OCR MEI Further Mechanics Minor 2019 June — Question 2 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeFind exponents with all unknowns
DifficultyStandard +0.3 This is a straightforward dimensional analysis question with standard steps: recall dimensions of pressure, unit conversion, equate dimensions to find exponents (with all unknowns given), and apply the formula. While it requires multiple parts and careful algebraic manipulation, it follows a well-practiced template with no novel insight required, making it slightly easier than average.
Spec6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions

2
  1. Write down the dimensions of pressure. The SI unit of pressure is the pascal (Pa). 15 Pa is equivalent to \(Q\) newtons per square centimetre.
  2. Find the value of \(Q\). Simon thinks the speed, \(v\), of sound in a gas is given by the formula \(v = k P ^ { x } d ^ { y } V ^ { z }\),
    where \(P\) is the pressure of the gas, \(d\) is the density of the gas, \(V\) is the volume of the gas, \(k\) is a dimensionless constant.
  3. Use dimensional analysis to

Question 2:
AnswerMarks Guidance
2(a) M Lโˆ’1 Tโˆ’2
[1]
AnswerMarks Guidance
(b)0.0015 B1
[1]
AnswerMarks Guidance
(c)LTโˆ’1 = (๐‘˜)(MLโˆ’1Tโˆ’2)๐‘ฅ(MLโˆ’3)๐‘ฆ(L3)๐‘ง M1
M11.1 Compare coeffs of M, L and T
1 1
๐‘ฅ = ๐‘ฆ = โˆ’
AnswerMarks Guidance
2 2A1ft 1.1
Show z = 0A1 1.1
[4]
AnswerMarks
(d)1
๐‘˜ = 340รทโˆš
AnswerMarks Guidance
1.29M1ft 3.4
Or speed = 340ร—โˆš
AnswerMarks Guidance
0.166= 386.165โ€ฆ
Speed = 948 (m sโˆ’1)A1ft 1.1
[2]
Question 2:
2 | (a) | M Lโˆ’1 Tโˆ’2 | B1 | 1.2
[1]
(b) | 0.0015 | B1 | 1.1
[1]
(c) | LTโˆ’1 = (๐‘˜)(MLโˆ’1Tโˆ’2)๐‘ฅ(MLโˆ’3)๐‘ฆ(L3)๐‘ง | M1 | 3.3 | Set up | M: 0 = ๐‘ฅ +๐‘ฆ
M1 | 1.1 | Compare coeffs of M, L and T | L: 1 = โˆ’๐‘ฅ โˆ’3๐‘ฆ+3๐‘ง
1 1
๐‘ฅ = ๐‘ฆ = โˆ’
2 2 | A1ft | 1.1 | T: โˆ’1 = โˆ’2๐‘ฅ
Show z = 0 | A1 | 1.1 | AG
[4]
(d) | 1
๐‘˜ = 340รทโˆš
1.29 | M1ft | 3.4 | 1.29
Or speed = 340ร—โˆš
0.166 | = 386.165โ€ฆ
Speed = 948 (m sโˆ’1) | A1ft | 1.1 | 947.8066
[2]
2
\begin{enumerate}[label=(\alph*)]
\item Write down the dimensions of pressure.

The SI unit of pressure is the pascal (Pa). 15 Pa is equivalent to $Q$ newtons per square centimetre.
\item Find the value of $Q$.

Simon thinks the speed, $v$, of sound in a gas is given by the formula\\
$v = k P ^ { x } d ^ { y } V ^ { z }$,\\
where $P$ is the pressure of the gas,\\
$d$ is the density of the gas,\\
$V$ is the volume of the gas,\\
$k$ is a dimensionless constant.
\item Use dimensional analysis to

\begin{itemize}
  \item find the values of $x$ and $y$ and
  \item show that $z = 0$.\\[0pt]
[4]\\
At normal atmospheric pressure the density of air at sea level is $1.29 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$. Under the same conditions the density of helium is $0.166 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$.
\item Given that the speed of sound in air under these conditions is $340 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, use Simon's formula to find the speed of sound in helium under the same conditions.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2019 Q2 [8]}}