AP Physics C: Mechanics · Topic 3.1

Topic 3.1: Translational Kinetic Energy

Q: What is the formula for translational kinetic energy in AP Physics C?

Translational kinetic energy is K = \frac{1}{2}mv^2, where m is the object's mass and v is its speed. Essential knowledge 3.1.A.1 of the AP Physics C: Mechanics course and exam description states it in exactly those terms, and the equation is printed on the AP Physics C: Mechanics equation sheet. The unit is the joule. The formula is the same in AP Physics 1, AP Physics 2 and both AP Physics C courses. What differs in the calculus-based course is that the speed you put into it is often something you have to produce first, by differentiating a position function or integrating an acceleration function.

Unit 3: Work, Energy, and Power15-25% of the multiple-choice section

Translational kinetic energy is one half the mass times the speed squared, and it is a scalar. AP Physics C states this topic in the same words as AP Physics 1. The calculus shows up one step earlier: the speed you square is often something you differentiate a position function to get.

AP Physics: Unit 3 (topics 3.1 Translational Kinetic Energy). Topic 3.1 of the current AP Physics C: Mechanics course and exam description (Unit 3, Work, Energy, and Power, weighted 15 to 25% of the multiple-choice section at about 12 to 17 class periods). One learning objective, 3.1.A, and three essential knowledge statements: 3.1.A.1 gives K = (1/2)mv^2, 3.1.A.2 states that translational kinetic energy is a scalar quantity, and 3.1.A.3 states that different observers may measure different values depending on the observer's frame of reference. No boundary statement. Suggested skills 1.C, 2.C, 3.B and 3.C. The required content is word for word identical to AP Physics 1's Topic 3.1, including all three essential knowledge statements and the absence of a boundary statement; the single framework difference is that AP Physics 1 suggests skill 2.B (calculate an unknown quantity) where AP Physics C: Mechanics suggests skill 2.C (compare quantities between scenarios). The calculus differentiator is upstream of the formula: the C course can require v to be obtained by differentiating a position function or integrating an acceleration function. The C: Mechanics sheet prints Delta x = integral of v_x dt and Delta v_x = integral of a_x dt in its translational block but does not print v_x = dx/dt, which appears in the framework at 1.3.A.4.i; the rotational block does print omega = d theta/dt and alpha = d omega/dt. Neither course's sheet prints K = p^2/(2m). Learning objective 3.1.A appears in two of the four sample free-response questions (Questions 1 and 3) and in none of the fifteen sample multiple-choice questions.

What Topic 3.1 requires, in full

Topic 3.1 of AP Physics C: Mechanics carries one learning objective and three essential knowledge statements. That is the entire required content, and it fits on this screen.

Learning objective 3.1.A: describe the translational kinetic energy of an object in terms of the object's mass and velocity.

Statement	What the CED says
3.1.A.1	An object's translational kinetic energy is given by the equation $K = \frac{1}{2}mv^2$
3.1.A.2	Translational kinetic energy is a scalar quantity
3.1.A.3	Different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference

The suggested skills listed for this topic are 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of the physical system), 2.C (compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario), 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim) and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

Topic 3.1 prints no boundary statement. Unit 3 prints two in total, and both sit elsewhere, under Topic 3.2 and Topic 3.4.

Three statements is the smallest required-content block in Unit 3. Topic 3.3 has eight numbered statements with seven sub-statements under them. That asymmetry is the honest shape of the unit: kinetic energy is one definition, and almost everything hard in Unit 3 lives in what changes it.

This topic is identical to AP Physics 1's. Say it plainly.

Set the two course and exam descriptions side by side at Topic 3.1 and the required content matches word for word. Same learning objective sentence. Same three essential knowledge statements, in the same order, with the same numbers. Same equation. Neither course prints a boundary statement here.

There is exactly one difference in the framework, and it is in the suggested skills:

	AP Physics 1 Topic 3.1	AP Physics C: Mechanics Topic 3.1
Learning objective 3.1.A	identical wording	identical wording
3.1.A.1, 3.1.A.2, 3.1.A.3	identical wording	identical wording
Equation on the course's sheet	$K = \frac{1}{2}mv^2$	$K = \frac{1}{2}mv^2$
Boundary statement	none	none
Suggested skills	1.C, 2.B, 3.B, 3.C	1.C, 2.C, 3.B, 3.C

Skill 2.B is "calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway." Skill 2.C is "compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario." So the algebra-based course suggests you compute a kinetic energy, and the calculus-based course suggests you compare two of them.

