M6 Quantities and units in mechanics — revision question pack

1 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section M6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M6.1 · Understand and use fundamental quantities and units in the S.I. system: length, time, mass; understand and use derived quantities and units: velocity, acceleration, force, weight, moment.

Explanation

The fundamental S.I. quantities are length in metres, time in seconds and mass in kilograms; velocity uses $\text{m s}^{-1}$ , acceleration $\text{m s}^{-2}$ , force and weight newtons, and moment $\text{N m}$ .
Convert all measurements to consistent S.I. units before substituting, using for example $1\,\text{km h}^{-1}=\dfrac{5}{18}\,\text{m s}^{-1}$ and $1\,\text{tonne}=1000\,\text{kg}$ .
Since $F=ma$ , $1\,\text{N}=1\,\text{kg m s}^{-2}$ ; multiplying a force by a perpendicular distance gives a moment in $\text{N m}$ .
A common error is to confuse mass with weight: mass is measured in kilograms, whereas weight is a force measured in newtons; a moment is not measured in newtons alone.

Worked example

A resultant force gives a $750\,\text{kg}$ car an acceleration of $1.6\,\text{m s}^{-2}$ . Find the force in newtons and write the newton in fundamental S.I. units.

1.Using $F=ma$ , $F=750\times1.6=1200\,\text{N}$ .
2.Since mass has unit $\text{kg}$ and acceleration has unit $\text{m s}^{-2}$ , the fundamental-unit form is $\text{kg m s}^{-2}$ .

Answer: $1200\,\text{N}$ ; $1\,\text{N}=1\,\text{kg m s}^{-2}$

Common mistakes

Don't convert centimetres to metres by dividing by $10$ rather than by $100$ .
Don't treat the newton as a fundamental unit rather than deriving it from mass times acceleration.

Exam tip

Carry units through the calculation and express derived force units as kilograms metres per second squared when requested.

Tier 1 · Easy

1.
Convert $90\,\text{km h}^{-1}$ to metres per second.
(2)
(Total for Question 1 is 2 marks)
2.
Express $0.42\,\text{kN}$ in newtons and in fundamental S.I. units.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A student states that velocity is measured in $\text{m s}^{-2}$ , acceleration in $\text{m s}^{-1}$ , force in $\text{kg m s}^{-2}$ and moment in $\text{kg m}^2\text{s}^{-2}$ . Identify the two incorrect statements and correct them.
(4)
(Total for Question 1 is 4 marks)
2.
A force $F$ , acting for time $t$ on an object of mass $m$ moving at speed $v$ , is used to form $Q=\dfrac{Ft}{mv}$ . Show that $Q$ has no units. Find $Q$ when $F=2.4\,\text{kN}$ , $t=0.35\,\text{s}$ , $m=700\,\text{kg}$ and $v=54\,\text{km h}^{-1}$ .
(4)
(Total for Question 2 is 4 marks)
3.
A force of $480\,\text{mN}$ acts perpendicular to a lever at a distance of $75\,\text{cm}$ from its pivot. Find the moment of the force about the pivot in $\text{N m}$ and express its unit in fundamental S.I. units.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A vehicle of mass $1.8$ tonnes increases its speed uniformly from $54\,\text{km h}^{-1}$ to $90\,\text{km h}^{-1}$ in $8.0\,\text{s}$ . Find the resultant force. This force has a perpendicular distance of $0.65\,\text{m}$ from a pivot; find its moment about the pivot.
(6)
(Total for Question 1 is 6 marks)
2.
A flat panel has area $450\,\text{cm}^2$ . Air of density $1.18\,\text{kg m}^{-3}$ meets the panel at $72\,\text{km h}^{-1}$ . A model gives the force normal to the panel as $F=\tfrac12\rho Av^2$ . The line of action of this force is $0.32\,\text{m}$ from a hinge. Find the force and its moment about the hinge, giving both answers to 3 significant figures. Use unrounded values in your working.
(6)
(Total for Question 2 is 6 marks)
3.
A model proposes that a moment $M$ is given by $M=c m v L^p t^q$ , where $c$ is dimensionless, $m$ is a mass, $v$ is a speed, $L$ is a length and $t$ is a time. Use dimensional consistency to find $p$ and $q$ . Hence calculate $M$ when $c=0.500$ , $m=350\,\text{g}$ , $v=72\,\text{km h}^{-1}$ , $L=40\,\text{cm}$ and $t=80\,\text{ms}$ , giving your answer in $\text{N m}$ to 3 significant figures.
(5)
(Total for Question 3 is 5 marks)
4.
A model for a force is $F=kAv^2$ , where $F$ is measured in newtons, $A$ in square metres and $v$ in metres per second. Determine the S.I. unit of $k$ . An engineer instead enters $A$ in $\text{cm}^2$ and $v$ in $\text{km h}^{-1}$ , while still obtaining $F$ in newtons, and writes the model as $F=KAv^2$ . Given that $k=0.648$ in S.I. units, find the exact numerical value of $K$ . Hence find $F$ when $A=400\,\text{cm}^2$ and $v=72\,\text{km h}^{-1}$ . Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
A proposed model for a time $T$ uses only a length $L$ and an acceleration $g$ . Three formulae are suggested: $T=L/g$ , $T=\sqrt{L/g}$ and $T=\sqrt{g/L}$ . Use S.I. units to determine which formula is dimensionally consistent. Using that formula, find the exact value of $T$ when $L=80\,\text{cm}$ and $g=9.8\,\text{m s}^{-2}$ . State how $L$ must change if $T$ is to double while $g$ remains unchanged, and find the new value of $L$ . Give all numerical answers exactly.
(5)
(Total for Question 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

