelastic collision. For the full audit of which quantities survive an impact, what is conserved in a collision works one collision through three endings.

Which one to reach for on a problem

The choice is usually forced by what the question gives you and what it asks for.

The problem gives you	The problem asks for	Use
Two objects interacting, no forces stated	Final velocities	Momentum conservation
A collision plus the word elastic	Both final velocities	Momentum and kinetic energy together
A contact time or a force-time graph	An average force	Impulse and momentum
A force and a distance	A final speed	Work and kinetic energy
A height, a spring, or a rough patch	A speed somewhere else	Energy conservation
An explosion from rest	The ratio of the two speeds	Momentum conservation

Two of those rows deserve a note.

An elastic collision is the case where you get both equations at once, and that is precisely why the label matters: it hands you a second independent equation and therefore lets you solve for two unknowns.

An explosion is the cleanest demonstration that the two quantities behave differently. A system at rest has zero momentum and zero kinetic energy. Let it blow apart and the momentum is still zero, because the fragments fly off with momenta that cancel, while the kinetic energy has gone from zero to something large, supplied by stored chemical or elastic energy. EK 4.1.A.3.iii defines an explosion as a model for an interaction in which forces internal to the system move objects within that system apart. Total momentum unchanged, total kinetic energy manufactured. No scalar could do that.

When the two rankings agree, and why that lulls you

For a single object speeding up or slowing down, momentum and kinetic energy rise and fall together, so nothing in everyday practice forces you to keep them apart. The same is true for any comparison between objects of equal mass: at equal mass, $K = p^2/2m$ makes the ordering by kinetic energy identical to the ordering by momentum magnitude.

That is the whole trap, because introductory problems are full of equal-mass carts on tracks.

Three more near coincidences are worth naming so you do not read them as sameness.

In an elastic collision both totals are conserved, so any check you run gives the same verdict twice and you never find out which conservation law you were using. Inelastic collisions are where the two separate.
Both quantities depend on the observer. Kinetic energy's frame dependence is stated outright at EK 3.1.A.3: different observers may measure different values of an object's translational kinetic energy depending on the observer's frame of reference. Momentum inherits the same dependence through $\vec{v}$ . So frame dependence is not a way to tell them apart.
Both are zero for an object at rest, and both are unchanged for an object coasting at constant velocity. The agreement holds right up to the moment a second object enters the problem.

The reflex worth building is to check the units of the answer you are about to write. If the question wanted $\text{kg}\cdot\text{m/s}$ and your line ends in joules, you solved a different problem.

Where the confusion costs a mark

Each of these is a scoring event rather than general advice.

Writing that kinetic energy is conserved because momentum is. The error this page exists to prevent. Momentum conservation follows from Newton's third law and needs only an isolated system. Kinetic energy conservation needs the collision to be elastic, which is an extra claim you have to be given or have to prove.
Adding kinetic energies as though they could cancel. Two carts approaching each other have kinetic energies that add, never subtract. Only the momenta get signs.
Dropping the sign on a momentum in a head-on problem. Declare a positive direction before the first line. A cart moving left at $3.0$ m/s contributes $-p$ , not $+p$ , and the system total is the difference, not the sum.
Reporting a negative kinetic energy. If a line produces one, a sign that belonged to a momentum has leaked into an energy. Kinetic energy has no direction to carry.
Assuming equal momentum means equal kinetic energy. It only does when the masses match. The first worked example shows a factor of four appearing from nowhere else.
Using $\Delta K = \sum W_i$ when the question asked how long. Work has no time in it. Reach for $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ instead, and the reverse when a question about distance tempts you toward impulse.
Quoting $K = p^2/2m$ as though it were printed. It is not on either algebra-based sheet. Derive it or avoid it; do not cite it as a given.
Calling a collision elastic because the objects separated. Separating is not elasticity. Only a kinetic-energy audit, before against after, settles the label, and EK 4.4.A.2 warns that in an elastic collision the individual objects' kinetic energies may still change even though the system total does not.

What the CED asks, and what the sheet prints

This comparison straddles two units of AP Physics 1.

Kinetic energy lives in Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. Topic 3.1, Translational Kinetic Energy, carries one learning objective, 3.1.A, describe the translational kinetic energy of an object in terms of the object's mass and velocity, supported by EK 3.1.A.1 through 3.1.A.3. Its suggested skills are 1.C, 2.B, 3.B and 3.C.

