AP Physics 1
Kinematikten salınımlara: kavram, formül ve hata teşhisli sorular.
AP Physics 1 Ünite İlerleme Haritası
Tamamlanan üniteler: konu anlatımı ve sorulara doğrudan bağlantılar.
AP Physics Sınav Mimarisi
Sayısal branş olarak hibrit dijital kurallara tabidir:
Çoktan Seçmeli (MCQ) • Bluebook Ekranı
• Grafik, veri tablosu ve çoklu öncüllü sorular.
• Bluebook'ta yerleşik formül sayfası ve bilimsel/grafik hesap makinesi mevcuttur.
• 40 soru / 80 dakika (Physics 1, 2025 itibarıyla güncel format).
Serbest Yanıtlı (FRQ) • El Yazılı Kitapçık
• Deney Tasarımı: Bağımsız, bağımlı ve kontrol değişkenlerini açıkça belirt.
• Vektör Çizimleri: Cisim üzerine etkiyen tüm kuvvet oklarını temas noktalarından başlat.
• Sorular ekrandan okunup fiziksel cevap kağıdına yazılır.
• 4 FRQ / 100 dakika: Mathematical Routines (MR), Translation Between Representations (TBR), Experimental Design and Analysis (LAB), Qualitative/Quantitative Translation (QQT).
Kritik Fizik Formülleri
Sınavda verilen resmî formül sayfasıyla aynı gösterim (cebir tabanlı; integral ve türev yok). Asıl sayfa: AP Physics 1 Exam (College Board).
Yalnızca ivme sabitken geçerli. Eğik atışta yatay ve düşey hareketi ayrı ayrı yaz; yatayda \(a_x = 0\).
Statik sürtünme bir eşitsizliktir; en büyük değerine yalnızca kaymanın eşiğinde ulaşır. Merkezcil kuvvet yeni bir kuvvet değildir, net kuvvetin merkeze doğru bileşenidir.
Korunumlu kuvvetler varken mekanik enerji korunur: \(\Delta K + \Delta U = 0\). Sürtünme sistemden mekanik enerji çıkarır.
Dış net kuvvet sıfırsa toplam momentum her çarpışmada korunur; kinetik enerji yalnızca esnek çarpışmada korunur. Kuvvet–zaman grafiğinin altındaki alan itmedir.
Dış net tork sıfırsa açısal momentum korunur: \(I_i \omega_i = I_f \omega_f\). Kaymadan yuvarlanmada \(v_{cm} = R\omega\).
Yayda periyot g'ye bağlı değildir; küçük açılı sarkaçta periyot kütleye bağlı değildir.
Kaldırma kuvvetinde \(\rho\) akışkanın yoğunluğu, \(V\) batan hacimdir. Boru daralınca hız artar, basınç düşer (Bernoulli).
Yüzeye yakın yerde \(\Delta U_g = m g \Delta y\) kullanılır; uzak mesafelerde \(U_G\) negatiftir ve sonsuzda sıfırdır.
Laboratuvar Tasarım Sorularında Tam Puan Alma Kuralları
College Board değerlendiricilerinin aradığı 5 temel kriter:
AP Physics 1 Core Concepts
Algebra-based laws, free-body analysis and qualitative reasoning the AP Physics 1 exam rewards (no calculus required):
Newton's Second Law: The acceleration of a system equals the net external force divided by the system's mass:
\(\vec{a}_{sys} = \dfrac{\sum \vec{F}_{ext}}{m_{sys}}\)
1. Represent the body as a distinct point mass or extended rigid object.
2. Draw force vectors starting exactly at their point of application (e.g., gravity starts at the Center of Mass; normal force starts at the contact surface).
3. Never draw component vectors (such as $mg \sin\theta$ or $mg \cos\theta$) on the official FBD! Draw only the primary physical forces ($mg$, $F_N$, $F_f$, $T$). Components belong only in your scratch working.
Work by a Constant Force: Only the force component parallel to the displacement does work:
\(W = F_\parallel d = F d \cos\theta\)
The Work-Energy Theorem: The net work done by all forces equals the change in kinetic energy: \(W_{net} = \Delta K = \tfrac{1}{2} m v_f^2 - \tfrac{1}{2} m v_i^2\). On a force–displacement graph, work is the area under the curve.
Linear Momentum & Impulse: \(\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}\) (area under a force–time graph). In any isolated system ($\sum \vec{F}_{ext} = 0$), total momentum is conserved in all collisions (elastic, inelastic, completely inelastic). Kinetic energy is conserved only in perfectly elastic collisions!
→ Full Unit 4 Linear Momentum masterclass (Topics 4.1–4.4) and 23-question station
Rotational Analogue of Newton's 2nd Law:
\(\tau = r F \sin\theta \qquad \alpha_{sys} = \dfrac{\tau_{net}}{I_{sys}}\)
Rotational inertia depends on how mass is distributed about the axis (Solid cylinder: $\tfrac{1}{2} M R^2$; Solid sphere: $\frac{2}{5} M R^2$; Rod about center: $\frac{1}{12} M L^2$). For point masses: $I = \sum m r^2$.
\(v_{cm} = R \omega \quad \text{and} \quad a_{cm} = R \alpha\)
Total kinetic energy: \(K_{total} = \tfrac{1}{2} M v_{cm}^2 + \tfrac{1}{2} I_{cm} \omega^2\).
Unit 1: Kinematics (1D & 2D Motion) — Conceptual & Mathematical Rigor
Kinematics mathematically describes motion without considering the forces that cause it. Below are the foundational vector definitions, constant acceleration equations, graphical analysis techniques, and experimental design methodologies required for AP 5/5 mastery:
Position ($\vec{x}$ or $\vec{r}$): The location of an object relative to an origin. A vector quantity possessing both magnitude and direction.
Displacement ($\Delta \vec{x} = \vec{x}_f - \vec{x}_i$): The net vector change in position. Path-independent!
Distance ($d$): The actual scalar path length traveled ($d \ge |\Delta \vec{x}|$).
• Average Velocity: $\vec{v}_{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t}$ (vector pointing in the direction of displacement).
• Average Speed: $s_{\text{avg}} = \frac{\text{Total Distance}}{\Delta t}$ (always scalar and non-negative).
• Instantaneous Velocity (Calculus): $\vec{v}(t) = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt}$.
• Instantaneous Speed: The absolute magnitude of instantaneous velocity: $s(t) = |\vec{v}(t)|$.
• An object is speeding up if and only if velocity and acceleration point in the SAME direction ($\vec{v} \cdot \vec{a} > 0$).
• An object is slowing down if and only if velocity and acceleration point in OPPOSITE directions ($\vec{v} \cdot \vec{a} < 0$).
Never assume negative acceleration means slowing down! If $v = -8$ m/s and $a = -3$ m/s², the object is moving in the $-x$ direction and speeding up!
When acceleration is strictly constant ($a = \text{const}$), integration yields the four canonical kinematics equations:
1.
v_f = v_0 + a·t (Velocity as a linear function of time)2.
Δx = v_0·t + (1/2)·a·t² (Displacement as a quadratic function of time)3.
v_f² = v_0² + 2·a·Δx (Timeless equation linking speeds to displacement)4.
Δx = [(v_0 + v_f) / 2] · t (Displacement as average velocity times time)
Calculus Connection (Physics C · ileri): When acceleration is NOT constant ($a(t) = f(t)$), you MUST use calculus integration:
v(t) = v_0 + ∫_0^t a(t') dt' and x(t) = x_0 + ∫_0^t v(t') dt'.
Over 35% of multiple-choice and FRQ questions require extracting physics from graphs without numbers:
• Position vs. Time ($x-t$):
- Slope of tangent line $= \frac{dx}{dt} = \text{Instantaneous Velocity } v(t)$.
- Concavity: Concave up ($d^2x/dt^2 > 0$) means $a > 0$; Concave down ($d^2x/dt^2 < 0$) means $a < 0$.
• Velocity vs. Time ($v-t$):
- Slope of tangent line $= \frac{dv}{dt} = \text{Instantaneous Acceleration } a(t)$.
- Area under the curve: $\int_{t_1}^{t_2} v(t) dt = \Delta x$ (Net Displacement).
- Signed areas: Area above $t$-axis is $+$, area below $t$-axis is $-$.