M6.1 · Understand and use fundamental quantities and units in the S.I. system: length, time, mass; understand and use derived quantities and units: velocity, acceleration, force, weight, moment.

Tier 1 · Easy

Mark scheme for M6.1 Tier 1 · Easy
Question	Scheme	Marks
1	$25\,\text{m s}^{-1}$	2
(2 marks)		2
Notes
$90\,\text{km h}^{-1}=90\times\dfrac{1000}{3600}\,\text{m s}^{-1}=90\times\dfrac{5}{18}\,\text{m s}^{-1}=25\,\text{m s}^{-1}$ .
2	$420\,\text{N}=420\,\text{kg m s}^{-2}$	2
(2 marks)		2
Notes
$1\,\text{kN}=1000\,\text{N}$ , so $0.42\,\text{kN}=420\,\text{N}$ . Since $1\,\text{N}=1\,\text{kg m s}^{-2}$ , this is $420\,\text{kg m s}^{-2}$ .

Tier 2 · Standard

Mark scheme for M6.1 Tier 2 · Standard
Question	Scheme	Marks
1	Velocity is measured in $\text{m s}^{-1}$ , not $\text{m s}^{-2}$ . Acceleration is measured in $\text{m s}^{-2}$ , not $\text{m s}^{-1}$ . The stated base units for force and moment are correct.	4
(4 marks)		4
Notes
Velocity is displacement change per unit time, so its unit is $\text{m s}^{-1}$ . Acceleration is velocity change per unit time, so its unit is $\text{m s}^{-2}$ . From $F=ma$ , the newton is $\text{kg m s}^{-2}$ , and multiplying force by perpendicular distance gives moment unit $\text{kg m}^2\text{s}^{-2}$ .
2	$Q$ has no units $Q=0.08$	4
(4 marks)		4
Notes
The units of $Ft$ are $(\text{kg m s}^{-2})(\text{s})=\text{kg m s}^{-1}$ , the same as the units of $mv$ , so their ratio has no units. Convert $F=2400\,\text{N}$ and $v=54\times5/18=15\,\text{m s}^{-1}$ . Hence $Q=2400(0.35)/(700(15))=840/10500=0.08$ .
3	Moment $=0.360\,\text{N m}=0.360\,\text{kg m}^2\text{s}^{-2}$	4
(4 marks)		4
Notes
Convert the force and distance to S.I. units: $480\,\text{mN}=0.480\,\text{N}$ and $75\,\text{cm}=0.75\,\text{m}$ . The moment is $0.480(0.75)=0.360\,\text{N m}$ . Since $1\,\text{N}=1\,\text{kg m s}^{-2}$ , $1\,\text{N m}=1\,\text{kg m}^2\text{s}^{-2}$ .

