| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of cylinder |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation, centripetal force analysis at the point of losing contact, and projectile motion. While it involves multiple parts and careful angle work, the techniques are all routine for this module—identifying the normal force condition (N=0), applying conservation of energy twice, and analyzing projectile trajectory. The 150° angle and final projectile calculation add some computational complexity but no novel insight is required. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02e Calculate KE and PE: using formulae6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| At \(B\): \(mg\cos 60\ (+R) = \frac{mv^2}{a}\) | M1A1 | |
| \(\frac{1}{2}g = \frac{v^2}{a}\), so \(v = \sqrt{\frac{ag}{2}}\) | A1 | Given/shown result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Energy \(A\) to \(B\): \(\frac{1}{2}mu^2 - \frac{1}{2}m\left(\frac{ag}{2}\right) = mga\sin 30\) | M1A1A1 | |
| \(u^2 = \frac{ag}{2} + 2ag\times\frac{1}{2}\) | ||
| \(u = \sqrt{\frac{3ag}{2}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Horiz speed \(= \sqrt{\frac{ag}{2}}\cos 60 = \frac{1}{2}\sqrt{\frac{ag}{2}}\) | M1A1 | |
| Initial vert speed \(= (-)\sqrt{\frac{ag}{2}}\sin 60 = (-){\frac{1}{2}}\sqrt{\frac{3ag}{2}}\) | M1 | |
| \(v^2 = \frac{1}{4}\times\frac{3ag}{2} + 2g\times\frac{a}{2}\) | M1A1 | |
| \(v^2 = \frac{11ag}{8}\) | ||
| \(\tan\theta = \frac{\text{vert}}{\text{horiz}} = \sqrt{\frac{11ag}{8}\times\frac{8}{ag}} = \sqrt{11}\) | M1 | |
| \(\theta = 73.22...= 73°\) | A1 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| At $B$: $mg\cos 60\ (+R) = \frac{mv^2}{a}$ | M1A1 | |
| $\frac{1}{2}g = \frac{v^2}{a}$, so $v = \sqrt{\frac{ag}{2}}$ | A1 | Given/shown result |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| Energy $A$ to $B$: $\frac{1}{2}mu^2 - \frac{1}{2}m\left(\frac{ag}{2}\right) = mga\sin 30$ | M1A1A1 | |
| $u^2 = \frac{ag}{2} + 2ag\times\frac{1}{2}$ | | |
| $u = \sqrt{\frac{3ag}{2}}$ | A1 | |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| Horiz speed $= \sqrt{\frac{ag}{2}}\cos 60 = \frac{1}{2}\sqrt{\frac{ag}{2}}$ | M1A1 | |
| Initial vert speed $= (-)\sqrt{\frac{ag}{2}}\sin 60 = (-){\frac{1}{2}}\sqrt{\frac{3ag}{2}}$ | M1 | |
| $v^2 = \frac{1}{4}\times\frac{3ag}{2} + 2g\times\frac{a}{2}$ | M1A1 | |
| $v^2 = \frac{11ag}{8}$ | | |
| $\tan\theta = \frac{\text{vert}}{\text{horiz}} = \sqrt{\frac{11ag}{8}\times\frac{8}{ag}} = \sqrt{11}$ | M1 | |
| $\theta = 73.22...= 73°$ | A1 | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d19c7390-0332-4cab-82e5-72976bd499a2-11_412_533_258_685}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A smooth hollow cylinder of internal radius $a$ is fixed with its axis horizontal. A particle $P$ moves on the inner surface of the cylinder in a vertical circle with radius $a$ and centre $O$, where $O$ lies on the axis of the cylinder. The particle is projected vertically downwards with speed $u$ from point $A$ on the circle, where $O A$ is horizontal. The particle first loses contact with the cylinder at the point $B$, where $\angle A O B = 150 ^ { \circ }$, as shown in Figure 3. Given that air resistance can be ignored,
\begin{enumerate}[label=(\alph*)]
\item show that the speed of $P$ at $B$ is $\sqrt { } \left( \frac { a g } { 2 } \right)$,
\item find $u$ in terms of $a$ and $g$.
After losing contact with the cylinder, $P$ crosses the diameter through $A$ at the point $D$. At $D$ the velocity of $P$ makes an angle $\theta ^ { \circ }$ with the horizontal.
\item Find the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q6 [14]}}