| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Radial and transverse acceleration |
| Difficulty | Standard +0.3 This is a straightforward circular motion question requiring standard differentiation and substitution. Students must find radial acceleration (r̈ - rθ̇²) and transverse acceleration (rθ̈ + 2ṙθ̇), but since r is constant (2m), the formulas simplify significantly. Part (i) is a 'show that' which guides students, and part (ii) follows the same method. The calculus is routine (differentiating 1 - cos 2t twice) and requires only direct application of memorized formulas—slightly easier than average for Further Maths. |
| Spec | 6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a_R = 2(\frac{d\theta}{dt})^2 = 2(2\sin 2t)^2 = 2(2\sin \frac{\pi}{3})^2\) | M1 | Verify radial acceleration \(a_R\) at \(t = \frac{\pi}{6}\) from \(r\dot{\theta}^2\) |
| \(= 2(\sqrt{3})^2 = 6\) [m s\(^{-2}\)] (AG) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a_T = 2\frac{d^2\theta}{dt^2} = 2(4\cos 2t) = 2(4\cos \frac{\pi}{3})\) | M1 | Find transverse acceleration \(a_T\) at \(t = \frac{\pi}{6}\) by differentiation |
| \(= 4\) [m s\(^{-2}\)] | A1 |
**Question 1:**
**Part (i):**
$a_R = 2(\frac{d\theta}{dt})^2 = 2(2\sin 2t)^2 = 2(2\sin \frac{\pi}{3})^2$ | M1 | Verify radial acceleration $a_R$ at $t = \frac{\pi}{6}$ from $r\dot{\theta}^2$
$= 2(\sqrt{3})^2 = 6$ [m s$^{-2}$] (AG) | A1 |
**Part (ii):**
$a_T = 2\frac{d^2\theta}{dt^2} = 2(4\cos 2t) = 2(4\cos \frac{\pi}{3})$ | M1 | Find transverse acceleration $a_T$ at $t = \frac{\pi}{6}$ by differentiation
$= 4$ [m s$^{-2}$] | A1 |
---1 A particle $P$ moves along an arc of a circle with centre $O$ and radius 2 m . At time $t$ seconds, the angle POA is $\theta$, where $\theta = 1 - \cos 2 t$, and $A$ is a fixed point on the arc of the circle.\\
(i) Show that the magnitude of the radial component of the acceleration of $P$ when $t = \frac { 1 } { 6 } \pi$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{2aaf3493-6509-4668-91a2-9f4708bbbb58-03_65_1573_488_324}\\
(ii) Find the magnitude of the transverse component of the acceleration of $P$ when $t = \frac { 1 } { 6 } \pi$.\\
\hfill \mbox{\textit{CAIE FP2 2019 Q1 [4]}}