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AP Physics 1 · ⁨AP 물리학 1⁩

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AP Physics 1은 대수 기반 역학입니다: 운동학, 힘, 일, 에너지와 출력, 운동량, 토크와 회전, 진동 및 유체입니다. 미적분이 필요 없습니다. 오히려 개념적으로 더 어렵습니다. 수치를 계산하는 것이 아니라 상황을 논리적으로 추론해야 하기 때문입니다.

자유 서술형 문제의 점수는 독립적으로 산정됩니다. 상황의 타당성을 설명하거나, 두 상황을 비교하거나, 결과를 글로 서술할 것을 요청합니다. 숫자만 맞고 추론이 없으면 매우 낮은 점수만 받습니다.

먼저 원리를 명시하세요. "운동량이 보존되고 수평 방향 외부 힘이 작용하지 않으므로…"은 채점 기준에서 원하는 문장 형식입니다. 모든 문제에 대해 자유체도(Free-body diagram)를 그리세요. 불필요해 보이는 경우에도 반드시 그려야 합니다. 이는 이 과목에서 가장 높은 가치를 주는 습관입니다.

아래의 노트는 운동학부터 단순 조진동까지 CED(교재 커리큘럼) 단원별로 구성되었으며, 각 단원에 풀이 예제가 포함되어 있습니다. released 자유 서술형 문제와 채점 기준은 라이브러리에서 확인할 수 있으며, 실험 설계 문제는 별도로 연습할 가치가 있습니다. 매년 출제되며, 공식보다는 명확한 절차를 rewarding합니다.

  • 1

    Kinematics

    Watch lesson · ⁨수업 보기⁩
    1.1

    Scalars and Vectors in One Dimension

    Syllabus
    English

    Learning Objective 1.1.A: Describe a scalar or vector quantity using magnitude and direction, as appropriate.

    • 1.1.A.1 Scalars are quantities described by magnitude only; vectors are quantities described by both magnitude and direction.
    • 1.1.A.2 Vectors can be visually modeled as arrows with appropriate direction and lengths proportional to their magnitude.
    • 1.1.A.3 Distance and speed are examples of scalar quantities, while position, displacement, velocity, and acceleration are examples of vector quantities.
      • 1.1.A.3.i Vectors are notated with an arrow above the symbol for that quantity.
        • Equation: $\vec{v} = \vec{v}_0 + \vec{a}t$
      • 1.1.A.3.ii Vector notation is not required for vector components along an axis. In one dimension, the sign of the component completely describes the direction of that component.
        • Derived equation: $v_x = v_{x0} + a_x t$

    Learning Objective 1.1.B: Describe a vector sum in one dimension.

    • 1.1.B.1 When determining a vector sum in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.
    한국어

    학습 목표 1.1.A: 크기와 방향을 사용하여 스칼라 또는 벡터 양을 적절히 기술하시오.

    • 1.1.A.1 스칼라는 크기만으로 서술되는 양이며, 벡터는 크기와 방향 모두로 서술되는 양입니다.
    • 1.1.A.2 벡터는 적절한 방향과 크기에 비례하는 길이를 가진 화살표로 시각적으로 모델링할 수 있습니다.
    • 1.1.A.3 거리와 속도는 스칼라 양의 예시이며, 위치, 변위, 속도, 가속도는 벡터 양의 예시입니다.
      • 1.1.A.3.i 벡터는 해당 양의 기호 위에 화살표를 붙여 표기합니다.
        • 식: $\vec{v} = \vec{v}_0 + \vec{a}t$
      • 1.1.A.3.ii 축을 따라 있는 벡터 성분에 대해 벡터 표기는 필수하지 않습니다. 1차원에서 성분의 부호는 해당 성분의 방향을 완전히 나타냅니다.
        • 파생 식: $v_x = v_{x0} + a_x t$

    학습 목표 1.1.B: 1차원의 벡터 합을 기술하시오.

    • 1.1.B.1 주어진 1차원 좌표계에서 벡터 합을 구할 때, 반대 방향은 서로 다른 부호로 표시합니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A car speedometer: speed is the magnitude of velocity — a scalar rate of motion
    A car speedometer: speed is the magnitude of velocity — a scalar rate of motion

    Kinematics 运动学 describes how objects move, without asking why. First, two kinds of quantity:

    • A scalar 标量 has only size (magnitude 大小): distance 距离, speed, time, mass.
    • A vector 矢量 has magnitude and direction: displacement, velocity, acceleration, force.

    The difference matters. Distance is the total path length travelled – a scalar that only grows. Displacement is the straight-line change in position, with a direction. Walk $3\ \text{m}$ east then $1\ \text{m}$ back west: the distance is $4\ \text{m}$, but the displacement is only $2\ \text{m}$ east.

    In one dimension, direction is just a sign (+ or −) along a chosen axis. Choosing the positive direction first is essential – every vector's sign depends on it. A velocity of $-5\ \text{m/s}$ does not mean "slow"; it means $5\ \text{m/s}$ in the negative direction.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Kinematics/ˌkɪnɪˈmætɪks/ 운동학
    scalar/ˈskeɪlə/ 스칼라
    magnitude/ˈmæɡnɪtjuːd/ 크기
    vector/ˈvektə/ 벡터
    distance/ˈdɪstəns/ 거리
    1.2

    Displacement, Velocity, and Acceleration

    Syllabus
    English

    Learning Objective 1.2.A: Describe a change in an object's position.

    • 1.2.A.1 When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.
    • 1.2.A.2 Displacement is the change in an object's position.
      • Equation: $\Delta x = x - x_0$

    Learning Objective 1.2.B: Describe the average velocity and acceleration of an object.

    • 1.2.B.1 Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.
    • 1.2.B.2 Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs.
      • Equation: $\vec{v}_{avg} = \dfrac{\Delta \vec{x}}{\Delta t}$
    • 1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
      • Equation: $\vec{a}_{avg} = \dfrac{\Delta \vec{v}}{\Delta t}$

    Learning Objective 1.2.B: Describe the velocity and acceleration of an object.

    • 1.2.B.4 An object is accelerating if the magnitude and/or direction of the object's velocity are changing.
    • 1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.
    한국어

    학습 목표 1.2.A: 물체의 위치 변화를 기술하시오.

    • 1.2.A.1 물체 모델을 사용할 때는 크기, 형태, 내부 구성을 무시합니다. 물체는 질량과 전하와 같은 광범위한 특성을 가진 단일 점으로 취급될 수 있습니다.
    • 1.2.A.2 변위는 물체의 위치 변화입니다.
      • 식: $\Delta x = x - x_0$

    학습 목표 1.2.B: 물체의 평균 속도와 평균 가속도를 기술하시오.

    • 1.2.B.1 속도와 가속도의 평균값은 시간 구간 동안 물체의 초기 상태와 최종 상태를 고려하여 계산됩니다.
    • 1.2.B.2 평균 속도는 해당 변위가 발생하는 시간 구간으로 나눈 물체의 변위입니다.
      • 식: $\vec{v}_{avg} = \dfrac{\Delta \vec{x}}{\Delta t}$
    • 1.2.B.3 평균 가속도는 속도 변화량을 해당 속도 변화가 발생하는 시간 구간으로 나눈 값입니다.
      • 식: $\vec{a}_{avg} = \dfrac{\Delta \vec{v}}{\Delta t}$

    학습 목표 1.2.B: 물체의 속도와 가속도를 기술하시오.

    • 1.2.B.4 물체의 속도의 크기와/또는 방향이 변하고 있다면 물체는 가속하고 있습니다.
    • 1.2.B.5 매우 짧은 시간 구간 동안 평균 속도 또는 평균 가속도를 계산하면 instantaneous velocity 또는 instantaneous acceleration에 매우 근접한 값을 얻게 됩니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A high-speed train: kinematics describes displacement, velocity and acceleration over time
    A high-speed train: kinematics describes displacement, velocity and acceleration over time
    Projectile motion is two independent motions

    Three linked vectors describe motion along a line:

    • Displacement 位移 $\Delta x$ is the change in position – a vector from start to end (not the total path length, which is distance).
    • Velocity 速度 is the rate of change of position, $v=\dfrac{\Delta x}{\Delta t}$. Its sign gives direction; its magnitude is speed 速率.
    • Acceleration 加速度 is the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$.

    Be careful to separate average velocity 平均速度 (total displacement over total time) from instantaneous velocity 瞬时速度 (the velocity at one instant, the slope of the position–time graph at that point). They are equal only when the velocity is constant.

    An object speeds up when $v$ and $a$ have the same sign, and slows down (deceleration 减速) when they have opposite signs. Note that a negative acceleration does not always mean slowing down – a ball falling faster and faster has negative velocity and negative acceleration.

    For constant acceleration, the four kinematic equations (often called SUVAT) apply:

    $$v=v_0+at,\qquad \Delta x=v_0 t+\tfrac{1}{2}at^2,\qquad v^2=v_0^2+2a\,\Delta x,\qquad \Delta x=\tfrac{1}{2}(v_0+v)\,t.$$
    Pick the equation that contains the three quantities you know plus the one you want, so only one unknown is left. They apply only while $a$ is constant.

    Worked example. A car starts from rest and accelerates uniformly at $2.0\ \text{m/s}^2$ for $6.0\ \text{s}$. Find its final velocity and the distance it travels.

    List what you know: $v_0=0$, $a=2.0\ \text{m/s}^2$, $t=6.0\ \text{s}$.

    $$v=v_0+at=0+2.0\times 6.0=12\ \text{m/s},$$
    $$\Delta x=v_0 t+\tfrac12 at^2=0+\tfrac12\times 2.0\times 6.0^2=36\ \text{m}.$$

    Worked example (free fall). A ball is thrown straight up at $15\ \text{m/s}$. Taking $g=9.8\ \text{m/s}^2$ and up as positive, how high does it rise, and how long is it in the air before returning to the thrower's hand?

    At the highest point the velocity is momentarily zero, and $a=-g=-9.8\ \text{m/s}^2$ throughout (this is free fall 自由落体, ignoring air resistance 空气阻力):

    $$v^2=v_0^2+2a\,\Delta x \;\Rightarrow\; 0=15^2+2(-9.8)\Delta x \;\Rightarrow\; \Delta x=\frac{225}{19.6}=11.5\ \text{m}.$$
    Time to the top: $0=15-9.8\,t \Rightarrow t=1.53\ \text{s}$. By symmetry the fall takes the same time, so the total is $3.1\ \text{s}$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Displacement/dɪˈspleɪsmənt/ 치환 반응
    Velocity/vəˈlɒsɪti/ 속도
    speed/spiːd/ 속력
    Acceleration/əkˌseləˈreɪʃn/ 가속도
    average velocity/ˈævrɪdʒ vəˈlɒsɪti/ 평균 속도
    instantaneous velocity/ˌɪnstənˈteɪnɪəs vəˈlɒsɪti/ 즉각 속도
    deceleration/dɪˌseləˈreɪʃn/ 감속도
    free fall/friː fɔːl/ 자유 낙하
    air resistance/eə rɪˈzɪstəns/ 공기 저항입니다
    1.3

    Representing Motion

    Syllabus
    English

    Learning Objective 1.3.A: Describe the position, velocity, and acceleration of an object using representations of that object's motion.

    • 1.3.A.1 Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
    • 1.3.A.2 For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension:
      • Equation: $v_x = v_{x0} + a_x t$
      • Equation: $x = x_0 + v_{x0}t + \dfrac{1}{2}a_x t^2$
      • Equation: $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$
      • Note: The equations above are written to indicate motion in the x-direction, but these equations can be used in any single dimension as appropriate.
    • 1.3.A.3 Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to $a_g = g \approx 10 \ m/s^2$.
    • 1.3.A.4 Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
      • 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
      • 1.3.A.4.ii An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
      • 1.3.A.4.iii The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
      • 1.3.A.4.iv The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.

    Boundary statement: AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration. However, students will be expected to be able to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.

    Boundary statement: For all situations in which a numerical quantity is required for $g$, the value $g \approx 10 \ m/s^2$ will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81 \ \text{m/s}^2$ or $g = 9.8 \ \text{m/s}^2$.

    한국어

    학습 목표 1.3.A: 운동 도식을 사용하여 물체의 위치, 속도, 가속도를 기술하시오.

    • 1.3.A.1 운동은 운동 도식, 그림, 그래프, 식, 서술적 설명으로 표현할 수 있습니다.
    • 1.3.A.2 등가속도인 경우 1차원의 순간 선형 운동을 기술하기 위해 세 가지 운동학 식을 사용할 수 있습니다:
      • 식: $v_x = v_{x0} + a_x t$
      • 식: $x = x_0 + v_{x0}t + \dfrac{1}{2}a_x t^2$
      • 식: $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$
      • 참고: 위 식들은 x방향의 운동을 나타내도록 작성되었으나, appropriate single dimension으로 사용 가능합니다.
    • 1.3.A.3 지구 표면 근처에서 중력에 의해 발생하는 수직 가속도는 아래를 향하며 일정하고, 측정된 값은 약 $a_g = g \approx 10 \ m/s^2$에 가깝습니다.
    • 1.3.A.4 시간의 함수로서 위치, 속도, 가속도의 그래프를 사용하여 이러한 양들 사이의 관계를 찾을 수 있습니다.
      • 1.3.A.4.i 물체의 순간 속도는 물체의 위치의 변화율로, 시간의 함수로서의 물체 위치 그래프 상 한 점에 접한 직선의 기울기와 같습니다.
      • 1.3.A.4.ii 물체의 순간 가속도는 물체의 속도의 변화율로, 시간의 함수로서의 물체 속도 그래프 상 한 점에 접한 직선의 기울기와 같습니다.
      • 1.3.A.4.iii 시간 구간 동안 물체의 변위는 시간의 함수로서의 물체 속도 그래프 하단 면적(즉, 해당 구간에 대한 함수와 수평축으로 둘러싸인 면적)과 같습니다.
      • 1.3.A.4.iv 시간 구간 동안 물체의 속도 변화량은 시간의 함수로서의 물체 가속도 그래프 하단 면적과 같습니다.

    경계 statement: AP Physics 1에서는 학생들에게 비균일 가속도의 정량적 분석을 기대하지 않습니다. 그러나 학생들은 비균일 가속도가 발생하는 상황을 정성적으로 분석하고 적절한 그래프를 스케치하며 논의할 수 있어야 합니다.

    경계 statement: $g$에 대해 수량 값이 필요한 모든 상황에서 $g \approx 10 \ m/s^2$ 값을 사용합니다. 그러나 학생들은 더 정밀하게 일반적으로 받아들여지는 $g = 9.81 \ \text{m/s}^2$ 또는 $g = 9.8 \ \text{m/s}^2$ 값을 올바르게 사용하는 것에 대해 감점되지 않습니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The same motion appears as a description, a graph, a table, or an equation, and you should move between them:

    Reading a distance-time graph: flat means at rest, a straight slope means constant speed
    Reading a distance-time graph: flat means at rest, a straight slope means constant speed
    • On a position–time graph, the slope is velocity (steeper = faster; a curve = changing velocity).
    • On a velocity–time graph, the slope is acceleration, and the area under the line is displacement.

    Reading slopes and areas off graphs is a core exam skill. To get displacement from a velocity–time graph, split the area into triangles and rectangles and add them up; area below the time axis counts as negative displacement (motion the other way).

    On a velocity-time graph the slope is the acceleration and the shaded area is the displacement
    On a velocity-time graph the slope is the acceleration and the shaded area is the displacement

    Worked example. A cyclist speeds up uniformly from rest to $8.0\ \text{m/s}$ in $4.0\ \text{s}$, then holds $8.0\ \text{m/s}$ for $6.0\ \text{s}$. Find the total distance from the velocity–time graph.

    The area is a triangle followed by a rectangle:

    $$\Delta x=\underbrace{\tfrac12\times 4.0\times 8.0}_{\text{triangle}}+\underbrace{6.0\times 8.0}_{\text{rectangle}}=16+48=64\ \text{m}.$$

    Explore · ⁨탐색하기⁩

    Explore the velocity–time graph · ⁨속도-시간 그래프 탐구하기⁩

    Change the start velocity $u$ and the acceleration $a$. The gradient (slope) of the line is the acceleration; the area between the line and the time axis is the displacement. · ⁨초기 속도 $u$와 가속도 $a$를 변경하십시오. 직선의 기울이(gradient)는 가속도이며, 직선과 시간 축 사이의 면적은 **변위(displacement)**입니다.⁩

    1.4

    Reference Frames and Relative Motion

    Syllabus
    English

    Learning Objective 1.4.A: Describe the reference frame of a given observer.

    • 1.4.A.1 The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.

    Learning Objective 1.4.B: Describe the motion of objects as measured by observers in different inertial reference frames.

    • 1.4.B.1 Measurements from a given reference frame may be converted to measurements from another reference frame.
    • 1.4.B.2 The observed velocity of an object results from the combination of the object's velocity and the velocity of the observer's reference frame.
      • 1.4.B.2.i Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
      • 1.4.B.2.ii The acceleration of any object is the same as measured from all inertial reference frames.

    Boundary statement: Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial.

    Boundary statement: Adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1.

    한국어

    학습 목표 1.4.A: 특정 관측자의 기준 좌표계를 기술하시오.

    • 1.4.A.1 기준 좌표계의 선택은 해당 좌표계에서 관측자가 측정하는 물리량의 방향과 크기를 결정한다.

    학습 목표 1.4.B: 서로 다른 관성 기준 좌표계에서 관측자가 측정하는 물체의 운동을 설명한다.

    • 1.4.B.1 특정 기준 좌표계에서 측정한 값은 다른 기준 좌표계에서의 측정값으로 변환할 수 있다.
    • 1.4.B.2 물체의 관측된 속도는 물체 자체의 속도와 관측자의 기준 좌표계의 속도의 합성 결과이다.
      • 1.4.B.2.i 주어진 기준 좌표계 내에서 물체의 운동과 관측자의 운동을 합성하기 위해서는 벡터의 가산 또는 감산을 수행해야 한다.
      • 1.4.B.2.ii 모든 관성 기준 좌표계에서 측정할 때 임의의 물체의 가속도는 동일하다.

    경계 조항: 별다른 명시가 없는 한 모든 문제의 기준 좌표계는 관성 좌표계라고 가정할 수 있다.

    경계 조항: AP Physics 1에서는 상대속도를 구하기 위한 벡터의 가감산运算을 1차원 운동에만 적용한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    All motion is measured against a reference frame 参考系. Velocities measured in different frames differ, and you combine them by vector addition. The velocity of A relative to C is

    $$\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}.$$
    A person walking on a moving train has one velocity relative to the train and another relative to the ground – this is relative motion 相对运动. A useful shortcut: the velocity of A relative to B is $\vec{v}_{A/B}=\vec{v}_A-\vec{v}_B$ (subtract B's velocity).

    A frame that is not accelerating is an inertial reference frame 惯性参考系: one in which a free object (no net force) obeys Newton's first law, staying at rest or moving at constant velocity. A frame that accelerates – a braking car – is non-inertial, where objects seem to speed up with no force acting on them.

    Worked example. A boat points straight across a river and moves at $3.0\ \text{m/s}$ relative to the water. The current flows at $4.0\ \text{m/s}$ along the river. Find the boat's speed and direction relative to the bank.

