📚 IB Physics: Typical Problems Explained | IB 物理：典型例题详解
IB Physics assessments challenge students to apply concepts to both familiar and unfamiliar situations. This article walks through a set of carefully selected worked examples spanning mechanics, waves, electricity, magnetism, thermal physics, and modern physics. Each problem is broken down into logical steps, with clear reasoning in English followed immediately by its Chinese translation, to support bilingual learners in mastering problem-solving techniques.
IB 物理考试要求学生将概念运用到熟悉和陌生的情境中。本文精选了一组典型例题，涵盖力学、波动、电学、磁学、热物理和近代物理。每道题都分解为清晰的解题步骤，先给出英文解析，紧接着是中文翻译，帮助双语学习者掌握解题技巧。
1. Projectile Motion | 抛体运动
A ball is projected from ground level with an initial speed of 20 m·s⁻¹ at an angle of 30° to the horizontal. Assuming g = 9.81 m·s⁻² and negligible air resistance, calculate (a) the time of flight and (b) the maximum height reached.
一个球从地面以 20 m·s⁻¹ 的初速度、与水平方向成 30° 角抛出。取 g = 9.81 m·s⁻²，忽略空气阻力，计算 (a) 飞行时间，(b) 最大高度。
Step 1: Resolve the initial velocity into horizontal and vertical components. The vertical component u_y = u sinθ = 20 × sin30° = 20 × 0.5 = 10 m·s⁻¹.
步骤 1：将初速度分解为水平和竖直分量。竖直分量 u_y = u sinθ = 20 × sin30° = 20 × 0.5 = 10 m·s⁻¹。
Step 2: Time of flight is determined by vertical motion. Using the fact that vertical displacement is zero when the ball returns to the ground, we apply s_y = u_y t – ½ g t² = 0. This gives t (u_y – ½ g t) = 0, so t = 2u_y / g.
步骤 2：飞行时间由竖直运动决定。利用球落回地面时竖直位移为零，应用 s_y = u_y t – ½ g t² = 0，得到 t (u_y – ½ g t) = 0，因此 t = 2u_y / g。
t = 2 × 10 / 9.81 ≈ 2.04 s
Step 3: Maximum height occurs when vertical velocity becomes zero: v_y = u_y – g t_up = 0, so t_up = u_y / g = 10 / 9.81 ≈ 1.02 s. Then maximum height h = u_y t_up – ½ g t_up² = u_y² / (2g).
步骤 3：最大高度出现在竖直速度为零的时刻：v_y = u_y – g t_up = 0，所以 t_up = u_y / g = 10 / 9.81 ≈ 1.02 s。最大高度 h = u_y t_up – ½ g t_up² = u_y² / (2g)。
h = (10)² / (2 × 9.81) ≈ 5.10 m
2. Connected Bodies and Newton’s Laws | 连接体与牛顿定律
Two blocks are connected by a light inextensible string over a frictionless pulley. Block A of mass 5.0 kg rests on a smooth horizontal table. Block B of mass 3.0 kg hangs freely. The system is released from rest. Determine the acceleration of the system and the tension in the string.
两个物块通过一根轻质不可伸长的绳子跨过一个无摩擦滑轮连接。物块 A 质量 5.0 kg 放在光滑水平桌面上，物块 B 质量 3.0 kg 自由悬挂。系统从静止释放。求系统的加速度和绳子的张力。
Step 1: Identify forces. For block B, weight W_B = m_B g = 3.0 × 9.81 = 29.43 N downwards; tension T upwards. For block A, tension T is the only horizontal force.
步骤 1：分析受力。对物块 B，重力 W_B = m_B g = 3.0 × 9.81 = 29.43 N 向下；张力 T 向上。对物块 A，张力 T 是唯一的水平力。
Step 2: Apply Newton’s second law to each block. For A: T = m_A a. For B: m_B g – T = m_B a.
步骤 2：对每个物块应用牛顿第二定律。对 A：T = m_A a。对 B：m_B g – T = m_B a。
Step 3: Combine equations to solve for acceleration. Substitute T = m_A a into the second equation: m_B g – m_A a = m_B a → a = m_B g / (m_A + m_B).
