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Ace Your AP Physics 1 Exam: Top 10 Formulas You Need to Know

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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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Preparing for the AP Physics 1 exam can be overwhelming, especially when it comes to memorizing all the necessary formulas. But what if you only needed a total of 15 equations to answer every question on the AP Exam? In this short article, we’ll cover the top 10 formulas you need to know to ace the AP Physics 1 exam. By mastering these formulas, you’ll be well on your way to understanding the underlying concepts and feeling confident on test day.

A Note on Derivation

On the AP Physics 1 Exam, memorizing formulas does NOT mean you will be able to solve questions.

Rather, formulas aid in seeing the relationship between two or more variables. College board often want to see if you can first derive a formula, then use it to reach a logical conclusion.

All of physics is deriving equations from fundamental equations and concepts. For example, in Unit 2.4, you learned how to use Newton’s Second Law to derive an equation for any force. This is exactly what college board is testing you for.

Kinematic Formulas

The first set of formulas you need to know for AP Physics 1 are the FOUR kinematics formulas, which relate to the motion of objects. These include formulas for distance, displacement, velocity, and acceleration.

Table showing the big 5 kinematic equations and the one missing variable in each equation.

Pro-tip: If you have more that four formulas memorized for Kinematics, you do not understand the underlying concepts. All other motion equations, are derived from the big four equations. And, you guessed it, the AP exam will test if you know how to derive the equations.

Here’s a simple question they might ask: If you double the angle of a projectile, by what factor does the range increase or decrease?

Newton’s Laws of Motion

Newton’s laws are crucial for understanding the behavior of objects in motion.

Be able to answer conceptual questions like the following:

  • Why do you feel like you are being pushed back when accleerating forward?
  • Do you need a force to jump off the ground? And if so, what is the force?
  • Can an object have velocity but zero net force?
  • When traveling at 30 m/s you throw a ball vertically. Where will the ball land?

Additionally, you must master Newton’s Second Law: [katex] F_{\text{net}} = ma [/katex].

This will allow you to solve ANY problem involving forces on the AP Physics 1 Exam. Practice the simple problem solving framework for all force problems.

If you need a speed review all of forces and Newton’s laws, skim through this article

Other useful formulas to memorize when it comes to forces:

The force of friction –

[katex] f = \mu N [/katex]

Centripetal acceleration –

[katex] a_c = \frac{v^2}{r} [/katex]

Work-Energy Theorem

[katex] W = Fd = \Delta KE [/katex]

This theorem relates the work done on an object to its change in kinetic energy. It is essential for understanding the concepts of work and energy.

Important note: The force and displacement must be parallel.

Conservation of Energy

[katex] E_i = E_f [/katex]

In other words, the sum of initial types of energy = the sum of the final types of energy. This is used to solve every problem dealing with energy.

This principle states that energy cannot be created or destroyed, only transferred or transformed. It is crucial for understanding the conservation of energy in physical systems.

Note that energy is NOT conserved in open systems (aka a system where there are external forces). Furthermore, note that energy is a scalar and has no direction.

Conservation of Momentum

[katex] p_i = p_f [/katex]

Simply put: Sum objects’ momentum before collision = Sum objects’ momentum after collision. This is used to solve every problem involving collisions in a closed systems.

Momentum is conserved in EVERY collision and explosion.

This principle states that the total momentum of a system remains constant if no external forces act on it. It is crucial for understanding the behavior of objects in collisions.

The AP Physics 1 Exam will test your knowledge on Linear and Rotational Momentum. This is generally the second most missed question on the AP exam, after circular motion.

Impulse Theorem

Impulse is used when momentum is not conserved.

[katex] I = \Delta p = m \Delta v = Ft [/katex]

In short, this theorem relates the force applied to an object to its resulting change in momentum. It is essential for understanding the concepts of impulse and momentum.

Important note: On the AP Exam the most common mistake is the +/- signs on velocity.

For example, what is ∆v if a ball hits a wall at speed v then rebounds at the same speed?

Show Answer

2v

Hooke’s Law

[katex] F_{\text{spring}} = -kx [/katex]

So k is the spring constant [how stiff a spring is] and x [how much the spring is compressed or stretched from the resting state]. Note the negative sign just means the spring force tends to act opposite to the direction stretched or compressed.

Since this is technically a part of Newton’s Second Law, we can set kx = ma.

Note the negative sign is to show us that it a restoring force (displacement is opposite to force). You can ignore it in most cases.

Torque and Rotational Motion

Torque is a measure of the turning force applied to an object. Technically this is also a part of Newton’s second law.

[katex] \Tau = I\alpha = Fd [/katex]

  • I = rotational inertia. Every object will have its own formula. Rotational inertia of a point mass = mr2
  • Force and distance have to be perpendicular

Also note that ALL linear formulas mentioned previously can be converted to rotational formulas. So you don’t have to memorize any other formula for this section.

Gravitational force

This is the force between ANY two object. Two planets. A mass on a planet. A satellite and planet. One important thing to note is that r is the distance measured from the centers‘ of the masses.

[katex] T_g = G \frac {m_1m_2}{r^2} [/katex]

This is another example of Newton’s second law (forces). It can be used in combination of other forces or even centripetal acceleration (in the case of satellites).

Simple Harmonic Motion

Period (measure in seconds) is the time taken to make one complete oscillation.

Frequency (measured in s-1 aka hertz) is the number of oscillations in one second.

[katex] T_{\text{spring}} = 2 \pi \sqrt{\frac{m}{k}} [/katex]
[katex] T_{\text{pendulum}} = 2 \pi \sqrt{\frac{L}{g}} [/katex]

Real-World Applications:

Beyond the AP Physics 1 exam, understanding these formulas has real-world applications. Physics is a fundamental science that helps us understand the world around us, from the motion of objects to the behavior of light. By understanding these formulas, you’ll be able to apply physics principles to solve real-world problems, from designing cars to launching satellites.

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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