| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Challenging +1.2 This is a multi-step mechanics problem requiring energy conservation, circular motion dynamics (normal reaction = 0 at leaving point), and projectile motion. Part (a) involves setting up energy equations and finding the angle at which the particle leaves, which is non-trivial but follows standard M3 techniques. Part (b) requires careful geometric reasoning and projectile equations. The 16 total marks and combination of energy, circular motion, and projectiles make this moderately challenging, but it's a recognizable M3 question type without requiring exceptional insight. |
| Spec | 3.02i Projectile motion: constant acceleration model6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mV^2 - \frac{1}{2}m\frac{2gx}{x} = mga(1 - \cos \theta)\) | M1 A1 A1 | |
| \(mg\cos \theta = m\frac{V^2}{a}\) | M1 A1 | |
| \(V = \sqrt{\frac{4ag}{5}}\) | DM1 A1 | Correct final answer with no errors in working |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos \theta = \frac{4}{5}\) | B1 | |
| \(t = \frac{a - a\sin \theta}{V \cos \theta} = \sqrt{\frac{5a}{16g}}\) | M1 A1 | |
| \(s = Vt\sin \theta + \frac{1}{2}gt^2\) | M1 | |
| \(= \sqrt{\frac{4ag}{5}}\sqrt{\frac{5a}{16g}}\frac{3}{5} + \frac{1}{2}g(\frac{5a}{16g})\) | M1 A1 | |
| \(= \frac{73a}{160}\) | A1 | |
| \(AX = a\cos \theta - \frac{73a}{160}\) | M1 | |
| \(= \frac{11a}{32}\) | A1 | Correct final answer |
| (9) | ||
| 16 |
## 4(a)
$\frac{1}{2}mV^2 - \frac{1}{2}m\frac{2gx}{x} = mga(1 - \cos \theta)$ | M1 A1 A1 |
$mg\cos \theta = m\frac{V^2}{a}$ | M1 A1 |
$V = \sqrt{\frac{4ag}{5}}$ | DM1 A1 | Correct final answer with no errors in working
| (7) |
## 4(b)
$\cos \theta = \frac{4}{5}$ | B1 |
$t = \frac{a - a\sin \theta}{V \cos \theta} = \sqrt{\frac{5a}{16g}}$ | M1 A1 |
$s = Vt\sin \theta + \frac{1}{2}gt^2$ | M1 |
$= \sqrt{\frac{4ag}{5}}\sqrt{\frac{5a}{16g}}\frac{3}{5} + \frac{1}{2}g(\frac{5a}{16g})$ | M1 A1 |
$= \frac{73a}{160}$ | A1 |
$AX = a\cos \theta - \frac{73a}{160}$ | M1 |
$= \frac{11a}{32}$ | A1 | Correct final answer
| (9) |
| 16 |
---\includegraphics{figure_2}
A smooth sphere of radius $a$ is fixed with a point $A$ of its surface in contact with a fixed vertical wall. A particle is placed on the highest point of the sphere and is projected towards the wall and perpendicular to the wall with horizontal speed $\sqrt{\frac{2ag}{5}}$, as shown in Figure 2.
The particle leaves the surface of the sphere with speed $V$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \sqrt{\frac{4ag}{5}}$ [7]
\end{enumerate}
The particle strikes the wall at the point $X$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the distance $AX$. [9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q4 [16]}}