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AP Physics 1 Unit 1C: Kinematics: Relative & Circular Motion — practice questions — page 3

Explore 92 free Unit 1C: Kinematics: Relative & Circular Motion 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 1C: Kinematics: Relative & Circular Motion questions and exercises with answers

Showing questions 61–90 of 92.

Two springs k1 and k2 in parallel: equivalent spring constant:Read the answer and worked solution →Two springs k1 and k2 in series: equivalent spring constant:Read the answer and worked solution →A spring of force constant k is cut into two equal halves. Spring constant of each half:Read the answer and worked solution →A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:Read the answer and worked solution →In SHM, when does kinetic energy equal potential energy?Read the answer and worked solution →A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:Read the answer and worked solution →For SHM x = A sin(ωt), find ratio of kinetic energy at t = T/4 to t = T/8:Read the answer and worked solution →A particle's SHM is described by x = 3 sin(ωt) + 4 cos(ωt). Amplitude of resulting SHM:Read the answer and worked solution →Time period of a simple pendulum is T. If mass is doubled, period becomes:Read the answer and worked solution →A particle of mass 0.5 kg has SHM with amplitude 4 cm and time period 0.2 s. Find maximum kinetic energy:Read the answer and worked solution →A simple pendulum of length 1 m on Earth (g = 10) has frequency:Read the answer and worked solution →For SHM, the displacement-time graph is:Read the answer and worked solution →A mass m on a vertical spring of constant k. The equilibrium extension is x0 = mg/k. The mass is set into SHM. Frequency depends on:Read the answer and worked solution →Two simple pendulums of lengths 1 m and 4 m start swinging in phase. They will be in phase again after:Read the answer and worked solution →A particle has SHM with x = A cos(ωt). At t = 0, the particle is at:Read the answer and worked solution →Phase difference between displacement and velocity in SHM:Read the answer and worked solution →The total energy of SHM is proportional to:Read the answer and worked solution →Linear speed v and angular speed omega relation for circular motion:Read the answer and worked solution →Torque tau acting on a body produces:Read the answer and worked solution →Angular momentum L of a body rotating with angular velocity omega:Read the answer and worked solution →A wheel starting from rest reaches 20 rad/s in 4 s. Angular acceleration alpha:Read the answer and worked solution →Moment of inertia of a uniform solid sphere of mass M and radius R about an axis through center:Read the answer and worked solution →Moment of inertia of a thin uniform disc (mass M, radius R) about axis perpendicular to disc through center:Read the answer and worked solution →Parallel axis theorem states I_parallel = I_cm + Md². Here d is:Read the answer and worked solution →For a thin rod of mass M and length L, I about axis through center perpendicular to rod:Read the answer and worked solution →Kinetic energy of a rotating body with moment of inertia I and angular velocity omega:Read the answer and worked solution →For a body rolling without slipping with velocity v of center of mass:Read the answer and worked solution →A solid cylinder and a hollow cylinder of same mass and radius roll down an incline. Which reaches bottom first?Read the answer and worked solution →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 →A figure skater pulls in her arms during a spin. Her moment of inertia decreases. By conservation of L (I × omega), her angular velocity:Read the answer and worked solution →

Worked practice questions

Two springs k1 and k2 in parallel: equivalent spring constant:

  1. k1 - k2
  2. √(k1 k2)
  3. k1 k2 / (k1 + k2)
  4. k1 + k2
Show answer and solution

Correct answer: D. k1 + k2

Springs in parallel (both stretching by same amount): forces add. **So k_eq = k1 + k2. Period decreases (system stiffer).**

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Two springs k1 and k2 in series: equivalent spring constant:

  1. (k1 + k2)/2
  2. Cannot be determined
  3. k1 k2 / (k1 + k2)
  4. k1 + k2
Show answer and solution

Correct answer: C. k1 k2 / (k1 + k2)

**Springs in series:** each carries same force, extensions add. 1/k_eq = 1/k1 + 1/k2, so k_eq = k1 k2 / (k1 + k2). System more compliant (less stiff).

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A spring of force constant k is cut into two equal halves. Spring constant of each half:

  1. k²
  2. k
  3. 2k
  4. k/2
Show answer and solution

Correct answer: C. 2k

Spring constant is inversely proportional to length. **Cutting into half:** k_half = 2k. So each half is stiffer (shorter springs are stiffer). Two halves in series give back original k.

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More Physics topics

Unit 8: Fluids: Density, Buoyancy and ContinuityUnit 1B: Kinematics: 2D Motion & ProjectilesUnit 2A: Newton's Laws and Translational EquilibriumUnit 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