What is M8.2: Newton's second law?

Newton's second law is =ma; in two dimensions it gives one scalar equation in each resolved direction.

How many marks are M8.2 questions worth?

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

Which past exam papers has M8.2 appeared in?

We have not yet indexed a verified past-paper appearance for M8.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.2

Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Draft — not yet indexed

Newton's second law

Worked answers and methods for M8.2 on Edexcel A-level Maths 9MA0.

Explanation

Newton's second law is $\sum\mathbf{F}=m\mathbf{a}$ ; in two dimensions it gives one scalar equation in each resolved direction.
Choose axes parallel and perpendicular to the motion where possible, resolve every force onto those axes, then apply $\sum F=ma$ separately.
For motion along a fixed smooth plane, perpendicular acceleration is zero, so the perpendicular force equation determines the normal reaction while the parallel equation determines acceleration.
A common error is to write $F=ma$ for one force instead of the resultant, or to reverse signs for only some components after choosing a positive direction.

Resolve forces in perpendicular directions, then apply

\sum\mathbf F=m\mathbf a

component by component.

Worked example

A $5\,\text{kg}$ particle is on a smooth plane inclined at $30^\circ$ to the horizontal. A force of $40\,\text{N}$ pulls it up the line of greatest slope. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Resolve up the plane.
2.The component of weight down the plane is $5(9.8)\sin30^\circ=24.5\,\text{N}$ .
3.Hence the resultant up the plane is $40-24.5=15.5\,\text{N}$ .
4.From $15.5=5a$ , $a=3.1\,\text{m s}^{-2}$ up the plane.

Answer: $3.1\,\text{m s}^{-2}$ up the plane

Common mistakes

Don't include the normal reaction in the equation parallel to an inclined plane.
Don't resolve weight using the wrong trigonometric component relative to the inclined plane.

Exam tip

Draw a force diagram, choose axes parallel and perpendicular to the plane, then resolve every force consistently.

Worked practice

Tier 1 · Easy

A constant resultant force of

12\,\text{N}

acts on a particle of mass

3\,\text{kg}

. Find its acceleration.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$4\,\text{m s}^{-2}$ in the direction of the force	2
Notes
Newton's second law gives $F=ma$ , so $a=F/m=12/3=4\,\text{m s}^{-2}$ , directed with the resultant force.

(2 marks)

Tier 2 · Standard

4\,\text{kg}

particle moves on a smooth horizontal surface. A force of

20\,\text{N}

acts at

30^\circ

above the horizontal. Using

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

, find the horizontal acceleration and the normal reaction.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Exact acceleration $=\dfrac{5\sqrt3}{2}\,\text{m s}^{-2}$ Acceleration to $3$ significant figures $=4.33\,\text{m s}^{-2}$ Normal reaction $=29.2\,\text{N}$	5
Notes
Horizontally, the resultant force is $20\cos30^\circ=10\sqrt3\,\text{N}$ , so $4a=10\sqrt3$ and $a=5\sqrt3/2\,\text{m s}^{-2}$ . Vertically there is no acceleration, so $R+20\sin30^\circ-4(9.8)=0$ . Hence $R=39.2-10=29.2\,\text{N}$ .

(5 marks)

Tier 3 · Hard

6\,\text{kg}

particle is pulled up a smooth plane inclined at

20^\circ

to the horizontal by a force of

50\,\text{N}

acting at

15^\circ

above the plane. Using

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

, find the normal reaction and the particle's acceleration up the plane.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Normal reaction $=42.3\,\text{N}$ Acceleration $=4.70\,\text{m s}^{-2}$ up the plane	7
Notes
Perpendicular to the plane, $R+50\sin15^\circ-6(9.8)\cos20^\circ=0$ , so $R=6(9.8)\cos20^\circ-50\sin15^\circ=42.3\,\text{N}$ . Parallel to the plane, $50\cos15^\circ-6(9.8)\sin20^\circ=6a$ . Therefore $a=[50\cos15^\circ-58.8\sin20^\circ]/6=4.70\,\text{m s}^{-2}$ .

(7 marks)

Tier 1 · Easy

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A resultant force

(9\mathbf{i}-12\mathbf{j})\,\text{N}

acts on a particle of mass

3\,\text{kg}

. Find its acceleration.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$(3\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}$	2
Notes
Newton's second law gives $\mathbf{a}=\mathbf{F}/m=(9\mathbf{i}-12\mathbf{j})/3=(3\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}$ .

(2 marks)

Tier 2 · Standard

6\,\text{kg}

sledge moves on a smooth horizontal surface. It is pulled by a force of

32\,\text{N}

acting at

25^\circ

above the horizontal, while a horizontal resistance of

5\,\text{N}

opposes the motion. Using

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

, find the acceleration and the normal reaction, giving both answers to 3 significant figures.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Acceleration $=4.00\,\text{m s}^{-2}$ Normal reaction $=45.3\,\text{N}$	5
Notes
Horizontally, $32\cos25^\circ-5=6a$ , so $a=4.00\,\text{m s}^{-2}$ to 3 significant figures. Vertically there is no acceleration, so $R+32\sin25^\circ-6(9.8)=0$ . Hence $R=58.8-32\sin25^\circ=45.3\,\text{N}$ to 3 significant figures.

