AS Physics: MCQ Quick-Kill Techniques | AS物理：选择题秒杀技巧

📚 AS Physics: MCQ Quick-Kill Techniques | AS物理：选择题秒杀技巧

In AS Physics multiple-choice questions, you often have only a minute or so per question. Mastering certain ‘quick-kill’ techniques can help you eliminate wrong options instantly and boost both your speed and accuracy. These methods are built on fundamental principles, not luck, and with practice they become second nature.

在AS物理选择题中，你通常每道题只有一分钟左右的时间。掌握一些“秒杀”技巧可以帮助你立即排除错误选项，从而提高做题速度和准确率。这些方法建立在基本原理之上，而非运气，通过练习它们会成为你的第二天性。

1. Dimensional Analysis | 量纲分析

Dimensional analysis checks whether the units (or dimensions) of an expression match the quantity it is supposed to represent. If an option gives a distance in kg·m·s⁻², it is physically impossible. Always verify that both sides of an equation have the same base dimensions.

量纲分析可以检验表达式的单位（或量纲）是否与它所代表的物理量匹配。如果一个选项给出的距离单位是 kg·m·s⁻²，这在物理上是不可能的。务必验证方程两边具有相同的基本量纲。

For example, the period of a pendulum T may be given as 2π√(L/g). The dimensions of L/g are [L]/[LT⁻²] = [T²], so the square root yields [T]. Any option without dimensions of time can be eliminated. A common distractor might have √(g/L) which gives [T⁻¹], plainly wrong for a period.

例如，单摆的周期 T 可能给出 T = 2π√(L/g)。L/g 的量纲是 [L]/[LT⁻²] = [T²]，开方后得到 [T]。任何不具有时间量纲的选项都可以被排除。一个常见的干扰项可能是 √(g/L)，其量纲为 [T⁻¹]，作为周期显然是错误的。

Quick check: substitute base units into a formula. If the formula for force is claimed as F = m × a², the units would be kg × (m s⁻²)² = kg m² s⁻⁴, not N (kg m s⁻²). Instantly reject. Similarly, an option for kinetic energy as mv instead of ½mv² fails the unit test.

快速检查：将基本单位代入公式。如果说力的公式是 F = m × a²，单位将是 kg × (m s⁻²)² = kg m² s⁻⁴，而不是牛顿 N (kg m s⁻²)。立即排除。同样地，若动能选项为 mv 而非 ½mv²，也通不过单位检验。

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2. Estimation and Order of Magnitude | 估算与数量级

Many MCQs require you to estimate physical quantities. Knowing typical values (e.g., mass of a car ~1000 kg, speed of sound ~340 m s⁻¹, Earth’s radius ~6.4×10⁶ m) allows rapid sanity checks. If a result for the height of a person comes out as 10⁴ m, you know it’s wrong.

许多选择题要求你估算物理量。了解典型值（如汽车质量 ~1000 kg，声速 ~340 m s⁻¹，地球半径 ~6.4×10⁶ m）让你快速进行合理性检查。如果一个人的身高计算结果为 10⁴ m，你立刻知道是错的。

If an option suggests that the wavelength of red light is 700 m, you immediately know it’s absurd – red light is around 7×10⁻⁷ m. Quick order-of-magnitude estimates can eliminate 2 or 3 choices without detailed calculation.

如果一个选项提示红光的波长为 700 m，你立刻知道这是荒谬的——红光波长大约为 7×10⁻⁷ m。快速的数量级估算可以省去详细计算，直接排除2到3个选项。

Practice rounding numbers to 1 significant figure and using powers of ten. For instance, the acceleration due to gravity, g ≈ 10 m s⁻², simplifies many calculations in multiple-choice settings. Use π² ≈ 10 for an even faster route in pendulum problems.

练习将数字四舍五入到 1 位有效数字，并使用 10 的幂。例如，重力加速度 g ≈ 10 m s⁻² 可以简化选择题中的许多计算。在单摆问题中将 π² 视为 10 则能更快求解。

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3. Graphical Analysis & Proportionality | 图像分析与正比关系

Many AS Physics questions involve graphs; you can often deduce the relationship without full calculations. If a graph is a straight line through the origin, the two variables are directly proportional. The gradient then equals the constant of proportionality, and you can quickly match it to a physical constant.

许多AS物理题涉及图像；你通常可以在不完全计算的情况下推断出关系。如果图像是一条过原点的直线，那么这两个变量成正比。斜率就等于比例常数，你可以快速将它与某个物理常数匹配起来。

For a graph of distance vs. time² for an object starting from rest, a straight line indicates constant acceleration. The gradient is ½a. Recognizing the form y = mx + c lets you read off physical quantities instantly. If the graph plots v² against x, a straight line shows a relation of the type v² = u² + 2ax.

