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AP Physics C: Mechanics Unit 6 practice questions

Try 10 multiple-choice questions from Unit 6 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    Two satellites orbit a planet in circular paths. Satellite A has an orbital radius R, and Satellite B has an orbital radius 4R. Which claim correctly compares the orbital period T_B of Satellite B to the period T_A of Satellite A?

    T_B = 8 T_A, because the square of the period is proportional to the cube of the radius

    T_B = 4 T_A, because the period is directly proportional to the orbital radius value

    T_B = 2 T_A, because the period is proportional to the square root of the radius value

    T_B = 64 T_A, because the period is proportional to the cube of the orbital radius

    Show answer

    Answer: (A)

    Applying Kepler's third law (T² ∝ r³) requires cubing the radius factor (4³ = 64) and taking the square root to find the period factor (√64 = 8).

  2. Question 2

    A planet orbits a star of mass M at a distance R with period T. Which claim correctly predicts the period of a planet orbiting a star of mass 2M at a distance 2R?

    2T, because the period is proportional to R^(3/2) and inversely proportional to M^(1/2)

    4T, because the period is proportional to R^(3/2) and inversely proportional to M^(1/2)

    T, because the effects of doubling both the mass and the radius cancel each other out

    T/2, because the period is proportional to M^(1/2) and inversely proportional to R^(3/2)

    Show answer

    Answer: (A)

    Using T ∝ √(R³/M), doubling R introduces a factor of √8 (approx 2.8) and doubling M introduces a factor of 1/√2. The combined factor is √(8/2) = √4 = 2.

  3. Question 3

    A hoop and a solid disk, both with mass MM and radius RR, roll without slipping down an incline starting from rest at height HH. How does the translational kinetic energy KtransK_{trans} of the disk's center of mass compare to that of the hoop at the bottom?

    The energy KtransK_{trans} is greater for the disk

    The energy KtransK_{trans} is greater for the hoop

    The energy KtransK_{trans} is equal for both objects

    The energy KtransK_{trans} depends on the incline angle

    Show answer

    Answer: (A)

    Both objects convert the same potential energy MgHMgH into total kinetic energy. The hoop has a larger rotational inertia (I=MR2I=MR^2) than the disk (I=1/2MR2I=1/2 MR^2), so it stores a larger fraction of energy in rotation. This leaves less energy for translation compared to the disk.

  4. Question 4

    A hoop and a solid disk of equal mass and radius roll down an incline without slipping. A student claims the disk reaches the bottom first. Which of the following correctly evaluates this claim with appropriate reasoning?

    Correct, because the disk has a smaller moment of inertia, requiring less energy for rotation and allowing more for translation.

    Correct, because the disk has a larger moment of inertia, which generates a greater torque and angular acceleration.

    Incorrect, because both objects have the same mass and radius, so they experience identical gravitational forces and accelerations.

    Incorrect, because the hoop has its mass concentrated at the rim, giving it more rotational momentum to overcome friction.

    Show answer

    Answer: (A)

    Justifying rotational motion requires energy analysis. The disk's mass is closer to the center, giving it a lower moment of inertia (II). This means less potential energy is 'spent' on rotation (1/2Iω21/2 I \omega^2), leaving more for translational speed.

  5. Question 5

    A spherical planet has uniform density ρ\rho and radius RR, with escape speed vescv_{esc}. If the planet's density is doubled to 2ρ2\rho while the radius remains RR, what is the new escape speed?

    2vesc\sqrt{2}v_{esc}

    2vesc2v_{esc}

    4vesc4v_{esc}

    vesc/2v_{esc}/\sqrt{2}

    Show answer

    Answer: (A)

    Connecting density to mass via M=ρVM = \rho V reveals that doubling density (with constant radius) doubles the mass. Since escape speed depends on mass as vesc∝Mv_{esc} \propto \sqrt{M}, the new speed increases by a factor of 2\sqrt{2}.

  6. Question 6

    A satellite orbits a planet in a circular path of radius RR with period TT. If the satellite is moved to a stable circular orbit of radius 4R4R, what is the new orbital period?

    2T2T

    4T4T

    8T8T

    16T16T

    Show answer

    Answer: (C)

    Applying Kepler's Third Law (T2∝R3T^2 \propto R^3) establishes the functional dependence T∝R3/2T \propto R^{3/2}. Increasing the radius by a factor of 4 results in a period change of 43/2=(4)3=23=84^{3/2} = (\sqrt{4})^3 = 2^3 = 8.

  7. Question 7

    Two satellites of equal mass mm orbit a planet. Satellite 1 orbits at radius RR and Satellite 2 orbits at radius 9R9R. How does the magnitude of the angular momentum L2L_2 of Satellite 2 compare to L1L_1 of Satellite 1?

    L2=L1L_2 = L_1

    L2=3L1L_2 = 3L_1

    L2=9L1L_2 = 9L_1

    L2=27L1L_2 = 27L_1

    Show answer

    Answer: (B)

    Angular momentum for a circular orbit is L=mvr=mGMrL = mvr = m\sqrt{GMr}, meaning L∝rL \propto \sqrt{r}. Increasing radius by a factor of 9 increases angular momentum by a factor of 9=3\sqrt{9}=3.

  8. Question 8

    Two satellites orbit a planet in circular paths. Satellite X has orbital speed vv and Satellite Y has orbital speed 2v2v. How does the orbital period TYT_Y of Satellite Y compare to the period TXT_X of Satellite X?

    TY=TX/8T_Y = T_X / 8

    TY=TX/4T_Y = T_X / 4

    TY=TX/2T_Y = T_X / 2

    TY=8TXT_Y = 8 T_X

    Show answer

    Answer: (A)

    Since v∝r−1/2v \propto r^{-1/2}, doubling speed implies the radius decreases by factor of 4 (rY=rX/4r_Y = r_X/4). Since T∝r3/2T \propto r^{3/2}, the period decreases by (1/4)3/2=1/8(1/4)^{3/2} = 1/8.

  9. Question 9

    A satellite orbits Earth in a circle of radius RR with total mechanical energy EE. If the satellite transfers to a circular orbit of radius 3R3R, what is the new total mechanical energy? (Assume EE is negative).

    Enew=3EE_{new} = 3E

    Enew=E/3E_{new} = E/3

    Enew=E/9E_{new} = E/9

    Enew=9EE_{new} = 9E

    Show answer

    Answer: (B)

    Total mechanical energy is E=−GMm/2rE = -GMm/2r, meaning E∝1/rE \propto 1/r. Tripling the radius divides the energy value by 3 (making it a smaller negative number).

  10. Question 10

    Two satellites of equal mass orbit a planet. The period of Satellite B is 8 times the period of Satellite A. How does the angular momentum LBL_B of Satellite B compare to LAL_A of Satellite A?

    LB=2LAL_B = 2L_A

    LB=4LAL_B = 4L_A

    LB=8LAL_B = 8L_A

    LB=16LAL_B = 16L_A

    Show answer

    Answer: (A)

    Since T∝r3/2T \propto r^{3/2}, an 8-fold period increase implies rr increases by factor of 4 (43/2=84^{3/2}=8). Since L∝rL \propto \sqrt{r}, the angular momentum increases by factor of 4=2\sqrt{4}=2.

What's on the AP Physics C: Mechanics Unit 6 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (40 minutes)
  • 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
  • 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

Questions are drawn from 1408 multiple-choice questions and 13 FRQs written for Unit 6, Energy and Momentum of Rotating Systems.

Practice AP Physics C: Mechanics Unit 6 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 6 before you start

Other AP Physics C: Mechanics unit tests