AP Physics C: Mechanics, AP exams

Unit 4: Linear Momentum

Unit 4 develops momentum, impulse as the integral of force over time, and conservation of momentum for systems. Center-of-mass motion is treated carefully, including the fact that the center of mass of an isolated system moves at constant velocity regardless of internal interactions. Collisions in two dimensions appear here.

Position: 4 of 7 units
Topics covered: 5 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics C: Mechanics. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Momentum and impulse: Momentum as mass times velocity, and impulse as the time integral of force or the area under a force-time graph.
Conservation of momentum: The conditions for conservation, and how the choice of system determines whether forces are internal or external.
Center of mass: Locating the center of mass for discrete and continuous distributions, and its motion under internal forces.
Collisions: Elastic, inelastic, and perfectly inelastic collisions, in one and two dimensions.
Systems with changing mass: Analyzing situations where mass enters or leaves the system, such as a chain or a stream of particles.

Study tips

How to practice Linear Momentum.

Section-specific guidance that follows from how this material is usually tested.

Compute the center of mass of a continuous rod or plate by integration a few times so the setup is familiar.
Work two-dimensional collisions strictly component by component, writing both equations before solving.
State what is conserved and why before any calculation; the justification carries points on free-response questions.

Where this fits

Unit 4: Linear Momentum is 4 of 7 units in the AP Physics C: Mechanics guide. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics C: Mechanics guide Official exam information

Sample Linear Momentum questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A net force F(t) = 6t, with F in newtons and t in seconds, acts on a 4.0 kg object that is at rest at t = 0. What is the speed of the object at t = 2.0 s?
1. A1.5 m/s
2. B3.0 m/s
3. C6.0 m/s
4. D12 m/s
Show answer
Answer: B. 3.0 m/s
The impulse is the integral of 6t dt from 0 to 2.0, which is 3t^2 evaluated at 2.0, or 12 N s, and that equals the momentum change, so v = 12/4.0 = 3.0 m/s. The 6.0 m/s choice treats the force as constant at its final value of 12 N and so doubles the impulse, and the 12 m/s choice reports the impulse without dividing by the mass.
Key insight: Impulse is the area under a force-versus-time graph and equals the change in momentum.
Question 2Foundational
Two particles lie on the x-axis: a 2.0 kg particle at x = 0 and a 6.0 kg particle at x = 4.0 m. Where is the center of mass of this two-particle system?
1. Ax = 1.0 m
2. Bx = 2.0 m
3. Cx = 3.0 m
4. Dx = 3.5 m
Show answer
Answer: C. x = 3.0 m
The center of mass is the mass-weighted average position: [(2.0)(0) + (6.0)(4.0)]/8.0 = 3.0 m. The x = 2.0 m choice is the midpoint, which would be correct only if the two masses were equal.
Key insight: The center of mass always sits closer to the heavier object, in inverse proportion to the masses.
Question 3Intermediate
On a frictionless horizontal surface a 2.0 kg puck moving east at 3.0 m/s collides with a 1.0 kg puck moving north at 6.0 m/s, and the two stick together. What is the speed of the combined object immediately after the collision?
1. A2.0 m/s
2. B2.8 m/s
3. C3.0 m/s
4. D4.0 m/s
Show answer
Answer: B. 2.8 m/s
Momentum is conserved component by component: 6.0 kg m/s east and 6.0 kg m/s north give a magnitude of sqrt(72) = 8.49 kg m/s, so the 3.0 kg combined mass moves at 8.49/3.0 = 2.8 m/s. The 4.0 m/s choice adds the two momentum components as if they were along the same line.
Key insight: Momentum is a vector, so two-dimensional collisions need two separate conservation equations.
Question 4Advanced
A thin rod lies along the x-axis from x = 0 to x = L and has linear mass density that increases as lambda(x) = cx, where c is a positive constant. Where is the rod's center of mass?
1. AL/3
2. BL/2
3. C2L/3
4. D3L/4
Show answer
Answer: C. 2L/3
The total mass is the integral of cx dx from 0 to L, which is cL^2/2, and the first moment is the integral of x(cx) dx, which is cL^3/3, so x_cm = (cL^3/3)/(cL^2/2) = 2L/3. The L/2 choice assumes uniform density and ignores that mass piles up toward the far end.
Key insight: For a continuous body the center of mass is the integral of x dm divided by the total mass.
Question 5Intermediate
A 1.0 kg ball moving at 4.0 m/s collides head-on and elastically with a stationary 3.0 kg ball on a frictionless surface. What is the velocity of the 1.0 kg ball immediately after the collision?
1. A1.0 m/s in the original direction
2. B2.0 m/s in the original direction
3. C2.0 m/s in the reverse direction
4. D4.0 m/s in the reverse direction
Show answer
Answer: C. 2.0 m/s in the reverse direction
For a one-dimensional elastic collision with a stationary target, v1' = [(m1 - m2)/(m1 + m2)]v1 = (-2/4)(4.0) = -2.0 m/s, so the light ball bounces back at 2.0 m/s while the 3.0 kg ball moves forward at 2.0 m/s. Checking both totals confirms 4.0 kg m/s of momentum and 8.0 J of kinetic energy before and after.
Key insight: A light object bouncing off a heavier one always reverses direction in an elastic collision.

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Unit 4: Linear Momentum