| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on inner surface of sphere/bowl |
| Difficulty | Standard +0.8 This is a multi-step circular motion problem requiring energy conservation, Newton's second law in polar form, and the critical condition for loss of contact (normal force = 0). Part (a) involves deriving a non-trivial relationship through combining these principles, worth 7 marks. Part (b) requires applying the result with specific values. While the techniques are standard M3 content, the problem demands careful setup and algebraic manipulation across multiple physics principles, placing it moderately above average difficulty. |
| Spec | 6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Energy: \(\frac{1}{2}m(U^2 - v^2) = mga(1 + \cos\alpha)\) | M1 A1 = A1 | |
| \((T +) mg\cos\alpha = \frac{mv^2}{a}\) | M1 A1 | |
| Leaves circle when \(T = 0\): \(g\cos\alpha = \frac{U^2 - 2ga - 2ga\cos\alpha}{a}\) Eliminating \(v\) | M1 | |
| Leading to \(U^2 = ag(2 + 3\cos\alpha)\) | A1 (cso) | 7 marks |
| Answer | Marks |
|---|---|
| \(\frac{1}{2}m(U^2 - W^2) = mga\) | M1 A1 = A1 |
| \(W^2 = U^2 - 2ag\) | |
| Using \(\cos\alpha = \frac{1}{\sqrt{3}}\): \(W^2 = ag\left(2 + \frac{3}{\sqrt{3}}\right) - 2ag = ag\sqrt{3}\) | M1 |
| A1 (cso) | 5 marks [12] |
| Answer | Marks | Guidance |
|---|---|---|
| \(V^2 = ag\cos\alpha = \frac{ga}{\sqrt{3}}\) | ||
| Energy: \(\frac{1}{2}m(W^2 - V^2) = mga\cos\alpha\) | M1 A1 | |
| \(\frac{1}{2}m\left(W^2 - \frac{1}{\sqrt{3}}ag\right) = mga \times \frac{1}{\sqrt{3}}\) | A1 | |
| Leading to \(W^2 = ag\sqrt{3}\) | M1 A1 (cso) | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(V^2 = ag\cos\alpha = \frac{ga}{\sqrt{3}}\) | ||
| \(W_y^2 = V^2\sin^2\alpha + 2ga\cos\alpha = \frac{1}{\sqrt{3}}ag\left(1 - \frac{1}{3}\right) + 2ga \times \frac{1}{\sqrt{3}} = \frac{8\sqrt{3}}{9}ag\) | M1 A1 | |
| \(V_x = V\cos\alpha\) | A1 | |
| \(W^2 = W_y^2 + V_x^2 = \frac{8\sqrt{3}}{9}ag + \frac{1}{3}ag\sqrt{3} \times \frac{1}{3} = ag\sqrt{3}\) | M1 A1 (cso) | 5 marks |
**(a)** Energy: $\frac{1}{2}m(U^2 - v^2) = mga(1 + \cos\alpha)$ | M1 A1 = A1 |
$(T +) mg\cos\alpha = \frac{mv^2}{a}$ | M1 A1 |
Leaves circle when $T = 0$: $g\cos\alpha = \frac{U^2 - 2ga - 2ga\cos\alpha}{a}$ Eliminating $v$ | M1 |
Leading to $U^2 = ag(2 + 3\cos\alpha)$ | A1 (cso) | 7 marks
**(b)** Using conservation of energy from the lowest point of the surface:
$\frac{1}{2}m(U^2 - W^2) = mga$ | M1 A1 = A1 |
$W^2 = U^2 - 2ag$ | |
Using $\cos\alpha = \frac{1}{\sqrt{3}}$: $W^2 = ag\left(2 + \frac{3}{\sqrt{3}}\right) - 2ag = ag\sqrt{3}$ | M1 |
| A1 (cso) | 5 marks [12]
**Alternative to (b) using conservation of energy from the point where P loses contact with surface:**
$V^2 = ag\cos\alpha = \frac{ga}{\sqrt{3}}$ | |
Energy: $\frac{1}{2}m(W^2 - V^2) = mga\cos\alpha$ | M1 A1 |
$\frac{1}{2}m\left(W^2 - \frac{1}{\sqrt{3}}ag\right) = mga \times \frac{1}{\sqrt{3}}$ | A1 |
Leading to $W^2 = ag\sqrt{3}$ | M1 A1 (cso) | 5 marks
**Alternative to (b) using projectile motion from the point where P loses contact with surface:**
$V^2 = ag\cos\alpha = \frac{ga}{\sqrt{3}}$ | |
$W_y^2 = V^2\sin^2\alpha + 2ga\cos\alpha = \frac{1}{\sqrt{3}}ag\left(1 - \frac{1}{3}\right) + 2ga \times \frac{1}{\sqrt{3}} = \frac{8\sqrt{3}}{9}ag$ | M1 A1 |
$V_x = V\cos\alpha$ | A1 |
$W^2 = W_y^2 + V_x^2 = \frac{8\sqrt{3}}{9}ag + \frac{1}{3}ag\sqrt{3} \times \frac{1}{3} = ag\sqrt{3}$ | M1 A1 (cso) | 5 marks
---A particle $P$ is free to move on the smooth inner surface of a fixed thin hollow sphere of internal radius $a$ and centre $O$. The particle passes through the lowest point of the spherical surface with speed $U$. The particle loses contact with the surface when $OP$ is inclined at an angle $\alpha$ to the upward vertical.
\begin{enumerate}[label=(\alph*)]
\item Show that $U^2 = ag(2 + 3\cos \alpha)$. [7]
\end{enumerate}
The particle has speed $W$ as it passes through the level of $O$. Given that $\cos \alpha = \frac{1}{\sqrt{3}}$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $W^2 = ag\sqrt{3}$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q6 [12]}}