| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Bungee jumping problems |
| Difficulty | Standard +0.8 This is a substantial multi-part mechanics question requiring Hooke's law, energy conservation, SHM differential equations, and interpretation. While it follows a structured path through standard M3 techniques (finding stiffness, energy methods, deriving SHM equation, finding period), the combination of elastic strings with SHM and the five-part structure requiring sustained reasoning across multiple concepts makes it moderately harder than average A-level questions. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02d Mechanical energy: KE and PE concepts6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At equilibrium: \(kx_0 = mg\), extension \(= 0.8\) m | M1 | Equilibrium condition |
| \(k = \frac{60 \times 9.8}{0.8} = 735\ \text{N m}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| B falls \(32 + x\) m total from O | B1 | Identifying distance fallen |
| Energy: \(mg(32+x) = \frac{1}{2}kx^2\) | M1 | Energy equation (KE=0) |
| \(60 \times 9.8(32+x) = \frac{1}{2}(735)x^2\) | M1 | Substituting values |
| \(588(32+x) = 367.5x^2\) | A1 | |
| Rearranging: \(x^2 - 1.6x - 51.2 = 0\) ✓ | A1 | Shown correctly |
| \(x = \frac{1.6 + \sqrt{2.56 + 204.8}}{2} = \frac{1.6 + \sqrt{207.36}}{2}\) | M1 | Quadratic formula |
| \(x \approx 8\), length \(= 32 + 8 = 40\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation of motion: \(m\ddot{x} = mg - kx\) | M1 | Newton's second law |
| \(60\ddot{x} = 60 \times 9.8 - 735x\) | A1 | Substituting values |
| \(\ddot{x} = 9.8 - 12.25x\) | M1 | Dividing through |
| \(\ddot{x} + 12.25x = 9.8\) ✓ | A1 | Correct result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| SHM about \(x = \frac{9.8}{12.25} = 0.8\), amplitude \(= 8 - 0.8 = 7.2\) | M1 | Identifying centre and amplitude |
| \(\omega^2 = 12.25\), so \(\omega = 3.5\) | B1 | |
| Time from equilibrium to lowest point \(= \frac{\pi}{2\omega} = \frac{\pi}{7}\) | M1 | Quarter period |
| \(t = \frac{\pi}{7} \approx 0.449\ \text{s}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At lowest point \(x = 8\): \(\ddot{x} = 9.8 - 12.25 \times 8 = 9.8 - 98 = -88.2\ \text{m s}^{-2}\) | M1 A1 | |
| Magnitude \(88.2\ \text{m s}^{-2}\), which is \(9g\) upward; this is a very large force (\(9 \times\) body weight) on Ben, which could be dangerous/harmful | A1 | Comment required |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| At equilibrium: $kx_0 = mg$, extension $= 0.8$ m | M1 | Equilibrium condition |
| $k = \frac{60 \times 9.8}{0.8} = 735\ \text{N m}^{-1}$ | A1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| B falls $32 + x$ m total from O | B1 | Identifying distance fallen |
| Energy: $mg(32+x) = \frac{1}{2}kx^2$ | M1 | Energy equation (KE=0) |
| $60 \times 9.8(32+x) = \frac{1}{2}(735)x^2$ | M1 | Substituting values |
| $588(32+x) = 367.5x^2$ | A1 | |
| Rearranging: $x^2 - 1.6x - 51.2 = 0$ ✓ | A1 | Shown correctly |
| $x = \frac{1.6 + \sqrt{2.56 + 204.8}}{2} = \frac{1.6 + \sqrt{207.36}}{2}$ | M1 | Quadratic formula |
| $x \approx 8$, length $= 32 + 8 = 40$ m | A1 | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation of motion: $m\ddot{x} = mg - kx$ | M1 | Newton's second law |
| $60\ddot{x} = 60 \times 9.8 - 735x$ | A1 | Substituting values |
| $\ddot{x} = 9.8 - 12.25x$ | M1 | Dividing through |
| $\ddot{x} + 12.25x = 9.8$ ✓ | A1 | Correct result shown |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| SHM about $x = \frac{9.8}{12.25} = 0.8$, amplitude $= 8 - 0.8 = 7.2$ | M1 | Identifying centre and amplitude |
| $\omega^2 = 12.25$, so $\omega = 3.5$ | B1 | |
| Time from equilibrium to lowest point $= \frac{\pi}{2\omega} = \frac{\pi}{7}$ | M1 | Quarter period |
| $t = \frac{\pi}{7} \approx 0.449\ \text{s}$ | A1 | |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| At lowest point $x = 8$: $\ddot{x} = 9.8 - 12.25 \times 8 = 9.8 - 98 = -88.2\ \text{m s}^{-2}$ | M1 A1 | |
| Magnitude $88.2\ \text{m s}^{-2}$, which is $9g$ upward; this is a very large force ($9 \times$ body weight) on Ben, which could be dangerous/harmful | A1 | Comment required |
---3 Ben has mass 60 kg and he is considering doing a bungee jump using an elastic rope with natural length 32 m . One end of the rope is attached to a fixed point O , and the other end is attached to Ben. When Ben is supported in equilibrium by the rope, the length of the rope is 32.8 m .
To predict what will happen, Ben is modelled as a particle B, the rope is assumed to be light, and air resistance is neglected. B is released from rest at O and falls vertically. When the rope becomes stretched, $x \mathrm {~m}$ denotes the extension of the rope.\\
(i) Find the stiffness of the rope.\\
(ii) Use an energy argument to show that, when B comes to rest instantaneously with the rope stretched,
$$x ^ { 2 } - 1.6 x - 51.2 = 0$$
Hence find the length of the rope when B is at its lowest point.\\
(iii) Show that, while the rope is stretched,
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 12.25 x = 9.8$$
where $t$ is the time measured in seconds.\\
(iv) Find the time taken for B to travel between the equilibrium position $( x = 0.8 )$ and the lowest point.\\
(v) Find the acceleration of $\mathbf { B }$ when it is at the lowest point, and comment on the implications for Ben.
\hfill \mbox{\textit{OCR MEI M3 2007 Q3 [18]}}