| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Standard +0.8 This is a classic vertical circular motion problem requiring energy conservation and Newton's second law in polar coordinates. While the derivation of the reaction force formula is standard Further Maths fare, the problem requires careful handling of multiple concepts (energy, centripetal force, projectile motion after leaving the sphere). The second part involving finding the impact velocity adds computational complexity. It's moderately challenging but follows established patterns for FP2 mechanics questions. |
| Spec | 3.02h Motion under gravity: vector form6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1-\cos\theta)\) | B1 | Conservation of energy (B0 for \(v^2 = \ldots\)) |
| \(R = mg\cos\theta - mv^2/a\) | B1 | Equate radial forces |
| \(R = mg(3\cos\theta - 2) - mu^2/a\) A.G. | M1 A1 | Eliminate \(v\) to find \(R\) |
| \(\cos\alpha = \frac{1}{3}(\frac{1}{4} + 2) = \frac{3}{4}\) | M1 A1 | Find \(\cos\alpha\) when \(R = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_1^2 = \frac{1}{4}ga + \frac{1}{2}ga = \frac{3}{4}ga\) | B1 | Find \(v^2\) at this point |
| \(v_2^2 = v_1^2\sin^2\alpha + 2ga(1 + \cos\alpha)\) | M1 A1 | Find vertical component of \(v_1\) at plane (using \(u\) in place of \(v_1\) can earn M1 A0) |
| \(= (21/64 + 7/2)\,ga\) | A1 | |
| \(v_2 = \sqrt{245ga/64}\) or \((7/8)\sqrt{5ga}\) or \(1.96\sqrt{ga}\) or \(6.19\sqrt{a}\) | A1 | AEF |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1-\cos\theta)$ | B1 | Conservation of energy (B0 for $v^2 = \ldots$) |
| $R = mg\cos\theta - mv^2/a$ | B1 | Equate radial forces |
| $R = mg(3\cos\theta - 2) - mu^2/a$ **A.G.** | M1 A1 | Eliminate $v$ to find $R$ |
| $\cos\alpha = \frac{1}{3}(\frac{1}{4} + 2) = \frac{3}{4}$ | M1 A1 | Find $\cos\alpha$ when $R = 0$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_1^2 = \frac{1}{4}ga + \frac{1}{2}ga = \frac{3}{4}ga$ | B1 | Find $v^2$ at this point |
| $v_2^2 = v_1^2\sin^2\alpha + 2ga(1 + \cos\alpha)$ | M1 A1 | Find vertical component of $v_1$ at plane (using $u$ in place of $v_1$ can earn M1 A0) |
| $= (21/64 + 7/2)\,ga$ | A1 | |
| $v_2 = \sqrt{245ga/64}$ or $(7/8)\sqrt{5ga}$ or $1.96\sqrt{ga}$ or $6.19\sqrt{a}$ | A1 | AEF |
**Total: 11 marks**
---4 A smooth sphere, with centre $O$ and radius $a$, has its lowest point fixed on a horizontal plane. A particle $P$ of mass $m$ is projected horizontally with speed $u$ from the highest point on the outer surface of the sphere. In the subsequent motion, $O P$ makes an angle $\theta$ with the upward vertical through $O$. Show that, while $P$ remains in contact with the sphere, the magnitude of the reaction of the sphere on\\
$P$ is $m g ( 3 \cos \theta - 2 ) - \frac { m u ^ { 2 } } { a }$.
The particle loses contact with the surface of the sphere when $\theta = \alpha$. Given that $u = \frac { 1 } { 2 } \sqrt { } ( g a )$, find\\
(i) $\cos \alpha$,\\
(ii) the vertical component of the velocity of $P$ as it strikes the horizontal plane.
\hfill \mbox{\textit{CAIE FP2 2012 Q4 [11]}}