| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: general symbolic/proof questions |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring energy conservation with elastic potential energy, force analysis at extreme positions, and finding maximum speed at equilibrium. While the setup is standard for FM1, it requires careful bookkeeping of EPE, GPE, and KE across three positions, plus understanding that maximum speed occurs at the equilibrium position. The algebra is moderately involved but systematic. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| From \(A\) to \(B\): EPE lost \(=\) GPE gained | M1 | Use conservation of energy with EPE \(= \frac{\lambda x^2}{2a}\). All three terms required. Must be dimensionally correct. Condone sign errors. |
| \(\dfrac{kmg \times 4a^2}{2a} - \dfrac{kmg \times \frac{a^2}{4}}{2a} = mg \times \dfrac{5a}{2}\) | A1 | Correct unsimplified equation in \(k\) |
| \(k = \dfrac{4}{3}\) * | A1* | Derive given result from correct working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(A\), equation of motion: | M1 | Use \(T = \frac{\lambda x}{a}\) and N2L to form equation of motion. All terms required. Dimensionally correct. Condone sign errors. |
| \((T - mg =)\dfrac{4mg \times 2a}{3a} - mg = m \times \text{acceleration}\) | A1 | Correct unsimplified equation |
| \(\Rightarrow \text{acceleration} = \dfrac{5g}{3}\) | A1 | Correct only ISW. Condone 1.7g or better. Accept \(+/-\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max speed at equilibrium position | M1 | Maximum speed at equilibrium seen or implied, and correct method to find \(e\) |
| \(\dfrac{4mge}{3a} = mg\), \(\quad e = \dfrac{3a}{4}\) | A1 | Correct \(e\). Alternative: form energy equation and differentiate \(v^2\) wrt \(h\), giving \(h = \frac{5a}{4}\) (M1, A1) |
| Forms equation using conservation of energy | M1 | Form energy equation for movement from \(A\) to equilibrium position. Need all 4 terms. Correct form for EPE. Dimensionally correct. Condone sign errors. Allow in \(a\), \(g\) and \(e\) (with \(e\) defined) |
| \(\dfrac{4mg \times 4a^2}{3 \times 2a} - \dfrac{4mg \times \frac{9a^2}{16}}{3 \times 2a} = \dfrac{1}{2}mv^2 + mg \times \dfrac{5a}{4}\) | A1ft A1ft | Unsimplified equation in their \(e\) with at most one error; Correct unsimplified equation (using their \(e\)) for \(v\) |
| \(v = \dfrac{5}{2}\sqrt{\dfrac{ga}{3}}\) | A1 | Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\) |
| SHM Alternative for 7(b) and 7(c): Equilibrium: \(\frac{4mge}{3a} = mg\), \(e = \frac{3a}{4}\); equation of motion gives \(\ddot{x} = -\frac{4g}{3a}x\), hence SHM. For (b): use \(x = \frac{5a}{4}\) and \(\omega^2\): \(\ddot{x} = -\frac{4g}{3a} \times \frac{5a}{4} = -\frac{5g}{3}\), \( | \ddot{x} | = \frac{5g}{3}\) (M1, A1). For (c): use \(v_{\max} = \omega a_{\text{amp}}\): \(v_{\max} = \sqrt{\frac{4g}{3a}} \times \frac{5a}{4} = \frac{5}{2}\sqrt{\frac{ga}{3}}\) (M1, A2ft, A1). |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| From $A$ to $B$: EPE lost $=$ GPE gained | M1 | Use conservation of energy with EPE $= \frac{\lambda x^2}{2a}$. All three terms required. Must be dimensionally correct. Condone sign errors. |
| $\dfrac{kmg \times 4a^2}{2a} - \dfrac{kmg \times \frac{a^2}{4}}{2a} = mg \times \dfrac{5a}{2}$ | A1 | Correct unsimplified equation in $k$ |
| $k = \dfrac{4}{3}$ * | A1* | Derive given result from correct working. |
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## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $A$, equation of motion: | M1 | Use $T = \frac{\lambda x}{a}$ and N2L to form equation of motion. All terms required. Dimensionally correct. Condone sign errors. |
| $(T - mg =)\dfrac{4mg \times 2a}{3a} - mg = m \times \text{acceleration}$ | A1 | Correct unsimplified equation |
| $\Rightarrow \text{acceleration} = \dfrac{5g}{3}$ | A1 | Correct only ISW. Condone 1.7g or better. Accept $+/-$ |
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## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max speed at equilibrium position | M1 | Maximum speed at equilibrium seen or implied, and correct method to find $e$ |
| $\dfrac{4mge}{3a} = mg$, $\quad e = \dfrac{3a}{4}$ | A1 | Correct $e$. Alternative: form energy equation and differentiate $v^2$ wrt $h$, giving $h = \frac{5a}{4}$ (M1, A1) |
| Forms equation using conservation of energy | M1 | Form energy equation for movement from $A$ to equilibrium position. Need all 4 terms. Correct form for EPE. Dimensionally correct. Condone sign errors. Allow in $a$, $g$ and $e$ (with $e$ defined) |
| $\dfrac{4mg \times 4a^2}{3 \times 2a} - \dfrac{4mg \times \frac{9a^2}{16}}{3 \times 2a} = \dfrac{1}{2}mv^2 + mg \times \dfrac{5a}{4}$ | A1ft A1ft | Unsimplified equation in their $e$ with at most one error; Correct unsimplified equation (using their $e$) for $v$ |
| $v = \dfrac{5}{2}\sqrt{\dfrac{ga}{3}}$ | A1 | Any equivalent form. Accept $1.44\sqrt{ag}$ or $1.4\sqrt{ag}$ |
**SHM Alternative for 7(b) and 7(c):** Equilibrium: $\frac{4mge}{3a} = mg$, $e = \frac{3a}{4}$; equation of motion gives $\ddot{x} = -\frac{4g}{3a}x$, hence SHM. For (b): use $x = \frac{5a}{4}$ and $\omega^2$: $\ddot{x} = -\frac{4g}{3a} \times \frac{5a}{4} = -\frac{5g}{3}$, $|\ddot{x}| = \frac{5g}{3}$ (M1, A1). For (c): use $v_{\max} = \omega a_{\text{amp}}$: $v_{\max} = \sqrt{\frac{4g}{3a}} \times \frac{5a}{4} = \frac{5}{2}\sqrt{\frac{ga}{3}}$ (M1, A2ft, A1).\begin{enumerate}
\item A particle $P$, of mass $m$, is attached to one end of a light elastic spring of natural length $a$ and modulus of elasticity kmg.
\end{enumerate}
The other end of the spring is attached to a fixed point $O$ on a ceiling.\\
The point $A$ is vertically below $O$ such that $O A = 3 a$\\
The point $B$ is vertically below $O$ such that $O B = \frac { 1 } { 2 } a$\\
The particle is held at rest at $A$, then released and first comes to instantaneous rest at the point $B$.\\
(a) Show that $k = \frac { 4 } { 3 }$\\
(b) Find, in terms of $g$, the acceleration of $P$ immediately after it is released from rest at $A$.\\
(c) Find, in terms of $g$ and $a$, the maximum speed attained by $P$ as it moves from $A$ to $B$.
\hfill \mbox{\textit{Edexcel FM1 2019 Q7 [12]}}