Tier 3 · Hard

Mark scheme for M6.1 Tier 3 · Hard
Question	Scheme	Marks
1	$2250\,\text{N}$ $1462.5\,\text{N m}$	6
(6 marks)		6
Notes
The mass is $1800\,\text{kg}$ . The speeds are $54\times5/18=15\,\text{m s}^{-1}$ and $90\times5/18=25\,\text{m s}^{-1}$ , so $a=(25-15)/8=1.25\,\text{m s}^{-2}$ . Hence $F=ma=1800(1.25)=2250\,\text{N}$ . The moment is $F d=2250(0.65)=1462.5\,\text{N m}$ .
2	Force $=10.6\,\text{N}$ Moment $=3.40\,\text{N m}$	6
(6 marks)		6
Notes
Convert $A=450/10000=0.0450\,\text{m}^2$ and $v=72\times5/18=20\,\text{m s}^{-1}$ . Then $F=\tfrac12(1.18)(0.0450)(20^2)=10.62\,\text{N}$ , so the force is $10.6\,\text{N}$ to 3 significant figures. The moment is $10.62(0.32)=3.3984\,\text{N m}$ , which is $3.40\,\text{N m}$ to 3 significant figures.
3	$p=1$ and $q=-1$ $M=17.5\,\text{N m}$	5
(5 marks)		5
Notes
A moment has units $\text{N m}=\text{kg m}^2\text{s}^{-2}$ . The units on the right are $\text{kg m}^{1+p}\text{s}^{-1+q}$ . Equating powers gives $1+p=2$ and $-1+q=-2$ , so $p=1$ and $q=-1$ . Convert $m=0.350\,\text{kg}$ , $v=20\,\text{m s}^{-1}$ , $L=0.40\,\text{m}$ and $t=0.080\,\text{s}$ . Hence $M=0.500(0.350)(20)(0.40)(0.080)^{-1}=17.5\,\text{N m}$ to 3 significant figures.
4	Unit of $k$ is $\text{kg m}^{-3}$ $K=5\times10^{-6}$ $F=10.368\,\text{N}$	6
(6 marks)		6
Notes
Since $[F]=\text{kg m s}^{-2}$ and $[Av^2]=\text{m}^2(\text{m}^2\text{s}^{-2})$ , $[k]=\text{kg m}^{-3}$ . If the engineer's numerical entries are $A_c$ and $v_k$ , then $A=10^{-4}A_c$ and $v=(5/18)v_k$ . Therefore $F=k(10^{-4}A_c)(5v_k/18)^2=(k/129600)A_cv_k^2$ , so $K=0.648/129600=5\times10^{-6}$ . Hence $F=(5\times10^{-6})(400)(72^2)=10.368\,\text{N}$ .
5	$T=\sqrt{L/g}$ $T=\dfrac27\,\text{s}$ $L$ must be multiplied by $4$ New length $=3.2\,\text{m}$	5
(5 marks)		5
Notes
The unit of $L/g$ is $\text{m}/(\text{m s}^{-2})=\text{s}^2$ , so only $\sqrt{L/g}$ has unit seconds. With $L=0.80\,\text{m}$ , $T=\sqrt{0.80/9.8}=\sqrt{4/49}=2/7\,\text{s}$ . Since $T$ is proportional to $\sqrt L$ for fixed $g$ , doubling $T$ requires $L$ to be multiplied by $2^2=4$ . The new length is $4(0.80)=3.2\,\text{m}$ .