| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on circular wire/arc |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in the radial direction. The question guides students through each step with clear 'show that' targets, making it slightly easier than average. While it involves multiple parts and requires careful application of mechanics principles, the techniques are routine for M3 students and no novel insight is needed. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| (a) P.E. gained = K.E. lost: \(mgr(1 - \cos\theta) = \frac{1}{2}m(3gr) - \frac{1}{2}mv^2\) | M1 M1 A1 |
| \(v^2 = 3gr - 2gr + 2gr\cos\theta = gr(1 + 2\cos\theta)\) | M1 A1 |
| (b) \(R - mg\cos\theta = \frac{mv^2}{r}\) | M1 A1 |
| \(R = \frac{m}{r}gr(1 + 2\cos\theta) + mg\cos\theta\) | M1 A1 |
| \(R = mg(1 + 3\cos\theta)\) | A1 |
| (c) (i) When \(v = 0\), \(\cos\theta = -\frac{1}{2}\) | B1 B1 |
| (ii) When \(R = 0\), \(\cos\theta = -\frac{1}{3}\) | |
| (d) \(h_1 = r + \frac{1}{2}r = \frac{3r}{2}\) | M1 A1 A1 |
| \(h_2 = r + \frac{1}{3}r = \frac{4r}{3}\) | |
| Ratio \(= \frac{3r}{2} : \frac{4r}{3} = \frac{9}{6} : \frac{8}{6} = 9:8\) | A1 |
**(a)** P.E. gained = K.E. lost: $mgr(1 - \cos\theta) = \frac{1}{2}m(3gr) - \frac{1}{2}mv^2$ | M1 M1 A1 |
$v^2 = 3gr - 2gr + 2gr\cos\theta = gr(1 + 2\cos\theta)$ | M1 A1 |
**(b)** $R - mg\cos\theta = \frac{mv^2}{r}$ | M1 A1 |
$R = \frac{m}{r}gr(1 + 2\cos\theta) + mg\cos\theta$ | M1 A1 |
$R = mg(1 + 3\cos\theta)$ | A1 |
**(c)** (i) When $v = 0$, $\cos\theta = -\frac{1}{2}$ | B1 B1 |
(ii) When $R = 0$, $\cos\theta = -\frac{1}{3}$ | |
**(d)** $h_1 = r + \frac{1}{2}r = \frac{3r}{2}$ | M1 A1 A1 |
$h_2 = r + \frac{1}{3}r = \frac{4r}{3}$ | |
Ratio $= \frac{3r}{2} : \frac{4r}{3} = \frac{9}{6} : \frac{8}{6} = 9:8$ | A1 |
**Total: 14 marks**A small bead $P$, of mass $m$ kg, can slide on a smooth circular ring, with centre $O$ and radius $r$ m, which is fixed in a vertical plane. $P$ is projected from the lowest point $L$ of the ring with speed $\sqrt{(3gr)}$ ms$^{-1}$. When $P$ has reached a position such that $OP$ makes an angle $\theta$ with the downward vertical, as shown, its speed is $v$ ms$^{-1}$.
\includegraphics{figure_5}
\begin{enumerate}[label=(\alph*)]
\item Show that $v^2 = gr(1 + 2 \cos \theta)$. [5 marks]
\item Show that the magnitude of the reaction $RN$ of the ring on the bead is given by
$$R = mg(1 + 3 \cos \theta).$$ [4 marks]
\item Find the values of $\cos \theta$ when
\begin{enumerate}[label=(\roman*)]
\item $P$ is instantaneously at rest,
\item the reaction $R$ is instantaneously zero. [2 marks]
\end{enumerate}
\item Hence show that the ratio of the heights of $P$ above $L$ in cases (i) and (ii) is $9:8$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [14]}}