Would an increase in mass of a sphere shaped plastic affect its terminal velocity? Application Techniques | 球状塑料质量增加是否影响其终端速度？应用题解题技巧

📚 Would an increase in mass of a sphere shaped plastic affect its terminal velocity? Application Techniques | 球状塑料质量增加是否影响其终端速度？应用题解题技巧

In many IB Physics problems, students are asked to predict how a change in one variable—such as mass—affects the terminal velocity of a falling object. A typical scenario involves a plastic sphere falling through air. Intuition might suggest that a heavier object falls faster, but the formal link between mass and terminal velocity must be anchored in balanced forces and the correct drag model. This article unpacks the physics, walks through a structured problem-solving technique, and clarifies common misconceptions using a real-world spherical plastic object.

在许多IB物理考题中，学生需要预测某一变量（如质量）的变化如何影响下落物体的终端速度。一个典型情景是塑料球在空气中下落。直觉可能告诉我们，物体越重落得越快，但质量与终端速度之间的正式关联必须建立在力的平衡和正确的阻力模型之上。本文将拆解其中的物理原理，演示一套结构化的解题方法，并结合现实中的球形塑料物体澄清常见误区。

1. Understanding Terminal Velocity | 理解终端速度

Terminal velocity is the constant maximum speed reached by an object when the net force acting on it becomes zero. For a sphere falling through a fluid, the downward weight is eventually balanced by the upward drag force and the buoyant force. Once this equilibrium is established, acceleration ceases and the object continues to fall at a steady speed.

终端速度是物体所受合力为零时达到的恒定最大速度。对于在流体中下落的球体，向下的重力最终会被向上的阻力和浮力所平衡。一旦达到这种平衡，加速度为零，物体将以稳定速度继续下落。

It is crucial to note that terminal velocity is not an intrinsic property of the object but rather the result of a dynamic balance between driving and resisting forces. Thus, any factor that shifts this balance—mass, cross-sectional area, fluid density, drag coefficient—will alter the terminal speed.

关键在于终端速度并非物体的固有属性，而是驱动力与阻力之间动态平衡的结果。因此，任何打破这一平衡的因素——质量、横截面积、流体密度、阻力系数——都会改变终端速度。

2. The Core Physics: Balanced Forces | 核心物理：力的平衡

Consider a plastic sphere of mass m, radius r, and density ρₛ, falling through air of density ρₐ. The forces are: weight W = mg downward; buoyancy B = ρₐ V g upward (where V = (4/3)πr³ is the sphere volume); and drag force F_d opposing motion, upward. At terminal velocity vₜ, the net force is zero: mg – ρₐVg – F_d = 0.

考虑一个质量为m、半径为r、密度为ρₛ的塑料球在密度为ρₐ的空气中下落。受力情况为：向下的重力W=mg；向上的浮力B=ρₐVg（其中V=(4/3)πr³为球的体积）；以及方向向上的阻力F_d。在终端速度vₜ时，合力为零：mg – ρₐVg – F_d = 0。

In many IB problems, buoyancy is negligible for a plastic sphere in air because ρₛ ≫ ρₐ. The force balance then simplifies to mg = F_d. This is the starting point for most application questions.

在许多IB问题中，由于塑料球密度远大于空气密度，浮力可以忽略。此时力的平衡简化为mg = F_d。这是大多数应用题的起点。

3. Air Resistance Models for a Sphere | 球形物体的空气阻力模型

The drag force on a sphere depends on the flow regime. At low Reynolds numbers (small, slow spheres), the Stokes’ law applies: F_d = 6πη r v, where η is the fluid viscosity. At higher speeds typical for plastic balls, the drag is proportional to v² and given by the quadratic drag equation: F_d = ½ ρₐ v² C_d A, where C_d is the drag coefficient (~0.5 for a smooth sphere) and A = πr² is the cross-sectional area.

