题目

Problem

One end of a light elastic string, of natural length $l$ and modulus of elasticity $\lambda$ , is fixed to a point $A$ on a smooth plane. The plane is inclined at $30^\circ$ to the horizontal.

A small ball $B$ of mass $m$ is attached to the other end of the elastic string. Initially, $B$ is held at rest at the point $C$ on the plane with the elastic string lying along a line of greatest slope of the plane.

The point $C$ is below $A$ and $AC=l$ , as shown in Figure 2.

The ball is released and comes to instantaneous rest at a point $D$ on the plane.

The points $A$ , $C$ and $D$ all lie along a line of greatest slope of the plane and $AD=\dfrac{5l}4$

The ball is modelled as a particle and air resistance is modelled as being negligible.

Using the model,

(a) show that $\lambda=4mg$

(4)

(b) find, in terms of $g$ and $l$ , the greatest speed of $B$ as it moves from $C$ to $D$

(7)

题目中文翻译

一根原长为 $l$ 、弹性模量为 $\lambda$ 的弹性轻绳的一端固定于光滑斜面上的 $A$ 点，斜面与水平面成 $30^\circ$ 角。

质量为 $m$ 的小球 $B$ 连接于弹性绳的另一端。最初将 $B$ 静止保持在斜面上的 $C$ 点，弹性绳沿斜面的最大坡度线。

如图 2 所示， $C$ 位于 $A$ 下方，且 $AC=l$ 。

释放小球后，它在斜面上的 $D$ 点瞬时静止。

$A$ 、 $C$ 、 $D$ 均位于斜面的同一条最大坡度线上，且 $AD=\dfrac{5l}4$ 。

将小球视为质点，并忽略空气阻力。

利用此模型，

(a) 证明 $\lambda=4mg$ ；

(4)

(b) 用 $g$ 和 $l$ 表示 $B$ 从 $C$ 到 $D$ 过程中的最大速率。

(7)

解答

(a)

解法一

思路

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在 $C,D$ 两点速率都为零。重力势能损失全部成为绳的弹性势能；下降高度是沿斜面路程乘 $\sin30^\circ$ 。

答题过程

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At $C$ the string has its natural length, so its elastic potential energy is zero. At $D$ its extension is

e_D=AD-l=\frac l4.

The vertical fall from $C$ to $D$ is

CD\sin30^\circ=\frac l4\cdot\frac12=\frac l8.

The plane is smooth and the ball is at rest at both ends. Conservation of energy gives

mg\frac l8=\frac{\lambda}{2l}\left(\frac l4\right)^2 =\frac{\lambda l}{32}.

Cancelling $l$ and multiplying by $32$ ,

\boxed{\lambda=4mg.}

(b)

解法一

思路

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沿坡向下的重力分量不变，张力随伸长增加。先加速后减速，合力为零处取得最大速率；再从 $C$ 写能量方程。

答题过程

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Let $e$ be the extension of the string. During the motion from $C$ to $D$ , $0\leq e\leq l/4$ . Resolving down the plane,

m\ddot e=mg\sin30^\circ-\frac{\lambda e}{l} =\frac{mg}{2}-\frac{4mg}{l}e.

The acceleration is positive for $e<l/8$ and negative for $e>l/8$ . Since the ball moves down the plane between $C$ and $D$ , its speed is greatest when

\frac{mg}{2}=\frac{4mg}{l}e, \qquad e=\frac l8.

At this position, conservation of energy from $C$ gives

mg\left(\frac l8\sin30^\circ\right) =\frac12mv_{\max}^2+ \frac{4mg}{2l}\left(\frac l8\right)^2.

Hence

\frac{mgl}{16}=\frac12mv_{\max}^2+\frac{mgl}{32}, \qquad v_{\max}^2=\frac{gl}{16}.

Taking the positive root,

\boxed{v_{\max}=\frac{\sqrt{gl}}4.}

解法二

思路

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把位移改为从沿坡平衡位置量起，恢复力正比于反向位移，便可用简谐运动的中心最大速率。

答题过程

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At the equilibrium position $E$ ,

\frac{4mg}{l}e_E=\frac{mg}{2}, \qquad e_E=\frac l8.

Let $y$ be displacement down the plane from $E$ . The extension is $l/8+y$ , so

m\ddot y=\frac{mg}{2}-\frac{4mg}{l}\left(\frac l8+y\right) =-\frac{4mg}{l}y.

Therefore

\ddot y=-\frac{4g}{l}y, \qquad \omega=2\sqrt{\frac gl}.

At $C$ the ball is released from rest at $y=-l/8$ . Its amplitude is $b=l/8$ . Throughout the motion to $D$ , the extension is non-negative, so the SHM equation applies. The greatest speed occurs at $E$ and is

\boxed{v_{\max}=b\omega=\frac l8\cdot2\sqrt{\frac gl} =\frac{\sqrt{gl}}4.}

M3 2022 June Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程