    The two velocities are perpendicular, so add them as a right triangle:

    $$v=\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s},\qquad \theta=\tan^{-1}\!\frac{4.0}{3.0}=53^{\circ}\ \text{downstream from straight across.}$$

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    reference frame/ˈrefrəns freɪm/ 기준 좌표계
    relative motion/ˈrelətɪv ˈməʊʃn/ 相대 운동
    inertial reference frame/ɪˈnɜːʃl ˈrefrəns freɪm/ 관성 기준 좌표계
    1.5

    Vectors and Motion in Two Dimensions

    Syllabus
    Learning ObjectiveEssential Knowledge

    1.5.A
    Describe the perpendicular components of a vector.

    • 1.5.A.1 Vectors can be mathematically modeled as the resultant of two perpendicular components.
    • 1.5.A.2 Vectors can be resolved into components using a chosen coordinate system.
    • 1.5.A.3 Vectors can be resolved into perpendicular components using trigonometric functions and relationships.
      • Equation: $\sin \theta = \dfrac{a}{c}$
      • Equation: $\cos \theta = \dfrac{b}{c}$
      • Equation: $\tan \theta = \dfrac{a}{b}$
      • Equation: $a^2 + b^2 = c^2$

    1.5.B
    Describe the motion of an object moving in two dimensions.

    • 1.5.B.1 Motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
    • 1.5.B.2 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    In two dimensions, resolve each vector into components 分量 along perpendicular axes ($x$ and $y$), handle each axis separately, then recombine. A velocity $v$ at angle $\theta$ to the horizontal has components $v_x=v\cos\theta$ and $v_y=v\sin\theta$.

    A velocity vector resolved into its horizontal and vertical components
    A velocity vector resolved into its horizontal and vertical components

    For projectile motion 抛体运动 (an object moving under gravity alone): the horizontal and vertical motions are independent. Horizontally, velocity is constant ($a_x=0$); vertically, acceleration is $-g$ (down). The two motions share only the time. So a projectile's path (its trajectory 轨迹) is a parabola, and you solve it as two one-dimensional problems joined by $t$.

    A projectile launched at an angle: the horizontal and vertical motions are independent
    A projectile launched at an angle: the horizontal and vertical motions are independent
    A dropped ball and a horizontally launched ball fall together – the vertical motions are identical
    A dropped ball and a horizontally launched ball fall together – the vertical motions are identical

    Worked example. A ball is kicked at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal. Taking $g=9.8\ \text{m/s}^2$, find the time of flight, the maximum height, and the horizontal range 射程 (assume it lands at launch height).

    Split the launch velocity into components:

    $$v_{0x}=20\cos 30^{\circ}=17.3\ \text{m/s},\qquad v_{0y}=20\sin 30^{\circ}=10\ \text{m/s}.$$
    Vertical motion sets the time. At the top $v_y=0$, so $0=10-9.8\,t \Rightarrow t_{\text{up}}=1.02\ \text{s}$, and the total flight is $2t_{\text{up}}=2.0\ \text{s}$. The maximum height is
    $$\Delta y=\frac{v_{0y}^2}{2g}=\frac{10^2}{19.6}=5.1\ \text{m}.$$
    Horizontal motion runs at constant $v_{0x}$ for the whole flight, so the range is
    $$R=v_{0x}\times t_{\text{flight}}=17.3\times 2.0=35\ \text{m}.$$

    A common trap: at the top of the flight the vertical velocity is zero, but the ball is not at rest – its horizontal velocity $v_{0x}$ never changes. The speed at the top equals $v_{0x}=17.3\ \text{m/s}$.

    Explore · ⁨탐색하기⁩

    Explore projectile motion · ⁨포물선 운동 탐구하기⁩

    Fire the ball, then change the angle and speed. The horizontal motion stays steady while gravity pulls it down — together they trace a parabola. Find the launch angle that gives the longest range, and try the Moon. · ⁨공을 발사한 후 각도와 속도를 변경하십시오. 수직 운동은 일정하게 유지되며 중력이 아래로 당깁니다—이들이 결합하여 포물선을 그리게 됩니다. 가장 긴 사거리를 주는 발사 각도를 찾고 달에서 시도해 보십시오.⁩

    Explore · ⁨탐색하기⁩

    Explore vectors and their components · ⁨벡터와 그 성분 탐구하기⁩

    Drag the vectors to change their $x$- and $y$-components. See how a single vector is built from independent horizontal and vertical parts, and how two vectors add tip-to-tail into a resultant. · ⁨벡터를 드래그하여 $x$ 및 $y$ 성분을 변경하십시오. 단일 벡터가 독립적인 수평 및 수직 성분으로 구성되는方式和, 두 벡터가 끝到尾로 합쳐져 **합성 벡터(resultant)**가 되는 것을 보십시오.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    components/kəmˈpəʊnənts/ 구성 요소
    projectile motion/prəˈdʒektaɪl ˈməʊʃn/ 포물선 운동
    trajectory/trəˈdʒektəri/ 궤적(trajectory)
    range/reɪndʒ/ 범위Unless range.
    1.5

    Exam tips

    • Choose the right kinematic equation by listing the three quantities you know plus the one you want, so only one unknown remains; the SUVAT equations apply only while acceleration is constant.
    • Fix a positive direction first — every displacement, velocity, and acceleration then carries a sign; a negative velocity means "moving the other way", not "slow".
    • Treat a projectile as two independent 1-D problems sharing only the time $t$: constant velocity horizontally, $a=-g$ vertically. At the top $v_y=0$ but $v_x$ is unchanged.
    • On a velocity–time graph the gradient is the acceleration and the area is the displacement (area below the axis is negative).
    • Distinguish distance (scalar, total path) from displacement (vector, start-to-end), and speed from velocity.
  • 2

    Force and Translational Dynamics

    Watch lesson · ⁨수업 보기⁩
    2.1

    Systems and Center of Mass

    Syllabus
    English

    Learning Objective 2.1.A: Describe the properties and interactions of a system.

    • 2.1.A.1 System properties are determined by the interactions between objects within the system.
    • 2.1.A.2 If the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object.
    • 2.1.A.3 Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
    • 2.1.A.4 Individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
    • 2.1.A.5 The internal structure of a system affects the analysis of that system.
    • 2.1.A.6 As variables external to a system are changed, the system's substructure may change.

    Learning Objective 2.1.B: Describe the location of a system's center of mass with respect to the system's constituent parts.

    • 2.1.B.1 For systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
    • 2.1.B.2 The location of a system's center of mass along a given axis can be calculated using the equation
      • Equation: $\vec{x}_{cm} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
    • 2.1.B.3 A system can be modeled as a singular object that is located at the system's center of mass.

    Boundary statement: AP Physics 1 only expects students to calculate the center of mass for systems of five or fewer particles arranged in a two-dimensional configuration or for systems that are highly symmetrical.

    한국어

    학습 목표 2.1.A: 계의 특성과 상호작용을 설명한다.

    • 2.1.A.1 계의 특성은 계 내 구성 요소 간의 상호작용에 의해 결정된다.
    • 2.1.A.2 계를 거시적 현상으로 모델링할 때 구성 요소의 특성이나 상호작용이 중요하지 않다면, 계 전체를 단일 물체로 취급할 수 있다.
    • 2.1.A.3 계는 구성 요소와 환경 간 상호작용을 허용하며, 이는 에너지나 물질의 이동을 초래할 수 있다.
    • 2.1.A.4 선택된 계 내 개별 물체들은 서로 그리고 계 전체와도 다르게 행동할 수 있다.
    • 2.1.A.5 계의 내부 구조는 해당 계의 분석에 영향을 미친다.
    • 2.1.A.6 계 외부 변수가 변화하면 계의 하위 구조가 변경될 수 있다.

    학습 목표 2.1.B: 계의 질량 중심 위치를 계의 구성 요소에 대해 설명한다.

    • 2.1.B.1 대칭적인 질량 분포를 가진 계의 경우, 질량 중심은 대칭축 위에 위치한다.
    • 2.1.B.2 계의 질량 중심 위치는 주어진 축 상에서 다음 식을 사용하여 계산할 수 있다.
      • 식: $\vec{x}_{cm} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
    • 2.1.B.3 계는 질량 중심에 위치하는 단일 물체로 모델링될 수 있다.

    경계 조항: AP Physics 1에서는 학생에게 2차원 평면 구조를 이루는 5개 이하의 입자로 구성된 계의 질량 중심 계산이나 매우 대칭적인 계의 질량 중심 계산만 요구한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A system 系统 is the object or group of objects you choose to analyze. A system can be treated as a single point at its center of mass 质心 – the average position of its mass. External forces change the motion of the center of mass; internal forces (between parts of the system) do not.

    This is why a wrench spinning across a table still has its center of mass move in a straight line: the spinning is internal, and only the (near-zero) external force matters for the center of mass. For two masses $m_1,m_2$ on a line at positions $x_1,x_2$, the center of mass sits at $x_{\text{cm}}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ – always closer to the heavier mass.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    system/ˈsɪstəm/ system
    center of mass/ˈsentə ɒv mæs/ center of mass
    2.2

    Forces and Free-Body Diagrams

    Syllabus
    English

    Learning Objective 2.2.A: Describe a force as an interaction between two objects or systems.

    • 2.2.A.1 Forces are vector quantities that describe the interactions between objects or systems.
      • 2.2.A.1.i A force exerted on an object or system is always due to the interaction of that object with another object or system.
      • 2.2.A.1.ii An object or system cannot exert a net force on itself.
    • 2.2.A.2 Contact forces describe the interaction of an object or system touching another object or system and are macroscopic effects of interatomic electric forces.

    Learning Objective 2.2.B: Describe the forces exerted on an object or system using a free-body diagram.

    • 2.2.B.1 Free-body diagrams are useful tools for visualizing forces being exerted on a single object or system and for determining the equations that represent a physical situation.
    • 2.2.B.2 The free-body diagram of an object or system shows each of the forces exerted on the object by the environment.
    • 2.2.B.3 Forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot. A system is treated as though all of its mass is located at the center of mass.
    • 2.2.B.4 A coordinate system with one axis parallel to the direction of acceleration of the object or system simplifies the translation from free-body diagram to algebraic representation. For example, in a free-body diagram of an object on an inclined plane, it is useful to set one axis parallel to the surface of the incline.

    Boundary statement: AP Physics 1 only expects students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping.

    한국어

    학습 목표 2.2.A: 힘을 두 물체 또는 계 사이의 상호작용으로 설명한다.

    • 2.2.A.1 힘은 물체나 계 간의 상호작용을 나타내는 벡터량이다.
      • 2.2.A.1.i 물체나 계에 가해지는 힘은 항상 해당 물체가 다른 물체나 계와 상호작용한 결과이다.
      • 2.2.A.1.ii 물체나 계는 자신에게서 순힘을 가할 수 없다.
    • 2.2.A.2 접촉력은 물체나 계가 다른 물체나 계에 닿아 상호작용하는 것을 기술하며, 이는 원자 간 전기력의 거시적 효과이다.

    학습 목표 2.2.B: 자유체도를 사용하여 물체나 계에 가해지는 힘을 설명한다.

    • 2.2.B.1 자유체도는 단일 물체나 계에 가해지는 힘을 시각화하고 물리적 상황을 나타내는 방정식을 도출하는 데 유용한 도구이다.
    • 2.2.B.2 자유체도는 환경으로부터 물체나 계에 가해지는 각 힘을 보여준다.
    • 2.2.B.3 물체나 계에 가해지는 힘은 질량 중심(예: 점)에서 시작하는 벡터로 표현된다. 계는 모든 질량이 질량 중심에 집중되어 있는 것처럼 취급된다.
    • 2.2.B.4 물체나 계의 가속도 방향과 평행한 축을 갖는 좌표계를 설정하면 자유체도를 알기식 표현으로 변환하는 것이 용이하다. 예를 들어, 경사면 위의 물체에 대한 자유체도에서는 축 중 하나를 경사면 표면에 평행하게 설정하는 것이 유리하다.

    경계 조항: AP Physics 1에서는 학생에게 자유체도에 힘의 성분을 표시하지 않고 오직 물체에 가해지는 힘만을 묘사하도록 요구한다. AP Physics 시험에서 자유체도에 표시된 개별 힘은 점에서 시작하여 힘의 방향으로 향하는 개별 직선 화살표로 그려져야 하며, 같은 방향의 개별 힘은 겹치지 않고 나란히 그려져야 한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A skydiver: free-body diagrams balance gravity and air resistance at terminal speed
    A skydiver: free-body diagrams balance gravity and air resistance at terminal speed
    A rocket launch: net force changes momentum — F_net = ma for translational dynamics
    A rocket launch: net force changes momentum — F_net = ma for translational dynamics

    A force 力 is a push or pull – a vector, measured in newtons (N). A free-body diagram 受力图 shows one object as a dot with arrows for every force acting on it (weight 重力, normal, tension 张力, friction, applied), each labelled and pointing the right way. Draw it before any dynamics problem; it is where most marks are won or lost.

    A free-body diagram shows every force acting on one object
    A free-body diagram shows every force acting on one object

    Two rules keep free-body diagrams honest: draw only forces acting on the chosen object (not forces it exerts on other things), and draw only real, physical forces (a rope, a surface, gravity, a hand) – never an "$ma$" arrow, which is the result of the forces, not a force itself.

    Explore · ⁨탐색하기⁩

    Balance the forces on a free-body diagram · ⁨자유체 도상의 힘 평형 맞추기⁩

    A free-body diagram shows every force on one object as an arrow. The object accelerates only if the forces don't cancel — the net force sets $a=F/m$. · ⁨**자유체 도상(free-body diagram)**은 하나의 물체에 작용하는 모든 힘을 화살표로 보여줍니다. 물체가 가속되려면 힘이 상쇄되지 않아야 하며, **순힘(net force)**이 $a=F/m$를 결정합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    force/fɔːs/ force
    free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ free-body diagram
    weight/weɪt/ weight
    tension/ˈtenʃn/ tension
    2.3

    Newton's Third Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.3.A
    Describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object.

    • 2.3.A.1 Newton's third law describes the interaction of two objects in terms of the paired forces that each exerts on the other.
      • Equation: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
    • 2.3.A.2 Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
    • 2.3.A.3 Tension is the macroscopic net result of forces that segments of a string, cable, chain, or similar system exert on each other in response to an external force.
      • 2.3.A.3.i An ideal string has negligible mass and does not stretch when under tension.
      • 2.3.A.3.ii The tension in an ideal string is the same at all points within the string.
      • 2.3.A.3.iii In a string with nonnegligible mass, tension may not be the same at all points within the string.
      • 2.3.A.3.iv An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

    Boundary statement: AP Physics 1 only expects students to describe tension qualitatively in a string, cable, chain, or similar system with mass. For example, students might note that the tension in a hanging chain is greater toward the top of the chain.

    Boundary statement: The interaction between objects or systems at a distance is limited to gravitational forces in AP Physics 1. In AP Physics 2, gravitational, electric, and magnetic forces may be considered.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's third law 牛顿第三定律: if object A pushes on object B, then B pushes back on A with a force equal in size and opposite in direction. These two forces act on different objects, so they never cancel each other. Identify third-law pairs by the "A on B / B on A" wording.

    A Newton's third-law pair: equal and opposite forces on two different objects
    A Newton's third-law pair: equal and opposite forces on two different objects

    A classic trap: the weight of a book and the normal force from the table are not a third-law pair – they act on the same object (the book). The partner of the book's weight is the pull the book exerts on the Earth; the partner of the table's push is the push the book makes on the table.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Newton's third law/ˈnjuːtnz θɜːd lɔː/ Newton's third law
    2.4

    Newton's First Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.4.A
    Describe the conditions under which a system's velocity remains constant.

    • 2.4.A.1 The net force on a system is the vector sum of all forces exerted on the system.
    • 2.4.A.2 Translational equilibrium is a configuration of forces such that the net force exerted on a system is zero.
      • Derived equation: $\sum_i \vec{F}_i = 0$
    • 2.4.A.3 Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
    • 2.4.A.4 Forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
    • 2.4.A.5 An inertial reference frame is one from which an observer would verify Newton's first law of motion.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's first law 牛顿第一定律 (the law of inertia 惯性): an object's velocity stays constant unless a net force 合力 acts on it. So zero net force means constant velocity (including rest) – the object is in translational equilibrium 平衡. Inertia is the tendency to resist changes in motion, measured by mass.

    Worked example. A $1200\ \text{kg}$ car cruises at a steady $25\ \text{m/s}$ on a level road. What is the net force on it? Because the velocity is constant, the acceleration is zero, so by the first law the net force is zero – the forward drive force exactly balances drag and friction. "Steady speed" always means balanced forces.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Newton's first law/ˈnjuːtnz fɜːst lɔː/ Newton's first law
    inertia/ɪˈnɜːʃə/ inertia
    net force/net fɔːs/ net force
    translational equilibrium/trænˈsleɪʃənl ˌiːkwɪˈlɪbrɪəm/ translational equilibrium
    2.5

    Newton's Second Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.5.A
    Describe the conditions under which a system's velocity changes.

    • 2.5.A.1 Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
    • 2.5.A.2 Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.
      • Equation: $\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}} = \dfrac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$
    • 2.5.A.3 The velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's second law 牛顿第二定律 relates net force to acceleration:

    $$\vec{a}=\frac{\sum \vec{F}}{m},\qquad\text{i.e.}\qquad \sum\vec{F}=m\vec{a}.$$
    Apply it one axis at a time: add the force components along each axis and set the sum equal to $ma$ for that axis. Acceleration points the same way as the net force.

    Worked example. A $4.0\ \text{kg}$ box is pulled along the floor by a horizontal force of $18\ \text{N}$. Friction on the box is $6.0\ \text{N}$. Find its acceleration. Along the direction of motion the net force is $18-6.0=12\ \text{N}$, so

    $$a=\frac{\sum F}{m}=\frac{12}{4.0}=3.0\ \text{m/s}^2.$$

    Worked example (incline 斜面). A block of mass $m$ slides down a frictionless ramp tilted at angle $\theta$. Find its acceleration. Resolve gravity into components along and perpendicular to the ramp; only the along-ramp part, $mg\sin\theta$, drives the motion, so

    $$a=\frac{mg\sin\theta}{m}=g\sin\theta.$$
    The steeper the ramp, the larger $\sin\theta$ and the faster it accelerates; at $\theta=90^{\circ}$ it is free fall.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Newton's second law/ˈnjuːtnz ˈsekənd lɔː/ Newton's second law
    incline/ɪnˈklaɪn/ incline
    2.6

    Gravitational Force

    Syllabus
    English

    Learning Objective 2.6.A: Describe the gravitational interaction between two objects or systems with mass.

    • 2.6.A.1 Newton's law of universal gravitation describes the gravitational force between two objects or systems as directly proportional to each of their masses and inversely proportional to the square of the distance between the systems' centers of mass.
      • Equation: $\left|\vec{F}_g\right| = G\dfrac{m_1 m_2}{r^2}$
      • 2.6.A.1.i The gravitational force is attractive.
      • 2.6.A.1.ii The gravitational force is always exerted along the line connecting the centers of mass of the two interacting systems.
      • 2.6.A.1.iii The gravitational force on a system can be considered to be exerted on the system's center of mass.
    • 2.6.A.2 A field models the effects of a noncontact force exerted on an object at various positions in space.
      • 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
        • Equation: $\left|\vec{g}\right| = \dfrac{\left|\vec{F}_g\right|}{m} = G\dfrac{M}{r^2}$
      • 2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in m/s$^2$) is numerically equal to the magnitude of the gravitational field strength (in N/kg) at that location.
    • 2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
      • Equation: $\text{Weight} = F_g = mg$

    Learning Objective 2.6.B: Describe situations in which the gravitational force can be considered constant.

