| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Simple harmonic motion with elastic string |
| Difficulty | Standard +0.3 This is a standard M3 SHM question with elastic strings requiring energy conservation (part a), differentiation to show SHM (part b), and period/amplitude calculations (part c). While it involves multiple steps and techniques, these are well-practiced procedures in M3 with no novel insights required. The question is slightly easier than average A-level due to its structured guidance and routine application of standard methods. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mU^2 - \frac{1}{2}mv^2 = \dfrac{2mgx^2}{2l}\) | M1 A1 A1 | Use conservation of energy with 2 KE terms and 1 EPE term. EPE of form \(\frac{1}{2}kx^2\). Dimensionally correct. A1 unsimplified with at most one error, A1 correct unsimplified. |
| \(v^2 = U^2 - \dfrac{2gx^2}{l}\) * | A1* | Given answer from complete and correct working. Must include a line of working before given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2v\dfrac{dv}{dx} = -\dfrac{4gx}{l}\) | M1A1 | Differentiate wrt \(x\). Powers of \(v\) and \(x\) reduce by 1 and \(\frac{dv}{dx}\) seen. M0 for approach not involving differentiation wrt \(x\) e.g. N2L. A1 correct differentiated equation. |
| \(\ddot{x} = -\dfrac{2g}{l}x\), SHM \(\left(\omega = \sqrt{\dfrac{2g}{l}}\right)\) | A1 | Correct SHM equation. Must use \(\ddot{x}\) for acceleration and conclude SHM. |
| Period \(= \dfrac{2\pi}{\omega} = \pi\sqrt{\dfrac{2l}{g}}\) * | DM1 A1* | DM1 dependent on previous M. Correct use of period \(= \frac{2\pi}{\omega}\). Given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{\dfrac{gl}{2}} = a\sqrt{\dfrac{2g}{l}}\) OR \(0 = \dfrac{gl}{2} - \dfrac{2ga^2}{l}\) | M1 | |
| \(a = \dfrac{1}{2}l\) | A1 | |
| Time from \(x=a\) to \(x=\frac{1}{4}l\): \(\frac{1}{4}l = \frac{1}{2}l\cos\sqrt{\dfrac{2g}{l}}\,t\) | M1 | |
| \(t = \dfrac{\pi}{3}\sqrt{\dfrac{l}{2g}}\) | A1 | |
| Time \(= \dfrac{1}{4}\text{period} + \text{time from } x=a \text{ to } x=\dfrac{1}{4}l\) | M1 | |
| \(= \dfrac{5\pi}{6}\sqrt{\dfrac{l}{2g}}\) | A1 | (6 marks total for part) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{period} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{2g}{l}}} = \pi\sqrt{\frac{2l}{g}}\) | A1* | Given answer correctly obtained. Must include a line of working between \(\ddot{x} = -\omega^2 x\) and the given answer. Or: \(\omega = \sqrt{\frac{2g}{l}}\), \(\text{period} = 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(U = a\omega\) OR energy equation with \(v = 0\) and \(x = a\) to find the amplitude | M1 | Method mark for finding amplitude |
| Correct amplitude \(= \frac{l}{2}\) | A1 | |
| Use of \(x = a\cos(\omega t)\) where \(\frac{1}{4}l = \frac{1}{2}l\cos\sqrt{\frac{2g}{l}}\,t\) to give a partial time, OR use of \(x = a\sin(\omega t)\) where \(\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{2g}{l}}\,t\) to give a partial time | M1 | Complete method to find partial time using their \(a\) and their \(\omega\) |
| Use of \(x = a\cos(\omega t) \Rightarrow t = \frac{1}{\omega}\frac{\pi}{3}\) | A1 | Correct partial time; or use of \(x = a\sin(\omega t) \Rightarrow t = \frac{1}{\omega}\frac{\pi}{6}\) or \(\frac{1}{\omega}\frac{5\pi}{6}\) |
| Using \(x = a\cos(\omega t)\): Total time \(= \frac{1}{4}\text{period} + \text{time from } x=a \text{ to } x=\frac{1}{4}l = \frac{1}{4}\pi\sqrt{\frac{2l}{g}} + \frac{\pi}{3}\sqrt{\frac{l}{2g}}\) | M1 | Complete method to find total time |
| Using \(x = a\sin(\omega t)\) with \(\frac{1}{\omega}\frac{5\pi}{6}\): Total time \(= \frac{1}{\omega}\frac{5\pi}{6}\) | M1 | Alternative method |
| Using \(x = a\sin(\omega t)\) with \(\frac{1}{\omega}\frac{\pi}{6}\): Total time \(= \frac{1}{2}\text{period} - \text{time from } x=0 \text{ to } x=\frac{1}{4}l = \frac{1}{2}\pi\sqrt{\frac{2l}{g}} - \frac{\pi}{6}\sqrt{\frac{l}{2g}}\) | M1 | Alternative method |
| \(\frac{5\pi}{6}\sqrt{\frac{l}{2g}} = \frac{5\pi}{3}\sqrt{\frac{l}{8g}} = \pi\sqrt{\frac{25l}{72g}}\) | A1 | Correct final answer |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mU^2 - \frac{1}{2}mv^2 = \dfrac{2mgx^2}{2l}$ | M1 A1 A1 | Use conservation of energy with 2 KE terms and 1 EPE term. EPE of form $\frac{1}{2}kx^2$. Dimensionally correct. A1 unsimplified with at most one error, A1 correct unsimplified. |
