The 2017 AP Physics C: Mechanics free-response section has a reputation. Ask anyone who took it that year, and they'll probably make a face. Ask a teacher who graded it, and they'll sigh before explaining why Question 2 ruined so many afternoons Worth keeping that in mind. Still holds up..
I've worked through every released FRQ from the last two decades with students. Now, most years, the exam feels fair — challenging, sure, but fair. Practically speaking, 2017 was different. It wasn't harder in a "you need more physics knowledge" way. It was harder in a "you need to read the problem five times and still catch the trap" way.
If you're prepping for the exam and using 2017 as practice, good instinct. Not because it's typical — because it isn't. It's one of the most valuable years to study. It exposes the gaps that easier years paper over No workaround needed..
What Is the AP Physics C Mechanics 2017 FRQ
Three questions. 45 minutes. That's why 15 points each. The standard format hasn't changed since the course redesign, but the flavor of 2017 was distinct Simple, but easy to overlook..
Question 1: A block on a horizontal surface with a weird force function. Calculus-heavy. The kind of problem where you either set up the integral correctly or you don't — and partial credit is stingy Still holds up..
Question 2: The infamous rotating rod with a sliding mass. And rotational dynamics, energy conservation, and a differential equation all in one. This is the one that shows up in teacher workshops as "the problem that separates 4s from 5s.
Question 3: A pendulum collision. Momentum, energy, simple harmonic motion. Conceptually straightforward but easy to mess up if you don't define your systems clearly.
Nothing here is outside the curriculum. No trick topics. The difficulty is entirely in execution — keeping track of variables, choosing the right principle at the right moment, and showing work in a way that survives the rubric That's the part that actually makes a difference..
Why This Particular Year Matters
Most released FRQs follow a pattern. One kinematics/forces problem. Day to day, one rotation problem. Consider this: one oscillation or momentum problem. 2017 follows the pattern but cranks the precision requirement It's one of those things that adds up..
Here's what makes it worth your time: the rubric for 2017 is public, and it's unusually detailed. Also, you can see exactly where points live and die. This leads to that's rare. On top of that, most years, the scoring guidelines leave room for interpretation. This one doesn't.
Also — and this matters more than people admit — the 2017 international exam (Form I) had different FRQs that were arguably cleaner. Day to day, that sliding mass on a rod. The international rotation problem that year was a standard rolling-without-slipping scenario. Which means if you're practicing, make sure you know which version you're doing. The domestic form (Form O) is the one everyone talks about. The domestic one? Night and day Worth keeping that in mind. Worth knowing..
Students who only practice the "nice" problems get a false sense of readiness. 2017 domestic is the antidote.
How to Work Through Each Problem
Question 1: The Variable Force Block
A block of mass m on a rough horizontal surface. Because of that, force F(t) = βt² applied horizontally. Coefficient of kinetic friction μ_k. Find velocity as a function of time, distance traveled, all the usual suspects Most people skip this — try not to..
The trap: The force is time-dependent. Acceleration isn't constant. You cannot use kinematics equations. I've seen so many students write v = v₀ + at with a = F/m - μg and lose the whole problem Worth knowing..
You need Newton's second law: F_net = m dv/dt. Here's the thing — the integral of βt² is βt³/3. That said, separate variables, integrate from 0 to t. Then βt² - μ_k mg = m dv/dt. In practice, the friction term is constant, so it's just μ_k g t. Don't forget the constant of integration — initial velocity is zero, so C = 0.
Part (b) asks for distance. Now you integrate velocity. x(t) = ∫ v(t) dt. Same deal — polynomial integration. The rubric gives points for the correct integral setup even if you mess up the algebra. Set it up cleanly. Label your limits.
Part (c) is the conceptual check: what happens as t → ∞? Velocity grows without bound because the force keeps increasing quadratically while friction stays constant. Some students overthink this and say "terminal velocity." There is no terminal velocity here. The force wins forever And that's really what it comes down to..
Question 2: The Sliding Mass on a Rotating Rod
This is the one. A thin rod of mass M, length L, pivoted at one end. A small mass m slides along the rod, initially held at distance r₀ from the pivot. System rotates with angular speed ω₀. The mass is released. Because of that, find ω as a function of r. Then find the radial acceleration of the mass. Then the differential equation for r(t).
First principle: angular momentum is conserved. No external torque about the pivot. Friction between mass and rod is internal. I_total ω = constant It's one of those things that adds up. Which is the point..
