Question 2 - A-Level Maths

OCR MEI Further Mechanics Minor 2024 June — Question 2 8 marks

Exam Board	OCR MEI
Module	Further Mechanics Minor (Further Mechanics Minor)
Year	2024
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Dimensional Analysis
Type	Show dimensions are equivalent
Difficulty	Moderate -0.5 This is a straightforward dimensional analysis question with routine conversions and standard dimensional manipulations. Part (a) is pure recall, parts (b) and (c) involve simple arithmetic with given conversion factors, and part (d) requires showing [ML^{-1}T^{-2}] equivalence using standard energy and pressure dimensions—all standard textbook exercises requiring no novel insight.
Spec	6.01a Dimensions: M, L, T notation 6.01b Units vs dimensions: relationship 6.01c Dimensional analysis: error checking

State the dimensions of force. Use the following metric-imperial conversion factors for the rest of this question.
A unit of force used in the imperial system is the pound-force (lbf). 1 lbf is defined as the gravitational force exerted on 1 lb on the surface of the Earth.
Show that 1 lbf is approximately equal to 4.45 N . The pascal (Pa) is a unit of pressure equivalent to 1 Newton per square metre. Pressure can also be measured in pound-force per square inch (psi). A diver, at a depth of 40 m , experiences a typical pressure of \(5 \times 10 ^ { 5 } \mathrm {~Pa}\).
Determine whether this is greater or less than the pressure in a bicycle tyre of 80 psi . In various physical contexts, energy density is the amount of energy stored in a given region of space per unit volume.
Show that energy density and pressure are dimensionally equivalent.

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	M  L  T − 2
[1]	1.2
(b)	9 .8  2 .2 = 4 .4 5 4 5 4 5  4 .4 5	B1
[1]	1.1	AG
(c)	5105 Pa

= 5×105 ÷4.45(45…)÷39.42 psi

Answer	Marks
 7 2 .3 psi which is less than the pressure in the tyre.	M1

M1

Answer	Marks
A1	1.1

1.1

Answer	Marks
2.2a	Correct method converting N to lbf.

Correct method converting m2 to

in2.

Conclusion clearly compares two

Answer	Marks
values.	72.4 if 4.45

used

Or 80 psi

Answer	Marks	Guidance
= 8 0  4 .4 5 4 5  3 9 .4 2 Pa	= 8 0  4 .4 5 4 5  3 9 .4 2 Pa	M1
M1	M1	Correct method converting lbf to N.
M1	Correct method converting in2 to

m2.

Answer	Marks	Guidance
 5 5 3 0 0 0 which is greater than the pressure on the diver.	A1	Conclusion clearly compares two

values.

[3]

Answer	Marks
(d)	 Pressure =MLT −2 L2 =ML −1T −2

 E n e r g y  =  12 m v 2  = M  ( L  T − 1 ) 2 = M  L 2  T − 2

or

[𝐸𝑛𝑒𝑟𝑔𝑦] = [𝐹𝑑] = 𝑀 ∙𝐿∙𝑇−2∙𝐿 = 𝑀∙𝐿2∙𝑇−2

So  E n e r g y d e n s i t y  = M  L 2  T − 2 L 3 = M  L − 1  T − 2 which is

Answer	Marks
the same as the dimensions of pressure.	B1

M1

A1

Answer	Marks
[3]	1.1

1.2

Answer	Marks
2.4	Using some known formula (e.g.

KE, GPE, EPE, work, to derive the

dimensions of energy).

Show dimensions are consistent.

Question 2:
2 | (a) | M  L  T − 2 | B1
[1] | 1.2
(b) | 9 .8  2 .2 = 4 .4 5 4 5 4 5  4 .4 5 | B1
[1] | 1.1 | AG
(c) | 5105 Pa
= 5×105 ÷4.45(45…)÷39.42 psi
 7 2 .3 psi which is less than the pressure in the tyre. | M1
M1
A1 | 1.1
1.1
2.2a | Correct method converting N to lbf.
Correct method converting m2 to
in2.
Conclusion clearly compares two
values. | 72.4 if 4.45
used
Or 80 psi
= 8 0  4 .4 5 4 5  3 9 .4 2 Pa | = 8 0  4 .4 5 4 5  3 9 .4 2 Pa | M1
M1 | M1 | Correct method converting lbf to N.
M1 | Correct method converting in2 to
m2.
 5 5 3 0 0 0 which is greater than the pressure on the diver. | A1 | Conclusion clearly compares two
values.
[3]
(d) |  Pressure =MLT −2 L2 =ML −1T −2
 E n e r g y  =  12 m v 2  = M  ( L  T − 1 ) 2 = M  L 2  T − 2
or
[𝐸𝑛𝑒𝑟𝑔𝑦] = [𝐹𝑑] = 𝑀 ∙𝐿∙𝑇−2∙𝐿 = 𝑀∙𝐿2∙𝑇−2
So  E n e r g y d e n s i t y  = M  L 2  T − 2 L 3 = M  L − 1  T − 2 which is
the same as the dimensions of pressure. | B1
M1
A1
[3] | 1.1
1.2
2.4 | Using some known formula (e.g.
KE, GPE, EPE, work, to derive the
dimensions of energy).
Show dimensions are consistent.

Show LaTeX source

2
\begin{enumerate}[label=(\alph*)]
\item State the dimensions of force.

Use the following metric-imperial conversion factors for the rest of this question.

\begin{itemize}
  \item $1 \mathrm {~kg} = 2.2 \mathrm { lb }$ (pounds)
  \item $1 \mathrm {~m} = 39.4 \mathrm { in }$ (inches)
\end{itemize}

A unit of force used in the imperial system is the pound-force (lbf). 1 lbf is defined as the gravitational force exerted on 1 lb on the surface of the Earth.
\item Show that 1 lbf is approximately equal to 4.45 N .

The pascal (Pa) is a unit of pressure equivalent to 1 Newton per square metre. Pressure can also be measured in pound-force per square inch (psi).

A diver, at a depth of 40 m , experiences a typical pressure of $5 \times 10 ^ { 5 } \mathrm {~Pa}$.
\item Determine whether this is greater or less than the pressure in a bicycle tyre of 80 psi .

In various physical contexts, energy density is the amount of energy stored in a given region of space per unit volume.
\item Show that energy density and pressure are dimensionally equivalent.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2024 Q2 [8]}}

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