AQA Further Paper 3 Mechanics 2019 June — Question 7 9 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2019
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – with string attached to vertex or fixed point
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring 3D spatial reasoning to find the geometry of the conical motion, then applying both Hooke's law and circular motion dynamics. Students must resolve forces in horizontal and vertical directions with two strings under different tension conditions, which is significantly harder than standard single-string conical pendulum problems.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
7 A particle of mass 2.5 kilograms is attached to one end of a light, inextensible string of length 75 cm . The other end of this string is attached to a point \(A\). The particle is also attached to one end of an elastic string of natural length 30 cm and modulus of elasticity \(\lambda \mathrm { N }\). The other end of this string is attached to a point \(B\), which is 60 cm vertically below \(A\). The particle is set in motion so that it describes a horizontal circle with centre \(B\). The angular speed of the particle is \(8 \mathrm { rad } \mathrm { s } { } ^ { - 1 }\) Find \(\lambda\), giving your answer in terms of \(g\).
Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sin\theta = \frac{60}{75} = \frac{4}{5}\), \(\cos\theta = \frac{3}{5}\)B1 Finds \(\sin\theta\) and \(\cos\theta\). PI by \(\theta = 36.9°\) or \(53.1°\)
\(r = 0.75\cos\theta = 0.45\)B1F Finds radius using trig values or Pythagoras. Accept 45 cm
\(T_1\sin\theta = 2.5g\)M1 Forms equation for tension in inextensible string. Allow \(\sin\theta\) or \(\cos\theta\)
\(T_1 = \frac{25g}{8}\)A1 Correct tension for inextensible string
\(T_2 + T_1\cos\theta = 2.5 \times 0.45 \times 8^2\)M1 Resolves towards centre involving both tensions and \(mr\omega^2\) with their radius. Allow \(T_2\cos\theta\) or \(T_2\sin\theta\)
\(T_2 = 72 - \frac{15g}{8}\)A1 Correct tension for elastic string
\(72 - \frac{15g}{8} = \frac{\lambda}{0.3} \times (0.45 - 0.3)\)M1 Uses Hooke's Law to find modulus of elasticity with their \(T_2\)
\(\lambda = 144 - \frac{15g}{4}\)A1 Correct modulus of elasticity 
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = \frac{60}{75} = \frac{4}{5}$, $\cos\theta = \frac{3}{5}$ | B1 | Finds $\sin\theta$ and $\cos\theta$. PI by $\theta = 36.9°$ or $53.1°$ |
| $r = 0.75\cos\theta = 0.45$ | B1F | Finds radius using trig values or Pythagoras. Accept 45 cm |
| $T_1\sin\theta = 2.5g$ | M1 | Forms equation for tension in inextensible string. Allow $\sin\theta$ or $\cos\theta$ |
| $T_1 = \frac{25g}{8}$ | A1 | Correct tension for inextensible string |
| $T_2 + T_1\cos\theta = 2.5 \times 0.45 \times 8^2$ | M1 | Resolves towards centre involving both tensions and $mr\omega^2$ with their radius. Allow $T_2\cos\theta$ or $T_2\sin\theta$ |
| $T_2 = 72 - \frac{15g}{8}$ | A1 | Correct tension for elastic string |
| $72 - \frac{15g}{8} = \frac{\lambda}{0.3} \times (0.45 - 0.3)$ | M1 | Uses Hooke's Law to find modulus of elasticity with their $T_2$ |
| $\lambda = 144 - \frac{15g}{4}$ | A1 | Correct modulus of elasticity |
7 A particle of mass 2.5 kilograms is attached to one end of a light, inextensible string of length 75 cm . The other end of this string is attached to a point $A$.

The particle is also attached to one end of an elastic string of natural length 30 cm and modulus of elasticity $\lambda \mathrm { N }$. The other end of this string is attached to a point $B$, which is 60 cm vertically below $A$.

The particle is set in motion so that it describes a horizontal circle with centre $B$. The angular speed of the particle is $8 \mathrm { rad } \mathrm { s } { } ^ { - 1 }$

Find $\lambda$, giving your answer in terms of $g$.

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2019 Q7 [9]}}
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