What is M8.1: Forces and Newton's first law?

A force is an interaction with magnitude and direction; common modelled forces include weight, normal reaction, tension, thrust, compression and resistance.

How many marks are M8.1 questions worth?

M8.1 practice drills on this page are worth 1–5 marks each; marks in a real exam vary by question.

Which past exam papers has M8.1 appeared in?

We have not yet indexed a verified past-paper appearance for M8.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.1

Understand the concept of a force; understand and use Newton's first law.

Draft — not yet indexed

Forces and Newton's first law

Worked answers and methods for M8.1 on Edexcel A-level Maths 9MA0.

Explanation

A force is an interaction with magnitude and direction; common modelled forces include weight, normal reaction, tension, thrust, compression and resistance.
Draw a force diagram for the chosen body, including only forces acting on that body, then resolve forces along useful perpendicular directions.
Newton's first law says that a particle remains at rest or moves with constant velocity when the resultant force on it is zero.
A common error is to infer that a moving particle must have a forward resultant force; constant non-zero velocity also means zero resultant force.

Worked example

A powered trolley moves in a straight line with constant velocity. Its motor exerts a forward force of $8\,\text{N}$ . Find the resistance force and explain your answer.

1.Constant velocity means zero acceleration and hence, by Newton's first law, zero resultant force.
2.The resistance must therefore balance the motor force, so it is $8\,\text{N}$ backwards.

Answer: Resistance $=8\,\text{N}$ opposite to the motion; Constant velocity means the resultant force is zero

Common mistakes

Don't balance only the horizontal forces and declare equilibrium while a vertical resultant remains.
Don't assume motion implies a resultant force even when the velocity is constant.

Exam tip

For constant velocity, invoke zero acceleration and balance the forces with equal magnitudes in opposite directions.

Worked practice

Tier 1 · Easy

A book rests on a horizontal table. Its weight is

35\,\text{N}

. State the magnitude and direction of the normal reaction on the book and the resultant force on it.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Normal reaction $=35\,\text{N}$ upwards Resultant force $=0\,\text{N}$	2
Notes
The stationary book has zero acceleration, so Newton's first law requires zero resultant force. The upward normal reaction therefore balances the $35\,\text{N}$ weight and has magnitude $35\,\text{N}$ .

(2 marks)

Tier 2 · Standard

A book of mass

2.5\,\text{kg}

rests on a horizontal table. Using

g=9.8\,\text{m s}^{-2}

, find the weight of the book and the normal force exerted by the table, explaining why the forces have equal magnitudes.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Weight $24.5\,\text{N}$ downward Normal force $24.5\,\text{N}$ upward They have equal magnitudes because the book has zero acceleration, so the resultant force is zero.	3
Notes
The weight is $mg=2.5(9.8)=24.5\,\text{N}$ downward. The book is at rest, so its acceleration and resultant force are zero. The only vertical forces are its weight and the table's normal force, so the normal force is $24.5\,\text{N}$ upward.

(3 marks)

Tier 3 · Hard

A particle moves with constant velocity under three coplanar forces. Two of the forces are

(6\mathbf{i}-8\mathbf{j})\,\text{N}

and

(-2\mathbf{i}+5\mathbf{j})\,\text{N}

. Find the third force, giving also its magnitude.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Third force $=(-4\mathbf{i}+3\mathbf{j})\,\text{N}$ Magnitude $=5\,\text{N}$	4
Notes
Constant velocity requires the vector resultant to be zero. The two known forces add to $4\mathbf{i}-3\mathbf{j}$ , so the third force is its negative, $-4\mathbf{i}+3\mathbf{j}$ . Its magnitude is $\sqrt{(-4)^2+3^2}=5\,\text{N}$ .

(4 marks)

Tier 1 · Easy

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A probe is moving through space with velocity

(5\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}

. No resultant force acts on it. State its velocity

12\,\text{s}

later.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$(5\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}$	1
Notes
By Newton's first law, zero resultant force means zero acceleration, so the probe continues with the same constant velocity.

(1 mark)

Tier 2 · Standard

A parachutist of weight

735\,\text{N}

descends vertically at constant speed. Find the magnitude and direction of the air resistance. Explain why the parachutist can be moving although the resultant force is zero.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Air resistance $=735\,\text{N}$ upwards Zero resultant force gives constant velocity, not necessarily zero velocity	3
Notes
Constant speed in a fixed downward direction means constant velocity, so the acceleration and resultant force are zero. Air resistance must therefore balance the $735\,\text{N}$ weight and act upwards. Newton's first law permits motion with any constant velocity when the resultant force is zero.

