| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 7 |
| 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 problem requiring conservation of energy with three terms (gravitational PE, elastic PE, kinetic energy). The setup is straightforward with clear initial/final conditions (both at rest), and the method is a textbook application. The algebra involves solving a quadratic, which is routine at this level. Slightly easier than average due to its standard structure and clear problem statement. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(EPE = \frac{\lambda x^2}{2l} = \frac{4mgx^2}{2a}\) | M1 A1 | |
| Condition of ME: \(mg(a+x) = \frac{4mgx^2}{2a}\) | M1 A1 | |
| \(\therefore a(a+x) = 2x^2\) giving \(2x^2 - ax - a^2 = 0\) | A1 | |
| \((2x+a)(x-a) = 0\) \(\therefore x = -\frac{a}{4}\) or \(a\) | M1 | |
| \(x > 0\) \(\therefore x = a\) so \(AB = 2a\) | A1 | (7) |
$EPE = \frac{\lambda x^2}{2l} = \frac{4mgx^2}{2a}$ | M1 A1 |
Condition of ME: $mg(a+x) = \frac{4mgx^2}{2a}$ | M1 A1 |
$\therefore a(a+x) = 2x^2$ giving $2x^2 - ax - a^2 = 0$ | A1 |
$(2x+a)(x-a) = 0$ $\therefore x = -\frac{a}{4}$ or $a$ | M1 |
$x > 0$ $\therefore x = a$ so $AB = 2a$ | A1 | (7)\begin{enumerate}
\item A light elastic string has natural length $a$ and modulus of elasticity 4 mg . One end of the string is attached to a fixed point $A$ and a particle of mass $m$ is attached to the other end.
\end{enumerate}
The particle is released from rest at $A$ and falls vertically until it comes to rest instantaneously at the point $B$.
Find the distance $A B$ in terms of $a$.\\
(7 marks)\\
\hfill \mbox{\textit{Edexcel M3 Q1 [7]}}