Elastic strings and springs and elastic energy

FM1-3 Elastic strings and springs and elastic energy — revision question pack

2 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FM1-3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FM1-3.1 · Elastic strings and springs. Hooke's law.

Explanation

Extension is stretched length minus natural length. A Hookean spring has tension magnitude $T=kx$ , where $k$ is the spring constant in $\text{N m}^{-1}$ .
A taut elastic string of natural length $l$ and modulus $\lambda$ obeys $T=\lambda x/l$ , where $\lambda$ is measured in newtons.
A string cannot push: when its extension is zero or negative it is slack and its tension is zero.
A spring may be extended or compressed, with force opposing the displacement.
Examiners expect natural length, actual length and extension to be distinguished, equilibrium forces to be resolved consistently, and the correct Hooke's-law form chosen for a spring or string.

The linear tension-extension graph for a Hookean spring.

Worked example

An elastic string has natural length $0.9\,\text{m}$ and modulus $54\,\text{N}$ . Find its tension when its length is $1.15\,\text{m}$ .

1.The extension is $x=1.15-0.90=0.25\,\text{m}$ .
2.Use the string form $T=\lambda x/l$ .
3. $T=54(0.25)/0.9=15\,\text{N}$ .

Answer: The tension is $15\,\text{N}$ .

Common mistakes

Don't substitute the stretched length for $x$ instead of calculating extension from the natural length.
Don't use $T=kx$ with the modulus of a string and omit division by natural length.
Don't assign a negative compressive tension to a slack string.

Exam tip

Write the extension explicitly before Hooke's law; this separates the length mark from the force calculation.

Tier 1 · Easy

1.
A light spring has spring constant $80\,\text{N m}^{-1}$ and is extended by $0.06\,\text{m}$ . Find its tension.
(2)
(Total for Question 1 is 2 marks)
2.
Two elastic strings have the same natural length, $1\,\text{m}$ , and the same modulus of elasticity. Their lengths are $1.25\,\text{m}$ and $1.4\,\text{m}$ . The tension in the first string is $20\,\text{N}$ . Find the tension in the second string.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
An elastic string has natural length $1.5\,\text{m}$ and modulus $120\,\text{N}$ . It is stretched to length $1.8\,\text{m}$ . Calculate its tension.
(3)
(Total for Question 1 is 3 marks)
2.
A particle of mass $3\,\text{kg}$ rests in equilibrium on the upper end of a light vertical spring. The spring has natural length $0.8\,\text{m}$ and spring constant $196\,\text{N m}^{-1}$ ; its lower end is fixed. Use $g=9.8\,\text{m s}^{-2}$ . Determine the thrust in the spring, its compression and its equilibrium length.
(4)
(Total for Question 2 is 4 marks)
3.
A $5\,\text{kg}$ particle hangs in equilibrium from two light vertical springs. The springs have natural lengths $0.6\,\text{m}$ and $0.8\,\text{m}$ , and spring constants $50\,\text{N m}^{-1}$ and $100\,\text{N m}^{-1}$ respectively. Their upper ends are fixed at the same level and their lower ends are attached to the particle. Taking $g=9.8\,\text{m s}^{-2}$ , find their common equilibrium length and the tension in each spring.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A $1.5\,\text{kg}$ particle hangs in equilibrium from a vertical elastic string of natural length $0.8\,\text{m}$ and modulus $49\,\text{N}$ . Taking $g=9.8\,\text{m s}^{-2}$ , determine the extension and the equilibrium length of the string.
(4)
(Total for Question 1 is 4 marks)
2.
Fixed points $A$ and $B$ are vertically $6\,\text{m}$ apart, with $A$ above $B$ . A particle $P$ of weight $W\,\text{N}$ , where $0<W<60$ , rests between them. It is attached to each point by an elastic string of natural length $2\,\text{m}$ . String $AP$ has modulus $30\,\text{N}$ and string $BP$ has modulus $15\,\text{N}$ . Determine the equilibrium position and state, for each value of $W$ , whether the lower string is taut or slack.
(7)
(Total for Question 2 is 7 marks)
3.
The upper end of the first light vertical spring is attached to a ceiling, its lower end is attached to a $1\,\text{kg}$ particle $A$ , and a second light vertical spring joins $A$ to a $2\,\text{kg}$ particle $B$ below it. The first spring has natural length $0.5\,\text{m}$ and spring constant $98\,\text{N m}^{-1}$ ; the second has natural length $0.4\,\text{m}$ and spring constant $49\,\text{N m}^{-1}$ . The particles rest in equilibrium. Taking $g=9.8\,\text{m s}^{-2}$ , find the tension, extension and length of each spring, and the distance from the ceiling to $B$ . Verify that both springs are extended.
(7)
(Total for Question 3 is 7 marks)
4.
A particle of weight $50\,\text{N}$ is held in equilibrium on a smooth plane inclined at an angle $\alpha$ , where $\sin\alpha=3/5$ , by a light spring parallel to a line of greatest slope. The spring has length $1.2\,\text{m}$ . The plane is now inclined at an angle $\beta$ , where $\sin\beta=4/5$ , and the same particle is held in equilibrium by the same spring, now of length $1.4\,\text{m}$ . In each state the spring is extended. Find the natural length and spring constant of the spring, and the normal reaction in each state.
(7)
(Total for Question 4 is 7 marks)
5.
A particle $P$ of weight $63\,\text{N}$ is held in equilibrium by two light elastic strings $PA$ and $PB$ . The unit directions from $P$ towards $A$ and $B$ are $(-4\mathbf i+3\mathbf j)/5$ and $(5\mathbf i+12\mathbf j)/13$ respectively, where $\mathbf j$ is vertically upwards. String $PA$ has natural length $1.2\,\text{m}$ and length $1.5\,\text{m}$ in equilibrium. String $PB$ has natural length $2\,\text{m}$ and length $2.5\,\text{m}$ in equilibrium. Find the tension and modulus of elasticity of each string, and verify the vertical equilibrium.
(8)
(Total for Question 5 is 8 marks)

