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AP Physics C: Mechanics Unit 4 practice questions

Try 10 multiple-choice questions from Unit 4 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

    A 4.0 kg object moves along the x-axis. A net force F(t) = 3t² (in Newtons) acts on the object from t = 0 to t = 2.0 s. A student claims that the change in momentum of the object during this interval is 24 kg·m/s. Which of the following correctly evaluates this claim with appropriate justification?

    Incorrect, because the impulse is the integral ∫F dt = 8.0 N·s, not the product of final force and time.

    Correct, because the impulse is the product F_final·Δt = 24 N·s, which applies to any time-dependent force.

    Incorrect, because the impulse is the average force F_avg = 6.0 N times the time, yielding 12 N·s.

    Correct, because the impulse is the integral ∫F dt = 24 N·s when calculated for this force function.

    Show answer

    Answer: (A)

    Integrating the force function F(t) = 3t² from t = 0 to t = 2.0 yields [t³] evaluated at limits, which is 8.0 N·s. The student incorrectly calculated impulse by multiplying the final force (12 N) by the time interval (2 s).

  2. Question 2

    Sand falls vertically at a rate of λ kg/s onto a conveyor belt moving horizontally at a constant speed v. A student claims that a horizontal external force of magnitude F = λv is required to maintain the belt's constant speed. Which of the following correctly evaluates this claim with appropriate justification?

    Correct, because this force provides the impulse rate v(dm/dt) needed to accelerate the sand horizontally.

    Incorrect, because the belt moves at constant velocity, so the net force must be zero by Newton's first law.

    Incorrect, because the sand only exerts a vertical normal force, so no horizontal force is required.

    Correct, because this force equals the product of the total accumulated mass on the belt and its velocity.

    Show answer

    Answer: (A)

    Applying the impulse-momentum theorem to a system with changing mass (F = dp/dt) shows that force equals velocity times the rate of mass addition. The sand enters with zero horizontal momentum and must be accelerated to velocity v.

  3. Question 3

    A 0.50 kg ball hits a wall with a velocity of 10 m/s to the right and rebounds with a velocity of 8.0 m/s to the left. The contact lasts for 0.10 s. A student claims that the magnitude of the average force exerted by the wall on the ball is 10 N. Which of the following correctly evaluates this claim with appropriate justification?

    Incorrect, because velocity is a vector, so the momentum change is 9.0 kg·m/s, yielding a force of 90 N.

    Incorrect, because the impulse-momentum theorem requires using velocity as a vector, so the momentum change is 18 kg·m/s, yielding a force of 180 N.

    Correct, because impulse equals the change in kinetic energy, and the kinetic energy change is 1.0 kJ, so the average force is 10 N.

    Incorrect, because the momentum change calculation requires treating velocity as a vector, but the average force should equal the magnitude of the net impulse divided by time, yielding 45 N.

    Show answer

    Answer: (A)

    Momentum is a vector quantity, so the change is final minus initial: (-8.0) - (10) = -18 m/s. The student likely treated speed as a scalar (10 - 8), ignoring the direction change.

  4. Question 4

    A 2.0 kg object starts from rest. A net force acts on it, modeled by a triangle graph: increasing linearly from 0 to 10 N at 2.0 s, then decreasing linearly to 0 at 4.0 s. A student claims the final velocity is 10 m/s. Which of the following correctly evaluates this claim with appropriate justification?

    Correct, because the impulse is the area under the curve (20 N·s), resulting in a velocity of 10 m/s.

    Incorrect, because the final velocity depends on the maximum force applied, which would yield 20 m/s.

    Correct, because the average force is 10 N, and applying Newton's law for 4.0 s yields 10 m/s.

    Incorrect, because the impulse is the product of average force (5 N) and half the time, yielding 5.0 m/s.

    Show answer

    Answer: (A)

    Impulse is the area under the force-time graph (0.5 × base × height = 0.5 × 4.0 s × 10 N = 20 N·s). Equating impulse to change in momentum (mΔv) gives 20 = 2.0(v_f), so v_f = 10 m/s.

  5. Question 5

    A block of mass mm is initially at rest on a frictionless surface. At time t=0t = 0, a net force F(t)=αtF(t) = \alpha t is applied to the block, where α\alpha is a positive constant. Which of the following expressions gives the speed of the block at time t=Tt = T?

