| Learning Objective | Essential Knowledge |
|---|---|
2.1.A |
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2.1.B |
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Force and Translational Dynamics
AP Physics C: Mechanics · Topic 2
2.1
Systems and Center of Mass
Syllabus
Source: College Board AP Course and Exam Description
A system 系统 is the object or objects you analyze, treated as a point at its center of mass 质心. For a set of masses, $x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$; for a continuous body, $x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$. Only external forces move the center of mass.
| English | Chinese | Pinyin |
|---|---|---|
| system | 系统 | xì tǒng |
| center of mass | 质心 | zhì xīn |
2.2
Forces and Free-Body Diagrams
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.2.A |
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2.2.B |
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect 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. |
Source: College Board AP Course and Exam Description
A force 力 is a push or pull (a vector, in newtons). A free-body diagram 受力图 draws one object with an arrow for every force on it – weight, normal, tension, friction, applied, drag. Draw it first; it sets up every dynamics equation.
A free-body diagram shows every force acting on one object
Balance the forces on a free-body diagram
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 $a=F/m$.
| English | Chinese | Pinyin |
|---|---|---|
| force | 力 | lì |
| free-body diagram | 受力图 | shòu lì tú |
2.3
Newton's Third Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.3.A |
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Source: College Board AP Course and Exam Description
Newton's third law 牛顿第三定律: A pushes B, B pushes back equally and oppositely. The pair acts on different objects, so it never cancels within one free-body diagram.
A Newton's third-law pair: equal and opposite forces on two different objects
| English | Chinese | Pinyin |
|---|---|---|
| Newton's third law | 牛顿第三定律 | niú dùn dì sān dìng lǜ |
2.4
Newton's First Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.4.A |
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Source: College Board AP Course and Exam Description
Newton's first law (inertia 惯性): with zero net force 合力, velocity stays constant – the object is in translational equilibrium 平动平衡.
| English | Chinese | Pinyin |
|---|---|---|
| inertia | 惯性 | guàn xìng |
| net force | 合力 | hé lì |
| translational equilibrium | 平动平衡 | píng dòng píng héng |
2.5
Newton's Second Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.5.A |
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Source: College Board AP Course and Exam Description
The general form uses momentum:
Worked example. A $2.0\ \text{kg}$ block slides down a frictionless incline 斜面 at $30^{\circ}$. Only the along-ramp component of gravity drives it, so $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$, independent of the mass.
Worked example (two bodies). A $6.0\ \text{kg}$ cart on a frictionless table is pulled by a string over a light pulley to a hanging $2.0\ \text{kg}$ mass. Treat the pair as one system: only the hanging weight drives it, so $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$. Then isolate the cart alone to find the string tension 张力: $T=6.0(2.45)\approx15\ \text{N}$. System first for $a$, single body for internal forces – that two-step is the standard pattern.
| English | Chinese | Pinyin |
|---|---|---|
| differential equation | 微分方程 | wēi fēn fāng chéng |
| incline | 斜面 | xié miàn |
| tension | 张力 | zhāng lì |
2.6
Gravitational Force
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.6.A |
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2.6.B |
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2.6.C |
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2.6.D |
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2.6.E |
Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem. |
Source: College Board AP Course and Exam Description
Near a surface, weight is $F_g=mg$. In general, Newton's law of gravitation 万有引力定律:
Two masses attract each other with equal, opposite, inverse-square forces along the line joining them
| English | Chinese | Pinyin |
|---|---|---|
| Newton's law of gravitation | 万有引力定律 | wàn yǒu yǐn lì dìng lǜ |
2.7
Kinetic and Static Friction
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.7.A |
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2.7.B |
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Source: College Board AP Course and Exam Description
Friction 摩擦力 opposes sliding along a surface: kinetic 动摩擦 $f_k=\mu_k N$ while sliding, and static 静摩擦 $f_s\le\mu_s N$ up to a maximum before sliding. $N$ is the normal force 法向力. Note the inequality: static friction is only as large as it needs to be, and $\mu_s N$ is its ceiling, not its value.
Worked example. The same block on the same $30^{\circ}$ incline, now with $\mu_k=0.20$. Perpendicular to the ramp: $N=mg\cos30^{\circ}$. Along the ramp: $ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$, so $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ – the mass cancels again.
