| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Bungee jumping problems |
| Difficulty | Standard +0.3 This is a standard M2 bungee jumping problem requiring application of Hooke's law, Newton's second law, and energy conservation. Part (a) involves equating weight to elastic force (routine). Part (b) uses energy conservation to derive a quadratic equation (shown result reduces difficulty) and solve it. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| At zero acceleration, tension \(=\) weight | M1 | Correct condition |
| \(\frac{\lambda x}{l} = mg \Rightarrow \frac{784x}{16} = 80(9.8)\) | M1 | Hooke's law applied |
| \(49x = 784 \Rightarrow x = 16\) m extension | A1 | Correct extension |
| Length of cord \(= 16 + 16 = 32\) m | A1 | Correct length |
| Answer | Marks | Guidance |
|---|---|---|
| Using energy conservation from start to rest: | M1 | Energy method |
| Loss in GPE \(=\) Gain in EPE | ||
| Distance fallen \(= 16 + x\) | ||
| \(mg(16+x) = \frac{\lambda x^2}{2l} = \frac{784x^2}{32}\) | M1 A1 | Correct energy equation |
| \(80(9.8)(16+x) = \frac{784x^2}{32}\) | ||
| \(784(16+x) = \frac{784x^2}{32}\) | ||
| \(32(16+x) = x^2\) | M1 | Simplification |
| \(512 + 32x = x^2\) | ||
| \(x^2 - 32x - 512 = 0\) | A1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 32x - 512 = 0 \Rightarrow (x-)(x+) = 0\) | M1 | Attempt to solve quadratic |
| \(x = \frac{32 \pm \sqrt{1024 + 2048}}{2} = \frac{32 \pm \sqrt{3072}}{2}\) | M1 | Correct use of formula |
| \(x = \frac{32 + 55.43}{2} = 43.7\) m (taking positive root) | A1 | Correct extension |
| Height above ground \(= 65 - (16 + 43.7) = 65 - 59.7 = 5.3\) m | M1 A1 | Correct final answer |
# Question 9:
## Part (a):
| At zero acceleration, tension $=$ weight | M1 | Correct condition |
|---|---|---|
| $\frac{\lambda x}{l} = mg \Rightarrow \frac{784x}{16} = 80(9.8)$ | M1 | Hooke's law applied |
| $49x = 784 \Rightarrow x = 16$ m extension | A1 | Correct extension |
| Length of cord $= 16 + 16 = 32$ m | A1 | Correct length |
## Part (b)(i):
| Using energy conservation from start to rest: | M1 | Energy method |
|---|---|---|
| Loss in GPE $=$ Gain in EPE | | |
| Distance fallen $= 16 + x$ | | |
| $mg(16+x) = \frac{\lambda x^2}{2l} = \frac{784x^2}{32}$ | M1 A1 | Correct energy equation |
| $80(9.8)(16+x) = \frac{784x^2}{32}$ | | |
| $784(16+x) = \frac{784x^2}{32}$ | | |
| $32(16+x) = x^2$ | M1 | Simplification |
| $512 + 32x = x^2$ | | |
| $x^2 - 32x - 512 = 0$ | A1 | Correctly shown |
## Part (b)(ii):
| $x^2 - 32x - 512 = 0 \Rightarrow (x-)(x+) = 0$ | M1 | Attempt to solve quadratic |
|---|---|---|
| $x = \frac{32 \pm \sqrt{1024 + 2048}}{2} = \frac{32 \pm \sqrt{3072}}{2}$ | M1 | Correct use of formula |
| $x = \frac{32 + 55.43}{2} = 43.7$ m (taking positive root) | A1 | Correct extension |
| Height above ground $= 65 - (16 + 43.7) = 65 - 59.7 = 5.3$ m | M1 A1 | Correct final answer |
These pages (7 and 8) from the AQA January 2009 MM2B paper both state **"There are no questions printed on this page"** — they are blank pages and contain no mark scheme content to extract.9 A bungee jumper, of mass 80 kg , is attached to one end of a light elastic cord, of natural length 16 metres and modulus of elasticity 784 N . The other end of the cord is attached to a horizontal platform, which is at a height of 65 metres above the ground.
The bungee jumper steps off the platform at the point where the cord is attached and falls vertically. The bungee jumper can be modelled as a particle. Hooke's law can be assumed to apply throughout the motion and air resistance can be assumed to be negligible.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the cord when the acceleration of the bungee jumper is zero.
\item The cord extends by $x$ metres beyond its natural length before the bungee jumper first comes to rest.
\begin{enumerate}[label=(\roman*)]
\item Show that $x ^ { 2 } - 32 x - 512 = 0$.
\item Find the distance above the ground at which the bungee jumper first comes to rest.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q9 [11]}}