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CIE 9709 2025 November Paper 41 Q2

A Level / CIE / M1

CIE 9709 2025 Nov Paper 41 Paper · Question 2

题目

Problem

Two particles, PP and QQ, of masses 3 kg3\,\mathrm{kg} and 5 kg5\,\mathrm{kg} respectively are at rest on a smooth horizontal plane. PP is projected at a speed of 4 m s−14\,\mathrm{m\,s^{-1}} directly towards QQ. After PP and QQ collide, PP has speed 1 m s−11\,\mathrm{m\,s^{-1}}.

(a) Find the two possible speeds of QQ after the collision.

(3)

It is given that λ J\lambda\,\mathrm{J} of kinetic energy, where λ>0\lambda>0, is lost during the collision.

(b) Find the value of λ\lambda.

(2)
题目中文翻译

质量分别为 3 kg3\,\mathrm{kg} 和 5 kg5\,\mathrm{kg} 的两个质点 PP 和 QQ 静止在光滑水平面上。PP 以 4 m s−14\,\mathrm{m\,s^{-1}} 的速率直接朝 QQ 发射。PP 与 QQ 碰撞后,PP 的速率为 1 m s−11\,\mathrm{m\,s^{-1}}。

(a) 求碰撞后 QQ 的两种可能速率。

(3)

已知碰撞过程中损失了 λ J\lambda\,\mathrm{J} 的动能,其中 λ>0\lambda>0。

(b) 求 λ\lambda 的值。

(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 (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

(a)

解法一

思路

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P碰后速率1但方向未知,分别把其速度取+1和−1列动量守恒。

答题过程

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Take PP‘s initial direction as positive. If PP continues forwards after impact, conservation of momentum gives

3(4)=3(1)+5v,v=1.8 m s−1.3(4)=3(1)+5v, \qquad v=1.8\text{ m s}^{-1}.

If PP rebounds,

3(4)=3(−1)+5v,v=3 m s−1.3(4)=3(-1)+5v, \qquad v=3\text{ m s}^{-1}.

Thus the two possible speeds of QQ are

1.8 m s−1and3 m s−1.\boxed{1.8\text{ m s}^{-1}\quad\text{and}\quad3\text{ m s}^{-1}}.

(b)

解法一

思路

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先检查两种情况的动能;题设损失严格为正,排除零损失情形。

答题过程

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The initial kinetic energy is

Ki=12(3)(4)2=24 J.K_i=\frac12(3)(4)^2=24\text{ J}.

If QQ has speed 3 m s−13\,\mathrm{m\,s^{-1}}, the final kinetic energy is

Kf=12(3)(1)2+12(5)(3)2=24 J,K_f=\frac12(3)(1)^2+\frac12(5)(3)^2=24\text{ J},

which gives zero loss and is excluded by λ>0\lambda>0.

For QQ‘s speed of 1.8 m s−11.8\,\mathrm{m\,s^{-1}},

Kf=12(3)(1)2+12(5)(1.8)2=9.6 J.K_f=\frac12(3)(1)^2+\frac12(5)(1.8)^2=9.6\text{ J}.

Hence

λ=24−9.6=14.4.\lambda=24-9.6=\boxed{14.4}.