| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring systematic checking of each term against [Power]. The given hint about derivatives simplifies the work, and the method is routine: identify dimensions of each variable, substitute into each term, and verify all match ML²T⁻³. While it requires careful bookkeeping across multiple terms, it involves no problem-solving insight or novel reasoning—just methodical application of a standard technique. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking |
| Answer | Marks |
|---|---|
| 5 | Correct use of dimensions using a suitable |
| Answer | Marks |
|---|---|
| consistent. | M1 |
| Answer | Marks |
|---|---|
| [6] | 2.1 |
| Answer | Marks |
|---|---|
| 2.2a | DR |
| Answer | Marks |
|---|---|
| Condone [] seen around M,L,T | Argument could be based around |
Question 5:
5 | Correct use of dimensions using a suitable
formula for at least one of the terms in the
equation
[P] = [Fv] = MLT–2 LT–1 = ML2T–3
(or [WD/t])
d v m v 2
1st term: k m v = = M (LT–1)2 T–1
1
d t t
= ML2T–2T–1 = ML2T–3
2nd term: [k mgv] = M LT–2 LT–1 = ML2T–3
2
3rd term: [k E] = ML2T–3 since a rate of energy
3
loss will have the same dimensions as a power
So, because the constants are dimensionless
and because quantities of the same
dimensions add to give quantities of that
dimension, LHS and RHS have same
dimensions so equation is dimensionally
consistent. | M1
A1
A1
A1
A1
A1
[6] | 2.1
3.4
2.2a
3.5a
2.2a | DR
May not be fully simplified.
Using a suitable relationship
involving power and quantities
with determinable dimensions to
derive [P]
Condone missing k . Derivation
1
must be convincing.
Condone missing k . Derivation
2
must be convincing.
Condone missing k . Can be
3
explained using rate of change
of energy or done as [P] above.
Explanation must deal with k’s
if these are not included in at
least one of the above.
Argument must include dealing
with + (ie different terms) and
must reach a clear conclusion.
May see the addition in a
formula.
Do not allow “all the parts have
the same dimension so you’re
able to add the terms” – needs to
conclude with dimensions of
sum
Condone [] seen around M,L,T | Argument could be based around
units. eg 1 W = 1 J/s = 1 Nm/s etc
until dimensions reached.
Need to get correctly to the answer5 In this question you may assume that if $x$ and $y$ are any physical quantities then $\left[ \frac { \mathrm { dy } } { \mathrm { dx } } \right] = \left[ \frac { \mathrm { y } } { \mathrm { x } } \right]$.
A machine drives a piston of mass $m$ into a vertical cylinder. The equation below is suggested to model the power developed by the machine, $P$, while it is not doing any other external work.
$$\mathrm { P } = \mathrm { k } _ { 1 } \mathrm { mv } \frac { \mathrm { dv } } { \mathrm { dt } } + \mathrm { k } _ { 2 } \mathrm { mgv } + \mathrm { k } _ { 3 } \mathrm { E }$$
in which
\begin{itemize}
\item $v$ is the velocity of the piston at a given time,
\item $g$ is the acceleration due to gravity,
\item $E$ is the rate at which heat energy is lost to the surroundings,
\item $k _ { 1 } , k _ { 2 }$ and $k _ { 3 }$ are dimensionless constants.
\end{itemize}
Determine whether the equation is dimensionally consistent. Show all the steps in your argument.
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q5 [6]}}