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AP Physics 1 Unit 2A: Newton's Laws and Translational Equilibrium — practice questions

Explore 92 free Unit 2A: Newton's Laws and Translational Equilibrium questions in AP Physics 1 Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 2A: Newton's Laws and Translational Equilibrium questions and exercises with answers

Showing questions 1–30 of 92.

Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:Read the answer and worked solution →For a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:Read the answer and worked solution →Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:Read the answer and worked solution →A 50 kg flywheel of radius 0.5 m rotates at 60 rev/min. Angular momentum (I = MR²/2):Read the answer and worked solution →A solid sphere rolls down without slipping. Fraction of KE that is rotational:Read the answer and worked solution →A wheel rotating at 50 rad/s decelerates uniformly to rest in 10 s. Total angular displacement:Read the answer and worked solution →Angular acceleration alpha = 4 rad/s² acts for 5 s on a wheel starting from rest. Number of revolutions:Read the answer and worked solution →A solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s. Total kinetic energy:Read the answer and worked solution →For a solid cylinder rolling without slipping, the friction force at the contact point:Read the answer and worked solution →Moment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis through center:Read the answer and worked solution →Stress is defined as:Read the answer and worked solution →Strain (longitudinal) is:Read the answer and worked solution →Young's modulus Y is ratio of:Read the answer and worked solution →Hydraulic pressure at depth h below liquid surface (density ρ):Read the answer and worked solution →Pascal's law states pressure applied to enclosed fluid is:Read the answer and worked solution →Bulk modulus B for fluid relates pressure change to:Read the answer and worked solution →Poisson's ratio σ for a stretched rod relates:Read the answer and worked solution →Bernoulli's equation: along streamline of ideal fluid:Read the answer and worked solution →Equation of continuity for incompressible fluid:Read the answer and worked solution →Stokes' law: viscous drag on sphere of radius r moving at v through fluid of viscosity η:Read the answer and worked solution →Terminal velocity of sphere (radius r, density ρ_s) falling in viscous fluid (density ρ_f, viscosity η):Read the answer and worked solution →Surface tension of liquid γ has SI unit:Read the answer and worked solution →Excess pressure inside spherical drop of radius r (surface tension γ):Read the answer and worked solution →Capillary rise h for liquid in tube of radius r:Read the answer and worked solution →Reynolds number R_e for tube flow:Read the answer and worked solution →Two wires of same material, same length, but cross-sections 1 mm² and 2 mm². Same load: ratio of extensions:Read the answer and worked solution →Energy stored per unit volume in stretched wire:Read the answer and worked solution →Two wires of same material (Y, ρ) and same length, one twice the diameter. Ratio of extensions under same load:Read the answer and worked solution →For elastic wire, breaking stress depends on:Read the answer and worked solution →For a metallic rod heated through ΔT with Y and α (thermal expansion) and rigidly held:Read the answer and worked solution →

Worked practice questions

Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:

  1. 12
  2. 24
  3. 48
  4. 36
Show answer and solution

Correct answer: B. 24

Each mass at distance 2 m from center. We have I = sum m × r². That is I = 2 × (3 × 4). So I = 24 kg m². Wait, recompute: 3 × 2² + 3 × 2² = 12 + 12 = 24. So I = 24. Hmm option B says 24 — I should select B.

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For a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:

  1. ML²/12
  2. ML²
  3. ML²/2
  4. ML²/3
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Correct answer: D. ML²/3

By parallel-axis theorem: I_end = I_cm + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = 4ML²/12 = ML²/3.

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Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:

  1. 4 rad/s²
  2. 100 rad/s²
  3. 20 rad/s²
  4. 5 rad/s²
Show answer and solution

Correct answer: A. 4 rad/s²

tau = I × alpha → alpha = tau/I = 20/5 = 4 rad/s².

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More Physics topics

Unit 8: Fluids: Density, Buoyancy and ContinuityUnit 1B: Kinematics: 2D Motion & ProjectilesUnit 1C: Kinematics: Relative & Circular MotionUnit 2B: Friction, Gravitational and Normal ForcesUnit 3A: Work and Kinetic Energy TheoremUnit 3B: Potential Energy & Conservation of EnergyUnit 4A: Linear Momentum and ImpulseUnit 4B: Conservation of Linear Momentum & Collisions