--- title: "AP Physics C: Mechanics Unit 4 Practice Test" description: "A timed AP Physics C: Mechanics Unit 4 (Linear Momentum) practice test in the AP format, drawn from 736 multiple-choice questions and 24 FRQs written for the unit. The standard version takes about 88 minutes." canonical: "https://fiveable.me/ap-physics-c-mechanics/unit-test/4" type: "unit-test" subject: "AP Physics C: Mechanics" unit: "Unit 4" --- # AP Physics C: Mechanics Unit 4 Practice Test ## Overview A timed AP Physics C: Mechanics Unit 4 (Linear Momentum) practice test in the AP format, drawn from 736 multiple-choice questions and 24 FRQs written for the unit. The standard version takes about 88 minutes. ## Test Format - 20 Multiple Choice Questions (40 minutes) - 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes) - 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes) ## Unit FRQs by Type - [FRQ 1 – Mathematical Routines](/ap-physics-c-mechanics/frqs/mathematical-routines?unit=4): 7 questions - [FRQ 2 – Translation Between Representations](/ap-physics-c-mechanics/frqs/translation-between-representations?unit=4): 5 questions - [FRQ 3 – Experimental Design](/ap-physics-c-mechanics/frqs/experimental-design?unit=4): 5 questions - [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-c-mechanics/frqs/qualitative-quantitative-translation?unit=4): 7 questions ## Practice Questions ### 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? - (A) Incorrect, because the impulse is the integral ∫F dt = 8.0 N·s, not the product of final force and time. - (B) Correct, because the impulse is the product F_final·Δt = 24 N·s, which applies to any time-dependent force. - (C) Incorrect, because the impulse is the average force F_avg = 6.0 N times the time, yielding 12 N·s. - (D) Correct, because the impulse is the integral ∫F dt = 24 N·s when calculated for this force function. **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). ### 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? - (A) Correct, because this force provides the impulse rate v(dm/dt) needed to accelerate the sand horizontally. - (B) Incorrect, because the belt moves at constant velocity, so the net force must be zero by Newton's first law. - (C) Incorrect, because the sand only exerts a vertical normal force, so no horizontal force is required. - (D) Correct, because this force equals the product of the total accumulated mass on the belt and its velocity. **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. ### 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? - (A) Incorrect, because velocity is a vector, so the momentum change is 9.0 kg·m/s, yielding a force of 90 N. - (B) 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. - (C) 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. - (D) 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. **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. ### 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? - (A) Correct, because the impulse is the area under the curve (20 N·s), resulting in a velocity of 10 m/s. - (B) Incorrect, because the final velocity depends on the maximum force applied, which would yield 20 m/s. - (C) Correct, because the average force is 10 N, and applying Newton's law for 4.0 s yields 10 m/s. - (D) Incorrect, because the impulse is the product of average force (5 N) and half the time, yielding 5.0 m/s. **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. ### Question 5 A block of mass $$m$$ is initially at rest on a frictionless surface. At time $$t = 0$$, a net force $$F(t) = \alpha t$$ is applied to the block, where $$\alpha$$ is a positive constant. Which of the following expressions gives the speed of the block at time $$t = T$$? - (A) $$\frac{\alpha T^2}{2m}$$ - (B) $$\frac{\alpha T^2}{m}$$ - (C) $$\frac{2\alpha T^2}{m}$$ - (D) $$\frac{\alpha T}{m}$$ **Answer: (A)** Integrating the force function $$F(t) = \alpha t$$ over time yields the impulse $$J = \int_0^T \alpha t \, dt = \frac{1}{2}\alpha T^2$$. Applying the impulse-momentum theorem $$\Delta