Question 10 EITHER - A-Level Maths

CAIE FP2 2011 June — Question 10 EITHER

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2011
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Simple Harmonic Motion
Type	Complete motion cycle with slack phase
Difficulty	Challenging +1.8 This is a challenging Further Maths mechanics problem requiring understanding of SHM with a slack phase. The first part is standard bookwork showing the period formula. The second part demands recognizing that motion splits into two phases (SHM while taut, then free fall under gravity when slack), finding the transition point, and combining times from both phases using energy considerations and SHM equations. This requires significant problem-solving insight beyond routine SHM questions.
Spec	4.10f Simple harmonic motion: x'' = -omega^2 x 6.02h Elastic PE: 1/2 k x^2 6.02i Conservation of energy: mechanical energy principle

One end of a light elastic string is attached to a fixed point \(O\). A particle of mass \(m\) is attached to the other end of the string and hangs freely under gravity. In the equilibrium position, the extension of the string is \(d\). Show that the period of small vertical oscillations about the equilibrium position is \(2 \pi \sqrt { } \left( \frac { d } { g } \right)\). The particle is now pulled down and released from rest at a distance \(2 d\) below the equilibrium position. Given that the particle does not reach \(O\) in the subsequent motion, show that the time taken until the particle first comes to instantaneous rest is \(\left( \sqrt { } 3 + \frac { 2 } { 3 } \pi \right) \sqrt { } \left( \frac { d } { g } \right)\).

Show mark scheme Show mark scheme source

Question 10:

Part (a)

Answer	Marks	Guidance
Answer	Mark	Guidance
Resolve vertically at equilibrium: \(\lambda d/a = mg\) \([\lambda/a = mg/d]\)	B1
Newton's Law at general point: \(m\,d^2x/dt^2 = mg - \lambda(d+x)/a\)
\([or\ -mg + \lambda(d-x)/a]\)	M1 A1
Simplify: \(d^2x/dt^2 = -(\lambda/ma)\,x\) or \(-(g/d)\,x\)	A1
Stating without derivation (max 3/5)	(B1)
Find period \(T\) using SHM \(\omega = \sqrt{g/d}\): \(T = [2\pi\sqrt{(ma/\lambda)}] = 2\pi\sqrt{(d/g)}\) A.G.	B1	5 marks total
Use SHM formula with amplitude \(2d\): \(x = 2d\cos(\omega t)\) \([or\ \sin]\)	M1
Find time \(t_1\) to string becoming slack: \(t_1 = (1/\omega)\cos^{-1}(-1/2)\)
or \(T/4 + (1/\omega)\sin^{-1}(1/2)\)	M1 A1
A.G. \(t_1 = (1/\omega)\,2\pi/3 = (2\pi/3)\sqrt{(d/g)}\)	A1
Find speed \(v\) when string slack: \(v = \omega\sqrt{(4d^2 - d^2)} = \omega d\sqrt{3}\) or \(\sqrt{(3dg)}\)	M1 A1
Find further time \(t_2\) to instantaneous rest: \(t_2 = v/g\)	B1
A.G. \(t_2 = \sqrt{(3dg)}/g = \sqrt{3}\sqrt{(d/g)}\)	M1 A1	9 marks total

Part (b)

Answer	Marks	Guidance
Answer	Mark	Guidance
Find mean and variance: \(262/200 = 1.31\) and \((586 - 262^2/200)/200 = 1.21\)	M1 A1
Values close, so distribution appropriate	B1	3 marks total; AEF, needs values approx correct

Part (b)(i)

Answer	Marks	Guidance
Answer	Mark	Guidance
A.G.: \(p = 200(1.31^2/2)e^{-1.31} = 46.304\)	B1
\(q = 200(1.31^3/6)e^{-1.31} = 20.2\)	B1	2 marks total; can use \(\sum E_i = 200\)

Part (b)(ii)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(H_0\): Poisson fits data	B1	State at least null hypothesis
Combine last 3 cells since exp. value \(< 5\): \(O: \ldots\ 5\)
\(E: \ldots\ 8.82\)	*M1 A1
\(\chi^2 = 5.54\)	M1 A1	Calculate \(\chi^2\) to 2 dp; A1 dep *M1
\(\chi^2_{3,\,0.9} = 6.251\)	M1 A1	Compare consistent tabular value to 2 dp
\([\chi^2_{4,\,0.9} = 7.779,\ \chi^2_{5,\,0.9} = 9.236]\)		A1 dep *M1
Accept \(H_0\) if \(\chi^2 <\) tabular value	M1	Valid method for conclusion
\(5.54 < 6.25\) so Poisson does fit	A1	9 marks total; A.E.F., needs correct values

