AQA Further Paper 3 Mechanics 2024 June — Question 6 10 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.3 This is a standard elastic string energy problem with straightforward applications of EPE = λx²/(2l), conservation of energy, and force comparison. All parts follow routine procedures taught in Further Maths mechanics with no novel insight required, making it slightly easier than average.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings

In this question use \(g = 9.8\) m s\(^{-2}\) A light elastic string has natural length 3 metres and modulus of elasticity 18 newtons. One end of the elastic string is attached to a particle of mass 0.25 kg The other end of the elastic string is attached to a fixed point \(O\) The particle is released from rest at a point \(A\), which is 4.5 metres vertically below \(O\)
  1. Calculate the elastic potential energy of the string when the particle is at \(A\) [2 marks]
  2. The point \(B\) is 3 metres vertically below \(O\) Calculate the gravitational potential energy gained by the particle as it moves from \(A\) to \(B\) [2 marks]
  3. Find the speed of the particle at \(B\) [3 marks]
  4. The point \(C\) is 3.6 metres vertically below \(O\) Explain, showing any calculations that you make, why the speed of the particle is increasing the first time that the particle is at \(C\) [3 marks]

Question 6:

AnswerMarks Guidance
6(a)Uses the EPE formula with their
extension.1.1a M1
EPE= =6.75 J
2×3
Obtains 6.75 or 6.8
AnswerMarks Guidance
Condone missing units.1.1b A1
Subtotal2
QMarking Instructions AO

AnswerMarks Guidance
6(b)Uses GPE formula with their AB 1.1a
= 3.675 J
= 3.7 J (to 2sf)
Obtains AWRT 3.7
AnswerMarks Guidance
Condone missing units.1.1b A1
Subtotal2
QMarking Instructions AO

AnswerMarks
6(c)Forms a three-term energy
equation using their EPE and GPE.
AnswerMarks Guidance
Condone sign errors.3.3 M1
6.75 – 3.675 = × 0.25v2
2
v2 = 24.6
v = 5.0 m s–1
Forms a correct equation using
AnswerMarks Guidance
their energies.1.1b A1F
Obtains 5 or 5.0
AnswerMarks Guidance
Must include units.3.2b A1
Subtotal3
QMarking Instructions AO

AnswerMarks Guidance
6(d)Obtains 3.6 for tension at C 1.1a
T= =3.6 N
3
Resultant force = T – mg
= 3.6 – 0.25 × 9.8
= 1.15 N
Resultant force is upwards and the
particle is moving upwards, so
speed is increasing.
Finds the resultant force or
acceleration of the particle at C or
compares the value of mg with the
AnswerMarks Guidance
tension.3.3 M1
Explains that this force is upwards
and so the speed is increasing or
an explanation that is consistent
AnswerMarks Guidance
with their calculations.2.4 R1
Subtotal3
Question total10
QMarking Instructions AO
Question 6:
--- 6(a) ---
6(a) | Uses the EPE formula with their
extension. | 1.1a | M1 | 18×1.52
EPE= =6.75 J
2×3
Obtains 6.75 or 6.8
Condone missing units. | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 6(b) ---
6(b) | Uses GPE formula with their AB | 1.1a | M1 | GPE = 0.25 × 9.8 × 1.5
= 3.675 J
= 3.7 J (to 2sf)
Obtains AWRT 3.7
Condone missing units. | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 6(c) ---
6(c) | Forms a three-term energy
equation using their EPE and GPE.
Condone sign errors. | 3.3 | M1 | 1
6.75 – 3.675 = × 0.25v2
2
v2 = 24.6
v = 5.0 m s–1
Forms a correct equation using
their energies. | 1.1b | A1F
Obtains 5 or 5.0
Must include units. | 3.2b | A1
Subtotal | 3
Q | Marking Instructions | AO | Marks | Typical Solution
--- 6(d) ---
6(d) | Obtains 3.6 for tension at C | 1.1a | B1 | 18×0.6
T= =3.6 N
3
Resultant force = T – mg
= 3.6 – 0.25 × 9.8
= 1.15 N
Resultant force is upwards and the
particle is moving upwards, so
speed is increasing.
Finds the resultant force or
acceleration of the particle at C or
compares the value of mg with the
tension. | 3.3 | M1
Explains that this force is upwards
and so the speed is increasing or
an explanation that is consistent
with their calculations. | 2.4 | R1
Subtotal | 3
Question total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
In this question use $g = 9.8$ m s$^{-2}$

A light elastic string has natural length 3 metres and modulus of elasticity 18 newtons.

One end of the elastic string is attached to a particle of mass 0.25 kg

The other end of the elastic string is attached to a fixed point $O$

The particle is released from rest at a point $A$, which is 4.5 metres vertically below $O$

\begin{enumerate}[label=(\alph*)]
\item Calculate the elastic potential energy of the string when the particle is at $A$
[2 marks]

\item The point $B$ is 3 metres vertically below $O$

Calculate the gravitational potential energy gained by the particle as it moves from $A$ to $B$
[2 marks]

\item Find the speed of the particle at $B$
[3 marks]

\item The point $C$ is 3.6 metres vertically below $O$

Explain, showing any calculations that you make, why the speed of the particle is increasing the first time that the particle is at $C$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q6 [10]}}