FM1-3.2 · Energy stored in an elastic string or spring.

Explanation

Elastic energy is the work done in stretching or compressing a Hookean object from zero extension. It is the area under the tension-extension graph, so a spring stores $\tfrac12kx^2$ and a taut elastic string stores $\dfrac{\lambda x^2}{2l}$ .
The factor $\tfrac12$ arises because tension increases linearly from zero to its final value.
In a work-energy problem, calculate extension from natural length at every position and include string energy only while the string is taut.
Gravitational and kinetic energy may appear in the same balance.
Examiners expect a clear initial-equals-final energy equation and reject final tension times extension as the stored energy.

Worked example

A $1.5\,\text{kg}$ particle hangs from a vertical elastic string of natural length $0.8\,\text{m}$ and modulus $48\,\text{N}$ . It is released from rest at natural length. Find its speed after descending $0.4\,\text{m}$ , taking $g=9.8\,\text{m s}^{-2}$ .

1.Loss of gravitational potential energy: $1.5(9.8)(0.4)=5.88\,\text{J}$ .
2.Elastic energy at extension $0.4$ : $\dfrac{48(0.4)^2}{2(0.8)}=4.8\,\text{J}$ .
3. $5.88=\tfrac12(1.5)v^2+4.8$ , so $0.75v^2=1.08$ .

Answer: $v=1.20\,\text{m s}^{-1}$ .

Common mistakes

Don't use $Tx$ for stored energy instead of the triangular area $\tfrac12Tx$ .
Don't measure extension from the release point rather than from natural length.
Don't include elastic-string energy at a position where the string is slack.

Exam tip

In a multi-energy question, label each term as kinetic, gravitational or elastic before simplifying.