Do not over-read that. The framework's exam-weighting page carries an explicit note: required course content, meaning the learning objectives and essential knowledge, can be assessed with any skill. The suggested skills describe how the content is usually approached in the classroom, not a fence around the exam.

But the pattern does hold up where it can be checked. On the multiple-choice section of the AP Physics C: Mechanics exam, skill 2.C is weighted 10 to 15% and skill 2.B is weighted 20 to 25%, so 2.C is the rarer of the two. And in the CED's own sample question set, learning objective 3.1.A appears in two of the four sample free-response questions and in none of the fifteen sample multiple-choice questions. On the evidence the CED provides about itself, Topic 3.1 reaches you through a derivation, not through a plug-in.

If you are in the algebra-based course, the page you want is AP Physics 1 Topic 3.1. Nothing on this page contradicts it. This page adds the calculus context that the C exam assumes around the same three sentences.

Where the calculus enters: the v you square

$K = \frac{1}{2}mv^2$ needs a speed. In an algebra-based problem that speed arrives already measured, or comes out of a constant-acceleration equation. In a calculus-based problem it is frequently something you have to produce.

Given a position function, differentiate:

v_x = \frac{dx}{dt} \qquad \text{then} \qquad K(t) = \tfrac{1}{2}m\left(\frac{dx}{dt}\right)^2

Given an acceleration function, integrate:

\Delta v_x = \int a_x(t)\, dt

A detail worth having right, because it is the kind of thing a page gets wrong from memory. On the AP Physics C: Mechanics equation sheet, the translational block prints the two integrals, $\Delta x = \int v_x(t)\, dt$ and $\Delta v_x = \int a_x(t)\, dt$ , and does not print $v_x = dx/dt$ . That derivative appears in the framework instead, as the relevant equation under essential knowledge 1.3.A.4.i, which says an object's instantaneous velocity is the rate of change of the object's position and is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time. The rotational block of the same sheet does print its derivatives, $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ . So the sheet is not uniform about this, and knowing that the linear derivative is a definition you supply rather than a line you copy is worth a few seconds on exam day.

None of the three statements in Topic 3.1 mentions calculus. That is the point. The calculus in this topic is entirely upstream of the formula, in getting $v$ , and it is why an AP Physics C question can hand you $x(t)$ as a polynomial and still be a Topic 3.1 question. Worked example 1 below is exactly that shape.

One more line that is not on either sheet: $K = \dfrac{p^2}{2m}$ . Both the AP Physics 1 and the AP Physics C: Mechanics sheets print $K = \frac{1}{2}mv^2$ and $\vec{p} = m\vec{v}$ separately, and neither prints the combination. It follows in one line, since $p = mv$ gives $p^2 = m^2v^2$ and so $\frac{p^2}{2m} = \frac{1}{2}mv^2$ , and it is genuinely useful in Unit 4 collision work. Derive it; do not go looking for it in the appendix.

Kinetic energy is a scalar (3.1.A.2), and Unit 3 leans on that

Statement 3.1.A.2 is one sentence, and it earns its place because almost everything else on the same equation sheet is a vector. Momentum is printed as $\vec{p} = m\vec{v}$ with an arrow. Force, impulse, and the position of the center of mass all carry arrows. Kinetic energy does not.

What that buys you:

Kinetic energies add as numbers, whatever the directions. Two 3.0 kg carts approaching each other at 4.0 m/s each have zero total momentum and 48 J of total kinetic energy. The momentum cancels because it is a vector sum; the kinetic energy cannot cancel because $v^2$ is never negative.
A sign convention cannot make $K$ negative. If you declare leftward negative and get a negative kinetic energy, the arithmetic is wrong. This is the cheapest self-check in the unit.
There is no "kinetic energy in the $x$ direction." You may resolve a velocity into components and you may write $K = \frac{1}{2}m(v_x^2 + v_y^2)$ , but the two terms are bookkeeping, not vector components of $K$ . Adding them is ordinary addition, not vector addition.
Work, which changes $K$ , is also a scalar, and it does carry a sign. That is not a contradiction. $K$ is a store and is never negative; work is a transfer and can go either way. Topic 3.2 states that at 3.2.A.2: work is a scalar quantity that may be positive, negative, or zero.