球体受到的阻力取决于流动状态。在低雷诺数下（小球、慢速），斯托克斯定律适用：F_d = 6πη r v，其中η为流体黏度。在塑料球常见的较高速下，阻力与v²成正比，由平方阻力公式给出：F_d = ½ ρₐ v² C_d A，其中C_d为阻力系数（光滑球体约为0.5），A = πr²为横截面积。

IB exam questions usually specify which drag regime to use or provide the appropriate formula. Always check the speed and size to confirm whether linear or quadratic drag is expected. For most everyday plastic balls, quadratic drag dominates.

IB考试通常会明确指出应使用哪种阻力模式或直接给出相应公式。务必通过速度与尺寸判断应该采用线性还是平方阻力。对于大多数日常塑料球，平方阻力占据主导。

4. Deriving the Terminal Velocity Equation | 推导终端速度方程

Ignoring buoyancy and using quadratic drag, set mg = ½ ρₐ vₜ² C_d A. Solving for terminal velocity gives:

vₜ = √(2mg / (ρₐ C_d A))

忽略浮力并采用平方阻力模型，由mg = ½ ρₐ vₜ² C_d A，解得终端速度：

vₜ = √(2mg / (ρₐ C_d A))

If the sphere is solid and its mass increases without changing its radius (e.g., by using a denser plastic), the cross-sectional area A = πr² stays constant. Then vₜ ∝ √m. This square-root dependence is critical for understanding how a change in mass influences terminal speed.

若球体为实心且质量增加而半径不变（例如使用密度更高的塑料），横截面积A = πr²保持不变，则vₜ ∝ √m。这一平方根关系对理解质量变化如何影响终端速度至关重要。

If the mass increase comes from a larger sphere of the same material, m ∝ r³ and A ∝ r², leading to vₜ ∝ √(r³/r²) = √r. So terminal velocity still increases, but the scaling law changes.

若质量增加源于用同种材料制成的更大球体，则m ∝ r³且A ∝ r²，得到vₜ ∝ √(r³/r²) = √r。因此终端速度依然增大，只是比例规律有所不同。

5. Effect of Mass on Terminal Velocity | 质量对终端速度的影响

Direct answer: Yes, an increase in mass of a sphere-shaped plastic object does increase its terminal velocity, provided the shape and size remain identical. Doubling the mass multiplies the terminal speed by a factor of √2 ≈ 1.41, not 2. This is a classic pitfall in multiple-choice questions.

直接回答：是的，若形状和尺寸保持不变，球状塑料物体质量的增加确实会使其终端速度增大。将质量加倍，终端速度将乘以√2≈1.41，而非2。这是选择题中的经典雷区。

This non-linear response arises because drag force grows with the square of velocity. A heavier sphere must fall faster to generate enough drag to balance its extra weight. Understanding this relationship avoids the misconception that terminal velocity is simply proportional to mass.

这种非线性响应源于阻力随速度平方增长。较重的球必须以更快的速度下落，才能产生足以平衡额外重力的阻力。理解这一关系可以避免终端速度简单地与质量成正比的误解。

6. Common Pitfall: Assuming Direct Proportionality | 常见误区：假设成正比

Many students instinctively write vₜ ∝ m, which is incorrect under quadratic drag. This mistake often appears when they memorise a terminal velocity formula without recognising its underlying assumptions. Remind yourself that the exponent ½ on m comes from solving v² ∝ m.

许多学生会本能地写出vₜ ∝ m，这在平方阻力模型下是错误的。这一错误常出现在死记终端速度公式而未理解其基本假设之时。要提醒自己，m的指数½源于求解v² ∝ m的过程。

Another subtle trap: if a question mentions a ‘plastic sphere’ and also gives the viscosity of air, students may incorrectly apply Stokes’ law, yielding vₜ ∝ m. Always check if the Reynolds number context is consistent with the chosen drag formula.

另一个隐蔽陷阱：如果题目提到“塑料球”并给出空气黏度，学生可能会错误地使用斯托克斯定律，得出vₜ ∝ m。务必检查雷诺数背景是否与所选阻力公式一致。

7. Step-by-Step Application Technique 1: Identify the System | 分段应用技巧1：识别系统

Begin by reading the problem carefully and highlight the known quantities: mass m, radius r, fluid density ρ, drag coefficient C_d, and whether buoyancy can be neglected. Sketch a free-body diagram showing weight and drag arrows for the sphere at terminal velocity.

