Edexcel A-Level Mathematics A2 M3.2 Elastic Strings and Springs Questions

Practise modelling elastic strings and springs with Hooke's law, equilibrium, acceleration and energy changes for particles released from rest.

Exam points

form Hooke's law tensions from extensions and use equilibrium to prove lengths
resolve forces along a string or plane to find acceleration after release from rest
use elastic energy with GPE and KE to find speeds or prove a modulus value

Covered syllabus points

M3.2.1 - Elastic strings and springs M3.2.2 - Energy stored in elastic strings and springs

[Maximum number: 9]

Figure 1

A light elastic spring has natural length l and modulus of elasticity $\lambda$ .
One end of the spring is attached to a point A on a smooth plane.
The plane is inclined at angle $\theta$ to the horizontal, where $\tan\theta=\frac5{12}$ .
A particle P of mass m is attached to the other end of the spring.
Initially P is held at the point B on the plane, where AB is a line of greatest slope of the plane.
The point B is lower than A and AB=2l, as shown in Figure 1.
The particle is released from rest at B and first comes to instantaneous rest at the point C on AB, where AC=0.7l.

(a)

Use the principle of conservation of mechanical energy to show that

$\lambda=\frac{100}{91}mg.$

[ 5 ]

Change in GPE
$=mg\cdot1.3l\sin\theta=0.5lmg$ $\mathrm{EPE}=\frac{\lambda l^2}{2l}\quad\text{or}\quad \frac{\lambda(0.3l)^2}{2l}$
Energy equation from B to A:
$\frac{\lambda l^2}{2l}-\frac{\lambda(0.3l)^2}{2l}=0.5lmg$ $\lambda=mg\frac1{1-0.09}=\frac{100}{91}mg$
A1*

(b)

Find the acceleration of P when it is released from rest at B.

[ 4 ]

Equation of motion:
$T-mg\sin\theta=ma$ $\frac{\lambda l}{l}-mg\sin\theta=ma$ $\frac{100}{91}mg-\frac5{13}mg=ma$ $a=\frac57g$

[Maximum number: 4]

The fixed point A is vertically above the fixed point B, with A B=3 l

A light elastic string has natural length l and modulus of elasticity 4 m g One end of the string is attached to A and the other end is attached to a particle P of mass m

A second light elastic string also has natural length l and modulus of elasticity 4 m g One end of this string is attached to P and the other end is attached to B.

Initially P rests in equilibrium at the point E, where AEB is a vertical straight line.

Show that $A E=\frac{13}{8} l$

mg+4mg\left(\frac{2l-AE}{l}\right)=4mg\left(\frac{AE-l}{l}\right)

13mg=\frac{8mg}{l}AE

AE=\frac{13l}{8}

A1*

AE=\frac{13}{8}l

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2