| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Standard +0.3 This is a standard energy conservation problem on a smooth curved surface requiring one application of conservation of energy (KE at bottom = PE at highest point) and basic trigonometry to find the height. The second part asks for a modeling assumption which is routine. Slightly above average difficulty due to being Further Maths content, but mechanically straightforward with no novel problem-solving required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial \(\text{KE} = \frac{1}{2} \times 3u^2\) | B1 | Used in solution. Accept \(1.5u^2\) |
| \(\text{PE when at rest} = 3g \times (3.2 - 3.2\cos 60°)\) | M1 | \(24g/5\) or \(47.04\) or \(4.8g\). Attempt to use \(mgh\) to find PE at instant \(P\) comes to rest. Use of suvat is M0. Assuming zero PE level at lowest point; otherwise this mark is for attempting to find the difference between PE when at rest and PE at the bottom |
| \(\frac{3}{2}u^2 = \frac{24g}{5}\) | M1 | Using energy conservation and their expressions to set up an equation in \(u^2\) |
| \(u = 5.6\) | A1 | No need to eliminate negative value explicitly here |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume that air resistance is negligible | B1 | Any sensible specific assumption e.g. no air resistance; \(P\) is a particle/has no dimensions; "No resistance to motion"; "No energy is lost to the surroundings"; "No other forces acting" are insufficient. Ignore references to assumptions stated in question e.g. no friction between \(P\) and the surface, "other forces", "does not come off the surface" |
| [1] |
## Question 2:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial $\text{KE} = \frac{1}{2} \times 3u^2$ | B1 | Used in solution. Accept $1.5u^2$ |
| $\text{PE when at rest} = 3g \times (3.2 - 3.2\cos 60°)$ | M1 | $24g/5$ or $47.04$ or $4.8g$. Attempt to use $mgh$ to find PE at instant $P$ comes to rest. Use of suvat is M0. Assuming zero PE level at lowest point; otherwise this mark is for attempting to find the difference between PE when at rest and PE at the bottom |
| $\frac{3}{2}u^2 = \frac{24g}{5}$ | M1 | Using energy conservation and their expressions to set up an equation in $u^2$ |
| $u = 5.6$ | A1 | No need to eliminate negative value explicitly here |
| **[4]** | | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume that air resistance is negligible | B1 | Any sensible specific assumption e.g. no air resistance; $P$ is a particle/has no dimensions; "No resistance to motion"; "No energy is lost to the surroundings"; "No other forces acting" are insufficient. Ignore references to assumptions stated in question e.g. no friction between $P$ and the surface, "other forces", "does not come off the surface" |
| **[1]** | | |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{b190b8c9-75b0-4ede-913f-cdecdb58180f-2_337_579_842_246}
A small body $P$ of mass 3 kg is at rest at the lowest point of the inside of a smooth hemispherical shell of radius 3.2 m and centre $O$.\\
$P$ is projected horizontally with a speed of $u \mathrm {~ms} ^ { - 1 }$. When $P$ first comes to instantaneous rest $O P$ makes an angle of $60 ^ { \circ }$ with the downward vertical through $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $u$.
\item State one assumption made in modelling the motion of $P$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2023 Q2 [5]}}