• Acceleration vs. Time ($a-t$):
- Area under the curve: $\int_{t_1}^{t_2} a(t) dt = \Delta v = v_f - v_i$ (Change in velocity).
The Independence Principle: Horizontal and vertical motions are strictly perpendicular and act completely independently of each other. The ONLY variable linking both dimensions is time ($t$).
• Horizontal Dimension ($x$): Constant velocity ($a_x = 0$):
\(v_x = v_0 \cdot cos(θ) = \text{const}, \quad x(t) = (v_0 \cdot cos(θ)) \cdot t\).
• Vertical Dimension ($y$): Free-fall acceleration ($a_y = -g$):
\(v_y(t) = v_0 \cdot sin(θ) - g \cdot t, \quad y(t) = y_0 + (v_0 \cdot sin(θ)) \cdot t - (1/2) \cdot g \cdot t^2\).
• Apex (Highest Point): $v_y = 0 \implies t_{\text{apex}} = \frac{v_0 \sin\theta}{g}$. Maximum height: $H = \frac{(v_0 \sin\theta)^2}{2g}$.
• Horizontal Range (Flat Ground): $R = v_x \cdot (2 t_{\text{apex}}) = \frac{v_0^2 \sin(2\theta)}{g}$ (Maximum at $\theta = 45^\circ$).
If Ball A is dropped from rest and Ball B is fired horizontally from the same height $h$, BOTH strike the flat ground at the exact same instant ($t = \sqrt{2h/g}$) because both have zero initial vertical velocity ($v_{y0} = 0$) and identical vertical acceleration ($-g$)!
Graph Linearization in Kinematics Experiments:
To experimentally determine $g$ from a free-fall experiment ($h = \frac{1}{2} g t^2$):
1. Never plot $h$ vs. $t$ (yields a curved parabola).
2. Linearize by setting $y = h$ and $x = t^2$. The resulting line is $y = m x$ where slope $m = \frac{g}{2}$.
3. Calculate experimental acceleration: $g = 2 \cdot (\text{Slope of best-fit line})$.
• Mathematical Routines (MR): symbolic and numerical derivations from first principles.
• Translation Between Representations (TBR): move between diagrams, graphs, equations, and words.
• Experimental Design and Analysis (LAB): procedure, linearization, slope interpretation, uncertainty.
• Qualitative/Quantitative Translation (QQT): a written argument that must agree with a derived equation.
(The old "Paragraph-Length Response (PLR)" label belongs to the pre-2025 exam and is no longer a separate FRQ type.)
1. State the fundamental physical principle: e.g., "In the absence of net external horizontal forces, horizontal velocity remains constant."
2. Connect principle to variables via mathematical proportionality: "Since $t = \sqrt{2h/g}$, flight time depends only on height $h$. Doubling $v_0$ doubles horizontal displacement $x = v_0 t$ because $t$ is unchanged."
3. Conclude with direct resolution of the prompt.
Unit 1 Kinematics: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 2: Force and Translational Dynamics (Newton's Laws) — Mechanical Rigor
Dynamics explains the physical causes of motion through interactions between bodies and their surroundings. Below are the foundational laws of motion, free-body diagram protocols, contact force models, coupled system mechanics, and differential fluid drag formulations required for AP Physics 5/5 mastery:
System Definition: An object or collection of objects treated as a single entity. Forces acting between constituents of the system are internal forces and cancel out in pairs; only external forces change the momentum of the center of mass.
1. Dot or Body Representation: Represent the object as a point mass (or shape if torque is considered).
2. Origin of Vectors: Every force vector must originate directly on the dot or point of contact and point strictly in the direction of the exerted force.
3. No Components on Official FBDs: NEVER draw broken-down components ($mg\sin\theta, mg\cos\theta$) on an official College Board FBD! Doing so costs the diagram point. Draw only fundamental forces ($F_g, F_N, F_f, F_T$). If you need components, draw a separate scrap diagram.
4. Relative Arrow Lengths: Vector lengths must accurately reflect relative magnitudes (e.g., if moving at constant velocity on a horizontal surface with friction, $|F_{\text{pull}}| = |f_k|$).
Newton's First Law (Law of Inertia): An object in an inertial reference frame remains at rest or continues at constant velocity in a straight line unless acted upon by a non-zero net external force ($\sum \vec{F}_{\text{ext}} = 0 \iff \vec{a} = 0$).
The translational acceleration of a system is directly proportional to the net external force acting on it and inversely proportional to the total system inertial mass:
$$\vec{a} = \frac{\sum \vec{F}_{\text{ext}}}{m_{\text{total}}} \iff \sum F_x = m a_x \quad \text{and} \quad \sum F_y = m a_y$$ Key AP Principle: Net force and acceleration are always collinear and codirectional. An object moving at constant speed along a curved path is NOT in equilibrium because the direction of velocity changes ($a_c = v^2/r \ne 0$).
A scale measures the normal contact force $F_N$ exerted by the scale on the person (apparent weight), NOT the actual gravitational force $mg$!
• Accelerating upward ($+a$): $F_N - mg = m(+a) \implies F_N = m(g + a)$ (Feels heavier).
• Accelerating downward ($-a$): $F_N - mg = m(-a) \implies F_N = m(g - a)$ (Feels lighter).
• Free-fall ($a = -g$): $F_N = m(g - g) = 0$ (Weightless sensation; no contact force).
Newton's Third Law: Whenever object A exerts a force on object B ($\vec{F}_{A \to B}$), object B simultaneously exerts an equal in magnitude and opposite in direction force on object A ($\vec{F}_{B \to A} = -\vec{F}_{A \to B}$).
1. They act on TWO DIFFERENT OBJECTS (Object A on B, Object B on A).
2. They are of the EXACT SAME PHYSICAL TYPE (both gravitational, both contact/normal, both magnetic, etc.).
3. They have IDENTICAL MAGNITUDES at all times, regardless of whether the system is accelerating, at rest, or of vastly different masses!
4. They act in OPPOSITE DIRECTIONS along the line joining the centers.
For a book resting on a table, $F_N$ (table pushing up on book) and $F_g$ (Earth pulling down on book) both act on the same single object (the book). They balance each other due to Newton's 2nd Law equilibrium ($\sum F_y = 0$).
• The true reaction to $F_g$ (Earth on book) is the gravitational force exerted by the book pulling UP on the Earth.
• The true reaction to $F_N$ (table on book) is the normal contact force exerted by the book pushing DOWN on the table.
Normal Force ($F_N$): The perpendicular contact force exerted by a surface on a body preventing interpenetration. It is an electromagnetic constraint force and self-adjusts to match external normal loads.
• Static Friction ($f_s$): Acts when there is no relative motion between surfaces at contact. It is an inequality:
$$0 \le f_s \le f_{s,\text{max}} = \mu_s F_N$$ Static friction is self-adjusting: it equals the applied parallel shear force until the maximum threshold $\mu_s F_N$ is exceeded!
• Kinetic Friction ($f_k$): Acts when surfaces slide relative to each other. It is constant and opposes relative motion:
$$f_k = \mu_k F_N \quad (\text{where typically } \mu_k < \mu_s)$$
For a crate resting stationary on an incline of angle $\theta$:
$\sum F_{\parallel} = mg\sin\theta - f_s = 0 \implies f_s = mg\sin\theta$.
Do NOT calculate $f_s = \mu_s mg\cos\theta$! The term $\mu_s mg\cos\theta$ is the maximum threshold before slipping occurs, not the actual friction force in equilibrium!
System Approach vs. Individual Free-Body Approach:
For connected masses (ideal massless, inextensible string over ideal frictionless pulley):
1. System Acceleration: Treat the entire multi-body setup as one system unfolding along the line of motion:
$$a = \frac{\sum F_{\text{external, driving}} - \sum F_{\text{external, resisting}}}{\sum m_{\text{total}}}$$
Internal tensions cancel out across the whole system.
2. Internal Tension ($T$): Isolate an individual mass and apply $\sum F = m_i a$.
• System acceleration: $a = \frac{m_2 g - f_{k1}}{m_1 + m_2}$ (where $f_{k1} = \mu_k m_1 g$).
• String Tension: Isolate hanging mass $m_2$: $m_2 g - T = m_2 a \implies T = m_2(g - a) < m_2 g$.
