| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Collision/impulse during circular motion |
| Difficulty | Standard +0.8 This is a sophisticated multi-stage mechanics problem requiring energy conservation for part (a), impulse-momentum for the collision in part (b), then energy conservation again for parts (c)(i) and (c)(ii). The collision during circular motion adds complexity beyond standard pendulum questions, and students must carefully track the change in mass and apply conservation principles across different stages. However, each individual step uses standard M3 techniques without requiring novel geometric insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| Energy \(\frac{1}{2}mv^2 = mga\cos\theta\); \(v^2 = 2ga\cos\theta\) | M1A1 |
| \(F = ma \leadsto T - mg\cos\theta = \frac{mv^2}{a}\) | M1A1 |
| Sub for \(\frac{v^2}{a}\): \(T = mg\cos\theta + 2mg\cos\theta\); \(\theta = 60° \therefore T = \frac{3}{2}mg\) | M1A1 (6) |
| Answer | Marks |
|---|---|
| Speed of \(P\) before impact \(= \sqrt{2ga}\) | B1 |
| \(\to \sqrt{2ga} \to 0 \to u\) | |
| PCLM: \(\bullet\) (m) \(\bullet\) (3m) \(\to\) \(\bullet\) (4m) \(\therefore u = \frac{\sqrt{2ga}}{4} = \sqrt{\frac{ga}{8}}\) | M1A1cso (3) |
| Answer | Marks |
|---|---|
| At A \(v = 0\) so conservation of energy gives: \(\frac{1}{2} \cdot 4mu^2 = 4mg a(1 - \cos\theta)\) | M1A1 |
| \(\frac{ga}{16} = ga(1 - \cos\theta)\) | M1 |
| \(\cos\theta = \frac{15}{16}\), \(\theta = 20°\) | A1 |
| Answer | Marks |
|---|---|
| At A \(T = 4mg\cos\theta = \frac{15mg}{4}\) (accept 3.75mg) | M1A1 (6) |
**Part (a):**
Energy $\frac{1}{2}mv^2 = mga\cos\theta$; $v^2 = 2ga\cos\theta$ | M1A1 |
$F = ma \leadsto T - mg\cos\theta = \frac{mv^2}{a}$ | M1A1 |
Sub for $\frac{v^2}{a}$: $T = mg\cos\theta + 2mg\cos\theta$; $\theta = 60° \therefore T = \frac{3}{2}mg$ | M1A1 (6) |
**Part (b):**
Speed of $P$ before impact $= \sqrt{2ga}$ | B1 |
$\to \sqrt{2ga} \to 0 \to u$ | |
PCLM: $\bullet$ (m) $\bullet$ (3m) $\to$ $\bullet$ (4m) $\therefore u = \frac{\sqrt{2ga}}{4} = \sqrt{\frac{ga}{8}}$ | M1A1cso (3) |
**Part (c)(i):**
At A $v = 0$ so conservation of energy gives: $\frac{1}{2} \cdot 4mu^2 = 4mg a(1 - \cos\theta)$ | M1A1 |
$\frac{ga}{16} = ga(1 - \cos\theta)$ | M1 |
$\cos\theta = \frac{15}{16}$, $\theta = 20°$ | A1 |
**Part (c)(ii):**
At A $T = 4mg\cos\theta = \frac{15mg}{4}$ (accept 3.75mg) | M1A1 (6) |
**Total: 15 Marks**
---\begin{enumerate}
\item A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is released from rest with the string taut and $O P$ horizontal.\\
(a) Find the tension in the string when $O P$ makes an angle of $60 ^ { \circ }$ with the downward vertical.
\end{enumerate}
A particle $Q$ of mass $3 m$ is at rest at a distance $a$ vertically below $O$. When $P$ strikes $Q$ the particles join together and the combined particle of mass $4 m$ starts to move in a vertical circle with initial speed $u$.\\
(b) Show that $u = \sqrt { } \left( \frac { g a } { 8 } \right)$.
The combined particle comes to instantaneous rest at $A$.\\
(c) Find\\
(i) the angle that the string makes with the downward vertical when the combined particle is at $A$,\\
(ii) the tension in the string when the combined particle is at $A$.\\
\section*{LU \\
$\_\_\_\_$}
\hfill \mbox{\textit{Edexcel M3 2008 Q5 [15]}}