| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question with two elastic strings. Part (a) is routine equilibrium using Hooke's law. Part (b) requires setting up forces and showing the SHM equation (standard technique). Parts (c) and (d) involve finding when strings become slack and timing calculations - all well-practiced M3 techniques with no novel insight required. The multi-part structure and need for careful algebra place it slightly above average difficulty. |
| 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 | Mark | Guidance |
| \((T=)\frac{20(1)}{2} = \frac{\lambda \times 0.8}{1.2}\) | M1A1 | Form equation equating 2 tensions found using Hooke's Law |
| \(\lambda = 15\)* | A1* | Correct answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either \(1.25\ddot{x} = \frac{15(0.8-x)}{1.2} - \frac{20(1+x)}{2}\) or \(1.25\ddot{x} = \frac{20(1-x)}{2} - \frac{15(0.8+x)}{1.2}\) | M1A1A1 | Equation of motion for \(P\); correct with at most one error; fully correct (acceleration may be \(a\)) |
| \(\ddot{x} = -18x\)* | A1* | Correct given equation, correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(10 = a\sqrt{18} \Rightarrow a = \frac{10}{\sqrt{18}}\) oe | B1 | Correct amplitude; equivalents: \(\frac{5\sqrt{2}}{3}\), \(\frac{\sqrt{50}}{3}\), \(2.4\) |
| When string \(PB\) slack: \(v^2 = 18\left(\left(\frac{10}{\sqrt{18}}\right)^2 - 0.8^2\right)\) | M1 | Use \(v^2 = \omega^2(a^2 - x^2)\) with \(x=0.8\) |
| \(v = 9.4063...\approx 9.4\) or \(9.41\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{20\times1^2}{2\times2}+\frac{1.25\times10^2}{2}+\frac{15\times0.8^2}{2\times1.2} = \frac{20\times1.8^2}{2\times2}+\frac{1.25\times v^2}{2}\) | M1, B1 | Dimensionally correct energy equation with 3 EPE and 2 KE terms |
| \(v = 9.4063...\approx 9.4\) or \(9.41\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8 = \frac{10}{\sqrt{18}}\sin\sqrt{18}\,t_1\) | M1A1 | Use \(x=0.8\) to find time until \(PB\) slack |
| \(t_1 = \frac{1}{\sqrt{18}}\sin^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right) = 0.0816...\) | A1 | Allow consistent use of degrees |
| \(PA\) becomes slack when \(x = -1\): \((\pm1) = \frac{10}{\sqrt{18}}\sin\sqrt{18}\,t_2\) | M1 | Use \(x=\pm1\) to find time until \(PA\) slack |
| \(t_2 = \frac{1}{\sqrt{18}}\sin^{-1}\!\left(\frac{\sqrt{18}}{10}\right) = 0.1032...\) | A1 | |
| \(T = 2(t_1+t_2) = 2\!\left(\frac{1}{\sqrt{18}}\sin^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right)+\frac{1}{\sqrt{18}}\sin^{-1}\!\left(\frac{\sqrt{18}}{10}\right)\right)\) | A1 | Complete to correct value of \(T\) |
| \(= 0.3697... \approx 0.37\) or \(0.370\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8 = \frac{10}{\sqrt{18}}\cos\sqrt{18}\,t_1\) | M1A1 | |
| \(t_1 = \frac{1}{\sqrt{18}}\cos^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right) = 0.2886...\) | A1 | |
| \(-1 = \frac{10}{\sqrt{18}}\cos\sqrt{18}\,t_2\) | M1 | |
| \(t_2 = \frac{1}{\sqrt{18}}\cos^{-1}\!\left(-\frac{\sqrt{18}}{10}\right) = 0.4735...\) | A1 | |
| \(T = 2(t_2 - t_1) = 0.3697... \approx 0.37\) or \(0.370\) | A1 |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(T=)\frac{20(1)}{2} = \frac{\lambda \times 0.8}{1.2}$ | M1A1 | Form equation equating 2 tensions found using Hooke's Law |
