| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: speed at specific point |
| Difficulty | Standard +0.3 This is a standard vertical circle problem requiring energy conservation in two parts: (i) finding speed at lowest point using PE→KE conversion, and (ii) finding angle where particle stops using energy balance. Both are routine applications of mechanical energy conservation with no novel insight required, making it slightly easier than average for Further Maths mechanics. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(KE = \frac{1}{2} \times m \times 1.2^2\) (= \(0.72m\)) | B1 | |
| \(PE\text{ difference} = mg \times 3.2(1 - \cos 15°)\) (=\(1.0685...m\)) | M1 | |
| \(\frac{1}{2} \times m \times v^2 = mg \times 3.2(1 - \cos 15°) + 0.72m\) | M1 | Conservation of energy (in 3 terms) (condone if \(m\) cancelled) |
| \(1.89\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg \times 3.2(1 - \cos\theta) = 1.7885...m\) | M1 | Conservation of energy with \(v=0\) (condone if \(m\) cancelled). Their non-zero \(\frac{1}{2}mu^2\) |
| \(\theta = 19.4\) | A1 | Allow 19.5 from correct working |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $KE = \frac{1}{2} \times m \times 1.2^2$ (= $0.72m$) | B1 | |
| $PE\text{ difference} = mg \times 3.2(1 - \cos 15°)$ (=$1.0685...m$) | M1 | |
| $\frac{1}{2} \times m \times v^2 = mg \times 3.2(1 - \cos 15°) + 0.72m$ | M1 | Conservation of energy (in 3 terms) (condone if $m$ cancelled) |
| $1.89$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg \times 3.2(1 - \cos\theta) = 1.7885...m$ | M1 | Conservation of energy with $v=0$ (condone if $m$ cancelled). Their non-zero $\frac{1}{2}mu^2$ |
| $\theta = 19.4$ | A1 | Allow 19.5 from correct working |
---1\\
\begin{tikzpicture}[>=latex, line join=round, line cap=round]
% Coordinates
\coordinate (O) at (0, 0);
\coordinate (V) at (0, -2.5); % End of dashed vertical line
\coordinate (P) at (-75:4.5); % Position of mass P
% Draw dashed vertical line
\draw[densely dashed, semithick] (O) -- (V);
% Draw pendulum string
\draw[thick] (O) -- (P) node[midway, right=4pt] {$3.2\mathrm{\,m}$};
% Draw pivot point label
\node[left=2pt] at (O) {$O$};
% Draw angle arc
\draw (0, -1.2) arc (-90:-75:1.2);
% Draw angle label and curved pointer
\node (angle_label) at (1.8, -0.8) {$15^\circ$};
% Use a Bezier curve to match the specific "under-hook" shape from the image
\draw[->, semithick] (1.7, -1.0) .. controls (1.7, -1.6) and (0.6, -1.6) .. (-80:1.2);
% Draw mass P
\fill (P) circle (0.15);
\node[right=4pt] at (P) {$P$};
% Draw velocity vector below P
% Using coordinates relative to P to maintain the correct slope and position
\coordinate (v_start) at ($(P) + (1.2, -0.9)$);
\coordinate (v_end) at ($(P) + (-0.5, -1.2)$);
\draw[->, semithick] (v_start) -- (v_end);
% Draw velocity label centered below the arrow
\node[below=2pt] at ($(v_start)!0.5!(v_end)$) {$1.2\mathrm{\,m\,s^{-1}}$};
\end{tikzpicture}
A particle $P$ of mass $m \mathrm {~kg}$ is attached to one end of a light inextensible string of length 3.2 m . The other end of the string is attached to a fixed point $O$. The particle is held at rest, with the string taut and making an angle of $15 ^ { \circ }$ with the vertical. It is then projected with velocity $1.2 \mathrm {~ms} ^ { - 1 }$ in a direction perpendicular to $O P$ and with a downwards component so that it begins to move in a vertical circle (see diagram). In the ensuing motion the string remains taut and the angle it makes with the downwards vertical through $O$ is denoted by $\theta ^ { \circ }$.\\
(i) Find the speed of $P$ at the point on its path vertically below $O$.\\
(ii) Find the value of $\theta$ at the point where $P$ first comes to instantaneous rest.
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q1 [6]}}