Question 7 - A-Level Maths

AQA Further AS Paper 2 Mechanics 2021 June — Question 7 8 marks

Exam Board	AQA
Module	Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year	2021
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hooke's law and elastic energy
Type	Elastic string on smooth inclined plane
Difficulty	Challenging +1.2 This is a multi-part mechanics question involving elastic strings on an inclined plane, requiring energy conservation and projectile motion analysis. While it tests several concepts (EPE formula, energy methods, kinematics on slopes), the techniques are standard for Further Maths mechanics with straightforward application of formulas and energy conservation. The main challenge is careful bookkeeping across three parts rather than novel problem-solving, placing it moderately above average difficulty.
Spec	6.02d Mechanical energy: KE and PE concepts 6.02h Elastic PE: 1/2 k x^2 6.02i Conservation of energy: mechanical energy principle

Use \(g\) as 9.81 m s\(^{-2}\) in this question. A light elastic string has one end attached to a fixed point A on a smooth plane inclined at 25° to the horizontal. The other end of the string is attached to a wooden block of mass 2.5 kg, which rests on the plane. The elastic string has natural length 3 metres and modulus of elasticity 125 newtons. The block is pulled down the line of greatest slope of the plane to a point 4.5 metres from A and then released.

Find the elastic potential energy of the string at the point when the block is released. [1 mark]
Calculate the speed of the block when the string becomes slack. [4 marks]
Determine whether the block reaches the point A in the subsequent motion, commenting on any assumptions that you make. [3 marks]

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks
7(a)	Recalls the formula for elastic

potential energy and calculates

the initial stored energy in the

stretched string – substituting

the appropriate values

Accept 47

Condone missing or incorrect

Answer	Marks	Guidance
units	1.1b	B1

2 2

𝜆𝜆𝑥𝑥 125(1.5)

2𝑙𝑙 = 6

J

== 4466..98 7J5

Answer	Marks	Guidance
Total	1
Q	Marking instructions	AO

Answer	Marks
7(b)	Finds the increase in vertical

height from the starting position

to the point when the string

Answer	Marks	Guidance
becomes slack	1.1b	B1

EPE lost = PE gained + KE gained

1 2

46.875 = (2.5)𝑣𝑣

2 0

+2.5g(1.5sin 25 )

2 𝑣𝑣 = 25.062…

v = 5.01 m s–1

Forms a conservation of energy

equation containing expressions

for EPE, KE and PE –

substituting the appropriate

Answer	Marks	Guidance
values	3.4	M1

Obtains a fully correct three

Answer	Marks	Guidance
term equation	1.1b	A1

Solves the equation correctly to

obtain the value of v

AWRT 5.0

FT their expression or value for

Answer	Marks	Guidance
the increase in height	1.1b	A1F
Total	4
Q	Marking instructions	AO

Answer	Marks
7(c)	Obtains or states two quantities

that can be used as a basis for a

comparison to reach a

Answer	Marks	Guidance
conclusion	3.4	M1

2.5 (4.5sin 25°) = 46.6 J

EPEg at start = 46.9 J

46.6 < 46.9

This model assumes no air

resistance

On that basis, the block will reach

A as there is enough initial energy

States at least one appropriate

assumption

For example:

• Assumes no air resistance

Answer	Marks	Guidance
• All energy is conserved	3.5a	E1

Makes an inference using the

quantities that have been

correctly calculated, in line with

any stated assumptions

Either

Infers that the block will reach A

as there is enough initial energy

and the block does not

experience any air resistance

Or

Infers that the block will not

reach A if air resistance is taken

into account

Must be a correct inference from

calculations and assumptions

Answer	Marks	Guidance
stated	2.2b	R1
Total	3
Question total	8
Q	Marking instructions	AO

Question 7:
--- 7(a) ---
7(a) | Recalls the formula for elastic
potential energy and calculates
the initial stored energy in the
stretched string – substituting
the appropriate values
Accept 47
Condone missing or incorrect
units | 1.1b | B1 | EPE =
2 2
𝜆𝜆𝑥𝑥 125(1.5)
2𝑙𝑙 = 6
J
== 4466..98 7J5
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Finds the increase in vertical
height from the starting position
to the point when the string
becomes slack | 1.1b | B1 | Increase in height = 1.5sin 250
EPE lost = PE gained + KE gained
1 2
46.875 = (2.5)𝑣𝑣
2 0
+2.5g(1.5sin 25 )
2
𝑣𝑣 = 25.062…
v = 5.01 m s–1
Forms a conservation of energy
equation containing expressions
for EPE, KE and PE –
substituting the appropriate
values | 3.4 | M1
Obtains a fully correct three
term equation | 1.1b | A1
Solves the equation correctly to
obtain the value of v
AWRT 5.0
FT their expression or value for
the increase in height | 1.1b | A1F
Total | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c) ---
7(c) | Obtains or states two quantities
that can be used as a basis for a
comparison to reach a
conclusion | 3.4 | M1 | PE gained from start to A =
2.5 (4.5sin 25°) = 46.6 J
EPEg at start = 46.9 J
46.6 < 46.9
This model assumes no air
resistance
On that basis, the block will reach
A as there is enough initial energy
States at least one appropriate
assumption
For example:
• Assumes no air resistance
• All energy is conserved | 3.5a | E1
Makes an inference using the
quantities that have been
correctly calculated, in line with
any stated assumptions
Either
Infers that the block will reach A
as there is enough initial energy
and the block does not
experience any air resistance
Or
Infers that the block will not
reach A if air resistance is taken
into account
Must be a correct inference from
calculations and assumptions
stated | 2.2b | R1
Total | 3
Question total | 8
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

Use $g$ as 9.81 m s$^{-2}$ in this question.

A light elastic string has one end attached to a fixed point A on a smooth plane inclined at 25° to the horizontal.

The other end of the string is attached to a wooden block of mass 2.5 kg, which rests on the plane.

The elastic string has natural length 3 metres and modulus of elasticity 125 newtons.

The block is pulled down the line of greatest slope of the plane to a point 4.5 metres from A and then released.

\begin{enumerate}[label=(\alph*)]
\item Find the elastic potential energy of the string at the point when the block is released. [1 mark]
\item Calculate the speed of the block when the string becomes slack. [4 marks]
\item Determine whether the block reaches the point A in the subsequent motion, commenting on any assumptions that you make. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2021 Q7 [8]}}

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