p = m \Delta v$$, the final speed is $$v = \frac{J}{m} = \frac{\alpha T^2}{2m}$$. ### Question 6 A tennis ball of mass $$m$$ strikes a wall with speed $$v$$ at an angle $$\theta$$ relative to the normal of the wall. It rebounds with the same speed $$v$$ at the same angle $$\theta$$ relative to the normal. What is the magnitude of the impulse exerted by the wall on the ball? - (A) $$2mv \cos\theta$$ - (B) $$2mv \sin\theta$$ - (C) $$mv \cos\theta$$ - (D) $$mv \sin\theta$$ **Answer: (A)** Calculating the change in momentum along the normal direction involves subtracting the initial momentum from the final momentum: $$mv \cos\theta - (-mv \cos\theta) = 2mv \cos\theta$$. The components parallel to the wall remain unchanged, so the net impulse is purely perpendicular. ### Question 7 A particle of mass $$m$$ is subject to a time-dependent force $$F(t) = F_0 e^{-t/\tau}$$, where $$F_0$$ and $$\tau$$ are positive constants. If the particle starts from rest at $$t=0$$, what is its speed as $$t$$ approaches infinity? - (A) $$\frac{F_0 \tau}{m}$$ - (B) $$F_0 \tau$$ - (C) $$\frac{F_0}{m}$$ - (D) $$\frac{F_0 \tau^2}{m}$$ **Answer: (A)** Integrating the force from $$0$$ to infinity gives the total impulse: $$J = \int_0^\infty F_0 e^{-t/\tau} dt = F_0 [-\tau e^{-t/\tau}]_0^\infty = F_0 \tau$$. Dividing by mass $$m$$ gives the final speed. ### Question 8 A horizontal stream of water with density $$\rho$$ and cross-sectional area $$A$$ hits a vertical wall with speed $$v$$ 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? - (A) $$\rho A v^2$$ - (B) $$\rho A v$$ - (C) $$\frac{1}{2}\rho A v^2$$ - (D) $$\rho A^2 v$$ **Answer: (A)** Determining the rate of mass hitting the wall as $$\frac{dm}{dt} = \rho A v$$ allows calculating the force as the rate of momentum change: $$F = \frac{dp}{dt} = v \frac{dm}{dt} = v(\rho A v) = \rho A v^2$$. ### Question 9 A net force $$F(t)$$ acts on an object of mass $$m$$ initially at rest. The force increases linearly from $$0$$ to $$F_{max}$$ over the time interval from $$t=0$$ to $$t=T$$. Which expression gives the final speed of the object at $$t=T$$? - (A) $$\frac{F_{max} T}{2m}$$ - (B) $$\frac{F_{max} T}{m}$$ - (C) $$\frac{2 F_{max} T}{m}$$ - (D) $$\frac{F_{max} T}{4m}$$ **Answer: (A)** Calculating the impulse as the area under the force-time graph (a triangle with base $$T$$ and height $$F_{max}$$) gives $$J = \frac{1}{2} F_{max} T$$. Dividing by mass $$m$$ yields the final speed. ### Question 10 Sand falls vertically from a stationary hopper at a rate of $$\sigma$$ (in kg/s) onto a horizontal conveyor belt moving at a constant speed $$v$$. What is the magnitude of the horizontal force required to keep the belt moving at this constant speed? - (A) $$\sigma v$$ - (B) $$\frac{1}{2}\sigma v$$ - (C) $$\sigma v^2$$ - (D) $$\sigma g$$ **Answer: (A)** Applying Newton's second law in the form $$F = \frac{dp}{dt}$$ where velocity is constant but mass changes gives $$F = v \frac{dm}{dt}$$. Substituting $$\frac{dm}{dt} = \sigma$$ yields $$F = \sigma v$$. ## Review the Unit - [4.1 Linear Momentum](/ap-physics-c-mechanics/unit-4/1-linear-momentum/study-guide/vUy5cuBKbgKou3xJ) - [4.2 Change in Momentum and Impulse](/ap-physics-c-mechanics/unit-4/2-change-in-momentum-and-impulse/study-guide/dj7haZ7RqOOTuCat) - [4.3 Conservation of Linear Momentum](/ap-physics-c-mechanics/unit-4/3-conservation-of-linear-momentum/study-guide/qWLd3tCmiQRSZKv0) - [4.4 Elastic and Inelastic Collisions](/ap-physics-c-mechanics/unit-4/4-elastic-and-inelastic-collisions/study-guide/JvjEzhTLprTtp64U) ## Other Unit Tests - [Unit 1: Kinematics](/ap-physics-c-mechanics/unit-test/1) - [Unit 2: Force and Translational Dynamics](/ap-physics-c-mechanics/unit-test/2) - [Unit 3: Work, Energy, and Power](/ap-physics-c-mechanics/unit-test/3) - [Unit 5: Torque and Rotational Dynamics](/ap-physics-c-mechanics/unit-test/5) - [Unit 6: Energy and Momentum of Rotating Systems](/ap-physics-c-mechanics/unit-test/6) - [Unit 7: Oscillations](/ap-physics-c-mechanics/unit-test/7) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)