Critical Truth: Tension in the cord is strictly less than the weight of the hanging mass ($T < m_2 g$) because $m_2$ must have a net downward force to accelerate!
When a body moves through a fluid (liquid or gas), it experiences a resistive drag force $\vec{F}_{\text{drag}}$ opposite to velocity:
• Equation of motion: $m \frac{dv}{dt} = mg - bv$.
• At terminal velocity, acceleration ceases ($a = 0$): $mg - bv_t = 0 \implies v_t = \frac{mg}{b}$.
• Exact velocity function via differential separation of variables:
$$\int_0^v \frac{dv'}{g - \frac{b}{m}v'} = \int_0^t dt' \implies v(t) = v_t \left( 1 - e^{-\frac{b}{m}t} \right)$$ • Quadratic Drag ($F_d = \frac{1}{2} C_D \rho A v^2$, high-speed turbulent flow in air):
$mg - \frac{1}{2} C_D \rho A v_t^2 = 0 \implies v_t = \sqrt{\frac{2mg}{C_D \rho A}}$.
Unit 2 Force & Dynamics: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 2 (cont.): Circular Motion & Gravitation — Orbital Mechanics & Planetary Dynamics
Kinematics of Uniform Circular Motion (UCM): When a particle travels in a circular path of radius $r$ at constant speed $v$, its velocity magnitude is constant, but its direction changes continuously. This generates a purely radial, inward-directed acceleration: $$\vec{a}_c = -\frac{v^2}{r} \hat{r} = -\omega^2 r \hat{r} = -\frac{4\pi^2 r}{T^2} \hat{r}$$ where $\omega = \frac{2\pi}{T} = 2\pi f$ is the angular speed and $T$ is the period.
If speed changes along the curve, the particle possesses both radial and tangential accelerations: $$\vec{a}_{\text{total}} = \vec{a}_c + \vec{a}_t = -\left(\frac{v^2}{r}\right)\hat{r} + \left(\frac{dv}{dt}\right)\hat{\theta} \implies |\vec{a}_{\text{total}}| = \sqrt{a_c^2 + a_t^2}$$
There is NO distinct physical interaction called "centripetal force" ($F_c$), and centrifugal force does NOT exist in an inertial reference frame!
• The Core Truth: $\sum F_r = m a_c = \frac{m v^2}{r}$ is Newton's Second Law applied along the radial axis. The required centripetal acceleration is provided by real physical forces (tension $F_T$, gravity $F_g$, normal force $F_N$, or static friction $f_s$).
• Official Rubric Penalty: If you draw a vector labeled "$F_c$" on an AP Free-Body Diagram, College Board graders automatically deduct the diagram point! Draw ONLY the actual physical agents ($F_T, F_g, F_N, f_s$).
Frictionless Banked Curve & Conical Pendulum Equilibrium:
A road banked at angle $\theta$ allows a vehicle of mass $m$ to negotiate a curve of radius $r$ at a specific design speed $v_{\text{design}}$ with zero lateral friction:
$$\text{Vertical: } F_N \cos\theta - mg = 0 \implies F_N \cos\theta = mg$$
$$\text{Horizontal (Radial): } F_N \sin\theta = \frac{m v^2}{r}$$
Dividing the radial equation by the vertical equation eliminates mass and normal force:
$$\tan\theta = \frac{v_{\text{design}}^2}{r g} \implies v_{\text{design}} = \sqrt{r g \tan\theta}$$
• Speed Variations with Friction: If $v > v_{\text{design}}$, the car tends to skid up the incline, requiring static friction directed down the bank. If $v < v_{\text{design}}$, static friction acts up the bank.
For an object of mass $m$ moving in a vertical circle of radius $R$:
1. At the Bottom: Radial net force points upward: $$\sum F_r = F_N - mg = \frac{m v_{\text{bot}}^2}{R} \implies F_N = mg + \frac{m v_{\text{bot}}^2}{R} \quad (\text{Maximum Apparent Weight!})$$ 2. At the Top: Both normal force and gravity point downward toward the center: $$\sum F_r = F_N + mg = \frac{m v_{\text{top}}^2}{R} \implies F_N = \frac{m v_{\text{top}}^2}{R} - mg$$ 3. Critical Speed at the Apex: To complete the loop without falling or losing contact, the track must exert at least an infinitesimal contact force ($F_N \ge 0$). At the threshold where $F_N \to 0$: $$mg = \frac{m v_{\text{crit}}^2}{R} \implies v_{\text{crit}} = \sqrt{g R}$$ 4. Minimum Release Height for a Frictionless Roller Coaster: By conservation of mechanical energy ($E_{\text{release}} = E_{\text{top}}$): $$m g h = m g (2R) + \frac{1}{2}m v_{\text{top}}^2 \implies g h = 2g R + \frac{1}{2}(g R) \implies h_{\text{min}} = \frac{5}{2}R = 2.5R$$
The Universal Inverse-Square Law: Every point mass attracts every other point mass with a force along the line connecting their centers: $$F_g = G \frac{m_1 m_2}{r^2} \quad \left(G = 6.674 \times 10^{-11} \frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}\right)$$ By Newton's Third Law, both masses experience forces of identical magnitude regardless of any difference in their masses ($\vec{F}_{1 \to 2} = -\vec{F}_{2 \to 1}$).
• Altitude $h$ above Surface ($r = R_p + h$): $g(h) = \frac{G M}{(R_p + h)^2} = g_0 \left(\frac{R_p}{R_p + h}\right)^2$.
• Interior Field for a Uniform Density Planet ($r < R_p$): By Newton's Shell Theorem, outer shells exert zero net gravity; only the enclosed mass ($M_{\text{enc}} = M \frac{r^3}{R_p^3}$) acts: $$g_{\text{interior}}(r) = \frac{G M_{\text{enc}}}{r^2} = \frac{G M r}{R_p^3} = g_0 \left(\frac{r}{R_p}\right) \quad (\text{Linear decrease to zero at the core!})$$
Circular Satellite Orbits: Gravity acts as the sole centripetal force holding satellite of mass $m$ in orbit around planet of mass $M$: $$\frac{G M m}{r^2} = \frac{m v_{\text{orb}}^2}{r} \implies v_{\text{orb}} = \sqrt{\frac{G M}{r}}$$ Key AP Takeaway: Orbital speed is completely independent of the satellite's mass $m$, and decreases inversely with the square root of the orbital radius ($v_{\text{orb}} \propto 1/\sqrt{r}$).
1. Kepler's 1st Law (Elliptical Orbits): Planetary orbits are ellipses with the central mass located at one focus.
2. Kepler's 2nd Law (Equal Areas in Equal Times): A line segment joining planet and star sweeps out equal areas during equal intervals of time ($dA/dt = \text{const}$).
• Conservation of Angular Momentum: Gravitational force is strictly a central force ($\vec{r} \parallel \vec{F}_g$), so the net gravitational torque is zero: $$\vec{\tau}_{\text{net}} = \vec{r} \times \vec{F}_g = 0 \implies \vec{L} = m (\vec{r} \times \vec{v}) = \text{constant}$$ $$\implies m r_p v_p = m r_a v_a \implies v_p r_p = v_a r_a$$ The planet moves fastest at perihelion ($r_p$, closest) and slowest at aphelion ($r_a$, farthest).
3. Kepler's 3rd Law (Harmonic Law Derivation): Using $v = \frac{2\pi r}{T}$ in the circular orbital equation: $$\frac{4\pi^2 r^2}{T^2} = \frac{G M}{r} \implies T^2 = \left(\frac{4\pi^2}{G M}\right) r^3 \iff \frac{T^2}{r^3} = \text{constant}$$
• Universal Gravitational Potential Energy: Zero reference chosen at infinite separation ($r \to \infty$): $$U_g(r) = -\frac{G M m}{r}$$ • Total Mechanical Energy in Circular Orbit: $$E_{\text{total}} = K + U_g = \frac{1}{2}m v_{\text{orb}}^2 - \frac{G M m}{r} = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r} = \frac{1}{2}U_g = -K$$ • Escape Speed ($v_{\text{esc}}$): The minimum launch speed from surface ($R_p$) to reach infinity with zero kinetic energy ($E_{\text{total}} = 0$): $$\frac{1}{2}m v_{\text{esc}}^2 - \frac{G M m}{R_p} = 0 \implies v_{\text{esc}} = \sqrt{\frac{2 G M}{R_p}} = \sqrt{2} \cdot v_{\text{orb, surface}}$$
Unit 2 (cont.) Circular Motion & Gravitation: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 3: Work, Energy, and Power — The Bookkeeping of Mechanics
Energy methods replace vector bookkeeping with scalar bookkeeping. Instead of tracking every force at every instant, you choose a system, identify how energy enters or leaves it (work), and track how it is stored (kinetic, gravitational, elastic, internal). Below are the definitions, theorems, graphical tools, and written-reasoning habits that the AP Physics 1 CED (Unit 3) expects (calculus extensions are marked "Physics C · ileri"):
Translational Kinetic Energy: $K = \frac{1}{2}mv^2$. It is a scalar, it is never negative, and it depends on the square of speed (doubling $v$ quadruples $K$). Its value depends on the reference frame.