步骤 3：联立方程求解加速度。将 T = m_A a 代入第二个方程：m_B g – m_A a = m_B a → a = m_B g / (m_A + m_B)。
a = (3.0 × 9.81) / (5.0 + 3.0) = 29.43 / 8.0 ≈ 3.68 m·s⁻²
Step 4: Find tension T = m_A a = 5.0 × 3.68 ≈ 18.4 N.
步骤 4：求张力 T = m_A a = 5.0 × 3.68 ≈ 18.4 N。
3. Circular Motion: Car on a Humpback Bridge | 圆周运动：拱桥上的汽车
A car of mass 1200 kg travels at a constant speed of 15 m·s⁻¹ over a humpback bridge with a radius of curvature 50 m at its peak. Calculate the normal reaction force exerted by the road on the car at the peak.
一辆质量为 1200 kg 的汽车以 15 m·s⁻¹ 的恒定速度驶过一座拱桥，桥顶的曲率半径为 50 m。计算桥顶处路面对汽车的法向反作用力。
Step 1: At the peak, the centripetal force is provided by the net downward force: mg – N = m v² / r. The car’s weight acts downwards, and the normal reaction N acts upwards.
步骤 1：在桥顶，向心力由向下的合力提供：mg – N = m v² / r。汽车的重力向下，法向反作用力 N 向上。
Step 2: Rearrange to find N: N = m(g – v² / r).
步骤 2：整理得 N = m(g – v² / r)。
N = 1200 × (9.81 – 15² / 50) = 1200 × (9.81 – 225/50) = 1200 × (9.81 – 4.5) = 1200 × 5.31 = 6372 N
Step 3: Interpretation: The reaction force is less than the car’s weight (mg = 11772 N) because part of the weight provides centripetal acceleration. If speed reaches √(gr) ≈ √(9.81×50) ≈ 22.1 m·s⁻¹, N becomes zero and the car loses contact.
步骤 3：解释：反作用力小于汽车的重力 (mg = 11772 N)，因为一部分重力提供了向心加速度。如果速度达到 √(gr) ≈ √(9.81×50) ≈ 22.1 m·s⁻¹，N 为零，汽车将脱离桥面。
4. Work, Energy and Springs | 功、能与弹簧
A spring of force constant k = 200 N·m⁻¹ is compressed by 0.15 m against a block of mass 0.50 kg on a frictionless horizontal surface. The spring is released, pushing the block away. Find the speed of the block just as it leaves the spring.
一个劲度系数 k = 200 N·m⁻¹ 的弹簧被压缩 0.15 m，顶住一个放在光滑水平面上的质量为 0.50 kg 的物块。弹簧释放后将物块推开。求物块刚脱离弹簧时的速度。
Step 1: The elastic potential energy stored in the spring is U = ½ k x², where x is the compression.
步骤 1：弹簧储存的弹性势能为 U = ½ k x²，其中 x 为压缩量。
Step 2: This energy is converted entirely into kinetic energy of the block: ½ k x² = ½ m v². Cancel the factor ½ and solve for v: v = x √(k / m).
步骤 2：该能量全部转化为物块的动能：½ k x² = ½ m v²。消去 ½，解出 v = x √(k / m)。
v = 0.15 × √(200 / 0.50) = 0.15 × √400 = 0.15 × 20 = 3.0 m·s⁻¹
5. Conservation of Momentum: Elastic Collision | 动量守恒：弹性碰撞
A cart of mass 2.0 kg moving at 4.0 m·s⁻¹ to the right collides head-on and elastically with a stationary cart of mass 2.0 kg on a frictionless track. Determine the velocities of both carts after the collision.
一辆质量为 2.0 kg 的小车以 4.0 m·s⁻¹ 的速度向右运动，与一辆静止在光滑轨道上的质量同样为 2.0 kg 的小车发生正碰且为弹性碰撞。求碰撞后两小车的速度。
Step 1: For an elastic head-on collision with equal masses, the moving cart stops and the stationary cart moves off with the initial velocity. This is a well-known result from conservation of momentum and kinetic energy.
步骤 1：对于等质量的正碰弹性碰撞，运动的车停止，静止的车以初速度运动。这是由动量守恒和动能守恒得出的熟知结论。
Step 2: Verify with equations: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, and for elasticity, u₁ – u₂ = v₂ – v₁. Here m₁=m₂=2.0 kg, u₁=4.0, u₂=0. Solving gives v₁=0, v₂=4.0 m·s⁻¹.
