The Ultimate AP Physics 1 Comprehensive Study Guide and Cheat Sheet

Kinematics (10–15%)

To master kinematics, follow a consistent procedure: sketch the axes, list all known variables, circle the unknowns you are solving for, select an equation containing those variables, and solve. Always remember that $\Delta = \text{final} - \text{initial}$ and watch signs closely. Use the following decision tree to pick equations: if you have $v_0$, $a$, and $t$, use $v = v_0 + at$ or $\Delta x = v_0t + \frac{1}{2}at^2$. If you have $v_0$, $v$, and $a$ but no $t$, use $v^2 = v_0^2 + 2a\Delta x$. If you have $t$ but no $a$, use $\Delta x = v_{\text{avg}} \times t = \frac{v_i+v_f}{2} \times t$. If one variable is missing, it is your unknown; if a variable is extra and unused, delete it from consideration.

Several common traps exist in kinematics. Trap 1 involves the misconception that "slowing" means negative acceleration. For a ball thrown upward, $v$ is positive but $a$ is negative (opposite directions). At the peak of flight, $v = 0$ but $a = -g$ (it is not zero). To fix this, plot a $v-t$ graph where the slope equals $a$. Trap 2 is that displacement $\Delta x$ does not equal distance; in a U-turn scenario, $\Delta x = 0$ while distance is greater than zero. Trap 3 concerns objects dropped vs objects thrown horizontally: objects at the "same height" have the same fall time, but not the same final speed. For a dropped object, $v_{0} = 0$, whereas for one thrown horizontal, $v_{0,y} = 0$ and $v_{0,x} \neq 0$. Only the vertical motion and fall time are identical; final speeds differ. Proportional reasoning dictates that if $a$ doubles, $v_f$ doubles (linear relationship to $a$). If $a$ doubles, $\Delta x$ quadruples (proportional to $a$ in the transcript). If $v_0$ doubles, stopping distance increases by a factor of $4$ (quadratic relationship to $v_0^2$). Time is independent of both.

For projectiles, motion must be split into $x$ and $y$ components. In $x$-motion, $v_x = v_{0,x}$ which is a constant (assuming no air resistance), and $a_x = 0$. Use the equation $\Delta x = v_x \times t$. In $y$-motion, $a_y = -g$ always. Use $v_y = v_{0,y} - gt$, $\Delta y = v_{0,y}t - \frac{1}{2}gt^2$, or $v_y^2 = v_{0,y}^2 - 2g\Delta y$. The strategy is to solve for the $y$-motion first (finding $t$ when the object lands based on $\Delta y$) and then use that $t$ to find the $x$ distance. At the peak of flight, $v_y = 0$, meaning the time to the peak is $t_{\text{up}} = \frac{v_{0,y}}{g}$. For a symmetric launch (level start and end), the range is maximized at $45^\circ$, and $t_{\text{up}} = t_{\text{down}}$. If landing on a ramp, the relationship is $y = -x \times \tan(\theta)$, which must be solved simultaneously with the $x$ and $y$ equations.

Graph reading is essential: the slope of an $x-t$ graph is $v$, the slope of a $v-t$ graph is $a$, and the area under a $v-t$ graph is $\Delta x$. A straight line on a $v-t$ graph indicates constant $a$, while a curved line on an $x-t$ graph indicates non-zero $a$. Note that peak velocity is purely horizontal because peak vertical velocity is zero ($v_y = 0$, $v_x = \text{const}$). Horizontal speed never changes in the air. For FRQ strategies, clearly state the equation, substitute values, and show algebra steps. To "justify," cite the kinematics equation and state why it applies. To "derive," begin with a fundamental equation like $v = v_0 + at$ and manipulate it algebraically.

Force & Dynamics (18–23%)

When creating a Free Body Diagram (FBD), always include Weight $mg$ (downward), Normal Force $N$ (perpendicular to surface, away from object), Tension $T$ (along a rope, pulling away), Friction $f$ (parallel to the surface, opposing motion), and Applied Force $F$ (direction based on the problem). Never draw reaction forces exerted from the object on others.

