📚 SAT2 Physics Formula Cheat Sheet with Worked Examples | SAT2物理必背公式清单及例题应用
The SAT Subject Test in Physics requires not only conceptual understanding but also rapid, accurate application of core formulas. This guide condenses every high-yield equation into a single revision sheet, paired with exam-style worked examples so you can see exactly how each formula is applied under time pressure.
SAT2物理考试不仅考查概念理解，更考查你在有限时间内快速准确运用核心公式的能力。本清单浓缩了所有高频必背公式，并配以贴近真题的例题，手把手教你如何在考场上高效套用。
1. Kinematics in One Dimension | 一维运动学
For motion with constant acceleration, the three essential equations are:
匀变速直线运动中，三个核心方程为：
v = v₀ + at
x = v₀t + ½at²
v² = v₀² + 2ax
Here x is displacement, v₀ is initial velocity, v is final velocity, a is acceleration, and t is time. The sign convention matters: choose a positive direction and stick to it.
其中 x 为位移，v₀ 为初速度，v 为末速度，a 为加速度，t 为时间。注意正方向约定，选定正方向后全程保持一致。
Example | 例题: A car accelerates from rest at 2 m/s² for 5 s. How far does it travel?
Solution | 解答: Since v₀ = 0 and a = 2 m/s², use x = v₀t + ½at² = 0 + ½ × 2 × 5² = 25 m.
因 v₀ = 0，a = 2 m/s²，使用 x = v₀t + ½at² = 0 + ½ × 2 × 5² = 25 m。
2. Projectile Motion | 抛体运动
Break motion into independent horizontal and vertical components. Horizontal velocity is constant (ignore air resistance); vertical motion has acceleration g = 9.8 m/s² downward.
将运动分解为相互独立的水平与竖直分量。水平方向速度恒定（忽略空气阻力）；竖直方向加速度为 g = 9.8 m/s² 向下。
x = vₓt, vₓ = v₀cosθ
y = v₀ᵧt − ½gt², v₀ᵧ = v₀sinθ
Example | 例题: A ball is launched at 20 m/s at 30° above horizontal. Find its flight time.
Solution | 解答: v₀ᵧ = 20sin30° = 10 m/s. Total flight time satisfies y = 0 at landing: 0 = 10t − 4.9t², so t = 10/4.9 ≈ 2.04 s.
v₀ᵧ = 20sin30° = 10 m/s。落地时 y = 0：0 = 10t − 4.9t²，解得 t = 10/4.9 ≈ 2.04 s。
3. Newton’s Laws and Friction | 牛顿定律与摩擦力
Newton’s second law links net force to acceleration. Friction is proportional to the normal force, with the coefficient depending on whether surfaces are static or kinetic.
牛顿第二定律将合力与加速度联系起来。摩擦力与正压力成正比，动、静摩擦系数不同。
Fₙₑₜ = ma
fₛ ≤ μₛN, fₖ = μₖN
Example | 例题: A 10 kg block is pushed with 50 N on a surface with μₖ = 0.2. Find acceleration.
Solution | 解答: N = mg = 98 N, fₖ = 0.2 × 98 = 19.6 N. Fₙₑₜ = 50 − 19.6 = 30.4 N. a = 30.4/10 = 3.04 m/s².
N = mg = 98 N，fₖ = 0.2 × 98 = 19.6 N。合力 Fₙₑₜ = 50 − 19.6 = 30.4 N。a = 30.4/10 = 3.04 m/s²。
4. Circular Motion and Gravitation | 圆周运动与万有引力
For uniform circular motion, centripetal force always points toward the center. Newton’s law of gravitation describes the force between two masses.
匀速圆周运动中，向心力始终指向圆心。牛顿万有引力定律描述两质点间的引力。
a꜀ = v²/r = ω²r
F꜀ = mv²/r
F = Gm₁m₂/r²
Example | 例题: A 2 kg object moves at 3 m/s in a circle of radius 1.5 m. Find centripetal force.
Solution | 解答: F꜀ = mv²/r = 2 × 3² / 1.5 = 12 N.
F꜀ = mv²/r = 2 × 3² / 1.5 = 12 N。
5. Work, Energy, and Power | 功、能量与功率
Work transfers energy. The work-energy theorem relates net work to change in kinetic energy. Power is the rate of doing work.
功传递能量。动能定理将合外力做功与动能变化联系起来。功率是做功的快慢。
W = Fdcosθ
KE = ½mv²
PE꜀ = mgh
P = W/t = Fv
Example | 例题: A 5 kg box is lifted 2 m at constant speed. How much work is done against gravity?
Solution | 解答: W = mgh = 5 × 9.8 × 2 = 98 J.
W = mgh = 5 × 9.8 × 2 = 98 J。
6. Momentum and Collisions | 动量与碰撞
Momentum is conserved in isolated systems. Impulse equals the change in momentum. Collisions may be elastic or inelastic.