Tier 1 · Easy

1.
A spring of constant $200\,\text{N m}^{-1}$ is compressed by $0.08\,\text{m}$ . Calculate the elastic energy stored.
(2)
(Total for Question 1 is 2 marks)
2.
A $0.5\,\text{kg}$ particle is held against a light spring on a smooth horizontal surface. The spring has natural length $0.8\,\text{m}$ and modulus $40\,\text{N}$ , and is compressed by $0.2\,\text{m}$ . The particle is released from rest. Find its speed when the spring reaches its natural length.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
An elastic string has natural length $1.2\,\text{m}$ and modulus $60\,\text{N}$ . Find the energy stored when its length is $1.5\,\text{m}$ .
(3)
(Total for Question 1 is 3 marks)
2.
A $1\,\text{kg}$ particle moves on a horizontal surface against a constant resistance of $2\,\text{N}$ . It is attached to an elastic string of natural length $2\,\text{m}$ and modulus $18\,\text{N}$ , and is released from rest at length $3\,\text{m}$ . Find its speed at natural length and the further distance travelled before it stops. State whether the string is taut or slack over this final stage.
(5)
(Total for Question 2 is 5 marks)
3.
The natural length of an elastic string is $2\,\text{m}$ . The work done in stretching the string from a length of $2.4\,\text{m}$ to a length of $2.8\,\text{m}$ is $4.2\,\text{J}$ . Find the modulus of elasticity of the string and its tension at length $2.8\,\text{m}$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A $2\,\text{kg}$ particle is attached to the lower end of a vertical elastic string with natural length $1\,\text{m}$ and modulus $49\,\text{N}$ . It is released from rest when the string is at natural length. Taking $g=9.8\,\text{m s}^{-2}$ , find its speed after descending $0.6\,\text{m}$ .
(5)
(Total for Question 1 is 5 marks)
2.
A smooth plane has angle of inclination $\alpha$ , with $\sin\alpha=3/5$ . A particle $P$ of mass $m$ is held a distance $a$ down the line of greatest slope from a fixed point $O$ . A light elastic string joins $P$ to $O$ and has natural length $a$ and modulus of elasticity $2mg$ , so the string is initially at its natural length. The particle is released from rest and moves down the plane. Find the greatest extension of the string and the speed of the particle when the extension is half its greatest value.
(7)
(Total for Question 2 is 7 marks)
3.
A $2\,\text{kg}$ particle $P$ moves on a smooth horizontal line and is attached to a fixed point $O$ by a light elastic string of natural length $1.2\,\text{m}$ and modulus of elasticity $48\,\text{N}$ . Initially $OP=0.6\,\text{m}$ and $P$ is projected directly away from $O$ at $6\,\text{m s}^{-1}$ . Using the principle of conservation of energy, find the speed of $P$ when $OP=1.5\,\text{m}$ and the greatest value of $OP$ . State whether the string is taut or slack at the initial position, at $OP=1.2\,\text{m}$ , at $OP=1.5\,\text{m}$ and at the turning point.
(7)
(Total for Question 3 is 7 marks)
4.
Two light springs lie along a smooth horizontal line between fixed points $A$ and $B$ , where $AB=3\,\text{m}$ . A $2\,\text{kg}$ particle $P$ is attached between them. Each spring has natural length $1\,\text{m}$ ; the spring joined to $A$ has spring constant $20\,\text{N m}^{-1}$ and the spring joined to $B$ has spring constant $40\,\text{N m}^{-1}$ . Initially $AP=0.5\,\text{m}$ and $P$ is released from rest. Find the speed and the resultant force on $P$ when $AP=1.5\,\text{m}$ . Find also the other value of $AP$ at which $P$ is instantaneously at rest.
(8)
(Total for Question 4 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FM1-3.1 · Elastic strings and springs. Hooke's law.