There are significant traps regarding Normal Force and Tension. Normal Force $N$ is not always equal to $mg$. On a ramp, $N = mg \times \cos(\theta)$ (the perpendicular component). In an elevator moving up, $N = m(g+a)$, and moving down, $N = m(g-a)$. At the top of a loop, N < mg where both forces point toward the center. In free fall, the object is weightless and $N = 0$. Regarding Tension, it does not always equal weight. For an Atwood machine, write $\Sigma F = ma$ for each mass separately: $m_1g - T = m_1a$ for the heavier mass moving down, and $T - m_2g = m_2a$ for the lighter mass moving up. Adding these gives $(m_1-m_2)g = (m_1+m_2)a$, after which you can solve for Tension with $T = m_1(g-a)$.

Friction follow specific rules: static friction is $f_s \leq \mu_s \times N$ and adjusts to applied force up to its maximum; kinetic friction is $f_k = \mu_k \times N$ and is constant regardless of speed. Note that \mu_s > \mu_k always. If moving, use kinetic; if at rest with no push, $f = 0$. Problem setup involves a decision tree. For an incline, tilt the axes parallel and perpendicular to the ramp. Parallel: $mg \times \sin(\theta) - f = ma$. Perpendicular: $N = mg \times \cos(\theta)$. For horizontal circular motion, the net inward force is $\frac{mv^2}{r}$ and centripetal acceleration is $a = \frac{v^2}{r}$ toward the center. For connected objects, find the total acceleration using $\frac{\Sigma F_{\text{total}}}{m_{\text{total}}}$ and backsolve for individual rope tensions.

Newton's 3rd Law states that equal and opposite pairs act on different objects and never cancel. Constant speed means $a = 0$ and the system is in equilibrium ($\Sigma F = 0$), but constant speed in a circle means $a \neq 0$ as there is centripetal acceleration toward the center. Distinguish between a constant magnitude of velocity and a zero acceleration vector. Generally, use Forces when you need $a$, $T$, $N$, or $f$. Use Energy when you need speed and path length does not matter.

Work, Energy & Power (18–23%)

Work is defined as $W = F \times d \times \cos(\theta)$, where $\theta$ is the angle between the force and motion. If force is perpendicular to motion, $W = 0$. Key energy formulas include Kinetic Energy $KE = \frac{1}{2}mv^2$, Gravitational Potential Energy $PE_{\text{grav}} = mgh$ (referenced to a convenient $h=0$), and Spring Potential Energy $PE_{\text{spring}} = \frac{1}{2}kx^2$ (measured from equilibrium, not natural length). Conservative forces like gravity and springs are path-irrelevant and have an associated PE. Non-conservative forces like friction and air drag are path-dependent and do not have a PE. Work done by friction is $W_f = -\mu \times N \times d$ and is always negative, representing energy lost to heat.

Proportional scaling and traps in energy involve the quadratic nature of speed: if $v$ doubles, $KE$ increases by a factor of $4$ ($KE \propto v^2$). Stopping distance similarly scales by $v^2$ (since $KE = \text{work against friction} = \mu mgd$). Halving the friction coefficient doubles the stopping distance. Friction removes energy as heat, which is irreversible and path-dependent; a longer path leads to more dissipation, reducing max height and final speed. In a spring-gravity system, equilibrium occurs at $\Delta x = \frac{mg}{k}$; the system oscillates around this point. Include both $PE_{\text{spring}} = \frac{1}{2}kx^2$ and $PE_{\text{gravity}} = -mgx$ when measuring from natural length.

Energy bar charts use columns for $KE$, $PE_{\text{grav}}$, $PE_{\text{spring}}$, and $E_{\text{friction lost}}$. The heights represent energy amounts, and initial and final states are shown side-by-side to illustrate energy flow and dissipation. Power is defined as $P = \frac{W}{\Delta t} = F \times v$ (instantaneous). It represents how fast work is done and is measured in Watts ($J/s$). Average power is $P_{\text{avg}} = \frac{\text{total } W}{\text{total } \Delta t}$. In systems with no friction, mechanical energy is conserved. If friction is present, $KE_i + PE_i = KE_f + PE_f + E_{\text{friction dissipated}}$. The Work-Energy Theorem states $W_{\text{net}} = \Delta KE$, and should be used when speed change is needed without tracking every individual force.