The word translational is also doing work in the title. It is there to leave room for Unit 6, where a rigid body spinning about its center of mass has rotational kinetic energy as well, and the total is the sum of the two. See Topic 6.1 for that. In Unit 3 every object is treated as a point or as a system whose center of mass carries the motion, so $\frac{1}{2}mv^2$ is the whole of the kinetic energy.

Two observers, two kinetic energies (3.1.A.3)

Statement 3.1.A.3 says different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference. It is the only statement in Topic 3.1 that is not a definition, and it is the one that generates questions.

The reason is that $v$ in $K = \frac{1}{2}mv^2$ is a velocity relative to something. Change the something and you change the number. Because the speed is squared, the change is not proportional: a frame shift that adds a modest amount to a speed can multiply the kinetic energy many times over. Worked example 2 puts numbers on that.

The exam constrains which frames are in play. The conventions box printed on the AP Physics C: Mechanics Table of Information has three bullets, and the first is: the frame of reference of any problem is assumed to be inertial unless otherwise stated. (The AP Physics 1 conventions box prints those same three bullets plus a fourth about fluids and pipes, which the C course does not need.) So unless a problem says otherwise, you are working in a frame where Newton's laws hold, and any second frame you introduce should be moving at constant velocity relative to the first.

Three consequences that show up in problems:

Kinetic energy is frame-dependent, and so is the change in it. If $K$ differs between frames, then $\Delta K$ can differ too, and by the work-energy theorem at 3.2.A.4 the net work done on the object differs as well. Two observers can disagree about how much work was done and both be right.
Zero kinetic energy is not an absolute state. A book on a table has $K = 0$ for you and a very large $K$ for an observer moving past at speed. Neither of you is making an error.
Conservation still works in every inertial frame. If mechanical energy is conserved for one inertial observer, it is conserved for every other inertial observer, at a different numerical total. Pick one frame, declare it, and stay in it for the whole problem. Switching frames mid-solution is a defect, not a shortcut.

The C course gives reference frames their own topic, Topic 1.4, Reference Frames and Relative Motion, so 3.1.A.3 is a callback rather than a new idea.

The traps, and where the marks actually go

$K$ goes as $v^2$ , not as $v$ . Double the speed and the kinetic energy quadruples. This single fact is the whole of sample multiple-choice question 11 in the CED, which is aligned to Topic 3.2 and skill 2.D but is answered by the $v^2$ dependence: two identical blocks are dropped from rest from different heights, Block 2 arrives at the floor with twice the speed of Block 1, and you are asked for the ratio of the work done by gravity. The published answer is $4W_1$ . If you reason with $v$ rather than $v^2$ you get $2W_1$ , which is one of the distractors.

Kinetic energy is not conserved on its own. "Conservation of energy" is about a total, not about $K$ . A pendulum bob has its kinetic energy vary continuously through every swing. Only in an elastic collision, and only for the collision itself, does the total kinetic energy come out unchanged.

Mass matters linearly, speed matters quadratically. Two objects with the same kinetic energy do not have the same momentum. Worked example 3 shows the ratio.

Where the marks are. Sample free-response Question 1, the Mathematical Routines question worth 10 points, aligns to learning objectives 3.1.A, 3.3.A and 3.4.B. Its part A gives a graph of the potential energy of a two-object system and releases Object A from rest, and one of the five scoring points in that part reads: for indicating that either the velocity of Object A or the kinetic energy of Object A is zero when Object A is at position $x = 3x_0$ . That is a full point of a 10-point question for recognising that released from rest means $K_0 = 0$ .

Sample free-response Question 3, the Experimental Design and Analysis question, also worth 10 points, aligns to 3.1.A alongside 2.1.B, 3.4.B, 3.4.C, 4.2.B and 4.4.A. So both of the sample free-response questions that touch this topic use it inside a longer argument rather than asking for a kinetic energy directly.

That is the realistic exam picture for Topic 3.1: you will almost never be asked "what is the kinetic energy." You will be asked for something else, and $\frac{1}{2}mv^2$ will be one line of your derivation.

