AP Physics 1 Practice Test — 20 Questions with Answer Explanations (2026)

Use this free AP Physics 1 practice test to check your understanding across all major topics. 20 questions, organized by unit, with a complete explanation for every answer.

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AP Physics 1 Exam Overview

The AP Physics 1 exam covers algebra-based physics — no calculus required. The five FRQ questions are worth 12 points each.

AP Physics 1 Practice Questions

Unit 1 — Kinematics

Question 1

A car starts from rest and accelerates uniformly at 4 m/s² for 5 seconds. What is its displacement?

A) 10 m B) 20 m C) 50 m D) 100 m

Answer: C — 50 m

Using x = v₀t + ½at² with v₀ = 0: x = 0 + ½(4)(25) = 50 m. You can verify with average velocity: final velocity v = at = 20 m/s, so average v = (0 + 20)/2 = 10 m/s × 5 s = 50 m. ✓

Question 2

A ball is thrown horizontally from a cliff at 10 m/s. After 2 seconds (ignoring air resistance), the ball's vertical velocity is:

A) 10 m/s downward B) 20 m/s downward C) 5 m/s downward D) 22 m/s downward at an angle

Answer: B — 20 m/s downward

Horizontal and vertical motion are independent. Vertically, the ball starts at rest and accelerates at g = 10 m/s². After 2 s: v_y = gt = 10 × 2 = 20 m/s downward. Horizontal velocity remains 10 m/s throughout.

Question 3

Which graph correctly represents an object moving at constant velocity?

A) A position-time graph with increasing slope B) A velocity-time graph showing a horizontal (flat) line C) An acceleration-time graph showing an increasing line D) A position-time graph showing a curved (parabolic) line

Answer: B — velocity-time graph, horizontal line

Constant velocity = velocity that does not change. On a v-t graph, this is a horizontal line. On a p-t graph, constant velocity appears as a straight diagonal line with constant slope. Acceleration = 0 throughout.

Unit 2 — Newton's Laws

Question 4

A 5 kg box is pulled across a frictionless surface by a 20 N horizontal force. Its acceleration is:

A) 0.25 m/s² B) 4 m/s² C) 20 m/s² D) 100 m/s²

Answer: B — 4 m/s²

Newton's second law: F = ma → a = F/m = 20/5 = 4 m/s². On a frictionless surface, net force equals applied force. If friction were present, you would subtract f_k = μₖmg before dividing.

Question 5

A 10 kg block rests on a flat table. What is the normal force from the table on the block?

A) 0 N B) 98 N (equals the block's weight) C) More than 98 N D) Less than 98 N

Answer: B — 98 N

No vertical acceleration → ΣF_y = 0 → N = mg = 10 × 9.8 ≈ 98 N. Normal force equals weight only when acceleration is zero. If the surface accelerates upward (like an elevator), N > mg.

Question 6

Newton's third law states that when object A exerts a force on object B:

A) Object B exerts an equal force in the same direction on A B) Object B exerts an equal and opposite force on A C) Object B exerts a larger force on A D) Object B exerts a smaller force on A

Answer: B — equal and opposite force on A

Action-reaction pairs are always equal in magnitude, opposite in direction, and act on different objects. These forces don't cancel because they act on different objects. A horse pulling a cart exerts 500 N forward — the cart pulls the horse 500 N backward.

Question 7

A 60 kg person stands on a scale in an elevator accelerating upward at 2 m/s². The scale reads: (g = 10 m/s²)

A) 480 N B) 600 N C) 720 N D) 60 N

Answer: C — 720 N

ΣF = ma → N − mg = ma → N = m(g + a) = 60(10 + 2) = 720 N. The person feels heavier because the floor must support their weight AND accelerate them upward. Downward acceleration at 2 m/s² would give N = 60(8) = 480 N.

Unit 3 — Work, Energy, Power

Question 8

A 2 kg ball is lifted 5 meters vertically. How much gravitational potential energy does it gain? (g = 10 m/s²)

A) 10 J B) 50 J C) 100 J D) 25 J

Answer: C — 100 J

ΔPE = mgh = 2 × 10 × 5 = 100 J. This energy came from the person lifting the ball. When released, all 100 J converts to kinetic energy, giving a final speed of v = √(2 × 100/2) = 10 m/s at the original height.

Question 9

A force of 50 N is applied at 60° above the horizontal to move a box 10 m horizontally. The work done is:

A) 500 J B) 250 J C) 433 J D) 0 J

Answer: B — 250 J

W = Fd·cos(θ) = 50 × 10 × cos(60°) = 500 × 0.5 = 250 J. Only the horizontal component of force does work. The vertical component (50 sin 60° ≈ 43.3 N) does zero work because it's perpendicular to the displacement.

Question 10

A machine does 600 J of work in 30 seconds. Its power output is:

A) 18,000 W B) 0.05 W C) 20 W D) 600 W

Answer: C — 20 W

Power = Work/time = 600 J / 30 s = 20 W. One watt = one joule per second. A human at moderate exercise outputs roughly 200–400 W; a light bulb uses 60–100 W.

