质点 $P$ 和 $Q$ 的质量分别为 $0.3\,\mathrm{kg}$ 和 $0.6\,\mathrm{kg}$。两质点被保持静止在粗糙斜面的一条最大坡度线上，相距 $2.75\,\mathrm{m}$。斜面与水平面的夹角为 $\alpha$，其中 $\sin\alpha=0.28$（见图）。每个质点与斜面之间的摩擦系数均为 $0.5$。

随后释放质点 $P$ 和 $Q$。

(a) 证明两个质点都不移动。

现在将质点 $P$ 以 $6\,\mathrm{m\,s^{-1}}$ 的速率直接朝 $Q$ 发射。

(b) 求 $P$ 在与 $Q$ 碰撞前一瞬间的速率。

(c) 碰撞后一瞬间，$P$ 开始以 $1\,\mathrm{m\,s^{-1}}$ 的速率沿斜面向上运动。

求碰撞后两质点都停止时它们之间的距离。

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 ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

静止摩擦力不是自动取最大值；比较维持静止所需的重力沿坡分量与最大静摩擦力。

Since $\cos\alpha=\sqrt{1-0.28^2}=0.96$, for either particle of mass $m$ kilograms,

The maximum available friction is

whereas the force required to prevent sliding is $mg\sin\alpha=0.28mg$. Since $0.28mg<0.48mg$, static friction can balance the downslope weight component for both particles. Neither particle moves.

也可把静止条件化为斜面角正切小于摩擦系数。

For equilibrium, friction must balance $mg\sin\alpha$ and must not exceed $\mu mg\cos\alpha$. Thus the condition is

Here

This condition is independent of mass, so both particles remain at rest.

P投射后下滑，摩擦取0.5R且向上，得到下坡加速度−2，再用路程2.75求碰前速度。

Taking down the plane as positive, for $P$,

Thus $a=2.8-4.8=-2\,\mathrm{m\,s^{-2}}$. Over $2.75\,\mathrm{m}$,

Hence the speed immediately before collision is $\boxed{5\text{ m s}^{-1}}$.

也可用初动能加下滑损失的势能，减去摩擦耗散求末动能。

For $P$, over $2.75\,\mathrm{m}$ down the plane,

Thus

The required speed is $\boxed{5\text{ m s}^{-1}}$.

碰撞先守恒动量，P反弹速度取负；之后P上行与Q下行的减速度不同，分别求路程再加。

Take down the plane as positive. Let $w$ be $Q$‘s velocity immediately after impact. Conservation of momentum gives

For $Q$ moving down the plane, $a_Q=-2\,\mathrm{m\,s^{-2}}$, as in (b). Thus its stopping distance is

For $P$ moving up the plane, gravity and friction both act down the plane. Taking upwards as positive,

so $a_P=-7.6\,\mathrm{m\,s^{-2}}$. Hence

The particles move in opposite directions from the collision point and remain at rest once stopped, by (a). Their final separation is

碰后各段也可独立用能量方程求停止距离，再相加。

Taking down the plane as positive, conservation of momentum gives

For $Q$ descending distance $d_Q$, the initial kinetic energy plus the loss of potential energy equals work against friction:

Thus $2.7+1.68d_Q=2.88d_Q$, giving $d_Q=9/4$ metres.

For $P$ ascending distance $d_P$, its initial kinetic energy is used to gain potential energy and overcome friction:

Thus $0.15=(0.84+1.44)d_P$, giving $d_P=5/76$ metres. By (a), each remains at rest once stopped. Their separation is