Comprehensive study guide for AP Physics 1 (Algebra-Based), aligned with the College Board Course and Exam Description. Covers mechanics and rotational motion with emphasis on conceptual understanding and problem-solving.

Quantity	Type	Definition	SI Unit
Displacement	Vector	Change in position, $\Delta \vec{r}$	m
Distance	Scalar	Total path length	m
Velocity	Vector	Rate of change of displacement	m/s
Speed	Scalar	Rate of change of distance	m/s
Acceleration	Vector	Rate of change of velocity	m/s$^2$

Constant acceleration equations:

$$v = v_0 + at$$ $$x = x_0 + v_0 t + \frac{1}{2}at^2$$ $$v^2 = v_0^2 + 2a(x - x_0)$$ $$x = x_0 + \frac{1}{2}(v_0 + v)t$$

In 2D, resolve all vectors into perpendicular components. The $x$ and $y$ components of motion are independent of each other (superposition principle).

For a projectile launched with initial speed $v_0$ at angle $\theta$ above the horizontal (ignoring air resistance):

$$v_{0x} = v_0 \cos\theta, \quad v_{0y} = v_0 \sin\theta$$

Horizontal (constant velocity):

$$x(t) = v_{0x}\, t$$

Vertical (constant acceleration $a_y = -g$):

$$y(t) = v_{0y}\, t - \frac{1}{2}gt^2$$ $$v_y(t) = v_{0y} - gt$$

Time of flight: $t = \dfrac{2v_{0y}}{g}$

Maximum height: $H = \dfrac{v_{0y}^2}{2g}$

Range: $R = \dfrac{v_0^2 \sin 2\theta}{g}$

An object in free fall experiences only the force of gravity. All objects in free fall have the same acceleration $g \approx 9.8\ \text{m/s}^2$ near Earth”s surface, regardless of mass.

• Position-time graph: Slope gives velocity
• Velocity-time graph: Slope gives acceleration; area under the curve gives displacement
• Acceleration-time graph: Area under the curve gives change in velocity

First Law (Inertia): An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net external force.

Second Law: The net force on an object equals its mass times its acceleration:

$$\vec{F}_{\text{net}} = m\vec{a}$$

This is a vector equation — apply separately in each direction.

Third Law: For every action force there is an equal and opposite reaction force. Forces always come in pairs acting on different objects.

1. Isolate the object of interest
2. Draw arrows for every force acting on the object
3. Forces act on the object; the object’s forces act on other things
4. Common forces: weight ($mg$), normal force ($N$), tension ($T$), friction ($f$), applied force ($F$)

Static friction: $f_s \leq \mu_s N$ (prevents motion; maximum value is $\mu_s N$)

Kinetic friction: $f_k = \mu_k N$ (opposes motion; constant magnitude)

where $\mu_s$ and $\mu_k$ are coefficients of static and kinetic friction respectively, and $N$ is the normal force.

On an incline at angle $\theta$ to the horizontal:

$$a = g(\sin\theta - \mu_k \cos\theta)$$

The component of weight parallel to the plane is $mg\sin\theta$; the component perpendicular (which equals the normal force) is $mg\cos\theta$.

Two masses $m_1$ and $m_2$ ($m_2 > m_1$) connected by a string over a frictionless pulley:

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$ $$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$

An object moving in a circle of radius $r$ at constant speed $v$ has a centripetal acceleration directed towards the centre:

$$a_c = \frac{v^2}{r}$$

The net inward force providing this acceleration is:

$$F_c = \frac{mv^2}{r} = m\omega^2 r$$

where $\omega$ is the angular velocity in rad/s.

Newton’s law of universal gravitation:

$$F_g = \frac{Gm_1 m_2}{r^2}$$

where $G = 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

The gravitational field strength at distance $r$ from mass $M$:

$$g = \frac{GM}{r^2}$$

Work done by a constant force:

$$W = Fd\cos\theta$$

where $\theta$ is the angle between the force and the displacement. Work is positive when the force has a component in the direction of displacement and negative when opposing displacement.

For a variable force:

$$W = \int_a^b F(x)\, dx$$

The net work done on an object equals its change in kinetic energy:

$$W_{\text{net}} = \Delta K = K_f - K_i$$$$K = \frac{1}{2}mv^2$$

Gravitational potential energy (near Earth’s surface):

$$U_g = mgh$$

Elastic (spring) potential energy:

$$U_s = \frac{1}{2}kx^2$$

where $k$ is the spring constant and $x$ is the displacement from equilibrium.

