| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Moderate -0.5 This is a straightforward dimensional analysis question requiring systematic application of standard techniques. Part (a) involves simple rearrangement of a formula to find dimensions of G, while part (b) requires equating powers of M, L, T—both are routine procedures covered in M3 with no novel insight needed. Slightly easier than average due to clear structure and standard method. |
| Spec | 6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(MLT^{-2} = \frac{[G]MM}{L^2}\) | M1 | |
| A1 | ||
| \([G] = L^3M^{-1}T^{-2}\) | A1F | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t = km^\alpha R^\beta G^\gamma\) | ||
| \(T = M^\alpha L^\beta M^{-\gamma} L^{3\gamma} T^{-2\gamma}\) | M1, A1F | L, M, T for \(G\) are needed to gain M1 |
| \(-2\gamma = 1 \Rightarrow \gamma = -\frac{1}{2}\) | m1 | Getting 3 equations |
| \(\alpha - \gamma = 0 \Rightarrow \alpha = -\frac{1}{2}\) | m1 | Solution |
| \(\beta + 3\gamma = 0 \Rightarrow \beta = \frac{3}{2}\) | A1F | Finding \(\alpha, \beta, \gamma\); Total: 5 marks |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $MLT^{-2} = \frac{[G]MM}{L^2}$ | M1 | |
| | A1 | |
| $[G] = L^3M^{-1}T^{-2}$ | A1F | Total: 3 marks |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t = km^\alpha R^\beta G^\gamma$ | | |
| $T = M^\alpha L^\beta M^{-\gamma} L^{3\gamma} T^{-2\gamma}$ | M1, A1F | L, M, T for $G$ are needed to gain M1 |
| $-2\gamma = 1 \Rightarrow \gamma = -\frac{1}{2}$ | m1 | Getting 3 equations |
| $\alpha - \gamma = 0 \Rightarrow \alpha = -\frac{1}{2}$ | m1 | Solution |
| $\beta + 3\gamma = 0 \Rightarrow \beta = \frac{3}{2}$ | A1F | Finding $\alpha, \beta, \gamma$; Total: 5 marks |
---1 The magnitude of the gravitational force, $F$, between two planets of masses $m _ { 1 }$ and $m _ { 2 }$ with centres at a distance $x$ apart is given by
$$F = \frac { G m _ { 1 } m _ { 2 } } { x ^ { 2 } }$$
where $G$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item By using dimensional analysis, find the dimensions of $G$.
\item The lifetime, $t$, of a planet is thought to depend on its mass, $m$, its initial radius, $R$, the constant $G$ and a dimensionless constant, $k$, so that
$$t = k m ^ { \alpha } R ^ { \beta } G ^ { \gamma }$$
where $\alpha , \beta$ and $\gamma$ are constants.\\
Find the values of $\alpha , \beta$ and $\gamma$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2007 Q1 [8]}}