Momentum lives in Unit 4, Linear Momentum, weighted at 10 to 15 percent over about 10 to 15 class periods. Topic 4.1, Linear Momentum, carries learning objective 4.1.A, describe the linear momentum of an object or system, supported by EK 4.1.A.1 through 4.1.A.3, with suggested skills 1.C, 2.B, 2.C and 3.B. A boundary statement adds that unless otherwise stated the general term momentum refers specifically to linear momentum. Topic 4.4, Elastic and Inelastic Collisions, carries learning objective 4.4.A and EK 4.4.A.1 through 4.4.A.5, and that is where the kinetic-energy test for the label is defined.

On the AP Physics 1 equation sheet, the Mechanics and Fluids panel prints $K = \frac{1}{2}mv^2$ , $\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$ , $\vec{p} = m\vec{v}$ , $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ , $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ and $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i}$ . Its symbol key lists $K$ as kinetic energy and $p$ as momentum. The same panel appears on the AP Physics 2 sheet. What neither sheet prints is a statement of momentum conservation, a statement of energy conservation, or the bridge $K = p^2/2m$ . Those three you supply.

Going further: conservation of momentum for the system-selection routine, the work-energy theorem for the energy route to a speed, and conservation of energy for the full accounting. The CED framing for each topic is on Topic 3.1 and Topic 4.1.

Equal momentum, unequal kinetic energy, and then the reverse

Part one: an $8.0$ kg object moves at $3.0$ m/s and a $2.0$ kg object moves at $12$ m/s, both in the same direction. Compare their momenta and their kinetic energies. Part two: now give an $8.0$ kg object and a $2.0$ kg object the same kinetic energy, $100$ J each. Compare their speeds and their momenta.

Part one, momenta. Heavy object: $p = mv = (8.0)(3.0) = 24\ \text{kg}\cdot\text{m/s}$ . Light object: $p = (2.0)(12) = 24\ \text{kg}\cdot\text{m/s}$ . Identical, and they point the same way, so nothing distinguishes them by momentum.
Part one, kinetic energies. Heavy: $K = \frac{1}{2}(8.0)(3.0)^2 = \frac{1}{2}(8.0)(9.0) = 36\ \text{J}$ . Light: $K = \frac{1}{2}(2.0)(12)^2 = \frac{1}{2}(2.0)(144) = 144\ \text{J}$ . The light object holds four times the kinetic energy of the heavy one while carrying exactly the same momentum.
Check both with the bridge. $K = p^2/2m$ gives $24^2/(2 \times 8.0) = 576/16 = 36\ \text{J}$ and $576/(2 \times 2.0) = 576/4 = 144\ \text{J}$ . The ratio $144/36 = 4$ is exactly the mass ratio $8.0/2.0$ turned upside down, which is what an inverse dependence on $m$ at fixed $p$ means.
Part two, speeds at equal kinetic energy. From $K = \frac{1}{2}mv^2$ , $v = \sqrt{2K/m}$ . Heavy: $v = \sqrt{2(100)/8.0} = \sqrt{25} = 5.0\ \text{m/s}$ . Light: $v = \sqrt{2(100)/2.0} = \sqrt{100} = 10\ \text{m/s}$ .
Part two, momenta. Heavy: $p = (8.0)(5.0) = 40\ \text{kg}\cdot\text{m/s}$ . Light: $p = (2.0)(10) = 20\ \text{kg}\cdot\text{m/s}$ . Now the heavy object carries twice the momentum from the same kinetic energy. Check with $p = \sqrt{2mK}$ : $\sqrt{2(8.0)(100)} = \sqrt{1600} = 40$ and $\sqrt{2(2.0)(100)} = \sqrt{400} = 20$ . The ratio is $\sqrt{4} = 2$ , the square root of the mass ratio.
Put the two parts side by side. The ranking reversed. At equal momentum the light object won on kinetic energy by a factor of four; at equal kinetic energy the heavy object won on momentum by a factor of two. Two quantities whose ordering of the same pair of objects can point in opposite directions are not two names for one idea.

At equal momentum $24\ \text{kg}\cdot\text{m/s}$ : the $8.0$ kg object has $36$ J and the $2.0$ kg object has $144$ J, four times more. At equal kinetic energy $100$ J: the $8.0$ kg object moves at $5.0$ m/s with $40\ \text{kg}\cdot\text{m/s}$ and the $2.0$ kg object moves at $10$ m/s with $20\ \text{kg}\cdot\text{m/s}$ , half as much. Fixing one quantity never fixes the other.

One head-on collision: momentum survives, kinetic energy may not

Two $2.0$ kg carts roll toward each other on a frictionless track, each at $3.0$ m/s. Take rightward as positive. Find the system's total momentum and total kinetic energy before the collision, then again for two endings: (a) the carts stick together, and (b) the collision is elastic and each cart rebounds at $3.0$ m/s.

