AP Physics 1 · Topic 2.9

Circular Motion Practice

Part of Force and Translational Dynamics.(TOP-2.I)

Practice questions

22

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Sample questions

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  1. Sample 1difficulty 1/5

    An astronaut has mass 70 kg70~\text{kg} on Earth. On the Moon (g1.6 m/s2g \approx 1.6~\text{m/s}^2), her <strong>mass</strong> is

    • A

      112 kg112~\text{kg}

    • B

      70 kg70~\text{kg}

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    • C

      11 kg11~\text{kg}

    • D

      700 kg700~\text{kg}

    Why

    Mass is a property of the object and does not change with gravity. Only weight (mgm g) changes.

  2. Sample 2difficulty 1/5

    A force is applied to a moving object opposite to its motion. Which must be true?

    • A

      Object speeds up.

    • B

      Object reverses immediately.

    • C

      Net force is zero.

    • D

      Object slows down.

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    Why

    Force opposite to motion produces deceleration; the object slows down (and may eventually reverse if force persists).

  3. Sample 3difficulty 2/5

    On a frictionless surface, blocks AA (2 kg2~\text{kg}) and BB (3 kg3~\text{kg}) are in contact. You push AA with a 20 N20~\text{N} horizontal force so both accelerate together. What is the contact force that AA exerts on BB?

    • A

      20 N20~\text{N}

    • B

      12 N12~\text{N}

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    • C

      8 N8~\text{N}

    • D

      4 N4~\text{N}

    Why

    System: a=20/5=4 m/s2a = 20/5 = 4~\text{m/s}^2. Isolating BB: contact force =mBa=(3)(4)=12 N= m_B a = (3)(4) = 12~\text{N}.

  4. Sample 4difficulty 2/5

    Two children pull on opposite ends of a rope, each with 200 N200~\text{N}. The rope is in equilibrium. What is the tension in the rope?

    • A

      200 N200~\text{N}

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    • B

      100 N100~\text{N}

    • C

      00

    • D

      400 N400~\text{N}

    Why

    Tension at any cross-section equals the force pulling at either end: T=200 NT = 200~\text{N}. (You don't add the two pulls.)

  5. Sample 5difficulty 2/5

    A car of mass 1200 kg1200~\text{kg} rounds a horizontal curve of radius 40 m40~\text{m} at 10 m/s10~\text{m/s}. What centripetal force does friction provide?

    • A

      12,000 N12{,}000~\text{N}

    • B

      1500 N1500~\text{N}

    • C

      3000 N3000~\text{N}

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    • D

      300 N300~\text{N}

    Why

    Fc=mv2/r=(1200)(100)/40=3000 NF_c = m v^2/r = (1200)(100)/40 = 3000~\text{N}.