题目

Problem

Two particles, $A$ and $B$ , are moving in opposite directions along the same straight line on a smooth horizontal surface when they collide directly. The mass of $A$ is $2m$ and the mass of $B$ is $3m$ .

Immediately after the collision, $A$ and $B$ are moving in opposite directions with the same speed $v$ .

In the collision, $A$ receives an impulse of magnitude $5mv$ .

(a) Find the coefficient of restitution between $A$ and $B$ .

(6)

After the collision with $A$ , particle $B$ strikes a smooth fixed vertical wall and rebounds. The wall is perpendicular to the direction of motion of the particles.

The coefficient of restitution between $B$ and the wall is $f$ .

As a result of its collision with $A$ and with the wall, the total kinetic energy lost by $B$ is $E$ . As a result of its collision with $B$ , the kinetic energy lost by $A$ is $2E$ .

(b) Find the value of $f$ .

(4)

题目中文翻译

质点 $A$ 、 $B$ 在光滑水平面上沿同一直线相向运动，发生正碰。 $A$ 的质量为 $2m$ ， $B$ 的质量为 $3m$ 。

碰撞后瞬间， $A$ 、 $B$ 以相同速率 $v$ 沿相反方向运动。

碰撞中， $A$ 受到大小为 $5mv$ 的冲量。

(a) 求 $A$ 与 $B$ 之间的恢复系数。

(6)

与 $A$ 碰撞后，质点 $B$ 撞上光滑固定竖直墙并反弹。墙面垂直于质点的运动方向。

$B$ 与墙之间的恢复系数为 $f$ 。

由于与 $A$ 及墙的碰撞， $B$ 损失的总动能为 $E$ 。由于与 $B$ 的碰撞， $A$ 损失的动能为 $2E$ 。

(b) 求 $f$ 的值。

(4)

解答

(a)

解法一

思路

展开

两物体碰后分离、速率同为 $v$ ，所以末速度符号相反。用已知冲量求各自碰前速率；两者受冲量大小相等。恢复系数分子为分离速率 $2v$ 。

答题过程

展开

Take the initial direction of motion of $A$ as positive. Let the initial velocities of $A$ and $B$ be $x$ and $-y$ , where $x,y>0$ . Immediately after collision their velocities are $-v$ and $v$ .

For $A$ , the impulse has magnitude $5mv$ , so

2m(x+v)=5mv,\qquad x=\frac32v.

The impulse on $B$ has the same magnitude and opposite direction. Thus

3m(v+y)=5mv,\qquad y=\frac23v.

The separation speed is $v-(-v)=2v$ , while the approach speed is $x+y$ . Hence

e=\frac{2v}{(3/2)v+(2/3)v}.

Therefore

\boxed{e=\frac{12}{13}}.

(b)

解法一

思路

展开

$B$ 与墙碰后速率为 $fv$ 。 $B$ 的总动能损失直接由首次碰撞前到撞墙后计算，不必把两次损失分别展开。注意题目给的是 $A$ 的损失为 $B$ 的两倍。

答题过程

展开

After rebounding from the fixed wall, $B$ has speed $fv$ .

The total kinetic energy lost by $B$ is

E=\frac12(3m)\left[\left(\frac{2v}{3}\right)^2-(fv)^2\right].

The kinetic energy lost by $A$ is

2E=\frac12(2m)\left[\left(\frac{3v}{2}\right)^2-v^2\right].

Consequently,

2\left[\frac32mv^2\left(\frac49-f^2\right)\right]=mv^2\left(\frac94-1\right).

Cancelling $mv^2$ ,

3\left(\frac49-f^2\right)=\frac54.

Thus

f^2=\frac49-\frac5{12}=\frac1{36}.

Since a coefficient of restitution is non-negative,

\boxed{f=\frac16}.

M2 2021 Oct Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程