Tier 1 · Easy

Mark scheme for FM1-3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$4.8\,\text{N}$	2
(2 marks)		2
Notes
Hooke's law for a spring gives $T=kx=80(0.06)=4.8\,\text{N}$ .
2	The first extension is $0.25\,\text{m}$ The common modulus is $80\,\text{N}$ The tension in the second string is $32\,\text{N}$	3
(3 marks)		3
Notes
For the first string, $20=\lambda(1.25-1)/1$ , so the common modulus is $\lambda=80\,\text{N}$ . The second extension is $1.4-1=0.4\,\text{m}$ , giving tension $T=80(0.4)/1=32\,\text{N}$ .

Tier 2 · Standard

Mark scheme for FM1-3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$24\,\text{N}$	3
(3 marks)		3
Notes
The extension is $x=1.8-1.5=0.3\,\text{m}$ . Therefore $T=\lambda x/l=120(0.3)/1.5=24\,\text{N}$ .
2	Vertical equilibrium gives a spring thrust equal to the weight, $T=3(9.8)$ Therefore the thrust is $29.4\,\text{N}$ Hooke's law gives $196x=29.4$ , so the compression is $x=0.15\,\text{m}$ The equilibrium length is $0.8-0.15=0.65\,\text{m}$	4
(4 marks)		4
Notes
Vertical equilibrium gives an upward thrust equal to the weight, $T=3(9.8)=29.4\,\text{N}$ . If the compression is $x$ , Hooke's law gives $196x=29.4$ , so $x=0.15\,\text{m}$ . The equilibrium length is therefore $0.8-0.15=0.65\,\text{m}$ .
3	At common length $L$ , the extensions are $L-0.6$ and $L-0.8\,\text{m}$ Equilibrium gives $50(L-0.6)+100(L-0.8)=49$ Solving gives $L=1.06\,\text{m}$ The tensions are $50(0.46)=23\,\text{N}$ and $100(0.26)=26\,\text{N}$	4
(4 marks)		4
Notes
If the common length is $L$ , Hooke's law gives tensions $50(L-0.6)$ and $100(L-0.8)$ . Vertical equilibrium requires their sum to equal $5(9.8)=49\,\text{N}$ , so $150L-110=49$ and $L=1.06\,\text{m}$ . The tensions are therefore $23\,\text{N}$ and $26\,\text{N}$ .

