📚 Worked Examples for IB and OCR Physics | IB OCR 物理：典型例题详解
This article presents carefully selected worked examples covering core topics in IB Physics (both SL and HL) and OCR A Level Physics. Each problem is solved step by step to reinforce conceptual understanding and problem-solving techniques.
本文精选了涵盖IB物理（SL和HL）及OCR A Level物理核心主题的典型例题。每个问题都按步骤求解，以强化概念理解和解题技巧。
1. Kinematics – Uniformly Accelerated Motion | 运动学 – 匀加速直线运动
A cyclist starts from rest and accelerates uniformly at 1.5 m s⁻² for 8.0 s. Calculate the final velocity and the distance covered.
一名自行车手从静止开始，以1.5 m s⁻² 的加速度匀加速运动8.0 s。计算末速度和所经过的距离。
v = u + at = 0 + 1.5 × 8.0 = 12 m s⁻¹
v = u + at = 0 + 1.5 × 8.0 = 12 m s⁻¹
s = ut + ½at² = 0 + ½ × 1.5 × (8.0)² = 48 m
s = ut + ½at² = 0 + ½ × 1.5 × (8.0)² = 48 m
2. Forces – Newton’s Second Law on an Incline | 力 – 斜面上的牛顿第二定律
A 5.0 kg block slides down a smooth incline inclined at 30° to the horizontal. Calculate the acceleration of the block. (g = 9.8 m s⁻²)
一个5.0 kg的物块沿倾角为30°的光滑斜面下滑。计算物块的加速度。(g = 9.8 m s⁻²)
Component of weight along incline: F = mg sin θ = 5.0 × 9.8 × sin 30° = 24.5 N. Since surface is smooth, net force = 24.5 N.
重力沿斜面的分量：F = mg sin θ = 5.0 × 9.8 × sin 30° = 24.5 N。由于斜面光滑，合力 = 24.5 N。
a = F/m = 24.5 / 5.0 = 4.9 m s⁻².
a = F/m = 24.5 / 5.0 = 4.9 m s⁻²。
3. Circular Motion – Centripetal Force | 圆周运动 – 向心力
A 0.20 kg mass is swung in a horizontal circle of radius 0.50 m at a constant speed of 2.0 m s⁻¹. Find the centripetal acceleration and the tension in the string.
一个0.20 kg的物体在水平面上以半径0.50 m做匀速圆周运动，线速度为2.0 m s⁻¹。求向心加速度和绳的张力。
Centripetal acceleration a = v²/r = (2.0)²/0.50 = 8.0 m s⁻².
向心加速度 a = v²/r = (2.0)²/0.50 = 8.0 m s⁻²。
Tension provides centripetal force: T = m a = 0.20 × 8.0 = 1.6 N.
张力提供向心力：T = m a = 0.20 × 8.0 = 1.6 N。
4. Work, Energy and Power – Conservation of Energy | 功、能量与功率 – 能量守恒
A 0.50 kg pendulum bob is released from rest at a height of 0.20 m above its lowest point. Neglecting air resistance, find its speed at the lowest point. (g = 9.8 m s⁻²)
一个0.50 kg的摆球从最低点上方0.20 m处静止释放。忽略空气阻力，求它在最低点时的速度。(g = 9.8 m s⁻²)
Gravitational potential energy lost = mgh = 0.50 × 9.8 × 0.20 = 0.98 J.
减少的重力势能 = mgh = 0.50 × 9.8 × 0.20 = 0.98 J。
Kinetic energy gained = ½mv² → ½ × 0.50 × v² = 0.98 → v² = 3.92 → v = 1.98 ≈ 2.0 m s⁻¹.
获得的动能 = ½mv² → ½ × 0.50 × v² = 0.98 → v² = 3.92 → v = 1.98 ≈ 2.0 m s⁻¹。
5. Momentum and Impulse – Collisions | 动量与冲量 – 碰撞
A 0.40 kg ball moving at 3.0 m s⁻¹ collides head-on with a stationary 0.60 kg ball. After an elastic collision, the first ball rebounds at 0.60 m s⁻¹. Determine the velocity of the second ball.
