| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Standard +0.3 This is a standard vertical circular motion problem requiring energy conservation and Newton's second law for circular motion. Part (a) is straightforward application of energy conservation, while part (b) requires recognizing that the particle leaves when normal reaction becomes zero. Both are well-practiced techniques in M2 with clear methodical approaches and no novel insight required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Using conservation of energy: \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - mgh\) | M1A1 | M1 for 3 terms, 2 KE and PE or 4 terms, 2 KE and 2 PE |
| \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - mga(1-\cos\theta)\) | M1A1 | M1A1 for finding \(h\) |
| \(v^2 = u^2 + 2ga(1-\cos\theta)\) | ||
| \(v = \left(u^2 + 2ga[1-\cos\theta]\right)^{\frac{1}{2}}\) | A1 | 5, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(F = ma\) radially: \(mg\cos\theta - N = \frac{mv^2}{a}\) | M1A1 | M1 correct 3 terms; A1 correct signs \((-N\) or \(+N)\) |
| Particle leaves surface when \(N = 0\) | B1 | |
| \(mg\cos\theta = \frac{m}{a}\left(u^2 + 2ga[1-\cos\theta]\right)\) | M1 | |
| \(\cos\theta = \frac{u^2}{ga} + 2 - 2\cos\theta\) | ||
| \(\cos\theta = \frac{1}{3}\left(\frac{u^2}{ga} + 2\right)\) | A1 | 5 |
## Question 7:
**Part (a):**
| Using conservation of energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - mgh$ | M1A1 | M1 for 3 terms, 2 KE and PE or 4 terms, 2 KE and 2 PE |
| $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - mga(1-\cos\theta)$ | M1A1 | M1A1 for finding $h$ |
| $v^2 = u^2 + 2ga(1-\cos\theta)$ | | |
| $v = \left(u^2 + 2ga[1-\cos\theta]\right)^{\frac{1}{2}}$ | A1 | 5, AG |
**Part (b):**
| Using $F = ma$ radially: $mg\cos\theta - N = \frac{mv^2}{a}$ | M1A1 | M1 correct 3 terms; A1 correct signs $(-N$ or $+N)$ |
| Particle leaves surface when $N = 0$ | B1 | |
| $mg\cos\theta = \frac{m}{a}\left(u^2 + 2ga[1-\cos\theta]\right)$ | M1 | |
| $\cos\theta = \frac{u^2}{ga} + 2 - 2\cos\theta$ | | |
| $\cos\theta = \frac{1}{3}\left(\frac{u^2}{ga} + 2\right)$ | A1 | 5 |
---7 A smooth hemisphere, of radius $a$ and centre $O$, is fixed with its plane face on a horizontal surface. A particle, of mass $m$, can move freely on the surface of the hemisphere.
The particle is placed at the point $A$, the highest point of the hemisphere, and is set in motion along the surface with speed $u$.
\begin{enumerate}[label=(\alph*)]
\item While the particle is in contact with the hemisphere at a point $P , O P$ makes an angle $\theta$ with the upward vertical.\\
\includegraphics[max width=\textwidth, alt={}, center]{06b431ca-d3a8-46d6-b9f8-bac08d3fd51e-5_366_1246_715_395}
Show that the speed of the particle at $P$ is
$$\left( u ^ { 2 } + 2 g a [ 1 - \cos \theta ] \right) ^ { \frac { 1 } { 2 } }$$
\item The particle leaves the surface of the hemisphere when $\theta = \alpha$.
Find $\cos \alpha$ in terms of $a , u$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q7 [10]}}