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CIE 9709 2023 November Paper 43 Q2

A Level / CIE / M1

CIE 9709 2023 Nov Paper 43 Paper · Question 2

题目

Problem

A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of a nail. The mass of the hammerhead is 1.2 kg1.2\,\mathrm{kg} and the mass of the nail is 0.004 kg0.004\,\mathrm{kg} (see diagram). The hammerhead hits the nail with speed v m s−1v\,\mathrm{m\,s^{-1}} and remains in contact with the nail after the impact. The combined hammerhead and nail move immediately after the impact with speed 40 m s−140\,\mathrm{m\,s^{-1}}.

(a) Calculate vv, giving your answer as an exact fraction.

(2)

(b) The nail is driven 4 cm4\,\mathrm{cm} into the wood.

Find the constant force resisting the motion.

(3)

题目中文翻译

一台将钉子打入木块的机器使锤头竖直下落,撞击钉子顶端。锤头的质量为 1.2 kg1.2\,\mathrm{kg},钉子的质量为 0.004 kg0.004\,\mathrm{kg}(见图)。锤头以 v m s−1v\,\mathrm{m\,s^{-1}} 的速率撞击钉子,并在撞击后与钉子保持接触。撞击后瞬间,锤头与钉子作为整体以 40 m s−140\,\mathrm{m\,s^{-1}} 的速率运动。

(a) 计算 vv,以精确分数形式给出答案。

(2)

(b) 钉子被打入木块 4 cm4\,\mathrm{cm}。

求阻碍运动的恒力。

(3)

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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碰撞前钉子静止,碰撞后锤头与钉子共同运动。按要求保留精确分数。

答题过程

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Taking downwards as positive, conservation of momentum gives

1.2v=(1.2+0.004)(40)=48.16.1.2v=(1.2+0.004)(40)=48.16.

Hence

v=48.161.2=60215 m s−1.\boxed{v=\frac{48.16}{1.2}=\frac{602}{15}\,\mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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先用4厘米减速到静止求加速度。重力沿向下,木块的阻力向上,不能漏掉重力。

答题过程

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The combined mass is 1.204 kg1.204\,\mathrm{kg} and the stopping distance is 0.04 m0.04\,\mathrm{m}. Taking downwards as positive,

0=402+2a(0.04),a=−20000 m s−2.0=40^2+2a(0.04),\qquad a=-20000\,\mathrm{m\,s^{-2}}.

If the resistance magnitude is RR, Newton’s second law gives

1.204(10)−R=1.204(−20000).1.204(10)-R=1.204(-20000).

Hence

R=24092.04 N≈24100 N(3 s.f.).R=24092.04\,\mathrm{N} \approx\boxed{24100\,\mathrm{N}}\quad\text{(3 s.f.)}.

解法二

思路

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也可用能量法,初动能加下降损失的势能,全部用于克服木块阻力。

答题过程

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During the 0.04 m0.04\,\mathrm{m} descent, the work done against resistance equals the initial kinetic energy plus the loss in potential energy:

0.04R= 12(1.204)(402) +1.204(10)(0.04).\begin{align*} 0.04R=&\,\frac12(1.204)(40^2)\\[4mm] &\,\hspace{2pt}+1.204(10)(0.04). \end{align*}

Thus

R=963.2+0.48160.04=24092.04 N,R=\frac{963.2+0.4816}{0.04}=24092.04\,\mathrm{N},

giving

R≈24100 N(3 s.f.).R\approx\boxed{24100\,\mathrm{N}}\quad\text{(3 s.f.)}.