| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 8 |
| Topic | Circular Motion 2 |
| Type | Vertical circle: speed at specific point |
| Difficulty | Standard +0.3 This is a standard Further Maths vertical circle problem requiring energy conservation and checking if the particle reaches horizontal level. The impulse-momentum start is routine, and the energy equation setup is straightforward. While it involves multiple steps and FM1 content, it follows a well-established template with no novel insights required, making it slightly easier than average for FM1 questions. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = \frac{9.3}{1.5}\) | M1 | Use of impulse = change of momentum |
| Initial KE \(= \frac{1}{2} \times 1.5 \times 6.2^2\) (= 28.83) | B1 | Need not be evaluated |
| PE gain \(= 1.5g \times 2.4(1 - \cos \frac{1}{6}\pi)\) \(= 4.72\ldots\) | M1, A1 | Attempt at \(1.5g \times\) height; allow trig errors |
| \(\frac{1}{2} \times 1.5 \times 6.2^2 = \frac{1}{2} \times 1.5v^2 + 4.72\ldots\) \(v = 5.67\) | M1, A1 [6] | Use of conservation of energy (3 terms); May be implied by correct answer www |
| Answer | Marks |
|---|---|
| Required gain in PE is \(1.5g \times 2.4 = 35.28\) \(35.28 > 28.83\), so not enough energy to reach the level | M1, E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times 1.5 \times 6.2^2 = 1.5g \times 2.4(1 - \cos\theta)\) \(\cos\theta = 0.1828\ldots < 1\) so comes to rest below horizontal | M1, E1 [2] | \(\theta \approx 1.39\) (79.5°) |
## (i)
$u = \frac{9.3}{1.5}$ | M1 | Use of impulse = change of momentum
Initial KE $= \frac{1}{2} \times 1.5 \times 6.2^2$ (= 28.83) | B1 | Need not be evaluated
PE gain $= 1.5g \times 2.4(1 - \cos \frac{1}{6}\pi)$ $= 4.72\ldots$ | M1, A1 | Attempt at $1.5g \times$ height; allow trig errors
$\frac{1}{2} \times 1.5 \times 6.2^2 = \frac{1}{2} \times 1.5v^2 + 4.72\ldots$ $v = 5.67$ | M1, A1 [6] | Use of conservation of energy (3 terms); May be implied by correct answer www
## (ii)
Required gain in PE is $1.5g \times 2.4 = 35.28$ $35.28 > 28.83$, so not enough energy to reach the level | M1, E1 |
**Alternative solution**
$\frac{1}{2} \times 1.5 \times 6.2^2 = 1.5g \times 2.4(1 - \cos\theta)$ $\cos\theta = 0.1828\ldots < 1$ so comes to rest below horizontal | M1, E1 [2] | $\theta \approx 1.39$ (79.5°)
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\includegraphics[max width=\textwidth, alt={}, center]{a1a43547-0a68-4346-884a-0c6d9302cf24-2_473_298_1037_884}
A particle $P$ of mass 1.5 kg is attached to one end of a light inextensible string of length 2.4 m . The other end of the string is attached to a fixed point $O$. The particle is initially at rest directly below $O$. A horizontal impulse of magnitude 9.3 Ns is applied to $P$. In the subsequent motion the string remains taut and makes an angle of $\theta$ radians with the downwards vertical at $O$, as shown in the diagram.\\
(i) Find the speed of $P$ when $\theta = \frac { 1 } { 6 } \pi$.\\
(ii) Determine whether $P$ will reach the same horizontal level as $O$.
\hfill \mbox{\textit{OCR FM1 AS 2017 Q3 [8]}}