| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis problem requiring students to equate dimensions of time with combinations of g, l, and m. Part (a) involves setting up and solving simple simultaneous equations from dimensional homogeneity. Parts (b)(i) and (b)(ii) are direct substitutions once the formula is found. While it requires systematic working, it's a standard textbook exercise with no novel insight needed, making it slightly easier than average. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking |
| Answer | Marks | Guidance |
|---|---|---|
| \([g] = LT^{-2}\) | B1 | Notation must be fully correct; M, L, T and nothing else; do not allow extra terms |
| \([T =] (LT^{-2})^\alpha L^\beta M^\gamma\) | M1 | Forming a dimensional equation between quantities with \([c] = 1\) soi and their \([g]\) |
| M: \(\gamma = 0\) | B1 | Allow \(\beta = 0\) if using \(MLT^{-2}\) for \([g]\); \(MLT^{-2}\) may also be seen (\(\Rightarrow \gamma = \frac{1}{2}\) later) |
| L: \(\alpha + \beta = 0\) and T: \(1 = -2\alpha\) | M1 | Equating their \(g^\alpha l^\beta m^\gamma\) to T (soi) and deriving equations for L and T; allow e.g. \(0 = L^\alpha + L^\beta\) if recovered by stating \(\alpha + \beta = 0\) |
| \(\alpha = -\frac{1}{2},\ \beta = \frac{1}{2}\) | A1 [5] | Could be embedded as e.g. \(\tau = c\sqrt{\frac{l}{g}}\); at least one of the two M marks should be scored; allow A1 if correct values obtained using \(MLT^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tau' = c\sqrt{\frac{4l}{g}} = 2c\sqrt{\frac{l}{g}}\) | M1 | Using the model to consider length quadrupling and all other parameters (except \(\tau\)) staying the same (must see \(\sqrt{4} = 2\) or equivalent for *their* \(\beta\)) |
| So the period (or \(\tau\)) is doubled | B1FT [2] | or just "so period doubles" oe; FT their value of \(\beta\); SC B1 for "no change" (with reason given) if using \(MT^{-2}\) for \([g]\) and consistent with their equations |
| Answer | Marks | Guidance |
|---|---|---|
| If mass quadruples, there is no change to the period (or \(\tau\)) since there is no dependency on \(m\) \((\gamma = 0)\) | B1 [1] | Answer must include both behaviour (no change to period) *and* reason (no dependency on \(m\)), i.e. not just 'no effect'; allow SC B1 for time doubling (with reason given) if using \(MT^{-2}\) or \(LMT^{-2}\) for \([g]\) and consistent with their equations |
# Question 3:
## Part (a)
$[g] = LT^{-2}$ | **B1** | Notation must be fully correct; M, L, T and nothing else; do not allow extra terms
$[T =] (LT^{-2})^\alpha L^\beta M^\gamma$ | **M1** | Forming a dimensional equation between quantities with $[c] = 1$ soi and their $[g]$
M: $\gamma = 0$ | **B1** | Allow $\beta = 0$ if using $MLT^{-2}$ for $[g]$; $MLT^{-2}$ may also be seen ($\Rightarrow \gamma = \frac{1}{2}$ later)
L: $\alpha + \beta = 0$ and T: $1 = -2\alpha$ | **M1** | Equating their $g^\alpha l^\beta m^\gamma$ to T (soi) and deriving equations for L and T; allow e.g. $0 = L^\alpha + L^\beta$ if recovered by stating $\alpha + \beta = 0$
$\alpha = -\frac{1}{2},\ \beta = \frac{1}{2}$ | **A1** [5] | Could be embedded as e.g. $\tau = c\sqrt{\frac{l}{g}}$; at least one of the two M marks should be scored; allow A1 if correct values obtained using $MLT^{-2}$
## Part (b)(i)
$\tau' = c\sqrt{\frac{4l}{g}} = 2c\sqrt{\frac{l}{g}}$ | **M1** | Using the model to consider length quadrupling and all other parameters (except $\tau$) staying the same (must see $\sqrt{4} = 2$ or equivalent for *their* $\beta$)
So the period (or $\tau$) is doubled | **B1FT** [2] | or just "so period doubles" oe; FT their value of $\beta$; SC B1 for "no change" (with reason given) if using $MT^{-2}$ for $[g]$ and consistent with their equations
## Part (b)(ii)
If mass quadruples, there is no change to the period (or $\tau$) since there is no dependency on $m$ $(\gamma = 0)$ | **B1** [1] | Answer must include **both** behaviour (no change to period) *and* reason (no dependency on $m$), i.e. not just 'no effect'; allow SC B1 for time doubling (with reason given) if using $MT^{-2}$ or $LMT^{-2}$ for $[g]$ and consistent with their equations3 A small object $P$ of mass $m$ is suspended from a fixed point by a light inextensible string of length l. When $P$ is displaced and released in a certain way, it oscillates in a vertical plane. The time taken for one complete oscillation is called the period and is denoted by $\tau$.
A student is carrying out experiments with $P$ and suggests the following formula to model the value of $\tau$.\\
$\tau = \mathrm { cg } \mathrm { a } ^ { \mathrm { a } } \mathrm { l } _ { \mathrm { m } } { } ^ { \gamma }$\\
in which
\begin{itemize}
\item $g$ is the acceleration due to gravity,
\item $C$ is a dimensionless constant.
\begin{enumerate}[label=(\alph*)]
\item Use dimensional analysis to determine the values of the constants $\alpha , \beta$ and $\gamma$.
\item \begin{enumerate}[label=(\roman*)]
\item Determine the effect on the period, according to the model, if the length of the string is then multiplied by 4, all other conditions being unchanged.
\item Determine the effect on the period, according to the model, if instead the mass of the object is multiplied by 4, all other conditions being unchanged.
\end{itemize}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2024 Q3 [8]}}