| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – conical pendulum (string inclined to vertical) |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (tension, weight) and application of Hooke's law with circular motion formula. The geometry is straightforward (given angle and radius), and the solution follows a routine two-part structure with no novel insight required. Slightly above average difficulty due to the 3D geometry and combining elastic strings with circular motion, but well within typical M3 scope. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T\cos 30 = 0.5g\) | M1A1 | Resolve vertically; tension must be resolved (cos or sin allowed); weight not resolved |
| \(\text{ext} = \dfrac{0.9}{\cos 60} - 1.2 = 0.6\text{ m}\) | M1A1 | Use trigonometry to calculate extension; must not use erroneous 1.2 m on vertical |
| \(T = \dfrac{\lambda x}{l} = \dfrac{\lambda \times ``0.6"}{1.2}\) | M1 | Use Hooke's Law with their extension |
| \(\dfrac{\lambda}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{g}{2}\), \(\lambda = \dfrac{2g}{\sqrt{3}} = 11.31\ldots = 11.3\) or \(11\) | dM1A1 [7] | Eliminate \(T\) and solve; depends on first and third M marks; correct answer 2 or 3 significant figures |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T\cos 60 = 0.9m\omega^2\) | ||
| \(T\cos 60 = 0.9 \times 0.5\omega^2\) | M1A1 | Equation of motion along radius; \(T\) must be resolved; acceleration \(0.9\omega^2\); mass to be 0.5 here or later |
| \(\dfrac{2g}{\sqrt{3}} \times \dfrac{0.6}{1.2} \times \dfrac{1}{2} = 0.9 \times 0.5\omega^2\) | dM1A1 | Eliminate \(T\) using Hooke's Law with their \(\lambda\) from (a) or using vertical equation; depends on first M in (b) |
| \(\omega = 2.507\ldots = 2.5\) or \(2.51\) | [4] | Correct value; must be 2 or 3 sig figs |
# Question 2(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T\cos 30 = 0.5g$ | M1A1 | Resolve vertically; tension must be resolved (cos or sin allowed); weight not resolved |
| $\text{ext} = \dfrac{0.9}{\cos 60} - 1.2 = 0.6\text{ m}$ | M1A1 | Use trigonometry to calculate extension; must not use erroneous 1.2 m on vertical |
| $T = \dfrac{\lambda x}{l} = \dfrac{\lambda \times ``0.6"}{1.2}$ | M1 | Use Hooke's Law with their extension |
| $\dfrac{\lambda}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{g}{2}$, $\lambda = \dfrac{2g}{\sqrt{3}} = 11.31\ldots = 11.3$ or $11$ | dM1A1 [7] | Eliminate $T$ and solve; depends on first and third M marks; correct answer 2 or 3 significant figures |
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# Question 2(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T\cos 60 = 0.9m\omega^2$ | | |
| $T\cos 60 = 0.9 \times 0.5\omega^2$ | M1A1 | Equation of motion along radius; $T$ must be resolved; acceleration $0.9\omega^2$; mass to be 0.5 here or later |
| $\dfrac{2g}{\sqrt{3}} \times \dfrac{0.6}{1.2} \times \dfrac{1}{2} = 0.9 \times 0.5\omega^2$ | dM1A1 | Eliminate $T$ using Hooke's Law with their $\lambda$ from (a) or using vertical equation; depends on first M in (b) |
| $\omega = 2.507\ldots = 2.5$ or $2.51$ | [4] | Correct value; must be 2 or 3 sig figs |
---2. A light elastic string has natural length 1.2 m and modulus of elasticity $\lambda$ newtons. One end of the string is attached to a fixed point $O$. A particle of mass 0.5 kg is attached to the other end of the string. The particle is moving with constant angular speed $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle with the string stretched. The circle has radius 0.9 m and its centre is vertically below $O$. The string is inclined at $60 ^ { \circ }$ to the horizontal.
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $\lambda$,
\item the value of $\omega$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q2 [11]}}