| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Detachment or peg impact mid-motion |
| Difficulty | Challenging +1.8 This M5 question requires calculating moment of inertia using parallel axis theorem for a composite body, then applying energy conservation and impulse-momentum principles to rotational motion. While systematic, it demands careful bookkeeping of multiple components, correct application of advanced mechanics theorems, and algebraic manipulation through several steps—significantly harder than typical A-level mechanics but standard for Further Maths M5. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \frac{1}{2}ma^2 + m(3a)^2 + \frac{1}{2}ma^2 + m(5a)^2\) | M1 A1 A1 | |
| \(\frac{1}{3}(3m)(4a)^2 + 3ma^2\) | M1 A1 | |
| \(= 54ma^2\) (*) | A1 | (6 marks) |
| \(\frac{1}{2} \times 54ma^2 \times \omega^2 = mg\times\frac{5a}{2} + 3mg\times\frac{a}{2} - mg\times\frac{3a}{2}\) | M1 A2, 1, 0 | |
| \(\Rightarrow \omega = \sqrt{\frac{5g}{54a}} = \frac{1}{3}\sqrt{\frac{5g}{6a}}\) | M1 A1 | |
| \(3a \times I = 54ma^2\left[\frac{1}{2}\sqrt{\frac{5g}{54a}} - \left(-\sqrt{\frac{5g}{54a}}\right)\right]\) | M1 A2, 1, 0 ft |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 9m\sqrt{\frac{5ga}{6}}\) | M1 A1 | (10 marks) |
## Question 7:
### Part (a)
$I = \frac{1}{2}ma^2 + m(3a)^2 + \frac{1}{2}ma^2 + m(5a)^2$ | M1 A1 A1 |
$\frac{1}{3}(3m)(4a)^2 + 3ma^2$ | M1 A1 |
$= 54ma^2$ (*) | A1 | (6 marks)
$\frac{1}{2} \times 54ma^2 \times \omega^2 = mg\times\frac{5a}{2} + 3mg\times\frac{a}{2} - mg\times\frac{3a}{2}$ | M1 A2, 1, 0 |
$\Rightarrow \omega = \sqrt{\frac{5g}{54a}} = \frac{1}{3}\sqrt{\frac{5g}{6a}}$ | M1 A1 |
$3a \times I = 54ma^2\left[\frac{1}{2}\sqrt{\frac{5g}{54a}} - \left(-\sqrt{\frac{5g}{54a}}\right)\right]$ | M1 A2, 1, 0 ft |
$I = \frac{54ma^2}{3a} \times \frac{3}{2}\sqrt{\frac{5g}{54a}}$
$= \frac{54ma^2}{3a} \times \frac{3}{2} \times \frac{1}{3}\sqrt{\frac{5g}{6a}}$
$= 9m\sqrt{\frac{5ga}{6}}$ | M1 A1 | (10 marks)7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{26fef791-e6fb-45a8-89e9-16c4b4a1a4e4-5_313_1443_317_356}
\end{center}
\end{figure}
A body consists of two uniform circular discs, each of mass $m$ and radius $a$, with a uniform rod. The centres of the discs are fixed to the ends $A$ and $B$ of the rod, which has mass $3 m$ and length 8a. The discs and the rod are coplanar, as shown in Fig. 2. The body is free to rotate in a vertical plane about a smooth fixed horizontal axis. The axis is perpendicular to the plane of the discs and passes through the point $O$ of the rod, where $A O = 3 a$.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the body about the axis is $54 m a ^ { 2 }$.
The body is held at rest with $A B$ horizontal and is then released. When the body has turned through an angle of $30 ^ { \circ }$, the rod $A B$ strikes a small fixed smooth peg $P$ where $O P = 3 a$. Given that the body rebounds from the peg with its angular speed halved by the impact,
\item show that the magnitude of the impulse exerted on the body by the peg at the impact is
$$9 m \sqrt { \left( \frac { 5 g a } { 6 } \right) } .$$
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2004 Q7 [16]}}