Slide a block down a slope with friction
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.
| English | Chinese | Pinyin |
|---|---|---|
| Friction | 摩擦力 | mó cā lì |
| kinetic | 动摩擦 | dòng mó cā |
| static | 静摩擦 | jìng mó cā |
| normal force | 法向力 | fǎ xiàng lì |
2.8
Spring Forces
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.8.A |
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2.8.B |
Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel. |
Source: College Board AP Course and Exam Description
An ideal spring obeys Hooke's law 胡克定律 $F_s=-kx$, a restoring force set by the spring constant 弹簧劲度系数 $k$. Its stored energy is $U=\tfrac12 kx^2$.
Hooke's law: extension is proportional to load up to the limit of proportionality
Combined springs act as one equivalent spring of constant $k_{\text{eq}}$. In parallel (side by side, sharing the load) the constants add, $k_{\text{eq}}=k_1+k_2$ – stiffer than either. In series (end to end) they combine as $\dfrac{1}{k_{\text{eq}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ – softer than the softest one. Two springs pulling a mass from opposite sides also act in parallel, giving a restoring constant $k_1+k_2$. Whichever you have, the SHM period uses $k_{\text{eq}}$: $T=2\pi\sqrt{m/k_{\text{eq}}}$.
Stretch a spring (Hooke's law)
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.
| English | Chinese | Pinyin |
|---|---|---|
| Hooke's law | 胡克定律 | hú kè dìng lǜ |
| spring constant | 弹簧劲度系数 | tán huáng jìn dù xì shù |
2.9
Resistive Forces
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.9.A |
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Source: College Board AP Course and Exam Description
A resistive force 阻力 (drag) opposes motion through a fluid and grows with speed, often modeled as $F=-bv$ or $F=-cv^2$. Newton's second law then gives a differential equation, e.g. $m\dfrac{dv}{dt}=mg-bv$ for a falling object. As speed rises, drag builds until it balances the driving force; the object then stops accelerating and moves at a constant terminal velocity 终极速度, found by setting the net force (and $dv/dt$) to zero.
Worked example. For a falling object with $m\dfrac{dv}{dt}=mg-bv$, the terminal velocity is where $\dfrac{dv}{dt}=0$: $mg=bv_T$, so $v_T=\dfrac{mg}{b}$. With $m=0.10\ \text{kg}$ and $b=0.50\ \text{kg/s}$, $v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$.
To get the full motion $v(t)$, separate the variables 分离变量 and integrate from rest:
An object falling through a fluid speeds up to a terminal velocity
| English | Chinese | Pinyin |
|---|---|---|
| resistive force | 阻力 | zǔ lì |
| terminal velocity | 终极速度 | zhōng jí sù dù |
| separate the variables | 分离变量 | fēn lí biàn liàng |
2.10
Circular Motion
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.10.A |
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2.10.B |
Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. |
Source: College Board AP Course and Exam Description
Uniform circular motion has a centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}$ pointing to the center, requiring a net inward force $F_c=\dfrac{mv^2}{r}$ supplied by a real force (tension, gravity, friction, normal). There is no outward force. For vertical circles and banked curves, decompose the real forces to find which provides the centripetal requirement.
Worked example. A $0.50\ \text{kg}$ ball is whirled on a $1.0\ \text{m}$ string in a horizontal circle at $3.0\ \text{m/s}$. The string tension supplies the whole centripetal force: $T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$.
Worked example (vertical circle). At the top of a vertical loop of radius $r$, gravity and the normal force both point toward the center: $N+mg=\dfrac{mv^2}{r}$. The slowest possible speed at the top is where the track pushes with nothing at all ($N=0$): $v_{\min}=\sqrt{gr}$. For $r=2.5\ \text{m}$: $v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$. Any slower and the cart leaves the track before the top.
The velocity points along the tangent; the centripetal force points to the centre
| English | Chinese | Pinyin |
|---|---|---|
| centripetal acceleration | 向心加速度 | xiàng xīn jiā sù dù |
2.10
Exam tips
- Locate the center of mass with $x_{cm}=\tfrac{1}{M}\int x\,dm$ (or $\tfrac{\sum m_i x_i}{\sum m_i}$ for point masses) and use $\lambda,\sigma,\rho$ for the mass element.
- The net external force moves the center of mass as if all mass sat there: $\vec F_{net}=M\vec a_{cm}$.
- Define your system clearly — internal forces cancel, so only external forces change its momentum.
- Exploit symmetry to shortcut a center-of-mass integral.
- Distinguish center of mass from center of gravity (identical in a uniform field).