Question 9 - A-Level Maths

AQA Further Paper 3 Mechanics 2021 June — Question 9 10 marks

Exam Board	AQA
Module	Further Paper 3 Mechanics (Further Paper 3 Mechanics)
Year	2021
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 1
Type	Elastic string – conical pendulum (string inclined to vertical)
Difficulty	Challenging +1.8 This is a challenging Further Maths mechanics problem requiring elastic string theory, circular motion, and resolving forces in 3D. Part (a) demands careful setup of Hooke's law with extension calculation, horizontal/vertical force resolution, and algebraic manipulation to reach a non-obvious target equation. Parts (b-c) are more routine once the equation is established. The multi-step reasoning, combination of elastic strings with conical pendulum dynamics, and the need to handle both tension components and centripetal acceleration place this well above average difficulty, though the structured parts provide scaffolding.
Spec	6.02j Conservation with elastics: springs and strings 6.05c Horizontal circles: conical pendulum, banked tracks

In this question use $g = 9.81 \text{ m s}^{-2}$ A conical pendulum is made from an elastic string and a sphere of mass 0.2 kg The string has natural length 1.6 metres and modulus of elasticity 200 N The sphere describes a horizontal circle of radius 0.5 metres at a speed of $v \text{ m s}^{-1}$ The angle between the elastic string and the vertical is $\alpha$

Show that $$62.5 - 200 \sin \alpha = 1.962 \tan \alpha$$ [5 marks]
Use your calculator to find $\alpha$ [1 mark]
Find the value of $v$ [4 marks]

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks
9(a)	Forms an equation that relates the

length of the string to the radius of

Answer	Marks	Guidance
the circle	3.3	M1

0 . 5

e 1 . 6 = −

s in 

0 . 5  

2 0 0 1 . 6 −

s in 

T =

1 . 6

6 2 . 5

2 0 0 = −

s in 

T c o s 0 . 2 g 1 . 9 6 2  = =

1 . 9 6 2

T =

c o s 

1.962 62.5

= −200

cos sin

1.962 62.5

sin= sin−200sin

cos sin

1.962tan= 62.5−200sin

Forms an expression for the

Answer	Marks	Guidance
tension using Hooke’s Law	1.1a	M1

Resolves tension vertically to form

Answer	Marks	Guidance
an equation	3.4	M1

Combines the three equations to

Answer	Marks	Guidance
eliminate tension and extension	1.1a	M1

Derives the required result using a

rigorous argument, must see

Answer	Marks	Guidance
justification for tan	2.1	R1
Total	5
Q	Marking Instructions	AO

Answer	Marks
9(b)	Obtains correct solution from

calculator or numerical method.

Answer	Marks	Guidance
AWRT 18	1.1b	B1
Total	1
Q	Marking Instructions	AO

Answer	Marks	Guidance
9(c)	Uses their angle to find an
expression for tension	1.1a	B1

T = = 2.0632

cos(18.02)

v 2

2 . 0 6 3 2 s in 0 . 2  =

0 . 5

2 . 0 6 3 2 ( s in 1 8 . 0 2  )

v =

0 . 4

= 1 . 2 6 3

= 1 . 2 6 m s 1 − t o 3 s f

Resolves horizontally to form an

Answer	Marks	Guidance
equation in v	3.4	M1
Obtains correct equation	1.1b	A1

Solves their equation to find the

Answer	Marks	Guidance
correct speed	3.4	A1
Total	4
Question total	10
Paper total	50

Question 9:
--- 9(a) ---
9(a) | Forms an equation that relates the
length of the string to the radius of
the circle | 3.3 | M1 | (1.6+e)sinα =0.5
0 . 5
e 1 . 6 = −
s in 
0 . 5  
2 0 0 1 . 6 −
s in 
T =
1 . 6
6 2 . 5
2 0 0 = −
s in 
T c o s 0 . 2 g 1 . 9 6 2  = =
1 . 9 6 2
T =
c o s 
1.962 62.5
= −200
cos sin
1.962 62.5
sin= sin−200sin
cos sin
1.962tan= 62.5−200sin
Forms an expression for the
tension using Hooke’s Law | 1.1a | M1
Resolves tension vertically to form
an equation | 3.4 | M1
Combines the three equations to
eliminate tension and extension | 1.1a | M1
Derives the required result using a
rigorous argument, must see
justification for tan | 2.1 | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
--- 9(b) ---
9(b) | Obtains correct solution from
calculator or numerical method.
AWRT 18 | 1.1b | B1 | 18.0
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 9(c) ---
9(c) | Uses their angle to find an
expression for tension | 1.1a | B1 | 1.962
T = = 2.0632
cos(18.02)
v 2
2 . 0 6 3 2 s in 0 . 2  =
0 . 5
2 . 0 6 3 2 ( s in 1 8 . 0 2  )
v =
0 . 4
= 1 . 2 6 3
= 1 . 2 6 m s 1 − t o 3 s f
Resolves horizontally to form an
equation in v | 3.4 | M1
Obtains correct equation | 1.1b | A1
Solves their equation to find the
correct speed | 3.4 | A1
Total | 4
Question total | 10
Paper total | 50

Show LaTeX source

In this question use $g = 9.81 \text{ m s}^{-2}$

A conical pendulum is made from an elastic string and a sphere of mass 0.2 kg

The string has natural length 1.6 metres and modulus of elasticity 200 N

The sphere describes a horizontal circle of radius 0.5 metres at a speed of $v \text{ m s}^{-1}$

The angle between the elastic string and the vertical is $\alpha$

\begin{enumerate}[label=(\alph*)]
\item Show that

$$62.5 - 200 \sin \alpha = 1.962 \tan \alpha$$
[5 marks]

\item Use your calculator to find $\alpha$
[1 mark]

\item Find the value of $v$
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2021 Q9 [10]}}

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