| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – no string (normal reaction only) |
| Difficulty | Standard +0.8 This is a standard Further Mechanics 2 conical pendulum problem requiring resolution of forces in two directions, friction consideration, and finding limiting conditions. It involves multiple steps (finding cone angle, resolving forces perpendicular and parallel to surface, applying friction law, solving for ω) but follows a well-established method taught in FM2. The geometry setup and friction limit make it moderately challenging but not exceptional for Further Maths. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| \(F\cos\theta + mg = R\sin\theta\) | A1 | Unsimplified equation with at most one error |
| \((4F + 5mg = 3R)\) | A1 | Correct unsimplified equation |
| Equation for motion towards centre | M1 | Need all terms. Condone sign errors and sin/cos confusion in \(R\) or their \(R\) |
| \(F\sin\theta + R\cos\theta = m\times 0.5\omega^2\) | A1 | Unsimplified equation with at most one error |
| \((3F + 4R = m\times 2.5\omega^2)\) | A1 | Correct unsimplified equation in \(R\) or their \(R\) |
| At max \(\omega\): \(R + 5mg = 3R \Rightarrow R = \frac{5mg}{2}\) | M1 | Use of \(F = \mu R\) to eliminate \(F\) or \(R\). Condone inequality |
| \(\frac{3}{4}R + 4R = m\times 2.5\omega^2 \Rightarrow 19R = 10m\omega^2\) | ||
| Solve for \(\omega\): \(19\times\frac{5mg}{2} = 10m\omega^2\) | DM1 | Complete method including substitution of trig values to obtain a value for \(\omega\) |
| \(\omega^2 = \frac{19g}{4} \Rightarrow \max\omega = 6.8\ (6.82)\) | A1 | 2 s.f or 3 s.f only; \(\sqrt{\frac{19g}{4}}\) is A0. Must be an equation |
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $F\cos\theta + mg = R\sin\theta$ | A1 | Unsimplified equation with at most one error |
| $(4F + 5mg = 3R)$ | A1 | Correct unsimplified equation |
| Equation for motion towards centre | M1 | Need all terms. Condone sign errors and sin/cos confusion in $R$ or their $R$ |
| $F\sin\theta + R\cos\theta = m\times 0.5\omega^2$ | A1 | Unsimplified equation with at most one error |
| $(3F + 4R = m\times 2.5\omega^2)$ | A1 | Correct unsimplified equation in $R$ or their $R$ |
| At max $\omega$: $R + 5mg = 3R \Rightarrow R = \frac{5mg}{2}$ | M1 | Use of $F = \mu R$ to eliminate $F$ or $R$. Condone inequality |
| $\frac{3}{4}R + 4R = m\times 2.5\omega^2 \Rightarrow 19R = 10m\omega^2$ | | |
| Solve for $\omega$: $19\times\frac{5mg}{2} = 10m\omega^2$ | DM1 | Complete method including substitution of trig values to obtain a value for $\omega$ |
| $\omega^2 = \frac{19g}{4} \Rightarrow \max\omega = 6.8\ (6.82)$ | A1 | 2 s.f or 3 s.f only; $\sqrt{\frac{19g}{4}}$ is A0. Must be an equation |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-20_611_782_210_660}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A hollow right circular cone, of internal base radius 0.6 m and height 0.8 m , is fixed with its axis vertical and its vertex $V$ pointing downwards, as shown in Figure 4.
A particle $P$ of mass $m \mathrm {~kg}$ moves in a horizontal circle of radius 0.5 m on the rough inner surface of the cone.
The particle $P$ moves with constant angular speed $\omega$ rads $^ { - 1 }$\\
The coefficient of friction between the particle $P$ and the inner surface of the cone is 0.25
Find the greatest possible value of $\omega$
\hfill \mbox{\textit{Edexcel FM2 2023 Q6 [9]}}