| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.8 This M3 circular motion question requires resolving forces in 3D geometry, applying Newton's second law for circular motion, and comparing two scenarios. Part (a) involves standard force resolution with geometry, part (b) applies centripetal force principles, but part (c) requires careful analysis of a tangential force scenario with inequality proof—demanding more sophisticated reasoning than typical M3 questions but still within standard mechanics techniques. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Reaction \(R\) acts on \(P\) towards centre of sphere, at \(\theta\) to vertical where \(\cos\theta = \frac{h}{r}\) | M1 | |
| \(R\cos\theta = mg\), so \(R = \frac{mgr}{h}\) | B1 M1 A1 | |
| (b) Resolve towards centre: \(R\sin\theta = \frac{mv^2}{r\sin\theta}\) | B1 | |
| \(v^2 = \frac{mgr\sin^2\theta}{h} - \frac{gr(r^2-h^2)}{h r^2}\) | \(v = \sqrt{\frac{g(r^2-h^2)}{h}}\) | M1 A1 A1 |
| (c) \(mg\sin\theta + S\cos\theta = mg\) | \(S = mg\frac{(1-\sin\theta)}{\cos\theta} = mg\left(1 - \frac{0 \cdot 866}{0 \cdot 5}\right)\) | M1 A1 M1 |
| \(= 0 \cdot 268mg\) | \(R = mg\frac{2h}{h} = 2mg\) | Hence \(S < \frac{R}{6}\) |
(a) Reaction $R$ acts on $P$ towards centre of sphere, at $\theta$ to vertical where $\cos\theta = \frac{h}{r}$ | M1
$R\cos\theta = mg$, so $R = \frac{mgr}{h}$ | B1 M1 A1
(b) Resolve towards centre: $R\sin\theta = \frac{mv^2}{r\sin\theta}$ | B1
$v^2 = \frac{mgr\sin^2\theta}{h} - \frac{gr(r^2-h^2)}{h r^2}$ | $v = \sqrt{\frac{g(r^2-h^2)}{h}}$ | M1 A1 A1
(c) $mg\sin\theta + S\cos\theta = mg$ | $S = mg\frac{(1-\sin\theta)}{\cos\theta} = mg\left(1 - \frac{0 \cdot 866}{0 \cdot 5}\right)$ | M1 A1 M1
$= 0 \cdot 268mg$ | $R = mg\frac{2h}{h} = 2mg$ | Hence $S < \frac{R}{6}$ | A1 A1
**Total: 13 marks**The diagram shows a particle $P$ of mass $m$ kg moving on the inner surface of a smooth fixed hemispherical bowl of radius $r$ m which is fixed with its axis vertical. $P$ moves at a constant speed in a horizontal circle, at a depth $h$ m below the top of the bowl.
\includegraphics{figure_6}
\begin{enumerate}[label=(\alph*)]
\item Show that the force $R$ exerted on $P$ by the bowl has magnitude $\frac{mgr}{h}$ N. [4 marks]
\item Find, in terms of $g$, $h$ and $r$, the constant speed of $P$. [4 marks]
\end{enumerate}
The bowl is now inverted and $P$ moves on the smooth outer surface at a height $h$ above the plane face under the action of a force of magnitude $mg$ applied tangentially as shown. The reaction of the surface of the sphere on $P$ now has magnitude $S$ N.
\includegraphics{figure_6b}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Given that $r = 2h$, prove that $S < \frac{1}{6}R$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}