AP Physics 1 practice problems: practice problems and exam questions

Practice for the AP Physics 1 exam topics: kinematics, forces and Newton’s laws, circular motion, energy, momentum and simple harmonic motion.

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Topics in this course

  1. Kinematicsspeeding up · braking · vertical throwcalculations
  2. Newton’s lawspush with friction · inclined plane · table and pulleycalculations
  3. Projectile motionhorizontal launch · launch from ground · launch from heightcalculations
  4. Circular motioncentripetal force · friction on a curve · vertical circlecalculations
  5. Work & energyfree fall · slope with friction · ramp, friction and springcalculations
  6. Momentum & collisionsperfectly inelastic · elastic · impulse and forcecalculations
  7. Springs & pendulumsmass on a spring · pendulum · speed, acceleration and energycalculations

Exam tips for AP Physics 1 practice problems

  • Start force problems with a free-body diagram and a chosen positive direction.
  • Use energy conservation when the question does not ask about time.
  • Check the units and the size of your answer before moving on.

Worked examples

KinematicsEasy
t (s)v (m/s)20 s10

A train slows down uniformly from 15 m/s to 10 m/s in 20 s. How far does it travel in that time? Find s.

v₀ = 15 m/sv = 10 m/st = 20 s

Find s (m)

Show hints
  1. With constant acceleration the average speed is the mean of the initial and final speeds.
  2. s = (v₀ + v)/2 · t. It is the area under the v–t graph.
  3. average speed = 12.5 m/s.
Show the answer and the step-by-step solution

Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.

Average speed

(v₀ + v)/2 = (15 + 10)/2 = 12.5 m/s

Distance

s = 12.5·20 = 250 m

Answer

s = 250 m
Newton’s lawsMedium
25 kg35°F = 140 Nμ = 0.25

A 25 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 140 N. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the sled.

m = 25 kgF = 140 Nθ = 35°μ = 0.25g = 9.81 m/s²

Find a (m/s²)

Show hints
  1. Split the pull into F·cos θ (forward) and F·sin θ (up).
  2. Vertically: N + F·sin θ = m·g. Friction is μ·N.
  3. N = 164.9 N.
Show the answer and the step-by-step solution

Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.

Normal force

N = m·g − F·sin θ = 245.3 − 140·sin 35° = 164.9 N

Horizontal forces

m·a = F·cos θ − μ·N = 114.7 − 41.24
a = 73.44/25 = 2.938 m/s²

Answer

a = 2.938 m/s²

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

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