| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Simple harmonic motion with elastic string |
| Difficulty | Challenging +1.8 This is a challenging M3 question requiring multiple sophisticated techniques: finding equilibrium with two elastic strings, proving SHM from first principles using Hooke's law, applying energy conservation across different phases of motion (both strings taut vs one slack), and determining when slack occurs. The multi-phase nature and need to track when strings become slack elevates this significantly above standard SHM questions. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = \frac{20e}{2} = \frac{15(1.8-e)}{1.2}\) | M1A1 | Attempting to obtain and equate tensions in the two parts of the string; correct equation |
| \(10e \times 1.2 = 15(1.8-e)\) | ||
| \(e = 1\) | A1 | Obtaining correct extension in either string (ext in \(BP = 0.8\) m) |
| \(AO = 3\) m | A1cso | cso Obtaining the correct given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(0.5\ddot{x} = \frac{20(1-x)}{2} - \frac{15(0.8+x)}{1.2}\) | M1, A1, A1 | Forming equation of motion at general point; both tensions with variable; deduct 1 for each error |
| \(\ddot{x} = -45x \therefore\) SHM | A1 cso | cso Correct equation in required form with concluding statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| String becomes slack when \(x = (-)0.8\) | B1 | For \(x = \pm 0.8\); need not be shown explicitly |
| \(v^2 = \omega^2(a^2 - x^2)\) | ||
| \(v^2 = 45(1 - 0.8^2)\ (= 16.2)\) | M1, A1ft | Using \(v^2 = \omega^2(a^2 - x^2)\) with their numerical \(\omega\) and their \(x\); equation with correct numbers |
| \(v = 4.024...\) m s\(^{-1}\) (4.0 or better) | A1ft | Correct value for \(v\), 2sf or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{2} \times \frac{20y^2}{2} - \frac{1}{2} \times \frac{20 \times 1.8^2}{2} = \frac{1}{2} \times 0.5 \times 16.2\) | M1, A1, A1ft | Attempting energy equation with 2 EPE terms and a KE term; 2 correct terms (may have \((1.8+x)\) instead of \(y\)); completely correct equation following through \(v\) from (c) |
| \(20y^2 - 64.8 = 16.2\) | ||
| \(y^2 = 4.05,\quad y = 2.012...\) | A1 | Correct value for distance travelled after \(PB\) became slack; \(x = 0.21\) |
| Distance \(DB = \ | 5 - 4.012...\ | = 0.988...\) m (accept 0.99 or better) |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = \frac{20e}{2} = \frac{15(1.8-e)}{1.2}$ | M1A1 | Attempting to obtain and equate tensions in the two parts of the string; correct equation |
| $10e \times 1.2 = 15(1.8-e)$ | | |
| $e = 1$ | A1 | Obtaining correct extension in either string (ext in $BP = 0.8$ m) |
| $AO = 3$ m | A1cso | **cso** Obtaining the correct given answer |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $0.5\ddot{x} = \frac{20(1-x)}{2} - \frac{15(0.8+x)}{1.2}$ | M1, A1, A1 | Forming equation of motion at general point; both tensions with variable; deduct 1 for each error |
| $\ddot{x} = -45x \therefore$ SHM | A1 cso | **cso** Correct equation in required form with concluding statement |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| String becomes slack when $x = (-)0.8$ | B1 | For $x = \pm 0.8$; need not be shown explicitly |
| $v^2 = \omega^2(a^2 - x^2)$ | | |
| $v^2 = 45(1 - 0.8^2)\ (= 16.2)$ | M1, A1ft | Using $v^2 = \omega^2(a^2 - x^2)$ with their numerical $\omega$ and their $x$; equation with correct numbers |
| $v = 4.024...$ m s$^{-1}$ (4.0 or better) | A1ft | Correct value for $v$, 2sf or better |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2} \times \frac{20y^2}{2} - \frac{1}{2} \times \frac{20 \times 1.8^2}{2} = \frac{1}{2} \times 0.5 \times 16.2$ | M1, A1, A1ft | Attempting energy equation with 2 EPE terms and a KE term; 2 correct terms (may have $(1.8+x)$ instead of $y$); completely correct equation following through $v$ from (c) |
| $20y^2 - 64.8 = 16.2$ | | |
| $y^2 = 4.05,\quad y = 2.012...$ | A1 | Correct value for distance travelled after $PB$ became slack; $x = 0.21$ |
| Distance $DB = \|5 - 4.012...\| = 0.988...$ m (accept 0.99 or better) | A1ft | Complete to distance $DB$; follow through their distance travelled after $PB$ became slack |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb73b211-7629-4ed7-9b71-91841c29bb85-16_193_931_269_520}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two fixed points $A$ and $B$ are 5 m apart on a smooth horizontal floor. A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string, of natural length 2 m and modulus of elasticity 20 N . The other end of the string is attached to $A$. A second light elastic string, of natural length 1.2 m and modulus of elasticity 15 N , has one end attached to $P$ and the other end attached to $B$.
Initially $P$ rests in equilibrium at the point $O$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that $A O = 3 \mathrm {~m}$.
The particle is now pulled towards $A$ and released from rest at the point $C$, where $A C B$ is a straight line and $O C = 1 \mathrm {~m}$.
\item Show that, while both strings are taut, $P$ moves with simple harmonic motion.
\item Find the speed of $P$ at the instant when the string $P B$ becomes slack.
The particle first comes to instantaneous rest at the point $D$.
\item Find the distance $D B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q5 [17]}}