首先仔细读题，标出已知量：质量m、半径r、流体密度ρ、阻力系数C_d，以及浮力是否可忽略。画出球体在终端速度下的受力图，标注重力与阻力箭头。

Decide which drag regime is appropriate: if the question provides a drag force equation or explicitly states ‘turbulent drag’, use the quadratic form. If it gives viscosity and describes the sphere as ‘very small’ or ‘slow’, linear drag may be intended. This decision determines the algebraic starting point.

判断应采用何种阻力模式：如果题目给出了阻力方程或明确说明“湍流阻力”，则使用平方形式；若提供了黏度并将球描述为“非常小”或“很慢”，则可能适用线性阻力。这一决定确定了代数推导的起点。

8. Step-by-Step Application Technique 2: Write the Force Balance | 分段应用技巧2：写出力平衡方程

Set up the equilibrium condition: ΣF = 0. For negligible buoyancy, this is simply mg = F_d. Substitute the appropriate expression for F_d, ensuring all variables are in SI units. Write the equation before inserting numbers; this symbolic approach reveals the physics relationships and guards against unit errors.

建立平衡条件：ΣF = 0。浮力可忽略时，即为mg = F_d。代入恰当的阻力表达式，确保所有变量均采用国际单位制。先写出方程再代入数值；这种符号化方法可以揭示物理关系，并防止单位错误。

For a quadratic drag problem: mg = ½ ρ C_d πr² vₜ². Rearranging gives vₜ² = 2mg/(ρ C_d πr²). This is your master equation. Notice that m appears in the numerator under the square root, providing the √m dependence.

对于平方阻力问题：mg = ½ ρ C_d πr² vₜ²。整理得vₜ² = 2mg/(ρ C_d πr²)。这就是主方程。注意m出现在分子上并在根号内，给出了√m的相关性。

9. Step-by-Step Application Technique 3: Solve Symbolically First | 分段应用技巧3：先符号求解

Before plugging in values, derive an expression for vₜ in terms of m and other constants. Then, to find how vₜ changes when mass varies, form a ratio: vₜ,new / vₜ,old = √(m_new/m_old). This eliminates the need to know C_d, ρ, or A, provided they remain unchanged.

在代入数值之前，先用m和其他常量推导出vₜ的表达式。然后，欲知质量变化时vₜ如何改变，可构成比值：vₜ,new / vₜ,old = √(m_new/m_old)。这可以免除对C_d、ρ或A的需求，只要它们保持不变。

This technique is incredibly powerful in application problems. Often the question asks ‘by what factor will the terminal velocity increase if the mass is tripled?’ Simply compute √3 ≈ 1.73. The ratio method works irrespective of the object’s size or drag coefficient, as long as shape and fluid are constant.

这一技巧在应用题中极为有效。题目常问“若质量增至三倍，终端速度将增大到原来的多少倍？”只需计算√3≈1.73。只要形状和流体不变，这种比值法无需知道物体尺寸或阻力系数。

10. Worked Example: Plastic Sphere with Increased Mass | 实例演示：塑料球质量增加

A solid plastic sphere of radius 4.0 cm and mass 0.15 kg is dropped from rest. Assume quadratic drag with air density ρ = 1.2 kg/m³ and C_d = 0.50. (a) Find its terminal velocity. (b) The sphere is replaced by another of identical size but made of a denser plastic, so its mass becomes 0.30 kg. Predict the new terminal velocity without recalculating all steps.

一个半径为4.0 cm、质量为0.15 kg的实心塑料球从静止下落。假设平方阻力，空气密度ρ=1.2 kg/m³，C_d=0.50。(a)求其终端速度。(b)换用尺寸相同但密度更大的塑料球，质量变为0.30 kg。不重复所有步骤，预测新的终端速度。

Solution (a): A = πr² = π(0.04)² = 5.03×10⁻³ m². Then vₜ = √(2mg/(ρ C_d A)) = √(2×0.15×9.8/(1.2×0.50×5.03×10⁻³)) ≈ √(2.94/(0.003018)) ≈ √974 ≈ 31.2 m/s.