    • 2.6.B.1 If the gravitational force between two systems' centers of mass has a negligible change as the relative position of the two systems changes, the gravitational force can be considered constant at all points between the initial and final positions of the systems.
    • 2.6.B.2 Near the surface of Earth, the strength of the gravitational field is $g \approx 10 \text{ N/kg}$

    Learning Objective 2.6.C: Describe the conditions under which the magnitude of a system's apparent weight is different from the magnitude of the gravitational force exerted on that system.

    • 2.6.C.1 The magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system.
    • 2.6.C.2 If the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on the system.
    • 2.6.C.3 A system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system.
    • 2.6.C.4 The equivalence principle states that an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field.

    Learning Objective 2.6.D: Describe inertial and gravitational mass.

    • 2.6.D.1 Objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object.
    • 2.6.D.2 Gravitational mass is related to the force of attraction between two systems with mass.
    • 2.6.D.3 Inertial mass and gravitational mass have been experimentally verified to be equivalent.
    한국어

    학습 목표 2.6.A: 질량을 가진 두 물체나 시스템 간의 중력 상호작용을 설명할 수 있다.

    • 2.6.A.1 만유인력 법칙은 두 물체나 시스템 간의 중력 힘을 각 질량의 곱에 직접 비례하고 두 시스템의 질량 중심 사이의 거리의 제곱에 반비례한다고 설명한다.
      • 식: $\left|\vec{F}_g\right| = G\dfrac{m_1 m_2}{r^2}$
      • 2.6.A.1.i 중력력은 인력(attractive)이다.
      • 2.6.A.1.ii 중력력은 항상 두 상호작용 시스템의 질량 중심을 잇는 직선 방향으로 작용한다.
      • 2.6.A.1.iii 시스템에 작용하는 중력력은 시스템의 질량 중심에 작용하는 것으로 간주할 수 있다.
    • 2.6.A.2 장은 공간상의 다양한 위치에서 물체에 가해지는 비접촉력의 효과를 모델링한다.
      • 2.6.A.2.i 질량 $M$인 시스템이 생성한 중력장의 세기는 특정 지점에서 시스템이 질량 $m$의 시험 물체에 가하는 중력힘과 시험 물체의 질량의 비율과 같다.
        • 식: $\left|\vec{g}\right| = \dfrac{\left|\vec{F}_g\right|}{m} = G\dfrac{M}{r^2}$
      • 2.6.A.2.ii 중력만이 물체에 작용한다면, 관측된 물체의 가속도(m/s$^2$)는 해당 지점의 중력장 세기(N/kg)와 수치적으로 같다.
    • 2.6.A.3 천체가 상대적으로 작은 근접 물체에 가하는 중력력을 무게(weight)라고 한다.
      • 식: $\text{Weight} = F_g = mg$

    학습 목표 2.6.B: 중력력을 일정하게 간주할 수 있는 상황을 설명할 수 있다.

    • 2.6.B.1 두 시스템의 질량 중심 간 중력력이 두 시스템의 상대적 위치 변화에 따라 무시할 만큼 변하지 않는다면, 초기 및 최종 위치 사이 모든 지점에서 중력력을 일정하다고 간주할 수 있다.
    • 2.6.B.2 지구 표면 근처에서 중력장의 세기는 $g \approx 10 \text{ N/kg}$이다.

    학습 목표 2.6.C: 시스템의 겉보기 무게(apparent weight) 크기가该系统에 가해지는 중력힘 크기와 다른 조건을 설명할 수 있다.

    • 2.6.C.1 시스템의 겉보기 무게 크기는该系统에 가해지는 법선력의 크기와 같다.
    • 2.6.C.2 시스템이 가속 중이라면该系统의 겉보기 무게 크기는该系统에 가해지는 중력힘 크기와 다르다.
    • 2.6.C.3该系统에 작용하는 힘이 없거나 중력만이该系统에 작용하는 경우该系统은 무중력 상태(weightlessness)로 보인다.
    • 2.6.C.4 등가원리는 비관성 기준계에 있는 관측자가 물체의 겉보기 무게와 중력장이 물체에 가하는 중력 사이의 차이를 구별할 수 없음을 의미한다.

    학습 목표 2.6.D: 관성 질량과 중력 질량을 설명하라.

    • 2.6.D.1 물체는 관성 질량, 즉 관성을 가지는데, 이는 다른 물체와 상호작용할 때 물체의 운동이 변하는 데 얼마나 저항하는지를 결정하는 성질이다.
    • 2.6.D.2 중력 질량은 질량을 가진 두 시스템 간의 인력他与 관련된다.
    • 2.6.D.3 관성 질량과 중력 질량은 실험적으로 등가임이 입증되었다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Orbital motion (Kepler's 2nd law)

    Near a planet's surface, the gravitational force (weight) is $F_g=mg$, directed down, where $g$ is the gravitational field strength 重力场强度. More generally, Newton's law of gravitation 万有引力定律 gives the attraction between any two masses:

    $$F_g=\frac{G m_1 m_2}{r^2},$$
    directed along the line joining them, weaker as the distance $r$ grows (an inverse-square law). Doubling the separation quarters the force.

    Two masses attract each other with equal, opposite, inverse-square forces along the line joining them
    Two masses attract each other with equal, opposite, inverse-square forces along the line joining them

    Worked example. A $2.0\ \text{kg}$ object weighs $19.6\ \text{N}$ on Earth ($g=9.8\ \text{m/s}^2$). On the Moon $g_{\text{Moon}}=1.6\ \text{m/s}^2$. Its mass is unchanged ($2.0\ \text{kg}$), but its weight becomes $F_g=mg=2.0\times1.6=3.2\ \text{N}$. Mass measures inertia; weight is a force that depends on where you are.

    Mass actually plays two distinct roles. Inertial mass 惯性质量 sets how strongly an object resists acceleration ($F=ma$); gravitational mass 引力质量 sets how strongly it attracts other masses ($F=Gm_1m_2/r^2$). Experiments confirm the two are equal – which is exactly why every object, heavy or light, falls with the same $g$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    gravitational field strength/ˌɡrævɪˈteɪʃənl fiːld streŋθ/ gravitational field strength
    Newton's law of gravitation/ˈnjuːtnz lɔː ɒv ˌɡrævɪˈteɪʃn/ Newton's law of gravitation
    Inertial mass/ɪˈnɜːʃl mæs/ Inertial mass
    gravitational mass/ˌɡrævɪˈteɪʃənl mæs/ gravitational mass
    2.7

    Kinetic and Static Friction

    Syllabus
    English

    Learning Objective 2.7.A: Describe kinetic friction between two surfaces

    • 2.7.A.1 Kinetic friction occurs when two surfaces in contact move relative to each other.
      • 2.7.A.1.i The kinetic friction force is exerted in a direction opposite to the motion of each surface relative to the other surface.
      • 2.7.A.1.ii The force of friction between two surfaces does not depend on the size of the surface area of contact.
    • 2.7.A.2 The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction.
      • Equation: $\left|\vec{F}_{f,k}\right| = \left|\mu_k \vec{F}_n\right|$
      • 2.7.A.2.i The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
      • 2.7.A.2.ii Normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.

    Learning Objective 2.7.B: Describe static friction between two surfaces.

    • 2.7.B.1 Static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.
    • 2.7.B.2 Static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface.
      • Equation: $\left|\vec{F}_{f,s}\right| \leq \left|\mu_s \vec{F}_n\right|$
      • 2.7.B.2.i Slipping and sliding refer to situations in which two surfaces are moving relative to each other.
      • 2.7.B.2.ii There exists a maximum value for which static friction will prevent an object from slipping on a given surface.
        • Equation: $F_{f,s,\max} = \mu_s F_n$
    • 2.7.B.3 The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.
    한국어

    학습 목표 2.7.A: 두 표면 사이의 운동 마찰력을 설명하라.

    • 2.7.A.1 접촉된 두 표면이 서로에 대해 상대 운동을 할 때 운동 마찰력이 발생한다.
      • 2.7.A.1.i 운동 마찰력의 힘은 각 표면이 다른 표면에 대해 이동하는 방향과 반대 방향으로 작용한다.
      • 2.7.A.1.ii 두 표면 사이의 마찰력은 접촉 면적의 크기에 의존하지 않는다.
    • 2.7.A.2 물체에 작용하는 운동 마찰력의 세기는 표면에 의해 물체에 가해지는 수직 반력(정반력)과 운동 마찰 계수의 곱으로 주어진다.
      • 식: $\left|\vec{F}_{f,k}\right| = \left|\mu_k \vec{F}_n\right|$
      • 2.7.A.2.i 운동 마찰 계수는 접촉된 표면의 재질 특성에 따라 달라진다.
      • 2.7.A.2.ii 수직 반력은 물체에接触的한 표면에 의해 물체에 가해지는 힘의 수직 성분이며, 표면으로부터外向로 향한다.

    학습 목표 2.7.B: 두 표면 사이의 정지 마찰력을 설명하라.

    • 2.7.B.1 정지 마찰력은 서로에 대해 정지해 있는 두 물체의 접촉면 사이에서 발생할 수 있다.
    • 2.7.B.2 정지 마찰력은 물체가表面上을 미끄러지거나 미끄러지는 것을 방지하기 위해 필요한 값과 방향을 취한다.
      • 식: $\left|\vec{F}_{f,s}\right| \leq \left|\mu_s \vec{F}_n\right|$
      • 2.7.B.2.i 미끄러짐(slipping)과 미끄러짐(sliding)은 두 표면이 서로에 대해 상대 운동을 하는 상황을 의미한다.
      • 2.7.B.2.ii 주어진表面上에서 물체의 미끄러짐을 방지하는 정지 마찰력에는 최대값이 존재한다.
        • 식: $F_{f,s,\max} = \mu_s F_n$
    • 2.7.B.3 특정 쌍의 표면에 대해 정지 마찰 계수는 일반적으로 운동 마찰 계수보다 크다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Friction 摩擦力 acts along a surface, opposing relative sliding (or the tendency to slide):

    • Kinetic friction 动摩擦 (while sliding): $f_k=\mu_k N$.
    • Static friction 静摩擦 (while not yet sliding): $f_s\le \mu_s N$ – it adjusts up to a maximum to prevent motion.

    Here $N$ is the normal force 法向力 (surface push, perpendicular to the surface) and $\mu$ is the coefficient of friction 摩擦系数.

    Worked example. A $5.0\ \text{kg}$ crate sits on a level floor with $\mu_s=0.40$. Will a horizontal push of $15\ \text{N}$ move it? On a level floor $N=mg=5.0\times9.8=49\ \text{N}$, so the largest static friction is $f_{s,\max}=\mu_s N=0.40\times49=19.6\ \text{N}$. The $15\ \text{N}$ push is smaller than $19.6\ \text{N}$, so friction rises to match it and the crate stays still.

    Explore · ⁨탐색하기⁩

    Slide a block down a slope with friction · ⁨마찰이 있는 경사면을 따라 블록 미끄러지기⁩

    Friction opposes motion up to a maximum $\mu N$. Tilt the slope until gravity's pull along it beats static friction and the block starts to slide. · ⁨마찰은 최대 $\mu N$까지 운동을 방해합니다. 경사면을 기울여 중력의 경사 방향 분력이 정지 마찰력을 넘어서도록 하면 블록이 미끄러지기 시작합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Friction/ˈfrɪkʃn/ Friction
    Kinetic friction/kɪˈnetɪk ˈfrɪkʃn/ Kinetic friction
    Static friction/ˈstætɪk ˈfrɪkʃn/ Static friction
    normal force/ˈnɔːml fɔːs/ normal force
    coefficient of friction/ˌkəʊɪˈfɪʃənt ɒv ˈfrɪkʃn/ coefficient of friction
    2.8

    Spring Forces

    Syllabus
    English

    Learning Objective 2.8.A: Describe the force exerted on an object by an ideal spring

    • 2.8.A.1 An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length.
    • 2.8.A.2 The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law:
      • Equation: $\vec{F}_s = -k\Delta\vec{x}$
    • 2.8.A.3 The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.
    한국어

    학습 목표 2.8.A: 이상적인 스프링이 물체에 가하는 힘을 설명하라.

    • 2.8.A.1 이상적인 스프링은 질량이 무시할 만큼 작으며, 이완된 길이로부터 측정된 길이의 변화량에 비례하는 힘을 가한다.
    • 2.8.A.2 이상적인 스프링이 물체에 가하는 힘의 세기는 훅의 법칙에 의해 주어진다:
      • 식: $\vec{F}_s = -k\Delta\vec{x}$
    • 2.8.A.3 스프링이 물체에 가하는 힘은 항상 물체-스프링 시스템의 평형 위치를 향한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Hooke's law & the elastic limit

    An ideal spring exerts a restoring force 回复力 proportional to its stretch or compression – Hooke's law 胡克定律:

    $$F_s=-kx,$$
    where $k$ is the spring constant 弹簧常数 (stiffness) and $x$ is the displacement from the spring's natural length. The minus sign means the force points back toward equilibrium.

    Hooke's law: extension is proportional to load up to the limit of proportionality
    Hooke's law: extension is proportional to load up to the limit of proportionality

    Worked example. A spring with $k=200\ \text{N/m}$ hangs vertically and a $0.50\ \text{kg}$ mass is hung on it. How far does it stretch at rest? At rest the spring force balances the weight, $kx=mg$, so

    $$x=\frac{mg}{k}=\frac{0.50\times9.8}{200}=0.025\ \text{m}=2.5\ \text{cm}.$$

    Explore · ⁨탐색하기⁩

    Stretch a spring (Hooke's law) · ⁨스프링 늘리기 (후크의 법칙)⁩

    A spring's force is proportional to its extension, $F=kx$ (Hooke's law). Pull harder and the extension grows in step — until the spring's limit. · ⁨스프링의 힘은 신도에 비례하며, $F=kx$(후크의 법칙)입니다. 더 세게 당길수록 신도가 비례하여 증가—but 스프링의 한계까지.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    restoring force/rɪˈstɔːrɪŋ fɔːs/ restoring force
    Hooke's law/hʊks lɔː/ Hooke's law
    spring constant/sprɪŋ ˈkɒnstənt/ spring constant
    2.9

    Circular Motion

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.9.A
    Describe the motion of an object traveling in a circular path.

    • 2.9.A.1 Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.
      • 2.9.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
        • Equation: $a_c = \dfrac{v^2}{r}$
      • 2.9.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
    • 2.9.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.
      • 2.9.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum speed, the gravitational force is the only force that causes the centripetal acceleration.
        • Equation: $v = \sqrt{gr}$
      • 2.9.A.2.ii Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.
      • 2.9.A.2.iii A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
    • 2.9.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
    • 2.9.A.4 The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
    • 2.9.A.5 The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.
      • 2.9.A.5.i The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, $T$.
      • 2.9.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
        • Equation: $T = \dfrac{1}{f}$
      • 2.9.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
        • Equation: $T = \dfrac{2\pi r}{v}$

    2.9.B
    Describe circular orbits using Kepler's third law.

    • 2.9.B.1 For a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. The period and radius of the circular orbit are related to the mass of the central body.
      • Equation: $T^2 = \dfrac{4\pi^2}{GM} R^3$

    Boundary statement: AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion. Analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions.

    Boundary statement: AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Uniform circular motion

    An object moving in a circle at constant speed still accelerates, because its velocity direction keeps changing. This centripetal acceleration 向心加速度 points toward the center:

    $$a_c=\frac{v^2}{r}.$$
    It is produced by a net inward (centripetal) force 向心力 $F_c=\dfrac{mv^2}{r}$ – supplied by whatever real force points inward (tension, gravity, friction, normal). The time for one full revolution is the period 周期 $T$, and the number of revolutions per second is the frequency 频率 $f$; they are reciprocals, $f = 1/T$, and the same definitions describe any rotation or oscillation.There is no separate outward force; "centrifugal" is only an apparent effect.

    The velocity points along the tangent; the centripetal force and acceleration point to the centre
    The velocity points along the tangent; the centripetal force and acceleration point to the centre

    Worked example. A $0.30\ \text{kg}$ ball on a string is whirled in a horizontal circle of radius $0.80\ \text{m}$ at $4.0\ \text{m/s}$. Find the tension in the string. The tension supplies the whole centripetal force:

    $$T=\frac{mv^2}{r}=\frac{0.30\times4.0^2}{0.80}=6.0\ \text{N}.$$
    If the string can take at most $6.0\ \text{N}$, this is the fastest the ball can go at that radius – spin any faster and the string breaks.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ centripetal acceleration
    net inward (centripetal) force/net ˈɪnwəd fɔːs/ net inward (centripetal) force
    period/ˈpɪərɪəd/ period
    frequency/ˈfriːkwənsi/ frequency
    2.9

    Exam tips

    • Always draw a free-body diagram first: only real forces on the chosen object (weight, normal, tension, friction, applied) — never an "$ma$" arrow.
    • Apply $\sum F = ma$ one axis at a time; on an incline resolve gravity into $mg\sin\theta$ (along) and $mg\cos\theta$ (perpendicular).
    • A Newton's third-law pair acts on two different objects — a book's weight and the table's normal force are not a pair (both act on the book).
    • Static friction adjusts up to $\mu_s N$ (use it to test whether motion starts); once sliding, use kinetic friction $f_k=\mu_k N$.
    • Circular motion needs a net inward (centripetal) force $mv^2/r$ supplied by a real force — there is no outward "centrifugal" force.
  • 3

    Work, Energy, and Power

    Watch lesson · ⁨수업 보기⁩
    3.1

    Translational Kinetic Energy · ⁨평동 운동 에너지⁩

    Syllabus
    English

    Learning Objective 3.1.A: Describe the translational kinetic energy of an object in terms of the object's mass and velocity.

    • 3.1.A.1 An object's translational kinetic energy is given by the equation
      • Equation: $K = \dfrac{1}{2}mv^2$
    • 3.1.A.2 Translational kinetic energy is a scalar quantity.
    • 3.1.A.3 Different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference.
    한국어

    학습 목표 3.1.A: 물체의 질량과 속도를 통해 물체의 병진 운동 에너지를 설명한다.

    • 3.1.A.1 물체의 병진 운동 에너지는 다음 식으로 주어진다.
      • 식: $K = \dfrac{1}{2}mv^2$
    • 3.1.A.2 병진 운동 에너지는 스칼라량이다.
    • 3.1.A.3 관측자의 기준계에 따라 물체의 병진 운동 에너지 측정값이 다를 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Energy 能量 is the capacity to do work, measured in joules 焦耳 (J). A moving object has kinetic energy 动能:

    $$K=\tfrac{1}{2}mv^2.$$
    It depends on the square of the speed, so doubling the speed quadruples the kinetic energy. Kinetic energy is a scalar and is never negative.

    Worked example. A $1500\ \text{kg}$ car travels at $20\ \text{m/s}$. Its kinetic energy is $K=\tfrac12\times1500\times20^2=3.0\times10^{5}\ \text{J}=300\ \text{kJ}$. If it speeds up to $40\ \text{m/s}$ (double), the kinetic energy becomes $4\times$ larger, $1200\ \text{kJ}$ – which is why stopping distance grows so fast with speed.

    한국어

    에너지는 일을 할 수 있는 능력이며, 단위는 줄(J)이다. 움직이는 물체는 운동 에너지를 가진다:

    $$K=\tfrac{1}{2}mv^2.$$
    속도의 제곱에 비례하므로 속도를 두 배로 하면 운동 에너지는 네 배가 된다. 운동 에너지는 스칼라 quantity이며 절대 음수가 될 수 없다.

