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CIE 9709 2023 June Paper 41 Q1

A Level / CIE / M1

CIE 9709 2023 June Paper 41 Paper · Question 1

题目

Problem

Two particles PP and QQ, of masses m kgm\,\mathrm{kg} and 0.3 kg0.3\,\mathrm{kg} respectively, are at rest on a smooth horizontal plane. PP is projected at a speed of 5 m s−15\,\mathrm{m\,s^{-1}} directly towards QQ. After PP and QQ collide, PP moves with a speed of 2 m s−12\,\mathrm{m\,s^{-1}} in the same direction as it was originally moving.

(a) Find, in terms of mm, the speed of QQ after the collision.

(2)

After this collision, QQ moves directly towards a third particle RR, of mass 0.6 kg0.6\,\mathrm{kg}, which is at rest on the plane. QQ is brought to rest in the collision with RR, and RR begins to move with a speed of 1.5 m s−11.5\,\mathrm{m\,s^{-1}}.

(b) Find the value of mm.

(2)

题目中文翻译

两个质点 PP 和 QQ 的质量分别为 m kgm\,\mathrm{kg} 和 0.3 kg0.3\,\mathrm{kg},静止在光滑水平面上。PP 以 5 m s−15\,\mathrm{m\,s^{-1}} 的速率直接向 QQ 抛出。PP 与 QQ 碰撞后,PP 沿原来的运动方向以 2 m s−12\,\mathrm{m\,s^{-1}} 的速率运动。

(a) 用 mm 表示碰撞后 QQ 的速率。

(2)

此次碰撞后,QQ 直接向第三个质点 RR 运动。RR 的质量为 0.6 kg0.6\,\mathrm{kg},静止在平面上。QQ 与 RR 碰撞后静止,RR 开始以 1.5 m s−11.5\,\mathrm{m\,s^{-1}} 的速率运动。

(b) 求 mm 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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第一次碰撞沿P初始方向守恒动量,P仍同方向运动,速度取正2。

答题过程

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Taking the initial direction of PP as positive, conservation of momentum gives

m(5)=m(2)+0.3v.m(5)=m(2)+0.3v.

Thus

v=3m0.3=10m m s−1.\boxed{v=\frac{3m}{0.3}=10m\,\mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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第二次碰撞后Q静止,全部初动量交给R。把(a)的速度表达式代入。

答题过程

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For the collision of QQ and RR, conservation of momentum gives

0.3(10m)=0.6(1.5).0.3(10m)=0.6(1.5).

Thus 3m=0.93m=0.9, so

m=0.3.\boxed{m=0.3}.