How this topic connects to the rest of Unit 3

Topic 3.1 defines the quantity; the other four topics of the unit are about changing it, storing it, and timing it.

Topic 3.2, Work supplies the work-energy theorem, $\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$ , which is printed on the sheet. This is the only equation in the course that changes a $K$ .
Topic 3.3, Potential Energy is where the calculus-based course pulls away from the algebra-based one hardest, with $F_x = -dU/dx$ and $\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$ both printed.
Topic 3.4, Conservation of Energy turns $K$ and $U$ into one conserved total, given the right choice of system.
Topic 3.5, Power is the rate at which any of this happens.

Unit 3 as a whole is weighted 15 to 25% of the multiple-choice section of the AP Physics C: Mechanics exam, at about 12 to 17 class periods. AP Physics 1's Unit 3, with the same five topic titles, is weighted 18 to 23% at about 22 to 27 class periods. The calculus-based course spends roughly half the classroom time on the same list of topic titles, which is the clearest single statement of how much prior fluency it assumes.

For the arithmetic itself, the kinetic energy calculator checks a $\frac{1}{2}mv^2$ and its inverse. For the routine that turns work into a speed, the work-energy theorem guide owns that procedure, and the conservation of energy guide owns the energy-accounting one. Both were written for the constant-force case that both courses share. The Unit 3 hub lists every equation in the unit and which of them the sheet prints.

Kinetic energy from a position function

A 4.0 kg object moves along the $x$ axis with position $x(t) = \alpha t^3 - \beta t$ , where $\alpha = 0.50\ \mathrm{m/s^3}$ , $\beta = 2.0\ \mathrm{m/s}$ and $t$ is in seconds. Find (a) the kinetic energy at $t = 0$ , (b) the time at which the kinetic energy is zero, (c) the kinetic energy at $t = 3.0$ s, (d) the net work done on the object between $t = 0$ and $t = 3.0$ s, and (e) another instant at which the kinetic energy equals its value at $t = 0$ .

Declare the convention: positive $x$ is the positive direction, and every velocity below is measured in the frame in which $x(t)$ was given. Statement 3.1.A.3 means that choice has to be stated once and then kept.
Differentiate to get the velocity, using $v_x = dx/dt$ : $v_x(t) = 3\alpha t^2 - \beta = 1.5t^2 - 2.0$ in metres per second.
(a) At $t = 0$ : $v_x = -2.0$ m/s. Then $K = \frac{1}{2}(4.0)(-2.0)^2 = \frac{1}{2}(4.0)(4.0) = 8.0$ J. The velocity is negative and the kinetic energy is positive, which is 3.1.A.2 in one line.
(b) $K = 0$ requires $v_x = 0$ : $1.5t^2 = 2.0$ , so $t^2 = 1.3333\ \mathrm{s^2}$ and $t = 1.1547$ s, or 1.15 s.
(c) At $t = 3.0$ s: $v_x = 1.5(9.0) - 2.0 = 13.5 - 2.0 = 11.5$ m/s. Then $K = \frac{1}{2}(4.0)(11.5)^2 = 2.0(132.25) = 264.5$ J, or 265 J to three figures.
(d) The work-energy theorem of 3.2.A.4 gives the net work as the change in kinetic energy: $W_{\text{net}} = 264.5 - 8.0 = 256.5$ J, or 257 J. No force function was needed, and none was given.
(e) $K$ returns to 8.0 J when $\lvert v_x \rvert = 2.0$ m/s again, this time positive: $1.5t^2 - 2.0 = +2.0$ gives $t^2 = 2.6667\ \mathrm{s^2}$ and $t = 1.6330$ s, or 1.63 s.
Check that last one: $1.5(2.6667) - 2.0 = 4.0 - 2.0 = 2.0$ m/s, and $\frac{1}{2}(4.0)(2.0)^2 = 8.0$ J. Two instants, 1.63 s apart, opposite velocities, one kinetic energy.
Note what an algebra-based course could not have posed here. The acceleration $a_x = dv_x/dt = 3.0t$ is not constant, so no kinematic equation of the $v^2 = v_0^2 + 2a\Delta x$ family applies at any point in this problem.

(a) $K = 8.0$ J. (b) $K = 0$ at $t = 1.15$ s. (c) $K = 265$ J at $t = 3.0$ s. (d) The net work done is 257 J. (e) $K = 8.0$ J again at $t = 1.63$ s, where the velocity is $+2.0$ m/s rather than $-2.0$ m/s.