步骤 2：用方程验证：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，弹性碰撞的恢复系数为 u₁ – u₂ = v₂ – v₁。这里 m₁=m₂=2.0 kg，u₁=4.0，u₂=0。解得 v₁=0，v₂=4.0 m·s⁻¹。
6. Simple Harmonic Motion: The Simple Pendulum | 简谐运动：单摆
A simple pendulum has a length of 1.20 m and a bob of mass 0.50 kg. It is displaced by a small angle and released. Calculate the period of oscillation. (Take g = 9.81 m·s⁻²)
一个单摆的摆长为 1.20 m，摆球质量为 0.50 kg。将其拉开一个小角度后释放。计算摆动周期。（取 g = 9.81 m·s⁻²）
Step 1: For small angles (less than about 10°), the period T is independent of amplitude and mass: T = 2π √(L / g).
步骤 1：对于小角度（小于约 10°），周期 T 与振幅和质量无关：T = 2π √(L / g)。
T = 2π × √(1.20 / 9.81) = 2π × √0.1223 ≈ 2π × 0.3497 ≈ 2.20 s
Step 2: Note that if the angle exceeds about 10°, the motion is no longer simple harmonic and the period begins to depend on amplitude.
步骤 2：注意如果角度超过约 10°，运动不再为简谐运动，周期开始依赖于振幅。
7. Electric Field from Point Charges | 点电荷的电场
Two point charges, +4.0 μC and +9.0 μC, are placed 0.50 m apart in vacuum. Locate the point along the line joining them where the resultant electric field is zero.
两个点电荷，+4.0 μC 和 +9.0 μC，在真空中相距 0.50 m。求在它们连线上合电场为零的位置。
Step 1: Let the distance from the +4.0 μC charge to the point be x. Then the distance from the +9.0 μC charge is (0.50 – x). Since both charges are positive, the zero-field point lies between them where the fields oppose each other.
步骤 1：设 +4.0 μC 电荷到该点的距离为 x，则 +9.0 μC 电荷到该点的距离为 (0.50 – x)。由于两电荷均为正，场强为零的点位于它们之间，两电荷产生的电场方向相反。
Step 2: Set magnitudes equal: k × 4.0 / x² = k × 9.0 / (0.50 – x)². Cancel k and take square roots: 2 / x = 3 / (0.50 – x).
步骤 2：令场强大小相等：k × 4.0 / x² = k × 9.0 / (0.50 – x)²。消去 k 并开方：2 / x = 3 / (0.50 – x)。
Step 3: Cross-multiply: 2(0.50 – x) = 3x → 1.0 – 2x = 3x → 1.0 = 5x → x = 0.20 m.
步骤 3：交叉相乘：2(0.50 – x) = 3x → 1.0 – 2x = 3x → 1.0 = 5x → x = 0.20 m。
Thus the zero-field point is 0.20 m from the +4.0 μC charge and 0.30 m from the +9.0 μC charge.
因此合电场为零的点距离 +4.0 μC 电荷 0.20 m，距离 +9.0 μC 电荷 0.30 m。
8. DC Circuits and Kirchhoff’s Laws | 直流电路与基尔霍夫定律
In the circuit shown, a 12 V battery with negligible internal resistance is connected to three resistors: R₁ = 6 Ω in series, and a parallel combination of R₂ = 4 Ω and R₃ = 12 Ω. Find the current through each resistor. (Note: diagram described in text.)
如图所示电路中，一个内阻可忽略的 12 V 电池连接三个电阻：R₁ = 6 Ω 串联，然后并联 R₂ = 4 Ω 和 R₃ = 12 Ω。求通过每个电阻的电流。（注：电路图以文字描述。）
Step 1: Calculate the equivalent resistance of the parallel section: 1/R_par = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so R_par = 3 Ω.
步骤 1：计算并联部分的等效电阻：1/R_par = 1/4 + 1/12 = 3/12 + 1/12 = 4/12，因此 R_par = 3 Ω。
Step 2: Total circuit resistance R_tot = R₁ + R_par = 6 + 3 = 9 Ω. Main current I_total = V / R_tot = 12 / 9 = 1.33 A. This current flows through R₁.
