| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – lower particle also moves in horizontal circle (conical pendulum below) |
| Difficulty | Challenging +1.2 This is a coupled circular motion problem requiring resolution of forces and application of circular motion equations. While it involves multiple particles and 3D geometry (string at angle), the setup is clearly described, the angle is given explicitly (not needing calculation), and the solution follows standard mechanics procedures: resolve tension components, apply F=mrω² for both particles, and solve simultaneous equations. More challenging than basic single-particle circular motion but still a structured multi-step problem without requiring novel insight. |
| 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 | Marks | Guidance |
| \(T\cos\theta = m_B g\) | M1* | Balancing vertical forces on \(B\) |
| \(T\sin\theta = m_B \times 0.6 \times \omega^2\) | M1* | NII for \(B\) with \(r = 0.6\) (could use \(v^2/0.6\)) |
| \(\tan\theta = (0.6\omega^2)/g\) | M1dep | Combining equations and eliminating \(T\) |
| \(\tan\theta = \frac{3}{4}\) oe | B1 | May be implied. Accept \(\theta = 36.9\) |
| \(\omega = 3.5\) | A1 | |
| \(t = \dfrac{2\pi}{3.5}\) | M1 | Their \(3.5\); or \(t = 2\pi r/v\) \((v = 2.1\text{ m/s})\) |
| Time for one revolution is \(1.8\) seconds | A1 | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 1.2 \times 0.6\omega^2\ (= 8.82)\) | M1 | NII for \(A\) and for realising that \(\omega\) is the same for \(A\) and \(B\). Could be seen in (i) |
| \(8.82\cos\theta = m_B g\) or \(8.82\sin\theta = m_B \times 0.6\omega^2\) | M1 | Substituting their \(T\) into either of their equations of motion for \(B\) |
| \(m_B = 0.72\) | A1 | |
| [3] |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\cos\theta = m_B g$ | M1* | Balancing vertical forces on $B$ |
| $T\sin\theta = m_B \times 0.6 \times \omega^2$ | M1* | NII for $B$ with $r = 0.6$ (could use $v^2/0.6$) |
| $\tan\theta = (0.6\omega^2)/g$ | M1dep | Combining equations and eliminating $T$ |
| $\tan\theta = \frac{3}{4}$ oe | B1 | May be implied. Accept $\theta = 36.9$ |
| $\omega = 3.5$ | A1 | |
| $t = \dfrac{2\pi}{3.5}$ | M1 | Their $3.5$; or $t = 2\pi r/v$ $(v = 2.1\text{ m/s})$ |
| Time for one revolution is $1.8$ seconds | A1 | |
| **[7]** | | |
---
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 1.2 \times 0.6\omega^2\ (= 8.82)$ | M1 | NII for $A$ and for realising that $\omega$ is the same for $A$ and $B$. Could be seen in (i) |
| $8.82\cos\theta = m_B g$ or $8.82\sin\theta = m_B \times 0.6\omega^2$ | M1 | Substituting their $T$ into either of their equations of motion for $B$ |
| $m_B = 0.72$ | A1 | |
| **[3]** | | |6 Two particles $A$ and $B$ are connected by a light inextensible string. Particle $A$ has mass 1.2 kg and moves on a smooth horizontal table in a circular path of radius 0.6 m and centre $O$. The string passes through a small smooth hole at $O$. Particle $B$ moves in a horizontal circle in such a way that it is always vertically below $A$. The angle that the portion of the string below the table makes with the downwards vertical through $O$ is $\theta$, where $\cos \theta = \frac { 4 } { 5 }$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{5960a9cf-2c51-4c07-9973-c29604762df7-4_519_803_484_632}\\
(i) Find the time taken for the particles to perform a complete revolution.\\
(ii) Find the mass of $B$.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q6 [10]}}