| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Standard +0.8 This is a multi-step energy conservation problem requiring: (1) finding equilibrium extension using Hooke's law, (2) determining the position where the particle comes to rest using energy conservation with elastic PE, gravitational PE, and KE, (3) solving a quadratic equation. While systematic, it requires careful bookkeeping of multiple energy forms and reference points, making it moderately challenging for FM1 students. |
| 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| GPE from \(E\) to instantaneous rest e.g. \(mgx\), \(mg(d-a)\) | M1 | Use of GPE for unknown distance from \(E\) to instantaneous rest. May be implied by a difference of 2 GPE terms. |
| Use of conservation of energy principle | M1 | Form equation with one KE term, one GPE term, two EPE terms. All terms required for A marks. Condone \(\pm\) sign errors. M0 if 'E' or similar used for unknown EPE unless recovered. |
| \(\frac{1}{2}m\frac{9ag}{4} + mgx = \frac{2mg(a+x)^2}{4a} - \frac{2mga^2}{4a}\) oe | A1 | All 4 terms present in energy equation with one unknown length. At most one error. A0 if energy term missing. |
| \(\frac{1}{2}m\frac{9ag}{4} + mg(d-a) = \frac{2mgd^2}{2(2a)} - \frac{2mga^2}{2(2a)}\) oe | A1 | Fully correct equation in one unknown length. |
| \(x = \frac{3a}{2}\) or \(d = \frac{5a}{2}\) | A1 | cao |
| Use of EPE formula | M1 | Use of EPE formula at least once. EPE must have form \(\frac{\lambda x^2}{ka}\) where \(\lambda\) is modulus of elasticity, \(k\) is a constant and \(x\) is extension. |
| \(\frac{25mga}{8}\) | A1 | Correct answer |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| GPE from $E$ to instantaneous rest e.g. $mgx$, $mg(d-a)$ | M1 | Use of GPE for unknown distance from $E$ to instantaneous rest. May be implied by a difference of 2 GPE terms. |
| Use of conservation of energy principle | M1 | Form equation with one KE term, one GPE term, two EPE terms. All terms required for A marks. Condone $\pm$ sign errors. M0 if 'E' or similar used for unknown EPE unless recovered. |
| $\frac{1}{2}m\frac{9ag}{4} + mgx = \frac{2mg(a+x)^2}{4a} - \frac{2mga^2}{4a}$ oe | A1 | All 4 terms present in energy equation with one unknown length. At most one error. A0 if energy term missing. |
| $\frac{1}{2}m\frac{9ag}{4} + mg(d-a) = \frac{2mgd^2}{2(2a)} - \frac{2mga^2}{2(2a)}$ oe | A1 | Fully correct equation in one unknown length. |
| $x = \frac{3a}{2}$ or $d = \frac{5a}{2}$ | A1 | cao |
| Use of EPE formula | M1 | Use of EPE formula at least once. EPE must have form $\frac{\lambda x^2}{ka}$ where $\lambda$ is modulus of elasticity, $k$ is a constant and $x$ is extension. |
| $\frac{25mga}{8}$ | A1 | Correct answer |
---\begin{enumerate}
\item A light elastic string has natural length $2 a$ and modulus of elasticity $2 m g$. One end of the string is attached to a fixed point $A$ on a horizontal ceiling. The other end is attached to a particle $P$ of mass $m$.
\end{enumerate}
The particle $P$ hangs in equilibrium at the point $E$, where $A E = 3 a$.\\
The particle $P$ is then projected vertically downwards from $E$ with speed $\frac { 3 } { 2 } \sqrt { a g }$\\
Air resistance is assumed to be negligible.\\
Find the elastic energy stored in the string, when $P$ first comes to instantaneous rest. Give your answer in the form kmga, where $k$ is a constant to be found.
\hfill \mbox{\textit{Edexcel FM1 2024 Q5 [7]}}