步骤 2：总电阻 R_tot = R₁ + R_par = 6 + 3 = 9 Ω。总电流 I_total = V / R_tot = 12 / 9 = 1.33 A。此电流流过 R₁。
Step 3: Voltage across the parallel branch: V_par = I_total × R_par = 1.33 × 3 = 4.0 V. Then I₂ = V_par / R₂ = 4.0 / 4 = 1.0 A, and I₃ = V_par / R₃ = 4.0 / 12 = 0.33 A. Check: I₂ + I₃ = 1.33 A.
步骤 3：并联支路两端电压：V_par = I_total × R_par = 1.33 × 3 = 4.0 V。则 I₂ = V_par / R₂ = 4.0 / 4 = 1.0 A，I₃ = V_par / R₃ = 4.0 / 12 = 0.33 A。验证：I₂ + I₃ = 1.33 A。
9. Electromagnetic Induction: Faraday’s Law | 电磁感应：法拉第定律
A coil of 200 turns and area 0.025 m² is placed perpendicular to a uniform magnetic field of strength 0.80 T. The field is reduced uniformly to zero in 0.50 s. Calculate the magnitude of the average induced emf in the coil.
一个匝数为 200、面积为 0.025 m² 的线圈，垂直于强度为 0.80 T 的均匀磁场放置。磁场在 0.50 s 内均匀减小到零。计算线圈中平均感应电动势的大小。
Step 1: Initial magnetic flux linkage Φ_initial = N B A = 200 × 0.80 × 0.025 = 4.0 Wb. Final flux linkage is zero. Change in flux linkage ΔΦ = 0 – 4.0 = -4.0 Wb.
步骤 1：初始磁链 Φ_initial = N B A = 200 × 0.80 × 0.025 = 4.0 Wb。末磁链为零。磁链变化量 ΔΦ = 0 – 4.0 = -4.0 Wb。
Step 2: By Faraday’s law, average induced emf magnitude |ε| = |ΔΦ / Δt| = 4.0 / 0.50 = 8.0 V.
步骤 2：根据法拉第定律，平均感应电动势的大小 |ε| = |ΔΦ / Δt| = 4.0 / 0.50 = 8.0 V。
10. Photoelectric Effect | 光电效应
Ultraviolet light of frequency 1.2 × 10¹⁵ Hz illuminates a metal surface with a work function of 4.14 eV. Determine (a) the maximum kinetic energy of the emitted electrons in joules, and (b) the cutoff frequency. (h = 6.63 × 10⁻³⁴ J·s, 1 eV = 1.60 × 10⁻¹⁹ J)
频率为 1.2 × 10¹⁵ Hz 的紫外光照射在逸出功为 4.14 eV 的金属表面上。计算 (a) 发射电子的最大动能（以焦耳为单位），(b) 截止频率。（h = 6.63 × 10⁻³⁴ J·s，1 eV = 1.60 × 10⁻¹⁹ J）
Step 1: Convert work function to joules: Φ = 4.14 eV × 1.60 × 10⁻¹⁹ J/eV = 6.624 × 10⁻¹⁹ J.
步骤 1：将逸出功换算为焦耳：Φ = 4.14 eV × 1.60 × 10⁻¹⁹ J/eV = 6.624 × 10⁻¹⁹ J。
Step 2: Photon energy E_photon = h f = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵ = 7.956 × 10⁻¹⁹ J.
步骤 2：光子能量 E_photon = h f = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵ = 7.956 × 10⁻¹⁹ J。
Step 3: Maximum kinetic energy E_k max = h f – Φ = 7.956 × 10⁻¹⁹ – 6.624 × 10⁻¹⁹ = 1.332 × 10⁻¹⁹ J (about 0.833 eV).
步骤 3：最大动能 E_k max = h f – Φ = 7.956 × 10⁻¹⁹ – 6.624 × 10⁻¹⁹ = 1.332 × 10⁻¹⁹ J（约 0.833 eV）。
Step 4: Cutoff frequency f₀ = Φ / h = 6.624 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ ≈ 1.00 × 10¹⁵ Hz.
步骤 4：截止频率 f₀ = Φ / h = 6.624 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ ≈ 1.00 × 10¹⁵ Hz。
Published by TutorHao | Physics Revision Series | aleveler.com
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