| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find dimensions of compound quantity |
| Difficulty | Standard +0.3 This is a structured dimensional analysis question with clear scaffolding through parts (a)-(e). Parts (a) and (b) are straightforward recall of definitions requiring minimal calculation. Part (c) is a standard dimensional analysis problem with guided setup. Parts (d) and (e) test understanding of the derived formula through simple substitution and conceptual reasoning. While dimensional analysis is a Further Maths topic, the execution here is methodical and well-guided, making it slightly easier than a typical A-level question. |
| 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 |
|---|---|---|
| 2 | (a) | [Strain] = [extension]/[length] = L/L = 1 or |
| L0 or M0L0T0 or dimensionless | B1 | 2.2a |
| Answer | Marks |
|---|---|
| and stated as dimensionless | Do not allow use of m (metres). Do |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | [Stress] = MLT–2 L–2 = ML–1T–2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [E] = [Stress] = ML–1T–2 | B1 | 2.2a |
| considering dimensions of strain | Allow clear [stress]/[strain] = |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | [V] = L3 and [] = ML–3 and [c] = LT–1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| LT–1 = (ML–1T–2)(L3)(ML–3) | M1* | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| T: –1 = –2 | M1dep | 1.1 |
| Answer | Marks |
|---|---|
| equation | M2 for correct consideration by |
| Answer | Marks | Guidance |
|---|---|---|
| = ½ => = –½ => = 0 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| | A1 | 1.1 |
| also | If correct answer seen from a |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | (i) | 5002 ms–1 or awrt 707 ms–1 |
| from part (c) | Numerical answers needed |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | (ii) | According to the model there is no |
| dependency on V so 500 ms–1 | B1 | 3.5a |
| Answer | Marks |
|---|---|
| matter” so unchanged | Needs to be an explanatory |
| Answer | Marks |
|---|---|
| (e) | There may be other (dimensionless) |
| Answer | Marks | Guidance |
|---|---|---|
| temperature) | B1 | 3.5b |
| Answer | Marks |
|---|---|
| that are dimensionally consistent | Do not allow “there could be a |
Question 2:
2 | (a) | [Strain] = [extension]/[length] = L/L = 1 or
L0 or M0L0T0 or dimensionless | B1 | 2.2a | AG. Condone = 0 (or =k) or [L]
provided that the ratio is correct
and stated as dimensionless | Do not allow use of m (metres). Do
not allow L : L
[1]
(b) | [Stress] = MLT–2 L–2 = ML–1T–2 | B1 | 1.1
So because “Strain is dimensionless”,
[E] = [Stress] = ML–1T–2 | B1 | 2.2a | oe e.g show divide by 1 when
considering dimensions of strain | Allow clear [stress]/[strain] =
[stress]
[2]
(c) | [V] = L3 and [] = ML–3 and [c] = LT–1 | B1 | 3.3 | All used in the solution | B1 correct dimensions all used in
solution
c = kEV (where k is dimensionless) =>
LT–1 = (ML–1T–2)(L3)(ML–3) | M1* | 3.3 | Setting up the model with their
dimensions (condone missing k
here). Must include ,, or
equivalent
M: 0 = +
L: 1 = – + 3 – 3
T: –1 = –2 | M1dep | 1.1 | Comparing dimensions on both
sides to derive three equations in
, and . Allow one incorrect
equation | M2 for correct consideration by
division with their dimensions and
manipulation of indices leading to
their dimensions for c
= ½ => = –½ => = 0 | A1 | 1.1 | Give this if = 0 seen in final
equation (i.e. no dependence on
V)
E
c k =
| A1 | 1.1 | k necessary here. Allow indices
also | If correct answer seen from a
division method using correct
dimensions allow A2. If k not
present A1.
[5]
(d) | (i) | 5002 ms–1 or awrt 707 ms–1 | B1FT | 3.4 | Follow through their model
from part (c) | Numerical answers needed
[1]
(d) | (ii) | According to the model there is no
dependency on V so 500 ms–1 | B1 | 3.5a | Some explanation involving (or
calculation using) the model
must be given with reference to
V. Allow “volume doesn’t
matter” so unchanged | Needs to be an explanatory
comment, not just “500”
No FT in this part
[1]
(e) | There may be other (dimensionless)
parameters which affect c but have not been,
or cannot be, taken into account (eg
temperature) | B1 | 3.5b | Allow:
can’t calculate/find value of
dimensionless constant
There may be additional terms
that are dimensionally consistent | Do not allow “there could be a
dimensionless constant”
B1 BOD can’t identify
[1]Materials have a measurable property known as the Young's Modulus, $E$.
If a force is applied to one face of a block of the material then the material is stretched by a distance called the extension. Young's modulus is defined as the ratio $\frac{\text{Stress}}{\text{Strain}}$ where Stress is defined as the force per unit area and Strain is the ratio of the extension of the block to the length of the block.
\begin{enumerate}[label=(\alph*)]
\item Show that Strain is a dimensionless quantity. [1]
\item By considering the dimensions of both Stress and Strain determine the dimensions of $E$. [2]
\end{enumerate}
It is suggested that the speed of sound in a material, $c$, depends only upon the value of Young's modulus for the material, $E$, the volume of the material, $V$, and the density (or mass per unit volume) of the material, $\rho$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use dimensional analysis to suggest a formula for $c$ in terms of $E$, $V$ and $\rho$. [5]
\item The speed of sound in a certain material is $500$ m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Use your formula from part (c) to predict the speed of sound in the material if the value of Young's modulus is doubled but all other conditions are unchanged. [1]
\item With reference to your formula from part (c), comment on the effect on the speed of sound in the material if the volume is doubled but all other conditions are unchanged. [1]
\end{enumerate}
\item Suggest one possible limitation caused by using dimensional analysis to set up the model in part (c). [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2023 Q2 [11]}}