| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with partial constraints |
| Difficulty | Standard +0.8 This is a multi-part dimensional analysis question requiring students to set up and solve simultaneous equations from dimensional consistency, then use physical reasoning (straight line graph, boundary conditions) to determine unknown exponents and derive a kinematic equation. While the mechanics topic is standard, the systematic approach to deriving a known formula through dimensional analysis and physical constraints requires more sophisticated reasoning than typical A-level questions. |
| 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 |
|---|---|---|
| \([v] = LT^{-1}\) | B1 (AO 1.2) | Used in solution; penalise wrong basic terms only once |
| \([u^\alpha a^\beta t^\gamma] = L^\alpha T^{-\alpha} L^\beta T^{-2\beta} T^\gamma\) | B1 (AO 3.3) | Correctly finding the dimensions of \(u^\alpha a^\beta t^\gamma\) in terms of \(\alpha\), \(\beta\) and \(\gamma\); allow unsimplified |
| \(1 = \alpha + \beta\) or \(-1 = -\alpha - 2\beta + \gamma\) | M1 (AO 3.4) | Equating their dimensions L and T |
| \(\Rightarrow \alpha = 1 - \beta\) | A1 (AO 1.1) | |
| \(\Rightarrow \gamma = \beta\) | A1 [5] (AO 1.1) | www; if extra term such as M is included, then B1B0M1A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| For straight line graph \(t^\gamma\) must be 1 (or constant or \(t^0\)) or \(t\) (or \(t^1\))... | M1 (AO 3.1b) | For clear understanding that the relationship must be of the form \(v = mt + c\) where both \(mt\) and \(c\) must take the form \([k]u^\alpha a^\beta t^\gamma\); or could see e.g. \(v = u^{1-\beta}a^\beta t^\beta\) with \(\beta = 1 \Rightarrow v = at\) and \(\beta = 0 \Rightarrow v = u\) |
| ...(so \(\gamma = 0\) or 1) so \(\beta = 0\) or 1 | A1 [2] (AO 1.1) | SC1 for \(\beta = 1\) using direct proportion or unsupported but www e.g. \(\beta = -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v\) must be the sum of terms like \(ku^\alpha a^\beta t^\gamma\) | M1 (AO 2.1) | AG (or \(k_1u + k_2at\) or \(mt + c\)); award if at least one term seen, must have \(k\), \(u\), \(a\) and \(t\) |
| \(v = k_1u + k_2at\) and \(v = u\) when \(t = 0 \Rightarrow k_1 = 1\) | A1 (AO 3.4) | |
| and \(v = u + a\) when \(t = 1 \Rightarrow k_1 = 1\) so \(v = u + at\) | A1 [3] (AO 2.2a) |
# Question 6:
## Part (a)
$[v] = LT^{-1}$ | B1 (AO 1.2) | Used in solution; penalise wrong basic terms only once
$[u^\alpha a^\beta t^\gamma] = L^\alpha T^{-\alpha} L^\beta T^{-2\beta} T^\gamma$ | B1 (AO 3.3) | Correctly finding the dimensions of $u^\alpha a^\beta t^\gamma$ in terms of $\alpha$, $\beta$ and $\gamma$; allow unsimplified
$1 = \alpha + \beta$ or $-1 = -\alpha - 2\beta + \gamma$ | M1 (AO 3.4) | Equating their dimensions L and T
$\Rightarrow \alpha = 1 - \beta$ | A1 (AO 1.1) |
$\Rightarrow \gamma = \beta$ | A1 [5] (AO 1.1) | www; if extra term such as M is included, then B1B0M1A0A0
## Part (b)
For straight line graph $t^\gamma$ must be 1 (or constant or $t^0$) or $t$ (or $t^1$)... | M1 (AO 3.1b) | For clear understanding that the relationship must be of the form $v = mt + c$ where both $mt$ and $c$ must take the form $[k]u^\alpha a^\beta t^\gamma$; or could see e.g. $v = u^{1-\beta}a^\beta t^\beta$ with $\beta = 1 \Rightarrow v = at$ and $\beta = 0 \Rightarrow v = u$
...(so $\gamma = 0$ or 1) so $\beta = 0$ or 1 | A1 [2] (AO 1.1) | SC1 for $\beta = 1$ using direct proportion or unsupported but www e.g. $\beta = -1$
## Part (c)
$v$ must be the sum of terms like $ku^\alpha a^\beta t^\gamma$ | M1 (AO 2.1) | AG (or $k_1u + k_2at$ or $mt + c$); award if at least one term seen, must have $k$, $u$, $a$ and $t$
$v = k_1u + k_2at$ and $v = u$ when $t = 0 \Rightarrow k_1 = 1$ | A1 (AO 3.4) |
and $v = u + a$ when $t = 1 \Rightarrow k_1 = 1$ so $v = u + at$ | A1 [3] (AO 2.2a) |
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6 A particle moves in a straight line with constant acceleration $a$. Its initial velocity is $u$ and at time $t$ its velocity is $v$.
It is assumed that $v$ depends only on $u , a$ and $t$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that this dependency is of the form $\mathrm { u } ^ { \alpha } \mathrm { a } ^ { \beta } \mathrm { t } ^ { \gamma }$, use dimensional analysis to find $\alpha$ and $\gamma$ in terms of $\beta$.
\item By noting that the graph of $v$ against $t$ must be a straight line, determine the possible values of $\beta$.
You may assume that the units of the given quantities are the corresponding SI units.
\item By considering $v$ when $t = 0$ seconds and when $t = 1$ second, derive the equation of motion $\mathrm { v } = \mathrm { u } + \mathrm { at }$, explaining your reasoning.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2022 Q6 [10]}}