What is M8.5: Resultant forces?

Forces add as vectors, so a resultant may be written in component form and then converted to magnitude-direction form.

How many marks are M8.5 questions worth?

M8.5 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M8.5 appeared in?

We have not yet indexed a verified past-paper appearance for M8.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.5

Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Draft — not yet indexed

Resultant forces

Worked answers and methods for M8.5 on Edexcel A-level Maths 9MA0.

Explanation

Forces add as vectors, so a resultant may be written in component form and then converted to magnitude-direction form.
Resolve each force along two perpendicular axes, add corresponding components, and apply $\sum\mathbf{F}=m\mathbf{a}$ to obtain the acceleration vector.
If the resultant is $X\mathbf{i}+Y\mathbf{j}$ , its magnitude is $\sqrt{X^2+Y^2}$ and its direction must be placed in the correct quadrant from the component signs.
A common error is to add force magnitudes without accounting for their directions, or to quote an inverse-tangent angle in the wrong quadrant.

The resultant force is the vector sum of the component forces.

Worked example

Two forces of magnitudes $12\,\text{N}$ and $10\,\text{N}$ act at an angle of $120^\circ$ to each other. Find the magnitude of their resultant and the angle the resultant makes with the $12\,\text{N}$ force.

1.Take the $12\,\text{N}$ force along the positive horizontal direction.
2.The other force has components $10\cos120^\circ=-5$ and $10\sin120^\circ=5\sqrt3$ .
3.The resultant is $7\mathbf{i}+5\sqrt3\mathbf{j}$ , with magnitude $\sqrt{49+75}=2\sqrt{31}\,\text{N}$ .
4.Its angle is $\tan^{-1}(5\sqrt3/7)=51.1^\circ$ .

Answer: Magnitude $=2\sqrt{31}\,\text{N}$ ; Angle $=51.1^\circ$ towards the $10\,\text{N}$ force

Common mistakes

Don't find a resultant magnitude using Pythagoras even though the force components are not perpendicular.
Don't add force magnitudes arithmetically despite a non-zero angle between them.

Exam tip

Resolve forces into perpendicular components or use the cosine rule, then find the resultant direction from its components.

Worked practice

Tier 1 · Easy

A force is

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

. Find its magnitude and its angle below the positive

\mathbf{i}

direction.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Magnitude $=10\,\text{N}$ Angle $=53.1^\circ$ below the positive $\mathbf{i}$ direction	3
Notes
The magnitude is $\sqrt{6^2+(-8)^2}=10\,\text{N}$ . Since the components place the force in the fourth quadrant, the angle below the positive $\mathbf{i}$ direction is $\tan^{-1}(8/6)=53.1^\circ$ .

(3 marks)

Tier 2 · Standard

A particle of mass

2\,\text{kg}

is initially at rest. Constant forces

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

and

(-2\mathbf{i}+10\mathbf{j})\,\text{N}

act on it. Find the resultant force, its acceleration, and its speed after

2\,\text{s}

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Resultant force $=(6\mathbf{i}+8\mathbf{j})\,\text{N}$ Acceleration $=(3\mathbf{i}+4\mathbf{j})\,\text{m s}^{-2}$ Speed after $2\,\text{s}$ is $10\,\text{m s}^{-1}$	5
Notes
Add the two forces component by component: $\mathbf F=(8-2)\mathbf i+(-2+10)\mathbf j=6\mathbf i+8\mathbf j$ . Newton's second law gives $\mathbf a=\mathbf F/m=3\mathbf i+4\mathbf j$ . From rest, after $2$ seconds $\mathbf v=2\mathbf a=6\mathbf i+8\mathbf j$ , whose magnitude is $\sqrt{6^2+8^2}=10\,\text{m s}^{-1}$ .

(5 marks)

Tier 3 · Hard

A particle of mass

5\,\text{kg}

has initial velocity

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

. Constant forces

(15\mathbf{i}+20\mathbf{j})\,\text{N}

and

(-5\mathbf{i}+10\mathbf{j})\,\text{N}

act on it. Find its acceleration, velocity after

3\,\text{s}

and displacement during those

3\,\text{s}

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Acceleration $=(2\mathbf{i}+6\mathbf{j})\,\text{m s}^{-2}$ Velocity $=(8\mathbf{i}+17\mathbf{j})\,\text{m s}^{-1}$ Displacement $=(15\mathbf{i}+24\mathbf{j})\,\text{m}$	7
Notes
The resultant force is $10\mathbf{i}+30\mathbf{j}$ , so $\mathbf{a}=\mathbf{F}/m=2\mathbf{i}+6\mathbf{j}$ . Then $\mathbf{v}=\mathbf{u}+\mathbf{a}t=(2\mathbf{i}-\mathbf{j})+3(2\mathbf{i}+6\mathbf{j})=8\mathbf{i}+17\mathbf{j}$ . Finally $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2=(6\mathbf{i}-3\mathbf{j})+(9\mathbf{i}+27\mathbf{j})=15\mathbf{i}+24\mathbf{j}$ .

