| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring routine application of standard techniques. Part (a) is simple rearrangement of a formula to find dimensions of G. Part (b) involves setting up dimensional equations and solving simultaneous equations for three unknowns—standard A-level Further Maths fare. Part (c) requires basic interpretation. The question is well-scaffolded with clear steps and uses familiar physical quantities. |
| Spec | 6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([r] = L,\ [m] = M\) and \([U] = LT^{-1}\) | B1 | Correct dimensions for other parameters (\(U\), \(r\) and \(m\)) soi |
| \([G] = \left[\frac{u^2 r}{m}\right]\) | M1 | Comparing dimensions, realising that 2 is dimensionless and rearranging. Could be done by dimensional analysis e.g. \([G] = L^\alpha M^\beta T^\gamma\) and equate indices using \(U = \sqrt{\frac{2Gm}{r}}\) |
| \(\therefore [G] = (LT^{-1})^2 LM^{-1} = L^3 M^{-1} T^{-2}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P] = (MLT^{-2}L)/T = ML^2T^{-3}\) | B1 | Using \(P = WD/t\) |
| Need \(LT^{-1} = M^\alpha L^{2\alpha} T^{-3\alpha} M^\beta L^\beta T^{-2\beta} T^\gamma\) | B1 | Realising condition for equation to be dimensionally correct and substituting in dimensions. ft errors in \([P]\) and/or \([W]\) here and in subsequent method marks provided M, L and T appear at least twice on the RHS |
| M: \(\alpha + \beta = 0\), L: \(1 = 2\alpha + \beta\) | M1 | Comparing to obtain equations in \(\alpha\) and \(\beta\) |
| \(\alpha = 1,\ \beta = -1\) | A1 | |
| T: \(-1 = -3\alpha - 2\beta + \gamma\) | M1 | Comparing to obtain equation in \(\gamma\) |
| \(\gamma = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Because \(\gamma = 0\), the modelled minimum launch speed \(V\) does not depend on the time \(t\) for which the engines operate | B1ft | i.e. the modified model predicts that \(V\) does not vary when \(t\) varies. Or appropriate comment from their result, e.g. if \(\gamma = -1\), then \(V\) is inversely proportional to \(t\) |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[r] = L,\ [m] = M$ and $[U] = LT^{-1}$ | B1 | Correct dimensions for other parameters ($U$, $r$ and $m$) soi |
| $[G] = \left[\frac{u^2 r}{m}\right]$ | M1 | Comparing dimensions, realising that 2 is dimensionless and rearranging. Could be done by dimensional analysis e.g. $[G] = L^\alpha M^\beta T^\gamma$ and equate indices using $U = \sqrt{\frac{2Gm}{r}}$ |
| $\therefore [G] = (LT^{-1})^2 LM^{-1} = L^3 M^{-1} T^{-2}$ | A1 | AG |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P] = (MLT^{-2}L)/T = ML^2T^{-3}$ | B1 | Using $P = WD/t$ |
| Need $LT^{-1} = M^\alpha L^{2\alpha} T^{-3\alpha} M^\beta L^\beta T^{-2\beta} T^\gamma$ | B1 | Realising condition for equation to be dimensionally correct and substituting in dimensions. ft errors in $[P]$ and/or $[W]$ here and in subsequent method marks provided M, L and T appear at least twice on the RHS |
| M: $\alpha + \beta = 0$, L: $1 = 2\alpha + \beta$ | M1 | Comparing to obtain equations in $\alpha$ and $\beta$ |
| $\alpha = 1,\ \beta = -1$ | A1 | |
| T: $-1 = -3\alpha - 2\beta + \gamma$ | M1 | Comparing to obtain equation in $\gamma$ |
| $\gamma = 0$ | A1 | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Because $\gamma = 0$, the modelled minimum launch speed $V$ does not depend on the time $t$ for which the engines operate | B1ft | i.e. the modified model predicts that $V$ does not vary when $t$ varies. Or appropriate comment from their result, e.g. if $\gamma = -1$, then $V$ is inversely proportional to $t$ |
---5 The escape speed of an unpowered object is the minimum speed at which it must be projected to escape the gravitational influence of the Earth if it is projected vertically upwards from the Earth's surface. A formula for the escape speed $U$ of an unpowered object of mass $m$ is $U = \sqrt { \frac { 2 G m } { r } }$ where $r$ is the radius of the Earth and $G$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the dimensions of $G$ are $\mathrm { M } ^ { - 1 } \mathrm {~L} ^ { 3 } \mathrm {~T} ^ { - 2 }$.
A rocket is a powered object. A rocket is launched with a given launch speed and is then powered by engines which apply a constant force for a period of time after the launch.
A student wishes to apply the formula given above to a rocket launch. They wish to model the minimum launch speed required for a rocket to escape the Earth's gravitational influence.
They realise that the given formula is for unpowered objects and so they include an extra term in the formula to obtain $V = \sqrt { \frac { 2 G m } { r } } - \mathrm { kP } ^ { \alpha } \mathrm { W } ^ { \beta } \mathrm { t } ^ { \gamma }$.
In their modified formula, $G$ and $r$ are the same as before. The other variables are defined as follows.
\begin{itemize}
\item $V$ is the required minimum launch speed of the rocket
\item $k , \alpha , \beta$ and $\gamma$ are dimensionless constants
\item $P$ is the power developed by the engines of the rocket
\item $m$ is the initial mass of the rocket
\item $W$ is the initial weight of the rocket
\item $t$ is the total time for which the engines of the rocket operate
\item Use dimensional analysis to determine the values of $\alpha , \beta$ and $\gamma$.
\item By considering the value of $\gamma$ found in part (b) explain the relationship between $t$ and $V$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2021 Q5 [10]}}