Work: Work is the transfer of energy into or out of a system by a force acting through a displacement. For a constant force $\vec{F}$ and a straight displacement $\vec{d}$:
$$W = \vec{F} \cdot \vec{d} = F d \cos\theta$$
where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$. Only the component of force parallel to the displacement does work. Unit: $1\text{ J} = 1\text{ N}\cdot\text{m}$.
• $0^\circ \le \theta < 90^\circ$: $W > 0$ — the force transfers energy into the object.
• $\theta = 90^\circ$: $W = 0$ — e.g., the normal force on a block sliding on a level floor, or the tension in a string for an object in uniform circular motion.
• $90^\circ < \theta \le 180^\circ$: $W < 0$ — the force removes energy (kinetic friction on a sliding block, gravity on a rising ball).
• No displacement of the point of application $\Rightarrow$ no work, however large the force (pushing on a wall that does not move).
When the force component along the displacement changes with position, the work is the signed area between the $F_x$–$x$ curve and the $x$-axis: $$W = \int_{x_i}^{x_f} F_x(x)\, dx$$ Area above the axis counts as positive work; area below the axis counts as negative work. In AP Physics 1 you compute areas of rectangles and triangles; in Physics C you integrate the function.
• The force exerted by a person stretching a spring from $x_1$ to $x_2$ is $F = +kx$, so the work done on the spring is the area of a trapezoid: $$W = \int_{x_1}^{x_2} kx\, dx = \frac{1}{2}k x_2^2 - \frac{1}{2}k x_1^2$$ • Note that this is not $\frac{1}{2}k(x_2 - x_1)^2$. Spring energy depends on the deformation measured from the natural length, not on the change in deformation.
For an object that can be modelled as a point particle, the net work done by all forces equals the change in its translational kinetic energy: $$W_{\text{net}} = \sum W_i = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$ Derivation (constant net force): $v_f^2 = v_i^2 + 2a\Delta x$ and $F_{\text{net}} = ma$ give $F_{\text{net}}\Delta x = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
$$-\mu_k m g\, d = 0 - \frac{1}{2}mv_0^2 \implies d = \frac{v_0^2}{2\mu_k g}$$ The mass cancels, and the stopping distance scales with $v_0^2$: doubling the speed quadruples the stopping distance.
Conservative force: the work it does on an object moving between two points is independent of the path, and the work around any closed path is zero (gravity, ideal spring force). Only conservative forces have an associated potential energy, defined by $\Delta U = -W_{\text{conservative}}$.
Nonconservative force: the work depends on the path (kinetic friction, air resistance, applied pushes). Kinetic friction around a closed path always does negative work, converting mechanical energy into internal (thermal) energy.
• Object–Earth system near the surface: $\Delta U_g = mg\Delta y$ (choose any convenient reference height $y = 0$).
• Spring–object system: $U_s = \frac{1}{2}kx^2$, with $x$ measured from the natural (unstretched) length.
• General gravitation (two masses, zero at infinite separation): $U_G = -\frac{Gm_1m_2}{r}$.
• Force from potential energy (Physics C · ileri): $F_x = -\frac{dU}{dx}$. Equilibrium where $\frac{dU}{dx} = 0$; stable at a minimum of $U$, unstable at a maximum.
General energy equation for a chosen system: $$K_i + U_i + W_{\text{ext}} = K_f + U_f + \Delta E_{\text{int}}$$ If the system is closed and isolated ($W_{\text{ext}} = 0$) and no nonconservative forces act inside it ($\Delta E_{\text{int}} = 0$), mechanical energy is conserved: $$K_i + U_{g,i} + U_{s,i} = K_f + U_{g,f} + U_{s,f}$$ Otherwise: $\Delta E_{\text{mech}} = W_{\text{nc}}$ (for example, $W_{\text{friction}} = -f_k d$).
1. Define the system in words (e.g., "block + spring + Earth").
2. Draw the initial-state bars ($K_i$, $U_{g,i}$, $U_{s,i}$), the energy-transfer bar ($W_{\text{ext}}$, can be negative), and the final-state bars ($K_f$, $U_{g,f}$, $U_{s,f}$, $\Delta E_{\text{int}}$).
3. The sum of the bar heights on the left must equal the sum on the right — bar charts are a visual version of the energy equation.
4. A bar of zero height is drawn as an empty column, not omitted.
| $K_i$ | $U_{g,i}$ | $U_{s,i}$ | $W_{\text{ext}}$ | $K_f$ | $U_{g,f}$ | $U_{s,f}$ | $\Delta E_{\text{int}}$ |
|---|---|---|---|---|---|---|---|
| 0 | 0 | ■■■■ | 0 | ■ | ■■ | 0 | ■ |
Average power: $P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$. Instantaneous power delivered by a force: $P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = Fv\cos\theta$. Unit: $1\text{ W} = 1\text{ J/s}$.
1. Using $\frac{1}{2}k(\Delta x)^2$ for the energy change of a spring that was already stretched.
2. Writing $f_k = \mu_k mg$ on an incline (correct: $\mu_k mg\cos\theta$, or $\mu_k F_N$ in general).
3. Claiming "energy is lost". Energy is transferred out of the system or converted into internal energy — name where it goes.
4. Treating speed as path-dependent on frictionless tracks. Final speed depends only on the height drop, not on the shape of the track or the launch direction.
5. Confusing power and energy (kW vs. kWh; "watts per second" is not a unit of power).
• Mathematical Routines: start every derivation from a fundamental equation ($K_i + U_i + W = K_f + U_f + \Delta E_{\text{int}}$), then substitute; do not start from a memorised special result.
• Translation Between Representations: make energy bar charts consistent with your equation term by term, and keep relative bar heights proportional.
• Experimental Design and Analysis: linearize, e.g. launch height $h$ versus compression $x$ satisfies $h = \frac{k}{2mg}x^2$, so plot $h$ against $x^2$ and use slope $= \frac{k}{2mg}$.
• Qualitative/Quantitative Translation: state the system, state whether external work or internal energy change is zero, and make your written claim agree with the derived expression (e.g., "since $d = v_0^2 / (2\mu_k g)$, doubling $v_0$ quadruples $d$, independent of mass").
Unit 3 Work, Energy, and Power: İnteraktif Soru İstasyonu
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AP Physics 1 — Unit 4: Linear Momentum — Interactions, Impulse, and Collisions
Energy tells you how much can happen; momentum tells you how objects share motion when they interact. Momentum is a vector, it is conserved for every isolated system, and it is the only reliable tool for the short, violent interactions (collisions, explosions, recoil) in which forces are unknown. Below are the definitions, theorems, graphical tools, and FRQ habits that the AP Physics 1 CED (Unit 4, Topics 4.1–4.4) expects (calculus extensions are marked "Physics C · ileri"):
Definition: The linear momentum of an object is the product of its mass and velocity:
$$\vec{p} = m\vec{v}$$
It is a vector with the same direction as the velocity. SI unit: $\text{kg}\cdot\text{m/s}$, which is identical to $\text{N}\cdot\text{s}$.
Momentum of a system: the vector sum of the momenta of its parts, $\vec{p}_{\text{sys}} = \sum m_i\vec{v}_i$. In one dimension, choose a positive direction and keep the signs; opposite motions can cancel to give zero total momentum even when every object is moving.
• $\vec{p} = m\vec{v}$ is a vector and scales linearly with speed; $K = \frac{1}{2}mv^2$ is a scalar and scales with the square of speed.
