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AP Physics C Mechanics Practice

AP Physics C — Mechanics Practice

16 MCQ practice problems. Select an answer, submit, and review the explanation.

Kinematics

Intuition: Kinematics is the geometry of motion. It describes how objects move (trajectory, velocity, acceleration) without worrying about why they move (forces).

Common Mistakes:

  • Confusing displacement (vector) with distance (scalar)
  • Using kinematic equations for non-constant acceleration
  • Forgetting that velocity and acceleration can have opposite signs

Free-Response Practice Problems

Problem 1: Projectile Motion with Air Resistance (15 points)

A projectile of mass mm is launched from the origin with initial speed v0v_0 at angle θ\theta above the horizontal. Assume air resistance is proportional to velocity: Fdrag=−bv\mathbf{F}_{\text{drag}} = -b\mathbf{v}.

(a) Write the differential equations of motion for the xx and yy components. (4 points)

(b) For the case b=0b = 0 (no air resistance), derive expressions for the time of flight, maximum height, and range. (6 points)

(c) Explain qualitatively how air resistance affects each of these quantities. (3 points)

(d) If bb is small, suggest an approximation method to find the corrected range. (2 points)

Solution:

(a) The equations of motion are:

mdvxdt=−bvx  ⟹  dvxdt=−bmvxm\frac{dv_x}{dt} = -bv_x \implies \frac{dv_x}{dt} = -\frac{b}{m}v_x mdvydt=−mg−bvy  ⟹  dvydt=−g−bmvym\frac{dv_y}{dt} = -mg - bv_y \implies \frac{dv_y}{dt} = -g - \frac{b}{m}v_y

(b) Without air resistance:

  • Time of flight: T=2v0sin⁡θgT = \frac{2v_0\sin\theta}{g}
  • Maximum height: H=v02sin⁡2θ2gH = \frac{v_0^2\sin^2\theta}{2g}
  • Range: R=v02sin⁡2θgR = \frac{v_0^2\sin 2\theta}{g}

(c) Air resistance:

  • Reduces time of flight (projectile decelerates faster)
  • Reduces maximum height (energy is lost to drag)
  • Reduces range (both horizontal and vertical motion are affected)

(d) Perturbation theory: expand the solution in powers of b/mb/m and solve order by order.

Scoring rubric:

  • Part (a): 2 points for each correct differential equation
  • Part (b): 2 points each for correct expressions
  • Part (c): 1 point for each correct qualitative effect
  • Part (d): 2 points for mentioning perturbation theory

Problem 2: Rotational Dynamics (15 points)

A uniform solid cylinder of mass MM and radius RR is placed on a rough inclined plane at angle θ\theta to the horizontal. The coefficient of static friction is μs\mu_s.

(a) Draw a free-body diagram showing all forces acting on the cylinder. (3 points)

(b) Write the equations for translational and rotational motion. (4 points)

(c) Derive the condition for the cylinder to roll without slipping. (4 points)

(d) Find the minimum angle θmin\theta_{\text{min}} at which the cylinder begins to slip. (4 points)

Solution:

(a) Forces: weight MgMg downward, normal force NN perpendicular to incline, friction ff up the incline (provides torque for rolling).

(b) Translation along incline: Mgsin⁡θ−f=MaMg\sin\theta - f = Ma

Translation perpendicular: N−Mgcos⁡θ=0N - Mg\cos\theta = 0

Rotation: τ=fR=Iα\tau = fR = I\alpha

Rolling condition: a=Rαa = R\alpha

(c) For a solid cylinder, I=12MR2I = \frac{1}{2}MR^2.

From torque equation: fR=12MR2⋅aR  ⟹  f=12MafR = \frac{1}{2}MR^2 \cdot \frac{a}{R} \implies f = \frac{1}{2}Ma

Substituting into translation equation:

Mgsin⁡θ−12Ma=Ma  ⟹  a=2gsin⁡θ3Mg\sin\theta - \frac{1}{2}Ma = Ma \implies a = \frac{2g\sin\theta}{3} f=12M⋅2gsin⁡θ3=Mgsin⁡θ3f = \frac{1}{2}M \cdot \frac{2g\sin\theta}{3} = \frac{Mg\sin\theta}{3}

Rolling without slipping requires f≤μsN=μsMgcos⁡θf \leq \mu_s N = \mu_s Mg\cos\theta:

Mgsin⁡θ3≤μsMgcos⁡θ  ⟹  tan⁡θ≤3μs\frac{Mg\sin\theta}{3} \leq \mu_s Mg\cos\theta \implies \tan\theta \leq 3\mu_s

(d) The cylinder begins to slip when tan⁡θmin=3μs\tan\theta_{\text{min}} = 3\mu_s, so:

θmin=tan⁡−1(3μs)\theta_{\text{min}} = \tan^{-1}(3\mu_s)

Scoring rubric:

  • Part (a): 1 point for each correct force and direction
  • Part (b): 1 point for each correct equation
  • Part (c): 2 points for correct moment of inertia, 2 points for deriving the condition
  • Part (d): 2 points for the final expression

Common Mistakes in AP Physics C Mechanics

  1. Confusing mass and weight. Mass is in kg; weight is in Newtons (W=mgW = mg).

  2. Forgetting that friction does negative work. When an object slides, friction opposes motion and removes mechanical energy from the system.

  3. Using v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x for non-constant acceleration. This equation only works for constant acceleration.

  4. Sign errors in rotational dynamics. Choose a consistent sign convention for torque and angular acceleration.

  5. Assuming all objects roll without slipping. Check whether the friction is sufficient to prevent slipping.

  6. Confusing center of mass with geometric center. For non-uniform objects, these may differ.

Exam Tips

  • Always draw a free-body diagram before writing equations
  • For rotational problems, specify the axis of rotation
  • Use conservation of energy when forces are conservative
  • Check that your answer has correct units and reasonable magnitude
  • For free-response problems, show all steps and box your final answer

Cross-References