| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.8 This is a substantial M3 circular motion problem requiring energy conservation, circular motion dynamics at multiple points, and vector analysis of normal reaction changes. While the individual techniques are standard (energy methods, N = mv²/r + mg cos θ), the multi-part nature, careful geometry with angles, and the conceptual part (e) about modeling refinements make this moderately challenging but still within typical M3 scope. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(B\) to \(C\): \(\tfrac{1}{2}mv^2 - \tfrac{1}{2}m(20)^2 = mg\times 50(1-\sin 30°)\) | M1 A1 | All M marks require correct number of terms with appropriate terms resolved |
| \(v = 30\ \text{ms}^{-1}\ (29.8)\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((\uparrow)\) at \(C\): \(R - mg = m\dfrac{890}{50}\) | M1 A1 ft | |
| \(R = 1900\ \text{N}\ (1930\ \text{N})\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(C\) to \(D\): \(\tfrac{1}{2}m(890) - \tfrac{1}{2}mw^2 = mg\times 50(1-\cos 30°)\) | M1 A1 ft | |
| \(w = 28\ \text{ms}^{-1}\ (27.5)\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Before: \(R = mg\cos\theta\) | B1 | |
| After: \(R = mg\cos\theta + m\dfrac{20^2}{50}\) | M1 A1 | |
| Change \(= 70\times\dfrac{20^2}{50} = 560\ \text{N}\) | A1 c.s.o | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Lower speed at \(C \Rightarrow R\) reduced | M1 A1 | (2 marks); (15 marks) |
## Question 6:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $B$ to $C$: $\tfrac{1}{2}mv^2 - \tfrac{1}{2}m(20)^2 = mg\times 50(1-\sin 30°)$ | M1 A1 | All M marks require correct number of terms with appropriate terms resolved |
| $v = 30\ \text{ms}^{-1}\ (29.8)$ | A1 | (3 marks) |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(\uparrow)$ at $C$: $R - mg = m\dfrac{890}{50}$ | M1 A1 ft | |
| $R = 1900\ \text{N}\ (1930\ \text{N})$ | A1 | (3 marks) |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $C$ to $D$: $\tfrac{1}{2}m(890) - \tfrac{1}{2}mw^2 = mg\times 50(1-\cos 30°)$ | M1 A1 ft | |
| $w = 28\ \text{ms}^{-1}\ (27.5)$ | A1 | (3 marks) |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Before: $R = mg\cos\theta$ | B1 | |
| After: $R = mg\cos\theta + m\dfrac{20^2}{50}$ | M1 A1 | |
| Change $= 70\times\dfrac{20^2}{50} = 560\ \text{N}$ | A1 c.s.o | (4 marks) |
### Part (e):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Lower speed at $C \Rightarrow R$ reduced | M1 A1 | (2 marks); **(15 marks)** |
---6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{b9e9b91c-7e6d-4b84-9f0e-180b626887c2-4_460_799_301_657}
\end{center}
\end{figure}
Figure 3 represents the path of a skier of mass 70 kg moving on a ski-slope $A B C D$. The path lies in a vertical plane. From $A$ to $B$, the path is modelled as a straight line inclined at $60 ^ { \circ }$ to the horizontal. From $B$ to $D$, the path is modelled as an arc of a vertical circle of radius 50 m . The lowest point of the $\operatorname { arc } B D$ is $C$.
At $B$, the skier is moving downwards with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At $D$, the path is inclined at $30 ^ { \circ }$ to the horizontal and the skier is moving upwards. By modelling the slope as smooth and the skier as a particle, find
\begin{enumerate}[label=(\alph*)]
\item the speed of the skier at $C$,
\item the normal reaction of the slope on the skier at $C$,
\item the speed of the skier at $D$,
\item the change in the normal reaction of the slope on the skier as she passes $B$.
The model is refined to allow for the influence of friction on the motion of the skier.
\item State briefly, with a reason, how the answer to part (b) would be affected by using such a model. (No further calculations are expected.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q6 [15]}}