题目

Problem

Three particles $A$ , $B$ and $C$ , each of mass $m$ , lie at rest in a straight line $L$ on a smooth horizontal surface, with $B$ between $A$ and $C$ .

Particles $A$ and $B$ are projected directly towards each other with speeds $5u$ and $4u$ respectively.

At the same instant, particle $C$ is projected directly away from $B$ with speed $3u$ .

In the subsequent motion, $A$ , $B$ and $C$ move along $L$ .

Particles $A$ and $B$ collide directly.

The coefficient of restitution between $A$ and $B$ is $e$ .

(a) Find

(i) the speed of $A$ immediately after the collision,

(ii) the speed of $B$ immediately after the collision.

(7)

Given that the direction of motion of $A$ is reversed in the collision between $A$ and $B$ , and that there is no collision between $B$ and $C$ ,

(b) find the set of possible values of $e$ .

(4)

题目中文翻译

三个质量均为 $m$ 的质点 $A$ 、 $B$ 、 $C$ 静止于光滑水平面上的直线 $L$ 上， $B$ 位于 $A$ 与 $C$ 之间。

将质点 $A$ 、 $B$ 分别以速率 $5u$ 、 $4u$ 沿直线相向发射。

在同一瞬间，将质点 $C$ 以速率 $3u$ 沿直线向远离 $B$ 的方向发射。

在随后的运动中， $A$ 、 $B$ 、 $C$ 均沿 $L$ 运动。

质点 $A$ 、 $B$ 发生正碰。

$A$ 、 $B$ 之间的恢复系数为 $e$ 。

(a) 求

(i) 碰撞后瞬间 $A$ 的速率；

(ii) 碰撞后瞬间 $B$ 的速率。

(7)

已知 $A$ 、 $B$ 碰撞使 $A$ 的运动方向反转，且 $B$ 、 $C$ 之间不发生碰撞，

(b) 求 $e$ 的可能取值集合。

(4)

解答

(a)

解法一

思路

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统一取 A 初运动方向为正，B 初速度为负。先求有符号末速度，再把 A 的速度取绝对值，因为题目要求速率。

答题过程

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Take the initial direction of $A$ as positive. Let the signed velocities of $A$ and $B$ after their collision be $v$ and $w$ . Conservation of momentum gives

m(5u)+m(-4u)=mv+mw,\qquad v+w=u.

The speed of approach is $5u-(-4u)=9u$ . Newton’s law of restitution gives

w-v=9eu.

Adding and subtracting these equations,

2w=u(1+9e),\qquad 2v=u(1-9e).

Hence

(i) the speed of $A$ is

\boxed{\frac u2|1-9e|};

(ii) since $w>0$ , the speed of $B$ is

\boxed{\frac u2(1+9e)}.

(b)

解法一

思路

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A 反向要求其有符号末速度严格为负。B 碰后仍在 C 后方，若速度不超过 C 就不会追上；等速时仍不会碰撞，所以上界可取等号。

答题过程

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For the direction of $A$ to be reversed,

v=\frac u2(1-9e)<0\quad\Rightarrow\quad e>\frac19.

Before the $A$ – $B$ collision, $B$ moves away from $C$ while $C$ moves to the right. Therefore $C$ is still ahead of $B$ when that collision occurs. Afterwards, $C$ continues at speed $3u$ . There is no $B$ – $C$ collision precisely when $B$ does not catch $C$ , requiring

w\leq3u,\qquad \frac u2(1+9e)\leq3u.

Thus $e\leq5/9$ . Equality gives equal speeds with a positive separation, so it is allowed. Combining both conditions,

\boxed{\frac19<e\leq\frac59}.

M2 2025 Oct A Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程