Declare the axis first and hold it: rightward is positive, so the left-moving cart contributes a negative momentum in every line below.
Before, momentum. $p_{\text{total}} = (2.0)(+3.0) + (2.0)(-3.0) = +6.0 - 6.0 = 0$ . The system is moving vigorously and its total momentum is zero, because momenta are vectors and these two cancel exactly.
Before, kinetic energy. $K_{\text{total}} = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(2.0)(3.0)^2 = 9.0 + 9.0 = 18\ \text{J}$ . Notice the second term was added, not subtracted. Kinetic energy is a scalar and the leftward cart's direction never entered the arithmetic.
Ending (a), the carts stick. Momentum conservation fixes the answer immediately: the total must still be zero, and a single combined object of mass $4.0$ kg with zero momentum has zero velocity. Both carts stop dead.
Ending (a), the energy audit. $K_{\text{after}} = \frac{1}{2}(4.0)(0)^2 = 0$ . All $18$ J of kinetic energy is gone, transformed by nonconservative forces into thermal energy, sound and deformation, exactly as EK 4.4.A.4 describes. This is the largest fraction of its kinetic energy any collision can lose, all of it, and it is available only because the system's total momentum was zero.
Ending (b), elastic. After: $p_{\text{total}} = (2.0)(-3.0) + (2.0)(+3.0) = 0$ , unchanged. $K_{\text{total}} = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(2.0)(3.0)^2 = 18\ \text{J}$ , also unchanged, which is what EK 4.4.A.1 requires of an elastic collision.
Now compare the two endings. The momentum answer was the same in both, $0$ before and $0$ after, and it was fixed before you knew anything about the collision. The kinetic energy answer differed by the whole $18$ J, and nothing but the label decided which one applied. One quantity was determined by conservation, the other by the physics of the impact.
One check worth running: could a collision have destroyed all the momentum instead? No. With no net external force the total is locked at zero, and it was already zero. Could a collision have destroyed all the kinetic energy of a system with nonzero total momentum? Also no, because that system has to keep moving, and a moving system has kinetic energy. The zero-momentum case is the only one where the kinetic energy can be taken all the way to zero.

Before: total momentum $0$ , total kinetic energy $18$ J. Sticking together: total momentum $0$ , kinetic energy $0$ , with all $18$ J transformed to other forms. Elastic rebound: total momentum $0$ , kinetic energy $18$ J. Momentum gave the same answer in both endings without being told the collision type; kinetic energy needed the label.

Same brakes, twice the speed: double the time, quadruple the distance

A $1200$ kg car brakes to a stop under a constant $6000$ N retarding force. Find the stopping time and the stopping distance from $15$ m/s, then from $30$ m/s. Take the direction of travel as positive.

At $15$ m/s, momentum. $p = mv = (1200)(15) = 18{,}000\ \text{kg}\cdot\text{m/s}$ . The car must lose all of it, so $\Delta p = -18{,}000\ \text{kg}\cdot\text{m/s}$ .
Stopping time from the impulse relation. $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ , so $\Delta t = \lvert \Delta p \rvert / F = 18{,}000/6000 = 3.0\ \text{s}$ .
At $15$ m/s, kinetic energy. $K = \frac{1}{2}(1200)(15)^2 = \frac{1}{2}(1200)(225) = 135{,}000\ \text{J}$ .
Stopping distance from the work-energy theorem. The friction force is antiparallel to the motion, so it does work $-Fd$ , and $\Delta K = -Fd$ gives $d = K/F = 135{,}000/6000 = 22.5\ \text{m}$ .
Now at $30$ m/s. $p = (1200)(30) = 36{,}000\ \text{kg}\cdot\text{m/s}$ , exactly double. $K = \frac{1}{2}(1200)(900) = 540{,}000\ \text{J}$ , exactly four times.
The two answers at $30$ m/s. Time: $\Delta t = 36{,}000/6000 = 6.0\ \text{s}$ , double the earlier $3.0$ s. Distance: $d = 540{,}000/6000 = 90\ \text{m}$ , four times the earlier $22.5$ m.
Read the pattern off the algebra rather than the numbers. $\Delta t = mv/F$ is linear in $v$ because momentum is. $d = mv^2/2F$ is quadratic in $v$ because kinetic energy is. Same car, same brakes, same stop, and the two questions scale differently because they are questions about two different quantities.
Sanity check with kinematics, which must agree. $a = F/m = 6000/1200 = 5.0\ \text{m/s}^2$ of deceleration. From $30$ m/s: $t = 30/5.0 = 6.0$ s and $d = v^2/2a = 900/10 = 90$ m. Both match.