    해설 예제. 질량 $1500\ \text{kg}$인 자동차가 속도 $20\ \text{m/s}$로 주행 중이다. 이 때 운동 에너지는 $K=\tfrac12\times1500\times20^2=3.0\times10^{5}\ \text{J}=300\ \text{kJ}$이다. 만약 속도가 $40\ \text{m/s}$(두 배)로 증가하면 운동 에너지는 $4\times$만큼 커지며, 이는 $1200\ \text{kJ}$ — 이것이 속도에 따라 정지 거리가 급격히 증가하는 이유이다.

    3.2

    Work · ⁨일⁩

    Syllabus
    English

    Learning Objective 3.2.A: Describe the work done on an object or system by a given force or collection of forces.

    • 3.2.A.1 Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance.
      • 3.2.A.1.i The work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system.
      • 3.2.A.1.ii The work done by a conservative force on a system—or the change in the potential energy of the system—will be zero if the system returns to its initial configuration.
      • 3.2.A.1.iii Potential energies are associated only with conservative forces.
      • 3.2.A.1.iv The work done by a nonconservative force is path-dependent.
      • 3.2.A.1.v Examples of nonconservative forces are friction and air resistance.
    • 3.2.A.2 Work is a scalar quantity that may be positive, negative, or zero.
    • 3.2.A.3 The amount of work done on a system by a constant force is related to the components of that force and the displacement of the point at which that force is exerted.
      • 3.2.A.3.i Only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy.
        • Equation: $W = F_{\parallel}d = Fd\cos\theta$
      • 3.2.A.3.ii The component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy.
    • 3.2.A.4 The work-energy theorem states that the change in an object's kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object.
      • Equation: $\Delta K = \sum_{i} W_i = \sum_{i} F_{\parallel,i}\,d$
      • 3.2.A.4.i An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system.
      • 3.2.A.4.ii If the system's center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system's kinetic energy can change.
      • 3.2.A.4.iii The energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted
        • Equation: $\Delta E_{\text{mech}} = F_f\,d\cos\theta$
    • 3.2.A.5 Work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement.

    Boundary statement: AP Physics 1 only expects students to analyze the transfer of mechanical energy (as defined in Unit 3, Topic 4: Conservation of Energy), although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound. In AP Physics 2, students will also study how thermal energy can be transferred between systems through heating or cooling.

    한국어

    학습 목표 3.2.A: 주어진 힘이나 힘의 집합에 의해 물체나 계에 가해진 일을 설명한다.

    • 3.2.A.1 일은 한 힘이 그系统进行 거리 동안 가할 때 시스템으로 들어오거나 나가는 에너지의 양이다.
      • 3.2.A.1.i 보존력에 의한 시스템에 대한 일은 경로와 무관하며 오직该系统의 초기 및 최종 구성에만 의존한다.
      • 3.2.A.1.ii 보존력에 의한 시스템에 대한 일, 혹은 시스템의ポテンシャル 에너지의 변화량은该系统가 초기 구성으로 돌아갈 경우 0이 된다.
      • 3.2.A.1.iii 포텐셜 에너지는 오직 보존력과 관련된다.
      • 3.2.A.1.iv 비보존력에 의한 일은 경로에 의존한다.
      • 3.2.A.1.v 비보존력의 예로는 마찰력과 공기 저항이 있다.
    • 3.2.A.2 일은 양수, 음수, 또는 0이 될 수 있는 스칼라량이다.
    • 3.2.A.3 상수력에 의한 시스템에 대한 일의 양은 해당 힘의 성분과 해당 힘이 가해지는 점의 변위와 관련된다.
      • 3.2.A.3.i 힘의 적용점의 변위에 평행한 힘의 성분만이 시스템의 총 에너지를 변경시킨다.
        • 식: $W = F_{\parallel}d = Fd\cos\theta$
      • 3.2.A.3.ii 시스템의 질량 중심의 변위 방향에 수직인 힘의 성분은 시스템의 운동 방향을 변경시킬 수 있으나 시스템의 운동 에너지는 변경시키지 못한다.
    • 3.2.A.4 운동 에너지 정리(Work-Energy Theorem)에 따르면 물체의 운동 에너지 변화량은 물체에 가해지는 모든 힘에 의해 이루어지는 일(순일)의 합과 같다.
      • 식: $\Delta K = \sum_{i} W_i = \sum_{i} F_{\parallel,i}\,d$
      • 3.2.A.4.i 외부 힘은 시스템의 구성을 변경시킬 수 있다. 변위에 평행한 외부 힘의 성분과 힘의 적용점의 변위를 곱하면 시스템의 운동 에너지 변화량을 구할 수 있다.
      • 3.2.A.4.ii 시스템의 질량 중심과 힘의 적용점이 같은 거리를 이동한다면,该系统는 물체로 모델링될 수 있으며 시스템의 운동 에너지만 변경될 수 있다.
      • 3.2.A.4.iii 마찰에 의해 소산되는 에너지는 일반적으로 마찰력 크기와 해당 힘이 가해진 경로의 길이의 곱과 동일시된다.
        • 식: $\Delta E_{\text{mech}} = F_f\,d\cos\theta$
    • 3.2.A.5 일은 $F_{\parallel}$를 변위의 함수로 나타낸 그래프 아래 면적과 같다.

    경계 설명: AP 물리학 1에서는 학생들에게 단원 3, 주제 4: 에너지 보존에서 정의된 대로 역학적 에너지(mechanical energy)의 전달만을 분석할 것을 기대한다. 다만 역학적 에너지가 열에너지나 소리 형태로 소산될 수 있다는 점을 인지하고 있어야 한다. AP 물리학 2에서는 학생들은 열에너지를 가열이나 냉각을 통해 시스템 간에 전달하는 방식도 연구하게 된다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Work 功 is energy transferred by a force acting over a displacement:

    $$W=F\,d\cos\theta,$$
    where $\theta$ is the angle between the force and the displacement. Work is positive when the force has a component along the motion (adds energy), negative when it opposes the motion (removes energy), and zero when the force is perpendicular. On a force–position graph, work is the area under the curve. The work–energy theorem 动能定理 states that the net work equals the change in kinetic energy: $W_{\text{net}}=\Delta K$.

    Worked example. A $2.0\ \text{kg}$ block moving at $3.0\ \text{m/s}$ on a frictionless floor is pushed by a $5.0\ \text{N}$ force over $4.0\ \text{m}$ in the direction of motion. Find its final speed. The net work is $W=Fd=5.0\times4.0=20\ \text{J}$, and by the work–energy theorem $W=\tfrac12 m(v^2-v_0^2)$:

    $$20=\tfrac12\times2.0\times(v^2-3.0^2)\;\Rightarrow\;v^2=29\;\Rightarrow\;v=5.4\ \text{m/s}.$$

    한국어

    일은 변위 동안 힘에 의해 전달된 에너지이다:

    $$W=F\,d\cos\theta,$$
    여기서 $\theta$는 힘과 변위 사이의 각도이다. 힘이 운동 방향으로 성분을 가질 때는 양수이며 에너지를 더하고, 운동과 반대 방향일 때는 음수이며 에너지를 빼며, 힘과 수직일 때는 영점이다. 힘-위치 그래프에서 일은 곡선 아래의 면적이다. 일-에너지 정리는 순일이 운동 에너지의 변화량과 같다고 말한다: $W_{\text{net}}=\Delta K$.

    변위 방향으로만 힘의 성분이 일을 한다
    변위 방향으로만 힘의 성분이 일을 한다

    해설 예제. 질량 $2.0\ \text{kg}$인 블록이 마찰 없는 바닥 위에서 속도 $3.0\ \text{m/s}$로 움직이고 있으며, 운동 방향과 동일한 $5.0\ \text{N}$의 힘에 의해 $4.0\ \text{m}$만큼 밀려나고 있다. 최종 속도를 구하시오. 순일은 $W=Fd=5.0\times4.0=20\ \text{J}$이며, 일-에너지 정리에 의하면 $W=\tfrac12 m(v^2-v_0^2)$:

    $$20=\tfrac12\times2.0\times(v^2-3.0^2)\;\Rightarrow\;v^2=29\;\Rightarrow\;v=5.4\ \text{m/s}.$$

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Work/wɜːk/ 일
    work–energy theorem/wɜːk ˈenədʒi ˈθɪərəm/ 일-에너지 정리
    Potential energy/pəˈtenʃl ˈenədʒi/ 위치 에너지
    Gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ 중력 포텐셜 에너지
    Elastic potential energy/ɪˈlæstɪk pəˈtenʃl ˈenədʒi/ 탄성Potential 에너지
    conservative forces/kənˈsɜːvətɪv ˈfɔːsɪz/ 보존력
    total mechanical energy/ˈtəʊtl mɪˈkænɪkl ˈenədʒi/ 총 역학적 에너지
    conserved/kənˈsɜːvd/ 보존된다
    thermal energy/ˈθɜːml ˈenədʒi/ 열에너지
    3.3

    Potential Energy · ⁨위치 에너지⁩

    Syllabus
    English

    Learning Objective 3.3.A: Describe the potential energy of a system.

    • 3.3.A.1 A system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces.
    • 3.3.A.2 Potential energy is a scalar quantity associated with the position of objects within a system.
    • 3.3.A.3 The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis.
    • 3.3.A.4 The potential energy of common physical systems can be described using the physical properties of that system.
      • 3.3.A.4.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
        • Equation: $U_s = \dfrac{1}{2}k(\Delta x)^2$
      • 3.3.A.4.ii The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (e.g., moons, planets or stars) is given by the equation
        • Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
      • 3.3.A.4.iii Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass $m$ and a planet with gravitational field of magnitude $g$ when the object is near the surface of the planet may be approximated by the equation
        • Equation: $\Delta U_g = mg\Delta y$
    • 3.3.A.5 The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system.
    한국어

    학습 목표 3.3.A: 시스템의 포텐셜 에너지를 설명한다.

    • 3.3.A.1 두 개 이상의 물체로 구성된 시스템은该系统 내의 물체들이 서로 보존력만으로 상호작용할 경우 포텐셜 에너지를 가진다.
    • 3.3.A.2 포텐셜 에너지는该系统 내의 물체의 위치와 관련된 스칼라량이다.
    • 3.3.A.3 특정 시스템에 대한 포텐셜 에너지의 0점 정의는 상황을 고려하여 분석을 단순화하거나 돕기 위해 관측자가 결정하는 사항이다.
    • 3.3.A.4 일반적인 물리 시스템의 포텐셜 에너지는该系统의 물리적 특성을 사용하여 설명할 수 있다.
      • 3.3.A.4.i 이상적인 스프링의 탄성 포텐셜 에너지는 다음 식으로 주어지며, 여기서 $\Delta x$는 스프링이 평형 길이로부터 늘어나거나 압축된 거리이다.
        • 식: $U_s = \dfrac{1}{2}k(\Delta x)^2$
      • 3.3.A.4.ii 대략적인 구형 질량 분포(예: 위성, 행성 또는 별)로 구성된 시스템의 중력 포텐셜 에너지의 일반 형태는 다음 식으로 주어진다.
        • 식: $U_g = -G\dfrac{m_1 m_2}{r}$
      • 3.3.A.4.iii 행성 표면 근처의 중력장은 거의 일정하므로, 질량이 $m$인 물체와 중력장 세기가 $g$인 행성이 있는 시스템에서 물체가 행성 표면 근처에 있을 때 중력 포텐셜 에너지의 변화량은 다음 식으로 근사할 수 있다.
        • 식: $\Delta U_g = mg\Delta y$
    • 3.3.A.5 두 개 이상의 물체를 포함하는 시스템의 총 포텐셜 에너지는该系统 내의 각 물체 쌍의 포텐셜 에너지의 합이다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Potential energy 势能 is stored energy that depends on position or configuration:

    • Gravitational potential energy 重力势能 near the surface: $U_g=mgh$ (height $h$ above a reference level).
    • Elastic potential energy 弹性势能 in a spring: $U_s=\tfrac{1}{2}kx^2$.

    Potential energy is defined only for conservative forces 保守力 (gravity, springs), for which the stored energy depends on position, not path. Only changes in potential energy matter, so you may put the zero level wherever is convenient.

    The $U_g=mgh$ form only holds near a surface where $g$ is roughly constant. The general form, for two spherical masses a distance $r$ apart, is

    $$U_g=-\frac{G m_1 m_2}{r}.$$
    It is negative and defined to be zero at infinite separation, so gravitational PE rises toward zero as the masses move apart. $mgh$ is simply its near-surface approximation. (This is the form you need for satellites and escape speed.)

    한국어

    위치 에너지는 위치 또는 구성에 따라 저장된 에너지이다:

    • 중력 위치 에너지 (표면 근처): $U_g=mgh$ (기준 수준으로부터 높이 $h$).
    • 탄성 위치 에너지 (스프링): $U_s=\tfrac{1}{2}kx^2$.

    위치 에너지는 보존력(중력, 스프링 등)에만 정의되며, 이때 저장된 에너지는 경로가 아닌 위치에 의존한다. 오직 위치 에너지의 변화량만이 중요하므로, 편하게 설정할 수 있는 어디든 영점을 둘 수 있다.

    $U_g=mgh$ 형태는 오직 표면에 가까운 곳에서만 유효하며, 이때 $g$은 대략 일정하다. 일반적인 형태는 거리 $r$만큼 떨어진 두 구형 질량에 대해 다음과 같다

    $$U_g=-\frac{G m_1 m_2}{r}.$$
    이것은 음수이며 무한한 거리에서의 분리를 기준으로 영점으로 정의되므로, 질량들이 멀어질수록 중력 PE는 영점 쪽으로 상승한다. $mgh$는 단순히 그 표면 근처 근사값이다. (위성과 탈출 속도에 필요한 형태는 이 것이다.)

    Explore · ⁨탐색하기⁩

    Store elastic potential energy in a spring · ⁨스프링에 탄성 위치에너지 저장하기⁩

    Stretching a spring stores elastic potential energy $\tfrac12 kx^2$ — the area under the force-extension line. Release it and that energy becomes kinetic. · ⁨스프링을 늘리면 탄력 Potential 에너지 $\tfrac12 kx^2$가 저장되는데 — 이는 힘-신도 선 아랫면적입니다. 놓아주면 이 에너지가 운동 에너지로 전환됩니다.⁩

    3.4

    Conservation of Energy · ⁨에너지 보존⁩

    Syllabus
    English

    Learning Objective 3.4.A: Describe the energies present in a system.

    • 3.4.A.1 A system composed of only a single object can only have kinetic energy.
    • 3.4.A.2 A system that contains objects that interact via conservative forces or that can change its shape reversibly may have both kinetic and potential energies.

    Learning Objective 3.4.B: Describe the behavior of a system using conservation of mechanical energy principles.

    • 3.4.B.1 Mechanical energy is the sum of a system's kinetic and potential energies.
    • 3.4.B.2 Any change to a type of energy within a system must be balanced by an equivalent change of other types of energies within the system or by a transfer of energy between the system and its surroundings.
    • 3.4.B.3 A system may be selected so that the total energy of that system is constant.
    • 3.4.B.4 If the total energy of a system changes, that change will be equivalent to the energy transferred into or out of the system.

    Learning Objective 3.4.C: Describe how the selection of a system determines whether the energy of that system changes.

    • 3.4.C.1 Energy is conserved in all interactions.
    • 3.4.C.2 If the work done on a selected system is zero and there are no nonconservative interactions within the system, the total mechanical energy of the system is constant.
    • 3.4.C.3 If the work done on a selected system is nonzero, energy is transferred between the system and the environment.

    Boundary statement: AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.

    한국어

    학습 목표 3.4.A: 시스템에 존재하는 에너지를 설명한다.

    • 3.4.A.1 단일 물체만으로 구성된 시스템은 운동 에너지만 가질 수 있다.
    • 3.4.A.2 보존력을 통해 상호작용하거나 형상을 가역적으로 변경할 수 있는 물체를 포함하는 시스템은 운동 에너지와 포텐셜 에너지를 모두 가질 수 있다.

    학습 목표 3.4.B: 역학적 에너지 보존 원리를 사용하여 시스템의 거동을 설명한다.

    • 3.4.B.1 역학적 에너지는 시스템의 운동에너지와 위치에너지의 합이다.
    • 3.4.B.2 시스템 내 한 형태의 에너지가 변화할 경우,该系统 내 다른 형태의 에너지에 동일한 크기의 변화가 일어나거나 시스템과 주변 환경 간에 에너지가 이동해야 한다.
    • 3.4.B.3 특정 시스템의 총 에너지를 일정하게 만들 수 있도록 시스템을 선택할 수 있다.
    • 3.4.B.4 시스템의 총 에너지가 변화할 경우, 그 변화량은 시스템으로 유입되거나 시스템에서 유출된 에너지와 동일하다.

    학습 목표 3.4.C: 시스템의 선택이 해당 시스템의 에너지 변화 여부에 미치는 영향을 설명하라.

    • 3.4.C.1 모든 상호작용에서 에너지는 보존된다.
    • 3.4.C.2 선택된 시스템에 가해진仕事が 0이며 시스템 내에 비보존적 상호작용이 없으면, 시스템의 총 역학적 에너지는 일정하다.
    • 3.4.C.3 선택한 물체에 가해진 일이 영이 아닌 경우, 에너지가 물체와 환경 사이를 이동한다.

    경계 선언: AP 물리학 1은 학생이 비보존적 힘에 의해 역학적 에너지가 열에너지나 소리로 산란될 수 있음을 알아야 한다고 기대한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English
    Energy conservation: KE ⇄ PE

    The total mechanical energy 机械能 is $E=K+U$. When only conservative forces do work, mechanical energy is conserved 守恒:

    $$K_1+U_1=K_2+U_2.$$
    When friction or other non-conservative forces act, they transfer mechanical energy to thermal energy 热能; then the general statement is that total energy (including thermal) is conserved. Energy bar charts are a good way to track where the energy goes.

    Worked example. A ball is released from rest at the top of a frictionless ramp $5.0\ \text{m}$ high. Find its speed at the bottom. All the gravitational potential energy becomes kinetic energy:

    $$mgh=\tfrac12 mv^2\;\Rightarrow\;v=\sqrt{2gh}=\sqrt{2\times9.8\times5.0}=9.9\ \text{m/s}.$$
    The mass cancels, so every object reaches the same speed – exactly the free-fall result, now got from energy. If instead $30\ \text{J}$ were lost to friction, you would subtract it: $mgh-30=\tfrac12 mv^2$.

    한국어
    에너지 보존: KE ⇄ PE
    하늘을 배경으로 한 높고 긴 루프가 있는 녹색 및 빨간색 레일의 롤러코스터
    롤러코스터는 에너지를 오고가며 교환합니다: 가장 높은 지점(최대 위치 에너지)과 가장 빠른 지점(최대 운동 에너지)에서 각각 최대값을 가집니다

    총 기계적 에너지는 $E=K+U$입니다. 비소멸력만 작용할 때 기계적 에너지는 보존됩니다:

    $$K_1+U_1=K_2+U_2.$$
    마찰력이나 다른 비보존력이 작용하면 기계적 에너지를 열에너지로 전달합니다. 이때 총 에너지(열에너지 포함)는 보존된다는 일반적인 진술이 성립합니다. 에너지 막대 차트는 에너지가 어디로 가는지를 추적하는 데 유용한 방법입니다.

