竖直的光滑墙面上有 A 和 B 两个钉子，二者处于同一水平高度，间距离为 l。有一原长为 l、劲度系...

创建于：2025年3月23日

问题

竖直的光滑墙面上有 A 和 B 两个钉子，二者处于同一水平高度，间距离为 l。有一原长为 l、劲度系数为 k 的轻橡皮筋，一端由 A 钉固定，另一端系有一质量为 m=kl/(4g) 的小球，其中 g 为重力加速度。钉子和小球都可视为质点，小球和任何物体的碰撞都是完全非弹性碰撞，且不发生粘连。现将小球水平向右拉伸到与 A 钉距离为 2l 的 C 点，（ C与A,B在同一条直线上, A，B处均有钉子把橡皮筋固定，但C处没有。）钉恰好处于橡皮筋下垂并始终与之光滑接触。初始时刻小球获得大小为 (√gl)/2，方向竖直向下的速度。试确定此后小球沿竖直方向的速度第一次为零的时刻。

思考

Analyzing the problem

OK, let’s break this down: we have a rubber band and a small mass celebrating inertia against a smooth wall. The goal? Find the exact moment when the mass stops vertically moving for the first time.

Assessing initial conditions

Hmm, I’m thinking about the initial setup: a rubber band, aligned to nail B, is stretched horizontally to twice its length, and the small mass starts with an initial downward velocity.Evaluating the motion

I'm figuring out when the small mass's vertical velocity hits zero. The rubber band, kept in place by nail B, likely causes intricate motion for the ball.

Analyzing energy dynamics

I'm considering energy and geometry to understand the time when the mass's vertical velocity hits zero. The tension in the rubber band and the system's constraints are key factors.

Assessing force dynamics

OK, let's see. The tension in the rubber band rivals the mass's weight, with B being a geometric focal point for the band. This tension will shape the band's formation.Evaluating the setup

OK, let me see: the rubber band is anchored at A and extended to 2l. Now, I'm considering the band's geometry as it reaches nail B and the mass at C.

Considering the band's elasticity

OK, let me see. The rubber band stretches beyond its natural length and touches nail B when the mass is at 2l from A, creating a frictionless loop around B.

Clarifying band stretch

It's interesting to see that the band stretches beyond its natural length due to the frictionless loop around B, despite the total rest length being l. This confirms B is an extra point of extension beyond l.Understanding tension dynamics

The tension in the rubber band increases with the distance from B to the mass. This relationship hinges on the mass and the band's stretching properties, as described in the problem.

Tracing motion

The mass at (2l, 0) moves downward due to its initial velocity, constrained by the rubber band loop around B. I’m thinking about the time when the vertical velocity hits zero.

Taking a closer look

The rubber band is anchored at A with the mass at one end and nail B positioned between them. It's interesting to see the band touches B from below.

Understanding extension tension

The band forms a broken line due to gravity, stretching to a total length L. The extension from the rest length l to L results in tension k(L - l).

Ensuring tautness

OK, let me see. The band is taut with no slack, consistently touching B from below. This setup aligns with the problem's constraints, confirming the tension path from A to B to the mass.Understanding band dynamics

With a 2l distance from A to C, the band stretches between B and the mass. This setup ensures the band remains taut, accommodating the required tension.

Understanding tension dynamics

The band wraps around B with tension T on both segments AB and B-ball. The tension is proportional to the distance from B to the ball, ensuring uniform tension throughout.

Analysing tension dynamics

With the origin at B, the x-axis extends right, and the y-axis moves up. The tension T is aligned with the distance from B to the mass, which is x.

Mapping out coordinates

OK, let's reframe things: setting B at (0,0) means A is at (-l,0) and the ball is at (2l,0), maintaining the 2l distance horizontally from A to the ball.

Pinpointing the ball's location

OK, let's figure out: the ball's at (l, 0), 1l away from B, with the band’s rest length. This solidifies the tension calculation from B to the ball.

Calculating net force

Hmm, I’m thinking about how the net horizontal force on the ball is T = k l, which is four times the gravitational force. The vertical force is zero since the tension line is horizontal.

Noting the ball's motion

The ball's initial vertical force is just gravity, giving an initial downward velocity of sqrt(g l)/2. The tension holds it to B horizontally, combining with gravity to influence acceleration.

Breaking down the forces

Net force combines tension from B and gravity, resulting in (-k x, -k y - mg). The mass of the ball is given by m = k l/(4 g).

Reworking equations

I’m mapping out the ball's equations of motion, determining: m · x" = -k x and m · y" = -k y - mg. Using ω^2 = k/m, I find ω = 2 √(g/l).

