| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 16 |
| 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 energy conservation with elastic potential energy. While it involves multiple parts and careful bookkeeping of energy terms (gravitational PE, kinetic energy, elastic PE), the approach is methodical and follows a well-practiced template. The algebra is straightforward, and finding maximum speed by differentiating the energy equation is a routine technique taught explicitly for this topic. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x \geq 22\), KE is \(\frac{1}{2}\times 49 \times v^2\) | ||
| EPE is \(\frac{1078(x-22)^2}{2\times 22}\) | M1A1 | M1 for any \(\frac{1078p^2}{2\times 22}\) |
| Change in PE is \(49 \times g \times x\) | ||
| Conservation of energy: \(\frac{1}{2}\times 49\times v^2 + \frac{1078(x-22)^2}{2\times 22} = 49\times g\times x\) | M1A1, A1 | M1 for 3 terms (PE, PE, EPE); A1 two terms correct; A1 all three terms correct |
| \(v^2 + (x-22)^2 = 19.6x\) | SC3: \(\frac{49}{2}v^2 + \frac{49}{2}e^2 = 49g(e+22)\) [could use \(x\) for \(e\)] | |
| \(5v^2 = 318x - 5x^2 - 2420\) | A1 | 6, AG |
| Answer | Marks | Guidance |
|---|---|---|
| If \(x\) is not greater than 22, cord is not stretched | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| At maximum value of \(x\), \(v = 0\), \(\therefore 5x^2 - 318x + 2420 = 0\) | M1 | |
| \(x = \frac{318 \pm \sqrt{318^2 - 4\times 5\times 2420}}{2\times 5}\) | m1 | dep on M1 above |
| \(x = 54.76..\) or \(8.84..\), \(= 54.8\) | A1, E1 | 4; A1 for either solution; needs reason for deletion of second root; both roots must be positive: one above 22, one below 22 |
| Answer | Marks | Guidance |
|---|---|---|
| When speed is maximum, \(a = 0\), tension \(=\) gravitational force | M1 | or \(\frac{d(5v^2)}{dx} = 318 - 10x = 0\) at maximum speed \(\Rightarrow 318 - 10x = 0\) |
| \(\frac{1078(x-22)}{22} = 49g\), so \(x - 22 = 9.8\), \(x = 31.8\) | A1, A1 | 3, AG |
| Answer | Marks | Guidance |
|---|---|---|
| From part (a): \(v^2 = 19.6\times 31.8 - 9.8^2\) | M1 | |
| \(v = 22.96\), maximum speed is \(23.0\text{ ms}^{-1}\) | A1 | 2 |
## Question 8:
**Part (a):**
| When $x \geq 22$, KE is $\frac{1}{2}\times 49 \times v^2$ | | |
| EPE is $\frac{1078(x-22)^2}{2\times 22}$ | M1A1 | M1 for any $\frac{1078p^2}{2\times 22}$ |
| Change in PE is $49 \times g \times x$ | | |
| Conservation of energy: $\frac{1}{2}\times 49\times v^2 + \frac{1078(x-22)^2}{2\times 22} = 49\times g\times x$ | M1A1, A1 | M1 for 3 terms (PE, PE, EPE); A1 two terms correct; A1 all three terms correct |
| $v^2 + (x-22)^2 = 19.6x$ | | SC3: $\frac{49}{2}v^2 + \frac{49}{2}e^2 = 49g(e+22)$ [could use $x$ for $e$] |
| $5v^2 = 318x - 5x^2 - 2420$ | A1 | 6, AG |
**Part (b):**
| If $x$ is not greater than 22, cord is not stretched | B1 | 1 |
**Part (c):**
| At maximum value of $x$, $v = 0$, $\therefore 5x^2 - 318x + 2420 = 0$ | M1 | |
| $x = \frac{318 \pm \sqrt{318^2 - 4\times 5\times 2420}}{2\times 5}$ | m1 | dep on M1 above |
| $x = 54.76..$ or $8.84..$, $= 54.8$ | A1, E1 | 4; A1 for either solution; needs reason for deletion of second root; both roots must be positive: one above 22, one below 22 |
**Part (d)(i):**
| When speed is maximum, $a = 0$, tension $=$ gravitational force | M1 | or $\frac{d(5v^2)}{dx} = 318 - 10x = 0$ at maximum speed $\Rightarrow 318 - 10x = 0$ |
| $\frac{1078(x-22)}{22} = 49g$, so $x - 22 = 9.8$, $x = 31.8$ | A1, A1 | 3, AG |
**Part (d)(ii):**
| From part (a): $v^2 = 19.6\times 31.8 - 9.8^2$ | M1 | |
| $v = 22.96$, maximum speed is $23.0\text{ ms}^{-1}$ | A1 | 2 |8 A bungee jumper, of mass 49 kg , is attached to one end of a light elastic cord of natural length 22 metres and modulus of elasticity 1078 newtons. The other end of the cord is attached to a horizontal platform, which is at a height of 60 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. Assume that Hooke's Law applies whilst the cord is taut and that air resistance is negligible throughout the motion.
When the bungee jumper has fallen $x$ metres, his speed is $v \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item By considering energy, show that, when $x$ is greater than 22,
$$5 v ^ { 2 } = 318 x - 5 x ^ { 2 } - 2420$$
\item Explain why $x$ must be greater than 22 for the equation in part (a) to be valid. ( 1 mark)
\item Find the maximum value of $x$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the speed of the bungee jumper is a maximum when $x = 31.8$.
\item Hence find the maximum speed of the bungee jumper.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q8 [16]}}