    흔들리는 진자는 중력포텐셜 에너지를 운동 에너지로 바꾸고 다시 되돌린다
    흔들리는 진자는 중력포텐셜 에너지를 운동 에너지로 바꾸고 다시 되돌린다

    해설 예제. 마찰 없는 경사 $5.0\ \text{m}$ 높이에서 정지 상태에서 공을 놓습니다. 바닥에서의 속도를 구하십시오. 중력ポ텐셜 에너지가 모두 운동 에너지로 변합니다:

    $$mgh=\tfrac12 mv^2\;\Rightarrow\;v=\sqrt{2gh}=\sqrt{2\times9.8\times5.0}=9.9\ \text{m/s}.$$
    질량이 소거되므로 모든 물체는 같은 속도에 도달합니다 – 이는 자유 낙하의 결과와 정확히 일치하며, 이제는 에너지를 통해 얻은 것입니다. 만약 $30\ \text{J}$가 마찰로 인해 손실된다면 이를 빼야 합니다: $mgh-30=\tfrac12 mv^2$.

    Explore · ⁨탐색하기⁩

    Watch energy convert as an object falls · ⁨物体 falling 시 에너지 변환 Watching⁩

    With no friction, mechanical energy is conserved: as an object falls, gravitational potential energy turns into kinetic energy while the total stays fixed. · ⁨마찰이 없을 때, 기계적 에너지 보존: Object falling 시 중력 **위치에너지(potential energy)**가 **운동에너지(kinetic energy)**로 전환되며 총량은 일정합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Energy/ˈenədʒi/ 에너지
    joules/dʒuːlz/ 줄
    kinetic energy/kɪˈnetɪk ˈenədʒi/ 운동 에너지Unless kinetic energy.
    3.5

    Power · ⁨비율(Power)⁩

    Syllabus
    English

    Learning Objective 3.5.A: Describe the transfer of energy into, out of, or within a system in terms of power.

    • 3.5.A.1 Power is the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system.
    • 3.5.A.2 Average power is the amount of energy being transferred or converted, divided by the time it took for that transfer or conversion to occur.
      • Equation: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
    • 3.5.A.3 Because work is the change in energy of an object or system due to a force, average power is the total work done, divided by the time during which that work was done.
      • Equation: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
    • 3.5.A.4 The instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity can be described with the derived equation.
      • Equation: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$
    한국어

    학습 목표 3.5.A: 출력을 사용하여 시스템 안으로, 밖으로, 혹은 내부로 에너지가 이동하는 과정을 설명하라.

    • 3.5.A.1 출력은 시간당 에너지의 변화율로, 시스템으로 유입되거나 유출되는 경우이거나 시스템 내에서 한 형태에서 다른 형태로 전환되는 경우를 의미한다.
    • 3.5.A.2 평균 출력은 이동하거나 전환된 에너지의 양을, 해당 이동이나 전환이发生的所需时间로 나눈 값이다.
      • 식: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
    • 3.5.A.3 일은 힘에 의해 물체나 시스템의 에너지가 변한 것이므로, 평균 출력은 행해진 총 일량을 그 일이 행해진 시간으로 나눈 값이다.
      • 식: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
    • 3.5.A.4 상수 힘의 성분이 물체의 속도 방향과 평행할 때 물체에 전달되는 순간 출력은 유도된 식으로 기술할 수 있다.
      • 식: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Power 功率 is the rate of doing work or transferring energy, measured in watts 瓦特 (W):

    $$P=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t},\qquad\text{and instantaneously}\qquad P=Fv.$$
    So the same job done faster requires more power. On an energy–time graph, power is the slope.

    Worked example. A motor lifts a $50\ \text{kg}$ load at a steady $2.0\ \text{m/s}$. Because it moves at constant speed, the lifting force equals the weight, so

    $$P=Fv=mgv=50\times9.8\times2.0=980\ \text{W}.$$

    Real machines waste some energy, so we quote efficiency 效率 – useful output power divided by total input power. If this motor draws $1400\ \text{W}$ of electrical power to deliver $980\ \text{W}$ of useful lifting, its efficiency is $980/1400=0.70$, or $70\%$; the other $30\%$ becomes heat and sound.

    한국어

    출력(Power) 은 일 또는 에너지 전달의 속도이며, 단위는 와트(W) 입니다:

    $$P=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t},\qquad\text{and instantaneously}\qquad P=Fv.$$
    따라서 동일한 일을 더 빠르게 수행하려면 더 많은 출력이 필요합니다. 에너지-시간 그래프에서 출력은 기울기입니다.

    Power is the slope of the work-time graph: the same work in less time means more power
    Power is the slope of the work-time graph: the same work in less time means more power

    해설 예제. 모터가 $50\ \text{kg}$ 하중을 일정한 $2.0\ \text{m/s}$ 속도로 들어 올립니다. 속도가 일정하므로 들어 올리는 힘은 무게와 같으므로,

    $$P=Fv=mgv=50\times9.8\times2.0=980\ \text{W}.$$

    실제 기계는 일부 에너지를 손실하므로 효율성을 제시합니다. 이는 유용한 출력력을 총 입력력으로 나눈 값입니다. 이 모터가 $1400\ \text{W}$의 전기입력을 받아 $980\ \text{W}$의 유효한 들어올림 출력을 제공한다면, 그 효율성은 $980/1400=0.70$, 즉 $70\%$이며, 나머지 $30\%$는 열과 소리로 변합니다.

    A waterfall: power is how quickly energy is transferred — the same drop in less time is more power
    A waterfall: power is how quickly energy is transferred — the same drop in less time is more power
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Power/ˈpaʊə/ 출력
    watts/wɒts/ 와트(watts)
    efficiency/ɪˈfɪʃənsi/ 효율(efficiency)
    3.5

    Exam tips · ⁨시험 팁⁩

    English
    • Use $W=Fd\cos\theta$: work is zero when the force is perpendicular to the motion, and negative when it opposes it.
    • Reach for the work–energy theorem ($W_{\text{net}}=\Delta K$) or energy conservation ($K_1+U_1=K_2+U_2$) instead of forces whenever the path is complicated.
    • When friction acts, mechanical energy is not conserved — subtract the energy lost to heat.
    • Remember $K\propto v^2$: doubling the speed quadruples the kinetic energy (and the stopping distance).
    • Use $P=Fv$ for power at a steady speed; at constant velocity the net force is zero but the power is not.
    한국어
    • $W=Fd\cos\theta$를 사용하십시오: 힘의 방향이 운동에 수직이면 일은 0이며, 반대方向이면 일은 음수입니다.
    • 경로가 복잡할 경우Force 대신 일-에너지 정리($W_{\text{net}}=\Delta K$) 또는 에너지 보존($K_1+U_1=K_2+U_2$)을 적용하십시오.
    • 마찰력이 작용하면 기계적 에너지는 보존되지 않습니다 – 열로 손실된 에너지를 빼십시오.
    • $K\propto v^2$를 기억하십시오: 속도를 두 배로 하면 운동 에너지는 네 배가 됩니다 (정지 거리도 네 배).
    • 일정한 속도에서의aces에 $P=Fv$를 사용하십시오; 일정 속도로 이동할 때 합力は 제로이지만aces는 아닙니다.
  • 4

    Linear Momentum

    Watch lesson · ⁨수업 보기⁩
    4.1

    Linear Momentum

    Syllabus
    English

    Learning Objective 4.1.A: Describe the linear momentum of an object or system.

    • 4.1.A.1 Linear momentum is defined by the equation $\vec{p} = m\vec{v}$.
      • Equation: $\vec{p} = m\vec{v}$
    • 4.1.A.2 Momentum is a vector quantity and has the same direction as the velocity.
    • 4.1.A.3 Momentum can be used to analyze collisions and explosions.
      • 4.1.A.3.i A collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.
      • 4.1.A.3.ii As only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions.
      • 4.1.A.3.iii An explosion is a model for an interaction in which forces internal to the system move objects within that system apart.

    Boundary statement: Unless otherwise stated, the general term "momentum" will refer specifically to linear momentum.

    한국어

    학습 목표 4.1.A: 물체나 시스템의 선형 운동량을 설명하라.

    • 4.1.A.1 선형 운동량은 식 $\vec{p} = m\vec{v}$에 의해 정의된다.
      • 식: $\vec{p} = m\vec{v}$
    • 4.1.A.2 운동량은 벡터량이며 속도와 같은 방향을 가진다.
    • 4.1.A.3 운동량은 충돌 및 폭발 현상을 분석하는 데 사용될 수 있다.
      • 4.1.A.3.i 충돌은 시스템 내 관련 물체들 사이에 작용하는 힘이 상호작용 중 물체에 가해지는 외부 합력에 비해 훨씬 큰 상호작용을 모델링하는 것이다.
      • 4.1.A.3.ii 충돌의 초기 상태와 최종 상태만 분석되므로, 물체 모델을 사용하여 충돌을 분석할 수 있다.
      • 4.1.A.3.iii 폭발은 시스템 내부의 힘들이 시스템 내 물체들을 서로 멀어지게 하는 상호작용을 모델링하는 것이다.

    경계 선언: 별다른 서식이 명시되지 않는 한, 일반 용어 "운동량"은 선형 운동량을 지칭한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Linear momentum 动量 is mass times velocity – a vector pointing the same way as the velocity:

    $$\vec{p}=m\vec{v}.$$
    It measures "how hard it is to stop" a moving object. A heavy slow truck and a light fast ball can have the same momentum. Its unit is $\text{kg m/s}$, the same as $\text{N s}$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Linear momentum/ˈlɪnɪə məʊˈmentəm/ Linear momentum
    4.2

    Change in Momentum and Impulse

    Syllabus
    Learning ObjectiveEssential Knowledge

    4.2.A
    Describe the impulse delivered to an object or system.

    • 4.2.A.1 The rate of change of momentum is equal to the net external force exerted on an object or system.
      • Equation: $\vec{F}_{\text{net}} = \dfrac{\Delta \vec{p}}{\Delta t}$
    • 4.2.A.2 Impulse is defined as the product of the average force exerted on a system and the time interval during which that force is exerted on the system.
      • Equation: $\vec{J} = \vec{F}_{\text{avg}} \Delta t$
    • 4.2.A.3 Impulse is a vector quantity and has the same direction as the net force exerted on the system.
    • 4.2.A.4 The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
    • 4.2.A.5 The net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.

    4.2.B
    Describe the relationship between the impulse exerted on an object or a system and the change in momentum of the object or system.

    • 4.2.B.1 Change in momentum is the difference between a system's final momentum and its initial momentum.
      • Equation: $\Delta \vec{p} = \vec{p} - \vec{p}_0$
    • 4.2.B.2 The impulse–momentum theorem relates the impulse exerted on a system and the system's change in momentum.
      • Equation: $\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}$
    • 4.2.B.3 Newton's second law of motion is a direct result of the impulse–momentum theorem applied to systems with constant mass.
      • Equation: $\vec{F}_{\text{net}} = \dfrac{\Delta \vec{p}}{\Delta t} = m\dfrac{\Delta \vec{v}}{\Delta t} = m\vec{a}$

    Boundary statement: AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A net force acting over time changes momentum. The impulse 冲量 delivered is

    $$\vec{J}=\vec{F}\,\Delta t=\Delta\vec{p}.$$
    This is the impulse–momentum theorem: impulse equals the change in momentum. On a force–time graph, impulse is the area under the curve. It explains why airbags and follow-through help – spreading the same momentum change over a longer time reduces the force.

    Impulse is the area under the force-time curve, equal to the average force times the contact time
    Impulse is the area under the force-time curve, equal to the average force times the contact time

    Worked example. A $0.15\ \text{kg}$ ball hits a wall at $20\ \text{m/s}$ and bounces straight back at $15\ \text{m/s}$. The contact lasts $0.020\ \text{s}$. Find the average force on the ball. Take the rebound direction as positive, so $u=-20\ \text{m/s}$ and $v=+15\ \text{m/s}$:

    $$\Delta p=m(v-u)=0.15\big(15-(-20)\big)=5.25\ \text{kg m/s},\qquad F=\frac{\Delta p}{\Delta t}=\frac{5.25}{0.020}=260\ \text{N}.$$
    The sign work matters: forgetting that the velocity reverses is the most common mistake here.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    impulse/ˈɪmpʌls/ 신호
    4.3

    Conservation of Linear Momentum

    Syllabus
    Learning ObjectiveEssential Knowledge

    4.3.A
    Describe the behavior of a system using conservation of linear momentum.

    • 4.3.A.1 A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
      • 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation $\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum m_i \vec{v}_i}{\sum m_i}$.
      • 4.3.A.1.ii The velocity of a system's center of mass is constant in the absence of a net external force.
    • 4.3.A.2 The total momentum of a system is the sum of the momenta of the system's constituent parts.
    • 4.3.A.3 In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.
      • 4.3.A.3.i The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law.
      • 4.3.A.3.ii A system may be selected so that the total momentum of that system is constant.
      • 4.3.A.3.iii If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system.
        • Equation: $\vec{J} = \Delta \vec{p}$
    • 4.3.A.4 Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.

    Boundary statement: AP Physics 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the AP Physics 1 Exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 includes a full treatment of conservation of momentum in two dimensions for problems that include one unknown final velocity.

    4.3.B
    Describe how the selection of a system determines whether the momentum of that system changes.

    • 4.3.B.1 Momentum is conserved in all interactions.
    • 4.3.B.2 If the net external force on the selected system is zero, the total momentum of the system is constant.
    • 4.3.B.3 If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    If the net external force on a system is zero, its total momentum is conserved 守恒:

    $$\sum \vec{p}_{\text{before}}=\sum \vec{p}_{\text{after}}.$$
    Internal forces (like the push between two colliding carts) come in third-law pairs and cancel, so they cannot change the total momentum. This is the key tool for collisions 碰撞 and explosions – apply it separately in the $x$ and $y$ directions.

    A head-on collision: total momentum before equals total momentum after
    A head-on collision: total momentum before equals total momentum after

    Worked example (recoil 反冲). A $60\ \text{kg}$ skater, initially at rest on frictionless ice, throws a $2.0\ \text{kg}$ ball at $8.0\ \text{m/s}$. Find her recoil speed. The total momentum starts at zero and stays zero:

    $$0=(60)v+(2.0)(8.0)\;\Rightarrow\;v=-\frac{16}{60}=-0.27\ \text{m/s},$$
    so she moves at $0.27\ \text{m/s}$ in the opposite direction to the ball – the principle behind rockets and guns.

    A whole collection of objects can be described by a single center-of-mass velocity 质心速度:

    $$\vec{v}_{\text{cm}}=\frac{\sum m_i\vec{v}_i}{\sum m_i}=\frac{\vec{p}_{\text{total}}}{M_{\text{total}}}.$$
    Since total momentum is conserved when no net external force acts, $\vec{v}_{\text{cm}}$ then stays constant – a $v_{\text{cm}}$-versus-time graph is flat unless an outside force acts. An internal collision or explosion never changes it, however violently the parts fly apart.

    Explore · ⁨탐색하기⁩

    Collide two carts and conserve momentum · ⁨두 카트 충돌 및 운동량 보존⁩

    In any collision the total momentum $\sum mv$ before equals the total after. Set the masses and speeds and check the momentum bookkeeping. · ⁨모든 충돌에서 충돌 전 총운동량 $\sum mv$는 충돌 후와 같습니다. 질량과 속도를 설정하여 운동량 계산이 맞는지 확인해 보십시오.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    conserved/kənˈsɜːvd/ 보존된다
    collisions/kəˈlɪʒnz/ 충돌
    center-of-mass velocity/ˈsentə ɒv mæs vəˈlɒsɪti/ 질량 중심 속도
    recoil/rɪˈkɔɪl/ 후진
    4.4

    Elastic and Inelastic Collisions

    Syllabus
    English

    Learning Objective 4.4.A: Describe whether an interaction between objects is elastic or inelastic.

    • 4.4.A.1 An elastic collision between objects is one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system.
    • 4.4.A.2 In an elastic collision, the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies.
    • 4.4.A.3 An inelastic collision between objects is one in which the total kinetic energy of the system decreases.
    • 4.4.A.4 In an inelastic collision, some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.
    • 4.4.A.5 In a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision.
    한국어

    학습 목표 4.4.A: 물체 간의 상호작용이 탄성인지 비탄성인지 서술한다.

    • 4.4.A.1 탄성 충돌이란 시스템의 초기 운동 에너지가 최종 운동 에너지와 동일한 충돌이다.
    • 4.4.A.2 탄성 충돌에서 시스템 내 각 물체의 최종 운동 에너지는 초기 운동 에너지와 다를 수 있다.
    • 4.4.A.3 비탄성 충돌이란 시스템의 총 운동 에너지가 감소하는 충돌이다.
    • 4.4.A.4 비탄성 충돌에서 일부 초기 운동 에너지는 운동 에너지로 회복되지 않고 비보존력에 의해 다른 형태의 에너지로 변환된다.
    • 4.4.A.5 완벽 비탄성 충돌에서는 물체가 서로 붙어 충돌 후 동일한 속도로 움직인다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Conservation of momentum in a collision

    Momentum is conserved in every collision (with no external force). Kinetic energy is not:

    A glancing collision, resolved along two perpendicular axes
    A glancing collision, resolved along two perpendicular axes
    • In an elastic collision 弹性碰撞, kinetic energy is also conserved (objects bounce apart cleanly).
    • In an inelastic collision 非弹性碰撞, some kinetic energy becomes heat or deformation. In a perfectly inelastic collision the objects stick together and move with one common velocity afterward.

    Strategy: always write momentum conservation; add energy conservation only if the collision is stated to be elastic. One elastic fact worth memorising: in a 1D elastic collision between equal masses, the two objects simply swap velocities (a moving ball striking an identical stationary one stops dead, and the target flies off at the incoming speed).

    Worked example. A $1000\ \text{kg}$ car moving at $20\ \text{m/s}$ runs into a stationary $1500\ \text{kg}$ car and they lock together. Find their common speed, and the kinetic energy lost. Momentum conservation gives

    $$1000\times20=(1000+1500)\,v\;\Rightarrow\;v=\frac{20000}{2500}=8.0\ \text{m/s}.$$
    Kinetic energy before is $\tfrac12(1000)(20^2)=2.0\times10^{5}\ \text{J}$; after is $\tfrac12(2500)(8.0^2)=8.0\times10^{4}\ \text{J}$. So $1.2\times10^{5}\ \text{J}$ (about $60\%$) is lost to crumpling and heat – momentum is still conserved, but kinetic energy is not.

    A Newton's cradle: five steel balls hanging in a row
    A Newton's cradle shows momentum and kinetic energy passing through a line of balls in a near-elastic collision
    Explore · ⁨탐색하기⁩

    Compare elastic and inelastic collisions · ⁨탄성 충돌과 비탄성 충돌 비교⁩

    Momentum is always conserved, but kinetic energy is only conserved in an elastic collision. In an inelastic one the carts stick and some energy becomes heat. · ⁨운동량은 항상 보존되지만, 운동 에너지는 탄성 충돌에서만 보존됩니다. 비탄성 충돌에서는 카트가 붙게 되며 일부 에너지가 열에너지로 변환됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    elastic collision/ɪˈlæstɪk kəˈlɪʒn/ 탄성 충돌
    inelastic collision/ɪnɪˈlæstɪk kəˈlɪʒn/ 비탄성 충돌
    4.4

    Exam tips

    • Momentum is a vector — assign $+$/$-$ signs before adding; a ball that rebounds reverses its velocity, giving a large $\Delta p$.
    • Momentum is conserved in every collision (no external force); kinetic energy is conserved only if the collision is stated to be elastic.
    • In a perfectly inelastic collision the objects stick and move with one common velocity.
    • Impulse $=F\,\Delta t=\Delta p$ = the area under a force–time graph; spreading a collision over a longer time reduces the force (airbags, bending knees).
    • For recoil/explosions, set the total momentum equal before and after (often zero before).
  • 5

    Torque and Rotational Dynamics

    Watch lesson · ⁨수업 보기⁩
    5.1

    Rotational Kinematics

    Syllabus
    English

    Learning Objective 5.1.A: Describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration.

