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CIE 9709 2024 November Paper 43 Q4

A Level / CIE / M1

CIE 9709 2024 Nov Paper 43 Paper · Question 4

题目

Problem

Two particles, AA and BB, of masses 3 kg3\,\mathrm{kg} and 6 kg6\,\mathrm{kg} respectively, lie on a smooth horizontal plane. Initially, BB is at rest and AA is moving towards BB with speed 8 m s−18\,\mathrm{m\,s^{-1}}. After AA and BB collide, AA moves with speed 2 m s−12\,\mathrm{m\,s^{-1}}.

Find the greater of the two possible total losses of kinetic energy due to the collision.

(6)

题目中文翻译

两个质点 AA 和 BB 的质量分别为 3 kg3\,\mathrm{kg} 和 6 kg6\,\mathrm{kg},位于一个光滑水平面上。最初,BB 静止,AA 以 8 m s−18\,\mathrm{m\,s^{-1}} 的速率向 BB 运动。AA 和 BB 碰撞后,AA 以 2 m s−12\,\mathrm{m\,s^{-1}} 的速率运动。

求碰撞造成的两种可能的总动能损失中较大的一个。

(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 (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

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A碰后的速率给出两个可能速度,分别沿原方向与反弹。逐一守恒动量并计算总动能损失,才能判断较大者。

答题过程

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Take the original direction of AA as positive. The initial kinetic energy is

E0=12(3)(82)=96 J.E_0=\frac12(3)(8^2)=96\,\mathrm{J}.

If AA continues forwards with velocity +2+2, conservation of momentum gives

3(8)=3(2)+6v,v=3.3(8)=3(2)+6v,\qquad v=3.

The final kinetic energy and loss are

E1=12(3)(22)+12(6)(32)=33 J,E_1=\frac12(3)(2^2)+\frac12(6)(3^2)=33\,\mathrm{J}, L1=96−33=63 J.L_1=96-33=63\,\mathrm{J}.

If AA rebounds with velocity −2-2, then

3(8)=3(−2)+6v,v=5.3(8)=3(-2)+6v,\qquad v=5.

Now

E2= 12(3)(22)+12(6)(52)=81 J,L2= 96−81=15 J.\begin{align*} E_2=&\,\frac12(3)(2^2)+\frac12(6)(5^2)=81\,\mathrm{J},\\[4mm] L_2=&\,96-81=15\,\mathrm{J}. \end{align*}

Hence the greater possible loss is

63 J.\boxed{63\,\mathrm{J}}.