| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string vertical motion |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question involving elastic strings, equilibrium, and energy conservation. Part (i) is straightforward Hooke's law application. Part (ii) requires resolving forces at equilibrium (standard technique). Parts (iii) and (iv) use energy methods with some geometric reasoning to find string lengths at different positions. While it requires multiple techniques and careful geometry, these are all standard M3 methods with no novel insight required, making it moderately above average difficulty for A-level. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02g 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 | Marks | Guidance |
| Extension \(= 1.3 - 1.2 = 0.1\ \text{m}\) | M1 | |
| Tension \(= 637 \times 0.1 = 63.7\ \text{N}\) | A1 | |
| Elastic energy \(= \frac{1}{2} \times 637 \times 0.1^2 = 3.185\ \text{J}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At equilibrium, \(AR = 1.3\ \text{m}\); horizontal distance \(= 1.2\ \text{m}\), so vertical drop \(= \sqrt{1.3^2 - 1.2^2} = 0.5\ \text{m}\) | M1 A1 | |
| Vertical component of tension \(= T\times\frac{0.5}{1.3} = 63.7 \times \frac{0.5}{1.3}\) | M1 | |
| \(mg = 24.5\ \text{N}\), \(m = 2.5\ \text{kg}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(AR = 1.5\ \text{m}\): vertical drop from A \(= \sqrt{1.5^2 - 1.2^2} = 0.9\ \text{m}\) | B1 | |
| R drops \(0.9 - 0.5 = 0.4\ \text{m}\) below equilibrium | B1 | |
| Elastic PE at \(AR=1.5\): \(\frac{1}{2}(637)(0.3)^2 = 28.665\ \text{J}\) | M1 | |
| Energy conservation: \(\frac{1}{2}mu^2 + EPE_{eq} + mg(0.4) = EPE_{1.5}\) | M1 A1 | |
| \(\frac{1}{2}(2.5)u^2 + 3.185 + 2.5(9.8)(0.4) = 28.665\) | A1 | |
| \(u = \sqrt{\frac{2(28.665 - 3.185 - 9.8)}{2.5}}\), \(u \approx 3.56\ \text{m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find energy at level of A (R is \(0.5\ \text{m}\) above equilibrium, \(AR = \sqrt{1.2^2+0^2}=1.2\ \text{m}\), natural length, so EPE \(= 0\)) | M1 | |
| Energy conservation from equilibrium upward: \(\frac{1}{2}mu^2 + EPE_{eq} = mg(0.5) + EPE_A + \frac{1}{2}mv_A^2\) | M1 | |
| KE at A level \(= \frac{1}{2}(2.5)u^2 + 3.185 - 2.5(9.8)(0.5) - 0\) | A1 | |
| KE \(> 0\), so ring rises above level of A | A1 |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Extension $= 1.3 - 1.2 = 0.1\ \text{m}$ | M1 | |
| Tension $= 637 \times 0.1 = 63.7\ \text{N}$ | A1 | |
| Elastic energy $= \frac{1}{2} \times 637 \times 0.1^2 = 3.185\ \text{J}$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At equilibrium, $AR = 1.3\ \text{m}$; horizontal distance $= 1.2\ \text{m}$, so vertical drop $= \sqrt{1.3^2 - 1.2^2} = 0.5\ \text{m}$ | M1 A1 | |
| Vertical component of tension $= T\times\frac{0.5}{1.3} = 63.7 \times \frac{0.5}{1.3}$ | M1 | |
| $mg = 24.5\ \text{N}$, $m = 2.5\ \text{kg}$ | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $AR = 1.5\ \text{m}$: vertical drop from A $= \sqrt{1.5^2 - 1.2^2} = 0.9\ \text{m}$ | B1 | |
| R drops $0.9 - 0.5 = 0.4\ \text{m}$ below equilibrium | B1 | |
| Elastic PE at $AR=1.5$: $\frac{1}{2}(637)(0.3)^2 = 28.665\ \text{J}$ | M1 | |
| Energy conservation: $\frac{1}{2}mu^2 + EPE_{eq} + mg(0.4) = EPE_{1.5}$ | M1 A1 | |
| $\frac{1}{2}(2.5)u^2 + 3.185 + 2.5(9.8)(0.4) = 28.665$ | A1 | |
| $u = \sqrt{\frac{2(28.665 - 3.185 - 9.8)}{2.5}}$, $u \approx 3.56\ \text{m s}^{-1}$ | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find energy at level of A (R is $0.5\ \text{m}$ above equilibrium, $AR = \sqrt{1.2^2+0^2}=1.2\ \text{m}$, natural length, so EPE $= 0$) | M1 | |
| Energy conservation from equilibrium upward: $\frac{1}{2}mu^2 + EPE_{eq} = mg(0.5) + EPE_A + \frac{1}{2}mv_A^2$ | M1 | |
| KE at A level $= \frac{1}{2}(2.5)u^2 + 3.185 - 2.5(9.8)(0.5) - 0$ | A1 | |
| KE $> 0$, so ring rises above level of A | A1 | |
---3 A light elastic string has natural length 1.2 m and stiffness $637 \mathrm { Nm } ^ { - 1 }$.\\
(i) The string is stretched to a length of 1.3 m . Find the tension in the string and the elastic energy stored in the string.
One end of this string is attached to a fixed point $A$. The other end is attached to a heavy ring $R$ which is free to move along a smooth vertical wire. The shortest distance from A to the wire is 1.2 m (see Fig. 3).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-4_357_337_669_863}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The ring is in equilibrium when the length of the string $A R$ is 1.3 m .\\
(ii) Show that the mass of the ring is 2.5 kg .
The ring is given an initial speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downwards from its equilibrium position. It first comes to rest, instantaneously, in the position where the length of AR is 1.5 m .\\
(iii) Find $u$.\\
(iv) Determine whether the ring will rise above the level of A .
\hfill \mbox{\textit{OCR MEI M3 2007 Q3 [18]}}