CAIE FP2 2010 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRotation about fixed axis: angular acceleration and velocity
DifficultyChallenging +1.2 This is a standard rotational dynamics problem requiring application of Newton's second law to the particle, torque equation to the disc, and the constraint that linear acceleration equals angular acceleration times radius. While it involves multiple connected concepts (moments of inertia, torque, constraint equations), it follows a well-established solution pattern taught in Further Maths mechanics with no novel insight required. The setup is clear and the method is routine for this topic.
Spec3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods

1 A uniform disc with centre \(O\) has mass \(m\) and radius \(a\). It is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through \(O\). One end of a light inextensible string is attached to a point on the circumference and is wrapped several times round the circumference. A particle \(P\), of mass \(2 m\), is attached to the free end of the string and the disc is held at rest with \(P\) hanging freely. The system is released from rest. Assuming that resistances may be neglected, find the acceleration of \(P\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation of motion for \(P\): \(2mg - T = 2mA\)M1
Equation of motion for disc: \(Ta = I_{disc}A/a\)M1
\((2mga = I_{disc}A/a\) can earn this M1; usually 1/5)
Substitute \(\frac{1}{2}ma^2\) for \(MI_{disc}\) of disc: \(T = \frac{1}{2}mA\)A1
Eliminate \(T\) to find accel. \(A\): \(2mg = \frac{5}{2}mA\)M1
\(A = 4g/5\) or \(8\)A1 Total: 5
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $P$: $2mg - T = 2mA$ | M1 | |
| Equation of motion for disc: $Ta = I_{disc}A/a$ | M1 | |
| $(2mga = I_{disc}A/a$ can earn this M1; usually 1/5) | | |
| Substitute $\frac{1}{2}ma^2$ for $MI_{disc}$ of disc: $T = \frac{1}{2}mA$ | A1 | |
| Eliminate $T$ to find accel. $A$: $2mg = \frac{5}{2}mA$ | M1 | |
| $A = 4g/5$ or $8$ | A1 | **Total: 5** |

---
1 A uniform disc with centre $O$ has mass $m$ and radius $a$. It is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through $O$. One end of a light inextensible string is attached to a point on the circumference and is wrapped several times round the circumference. A particle $P$, of mass $2 m$, is attached to the free end of the string and the disc is held at rest with $P$ hanging freely. The system is released from rest. Assuming that resistances may be neglected, find the acceleration of $P$.

\hfill \mbox{\textit{CAIE FP2 2010 Q1 [5]}}