AQA Further Paper 3 Mechanics 2021 June — Question 7 9 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2021
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle with peg/obstacle
DifficultyChallenging +1.8 This Further Maths mechanics problem requires understanding of energy conservation and geometry of a string wrapping around a peg. Part (a) tests diagram interpretation (straightforward), while part (c) demands setting up energy equations with the constraint that the string goes slack at a specific angle, requiring careful geometric analysis and solving a transcendental equation. The multi-step reasoning and geometric insight needed elevate this above standard mechanics questions, though the techniques themselves are well-established.
Spec6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration

A light string has length 1.5 metres. A small sphere is attached to one end of the string. The other end of the string is attached to a fixed point O A thin horizontal bar is positioned 0.9 metres directly below O The bar is perpendicular to the plane in which the sphere moves. The sphere is released from rest with the string taut and at an angle \(\alpha\) to the downward vertical through O The string becomes slack when the angle between the two sections of the string is 60° Ben draws the diagram below to show the initial position of the sphere, the bar and the path of the sphere. \includegraphics{figure_7}
  1. State two reasons why Ben's diagram is not a good representation of the situation. [2 marks]
  2. Using your answer to part (a), sketch an improved diagram. [1 mark]
  3. Find \(\alpha\), giving your answer to the nearest degree. [6 marks]

Question 7:

AnswerMarks Guidance
7(a)States that the 60 angle is too
small2.3 E1
The sphere must start at a higher
level than where the string
becomes slack
States that the initial position must
be higher than the point at which
AnswerMarks Guidance
the string becomes slack2.3 E1
Total2
QMarking Instructions AO

AnswerMarks
7(b)Draws a diagram that shows the
initial position more accurately,
where the sphere starts at higher
AnswerMarks Guidance
level than the final level shown.2.3 B1
Total1
QMarking Instructions AO

AnswerMarks Guidance
7(c)Introduces a variable for the mass
and uses it in an equation1.1b B1
string becomes slack = v
m v 2
m g c o s 6 0  =
0 . 6
v 2 = 0 . 6 g c o s 6 0 
Take GPE as zero at the bar
1
mgh= m0.6gcos60+mg0.6cos60
2
h=0.45
0.45
cos= =0.3
1.5
= 73
Resolves radially at the point where
the string becomes slack and uses
AnswerMarks Guidance
Newton’s second law.3.3 M1
Obtains correct expression for v or
AnswerMarks Guidance
v 21.1b A1
Forms a three term energy
equation and substitutes their value
AnswerMarks Guidance
for v3.4 M1
Obtains a correct energy equation1.1b A1
Deduces the correct angle2.2a A1
Total6
Question total9
QMarking Instructions AO
Question 7:
--- 7(a) ---
7(a) | States that the 60 angle is too
small | 2.3 | E1 | The 60 angle is too small
The sphere must start at a higher
level than where the string
becomes slack
States that the initial position must
be higher than the point at which
the string becomes slack | 2.3 | E1
Total | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 7(b) ---
7(b) | Draws a diagram that shows the
initial position more accurately,
where the sphere starts at higher
level than the final level shown. | 2.3 | B1
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 7(c) ---
7(c) | Introduces a variable for the mass
and uses it in an equation | 1.1b | B1 | Let mass = m and speed when the
string becomes slack = v
m v 2
m g c o s 6 0  =
0 . 6
v 2 = 0 . 6 g c o s 6 0 
Take GPE as zero at the bar
1
mgh= m0.6gcos60+mg0.6cos60
2
h=0.45
0.45
cos= =0.3
1.5
= 73
Resolves radially at the point where
the string becomes slack and uses
Newton’s second law. | 3.3 | M1
Obtains correct expression for v or
v 2 | 1.1b | A1
Forms a three term energy
equation and substitutes their value
for v | 3.4 | M1
Obtains a correct energy equation | 1.1b | A1
Deduces the correct angle | 2.2a | A1
Total | 6
Question total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
A light string has length 1.5 metres.

A small sphere is attached to one end of the string.

The other end of the string is attached to a fixed point O

A thin horizontal bar is positioned 0.9 metres directly below O

The bar is perpendicular to the plane in which the sphere moves.

The sphere is released from rest with the string taut and at an angle $\alpha$ to the downward vertical through O

The string becomes slack when the angle between the two sections of the string is 60°

Ben draws the diagram below to show the initial position of the sphere, the bar and the path of the sphere.

\includegraphics{figure_7}

\begin{enumerate}[label=(\alph*)]
\item State two reasons why Ben's diagram is not a good representation of the situation.
[2 marks]

\item Using your answer to part (a), sketch an improved diagram.
[1 mark]

\item Find $\alpha$, giving your answer to the nearest degree.
[6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2021 Q7 [9]}}