(5 marks)

Tier 3 · Hard

3\,\text{kg}

particle lies on a smooth plane inclined at angle

\alpha

to the horizontal, where

\tan\alpha=\dfrac34

. A horizontal force of magnitude

P\,\text{N}

acts on the particle towards the upward side of the plane. The particle accelerates up the plane at

1.5\,\text{m s}^{-2}

. Using

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

, find

P

and the normal reaction, giving both answers to 3 significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$P=27.7\,\text{N}$ Normal reaction $=40.1\,\text{N}$	6
Notes
Here $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Resolving up the plane gives $(4/5)P-3(9.8)(3/5)=3(1.5)$ , so $P=27.675\,\text{N}=27.7\,\text{N}$ to 3 significant figures. Perpendicular to the plane, the horizontal force presses into the plane, so $R=3(9.8)(4/5)+(3/5)P=40.125\,\text{N}=40.1\,\text{N}$ to 3 significant figures.

(6 marks)

Tier 2 · Standard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. Constant forces

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

and

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

act on a particle. During

4\,\text{s}

its velocity changes by

(4\mathbf{i}+6\mathbf{j})\,\text{m s}^{-1}

. Find the mass of the particle.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Mass $=8\,\text{kg}$	4
Notes
The resultant force is $(8\mathbf{i}+12\mathbf{j})\,\text{N}$ . The acceleration is the velocity change divided by time, so $\mathbf{a}=(\mathbf{i}+1.5\mathbf{j})\,\text{m s}^{-2}$ . From $\mathbf{F}=m\mathbf{a}$ , either component gives $m=8/1=12/1.5=8\,\text{kg}$ .

(4 marks)

Tier 3 · Hard

A particle moves on a smooth horizontal plane under two forces. One force has magnitude

10\,\text{N}

and acts due east. The other has magnitude

P\,\text{N}

and acts at

120^\circ

anticlockwise from east. The particle's acceleration has magnitude

2\,\text{m s}^{-2}

and direction

60^\circ

north of east. Find

P

and the mass of the particle.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$P=10\,\text{N}$ Mass $=5\,\text{kg}$	5
Notes
The resultant force has components $(10-\tfrac12P)\,\text{N}$ east and $(\tfrac{\sqrt3}{2}P)\,\text{N}$ north. Its direction is $60^\circ$ north of east, so $\tfrac{\sqrt3}{2}P=\sqrt3(10-\tfrac12P)$ . Hence $P=10\,\text{N}$ . The resultant components are then $5$ and $5\sqrt3$ , so its magnitude is $10\,\text{N}$ . From $F=ma$ , $10=2m$ and $m=5\,\text{kg}$ .

(5 marks)

Tier 3 · Hard

5\,\text{kg}

particle rests on a smooth horizontal plane and is pulled by a force of magnitude

P\,\text{N}

acting upwards and forwards at an angle

\alpha

to the horizontal, where

\sin\alpha=3/5

. Take

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

. While the particle remains in contact with the plane, its horizontal acceleration is

6\,\text{m s}^{-2}

. Find

P

and the normal reaction. The force is then increased, its direction unchanged. Find the exact value of

P

at which contact with the plane is just lost and the horizontal acceleration at that instant. Give all numerical answers exactly.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Initially $P=37.5\,\text{N}$ Initially $R=26.5\,\text{N}$ Contact is lost when $P=\dfrac{245}{3}\,\text{N}$ Horizontal acceleration then $=\dfrac{196}{15}\,\text{m s}^{-2}$	7
Notes
Here $\cos\alpha=4/5$ . Horizontally, $(4/5)P=5(6)$ , so $P=37.5\,\text{N}$ . Vertically, $R+(3/5)P=5(9.8)$ , giving $R=49-22.5=26.5\,\text{N}$ . Contact is just lost when $R=0$ , so $(3/5)P=49$ and $P=245/3\,\text{N}$ . The horizontal acceleration is then $[(4/5)(245/3)]/5=196/15\,\text{m s}^{-2}$ .

(7 marks)

Q10

Tier 3 · Hard

10.

4\,\text{kg}

particle moves upwards while remaining in contact with a smooth vertical wall. A force of magnitude

P\,\text{N}

acts upwards and towards the wall at an angle

\alpha

above the horizontal, where

\tan\alpha=3/4

. The particle's acceleration is

2\,\text{m s}^{-2}

upwards. Take

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

. Find the exact value of

P

and the normal reaction from the wall. If the particle starts from rest, find its speed after

3\,\text{s}

. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=\dfrac{236}{3}\,\text{N}$ Normal reaction $=\dfrac{944}{15}\,\text{N}$ Speed after $3\,\text{s}$ is $6\,\text{m s}^{-1}$	6
Notes
Since $\tan\alpha=3/4$ , $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Vertically, $(3/5)P-4(9.8)=4(2)$ , so $(3/5)P=47.2$ and $P=236/3\,\text{N}$ . Perpendicular to the wall there is no acceleration, so the normal reaction balances the horizontal component: $R=(4/5)P=944/15\,\text{N}$ . From rest with constant acceleration $2\,\text{m s}^{-2}$ , the speed after $3\,\text{s}$ is $v=2(3)=6\,\text{m s}^{-1}$ .

(6 marks)

Verified exam appearances

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Other points in M8 Forces and Newton's laws

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