AQA Further Paper 3 Mechanics Specimen — Question 5 6 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyModerate -0.5 This is a standard conical pendulum problem requiring resolution of forces and application of circular motion equations. Part (a) is a 'show that' using vertical equilibrium (T cos 30° = mg), part (b) applies horizontal circular motion (T sin 30° = mrω²), and part (c) tests understanding of modelling assumptions. While it's Further Maths content, the problem-solving is routine and methodical with no novel insight required, making it slightly easier than average.
Spec3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks

5 In this question use \(\boldsymbol { g } = 9.8 \mathbf { m ~ s } ^ { \mathbf { - 2 } }\).
A conical pendulum consists of a string of length 60 cm and a particle of mass 400 g . The string is at an angle of \(30 ^ { \circ }\) to the vertical, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-08_501_606_644_854} 5
  1. Show that the tension in the string is 4.5 N . 5
  2. Find the angular speed of the particle.
    [0pt] [3 marks]
    5
  3. State two assumptions that you have made about the string.

Question 5:
Part (a):
AnswerMarks Guidance
\(0.4 \times 9.8 = T\cos 30°\)M1 Resolves vertically to form equation for tension
\(T = 4.5\text{ N}\)A1 Correct tension
Part (b):
AnswerMarks Guidance
\(r = 0.6\sin 30° = 0.3\)B1 Finds radius
\(T\sin 30° = 0.4 \times r\omega^2\)M1 Forms equation using their radius and tension
\(\omega = \sqrt{\dfrac{4.5\sin 30°}{0.4 \times 0.3}} = 4.3\text{ rad s}^{-1}\)A1F Correct angular speed to 2 sf; FT if both M1 marks awarded
Part (c):
AnswerMarks Guidance
Light and inextensibleB1 Two appropriate assumptions stated
## Question 5:

### Part (a):
| $0.4 \times 9.8 = T\cos 30°$ | M1 | Resolves vertically to form equation for tension |
| $T = 4.5\text{ N}$ | A1 | Correct tension |

### Part (b):
| $r = 0.6\sin 30° = 0.3$ | B1 | Finds radius |
| $T\sin 30° = 0.4 \times r\omega^2$ | M1 | Forms equation using their radius and tension |
| $\omega = \sqrt{\dfrac{4.5\sin 30°}{0.4 \times 0.3}} = 4.3\text{ rad s}^{-1}$ | A1F | Correct angular speed to 2 sf; FT if both M1 marks awarded |

### Part (c):
| Light and inextensible | B1 | Two appropriate assumptions stated |

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5 In this question use $\boldsymbol { g } = 9.8 \mathbf { m ~ s } ^ { \mathbf { - 2 } }$.\\
A conical pendulum consists of a string of length 60 cm and a particle of mass 400 g . The string is at an angle of $30 ^ { \circ }$ to the vertical, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-08_501_606_644_854}

5
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string is 4.5 N .

5
\item Find the angular speed of the particle.\\[0pt]
[3 marks]\\

5
\item State two assumptions that you have made about the string.
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics  Q5 [6]}}