Edexcel M5 2006 January — Question 5 10 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2006
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRotation about fixed axis: impulsive impact and subsequent motion
DifficultyChallenging +1.8 This M5 question requires multiple sophisticated steps: applying conservation of angular momentum for an off-center collision, using the parallel axis theorem for moment of inertia, setting up energy conservation with changing gravitational PE for both disc and particle, and solving the resulting equation. While the individual techniques are standard for Further Maths mechanics, the combination and careful tracking of geometry through 90° rotation makes this substantially harder than typical A-level questions.
Spec1.10g Problem solving with vectors: in geometry3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

5. A uniform circular disc has mass \(m\) and radius \(a\). The disc can rotate freely about an axis that is in the same plane as the disc and tangential to the disc at a point \(A\) on its circumference. The disc hangs at rest in equilibrium with its centre \(O\) vertically below \(A\). A particle \(P\) of mass \(m\) is moving horizontally and perpendicular to the disc with speed \(\sqrt { } ( k g a )\), where \(k\) is a constant. The particle then strikes the disc at \(O\) and adheres to it at \(O\). Given that the disc rotates through an angle of \(90 ^ { \circ }\) before first coming to instantaneous rest, find the value of \(k\).
(Total 10 marks)

AnswerMarks
MI of disc all diameter = \(\frac{1}{4}ma^2\)M1
MI of disc about axis = \(\frac{4}{9}ma^2 + ma^2 = \frac{13}{9}ma^2\)M1, A1
CAM: \(m.a.\sqrt{kg.a} = \left(\frac{13}{9}ma^2 + ma^2\right)\omega\)M1, A1
\(\omega = \frac{4}{9}\sqrt{\frac{kg}{a}}\)A1
Energy: \(\frac{1}{2}.\frac{9ma^2}{4}.16kg\frac{9g.a}{10a} = 2mg.a\)M1, A1, A1
\(k = 9\)A1 (10)
MI of disc all diameter = $\frac{1}{4}ma^2$ | M1 |
MI of disc about axis = $\frac{4}{9}ma^2 + ma^2 = \frac{13}{9}ma^2$ | M1, A1 |
CAM: $m.a.\sqrt{kg.a} = \left(\frac{13}{9}ma^2 + ma^2\right)\omega$ | M1, A1 |
$\omega = \frac{4}{9}\sqrt{\frac{kg}{a}}$ | A1 |
Energy: $\frac{1}{2}.\frac{9ma^2}{4}.16kg\frac{9g.a}{10a} = 2mg.a$ | M1, A1, A1 |
$k = 9$ | A1 (10) |
5. A uniform circular disc has mass $m$ and radius $a$. The disc can rotate freely about an axis that is in the same plane as the disc and tangential to the disc at a point $A$ on its circumference. The disc hangs at rest in equilibrium with its centre $O$ vertically below $A$. A particle $P$ of mass $m$ is moving horizontally and perpendicular to the disc with speed $\sqrt { } ( k g a )$, where $k$ is a constant. The particle then strikes the disc at $O$ and adheres to it at $O$. Given that the disc rotates through an angle of $90 ^ { \circ }$ before first coming to instantaneous rest, find the value of $k$.\\
(Total 10 marks)\\

\hfill \mbox{\textit{Edexcel M5 2006 Q5 [10]}}