(7 marks)

Tier 1 · Easy

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. Forces

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

and

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

act on a particle. Find their resultant.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$(4\mathbf{i}+7\mathbf{j})\,\text{N}$	2
Notes
Add corresponding components: $(7-3)\mathbf{i}+(2+5)\mathbf{j}=4\mathbf{i}+7\mathbf{j}$ .

(2 marks)

Tier 2 · Standard

A particle of mass

4\,\text{kg}

is acted on by a force of

18\,\text{N}

due east and a force of

12\,\text{N}

120^\circ

anticlockwise from east. Find the magnitude and direction of its acceleration. Give the direction to 3 significant figures.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Acceleration magnitude $=\dfrac{3\sqrt7}{2}\,\text{m s}^{-2}$ Direction $=40.9^\circ$ north of east	5
Notes
The resultant is $(18+12\cos120^\circ)\mathbf{i}+12\sin120^\circ\mathbf{j}=12\mathbf{i}+6\sqrt3\mathbf{j}$ . Its magnitude is $\sqrt{12^2+(6\sqrt3)^2}=6\sqrt7\,\text{N}$ , so the acceleration magnitude is $6\sqrt7/4=3\sqrt7/2\,\text{m s}^{-2}$ . Its direction is $\tan^{-1}(6\sqrt3/12)=40.9^\circ$ north of east.

(5 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle of mass

3\,\text{kg}

has initial velocity

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

. Constant forces

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

and

(p\mathbf{i}+12\mathbf{j})\,\text{N}

act on it. After

2\,\text{s}

its velocity is parallel to

\mathbf{i}+2\mathbf{j}

and has a positive

\mathbf{i}

component. Find

p

, the speed after

2\,\text{s}

and the displacement during these

2\,\text{s}

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$p=-\dfrac{33}{4}$ Speed $=\dfrac{5\sqrt5}{2}\,\text{m s}^{-1}$ Displacement $=(\dfrac92\mathbf{i}+4\mathbf{j})\,\text{m}$	6
Notes
The acceleration is $[(9+p)/3]\mathbf{i}+3\mathbf{j}$ . After $2\,\text{s}$ , the velocity is $[2+2(9+p)/3]\mathbf{i}+5\mathbf{j}$ . Parallel to $\mathbf{i}+2\mathbf{j}$ with positive components requires the $\mathbf{j}$ component to be twice the $\mathbf{i}$ component, so $5=2[2+2(9+p)/3]$ , giving $p=-33/4$ . Thus $\mathbf{a}=\tfrac14\mathbf{i}+3\mathbf{j}$ and $\mathbf{v}=\tfrac52\mathbf{i}+5\mathbf{j}$ , whose magnitude is $5\sqrt5/2$ . Finally $\mathbf{s}=2\mathbf{u}+\tfrac12\mathbf{a}(2^2)=(4,-2)+(1/2,6)=(9/2,4)$ metres.

(6 marks)

Tier 2 · Standard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. One of two forces acting on a

2\,\text{kg}

particle is

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

. Their resultant is

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

. Find the second force, its exact magnitude and the acceleration of the particle.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Second force $=(3\mathbf{i}+9\mathbf{j})\,\text{N}$ Its magnitude is $3\sqrt{10}\,\text{N}$ Acceleration $=(5\mathbf{i}+2.5\mathbf{j})\,\text{m s}^{-2}$	5
Notes
The second force is the resultant minus the first force: $(10\mathbf{i}+5\mathbf{j})-(7\mathbf{i}-4\mathbf{j})=3\mathbf{i}+9\mathbf{j}$ . Its magnitude is $\sqrt{3^2+9^2}=3\sqrt{10}\,\text{N}$ . Newton's second law gives $\mathbf{a}=\mathbf{F}/m=(10\mathbf{i}+5\mathbf{j})/2=5\mathbf{i}+2.5\mathbf{j}\,\text{m s}^{-2}$ .

