| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Standard +0.3 This is a standard dimensional analysis question requiring systematic application of a well-defined technique. Part (a) is straightforward rearrangement of dimensions from Newton's law of gravitation. Part (b) involves setting up and solving three simultaneous equations from dimensional homogeneity—routine for Further Maths students who have learned this method, though slightly above average difficulty due to the algebraic manipulation required. |
| Spec | 6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([F] = MLT^{-2}\), \([d] = L\), \([m_1] = [m_2] = M\) | M1 | Recalls dimensions for force, distance and mass to form equation for dimensional consistency |
| \([G] = \frac{MLT^{-2} \times L^2}{M^2} = M^{-1}L^3T^{-2}\) | R1 | Completes rigorous argument using both dimensions for force, distance and mass to verify dimensions of \(G\) are \(M^{-1}L^3T^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([m^a r^b G^c] = (M)^a(L)^b(M^{-1}L^3T^{-2})^c = M^{a-c}L^{b+3c}T^{-2c}\) | M1 | Uses dimensions to form correct expression for dimensions of \(\left[m^a r^b G^c\right]\) |
| Forms three simultaneous equations: \(a - c = 0\), \(b + 3c = 0\), \(-2c = 1\) | M1 | PI by correct values of \(a\), \(b\), \(c\) |
| \(c = -0.5\), \(b = 1.5\), \(a = -0.5\) | A1 | CAO |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[F] = MLT^{-2}$, $[d] = L$, $[m_1] = [m_2] = M$ | M1 | Recalls dimensions for force, distance and mass to form equation for dimensional consistency |
| $[G] = \frac{MLT^{-2} \times L^2}{M^2} = M^{-1}L^3T^{-2}$ | R1 | Completes rigorous argument using both dimensions for force, distance and mass to verify dimensions of $G$ are $M^{-1}L^3T^{-2}$ |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[m^a r^b G^c] = (M)^a(L)^b(M^{-1}L^3T^{-2})^c = M^{a-c}L^{b+3c}T^{-2c}$ | M1 | Uses dimensions to form correct expression for dimensions of $\left[m^a r^b G^c\right]$ |
| Forms three simultaneous equations: $a - c = 0$, $b + 3c = 0$, $-2c = 1$ | M1 | PI by correct values of $a$, $b$, $c$ |
| $c = -0.5$, $b = 1.5$, $a = -0.5$ | A1 | CAO |
---6 The magnitude of the gravitational force $F$ between two planets of masses $m _ { 1 }$ and $m _ { 2 }$ with centres at a distance $d$ apart is given by
$$F = \frac { G m _ { 1 } m _ { 2 } } { d ^ { 2 } }$$
where $G$ is a constant.\\
6
\begin{enumerate}[label=(\alph*)]
\item Show that $G$ must have dimensions $L ^ { 3 } M ^ { - 1 } T ^ { - 2 }$, where $L$ represents length, $M$ represents mass and $T$ represents time.\\
6
\item The lifetime $t$ of a planet is thought to depend on its mass $m$, its radius $r$, the constant $G$ and a dimensionless constant $k$ such that
$$t = k m ^ { a } r ^ { b } G ^ { c }$$
where $a , b$ and $c$ are constants.\\
Determine the values of $a , b$ and $c$.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2020 Q6 [5]}}