解(a)：A = πr² = π(0.04)² = 5.03×10⁻³ m²。代入得vₜ = √(2×0.15×9.8/(1.2×0.50×5.03×10⁻³)) ≈ √(2.94/0.003018) ≈ √974 ≈ 31.2 m/s。

(b) Since only mass changes, vₜ,new = vₜ,old × √(0.30/0.15) = 31.2 × √2 ≈ 31.2 × 1.414 = 44.1 m/s. The ratio method confidently gives the answer without repeating the full computation; it also makes clear that doubling mass raises terminal speed by about 41%, not 100%.

(b)由于只有质量变化，vₜ,新 = vₜ,旧 × √(0.30/0.15) = 31.2 × √2 ≈ 31.2 × 1.414 = 44.1 m/s。运用比值法无需全面重复计算即可得出答案，同时明确显示：质量加倍使终端速度增加约41%，而非100%。

11. Graph Interpretation and Exam Tips | 图表解读与考试提示

Velocity-time graphs for a falling sphere typically show a steep initial rise that levels off asymptotically at vₜ. If asked to sketch the effect of increased mass on the same axes, draw a new curve that reaches a higher asymptote. The initial slope (acceleration) is still g for both because drag is initially zero; the curves separate as speed builds up.

下落球体的速度-时间图像通常显示初段陡升，然后逐渐趋向vₜ。若要求在相同坐标轴上画出质量增大后的效果，应绘制一条到达更高渐近线的新曲线。两条曲线的初始斜率（加速度）均为g，因为初始时阻力为零；随着速度增长，曲线将分开。

Exam marking points often reward: identifying the force balance at terminal velocity, using the correct drag formula, showing the square-root relationship, and interpreting results as a factor change rather than absolute values. Always write assumptions clearly, e.g., ‘neglecting buoyancy’ or ‘assuming constant C_d‘, to demonstrate thorough scientific thinking.

阅卷采分点常在于：指出终端速度时的力平衡、采用正确的阻力公式、展示平方根关系，以及将结果解释为倍率变化而非绝对值。务必清晰写明假设，例如“忽略浮力”或“假设C_d恒定”，以展现严谨的科学思维。

When you encounter a question phrased as ‘Would an increase in mass affect terminal velocity?’, do not simply answer ‘yes’. Justify using the equation vₜ ∝ √m, and state the condition: provided cross-sectional area and drag coefficient remain constant. This precision distinguishes a high-level response.

遇到“质量增加会影响终端速度吗？”这类问题时，不要仅回答“是”。要用方程vₜ ∝ √m进行论证，并说明条件：在横截面积和阻力系数保持不变的情况下。这种精确性是高分答案的标志。

12. Conclusion: Application Wisdom | 结论：应用题智慧

IB Physics application problems on terminal velocity reward a systematic, symbolic approach. Instead of memorising outcomes for every scenario, students should master the method: draw forces, write balance equations, choose the drag model, solve symbolically, and finally plug in numbers. This technique seamlessly handles variations in mass, shape, or fluid properties.

IB物理关于终端速度的应用题青睐系统性的符号化方法。学生不应死记每种情景的结果，而应掌握这套方法：画受力、列平衡方程、选择阻力模型、符号求解，最后代入数值。这一技巧能无缝应对质量、形状或流体性质的各种变化。

For the specific case of a plastic sphere with increased mass, the terminal velocity does increase, scaling with the square root of mass if radius is unchanged. Using the ratio technique and justifying every step will build confidence and earn full marks on even the trickiest application questions.

就塑料球质量增加这一具体情形而言，若半径不变，终端速度确实会增大，且与质量的平方根成正比。使用比值法并为每一步提供论证，将帮助学生在最复杂的应用题中建立信心并获取满分。

Published by TutorHao | IB Physics Revision Series | aleveler.com

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