一颗质量为 $0.1\,\mathrm{kg}$ 的珠子 $A$ 穿在一根长直刚性金属丝上，金属丝与水平面的夹角为 $\sin^{-1}\big(\frac7{25}\big)$。$A$ 从静止释放并沿金属丝向下运动。$A$ 与金属丝之间的摩擦系数为 $\mu$。当 $A$ 沿金属丝向下移动 $0.45\,\mathrm{m}$ 时，其速率为 $0.6\,\mathrm{m\,s^{-1}}$。

(a) 证明 $\mu=0.25$。

另一颗质量为 $0.5\,\mathrm{kg}$ 的珠子 $B$ 也穿在该金属丝上。$A$ 沿金属丝向下移动 $0.45\,\mathrm{m}$ 时，撞到在金属丝上瞬时静止的 $B$。碰撞使 $A$ 瞬时静止。$B$ 与金属丝之间的摩擦系数为 $0.275$。

(b) 求从此次碰撞发生起，直到 $A$ 再次与 $B$ 碰撞为止的时间。

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 $10\,\mathrm{m\,s^{-2}}$。

先由位移和末速求恒加速度，再用沿线方向的牛顿第二定律求摩擦力及摩擦系数。

Let the inclination be $\theta$. Since $\sin\theta=7/25$ and the wire slopes above the horizontal,

The acceleration of $A$ is constant. Using $v^2=u^2+2as$,

The normal reaction is

Resolving down the wire,

Thus $0.96\mu=0.28-0.04=0.24$, giving $\boxed{\mu=0.25}$, as required.

也可用功—能法，把重力势能减少量分为动能增加与克服摩擦的功。

As $A$ moves $0.45\,\mathrm{m}$ down the wire, the loss of gravitational potential energy and gain in kinetic energy are

The normal reaction is $R=0.1(10)(24/25)=0.96\,\mathrm{N}$. Hence the work-energy equation gives

Thus $0.432\mu=0.108$, so $\boxed{\mu=0.25}$, as required.

碰撞时沿线方向动量守恒求B的初速。碰后A重新由静止下滑，两者加速度不同；从碰撞点写位移并令其相等。

Taking down the wire as positive, conservation of momentum at the first collision gives

Hence $u_B=0.12\,\mathrm{m\,s^{-1}}$.

After the collision, $A$ again accelerates down the wire at

For $B$, Newton’s second law gives

so $a_B=0.16\,\mathrm{m\,s^{-2}}$.

Let $t$ be the time since the collision, with displacement measured down the wire from the collision point. Then

For the next collision,

The root $t=0$ is the first collision, so the required elapsed time is $\boxed{1\,\mathrm{s}}$.