| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law at the highest point. Part (a) uses basic energy methods (KE + PE), while part (b) adds the condition that centripetal force equals weight. Both are textbook exercises with well-rehearsed techniques, making this slightly easier than average for A-level. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh\), \(v=0\), \(h=2r\): \(\frac{1}{2}u^2 = 2gr\) | \(u = 2\sqrt{gr}\) | M1 A1 A1 |
| (b) At top, force towards centre = \(\frac{mu^2}{r} = mg\) as \(R=0\) | M1 A1 | |
| Thus \(v^2 = gr\), so \(mu^2 = mv^2 + 4mgr = 5mgr\) | \(u = \sqrt{5gr}\) | M1 A1 |
(a) $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh$, $v=0$, $h=2r$: $\frac{1}{2}u^2 = 2gr$ | $u = 2\sqrt{gr}$ | M1 A1 A1
(b) At top, force towards centre = $\frac{mu^2}{r} = mg$ as $R=0$ | M1 A1
Thus $v^2 = gr$, so $mu^2 = mv^2 + 4mgr = 5mgr$ | $u = \sqrt{5gr}$ | M1 A1
**Total: 7 marks**A small bead is threaded onto a smooth circular hoop, of radius $r$ m, fixed in a vertical plane. It is projected with speed $u$ ms$^{-1}$ from the lowest point of the hoop. Find $u$ in terms of $g$ and $r$ if
\begin{enumerate}[label=(\alph*)]
\item the bead just reaches the highest point of the hoop, [3 marks]
\item the reaction on the bead is zero when it is at the highest point of the hoop. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q1 [7]}}