| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rod or pendulum circular motion energy |
| Difficulty | Standard +0.3 This is a standard vertical circular motion problem using energy conservation. Part (a) requires straightforward application of energy conservation to find speed at a given angle. Part (b) extends this to find when the particle comes to rest. Both parts follow routine procedures taught in Further Mechanics, with clear setup and no novel insights required—slightly easier than average A-level standard. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial Energy \(= \frac{1}{2} \times 5.6 \times 5^2\) | B1 | Using \(u\) to find \(P\)'s initial (kinetic) energy \((=70\) J\()\) |
| When \(\theta = \frac{1}{4}\pi\), \(P\)'s PE \(= 5.6g \times \left(2.1 - 2.1\cos\frac{1}{4}\pi\right)\) | *M1 | Finding \(P\)'s PE when \(\theta = \frac{1}{4}\pi\) \((= 33.8\) J\()\); allow 1 slip, but PE must not become negative |
| So conservation of energy: \(\frac{1}{2} \times 5.6v^2 + 5.6g \times \left(2.1 - 2.1\cos\frac{1}{4}\pi\right) = 70\) | M1 dep | Finding expression for \(P\)'s energy \((KE + PE)\) and equating to initial energy |
| \(v =\) awrt \(3.6\) | A1 | Final speed must be \(< 3.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(v = 0\), \(P\)'s energy \(= 5.6g \times (2.1 - 2.1\cos\theta) = 70\) | M1 | Finding expression for \(P\)'s final (potential) energy and equating to initial energy |
| \(\theta =\) awrt \(1.17\) rads | A1 | cao; allow in degrees awrt \(66.9°\) |
## Question 3:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial Energy $= \frac{1}{2} \times 5.6 \times 5^2$ | B1 | Using $u$ to find $P$'s initial (kinetic) energy $(=70$ J$)$ |
| When $\theta = \frac{1}{4}\pi$, $P$'s PE $= 5.6g \times \left(2.1 - 2.1\cos\frac{1}{4}\pi\right)$ | *M1 | Finding $P$'s PE when $\theta = \frac{1}{4}\pi$ $(= 33.8$ J$)$; allow 1 slip, but PE must not become negative |
| So conservation of energy: $\frac{1}{2} \times 5.6v^2 + 5.6g \times \left(2.1 - 2.1\cos\frac{1}{4}\pi\right) = 70$ | M1 dep | Finding expression for $P$'s energy $(KE + PE)$ and equating to initial energy |
| $v =$ awrt $3.6$ | A1 | Final speed must be $< 3.6$ |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $v = 0$, $P$'s energy $= 5.6g \times (2.1 - 2.1\cos\theta) = 70$ | M1 | Finding expression for $P$'s final (potential) energy and equating to initial energy |
| $\theta =$ awrt $1.17$ rads | A1 | cao; allow in degrees awrt $66.9°$ |
---3 A particle $P$ of mass 5.6 kg is attached to one end of a light rod of length 2.1 m . The other end of the rod is freely hinged to a fixed point $O$.
The particle is initially at rest directly below $O$. It is then projected horizontally with speed $5 \mathrm {~ms} ^ { - 1 }$. In the subsequent motion, the angle between the rod and the downward vertical at $O$ is denoted by $\theta$ radians, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{0501e5a4-2137-4e7d-98ff-2ee81941cbf3-2_499_312_1905_244}
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when $\theta = \frac { 1 } { 4 } \pi$.
\item Find the value of $\theta$ when $P$ first comes to instantaneous rest.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2020 Q3 [6]}}