| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.3 This is a standard M3 elastic string energy conservation problem with straightforward application of Hooke's law and energy principles. Part (a) requires setting up energy conservation at maximum extension (a routine 'show that' proof), while part (b) applies the same method at an intermediate position. Both parts follow textbook procedures with no novel insight required, making this slightly easier than average for M3 material. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks |
|---|---|
| (a) At greatest depth, gravitational P.E. lost = elastic P.E. gained | M1 M1 A1 |
| \(mg(2l) = \frac{\lambda l^2}{2l}\) | |
| \(\lambda = 4mg\) | |
| (b) Gravitational P.E. lost = elastic P.E. gained + K.E. gained | M1 A1 A1 |
| \(mg \frac{4l}{4} = \frac{4mg \cdot l}{2l \cdot 16} + \frac{1}{2}mv^2\) | |
| \(\frac{5gl}{4} = \frac{gl}{8} + \frac{1}{2}v^2\) | |
| \(v^2 = \frac{9g}{4}\) | M1 A1 A1 |
| \(v = \frac{3}{2}\sqrt{gl} \, \text{ms}^{-1}\) | |
| Total: 9 marks |
**(a)** At greatest depth, gravitational P.E. lost = elastic P.E. gained | M1 M1 A1 |
$mg(2l) = \frac{\lambda l^2}{2l}$ | |
$\lambda = 4mg$ | |
**(b)** Gravitational P.E. lost = elastic P.E. gained + K.E. gained | M1 A1 A1 |
$mg \frac{4l}{4} = \frac{4mg \cdot l}{2l \cdot 16} + \frac{1}{2}mv^2$ | |
$\frac{5gl}{4} = \frac{gl}{8} + \frac{1}{2}v^2$ | |
$v^2 = \frac{9g}{4}$ | M1 A1 A1 |
$v = \frac{3}{2}\sqrt{gl} \, \text{ms}^{-1}$ | |
| | **Total: 9 marks** |A particle $P$ of mass $m$ kg is attached to one end of a light elastic string of natural length $l$ m and modulus of elasticity $\lambda$ N. The other end of the string is attached to a fixed point $O$. $P$ is released from rest at $O$ and falls vertically downwards under gravity. The greatest distance below $O$ reached by $P$ is $2l$ m.
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 4mg$. [3 marks]
\item Find, in terms of $l$ and $g$, the speed with which $P$ passes through the point $A$, where $OA = \frac{5l}{4}$ m. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [9]}}