In the absence of non-conservative forces (e.g., friction):

$$K_i + U_i = K_f + U_f$$

When friction is present:

$$K_i + U_i = K_f + U_f + E_{\text{lost to friction}}$$

where $E_{\text{lost to friction}} = f_k d$.

Average power is the rate at which work is done:

$$P = \frac{W}{\Delta t}$$

Instantaneous power:

$$P = Fv\cos\theta$$

SI unit: watt (W), where $1\ \text{W} = 1\ \text{J/s}$.

Momentum:

$$\vec{p} = m\vec{v}$$

Impulse:

$$\vec{J} = \vec{F}_{\text{avg}}\,\Delta t = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$$

The impulse-momentum theorem states that the impulse on an object equals its change in momentum. Graphically, impulse equals the area under a force-time graph.

When the net external force on a system is zero, the total momentum of the system is conserved:

$$m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}$$

This applies independently to each component direction.

Elastic collision: Both momentum and kinetic energy are conserved.

For a 1D elastic collision between masses $m_1$ and $m_2$:

$$v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2 v_{2i}}{m_1 + m_2}$$ $$v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1 v_{1i}}{m_1 + m_2}$$

Inelastic collision: Momentum is conserved but kinetic energy is not.

Perfectly inelastic collision: The objects stick together after collision.

$$(m_1 + m_2)v_f = m_1 v_{1i} + m_2 v_{2i}$$

The centre of mass of a system of particles:

$$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots}$$

The centre of mass of a system moves as if all external forces act on a single particle of total mass located at the centre of mass.

Torque is the rotational analogue of force:

$$\tau = rF\sin\theta = r_\perp F$$

where $r_\perp = r\sin\theta$ is the perpendicular distance from the axis of rotation to the line of action of the force (the moment arm). Positive torque causes counter-clockwise rotation; negative causes clockwise.

Net torque and angular acceleration:

$$\tau_{\text{net}} = I\alpha$$

where $I$ is the rotational inertia and $\alpha$ is the angular acceleration.

$$I = \sum m_i r_i^2$$

For continuous objects: $I = \int r^2\, dm$

Common moments of inertia:

Object	Rotational Inertia
Solid disc/cylinder (axis through centre)	$\dfrac{1}{2}MR^2$
Hoop/ring (axis through centre)	$MR^2$
Solid sphere	$\dfrac{2}{5}MR^2$
Thin rod (axis through centre)	$\dfrac{1}{12}ML^2$
Thin rod (axis through end)	$\dfrac{1}{3}ML^2$
Point mass at distance $r$	$mr^2$

$$L = I\omega$$

For a point mass: $L = mvr_\perp$

Conservation of angular momentum: When no net external torque acts on a system, $I\omega$ is constant:

$$I_i\omega_i = I_f\omega_f$$$$K_r = \frac{1}{2}I\omega^2$$

Total kinetic energy for a rolling object:

$$K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For an object rolling without slipping:

$$v_{\text{cm}} = R\omega$$ $$a_{\text{cm}} = R\alpha$$

Acceleration of a solid sphere rolling down an incline of angle $\theta$:

$$a = \frac{5}{7}g\sin\theta$$

An object undergoes SHM when a restoring force is proportional to the displacement from equilibrium:

$$F = -kx$$

This yields sinusoidal motion:

$$x(t) = A\cos(\omega t + \phi)$$

where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

$$\omega = \sqrt{\frac{k}{m}}, \quad T = 2\pi\sqrt{\frac{m}{k}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

where $T$ is the period (time for one complete oscillation) and $f$ is the frequency.

For small angles ($\theta < 15°$), a simple pendulum approximates SHM:

$$\omega = \sqrt{\frac{g}{L}}, \quad T = 2\pi\sqrt{\frac{L}{g}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$$

Note: the period of a pendulum is independent of mass.

