| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Moderate -0.3 This is a straightforward dimensional analysis question requiring systematic application of standard techniques: verifying dimensions of a given equation, identifying dimensions of constants from context, and basic differentiation. While it spans multiple parts, each step follows directly from the previous with no novel insight required, 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 checking6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | [F] = MLT–2 |
| Answer | Marks |
|---|---|
| dx x L | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.1 | Correctly finding the dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | Only quantities with the same dimensions can |
| Answer | Marks | Guidance |
|---|---|---|
| means that [a] = [x]] | B1 | |
| [1] | 2.4 | |
| 3 | (c) | [ k ] M −1 2 ( L2) 1 2 =LT −1 |
| [k]=M 1 2T −1 | M1 | |
| A1 | 2.2a | |
| 1.1 | Use of formula for v to derive |
| Answer | Marks | Guidance |
|---|---|---|
| Alternative solution | M1 | Use of formula for v to derive |
| Answer | Marks |
|---|---|
| [k]=M 1 2T −1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | dv −1 1 −1 |
| Answer | Marks |
|---|---|
| ∴F =−k2x | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Use of chain rule to differentiate |
| Answer | Marks |
|---|---|
| dx | dv −1 −1 |
Question 3:
3 | (a) | [F] = MLT–2
dv [ m ][ v ][ v ] ML2T −2
mv = = =MLT −2
[ ]
dx x L | B1
B1
[2] | 1.1
2.1 | Correctly finding the dimensions
of both sides is sufficient for
B1B1; an explicit conclusion is
not necessary.
3 | (b) | Only quantities with the same dimensions can
be added (or subtracted) [so [a2] = [x2] which
means that [a] = [x]] | B1
[1] | 2.4
3 | (c) | [ k ] M −1 2 ( L2) 1 2 =LT −1
[k]=M 1 2T −1 | M1
A1 | 2.2a
1.1 | Use of formula for v to derive
dimensional equation for [k]
Alternative solution | M1 | Use of formula for v to derive
units of k.
1
−1 vm2
v=km 2 a2−x2 ⇒k = so the
a2−x2
units of k are kg 1 2s −1
[k]=M 1 2T −1 | A1
[2]
3 | (d) | dv −1 1 −1
=km 2(−2x) (a2−x2) 2
dx 2
dv
∴F =mv
dx
−1 1 −1 1 −1
=m×km 2(a2−x2) 2km 2(−2x) (a2−x2) 2
2
∴F =−k2x | M1
M1
A1
[3] | 1.1
1.1
1.1 | Use of chain rule to differentiate
v wrt x
Use of formula for F with m, v
dv
and their substituted in.
dx | dv −1 −1
=−km 2x(a2−x2) 2
dx
Use of formula for v to derive
units of k.3 A particle $P$ of mass $m$ moves on the $x$-axis under the action of a force $F$ directed along the axis. When the displacement of $P$ from the origin is $x$ its velocity is $v$.
\begin{enumerate}[label=(\alph*)]
\item By using the fact that the dimensions of the derivative $\frac { d v } { d x }$ are the same as those of $\frac { v } { x }$, verify that the equation $\mathrm { F } = \mathrm { mv } \frac { \mathrm { dv } } { \mathrm { dx } }$ is dimensionally consistent.
It is given that $\mathrm { v } = \mathrm { km } ^ { - \frac { 1 } { 2 } } \sqrt { \mathrm { a } ^ { 2 } - \mathrm { x } ^ { 2 } }$ where $a$ and $k$ are constants.
\item Explain why $[ a ]$ must be the same as $[ x ]$.
\item Deduce the dimensions of $k$.
\item Find an expression for $F$ in terms of $x$ and $k$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q3 [8]}}