题目

Problem

A ball of mass $2\text{ kg}$ is projected vertically downwards with speed $5\text{ m s}^{-1}$ through a liquid. At time $t\text{ s}$ after projection, the velocity of the ball is $v\text{ m s}^{-1}$ and its displacement from its starting point is $x\text{ m}$ . The forces acting on the ball are its weight and a resistive force of magnitude $0.2v^2\text{ N}$ .

(a) Find an expression for $v$ in terms of $t$ .

(6)

(b) Deduce what happens to $v$ for large values of $t$ .

(1)

题目中文翻译

一个质量为 $2\text{ kg}$ 的球以 $5\text{ m s}^{-1}$ 的速率竖直向下抛入液体中。抛出后 $t\text{ s}$ 时，球的速度为 $v\text{ m s}^{-1}$ ，相对于出发点的位移为 $x\text{ m}$ 。作用在球上的力为重力和大小为 $0.2v^2\text{ N}$ 的阻力。

(a) 求 $v$ 关于 $t$ 的表达式。

(6)

(b) 由此推断，当 $t$ 很大时， $v$ 会如何变化。

(1)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ( $g$ ) is needed, use $10\,\mathrm{m\,s^{-2}}$ .

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

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取向下为正，用重力减去二次阻力得到速度的微分方程。分离变量后，将 $100-v^2$ 因式分解。

答题过程

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Take downwards as positive. Newton’s second law gives

\begin{align*} 2\frac{\mathrm{d}v}{\mathrm{d}t}=&\,20-0.2v^2,\\[4mm] \frac{\mathrm{d}v}{\mathrm{d}t}=&\,\frac{100-v^2}{10}. \end{align*}

On the branch starting at $v=5$ , the speed increases towards the terminal speed $10$ , so $5\le v<10$ . The logarithm arguments below are therefore positive. Using partial fractions,

\begin{align*} \frac{10}{100-s^2}=&\,\frac12\bigg(\frac1{10+s}+\frac1{10-s}\bigg),\\[4mm] \int_5^v\frac{10\,\mathrm{d}s}{100-s^2}=&\,\int_0^t1\,\mathrm{d}\tau,\\[4mm] \frac12\big[\ln(10+s)-\ln(10-s)\big]_5^v=&\,t,\\[4mm] \frac12\ln\bigg(\frac{10+v}{3(10-v)}\bigg)=&\,t. \end{align*}

Thus $(10+v)/(10-v)=3e^{2t}$ , so $10+v=30e^{2t}-3ve^{2t}$ and $v(1+3e^{2t})=30e^{2t}-10$ . Solving for $v$ and multiplying numerator and denominator by $e^{-2t}$ gives

v=10\frac{3-e^{-2t}}{3+e^{-2t}}.

(b)

解法一

思路

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考察 $t$ 很大时指数项的极限，即可得到终端速度。

答题过程

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As $t\to\infty$ , $e^{-2t}\to0$ . Thus $v\to10\,\mathrm{m\,s^{-1}}$ downwards. At this speed the weight and resistance balance.

CIE 9231 2023 November Paper 31 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程