Refining the equations

Interestingly, the x equation simplifies to a simple harmonic oscillator: $\ddot{x} = - 4g/l \, x$ . Meanwhile, the y equation becomes: $\ddot{y} = - 4g/l \, y - g$ , indicating a shifted harmonic oscillator.Examining initial conditions

Starting with x(0) = l and y(0) = 0, I’ve already noted x(0) condition. For y, $\dot{y}(0) = - \sqrt{g l}/2$ .

Solving for zero vertical velocity

I’m thinking through the system's equations to find the time when the vertical velocity first hits zero. Using homogeneous and particular solutions to y(t), I’m working towards this goal.

Setting initial conditions

I began by applying the initial conditions to determine coefficients $A$ and $B$ in the general solution $y(t) = A \cos(\omega t) + B \sin(\omega t) - \frac{g}{\omega^2}$ .

Revisiting dimensional dimensions

I confirmed that $\omega$ and coefficients $A$ and $B$ align dimensionally with the system's parameters. Then, I revisited the homogeneous and particular solutions.

Determining coefficients

I meticulously verified $A$ and $B$ by ensuring their values accurately reflected the system's parameters and matched the initial conditions, confirming $y(t)$ as $l/4 \cos(\omega t) - l/4 \sin(\omega t) - l/4$ .

Ensuring sign accuracy

I meticulously calculated $\dot{y}(t)\ by carefully assessing the derivatives of the terms in the expression for \(y(t)$ , ensuring the accuracy of each sign in the result.

Identifying the time

I established that the first positive time when $\dot{y}(t)$ is zero, given $\omega = 2 \sqrt{g/l}$ , is $t_1 = \frac{3 \pi}{8} \sqrt{\frac{l}{g}}$ .

Ensuring collision-free motion

I'm piecing together the ball's trajectory, especially its vertical velocity and potential collisions, to confirm the motion remains collision-free, using equations for vertical and horizontal movements.

Establishing trajectory

I determined that $x(t) = l \cos(\omega t)$ with $\omega = 2 \sqrt{g/l}$ , and confirmed that the first time $x(t)$ is zero is later than $t_1$ , ensuring no collision with point B.

Confirming intersection

I verified that the ball intersects point B at $x = 0$ at time $t_B = \frac{\pi}{8} \sqrt{\frac{l}{g}}$ , and I determined that $y(t_B) = 0$ , ensuring no collision.Mapping out the intersection

I determined that the ball passes the vertical line $x = 0$ at $y = -\frac{l}{2}$ , which is half a ball's diameter below point B.

Determining collision

I concluded that there is no collision with point B, as the ball's trajectory does not intersect at $(0,0)$ . The problem details the wall and nails' positions.

Clarifying spatial setup

I’m rethinking the spatial context: the wall is vertical, but the nails could be on a horizontal line at the same height, maintaining the distance $l$ .

Mapping out trajectories

I’m piecing together the ball's trajectory, noting the initial horizontal position and ensuring no collision with nails or the ground by using a clear coordinate system.

Aligning the system

I’m adjusting the coordinate system for consistency, aligning B to $x=0$ and A to $x=-l$ , with the ball at $x=l$ . This setup simplifies equation management and ensures clarity.

Clarifying forces

I’m the problem by defining the position vector $\vec{r}$ from B, calculating forces from the spring and gravity, and ensuring vector components are accurately addressed.

Clarifying tension

I'm working through the tension on the mass from B to M, ensuring it's accurately defined as $k \times \text{extension}$ in the direction of $-\vec{r}/r$ .

Balancing forces

I’m piecing together the net force by combining the spring force $- k \vec{r}$ with the downward gravitational force $\vec{g} = (0, -g)$ . This ensures the system remains consistent.

Formulating system dynamics

I'm noting the system includes uncoupled equations for $x$ and $y$ , with constants $\omega^2 = k/m$ and the gravitational effect only in $y$ . Initial conditions define the mass's starting point.

Determining point locations

I'm defining the coordinate of point C, ensuring AC equals $2l$ , and aligning it appropriately with points A and B, optimizing the given constraints and distances.

Revisiting point locations

I’m re-evaluating the positions of A, B, and C, ensuring B is at $x=l$ and C at $x=2l$ , establishing a distance of $l$ between B and C.

Confirming alignment

I’m working through the positions of the mass and B, ensuring the distance aligns with the extension $l$ , and verifying the acceleration due to gravity and initial conditions as $x(0) = l$ .

Mapping coordinates

I setting the origin at B, establishing A at $-l$ , and determining C at $x=l$ to maintain a $2l$ distance from A. This arrangement ensures the mass's coordinate relative to B is positive.

