AQA Further Paper 3 Mechanics 2024 June — Question 5 4 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeFind exponents with partial constraints
DifficultyStandard +0.3 This is a straightforward dimensional analysis question requiring students to equate dimensions on both sides of an equation. Part (a) is trivial recall (force = MLT^-2), and part (b) involves setting up and solving simple simultaneous equations from dimension matching. While it's a Further Maths mechanics question, dimensional analysis is a standard technique with no conceptual subtlety or problem-solving insight required—just methodical application of a learned procedure.
Spec6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions

When a sphere of radius \(r\) metres is falling at \(v\) m s\(^{-1}\) it experiences an air resistance force \(F\) newtons. The force is to be modelled as $$F = kr^\alpha v^\beta$$ where \(k\) is a constant with units kg m\(^{-2}\)
  1. State the dimensions of \(F\) [1 mark]
  2. Use dimensional analysis to find the value of \(\alpha\) and the value of \(\beta\) [3 marks]

Question 5:

AnswerMarks Guidance
5(a)States MLT–2 1.2
Subtotal1
QMarking Instructions AO

AnswerMarks
5(b)Uses dimensional analysis notation
to form an equation.
AnswerMarks Guidance
Must not use units.3.3 M1
MLT–2 = ML–2 × Lα × LβT–β
1 = –2 + α + β
α + β = 3
–2 = –β
β = 2
α = 1
Obtains either α = 1 or β = 2
AnswerMarks Guidance
Condone use of units.1.1a M1
Completes a reasoned argument
using dimensions to show that
AnswerMarks Guidance
α = 1 and β = 22.2a R1
Subtotal3
Question total4
QMarking Instructions AO
Question 5:
--- 5(a) ---
5(a) | States MLT–2 | 1.2 | B1 | MLT–2
Subtotal | 1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 5(b) ---
5(b) | Uses dimensional analysis notation
to form an equation.
Must not use units. | 3.3 | M1 | [F] = [k][r]α[v]β
MLT–2 = ML–2 × Lα × LβT–β
1 = –2 + α + β
α + β = 3
–2 = –β
β = 2
α = 1
Obtains either α = 1 or β = 2
Condone use of units. | 1.1a | M1
Completes a reasoned argument
using dimensions to show that
α = 1 and β = 2 | 2.2a | R1
Subtotal | 3
Question total | 4
Q | Marking Instructions | AO | Marks | Typical Solution
When a sphere of radius $r$ metres is falling at $v$ m s$^{-1}$ it experiences an air resistance force $F$ newtons.

The force is to be modelled as
$$F = kr^\alpha v^\beta$$
where $k$ is a constant with units kg m$^{-2}$

\begin{enumerate}[label=(\alph*)]
\item State the dimensions of $F$
[1 mark]

\item Use dimensional analysis to find the value of $\alpha$ and the value of $\beta$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q5 [4]}}