| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a substantial M3 circular motion problem requiring energy conservation, Newton's second law in circular motion, and projectile motion after string breaks. While the techniques are standard for M3 (energy methods, tension formula, projectile trajectory), it requires careful multi-step reasoning across three connected parts with 15 total marks, placing it moderately above average difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| (a) Energy: \(\frac{1}{2}(0.4)(1.4)^2 = 0.4 \times 9.8 \times 0.2(1-\cos\theta) + \frac{1}{2} \times 0.4v^2\) | M1 A1 A1 |
| \(v^2 = 1.96 - 3.92(1-\cos\theta) = 3.92\cos\theta - 1.96\) | A1 |
| \(v^2 \geq 0\), so \(\cos\theta \geq \frac{1}{2}\) | M1 A1 |
| \(\theta \leq 60°\) | |
| (b) \(T - mg\cos\theta = \frac{mv^2}{r}\) | B1 M1 A1 |
| \(T = 0.4 \times 9.8\cos\theta + 2(3.92\cos\theta - 1.96)\) | |
| \(T = 3.92(3\cos\theta - 1)\) | A1 |
| (c) \(u^2 = 3.92(0.6) - 1.96 = 0.392\) | M1 A1 |
| Energy: \(\frac{1}{2}m(0.392)^2 = mgh\) | M1 A1 A1 |
| \(h = 0.00784\) | |
| Greatest height = 0.08 + 0.00784 = 0.0878 m | 15 marks total |
**(a)** Energy: $\frac{1}{2}(0.4)(1.4)^2 = 0.4 \times 9.8 \times 0.2(1-\cos\theta) + \frac{1}{2} \times 0.4v^2$ | M1 A1 A1
$v^2 = 1.96 - 3.92(1-\cos\theta) = 3.92\cos\theta - 1.96$ | A1
$v^2 \geq 0$, so $\cos\theta \geq \frac{1}{2}$ | M1 A1
$\theta \leq 60°$ |
**(b)** $T - mg\cos\theta = \frac{mv^2}{r}$ | B1 M1 A1
$T = 0.4 \times 9.8\cos\theta + 2(3.92\cos\theta - 1.96)$ |
$T = 3.92(3\cos\theta - 1)$ | A1
**(c)** $u^2 = 3.92(0.6) - 1.96 = 0.392$ | M1 A1
Energy: $\frac{1}{2}m(0.392)^2 = mgh$ | M1 A1 A1
$h = 0.00784$ |
Greatest height = 0.08 + 0.00784 = 0.0878 m | 15 marks totalA particle $P$ of mass 0.4 kg hangs by a light, inextensible string of length 20 cm whose other end is attached to a fixed point $O$. It is given a horizontal velocity of 1.4 ms$^{-1}$ so that it begins to move in a vertical circle. If in the ensuing motion the string makes an angle of $\theta$ with the downward vertical through $O$, show that
\begin{enumerate}[label=(\alph*)]
\item $\theta$ cannot exceed 60°, [6 marks]
\item the tension, $T$ N, in the string is given by $T = 3.92(3 \cos \theta - 1)$. [4 marks]
\end{enumerate}
If the string breaks when $\cos \theta = \frac{3}{5}$ and $P$ is ascending,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the greatest height reached by $P$ above the initial point of projection. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [15]}}