孤立系统内动量守恒。冲量等于动量变化。碰撞分为弹性碰撞与非弹性碰撞。
p = mv
J = FΔt = Δp
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
Example | 例题: A 2 kg ball moving at 4 m/s hits a stationary 3 kg ball and sticks. Find final speed.
Solution | 解答: By conservation of momentum: 2 × 4 = (2 + 3)v, so v = 8/5 = 1.6 m/s.
由动量守恒：2 × 4 = (2 + 3)v，解得 v = 8/5 = 1.6 m/s。
7. Simple Harmonic Motion and Waves | 简谐运动与波动
Simple harmonic motion involves a restoring force proportional to displacement. Wave speed depends on wavelength and frequency.
简谐运动的回复力与位移成正比。波速由波长与频率共同决定。
F = −kx
T = 2π√(m/k)
v = fλ
Example | 例题: A wave has frequency 50 Hz and wavelength 0.2 m. Find wave speed.
Solution | 解答: v = fλ = 50 × 0.2 = 10 m/s.
v = fλ = 50 × 0.2 = 10 m/s。
8. Electrostatics and Coulomb’s Law | 静电学与库仑定律
Charges exert forces on each other according to Coulomb’s law. Electric field is force per unit charge.
电荷间遵循库仑定律相互作用。电场强度为单位电荷所受的力。
F = k|q₁q₂|/r²
E = F/q
E = kQ/r²
Example | 例题: Two charges, 2 μC and 3 μC, are 0.1 m apart. Find force (k = 9 × 10⁹ N·m²/C²).
Solution | 解答: F = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.1)² = 5.4 N.
F = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.1)² = 5.4 N。
9. Electric Circuits | 直流电路
Ohm’s law relates voltage, current, and resistance. Electric power can be expressed in multiple equivalent forms. Resistors combine differently in series and parallel.
欧姆定律联系电压、电流与电阻。电功率有多种等价表达形式。电阻串联与并联的计算方式不同。
V = IR
P = IV = I²R = V²/R
Rₛ = R₁ + R₂ + …
1/Rₚ = 1/R₁ + 1/R₂ + …
Example | 例题: A 12 V battery drives 3 A through a resistor. Find the resistance and power.
Solution | 解答: R = V/I = 12/3 = 4 Ω. P = IV = 3 × 12 = 36 W.
R = V/I = 12/3 = 4 Ω。P = IV = 3 × 12 = 36 W。
10. Magnetism and Electromagnetic Induction | 磁场与电磁感应
A moving charge experiences a magnetic force. Faraday’s law states that a changing magnetic flux induces an electromotive force.
运动电荷在磁场中受力。法拉第定律表明变化的磁通量会产生感应电动势。
F = qvBsinθ
ε = −N ΔΦ/Δt
Example | 例题: A 200-turn coil experiences a flux change of 0.05 Wb in 0.2 s. Find induced emf.
Solution | 解答: ε = N ΔΦ/Δt = 200 × 0.05 / 0.2 = 50 V.
ε = N ΔΦ/Δt = 200 × 0.05 / 0.2 = 50 V。
11. Thermodynamics and Ideal Gas | 热学与理想气体
The ideal gas law connects pressure, volume, temperature, and mole number. The first law of thermodynamics governs energy conservation in thermal processes.
理想气体状态方程联系压强、体积、温度与物质的量。热力学第一定律描述热过程中的能量守恒。
PV = nRT
ΔU = Q − W
Example | 例题: 2 mol of gas at 300 K occupies 0.05 m³. Find pressure (R = 8.31 J/(mol·K)).
Solution | 解答: P = nRT/V = 2 × 8.31 × 300 / 0.05 = 9.97 × 10⁴ Pa ≈ 1.0 × 10⁵ Pa.
P = nRT/V = 2 × 8.31 × 300 / 0.05 = 9.97 × 10⁴ Pa ≈ 1.0 × 10⁵ Pa。
12. Modern Physics Essentials | 现代物理核心公式
Photons carry energy proportional to frequency. Einstein’s mass-energy relation and the photoelectric effect equation are standard SAT2 topics.
光子能量与频率成正比。爱因斯坦质能方程和光电效应方程是SAT2的常考内容。
E = hf
E = mc²
KEₘₐₓ = hf − φ
Example | 例题: Light of frequency 6 × 10¹⁴ Hz shines on a metal with work function 1.5 eV. Find maximum kinetic energy in eV (h = 4.14 × 10⁻¹⁵ eV·s).
Solution | 解答: KEₘₐₓ = hf − φ = (4.14 × 10⁻¹⁵)(6 × 10¹⁴) − 1.5 = 2.484 − 1.5 = 0.98 eV.
KEₘₐₓ = hf − φ = (4.14 × 10⁻¹⁵)(6 × 10¹⁴) − 1.5 = 2.484 − 1.5 = 0.98 eV。
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