Linear Momentum (10–15%)

Momentum is the vector $p = m \times v$. Impulse is $J = F \times \Delta t = \Delta p$. A longer collision duration results in a smaller average force. In an elastic collision, both $p$ and $KE$ are conserved. In an inelastic collision, $p$ is conserved but $KE$ decreases due to heat or deformation. In an explosion, $p$ is conserved while $KE$ increases as internal energy is released.

In 1D collision methods, assigning signs is critical: $m_1v_1 + m_2v_2 = (m_1+m_2)v_{\text{final}}$. If objects move in opposite directions, use opposite signs rather than just adding magnitudes. In a perfectly inelastic collision, objects stick together to move with one final velocity. Energy lost to heat, sound, or deformation is calculated as $\Delta KE_{\text{lost}} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 - \frac{1}{2}(m_1+m_2)v_{\text{final}}^2$. For recoil where $p_{\text{initial}} = 0$, the relation is $m_1v_1 + m_2v_2 = 0$, leading to $v_1 = -\frac{m_2}{m_1}v_2$. Here, the lighter object moves much faster ($v \propto \frac{1}{m}$); gun recoil is proportional to the bullet mass divided by the gun mass.

Advanced topics include 2D collisions and the ballistic pendulum. In 2D, conserve $p_x$ and $p_y$ separately if there are no external forces in those directions. Find the final velocity using $v_f = \sqrt{v_{fx}^2 + v_{fy}^2}$ and the angle using $\arctan(\frac{v_{fy}}{v_{fx}})$. The ballistic pendulum involves two steps: first, a momentum collision ($m_{\text{bullet}} \times v_b = (m_b + m_{\text{block}}) \times v_{\text{after}}$), then an energy swing ($\frac{1}{2}(m_b + m_{\text{block}}) \times v_{\text{after}}^2 = (m_b + m_{\text{block}}) \times g \times h$) to backsolve for $v_b$. Center of mass velocity $v_{\text{cm}}$ is constant if $\Sigma F_{\text{ext}} = 0$. Even if pieces separate in an explosion, the center of mass velocity remains unchanged.

Momentum is a vector; never add magnitudes in multi-directional problems. Proportional reasoning shows that doubling $m$ or $v$ doubles $p$, but collision impulse is proportional to $\Delta v$, not $\Delta p^2$. A longer contact time leads to a lower average force. On an impulse graph, the area under the $F$-vs-$t$ curve is impulse $\Delta p$. For a triangular force, $J = \frac{1}{2} \times F_{\text{max}} \times \Delta t$, while for a constant force, $J = F \times \Delta t$. Choosing a method: use Momentum for collisions and explosions, Energy for speed and height, and Forces for acceleration and tension. Many FRQs will require these methods sequentially.

Torque & Rotation (10–15%)

Torque is calculated as $\tau = r \times F \times \sin(\theta) = r_{\text{perp}} \times F$, where $r_{\text{perp}}$ is the perpendicular distance from the pivot to the line of force. Careful geometry drawing is required. Following sign convention, counterclockwise is positive and clockwise is negative. A smart pivot choice is to pick the pivot at the location of an unknown force so that its torque is zero, simplifying the equation. For example, in a ladder leaning on a wall, picking the pivot at the wall makes the wall force torque vanish.

Static equilibrium requires both $\Sigma F = 0$ and $\Sigma \tau = 0$. For instance, for a plank on two supports, $\Sigma F_y = 0$ finds the total support force, while $\Sigma \tau = 0$ finds the distribution of force (which is unequal if the load is off-center). Gravity acts at the Center of Mass (CM). For uniform objects, the CM is the geometric center. In non-uniform objects, find the CM first then apply weight there. Note that $\Sigma \tau_{\text{COM}}$ is not always zero if the object is rotating.

