题目

Problem

A particle $P$ of mass $3\,\mathrm{kg}$ is projected with a speed of $8\,\mathrm{m\,s^{-1}}$ up a line of greatest slope of a rough plane inclined at $30^\circ$ to the horizontal. $P$ is projected from a point $A$ on the plane and comes to instantaneous rest at a point $B$ on the plane. $P$ then slides back down the plane. The coefficient of friction between $P$ and the plane is $\frac{1}{12}\sqrt{3}$ .

Using an energy method throughout, find the speed of $P$ at the instant it returns to $A$ .

(6)

题目中文翻译

一个质量为 $3\,\mathrm{kg}$ 的质点 $P$ ，以 $8\,\mathrm{m\,s^{-1}}$ 的速率沿粗糙斜面的一条最大坡度线向上发射，斜面与水平面成 $30^\circ$ 角。 $P$ 从斜面上的点 $A$ 发射，在斜面上的点 $B$ 瞬时静止，随后沿斜面滑回。 $P$ 与斜面之间的摩擦系数为 $\frac{1}{12}\sqrt{3}$ 。

全程使用能量方法，求 $P$ 返回 $A$ 的瞬间的速率。

(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

解法一

思路

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全程指定能量法。上行先求AB及单程摩擦耗散；完整往返势能净变化为零，但摩擦耗散两次。

答题过程

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The normal contact force is

R=3(10)\cos30^\circ=15\sqrt3\text{ N},

so friction has magnitude

F=\frac{\sqrt3}{12}R=\frac{15}{4}\text{ N}.

Let $AB=d$ metres. On the upward journey, the initial kinetic energy equals the potential energy gained plus work against friction:

\frac12(3)(8)^2=3(10)d\sin30^\circ+\frac{15}{4}d,

96=\frac{75}{4}d, \qquad d=\frac{128}{25}=5.12\text{ m}.

The work against friction on each leg is $(15/4)(5.12)=19.2\,\mathrm{J}$ . Over the complete return journey, the potential energy has no net change, so

\frac12(3)v^2=96-2(19.2)=57.6.

Therefore

\begin{align*} v^2=&\,38.4,\\[4mm] v=&\,\frac{8\sqrt{15}}5\\[4mm] \approx&\,\boxed{6.20\text{ m s}^{-1}}\text{ (3 s.f.)}. \end{align*}

CIE 9709 2025 June Paper 42 Q7

题目

解答

解法一

思路

答题过程