(5 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle of mass

3\,\text{kg}

has velocity

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

at time

t=0

and velocity

(8\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}

at time

t=4\,\text{s}

. A constant resultant force acts throughout. Find this force, its magnitude and its exact direction relative to the positive

\mathbf{i}

direction. Find also the displacement of the particle during the

4\,\text{s}

and the exact distance between its initial and final positions.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Resultant force $=(4.5\mathbf{i}+6\mathbf{j})\,\text{N}$ Magnitude $=7.5\,\text{N}$ Direction $=\tan^{-1}(4/3)$ above the positive $\mathbf{i}$ direction Displacement $=(20\mathbf{i}+12\mathbf{j})\,\text{m}$ Distance between the positions $=4\sqrt{34}\,\text{m}$	6
Notes
The acceleration is the change in velocity divided by time: $\mathbf{a}=[(8-2)\mathbf{i}+(7-(-1))\mathbf{j}]/4=1.5\mathbf{i}+2\mathbf{j}$ . Hence $\mathbf{F}=3\mathbf{a}=4.5\mathbf{i}+6\mathbf{j}\,\text{N}$ . Its magnitude is $\sqrt{4.5^2+6^2}=7.5\,\text{N}$ and its direction is $\tan^{-1}(6/4.5)=\tan^{-1}(4/3)$ above $\mathbf{i}$ . Constant acceleration gives displacement $\tfrac12(\mathbf{u}+\mathbf{v})t=\tfrac12(10\mathbf{i}+6\mathbf{j})(4)=20\mathbf{i}+12\mathbf{j}\,\text{m}$ , whose magnitude is $\sqrt{20^2+12^2}=4\sqrt{34}\,\text{m}$ .

(6 marks)

Tier 3 · Hard

Two forces of magnitudes

13\,\text{N}

and

15\,\text{N}

act on a particle, and their resultant has magnitude

14\,\text{N}

. Find the exact cosine of the angle between the two forces. Find also the exact cosine of the acute angle between the resultant and the

13\,\text{N}

force. If the particle has mass

7\,\text{kg}

, find the magnitude of its acceleration. Give all numerical answers exactly.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Cosine of angle between forces $=-\dfrac{33}{65}$ Cosine of angle between resultant and $13\,\text{N}$ force $=\dfrac5{13}$ Acceleration magnitude $=2\,\text{m s}^{-2}$	5
Notes
If $\theta$ is the angle between the forces, $14^2=13^2+15^2+2(13)(15)\cos\theta$ , so $\cos\theta=(196-169-225)/390=-33/65$ . In the force triangle, if $\phi$ is the angle between the resultant and the $13\,\text{N}$ force, the side opposite $\phi$ has length $15$ , so $15^2=13^2+14^2-2(13)(14)\cos\phi$ . Hence $\cos\phi=5/13$ . Newton's second law gives acceleration magnitude $14/7=2\,\text{m s}^{-2}$ .

(5 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A

3\,\text{kg}

particle has initial velocity

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

. A known constant force

(2\mathbf{i}-1.5\mathbf{j})\,\text{N}

and an unknown constant force act on it. During the next

4\,\text{s}

its displacement is

(24\mathbf{i}+8\mathbf{j})\,\text{m}

. Find the unknown force, its exact magnitude and direction relative to the positive

\mathbf{i}

direction. Find also the particle's velocity after

4\,\text{s}

. Give all numerical answers exactly.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Unknown force $=(4\mathbf{i}+6\mathbf{j})\,\text{N}$ Magnitude $=2\sqrt{13}\,\text{N}$ Direction $=\tan^{-1}(3/2)$ above the positive $\mathbf{i}$ direction Velocity after $4\,\text{s}$ is $(10\mathbf{i}+5\mathbf{j})\,\text{m s}^{-1}$	7
Notes
Using $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $(24,8)=4(2,-1)+8\mathbf a$ , so $8\mathbf a=(16,12)$ and $\mathbf a=(2,3/2)\,\text{m s}^{-2}$ . The resultant force is $3\mathbf a=(6,9/2)\,\text{N}$ . Subtracting the known force $(2,-3/2)$ gives the unknown force $(4,6)\,\text{N}$ . Its magnitude is $2\sqrt{13}$ and its direction is $\tan^{-1}(6/4)=\tan^{-1}(3/2)$ above $\mathbf i$ . Finally $\mathbf v=\mathbf u+4\mathbf a=(10,5)\,\text{m s}^{-1}$ .

(7 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for M8.5. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in M8 Forces and Newton's laws

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