• They are linked by $K = \frac{p^2}{2m}$, i.e. $p = \sqrt{2mK}$.
• Two objects with equal momentum magnitudes: the lighter one has more kinetic energy ($K \propto 1/m$).
• Two objects with equal kinetic energies: the heavier one has more momentum ($p \propto \sqrt{m}$).
Newton's second law in momentum form: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$. The net force is the rate of change of momentum — the slope of a $p$–$t$ graph.
Impulse: the impulse delivered by a force over a time interval is
$$\vec{J} = \int_{t_i}^{t_f} \vec{F}\,dt = \vec{F}_{\text{avg}}\,\Delta t$$
Impulse–Momentum Theorem: the net impulse on an object equals its change in momentum:
$$\vec{J}_{\text{net}} = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$$
• Area under a force–time graph = impulse = $\Delta p$. Area above the $t$-axis is positive; area below is negative.
• Slope of a momentum–time graph = net force.
• Example: a bat–ball force rises linearly from $0$ to $600\text{ N}$ and back to $0$ over $0.010\text{ s}$. Impulse = triangle area $= \frac{1}{2}(0.010)(600) = 3.0\text{ N}\cdot\text{s}$. A $0.15\text{ kg}$ ball at rest leaves at $3.0/0.15 = 20\text{ m/s}$.
| Graph | Slope means | Area means |
|---|---|---|
| $F$ vs. $t$ | (not used) | Impulse $J = \Delta p$ |
| $p$ vs. $t$ | Net force $F_{\text{net}}$ | (not used) |
| $F$ vs. $x$ (Unit 3) | (not used) | Work $W = \Delta K$ |
Law of Conservation of Linear Momentum: if the net external force on a system is zero, the total momentum of the system is constant: $$\sum\vec{F}_{\text{ext}} = 0 \implies \vec{p}_{\text{sys},i} = \vec{p}_{\text{sys},f}$$ Internal forces (between parts of the system) come in Newton's-third-law pairs, deliver equal and opposite impulses, and cancel. They can transfer momentum between parts of the system but cannot change the total.
• Include every object that exerts a significant force during the interaction. For a collision of two carts, the system is "cart A + cart B".
• For a ball falling to the ground, the ball alone is not isolated (gravity is external); the ball–Earth system is.
• Approximately isolated: during a very short collision, external forces such as friction or gravity deliver negligible impulse compared with the large internal forces, so momentum is conserved "just before" to "just after".
• Component conservation: if external forces act only vertically (e.g., gravity and normal force), horizontal momentum is still conserved.
• A shell that explodes in mid-air: its center of mass continues along the original parabola, because the explosion forces are internal and the only external force is gravity.
Momentum is conserved in every collision of an isolated system. What distinguishes collision types is what happens to the total kinetic energy of the system:
1. Confirm $\vec{p}_i = \vec{p}_f$ (if not, the system is not isolated or the data are wrong).
2. Compute $K_i = \sum\frac{1}{2}m v_i^2$ and $K_f = \sum\frac{1}{2}m v_f^2$.
• $K_f = K_i$: elastic (e.g., hard steel spheres, magnetic bumpers, atomic collisions).
• $K_f < K_i$, objects separate: inelastic; the missing kinetic energy becomes internal energy (thermal energy, sound, permanent deformation).
• $K_f < K_i$, objects move together: perfectly (completely) inelastic; this is the maximum possible kinetic energy loss consistent with momentum conservation.
• $K_f > K_i$: an internal energy source (spring, explosion) did positive work — an "explosion" or "release".
• $m_1 \ll m_2$: the light object bounces back with nearly its original speed.
• $m_1 \gg m_2$: the heavy object continues almost unchanged; the light target leaves at nearly $2v_1$.
• Relative speed is preserved in 1D elastic collisions: $v_2' - v_1' = v_1 - v_2$.
• Stage 1 (bullet embeds in block): momentum conserved, kinetic energy not conserved: $mv = (m + M)V$.
• Stage 2 (block swings up): mechanical energy conserved: $\frac{1}{2}(m + M)V^2 = (m + M)gh$.
• Combined: $v = \frac{m + M}{m}\sqrt{2gh}$. Applying energy conservation across the collision is the most common error.
1. Define the system in words ("System: cart A + cart B") and justify why it is isolated, or isolated in one direction ("the track is frictionless and the vertical forces cancel, so no net external horizontal force acts").
2. Draw a before/after diagram: a sketch of the system just before and just after the interaction, with a labelled velocity arrow and symbol for every object ($v_{1i}$, $v_{2i}$, $v_{1f}$, $v_{2f}$) and an explicit positive direction.
3. Start from the fundamental equation: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$, then substitute with signs.
4. Check kinetic energy if the question asks about collision type or energy dissipated.
5. Interpret: state the direction of the answer in words and check limiting cases (does it reduce correctly when $m_2 \to 0$ or $m_1 = m_2$?).
• Mathematical Routines: derive symbolically first ($v_f = \frac{m_1v_1}{m_1 + m_2}$), then substitute numbers.
• Translation Between Representations: sketch $F$–$t$ graphs whose areas match the impulses; sketch $p$–$t$ graphs whose slopes match the forces; draw momentum bar charts for each object before and after.
• Experimental Design and Analysis: name the instruments (motion detectors or photogates for velocities, an electronic balance for masses, a force sensor for $F(t)$). Linearize: for sticking collisions, $\frac{1}{v_f} = \frac{1}{mv_0}M + \frac{1}{v_0}$, so plot $1/v_f$ against $M$ and use the intercept to find $v_0$.
• Qualitative/Quantitative Translation: reasoning such as "the airbag increases $\Delta t$; since $F_{\text{avg}} = \Delta p/\Delta t$ and $\Delta p$ is fixed by the initial and final speeds, the average force decreases" must agree with your equation.
1. Dropping the minus sign of a velocity that reverses direction.
2. Using energy conservation across a collision that is not stated to be elastic.
3. Claiming momentum is conserved for a single object acted on by an external force (e.g., the ball alone during a bounce).
4. Adding 2D momenta as scalars instead of by components.
5. Saying "momentum is lost" in an inelastic collision. Kinetic energy is converted; momentum of the isolated system is unchanged.
Unit 4 Linear Momentum: İnteraktif Soru İstasyonu
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AP Physics 1 — Unit 5: Torque and Rotational Dynamics — Rigid Bodies, Lever Arms, and Angular Acceleration
Units 1–4 treated every object as a point. Real objects have size, and a force applied off-centre can make them turn. Unit 5 builds a complete rotational copy of Newtonian mechanics: angle, angular velocity, and angular acceleration replace position, velocity, and acceleration; torque replaces force; rotational inertia replaces mass. Below are the definitions, models, and FRQ habits expected by the current AP Physics 1 CED (Unit 5, Topics 5.1–5.6). Rotational energy and angular momentum follow in Unit 6.
Rigid-body model: every point of a rigid object rotating about a fixed axis turns through the same angle in the same time. We therefore describe the whole object with one set of angular quantities: $$\Delta\theta \;(\text{rad}), \qquad \omega = \frac{d\theta}{dt} \;(\text{rad/s}), \qquad \alpha = \frac{d\omega}{dt} \;(\text{rad/s}^2)$$ Choose a positive sense of rotation (usually counterclockwise) and keep the signs, exactly as in one-dimensional kinematics. One revolution is $2\pi\text{ rad} = 360^\circ$.
• Graphs work exactly as in Unit 1: the slope of $\theta$–$t$ is $\omega$, the slope of $\omega$–$t$ is $\alpha$, and the area under $\omega$–$t$ is $\Delta\theta$.
A point at distance $r$ from the axis travels along an arc. Its linear quantities follow from the angular ones (angles in radians): $$s = r\,\Delta\theta, \qquad v_t = r\omega, \qquad a_t = r\alpha, \qquad a_c = \frac{v_t^2}{r} = \omega^2 r$$ All points share $\omega$ and $\alpha$, but points farther from the axis have larger tangential speed and larger accelerations.
• A light string that unwinds from a drum of radius $R$ without slipping moves with $v = R\omega$ and $a = R\alpha$. This constraint equation links a hanging block to a rotating pulley.
• Example: a bucket descends with $a = 1.2\text{ m/s}^2$ on a rope wound around a drum of radius $0.20\text{ m}$. Then $\alpha = \frac{a}{R} = 6.0\text{ rad/s}^2$.
