Natural Sciences Admissions Assessment 2018 Section 2 Answer Guide | 2018 年剑桥自然科学入学评估第二部分答案解析

📚 Natural Sciences Admissions Assessment 2018 Section 2 Answer Guide | 2018 年剑桥自然科学入学评估第二部分答案解析

The NSAA Section 2 tests your ability to apply scientific principles across physics, chemistry, biology and mathematics. This guide provides clear, step‑by‑step answers and explanations for key questions from the 2018 paper, helping you understand the reasoning behind each correct option.

NSAA 第二部分考查你跨物理、化学、生物和数学运用科学原理的能力。这份指南针对 2018 年真题中的典型问题，提供了清晰的逐步答案与解析，帮助你理解每个正确答案背后的推理。

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1. Projectile Motion | 问题1：抛体运动

Answer: 34.6 m (approximately 35 m)

答案：34.6 米（约 35 米）

Resolve the initial velocity. The horizontal component uₓ = u cos θ = 20 cos 30° = 20 × (√3) ≈ 17.32 m s⁻¹. The vertical component uᵧ = u sin θ = 20 sin 30° = 20 × 0.5 = 10.0 m s⁻¹.

步骤1：分解初速度。水平分量 uₓ = u cos θ = 20 cos 30° = 20 × (√3/2) ≈ 17.32 m s⁻¹。垂直分量 uᵧ = u sin θ = 20 sin 30° = 10.0 m s⁻¹。

is gives t (uᵧ − ½ g t) = 0, hence total time t = (2 u sin θ) / g = (2 × 10) / 10 = 2.0 s.

步骤2：计算飞行时间。垂直于运动方程 s = uᵧ t − ½ g t²，落回地面时 s = 0，得 t (uᵧ − ½ g t) = 0，因此总时间 t = (2 u sin θ) / g = 20 / 10 = 2.0 s。

tion is constant: R = uₓ × t = 17.32 × 2.0 ≈ 34.6 m. You can also use the derived formula:

步骤3：计算水平射程。水平匀速运动：R = uₓ × t = 17.32 × 2.0 ≈ 34.6 m。也可以使用推导公式：

R = (u² sin 2θ) / g

Substituting u = 20 m s⁻¹, θ = 30° (sin 60° = √3/2), g = 10 m s⁻² yields R = (400 × √3/2) / 10 ≈ 34.6 m.

代入 u = 20 m s⁻¹，θ = 30°（sin 60° = √3/2），g = 10 m s⁻²，得 R = (400 × √3/2) / 10 ≈ 34.6 m。

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2. Electrical Circuits | 问题2：电路

Answer: 1.2 A

答案：1.2 A

The circuit consists of a 12 V battery connected to three resistors in series: 2 Ω, 3 Ω and 5 Ω. The total resistance in series is the sum: R_total = 2 + 3 + 5 = 10 Ω.

电路由 12 V 电池和三个串联电阻组成：2 Ω、3 Ω 和 5 Ω。串联总电阻为各个电阻之和：R_total = 2 + 3 + 5 = 10 Ω。

Apply Ohm's Law: V = I R, so I = V / R_total = 12 V / 10 Ω = 1.2 A. The same 1.2 A flows through each resistor.

应用欧姆定律：V = I R，所以 I = V / R_total = 12 V / 10 Ω = 1.2 A。每个电阻中流过的电流均为 1.2 A。

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3. Stoichiometry and the Mole | 问题3：化学计量与摩尔

Answer: 4.40 g of CO₂

答案：4.40 g CO₂

+ 12.0 + (16.0 × 3) = 100.1 g mol⁻¹, typically rounded to 100 g mol⁻¹ for the calculation.

热分解反应为：CaCO₃ → CaO + CO₂。CaCO₃ 的摩尔质量 = 100 g mol⁻¹（计算时常用取整值）。

0.100 mol. The balanced equation shows a 1:1 mole ratio, so 0.100 mol of CO₂ is produced.

CaCO₃ 的物质的量 = 10.0 g / 100 g mol⁻¹ = 0.100 mol。配平方程显示 1:1 的计量比，因此生成 0.100 mol CO₂。

Mass of CO₂ = moles × molar mass of CO₂ = 0.100 mol × 44.0 g mol⁻¹ = 4.40 g.

CO₂ 的质量 = 0.100 mol × 44.0 g mol⁻¹ = 4.40 g。

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4. Equilibrium Constant Expression | 问题4：平衡常数表达式

Answer: Kc = [NH₃]² / ([N₂][H₂]³); units = mol⁻² dm⁶

答案：Kc = [NH₃]² / ([N₂][H₂]³)；单位为 mol⁻² dm⁶

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant Kc is written as the ratio of product and reactant concentrations raised to their stoichiometric coefficients.

对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，平衡常数 Kc 表示为生成物浓度与反应物浓度以其化学计量系数为幂次的比值。

Kc = [NH₃]² / ([N₂][H₂]³)

dm⁻³): (mol dm⁻³)² / [ (mol dm⁻³) × (mol dm⁻³)³ ] = mol² dm⁻⁶ / (mol⁴ dm⁻¹²) = mol⁻² dm⁶.