| $\lambda = 15$* | A1* | Correct answer correctly obtained |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Either $1.25\ddot{x} = \frac{15(0.8-x)}{1.2} - \frac{20(1+x)}{2}$ or $1.25\ddot{x} = \frac{20(1-x)}{2} - \frac{15(0.8+x)}{1.2}$ | M1A1A1 | Equation of motion for $P$; correct with at most one error; fully correct (acceleration may be $a$) |
| $\ddot{x} = -18x$* | A1* | Correct given equation, correctly obtained |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $10 = a\sqrt{18} \Rightarrow a = \frac{10}{\sqrt{18}}$ oe | B1 | Correct amplitude; equivalents: $\frac{5\sqrt{2}}{3}$, $\frac{\sqrt{50}}{3}$, $2.4$ |
| When string $PB$ slack: $v^2 = 18\left(\left(\frac{10}{\sqrt{18}}\right)^2 - 0.8^2\right)$ | M1 | Use $v^2 = \omega^2(a^2 - x^2)$ with $x=0.8$ |
| $v = 9.4063...\approx 9.4$ or $9.41\ \text{ms}^{-1}$ | A1 | |
**ALT (c) Conservation of Energy, O to slack:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{20\times1^2}{2\times2}+\frac{1.25\times10^2}{2}+\frac{15\times0.8^2}{2\times1.2} = \frac{20\times1.8^2}{2\times2}+\frac{1.25\times v^2}{2}$ | M1, B1 | Dimensionally correct energy equation with 3 EPE and 2 KE terms |
| $v = 9.4063...\approx 9.4$ or $9.41\ \text{ms}^{-1}$ | A1 | |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8 = \frac{10}{\sqrt{18}}\sin\sqrt{18}\,t_1$ | M1A1 | Use $x=0.8$ to find time until $PB$ slack |
| $t_1 = \frac{1}{\sqrt{18}}\sin^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right) = 0.0816...$ | A1 | Allow consistent use of degrees |
| $PA$ becomes slack when $x = -1$: $(\pm1) = \frac{10}{\sqrt{18}}\sin\sqrt{18}\,t_2$ | M1 | Use $x=\pm1$ to find time until $PA$ slack |
| $t_2 = \frac{1}{\sqrt{18}}\sin^{-1}\!\left(\frac{\sqrt{18}}{10}\right) = 0.1032...$ | A1 | |
| $T = 2(t_1+t_2) = 2\!\left(\frac{1}{\sqrt{18}}\sin^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right)+\frac{1}{\sqrt{18}}\sin^{-1}\!\left(\frac{\sqrt{18}}{10}\right)\right)$ | A1 | Complete to correct value of $T$ |
| $= 0.3697... \approx 0.37$ or $0.370$ | | |
**ALT 7(d) Using cos:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8 = \frac{10}{\sqrt{18}}\cos\sqrt{18}\,t_1$ | M1A1 | |
| $t_1 = \frac{1}{\sqrt{18}}\cos^{-1}\!\left(0.8\frac{\sqrt{18}}{10}\right) = 0.2886...$ | A1 | |
| $-1 = \frac{10}{\sqrt{18}}\cos\sqrt{18}\,t_2$ | M1 | |
| $t_2 = \frac{1}{\sqrt{18}}\cos^{-1}\!\left(-\frac{\sqrt{18}}{10}\right) = 0.4735...$ | A1 | |
| $T = 2(t_2 - t_1) = 0.3697... \approx 0.37$ or $0.370$ | A1 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-24_165_1392_258_338}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows two fixed points, $A$ and $B$, which are 5 m apart on a smooth horizontal floor.
A particle $P$ of mass 1.25 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 $\lambda$ newtons, has one end attached to $P$ and the other end attached to $B$
Initially $P$ rests in equilibrium at the point $O$, where $A O = 3 \mathrm {~m}$
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 15$
The particle is now projected along the floor towards $B$\\
At time $t$ seconds, $P$ is a displacement $x$ metres from $O$ in the direction $O B$
\item Show that, while both strings are taut, $P$ moves with simple harmonic motion where $\ddot { x } = - 18 x$
The initial speed of $P$ is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\item Find the speed of $P$ at the instant when the string $P B$ becomes slack.
Both strings are taut for $T$ seconds during one complete oscillation.
\item Find the value of $T$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q7 [16]}}