| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Show quantity independence |
| Difficulty | Moderate -0.5 This is a straightforward dimensional analysis problem requiring students to set up equations with powers and solve simultaneously. While it involves algebraic manipulation across multiple steps, the technique is standard and methodical with no conceptual surprises—slightly easier than average for M3 level. |
| Spec | 6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| \(T^1 = L^a \times M^b \times (LT^{-2})^c\) | M1A1E1 | There is no M on the left, so \(b = 0\) |
| \(a = \frac{1}{2}, c = -\frac{1}{2}\) | m1 | Solution |
| \(\therefore \text{Period} = kl^{\frac{1}{2}}g^{-\frac{1}{2}}\) | A1F | Constant needed |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T^1 = L^{a+c} \times M^b \times T^{-2}\) | M1 | |
| \(-2c = 1\) | m1 | Equating corresponding indices |
| \(a + c = 0\) | m1 | |
| \(a = \frac{1}{2}, c = -\frac{1}{2}\) | m1 | Solution |
| \(\therefore \text{Period} = kl^{\frac{1}{2}}g^{-\frac{1}{2}}\) | A1F | Constant needed |
| 4 | ||
| 7 | Total |
### Part (i)
$T^1 = L^a \times M^b \times (LT^{-2})^c$ | M1A1E1 | There is no M on the left, so $b = 0$
$a = \frac{1}{2}, c = -\frac{1}{2}$ | m1 | Solution
$\therefore \text{Period} = kl^{\frac{1}{2}}g^{-\frac{1}{2}}$ | A1F | Constant needed
| | 3 |
### Part (ii)
$T^1 = L^{a+c} \times M^b \times T^{-2}$ | M1 |
$-2c = 1$ | m1 | Equating corresponding indices
$a + c = 0$ | m1 |
$a = \frac{1}{2}, c = -\frac{1}{2}$ | m1 | Solution
$\therefore \text{Period} = kl^{\frac{1}{2}}g^{-\frac{1}{2}}$ | A1F | Constant needed
| | 4 |
| | **7** | **Total** |1 The time $T$ taken for a simple pendulum to make a single small oscillation is thought to depend only on its length $l$, its mass $m$ and the acceleration due to gravity $g$.
By using dimensional analysis:
\begin{enumerate}[label=(\alph*)]
\item show that $T$ does not depend on $m$;
\item express $T$ in terms of $l , g$ and $k$, where $k$ is a dimensionless constant.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2006 Q1 [7]}}