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CIE 9231 2024 June Paper 33 Q6

A Level / CIE / FM

CIE 9231 2024 June Paper 33 Paper · Question 6

题目

Problem

A particle PP is projected with speed u m s−1u\text{ m s}^{-1} at an angle θ\theta above the horizontal from a point OO and moves freely under gravity. After 5 seconds the speed of PP is 34u\frac34u.

(a) Show that 716u2−100usin⁡θ+2500=0\frac{7}{16}u^2-100u\sin\theta+2500=0.

(3)

(b) It is given that the velocity of PP after 5 seconds is perpendicular to the initial velocity.

Find, in either order, the value of uu and the value of sin⁡θ\sin\theta.

(5)
题目中文翻译

质点 PP 从点 OO 以速率 u m s−1u\text{ m s}^{-1}、与水平方向成角 θ\theta 向上抛出，并仅在重力作用下自由运动。5 秒后，PP 的速率为 34u\frac34u。

(a) 证明 716u2−100usin⁡θ+2500=0\frac{7}{16}u^2-100u\sin\theta+2500=0。

(3)

(b) 已知 PP 在 5 秒后的速度与初速度垂直。

求 uu 和 sin⁡θ\sin\theta 的值，求解顺序不限。

(5)
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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时，取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

(a)

解法一

思路

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重力只改变竖直速度分量。将第 55 秒时的两个速度分量平方相加，并与 (3u/4)2(3u/4)^2 相等。

答题过程

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At t=5t=5, the velocity is (ucos⁡θ,usin⁡θ−50)(u\cos\theta,u\sin\theta-50). The square of its speed is

 (ucos⁡θ)2+(usin⁡θ−50)2= u2cos⁡2θ+u2sin⁡2θ−100usin⁡θ+2500= u2−100usin⁡θ+2500.\begin{align*} &\,(u\cos\theta)^2+(u\sin\theta-50)^2\\[4mm] =&\,u^2\cos^2\theta+u^2\sin^2\theta-100u\sin\theta+2500\\[4mm] =&\,u^2-100u\sin\theta+2500. \end{align*}

Equating this to 9u2/169u^2/16 and rearranging gives 716u2−100usin⁡θ+2500=0\frac7{16}u^2-100u\sin\theta+2500=0.

(b)

解法一

思路

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用两个运动方向的斜率乘积为 −1-1 求出 uu 与 sin⁡θ\sin\theta 的关系，再代入 (a)；也可等价地使用两个速度向量的点积为零。

答题过程

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Let α\alpha be the direction angle at t=5t=5. Then tan⁡α=(usin⁡θ−50)/(ucos⁡θ)\tan\alpha=(u\sin\theta-50)/(u\cos\theta). Since the directions are perpendicular, tan⁡αtan⁡θ=−1\tan\alpha\tan\theta=-1, so

usin⁡2θ−50sin⁡θ= −ucos⁡2θ,u= 50sin⁡θ.\begin{align*} u\sin^2\theta-50\sin\theta=&\,-u\cos^2\theta,\\[4mm] u=&\,50\sin\theta. \end{align*}

Equivalently, the initial velocity is (ucos⁡θ,usin⁡θ)(u\cos\theta,u\sin\theta), whose dot product with the velocity at t=5t=5 is

 u2cos⁡2θ+usin⁡θ(usin⁡θ−50)= u2−50usin⁡θ.\begin{align*} &\,u^2\cos^2\theta+u\sin\theta(u\sin\theta-50)\\[4mm] =&\,u^2-50u\sin\theta. \end{align*}

This is zero, which again gives u=50sin⁡θu=50\sin\theta since u>0u>0. Substituting sin⁡θ=u/50\sin\theta=u/50 into part (a) gives

716u2−2u2+2500= 0,2516u2= 2500,u2= 1600.\begin{align*} \frac7{16}u^2-2u^2+2500=&\,0,\\[4mm] \frac{25}{16}u^2=&\,2500,\\[4mm] u^2=&\,1600. \end{align*}

Taking the positive speed gives u=40 m s−1u=40\,\mathrm{m\,s^{-1}} and sin⁡θ=4/5\sin\theta=4/5.

解法二

思路

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也可直接把初速度的方向顺时针转 90∘90^\circ，得到第 55 秒速度的方向。比较两个速度向量的水平分量及竖直分量。

答题过程

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The later velocity still points forwards horizontally and is perpendicular to the initial velocity, so its components are (34usin⁡θ,−34ucos⁡θ)(\frac34u\sin\theta,-\frac34u\cos\theta). The horizontal component is unchanged, so

34usin⁡θ= ucos⁡θ,tan⁡θ= 43.\begin{align*} \frac34u\sin\theta=&\,u\cos\theta,\\[4mm] \tan\theta=&\,\frac43. \end{align*}

Thus sin⁡θ=4/5\sin\theta=4/5 and cos⁡θ=3/5\cos\theta=3/5. Comparing the vertical components gives

usin⁡θ−50= −34ucos⁡θ,50= u(45+920)=54u.\begin{align*} u\sin\theta-50=&\,-\frac34u\cos\theta,\\[4mm] 50=&\,u\bigg(\frac45+\frac9{20}\bigg)=\frac54u. \end{align*}

Hence u=40 m s−1u=40\,\mathrm{m\,s^{-1}}.

解法三

思路

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从水平方向先求出 sin⁡θ\sin\theta 后，也可以代入 (a) 解二次方程。要用第 55 秒速度的竖直方向判断哪一个根符合垂直条件。

答题过程

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Perpendicularity and the common positive horizontal component give ucos⁡θ=(3u/4)sin⁡θu\cos\theta=(3u/4)\sin\theta, so sin⁡θ=4/5\sin\theta=4/5. Part (a) then becomes

7u2−1280u+40000=(u−40)(7u−1000)=0.7u^2-1280u+40000=(u-40)(7u-1000)=0.

The later velocity must point downwards, so 4u/5−50<04u/5-50<0. This excludes u=1000/7u=1000/7, leaving u=40 m s−1u=40\,\mathrm{m\,s^{-1}}.