IB Physics Worked Examples Explained in Detail | IB 物理典型例题详解

📚 IB Physics Worked Examples Explained in Detail | IB 物理典型例题详解

Mastering IB Physics requires a deep understanding of core concepts and the ability to apply them to solve typical exam-style problems. This article presents a selection of worked examples drawn from key topics, following the rigorous problem-solving style found in IB and Edexcel-style examinations. Each example is solved step-by-step with clear explanations in both English and Chinese, helping you build confidence and strengthen your exam technique.

掌握 IB 物理需要深刻理解核心概念，并能熟练应用于典型考题。本文精选了来自核心知识点的例题，风格接近 IB 与 Edexcel 考试题目。每个例题都配有中英双语逐步解析，帮助巩固知识、提升应试技巧。

1. Projectile Motion | 抛体运动

A ball is kicked from the ground with an initial speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Assuming air resistance is negligible, calculate (a) the maximum height reached, (b) the time of flight, and (c) the horizontal range. Take g = 9.8 m s⁻².

一足球以 20 m s⁻¹ 的初速度与水平方向成 30° 角踢出。忽略空气阻力，计算 (a) 最大高度, (b) 飞行时间, (c) 水平射程。取 g = 9.8 m s⁻²。

First, resolve the initial velocity into horizontal and vertical components.

u_x = u cosθ = 20 cos30° = 17.3 m s⁻¹

u_y = u sinθ = 20 sin30° = 10 m s⁻¹

首先将初速度分解为水平和竖直分量。水平分量 u_x = 20 cos30° = 17.3 m s⁻¹；竖直分量 u_y = 20 sin30° = 10 m s⁻¹。

(a) At maximum height, the vertical velocity becomes zero. Use v_y² = u_y² – 2gh, setting v_y = 0:

0 = (10)² – 2 × 9.8 × h → h = 100 / 19.6 = 5.10 m

(a) 在最大高度时，竖直速度为零。利用 v_y² = u_y² – 2gh，令 v_y = 0，解得 h = 100 / 19.6 = 5.10 m。

(b) The time of flight is found from the vertical motion. With s_y = 0 when the ball lands, s_y = u_y t – ½ g t²:

0 = 10 t – 4.9 t² → t(10 – 4.9 t) = 0

Ignoring the t = 0 solution, t = 10 / 4.9 ≈ 2.04 s.

(b) 飞行时间由竖直运动求得。落回地面时 s_y = 0，由 s_y = u_y t – ½ g t² 得 0 = 10 t – 4.9 t²。舍去 t = 0，解得 t = 10 / 4.9 ≈ 2.04 s。

R = 17.3 × 2.04 ≈ 35.3 m

2. Newton’s Laws and the Inclined Plane | 牛顿定律与斜面

A 5.0 kg box is pulled at constant speed up a rough incline of 30° by a force F parallel to the slope. The coefficient of kinetic friction between the box and the incline is 0.30. Calculate the magnitude of F. Take g = 9.8 m s⁻².

一个 5.0 kg 的箱子在沿斜面方向的拉力 F 作用下，沿粗糙的 30° 斜面匀速上升。箱子与斜面间的动摩擦系数为 0.30。求 F 的大小。取 g = 9.8 m s⁻²。

Since the box moves at constant speed, the net force along the incline is zero. The forces acting are: component of weight down the slope mg sinθ, kinetic friction f_k opposite to motion, and the applied force F up the slope.

箱子匀速运动，故沿斜面方向的合外力为零。受力分析：重力沿斜面分量 mg sinθ 向下，动摩擦力 f_k 沿斜面向下，拉力 F 沿斜面向上。

The normal reaction N = mg cosθ. Hence the kinetic friction f_k = μ_k N = μ_k mg cosθ.

支持力 N = mg cosθ，因此动摩擦力 f_k = μ_k N = μ_k mg cosθ。

Applying Newton’s first law along the slope:

F – mg sinθ – f_k = 0 → F = mg (sinθ + μ_k cosθ)

Substituting the values:

F = 5.0 × 9.8 × (sin30° + 0.30 × cos30°) = 49 × (0.5 + 0.30 × 0.866) = 49 × (0.5 + 0.260) = 49 × 0.760 ≈ 37.2 N.

沿斜面应用牛顿第一定律：F – mg sinθ – f_k = 0，所以 F = mg (sinθ + μ_k cosθ)。代入数值：F = 5.0×9.8×(0.5+0.30×0.866)=49×0.760≈37.2 N。

3. Energy Conservation in Vertical Circular Motion | 竖直圆周运动中的能量守恒

A 0.50 kg stone is tied to a 1.0 m string and whirled in a vertical circle. At the lowest point of the circle the stone has a speed of 8.0 m s⁻¹. Using energy conservation, find the speed of the stone at the highest point and the tension in the string at that point. Take g = 9.8 m s⁻².

一个 0.50 kg 的小石块系在 1.0 m 长的绳子一端，在竖直平面内做圆周运动。在最低点石块的速率为 8.0 m s⁻¹。利用能量守恒求最高点石块的速度以及该处绳的张力。取 g = 9.8 m s⁻²。

Take the lowest point as the reference level for gravitational potential energy. At the bottom, the energy is purely kinetic: E_bottom = ½ m v_bottom². At the top, the stone has risen by a height of 2L = 2.0 m, so its total energy is ½ m v_top² + mg(2L).

取最低点为零重力势能参考面。在最低点，能量全部为动能：E_bottom = ½ m v_bottom²。在最高点，石块上升了 2L = 2.0 m，总能量为 ½ m v_top² + mg(2L)。

By conservation of mechanical energy:

½ m v_bottom² = ½ m v_top² + 2mgL

Cancel m and solve for v_top²: v_top² = v_bottom² – 4gL = 64 – 4×9.8×1.0 = 64 – 39.2 = 24.8. Thus v_top = √24.8 ≈ 4.98 m s⁻¹.

由机械能守恒：½ m v_bottom² = ½ m v_top² + 2mgL。消去 m 解得 v_top² = 64 – 39.2 = 24.8，v_top ≈ 4.98 m s⁻¹。

At the top of the circle, the net force toward the centre provides the centripetal force: T

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IB Physics Worked Examples Explained in Detail | IB 物理典型例题详解