| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle in circular tube or on wire |
| Difficulty | Challenging +1.8 This M3 question requires energy conservation with geometric constraints (string over cylinder), finding normal force using circular motion dynamics, and solving a transcendental equation. It demands careful geometric analysis of the string configuration, multi-step energy calculations, and understanding of contact forces. The transcendental equation in part (ii) is sophisticated but the overall approach follows standard M3 techniques for connected particles over pulleys/cylinders. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Conservation of energy | M1, M1 | Need 4 terms; allow sign & trig errors; Both KE or both PE correct |
| \(\frac{1}{2}(0.4)v^2 + \frac{1}{2}(0.6)v^2 + 0.4ga\sin\theta - 0.6ga\theta = 0\) | A1 | Completely correct |
| \(v^2 = 3.92a(3\theta - 2\sin\theta)\) | M1, A1 | Attempt to find \(v^2\) dep both earlier M1s AG; Allow with sign and trig errors; No errors |
| \(F = ma\) radially for \(P\) | M1* | Allow sign and trig errors |
| \(0.4g\sin\theta - R = \frac{0.4v^2}{a}\) | A1 | |
| \(R = -4.704\theta + 7.056\sin\theta\) | *M1, A1 [9] | Manipulation attempted, leading to \(a\theta + b\sin\theta\); Allow sign and trig errors; \(2.352(-2\theta + 3\sin\theta)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(R = 0\) | M1 | \(0 = -4.704\theta + 7.056\sin\theta\) |
| \((k =)\ \frac{2}{3}\) | A1 [2] | Must be from correct expression in (i) |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Conservation of energy | M1, M1 | Need 4 terms; allow sign & trig errors; Both KE or both PE correct |
| $\frac{1}{2}(0.4)v^2 + \frac{1}{2}(0.6)v^2 + 0.4ga\sin\theta - 0.6ga\theta = 0$ | A1 | Completely correct |
| $v^2 = 3.92a(3\theta - 2\sin\theta)$ | M1, A1 | Attempt to find $v^2$ dep both earlier M1s **AG**; Allow with sign and trig errors; No errors |
| $F = ma$ radially for $P$ | M1* | Allow sign and trig errors |
| $0.4g\sin\theta - R = \frac{0.4v^2}{a}$ | A1 | |
| $R = -4.704\theta + 7.056\sin\theta$ | *M1, A1 **[9]** | Manipulation attempted, leading to $a\theta + b\sin\theta$; Allow sign and trig errors; $2.352(-2\theta + 3\sin\theta)$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $R = 0$ | M1 | $0 = -4.704\theta + 7.056\sin\theta$ |
| $(k =)\ \frac{2}{3}$ | A1 **[2]** | Must be from correct expression in (i) |
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4 A smooth cylinder of radius $a \mathrm {~m}$ is fixed with its axis horizontal and $O$ is the centre of a cross-section. Particle $P$, of mass 0.4 kg , and particle $Q$, of mass 0.6 kg , are connected by a light inextensible string of length $\pi a \mathrm {~m}$. The string is held at rest with $P$ and $Q$ at opposite ends of the horizontal diameter of the crosssection through $O$ (see Fig. 1). The string is released and $Q$ begins to descend. When $O P$ has rotated through $\theta$ radians, with $P$ remaining in contact with the cylinder, the speed of each particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see Fig. 2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_365_433_520_424}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_396_643_484_1000}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(i) Show that $v ^ { 2 } = 3.92 a ( 3 \theta - 2 \sin \theta )$ and find an expression in terms of $\theta$ for the normal force of the cylinder on $P$ at this time.\\
(ii) Given that $P$ leaves the surface of the cylinder when $\theta = \alpha$, show that $\sin \alpha = k \alpha$ where $k$ is a constant to be found.
\hfill \mbox{\textit{OCR M3 2013 Q4 [11]}}