Tier 3 · Hard

Mark scheme for FM1-3.1 Tier 3 · Hard
Question	Scheme	Marks
1	Extension $0.24\,\text{m}$ Equilibrium length $1.04\,\text{m}$	4
(4 marks)		4
Notes
Equilibrium gives $T=mg=1.5(9.8)=14.7\,\text{N}$ . Hooke's law gives $14.7=49x/0.8$ , so $x=0.24\,\text{m}$ . Adding the natural length gives $0.8+0.24=1.04\,\text{m}$ .
2	Writing $AP=x$ , the upper tension is $T_{AP}=15(x-2)\,\text{N}$ Since $BP=6-x$ , the lower tension when taut is $T_{BP}=7.5(4-x)\,\text{N}$ With both strings taut, equilibrium gives $15(x-2)=W+7.5(4-x)$ Solving gives $x=2(W+60)/45\,\text{m}$ The lower string is taut for $W<30$ ; at $W=30$ , $x=4$ and it has zero tension For $30<W<60$ it is slack, and $15(x-2)=W$ gives $x=2+W/15\,\text{m}$ This slack-branch value has $4<x<6$ because $30<W<60$ , so the particle remains between $A$ and $B$	7
(7 marks)		7
Notes
Write $AP=x$ , so $BP=6-x$ . When both strings are taut, the upper and lower tensions are $15(x-2)$ and $7.5(4-x)$ newtons. Equilibrium gives $15(x-2)=W+7.5(4-x)$ , hence $x=2(W+60)/45$ . The lower string is stretched exactly when $6-x>2$ , which reduces to $W<30$ ; the upper string is stretched throughout this range. At $W=30$ , $x=4$ , so the straight lower string is just taut at natural length with zero tension. For $W>30$ take the lower string as slack: $15(x-2)=W$ , so $x=2+W/15$ . This gives $x>4$ , confirming slackness, and $x<6$ because $W<60$ .
3	Equilibrium of $B$ gives the tension in spring 2 as $2(9.8)=19.6\,\text{N}$ The extension of spring 2 is $19.6/49=0.4\,\text{m}$ The length of spring 2 is $0.4+0.4=0.8\,\text{m}$ Equilibrium of $A$ gives the tension in spring 1 as $19.6+9.8=29.4\,\text{N}$ The extension of spring 1 is $29.4/98=0.3\,\text{m}$ The length of spring 1 is $0.5+0.3=0.8\,\text{m}$ The ceiling-to- $B$ distance is $1.6\,\text{m}$ , and both extensions are positive	7
(7 marks)		7
Notes
For particle $B$ , equilibrium gives $T_2=2g=19.6\,\text{N}$ , so spring 2 extends by $T_2/49=0.4\,\text{m}$ and has length $0.8\,\text{m}$ . Particle $A$ is pulled down by its own weight and by spring 2, so $T_1=9.8+19.6=29.4\,\text{N}$ . Spring 1 therefore extends by $29.4/98=0.3\,\text{m}$ and also has length $0.8\,\text{m}$ . The total distance is $1.6\,\text{m}$ ; both calculated extensions are positive.
4	Resolving along the first plane gives tension $T_1=50(3/5)=30\,\text{N}$ Resolving along the second plane gives tension $T_2=50(4/5)=40\,\text{N}$ If the natural length is $l\,\text{m}$ and the spring constant is $k\,\text{N m}^{-1}$ , then $30=k(1.2-l)$ and $40=k(1.4-l)$ Subtracting the equations gives $10=0.2k$ , so $k=50\,\text{N m}^{-1}$ Substitution gives $l=0.6\,\text{m}$ The first normal reaction is $50\cos\alpha=40\,\text{N}$ The second normal reaction is $50\cos\beta=30\,\text{N}$	7
(7 marks)		7
Notes
Resolving parallel to the plane gives spring tensions $30\,\text{N}$ and $40\,\text{N}$ in the two states. If $l$ and $k$ are the natural length and spring constant, Hooke's law gives $30=k(1.2-l)$ and $40=k(1.4-l)$ . Subtraction gives $k=50\,\text{N m}^{-1}$ , and substitution gives $l=0.6\,\text{m}$ . Resolving perpendicular to the plane gives reactions $50(4/5)=40\,\text{N}$ and $50(3/5)=30\,\text{N}$ .
5	Horizontal equilibrium gives $-4T_A/5+5T_B/13=0$ Hence $T_B=52T_A/25$ Vertical equilibrium gives $3T_A/5+12T_B/13=63$ Substitution gives $63T_A/25=63$ , so $T_A=25\,\text{N}$ Hence $T_B=52\,\text{N}$ For $PA$ , $25=\lambda_A(0.3)/1.2$ , so $\lambda_A=100\,\text{N}$ For $PB$ , $52=\lambda_B(0.5)/2$ , so $\lambda_B=208\,\text{N}$ The upward components are $15\,\text{N}$ and $48\,\text{N}$ , whose sum equals the $63\,\text{N}$ weight	8
(8 marks)		8
Notes
Resolve the two tensions horizontally and vertically. The horizontal equation gives $T_B=52T_A/25$ . Vertical equilibrium gives $3T_A/5+12T_B/13=63$ , so $63T_A/25=63$ and hence $T_A=25$ and $T_B=52\,\text{N}$ . The extensions are $0.3$ and $0.5\,\text{m}$ , so Hooke's law for a string gives moduli $100$ and $208\,\text{N}$ . The two upward components are $15$ and $48\,\text{N}$ , confirming equilibrium.

