CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 41 Q2

A Level / CIE / M1

CIE 9709 2025 June Paper 41 Paper · Question 2

题目

Problem

Three coplanar forces of magnitudes P NP\,\mathrm{N}, 5 N5\,\mathrm{N} and 10 N10\,\mathrm{N} act at a point OO, as shown in the diagram. The resultant of the three forces has magnitude Q NQ\,\mathrm{N} and acts in a direction perpendicular to the force of magnitude P NP\,\mathrm{N}.

Find the value of PP and the value of QQ.

(4)
题目中文翻译

三个大小分别为 P NP\,\mathrm{N}、5 N5\,\mathrm{N} 和 10 N10\,\mathrm{N} 的共面力作用于点 OO，如图所示。这三个力的合力大小为 Q NQ\,\mathrm{N}，方向垂直于大小为 P NP\,\mathrm{N} 的力。

求 PP 和 QQ 的值。

(4)
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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时，取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

解法一

思路

展开

P力竖直，合力垂直它即水平，先令竖直合分量为零，再求水平合力的大小。

答题过程

展开

Since the resultant is perpendicular to the vertical force PP, its vertical component is zero:

P+5sin⁡60∘−10sin⁡30∘=0.P+5\sin60^\circ-10\sin30^\circ=0.

Thus

P=5−532≈0.670 (3 s.f.).P=5-\frac{5\sqrt3}{2}\approx\boxed{0.670}\text{ (3 s.f.)}.

The horizontal resultant acts to the left, with magnitude

Q= 10cos⁡30∘−5cos⁡60∘= 53−52≈ 6.16 (3 s.f.).\begin{align*} Q=&\,10\cos30^\circ-5\cos60^\circ\\[4mm] =&\,5\sqrt3-\frac52\\[4mm] \approx&\,\boxed{6.16}\text{ (3 s.f.)}. \end{align*}