AP Physics 2 2019 FRQ Solutions and Analysis | AP物理2 2019年自由回答题解析

📚 AP Physics 2 2019 FRQ Solutions and Analysis | AP物理2 2019年自由回答题解析

The 2019 AP Physics 2 free-response section featured four questions that tested core concepts in thermodynamics, DC circuits, geometric optics, and modern physics. This article provides detailed solutions and analysis, highlighting key strategies, common pitfalls, and the underlying physical principles for each problem. Students will gain a deeper understanding of how to apply ideal gas laws, circuit analysis with internal resistance, total internal reflection in fiber optics, and the photoelectric effect.

2019年AP物理2自由回答部分共有四道试题，分别考查了热力学、直流电路、几何光学和现代物理的核心知识。本文给出了这些真题的详细解答与深度解析，逐一分析每道题的解题思路、常见错误以及背后的物理原理。通过阅读本文，学生将深入理解如何运用理想气体定律、含内阻的电路分析、光纤中的全反射现象以及光电效应。

1. Overview and Exam Strategies | 总体概述与考试策略

Before diving into the solutions, it is helpful to quickly scan the four questions. The thermodynamics problem involves a gas in a cylinder undergoing an isobaric expansion followed by an isochoric cooling. The circuit question requires comparing bulb brightness when a switch is closed, emphasizing the effect of internal resistance. The optics problem deals with refraction at a fiber entrance and total internal reflection at the core-cladding boundary. The modern physics question asks for an experimental design to determine Planck’s constant using the photoelectric effect. Time management is crucial: allocate about 20 minutes per question and always show clear, logical reasoning.

在深入解析之前，快速浏览四道题的整体布局会很有帮助。热力学题涉及气缸内气体经历等压膨胀后再等容冷却的过程；电路题要求比较开关闭合前后灯泡的亮度，重点考察电源内阻的影响；光学题讨论光在光纤端面的折射以及在芯-包界面的全反射；现代物理题则让学生设计实验，利用光电效应测定普朗克常数。考试时间分配十分关键，建议每道题留出约20分钟，并且解题步骤一定要清晰完整。

2. Thermodynamics FRQ – Work Done by Gas | 热力学题 – 气体做功

The problem describes a sample of an ideal gas in a cylinder with a movable piston. Initially, the gas has pressure P₀, volume V₀, and temperature T₁. The gas is heated slowly while the piston is allowed to move, maintaining a constant pressure P₀ until the volume reaches 3V₀ (state 2). For an isobaric process, the work done by the gas on the piston is given by W = P × ΔV. Therefore, the work done in this step is W₁₂ = P₀ × (3V₀ – V₀) = 2P₀V₀. It is essential to note that the work is positive because the gas expands.

题目描述一个理想气体样品装在带有可移动活塞的气缸中。气体的初始压强为 P₀、体积为 V₀、温度为 T₁。缓慢加热气体并允许活塞移动，使压强保持 P₀ 不变，直到体积膨胀为 3V₀（状态2）。由于这是一个等压过程，气体对活塞做的功为 W = P × ΔV。因此，这一步中气体做功为 W₁₂ = P₀ × (3V₀ – V₀) = 2P₀V₀。注意气体膨胀对外做正功，所以功的符号为正。

3. Thermodynamics FRQ – First Law and Heat | 热力学题 – 第一定律与热量

Next, the piston is locked in place so the volume remains constant at 3V₀, and the gas cools until its temperature returns to the initial temperature T₁ (state 3). Because the initial and final states of the whole two-step process have the same temperature, the total change in internal energy ΔU is

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