What is M6.1: SI units and quantities?

The fundamental S.I. quantities are length in metres, time in seconds and mass in kilograms; velocity uses m s^-1, acceleration m s^-2, force and weight newtons, and moment N m.

How many marks are M6.1 questions worth?

M6.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has M6.1 appeared in?

We have not yet indexed a verified past-paper appearance for M6.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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.

Draft — not yet indexed

SI units and quantities

Worked answers and methods for M6.1 on Edexcel A-level Maths 9MA0.

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.

Worked practice

Tier 1 · Easy

Convert

90\,\text{km h}^{-1}

to metres per second.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$25\,\text{m s}^{-1}$	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 marks)

Tier 2 · Standard

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 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	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
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 $\text{kg m}^2\text{s}^{-2}$ .

(4 marks)

Tier 3 · Hard

A vehicle of mass

1.8

tonnes increases its speed uniformly from

54\,\text{km h}^{-1}

90\,\text{km h}^{-1}

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 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$2250\,\text{N}$ $1462.5\,\text{N m}$	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 $F d=2250(0.65)=1462.5\,\text{N m}$ .

(6 marks)

Tier 1 · Easy

Express

0.42\,\text{kN}

in newtons and in fundamental S.I. units.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$420\,\text{N}=420\,\text{kg m s}^{-2}$	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 $420\,\text{kg m s}^{-2}$ .

(2 marks)

Tier 2 · Standard

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 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$Q$ has no units $Q=0.08$	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 $Q=2400(0.35)/(700(15))=840/10500=0.08$ .

(4 marks)

Tier 3 · Hard

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 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Force $=10.6\,\text{N}$ Moment $=3.40\,\text{N m}$	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.

(6 marks)

Tier 2 · Standard

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 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Moment $=0.360\,\text{N m}=0.360\,\text{kg m}^2\text{s}^{-2}$	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 $1\,\text{N}=1\,\text{kg m s}^{-2}$ , $1\,\text{N m}=1\,\text{kg m}^2\text{s}^{-2}$ .

(4 marks)

Tier 3 · Hard

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 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$p=1$ and $q=-1$ $M=17.5\,\text{N m}$	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.

(5 marks)

Tier 3 · Hard

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

\text{cm}^2

and

v

\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 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Unit of $k$ is $\text{kg m}^{-3}$ $K=5\times10^{-6}$ $F=10.368\,\text{N}$	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 $F=(5\times10^{-6})(400)(72^2)=10.368\,\text{N}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

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 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$T=\sqrt{L/g}$ $T=\dfrac27\,\text{s}$ $L$ must be multiplied by $4$ New length $=3.2\,\text{m}$	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 $4(0.80)=3.2\,\text{m}$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for M6.1. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Want help turning this into marks?

Bring M6.1 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call