    • 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
      • Equation: $\Delta\theta = \theta - \theta_0$
      • 5.1.A.1.i A rigid system is one that holds its shape but in which different points on the system move in different directions during rotation. A rigid system cannot be modeled as an object.
      • 5.1.A.1.ii One direction of angular displacement about an axis of rotation—clockwise or counterclockwise—is typically indicated as mathematically positive, with the other direction becoming mathematically negative.
      • 5.1.A.1.iii If the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object. For example, the rotation of Earth about its axis may be considered negligible when considering the revolution of Earth about the center of mass of the Earth–Sun system.
    • 5.1.A.2 Average angular velocity is the average rate at which angular position changes with respect to time.
      • Equation: $\omega_{\text{avg}} = \dfrac{\Delta\theta}{\Delta t}$
    • 5.1.A.3 Average angular acceleration is the average rate at which the angular velocity changes with respect to time.
      • Equation: $\alpha_{\text{avg}} = \dfrac{\Delta\omega}{\Delta t}$
    • 5.1.A.4 Angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.
      • 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
        • Equation: $\omega = \omega_0 + \alpha t$
        • Equation: $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
        • Equation: $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
      • 5.1.A.4.ii Graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities.

    Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.

    한국어

    학습 목표 5.1.A: 각변위, 각속도 및 각가속도를 사용하여 시스템의 회전을 시간에 대해 서술한다.

    • 5.1.A.1 각변위는 경질 시스템 상의 한 점이 특정 축을 중심으로 회전하는 각도의 라디안 단위로 측정된 값이다.
      • 식: $\Delta\theta = \theta - \theta_0$
      • 5.1.A.1.i 경질 시스템은 형태를 유지하지만 회전 중 시스템의 다른 점들이 서로 다른 방향으로 움직이는 시스템이다. 경질 시스템은 단일 물체로 모델링할 수 없다.
      • 5.1.A.1.ii 회전축에 대한 각변위의 한 방향(시계 방향 또는 반시계 방향)은 일반적으로 수학적으로 양수(+), 다른 방향은 음수(-)로 표시된다.
      • 5.1.A.1.iii 시스템의 축에 대한 회전이 시스템의 질량 중심의 운동으로 잘 묘사될 수 있다면, 시스템을 단일 물체로 취급할 수 있다. 예를 들어, 지구-태양 질량 중심에 대한 지구의 공전을 고려할 때 지구의 자전은 무시할 수 있다.
    • 5.1.A.2 평균 각속도는 각 위치가 시간에 따라 변하는 평균율이다.
      • 식: $\omega_{\text{avg}} = \dfrac{\Delta\theta}{\Delta t}$
    • 5.1.A.3 평균 각가속도는 각속도가 시간에 따라 변하는 평균율이다.
      • 식: $\alpha_{\text{avg}} = \dfrac{\Delta\omega}{\Delta t}$
    • 5.1.A.4 한 축에 대한 각변위, 각속도, 각가속도는 한 차원의 선형 변위, 속도, 가속도와 유사하며 동일한 수학적 관계를 나타낸다.
      • 5.1.A.4.i 일정 각가속도일 때, 각변위, 각속도, 각가속도 사이의 수학적 관계는 다음 식으로 설명할 수 있다:
        • 식: $\omega = \omega_0 + \alpha t$
        • 식: $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
        • 식: $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
      • 5.1.A.4.ii 시간 함수로서의 각변위, 각속도, 각가속도의 그래프를 사용하여 해당 물리량 간의 관계를 찾을 수 있다.

    범위 설명: 점이나 물체의 회전 방향에 대한 설명은 특정 회전축에 대해 시계 방향과 반시계 방향으로 제한된다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A crash-test collision: large forces and torques produce rapid changes in linear and angular motion
    A crash-test collision: large forces and torques produce rapid changes in linear and angular motion
    A spring under load: Hooke's law links force and extension — useful context for energy in rotating systems
    A spring under load: Hooke's law links force and extension — useful context for energy in rotating systems

    Rotation is described by angular quantities that mirror the linear ones:

    One radian is the angle whose arc length equals the radius
    One radian is the angle whose arc length equals the radius
    • angular displacement 角位移 $\theta$ (in radians 弧度),
    • angular velocity 角速度 $\omega=\dfrac{\Delta\theta}{\Delta t}$,
    • angular acceleration 角加速度 $\alpha=\dfrac{\Delta\omega}{\Delta t}$.

    For constant $\alpha$, the rotational kinematic equations have the same form as the linear ones, with $\theta,\omega,\alpha$ replacing $x,v,a$: $\omega=\omega_0+\alpha t$, $\theta=\omega_0 t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.

    Worked example. A wheel starts from rest and speeds up uniformly to $30\ \text{rad/s}$ in $6.0\ \text{s}$. Find its angular acceleration and the total angle turned:

    $$\alpha=\frac{\Delta\omega}{\Delta t}=\frac{30}{6.0}=5.0\ \text{rad/s}^2,\qquad \theta=\tfrac12\alpha t^2=\tfrac12\times5.0\times6.0^2=90\ \text{rad}.$$

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    angular displacement/ˈæŋɡjʊlə dɪˈspleɪsmənt/ 각변위
    radians/ˈreɪdɪənz/ 라디안(radians)
    angular velocity/ˈæŋɡjʊlə vəˈlɒsɪti/ 각속도
    angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ 각가속도
    5.2

    Connecting Linear and Rotational Motion

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.2.A
    Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.

    • 5.2.A.1 For a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
    • 5.2.A.2 Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:
      • Equation: $s = r\theta$
      • Equation: $v = r\omega$
      • Equation: $a_T = r\alpha$
    • 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.

    Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A point at radius $r$ from the axis has linear quantities tied to the angular ones:

    $$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha.$$
    Points farther from the axis move faster. This link lets you switch between "how fast the wheel spins" and "how fast a point on its rim moves."

    As the radius turns through an angle, a point moves along an arc at speed v
    As the radius turns through an angle, a point moves along an arc at speed v

    Worked example. A bicycle wheel of radius $0.35\ \text{m}$ spins at $12\ \text{rad/s}$. A point on the rim (and so the bike) moves at $v=r\omega=0.35\times12=4.2\ \text{m/s}$. A point halfway to the axis moves at half that speed.

    5.3

    Torque

    Syllabus
    English

    Learning Objective 5.3.A: Identify the torques exerted on a rigid system.

    • 5.3.A.1 Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.
    • 5.3.A.2 The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.

    Learning Objective 5.3.B: Describe the torques exerted on a rigid system.

    • 5.3.B.1 Torques can be described using force diagrams.
      • 5.3.B.1.i Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.
      • 5.3.B.1.ii Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.
    • 5.3.B.2 The magnitude of the torque exerted on a rigid system by a force is described by the following equation, where $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force.
      • Equation: $\tau = rF_{\perp} = rF\sin\theta$

    Boundary statement: While AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course.

    한국어

    학습 목표 5.3.A: 경질 시스템에 작용하는 토크를 식별한다.

    • 5.3.A.1 토크는 회전축에서 힘의 작용점까지의 위치 벡터에 수직인 힘 성분에서만 발생한다.
    • 5.3.A.2 레버암은 회전축에서 가해진 힘의 작용선까지의 수직 거리이다.

    학습 목표 5.3.B: 강체 시스템에 작용하는 토크를 설명하시오.

    • 5.3.B.1 토크는 힘 도표를 사용하여 설명할 수 있습니다.
      • 5.3.B.1.i 힘 도표는 자유체도(Free-body diagram)와 유사하며, 강체 시스템에 작용하는 토크를 분석하는 데 사용됩니다.
      • 5.3.B.1.ii 자유체도와 마찬가지로 힘 도표는 강체 시스템에 작용하는 힘의 상대적 크기와 방향을 나타냅니다. 또한 힘 도표는 회전 축에 대한 각 힘이 작용하는 위치도 함께 묘사합니다.
    • 5.3.B.2 힘에 의해 강체 시스템에 가해지는 토크의 크기는 다음 식으로 설명됩니다. 여기서 $\theta$는 힘 벡터와 회전축에서 힘의 작용점까지의 위치 벡터 사이의 각도입니다.
      • 식: $\tau = rF_{\perp} = rF\sin\theta$

    범위 문장: AP Physics 1은 학생들이 벡터 규약에 따라 토의 크기를 수학적으로 변형하도록 요구하지만, 토크의 방향은 과정 범위를 벗어납니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The principle of moments (torque)

    Torque 力矩 is the rotational effect of a force – how effectively it turns an object about an axis:

    $$\tau=r F\sin\theta = F\cdot r_\perp,$$
    where $r_\perp$ is the moment arm 力臂 (the perpendicular distance from the axis to the force's line of action). A force applied farther out, or more perpendicular, produces more torque. Torque has a sign (clockwise vs counterclockwise).

    The moment of a force depends on the perpendicular distance from the pivot
    The moment of a force depends on the perpendicular distance from the pivot

    Worked example. You push with $20\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench. Perpendicular to the wrench the torque is $\tau=rF=0.30\times20=6.0\ \text{N m}$. If you push at $60^{\circ}$ to the wrench instead, only the perpendicular part counts: $\tau=rF\sin 60^{\circ}=0.30\times20\times0.87=5.2\ \text{N m}$ – which is why you push at a right angle for the most turning effect.

    Explore · ⁨탐색하기⁩

    Balance torques on a beam · ⁨빔에 작용하는 토크 균형⁩

    Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal. · ⁨토크는 힘과 수직 거리의 곱으로 정의되며, $\tau=Fd$입니다. 양쪽의 토크가 같을 때 빔은 회전 평형 상태가 됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Torque/tɔːk/ 토크
    moment arm/ˈməʊmənt ɑːm/ 모멘트 암
    5.4

    Rotational Inertia

    Syllabus
    English

    Learning Objective 5.4.A: Describe the rotational inertia of a rigid system relative to a given axis of rotation.

    • 5.4.A.1 Rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.
    • 5.4.A.2 The rotational inertia of an object rotating a perpendicular distance $r$ from an axis is described by the equation
      • Equation: $I = mr^2$
    • 5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis:
      • Equation: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$

    Learning Objective 5.4.B: Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.

    • 5.4.B.1 A rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.
    • 5.4.B.2 The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:
      • Equation: $I' = I_{\text{cm}} + Md^2$

    Boundary statement: AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration.

    Boundary statement: Students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius.

    한국어

    학습 목표 5.4.A: 주어진 회전 축에 대해 강체 시스템의 회전 관성을 설명하시오.

    • 5.4.A.1 회전 관성은 강체 시스템이 회전 상태 변화에 저항하는 정도를 측정하며, 시스템의 질량과 질량이 회전 축에 대해 어떻게 분포되어 있는지에 관련되어 있습니다.
    • 5.4.A.2 회전축으로부터 수직 거리 $r$로 회전하는 물체의 회전 관성은 다음 식으로 설명됩니다.
      • 식: $I = mr^2$
    • 5.4.A.3 축에 대한 여러 물체 집합의 총 회전 관성은 각 물체가 해당 축에 대해 가지는 회전 관성의 합입니다:
      • 식: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$

    학습 목표 5.4.B: 중심的方法来 passes through the system's center of mass does not pass through the system's center of mass. (Original: Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.)

    • 5.4.B.1 강체 시스템의 특정 평면에서의 회전 관성은 회전 축이 시스템의 중심을 통과할 때 최소값을 가집니다.
    • 5.4.B.2 평행축 정리는 중심을 통과하는 축과 평행한 임의의 축에 대한 강체 시스템의 회전 관성을 연결하기 위해 다음 식을 사용합니다:
      • 식: $I' = I_{\text{cm}} + Md^2$

    범위 문장: AP Physics 1은 학생들이 2차원 구성으로 배열된 5개 이하의 물체 시스템에 대한 회전 관성만 계산하도록 요구합니다.

    범위 문장: 학생들은 확장된 강체 시스템의 회전 관성을 알 필요가 없으며, 이러한 값은 시험 내에서 제공될 것입니다. 학생들은 회전 관성에 영향을 미치는 요인에 대해 정성적인 이해를 가져야 합니다; 예를 들어, 질량이 회전 축에서 더 멀어질수록 회전 관성이 커지므로, 같은 질량과 반지름을 가진 고리(hoop)가 단단한 원반(solid disk)보다 더 큰 회전 관성을 가지는 이유입니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Rotational inertia 转动惯量 (moment of inertia) $I$ measures how hard it is to change an object's rotation – the rotational version of mass. It depends on both the mass and how far that mass sits from the axis: mass spread farther out gives a larger $I$. For a point mass, $I=mr^2$; for extended bodies, standard formulas are provided (a hoop is $mR^2$, a solid disk $\tfrac12 mR^2$). This is why a figure skater spins faster when she pulls her arms in – she reduces $I$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ 회전 관성
    5.5

    Rotational Equilibrium

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.5.A
    Describe the conditions under which a system's angular velocity remains constant.

    • 5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
      • 5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
      • 5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.
        • Equation: $\sum \tau_i = 0$
      • 5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
    • 5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.

    Boundary statement: AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    An object is in rotational equilibrium 转动平衡 when the net torque is zero, so its angular velocity stays constant. For a balanced (static) object, both the net force and the net torque are zero. Choosing the axis at an unknown force's location removes it from the torque equation – a useful trick for beam and ladder problems.

    At balance the clockwise and anticlockwise moments about the pivot are equal
    At balance the clockwise and anticlockwise moments about the pivot are equal

    Worked example. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from the pivot of a seesaw. Where must a $40\ \text{kg}$ child sit on the other side to balance it? Set the clockwise torque equal to the anticlockwise torque (the $g$'s cancel):

    $$30\times2.0=40\times d\;\Rightarrow\;d=\frac{60}{40}=1.5\ \text{m}.$$
    The heavier child sits closer to the pivot – less distance, same torque.

    Explore · ⁨탐색하기⁩

    Find the balance point · ⁨균형점 balance point 찾기⁩

    For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances. · ⁨회전 평형이 되기 위해서는 총 시계 방향 토크와 총 반시계 방향 토크가 서로 같아야 합니다. 빔이 균형을 이루도록 힘과 거리를 이동시키세요.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    rotational equilibrium/rəʊˈteɪʃənl ˌiːkwɪˈlɪbrɪəm/ 회전 평형
    5.6

    Newton's Second Law in Rotational Form

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.6.A
    Describe the conditions under which a system's angular velocity changes.

    • 5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
    • 5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system.
      • Equation: $\alpha_{\text{sys}} = \dfrac{\sum \tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
    • 5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Net torque produces angular acceleration, in direct analogy with $F=ma$:

    $$\sum\tau = I\alpha.$$
    So a larger net torque, or a smaller rotational inertia, gives a larger angular acceleration. Solve rotation problems just like translation problems, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, and $\alpha\leftrightarrow a$.

    Worked example. A net torque of $12\ \text{N m}$ acts on a wheel with rotational inertia $I=3.0\ \text{kg m}^2$. Its angular acceleration is $\alpha=\tau/I=12/3.0=4.0\ \text{rad/s}^2$ – the exact rotational twin of $a=F/m$.

    5.6

    Exam tips

    • Torque $\tau=Fr_\perp$ uses the perpendicular distance from the pivot; a force through the pivot gives zero torque.
    • For balance, set clockwise torque = anticlockwise torque; choosing the pivot at an unknown force removes it from the equation.
    • Use the rotational analogues: $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$, so $\sum\tau=I\alpha$ mirrors $\sum F=ma$.
    • Rotational inertia depends on how far the mass sits from the axis, not just its amount — a hoop resists spinning more than a disc of equal mass.
    • Convert angles to radians and link linear to angular with $v=r\omega$, $a_t=r\alpha$.
  • 6

    Energy and Momentum of Rotating Systems

    Watch lesson · ⁨수업 보기⁩
    6.1

    Rotational Kinetic Energy

    Syllabus
    English

    Learning Objective 6.1.A: Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.

    • 6.1.A.1 The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation $K = \frac{1}{2} I \omega^2$.
      • 6.1.A.1.i The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.
      • 6.1.A.1.ii The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.
    • 6.1.A.2 A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.
    • 6.1.A.3 Rotational kinetic energy is a scalar quantity.
    한국어

    학습 목표 6.1.A: 강체 시스템의 회전 운동 에너지를该系统의회전 관성과 각속도를 통해 설명하시오.

    • 6.1.A.1 물체나 강체 시스템의 회전 운동 에너지는该系统的回전 관성과 각속도에 관련되며, 다음 식으로 주어집니다: $K = \frac{1}{2} I \omega^2$.
      • 6.1.A.1.i 고정 축에 대한 물체의 회전 관성을 사용하면,该系统의回전 운동 에너지가该系统的平动 운동 에너지(총 운동 에너지)와 동등함을 보여줄 수 있습니다.
      • 6.1.A.1.ii 강체 시스템의 총 운동 에너지는该系统的center of mass에 대한 회전 운동 에너지와该系统的center of mass의 선형 운동에 의한 평동 운동 에너지의 합입니다.
    • 6.1.A.2 강체 시스템은该系统的center of mass가 정지해 있더라도该系统의internal points가 선형 속도를 가지므로 운동 에너지를 가지고 있어回전 운동 에너지를 가질 수 있습니다.
    • 6.1.A.3回전 운동 에너지는 스칼라 양입니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A spinning object has rotational kinetic energy 转动动能, the rotational twin of $\tfrac12 mv^2$:

    $$K_{\text{rot}}=\tfrac{1}{2}I\omega^2.$$
    An object that both moves and spins (like a rolling ball) has both translational and rotational kinetic energy, and its total is $K=\tfrac12 mv^2+\tfrac12 I\omega^2$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    rotational kinetic energy/rəʊˈteɪʃənl kɪˈnetɪk ˈenədʒi/ 회전 운동 에너지
    6.2

    Torque and Work

    Syllabus
    English

    Learning Objective 6.2.A: Describe the work done on a rigid system by a given torque or collection of torques.

    • 6.2.A.1 A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.
    • 6.2.A.2 The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted.
      • Equation: $W = \tau \Delta\theta$
    • 6.2.A.3 Work done on a rigid system by a given torque can be found from the area under the curve of a graph of torque as a function of angular position.
    한국어

    학습 목표 6.2.A: 주어진 토크 또는 일련의 토크가 강체 시스템에 가하는 일을 설명하시오.

    • 6.2.A.1 토크가 각 변위 동안 작용할 경우, 그 토크는 물체나 강체 시스템에 에너지를 전달하거나 빼낼 수 있다.
    • 6.2.A.2 토크에 의해 강체 시스템에 가해지는일의 양은 해당 토크의 크기와 토크가 작용하는 구간 동안 강체 시스템이 회전하는 각 변위에 관련되어 있다.
      • 식: $W = \tau \Delta\theta$
    • 6.2.A.3 주어진 토크에 의해 강체 시스템에 가해지는 일은 토크를 각 위치의 함수로 나타낸 그래프 아랫면적을 통해 구할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A torque acting through an angular displacement does work, changing rotational kinetic energy:

    $$W=\tau\,\Delta\theta,\qquad P=\tau\omega.$$
    This is the rotational form of $W=Fd$ and $P=Fv$, and it extends the work–energy theorem to rotation.

