# Questionnaire 1.2.3 — Elementary Classical Mechanics > Five questions on momentum and energy: momentum as a vector, kinetic vs. potential energy as > operational definitions, why kinetic energy is never negative, what actually raises potential > energy, and which quantities grow at a constant rate under constant acceleration. ## Key Ideas **Momentum** is the vector $\vec{p} = m\vec{v}$ — it carries a magnitude _and_ a direction. Any change of the velocity vector changes it, including a pure change of direction at constant speed: a body in uniform circular motion has constant $\lvert\vec{p}\rvert$ but a momentum that never stops changing, which is exactly why circular motion requires a force ($\vec{F} = d\vec{p}/dt$, Newton's second law in its momentum form). **Kinetic energy** is the energy a body has because it is moving — operationally, the energy you could extract from it by bringing it to rest: $$ E_k = \tfrac{1}{2}mv^2 = \frac{p^2}{2m} $$ It is a scalar built from $v^2 = \vec{v}\cdot\vec{v} \ge 0$, so in classical mechanics kinetic energy can never be negative, whatever the direction of motion. **Potential energy** is energy stored in a body's _position or configuration_ inside a force field, not in its motion. Near the Earth's surface, $\Delta E_p = mg\,\Delta h$, where $\Delta h$ is the net rise of the body's **center of mass**. Potential energy is a _state function_: it depends only on where you end up relative to where you started, not on the path taken — climbing up and coming back down leaves it unchanged. The **work–energy theorem** says a net force changes kinetic energy at a constant rate _per unit distance_: $\Delta E_k = F\,\Delta x$. Per unit _time_ the rate is the **power** $P = dE_k/dt = Fv$, which grows as the body speeds up. Under constant acceleration, momentum grows linearly in time ($dp/dt = F = \text{const}$) while kinetic energy grows quadratically ($E_k = \tfrac{1}{2}m(at)^2$ from rest) — "constant rate" is true for momentum per second and for energy per metre, but not for energy per second. ## Exercises **E1 (easy).** True or false: to change my momentum, I must either slow down or speed up (assuming my mass stays constant).

Solution

**Answer: False.** **Why, step by step:** 1. Momentum is the vector $\vec{p} = m\vec{v}$, so it changes whenever the velocity _vector_ changes — in magnitude **or** in direction. 2. A car rounding a bend at a steady 50 km/h neither slows down nor speeds up, yet its momentum changes continuously because $\vec{v}$ keeps rotating. 3. Newton's second law makes the same point dynamically: $\vec{F} = d\vec{p}/dt$. Uniform circular motion needs a (centripetal) force, and a force means momentum is changing — at constant speed. **Why the tempting reading fails:** - "True" treats momentum as $m \times \text{speed}$, a scalar. That silently discards the direction, which is half of what momentum is. **See it:** the body below never speeds up or slows down, yet drag the snapshot separation apart and $\lvert\Delta\vec{p}\rvert$ grows anyway — direction change alone does it. ```qm-demo circular-momentum ```

**E2 (easy).** True or false: potential energy is the amount of energy we could get out of a moving body by stopping it.

Solution

**Answer: False.** **Why, step by step:** 1. "Energy extracted by stopping a moving body" is the operational definition of **kinetic** energy: the work–energy theorem gives exactly $\tfrac{1}{2}mv^2$ when a body of speed $v$ is brought to rest. 2. Potential energy has nothing to do with the body currently moving — it is energy stored in the body's _position or configuration_ in a force field (height in gravity, stretch of a spring, separation of charges). 3. The operational test for potential energy is the opposite one: hold the body still, _release_ it, and see how much work the field does on it as it moves toward lower potential. **Why the tempting reading fails:** - The statement is word-for-word a correct definition — of the _other_ energy. The quiz swaps the labels and checks whether you classify energy by where it is stored (motion vs. position), not by formula memorization. **See it:** classify any energy bookkeeping question by asking where the energy sits right now. ```mermaid flowchart TD q["Where is the energy stored?"] --> motion["In the motion: v ≠ 0"] q --> config["In the position/configuration
(height, spring stretch, charge separation)"] motion --> ke["KINETIC energy ½mv²
extract it by STOPPING the body"] config --> pe["POTENTIAL energy (e.g. mgh)
extract it by RELEASING the body"] ke -.->|"the statement describes this…"| trap["…but names it 'potential' ✗"] ```

