| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – particle on horizontal surface |
| Difficulty | Moderate -0.8 This is a straightforward application of standard circular motion formulas (v²/r for acceleration, v=rω, T=mv²/r) followed by basic energy conservation. All parts require direct substitution into well-known equations with no problem-solving insight or geometric complexity. Easier than average A-level mechanics. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| \(a = \frac{v^2}{r} = \frac{12^2}{1.8} = 80\) (ms\(^{-2}\)) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Towards \(O\) | B1FT [1] | Acceleration in part (a) must be \(> 0\) for this mark |
| Answer | Marks |
|---|---|
| \(v = r\omega = 1.8 \times 8 = 14.4\) (ms\(^{-1}\)) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = r\omega^2 = 1.8 \times 8^2\ (= 115.2)\) | M1 | Using formula for radial acceleration and \(F = ma\) with tension as only force |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 46.1\) (N) | A1 [2] | 46.08 |
| Answer | Marks | Guidance |
|---|---|---|
| \(KE = \frac{1}{2} \times 0.4 \times \text{"14.4"}^2\) J \((= 41.472\) J) | M1 | Finding 'initial' energy |
| "\(41.472\)" \(= 0.4gh\) | M1 | ...and equating to final PE; or \(\ldots = 0.4g \times 20\sin\theta\); allow sin/cos confusion if expressing \(h\) in terms of \(\theta\) |
| \(\Rightarrow h = 10.579\ldots\) so \(\theta = \sin^{-1}(10.579\ldots/20)\) = awrt \(31.9(°)\) | A1 [3] | NB degrees specified in question; answer in radians (e.g. 0.557) scores A0; SCB2 for \(\theta = 31.9°\) from non-energy method e.g. suvat: \(14.4^2 = 2 \times 9.8 \times h \Rightarrow h = 10.579\ldots\) |
# Question 2:
## Part (a)(i)
$a = \frac{v^2}{r} = \frac{12^2}{1.8} = 80$ (ms$^{-2}$) | **B1** [1] |
## Part (a)(ii)
Towards $O$ | **B1FT** [1] | Acceleration in part (a) must be $> 0$ for this mark
## Part (b)(i)
$v = r\omega = 1.8 \times 8 = 14.4$ (ms$^{-1}$) | **B1** [1] |
## Part (b)(ii)
$a = r\omega^2 = 1.8 \times 8^2\ (= 115.2)$ | **M1** | Using formula for radial acceleration and $F = ma$ with tension as only force
$\Rightarrow T = ma = 0.4 \times$ "115.2"
$= 46.1$ (N) | **A1** [2] | 46.08
## Part (c)
$KE = \frac{1}{2} \times 0.4 \times \text{"14.4"}^2$ J $(= 41.472$ J) | **M1** | Finding 'initial' energy
"$41.472$" $= 0.4gh$ | **M1** | ...and equating to final PE; or $\ldots = 0.4g \times 20\sin\theta$; allow sin/cos confusion if expressing $h$ in terms of $\theta$
$\Rightarrow h = 10.579\ldots$ so $\theta = \sin^{-1}(10.579\ldots/20)$ = awrt $31.9(°)$ | **A1** [3] | NB degrees specified in question; answer in radians (e.g. 0.557) scores A0; **SCB2** for $\theta = 31.9°$ from non-energy method e.g. suvat: $14.4^2 = 2 \times 9.8 \times h \Rightarrow h = 10.579\ldots$
---2 A particle $P$ of mass 0.4 kg is attached to one end of a light inextensible string of length 1.8 m . The other end of the string is attached to a fixed point, $O$, on a smooth horizontal plane. Initially, $P$ is moving with a constant speed of $12 \mathrm {~ms} ^ { - 1 }$ in a horizontal circle with $O$ as its centre.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the magnitude of the acceleration of $P$.
\item State the direction of the acceleration of $P$.
A force is now applied to $P$ in such a way that its angular velocity increases. At the instant that the angular velocity reaches $8 \mathrm { rad } \mathrm { s } ^ { - 1 }$, the string breaks.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the speed with which $P$ is moving at the instant that the string breaks.
\item Find the tension in the string at the instant that the string breaks.
After the string has broken $P$ starts to move directly up a smooth slope which is fixed to the plane and inclined at an angle $\theta ^ { \circ }$ above the horizontal. Particle $P$ moves a distance of 20 m up the slope before coming to instantaneous rest.
\end{enumerate}\item Use an energy method to determine the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2024 Q2 [8]}}