• Tangential acceleration $a_t = r\alpha$ changes the speed; centripetal acceleration $a_c = \omega^2 r$ changes the direction. For a point on a speeding-up wheel, both are present and perpendicular.
Torque measures how effectively a force changes an object's rotation about a chosen axis: $$\tau = rF\sin\theta = r_\perp F = r F_\perp$$ Here $r$ is the distance from the axis to the point where the force is applied, and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. The lever arm $r_\perp = r\sin\theta$ is the perpendicular distance from the axis to the line of action of the force. SI unit: $\text{N}\cdot\text{m}$ (not a joule).
• Perpendicular component of the force times the full distance: $\tau = rF_\perp$.
• Full force times the lever arm: $\tau = r_\perp F$.
• Example: a $40\text{ N}$ force on a $0.25\text{ m}$ wrench at $60^\circ$ to the handle gives $\tau = (0.25)(40)\sin 60^\circ \approx 8.7\text{ N}\cdot\text{m}$.
• A force whose line of action passes through the axis (pushing a door toward its hinge) produces zero torque however large it is.
Rotational inertia (moment of inertia) is an object's resistance to changes in its rotation. It depends on the total mass and on how that mass is distributed relative to the axis: $$I = \sum m_i r_i^2 \qquad (\text{point masses}), \qquad I = \int r^2\,dm \qquad (\text{continuous objects})$$ The same object has different rotational inertias about different axes. Mass far from the axis contributes much more, because $r$ is squared.
• Thin hoop or thin-walled cylinder about its central axis: $I = MR^2$.
• Solid disk or solid cylinder about its central axis: $I = \frac{1}{2}MR^2$.
• Solid sphere about a diameter: $I = \frac{2}{5}MR^2$.
• Uniform thin rod about its centre: $I = \frac{1}{12}ML^2$; about one end: $I = \frac{1}{3}ML^2$.
• For equal $M$ and $R$: $I_{\text{hoop}} > I_{\text{disk}} > I_{\text{sphere}}$, because the hoop places all of its mass at the maximum distance.
• Solid disk about a point on its rim: $\frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
• $I_{cm}$ is always the smallest rotational inertia among all parallel axes; $d$ is measured from the center of mass, not from an end.
Newton's first law in rotational form: if the net torque on a rigid object is zero, its angular velocity is constant (it stays at rest or keeps rotating at the same rate). An object in static equilibrium satisfies two independent conditions: $$\sum \vec{F} = 0 \qquad \text{and} \qquad \sum \tau = 0 \;\text{ about any axis}$$ Because the net torque is zero about every axis when both conditions hold, you are free to choose the most convenient pivot.
1. Draw a force diagram of the extended object, placing each force at its point of application (weight at the center of mass).
2. Choose the pivot at the location of an unknown force; that force then has zero lever arm and drops out of the torque equation.
3. Write $\sum\tau = 0$ with signs, then $\sum F_x = 0$ and $\sum F_y = 0$ for the remaining unknowns.
• Example: a $4.0\text{ m}$, $20\text{ kg}$ uniform plank on two end supports with a $60\text{ kg}$ person $1.0\text{ m}$ from the left end ($g = 10\text{ m/s}^2$). Torques about the left end: $4.0F_R = (200)(2.0) + (600)(1.0) \Rightarrow F_R = 250\text{ N}$; then $F_L = 800 - 250 = 550\text{ N}$.
Newton's second law for rotation about a fixed axis: $$\alpha = \frac{\tau_{\text{net}}}{I} \qquad \Longleftrightarrow \qquad \tau_{\text{net}} = I\alpha$$ The angular acceleration is proportional to the net torque and inversely proportional to the rotational inertia. A large torque on a small-$I$ object gives a large $\alpha$.
• Block (translation, $+$ downward): $mg - T = ma$.
• Pulley (rotation): $TR = I\alpha = \frac{1}{2}MR^2\alpha$.
• Constraint: $a = R\alpha$, so $T = \frac{1}{2}Ma$.
• Combine: $$a = \frac{mg}{m + \frac{1}{2}M}, \qquad T = \frac{\frac{1}{2}M\,mg}{m + \frac{1}{2}M} < mg$$ • Example ($g = 9.8\text{ m/s}^2$): $m = 2.0\text{ kg}$, $M = 4.0\text{ kg}$: $a = \frac{19.6}{2.0 + 2.0} = 4.9\text{ m/s}^2$.
• In a two-block Atwood machine with a massive pulley, the tensions on the two sides are different; their difference $(T_1 - T_2)R$ is the net torque that angularly accelerates the pulley. Only an ideal massless pulley has equal tensions.
1. Draw an extended force diagram (not a dot): an outline of the object with every force drawn at its point of application, labelled with a distinct symbol, and the axis of rotation marked.
2. State the pivot and the positive rotation sense ("torques about the hinge, counterclockwise positive").
3. Start from the fundamental equations: $\sum\tau = I\alpha$ for each rotating object, $\sum F = ma$ for each translating object, and a constraint $a = R\alpha$ linking them.
4. Derive symbolically, then substitute, then check limits: as $M_{\text{pulley}} \to 0$ the result must reduce to the massless-pulley answer.
• Mathematical Routines: keep $I$ symbolic (e.g. $\beta MR^2$) until the last step so that the hoop/disk comparison is visible.
• Translation Between Representations: sketch $\omega$–$t$ graphs whose slopes equal $\tau_{\text{net}}/I$; draw torque arrows with lengths proportional to $r_\perp F$.
• Experimental Design and Analysis: to measure the rotational inertia of a pulley, hang different masses, measure $a$ with a motion detector or photogate, and linearize: $\frac{1}{a} = \frac{1}{g} + \frac{I}{R^2 g}\cdot\frac{1}{m}$; plot $\frac{1}{a}$ against $\frac{1}{m}$ and obtain $I$ from the slope.
• Qualitative/Quantitative Translation: "moving the same mass farther from the axis increases $I$, so the same torque produces a smaller $\alpha$" must agree with $\alpha = \tau/I$ and $I = \sum mr^2$.
1. Using the full distance $r$ instead of the lever arm $r_\perp$ when the force is not perpendicular.
2. Assuming equal tensions on both sides of a pulley that has rotational inertia.
3. Measuring $d$ in the parallel axis theorem from an end instead of from the center of mass.
4. Mixing degrees or revolutions with radians in $v = r\omega$ and the kinematic equations.
5. Writing $\sum\tau = 0$ for an object that is speeding up its rotation; equilibrium requires $\alpha = 0$.
Unit 5 Torque and Rotational Dynamics: İnteraktif Soru İstasyonu
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AP Physics 1 — Unit 6: Energy and Momentum of Rotating Systems — Rotational Energy, Angular Momentum, Rolling, and Orbits
Unit 5 answered the question "how does a torque change rotation?" with $\tau_{\text{net}} = I\alpha$. Unit 6 answers the same question with the two great bookkeeping tools of mechanics: energy (Unit 3) and momentum (Unit 4), now written for rotating objects. The current AP Physics 1 CED (Unit 6, Topics 6.1–6.6; exam weighting 5–8%) covers rotational kinetic energy, torque and work, angular momentum and angular impulse, conservation of angular momentum, rolling, and the motion of orbiting satellites. Rotational inertia values ($I = \beta MR^2$) and the parallel axis theorem are taught in Unit 5, Topic 5.4; here they are used, not re-derived.
Every particle of a rotating rigid object moves with speed $v_i = r_i\omega$. Adding their kinetic energies gives $$K_{\text{rot}} = \sum \tfrac{1}{2} m_i (r_i\omega)^2 = \tfrac{1}{2}\Big(\sum m_i r_i^2\Big)\omega^2 = \tfrac{1}{2} I\omega^2$$ Rotational kinetic energy is not a new kind of energy: it is ordinary kinetic energy of the particles, organised by the axis. It is a scalar, measured in joules, and is always positive.
• $I = \frac{2}{5}(2.0)(0.10)^2 = 8.0\times 10^{-3}\text{ kg}\cdot\text{m}^2$.
• $K_{\text{rot}} = \frac{1}{2}(8.0\times 10^{-3})(50)^2 = 10\text{ J}$.
