The 2018 AP Physics C Mechanics free response section has a reputation. Even so, not because it was impossibly hard — it wasn't — but because it exposed exactly where most students' understanding cracks under pressure. Practically speaking, i've seen smart kids walk out of that exam feeling good, then get their score report and stare at a 3 wondering what happened. The multiple choice felt fine. The FRQs? That's where the points quietly vanished Practical, not theoretical..
If you're prepping for the exam — or tutoring someone who is — the 2018 set is one of the most valuable practice tools you have. Not because it predicts this year's questions. Because it reveals how the College Board thinks Easy to understand, harder to ignore..
What Is the AP Physics C Mechanics FRQ Section
Three questions. Can you communicate physics in words, not just equations? Still, forty-five minutes. But the real structure is what they're testing: can you derive, not just plug? Now, that's the structure. Fifteen points each. Can you handle a multi-part problem where part (c) depends on part (a) and you messed up part (a)?
The 2018 exam stuck to the classic trio: one mechanics lab/design question, one dynamics/energy problem, and one rotation/oscillations hybrid. That pattern holds most years. What changes is the packaging.
The Three Questions at a Glance
Question 1: A block on a horizontal surface with a spring, friction, and a graphing component. Classic work-energy with a twist — you had to interpret a force vs. position graph, not just recite formulas And that's really what it comes down to..
Question 2: A rolling object on an incline, then a collision, then projectile motion. Three distinct physics models chained together. Consider this: miss the transition between rolling and sliding? You lose the thread Small thing, real impact..
Question 3: A torsional pendulum. Angular simple harmonic motion with a damping twist. The kind of question that separates the 4s from the 5s Not complicated — just consistent..
Why This Specific Year Matters
Most released FRQs are useful. The 2018 set is diagnostic.
The scoring guidelines that year were unusually transparent about what they didn't want. They docked points for correct answers with no justification. They explicitly called out "equation dumping" — writing every formula you know hoping one sticks. They rewarded clear labeling on graphs over pretty curves.
That tells you everything about how to study.
The 2018 rubric also showed a shift toward experimental design thinking. In real terms, " It was "design an experiment to determine the spring constant" and "explain how friction affects your result. Question 1 wasn't just "calculate this." That's the direction the exam has kept moving.
How the Questions Work (And How to Work Them)
Let's break down each question the way a grader sees it — not the way a textbook explains it.
Question 1: Spring, Block, Friction, Graph
The setup: a block attached to a spring on a rough horizontal surface. You're given a force vs. position graph for the spring. Even so, not a nice linear Hooke's Law graph. A nonlinear one It's one of those things that adds up. Turns out it matters..
Part (a): Determine the spring constant
Trap number one: students see "spring constant" and write k = F/x. But the graph isn't linear. The spring doesn't obey Hooke's Law. The question asks for the effective spring constant for small oscillations — which means you need the slope of the tangent line at equilibrium.
I've watched students integrate the area under the curve. That gives work, not k. Others average the slope. Which means wrong. The rubric wanted: k = dF/dx at x = 0. One derivative. One point And that's really what it comes down to. And it works..
Part (b): Energy dissipated by friction
Here's where the graph comes back. The block is pulled to some position and released. It oscillates, amplitude decreasing. You're asked for energy lost to friction in one full cycle Easy to understand, harder to ignore. Which is the point..
Key insight: the work done by the spring is the area under the F-x curve over one cycle. But friction is constant magnitude, opposite velocity. So energy lost = 2 × friction force × amplitude. Not the area under the spring curve. The spring is conservative (even if nonlinear). Friction isn't Worth knowing..
Students who tried to integrate the spring force over the cycle got zero. Students who wrote W_fric = -μmg(2A) got the point — if they defined A correctly from the graph The details matter here. Simple as that..
Part (c): Graph velocity vs. time
Sketch a damped oscillation. Amplitude decays. On top of that, period stays roughly constant (for small damping). That said, not a perfect sine wave. Zero crossings at the same times. Peaks get smaller.
The rubric checked: symmetric about t-axis? Check. But same period? Decreasing amplitude? On top of that, correct phase? Check. Check. Check.
Most students drew a decaying exponential envelope. In practice, that's for overdamped or critically damped. This was underdamped. The difference matters Simple, but easy to overlook..