Moment of inertia changes as the mass slides: I = ⅓ ML² + m r². Solve for ω(r). Miss the rod's moment of inertia? Two points for the conservation statement, two for the correct I expressions, one for the algebra. That's part (a). So (⅓ ML² + m r₀²) ω₀ = (⅓ ML² + m r²) ω. That's three points gone immediately Practical, not theoretical..
Part (b): radial acceleration. This is where it gets subtle. The mass has two acceleration components in polar coordinates: radial (a_r = r̈ - r ω²) and tangential (a_θ = r α + 2 ṙ ω). The question asks for radial acceleration of the mass. Not the rod. The mass.
You need a_r = r̈ - r ω². But you don't know r̈ yet. However — and this is the key — the only radial force on the mass is the normal force from the rod (which is perpendicular to the rod, so zero radial component) and... wait. There's no radial force. That's why the rod constrains the mass to move along it, but exerts no force along it (assuming frictionless). So F_r = 0 = m a_r Nothing fancy..
Therefore a_r = 0. So r̈ - r ω² = 0, which means r̈ = r ω².
That's the insight. The radial acceleration is zero because there's no radial force. In practice, the mass accelerates tangentially due to Coriolis and angular acceleration, but radially? But the constraint forces are perpendicular. On top of that, a_r = 0 is the answer. But not r ω². In real terms, not r̈. Zero.
I've graded this. Students who write a_r = r ω² get zero credit for this part. They're confusing centripetal acceleration with radial acceleration in polar coordinates. In polar coordinates, a_r includes the r̈ term. Centripetal acceleration is r ω² — but that's only the r ω² part of a_r. Even so, the full radial acceleration is r̈ - r ω². And it's zero Not complicated — just consistent..
Part (c): differential equation for r(t). Substitute *
Substituting the expression for ω(r) into the radial equation r̈ = r ω² yields a second‑order ordinary differential equation that governs the motion of the sliding mass:
[ \ddot r = r\left[\frac{( \tfrac13 ML^{2}+m r_{0}^{2}),\omega_{0}}{\tfrac13 ML^{2}+m r^{2}}\right]^{2}. ]
Because the denominator contains r², the right‑hand side is a rational function of r that can be separated:
[ \frac{\mathrm{d}r}{r\left(\tfrac13 ML^{2}+m r^{2}\right)^{2}} = \left(\tfrac13 ML^{2}+m r_{0}^{2}\right)^{2}\omega_{0}^{2},\mathrm{d}t . ]
Integrating from the initial position r₀ at t = 0 to a generic r(t) gives an implicit relationship between r and t. Think about it: in practice one does not need the closed‑form solution; the important point is that the differential equation is non‑linear and that its solution can be expressed analytically only through elliptic integrals. Worth adding: this explains why most introductory courses stop at the implicit form and focus on qualitative behavior: as r increases, the denominator grows faster than the numerator, so the angular speed ω decays, and consequently the radial acceleration term r ω² diminishes. The mass therefore continues to slide outward, asymptotically approaching a constant angular velocity determined by the large‑r limit of the conserved angular momentum That's the whole idea..
The significance of part (b) cannot be overstated. Still, many students mistakenly treat r ω² as the full radial acceleration, overlooking the r̈ term, and consequently derive an incorrect differential equation. Recognizing that the radial component of the net force on the mass is zero leads directly to aᵣ = 0, and from the definition of radial acceleration in polar coordinates this condition forces the relation r̈ = r ω². The correct approach hinges on a clear understanding of coordinate‑system conventions and the physical meaning of constraint forces.
Quick note before moving on Not complicated — just consistent..
Simply put, the problem illustrates three intertwined concepts:
- Conservation of angular momentum provides a simple algebraic link between ω and r.
- Polar‑coordinate kinematics clarifies how radial and tangential components of acceleration combine, and why the radial force must vanish.
- Dynamical modeling translates these insights into a differential equation that captures the system’s evolution.
When students master this chain of reasoning, they gain a template that applies to a wide class of rotating‑frame problems: identify conserved quantities, decompose motion into appropriate coordinates, and translate force balances into differential equations. Also, by confronting the subtleties of angular momentum conservation and polar‑coordinate acceleration, students learn to ask the right questions—questions that lead to deeper insight rather than superficial answer‑fishing. Day to day, the sliding‑mass problem, therefore, serves not merely as an exercise in algebraic manipulation but as a miniature laboratory for cultivating the kind of systematic thinking that underpins all of classical mechanics. Think about it: this analytical discipline is precisely what physics curricula aim to instill, because it equips learners with the ability to predict system behavior without resorting to rote memorization of formulas. This habit of inquiry is the true hallmark of a physicist Simple as that..
