| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Coalescing particles collision |
| Difficulty | Challenging +1.8 This M5 question requires applying conservation of angular momentum about a parallel axis (not through the center of mass), using the parallel axis theorem for moment of inertia, then analyzing forces immediately after impact including the reaction force at the pivot. It combines rotational dynamics with impulse-momentum principles in a non-trivial geometry, requiring careful setup and multiple connected steps, making it significantly harder than average A-level questions but standard for Further Mechanics. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| MI of disc about \(L = \frac{1}{4}m(2a)^2 + m(2a)^2 = 5ma^2\) | M1 A1 | |
| CAM: \(m3\sqrt{ag} \cdot 2a = (5ma^2 + m(2a)^2)\omega\) | M1 A1 ft | |
| \(\omega = \frac{2}{3}\sqrt{\frac{g}{a}}\) | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(M(A),\ 0 = I\ddot{\theta}\), therefore \(\ddot{\theta} = 0\) | B1 | |
| \(R(\leftarrow),\ X = 2m \cdot 2a\ddot{\theta} = 0\) | B1 | |
| \(R(\uparrow),\ Y - 2mg = 2m \cdot 2a\dot{\theta}^2\) | M1 A1 | |
| \(Y = 2mg + 4ma \cdot \frac{4g}{9a}\) | DM1 | |
| \(= \frac{34mg}{9}\) | A1 | |
| (6) | ||
| Total: 11 |
## Question 6:
### Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| MI of disc about $L = \frac{1}{4}m(2a)^2 + m(2a)^2 = 5ma^2$ | M1 A1 | |
| CAM: $m3\sqrt{ag} \cdot 2a = (5ma^2 + m(2a)^2)\omega$ | M1 A1 ft | |
| $\omega = \frac{2}{3}\sqrt{\frac{g}{a}}$ | A1 | |
| | **(5)** | |
### Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $M(A),\ 0 = I\ddot{\theta}$, therefore $\ddot{\theta} = 0$ | B1 | |
| $R(\leftarrow),\ X = 2m \cdot 2a\ddot{\theta} = 0$ | B1 | |
| $R(\uparrow),\ Y - 2mg = 2m \cdot 2a\dot{\theta}^2$ | M1 A1 | |
| $Y = 2mg + 4ma \cdot \frac{4g}{9a}$ | DM1 | |
| $= \frac{34mg}{9}$ | A1 | |
| | **(6)** | |
| | **Total: 11** | |\begin{enumerate}
\item A uniform circular disc has mass $m$, centre $O$ and radius $2 a$. It is free to rotate about a fixed smooth horizontal axis $L$ which lies in the same plane as the disc and which is tangential to the disc at the point $A$. The disc is hanging at rest in equilibrium with $O$ vertically below $A$ when it is struck at $O$ by a particle of mass $m$. Immediately before the impact the particle is moving perpendicular to the plane of the disc with speed $3 \sqrt { } ( a g )$. The particle adheres to the disc at $O$.\\
(a) Find the angular speed of the disc immediately after the impact.\\
(b) Find the magnitude of the force exerted on the disc by the axis immediately after the impact.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2010 Q6 [11]}}