| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2002 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rotation about fixed axis: impulsive impact and subsequent motion |
| Difficulty | Challenging +1.8 This M5 question requires conservation of angular momentum with parallel axis theorem, energy methods for oscillation, and small angle approximation for SHM—multiple advanced mechanics concepts combined. The 'show that' part demands precise moment of inertia calculations, while part (c) requires recognizing when to apply small oscillation theory, making it significantly harder than standard M1/M2 problems but accessible to well-prepared Further Maths students. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(I_B = \frac{1}{4}ma^2 + ma^2 = \frac{5ma^2}{4}\) | M1 A1 | |
| \(mua = \left(\frac{5ma^2}{4} + ma^2\right)\omega\) | M1 A1 A1↑ | |
| \(\omega = \frac{4u}{9a}\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{1}{2}\cdot\frac{9ma^2}{4}\left(\frac{4u}{9a}\right)^2 = 2mga(1-\cos\theta)\) | M1 A1 A1↑ | |
| \(\cos\theta = \frac{89}{90}\) | M1 | |
| \(\theta \approx 8.5° \approx 0.149^c\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(-2mga\sin\theta = \frac{9ma^2}{4}\ddot{\theta}\) | M1 A1 A1↑ | |
| \(\ddot{\theta} = -\frac{8g}{9a}\theta \quad (\theta \leq 9°)\) | M1 B1 | (SHM condition) |
| \(T = \pi\sqrt{\frac{9a}{8g}}\) | ||
| \(= \frac{3\pi}{2}\sqrt{\frac{a}{2g}}\) | A1 c.s.o. | (6) |
## Question 7(a):
| Working | Marks | Notes |
|---------|-------|-------|
| $I_B = \frac{1}{4}ma^2 + ma^2 = \frac{5ma^2}{4}$ | M1 A1 | |
| $mua = \left(\frac{5ma^2}{4} + ma^2\right)\omega$ | M1 A1 A1↑ | |
| $\omega = \frac{4u}{9a}$ | A1 | **(6)** |
---
## Question 7(b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}\cdot\frac{9ma^2}{4}\left(\frac{4u}{9a}\right)^2 = 2mga(1-\cos\theta)$ | M1 A1 A1↑ | |
| $\cos\theta = \frac{89}{90}$ | M1 | |
| $\theta \approx 8.5° \approx 0.149^c$ | A1 | **(5)** |
---
## Question 7(c):
| Working | Marks | Notes |
|---------|-------|-------|
| $-2mga\sin\theta = \frac{9ma^2}{4}\ddot{\theta}$ | M1 A1 A1↑ | |
| $\ddot{\theta} = -\frac{8g}{9a}\theta \quad (\theta \leq 9°)$ | M1 B1 | (SHM condition) |
| $T = \pi\sqrt{\frac{9a}{8g}}$ | | |
| $= \frac{3\pi}{2}\sqrt{\frac{a}{2g}}$ | A1 c.s.o. | **(6)** |7. A uniform plane circular disc, of mass $m$ and radius $a$, hangs in equilibrium from a point $B$ on its circumference. The disc is free to rotate about a fixed smooth horizontal axis which is in the plane of the disc and tangential to the disc at $B$. A particle $P$, of mass $m$, is moving horizontally with speed $u$ in a direction which is perpendicular to the plane of the disc. At time $t = 0 , P$ strikes the disc at its centre and adheres to the disc.
\begin{enumerate}[label=(\alph*)]
\item Show that the angular speed of the disc immediately after it has been struck by $P$ is $\frac { 4 u } { 9 a }$.\\
(6)
It is given that $u ^ { 2 } = \frac { 1 } { 10 } a g$, and that air resistance is negligible.
\item Find the angle through which the disc turns before it first comes to instantaneous rest.
The disc first returns to its initial position at time $t = T$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down an equation of motion for the disc.
\item Hence find $T$ in terms of $a , g$ and $m$, using a suitable approximation which should be justified.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2002 Q7 [17]}}