FM1-3.2 · Energy stored in an elastic string or spring.

Tier 1 · Easy

Mark scheme for FM1-3.2 Tier 1 · Easy
Question	Scheme	Marks
1	$0.64\,\text{J}$	2
(2 marks)		2
Notes
$E=\tfrac12kx^2=\tfrac12(200)(0.08)^2=0.64\,\text{J}$ .
2	The stored elastic energy is $40(0.2)^2/[2(0.8)]=1\,\text{J}$ Energy conservation gives $1=\tfrac12(0.5)v^2$ Hence $v=2\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
The stored elastic energy is $40(0.2)^2/[2(0.8)]=1\,\text{J}$ . On the smooth surface this becomes kinetic energy, so $1=\tfrac12(0.5)v^2$ . Therefore $v=2\,\text{m s}^{-1}$ .

Tier 2 · Standard

Mark scheme for FM1-3.2 Tier 2 · Standard
Question	Scheme	Marks
1	$2.25\,\text{J}$	3
(3 marks)		3
Notes
The extension is $x=1.5-1.2=0.3\,\text{m}$ . Thus $E=\lambda x^2/(2l)=60(0.3)^2/[2(1.2)]=2.25\,\text{J}$ .
2	The initial elastic energy is $4.5\,\text{J}$ Resistance does $2\,\text{J}$ of work before natural length The speed at natural length is $\sqrt5\,\text{m s}^{-1}$ The further distance travelled is $1.25\,\text{m}$ The string is slack over this final stage	5
(5 marks)		5
Notes
The initial elastic energy is $18(1)^2/[2(2)]=4.5\,\text{J}$ . Resistance does $2(1)=2\,\text{J}$ of work before natural length, leaving kinetic energy $2.5\,\text{J}$ ; hence $\tfrac12v^2=2.5$ and $v=\sqrt5$ . Beyond natural length the particle moves towards the fixed end, so the string is slack. Resistance alone removes the remaining $2.5\,\text{J}$ , giving a further distance $2.5/2=1.25\,\text{m}$ .
3	The initial extension is $0.4\,\text{m}$ The final extension is $0.8\,\text{m}$ The work done is the elastic-energy increase $\lambda(0.8^2-0.4^2)/[2(2)]=0.12\lambda$ $0.12\lambda=4.2$ , so $\lambda=35\,\text{N}$ The final tension is $35(0.8)/2=14\,\text{N}$	5
(5 marks)		5
Notes
The extensions are $0.4\,\text{m}$ and $0.8\,\text{m}$ . The work done is the increase in elastic energy, so $4.2=\lambda(0.8^2-0.4^2)/[2(2)]=0.12\lambda$ . Hence $\lambda=35\,\text{N}$ . Hooke's law then gives final tension $35(0.8)/2=14\,\text{N}$ .

