| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Challenging +1.2 This is a standard M3 projectiles-on-sphere problem requiring circular motion (finding leaving point) and energy conservation (finding impact speed). While it involves multiple steps and two distinct mechanics principles, the techniques are well-practiced in M3 and follow predictable patterns. The 7+7 mark allocation reflects computational length rather than conceptual novelty, placing it moderately above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Energy: \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(1-\cos\theta)\) | M1 A1 A1 | |
| Radial: \((±R) + mg\cos\theta = \frac{mv^2}{a}\) | M1 A1 | |
| Eliminating \(v\) and finding \(\cos\theta = \frac{u^2 + 2ga}{3ga}\) | M1, A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Energy (C and ground): \(\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mv^2 = mga(1-\cos\theta)\) | M1 A1 | |
| Eliminating \(v\): \(\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mag\cos\theta = mga(1+\cos\theta)\) | M1 A1 | |
| \(\cos\theta = \frac{5}{6}\) | M1 A1 ft | |
| \(\theta = 34°\) | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Or energy (A and ground): \(\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mu^2 = 2mga\) | M1 A1 | |
| \(u^2 = \frac{1}{2}ga\) | M1 A1 | |
| Using with (a) to find \(\cos\theta = \frac{5}{6}\); \(\theta = 34°\) | M1 A1; A1 | (7 marks) |
| Answer | Marks |
|---|---|
| Projectile approach: \(V_x = v\cos\theta\); \(V_y^2 = (v\sin\theta)^2 + 2ga(1+\cos\theta)\) | |
| \(\left(\frac{9ag}{2}\right) = V_x^2 + V_y^2 \Rightarrow \left(\frac{9ag}{2}\right) - v^2 = 2ga(1+\cos\theta)\) – M1 A1, then scheme |
## (a)
Energy: $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(1-\cos\theta)$ | M1 A1 A1 |
Radial: $(±R) + mg\cos\theta = \frac{mv^2}{a}$ | M1 A1 |
Eliminating $v$ and finding $\cos\theta = \frac{u^2 + 2ga}{3ga}$ | M1, A1 | (7 marks)
## (b)
Energy (C and ground): $\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mv^2 = mga(1-\cos\theta)$ | M1 A1 |
Eliminating $v$: $\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mag\cos\theta = mga(1+\cos\theta)$ | M1 A1 |
$\cos\theta = \frac{5}{6}$ | M1 A1 ft |
$\theta = 34°$ | A1 | (7 marks)
**(14 marks total)**
## Alt (b)
Or energy (A and ground): $\frac{1}{2}m\left(\frac{9ag}{2}\right) - \frac{1}{2}mu^2 = 2mga$ | M1 A1 |
$u^2 = \frac{1}{2}ga$ | M1 A1 |
Using with (a) to find $\cos\theta = \frac{5}{6}$; $\theta = 34°$ | M1 A1; A1 | (7 marks)
## Alt
Projectile approach: $V_x = v\cos\theta$; $V_y^2 = (v\sin\theta)^2 + 2ga(1+\cos\theta)$ |
$\left(\frac{9ag}{2}\right) = V_x^2 + V_y^2 \Rightarrow \left(\frac{9ag}{2}\right) - v^2 = 2ga(1+\cos\theta)$ – M1 A1, then scheme |
---\includegraphics{figure_2}
A particle is at the highest point $A$ on the outer surface of a fixed smooth sphere of radius $a$ and centre $O$. The lowest point $B$ of the sphere is fixed to a horizontal plane. The particle is projected horizontally from $A$ with speed $u$, where $u < \sqrt{ag}$. The particle leaves the sphere at the point $C$, where $OC$ makes an angle $\theta$ with the upward vertical, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\cos \theta$ in terms of $u$, $g$ and $a$. [7]
\end{enumerate}
The particle strikes the plane with speed $\sqrt{\frac{9ag}{2}}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, to the nearest degree, the value of $\theta$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q6 [14]}}