| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a multi-part vertical circle problem requiring energy conservation, collision mechanics, and finding when tension becomes zero. While it involves several steps and combines multiple mechanics concepts (circular motion, energy, momentum conservation), each individual part follows standard Further Maths mechanics techniques without requiring novel insight. The 'show that' in part (i) and the sequential build-up through parts (ii)-(iv) are typical of FM2 circular motion questions, making it moderately above average difficulty but well within expected syllabus material. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.05d Variable speed circles: energy methods |
| Answer | Marks |
|---|---|
| Find tension at top from \(F = ma\) vertically: \(T = \frac{mu^2}{a} - mg\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Find \(T\) at top by taking \(\theta = 0\): \(T = \frac{mu^2}{a} - mg\) | (B1) | |
| Find \(u_{\min}\) by requiring \(T \geq 0\) at top [or \(T > 0\)]: \(\frac{u^2}{a} - g \geq 0\) so \(u_{\min} = \sqrt{ag}\) | B1 | A.G. |
| Answer | Marks |
|---|---|
| \(v^2 = ag + 4ag,\quad v = \sqrt{5ag}\) | M1, A1 |
| Answer | Marks |
|---|---|
| *or* \(\frac{4}{5}\sqrt{5ag}\) AEF | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([w^2 = ag(6/5 - 2\cos\theta)]\) | M1 A1 | (condone \(m\) instead of \(m'\) here since cancels out) |
| Answer | Marks |
|---|---|
| \(\frac{1}{2}m'w^2 = \frac{1}{2}mu^2 + mga(1 - \cos\theta) - \frac{1}{4}mga(1 + \cos\theta)\) | (B1) |
| Answer | Marks |
|---|---|
| Find tension \(T'\) there by using \(F = ma\) radially: \(T' = \frac{m'w^2}{a} - m'g\cos\theta\) | B1 |
| Eliminate \(w^2\): \(= \frac{m'V^2}{a} - m'g(2 + 3\cos\theta)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| *or* \(\frac{3mg}{2} - \frac{15}{4}mg\cos\theta\) | A1 | AEF |
| Answer | Marks | Guidance |
|---|---|---|
| Find \(\cos\theta\) when string becomes slack from \(T' = 0\): \(\cos\theta = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}\) *or* \(0.4\) | M1 A1 | S.R. Allow if found from \(T' = mg(6/5 - 3\cos\theta)\) |
## Question 4:
**Part 1 - Tension at top (EITHER method):**
Find tension at top from $F = ma$ vertically: $T = \frac{mu^2}{a} - mg$ | B1 |
**OR method:**
Use energy at angle $\theta$ to upward vertical: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1 - \cos\theta)$
Find tension $T$ by using $F = ma$ radially: $T = \frac{mv^2}{a} - mg\cos\theta$
Eliminate $v^2$: $= \frac{mu^2}{a} + mg(2 - 3\cos\theta)$
Find $T$ at top by taking $\theta = 0$: $T = \frac{mu^2}{a} - mg$ | (B1) |
Find $u_{\min}$ by requiring $T \geq 0$ at top [or $T > 0$]: $\frac{u^2}{a} - g \geq 0$ so $u_{\min} = \sqrt{ag}$ | B1 | A.G.
**Total: 2 marks**
---
**Part 2 - Speed at bottom:**
Find $v$ at bottom from conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mg \times 2a$
$v^2 = ag + 4ag,\quad v = \sqrt{5ag}$ | M1, A1 |
**Total: 2 marks**
---
**Part 3 - New speed after mass added:**
Find new speed $V$ from conservation of momentum: $m'V = mv$ with $m' = m + \frac{1}{4}m$
$V = \frac{4v}{5} = \frac{4\sqrt{ag/5}}{1}$
*or* $\frac{4}{5}\sqrt{5ag}$ AEF | M1, A1 |
**Total: 2 marks**
---
**Part 4 - Speed at angle $\theta$:**
Find $w^2$ at angle $\theta$ from conservation of energy: $\frac{1}{2}m'w^2 = \frac{1}{2}m'V^2 - m'ga(1 + \cos\theta)$
$[w^2 = ag(6/5 - 2\cos\theta)]$ | M1 A1 | (condone $m$ instead of $m'$ here since cancels out)
**Total: 2 marks**
---
**S.R. Invalid energy method (max 2/5):**
[gives $T' = \frac{5mg}{4}(2 - 3\cos\theta)$]
$\frac{1}{2}m'w^2 = \frac{1}{2}mu^2 + mga(1 - \cos\theta) - \frac{1}{4}mga(1 + \cos\theta)$ | (B1) |
---
**Part 5 - Tension $T'$:**
Find tension $T'$ there by using $F = ma$ radially: $T' = \frac{m'w^2}{a} - m'g\cos\theta$ | B1 |
Eliminate $w^2$: $= \frac{m'V^2}{a} - m'g(2 + 3\cos\theta)$ | A1 |
Substitute for $m'$ and $V$: $= \frac{5mg}{4}(6/5 - 3\cos\theta)$
*or* $\frac{3mg}{2} - \frac{15}{4}mg\cos\theta$ | A1 | AEF
**Total: 5 marks**
---
**Part 6 - String becomes slack:**
Find $\cos\theta$ when string becomes slack from $T' = 0$: $\cos\theta = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$ *or* $0.4$ | M1 A1 | **S.R.** Allow if found from $T' = mg(6/5 - 3\cos\theta)$
**Total: 2 marks**
---4 A 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 $O$. When $P$ is hanging at rest vertically below $O$, it is projected horizontally. In the subsequent motion $P$ completes a vertical circle. The speed of $P$ when it is at its highest point is $u$.\\
(i) Show that the least possible value of $u$ is $\sqrt { } ( a g )$.\\
It is now given that $u = \sqrt { } ( a g )$. When $P$ passes through the lowest point of its path, it collides with, and coalesces with, a stationary particle of mass $\frac { 1 } { 4 } m$.\\
(ii) Find the speed of the combined particle immediately after the collision.\\
In the subsequent motion, when $O P$ makes an angle $\theta$ with the upward vertical the tension in the string is $T$.\\
(iii) Find an expression for $T$ in terms of $m , g$ and $\theta$.\\
(iv) Find the value of $\cos \theta$ when the string becomes slack.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q4 [13]}}