对于从静止开始的物体，距离-时间² 图像如果是直线，表明加速度恒定。斜率为 ½a。识别出 y = mx + c 的形式可以让你立即读出物理量。如果图像是 v² 对 x 的直线，则表明存在 v² = u² + 2ax 类型的关系。

When a graph is curved, check if squaring, rooting, or taking a reciprocal of one axis would linearise it. For example, if P ∝ 1/V, a plot of P against 1/V yields a straight line through origin. Inversely, a plot of PV vs P for a fixed amount of ideal gas gives a horizontal line.

当图像是曲线时，检查是否通过对某一轴平方、开方或取倒数可以使其直线化。例如，如果 P ∝ 1/V，则以 P 对 1/V 作图会得到一条过原点的直线。反过来，对于一定量的理想气体，PV-P 图像则是一条水平线。

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4. Special & Limiting Cases | 特殊值与极限法

Plug in extreme or special values (like 0, 90°, or infinity) to test a formula. If a formula for the period of a pendulum includes sin θ and the question says small angles, the option that diverges at θ=0 is wrong. The correct formula should give the well-known T = 2π√(L/g) when θ → 0.

代入极端或特殊值（如 0、90° 或无穷大）来检验一个公式。如果某单摆周期公式包含 sin θ，而题目说的是小角度，那么在 θ=0 时发散的选项就是错误的。正确的公式应在 θ→0 时给出众所周知的 T = 2π√(L/g)。

Consider the limit when a mass becomes very large or friction zero. In a collision problem, if one mass is infinitely heavy, the light object should bounce back with the same speed (elastic) or stick? Momentum conservation still holds. Use such limits to test answers. If you let m₂ → ∞, the final velocity of m₁ should become -u in a perfectly elastic head-on collision.

考虑质量变得非常大或摩擦力为零时的极限情况。在碰撞问题中，如果一个物体质量无限大，轻物体在完全弹性碰撞中应以与入射速率相同的速率反弹。用这种极限来检验答案。当 m₂ → ∞ 时，m₁ 的末速度应为 -u。

In projectile motion, setting the angle to 90° should give vertical motion only, and range zero. Check if the option satisfies this. Plugging θ=90° into a range formula R = (u² sin 2θ)/g gives 0, while a wrong formula might give a non-zero value.

在抛体运动中，将角度设为90°应只得到竖直运动，射程为零。检查选项是否满足这点。将 θ=90° 代入射程公式 R = (u² sin 2θ)/g 得 0，而一个错误公式可能会给出非零值。

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5. Unit Conversion Tricks | 单位换算技巧

AS Physics often has questions requiring unit conversions (e.g., cm² to m², km h⁻¹ to m s⁻¹). A fast method is to multiply by conversion factors written as fractions. 1 km h⁻¹ = (1000 m)/(3600 s) = 5/18 m s⁻¹. Memorising this factor saves precious seconds.

AS物理中经常有需要单位换算的题（如 cm² 到 m²，km h⁻¹ 到 m s⁻¹）。一个快速方法是将转换因子写成分数相乘。1 km h⁻¹ = (1000 m)/(3600 s) = 5/18 m s⁻¹。记住这个因子可以节省宝贵的时间。

For areas, remember that 1 m² = 10⁴ cm², not 100 cm². A quick way to avoid mistakes: write 1 cm = 10⁻² m, then (1 cm)² = (10⁻² m)² = 10⁻⁴ m². When converting volumes, 1 m³ = 10⁶ cm³.

对于面积，记住 1 m² = 10⁴ cm²，而不是 100 cm²。避免错误的一个快捷方法是：写出 1 cm = 10⁻² m，那么 (1 cm)² = (10⁻² m)² = 10⁻⁴ m²。在体积换算中，1 m³ = 10⁶ cm³。

When dealing with density, mass in g and volume in cm³ give density in g cm⁻³. To convert to kg m⁻³, multiply by 1000. Dimensional moves: multiply by (1 kg/1000 g) and (10⁶ cm³/1 m³). Mastering this prevents careless errors.

当处理密度时，质量用克，体积用 cm³，密度单位是 g cm⁻³。要转换为 kg m⁻³，乘以 1000。量纲转换法：乘以 (1 kg/1000 g) 和 (10⁶ cm³/1 m³)。掌握这一点可以避免粗心错误。

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6. Vector Shortcuts | 矢量捷径

Adding vectors at right angles: use Pythagoras. But if the question gives components and asks for direction, the tangent of the angle is opposite/adjacent. If the angle is 45°, the two perpendicular components must be equal. Quickly spot options where they are not.

直角矢量相加：使用毕达哥拉斯定理。但如果题目给出分量并要求方向，角度的正切是对边/邻边。若角度为45°，两个垂直分量必须相等。快速找出不相等的选项。

When resolving forces on an inclined plane, the component of weight down the slope is mg sin θ, and into the slope is mg cos θ. A common distractor swaps sin

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