Two observers disagree, and both are right

A train travels in a straight line at a constant 25 m/s relative to the ground. A 60 kg passenger walks along the aisle at 1.5 m/s relative to the train, in the direction the train is moving. Find the passenger's kinetic energy (a) in the train's frame and (b) in the ground frame. (c) The passenger turns around and walks at 1.5 m/s toward the back of the train. Find the change in kinetic energy in each frame, and say what that implies about the net work done on the passenger.

Both frames are inertial: the train's velocity is constant, so the conventions box on the equation sheet is satisfied and Newton's laws hold in both.
(a) In the train's frame, $v = 1.5$ m/s. $K_{\text{train}} = \frac{1}{2}(60)(1.5)^2 = 30(2.25) = 67.5$ J.
(b) In the ground frame, velocities add: $v = 25 + 1.5 = 26.5$ m/s. $K_{\text{ground}} = \frac{1}{2}(60)(26.5)^2 = 30(702.25) = 21067.5$ J, or $2.11 \times 10^4$ J.
The ratio is $21067.5 / 67.5 = 312.1$ . A 6% change in speed produced a factor of about 312 in kinetic energy, because the frame shift changed what is being squared.
(c) Walking toward the back: in the train's frame $v = -1.5$ m/s, so $K_{\text{train}} = 67.5$ J again and $\Delta K_{\text{train}} = 0$ .
In the ground frame $v = 25 - 1.5 = 23.5$ m/s, so $K_{\text{ground}} = 30(552.25) = 16567.5$ J and $\Delta K_{\text{ground}} = 16567.5 - 21067.5 = -4500$ J.
By the work-energy theorem, the net work done on the passenger is zero according to the train observer and $-4500$ J according to the ground observer. Neither has made a mistake. Work is the transfer of energy across a frame-dependent displacement, so it inherits the frame dependence of 3.1.A.3.
The practical rule that follows: choose one frame at the start of a problem, state it, and do every energy in that frame. Mixing a speed measured in one frame with a displacement measured in another is the error 3.1.A.3 exists to warn you about.

(a) 67.5 J in the train's frame. (b) $2.11 \times 10^4$ J in the ground frame, about 312 times larger. (c) Turning around changes the kinetic energy by 0 in the train's frame and by $-4500$ J in the ground frame, so the two observers also disagree about the net work done, and both are correct within their own inertial frame.

Same kinetic energy, different momentum

A 2.0 kg cart and an 8.0 kg cart each have 36 J of translational kinetic energy and both move in the $+x$ direction. Find each speed and each momentum, and give the ratio of the momenta as a formula in the two masses.

Invert $K = \frac{1}{2}mv^2$ : $v = \sqrt{2K/m}$ .
Light cart: $v_1 = \sqrt{2(36)/2.0} = \sqrt{36} = 6.0$ m/s.
Heavy cart: $v_2 = \sqrt{2(36)/8.0} = \sqrt{9.0} = 3.0$ m/s.
Momenta, from $\vec{p} = m\vec{v}$ : $p_1 = (2.0)(6.0) = 12$ kg m/s and $p_2 = (8.0)(3.0) = 24$ kg m/s, both in the $+x$ direction.
Symbolically, $p = mv = m\sqrt{2K/m} = \sqrt{2Km}$ , so at equal $K$ the momenta go as the square root of the mass: $p_2/p_1 = \sqrt{m_2/m_1} = \sqrt{8.0/2.0} = 2$ , which matches $24/12$ .
That relation, $K = p^2/2m$ , is not printed on the AP Physics C: Mechanics sheet or on the AP Physics 1 sheet. Both print $K = \frac{1}{2}mv^2$ and $\vec{p} = m\vec{v}$ and leave the combination to you.
The physical reading: equal kinetic energy does not mean equal momentum, and the heavier object always carries more momentum at equal kinetic energy. That is why a collision question and an energy question about the same pair of carts can have opposite-looking answers.

The 2.0 kg cart moves at 6.0 m/s with 12 kg m/s of momentum; the 8.0 kg cart moves at 3.0 m/s with 24 kg m/s. At equal kinetic energy, $p = \sqrt{2Km}$ , so the momenta are in the ratio $\sqrt{m_2/m_1} = 2$ .

