AP Physics C: Mechanics practice questions and exam guide

A solid cylinder rolls without slipping down an incline angle θ. Acceleration?

1. g sin θ
2. g sin θ / 2
3. (2/3) g sin θ
4. (1/2) g sin θ

Answer: C — (2/3) g sin θ

C) a = g sin θ /(1 + I/(MR²)) = g sin θ /(1 + 1/2) = (2/3) g sin θ. A) Frictionless slide. B/D) Wrong fraction.

The center of mass of a system moves as if:

1. All mass is at the geometric center
2. ALL mass is concentrated at the center of mass, and net external force acts there; internal forces don't move the CoM
3. CoM is fixed
4. CoM rotates
5. CoM has no momentum

Answer: B — ALL mass is concentrated at the center of mass, and net external force acts there; internal forces don't move the CoM

F⃗_ext = M·a⃗_CoM (Newton's 2nd law for systems). Even a complex tumbling system has a CoM moving in a simple trajectory (e.g., dropped wrench: CoM falls in straight line). Foundation for AP Physics C rigid body chapters.

A 2 kg block experiences F(x) = 3x² N. Acceleration at x = 2 m?

1. 12 m/s²
2. 24 m/s²
3. 3 m/s²
4. 6 m/s²

Answer: D — 6 m/s²

D) a = F/m = 3(2)²/2 = 12/2 = 6 m/s². A) Forgot dividing by mass. B) Wrong. C) Coefficient only.

A 0.05 kg bullet at 400 m/s embeds in a 2 kg block at rest on frictionless surface. Speed after embedding?

1. 9.76 m/s
2. 8 m/s
3. 10 m/s
4. 400 m/s

Answer: A — 9.76 m/s

A) v = (0.05·400)/(2.05) ≈ 9.76 m/s. B) Used wrong total mass. C) Wrong. D) No collision.

The center of mass of a system of particles is:

1. r_cm = (Σ m_i · r_i) / Σ m_i
2. The position of the heaviest particle
3. The midpoint of the system's extent
4. The point with the highest potential energy

Answer: A — r_cm = (Σ m_i · r_i) / Σ m_i

A) Standard formula. B/C/D) Each contradicts the definition.

Velocity is defined as:

1. The derivative of position with respect to time, v = dx/dt
2. The product of mass and acceleration
3. The integral of force with respect to position
4. The same as speed

Answer: A — The derivative of position with respect to time, v = dx/dt

A) Standard calculus-based definition. B) That's net force (Newton's second). C) Work. D) Speed is the scalar magnitude.

In an elastic 1D collision of equal masses where m₂ is at rest, after collision:

1. Both move with v₁/2
2. Both reverse direction
3. m₁ stops; m₂ moves off with v₁
4. Both stop

Answer: C — m₁ stops; m₂ moves off with v₁

C) Standard result for elastic equal-mass collision. A) That's perfectly inelastic. B) Sign error. D) Violates momentum.

Kepler's third law for planetary motion states:

1. Force is proportional to distance
2. Velocity is constant
3. The square of the orbital period is proportional to the cube of the semi-major axis (T² ∝ a³)
4. Acceleration is zero in orbit

Answer: C — The square of the orbital period is proportional to the cube of the semi-major axis (T² ∝ a³)

C) Kepler's third law. A/B/D) Each is incorrect.

Conservative force F satisfies:

1. It depends on path
2. Path-independent work: F = −dU/dx (1D) or F⃗ = −∇U (3D) — has a potential energy function; mechanical energy conserved
3. F is constant
4. F is friction-like
5. F can do negative work only

Answer: B — Path-independent work: F = −dU/dx (1D) or F⃗ = −∇U (3D) — has a potential energy function; mechanical energy conserved

Conservative forces (gravity, springs, electrostatic): path-independent work, mechanical energy conserved. Non-conservative (friction, drag): path-dependent. Relationship F = −dU/dx is essential. AP Physics C-defining.

A particle moves under F = (3x²)x̂ + (2y)ŷ. Work along path from (0,0) to (1,1) along x-axis then y-axis?

1. 0
2. 5
3. 3
4. ∫₀¹ 3x² dx + ∫₀¹ 2y dy = 1 + 1 = 2 J

Answer: D — ∫₀¹ 3x² dx + ∫₀¹ 2y dy = 1 + 1 = 2 J

D) Path integral splits cleanly along axes. Other choices ignore vector components.

Two blocks (3 kg, 5 kg) connected by string over frictionless pulley. Acceleration of system (g = 10):

1. g/8 = 1.25 m/s²
2. g = 10 m/s²
3. 5 m/s²
4. a = (5 − 3)g/(5 + 3) = 2.5 m/s²

Answer: D — a = (5 − 3)g/(5 + 3) = 2.5 m/s²

D) Atwood machine: a = (Δm)g/Σm. A) Wrong fraction. B) Ignores constraint. C) Bad denominator.

A force F(x) = 6x² N moves an object from x = 1 to x = 3 m. Work done?

1. 12 J
2. 26 J
3. 52 J
4. 54 J

Answer: C — 52 J

C) W = ∫₁³ 6x² dx = [2x³]₁³ = 54 − 2 = 52 J. A) Wrong. B) Off by 26. D) Used 0 lower bound.

Difference between Physics 1 and Physics C: Mechanics?

Physics 1: algebra-based, broader topics. Physics C: calculus-based, deeper into mechanics only.

Resonance?

Driving force at natural frequency → large amplitude oscillation.

Friction's role in rolling without slipping?

Static friction provides torque for angular acceleration. Often does no work.

Moment of inertia (point masses)?

I = Σm_i r_i².

If x(t) = t³ - 6t² + 9t, find v(t).

v(t) = 3t² - 12t + 9.

Newton's 2nd Law (rigorous)?

F_net = ma. For systems with changing mass: F = dp/dt.