CAIE FP2 2013 November — Question 3 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeSmall oscillations: rigid body compound pendulum
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem requiring application of rotational dynamics (moment of inertia, torque) and energy methods. While it involves multiple concepts (rotation, tension, resistance), the approach is methodical: use energy conservation or equations of motion with known formulas for disc MI. The given final condition makes it straightforward to set up equations. More challenging than typical A-level due to rotational mechanics, but follows standard FM2 patterns without requiring novel insight.
Spec6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work
3 \includegraphics[max width=\textwidth, alt={}, center]{b486decd-75b8-44bd-889f-2472f1163871-2_570_419_1539_863} A uniform disc, of mass 2 kg and radius 0.2 m , is free to rotate in a vertical plane about a smooth horizontal axis through its centre. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a small block of mass 4 kg , which hangs freely (see diagram). The system is released from rest. During the subsequent motion, the block experiences a constant resistance to its motion, of magnitude \(R \mathrm {~N}\). Given that the angular speed of the disc after it has turned through 2 radians is \(5 \mathrm { rad } \mathrm { s } ^ { - 1 }\), find \(R\) and the tension in the string.
[0pt] [9]
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(5^2 = 2 \times \frac{d\omega}{dt} \times 2\)M1 A1 Use \(\omega^2 = 2\frac{d\omega}{dt}\theta\) to find \(\frac{d\omega}{dt}\)
\(\frac{d\omega}{dt} = 25/4\) or \(6.25\) \([a = 1.25]\)A1
\(T \times 0.2 = (\frac{1}{2} \times 2 \times 0.2^2)\frac{d\omega}{dt}\)M1 A1 Find eqn of motion for disc
\(T = 0.2\frac{d\omega}{dt} = 5/4\) or \(1.25\)A1 Substitute to find \(T\)
\(4g - R - T = 4 \times 0.2\frac{d\omega}{dt}\) or \(4a\)M1 A1 Find eqn of motion for block
\(R = 4g - T - 0.8\frac{d\omega}{dt} = 135/4\) or \(33.7_{[5]}\)A1 Substitute to find \(R\)
S.R.: B1 only for \((4g-R) \times 0.2 = 0.2^2\frac{d\omega}{dt}\)
*OR* (energy method): \(\frac{1}{2}(0.2)^2\omega^2 + \frac{1}{2}(4)(0.2\omega)^2 = (4g-R) \times 0.2\theta\)(M1 A1) Use energy to find \(R\) without \(\frac{d\omega}{dt}\)
\(\frac{1}{2} + 2 = (4g-R) \times 0.4\)
\(R = 4g - 5/4 - 5 = 135/4\) or \(33.7_{[5]}\)(A1)
Find 2 eqns of motion (or energy for block), eliminate \(\frac{d\omega}{dt}\) to find \(T\):\((2 \times\) M1 A1)
\(4g - R - T = 4T\), \(T = 1.25\)(M1 A1)
Total: 9 + [9] marks 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $5^2 = 2 \times \frac{d\omega}{dt} \times 2$ | M1 A1 | Use $\omega^2 = 2\frac{d\omega}{dt}\theta$ to find $\frac{d\omega}{dt}$ |
| $\frac{d\omega}{dt} = 25/4$ or $6.25$ $[a = 1.25]$ | A1 | |
| $T \times 0.2 = (\frac{1}{2} \times 2 \times 0.2^2)\frac{d\omega}{dt}$ | M1 A1 | Find eqn of motion for disc |
| $T = 0.2\frac{d\omega}{dt} = 5/4$ or $1.25$ | A1 | Substitute to find $T$ |
| $4g - R - T = 4 \times 0.2\frac{d\omega}{dt}$ or $4a$ | M1 A1 | Find eqn of motion for block |
| $R = 4g - T - 0.8\frac{d\omega}{dt} = 135/4$ or $33.7_{[5]}$ | A1 | Substitute to find $R$ |
| **S.R.:** B1 only for $(4g-R) \times 0.2 = 0.2^2\frac{d\omega}{dt}$ | | |
| *OR* (energy method): $\frac{1}{2}(0.2)^2\omega^2 + \frac{1}{2}(4)(0.2\omega)^2 = (4g-R) \times 0.2\theta$ | (M1 A1) | Use energy to find $R$ without $\frac{d\omega}{dt}$ |
| $\frac{1}{2} + 2 = (4g-R) \times 0.4$ | | |
| $R = 4g - 5/4 - 5 = 135/4$ or $33.7_{[5]}$ | (A1) | |
| Find 2 eqns of motion (or energy for block), eliminate $\frac{d\omega}{dt}$ to find $T$: | $(2 \times$ M1 A1) | |
| $4g - R - T = 4T$, $T = 1.25$ | (M1 A1) | |
| **Total: 9 + [9] marks** | | |

---
3\\
\includegraphics[max width=\textwidth, alt={}, center]{b486decd-75b8-44bd-889f-2472f1163871-2_570_419_1539_863}

A uniform disc, of mass 2 kg and radius 0.2 m , is free to rotate in a vertical plane about a smooth horizontal axis through its centre. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a small block of mass 4 kg , which hangs freely (see diagram). The system is released from rest. During the subsequent motion, the block experiences a constant resistance to its motion, of magnitude $R \mathrm {~N}$. Given that the angular speed of the disc after it has turned through 2 radians is $5 \mathrm { rad } \mathrm { s } ^ { - 1 }$, find $R$ and the tension in the string.\\[0pt]
[9]

\hfill \mbox{\textit{CAIE FP2 2013 Q3 [9]}}
This paper (10 questions)
View full paper
Q1 6 Q2 8 Q3 9 Q4 10 Q5 11 Q6 6 Q7 7 Q9 11 Q11 EITHER Q11 OR