    αT22m\frac{\alpha T^2}{2m}

    αT2m\frac{\alpha T^2}{m}

    2αT2m\frac{2\alpha T^2}{m}

    αTm\frac{\alpha T}{m}

    Show answer

    Answer: (A)

    Integrating the force function F(t)=αtF(t) = \alpha t over time yields the impulse J=∫0Tαt dt=12αT2J = \int_0^T \alpha t \, dt = \frac{1}{2}\alpha T^2. Applying the impulse-momentum theorem Δp=mΔv\Delta p = m \Delta v, the final speed is v=Jm=αT22mv = \frac{J}{m} = \frac{\alpha T^2}{2m}.

  6. Question 6

    A tennis ball of mass mm strikes a wall with speed vv at an angle θ\theta relative to the normal of the wall. It rebounds with the same speed vv at the same angle θ\theta relative to the normal. What is the magnitude of the impulse exerted by the wall on the ball?

    2mvcos⁡θ2mv \cos\theta

    2mvsin⁡θ2mv \sin\theta

    mvcos⁡θmv \cos\theta

    mvsin⁡θmv \sin\theta

    Show answer

    Answer: (A)

    Calculating the change in momentum along the normal direction involves subtracting the initial momentum from the final momentum: mvcos⁡θ−(−mvcos⁡θ)=2mvcos⁡θmv \cos\theta - (-mv \cos\theta) = 2mv \cos\theta. The components parallel to the wall remain unchanged, so the net impulse is purely perpendicular.

  7. Question 7

    A particle of mass mm is subject to a time-dependent force F(t)=F0e−t/τF(t) = F_0 e^{-t/\tau}, where F0F_0 and τ\tau are positive constants. If the particle starts from rest at t=0t=0, what is its speed as tt approaches infinity?

    F0τm\frac{F_0 \tau}{m}

    F0τF_0 \tau

    F0m\frac{F_0}{m}

    F0τ2m\frac{F_0 \tau^2}{m}

    Show answer

    Answer: (A)

    Integrating the force from 00 to infinity gives the total impulse: J=∫0∞F0e−t/τdt=F0[−τe−t/τ]0∞=F0τJ = \int_0^\infty F_0 e^{-t/\tau} dt = F_0 [-\tau e^{-t/\tau}]_0^\infty = F_0 \tau. Dividing by mass mm gives the final speed.

  8. Question 8

    A horizontal stream of water with density ρ\rho and cross-sectional area AA hits a vertical wall with speed vv and splashes off parallel to the wall (losing all horizontal momentum). What is the magnitude of the average force exerted by the water on the wall?

    ρAv2\rho A v^2

    ρAv\rho A v

    12ρAv2\frac{1}{2}\rho A v^2

    ρA2v\rho A^2 v

    Show answer

    Answer: (A)

    Determining the rate of mass hitting the wall as dmdt=ρAv\frac{dm}{dt} = \rho A v allows calculating the force as the rate of momentum change: F=dpdt=vdmdt=v(ρAv)=ρAv2F = \frac{dp}{dt} = v \frac{dm}{dt} = v(\rho A v) = \rho A v^2.

  9. Question 9

    A net force F(t)F(t) acts on an object of mass mm initially at rest. The force increases linearly from 00 to FmaxF_{max} over the time interval from t=0t=0 to t=Tt=T. Which expression gives the final speed of the object at t=Tt=T?

    FmaxT2m\frac{F_{max} T}{2m}

    FmaxTm\frac{F_{max} T}{m}

    2FmaxTm\frac{2 F_{max} T}{m}

    FmaxT4m\frac{F_{max} T}{4m}

    Show answer

    Answer: (A)

    Calculating the impulse as the area under the force-time graph (a triangle with base TT and height FmaxF_{max}) gives J=12FmaxTJ = \frac{1}{2} F_{max} T. Dividing by mass mm yields the final speed.

  10. Question 10

    Sand falls vertically from a stationary hopper at a rate of σ\sigma (in kg/s) onto a horizontal conveyor belt moving at a constant speed vv. What is the magnitude of the horizontal force required to keep the belt moving at this constant speed?

    σv\sigma v

    12σv\frac{1}{2}\sigma v

    σv2\sigma v^2

    σg\sigma g

    Show answer

    Answer: (A)

    Applying Newton's second law in the form F=dpdtF = \frac{dp}{dt} where velocity is constant but mass changes gives F=vdmdtF = v \frac{dm}{dt}. Substituting dmdt=σ\frac{dm}{dt} = \sigma yields F=σvF = \sigma v.

What's on the AP Physics C: Mechanics Unit 4 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 736 multiple-choice questions and 24 FRQs written for Unit 4, Linear Momentum.

Practice AP Physics C: Mechanics Unit 4 FRQs

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

Review Unit 4 before you start

Other AP Physics C: Mechanics unit tests