Confirming orientation

I’m mapping out the initial conditions: B at 0, A at $-l$ , and C at $+l$ , the mass at $x = +l$ with an initial downward velocity of $\sqrt{gl}/2$ .

Setting initial parameters

I’m establishing the initial conditions for the equations of motion: $x(0) = l$ , $\dot{x}(0) = 0$ , $y(0) = 0$ , and $\dot{y}(0) = -\sqrt{gl}/2$ , defining $C$ and $D$ for the vertical motion.

Reaching first zero

I’m pulling together the expression for $y(t)$ and identifying when $\dot{y}(t) = 0$ . This leads to the finding $\omega t = 3\pi/4$ , giving the initial time $t_1$ .

Calculating collision time

OK, let me see. The mass hits nail B at $t_B = \frac{\pi}{4} \sqrt{\frac{l}{g}}$ , which is before the initial mass condition time $t_1 = \frac{3\pi}{8} \sqrt{\frac{l}{g}}$ .Assessing collision

I'm checking if the mass hits nail B. Since it's at (0, -l/2) and B's at (0,0), it feels like they're not intersecting. The mass likely meets point A next.

Pinpointing collision

I’m digging into when the mass hits point A. Calculating $\frac{\pi}{\omega}$ shows it’s after the initial mass's trajectory, so no collision occurs earlier.

