Questionnaire 1.2.3 — Elementary Classical Mechanics

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.

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.

flowchart TD
    q["Where is the energy stored?"] --> motion["In the motion: v ≠ 0"]
    q --> config["In the position/configuration<br/>(height, spring stretch, charge separation)"]
    motion --> ke["KINETIC energy ½mv²<br/>extract it by STOPPING the body"]
    config --> pe["POTENTIAL energy (e.g. mgh)<br/>extract it by RELEASING the body"]
    ke -.->|"the statement describes this…"| trap["…but names it 'potential' ✗"]

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.

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.

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.