题目

Problem

A particle $P$ is projected from a point $A$ with a speed of $4\,\mathrm{m\,s^{-1}}$ up a line of greatest slope of a rough plane inclined at an angle $\theta$ to the horizontal, where $\tan\theta=\frac43$ .

$P$ comes to instantaneous rest at a point $B$ on the plane (see diagram).

The coefficient of friction between $P$ and the plane is $\frac13$ .

(a) Using an energy method, show that the distance $AB$ is $0.8\,\mathrm{m}$ .

(4)

(b) After coming to instantaneous rest at $B$ , the particle slides down the plane.

Find the total time from the instant at which $P$ is projected from $A$ until it returns to $A$ .

(5)

题目中文翻译

质点 $P$ 从点 $A$ 以 $4\,\mathrm{m\,s^{-1}}$ 的速率沿粗糙斜面的一条最大坡度线向上发射。斜面与水平面的夹角为 $\theta$ ，其中 $\tan\theta=\frac43$ 。

质点在斜面上的点 $B$ 瞬时静止（见图）。

$P$ 与斜面之间的摩擦系数为 $\frac13$ 。

(a) 使用能量方法，证明距离 $AB$ 为 $0.8\,\mathrm{m}$ 。

(4)

(b) 在 $B$ 处瞬时静止后，质点沿斜面向下滑动。

求从 $P$ 自 $A$ 发射起至返回 $A$ 所经历的总时间。

(5)

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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题目指定能量法。设质量m，不擅自令m=1；先求正压力与摩擦，再写能量平衡。

答题过程

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Let the mass be $m\,\mathrm{kg}$ and $AB=d$ metres. From $\tan\theta=4/3$ ,

\sin\theta=\frac45,\qquad \cos\theta=\frac35.

The normal contact force and friction are

R=mg\cos\theta=6m, \qquad F=\frac13R=2m.

On the upward journey, the initial kinetic energy is used to increase potential energy and do work against friction:

\frac12m(4)^2=mgd\sin\theta+Fd,

8m=8md+2md=10md.

Since $m>0$ , cancellation gives

\boxed{AB=d=0.8\text{ m}},

as required.

(b)

解法一

思路

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上行由平均速度求时间，下行摩擦方向反向，重算下行加速度再求时间。

答题过程

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The upward journey has initial speed 4 and final speed 0, so

0.8=\frac{4+0}{2}t_{\rm up}, \qquad t_{\rm up}=0.4\text{ s}.

On the downward journey, friction acts up the plane. Resolving downwards,

ma=mg\sin\theta-\frac13mg\cos\theta=8m-2m,

so $a=6\,\mathrm{m\,s^{-2}}$ . Starting from rest at $B$ ,

\begin{align*} 0.8=&\,\frac12(6)t_{\rm down}^2,\\[4mm] t_{\rm down}=&\,\sqrt{\frac4{15}}=0.516397779\ldots\text{ s}. \end{align*}

Thus the total time is

0.4+0.516397779\ldots\approx\boxed{0.916\text{ s}}.

解法二

思路

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下行也可用能量法求返回A的速率，再由平均速度求下行时间。

答题过程

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The upward time is

t_{\rm up}=\frac{2(0.8)}{4+0}=0.4\text{ s}.

On descending from $B$ to $A$ , the loss of potential energy minus the work against friction gives

mg(0.8)\frac45-(2m)(0.8)=\frac12mv^2.

Thus $6.4m-1.6m=\frac12mv^2$ , so $v^2=9.6$ . Since the downward acceleration is constant,

t_{\rm down}=\frac{2(0.8)}{0+\sqrt{9.6}}=0.516397779\ldots\text{ s}.

Hence the total time is

t_{\rm up}+t_{\rm down}\approx\boxed{0.916\text{ s}}\text{ (3 s.f.)}.

CIE 9709 2026 June Paper 42 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

解法二

思路

答题过程