题目

Problem

Particles $X$ and $Y$ move in a straight line through points $A$ and $B$ . Particle $X$ starts from rest at $A$ and moves towards $B$ . At the same instant, $Y$ starts from rest at $B$ .

At time $t$ seconds after the particles start moving

the acceleration of $X$ in the direction $AB$ is given by $(12t+12)\,\mathrm{m\,s^{-2}}$ ,
the acceleration of $Y$ in the direction $AB$ is given by $(24t-8)\,\mathrm{m\,s^{-2}}$ .

(a) It is given that the velocities of $X$ and $Y$ are equal when they collide.

Calculate the distance $AB$ .

(6)

(b) It is given instead that $AB=36\,\mathrm{m}$ .

Verify that $X$ and $Y$ collide after $3\,\mathrm{s}$ .

(2)

题目中文翻译

质点 $X$ 和 $Y$ 沿经过 $A$ 、 $B$ 两点的直线运动。质点 $X$ 从 $A$ 点由静止出发朝 $B$ 运动。同一时刻， $Y$ 从 $B$ 点由静止开始运动。

两质点开始运动后 $t$ 秒时，

$X$ 沿 $AB$ 方向的加速度为 $(12t+12)\,\mathrm{m\,s^{-2}}$ ，
$Y$ 沿 $AB$ 方向的加速度为 $(24t-8)\,\mathrm{m\,s^{-2}}$ 。

(a) 已知 $X$ 和 $Y$ 碰撞时速度相等。

计算距离 $AB$ 。

(6)

(b) 现改为已知 $AB=36\,\mathrm{m}$ 。

验证 $X$ 和 $Y$ 在 $3\,\mathrm{s}$ 后碰撞。

(2)

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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两次积分分别求相对各自起点的速度、位移。碰撞时速度相同给出正时间，再由位置相等求初始间距；Y早期会反向，必须用位移而非路程。

答题过程

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Take direction $AB$ as positive. Since both particles start from rest, integrating their accelerations gives

\begin{align*} v_X=&\,6t^2+12t,\\[4mm] v_Y=&\,12t^2-8t. \end{align*}

At the collision their velocities are equal:

6t^2+12t=12t^2-8t, \qquad2t(3t-10)=0.

The nonzero collision time is $t=10/3$ . Their displacements from their respective starting points are

\begin{align*} s_X=&\,2t^3+6t^2,\\[4mm] s_Y=&\,4t^3-4t^2, \end{align*}

where each displacement is zero at $t=0$ . At collision, $s_X=AB+s_Y$ . Hence

\begin{align*} AB=&\,s_X\bigg(\frac{10}{3}\bigg)-s_Y\bigg(\frac{10}{3}\bigg)\\[4mm] =&\,\frac{3800}{27}-\frac{2800}{27}\\[4mm] =&\,\boxed{\frac{1000}{27}\,\mathrm{m}} \approx37.0\,\mathrm{m}. \end{align*}

(b)

解法一

思路

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用(a)得到的位移表达式，但改用新的初始间距36；实际代入3秒验证位置相等。

答题过程

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Using the displacement expressions from (a), the difference is

s_X-s_Y=-2t^3+10t^2.

At $t=3$ ,

\begin{align*} s_X-s_Y=&\,-2(3^3)+10(3^2)\\[4mm] =&\,-54+90\\[4mm] =&\,36=AB. \end{align*}

Thus $s_X=AB+s_Y$ at that time, verifying that they collide after $\boxed{3\,\mathrm{s}}$ .

CIE 9709 2022 November Paper 43 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程