| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2003 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rotation about fixed axis: impulsive impact and subsequent motion |
| Difficulty | Challenging +1.2 This is a standard M5 rigid body dynamics question requiring conservation of angular momentum for impact, energy conservation for rotation, and force analysis. While it involves multiple parts and careful bookkeeping of moments of inertia (rod about end: ma²/3, particle at distance 2a), the techniques are routine for Further Maths M5 students. The 'show that' in part (a) provides scaffolding, and parts (b) and (c) follow standard procedures without requiring novel insight. |
| Spec | 3.04a Calculate moments: about a point6.02i 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 |
|---|---|
| (a) \(I_A = \left(\frac{4}{3}ma^2 + m(2a)^2\right)\) | M1 A1 |
| \(mv(2a) = I_A\omega = \frac{16ma^2}{3}\omega\) | M1 A1 ft |
| \(\omega = \frac{3v}{8a}\) * | A1 cso (5) |
| Gain in PE = \(mg \cdot 3a(1 + \cos 60°)\) | M1 A1 |
| Attempt at \(\frac{1}{2}I_A\omega^2 =\) gain in PE | M1 |
| \(\frac{1}{2}\left(\frac{16ma^2}{3}\right)\left(\frac{3v}{8a}\right)^2 = mg \cdot 3a(1 + \cos 60°)\) | A1 ft |
| Finding \(v\): \(v = \sqrt{12ga}\) | M1 A1 (6) |
| (c) Acceleration of \(C\) of \(G = \left(\frac{3}{2}a\omega^2\right)\) | B1 |
| \(R - 2mg = mr\omega^2 = 2m\left(\frac{3}{2}a\omega^2\right)\) | M1 A1 |
| Substitution of \(\omega\) and \(v\) and finding \(R = \ldots\) | M1 |
| \(R = \frac{113}{16}mg\) | A1 (5) |
**(a)** $I_A = \left(\frac{4}{3}ma^2 + m(2a)^2\right)$ | M1 A1 |
$mv(2a) = I_A\omega = \frac{16ma^2}{3}\omega$ | M1 A1 ft |
$\omega = \frac{3v}{8a}$ * | A1 cso (5) |
Gain in PE = $mg \cdot 3a(1 + \cos 60°)$ | M1 A1 |
Attempt at $\frac{1}{2}I_A\omega^2 =$ gain in PE | M1 |
$\frac{1}{2}\left(\frac{16ma^2}{3}\right)\left(\frac{3v}{8a}\right)^2 = mg \cdot 3a(1 + \cos 60°)$ | A1 ft |
Finding $v$: $v = \sqrt{12ga}$ | M1 A1 (6) |
**(c)** Acceleration of $C$ of $G = \left(\frac{3}{2}a\omega^2\right)$ | B1 |
$R - 2mg = mr\omega^2 = 2m\left(\frac{3}{2}a\omega^2\right)$ | M1 A1 |
Substitution of $\omega$ and $v$ and finding $R = \ldots$ | M1 |
$R = \frac{113}{16}mg$ | A1 (5) |
**Total: (16 marks)**
---5. A uniform rod $A B$, of mass $m$ and length $2 a$, is free to rotate in a vertical plane about a fixed smooth horizontal axis through $A$. The rod is hanging in equilibrium with $B$ below $A$ when it is hit by a particle of mass $m$ moving horizontally with speed $v$ in a vertical plane perpendicular to the axis. The particle strikes the rod at $B$ and immediately adheres to it.
\begin{enumerate}[label=(\alph*)]
\item Show that the angular speed of the rod immediately after the impact is $\frac { 3 v } { 8 a }$.
Given that the rod rotates through $120 ^ { \circ }$ before first coming to instantaneous rest,
\item find $v$ in terms of $a$ and $g$.
\item find, in terms of $m$ and $g$, the magnitude of the vertical component of the force acting on the $\operatorname { rod }$ at $A$ immediately after the impact.\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2003 Q5 [16]}}