WJEC Further Unit 3 Specimen — Question 5 6 marks

Exam BoardWJEC
ModuleFurther Unit 3 (Further Unit 3)
SessionSpecimen
Marks6
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyStandard +0.3 This is a standard conical pendulum problem requiring resolution of forces (tension components) and circular motion (centripetal force). While it involves multiple steps (resolving horizontally and vertically, then combining equations), the setup is entirely routine for Further Maths mechanics with no novel insight required. The 6 marks reflect the working steps rather than conceptual difficulty, making it slightly easier than average overall.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

A particle of mass \(m\) kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(O\). The particle is set in motion such that it moves in a horizontal circle of radius 2 m with constant speed 4.8 ms\(^{-1}\). Calculate the angle the string makes with the vertical. [6]

AnswerMarks Guidance
Resolve vertically: \(T\cos\theta = mg\)M1 A1 AO3 AO2
N2L towards centre: \(T\sin\theta = \frac{mv^2}{r}\)M1 AO3
\(T\sin\theta = \frac{m \times 4.8^2}{2}\)A1 AO2
\(\tan\theta = \frac{4.8^2}{2 \times 9.8}\)m1 AO1
\(\theta = 49.61(2371...)°\)A1 AO1
Total: [6]
Resolve vertically: $T\cos\theta = mg$ | M1 A1 | AO3 AO2

N2L towards centre: $T\sin\theta = \frac{mv^2}{r}$ | M1 | AO3

$T\sin\theta = \frac{m \times 4.8^2}{2}$ | A1 | AO2

$\tan\theta = \frac{4.8^2}{2 \times 9.8}$ | m1 | AO1

$\theta = 49.61(2371...)°$ | A1 | AO1

**Total: [6]**

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A particle of mass $m$ kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $O$. The particle is set in motion such that it moves in a horizontal circle of radius 2 m with constant speed 4.8 ms$^{-1}$. Calculate the angle the string makes with the vertical. [6]

\hfill \mbox{\textit{WJEC Further Unit 3  Q5 [6]}}