    Worked example. A motor applies a steady torque of $8.0\ \text{N m}$ to a flywheel while it turns through $10\ \text{rad}$. The work done is $W=\tau\,\Delta\theta=8.0\times10=80\ \text{J}$, and if the flywheel started from rest this all becomes rotational kinetic energy.

    Explore · ⁨탐색하기⁩

    Balance torques on a seesaw

    Torque is force times perpendicular distance, $\tau = Fd$. The beam balances when the torques on each side are equal — move the forces and distances to find the balance point.

    6.3

    Angular Momentum and Angular Impulse

    Syllabus
    Learning ObjectiveEssential Knowledge

    6.3.A
    Describe the angular momentum of an object or rigid system.

    • 6.3.A.1 The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$.
    • 6.3.A.2 The magnitude of the angular momentum of an object about a given point is $L = rmv \sin\theta$.
      • 6.3.A.2.i The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
      • 6.3.A.2.ii The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.

    6.3.B
    Describe the angular impulse delivered to an object or rigid system by a torque.

    • 6.3.B.1 Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted.
      • Equation: $\text{angular impulse} = \tau \Delta t$
    • 6.3.B.2 Angular impulse has the same direction as the torque exerted on the object or system.
    • 6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.

    6.3.C
    Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.

    • 6.3.C.1 The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial angular momenta of the object or rigid system: $\Delta L = L - L_0$
    • 6.3.C.2 A rotational form of the impulse–momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
      • 6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
        • Equation: $\Delta L = \tau \Delta t$
      • 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of the rotational form of Newton's second law of motion for cases in which rotational inertia is constant: $\tau_{\text{net}} = \dfrac{\Delta L}{\Delta t} = I \dfrac{\Delta\omega}{\Delta t} = I\alpha$
    • 6.3.C.3 The net torque exerted on an object is equal to the slope of the graph of the angular momentum of an object as a function of time.
    • 6.3.C.4 The angular impulse delivered to an object is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.

    Boundary statement: While AP Physics 1 expects that students can mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Angular momentum 角动量 is the rotational version of linear momentum:

    $$L=I\omega.$$
    A net torque acting over time delivers an angular impulse 角冲量 that changes it: $\tau\,\Delta t=\Delta L$ – the rotational impulse–momentum theorem.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Angular momentum/ˈæŋɡjʊlə məʊˈmentəm/ 각운동량
    angular impulse/ˈæŋɡjʊlə ˈɪmpʌls/ 각 충량
    6.4

    Conservation of Angular Momentum

    Syllabus
    Learning ObjectiveEssential Knowledge

    6.4.A
    Describe the behavior of a system using conservation of angular momentum.

    • 6.4.A.1 The total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that axis.
    • 6.4.A.2 Any change to a system's angular momentum must be due to an interaction between the system and its surroundings.
      • 6.4.A.2.i The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. This is a direct result of Newton's third law.
      • 6.4.A.2.ii A system may be selected so that the total angular momentum of that system is constant.
      • 6.4.A.2.iii The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis.
      • 6.4.A.2.iv If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system.

    6.4.B
    Describe how the selection of a system determines whether the angular momentum of that system changes.

    • 6.4.B.1 Angular momentum is conserved in all interactions.
    • 6.4.B.2 If the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.
    • 6.4.B.3 If the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    If the net external torque on a system is zero, its total angular momentum is conserved 守恒:

    $$I_1\omega_1=I_2\omega_2.$$
    So if $I$ decreases, $\omega$ increases to keep $L$ constant – this is why a spinning skater speeds up when pulling their arms in. It applies to collisions and explosions of rotating systems too.

    Worked example. A skater spins at $2.0\ \text{rev/s}$ with rotational inertia $I_1=4.0\ \text{kg m}^2$. She pulls her arms in, dropping her rotational inertia to $I_2=1.6\ \text{kg m}^2$. With no external torque, angular momentum is conserved:

    $$\omega_2=\frac{I_1}{I_2}\,\omega_1=\frac{4.0}{1.6}\times2.0=5.0\ \text{rev/s}.$$
    Her kinetic energy actually rises – the extra energy comes from the work her muscles do pulling her arms in against the outward pull.

    Pulling mass inward lowers I, so ω rises to conserve L = Iω
    Pulling mass inward lowers I, so ω rises to conserve L = Iω
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    conserved/kənˈsɜːvd/ 보존된다
    6.5

    Rolling

    Syllabus
    English

    Learning Objective 6.5.A: Describe the kinetic energy of a system that has translational and rotational motion.

    • 6.5.A.1 The total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies.
      • Equation: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

    Learning Objective 6.5.B: Describe the motion of a system that is rolling without slipping.

    • 6.5.B.1 While rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with the equations:
      • Equation: $\Delta x_{\text{cm}} = r\Delta\theta$
      • Equation: $v_{\text{cm}} = r\omega$
      • Equation: $a_{\text{cm}} = r\alpha$
    • 6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.

    Learning Objective 6.5.C: Describe the motion of a system that is rolling while slipping.

    • 6.5.C.1 When slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.
    • 6.5.C.2 When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.

    Boundary statement: Rolling friction is beyond the scope of AP Physics 1.

    Boundary statement: The precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping.

    한국어

    학습 목표 6.5.A: 병진 운동과 회전 운동을 모두 가지는 시스템의 운동 에너지를 설명하시오。

    • 6.5.A.1 시스템의 총 운동 에너지는该系统的平动动能和转动动能之和。
      • 식: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

    학습 목표 6.5.B: 미끄러짐 없이 굴리는 시스템의 운동을 설명하시오。

    • 6.5.B.1 미끄러짐 없이 굴러갈 때, 시스템의 질량 중심의 병진 운동은 다음 식을 통해 시스템 자체의 회전 운동과 관련된다:
      • 식: $\Delta x_{\text{cm}} = r\Delta\theta$
      • 식: $v_{\text{cm}} = r\omega$
      • 식: $a_{\text{cm}} = r\alpha$
    • 6.5.B.2 이상적인 경우, 미끄러짐 없이 굴러가는 것은 마찰력이 굴리는 시스템으로부터 에너지를 소산시키지 않음을 의미한다.

    학습 목표 6.5.C: 미끄러지며 굴러가는 시스템의 운동을 설명하시오.

    • 6.5.C.1 미끄러질 때, 시스템의 질량 중심의 운동과 시스템의 회전 운동은 직접적으로 관련될 수 없다.
    • 6.5.C.2 회전하는 시스템이 다른 표면相对于로 미끄러질 때, 운동 마찰력의 작용점에Applied된 힘은 표면에 대해 이동하므로 운동 마찰력은 시스템으로부터 에너지를 소산시킨다.

    경계 문구: 구동 마찰력은 AP 물리학 1의 범위를 벗어난다.

    경계 문구: 강체가 미끄러지며 굴러갈 때 선량과 각량의 정확한 수학적 관계는 AP 물리학 1 및 2의 범위를 벗어난다. 학생들은 이러한 관계를 정량적으로 모델링할 것으로 기대되지 않는다. 그러나 강체가 미끄러지며 굴러갈 때 선량과 각량에发生改变되는 것을 정성적으로 설명할 것으로 기대된다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Rolling without slipping 纯滚动 links the translational and rotational motions: the contact point is momentarily at rest, so

    $$v=r\omega \qquad\text{and}\qquad a=r\alpha.$$
    A rolling object's energy splits between translation and rotation, so on an incline it accelerates more slowly than a frictionless sliding object – some energy goes into spin.

    Worked example. For a solid disk ($I=\tfrac12 mR^2$) that rolls without slipping, what fraction of its kinetic energy is rotational? Using $v=R\omega$, the rotational part is $\tfrac12 I\omega^2=\tfrac12(\tfrac12 mR^2)\omega^2=\tfrac14 mv^2$, while the translational part is $\tfrac12 mv^2$. So the total is $\tfrac34 mv^2$ and the rotational share is $\tfrac{1/4}{3/4}=\tfrac13$. A hoop, with its mass farther out, stores half its energy in spin and rolls down a ramp even more slowly.

    In rolling without slipping the contact point is at rest, so v = rω
    In rolling without slipping the contact point is at rest, so v = rω
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Rolling without slipping/ˈrəʊlɪŋ wɪˈðaʊt ˈslɪpɪŋ/ 미끄러짐 없는 굴림
    6.6

    Motion of Orbiting Satellites

    Syllabus
    English

    Learning Objective 6.6.A: Describe the motions of a system consisting of two objects interacting only via gravitational forces.

    • 6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, the motion of the central object itself is negligible.
    • 6.6.A.2 The motion of satellites in orbits is constrained by conservation laws.
      • 6.6.A.2.i In circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant.
      • 6.6.A.2.ii In elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change.
      • 6.6.A.2.iii The gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object.
        • Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
    • 6.6.A.3 The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite–central-object system is equal to zero.
      • 6.6.A.3.i When the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.
      • 6.6.A.3.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
        • Equation (derived): $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
    한국어

    학습 목표 6.6.A: 중력 상호작용만 있는 두 개 objekts가 이루어진 시스템의 운동을 설명하시오.

    • 6.6.A.1 질량이 매우 큰 중심 천체와 그 질량에 비해 질량이 무시할 수 있는 궤도 위 인공위성으로만 구성된 시스템에서, 중심 천체 자체의 운동은 무시할 수 있다.
    • 6.6.A.2 궤도 위 인공위성의 운동은 보존 법칙에 의해 제한받는다.
      • 6.6.A.2.i 원형 궤도에서 시스템의 총 역학적 에너지, 시스템의 중력ポテン셜에너지, 인공위성의 각운동량 및 운동에너지는 일정하다.
      • 6.6.A.2.ii 타원 궤도에서 시스템의 총 역학적 에너지와 인공위성의 각운동량은 일정하지만, 시스템의 중력ポテン셜에너지와 인공위성의 운동에너지는 각각 변할 수 있다.
      • 6.6.A.2.iii 인공위성과 질량이 매우 큰 중심 천체로 구성된 시스템의 중력ポテン셜에너지는 인공위성이 중심 천체로부터 무한한 거리에 있을 때 0으로 정의한다.
        • 식: $U_g = -G\dfrac{m_1 m_2}{r}$
    • 6.6.A.3 인공위성의 탈출속도는 인공위성-중심 천체 시스템의 역학적 에너지가 0이 되는 인공위성의 속도이다.
      • 6.6.A.3.i 중심 천체로부터의 중력만이 인공위성에 작용할 때, 탈출속도에 도달한 인공위성은 중심 천체로부터 무한한 거리에서 속도가 0이 될 때까지 중심 천체로부터 멀어진다.
      • 6.6.A.3.ii 질량 $M$인 중심 천체로부터 인공위성의 탈출속도는 에너지 보존 법칙을 사용하여 유도할 수 있다.
        • 식(유도): $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Orbital motion (Kepler's 2nd law)

    A satellite 卫星 in orbit is in free fall: gravity provides the exact centripetal force needed to curve its path into an orbit. Setting gravity equal to the centripetal requirement,

    $$\frac{GMm}{r^2}=\frac{mv^2}{r}\ \Rightarrow\ v=\sqrt{\frac{GM}{r}}.$$
    So a larger orbit means a slower speed. For a circular orbit, angular momentum and mechanical energy are both constant; for an elliptical orbit, angular momentum is conserved (no torque about the planet) while speed varies – fastest when closest.

    Gravity provides the centripetal force that keeps a satellite in orbit
    Gravity provides the centripetal force that keeps a satellite in orbit

    Worked example. Find the speed of a satellite in a low orbit just above the Earth, radius $r=6.4\times10^{6}\ \text{m}$, with $GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$:

    $$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4.0\times10^{14}}{6.4\times10^{6}}}=\sqrt{6.25\times10^{7}}\approx 7.9\times10^{3}\ \text{m/s},$$
    about $7.9\ \text{km/s}$ – and notice that the satellite's mass cancels, so all low orbits share this speed.

    A photograph of the International Space Station orbiting Earth
    The International Space Station: a real satellite held in orbit by gravity alone

    Escape velocity 逃逸速度 is the launch speed that just lets an object leave for good: its total mechanical energy is exactly zero, so it slows to zero speed only at infinite distance. Setting $\tfrac12 mv^2 - \dfrac{GMm}{r}=0$ (using the general gravitational PE $U_g=-GMm/r$) and solving for $v$,

    $$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$
    It is $\sqrt2$ times the circular-orbit speed at the same radius, and (like orbital speed) does not depend on the escaping object's mass.

    Explore · ⁨탐색하기⁩

    Compare orbits at different radii

    A satellite is in free fall, its gravity supplying the centripetal force. A larger orbit means a slower speed and a longer period — Kepler's third law.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    satellite/ˈsætəlaɪt/ 위성
    Escape velocity/eˈskeɪp vəˈlɒsɪti/ 탈출 속도
    6.6

    Exam tips

    • Conserve angular momentum $L=I\omega$ when no external torque acts: a smaller $I$ (arms pulled in) gives a larger $\omega$.
    • A rolling object splits its energy between $\tfrac12 mv^2$ and $\tfrac12 I\omega^2$, linked by $v=r\omega$ — so it accelerates down a ramp more slowly than a sliding one.
    • For a circular orbit set gravity equal to the centripetal requirement: $v=\sqrt{GM/r}$, so a larger orbit is slower and the satellite's mass cancels.
    • Pulling in raises the spin and the kinetic energy — the extra energy comes from the work done pulling inward; $L$ is unchanged.
    • Watch which rotational quantity is conserved: $L$ (no torque) versus energy (no friction) are different conditions.
  • 7

    Oscillations

    Watch lesson · ⁨수업 보기⁩
    7.1

    Defining Simple Harmonic Motion

    Syllabus
    English

    Learning Objective 7.1.A: Describe simple harmonic motion.

    • 7.1.A.1 Simple harmonic motion is a special case of periodic motion.
    • 7.1.A.2 SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position.
      • Equation: $ma_x = -k\Delta x$
      • 7.1.A.2.i A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position.
      • 7.1.A.2.ii An equilibrium position is a location at which the net force exerted on an object or system is zero.
      • 7.1.A.2.iii The motion of a pendulum with a small angular displacement can be modeled as simple harmonic motion because the restoring torque is proportional to the angular displacement.
    한국어

    학습 목표 7.1.A: 단순조화운동을 설명하시오.

    • 7.1.A.1 단순조화운동은 주기운동의 특별한 경우이다.
    • 7.1.A.2 평형 위치에서의 변위에 비례하는 복원력의 크기가 작용할 때 단순조화운동이 발생한다.
      • 식: $ma_x = -k\Delta x$
      • 7.1.A.2.i 복원력은 평형 위치에서의 변위와 반대 방향으로 작용하는 힘이다.
      • 7.1.A.2.ii 평형 위치란 물체나 시스템에 작용하는 합력이 0인 위치를 말한다.
      • 7.1.A.2.iii 작은 각변위를 가지는 단자의 운동은 복원 모멘트가 각변위에 비례하므로 단순조화운동으로 모델링할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Simple harmonic motion

    Simple harmonic motion 简谐运动 (SHM) is a back-and-forth oscillation 振动 caused by a restoring force 回复力 that is proportional to the displacement from equilibrium and always points back toward it: $F=-kx$. A mass on a spring and (for small angles) a pendulum are the standard examples. Because the force grows with displacement, the motion is smooth and repeating, tracing a sine curve in time.

    In SHM the acceleration always points back towards equilibrium, opposite the displacement
    In SHM the acceleration always points back towards equilibrium, opposite the displacement

    The test for SHM is exactly this: acceleration proportional to displacement and opposite in direction, $a=-\dfrac{k}{m}x$. A pendulum only obeys it for small swings, where $\sin\theta\approx\theta$; large swings are not quite SHM.

    A spring under load: Hooke's law F = −kx is the restoring force that produces SHM
    A spring under load: Hooke's law F = −kx is the restoring force that produces SHM
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Simple harmonic motion/ˈsɪmpl hɑːˈmɒnɪk ˈməʊʃn/ 간단 진동
    oscillation/ˌɒsɪˈleɪʃn/ 진동
    restoring force/rɪˈstɔːrɪŋ fɔːs/ 복원력
    7.2

    Frequency and Period of SHM

    Syllabus
    English

    Learning Objective 7.2.A: Describe the frequency and period of an object exhibiting SHM.

    • 7.2.A.1 The period of SHM is related to the frequency $f$ of the object's motion by the following equation:
      • Equation: $T = \dfrac{1}{f}$
      • 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation $T_s = 2\pi\sqrt{\dfrac{m}{k}}$.
      • 7.2.A.1.ii The period of a simple pendulum displaced by a small angle is given by the equation $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$.
    한국어

    학습 목표 7.2.A: SHM을 나타내는 물체의 주파수와 주기를 설명하시오.

    • 7.2.A.1 SHM의 주기는 물체의 운동의 주파수 $f$와 다음 식으로 관련된다:
      • 식: $T = \dfrac{1}{f}$
      • 7.2.A.1.i 물체-이상적 스프링 진동자의 주기는 다음 식으로 주어진다: $T_s = 2\pi\sqrt{\dfrac{m}{k}}$.
      • 7.2.A.1.ii 작은 각도로 변위된 단순 단자의 주기는 다음 식으로 주어진다: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    • The period 周期 $T$ is the time for one full cycle.
    • The frequency 频率 $f=\dfrac{1}{T}$ is cycles per second (hertz).

    For SHM these depend only on the system, not on the amplitude:

    $$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{L}{g}}.$$
    So a stiffer spring or smaller mass oscillates faster; a longer pendulum swings slower.

    Worked example. A $0.25\ \text{kg}$ mass hangs on a spring of stiffness $k=100\ \text{N/m}$. Its period is

    $$T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.25}{100}}=0.31\ \text{s},\qquad f=\frac{1}{T}=3.2\ \text{Hz}.$$

    Worked example. A pendulum clock ticks with a period of exactly $2.0\ \text{s}$. How long is it? Rearranging $T=2\pi\sqrt{L/g}$,

    $$L=g\left(\frac{T}{2\pi}\right)^2=9.8\times\left(\frac{2.0}{2\pi}\right)^2=0.99\ \text{m}.$$
    Notice the amplitude never entered – a wide or narrow swing keeps the same time, which is what makes pendulums good clocks.

    A set of tuning forks of different sizes
    Tuning forks: a longer, heavier fork vibrates more slowly, giving a lower frequency
    Explore · ⁨탐색하기⁩

    Time a pendulum's swing · ⁨진자의 진동 시간 측정⁩

    A pendulum's period depends on its length and gravity, not its mass or (small) amplitude: $T=2\pi\sqrt{L/g}$. Lengthen it and each swing takes longer. · ⁨진자의 주기는 길이에 중력加速度(중력)에 의존하며 질량이나 (작은) 진폭에는 의하지 않습니다: $T=2\pi\sqrt{L/g}$. 길이를 늘리면 한 번의 진동에 더 많은 시간이 걸립니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    period/ˈpɪərɪəd/ 주기
    frequency/ˈfriːkwənsi/ 주파수
    7.3

    Representing and Analyzing SHM

    Syllabus
    English

    Learning Objective 7.3.A: Describe the displacement, velocity, and acceleration of an object exhibiting SHM.

    • 7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$.
      • 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
      • 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
    • 7.3.A.2 Changing the amplitude of a system exhibiting SHM will not change the period of that system.
    • 7.3.A.3 Properties of SHM can be determined and analyzed using graphical representations.
    한국어

    학습 목표 7.3.A: SHM을 나타내는 물체의 변위, 속도 및 가속도를 설명하시오.

