| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Reject model using dimensions |
| Difficulty | Moderate -0.3 This is a straightforward dimensional analysis question requiring students to set up dimension equations and solve for powers. While it involves mechanics context and requires systematic working, the technique is standard and well-practiced in Further Maths mechanics modules. The two-part structure guides students through rejecting one model then finding constants in the other, making it slightly easier than average. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| \(LT^{-1} = (LT^{-2})^a \times L^b \times (MT^{-3})^c\), therefore \(0 = c\) | M1 | Shows that \(c = 0\) by considering the dimensions of mass and deduces that speed does not depend on density |
| Since \(c = 0\), \(v\) does not depend on the density of the liquid. So David's model is incorrect. | A1 | Rejects David's model because speed is shown not to depend on density |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{1}{2}\) | M1 | Uses dimensions to form an equation for dimensional consistency |
| \(b = \frac{1}{2}\) | A1 | Obtains correct values for \(a\) and \(b\) |
## Question 3(a):
$LT^{-1} = (LT^{-2})^a \times L^b \times (MT^{-3})^c$, therefore $0 = c$ | M1 | Shows that $c = 0$ by considering the dimensions of mass and deduces that speed does not depend on density
Since $c = 0$, $v$ does not depend on the density of the liquid. So David's model is incorrect. | A1 | Rejects David's model because speed is shown not to depend on density
---
## Question 3(b):
$LT^{-1} = (LT^{-2})^a \times L^b$
$1 = a + b$
$-1 = -2a$
$a = \frac{1}{2}$ | M1 | Uses dimensions to form an equation for dimensional consistency
$b = \frac{1}{2}$ | A1 | Obtains correct values for $a$ and $b$
---3 A tank full of liquid has a hole made in its base.\\
Two students, Sarah and David, propose two different models for the speed, $v$, at which liquid exits the tank.
David thinks that $v$ will depend on the height of the liquid in the tank, $h$, the acceleration due to gravity, $g$, and the density of the liquid, $\rho$, such that $v \propto g ^ { a } h ^ { b } \rho ^ { c }$ where $a$, $b$ and $c$ are constants.
Sarah thinks that $v$ will not depend on the density of the liquid and suggests the model $v \propto g ^ { a } h ^ { b }$
3
\begin{enumerate}[label=(\alph*)]
\item By considering dimensions, explain which student's model should be rejected.\\[0pt]
[2 marks]\\
3
\item Find the values of the constants in order for the model that you did not reject in part (a) to be dimensionally consistent.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics Q3 [4]}}