Question 4 - A-Level Maths

CAIE FP2 2011 June — Question 4 12 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2011
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 2
Type	Vertical circle with peg/obstacle
Difficulty	Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and critical analysis of motion constraints. The first part (showing the tension formula) is standard A-level mechanics using energy and centripetal force. However, the second part requires insight: recognizing that the critical condition occurs when tension becomes zero at the highest point of the new circle about Q, then working backwards through energy conservation with a changing radius. This multi-stage reasoning with a non-obvious constraint pushes it well above average difficulty.
Spec	6.02i Conservation of energy: mechanical energy principle 6.05d Variable speed circles: energy methods 6.05f Vertical circle: motion including free fall

4 A particle $P$ of mass $m$ is suspended from a fixed point $O$ by a light inextensible string of length $a$. When hanging at rest under gravity, $P$ is given a horizontal velocity of magnitude $\sqrt { } ( 3 a g )$ and subsequently moves freely in a vertical circle. Show that the tension $T$ in the string when $O P$ makes an angle $\theta$ with the downward vertical is given by $$T = m g ( 1 + 3 \cos \theta )$$ When the string is horizontal, it comes into contact with a small smooth peg $Q$ which is at the same horizontal level as $O$ and at a distance $x$ from $O$, where $x < a$. Given that $P$ completes a vertical circle about $Q$, find the least possible value of $x$.

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Question 4:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Use conservation of energy at general point: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 - \cos\theta)$	B1
Equate radial forces to find tension $T$: $T = mg\cos\theta + \frac{mv^2}{a}$	B1
Eliminate $v^2$, replace $u^2$ by $3ag$ and simplify: $T = mg(1 + 3\cos\theta)$	M1 A1	A.G.
Use energy to find speed $v$ when $PQ$ horizontal: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga,\ v^2 = ga$	M1 A1
Use energy to find speed $w$ when $P$ above $Q$: $\frac{1}{2}mw^2 = \frac{1}{2}mv^2 - mg(a-x)$	M1 A1
(note that $v$ need not be found): $\left[mw^2 = mg(2x-a)\right]$
EITHER:
Consider tension to find required condition: $T = \frac{mw^2}{(a-x)} - mg \geq (\text{or} >) 0$	M1 A1
Combine to find least value of $x$: $\frac{mg(3x-2a)}{(a-x)} \geq 0$
$x \geq \frac{2a}{3},\ x_{min} = \frac{2a}{3}$	M1 A1
OR:
Find $x$ for which $T$ becomes zero: $\frac{mw^2}{(a-x)} = mg,\ x = \frac{2a}{3}$	(M1 A1)
Show this is least possible value of $x$: $T = \frac{mg(3x-2a)}{(a-x)} \geq 0$ if $x \geq \frac{2a}{3}$	(M1 A1)	[Subtotal: 8]

Total: 12

## Question 4:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use conservation of energy at general point: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 - \cos\theta)$ | B1 | |
| Equate radial forces to find tension $T$: $T = mg\cos\theta + \frac{mv^2}{a}$ | B1 | |
| Eliminate $v^2$, replace $u^2$ by $3ag$ and simplify: $T = mg(1 + 3\cos\theta)$ | M1 A1 | **A.G.** | [Subtotal: 4] |
| Use energy to find speed $v$ when $PQ$ horizontal: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga,\ v^2 = ga$ | M1 A1 | |
| Use energy to find speed $w$ when $P$ above $Q$: $\frac{1}{2}mw^2 = \frac{1}{2}mv^2 - mg(a-x)$ | M1 A1 | |
| (note that $v$ need not be found): $\left[mw^2 = mg(2x-a)\right]$ | | |
| **EITHER:** | | |
| Consider tension to find required condition: $T = \frac{mw^2}{(a-x)} - mg \geq (\text{or} >) 0$ | M1 A1 | |
| Combine to find least value of $x$: $\frac{mg(3x-2a)}{(a-x)} \geq 0$ | | |
| $x \geq \frac{2a}{3},\ x_{min} = \frac{2a}{3}$ | M1 A1 | |
| **OR:** | | |
| Find $x$ for which $T$ becomes zero: $\frac{mw^2}{(a-x)} = mg,\ x = \frac{2a}{3}$ | (M1 A1) | |
| Show this is least possible value of $x$: $T = \frac{mg(3x-2a)}{(a-x)} \geq 0$ if $x \geq \frac{2a}{3}$ | (M1 A1) | [Subtotal: 8] |

**Total: 12**

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Show LaTeX source

4 A particle $P$ of mass $m$ is suspended from a fixed point $O$ by a light inextensible string of length $a$. When hanging at rest under gravity, $P$ is given a horizontal velocity of magnitude $\sqrt { } ( 3 a g )$ and subsequently moves freely in a vertical circle. Show that the tension $T$ in the string when $O P$ makes an angle $\theta$ with the downward vertical is given by

$$T = m g ( 1 + 3 \cos \theta )$$

When the string is horizontal, it comes into contact with a small smooth peg $Q$ which is at the same horizontal level as $O$ and at a distance $x$ from $O$, where $x < a$. Given that $P$ completes a vertical circle about $Q$, find the least possible value of $x$.

\hfill \mbox{\textit{CAIE FP2 2011 Q4 [12]}}

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Q1 8 Q2 11 Q3 12 Q4 12 Q5 6 Q6 7 Q7 8 Q8 11 Q9 11 Q10 EITHER Q10 OR