| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with partial constraints |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring systematic application of a standard technique. Part (a) involves basic force dimensions to find [k] = MT^{-1}, while part (b) requires equating dimensions of time on both sides to solve three simultaneous equations. Though it's a Further Maths topic with multiple steps (7 marks total), the method is mechanical with no conceptual insight or problem-solving required—students either know the dimensional analysis algorithm or they don't. |
| Spec | 6.01b Units vs dimensions: relationship6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | F = M L T − 2 or v = L T − 1 |
| Answer | Marks | Guidance |
|---|---|---|
| L T 1 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| [ k ] = M T − 1 | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (b) | [T]=k2muF |
| Answer | Marks | Guidance |
|---|---|---|
| | M1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| T=(MT −1)2M (LT −1) (MLT −2) | M1dep* | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 + = | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 , 3 , 3 = = = − | A1 | 2.2a |
Question 5:
5 | (a) | F = M L T − 2 or v = L T − 1 | B1 | 1.2 | For either the dimensions of force or velocity correct
(possibly seen in an expression for the dimensions of
k)
M L T− − 2
k =
L T 1 | M1 | 1.1 | Use of F = k v with at least one of [F] or [v]
correct
[ k ] = M T − 1 | A1 | 2.2a
[3]
(b) | [T]=k2muF
e.g.
| M1* | 2.1 | For correctly equating the dimensions of T (or either
of the other two terms on the rhs) to the dimensions
1
k2muF
of - this mark can be implied by setting
3
up an equation in M, L and T – if k is missing or
assumed to be dimensionless then no marks in this
part. Must be powers of k, m, u and F and no other
(incorrect) terms
T=(MT −1)2M (LT −1) (MLT −2) | M1dep* | 1.1 | Setting up an equation in M, L and T using their [k]
and at least one of [F] or [v] correct
2 3 + = −
2 + = −
0 + = | M1 | 1.1 | Setting up three equations in , and from their
MLT (dependent on both previous M marks) – allow
one slip from their MLT equation
1 , 3 , 3 = = = − | A1 | 2.2a
[4]A particle P of mass $m$ kg is projected with speed $u \text{ m s}^{-1}$ along a rough horizontal surface. During the motion of P, a constant frictional force of magnitude $F$ N acts on P. When the velocity of P is $v \text{ m s}^{-1}$, it experiences a force of magnitude $kv$ N due to air resistance, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Determine the dimensions of $k$. [3]
\end{enumerate}
At time $T$ s after projection P comes to rest. A formula approximating the value of $T$ is
$$T = \frac{mu}{F} - \frac{kmu^2}{2F^2} + \frac{1}{3}k^2m^{\alpha}u^{\beta}F^{\gamma}.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use dimensional analysis to find $\alpha$, $\beta$ and $\gamma$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q5 [7]}}