# Question 10:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolve vertically at equilibrium: $\lambda d/a = mg$ $[\lambda/a = mg/d]$ | B1 | |
| Newton's Law at general point: $m\,d^2x/dt^2 = mg - \lambda(d+x)/a$ | | |
| $[or\ -mg + \lambda(d-x)/a]$ | M1 A1 | |
| Simplify: $d^2x/dt^2 = -(\lambda/ma)\,x$ *or* $-(g/d)\,x$ | A1 | |
| Stating without derivation (max 3/5) | (B1) | |
| Find period $T$ using SHM $\omega = \sqrt{g/d}$: $T = [2\pi\sqrt{(ma/\lambda)}] = 2\pi\sqrt{(d/g)}$ **A.G.** | B1 | 5 marks total |
| Use SHM formula with amplitude $2d$: $x = 2d\cos(\omega t)$ $[or\ \sin]$ | M1 | |
| Find time $t_1$ to string becoming slack: $t_1 = (1/\omega)\cos^{-1}(-1/2)$ | | |
| *or* $T/4 + (1/\omega)\sin^{-1}(1/2)$ | M1 A1 | |
| **A.G.** $t_1 = (1/\omega)\,2\pi/3 = (2\pi/3)\sqrt{(d/g)}$ | A1 | |
| Find speed $v$ when string slack: $v = \omega\sqrt{(4d^2 - d^2)} = \omega d\sqrt{3}$ *or* $\sqrt{(3dg)}$ | M1 A1 | |
| Find further time $t_2$ to instantaneous rest: $t_2 = v/g$ | B1 | |
| **A.G.** $t_2 = \sqrt{(3dg)}/g = \sqrt{3}\sqrt{(d/g)}$ | M1 A1 | 9 marks total |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Find mean and variance: $262/200 = 1.31$ *and* $(586 - 262^2/200)/200 = 1.21$ | M1 A1 | |
| Values close, so distribution appropriate | B1 | 3 marks total; AEF, needs values approx correct |

## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| **A.G.:** $p = 200(1.31^2/2)e^{-1.31} = 46.304$ | B1 | |
| $q = 200(1.31^3/6)e^{-1.31} = 20.2$ | B1 | 2 marks total; can use $\sum E_i = 200$ |

## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Poisson fits data | B1 | State at least null hypothesis |
| Combine last 3 cells since exp. value $< 5$: $O: \ldots\ 5$ | | |
| $E: \ldots\ 8.82$ | *M1 A1 | |
| $\chi^2 = 5.54$ | M1 A1 | Calculate $\chi^2$ to 2 dp; A1 dep *M1 |
| $\chi^2_{3,\,0.9} = 6.251$ | M1 A1 | Compare consistent tabular value to 2 dp |
| $[\chi^2_{4,\,0.9} = 7.779,\ \chi^2_{5,\,0.9} = 9.236]$ | | A1 dep *M1 |
| Accept $H_0$ if $\chi^2 <$ tabular value | M1 | Valid method for conclusion |
| $5.54 < 6.25$ so Poisson does fit | A1 | 9 marks total; A.E.F., needs correct values |

Show LaTeX source

One end of a light elastic string is attached to a fixed point $O$. A particle of mass $m$ is attached to the other end of the string and hangs freely under gravity. In the equilibrium position, the extension of the string is $d$. Show that the period of small vertical oscillations about the equilibrium position is $2 \pi \sqrt { } \left( \frac { d } { g } \right)$.

The particle is now pulled down and released from rest at a distance $2 d$ below the equilibrium position. Given that the particle does not reach $O$ in the subsequent motion, show that the time taken until the particle first comes to instantaneous rest is $\left( \sqrt { } 3 + \frac { 2 } { 3 } \pi \right) \sqrt { } \left( \frac { d } { g } \right)$.

\hfill \mbox{\textit{CAIE FP2 2011 Q10 EITHER}}

This paper (11 questions)

View full paper

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