Frequently asked questions

What is the formula for translational kinetic energy in AP Physics C?

Translational kinetic energy is $K = \frac{1}{2}mv^2$ , where $m$ is the object's mass and $v$ is its speed. Essential knowledge 3.1.A.1 of the AP Physics C: Mechanics course and exam description states it in exactly those terms, and the equation is printed on the AP Physics C: Mechanics equation sheet. The unit is the joule. The formula is the same in AP Physics 1, AP Physics 2 and both AP Physics C courses. What differs in the calculus-based course is that the speed you put into it is often something you have to produce first, by differentiating a position function or integrating an acceleration function.

Is AP Physics C Topic 3.1 different from AP Physics 1 Topic 3.1?

The required content is identical. The learning objective sentence and all three essential knowledge statements, 3.1.A.1 on the equation, 3.1.A.2 on kinetic energy being a scalar, and 3.1.A.3 on frame dependence, are word for word the same in both course and exam descriptions, and neither course prints a boundary statement for this topic. The one difference in the framework is in the suggested skills: AP Physics 1 lists skill 2.B, calculate an unknown quantity, while AP Physics C: Mechanics lists skill 2.C, compare quantities between scenarios. In the CED's own sample questions, learning objective 3.1.A appears in two of the four sample free-response questions and in none of the fifteen sample multiple-choice questions.

Is kinetic energy a vector or a scalar?

Kinetic energy is a scalar. Essential knowledge 3.1.A.2 states this directly. It has a magnitude and a unit but no direction, so it is never resolved into components and it is never negative, because the speed is squared. Two objects moving in opposite directions have kinetic energies that add rather than cancel, which is the opposite of what happens to their momenta: momentum is a vector and can cancel to zero for the same pair. If a sign convention ever produces a negative kinetic energy, the arithmetic is wrong. Work, the quantity that changes kinetic energy, is also a scalar but does carry a sign, because it is a transfer rather than a store.

Why do different observers measure different kinetic energies?

Because speed is measured relative to a frame of reference, and kinetic energy depends on the square of that speed. Essential knowledge 3.1.A.3 states that different observers may measure different values of an object's translational kinetic energy depending on the observer's frame of reference. A passenger walking at 1.5 m/s along a train that moves at 25 m/s has a small kinetic energy in the train's frame and a very large one in the ground frame, because 26.5 squared is far larger than 1.5 squared. The change in kinetic energy is frame-dependent too, which means two inertial observers can disagree about how much net work was done and both be correct. The AP Physics C exam conventions box assumes every frame is inertial unless a problem states otherwise.

Do I need calculus for kinetic energy in AP Physics C?

Not for the formula itself, which is the same one half m v squared as in the algebra-based course. You need calculus for the step before it. AP Physics C: Mechanics can give you position as a function of time and expect you to differentiate to get the velocity, or give you acceleration as a function of time and expect you to integrate. The equation sheet supports the second route directly, printing the change in velocity as the integral of acceleration over time, while the derivative form for velocity appears in the framework at essential knowledge 1.3.A.4.i rather than on the sheet. Once you have the speed, the kinetic energy step is ordinary arithmetic.

What happens to kinetic energy if you double an object's speed?

It quadruples, because kinetic energy depends on the square of the speed. Halving the speed leaves one quarter of the kinetic energy, and tripling it gives nine times as much. The CED's sample multiple-choice question 11 is built on exactly this dependence: two identical blocks are dropped from rest from different heights, the second arrives with twice the speed of the first, and the published answer for the work done by gravity on the second block is four times the work done on the first. Mass, by contrast, enters linearly, so doubling the mass at fixed speed only doubles the kinetic energy.

Is K = p squared over 2m on the AP Physics C equation sheet?

No. Neither the AP Physics C: Mechanics sheet nor the AP Physics 1 sheet prints that combination. Both print the kinetic energy as one half m v squared and the momentum as mass times velocity on separate lines, and the relation between them is a one-line derivation you are expected to perform: since p equals m v, p squared equals m squared v squared, so p squared divided by 2m equals one half m v squared. It is worth knowing because it turns a momentum-conservation result straight into an energy comparison, which is a common shape for a question that spans Unit 3 and Unit 4.