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M1 2022 Oct Q1

A Level / Edexcel / M1

IAL 2022 Oct Paper · Question 1

题目

Problem

1.

A railway truck SS of mass 2020 tonnes is moving along a straight horizontal track when it collides with another railway truck TT of mass 3030 tonnes which is at rest. Immediately before the collision the speed of SS is 4 m s−14\,\mathrm{m\,s^{-1}}.

As a result of the collision, the two railway trucks join together.

Find

(a) the common speed of the railway trucks immediately after the collision,

(2)

(b) the magnitude of the impulse exerted on SS in the collision, stating the units of your answer.

(3)
题目中文翻译

一节质量为 2020 吨的货车厢 SS 沿水平直轨道运动,与另一节静止、质量为 3030 吨的货车厢 TT 发生碰撞。碰撞前一瞬间,SS 的速率为 4 m s−14\,\mathrm{m\,s^{-1}}。

碰撞后,两节货车厢连接在一起。

求

(a) 碰撞后瞬间两节货车厢的共同速率,

(2)

(b) 碰撞过程中 SS 所受冲量的大小,并写出答案的单位。

(3)

解答

(a)

解法一

思路

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两车碰后连在一起,具有共同速度。以原来运动方向为正,对整个系统用动量守恒,质量单位统一为千克。

答题过程

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The truck masses are 20000 kg and 30000 kg. Let their common velocity after the collision be vv, taking the original direction of SS as positive.

Conservation of momentum gives

20000(4)+30000(0)=(20000+30000)v.20000(4)+30000(0)=(20000+30000)v.

Hence the common speed is v=1.6 m s−1\boxed{v=1.6\ \mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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冲量是单个车厢的动量变化,必须用千克计算才能得到牛顿秒。求大小时把负的冲量取绝对值。

答题过程

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The impulse on SS is

IS=20000(1.6−4)=−48000 N s.I_S=20000(1.6-4)=-48000\ \mathrm{N\,s}.

Therefore its magnitude is 48000 N s\boxed{48000\ \mathrm{N\,s}}.

解法二

思路

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也可先求静止车厢 TT 的动量变化,再用两个车厢受到的碰撞冲量等大反向。

答题过程

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The impulse on TT is

IT=30000(1.6−0)=48000 N s.I_T=30000(1.6-0)=48000\ \mathrm{N\,s}.

The impulse on SS is equal and opposite, so its magnitude is 48000 N s\boxed{48000\ \mathrm{N\,s}}.