| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring application of F=mrω² and Hooke's law, followed by a straightforward equilibrium calculation. The two-part structure and clear setup make it slightly easier than average, though it requires careful handling of units and the modulus of elasticity formula. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| (a) \(T = 0.2(1-15)n^2\), \(T = \frac{\lambda}{l}(0-15)\), so \(\lambda = \frac{(0-2)(1-15)n^2}{0.15} = 15.1\) N | M1 A1 M1 A1 M1 A1 |
| (b) \(T_1 = mg = 0.2(9.8)\) | B1 |
| \(1.96 = \frac{\lambda}{l}x\) | M1 A1 |
| \(x = 0.1298\) m \(\approx 13\) cm | A1 |
| Total: 9 marks |
(a) $T = 0.2(1-15)n^2$, $T = \frac{\lambda}{l}(0-15)$, so $\lambda = \frac{(0-2)(1-15)n^2}{0.15} = 15.1$ N | M1 A1 M1 A1 M1 A1 |
(b) $T_1 = mg = 0.2(9.8)$ | B1 |
$1.96 = \frac{\lambda}{l}x$ | M1 A1 |
$x = 0.1298$ m $\approx 13$ cm | A1 |
| **Total: 9 marks** |A particle $P$ of mass 0.2 kg moves in a horizontal circle on one end of an elastic string whose other end is attached to a fixed point $O$. The angular velocity of $P$ is $\pi$ rad s$^{-1}$. The natural length of the string is 1 m and, while $P$ is in motion, the distance $OP = 1.15$ m.
\begin{enumerate}[label=(\alph*)]
\item Calculate, to 3 significant figures, the modulus of elasticity of the string. [6 marks]
\end{enumerate}
The motion now ceases and $P$ hangs at rest vertically below $O$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the extension in the string in this position is about 13 cm. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [9]}}