    • 7.3.A.1 SHM을 나타내는 물체의 변위는 평형 위치로부터 측정하여 다음 식 $x = A\cos(2\pi ft)$ 또는 $x = A\sin(2\pi ft)$로 표현할 수 있다.
      • 7.3.A.1.i 변위, 속도 및 가속도의 극소값, 극대값 및 영점은 조화운동의 특징이다.
      • 7.3.A.1.ii SHM의 변위, 속도 및 가속도가 극값이나 영점을 가지는 위치나 시간을 인지하면 운동의 거동을 정성적으로 설명하는 데 도움이 된다.
    • 7.3.A.2 SHM을 나타내는 시스템의 진폭을 변경则该系统的周期不会改变。
    • 7.3.A.3 SHM의 성질은 그래프 표현을 통해 결정하고 분석할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The displacement varies sinusoidally: $x(t)=A\cos(\omega t)$ (or sine), where $A$ is the amplitude 振幅 (maximum displacement) and $\omega=2\pi f$ is the angular frequency 角频率. Reading the motion:

    Displacement varies sinusoidally with time in simple harmonic motion
    Displacement varies sinusoidally with time in simple harmonic motion
    • At the extremes ($x=\pm A$): displacement and restoring force are maximum, so acceleration is maximum, but velocity is zero.
    • At equilibrium 平衡 ($x=0$): force and acceleration are zero, but speed is maximum.

    Velocity and acceleration are also sinusoidal, shifted in phase 相位 from the displacement – velocity leads displacement by a quarter cycle, and acceleration is exactly opposite to displacement.

    Displacement, velocity, and acceleration in SHM, each a quarter-cycle apart
    Displacement, velocity, and acceleration in SHM, each a quarter-cycle apart

    Read the figure as a story: where $x$ is largest (the turning point), $v$ has just fallen to zero and $a$ is at its most negative, hauling the mass back; a quarter-cycle later the mass races through the middle at top speed with zero acceleration.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    amplitude/ˈæmplɪtjuːd/ 진폭
    angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ 각주파수
    phase/feɪz/ 위상
    equilibrium/ˌiːkwɪˈlɪbrɪəm/ 평형
    7.4

    Energy of Simple Harmonic Oscillators

    Syllabus
    English

    Learning Objective 7.4.A: Describe the mechanical energy of a system exhibiting SHM.

    • 7.4.A.1 The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies.
      • Equation: $E_{\text{total}} = U + K$
    • 7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
    • 7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
    • 7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
      • 7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
      • 7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system.
      • Relevant equation for a spring–object system: $E_{\text{total}} = \dfrac{1}{2}kA^2$
    한국어

    학습 목표 7.4.A: SHM을 나타내는 시스템의 역학적 에너지를 설명하시오.

    • 7.4.A.1 SHM을 나타내는 시스템의 총 에너지는 시스템의 운동에너지와 포텐셜에너지의 합이다.
      • 식: $E_{\text{total}} = U + K$
    • 7.4.A.2 에너지 보존에 따라 SHM을 나타내는 시스템의 총 에너지는 일정하다.
    • 7.4.A.3 SHM을 나타내는 시스템의 운동에너지는 시스템의 포텐셜에너지가 최소일 때 최대가 된다.
    • 7.4.A.4 진동계가 단순조화운동(SHM)을 할 때, 운동 에너지가 최소일 때 시스템의ポテン셜 에너지는 최대가 된다.
      • 7.4.A.4.i 단순조화운동을 하는 시스템의 운동 에너지의 최소값은 0이다.
      • 7.4.A.4.ii 단순조화운동을 하는 시스템의 진폭을 변경하면 시스템의 최대 포텐셜 에너지와 따라서 시스템의 총 에너지가 바뀐다.
      • 스프링-물체 시스템에 대한 관련 식: $E_{\text{total}} = \dfrac{1}{2}kA^2$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Energy conservation: KE ⇄ PE

    Energy sloshes between kinetic and potential while the total stays constant (no friction):

    $$E=\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2+\tfrac{1}{2}mv^2.$$
    At the extremes it is all potential; at equilibrium it is all kinetic (maximum speed). Because $E\propto A^2$, doubling the amplitude quadruples the energy.

    Kinetic and potential energy swap over a cycle while the total energy stays constant
    Kinetic and potential energy swap over a cycle while the total energy stays constant

    Worked example. A $0.50\ \text{kg}$ mass on a spring of stiffness $k=200\ \text{N/m}$ oscillates with amplitude $A=0.10\ \text{m}$. Find its maximum speed. All the energy is kinetic at the equilibrium point, so $\tfrac12 kA^2=\tfrac12 mv_{\max}^2$:

    $$v_{\max}=A\sqrt{\frac{k}{m}}=0.10\times\sqrt{\frac{200}{0.50}}=0.10\times20=2.0\ \text{m/s}.$$
    The energy method is faster than tracking the sine functions when you only need the greatest speed.

    Explore · ⁨탐색하기⁩

    Trade kinetic and potential energy in SHM · ⁨SHM에서의 운동 에너지와 위치 에너지 교환⁩

    In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant. · ⁨**简单简谐运动(SHM)**에서 에너지는 중심부에서 가장 빠르고 극단에서 가장 큰 위치 에너지와 운동 에너지 사이를 오가며, 전체 에너지는 일정하게 유지됩니다.⁩

    7.4

    Exam tips

    • Test for SHM: the acceleration must be proportional to the displacement and directed back toward the middle ($a=-\tfrac{k}{m}x$).
    • The period does not depend on amplitude — use $T=2\pi\sqrt{m/k}$ (spring) or $T=2\pi\sqrt{L/g}$ (pendulum, small angles).
    • Speed is maximum at the middle (all kinetic) and zero at the extremes (all potential); acceleration is largest at the extremes.
    • Use energy ($\tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2$) to find the maximum speed quickly: $v_{\max}=A\sqrt{k/m}$.
    • Total energy $\propto A^2$, so doubling the amplitude quadruples the energy.
  • 8

    Fluids

    Watch lesson · ⁨수업 보기⁩
    8.1

    Internal Structure and Density

    Syllabus
    English

    Learning Objective 8.1.A: Describe the properties of a fluid.

    • 8.1.A.1 Distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules.
    • 8.1.A.2 A fluid is a substance that has no fixed shape.
    • 8.1.A.3 Fluids can be characterized by their density. Density is defined as a ratio of mass to volume.
      • Equation: $\rho = \dfrac{m}{V}$
    • 8.1.A.4 An ideal fluid is incompressible and has no viscosity.
    한국어

    학습 목표 8.1.A: 유체의 성질을 설명하라.

    • 8.1.A.1 고체, 액체, 기체의 구별 가능한 성질은 원자와 분자 사이의 상호작용이 다르게 나타남에서 기인한다.
    • 8.1.A.2 유체는 고정된 형태를 가지지 않는 물질이다.
    • 8.1.A.3 유체는 밀도로 특징지어질 수 있다. 밀도는 질량과 부피의 비로 정의된다.
      • 식: $\rho = \dfrac{m}{V}$
    • 8.1.A.4 이상 유체는 압축 불가능하며 점성이 없다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The differences between solids, liquids, and gases come from how strongly their particles interact. A fluid 流体 (liquid or gas) has no fixed shape – it flows because its particles move past one another. Density 密度 is mass per unit volume:

    $$\rho=\frac{m}{V}.$$
    It depends on the substance and its state. An object sinks or floats depending on how its density compares with the surrounding fluid's – less dense floats, more dense sinks. An ideal fluid 理想流体 is incompressible 不可压缩 (constant density, whatever the pressure) and has no viscosity 黏度 (internal friction) – the model AP uses throughout.

    Explore · ⁨탐색하기⁩

    Float or sink by density · ⁨밀도에 따른 부양 또는 침수⁩

    An object floats if it is less dense than the fluid. Change the density and watch it ride higher or sink, displacing its own weight of fluid. · ⁨物体가 유체보다 밀도가 낮으면 뜨게 됩니다. 밀도를 변화시키면 물체가 더 위로 떠 있거나 가라앉으며, 자신의 무게만큼의 유체를 밀어냅니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    fluid/ˈfluːɪd/ 유체로 treat 한다
    Density/ˈdensɪti/ 밀도
    ideal fluid/aɪˈdɪəl ˈfluːɪd/ 이상 유체
    incompressible/ˌɪnkəmˈpresəbl/ 압축 불가능한
    viscosity/vɪˈskɒsɪti/ 점성
    8.2

    Pressure

    Syllabus
    English

    Learning Objective 8.2.A: Describe the pressure exerted on a surface by a given force.

    • 8.2.A.1 Pressure is defined as the magnitude of the perpendicular force component exerted per unit area over a given surface area, as described by the equation
      • Equation: $P = \dfrac{F_\perp}{A}$
    • 8.2.A.2 Pressure is a scalar quantity.
    • 8.2.A.3 The volume and density of a given amount of an incompressible fluid is constant regardless of the pressure exerted on that fluid.

    Learning Objective 8.2.B: Describe the pressure exerted by a fluid.

    • 8.2.B.1 The pressure exerted by a fluid is the result of the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact.
    • 8.2.B.2 The absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$, such as the atmospheric pressure $P_{\text{atm}}$, and the gauge pressure $P_{\text{gauge}}$.
      • Equation: $P = P_0 + \rho g h$
    • 8.2.B.3 The gauge pressure of a vertical column of fluid is described by the equation
      • Equation: $P_{\text{gauge}} = \rho g h$
    한국어

    학습 목표 8.2.A: 주어진 힘이 표면에 가하는 압력을 설명하라.

    • 8.2.A.1 압력은 다음 식으로 기술되듯이, 특정 면적에 대해 단위 면적당 가해지는 수직 힘 성분의 크기로 정의된다.
      • 식: $P = \dfrac{F_\perp}{A}$
    • 8.2.A.2 압력은 스칼라 양이다.
    • 8.2.A.3 압축 불가능한 유체의 일정한 양에 대해 그 유체에 가해지는 압력에 상관없이 부피와 밀도는 일정하다.

    학습 목표 8.2.B: 유체가 가하는 압력을 설명하라.

    • 8.2.B.1 유체가 가하는 압력은 유체를 구성하는 입자와 그 입자가 상호작용하는 표면 간의 모든 상호작용의 결과이다.
    • 8.2.B.2 특정 지점에서의 유체의 절대 압력은 기준 압력 $P_0$(예: 대기압 $P_{\text{atm}}$)와 게이지 압력 $P_{\text{gauge}}$의 합으로 정의된다.
      • 식: $P = P_0 + \rho g h$
    • 8.2.B.3 수직 유체 기둥의 게이지 압력은 다음 식으로 기술된다.
      • 식: $P_{\text{gauge}} = \rho g h$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Pressure 压强 is the perpendicular force per unit area, a scalar 标量 measured in pascals (Pa):

    $$P=\frac{F_\perp}{A}.$$
    In a fluid at rest, pressure increases with depth 深度 because of the weight of fluid above:
    $$P=P_0+\rho g h,$$
    where $P_0$ is the pressure at the surface and $h$ is the depth. Pressure acts equally in all directions at a point and pushes perpendicular to any surface.

    Two hydraulic bottle jacks standing side by side, each with a narrow pump lever and a wide lifting ram
    Hydraulic jacks: pressure applied to a small piston is transmitted undiminished through the fluid to a large one, so a small force lifts a car

    Distinguish the two pressures AP asks about: the gauge pressure 表压 is the extra pressure the fluid column adds, $P_{\text{gauge}}=\rho g h$, while the absolute pressure 绝对压强 is the total, $P=P_0+P_{\text{gauge}}$ (with $P_0$ usually atmospheric). A tyre gauge reading "$200\ \text{kPa}$" is gauge pressure; the air inside is really at about $300\ \text{kPa}$ absolute.

    The weight of a liquid column sets the extra pressure at a depth
    The weight of a liquid column sets the extra pressure at a depth

    Worked example. Find the total pressure on a diver $10\ \text{m}$ below the surface of water ($\rho=1000\ \text{kg/m}^3$, surface pressure $P_0=1.0\times10^{5}\ \text{Pa}$):

    $$P=P_0+\rho g h=1.0\times10^{5}+1000\times9.8\times10=1.98\times10^{5}\ \text{Pa}.$$
    Every $10\ \text{m}$ of water adds roughly one extra atmosphere of pressure. Notice the pressure does not depend on the shape or width of the container, only on the depth.

    An aneroid barometer with a circular dial
    An aneroid barometer measures air pressure with a sealed metal box that flexes as the pressure outside changes
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Pressure/ˈpreʃə/ 압력
    scalar/ˈskeɪlə/ 스칼라
    depth/depθ/ 깊이
    gauge pressure/ɡeɪdʒ ˈpreʃə/ 표압
    absolute pressure/ˈæbsəluːt ˈpreʃə/ 절대 압력
    8.3

    Fluids and Newton's Laws

    Syllabus
    English

    Learning Objective 8.3.A: Describe the conditions under which a fluid's velocity changes.

    • 8.3.A.1 Newton's laws can be used to describe the motion of particles within a fluid.
    • 8.3.A.2 The macroscopic behavior of a fluid is a result of the internal interactions between the fluid's constituent particles and external forces exerted on the fluid.

    Learning Objective 8.3.B: Describe the buoyant force exerted on an object interacting with a fluid.

    • 8.3.B.1 The buoyant force is a net upward force exerted on an object by a fluid.
    • 8.3.B.2 The buoyant force exerted on an object by a fluid is a result of the collective forces exerted on the object by the particles making up the fluid.
    • 8.3.B.3 The magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object.
      • Equation: $F_b = \rho V g$
    한국어

    학습 목표 8.3.A: 유체의 속도가 변하는 조건을 설명하라.

    • 8.3.A.1 뉴턴의 법칙을 사용하여 유체 내 입자의 운동을 설명할 수 있다.
    • 8.3.A.2 유체의 거시적 행동은 유체를 구성하는 입자 간의 내부 상호작용과 유체에 가해지는 외부 힘의 결과이다.

    학습 목표 8.3.B: 유체와 상호작용하는 물체에 가해지는 부력을 설명하라.

    • 8.3.B.1 부력은 유체가 물체에 가하는 순상승 힘이다.
    • 8.3.B.2 유체가 물체에 가하는 부력은 유체를 구성하는 입자들이 물체에 가하는 힘들의 합계的结果이다.
    • 8.3.B.3 유체가 물체에 가하는 부력의 크기는 물체가 배제한 유체의 무게와 같다.
      • 식: $F_b = \rho V g$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Buoyancy: float or sink

    An object in a fluid feels an upward buoyant force 浮力 equal to the weight of the fluid it displaces – Archimedes' principle 阿基米德原理:

    $$F_b=\rho_{\text{fluid}}\,g\,V_{\text{displaced}}.$$
    Combine this with Newton's laws: the object floats when buoyancy balances weight, sinks when weight wins, and rises when buoyancy wins. A floating object displaces exactly its own weight of fluid.

    Worked example. A block of density $600\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction is under the surface? For floating, the buoyant force equals the weight, so $\rho_{\text{fluid}}\,g\,V_{\text{sub}}=\rho_{\text{object}}\,g\,V$:

    $$\frac{V_{\text{sub}}}{V}=\frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}=\frac{600}{1000}=0.60.$$
    So $60\%$ sits below the water – the same reason most of an iceberg (density $\approx 900$) hides underwater.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    buoyant force/ˈbɔɪənt fɔːs/ 부력
    Archimedes' principle/ˌɑːkɪˈmiːdiːz ˈprɪnsɪpl/ 아킴데스의 원리
    8.4

    Fluids and Conservation Laws

    Syllabus
    English

    Learning Objective 8.4.A: Describe the flow of an incompressible fluid through a cross-sectional area by using mass conservation.

    • 8.4.A.1 A difference in pressure between two locations causes a fluid to flow.
      • 8.4.A.1.i The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.
      • 8.4.A.1.ii The rate at which matter flows into a location is proportional to the cross-sectional area of the flow and the speed at which the fluid flows.
        • Derived equation: $\dfrac{V}{t} = A v$
    • 8.4.A.2 The continuity equation for fluid flow describes conservation of mass flow rate in incompressible fluids.
      • Equation: $A_1 v_1 = A_2 v_2$
    한국어

    학습 목표 8.4.A: 질량 보존을 사용하여 압축 불가능한 유체의 단면적을 통한 흐름을 설명하라.

    • 8.4.A.1 두 위치 간의 압력 차이는 유체의 흐름을 유발한다.
      • 8.4.A.1.i 양쪽 끝이 열린 관에 들어오는 물질의 유속은 관을 빠져나가는 물질의 유속과 같아야 한다.
      • 8.4.A.1.ii 어떤 지점으로 유입되는 물질의 유속은 흐름의 단면적과 유체가 흐르는 속도에 비례한다.
        • 파생 식: $\dfrac{V}{t} = A v$
    • 8.4.A.2 유체 흐름의 연속성 방정식은 압축 불가능한 유체에서의 질량 유속 보존을 기술한다.
      • 식: $A_1 v_1 = A_2 v_2$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    For an ideal fluid flowing steadily, two conservation ideas apply:

    Upthrust arises because the pressure on the bottom of an object exceeds that on the top
    Upthrust arises because the pressure on the bottom of an object exceeds that on the top
    • Continuity 连续性 (conservation of mass): the volume flow rate is constant, so $A_1 v_1 = A_2 v_2$. A narrower pipe forces faster flow.
    • Bernoulli's equation 伯努利方程 (conservation of energy per volume): along a streamline,
      $$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$

    Together they explain why fluid speeds up and its pressure drops where a pipe narrows or where flow is fastest.

    Where the pipe narrows the fluid speeds up (continuity) and its pressure drops (Bernoulli)
    Where the pipe narrows the fluid speeds up (continuity) and its pressure drops (Bernoulli)

    Worked example. Water flows at $2.0\ \text{m/s}$ through a pipe of cross-section $0.010\ \text{m}^2$, then enters a narrower section of $0.0040\ \text{m}^2$. By continuity the speed there is

    $$v_2=\frac{A_1 v_1}{A_2}=\frac{0.010\times2.0}{0.0040}=5.0\ \text{m/s}.$$
    By Bernoulli's equation this faster stream is at lower pressure – the effect that lifts an aeroplane wing and pulls two passing ships together.

    Exam skill. Choose the right law by what changes. If the pipe changes width, start with continuity ($A_1v_1=A_2v_2$) to get the speeds; if you then need a pressure, feed those speeds into Bernoulli. Watch the height term $\rho g y$ only when the pipe also changes level.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Continuity/kɒntɪˈnjuːɪti/ 연속성
    Bernoulli's equation/ˈbɜːnaʊliz ɪˈkweɪʒn/ 베를누이 방정식
    8.4

    Exam tips

    • Pressure with depth is $P=P_0+\rho g h$ — it depends on depth only, not the container's shape or width.
    • Buoyant force = weight of fluid displaced ($\rho_{\text{fluid}}\,gV_{\text{disp}}$); a floating object displaces its own weight, so the fraction submerged is $\rho_{\text{object}}/\rho_{\text{fluid}}$.
    • Compare densities to predict floating vs sinking; a floating object is in equilibrium (buoyancy = weight), not weightless.
    • Use continuity $A_1v_1=A_2v_2$: a narrower pipe means faster flow.
    • Bernoulli: where a fluid flows faster its pressure is lower (wing lift, spray).

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IGCSE, A-Level & AP