题目

Problem

Three particles $A$ , $B$ and $C$ have masses $3\,\mathrm{kg}$ , $1.5\,\mathrm{kg}$ and $m\,\mathrm{kg}$ respectively, where $m>1.5$ . Particles $A$ and $B$ are attached to the ends of a light inextensible string that passes over a smooth fixed pulley.

Particle $C$ is attached to particle $B$ . The system is held at rest with the string taut and the hanging parts of the string vertical (see diagram).

The system is released from rest and while the three particles are all attached as described, the tension in the string is $37.5\,\mathrm{N}$ .

(a) Find the value of $m$ .

(4)

At the instant when $A$ has moved a distance $0.8\,\mathrm{m}$ upwards from its initial position, $C$ separates from $B$ and falls away. In the subsequent motion, $A$ and $B$ continue to move and $A$ does not reach the pulley.

(b) Find the total time from when the system is released until $A$ first comes to instantaneous rest.

(6)

题目中文翻译

三个质点 $A$ 、 $B$ 和 $C$ 的质量分别为 $3\,\mathrm{kg}$ 、 $1.5\,\mathrm{kg}$ 和 $m\,\mathrm{kg}$ ，其中 $m>1.5$ 。质点 $A$ 和 $B$ 分别连接在一根轻且不可伸长的绳子的两端，绳子跨过一个光滑的定滑轮。

质点 $C$ 与质点 $B$ 相连。系统被保持静止，绳子拉紧，悬垂的绳段竖直（见图）。

系统从静止释放。在三个质点仍按上述方式全部连接着时，绳中的张力为 $37.5\,\mathrm{N}$ 。

(a) 求 $m$ 的值。

(4)

当 $A$ 从初始位置向上移动了 $0.8\,\mathrm{m}$ 的瞬间， $C$ 与 $B$ 分离并掉落。之后 $A$ 和 $B$ 继续运动，且 $A$ 不会到达滑轮。

(b) 求从系统释放到 $A$ 第一次瞬时静止所经过的总时间。

(6)

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

展开

由m>1.5知BC合质量较大，A向上加速；先在A上求加速度，再用于BC。

答题过程

展开

Since $m+1.5>3$ , $A$ accelerates upwards and $BC$ downwards. For $A$ , using $g=10$ ,

37.5-3(10)=3a, \qquad a=2.5\text{ m s}^{-2}.

For the combined particles $B$ and $C$ ,

(m+1.5)(10)-37.5=(m+1.5)(2.5).

Thus

\begin{align*} 7.5(m+1.5)=&\,37.5,\\[4mm] m+1.5=&\,5,\\[4mm] m=&\,\boxed{3.5}. \end{align*}

(b)

解法一

思路

展开

分离前后分两段匀加速运动；分离瞬间速度连续，但之后加速度反向。

答题过程

展开

Before separation, $A$ starts from rest and accelerates upwards at $2.5\,\mathrm{m\,s^{-2}}$ through $0.8\,\mathrm{m}$ . Its speed at separation is

v^2=0^2+2(2.5)(0.8)=4, \qquad v=2\text{ m s}^{-1}.

The time for this first stage is

t_1=\frac{2}{2.5}=0.8\text{ s}.

After $C$ falls away, let $a_2$ be $A$ ‘s upward acceleration and $T_2$ the tension. Then

T_2-30=3a_2,\qquad 15-T_2=1.5a_2.

Adding gives

-15=4.5a_2, \qquad a_2=-\frac{10}{3}\text{ m s}^{-2}.

Starting this stage with upward speed $2\,\mathrm{m\,s^{-1}}$ , $A$ first comes to rest when

0=2-\frac{10}{3}t_2, \qquad t_2=0.6\text{ s}.

The total time is

t_1+t_2=0.8+0.6=\boxed{1.4\text{ s}}.

CIE 9709 2026 June Paper 41 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程