| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Moderate -0.3 This is a straightforward application of Hooke's law and elastic potential energy formulas. Part (a) requires direct substitution into F = λx/l to find λ, and part (b) uses the standard EPE formula with given values. Both parts are routine calculations with no problem-solving insight required, making it slightly easier than average for M3 level. |
| Spec | 6.02c Work by variable force: using integration6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(T = \frac{\lambda x}{l}\) \(\therefore 4.5 = \frac{0.064}{0.15}\) giving \(\lambda = 11.25 \text{ N}\) | M1 A1; A1 | |
| (b) work done = change in EPE = \(\frac{\lambda}{2l}(x_2^2 - x_1^2)\) = \(\frac{11.25}{2 \times 0.15}(0.1^2 - 0.06^2) = 0.24 \text{ J}\) | M1 A1; M1 A1 | (7) |
(a) $T = \frac{\lambda x}{l}$ $\therefore 4.5 = \frac{0.064}{0.15}$ giving $\lambda = 11.25 \text{ N}$ | M1 A1; A1 |
(b) work done = change in EPE = $\frac{\lambda}{2l}(x_2^2 - x_1^2)$ = $\frac{11.25}{2 \times 0.15}(0.1^2 - 0.06^2) = 0.24 \text{ J}$ | M1 A1; M1 A1 | (7)
---\begin{enumerate}
\item The mechanism for releasing the ball on a pinball machine contains a light elastic spring of natural length 15 cm and modulus of elasticity $\lambda$.
\end{enumerate}
The spring is held compressed to a length of 9 cm by a force of 4.5 N .\\
(a) Find $\lambda$.\\
(b) Find the work done in compressing the spring from a length of 9 cm to a length of 5 cm .\\
(4 marks)\\
\hfill \mbox{\textit{Edexcel M3 Q1 [7]}}