From $15$ m/s the car stops in $3.0$ s over $22.5$ m. From $30$ m/s it stops in $6.0$ s over $90$ m. Doubling the speed doubles the momentum and therefore the stopping time, and quadruples the kinetic energy and therefore the stopping distance.

Frequently asked questions

What is the difference between momentum and kinetic energy?

Momentum is a vector equal to mass times velocity, measured in kilogram meters per second, and it points along the velocity. Kinetic energy is a scalar equal to one half mass times speed squared, measured in joules, and it has no direction. Two consequences follow. First, momentum is linear in speed while kinetic energy is quadratic, so doubling an object's speed doubles its momentum and quadruples its kinetic energy. Second, momenta can cancel within a system while kinetic energies can only add, which is why a system of two objects moving toward each other can have zero total momentum and a large total kinetic energy at the same time.

Can two objects have the same momentum but different kinetic energies?

Yes, and they always do unless their masses are equal. An 8.0 kg object at 3.0 m/s and a 2.0 kg object at 12 m/s both carry 24 kilogram meters per second of momentum, but the first has 36 J of kinetic energy and the second has 144 J, four times more. The relation behind it is that kinetic energy equals momentum squared divided by twice the mass, so at fixed momentum the lighter object always holds more kinetic energy. Running it the other way, two objects with the same kinetic energy have different momenta, and there the heavier object carries more, in proportion to the square root of its mass.

Is kinetic energy conserved in a collision?

Only if the collision is elastic. The AP Physics 1 CED defines an elastic collision at essential knowledge 4.4.A.1 as one in which the initial kinetic energy of the system equals the final kinetic energy, and defines an inelastic collision at 4.4.A.3 as one in which the total kinetic energy of the system decreases. Momentum is different: EK 4.3.B.1 states that momentum is conserved in all interactions, and EK 4.3.B.2 gives the working condition, that the total momentum of a chosen system is constant when the net external force on it is zero. So momentum conservation comes free with an isolated system, while kinetic energy conservation is an extra claim you must be given or must verify by totalling the kinetic energy before and after.

Why is momentum a vector and kinetic energy a scalar?

Because of how each is built from the velocity. Momentum multiplies the mass by the velocity itself, and velocity is a vector, so momentum inherits its direction. The AP Physics 1 CED states this at essential knowledge 4.1.A.2: momentum is a vector quantity and has the same direction as the velocity. Kinetic energy multiplies the mass by the square of the speed, and squaring destroys the direction and the sign, so what comes out is a plain positive number. EK 3.1.A.2 states that translational kinetic energy is a scalar quantity. This is why reversing an object's velocity flips the sign of its momentum and leaves its kinetic energy untouched.

Does doubling the speed double both the momentum and the kinetic energy?

No. Doubling the speed doubles the momentum but multiplies the kinetic energy by four, because momentum is linear in speed and kinetic energy depends on speed squared. The everyday version is braking. Under the same braking force, a car at twice the speed takes twice as long to stop, because stopping time comes from the impulse relation and therefore from momentum, but it travels four times as far while stopping, because stopping distance comes from the work-energy theorem and therefore from kinetic energy. A 1200 kg car braking at 6000 N stops from 15 m/s in 3.0 s over 22.5 m, and from 30 m/s in 6.0 s over 90 m.

Is the formula for kinetic energy in terms of momentum on the AP equation sheet?

No. Neither the AP Physics 1 sheet nor the AP Physics 2 sheet prints kinetic energy as momentum squared over twice the mass. The Mechanics and Fluids panel prints the two definitions separately, momentum as mass times velocity and kinetic energy as one half mass times speed squared, and leaves the combination to you. The derivation takes two lines: solve the momentum equation for the velocity, substitute it into the kinetic energy equation, and the masses combine to leave momentum squared over twice the mass. Write it in the margin rather than trusting a half-remembered version, because getting the factor of two or the position of the mass wrong is a common way to lose an otherwise correct answer.

Should I use momentum or energy to solve a collision problem?

Start with momentum, because it is conserved whatever kind of collision you are looking at, provided the net external force on your chosen system is negligible. That single equation is enough to find one unknown final velocity, and it is enough for a perfectly inelastic collision where the objects share one final velocity. Bring kinetic energy in only when you are told the collision is elastic, because then you get a second independent equation and can solve for two unknowns. If the question asks about the force during the impact or its duration, switch to the impulse relation instead, and if it asks about a distance travelled before or after the collision, switch to work and energy.