| $v^2 = U^2 - \dfrac{2gx^2}{l}$ * | A1* | Given answer from complete and correct working. Must include a line of working before given answer. |
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2v\dfrac{dv}{dx} = -\dfrac{4gx}{l}$ | M1A1 | Differentiate wrt $x$. Powers of $v$ and $x$ reduce by 1 and $\frac{dv}{dx}$ seen. M0 for approach not involving differentiation wrt $x$ e.g. N2L. A1 correct differentiated equation. |
| $\ddot{x} = -\dfrac{2g}{l}x$, SHM $\left(\omega = \sqrt{\dfrac{2g}{l}}\right)$ | A1 | Correct SHM equation. Must use $\ddot{x}$ for acceleration and conclude SHM. |
| Period $= \dfrac{2\pi}{\omega} = \pi\sqrt{\dfrac{2l}{g}}$ * | DM1 A1* | DM1 dependent on previous M. Correct use of period $= \frac{2\pi}{\omega}$. Given answer. |
### Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{\dfrac{gl}{2}} = a\sqrt{\dfrac{2g}{l}}$ **OR** $0 = \dfrac{gl}{2} - \dfrac{2ga^2}{l}$ | M1 | |
| $a = \dfrac{1}{2}l$ | A1 | |
| Time from $x=a$ to $x=\frac{1}{4}l$: $\frac{1}{4}l = \frac{1}{2}l\cos\sqrt{\dfrac{2g}{l}}\,t$ | M1 | |
| $t = \dfrac{\pi}{3}\sqrt{\dfrac{l}{2g}}$ | A1 | |
| Time $= \dfrac{1}{4}\text{period} + \text{time from } x=a \text{ to } x=\dfrac{1}{4}l$ | M1 | |
| $= \dfrac{5\pi}{6}\sqrt{\dfrac{l}{2g}}$ | A1 | (6 marks total for part) |
## Question A1 (Period):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{period} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{2g}{l}}} = \pi\sqrt{\frac{2l}{g}}$ | A1* | Given answer correctly obtained. Must include a line of working between $\ddot{x} = -\omega^2 x$ and the given answer. Or: $\omega = \sqrt{\frac{2g}{l}}$, $\text{period} = 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}$ |
---
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $U = a\omega$ OR energy equation with $v = 0$ and $x = a$ to find the amplitude | M1 | Method mark for finding amplitude |
| Correct amplitude $= \frac{l}{2}$ | A1 | |
| Use of $x = a\cos(\omega t)$ where $\frac{1}{4}l = \frac{1}{2}l\cos\sqrt{\frac{2g}{l}}\,t$ to give a partial time, OR use of $x = a\sin(\omega t)$ where $\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{2g}{l}}\,t$ to give a partial time | M1 | Complete method to find partial time using their $a$ and their $\omega$ |
| Use of $x = a\cos(\omega t) \Rightarrow t = \frac{1}{\omega}\frac{\pi}{3}$ | A1 | Correct partial time; or use of $x = a\sin(\omega t) \Rightarrow t = \frac{1}{\omega}\frac{\pi}{6}$ or $\frac{1}{\omega}\frac{5\pi}{6}$ |
| Using $x = a\cos(\omega t)$: Total time $= \frac{1}{4}\text{period} + \text{time from } x=a \text{ to } x=\frac{1}{4}l = \frac{1}{4}\pi\sqrt{\frac{2l}{g}} + \frac{\pi}{3}\sqrt{\frac{l}{2g}}$ | M1 | Complete method to find total time |
| Using $x = a\sin(\omega t)$ with $\frac{1}{\omega}\frac{5\pi}{6}$: Total time $= \frac{1}{\omega}\frac{5\pi}{6}$ | M1 | Alternative method |
| Using $x = a\sin(\omega t)$ with $\frac{1}{\omega}\frac{\pi}{6}$: Total time $= \frac{1}{2}\text{period} - \text{time from } x=0 \text{ to } x=\frac{1}{4}l = \frac{1}{2}\pi\sqrt{\frac{2l}{g}} - \frac{\pi}{6}\sqrt{\frac{l}{2g}}$ | M1 | Alternative method |
| $\frac{5\pi}{6}\sqrt{\frac{l}{2g}} = \frac{5\pi}{3}\sqrt{\frac{l}{8g}} = \pi\sqrt{\frac{25l}{72g}}$ | A1 | Correct final answer |\begin{enumerate}
\item A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $l$ and modulus of elasticity 2 mg . The other end of the string is attached to a fixed point $A$ on a smooth horizontal table. The particle $P$ is at rest at the point $B$ on the table, where $A B = l$.
\end{enumerate}
At time $t = 0 , P$ is projected along the table with speed $U$ in the direction $A B$.\\
At time $t$
\begin{itemize}
\item the elastic string has not gone slack
\item $B P = x$
\item the speed of $P$ is $v$\\
(a) Show that
\end{itemize}
$$v ^ { 2 } = U ^ { 2 } - \frac { 2 g x ^ { 2 } } { l }$$
(b) By differentiating this equation with respect to $x$, prove that, before the elastic string goes slack, $P$ moves with simple harmonic motion with period $\pi \sqrt { \frac { 2 l } { g } }$
Given that $U = \sqrt { \frac { g l } { 2 } }$\\
(c) find, in terms of $l$ and $g$, the exact total time, from the instant it is projected from $B$, that it takes $P$ to travel a total distance of $\frac { 3 } { 4 } l$ along the table.
\hfill \mbox{\textit{Edexcel M3 2024 Q7 [15]}}