| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Bungee jumping problems |
| Difficulty | Challenging +1.2 This is a standard M3 mechanics question involving energy methods and elastic strings/ropes. Part (i) requires recognizing maximum speed occurs when forces balance (at natural length), part (ii) is straightforward energy conservation, and part (iii) uses energy at lowest point. The second problem involves energy conservation with circular motion and finding maximum contact force via differentiation. While multi-step and requiring careful setup, these are well-practiced M3 techniques without novel insight required. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = 1470x/30\) | B1 | |
| \([49x = 70 \times 9.8]\) | M1 | For using \(T = mg\) |
| \(x = 14\) | A1 | |
| Distance fallen is \(44\text{m}\) | A1ft | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE loss \(= 70g(30 + 14)\) | B1ft | |
| EE gain \(= 1470 \times 14^2/(2 \times 30)\) | B1ft | |
| \([\frac{1}{2} \times 70v^2 = 30184 - 4802]\) | M1 | For a linear equation with terms representing KE, PE and EE changes |
| Speed is \(26.9\text{ms}^{-1}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.5v^2 = 14g - 68.6 + 30g]\) | M1 | For using Newton's 2nd law \((vdv/dx = g - 0.7x)\), integrating \((0.5v^2 = gx - 0.35x^2 + k)\), using \(v(0)^2 = 60g \Rightarrow k = 30g\), and substituting \(x = 14\) |
| For \(14g + 30g\) | B1ft | |
| For \(\mp 68.6\) | B1ft | |
| Speed is \(26.9\text{ms}^{-1}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE loss \(= 70g(30 + x)\) | B1ft | |
| EE gain \(= 1470x^2/(2 \times 30)\) | B1ft | |
| \([x^2 - 28x - 840 = 0]\) | M1 | For using PE loss \(=\) KE gain to obtain a 3 term quadratic equation |
| Extension is \(46.2\text{m}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For identifying SHM with \(n^2 = 1470/(70 \times 30)\) |
| (implied method) | M1 | For using \(v_{\max} = An\) |
| \(A = 26.9/\sqrt{0.7}\) | A1 | |
| Extension is \(46.2\text{m}\) | A1 | 4 |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = 1470x/30$ | B1 | |
| $[49x = 70 \times 9.8]$ | M1 | For using $T = mg$ |
| $x = 14$ | A1 | |
| Distance fallen is $44\text{m}$ | A1ft | 4 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE loss $= 70g(30 + 14)$ | B1ft | |
| EE gain $= 1470 \times 14^2/(2 \times 30)$ | B1ft | |
| $[\frac{1}{2} \times 70v^2 = 30184 - 4802]$ | M1 | For a linear equation with terms representing KE, PE and EE changes |
| Speed is $26.9\text{ms}^{-1}$ | A1 | 4 | AG |
**OR (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.5v^2 = 14g - 68.6 + 30g]$ | M1 | For using Newton's 2nd law $(vdv/dx = g - 0.7x)$, integrating $(0.5v^2 = gx - 0.35x^2 + k)$, using $v(0)^2 = 60g \Rightarrow k = 30g$, and substituting $x = 14$ |
| For $14g + 30g$ | B1ft | |
| For $\mp 68.6$ | B1ft | |
| Speed is $26.9\text{ms}^{-1}$ | A1 | 4 | AG; accept in unsimplified form |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE loss $= 70g(30 + x)$ | B1ft | |
| EE gain $= 1470x^2/(2 \times 30)$ | B1ft | |
| $[x^2 - 28x - 840 = 0]$ | M1 | For using PE loss $=$ KE gain to obtain a 3 term quadratic equation |
| Extension is $46.2\text{m}$ | A1 | 4 | |
**OR (iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For identifying SHM with $n^2 = 1470/(70 \times 30)$ |
| (implied method) | M1 | For using $v_{\max} = An$ |
| $A = 26.9/\sqrt{0.7}$ | A1 | |
| Extension is $46.2\text{m}$ | A1 | 4 | |
---6 A bungee jumper of mass 70 kg is joined to a fixed point $O$ by a light elastic rope of natural length 30 m and modulus of elasticity 1470 N . The jumper starts from rest at $O$ and falls vertically. The jumper is modelled as a particle and air resistance is ignored.\\
(i) Find the distance fallen by the jumper when maximum speed is reached.\\
(ii) Show that this maximum speed is $26.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures.\\
(iii) Find the extension of the rope when the jumper is at the lowest position.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09d3e8ca-0062-4f62-8453-7acaff591db5-4_543_616_310_301}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09d3e8ca-0062-4f62-8453-7acaff591db5-4_668_709_267_1135}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A smooth horizontal cylinder of radius 0.6 m is fixed with its axis horizontal and passing through a fixed point $O$. A light inextensible string of length $0.6 \pi \mathrm {~m}$ has particles $P$ and $Q$, of masses 0.3 kg and 0.4 kg respectively, attached at its ends. The string passes over the cylinder and is held at rest with $P , O$ and $Q$ in a straight horizontal line (see Fig. 1). The string is released and $Q$ begins to descend. When the line $O P$ makes an angle $\theta$ radians, $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$, with the horizontal, the particles have speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see Fig. 2).\\
(i) By considering the total energy of the system, or otherwise, show that
$$v ^ { 2 } = 6.72 \theta - 5.04 \sin \theta .$$
(ii) Show that the magnitude of the contact force between $P$ and the cylinder is
$$( 5.46 \sin \theta - 3.36 \theta ) \text { newtons. }$$
Hence find the value of $\theta$ for which the magnitude of the contact force is greatest.\\
(iii) Find the transverse component of the acceleration of $P$ in terms of $\theta$.
\hfill \mbox{\textit{OCR M3 2006 Q6 [12]}}