• For the same $M$, $R$, and $\omega$, a hoop ($\beta = 1$) stores $2.5$ times as much energy as the sphere ($\beta = \frac{2}{5}$), because $K_{\text{rot}} = \frac{1}{2}\beta MR^2\omega^2$ is proportional to $\beta$.
A force that produces a torque about a fixed axis does work as the object turns. For a constant torque acting through an angular displacement $\Delta\theta$ (in radians): $$W = \tau\,\Delta\theta, \qquad \text{and for a varying torque} \qquad W = \int \tau\,d\theta = \text{area under the } \tau\text{–}\theta \text{ graph}$$ The rotational work–energy theorem then reads $W_{\text{net}} = \Delta K_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$. The rate of doing work is the rotational power $P = \tau\omega$.
• $W = (5.0)(20) = 100\text{ J}$, so the grindstone gains $100\text{ J}$ of rotational kinetic energy (frictionless axle).
• If the angular displacement is given as $3.0$ revolutions, convert first: $\Delta\theta = 3.0 \times 2\pi \approx 18.8\text{ rad}$.
• A motor that supplies $8.0\text{ N}\cdot\text{m}$ while turning at a steady $150\text{ rad/s}$ delivers $P = \tau\omega = 1200\text{ W}$.
Angular momentum is the rotational analogue of linear momentum. Its magnitude depends on the chosen axis: $$L = I\omega \;\;(\text{rigid object about a fixed axis}), \qquad L = m v r_\perp = mvr\sin\theta \;\;(\text{point particle})$$ Here $r_\perp$ is the perpendicular distance from the axis to the particle's line of motion. A particle moving in a straight line still has angular momentum about any point that is not on that line. SI unit: $\text{kg}\cdot\text{m}^2/\text{s}$.
• On an $L$–$t$ graph the slope is the net torque; on a $\tau$–$t$ graph the area is the change in angular momentum.
If the net external torque on a system is zero, the total angular momentum of the system is constant. Internal torques (muscles, friction between two parts of the system, a child grabbing a merry-go-round) transfer angular momentum between parts but cannot change the total: $$\tau_{\text{ext}} = 0 \;\Longrightarrow\; L_i = L_f \;\Longrightarrow\; I_i\omega_i = I_f\omega_f$$ Angular momentum may be conserved while kinetic energy is not, exactly as in an inelastic collision in Unit 4.
• $\omega_f = \frac{(3.0)(2.0)}{1.8} \approx 3.3\text{ rad/s}$.
• $K_i = \frac{1}{2}(3.0)(2.0)^2 = 6.0\text{ J}$; $K_f = \frac{L^2}{2I_f} = \frac{(6.0)^2}{2(1.8)} = 10\text{ J}$. The extra $4.0\text{ J}$ comes from the work done by the student's arms (internal chemical energy).
• A disk ($I_A$, $\omega_0$) and a non-rotating coaxial disk ($I_B$) stick together: $\omega_f = \frac{I_A}{I_A + I_B}\omega_0$, and the fraction of kinetic energy kept is $\frac{I_A}{I_A + I_B}$.
• A child running tangentially onto the rim of a merry-go-round: $m v R = \left(I_{\text{disk}} + mR^2\right)\omega_f$. The child's linear motion carries angular momentum $mvR$ about the axle.
Rolling without slipping: the point of the object in contact with the surface is instantaneously at rest relative to the surface. This gives the constraint $$v_{cm} = R\omega, \qquad a_{cm} = R\alpha$$ Relative to the ground, the contact point has speed $0$, the centre moves at $v_{cm}$, and the top of the wheel moves at $2v_{cm}$. Because the contact point does not slide, static friction does no work on an object that rolls without slipping, so mechanical energy is conserved.
• Solid disk or cylinder ($\beta = \frac{1}{2}$): $\frac{1}{3}$ rotational, $\frac{2}{3}$ translational.
• Solid sphere ($\beta = \frac{2}{5}$): $\frac{2}{7}$ rotational, $\frac{5}{7}$ translational.
• The result is independent of $M$ and $R$. In a race, the object with the smallest $\beta$ wins: sphere, then disk, then hoop.
For a satellite of mass $m$ orbiting a much more massive central object $M$, gravity always points toward the centre of $M$, so it exerts zero torque about that centre. In the satellite–planet system, both the satellite's angular momentum about the planet and the total mechanical energy are conserved: $$L = m v r_\perp = \text{constant}, \qquad E = \tfrac{1}{2}mv^2 - \frac{GMm}{r} = \text{constant}$$
• At the closest and farthest points $\vec{v} \perp \vec{r}$, so $r_p v_p = r_a v_a$. Example ($GM = 4.0\times 10^{14}\text{ m}^3/\text{s}^2$, $r_p = 7.0\times 10^{6}\text{ m}$, $r_a = 1.4\times 10^{7}\text{ m}$): $v_p \approx 8.73\text{ km/s}$ and $v_a = \frac{r_p}{r_a}v_p \approx 4.36\text{ km/s}$; both points give the same specific energy $\frac{E}{m} \approx -1.9\times 10^{7}\text{ J/kg}$.
• Escape speed is the launch speed for which $E = 0$: $\frac{1}{2}mv_{esc}^2 - \frac{GMm}{R} = 0 \Rightarrow v_{esc} = \sqrt{\frac{2GM}{R}}$. For Earth ($M = 6.0\times 10^{24}\text{ kg}$, $R = 6.4\times 10^{6}\text{ m}$), $v_{esc} \approx 11.2\text{ km/s}$.
• A typical prompt: "A hoop and a solid disk of equal mass and radius roll from rest down the same incline. Without equations, explain which reaches the bottom first. Then derive $v$ for each and show that your derivation agrees with your explanation."
• Qualitative point: the disk has less of its mass far from the axis, so a smaller share of $Mgh$ goes into rotation and more into translation; the disk is faster.
• Quantitative point: $v = \sqrt{\frac{2gh}{1+\beta}}$ with $\beta_{\text{disk}} = \frac{1}{2} < \beta_{\text{hoop}} = 1$.
• Consistency point: you must explicitly connect the two ("the smaller $\beta$ in the denominator gives a larger $v$, as argued in part (a)"). Graders award this point separately.
1. Release the object from rest at several measured heights $h$ (metre stick) and measure the speed at the bottom with a photogate (object diameter divided by blocking time).
2. Repeat each height several times and average to reduce random error.
3. Linearize: $v^2 = \frac{2g}{1+\beta}\,h$. Plot $v^2$ on the vertical axis against $h$; the best-fit line should pass near the origin.
4. From the slope $k$: $\beta = \frac{2g}{k} - 1$. Example: a slope of $14.0\text{ m/s}^2$ with $g = 9.8\text{ m/s}^2$ gives $\beta = 0.40$, consistent with a solid sphere.
5. Name a real source of error: slipping at steep angles or rolling friction reduces $v$ and makes $\beta$ appear too large.
• Mathematical Routines: state the conserved quantity and the reason ("no external torque about the axle, so $L$ is conserved") before writing the equation.
• Translation Between Representations: draw energy bar charts with separate bars for $K_{\text{trans}}$, $K_{\text{rot}}$, and $U_g$; for a rolling disk the final $K_{\text{rot}}$ bar is exactly half the $K_{\text{trans}}$ bar.
• Experimental Design and Analysis: always say which quantity goes on each axis and how the slope gives the target quantity.
• QQT: justify every "increases/decreases/stays the same" with a named principle (conservation of $L$, conservation of $E$, rotational work–energy theorem).
1. Conserving kinetic energy in a rotational collision or when a skater pulls her arms in; only $L$ is conserved.
2. Forgetting $K_{\text{rot}}$ for a rolling object and writing $v = \sqrt{2gh}$.
3. Using revolutions instead of radians in $W = \tau\Delta\theta$ or $L = I\omega$.
4. Claiming that static friction removes energy from an object that rolls without slipping.
5. Treating orbital total energy as positive, or using $mgh$ for satellite altitude changes.
Unit 6 Energy and Momentum of Rotating Systems: İnteraktif Soru İstasyonu
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AP Physics 1 — Unit 7: Oscillations — Simple Harmonic Motion, Period, Graphs, and Energy
Units 2 and 3 described motion caused by forces and tracked with energy. Unit 7 applies both tools to one special and very common kind of motion: an object that moves back and forth about a stable equilibrium position. The current AP Physics 1 CED (Unit 7, Topics 7.1–7.4; exam weighting 5–8%) covers the definition of simple harmonic motion (SHM), the period and frequency of the spring–object oscillator and the simple pendulum, graphical and equation representations of SHM, and the energy of simple harmonic oscillators. Hooke's law ($F_s = -kx$) and elastic potential energy ($U_s = \frac{1}{2}kx^2$) are taught in Unit 3; here they are used, not re-derived.
