| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – no string (normal reaction only) |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem with clear geometric setup. Part (i) requires resolving forces on a cone (routine technique), part (ii) involves two normal reactions with horizontal circular motion (standard but slightly more complex), and part (iii) is straightforward energy calculation. The geometry is given explicitly, and all parts follow well-established M2 methods without requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| i | \(R\sin 30 = 0.3g\) | M1 |
| \(R\cos 30 = 0.3\omega^2 x \times 0.12\) | M1 | |
| \(\omega = 11.9 \text{ rads}^{-1}\) | A1 | \(R = 5.88\) or \(0.6g\); accept \(v^2/0.12\) for acceleration |
| A1 | cao | |
| A1 | ||
| [5] | ||
| ii | \(S + R\cos 30 = 0.3 \times 2.1^2/0.2\) | A1 |
| \(R = 5.88\) | B1ft | \(ft cv(R)\) from (i) |
| \(S = 1.52 \text{ N}\) | A1 | |
| [4] | ||
| iii | \(v_P = 11.9 \times 0.12\), or \(h = 0.2/\tan 30\) or \(0.12/\tan 30\) or \(0.08/\tan 30\) | B1 |
| \(+/-(Q - P) =\) | M1 | KE difference (\(ft\) on \(cv(\omega)\)) or PE difference |
| \(0.5 \times 0.3(2.1^2 - (11.9 \times 0.12)^2)\) | A2ft | \(Q - P = +/-(0.3556 + 0.4074)\) |
| \(+ (0.2/\tan 30 - 0.12/\tan 30) \times 0.3g\) | A1 | |
| \(Q - P = +/- 0.763 \text{ J}\) | [5] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| i | $R\sin 30 = 0.3g$ | M1 | |
| | $R\cos 30 = 0.3\omega^2 x \times 0.12$ | M1 | |
| | $\omega = 11.9 \text{ rads}^{-1}$ | A1 | $R = 5.88$ or $0.6g$; accept $v^2/0.12$ for acceleration |
| | | A1 | cao |
| | | A1 | |
| | | [5] | |
| ii | $S + R\cos 30 = 0.3 \times 2.1^2/0.2$ | A1 | Resolve and use N2L on sphere Q, 3 terms needed |
| | $R = 5.88$ | B1ft | $ft cv(R)$ from (i) |
| | $S = 1.52 \text{ N}$ | A1 | |
| | | [4] | |
| iii | $v_P = 11.9 \times 0.12$, or $h = 0.2/\tan 30$ or $0.12/\tan 30$ or $0.08/\tan 30$ | B1 | $cv(\omega)$ from (i). Attempt to calculate KE or PE for both particles |
| | $+/-(Q - P) =$ | M1 | KE difference ($ft$ on $cv(\omega)$) or PE difference |
| | $0.5 \times 0.3(2.1^2 - (11.9 \times 0.12)^2)$ | A2ft | $Q - P = +/-(0.3556 + 0.4074)$ |
| | $+ (0.2/\tan 30 - 0.12/\tan 30) \times 0.3g$ | A1 | |
| | $Q - P = +/- 0.763 \text{ J}$ | [5] | |6
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{65c47bd2-eace-4fec-b1e6-a0c904c4ec3f-3_538_478_758_836}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A container is constructed from a hollow cylindrical shell and a hollow cone which are joined along their circumferences. The cylindrical shell has radius 0.2 m , and the cone has semi-vertical angle $30 ^ { \circ }$. Two identical small spheres $P$ and $Q$ move independently in horizontal circles on the smooth inner surface of the container (see Fig. 1). Each sphere has mass 0.3 kg .\\
(i) $P$ moves in a circle of radius 0.12 m and is in contact with only the conical part of the container. Calculate the angular speed of $P$.\\
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{65c47bd2-eace-4fec-b1e6-a0c904c4ec3f-3_278_209_1845_1009}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
$Q$ moves with speed $2.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is in contact with both the cylindrical and conical surfaces of the container (see Fig. 2). Calculate the magnitude of the force which the cylindrical shell exerts on the sphere.\\
(iii) Calculate the difference between the mechanical energy of $P$ and of $Q$.
\section*{[Question 7 is printed overleaf.]}
\hfill \mbox{\textit{OCR M2 2011 Q6 [14]}}