| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 1 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Multiple choice dimension question |
| Difficulty | Easy -1.2 This is a straightforward dimensional analysis question requiring students to check dimensions of four given formulae. It involves only basic dimensional substitutions (force=MLT^-2, velocity=LT^-1, acceleration=LT^-2) and simple algebraic manipulation. While it's a Further Maths mechanics question, the technique is routine and requires no problem-solving insight—just systematic checking of each option. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship |
| \(a\) | represents acceleration, |
| \(T\) | represents time, |
| \(l\) | represents length, |
| \(m\) | represents mass, |
| \(v\) | represents velocity, |
| \(F\) | represents force. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Fl = mv^2\) | B1 | Circles correct answer |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Fl = mv^2$ | B1 | Circles correct answer |
**Total: 1 mark**
---2 Ns\\
2.4 N s
2 In this question
\begin{center}
\begin{tabular}{ c l }
$a$ & represents acceleration, \\
$T$ & represents time, \\
$l$ & represents length, \\
$m$ & represents mass, \\
$v$ & represents velocity, \\
$F$ & represents force. \\
\end{tabular}
\end{center}
One of these formulae is dimensionally consistent.\\
Circle your answer.\\[0pt]
[1 mark]
$$T = 2 \pi \sqrt { \frac { a } { l } } \quad v ^ { 2 } = \frac { 2 a l } { T } \quad F l = m v ^ { 2 } \quad F T = m \sqrt { a }$$
Turn over for the next question
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q2 [1]}}