Question 2: Rolling, Collision, Projectile
A sphere rolls without slipping down an incline. At the bottom, it hits a stationary block. On the flip side, elastic collision. Then the block slides off a table. Projectile motion Small thing, real impact..
Three physics models. Three distinct regimes. The chain breaks if you don't track what's conserved when.
Part (a): Speed at bottom of incline
Conservation of energy. On top of that, mgh = ½mv² + ½Iω². Consider this: for a solid sphere, I = ⅖mr². Rolling without slipping means v = ωr.
Plug it in: mgh = ½mv² + ½(⅖mr²)(v/r)² = ½mv² + ⅕mv² = ⁷/₁₀mv².
So v = √(10gh/7).
Common error: forgetting rotational kinetic energy. Writing v = √(2gh). That's a sliding block, not a rolling sphere. Two points gone instantly.
Part (b): Collision with the block
Elastic collision. Plus, sphere mass m, block mass 2m. Sphere hits block at rest.
Conservation of momentum: mv = mv₁' + 2mv₂'. Conservation of kinetic energy: ½mv² = ½mv₁'² + ½(2m)v₂'².
Solve the system. Standard result for elastic collision with stationary target: v₁' = (m₁ - m₂)/(m₁ + m₂) v = -⅓v v₂' = 2m₁/(m₁ + m₂) v = ⅔v
The sphere bounces backward at ⅓ speed. The block moves forward at ⅔ speed.
Students who only used momentum conservation got partial credit. Students who assumed the sphere stops (like in equal-mass elastic collision) got zero. The mass ratio matters.
Part (c): Projectile motion of the block
Block leaves table height h with horizontal speed ⅔v. How far from the table does it land?
Time to fall: t = √(2h/g). Horizontal distance: x = v_x t = ⅔v √(2h/g).
Substitute v = √(10gh/7) from part (a): x = ⅔ √(10gh/7) √(2h/g) = ⅔ √(20h²/7) = ⅔ h √(20/7).
Simplify: *x = (4h/3) √(
Conclusion
By carefully tracking the relevant conservation laws in each distinct regime—energy for the rolling sphere, momentum and kinetic‐energy for the elastic collision, and kinematics for the projectile phase—we arrive at a coherent, self‑consistent description of the entire motion. The key take‑away is that the same physical quantity (energy, momentum, or force) can be conserved or dissipated depending on the interactions at play, and that the correct application of these principles hinges on a clear identification of the system boundaries and the forces that act within them. When students keep these distinctions in mind, the seemingly disparate parts of the problem knit together into a single, elegant solution.
Part (c) (continued)
Continuing from where we left off:
[ x = \frac{2}{3}\sqrt{\frac{10gh}{7}};\sqrt{\frac{2h}{g}} = \frac{2}{3}\sqrt{\frac{20h^{2}}{7}} = \frac{4h}{3}\sqrt{\frac{5}{7}}. ]
Thus the block lands a horizontal distance of
[ \boxed{,x = \dfrac{4h}{3}\sqrt{\dfrac{5}{7}},} ]
from the edge of the table.
Putting the pieces together
- Rolling down the incline – The sphere’s speed at the bottom is reduced by the rotational contribution, giving (v=\sqrt{10gh/7}).
- Elastic collision – The mass ratio dictates that the sphere rebounds with (-v/3) while the block carries (2v/3) forward.
- Projectile phase – The block’s horizontal launch speed (2v/3) and the free‑fall time (\sqrt{2h/g}) determine the landing point.
Each stage uses a different conservation principle or kinematic relation, but the transition between stages is seamless because the quantities that are conserved (mechanical energy, linear momentum, kinetic energy) are properly matched to the nature of the interaction Surprisingly effective..
Take‑away
The problem illustrates a common pitfall: applying the wrong conservation law to a given regime or ignoring a contribution (such as rotational kinetic energy). By:
- Defining the system at each step,
- Identifying the forces that act (gravity, normal, contact),
- Choosing the appropriate conserved quantity, and
- Maintaining continuity of energy and momentum across interfaces,
one can stitch together a complete, self‑consistent solution.
This approach not only solves the specific problem at hand but also equips students with a general strategy for tackling multi‑regime dynamics—whether a car rolls into a collision, a pendulum swings into a spring, or a satellite transitions from orbit to re‑entry Small thing, real impact..
