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Linear momentum practice problems AP Physics C: Mechanics, Unit 4

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

Impulse as the integral of force, conservation of momentum, elastic and inelastic collisions, and centre of mass. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for linear momentum

p = mvmomentum
J = ∫ F dt = Δpimpulse
xcm = ∫ x dm / ∫ dmcentre of mass
ΣFext = M acmmotion of the centre of mass

Use g = 9.8 m/s², as on the AP reference sheet. You can type answers like 4.9, 1/3 or 3.6e2. The full sheet is in our AP Physics C equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
Find the momentum of a 0.15 kg ball moving at 40 m/s.
kg·m/s

Hint

p = mv.

Worked solution
  1. p = 0.15 × 40 = 6 kg·m/s.
Q2Warm-up
A constant 10 N force acts on an object for 0.30 s. Find the impulse.
N·s

Hint

J = FΔt for a constant force.

Worked solution
  1. J = 10 × 0.30 = 3 N·s.
Q3Warm-up
In a perfectly elastic collision between two objects:

Hint

Momentum is conserved in every collision with no external impulse.

Worked solution
  1. Momentum is always conserved here; elastic means kinetic energy is conserved too.
  2. The answer is C.

Exam-style questions

Questions at the level of the AP Physics C: Mechanics exam. The ones marked Challenge combine more than one idea.

Q4Exam-style
A 2.0 kg cart moving at 3.0 m/s collides with and sticks to a 1.0 kg cart at rest. Find their common speed.
m/s

Hint

A perfectly inelastic collision: one final velocity.

Worked solution
  1. 2.0 × 3.0 = 3.0v.
  2. v = 2 m/s.
Q5Exam-style
A bat exerts a force F(t) = 2000t(0.20 − t) N on a ball for 0 ≤ t ≤ 0.20 s. Find the impulse.
N·s

Hint

Integrate F(t) over the contact time.

Worked solution
  1. J = ∫00.20 2000(0.20t − t2) dt = 2000[0.10t2 − t3/3]00.20.
  2. = 2000(0.0040 − 0.002667).
  3. J = 2.67 N·s.
Q6Exam-style
A rod lies along the x-axis from x = 0 to x = 3 m, with linear density λ = 2x kg/m. Find the position of its centre of mass.
m

Hint

xcm = ∫ xλ dx / ∫ λ dx.

Worked solution
  1. Mass: ∫03 2x dx = 9 kg.
  2. ∫03 2x2 dx = 18 kg·m.
  3. xcm = 18/9 = 2 m.
Q7Exam-style
A system has no net external force on it. Its centre of mass:

Hint

ΣFext = Macm.

Worked solution
  1. With no external force, acm = 0.
  2. So the centre of mass moves with constant velocity (which may be zero): B.
Q8Challenge
A 1.0 kg ball moving at 6.0 m/s hits a 2.0 kg ball at rest in a head-on, perfectly elastic collision. Find the speed of the 2.0 kg ball afterwards.
m/s

Hint

Use conservation of momentum and of kinetic energy, or the elastic collision result v2 = 2m1u/(m1 + m2).

Worked solution
  1. Momentum: 6 = v1 + 2v2. Elastic, so relative speed reverses: v2 − v1 = 6.
  2. Adding: 3v2 = 12.
  3. v2 = 4 m/s (and the 1.0 kg ball rebounds at 2 m/s).

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Translation between representations style · 6 points · about 20 minutes

A 0.010 kg bullet is fired horizontally into a 1.99 kg wooden block hanging from a light string, and stays inside it. The block and bullet swing up to a maximum height of 0.20 m above the starting point.

  1. (a) Find the speed of the block and bullet just after the collision.
  2. (b) Find the speed of the bullet before it hit the block.
  3. (c) Explain why momentum, but not mechanical energy, is conserved during the collision.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

½(M + m)V2 = (M + m)gh, so V = √(2 × 9.8 × 0.20) = 1.98 m/s.

Part (b)

Momentum: 0.010u = 2.00 × 1.98, so u = 396 m/s.

Part (c)

No external horizontal impulse acts during the brief collision, but kinetic energy is converted to thermal energy and deformation as the bullet embeds.

Your mark: 0 of 6

Common mistakes in linear momentum

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics C: Mechanics past papers, with official scoring guidelines. · All AP Physics C: Mechanics units