OCR MEI Further Mechanics Minor Specimen — Question 3 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeFind exponents with all unknowns
DifficultyStandard +0.3 This is a standard dimensional analysis question with straightforward setup. Part (i) requires basic recall of dimensions, part (ii) involves solving a simple system of three linear equations (standard technique), and part (iii) is direct substitution with percentage changes. While it requires multiple steps, each component is routine for Further Maths students and follows a well-practiced method.
Spec6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions

3
  1. Find the dimensions of
    • density and
    • pressure (force per unit area).
    The frequency, \(f\), of the note emitted by an air horn is modelled as \(f = k s ^ { \alpha } p ^ { \beta } d ^ { \gamma }\), where
    • \(s\) is the length of the horn,
    • \(\quad p\) is the air pressure,
    • \(d\) is the air density,
    • \(k\) is a dimensionless constant.
    • Determine the values of \(\alpha , \beta\) and \(\gamma\).
    A particular air horn emits a note at a frequency of 512 Hz and the air pressure and air density are recorded. At another time it is found that the air pressure has fallen by \(2 \%\) and the air density has risen by \(1 \%\). The length of the horn is unchanged.
  2. Calculate the new frequency predicted by the model.

Question 3:
AnswerMarks Guidance
3(i) [density] = M L -3
[pressure] = M L T -2 / L-2 = M L -1 T -2B1
B1
AnswerMarks
[2]1.2
1.1
AnswerMarks Guidance
3(ii) T(cid:16)1 (cid:32)L(cid:68) (M L(cid:16)1 T(cid:16)2)(cid:69) (M L(cid:16)3)(cid:74)
α = – 1
1
β =
2
(cid:16)1
γ =
AnswerMarks
2M1
M1
A1
A1
AnswerMarks
[4]3.3
1.1
1.1
AnswerMarks
2.2aNSet up
compare coefficients of M, L and T
One correct
E
All correct
AnswerMarks Guidance
3(iii) Suppose the values for f = 512 are s , p and d
0 0 0
either
0.98p
ks 0
f ' 0 1.01d 0.98 C
(cid:32) 0 (cid:32)
512 p 1.01
ks 0
0 d
0
AnswerMarks Guidance
orIM1 M
3.1bAny complete method
E
p
Find k using 512(cid:32)ks 0
0 d
0
P
512 d 0.98p
so substituting for k, f '(cid:32) 0 (cid:117)s 0
s p 0 1.01d
AnswerMarks Guidance
0 0 0M1 oe Method must have k eliminated
S
AnswerMarks
so f ' = 504.33… so 504Hz (3 s.f.)A1
[2]1.1
Question 3:
3 | (i) | [density] = M L -3
[pressure] = M L T -2 / L-2 = M L -1 T -2 | B1
B1
[2] | 1.2
1.1
3 | (ii) | T(cid:16)1 (cid:32)L(cid:68) (M L(cid:16)1 T(cid:16)2)(cid:69) (M L(cid:16)3)(cid:74)
α = – 1
1
β =
2
(cid:16)1
γ =
2 | M1
M1
A1
A1
[4] | 3.3
1.1
1.1
2.2a | NSet up
compare coefficients of M, L and T
One correct
E
All correct
3 | (iii) | Suppose the values for f = 512 are s , p and d
0 0 0
either
0.98p
ks 0
f ' 0 1.01d 0.98 C
(cid:32) 0 (cid:32)
512 p 1.01
ks 0
0 d
0
or | IM1 | M
3.1b | Any complete method
E
p
Find k using 512(cid:32)ks 0
0 d
0
P
512 d 0.98p
so substituting for k, f '(cid:32) 0 (cid:117)s 0
s p 0 1.01d
0 0 0 | M1 | oe Method must have k eliminated
S
so f ' = 504.33… so 504Hz (3 s.f.) | A1
[2] | 1.1
3 (i) Find the dimensions of

\begin{itemize}
  \item density and
  \item pressure (force per unit area).
\end{itemize}

The frequency, $f$, of the note emitted by an air horn is modelled as $f = k s ^ { \alpha } p ^ { \beta } d ^ { \gamma }$, where

\begin{itemize}
  \item $s$ is the length of the horn,
  \item $\quad p$ is the air pressure,
  \item $d$ is the air density,
  \item $k$ is a dimensionless constant.\\
(ii) Determine the values of $\alpha , \beta$ and $\gamma$.
\end{itemize}

A particular air horn emits a note at a frequency of 512 Hz and the air pressure and air density are recorded. At another time it is found that the air pressure has fallen by $2 \%$ and the air density has risen by $1 \%$. The length of the horn is unchanged.\\
(iii) Calculate the new frequency predicted by the model.

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor  Q3 [8]}}