Finished reasoning

回答

思路梗概：

确立坐标系与弹力表达：
由于题目叙述中，橡皮筋始终绕过钉子 $B$ 与小球相连，且弹簧原长正好等于 $AB$ 的距离 $l$ ，因此从几何上可以看出：
- 橡皮筋的“ $A \to B$ ”这段正好是它的原长 $l$ ，无论如何都不再伸长;
- 真正发生弹性伸长(并提供弹力)的部分是“ $B \to \text{小球}$ ”这段。
  由于 $B$ 与小球之间的距离记作 $\rho$ ，则该段的伸长量就是 $\rho$ ，故弹力大小为
$T \;=\; k\,\rho.$
方向由小球指向钉子 $B$ 。
简化为质点的受力模型：
若取钉子 $B$ 作为坐标原点，建立如图所示的平面直角坐标系 $(x,y)$ ，其中 $x$ 轴水平向右， $y$ 轴竖直向上。则小球位置矢量为
$\mathbf{r}(t)\;=\;\bigl(x(t),\,y(t)\bigr).$
弹力为
$\mathbf{F}_{\text{弹}} \;=\;-\,k\,\rho\,\hat{\rho} \;=\;-\,k\,\bigl|\mathbf{r}\bigr|\;\frac{\mathbf{r}}{\bigl|\mathbf{r}\bigr|} \;=\;-\,k\,\mathbf{r}.$
由此可见，弹力在 $x$ 方向和 $y$ 方向上分别就是
$F_x = -\,k\,x,\quad F_y = -\,k\,y.$
再加上重力 $\,(0,\,-mg)$ ，小球总受力为
$\mathbf{F}_{\text{总}} \;=\;\bigl(-k\,x,\;-\,k\,y\;-\;m\,g\bigr).$
因此牛顿第二定律给出在 $x$ 和 $y$ 两个方向的运动方程：
$m\,\ddot{x} \;=\;-\,k\,x, \quad\quad m\,\ddot{y} \;=\;-\,k\,y \;-\;m\,g.$
确定系统固有角频率 $\omega$ ：
由题目给出
$m \;=\; \frac{k\,l}{4\,g},$
则
$\omega^2 \;=\;\frac{k}{m} \;=\;\frac{k}{\,k\,l/(4g)\,} \;=\;\frac{4\,g}{\,l\,} \;\;\Longrightarrow\;\; \omega \;=\;2\,\sqrt{\tfrac{g}{l}}.$
因而方程可写成
$\ddot{x} \;+\; \omega^2\,x \;=\;0, \quad\quad \ddot{y} \;+\; \omega^2\,y \;=\;-\,g.$
初始条件：
- 以 $B$ 为原点时，小球最初被拉到与 $B$ 相距 $l$ 的点(题目中说「与 $A$ 距离 $2l$ ，而 $B$ 与 $A$ 距离 $l$ ，可知 $B$ 到小球也是 $l$ )，且这时小球与 $B$ 在同一水平线上，所以 $x(0) \;=\;l,\quad y(0)\;=\;0.$
- 初速度只有竖直向下分量，大小为 $\tfrac{\sqrt{g\,l}}{2}$ ，方向向下即 $y$ 轴负向，故 $\dot{x}(0)\;=\;0,\quad \dot{y}(0)\;=\;-\,\frac{\sqrt{g\,l}}{\,2\,}.$
分别求解 $x(t)$ 和 $y(t)$ ：
- (i) $x$ 方向简谐振动：
  $\ddot{x} + \omega^2\,x = 0 \quad\Longrightarrow\quad x(t) = C\cos(\omega t)+D\sin(\omega t).$
  用 $x(0)=l,\;\dot{x}(0)=0$ 即可得
  $x(0)=l \;\Rightarrow\; C=l,\quad \dot{x}(0)=\omega\,D=0 \;\Rightarrow\; D=0.$
  因而
  $x(t) \;=\;l\,\cos(\omega t).$
- (ii) $y$ 方向简谐振动+恒力项：
  $\ddot{y} + \omega^2\,y \;=\;-\,g.$
  其通解为「齐次解 + particular解」：
  $y(t) \;=\;C'\cos(\omega t)\;+\;D'\sin(\omega t)\;-\;\frac{g}{\,\omega^2\,}.$
  其中 $\displaystyle -\,\dfrac{g}{\omega^2}$ 是常数特解(使左边 $\ddot{y}+\omega^2\,y$ 恰好等于 $-g$ )。
  用初值
  $y(0)=0 \;\;\Longrightarrow\;\; C' - \frac{g}{\omega^2}=0 \;\;\Longrightarrow\;\; C' = \frac{g}{\omega^2},$ $\dot{y}(0)= -\,\frac{\sqrt{g\,l}}{2} \;\;\Longrightarrow\;\; \dot{y}(0)=\omega\,D' \;=\; -\,\frac{\sqrt{g\,l}}{2} \;\;\Longrightarrow\;\; D'=-\,\frac{\sqrt{g\,l}}{\,2\,\omega\,}.$
  由于 (\displaystyle \omega^2=\frac{4,g}{l} ;\Longrightarrow; \omega=2\sqrt{\tfrac{g}{l}},) 可以验证 (\displaystyle \frac{g}{\omega^2}=\frac{l}{4},;; -,\frac{\sqrt{g,l}}{2,\omega}=-,\frac{l}{4}. ) 因而
  $y(t) = \frac{l}{4}\cos(\omega t) \;-\;\frac{l}{4}\,\sin(\omega t) \;-\;\frac{l}{4} \;=\; \frac{l}{4}\Bigl[\cos(\omega t)-\sin(\omega t)-1\Bigr].$
求竖直速度 $\dot{y}$ 第一次为零的时刻：
$\dot{y}(t) = \frac{l}{4}\,\Bigl[-\,\omega\,\sin(\omega t)\;-\;\omega\,\cos(\omega t)\Bigr] = -\,\frac{\omega\,l}{4}\,\Bigl[\sin(\omega t)+\cos(\omega t)\Bigr].$
令 $\dot{y}(t)=0$ 等价于
$\sin(\omega t)\;+\;\cos(\omega t)\;=\;0 \;\;\Longrightarrow\;\; \tan(\omega t) \;=\;-\,1 \;\;\Longrightarrow\;\; \omega t \;=\;-\tfrac{\pi}{4}+n\pi.$
寻找 $t>0$ 的最小解：令 $n=1$ 得
$\omega\,t = -\,\frac{\pi}{4} + \pi = \frac{3\,\pi}{4} \;\;\Longrightarrow\;\; t \;=\; \frac{3\,\pi}{\,4\,\omega\,}.$
又因为 $\,\omega=2\sqrt{\frac{g}{l}}$ , 故
$t \;=\; \frac{3\pi}{4\,\bigl(2\sqrt{g/l}\bigr)} \;=\; \frac{3\pi}{8}\,\sqrt{\frac{\,l\,}{\,g\,}}.$
这就是竖直分速度第一次归零的时刻。
碰撞(约束切换)是否会提前打断运动？
- 小球若要与钉子 $B$ 碰撞，必须同时满足
  $\,x(t)=0,\;y(t)=0.$
  但从上面可算出，当 $x(t)$ 首次为 0 时(即 $\cos(\omega t)=0$ )，那一刻 $y(t)$ 为 $-\,\tfrac{l}{2}$ ，并不在同一垂直高度，因此并未真正“撞”到 $B$ 。
- 小球也不会去撞到 $A$ (那要 $x=-l,y=0$ )，更不会“黏”在其它地方。
因此在第一次出现 $\dot{y}=0$ 之前，并无任何碰撞(完全非弹性碰撞)发生来改变其速度，故上面算出的时刻依然有效。

最终答案

根据上述分析，小球在此后的运动中，竖直方向速度首次变为零的时刻为

\boxed{\;t \;=\; \frac{3\pi}{8}\,\sqrt{\frac{l}{\,g\,}}\,. }

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