📚 Calculation Problems Practice for GCSE CIE Physics | GCSE CIE 物理：计算题专项训练
Calculations make up a significant portion of the GCSE CIE Physics exam. Students must be able to recall formulae, substitute values correctly, and present final answers with proper units. This guide provides targeted practice for all key topics, with step-by-step solutions showing how to approach numerical problems efficiently.
计算题在 GCSE CIE 物理考试中占很大比重。学生必须能够回忆公式、正确代入数值并给出带单位的最终答案。本指南针对所有重要主题提供专项训练，并配有逐步求解过程，展示如何高效处理数值题。
1. Motion and Kinematics | 运动学计算
the SUVAT equations. The basic relations are: average speed v = s / t, acceleration a = (v – u) / t. The full set is v = u + a t, s = u t + ½ a t², s = v t – and time t. Pick the equation that contains the known quantities and the unknown you need.
匀加速运动由 SUVAT 方程描述。基本关系为：平均速度 v = s / t，加速度 a = (v – u) / t。全套方程为 v = u + a t，s = u t + ½ a t²，s = v t – ½ a t²，s = ½ (u + v) t 以及 v² = u² + 2 a s。请始终确定初速度 u、末速度 v、加速度 a、位移 s 和时间 t。选择同时包含已知量和所求未知量的方程。
Example: A cyclist accelerates uniformly from rest to 12 m/s in 6.0 s. Calculate the acceleration and the distance travelled during this time.
例题：一名骑自行车的人从静止匀加速到 12 m/s，用时 6.0 s。计算加速度和这段时间内的行驶距离。
Solution: (a) u = 0, v = 12 m/s, t = 6 s. a = (v – u) / t = (12 – 0) / 6 = 2.0 m/s².
解答：(a) u = 0，v = 12 m/s，t = 6 s。a = (v – u) / t = (12 – 0) / 6 = 2.0 m/s²。
0 × (6)² = 0 + 0.5 × 2 × 36 = 36 m. Alternatively, v² = u² + 2 a s → 144 = 4 s → s = 36 m.
(b) 使用 s = u t + ½ a t²：s = 0 × 6 + 0.5 × 2.0 × (6)² = 36 m。也可用 v² = u² + 2 a s → 144 = 4 s → s = 36 m。
2. Forces, Mass and Weight | 力、质量和重量
Newton' is often taken as 10 m/s², but sometimes 9.8 m/s²; use the value given in the question. Free-body diagrams help resolve forces and find net force.
牛顿第二定律给出合力 F = m a，其中 m 是质量，a 是加速度。重量是作用在物体上的重力：W = m g。在 CIE 考试中，g 通常取 10 m/s²，但有时是 9.8 m/s²；请使用题目给定的数值。受力示意图有助于分解力并求出净力。
e acceleration. If a friction force of 4.0 N acts on the box, what is the new acceleration? Determine the weight of the box (g = 9.8 m/s²).
例题：一个 5.0 kg 的箱子在光滑水平面上受到 20 N 的水平拉力。求加速度。如果箱子受到 4.0 N 的摩擦力，新的加速度是多少？计算箱子的重量（g = 9.8 m/s²）。
Solution: Without friction, resultant force = 20 N, so a = F / m = 20 / 5.0 = 4.0 m/s².
解答：无摩擦时，合力 = 20 N，故 a = F / m = 20 / 5.0 = 4.0 m/s²。
With friction, resultant force = 20 N – 4.0 N = 16 N, a = 16 / 5.0 = 3.2 m/s².
有摩擦时，合力 = 20 N – 4.0 N = 16 N，a = 16 / 5.0 = 3.2 m/s²。
Weight: W = m g = 5.0 × 9.8 = 49 N.
重量：W = m g = 5.0 × 9.8 = 49 N。
3. Momentum and Impulse | 动量和冲量
Momentum p = m v. Impulse = resultant force × time = F t = Δ p = m v – external resultant force acts. Remember momentum is a vector; define a positive direction.
动量 p = m v。冲量 = 合力 × 时间 = F t = Δ p = m v – m u。在没有外合力作用时，碰撞和爆炸中动量守恒。记住动量是矢量，请定义正方向。
Example: A 0.50 kg ball moves at 4.0 m/s towards a wall, hits it and rebounds at 3.0 m/s. The contact time is 0.20 s. Calculate the change in momentum of the ball and the average force exerted by the wall on the ball.
例题：一个 0.50 kg 的球以 4.0 m/s 的速度向墙运动，撞击后以 3.0 m/s 的速度反弹。接触时间为 0.20 s。计算球的动量变化量以及墙对球的平均作用力。
s wall as positive. Initial momentum pᵢ = m u = 0.50 × 4.0 = 2.0 kg m/s. Final velocity is –3.0 m/s, so final momentum pf = m v = 0.50 × (–3.0) = –1.5 kg m/s. Change in momentum Δp = pf – pᵢ = –1.5 – 2.0 = –3.5 kg m/s. The magnitude of change is 3.5 kg m/s.
