| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| 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 with elastic string requiring resolution of forces and application of Hooke's law. Part (a) involves straightforward vertical equilibrium (tension balances weight), and part (b) applies the horizontal component to circular motion formula. The 'show that' in part (a) guides students significantly, and the setup is a textbook example with clean numbers that work out nicely (extension = half natural length). Slightly above average difficulty due to 3D geometry and combining multiple mechanics concepts, but follows a well-practiced method. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\cos 60° = 0.5g\), \(T = g\) | M1, A1 | |
| Extension \(= x\), \(T = \frac{\lambda x}{a} = \frac{19.6x}{0.8}\) | B1 | |
| \(g = 24.5x\), \(x = 0.4 \text{ m}\) | M1, A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\sin 60° = 0.5 \times r \times \omega^2\) | M1 A1 | |
| \(g\sin 60° = 0.5 \times (0.8 + 0.4)\sin 60°\cdot\omega^2\) | M1 A1 | |
| \(\omega^2 = \frac{2g}{1.2}\), \(\omega = \sqrt{\frac{5g}{3}}\) (4.04 or 4.0) | A1 | (5) Total: 10 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\cos 60° = 0.5g$, $T = g$ | M1, A1 | |
| Extension $= x$, $T = \frac{\lambda x}{a} = \frac{19.6x}{0.8}$ | B1 | |
| $g = 24.5x$, $x = 0.4 \text{ m}$ | M1, A1 | (5) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\sin 60° = 0.5 \times r \times \omega^2$ | M1 A1 | |
| $g\sin 60° = 0.5 \times (0.8 + 0.4)\sin 60°\cdot\omega^2$ | M1 A1 | |
| $\omega^2 = \frac{2g}{1.2}$, $\omega = \sqrt{\frac{5g}{3}}$ (4.04 or 4.0) | A1 | (5) Total: 10 |
---4. A light elastic string $A B$ has natural length 0.8 m and modulus of elasticity 19.6 N . The end $A$ is attached to a fixed point. A particle of mass 0.5 kg is attached to the end $B$. The particle is moving with constant angular speed $\omega$ rad s $^ { - 1 }$ in a horizontal circle whose centre is vertically below $A$. The string is inclined at $60 ^ { \circ }$ to the vertical.
\begin{enumerate}[label=(\alph*)]
\item Show that the extension of the string is 0.4 m .
\item Find the value of $\omega$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2012 Q4 [10]}}