求单位时，代入浓度单位 mol dm⁻³：(mol dm⁻³)² / [ (mol dm⁻³) × (mol dm⁻³)³ ] = mol⁻² dm⁶。

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5. Enzyme Kinetics (Michaelis–Menten) | 问题5：酶动力学（米氏方程）

Answer: V = Vmax / 2

答案：V = Vmax / 2

The Michaelis– + [S]). When the substrate concentration [S] equals the Michaelis constant Kₘ, the equation becomes:

米氏方程为 V = (Vmax [S]) / (Kₘ + [S])。当底物浓度 [S] 等于米氏常数 Kₘ 时：

V = (Vmax × Kₘ) / (Kₘ + Kₘ) = (Vmax Kₘ) / (2 Kₘ) = Vmax / 2.

V = (Vmax × Kₘ) / (Kₘ + Kₘ) = (Vmax Kₘ) / (2 Kₘ) = Vmax / 2。

tly half of the maximum rate when [S] = Kₘ, which is a key diagnostic feature of Michaelis–Menten kinetics.

因此，当 [S] = Kₘ 时，反应速率恰好为最大速率的一半，这是米氏动力学的一个关键诊断特征。

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6. Genetics: ABO Blood Group Probability | 问题6：遗传学：ABO 血型概率

Answer: 1/4 (25%)

答案：1/4（25%）

Parental genotypes are IAi (blood group A) and IBi (blood group B). Each parent can produce gametes carrying either the dominant allele or the recessive i allele with equal probability.

亲本基因型为 IAi（A 型）和 IBi（B 型）。每个亲本均能产生携带显性等位基因或隐性 i 的配子，且概率相等。

The Punnett square illustrates the offspring genotypes:

庞纳特方格展示后代基因型：

	IA	i
IB	IAIB	IBi
i	IAi	ii

Only the ii genotype gives blood group O. There is one ii square out of four equally likely combinations, so the probability is 1/4.

只有 ii 基因型表现为 O 型血。四个等可能组合中出现一个 ii，因此概率为 1/4。

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7. Ideal Gas Law Calculation | 问题7：理想气体计算

Answer: n ≈ 4.0 mol

答案：n ≈ 4.0 mol

Use the ideal gas equation PV = nRT. Rearrange to n = PV / (RT).

使用理想气体状态方程 PV = nRT，变形得 n = PV / (RT)。

Given P = 2.0 × 10⁵ Pa, V = 0.050 m³, T = 300 K, and R = 8.31 J mol⁻¹ K⁻¹:

已知 P = 2.0 × 10⁵ Pa，V = 0.050 m³，T = 300 K，R = 8.31 J mol⁻¹ K⁻¹：

¹ × 300 K) = 1.0 × 10⁴ / 2493 ≈ 4.01 mol. Rounded to two significant figures, n = 4.0 mol.

n = (2.0 × 10⁵ Pa × 0.050 m³) / (8.31 J mol⁻¹ K⁻¹ × 300 K) = 1.0 × 10⁴ / 2493 ≈ 4.01 mol。保留两位有效数字为 4.0 mol。

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8. Organic Reaction: Electrophilic Addition | 问题8：有机反应：亲电加成

Answer: 1,2‑dibromoethane; addition reaction

答案：1,2‑二溴乙烷；加成反应

Ethene (CH₂=CH₂) reacts with bromine (Br₂) at room temperature to give 1,2‑ sm is electrophilic addition, where the polarisable Br₂ molecule acts as the electrophile.

乙烯（CH₂=CH₂）与溴在室温下反应生成 1,2‑二溴乙烷。反应机理为亲电加成，可极化的 Br₂ 分子作为亲电试剂。

The π‑ mine atom, forming a cyclic bromonium ion intermediate. The released Br⁻ then attacks from the opposite side, giving anti‑addition and the product BrCH₂CH₂Br.

C=C 双键的 π 电子进攻一个溴原子，形成环状溴鎓离子中间体。释出的 Br⁻ 从背面进攻，完成反式加成，得到产物 BrCH₂CH₂Br。

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9. Photosynthesis: Light‑Dependent Reactions | 问题9：光合作用：光反应

Answer: ATP, NADPH and O₂

答案：ATP、NADPH 和 O₂

The light‑dependent reactions occur in the thylakoid membranes. Light energy splits water (photolysis), releasing O₂ as a by‑product. The energy is used to generate ATP via photophosphorylation and to reduce NADP⁺ to NADPH.

光反应发生在类囊体膜上。光能分解水（光解），释放 O₂ 作为副产物。能量用于通过光合磷酸化生成 ATP，并将 NADP⁺ 还原为 NADPH。

Therefore, the three key products are ATP, NADPH and O₂. These are essential for the Calvin cycle, which fixes CO₂ into sugars.

因此，三个关键产物是 ATP、NADPH 和 O₂。它们是卡尔文循环固定 CO₂ 合成糖类所必需的原料。

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10. Differentiation: Rate of Change | 问题10：微分：变化率

Answer: 14

答案：14

+ 2x, differentiate term by term using the power rule: dy/dx = 3 × 2x¹ + 2 × 1x⁰ = 6x + 2.

对 y = 3x² + 2x 逐项使用幂法则求导：dy/dx = 3 × 2x¹ + 2 × 1x⁰ = 6x + 2。

14. This represents the instantaneous rate of change of y with respect to x at that point.

代入 x = 2：dy/dx = 6(2) + 2 = 12 + 2 = 14。该值代表 y 关于 x 在该点的瞬时变化率。

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