解答：取向墙的方向为正。初动量 pᵢ = m u = 0.50 × 4.0 = 2.0 kg m/s。末速度为 –3.0 m/s，因此末动量 pf = 0.50 × (–3.0) = –1.5 kg m/s。动量变化量 Δp = pf – pᵢ = –1.5 – 2.0 = –3.5 kg m/s。变化量的大小为 3.5 kg m/s。
Average force: F = Δp / t = 3.5 / 0.20 = 17.5 N (direction opposite to initial motion).
平均力：F = Δp / t = 3.5 / 0.20 = 17.5 N（方向与初运动相反）。
4. Energy, Work and Power | 能量、功和功率
= energy transferred / time. Efficiency = (useful power output / total power input) × 100% or (useful energy output / total energy input) × 100%.
动能 KE = ½ m v²，重力势能 GPE = m g h。做功 = 力 × 沿力方向移动的距离。功率 P = 做功 / 时间 = 能量转移 / 时间。效率 = (有用输出功率 / 总输入功率) × 100% 或 (有用输出能量 / 总输入能量) × 100%。
m/s². Calculate the gain in GPE, the useful power output, and the efficiency of the crane.
例题：起重机在 10 s 内将 200 kg 的重物提升 15 m。电动机输入功率为 3500 W。取 g = 10 m/s²。计算增加的重力势能、有用输出功率和起重机的效率。
Solution: Gain in GPE = m g h = 200 × 10 × 15 = 30 000 J (30 kJ).
解答：增加的重力势能 = m g h = 200 × 10 × 15 = 30 000 J（30 kJ）。
Useful power output = work done / time = 30 000 J / 10 s = 3000 W.
有用输出功率 = 做功 / 时间 = 30 000 J / 10 s = 3000 W。
Efficiency = (useful power output / input power) × 100% = (3000 / 3500) × 100% ≈ 85.7%.
效率 = (有用输出功率 / 输入功率) × 100% = (3000 / 3500) × 100% ≈ 85.7%。
5. Density and Pressure | 密度和压强
pth of the liquid. This assumes the liquid is incompressible and the gravitational field is uniform.
密度 ρ = 质量 / 体积，ρ = m / V（单位：kg/m³）。压强 p = 力 / 面积，p = F / A（单位：Pa 或 N/m²）。液柱压强：p = ρ g h，其中 h 是液体深度。该公式假设液体不可压缩且重力场均匀。
its density and the pressure it exerts on a table when standing on one face (g = 10 N/kg).
例题：一个边长为 0.10 m 的立方体质量为 0.80 kg。计算其密度以及当它以一个面立在桌上时对桌面的压强（g = 10 N/kg）。
Solution: Volume = (0.10)³ = 0.0010 m³. Density ρ = m / V = 0.80 / 0.0010 = 800 kg/m³.
解答：体积 = (0.10)³ = 0.0010 m³。密度 ρ = 0.80 / 0.0010 = 800 kg/m³。
10 = 8.0 N. Contact area = (0.10)² = 0.010 m². Pressure p = F / A = 8.0 / 0.010 = 800 Pa.
重量 = 0.80 × 10 = 8.0 N。接触面积 = (0.10)² = 0.010 m²。压强 p = 8.0 / 0.010 = 800 Pa。
Additional example: What is the pressure at the bottom of a 5.0 m deep fresh water lake? (ρ_water = 1000 kg/m³, g = 10 m/s²) p = ρ g h = 1000 × 10 × 5.0 = 50 000 Pa (50 kPa).
附加例题：一个 5.0 m 深的淡水湖湖底的压强是多少？（ρ_水 = 1000 kg/m³，g = 10 m/s²）p = ρ g h = 1000 × 10 × 5.0 = 50 000 Pa（50 kPa）。
6. Waves: Speed, Frequency and Wavelength | 波：速度、频率和波长
/ f. You can combine these to solve for any unknown when two quantities are given. Remember that frequency remains constant when a wave passes from one medium to another.
波动方程为 v = f λ，其中 v 是波速（m/s），f 是频率（Hz），λ 是波长（m）。周期 T = 1 / f。当已知两个量时，可联立求解未知量。记住波从一种介质进入另一种介质时频率保持不变。
Example: A water wave has a frequency of 5.0 Hz and a wavelength of 0.30 m. Calculate its speed and its period.
例题：一个水波的频率为 5.0 Hz，波长为 0.30 m。计算其波速和周期。
Solution: v = f λ = 5.0 × 0.30 = 1.5 m/s.
解答：v = f λ = 5.0 × 0.30 = 1.5 m/s。
Period T = 1 / f = 1 /
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