题目

Problem

Two uniform smooth spheres $A$ and $B$ of equal radii have masses $2m$ and $m$ respectively. Sphere $A$ is moving in a straight horizontal line with speed $u$ , and sphere $B$ is stationary. Sphere $A$ collides directly with $B$ , and they both then move in the same direction with speeds $v_A$ and $v_B$ respectively. After the collision, the kinetic energy of $B$ is $\frac92$ times the kinetic energy of $A$ .

(a) Show that $v_B=\frac65u$ .

(3)

Sphere $B$ then collides with a fixed vertical barrier. Immediately before the collision, the direction of motion of $B$ makes an angle $\alpha$ with the barrier. Immediately after the collision, the direction of motion of $B$ makes an angle $\beta$ with the barrier. The coefficient of restitution between $B$ and the barrier is $\frac45$ . As a result of the collision, the velocity of $B$ is reduced to $\frac{12}{25}\sqrt5u$ .

(b) Find the value of $\sin(\alpha+\beta)$ .

(6)

题目中文翻译

两个半径相等的均匀光滑球 $A$ 和 $B$ 的质量分别为 $2m$ 和 $m$ 。球 $A$ 以速率 $u$ 沿水平直线运动，球 $B$ 静止。球 $A$ 与 $B$ 发生正碰，碰后两球沿相同方向运动，速率分别为 $v_A$ 和 $v_B$ 。碰后， $B$ 的动能是 $A$ 的动能的 $\frac92$ 倍。

(a) 证明 $v_B=\frac65u$ 。

(3)

随后，球 $B$ 与一个固定的竖直挡板发生碰撞。碰撞前瞬间， $B$ 的运动方向与挡板成 $\alpha$ 角；碰撞后瞬间， $B$ 的运动方向与挡板成 $\beta$ 角。 $B$ 与挡板之间的恢复系数为 $\frac45$ 。由于此次碰撞， $B$ 的速度减小至 $\frac{12}{25}\sqrt5u$ 。

(b) 求 $\sin(\alpha+\beta)$ 的值。

(6)

原卷通用作答要求

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 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

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把两球碰后的动能比换成速率比，再使用第一次碰撞的动量守恒。

答题过程

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The kinetic energy condition gives

\begin{align*} \frac12 mv_B^2=&\,\frac92\bigg(\frac12(2m)v_A^2\bigg),\\[4mm] v_B^2=&\,9v_A^2,\\[4mm] v_B=&\,3v_A \end{align*}

because the speeds are positive. The spheres move in the same direction, so conservation of momentum gives

2mu=2mv_A+mv_B=5mv_A.

Hence $v_A=2u/5$ and $v_B=6u/5$ .

(b)

解法一

思路

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题给角度是速度与墙面的夹角。平行墙面的分量不变，垂直分量反向且大小乘恢复系数。用碰后速率求出两个角的三角比。

答题过程

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Write $w=\frac{12}{25}\sqrt5\,u$ for the speed after the wall collision. The parallel component is unchanged, and the magnitude of the normal component is multiplied by $4/5$ , giving

\begin{align*} w\cos\beta=&\,v_B\cos\alpha,\\[4mm] w\sin\beta=&\,\frac45v_B\sin\alpha. \end{align*}

Since $w/v_B=(12\sqrt5/25)/(6/5)=2/\sqrt5$ , squaring, adding and dividing by $v_B^2$ gives

\begin{align*} \frac45=&\,\cos^2\alpha+\frac{16}{25}\sin^2\alpha\\[4mm] =&\,1-\frac9{25}\sin^2\alpha,\\[4mm] \frac9{25}\sin^2\alpha=&\,\frac15,\\[4mm] \sin^2\alpha=&\,\frac59. \end{align*}

Thus $\sin\alpha=\sqrt5/3$ and $\cos\alpha=2/3$ . Since $w/v_B=2/\sqrt5$ , we obtain $\cos\beta=\sqrt5/3$ and $\sin\beta=2/3$ . Therefore

\begin{align*} \sin(\alpha+\beta)=&\,\sin\alpha\cos\beta+\cos\alpha\sin\beta\\[4mm] =&\,\frac{\sqrt5}{3}\frac{\sqrt5}{3}+\frac23\frac23\\[4mm] =&\,\frac59+\frac49=\boxed1. \end{align*}

CIE 9231 2025 June Paper 31 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