CAIE A-Level Further Math A2 3.4 Hookes Law Questions

Practise modelling elastic tension from extension and natural length, then combining Hooke's law with equilibrium, Newton's second law or energy conservation.

Exam points

calculate tension from modulus, extension and natural length before applying the mechanics condition
combine Hooke's law with force balance or radial motion to determine extension or acceleration
include elastic potential energy with kinetic and gravitational terms between two configurations

Covered syllabus points

3.4.1—Hooke's law 3.4.2—Elastic energy 3.4.3—Problems involving forces due to elastic

[Maximum number: 5]

One end of a light elastic string, of natural length a and modulus of elasticity 3 m g, is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The string hangs with P vertically below O. The particle P is pulled vertically downwards so that the extension of the string is 2 a. The particle P is then released from rest.

(a)

Find the speed of P when it is at a distance $\frac{3}{4} a$ below O.

[ 3 ]

$\frac{3 m g}{2 a}(2 a)^{2}$
Correct EPE term seen

$\frac{1}{2} m v^{2}+m g \times\left(3 a-\frac{3}{4} a\right)=\frac{3 m g}{2 a}(2 a)^{2}$
Dimensionally correct energy equation. Must have
one KE, one EPE term and at least one GPE.
Allow sign errors.

$v=\sqrt{\frac{15}{2} a g} \quad[2.74 \sqrt{a g}]$
AEF

3

(b)

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

[ 2 ]

T-m g=m A and $T=\frac{3 m g}{a} \times 2 a$
N2L and Hooke's law

Acceleration =5 g [upwards]
Allow $\pm 50$ or $\pm 5 g$

2

[Maximum number: 8]

A light elastic string has natural length a and modulus of elasticity 24 m g. One end of the string is attached to a fixed point A. The other end of the string is attached to a particle of mass 2 m.

(a)

Find, in terms of a, the extension of the string when the particle hangs freely in equilibrium below A.

[ 2 ]

In equilibrium, $2 m g=24 m g \cdot \frac{x}{a}$

1
Use Hooke's law

$x=\frac{a}{12}$

1
2

(b)

The particle is released from rest at A.

Find, in terms of a, the distance of the particle below A when it first comes to instantaneous rest.

[ 6 ]

Let d be the distance below A Loss in gravitational potential energy (GPE) of particle =2 m g d

1
Gain in elastic potential energy (EPE) of string
$=\frac{1}{2} \times \frac{24 m g}{a} \times(d-a)^{2}$

1
No change in kinetic energy (KE), so
$2 m g d=\frac{1}{2} \times \frac{24 m g}{a} \times(d-a)^{2}$

1
Equate energy loss to energy gain

$6(d-a)^{2}=a d$

1
Attempt to solve

$d=\frac{3}{2} a$ or $\frac{2}{3} a$

1
Since d>a, distance below $A=\frac{3}{2} a$

1
6

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)