| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Show quantity is dimensionless |
| Difficulty | Moderate -0.8 This is a straightforward dimensional analysis question requiring basic manipulation of dimensions. Part (a) involves simple rearrangement to show k is dimensionless, while part (b) requires setting up and solving simultaneous equations from dimension matching - both are standard textbook exercises with no novel insight required. |
| Spec | 6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([t] = T\), \([g] = LT^{-2}\), \([w] = L\) | M1 | Substitutes dimensions for length, time and acceleration into both sides. Do not condone use of units. |
| \([k] = \frac{[gt^2]}{[w]} = \frac{LT^{-2}T^2}{L} = 1\), therefore \(k\) is a dimensionless constant | R1 | Completes a rigorous argument using dimensions to verify that \(k\) is a dimensionless constant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([(gd)^\alpha t^\beta] = (LT^{-2}L)^\alpha(T)^\beta = (L)^{2\alpha}(T)^{-2\alpha+\beta}\) | B1 | Uses dimensions to form a correct expression for the dimensions of \([(gd)^\alpha t^\beta]\). Need not be simplified. Do not condone use of units. |
| \(2\alpha = 1\) and \(-2\alpha + \beta = 0\) | M1 | Forms two simultaneous equations in \(\alpha\) and \(\beta\) consistent with their simplified \([(gd)^\alpha t^\beta]\). PI by a correct value of \(\alpha\) or \(\beta\) |
| \(\alpha = \frac{1}{2}\), \(\beta = 1\) | A1 | Obtains correct values for \(\alpha\) and \(\beta\) |
## Question 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[t] = T$, $[g] = LT^{-2}$, $[w] = L$ | M1 | Substitutes dimensions for length, time and acceleration into both sides. Do not condone use of units. |
| $[k] = \frac{[gt^2]}{[w]} = \frac{LT^{-2}T^2}{L} = 1$, therefore $k$ is a dimensionless constant | R1 | Completes a rigorous argument using dimensions to verify that $k$ is a dimensionless constant |
## Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[(gd)^\alpha t^\beta] = (LT^{-2}L)^\alpha(T)^\beta = (L)^{2\alpha}(T)^{-2\alpha+\beta}$ | B1 | Uses dimensions to form a correct expression for the dimensions of $[(gd)^\alpha t^\beta]$. Need not be simplified. Do not condone use of units. |
| $2\alpha = 1$ and $-2\alpha + \beta = 0$ | M1 | Forms two simultaneous equations in $\alpha$ and $\beta$ consistent with their simplified $[(gd)^\alpha t^\beta]$. PI by a correct value of $\alpha$ or $\beta$ |
| $\alpha = \frac{1}{2}$, $\beta = 1$ | A1 | Obtains correct values for $\alpha$ and $\beta$ |4 Wavelength is defined as the distance from the highest point on one wave to the highest point on the next wave.
Surfers classify waves into one of several types related to their wavelengths.\\
Two of these classifications are deep water waves and shallow water waves.\\
4
\begin{enumerate}[label=(\alph*)]
\item The wavelength $w$ of a deep water wave is given by
$$w = \frac { g t ^ { 2 } } { k }$$
where $g$ is the acceleration due to gravity and $t$ is the time period between consecutive waves.
Given that the formula for a deep water wave is dimensionally consistent, show that $k$ is a dimensionless constant.
4
\item The wavelength $w$ of a shallow water wave is given by
$$w = ( g d ) ^ { \alpha } t ^ { \beta }$$
where $g$ is the acceleration due to gravity, $d$ is the depth of water and $t$ is the time period between consecutive waves.
Use dimensional analysis to find the values of $\alpha$ and $\beta$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2022 Q4 [5]}}