| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Collision/impulse during circular motion |
| Difficulty | Challenging +1.8 This is a challenging M5 question requiring multiple mechanics concepts: moment of inertia calculation, conservation of angular momentum at collision, energy conservation for rotation, and solving a transcendental equation. Part (b) involves substantial algebraic manipulation and physical insight about the system's behavior after an inelastic collision, making it significantly harder than routine mechanics problems but still within the scope of well-prepared M5 students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_P = \frac{2}{3}m\left(\frac{a}{2}\right)^2 + 3m(2a)^2 = 15ma^2\) | M1 A1 | PRINTED ANSWER |
| OR \(\frac{1}{3}m\left(\frac{a}{2}\right)^2 + m\left(\frac{a}{2}\right)^2 + 3m(2a)^2 = 15ma^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| KE gain = PE loss | M1 | |
| \(\frac{2}{3}3mv^2 = 3mg \cdot 2a\) | M1 | |
| \(v = 2\sqrt{ag}\) | A1 | |
| CAM: \(3mv \cdot 2a = 15ma^2\omega\) | M1 A1 | |
| OR CAM: \((12ma^2)\Omega^2 = 3mg \cdot 2a\) | M1 A1 | |
| \(\Omega = \sqrt{\frac{g}{a}}\) | A1 | |
| \(\omega = \frac{2av}{5a^2} = \frac{4}{5}\sqrt{\frac{g}{a}}\) | A1 | |
| KE loss = PE gain | M1 A1 A1 | |
| \(\frac{1}{2}15ma^2\omega^2 = mg\frac{3a}{2}(1 - \cos\theta) + 3mg \cdot 2a(1 - \cos\theta)\) | ||
| \(\cos\theta = \frac{9}{25}\), i.e. \(\theta = \cos^{-1}\left(\frac{9}{25}\right)\) | M1 A1 | PRINTED ANSWER |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}15ma^2\omega^2 = 4mg\frac{15a}{8}(1 - \cos\theta)\) | M1 A1 A1 | |
| \(\cos\theta = \frac{9}{25}\), i.e. \(\theta = \cos^{-1}\left(\frac{9}{25}\right)\) | M1 A1 | PRINTED ANSWER |
## (a)
$I_P = \frac{2}{3}m\left(\frac{a}{2}\right)^2 + 3m(2a)^2 = 15ma^2$ | M1 A1 | PRINTED ANSWER |
OR $\frac{1}{3}m\left(\frac{a}{2}\right)^2 + m\left(\frac{a}{2}\right)^2 + 3m(2a)^2 = 15ma^2$ |
## (b)
KE gain = PE loss | M1 |
$\frac{2}{3}3mv^2 = 3mg \cdot 2a$ | M1 |
$v = 2\sqrt{ag}$ | A1 |
CAM: $3mv \cdot 2a = 15ma^2\omega$ | M1 A1 |
OR CAM: $(12ma^2)\Omega^2 = 3mg \cdot 2a$ | M1 A1 |
$\Omega = \sqrt{\frac{g}{a}}$ | A1 |
$\omega = \frac{2av}{5a^2} = \frac{4}{5}\sqrt{\frac{g}{a}}$ | A1 |
KE loss = PE gain | M1 A1 A1 |
$\frac{1}{2}15ma^2\omega^2 = mg\frac{3a}{2}(1 - \cos\theta) + 3mg \cdot 2a(1 - \cos\theta)$ |
$\cos\theta = \frac{9}{25}$, i.e. $\theta = \cos^{-1}\left(\frac{9}{25}\right)$ | M1 A1 | PRINTED ANSWER |
OR
$\frac{1}{2}15ma^2\omega^2 = 4mg\frac{15a}{8}(1 - \cos\theta)$ | M1 A1 A1 |
$\cos\theta = \frac{9}{25}$, i.e. $\theta = \cos^{-1}\left(\frac{9}{25}\right)$ | M1 A1 | PRINTED ANSWER |A uniform rod $PQ$ of mass $m$ and length $3a$, is free to rotate about a fixed smooth horizontal axis $L$, which passes through the end $P$ of the rod and is perpendicular to the rod. The rod hangs at rest in equilibrium with $Q$ vertically below $P$. One end of a light inextensible string of length $2a$ is attached to the rod at $P$ and the other end is attached to a particle of mass $3m$. The particle is held with the string taut, and horizontal and perpendicular to $L$, and is then released. After colliding, the particle sticks to the rod forming a body $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of $B$ about $L$ is $15ma^2$.
[2]
\item Show that $B$ first comes to instantaneous rest after it has turned through an angle $\arccos\left(\frac{9}{25}\right)$.
[10]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2012 Q3 [12]}}