Question 16 - A-Level Maths

AQA Paper 2 Specimen — Question 16 12 marks

Exam Board	AQA
Module	Paper 2 (Paper 2)
Session	Specimen
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Motion on a slope
Type	String at angle to slope
Difficulty	Standard +0.3 This is a standard A-level mechanics problem involving resolving forces in equilibrium (part a) and with acceleration (part b). Part (a) requires resolving horizontally and vertically with friction, which is routine. Part (b) adds an incline and acceleration but follows the same systematic approach. The 12 total marks reflect length rather than conceptual difficulty—all techniques are standard for A-level mechanics with no novel insight required.
Spec	3.03c Newton's second law: F=ma one dimension 3.03d Newton's second law: 2D vectors 3.03e Resolve forces: two dimensions 3.03f Weight: W=mg 3.03g Gravitational acceleration 3.03i Normal reaction force 3.03m Equilibrium: sum of resolved forces = 0 3.03r Friction: concept and vector form 3.03s Contact force components: normal and frictional 3.03t Coefficient of friction: F <= mu*R model 3.03v Motion on rough surface: including inclined planes

In this question use \(g = 9.8\) m s\(^{-2}\). The diagram shows a box, of mass 8.0 kg, being pulled by a string so that the box moves at a constant speed along a rough horizontal wooden board. The string is at an angle of 40° to the horizontal. The tension in the string is 50 newtons. \includegraphics{figure_16a} The coefficient of friction between the box and the board is \(\mu\) Model the box as a particle.

Show that \(\mu = 0.83\) [4 marks]
One end of the board is lifted up so that the board is now inclined at an angle of 5° to the horizontal. The box is pulled up the inclined board. The string remains at an angle of 40° to the board. The tension in the string is increased so that the box accelerates up the board at 3 m s\(^{-2}\) \includegraphics{figure_16b}
1. Draw a diagram to show the forces acting on the box as it moves. [1 mark]
2. Find the tension in the string as the box accelerates up the slope at 3 m s\(^{-2}\). [7 marks]

Show mark scheme Show mark scheme source

Question 16:

Answer	Marks
16(a)	Resolves horizontally and

vertically to obtain expressions

for F and R

(allow consistent mixing of sin

Answer	Marks	Guidance
and cos)	AO3.4	M1

R = 8 × 9.8 – 50sin 40°

Resolving horizontally

F = 50cos 40°

F R

50cos40(89.850sin40)

50cos40



89.850sin40

μ = 0.8279660445 = 0.83 (2 sf)

AG

Obtains correct expressions for

Answer	Marks	Guidance
R and F	AO1.1b	A1

States friction model F = μR with

Answer	Marks	Guidance
‘their’ values for F and R	AO1.2	B1

Completes a rigorous argument

that results with correct μ AG

Only award if they have a

completely correct solution,

which is clear, easy to follow

Answer	Marks	Guidance
and contains no slips	AO2.1	R1
(b)(i)	Draws correct diagram with

exactly four forces showing

arrow heads and labels

Answer	Marks	Guidance
Can use Mg or 8g or 78.4 for W	AO3.3	B1

R

F or μR

W

Answer	Marks	Guidance
Q	Marking Instructions	AO

Answer	Marks
16(b)(ii)	Resolves perpendicular to the

plane resulting in a three term

equation containing; R, 8gcos 5°

and Tsin 40° (or Tcos 50°)

OR

Resolves horizontally and

vertically to obtain equations of

motion in horizontal and vertical

Answer	Marks	Guidance
directions	AO3.1b	M1

= 78.1 – Tsin 40°

Tcos 40° – 8  9.8sin 5°– F = 8  3

Tcos 40° – 8  9.8sin 5°

– 0.827…×(8 × 9.8cos 5° –

Tsin 40°) = 24

2489.8sin 5o0.827...89.8cos 5o

T 

cos 40o0.827...sin 40o

T = 73.55939193 = 74 (2 sf)

Obtains correct expression for R

OR

Obtains correct horizontal and

Answer	Marks	Guidance
vertical equations	AO1.1b	A1

Forms a four term equation of

motion parallel to the plane with

correct terms

(allow sign errors)

OR

Answer	Marks	Guidance
Eliminates R to solve for T	AO3.1b	M1

Obtains correct equation of motion

OR

Obtains correct expression(s)

Answer	Marks	Guidance
without R	AO1.1b	A1

Uses the friction model in the four

term equation of motion where R is

in the form a – bT and a and b are

positive constants

OR

Uses the friction model in the

horizontal and vertical equations

Answer	Marks	Guidance
Dependent on previous M1	AO3.1b	dM1
Solves for T	AO1.1b	A1

Obtains correct value of T with 2 sf

accuracy.