Unit 4 — Momentum and Impulse

Question 11

A 1000 kg car moving at 20 m/s collides with and sticks to a stationary 1000 kg car. Their combined velocity after the collision is:

A) 20 m/s B) 10 m/s C) 5 m/s D) 0 m/s

Answer: B — 10 m/s

Conservation of momentum: m₁v₁ = (m₁ + m₂)v_f → 1000(20) = 2000·v_f → v_f = 10 m/s. This perfectly inelastic collision loses kinetic energy: KE before = ½(1000)(400) = 200,000 J; KE after = ½(2000)(100) = 100,000 J. Half the kinetic energy is lost to heat and deformation.

Question 12

The impulse-momentum theorem states that impulse equals:

A) The net force times distance B) The change in the object's momentum C) The object's mass times velocity D) The average force times acceleration

Answer: B — the change in momentum

Impulse = FΔt = Δp = mΔv. This is why airbags and crumple zones save lives: the same Δp (car stopping) over a longer time means smaller average force on the occupants.

Unit 5 — Rotation

Question 13

A disk rotating at 10 rad/s has its angular velocity doubled. By what factor does its rotational kinetic energy change?

A) 2 B) 4 C) 8 D) ½

Answer: B — 4

KE_rot = ½Iω². Doubling ω: KE_new = ½I(2ω)² = 4 × ½Iω². Rotational kinetic energy quadruples — same relationship as translational KE ∝ v².

Question 14

A figure skater pulls her arms inward while spinning. Her angular speed:

A) Decreases because her moment of inertia decreases B) Increases because angular momentum is conserved C) Stays the same because no external torque acts D) Decreases because she loses kinetic energy

Answer: B — increases

Conservation of angular momentum: L = Iω = constant (no external torque). Pulling arms inward reduces I, so ω must increase. Her rotational KE actually increases because she does work pulling her arms against the centrifugal tendency.

Unit 6 — Simple Harmonic Motion

Question 15

A mass on a spring oscillates with period T. If the mass is quadrupled while k remains constant, the new period is:

A) T/4 B) T/2 C) 2T D) 4T

Answer: C — 2T

T = 2π√(m/k). If m → 4m: T_new = 2π√(4m/k) = 2 × 2π√(m/k) = 2T. Period is proportional to √m. Importantly, period is independent of amplitude — doubling the amplitude doesn't change T.

Question 16

At what point in a pendulum's swing is kinetic energy maximum?

A) At the highest point B) At the lowest point C) At 45° from vertical D) Kinetic energy is constant throughout

Answer: B — at the lowest point

At the lowest point, all gravitational PE has converted to KE — maximum speed. At the highest points, the pendulum momentarily stops (KE = 0, PE maximum). Energy conservation: KE_max = mgh where h is the height above the lowest point.

Unit 7 — Waves

Question 17

A wave has frequency 5 Hz and wavelength 4 m. Its wave speed is:

A) 1.25 m/s B) 0.8 m/s C) 20 m/s D) 9 m/s

Answer: C — 20 m/s

v = fλ = 5 × 4 = 20 m/s. This fundamental wave equation applies to all wave types — sound, light, water waves, electromagnetic radiation.

Question 18

Two speakers emit sound in phase at the same frequency. At a point where path length difference = 1 full wavelength, the result is:

A) Destructive interference (silence) B) Constructive interference (loud sound) C) No interference D) 90° phase shift

Answer: B — constructive interference

Constructive interference: path difference = nλ (waves arrive in phase, amplitudes add). Destructive interference: path difference = (n + ½)λ (waves arrive 180° out of phase, amplitudes cancel).

Unit 8 — Electric Charge and Circuits

Question 19

Three 6 Ω resistors are connected in parallel. The equivalent resistance is:

A) 18 Ω B) 6 Ω C) 3 Ω D) 2 Ω

Answer: D — 2 Ω

1/R_eq = 1/6 + 1/6 + 1/6 = 3/6 → R_eq = 2 Ω. For n identical resistors in parallel: R_eq = R/n = 6/3 = 2 Ω. Parallel combinations always give lower resistance than any individual resistor.

Question 20

The gravitational force between two objects is F. If the distance between them is tripled, the new gravitational force is:

A) F/3 B) F/6 C) F/9 D) 3F

Answer: C — F/9

Newton's Law of Gravitation: F ∝ 1/r². Tripling r: F_new = F/3² = F/9. This inverse-square relationship applies to both gravitational and electrostatic forces.

Score Interpretation

These are approximate. See our AP Physics 1 Score Calculator for a full composite score estimate combining MC and FRQ.

What to Study Next

Missed kinematics questions? Review projectile motion, kinematic equations, and motion graphs.

Missed Newton's law questions? Practice drawing free-body diagrams for every scenario — elevator, inclined plane, pulley, circular motion.

Missed energy questions? Energy conservation problems are on every exam. Practice identifying which energies are present at each point in a system.

Missed momentum questions? Review elastic vs. inelastic collision types, and practice impulse-momentum theorem calculations.

Missed rotation questions? Rotational kinematics mirrors linear kinematics (θ, ω, α instead of x, v, a). Practice conservation of angular momentum for spinning objects.

Missed waves or circuits? Waves: memorize v = fλ and the interference conditions. Circuits: practice series vs. parallel calculations with Ohm's law.

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