| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.2 This is a standard M3 SHM question with elastic strings covering equilibrium, SHM verification, and motion with a slack phase. While it requires multiple techniques (Hooke's law, energy methods, projectile motion), each part follows predictable patterns taught in M3. The slack phase adds mild complexity but is a routine extension. More challenging than basic C3 calculus but typical for Further Maths mechanics. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.5g = \dfrac{29.4\times(1.4-l)}{l}\) OR \(0.5g = \dfrac{29.4\times x}{l}\), giving \(x=\dfrac{l}{6}\), \(\dfrac{7l}{6}=1.4\) | M1A1 | Use Hooke's Law to find extension at \(B\); correct equation |
| \(l=1.2\) ✱ | A1cso (3) | Obtain given value for \(l\) with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.5g - T = 0.5\ddot{x}\) | ||
| \(0.5g - \dfrac{29.4(x+0.2)}{1.2} = 0.5\ddot{x}\) | M1A1 | Attempt equation of motion using Hooke's law for tension when extension is \(x+0.2\); \(m\) or \(0.5\) allowed; acceleration can be \(\ddot{x}\) or \(a\) |
| \(\ddot{x} = -\dfrac{29.4}{1.2\times0.5}x\), \(\ddot{x}=-49x\) \(\therefore\) SHM | dM1A1 (4) | Re-arrange to form \(\ddot{x}=(\pm)\omega^2 x\); must be \(\ddot{x}\); correct equation and conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v^2 = 49\left(0.4^2-(\pm0.2)^2\right)\) | M1A1ft | Use \(v^2=\omega^2(a^2-x^2)\) with their \(\omega^2\); amplitude \(=0.4\); \(x=\pm0.2\) |
| \(v = 2.42487\ldots = 2.4\) or \(2.42\ \text{ms}^{-1}\) | A1 (3) | Correct speed at instant string becomes slack; must be 2 or 3 significant figures |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Motion under gravity: \(0 = 7''\sqrt{0.4^2-0.2^2}''-gt\) | M1A1ft | Use SUVAT for motion under gravity with speed from (c) to find time from \(D\) to string becoming taut again; correct numbers used following through speed from (c) |
| \(t = \left(7\sqrt{0.4^2-0.2^2}\right)\div g = 0.24743\ldots\) | A1 | Correct time, shown explicitly or implied by correct final answer |
| SHM: \(-0.2 = 0.4\cos 7t\) | M1 | \((\pm)0.2=0.4\cos''7''t\) or \(0.4\sin''7''t\) with their \(\omega\) |
| \(t = \dfrac{1}{7}\cos^{-1}(-0.5) = 0.29919\ldots\) | dM1A1 | Solve for time to \(C\); must be radians; depends on previous M mark; correct time |
| Total time \(= 0.29919\ldots+0.24743\ldots = 0.5466\ldots = 0.55\) or \(0.547\ \text{s}\) | A1 cao (7) [17] | Add times for 2 parts of motion; must be 2 or 3 significant figures |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.5g = \dfrac{29.4\times(1.4-l)}{l}$ OR $0.5g = \dfrac{29.4\times x}{l}$, giving $x=\dfrac{l}{6}$, $\dfrac{7l}{6}=1.4$ | M1A1 | Use Hooke's Law to find extension at $B$; correct equation |
| $l=1.2$ ✱ | A1cso (3) | Obtain given value for $l$ with no errors seen |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.5g - T = 0.5\ddot{x}$ | | |
| $0.5g - \dfrac{29.4(x+0.2)}{1.2} = 0.5\ddot{x}$ | M1A1 | Attempt equation of motion using Hooke's law for tension when extension is $x+0.2$; $m$ or $0.5$ allowed; acceleration can be $\ddot{x}$ or $a$ |
| $\ddot{x} = -\dfrac{29.4}{1.2\times0.5}x$, $\ddot{x}=-49x$ $\therefore$ SHM | dM1A1 (4) | Re-arrange to form $\ddot{x}=(\pm)\omega^2 x$; must be $\ddot{x}$; correct equation and conclusion stated |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $v^2 = 49\left(0.4^2-(\pm0.2)^2\right)$ | M1A1ft | Use $v^2=\omega^2(a^2-x^2)$ with their $\omega^2$; amplitude $=0.4$; $x=\pm0.2$ |
| $v = 2.42487\ldots = 2.4$ or $2.42\ \text{ms}^{-1}$ | A1 (3) | Correct speed at instant string becomes slack; must be 2 or 3 significant figures |
## Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Motion under gravity: $0 = 7''\sqrt{0.4^2-0.2^2}''-gt$ | M1A1ft | Use SUVAT for motion under gravity with speed from (c) to find time from $D$ to string becoming taut again; correct numbers used following through speed from (c) |
| $t = \left(7\sqrt{0.4^2-0.2^2}\right)\div g = 0.24743\ldots$ | A1 | Correct time, shown explicitly or implied by correct final answer |
| SHM: $-0.2 = 0.4\cos 7t$ | M1 | $(\pm)0.2=0.4\cos''7''t$ or $0.4\sin''7''t$ with their $\omega$ |
| $t = \dfrac{1}{7}\cos^{-1}(-0.5) = 0.29919\ldots$ | dM1A1 | Solve for time to $C$; must be radians; depends on previous M mark; correct time |
| Total time $= 0.29919\ldots+0.24743\ldots = 0.5466\ldots = 0.55$ or $0.547\ \text{s}$ | A1 cao (7) [17] | Add times for 2 parts of motion; must be 2 or 3 significant figures |
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There is **no mark scheme content visible** on this page to extract. This appears to be a blank back page of an exam paper or mark scheme document.
If you have other pages with actual mark scheme content, please share those and I'll be happy to extract and format the information for you.7. A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string. The string has natural length $l$ metres and modulus of elasticity 29.4 N . The other end of the string is attached to a fixed point $A$. The particle hangs freely in equilibrium at the point $B$, where $B$ is vertically below $A$ and $A B = 1.4 \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $l = 1.2$
The point $C$ is vertically below $A$ and $A C = 1.8 \mathrm {~m}$. The particle is pulled down to $C$ and released from rest.
\item Show that, while the string is taut, $P$ moves with simple harmonic motion.
\item Calculate the speed of $P$ at the instant when the string first becomes slack.
The particle first comes to instantaneous rest at the point $D$.
\item Find the time taken by $P$ to return directly from $D$ to $C$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q7 [17]}}