Moment of Inertia $I = \Sigma m \times r^2$. Standard shapes include a point mass ($mr^2$), disk or cylinder ($\frac{1}{2}mr^2$), solid sphere ($\frac{2}{5}mr^2$), hollow sphere ($\frac{2}{3}mr^2$), rod from the end ($\frac{1}{3}mL^2$), and rod from the center ($\frac{1}{12}mL^2$). Getting farther from the axis increases $I$ (specifically $I \propto r^2$). Rotational dynamics follows linear analogs: $x$ to $\theta$, $v$ to $\omega$, $a$ to $\alpha$, $F$ to $\tau$, and $m$ to $I$. Kinematics equations are identical in form: $\omega = \omega_0 + \alpha t$, $\theta = \omega_0t + \frac{1}{2}\alpha t^2$, and $\omega^2 = \omega_0^2 + 2\alpha\theta$. In rolling without slipping, $v_{\text{cm}} = r \times \omega$ and $a_{\text{cm}} = r \times \alpha$. Friction is static and does no net work. Massive pulleys result in different tensions ($(T_1 - T_2)R = I\alpha$), whereas massless pulleys have $T_1 = T_2$. The Parallel Axis Theorem states $I_{\text{parallel}} = I_{\text{COM}} + M \times d^2$, where $d$ is the distance from the COM to the new axis.

Rotational Energy & Angular Momentum (5–8%)

Angular momentum $L = I\omega$ is conserved when $\Sigma \tau_{\text{ext}} = 0$. In rolling motion, the total kinetic energy is $KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Using the $v = r\omega$ constraint, this becomes $KE = \frac{1}{2}mv^2(1 + \frac{I}{mr^2})$. The shape factor determines final speed, not mass or radius. For a sphere where $\frac{I}{mr^2} = \frac{2}{5}$, the total $KE = \frac{1}{2}mv^2 \times 1.4$. In a ramp race, the order from fastest to slowest is: Solid sphere (1.4 factor), Solid cylinder (1.5), Hollow sphere (1.67), and Hollow cylinder (2). Final speed is proportional to $\sqrt{\frac{1}{1 + \frac{I}{mr^2}}}$. Energy conservation in rolling means $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. If sliding (frictionless), the final speed is $v = \sqrt{2gh}$; in rolling, speed is slower: $v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^2}}}$. Static friction in rolling does no net work and reorients momentum rather than dissipating energy.

When $\Sigma \tau_{\text{ext}} = 0$, $L$ is conserved such that $I_i\omega_i = I_f\omega_f$. In an ice skater scenario, pulling arms in decreases $I$ and increases $\omega$. While $L$ is constant, the muscles do positive work, increasing $KE$. In a turntable or disc collision, angular momentum is constant: $I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_{\text{final}}$. Energy is lost in this inelastic process. Proportionality shows that doubling $\omega$ or $I$ doubles $L$. For a fixed $L$, if $I$ doubles, $\omega$ is halved. If $I$ doubles and $\omega$ is halved, the rotational energy $KE_{\text{rot}} = \frac{1}{2}I\omega^2$ remains the same. A key trap is the skater spinning faster: $KE$ increases because muscles perform work against centrifugal effects.

Simple Harmonic Motion (SHM) (5–8%)

The period of a spring is $T = 2\pi\sqrt{\frac{m}{k}}$ and only depends on $m$ and $k$. Doubling $m$ results in $T \times \sqrt{2}$, while doubling $k$ leads to $T \div \sqrt{2}$. Note that Amplitude $A$ and gravity do not affect the period. Halving a spring length doubles $k$, leading to $T \div \sqrt{2}$. For a pendulum, $T = 2\pi\sqrt{\frac{L}{g}}$ (at small angles < 15^\circ). Mass and amplitude do not matter. A longer string increases $T$, and the period is proportional to $\sqrt{g}$. A trap is the independence of amplitude; this holds for small angles but fails for large angles where dynamics become nonlinear.

