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M3 2025 Jan Q3

A Level / Edexcel / M3

IAL 2025 Jan Paper · Question 3

题目

Problem

3.

A light elastic string has natural length aa and modulus of elasticity mgmg.

One end of the elastic string is attached to a fixed point OO.

The other end is attached to a small smooth ring of mass mm.

The ring is threaded on a fixed smooth vertical pole which is a distance 32a\dfrac32a from OO.

Initially, the ring is held at the point AA on the pole with the elastic string horizontal.

The ring is then projected vertically downwards with speed ag\sqrt{ag} and reaches the point BB, where AB=2aAB=2a, with speed VV, as shown in Figure 2.

Air resistance is assumed to be negligible.

Find VV in terms of aa and gg.

(7)
题目中文翻译

一根弹性轻绳的原长为 aa,弹性模量为 mgmg。

绳的一端连接于固定点 OO,另一端连接于质量为 mm 的光滑小环。

小环套在一根固定的光滑竖直杆上,杆到 OO 的距离为 32a\dfrac32a。

最初将小环保持在杆上的 AA 点,弹性绳水平。

如图 2 所示,随后将小环以速率 ag\sqrt{ag} 竖直向下发射,它以速率 VV 到达 BB 点,其中 AB=2aAB=2a。

假设空气阻力可忽略。

用 aa 和 gg 表示 VV。

(7)

解答

解法一

思路

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环沿竖直光滑杆运动,杆的支持力不做功。初始绳已经伸长 a/2a/2,末状态伸长要先由直角三角形求绳长,不能把下降距离直接当伸长。

答题过程

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Initially the string length is OA=3a/2OA=3a/2, so its extension is

eA=3a2−a=a2.e_A=\frac{3a}{2}-a=\frac a2.

At BB, Pythagoras gives

OB=(3a2)2+(2a)2=5a2.OB=\sqrt{\left(\frac{3a}{2}\right)^2+(2a)^2} =\frac{5a}{2}.

Thus the final extension is

eB=5a2−a=3a2.e_B=\frac{5a}{2}-a=\frac{3a}{2}.

The ring falls through 2a2a. The reaction from the smooth vertical pole is horizontal and does no work. Conservation of mechanical energy gives

12m(ag)+mg(2a)+mg2a(a2)2=12mV2+mg2a(3a2)2.\frac12m(ag)+mg(2a)+\frac{mg}{2a}\left(\frac a2\right)^2 =\frac12mV^2+\frac{mg}{2a}\left(\frac{3a}{2}\right)^2.

Therefore

12mag+2mag+18mag=12mV2+98mag,\frac12mag+2mag+\frac18mag =\frac12mV^2+\frac98mag,

so

12mV2=32mag,V2=3ag.\frac12mV^2=\frac32mag, \qquad V^2=3ag.

Taking the positive root,

V=3ag.\boxed{V=\sqrt{3ag}.}