| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 9 |
| 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 multi-part bungee jumping problem requiring energy conservation with elastic potential energy, which is standard M3 content but involves careful bookkeeping across multiple stages (free fall then elastic extension). Part (i) requires recognizing maximum tension occurs at lowest point; part (ii) involves setting up energy equations with gravitational and elastic PE. More demanding than routine mechanics but follows established M3 patterns without requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) Greatest tension \(= \frac{1320 \times 35}{25} = 1848\) N | M1 | For use of \(\frac{\lambda x}{l}\) at lowest point |
| A1 [2] | For correct answer \(1848\) | |
| (ii)(a) \(mg \times 60 = \frac{1320}{2\times25}(60-25)^2\) | M1 | For use of correct EPE formula \(\frac{\lambda x^2}{2l}\) |
| A1 | For correct unsimplified expression for EPE | |
| Hence the girl's mass is \(55\) kg | M1 | For use of equation involving EPE and GPE |
| A1 [4] | For correct answer \(55\) | |
| (b) \(55g\times30 = \frac{1}{2}\times55v^2 + \frac{1320}{2\times25}\times(30-25)^2\) | M1 | For energy equation with KE, GPE and EPE |
| A1\(\checkmark\) | For equation with all terms correct | |
| So \(v^2 = 564\), hence speed is \(23.7 \text{ m s}^{-1}\) | A1 [3] | For correct answer \(24.3\) |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** Greatest tension $= \frac{1320 \times 35}{25} = 1848$ N | M1 | For use of $\frac{\lambda x}{l}$ at lowest point |
| | A1 **[2]** | For correct answer $1848$ |
| **(ii)(a)** $mg \times 60 = \frac{1320}{2\times25}(60-25)^2$ | M1 | For use of correct EPE formula $\frac{\lambda x^2}{2l}$ |
| | A1 | For correct unsimplified expression for EPE |
| Hence the girl's mass is $55$ kg | M1 | For use of equation involving EPE and GPE |
| | A1 **[4]** | For correct answer $55$ |
| **(b)** $55g\times30 = \frac{1}{2}\times55v^2 + \frac{1320}{2\times25}\times(30-25)^2$ | M1 | For energy equation with KE, GPE and EPE |
| | A1$\checkmark$ | For equation with all terms correct |
| So $v^2 = 564$, hence speed is $23.7 \text{ m s}^{-1}$ | A1 **[3]** | For correct answer $24.3$ |
---4 For a bungee jump, a girl is joined to a fixed point $O$ of a bridge by an elastic rope of natural length 25 m and modulus of elasticity 1320 N . The girl starts from rest at $O$ and falls vertically. The lowest point reached by the girl is 60 m vertically below $O$. The girl is modelled as a particle, the rope is assumed to be light, and air resistance is neglected.\\
(i) Find the greatest tension in the rope during the girl's jump.\\
(ii) Use energy considerations to find
\begin{enumerate}[label=(\alph*)]
\item the mass of the girl,
\item the speed of the girl when she has fallen half way to the lowest point.
\end{enumerate}
\hfill \mbox{\textit{OCR M3 Q4 [9]}}