📚 Year 12 Cambridge Physics Mock Unit Test Walkthrough | 剑桥物理12年级单元模拟卷精讲
This walkthrough dissects a representative Year 12 Cambridge Physics unit test, offering clear, step-by-step solutions to common question types. Each section pairs a worked problem with the essential physics principles, helping you reinforce both your problem-solving technique and conceptual understanding. Topics span kinematics, Newton’s laws, energy, momentum, circular motion, waves, superposition, electric fields, DC circuits, and particle physics.
本精讲深入解析一份典型的剑桥物理12年级单元模拟卷，以清晰的逐步解法展示常见题型。每个部分都将一道典型题目与核心物理原理相结合，帮助你同时强化解题技巧与概念理解。内容涵盖运动学、牛顿定律、能量、动量、圆周运动、波、叠加、电场、直流电路和粒子物理。
1. Kinematics: Projectile Range Calculation | 运动学：抛体射程计算
t an angle of 35° to the horizontal. Calculate the horizontal range. (Take g = 9.8 m s⁻²).
一个足球以初速度22 m s⁻¹、与水平面成35°角从地面踢出。计算水平射程。（取g = 9.8 m s⁻²）。
θ = 22 cos35° ≈ 18.0 m s⁻¹, and the vertical component is u sinθ = 22 sin35° ≈ 12.6 m s⁻¹.
将初速度分解为水平和竖直分量。水平分量为 u cosθ = 22 cos35° ≈ 18.0 m s⁻¹，竖直分量为 u sinθ = 22 sin35° ≈ 12.6 m s⁻¹。
ment in the vertical direction is zero when the ball returns to the ground: 0 = (u sinθ) t – ½ g t². Solving for t gives t = 2u sinθ / g = (2 × 12.6) / 9.8 ≈ 2.57 s.
利用竖直运动求飞行时间。足球回到地面时竖直位移为零：0 = (u sinθ) t – ½ g t²。解得 t = 2u sinθ / g = (2 × 12.6) / 9.8 ≈ 2.57 s。
= 18.0 × 2.57 ≈ 46.3 m. This result can also be obtained directly using the range formula.
水平射程 R 等于水平速度乘以飞行时间：R = (u cosθ) × t = 18.0 × 2.57 ≈ 46.3 m。该结果也可直接用射程公式求得。
R = u² sin(2θ) / g
Remember that air resistance is neglected, so the actual range would be shorter. The key steps are always resolving the velocity and treating the two perpendicular components independently.
注意此处忽略了空气阻力，因此实际射程会更短。解题关键始终是分解速度并独立处理两个垂直分量。
2. Newton’s Laws: Connected Bodies on a Frictionless Plane | 牛顿定律：光滑平面上连接体的动力学
Two blocks of masses 3.0 kg and 2.0 kg are joined by a light, inextensible string that passes over a smooth pulley. The 3.0 kg block rests on a frictionless horizontal table, and the 2.0 kg block hangs vertically. Determine the acceleration of the system and the tension in the string.
两个质量分别为3.0 kg和2.0 kg的木块由一根轻质不可伸长的绳子连接，绳子跨过一个光滑滑轮。3.0 kg的木块放在光滑水平桌面上，2.0 kg的木块竖直悬挂。求系统的加速度及绳子的张力。
tension T: T = 3.0a. For the hanging 2.0 kg block, the forces are weight and tension: 2.0g – T = 2.0a.
分别对每个质量应用牛顿第二定律。对于桌面上的3.0 kg木块，唯一的水平力是张力T：T = 3.0a。对于悬挂的2.0 kg木块，力有重力和张力：2.0g – T = 2.0a。
g / 5.0 = (2.0 × 9.8) / 5.0 ≈ 3.92 m s⁻². Substitute back to find T = 3.0 × 3.92 ≈ 11.8 N.
将两式相加消去T：2.0g = 5.0a，得 a = 2.0g / 5.0 = (2.0 × 9.8) / 5.0 ≈ 3.92 m s⁻²。代回可得 T = 3.0 × 3.92 ≈ 11.8 N。
a = m₂g / (m₁ + m₂) , T = m₁m₂g / (m₁ + m₂)
It is vital to consider the direction of acceleration consistently: the table block accelerates to the right, the hanging block accelerates downwards, and the string transmits the same tension to both.
务必保持加速度方向的一致性：桌上木块向右加速，悬挂木块向下加速，且绳子对两者的张力大小相等。
3. Energy: Conservation of Mechanical Energy with a Spring | 能量：涉及弹簧的机械能守恒
ssing it by 0.10 m. The block is then released. Find the speed of the block when it leaves the spring.
一个0.50 kg的木块置于光滑水平面上，紧靠劲度系数 k = 200 N m⁻¹ 的弹簧，并将其压缩0.10 m。然后释放木块。求木块离开弹簧时的速度。
². As the spring expands, this energy converts entirely into kinetic energy of the block, assuming no friction.
在释放瞬间，所有机械能以弹性势能储存在弹簧中：Eₚ = ½ k x²。随着弹簧伸长，该能量完全转化为木块的动能，假设无摩擦。
Set ½ k x² = ½ m v². Cancelling the factor ½ and solving for v gives v = x √k / m. Substitute the values: v
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