| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard circular motion problem requiring energy conservation and the condition for string becoming slack (tension = 0). Part (i) involves routine application of these principles with straightforward algebra to reach the given answer. Part (ii) requires projectile motion after the string slacks, which is a common extension. While it requires multiple steps and careful setup, the techniques are well-practiced in Further Maths and the question provides significant scaffolding by giving the answer to show in part (i). |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | 1mv2 = 1mu2 – mga cos θ | |
| 2 2 | M1 | Use conservation of energy to slack point P |
| Answer | Marks |
|---|---|
| mv2/a – mg cos θ = 0 | M1 |
| A1 | Equate tension at P to 0 by using F = ma |
| Answer | Marks | Guidance |
|---|---|---|
| v2 = 2ag – 2ag cos θ = ag cos θ | M1 | Combine to verify cos θ using u = √(2ag) |
| cos θ = 2/3 | A1 |
| Answer | Marks |
|---|---|
| 4(ii) | v = v sin θ = √(2ag/3) (√5/3) |
| Answer | Marks | Guidance |
|---|---|---|
| V | M1 | Find vertical speed v at P |
| Answer | Marks | Guidance |
|---|---|---|
| or 0⋅185 a | M1 A1 | Find height risen above P by considering vertical motion |
| Answer | Marks | Guidance |
|---|---|---|
| or 0⋅852 a | A1 | Find total height risen above level of O |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | 1mv2 = 1mu2 – mga cos θ
2 2 | M1 | Use conservation of energy to slack point P
1
mv2/a – mg cos θ = 0 | M1
A1 | Equate tension at P to 0 by using F = ma
1
A1 if both eqns correct, with m included. AG
v2 = 2ag – 2ag cos θ = ag cos θ | M1 | Combine to verify cos θ using u = √(2ag)
cos θ = 2/3 | A1
5
--- 4(ii) ---
4(ii) | v = v sin θ = √(2ag/3) (√5/3)
V
or v 2 = (10/27) ag
V | M1 | Find vertical speed v at P
V 1
h = v 2/2g = (5/27) a
V
or 0⋅185 a | M1 A1 | Find height risen above P by considering vertical motion
1
h + a cos θ = (23/27) a
or 0⋅852 a | A1 | Find total height risen above level of O
4
Question | Answer | Marks | GuidanceA particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$ and $P$ is held with the string taut and horizontal. The particle $P$ is projected vertically downwards with speed $\sqrt{(2ag)}$ so that it begins to move along a circular path. The string becomes slack when $OP$ makes an angle $\theta$ with the upward vertical through $O$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\cos \theta = \frac{2}{3}$.
[5]
\item Find the greatest height, above the horizontal through $O$, reached by $P$ in its subsequent motion.
[4]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2019 Q4 [9]}}