**E3 (easy).** True or false: in classical mechanics, kinetic energy can never be negative.

Solution

**Answer: True.** **Why, step by step:** 1. $E_k = \tfrac{1}{2}mv^2$ with $m > 0$ and $v^2 = \vec{v}\cdot\vec{v} \ge 0$ — the square of a real speed cannot be negative, so $E_k \ge 0$ always. 2. Direction is irrelevant: a body moving at $-5\ \text{m/s}$ (leftward) has $v^2 = 25\ \text{m}^2/\text{s}^2$, the same kinetic energy as one moving rightward. 3. Slowing down means kinetic energy _decreases toward_ zero — the change $\Delta E_k$ can be negative, but the value itself bottoms out at rest, $E_k = 0$. **Why the tempting reading fails:** - Negative velocities and decelerations invite "negative energy" intuitions, but squaring kills the sign, and a _decreasing_ positive quantity is still positive. - The qualifier "in classical mechanics" is there for a reason: later, in quantum mechanics, a particle can be found in regions where $E < V$ — where a naive "local kinetic energy" $E - V$ would be negative. Classically that region is simply forbidden. **See it:** drag $v$ leftward through zero. The body reverses and the marker climbs the far branch to the same height — then watch it refuse to cross the floor no matter what you do. ```qm-demo kinetic-energy-floor ```

**E4 (medium).** Which of the following will increase my potential energy? (Choose all correct answers) a) I am pulled across the room from one side to the other. b) I climb up the stairs and back down again. c) I jump into a swimming pool from a diving board above the pool. d) While otherwise standing still, I raise my hand to wave at my friend.

Solution

**Answer: d) only.** **Why, step by step:** 1. Near the Earth's surface, gravitational potential energy is $E_p = mgh$ evaluated at the **center of mass**, so the only question that matters is: _did my center of mass end up higher than it started?_ ($\Delta E_p = mg\,\Delta h_{\text{cm}}$.) 2. **a)** Being pulled across the room is horizontal: $\Delta h = 0$, so $\Delta E_p = 0$. Whatever work the pulling does goes into kinetic energy and friction, not storage. 3. **b)** Up the stairs and back down ends at the starting height: $\Delta h = 0$ net. Potential energy is a state function — it forgets the path, so the climb's gain is exactly repaid on the way down. 4. **c)** Jumping from a board _above_ the pool into the water moves the center of mass _down_: $\Delta h < 0$, so potential energy **decreases** (it converts to kinetic energy on the way down). 5. **d)** Raising an arm while otherwise standing still lifts part of your mass, so the center of mass rises slightly — $\Delta h_{\text{cm}} > 0$, hence $\Delta E_p > 0$. Small (a few-kilogram arm raised half a metre stores ~$10$–$20$ J), but unambiguously an increase. **Why the tempting options fail:** - **a)** feels like work is being done on you — it is, but work done _against friction or into motion_ is not stored as gravitational potential energy; only net height gain is. - **b)** tempts via "I burned energy climbing" — your muscles did, but $E_p$ tracks the state, not the effort; ending where you started means no net change. - **c)** plants the word "above the pool" to make height salient — but the _change_ is a drop, and the sign of $\Delta h$ is all that counts. **See it:** step the slider through all four paths and watch where each one leaves the dashed start line — the climb hands its joules straight back, and only the hand-wave ends above it. ```qm-demo center-of-mass-height ```

**E5 (medium).** I accelerate with constant acceleration (or, equivalently, with a constant rate of acceleration). (Here, by "accelerate" we mean we keep getting faster, going in a particular direction, just like accelerating in a car on a straight road.) Which statement is true? a) Momentum keeps increasing at a constant rate b) Kinetic energy keeps increasing at a constant rate c) Both (a) and (b) are TRUE d) Both (a) and (b) are FALSE

Solution

**Answer: a) Momentum keeps increasing at a constant rate.** **Why, step by step:** 1. Constant acceleration with constant mass means $\dfrac{dp}{dt} = m\dfrac{dv}{dt} = ma = F = \text{const}$ — momentum gains the same amount every second. Statement (a) is true. 2. Kinetic energy from rest is $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m(at)^2$ — quadratic in time. Its time rate is the power, $\dfrac{dE_k}{dt} = Fv$, which keeps growing because $v$ does. Each second adds more energy than the last, so (b) is false. 3. Concretely, with $m = 2\ \text{kg}$ and $a = 1\ \text{m/s}^2$: momentum steps $2, 2, 2, \ldots\ \text{kg·m/s}$ each second, while kinetic energy steps $1, 3, 5, 7, \ldots\ \text{J}$. 4. With (a) true and (b) false, options (c) and (d) fall automatically. **Why the tempting options fail:** - **b)** is the work–energy theorem misread: $\Delta E_k = F\,\Delta x$ does grow at a constant rate — _per metre_. But at constant acceleration you cover more metres each second, so the per-second energy gain grows. An unqualified "rate" means per unit time. - **c)** follows if you accept both readings without checking units of "rate"; **d)** usually comes from doubting (a) because the speed keeps changing — but a steadily _changing_ $v$ is exactly what a constant $dp/dt$ looks like. **See it:** drag the elapsed time and compare the two "last second" brackets — the momentum step never changes, the energy step never stops growing. ```qm-demo constant-acceleration-energy ```