| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.3 This is a standard M3 particle-on-sphere problem requiring energy conservation and circular motion equations. Part (a) is straightforward application of energy conservation, and part (b) requires setting normal reaction to zero—both are textbook exercises with well-rehearsed methods. Slightly easier than average due to the 'show that' structure and standard setup. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) con. of ME: \(\frac{1}{2}mv^2 = mgh\) | M1 | |
| giving \(v^2 = 2g(r - r\cos\theta) = 2 \times 9.8 \times 1.25(1-\cos\theta) = 24.5(1-\cos\theta)\) | M1 A1 | |
| (b) resolve \(\curvearrowright\): \(mg\cos\theta - R = \frac{mv^2}{r} = \frac{mv^2}{1.25}\) | M1 A1 | |
| leaves surface when \(R = 0 \therefore v^2 = 1.25g\cos\theta\) | M1 | |
| combining, \(24.5(1-\cos\theta) = 1.25g\cos\theta\) | M1 | |
| \(\therefore 24.5 = \cos\theta(1.25 \times 9.8 + 24.5)\) | A1 | |
| giving \(\cos\theta = \frac{24.5}{36.75} = \frac{2}{3}\) | A1 | (8) |
**(a)** con. of ME: $\frac{1}{2}mv^2 = mgh$ | M1 |
giving $v^2 = 2g(r - r\cos\theta) = 2 \times 9.8 \times 1.25(1-\cos\theta) = 24.5(1-\cos\theta)$ | M1 A1 |
**(b)** resolve $\curvearrowright$: $mg\cos\theta - R = \frac{mv^2}{r} = \frac{mv^2}{1.25}$ | M1 A1 |
leaves surface when $R = 0 \therefore v^2 = 1.25g\cos\theta$ | M1 |
combining, $24.5(1-\cos\theta) = 1.25g\cos\theta$ | M1 |
$\therefore 24.5 = \cos\theta(1.25 \times 9.8 + 24.5)$ | A1 |
giving $\cos\theta = \frac{24.5}{36.75} = \frac{2}{3}$ | A1 | **(8)**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0668f31-4b72-4dfd-9cf7-470acef0bfdb-2_469_465_776_680}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A particle $P$ of mass 0.5 kg is at rest at the highest point $A$ of a smooth sphere, centre $O$, of radius 1.25 m which is fixed to a horizontal surface.
When $P$ is slightly disturbed it slides along the surface of the sphere. Whilst $P$ is in contact with the sphere it has speed $v \mathrm {~ms} ^ { - 1 }$ when $\angle A O P = \theta$ as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = 24.5 ( 1 - \cos \theta )$.
\item Find the value of $\cos \theta$ when $P$ leaves the surface of the sphere.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [8]}}