题目

Problem

Two particles, $P$ of mass $9m$ and $Q$ of mass $3m$ , are at rest on a smooth horizontal surface.

Particle $P$ is projected with speed $4u$ towards $Q$ and $P$ collides with $Q$ .

The coefficient of restitution between the particles is $e$ .

(a) Show that the speed of $Q$ immediately after the collision is $3u(1+e)$ .

(5)

The total kinetic energy lost in the collision between $P$ and $Q$ is $kmu^2$

Given that $e=\dfrac23$

(b) find the value of $k$ .

(4)

After being struck by $P$ , particle $Q$ goes on to hit a fixed vertical wall. The wall is perpendicular to the direction of motion of $Q$ .

In the collision with the wall, $Q$ receives an impulse of magnitude $21mu$ .

(3)

题目中文翻译

两个质点 $P$ 、 $Q$ 静止在光滑水平面上，质量分别为 $9m$ 、 $3m$ 。

将质点 $P$ 以速率 $4u$ 向 $Q$ 发射， $P$ 与 $Q$ 发生碰撞。

质点之间的恢复系数为 $e$ 。

(a) 证明碰撞后瞬间 $Q$ 的速率为 $3u(1+e)$ 。

(5)

$P$ 、 $Q$ 碰撞中损失的总动能为 $kmu^2$ 。

已知 $e=\dfrac23$ ，

(b) 求 $k$ 的值。

(4)

受到 $P$ 撞击后，质点 $Q$ 继续运动并撞向固定竖直墙。墙面垂直于 $Q$ 的运动方向。

与墙碰撞中， $Q$ 受到大小为 $21mu$ 的冲量。

(3)

解答

(a)

解法一

思路

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碰撞中系统动量守恒，恢复定律给出末相对分离速度。两式联立求 Q 的速度，并说明其为正所以也是速率。

答题过程

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Take the initial direction of $P$ as positive, and let the velocities after collision be $v_P$ and $v_Q$ . Conservation of momentum gives

9m(4u)=9mv_P+3mv_Q,\qquad 12u=3v_P+v_Q.

Newton’s law of restitution gives

v_Q-v_P=e(4u),\qquad v_P=v_Q-4eu.

Substituting,

12u=3(v_Q-4eu)+v_Q,

4v_Q=12u(1+e),\qquad \boxed{v_Q=3u(1+e)}.

Since $u>0$ and $0\leq e\leq1$ , this is positive and is also the speed of $Q$ .

(b)

解法一

思路

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损失的动能是碰前总动能减碰后两粒子的总动能，不能遗漏 P 的末动能。

答题过程

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With $e=2/3$ ,

v_Q=3u\left(1+\frac23\right)=5u,

v_P=v_Q-4eu=5u-\frac83u=\frac73u.

The kinetic energy lost is

\Delta E=\frac12(9m)(4u)^2-\frac12(9m)\left(\frac73u\right)^2-\frac12(3m)(5u)^2.

Hence

\Delta E=\left(72-\frac{49}{2}-\frac{75}{2}\right)mu^2=10mu^2.

Comparing with $kmu^2$ ,

\boxed{k=10}.

(c)

解法一

思路

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墙使 Q 反弹，所以碰后速度与碰前反向。用冲量的有符号动量变化求反弹速率，再除以入射速率。

答题过程

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Let $w>0$ be the rebound speed. Taking motion towards the wall as positive, the velocities of $Q$ immediately before and after impact are $5u$ and $-w$ . The impulse is directed away from the wall, so

-21mu=3m(-w-5u).

Thus

w+5u=7u,\qquad w=2u.

As the wall is fixed, its coefficient of restitution with $Q$ is

\boxed{\frac{w}{5u}=\frac25}.

M2 2025 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

(c)

解法一

思路

答题过程