Tier 3 · Hard

Mark scheme for FM1-3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\sqrt{2.94}\,\text{m s}^{-1}\approx1.71\,\text{m s}^{-1}$	5
(5 marks)		5
Notes
The loss of gravitational potential energy is $2(9.8)(0.6)=11.76\,\text{J}$ . At extension $0.6\,\text{m}$ , elastic energy is $49(0.6)^2/[2(1)]=8.82\,\text{J}$ . Conservation of mechanical energy gives $11.76=\tfrac12(2)v^2+8.82$ , so $v^2=2.94$ and $v=\sqrt{2.94}\,\text{m s}^{-1}$ .
2	At extension $x$ , the loss of gravitational potential energy is $3mgx/5$ The elastic energy is $(2mg)x^2/(2a)=mgx^2/a$ Energy conservation gives $3mgx/5=\tfrac12mv^2+mgx^2/a$ At greatest extension, $v=0$ , so $x(3a/5-x)=0$ The non-zero turning point gives a greatest extension of $3a/5$ Half the greatest extension is $x=3a/10$ Substitution gives $v^2=9ag/50$ , so $v=3\sqrt{2ag}/10$	7
(7 marks)		7
Notes
Let $x$ be the extension after the particle has moved $x$ down the plane. The loss of gravitational potential energy is $mgx\sin\alpha=3mgx/5$ , while the string stores $(2mg)x^2/(2a)=mgx^2/a$ . Hence $3mgx/5=\tfrac12mv^2+mgx^2/a$ . At greatest extension $v=0$ , so $x(3a/5-x)=0$ . The root $x=0$ is the release point, leaving greatest extension $3a/5$ . At half this extension, $x=3a/10$ , the energy equation gives $\tfrac12mv^2=9mga/100$ , so $v^2=9ag/50$ and $v=3\sqrt{2ag}/10$ .
3	The string is slack initially and remains slack until $P$ has travelled $0.6\,\text{m}$ to $OP=1.2\,\text{m}$ At $OP=1.2\,\text{m}$ it is just taut at natural length with zero tension, and the kinetic energy is $\tfrac12(2)(6^2)=36\,\text{J}$ At $OP=1.5\,\text{m}$ the string is taut with extension $0.3\,\text{m}$ Its elastic energy there is $48(0.3)^2/[2(1.2)]=1.8\,\text{J}$ Energy conservation gives $v^2=36-1.8=171/5$ , so $v=\sqrt{171/5}\,\text{m s}^{-1}$ At the turning point the taut string satisfies $48x^2/[2(1.2)]=36$ , so $x=3/\sqrt5\,\text{m}$ The greatest value of $OP$ is $1.2+3/\sqrt5\,\text{m}$ , where the string is taut	7
(7 marks)		7
Notes
The particle travels from $OP=0.6$ to $OP=1.2\,\text{m}$ with the string slack, so its speed is still $6\,\text{m s}^{-1}$ when the string reaches natural length. The conserved energy is therefore $36\,\text{J}$ . At $OP=1.5$ , the extension is $0.3$ and the elastic energy is $1.8\,\text{J}$ , leaving kinetic energy $34.2\,\text{J}$ ; since $m=2$ , $v^2=34.2=171/5$ . At the turning point all $36\,\text{J}$ is elastic, so $20x^2=36$ and $x=3/\sqrt5$ . The greatest distance is the natural length plus this positive extension.
4	Initially the left spring is compressed by $0.5\,\text{m}$ and the right spring is extended by $1.5\,\text{m}$ The initial elastic energy is $10(0.5)^2+20(1.5)^2=47.5\,\text{J}$ At $AP=1.5\,\text{m}$ , both springs are extended by $0.5\,\text{m}$ Their total elastic energy there is $10(0.5)^2+20(0.5)^2=7.5\,\text{J}$ Thus the kinetic energy is $40\,\text{J}$ , so $v=2\sqrt{10}\,\text{m s}^{-1}$ The spring forces there are $10\,\text{N}$ towards $A$ and $20\,\text{N}$ towards $B$ , giving resultant $10\,\text{N}$ towards $B$ At a turning point $x=AP$ satisfies $10(x-1)^2+20(2-x)^2=47.5$ The roots are $x=0.5$ and $x=17/6$ , so the other rest position is $AP=17/6\,\text{m}$	8
(8 marks)		8
Notes
Elastic energy is $10(x-1)^2+20(2-x)^2$ when $AP=x$ . It is $47.5\,\text{J}$ at release and $7.5\,\text{J}$ at $x=1.5$ , so the kinetic energy there is $40\,\text{J}$ . With mass $2\,\text{kg}$ this gives speed $2\sqrt{10}$ . The two tensions at that point are $10$ and $20\,\text{N}$ in opposite directions. Equating the elastic energy at a later rest position to $47.5$ gives roots $1/2$ and $17/6$ .