| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.8 This is a challenging M3 SHM problem requiring energy conservation across slack/taut phases, identification of SHM parameters from elastic forces, and careful analysis of motion phases to find when the string goes slack. It demands strong conceptual understanding of the transition between different motion regimes and multi-step reasoning, placing it well above average difficulty but within reach of well-prepared Further Maths students. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g 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 |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{\lambda(D-l)}{l} = mg\) | M1A1 | M1: Use Hooke's law in \(D\) and equate to \(mg\); A1: Correct equation |
| \(\frac{\lambda(2l)^2}{2l} = mg \times 3l\) | M1A1A1 | M1: Energy equation with correct no. of terms, EPE of form \(\frac{\lambda x^2}{kl}, k\neq 1\); A1: Equation with at most one error; A1: Correct equation |
| \(D = \frac{5l}{3}\) * | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(mg - T = m\ddot{x}\) or \(T - mg = m\ddot{x}\) | M1 | Equation of motion in a *general* position, allow \(a\) for acceleration, correct no. of terms, condone sign errors |
| \(mg - \frac{3mg}{2l}\left(\frac{2l}{3}+x\right) = m\ddot{x}\) or \(\frac{3mg}{2l}\left(\frac{2l}{3}-x\right) - mg = m\ddot{x}\) | dM1A1 | dM1: Use Hooke's Law to sub for tension with extension measured from equilibrium position; A1: Correct unsimplified equation |
| \(-\frac{3g}{2l}x = \ddot{x}\) hence SHM | A1 | Correct SHM equation, must use \(\ddot{x}\) and conclude SHM |
| \(\text{period} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{3g}{2l}}}\) \(\left\{\omega = \sqrt{\frac{3g}{2l}}\right\}\) | M1 | Use of \(\frac{2\pi}{\omega}\) where \(\omega\) has come from attempt at N2L at general point |
| \(= 2\pi\sqrt{\frac{2l}{3g}}\) * | A1* | Must follow from fully correct working; at least one line between \(\ddot{x}=-\omega^2 x\) and given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(-\frac{2l}{3} = \frac{4l}{3}\cos\sqrt{\frac{3g}{2l}}\,t\) | M1A1A1 | M1: Complete method, correct \(\omega\) must be used; A1: Equation with at most one error; A1: Correct equation |
| \(t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(t = \frac{1}{4} \cdot 2\pi\sqrt{\frac{2l}{3g}} + t_1\) where \(\frac{2l}{3} = \frac{4l}{3}\sin\sqrt{\frac{3g}{2l}}\,t_1\) | M1A1A1 | Must include \(\frac{1}{4}T\) + their \(t\) value for M1 |
| \(t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(t = \frac{1}{2} \cdot 2\pi\sqrt{\frac{2l}{3g}} - t_1\) where \(\frac{2l}{3} = \frac{4l}{3}\cos\sqrt{\frac{3g}{2l}}\,t_1\) | M1A1A1 | Must include \(\frac{1}{2}T\) − their \(t\) value for M1 |
| \(t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}\) | A1 | or equivalent exact form |
## Question 7:
---
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{\lambda(D-l)}{l} = mg$ | M1A1 | M1: Use Hooke's law in $D$ and equate to $mg$; A1: Correct equation |
| $\frac{\lambda(2l)^2}{2l} = mg \times 3l$ | M1A1A1 | M1: Energy equation with correct no. of terms, EPE of form $\frac{\lambda x^2}{kl}, k\neq 1$; A1: Equation with at most one error; A1: Correct equation |
| $D = \frac{5l}{3}$ * | A1* | Given answer correctly obtained |
**Total: 6 marks**
---
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $mg - T = m\ddot{x}$ or $T - mg = m\ddot{x}$ | M1 | Equation of motion in a *general* position, allow $a$ for acceleration, correct no. of terms, condone sign errors |
| $mg - \frac{3mg}{2l}\left(\frac{2l}{3}+x\right) = m\ddot{x}$ or $\frac{3mg}{2l}\left(\frac{2l}{3}-x\right) - mg = m\ddot{x}$ | dM1A1 | dM1: Use Hooke's Law to sub for tension with extension measured from equilibrium position; A1: Correct unsimplified equation |
| $-\frac{3g}{2l}x = \ddot{x}$ hence SHM | A1 | Correct SHM equation, must use $\ddot{x}$ and conclude SHM |
| $\text{period} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{3g}{2l}}}$ $\left\{\omega = \sqrt{\frac{3g}{2l}}\right\}$ | M1 | Use of $\frac{2\pi}{\omega}$ where $\omega$ has come from attempt at N2L at general point |
| $= 2\pi\sqrt{\frac{2l}{3g}}$ * | A1* | Must follow from fully correct working; at least one line between $\ddot{x}=-\omega^2 x$ and given answer |
**Total: 6 marks**
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### Part (c):
**Method 1 (cosine):**
| Working | Marks | Notes |
|---------|-------|-------|
| $-\frac{2l}{3} = \frac{4l}{3}\cos\sqrt{\frac{3g}{2l}}\,t$ | M1A1A1 | M1: Complete method, correct $\omega$ must be used; A1: Equation with at most one error; A1: Correct equation |
| $t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}$ | A1 | cao |
**OR Method 2 (sine with $\frac{1}{4}T$):**
| Working | Marks | Notes |
|---------|-------|-------|
| $t = \frac{1}{4} \cdot 2\pi\sqrt{\frac{2l}{3g}} + t_1$ where $\frac{2l}{3} = \frac{4l}{3}\sin\sqrt{\frac{3g}{2l}}\,t_1$ | M1A1A1 | Must include $\frac{1}{4}T$ + their $t$ value for M1 |
| $t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}$ | A1 | oe |
**OR Method 3 (cosine with $\frac{1}{2}T$):**
| Working | Marks | Notes |
|---------|-------|-------|
| $t = \frac{1}{2} \cdot 2\pi\sqrt{\frac{2l}{3g}} - t_1$ where $\frac{2l}{3} = \frac{4l}{3}\cos\sqrt{\frac{3g}{2l}}\,t_1$ | M1A1A1 | Must include $\frac{1}{2}T$ − their $t$ value for M1 |
| $t = \frac{2\pi}{3}\sqrt{\frac{2l}{3g}}$ | A1 | or equivalent exact form |
**Total: 4 marks**
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**Overall Total: 16 marks**\begin{enumerate}
\item A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $l$. The other end of the string is attached to a fixed point on a ceiling. The particle $P$ hangs in equilibrium at a distance $D$ below the ceiling.
\end{enumerate}
The particle $P$ is now pulled vertically downwards until it is a distance $3 l$ below the ceiling and released from rest.
Given that $P$ comes to instantaneous rest just before it reaches the ceiling,\\
(a) show that $D = \frac { 5 l } { 3 }$\\
(b) Show that, while the elastic string is stretched, $P$ moves with simple harmonic motion, with period $2 \pi \sqrt { \frac { 2 l } { 3 g } }$\\
(c) Find, in terms of $g$ and $l$, the exact time from the instant when $P$ is released to the instant when the elastic string first goes slack.
\hfill \mbox{\textit{Edexcel M3 2023 Q7 [16]}}