In SHM, energy is exchanged. At equilibrium ($x = 0$), $v = v_{\text{max}}$, $a = 0$, $KE$ is at its maximum, and $PE = 0$. At the peak ($x = \pm A$), $v = 0$, $a = a_{\text{max}}$, $KE = 0$, and $PE$ is at its maximum. Total energy is $E_{\text{total}} = \frac{1}{2}kA^2$ and is constant without friction. Doubling amplitude $A$ results in $E_{\text{total}} \times 4$ ($E \propto A^2$). Doubling $k$ doubles $E$, but doubling $m$ leaves $E$ unchanged for a fixed $A$. Kinematic positions are: $x = A\cos(\omega t + \phi)$, $v = -A\omega\sin(...)$, and $a = -A\omega^2\cos(...)$ with $v_{\text{max}} = A\omega = A\sqrt{\frac{k}{m}}$. Acceleration leads position by $180^\circ$ and velocity leads position by $90^\circ$.

When a spring hangs vertically, the equilibrium shifts down by $\Delta x = \frac{mg}{k}$; the period $T$ is unchanged and the system oscillates around this new equilibrium. For spring combinations: series springs follow $\frac{1}{k_{\text{total}}} = \frac{1}{k_1} + \frac{1}{k_2}$ (weaker), while parallel springs follow $k_{\text{total}} = k_1 + k_2$ (stiffer). In damped SHM, friction removes energy, causing amplitude to decay according to $A(t) = A_0e^{-bt}$; the frequency also decreases slightly. In proportional reasoning, adding mass increases $T$ ($T \propto \sqrt{m}$). Cutting a spring in half doubles $k$. Doubling amplitude increases $E \times 4$ but only doubles $v_{\text{max}}$.

Fluids (10–15%)

Pressure at depth depends only on depth $h$ and not on container shape or object size according to $P = P_{\text{atm}} + \rho gh$. Gauge pressure is $P = \rho gh$ (above atmosphere), while Absolute pressure includes the atmosphere. Pressure is identical at the same depth regardless of the width of the vessel. Buoyancy, or Archimedes' principle, states $F_b = \rho_{\text{fluid}} \times V_{\text{submerged}} \times g$, representing the weight of the displaced fluid. This force acts upward from the center of buoyancy. In floating equilibrium, $F_b = mg$. The fraction submerged is $\frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$. An object floats if \rho_{\text{obj}} < \rho_{\text{fluid}}, is neutrally buoyant if they are equal, and sinks if \rho_{\text{obj}} > \rho_{\text{fluid}}. FBD analysis for fluids involves $F_b$ pointing up and weight $mg$ pointing down. A trap to avoid is that $F_b$ depends solely on the submerged volume, not the total volume.

Fluid flow is governed by the Continuity equation $A_1v_1 = A_2v_2 = Q$, where $Q$ is the constant volumetric flow rate. Narrow sections have faster flow ($v \propto \frac{1}{A}$). Bernoulli’s equation states $P + \frac{1}{2}\rho v^2 + \rho gh = \text{const}$ along a streamline. Faster flow results in lower pressure, and higher elevation results in lower pressure, representing a trade-off between pressure and kinetic energy. Torricelli's theorem finds the speed exiting a hole at depth $h$ to be $v = \sqrt{2gh}$, identical to free-fall speed from that same depth. Gauge pressure drives this flow. Flow rate is $Q = A \times v = A \times \sqrt{2gh}$. Bernoulli assumes ideal fluids which are incompressible, non-viscous, steady, and laminar. Real fluids possess viscosity and turbulence.

Exam Format & Problem-Solving Strategies

The AP Physics 1 exam consists of $40$ Multiple Choice Questions (MCQ) over $80$ minutes and $4$ Free Response Questions (FRQ) over another $80$ minutes. There is no penalty for guessing. You have approximately $2$ minutes per MCQ. Calculators and a reference sheet are provided. When an FRQ asks you to "Justify," you must provide the physical principle and an explanation of why it applies. To "Derive" means to show the algebra starting from fundamental principles.