FT incorrect value found for T

provided both M1 marks and dM1

Answer	Marks	Guidance
mark have been awarded	AO3.2a	A1F
Total	12
Q	Marking Instructions	AO

Question 16:
--- 16(a) ---
16(a) | Resolves horizontally and
vertically to obtain expressions
for F and R
(allow consistent mixing of sin
and cos) | AO3.4 | M1 | Resolving vertically
R = 8 × 9.8 – 50sin 40°
Resolving horizontally
F = 50cos 40°
F R
50cos40(89.850sin40)
50cos40

89.850sin40
μ = 0.8279660445 = 0.83 (2 sf)
AG
Obtains correct expressions for
R and F | AO1.1b | A1
States friction model F = μR with
‘their’ values for F and R | AO1.2 | B1
Completes a rigorous argument
that results with correct μ AG
Only award if they have a
completely correct solution,
which is clear, easy to follow
and contains no slips | AO2.1 | R1
(b)(i) | Draws correct diagram with
exactly four forces showing
arrow heads and labels
Can use Mg or 8g or 78.4 for W | AO3.3 | B1 | T
R
F or μR
W
Q | Marking Instructions | AO | Marks | Typical Solution
--- 16(b)(ii) ---
16(b)(ii) | Resolves perpendicular to the
plane resulting in a three term
equation containing; R, 8gcos 5°
and Tsin 40° (or Tcos 50°)
OR
Resolves horizontally and
vertically to obtain equations of
motion in horizontal and vertical
directions | AO3.1b | M1 | R = 8  9.8cos 5° – Tsin 40°
= 78.1 – Tsin 40°
Tcos 40° – 8  9.8sin 5°– F = 8  3
Tcos 40° – 8  9.8sin 5°
– 0.827…×(8 × 9.8cos 5° –
Tsin 40°) = 24
2489.8sin 5o0.827...89.8cos 5o
T 
cos 40o0.827...sin 40o
T = 73.55939193 = 74 (2 sf)
Obtains correct expression for R
OR
Obtains correct horizontal and
vertical equations | AO1.1b | A1
Forms a four term equation of
motion parallel to the plane with
correct terms
(allow sign errors)
OR
Eliminates R to solve for T | AO3.1b | M1
Obtains correct equation of motion
OR
Obtains correct expression(s)
without R | AO1.1b | A1
Uses the friction model in the four
term equation of motion where R is
in the form a – bT and a and b are
positive constants
OR
Uses the friction model in the
horizontal and vertical equations
Dependent on previous M1 | AO3.1b | dM1
Solves for T | AO1.1b | A1
Obtains correct value of T with 2 sf
accuracy.
FT incorrect value found for T
provided both M1 marks and dM1
mark have been awarded | AO3.2a | A1F
Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution

Show LaTeX source

In this question use $g = 9.8$ m s$^{-2}$.

The diagram shows a box, of mass 8.0 kg, being pulled by a string so that the box moves at a constant speed along a rough horizontal wooden board.

The string is at an angle of 40° to the horizontal.

The tension in the string is 50 newtons.

\includegraphics{figure_16a}

The coefficient of friction between the box and the board is $\mu$

Model the box as a particle.

\begin{enumerate}[label=(\alph*)]
\item Show that $\mu = 0.83$
[4 marks]

\item One end of the board is lifted up so that the board is now inclined at an angle of 5° to the horizontal.

The box is pulled up the inclined board.

The string remains at an angle of 40° to the board.

The tension in the string is increased so that the box accelerates up the board at 3 m s$^{-2}$

\includegraphics{figure_16b}

\begin{enumerate}[label=(\roman*)]
\item Draw a diagram to show the forces acting on the box as it moves.
[1 mark]

\item Find the tension in the string as the box accelerates up the slope at 3 m s$^{-2}$.
[7 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2  Q16 [12]}}

This paper (17 questions)

View full paper

Q1 1 Q2 1 Q3 6 Q4 6 Q5 9 Q6 5 Q7 4 Q8 8 Q9 10 Q10 1 Q11 2 Q12 4 Q13 5 Q14 7 Q15 11 Q16 12 Q17 8