| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Challenging +1.2 This is a substantial multi-part mechanics question requiring energy methods, SHM recognition, and careful bookkeeping of string extensions. While it involves multiple techniques (equilibrium, energy conservation, SHM equations), each part follows standard M3 procedures without requiring novel insight. The length and coordination across six parts elevates it above average difficulty, but it remains a typical Further Maths mechanics question testing standard methods rather than problem-solving creativity. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(BC = y\), \(12y = 30(3.5 - y)\) | M1 A1 | Using stiffness \(\times\) extension to find tension in both strings; Correct equation for a distance; Must use extension. Condone use of modulus for full marks |
| Distance BC is \(2.5\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(30(1.0-x) - 12(2.5+x) = 10.5\frac{d^2x}{dt^2}\) | B1 M1 A1 | For \(30(1.0-x)\) or \(12(2.5+x)\); Equation of motion; Allow \((\pm)10.5a\) on RHS; FT if BC is wrong; Two forces in terms of \(x\) and acc'n; When necessary, replacing \(a\) by \(-\frac{d^2x}{dt^2}\) requires some explanation |
| \(\frac{d^2x}{dt^2} = -4x\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum speed is \(A\omega\) | M1 | OR Using energy to obtain an equation for \(v\) at C; \(\frac{1}{2}(30)(3.5^2) = \frac{1}{2}(30)(1.0^2) + \frac{1}{2}(12)(2.5^2) + \frac{1}{2}(10.5)v^2\) |
| Maximum speed is \(5 \text{ ms}^{-1}\) | A1 | FT is \(2 \times BC\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4.8^2 = 4(2.5^2 - x^2)\) | M1 | Using \(v^2 = \omega^2(A^2 - x^2)\); OR Using energy to obtain an equation for a distance |
| Distance from C is \(0.7\) m | A1 | CAO; Condone \(\pm 0.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = -2.5\cos 2t\) | B1 | \(\pm 2.5\sin 2t\) or \(\pm 2.5\cos 2t\); FT if BC is wrong (or \(A\) or \(\omega\)) |
| \(1.0 = -2.5\cos 2t\) | M1 | Using \(x = 1.0\) to obtain a time |
| Time is \(0.991\) s (3 sf) | M1 A1 | Fully correct strategy for finding the required time; CAO; e.g. \(0.2058 + \frac{1}{4} \times \frac{2\pi}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}(30)(3.5^2) = \frac{1}{2}(12)(BE^2)\) | M1 A1 | Using change of elastic energy to obtain an equation for a distance; OR Fully correct strategy using two stages of SHM; Correct (FT) equation for a distance |
| \(BE = 5.534\); Distance DE is \(2.03\) m (3 sf) | A1 | CAO |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $BC = y$, $12y = 30(3.5 - y)$ | M1 A1 | Using stiffness $\times$ extension to find tension in both strings; Correct equation for a distance; Must use extension. Condone use of modulus for full marks |
| Distance BC is $2.5$ m | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $30(1.0-x) - 12(2.5+x) = 10.5\frac{d^2x}{dt^2}$ | B1 M1 A1 | For $30(1.0-x)$ or $12(2.5+x)$; Equation of motion; Allow $(\pm)10.5a$ on RHS; FT if BC is wrong; Two forces in terms of $x$ and acc'n; When necessary, replacing $a$ by $-\frac{d^2x}{dt^2}$ requires some explanation |
| $\frac{d^2x}{dt^2} = -4x$ | E1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum speed is $A\omega$ | M1 | OR Using energy to obtain an equation for $v$ at C; $\frac{1}{2}(30)(3.5^2) = \frac{1}{2}(30)(1.0^2) + \frac{1}{2}(12)(2.5^2) + \frac{1}{2}(10.5)v^2$ |
| Maximum speed is $5 \text{ ms}^{-1}$ | A1 | FT is $2 \times BC$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4.8^2 = 4(2.5^2 - x^2)$ | M1 | Using $v^2 = \omega^2(A^2 - x^2)$; OR Using energy to obtain an equation for a distance |
| Distance from C is $0.7$ m | A1 | CAO; Condone $\pm 0.7$ |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -2.5\cos 2t$ | B1 | $\pm 2.5\sin 2t$ or $\pm 2.5\cos 2t$; FT if BC is wrong (or $A$ or $\omega$) |
| $1.0 = -2.5\cos 2t$ | M1 | Using $x = 1.0$ to obtain a time |
| Time is $0.991$ s (3 sf) | M1 A1 | Fully correct strategy for finding the required time; CAO; e.g. $0.2058 + \frac{1}{4} \times \frac{2\pi}{2}$ |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}(30)(3.5^2) = \frac{1}{2}(12)(BE^2)$ | M1 A1 | Using change of elastic energy to obtain an equation for a distance; OR Fully correct strategy using two stages of SHM; Correct (FT) equation for a distance |
| $BE = 5.534$; Distance DE is $2.03$ m (3 sf) | A1 | CAO |
---3 Fig. 3 shows the fixed points A and F which are 9.5 m apart on a smooth horizontal surface and points B and D on the line AF such that $\mathrm { AB } = \mathrm { DF } = 3.0 \mathrm {~m}$. A small block of mass 10.5 kg is joined to A by a light elastic string of natural length 3.0 m and stiffness $12 \mathrm { Nm } ^ { - 1 }$; the block is joined to F by a light elastic string of natural length 3.0 m and stiffness $30 \mathrm { Nm } ^ { - 1 }$. The block is released from rest at B and then slides along part of the line AF . The block has zero acceleration when it is at a point C , and it comes to instantaneous rest at a point E .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{70a2c3ce-7bdb-4ddd-92fc-f7dcbdfdcfaf-4_221_1082_536_502}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Find the distance BC .
At time $t \mathrm {~s}$ the displacement of the block from C is $x \mathrm {~m}$, measured in the direction AF .\\
(ii) Show that, when the block is between B and $\mathrm { D } , \frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 4 x$.\\
(iii) Find the maximum speed of the block.\\
(iv) Find the distance of the block from C when its speed is $4.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(v) Find the time taken for the block to travel from B to D.\\
(vi) Find the distance DE .
\hfill \mbox{\textit{OCR MEI M3 2015 Q3 [18]}}