| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring resolution of forces (tension, weight, normal reaction), application of F=mrω², and geometric constraints. The 'show that' format with inequality requires finding the critical case (N=0 at limiting speed), but the setup and method are textbook-standard for this module. Slightly above average due to the inequality aspect and need to identify the limiting condition, but well within expected M3 difficulty. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((4a)^2 + r^2 = (8a-r)^2\), Radius \(= 3a\) | M1, A1 | M1: attempt to obtain radius, probably using Pythagoras (can justify via 3,4,5 triangle); A1: correct radius seen or used (providing M mark awarded) |
| \(T\cos\theta = mg - R\) | M1A1 | M1: resolve vertically, 3 forces with \(T\) resolved; A1: correct equation with \(\cos\theta\) or \(\frac{4}{5}\) |
| \(T + T\sin\theta = m \times \text{rad} \times \omega^2\) | M1A1A1 | M1: equation of motion along horizontal radius, \(T\) resolved, radius or \(r\) allowed; A1: forces correct with \(\sin\theta\) or \(\frac{3}{5}\), both tensions same; A1: acceleration correct in form shown |
| \(\frac{8}{5}T = 3ma\omega^2\) | ||
| \(2mg - 2R = 3ma\omega^2\) | DM1 | Eliminate \(T\) and replace trig functions with values (depends on 2nd and 3rd M marks but not the first) |
| \(R \geq 0 \Rightarrow 2mg - 3ma\omega^2 \geq 0 \Rightarrow \omega^2 \leq \frac{2g}{3a}\) | M1A1 | M1: use \(R \geq 0\) to obtain inequality for \(\omega^2\); A1: correct inequality, must use \(r = 3a\) |
| \(S \geq 2\pi\sqrt{\frac{3a}{2g}} = \pi\sqrt{\frac{6a}{g}}\) | DM1, A1cso | DM1: use period \(= \frac{2\pi}{\omega}\) to obtain inequality for \(S\), direction must change; A1cso: correct inequality as shown. NB: candidates assuming 3,4,5 triangle without justification lose first 2 marks and this mark |
| Total: [12] |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4a)^2 + r^2 = (8a-r)^2$, Radius $= 3a$ | M1, A1 | M1: attempt to obtain radius, probably using Pythagoras (can justify via 3,4,5 triangle); A1: correct radius seen or used (providing M mark awarded) |
| $T\cos\theta = mg - R$ | M1A1 | M1: resolve vertically, 3 forces with $T$ resolved; A1: correct equation with $\cos\theta$ or $\frac{4}{5}$ |
| $T + T\sin\theta = m \times \text{rad} \times \omega^2$ | M1A1A1 | M1: equation of motion along horizontal radius, $T$ resolved, radius or $r$ allowed; A1: forces correct with $\sin\theta$ or $\frac{3}{5}$, both tensions same; A1: acceleration correct in form shown |
| $\frac{8}{5}T = 3ma\omega^2$ | | |
| $2mg - 2R = 3ma\omega^2$ | DM1 | Eliminate $T$ and replace trig functions with values (depends on 2nd and 3rd M marks but **not** the first) |
| $R \geq 0 \Rightarrow 2mg - 3ma\omega^2 \geq 0 \Rightarrow \omega^2 \leq \frac{2g}{3a}$ | M1A1 | M1: use $R \geq 0$ to obtain inequality for $\omega^2$; A1: correct inequality, must use $r = 3a$ |
| $S \geq 2\pi\sqrt{\frac{3a}{2g}} = \pi\sqrt{\frac{6a}{g}}$ | DM1, A1cso | DM1: use period $= \frac{2\pi}{\omega}$ to obtain inequality for $S$, direction must change; A1cso: correct inequality as shown. **NB:** candidates assuming 3,4,5 triangle without justification lose first 2 marks and this mark |
| **Total: [12]** | | |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d93ae982-9395-4311-9972-be727b3ce954-10_547_841_244_555}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small smooth bead $P$ is threaded on a light inextensible string of length $8 a$. One end of the string is attached to a fixed point $A$ on a smooth horizontal table. The other end of the string is attached to the fixed point $B$, where $B$ is vertically above $A$ and $A B = 4 a$, as shown in Figure 2. The bead moves with constant angular speed, in a horizontal circle, centre $A$, with $A P$ horizontal. The bead remains in contact with the table and both parts of the string, $A P$ and $B P$, are taut. The time for $P$ to complete one revolution is $S$.
Show that $\quad S \geqslant \pi \sqrt { \frac { 6 a } { g } }$
\hfill \mbox{\textit{Edexcel M3 2018 Q4 [12]}}