Periodic motion repeats itself after a fixed time interval. Simple harmonic motion is a special case of periodic motion. Two definitions are needed:
• An equilibrium position is a location at which the net force exerted on the object (or system) is zero.
• A restoring force is a force exerted in the direction opposite to the object's displacement from that equilibrium position.
SHM results when the magnitude of the restoring force is proportional to the displacement from equilibrium: $$F_{\text{net}} = -kx \qquad \Longrightarrow \qquad a = -\frac{k}{m}x$$ The minus sign carries the physics: the force (and therefore the acceleration) always points back toward equilibrium, and it grows linearly as the object moves farther away.
• $F_x = -kx = -(200)(0.15) = -30\text{ N}$: a $30\text{ N}$ force directed to the left, toward equilibrium.
• At $x = -0.15\text{ m}$ the force is $+30\text{ N}$; at $x = 0$ it is zero. A graph of $F$ versus $x$ is a straight line through the origin with slope $-k$.
• A block of mass $m$ hanging from a spring has its equilibrium position where the spring is stretched by $\Delta y_0 = \frac{mg}{k}$, not at the spring's natural length. Measured from this new equilibrium, the net force is still $-ky$, so the motion is SHM with the same $k$.
• For a simple pendulum, the gravitational force exerts a restoring torque about the pivot of magnitude $mgL\sin\theta$. For small angles $\sin\theta \approx \theta$ (in radians), so the restoring torque is approximately proportional to the angular displacement and the motion can be modelled as SHM. At $10^\circ$, $\sin\theta = 0.1736$ while $\theta = 0.1745\text{ rad}$: an error of only about $0.5\%$.
The period $T$ is the time for one complete cycle; the frequency $f$ is the number of cycles per second (hertz). They are reciprocals: $$T = \frac{1}{f}$$ For the two standard oscillators on the AP Physics 1 equation sheet: $$T_{s} = 2\pi\sqrt{\frac{m}{k}} \;\;(\text{object–ideal-spring}), \qquad T_{p} = 2\pi\sqrt{\frac{L}{g}} \;\;(\text{simple pendulum, small angle})$$
• Spring: $m = 0.50\text{ kg}$, $k = 200\text{ N/m}$ gives $T = 2\pi\sqrt{0.0025} = 2\pi(0.050) \approx 0.31\text{ s}$ and $f = \frac{1}{T} \approx 3.2\text{ Hz}$.
• Pendulum: $L = 1.0\text{ m}$, $g = 9.8\text{ m/s}^2$ gives $T = 2\pi\sqrt{\frac{1.0}{9.8}} \approx 2.0\text{ s}$. A pendulum that ticks once per second each way (a "seconds pendulum", $T = 2.0\text{ s}$) needs $L = \frac{g}{\pi^2} \approx 0.99\text{ m}$.
• Spring: $T \propto \sqrt{m}$ and $T \propto \frac{1}{\sqrt{k}}$. Quadrupling the mass doubles the period; quadrupling $k$ halves it. The spring period does not depend on $g$, so a vertical spring has the same period as a horizontal one, on Earth or on the Moon.
• Pendulum: $T \propto \sqrt{L}$ and $T \propto \frac{1}{\sqrt{g}}$. The pendulum period does not depend on the bob's mass, because both the restoring force and the inertia are proportional to $m$.
• Neither period depends on the amplitude (as long as the pendulum angle stays small).
The displacement of an object in SHM, measured from equilibrium, is a sinusoidal function of time: $$x = A\cos(2\pi f t) \quad \text{(released from rest at } x = A \text{ when } t = 0\text{)}, \qquad x = A\sin(2\pi f t) \quad \text{(passing } x = 0 \text{ when } t = 0\text{)}$$ $A$ is the amplitude, the maximum displacement from equilibrium (half of the total peak-to-peak distance). The angle $2\pi f t$ is in radians.
| Location | Displacement $x$ | Speed $|v|$ | Acceleration $|a|$ | Net force |
|---|---|---|---|---|
| Turning points $x = \pm A$ | maximum magnitude | zero | maximum | maximum, toward equilibrium |
| Equilibrium $x = 0$ | zero | maximum | zero | zero |
• Acceleration is always opposite to displacement ($a = -\frac{k}{m}x$), so the $a$–$t$ graph is the $x$–$t$ graph flipped upside down and scaled.
• Compare with $A\cos(2\pi f t)$: $A = 0.12\text{ m}$, $2\pi f = 4\pi \Rightarrow f = 2.0\text{ Hz}$, $T = 0.50\text{ s}$.
• At $t = 0.125\text{ s} = \frac{T}{4}$: $x = 0.12\cos\left(\frac{\pi}{2}\right) = 0$, speed maximum, acceleration zero. At $t = 0.25\text{ s} = \frac{T}{2}$: $x = -0.12\text{ m}$, momentarily at rest, acceleration maximum in the $+x$ direction.
• The maximum acceleration magnitude follows from Newton's second law at a turning point: $a_{\max} = \frac{kA}{m}$.
The total mechanical energy of a system in SHM is the sum of its kinetic and potential energies, and with no dissipative forces it is constant: $$E_{\text{total}} = K + U = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2$$ At the turning points $K = 0$ and $U$ is maximum; at equilibrium $U$ is minimum and $K$ is maximum. The minimum kinetic energy is zero. Because $E_{\text{total}} = \frac{1}{2}kA^2$, changing the amplitude changes the total energy: doubling $A$ quadruples $E$.
• $E = \frac{1}{2}(50)(0.20)^2 = 1.0\text{ J}$.
• At $x = 0$: $K = 1.0\text{ J}$, so $v_{\max} = \sqrt{\frac{2(1.0)}{0.50}} = 2.0\text{ m/s}$. Equivalently, $v_{\max} = A\sqrt{\frac{k}{m}}$.
• At $x = 0.10\text{ m} = \frac{A}{2}$: $U = \frac{1}{2}(50)(0.10)^2 = 0.25\text{ J}$, so $K = 0.75\text{ J}$ and $v = \sqrt{3} \approx 1.7\text{ m/s}$. Half the amplitude stores only one quarter of the energy as potential energy.
• Kinetic and potential energy are equal where $\frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right)$, that is at $x = \pm\frac{A}{\sqrt{2}} \approx \pm 0.71A$.
1. Static method: hang known masses, measure each stretch $\Delta y$ with a metre stick, plot $mg$ (vertical) against $\Delta y$ (horizontal); the slope is $k$.
2. Dynamic method: for each known mass, time $10$ oscillations with a stopwatch and compute $T$. Plot $T^2$ (vertical) against $m$ (horizontal); the slope is $\frac{4\pi^2}{k}$, so $k = \frac{4\pi^2}{\text{slope}}$.
3. Name a real source of error: the spring's own mass adds to the oscillating inertia, so the $T^2$–$m$ line has a small positive intercept; the slope is still valid.
• Sketch $x$–$t$, $v$–$t$, and $a$–$t$ on aligned time axes. Each zero of $x$ must line up with an extremum of $v$, and each extremum of $x$ with a zero of $v$ and an opposite-sign extremum of $a$.
• Energy bar charts: at a turning point draw only a $U$ bar; at equilibrium only a $K$ bar of equal height; at $x = \frac{A}{2}$ draw $U$ at one quarter and $K$ at three quarters of the total.
1. Claiming that a larger amplitude gives a longer period.
2. Saying the acceleration is zero at the turning points because the object is momentarily at rest.
3. Using degrees inside $\cos(2\pi f t)$ or inverting the ratio inside the square root.
4. Treating energy as linear in $x$ (for example, $U = \frac{E}{2}$ at $x = \frac{A}{2}$).
5. Placing the equilibrium of a hanging spring at its natural length instead of at a stretch of $\frac{mg}{k}$.
Unit 7 Oscillations: İnteraktif Soru İstasyonu
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Unit 8 Fluids: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.