(3 marks)

Tier 3 · Hard

A maintenance platform moves with constant velocity under two cable forces and a resistance force. One cable pulls with force

40\,\text{N}

due east. The other pulls with force

30\,\text{N}

60^\circ

north of east. Take

\mathbf{i}

and

\mathbf{j}

to be unit vectors due east and due north respectively. Find the resistance force as a vector in terms of

\mathbf{i}

and

\mathbf{j}

, its exact magnitude, and its direction to 3 significant figures.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Resistance $=(-55\mathbf{i}-15\sqrt3\mathbf{j})\,\text{N}$ Magnitude $=10\sqrt{37}\,\text{N}$ Direction $25.3^\circ$ south of west	5
Notes
The cable resultant is $40\mathbf{i}+30(\cos60^\circ\mathbf{i}+\sin60^\circ\mathbf{j})=55\mathbf{i}+15\sqrt3\mathbf{j}$ . Constant velocity requires zero resultant force, so resistance is its negative. Its magnitude is $\sqrt{55^2+(15\sqrt3)^2}=10\sqrt{37}\,\text{N}$ . The acute angle from west is $\tan^{-1}(15\sqrt3/55)=25.3^\circ$ , towards the south.

(5 marks)

Tier 2 · Standard

At one instant a train is stationary at a signal. A student concludes that the resultant force on the train is zero because its velocity is zero at that instant. Explain why the observation is not sufficient to support this conclusion. State the possible forms of motion for an object on which the resultant force is zero.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Zero velocity at one instant does not imply zero acceleration, so the train could have a non-zero resultant force A zero resultant force would require the train to remain at rest or to move with constant velocity	3
Notes
An object can be instantaneously at rest while its velocity is changing, so the single observation does not establish that its acceleration or resultant force is zero. By Newton's first law, a zero resultant force makes the velocity constant; the train would remain at rest or continue with an unchanged non-zero velocity.

(3 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. Three constant forces

(4\mathbf{i}+7\mathbf{j})\,\text{N}

(-9\mathbf{i}+2\mathbf{j})\,\text{N}

and

(5\mathbf{i}-9\mathbf{j})\,\text{N}

act on a particle. Initially its position vector is

(2\mathbf{i}-5\mathbf{j})\,\text{m}

and its velocity is

(3\mathbf{i}+2\mathbf{j})\,\text{m s}^{-1}

. Show that the particle moves with constant velocity. Find its position vector at time

t

and the point at which its path crosses the line

y=7

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The resultant force is $\mathbf{0}$ , so the velocity remains $(3\mathbf{i}+2\mathbf{j})\,\text{m s}^{-1}$ $\mathbf{r}=(2+3t)\mathbf{i}+(-5+2t)\mathbf{j}\,\text{m}$ The path crosses $y=7$ at $(20,7)$	5
Notes
Adding the forces gives $(4-9+5)\mathbf{i}+(7+2-9)\mathbf{j}=\mathbf{0}$ . Newton's first law therefore gives constant velocity. Hence $\mathbf{r}=(2\mathbf{i}-5\mathbf{j})+t(3\mathbf{i}+2\mathbf{j})$ . On the line $y=7$ , $-5+2t=7$ , so $t=6$ and $x=2+3(6)=20$ .

(5 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A platform moves with constant velocity while forces

(7\mathbf{i}+3\mathbf{j})\,\text{N}

and

(-3\mathbf{i}+5\mathbf{j})\,\text{N}

act on it. A thruster exerts

P\mathbf{i}\,\text{N}

, where

P>0

, and a cable exerts

-T(\tfrac35\mathbf{i}+\tfrac45\mathbf{j})\,\text{N}

, where

T>0

. Find

P

and

T

, and explain why the platform's non-zero velocity can remain unchanged. Give all numerical answers exactly.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$T=10\,\text{N}$ $P=2\,\text{N}$ The forces have zero resultant, so Newton's first law gives constant velocity	5
Notes
Constant velocity requires zero resultant force. Resolving in the $\mathbf j$ direction gives $3+5-4T/5=0$ , so $T=10$ . Resolving in the $\mathbf i$ direction then gives $7-3+P-3(10)/5=0$ , hence $P=2$ . A zero resultant produces zero acceleration, not necessarily zero velocity, so the existing non-zero velocity remains constant.

(5 marks)

Q10

Tier 3 · Hard

10.

A raft moves with constant velocity. One engine force has magnitude

10\,\text{N}

and acts at

60^\circ

north of east. A second engine force has magnitude

P\,\text{N}

and acts at

30^\circ

south of east. The water resistance acts due west. Find the exact value of

P

and the magnitude of the water resistance. Give all numerical answers exactly.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=10\sqrt3\,\text{N}$ Water resistance $=20\,\text{N}$ due west	5
Notes
Constant velocity requires the north and east force components to balance. Resolving north, $10\sin60^\circ=P\sin30^\circ$ , so $5\sqrt3=P/2$ and $P=10\sqrt3\,\text{N}$ . The total eastward component of the engine forces is $10\cos60^\circ+10\sqrt3\cos30^\circ=5+15=20\,\text{N}$ . The resistance must therefore be $20\,\text{N}$ due west.

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for M8.1. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in M8 Forces and Newton's laws

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