| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle in circular tube or on wire |
| Difficulty | Standard +0.3 This is a standard vertical circular motion energy conservation problem requiring two straightforward applications of energy conservation. Part (a) uses conservation of energy with a given angle, and part (b) checks if the bead reaches the top. Both are routine calculations 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Initial (kinetic) energy \(= \frac{1}{2} \times m \times 8.4^2\) | B1 | \(35.28m\); \(m\) may be implied |
| Energy at \(0.8\) rad \(= \frac{1}{2}mv^2 + m \times 9.8 \times 2.5(1 - \cos 0.8)\) = Initial energy | M1 | Attempt to find \(KE + PE\) at \(0.8\) rad (or \(45.8°\)) and equate to initial kinetic energy (KE must use correct formula); NB \(\Delta h = 0.7582...\); \(\frac{1}{2}mv^2 + 7.4306...m\); Or subtract PE from initial KE (to give final KE) (Final KE is \(27.849...m\)) |
| \(v^2 = 55.698... \Rightarrow\) speed is \(7.46 \text{ m s}^{-1}\) | A1 | SC1 for use of constant acceleration without justification |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Minimum energy to reach top \(= m \times 9.8 \times (2 \times 2.5)\) | M1 | Or attempt to find angle when \(v = 0\): \(35.28m = 24.5m(1 - \cos\theta) + \frac{1}{2}m(0)^2\); Condone missing \(m\); Or attempt to find \(h\) when \(v = 0\) \(\left(h = \frac{35.28}{g}\right)\) |
| \(= 49m\) | A1 | \(\theta = 2.03\) rad or \(116°\); \(h = 3.6\) |
| \(49m > 35.28m\) so insufficient energy to reach top | A1ft | Comparison between their numerical multiples of \(m\) (\(m\) could be missing); Allow \(\neq\); and consistent ft conclusion; or comparison of their angle with \(2\pi\) or \(180°\); Or show that \(h = 3.6 < 5\); or show that \(v^2 = -27.44 < 0\) (is not valid) |
| [3] |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Initial (kinetic) energy $= \frac{1}{2} \times m \times 8.4^2$ | **B1** | $35.28m$; $m$ may be implied |
| Energy at $0.8$ rad $= \frac{1}{2}mv^2 + m \times 9.8 \times 2.5(1 - \cos 0.8)$ = Initial energy | **M1** | Attempt to find $KE + PE$ at $0.8$ rad (or $45.8°$) and equate to initial kinetic energy (KE must use correct formula); NB $\Delta h = 0.7582...$; $\frac{1}{2}mv^2 + 7.4306...m$; Or subtract PE from initial KE (to give final KE) (Final KE is $27.849...m$) |
| $v^2 = 55.698... \Rightarrow$ speed is $7.46 \text{ m s}^{-1}$ | **A1** | SC1 for use of constant acceleration without justification |
| **[3]** | | |
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### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Minimum energy to reach top $= m \times 9.8 \times (2 \times 2.5)$ | **M1** | Or attempt to find angle when $v = 0$: $35.28m = 24.5m(1 - \cos\theta) + \frac{1}{2}m(0)^2$; Condone missing $m$; Or attempt to find $h$ when $v = 0$ $\left(h = \frac{35.28}{g}\right)$ |
| $= 49m$ | **A1** | $\theta = 2.03$ rad or $116°$; $h = 3.6$ |
| $49m > 35.28m$ so insufficient energy to reach top | **A1ft** | Comparison between their numerical multiples of $m$ ($m$ could be missing); Allow $\neq$; and consistent ft conclusion; or comparison of their angle with $2\pi$ or $180°$; Or show that $h = 3.6 < 5$; or show that $v^2 = -27.44 < 0$ (is not valid) |
| **[3]** | | |1\\
\includegraphics[max width=\textwidth, alt={}, center]{74bada9e-60cf-4ed4-abd0-4be155b7cf81-2_533_424_402_246}
A smooth wire is shaped into a circle of radius 2.5 m which is fixed in a vertical plane with its centre at a point $O$. A small bead $B$ is threaded onto the wire. $B$ is held with $O B$ vertical and is then projected horizontally with an initial speed of $8.4 \mathrm {~ms} ^ { - 1 }$ (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$ at the instant when $O B$ makes an angle of 0.8 radians with the downward vertical through $O$.
\item Determine whether $B$ has sufficient energy to reach the point on the wire vertically above $O$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2019 Q1 [6]}}