| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Maximum or minimum speed problems |
| Difficulty | Standard +0.8 This M3 SHM question requires students to model tidal motion, extract period and amplitude from context, apply SHM velocity formula v = ±ω√(a²-x²), and solve for time intervals using inverse trig. While the setup is multi-step and contextual, the techniques are standard M3 material with straightforward arithmetic—moderately challenging but not requiring novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Period \(= 12.25\) hours | B1 | Correct period for the SHM (Allow 44100 s) |
| \(\frac{2\pi}{\omega} = 12.25\) | ||
| \(\omega = \frac{2\pi}{12.25}\) oe eg \(\omega = \frac{8\pi}{49}\) | M1A1ft | A correct expression for \(\omega\) follow through their period, inc with units changed. |
| Amplitude \(= 5\) m | B1 | Correct amplitude seen explicitly or used |
| \(v^2 = \omega^2(a^2 - x^2)\) | ||
| \(x = 4 \Rightarrow v^2 = \left(\frac{2\pi}{12.25}\right)^2(5^2 - 4^2)\) | M1A1 | |
| \(v = \left(\frac{2\pi}{12.25}\right) \times 3 = 1.53873... = 1.5\) m h\(^{-1}\) (1.5 or better or \(\frac{24\pi}{49}\)) | A1cao | (7) Correct value of \(v\), 2 sf or better Units must be m h\(^{-1}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 4\) \(4 = 5\cos\omega t\) | M1 | Using \(x = a\cos\omega t\) or \(a\sin\omega t\) with \(x = \pm\)(their amp \(-1\)), their \(\omega\) and amplitude |
| \(t = \frac{12.25}{2\pi}\cos^{-1}0.8\) (= 1.2546...) | M1A1 | Finding a correct time, either from the end point to reaching the bottom step or from the bottom step to the centre of the oscillation. (in hours or seconds) |
| Time ladder in the water \(= 12.25 - 2 \times \frac{12.25}{2\pi}\cos^{-1}0.8\) | DM1 | Using their time to obtain an expression for the required time |
| \(= 9.7407... = 9.7\) hours (9.7 or better) (9hr 44 min or 35 066 875 seconds) | A1 cao | (4) Time \(= 9.7\) hrs or better [11] |
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Period $= 12.25$ hours | B1 | Correct period for the SHM (Allow 44100 s) |
| $\frac{2\pi}{\omega} = 12.25$ | | |
| $\omega = \frac{2\pi}{12.25}$ oe eg $\omega = \frac{8\pi}{49}$ | M1A1ft | A correct expression for $\omega$ follow through their period, inc with units changed. |
| Amplitude $= 5$ m | B1 | Correct amplitude seen explicitly or used |
| $v^2 = \omega^2(a^2 - x^2)$ | | |
| $x = 4 \Rightarrow v^2 = \left(\frac{2\pi}{12.25}\right)^2(5^2 - 4^2)$ | M1A1 | |
| $v = \left(\frac{2\pi}{12.25}\right) \times 3 = 1.53873... = 1.5$ m h$^{-1}$ (1.5 or better or $\frac{24\pi}{49}$) | A1cao | (7) Correct value of $v$, 2 sf or better Units must be m h$^{-1}$. |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 4$ $4 = 5\cos\omega t$ | M1 | Using $x = a\cos\omega t$ or $a\sin\omega t$ with $x = \pm$(their amp $-1$), their $\omega$ and amplitude |
| $t = \frac{12.25}{2\pi}\cos^{-1}0.8$ (= 1.2546...) | M1A1 | Finding a correct time, either from the end point to reaching the bottom step or from the bottom step to the centre of the oscillation. (in hours or seconds) |
| Time ladder in the water $= 12.25 - 2 \times \frac{12.25}{2\pi}\cos^{-1}0.8$ | DM1 | Using their time to obtain an expression for the required time |
| $= 9.7407... = 9.7$ hours (9.7 or better) (9hr 44 min or 35 066 875 seconds) | A1 cao | (4) Time $= 9.7$ hrs or better [11] |
**Key Notes:**
- (a)B1: Correct period for the SHM (Allow 44100 s)
- (a)M1: Using period $= \frac{2\pi}{\omega}$ to obtain an expression for $\omega$ or any other complete method
- (a)A1ft: A correct expression for $\omega$ follow through their period, inc with units changed.
- (a)B1: Correct amplitude seen explicitly or used
- (a)M1: Using $v^2 = \omega^2(a^2 - x^2)$ with their $\omega$, amplitude and $x = $(their amp $-1$)
- (a)A1: Correct expression for $v^2$
- (a)A1cao: Correct value of $v$, 2 sf or better Units must be m h$^{-1}$.
- (b)M1: Using $x = a\cos\omega t$ or $a\sin\omega t$ with $x = \pm$(their amp $-1$), their $\omega$ and amplitude
- (b)A1: Finding a correct time, either from the end point to reaching the bottom step or from the bottom step to the centre of the oscillation. (in hours or seconds)
- (b)DM1: Using their time to obtain an expression for the required time
- (b)A1cao: Time $= 9.7$ hrs or better
- Sine used: Time $= 2 \times \frac{\text{period}}{4} + 2 \times$ time found
---A vertical ladder is fixed to a wall in a harbour. On a particular day the minimum depth of water in the harbour occurs at 0900 hours. The next time the water is at its minimum depth is 2115 hours on the same day. The bottom step of the ladder is 1 m above the lowest level of the water and 9 m below the highest level of the water. The rise and fall of the water level can be modelled as simple harmonic motion and the thickness of the step can be assumed to be negligible.
Find
\begin{enumerate}[label=(\alph*)]
\item the speed, in metres per hour, at which the water level is moving when it reaches the bottom step of the ladder, [7]
\item the length of time, on this day, between the water reaching the bottom step of the ladder and the ladder being totally out of the water once more. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q5 [11]}}