| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on inner surface of sphere/bowl |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in the radial direction. Parts (a)-(c) are routine applications of standard techniques (energy equation, resolving forces, finding limiting case when N=0). Part (d) requires projectile motion after leaving the bowl, adding modest complexity but still following a predictable solution path. The 120° angle and multi-part structure make it slightly above average difficulty, but it remains a textbook-style question testing well-practiced methods rather than requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga(1+\cos 60°)\) | M1 A1 | |
| \(v^2 = u^2 - 3ga\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(R + mg\cos 60° = \frac{mv^2}{a}\) | M1 A1 | |
| \(R = \frac{m}{a}(6ga - 3ga) - \frac{mg}{2}\) | ||
| \(= \frac{5mg}{2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(R=0\) at \(B \Rightarrow \frac{mg}{2} = \frac{mv^2}{a} \Rightarrow v^2 = \frac{1}{2}ag\) | M1 | |
| \(\Rightarrow u^2 = \frac{7ga}{2} \Rightarrow u = \sqrt{\frac{7ga}{2}}\) | M1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \((\to)\) B to C: \(v\cos 60°\times t = a\sqrt{3}\) | M1 A1 | |
| \(t = \frac{2a\sqrt{3}}{v}\) | ||
| \((\uparrow)\) B to C: \(0 = v\sin 60t - \frac{1}{2}gt^2\) | M1 A1 | |
| \(\Rightarrow t = \frac{2v\sin 60°}{g} = \frac{v\sqrt{3}}{g}\) | ||
| \(\therefore \frac{2a\sqrt{3}}{v} = \frac{v\sqrt{3}}{g} \Rightarrow v^2 = 2ga\) | M1 A1 | |
| \(\Rightarrow u^2 = 5ga\) | ||
| \(\Rightarrow u = \sqrt{5ga}\) | A1 | (7) |
# Question 7:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga(1+\cos 60°)$ | M1 A1 | |
| $v^2 = u^2 - 3ga$ | A1 | (3) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $R + mg\cos 60° = \frac{mv^2}{a}$ | M1 A1 | |
| $R = \frac{m}{a}(6ga - 3ga) - \frac{mg}{2}$ | | |
| $= \frac{5mg}{2}$ | A1 | (3) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $R=0$ at $B \Rightarrow \frac{mg}{2} = \frac{mv^2}{a} \Rightarrow v^2 = \frac{1}{2}ag$ | M1 | |
| $\Rightarrow u^2 = \frac{7ga}{2} \Rightarrow u = \sqrt{\frac{7ga}{2}}$ | M1 A1 | (3) |
## Part (d):
| Working | Marks | Notes |
|---------|-------|-------|
| $(\to)$ B to C: $v\cos 60°\times t = a\sqrt{3}$ | M1 A1 | |
| $t = \frac{2a\sqrt{3}}{v}$ | | |
| $(\uparrow)$ B to C: $0 = v\sin 60t - \frac{1}{2}gt^2$ | M1 A1 | |
| $\Rightarrow t = \frac{2v\sin 60°}{g} = \frac{v\sqrt{3}}{g}$ | | |
| $\therefore \frac{2a\sqrt{3}}{v} = \frac{v\sqrt{3}}{g} \Rightarrow v^2 = 2ga$ | M1 A1 | |
| $\Rightarrow u^2 = 5ga$ | | |
| $\Rightarrow u = \sqrt{5ga}$ | A1 | (7) |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\includegraphics[alt={},max width=\textwidth]{044c5866-0a12-4309-8ced-b463e1615fb0-5_604_596_391_760}
\end{center}
\end{figure}
Part of a hollow spherical shell, centre $O$ and radius $a$, is removed to form a bowl with a plane circular rim. The bowl is fixed with the circular rim uppermost and horizontal. The point $A$ is the lowest point of the bowl. The point $B$ is on the rim of the bowl and $\angle A O B = 120 ^ { \circ }$, as shown in Fig. 4. A smooth small marble of mass $m$ is placed inside the bowl at $A$ and given an initial horizontal speed $u$. The direction of motion of the marble lies in the vertical plane $A O B$. The marble stays in contact with the bowl until it reaches $B$. When the marble reaches $B$, its speed is $v$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v ^ { 2 }$.
\item For the case when $u ^ { 2 } = 6 g a$, find the normal reaction of the bowl on the marble as the marble reaches $B$.
\item Find the least possible value of $u$ for the marble to reach $B$.
The point $C$ is the other point on the rim of the bowl lying in the vertical plane $O A B$.
\item Find the value of $u$ which will enable the marble to leave the bowl at $B$ and meet it again at the point $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q7 [16]}}