Edexcel M3 Specimen — Question 4 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeBungee jumping problems
DifficultyStandard +0.8 This is a multi-stage energy problem requiring careful identification of when the elastic rope becomes taut, application of energy conservation with elastic potential energy (EPE = λx²/2l), and handling two different positions. While methodical, it demands more problem-solving than routine M3 questions and involves multiple energy transfers across extended vertical distances.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

4. A man of mass 75 kg is attached to one end of a light elastic rope of natural length 12 m . The other end of the rope is attached to a point on the edge of a horizontal ledge 19 m above the ground. The man steps off the ledge and falls vertically under gravity. The man is modelled as a particle falling from rest. He is brought to instantaneous rest by the rope when he is 1 m above the ground.
Find
  1. the modulus of elasticity of the rope,
    (5)
  2. the speed of the man when he is 2 m above the ground, giving your answer in \(\mathrm { m } \mathrm { s } ^ { - 1 }\) to 3 significant figures.
    (5)

Question 4:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Notes
Elastic energy gained \(= \dfrac{\lambda x^2}{2l}\)M1
\(\dfrac{\lambda \cdot 6^2}{2 \times 12} = \text{PE lost} = 75 \times 9.8 \times 18\)M1 A1
\(\lambda = 8820\) NM1 A1 ft (5)
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Notes
At 2 m off ground: \(\dfrac{1}{2}\times 75 \times v^2 = 75\times 9.8\times 17 - \dfrac{1}{2}\times\dfrac{8820\times 5^2}{12}\)M1 A1 A1ft
\(v^2 = 88.2\)
\(v \approx 9.39\) ms\(^{-1}\)M1 A1 (5)
# Question 4:

## Part (a):

| Working/Answer | Marks | Notes |
|---|---|---|
| Elastic energy gained $= \dfrac{\lambda x^2}{2l}$ | M1 | |
| $\dfrac{\lambda \cdot 6^2}{2 \times 12} = \text{PE lost} = 75 \times 9.8 \times 18$ | M1 A1 | |
| $\lambda = 8820$ N | M1 A1 ft | (5) |

## Part (b):

| Working/Answer | Marks | Notes |
|---|---|---|
| At 2 m off ground: $\dfrac{1}{2}\times 75 \times v^2 = 75\times 9.8\times 17 - \dfrac{1}{2}\times\dfrac{8820\times 5^2}{12}$ | M1 A1 A1ft | |
| $v^2 = 88.2$ | | |
| $v \approx 9.39$ ms$^{-1}$ | M1 A1 | (5) |

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4. A man of mass 75 kg is attached to one end of a light elastic rope of natural length 12 m . The other end of the rope is attached to a point on the edge of a horizontal ledge 19 m above the ground. The man steps off the ledge and falls vertically under gravity. The man is modelled as a particle falling from rest. He is brought to instantaneous rest by the rope when he is 1 m above the ground.\\
Find
\begin{enumerate}[label=(\alph*)]
\item the modulus of elasticity of the rope,\\
(5)
\item the speed of the man when he is 2 m above the ground, giving your answer in $\mathrm { m } \mathrm { s } ^ { - 1 }$ to 3 significant figures.\\
(5)
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q4 [10]}}