一个0.40 kg的球以3.0 m s⁻¹的速度与一个静止的0.60 kg的球发生正碰。若为弹性碰撞，且第一个球以0.60 m s⁻¹反弹，求第二个球的速度。
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ → 0.40×3.0 + 0 = 0.40×(-0.60) + 0.60×v₂.
动量守恒：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ → 0.40×3.0 + 0 = 0.40×(-0.60) + 0.60×v₂。
1.2 = -0.24 + 0.60 v₂ → 0.60 v₂ = 1.44 → v₂ = 2.4 m s⁻¹ (in the original direction of the first ball).
1.2 = -0.24 + 0.60 v₂ → 0.60 v₂ = 1.44 → v₂ = 2.4 m s⁻¹（方向与第一个球原方向相同）。
6. Electric Fields – Point Charges | 电场 – 点电荷
Two point charges, +4.0 μC and -2.0 μC, are placed 0.30 m apart in a vacuum. Calculate the magnitude and direction of the electric field at the midpoint between them. (k = 8.99×10⁹ N m² C⁻²)
两个点电荷，+4.0 μC 和 -2.0 μC，在真空中相距0.30 m。计算它们连线中点处的电场大小和方向。(k = 8.99×10⁹ N m² C⁻²)
Distance from each charge to midpoint: r = 0.15 m. E from +4.0 μC: E₁ = k|Q|/r² = (8.99×10⁹)×(4.0×10⁻⁶)/(0.15)² = 1.598×10⁶ N C⁻¹, directed away from the positive charge.
每个电荷到中点的距离：r = 0.15 m。+4.0 μC产生的场强：E₁ = k|Q|/r² = (8.99×10⁹)×(4.0×10⁻⁶)/(0.15)² = 1.598×10⁶ N C⁻¹，方向背离正电荷。
E from -2.0 μC: E₂ = k|Q|/r² = (8.99×10⁹)×(2.0×10⁻⁶)/(0.15)² = 0.799×10⁶ N C⁻¹, directed towards the negative charge. Both fields point in the same direction (towards the negative charge).
-2.0 μC产生的场强：E₂ = k|Q|/r² = (8.99×10⁹)×(2.0×10⁻⁶)/(0.15)² = 0.799×10⁶ N C⁻¹，方向指向负电荷。两电场同向（都指向负电荷）。
Total E = E₁ + E₂ = 2.397×10⁶ N C⁻¹ ≈ 2.40×10⁶ N C⁻¹, towards the negative charge.
总场强 E = E₁ + E₂ = 2.397×10⁶ N C⁻¹ ≈ 2.40×10⁶ N C⁻¹，方向指向负电荷。
7. Circuits – Internal Resistance and EMF | 电路 – 内阻和电动势
A battery has an emf of 1.50 V. When connected to a 4.0 Ω resistor, the terminal voltage is 1.20 V. Determine the internal resistance of the battery and the current in the circuit.
一个电池的电动势为1.50 V。当它连接到一个4.0 Ω的电阻器时，端电压为1.20 V。求电池的内阻和电路中的电流。
Current: I = V_R / R = 1.20 / 4.0 = 0.30 A.
电流：I = V_R / R = 1.20 / 4.0 = 0.30 A。
Lost volts = emf – terminal p.d. = 1.50 – 1.20 = 0.30 V. Internal resistance r = lost volts / I = 0.30 / 0.30 = 1.0 Ω.
损失的电压 = 电动势 – 端电压 = 1.50 – 1.20 = 0.30 V。内阻 r = 损失的电压 / I = 0.30 / 0.30 = 1.0 Ω。
8. Waves – Double-Slit Interference | 波 –
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