| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Moderate -0.8 This is a straightforward dimensional analysis question requiring routine application of standard techniques. Part (i) involves simple algebraic manipulation of dimensions, part (ii) requires basic graph interpretation, and part (iii) involves recognizing that dimensionless quantities cannot determine exponents. All steps are standard textbook exercises with no novel problem-solving required. |
| Spec | 6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | p |
| Answer | Marks |
|---|---|
| L2T(cid:16)1 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 3.3 | Correct dimensions for R and one |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (ii) | E.g. Most of the points seem to lie on a straight line, |
| Answer | Marks |
|---|---|
| [with one point discarded] is consistent with the model. | B1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 3.5a | For discussing a possible linear |
| Answer | Marks |
|---|---|
| the experimental data. | This mark is for a comment |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. The result at v = 4 looks inconsistent with the | E1 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. The experiment is not consistent with the model. | A1 | i |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (iii) | (A) |
| Answer | Marks |
|---|---|
| (cid:171) (cid:172) (cid:75) (cid:187) (cid:188) ML(cid:16)1T(cid:16)1 p | c |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Find the dimensions of the |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (iii) | (B) |
| Answer | Marks | Guidance |
|---|---|---|
| dimensions of the term | E1 | |
| [1] | 2.4 | Could say M0L0T0 instead of |
Question 3:
3 | (i) | p
MLT(cid:16)2 (cid:32)L(cid:117)LT(cid:16)1(cid:117)[(cid:75)]
S
MLT(cid:16)2
so (cid:62)(cid:75)(cid:64)(cid:32) =ML(cid:16)1T(cid:16)1 AG
L2T(cid:16)1 | B1
M1
E1
[3] | 1.1
1.1
3.3 | Correct dimensions for R and one
other
(cid:62)(cid:75)(cid:64)
Equating dimensions with .
Clearly shown
3 | (ii) | E.g. Most of the points seem to lie on a straight line,
which passes through or near the origin.
either
E.g. The result for v = 4 seems inconsistent with the
others, perhaps some experimental error, so discard it.
E.g. Model suggests a proportional relationship
between R and v [with r and fixed], so experiment
[with one point discarded] is consistent with the model. | B1
E1
A1 | 1.1
2.2b
3.5a | For discussing a possible linear
model
Forn commenting on the
experimental data
eFor a conclusion which is
consistent with their comments on
the experimental data. | This mark is for a comment
about the experimental data
suggesting a proportional
relationship.
A1 can only be awarded for
claiming proportionality if the
line was claimed to pass through
the origin.
or
E.g. The result at v = 4 looks inconsistent with the | E1 | E1 | m | m
others, so possibly the linear relationship is not correct.
E.g. The experiment is not consistent with the model. | A1 | i
[3]
3 | (iii) | (A) | e
(cid:170)(cid:85)rv(cid:186) ML(cid:16)3(cid:117)L(cid:117)LT(cid:16)1
(cid:32) (cid:32)1
(cid:171) (cid:172) (cid:75) (cid:187) (cid:188) ML(cid:16)1T(cid:16)1 p | c
B1
B1
[2] | 1.1
1.1 | Find the dimensions of the
expression
Dimensions of density
3 | (iii) | (B) | S
Dimensionless so raising to power α will not affect the
dimensions of the term | E1
[1] | 2.4 | Could say M0L0T0 instead of
dimensionless.
3 Solid toy aeroplane nose cones of various sizes are made in the shape shown in Fig. 3.1, where OA is its line of symmetry.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-3_364_432_395_845}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
The air resistance against the nose cone as the aeroplane flies through the air is initially modelled by $R = k r v \eta$, where $R$ is the air resistance, $r$ is the radius of the circular flat end of the nose cone, $v$ is the velocity of the nose cone, $\eta$ is the viscosity of the air and $k$ is a dimensionless constant.
\begin{enumerate}[label=(\roman*)]
\item Use dimensional analysis to show that the dimensions of $\eta$ are $\mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 1 }$.
In an experiment conducted on a particular nose cone, measurements of air resistance are taken for different velocities. The viscosity of the air does not vary during the experiment. The graph in Fig. 3.2 shows the results. Measurements are given using the appropriate S.I. units.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-3_794_1166_1411_427}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\item Comment on whether the results of this experiment are consistent with the initial model.
It is now suggested that a better model for the air resistance is $R = K r v \left( \frac { \rho r v } { \eta } \right) ^ { \alpha }$, where $\rho$ is the density of the air, $K$ is a dimensionless constant and $R , r , v$ and $\eta$ are as before.
\item (A) Find the dimensions of $\frac { \rho r v } { \eta }$.\\
(B) Explain why you cannot use dimensional analysis to find the value of $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS Q3 [9]}}