| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | January |
| Marks | 5 |
| 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 vertical elastic string problem requiring energy conservation with gravitational PE and elastic PE. The setup is straightforward (release from natural length position), and the method is routine: equate initial and final energy states. While it involves multiple energy terms and solving a quadratic, this is a textbook application of a well-practiced technique with no novel insight required. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation6.02g 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{29.4(y-1.2)^2}{2 \times 1.2} = 0.9 \times 9.8y\) | M1A1A1 | M1: energy equation with GPE term and single EPE term of form \(k\frac{\lambda x^2}{l}\); A1 either term correct; A1 both terms correct |
| \(y^2 - 3.12y + 1.44 = 0\) | ||
| \(y = \frac{3.12 \pm \sqrt{3.12^2 - 4 \times 1.44}}{2}\), \(\quad y = 2.556... = 2.6\) or \(2.56\) (m) | DM1A1 | DM1: obtain 3-term quadratic and attempt solution (formula must be correct if shown; calculator only if final answer correct); A1cao: correct distance \(AB\), second solution need not be shown |
| ALT: \(\frac{29.4x^2}{2\times1.2} = 0.9\times9.8(x+1.2)\) (solves to \(1.356...\)) | M1A1A1 | Second M mark requires completion to distance \(AB\) |
| Total: [5] |
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{29.4(y-1.2)^2}{2 \times 1.2} = 0.9 \times 9.8y$ | M1A1A1 | M1: energy equation with GPE term and single EPE term of form $k\frac{\lambda x^2}{l}$; A1 either term correct; A1 both terms correct |
| $y^2 - 3.12y + 1.44 = 0$ | | |
| $y = \frac{3.12 \pm \sqrt{3.12^2 - 4 \times 1.44}}{2}$, $\quad y = 2.556... = 2.6$ or $2.56$ (m) | DM1A1 | DM1: obtain 3-term quadratic and attempt solution (formula must be correct if shown; calculator only if final answer correct); A1cao: correct distance $AB$, second solution need not be shown |
| **ALT:** $\frac{29.4x^2}{2\times1.2} = 0.9\times9.8(x+1.2)$ (solves to $1.356...$) | M1A1A1 | Second M mark requires completion to distance $AB$ |
| **Total: [5]** | | |
---\begin{enumerate}
\item A particle of mass 0.9 kg is attached to one end of a light elastic string, of natural length 1.2 m and modulus of elasticity 29.4 N . The other end of the string is attached to a fixed point $A$ on a ceiling.
\end{enumerate}
The particle is held at $A$ and then released from rest. The particle first comes to instantaneous rest at the point $B$.
Find the distance $A B$.\\
(5)
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\hfill \mbox{\textit{Edexcel M3 2018 Q2 [5]}}