| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle on sphere or circular surface |
| Difficulty | Challenging +1.2 This is a standard circular motion problem requiring energy conservation and the condition for losing contact (normal force = 0). Part (i) involves setting up two equations (energy and centripetal force) and solving simultaneously—a routine technique for this topic. Part (ii) requires projectile motion after leaving the sphere, which is straightforward once the velocity at separation is known. While it requires multiple steps and careful application of mechanics principles, the approach is well-established and commonly practiced in Further Maths mechanics. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of energy: \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 + \cos\theta)\), \([v^2 = 2ag(1 - \cos\theta)]\) | B1 | |
| Equate radial forces [may imply \(R = 0\)]: \(mv^2/a = mg\cos\theta + R\) | M1 | |
| Take \(R = 0\) when contact lost: \(mv^2/a = mg\cos\theta\), \([v^2 = ag\cos\theta]\) | A1 | |
| Eliminate \(v^2\), replace \(u^2\) by \(4ag\): \(4mg - 2mg(1 + \cos\theta) = mg\cos\theta\) | M1 | |
| Solve: \(\cos\theta = 2/3\) | A1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find further height \(h_2\) risen: \(h_2 = v^2\sin^2\theta/2g\) | M1 | |
| Substitute for \(v\) and \(\theta\): \(= (2ag/3)(5/9)/2g = 5a/27\) | M1 A1 | |
| Total height above centre \(O\): \(a\cos\theta + h_2 = 23a/27\) | B1 | |
| Subtotal: 4 marks | Total: 9 marks |
## Question 3(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 + \cos\theta)$, $[v^2 = 2ag(1 - \cos\theta)]$ | B1 | |
| Equate radial forces [may imply $R = 0$]: $mv^2/a = mg\cos\theta + R$ | M1 | |
| Take $R = 0$ when contact lost: $mv^2/a = mg\cos\theta$, $[v^2 = ag\cos\theta]$ | A1 | |
| Eliminate $v^2$, replace $u^2$ by $4ag$: $4mg - 2mg(1 + \cos\theta) = mg\cos\theta$ | M1 | |
| Solve: $\cos\theta = 2/3$ | A1 | A.G. |
**Subtotal: 5 marks**
## Question 3(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find further height $h_2$ risen: $h_2 = v^2\sin^2\theta/2g$ | M1 | |
| Substitute for $v$ and $\theta$: $= (2ag/3)(5/9)/2g = 5a/27$ | M1 A1 | |
| Total height above centre $O$: $a\cos\theta + h_2 = 23a/27$ | B1 | |
**Subtotal: 4 marks | Total: 9 marks**
---3 A fixed hollow sphere with centre $O$ has a smooth inner surface of radius $a$. A particle $P$ of mass $m$ is projected horizontally with speed $2 \sqrt { } ( a g )$ from the lowest point of the inner surface of the sphere. The particle loses contact with the inner surface of the sphere when $O P$ makes an angle $\theta$ with the upward vertical.\\
(i) Show that $\cos \theta = \frac { 2 } { 3 }$.\\
(ii) Find the greatest height that $P$ reaches above the level of $O$.
\hfill \mbox{\textit{CAIE FP2 2011 Q3 [9]}}