Total mechanical energy is conserved and constant:

$$E_{\text{total}} = \frac{1}{2}kA^2 = K_{\text{max}} = U_{\text{max}}$$

At any position $x$:

$$K = \frac{1}{2}k(A^2 - x^2), \quad U = \frac{1}{2}kx^2$$$$T = \frac{1}{f}, \quad f = \frac{1}{T}, \quad \omega = 2\pi f = \frac{2\pi}{T}$$$$v = v_0 + at \quad x = x_0 + v_0 t + \tfrac{1}{2}at^2 \quad v^2 = v_0^2 + 2a\Delta x$$ $$x_{\text{max}} = \frac{v_0^2\sin^2\theta}{2g} \quad R = \frac{v_0^2 \sin 2\theta}{g}$$$$F_{\text{net}} = ma \quad f_k = \mu_k N \quad F_c = \frac{mv^2}{r}$$$$K = \tfrac{1}{2}mv^2 \quad U_g = mgh \quad U_s = \tfrac{1}{2}kx^2$$ $$W = Fd\cos\theta \quad P = \frac{W}{\Delta t} = Fv\cos\theta$$$$p = mv \quad J = F_{\text{avg}}\,\Delta t = \Delta p$$$$\tau = I\alpha \quad L = I\omega \quad K_r = \tfrac{1}{2}I\omega^2$$$$T_{\text{spring}} = 2\pi\sqrt{\frac{m}{k}} \quad T_{\text{pendulum}} = 2\pi\sqrt{\frac{L}{g}}$$

1. Draw a free-body diagram for every problem. It earns points, organises your thinking, and helps identify all forces. Never skip this step on the free-response section.
2. Check dimensions and units at every step. If your answer has units of kg·m/s when you expect joules, there is an error. Dimensional analysis catches many mistakes quickly.
3. Define your coordinate system. State which direction is positive and where the origin is. This prevents sign errors and makes your reasoning clear to the grader.
4. Distinguish between vector and scalar quantities. Displacement is not distance; velocity is not speed. The AP exam frequently tests this distinction.
5. Apply conservation laws first. Before writing out Newton’s second law for every object, check whether conservation of energy or momentum solves the problem more efficiently.
6. Explain your reasoning in words. The AP Physics 1 exam emphasises conceptual understanding. After writing equations, explain what each term represents and why the equation applies.
7. Use graphs to support your answers. The exam often asks you to sketch or interpret graphs. Practise translating between equations, graphs, and physical descriptions.

1. Confusing mass and weight. Mass is an intrinsic property measured in kg; weight is the gravitational force $mg$ measured in newtons. An object has the same mass on the Moon but different weight.
2. Including internal forces in $\vec{F}_{\text{net}}$. Only external forces contribute to the net force on a system. The tension in a rope between two parts of the same system is an internal force and cancels out.
3. Forgetting the normal force is not always $mg$. On an incline, $N = mg\cos\theta$. In an accelerating lift, $N = m(g \pm a)$. Always derive $N$ from Newton’s second law.
4. Mixing up centripetal and centrifugal forces. “Centrifugal force” is a fictitious force that appears in a rotating reference frame. In an inertial frame, only the centripetal force (directed towards the centre) acts on the object.
5. Using the wrong collision formula. Only use elastic collision equations when kinetic energy is conserved. For perfectly inelastic collisions, use conservation of momentum alone and set final velocities equal.
6. Incorrect rotational inertia. Do not assume $I = MR^2$ for all objects. A solid disc has $I = \tfrac{1}{2}MR^2$; a solid sphere has $I = \tfrac{2}{5}MR^2$. Know the common values.
7. Applying the small-angle approximation outside its range. The pendulum formula $T = 2\pi\sqrt{L/g}$ only holds for angles below roughly 15°. For larger amplitudes, the period is longer.

Topic	Key Ideas
Kinematics	Displacement, velocity, acceleration, projectile motion, free fall, motion graphs
Dynamics	Newton’s three laws, free-body diagrams, friction, inclined planes, circular motion
Work and Energy	Work, work-energy theorem, kinetic/potential energy, conservation of energy, power
Momentum	Impulse-momentum theorem, conservation of momentum, elastic and inelastic collisions
Rotational Motion	Torque, rotational inertia, angular momentum, conservation of angular momentum, rolling
Oscillations	SHM, mass-spring systems, pendulums, energy in oscillations

AP Physics 1 rewards deep conceptual understanding over memorisation. Focus on building physical intuition — ask yourself why an object behaves as it does before reaching for an equation. Practise explaining your reasoning evidently in writing, as the free-response section heavily weights qualitative explanations alongside quantitative solutions.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.